Samacheer Kalvi 12th Business Maths Guide Chapter 7 Probability Distributions Ex 7.1

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 7 Probability Distributions Ex 7.1 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.1

Question 1.
Define Biomial distribution.
Solution:
Binomial distribution was discovered by James Bernoulli (1654-1705) in the year 1700.
A random variable X is said to follow binomial distribution with parameter n and p, if it assumes only non-negative value and its probability mass function in given by
Samacheer Kalvi 12th Business Maths Guide Chapter 7 Probability Distributions Ex 7.1 1

Samacheer Kalvi 12th Business Maths Guide Chapter 7 Probability Distributions Ex 7.1

Question 2.
Define Bernouli trials.
Solution:
A random experiment whose outcomes are of two types namely success S and failure F, occurring with probabilities p and q respectively, is called a Bernoulli trial.
Some examples of Bernoulli trials are:
(i) Tossing of a coin (Head or tail)
(ii) Throwing of a die (getting even or odd number)

Question 3.
Derive the mean and variance of bionomial distribution
Solution:
Derivation of the Mean and Variance of Binomial distribution:
The mean of the binomial distribution
Samacheer Kalvi 12th Business Maths Guide Chapter 7 Probability Distributions Ex 7.1 2
= np(q + p)n – 1 [since p + q = 1]
= np
E(X) = np
The mean of the binomial distribution is np.
Var(X) = E(X²) – E(X²)
Samacheer Kalvi 12th Business Maths Guide Chapter 7 Probability Distributions Ex 7.1 3
= n(n – 1)p²(q + p)(n – 2) + np
n(n – 1 )p² + np
Variance = E(X²) – [E(X)]²
= n²p² – np² + np – n²p²
= np(1 – p) = npq
Hence, mean of the BD is np and the Variance is npq.

Question 4.
Write down the condition for which the bionomial distribution can be used.
Solution:
The Binomial distribution can be used under the following conditions:
1. The number of trials ‘n finite
2. The trials are independent of each other.
3. The probability of success ‘p’ is constant for each trial.
4. In every trial there are only two possible outcomes – success or failure.

Question 5.
Mention the properties of bionomial distribution.
Solution:
Properties of Binomial distribution
1. Binomial distribution is symmetrical if p = q = 0.5, It is skew symmetric if p ≠ q. It is positively skewed if p < 0.5 and it is negatively skewed it p > 0.5.
2. For Binomial distribution, variance is less than mean
Variance npq = (np) q < np

Samacheer Kalvi 12th Business Maths Guide Chapter 7 Probability Distributions Ex 7.1

Question 6.
If 5% of the items produced turn out to be defective, then find out the probability that out of 10 items selected at random there are
(i) exactly three defectives
(ii) atleast two defectives
(iii) exactly 4 defectives
(iv) find the mean and variance.
Solution:
Probability of getting a defective item
Samacheer Kalvi 12th Business Maths Guide Chapter 7 Probability Distributions Ex 7.1 4
In binomial distribution
p(X = x) = nCxpxqn-x
Samacheer Kalvi 12th Business Maths Guide Chapter 7 Probability Distributions Ex 7.1 5
Let x = (19)²
log x = 7 log(19)7
= 7 log 19
= 7 × 1.2788
= 8.9516
x = Antilog 8.956
= 8.945 × 108
ut y = (20)10
log y = (10) log 20
= 10 × 1.3010
= 13.010
y = Anti log 13.010
= 1.023 × 10³

(ii) p(atleast two defectives)
= p(x ≥ 2)
= p(x = 2) = p(x = 3) + ………….. + p(x = 10)
= 1 – p(x < 2)
= 1 – {p(x = 0) + p(x = 1)}
Samacheer Kalvi 12th Business Maths Guide Chapter 7 Probability Distributions Ex 7.1 6
Let x = (19)9
log x = 9 log 19
= 9 × 1.2788
= 11.5092
x = Antilog 11.5092
= 3.229 × 1011

(iii) p(extactly 4 defectives) = p(X = 4)
Samacheer Kalvi 12th Business Maths Guide Chapter 7 Probability Distributions Ex 7.1 7
Let x = (19)6
log x = 6 log 19
= 6 × 1.2788
log x = 7.66728
Antilog (7.66728)
x = 4.648 × 107

(iv) mean E(x) = np
= 10 × \(\frac { 1 }{20}\) = \(\frac { 1 }{2}\) = 0.5
Varaince = npq
= 10 × \(\frac { 1 }{20}\) = \(\frac { 19 }{20}\) = \(\frac { 19 }{40}\) = 0.475

Samacheer Kalvi 12th Business Maths Guide Chapter 7 Probability Distributions Ex 7.1

Question 7.
In a particular university 40% of the students are having news paper reading habit. Nine university students are selected to find their views on reading habit. Find the probability that
(i) none of those selected have news paper reading habit
(ii) all those selected have news paper reading habit
(iii) atleast two third have news paper reading habit.
Solution:
let p to the probability of having newspaper reading habit
p = \(\frac { 40 }{100}\) = \(\frac { 2 }{5}\)
q = 1 – p = 1\(\frac { 2 }{5}\) = \(\frac { 5-2 }{5}\) = \(\frac { 3 }{5}\) and n = 9
In the binomial distribution p(x = 4) = ncx pxqn-r
The binomial distribution P(x) = 9cx (\(\frac { 2 }{5}\))x (\(\frac { 3 }{5}\))9-x

(i) p(none of those selected have newspaper reading
Samacheer Kalvi 12th Business Maths Guide Chapter 7 Probability Distributions Ex 7.1 8

(ii) P(all those selected have newspaper reading habit)
Samacheer Kalvi 12th Business Maths Guide Chapter 7 Probability Distributions Ex 7.1 9

(iii) p(at least two third have newspaper reading habit)
Samacheer Kalvi 12th Business Maths Guide Chapter 7 Probability Distributions Ex 7.1 10

Question 8.
In a family of 3 children, what is the probability that there will be exactly 2 girls?
Solution:
let p be the probability of getting a girl child= 1/2 q = 1 – p ⇒ = 1 – 1/2
∴ q = \(\frac { 1 }{2}\) and n = 3
In a binomial distibution
p(x = x) ncx pxqn-x
p(exactly 2 girls) p(x = 2)
Samacheer Kalvi 12th Business Maths Guide Chapter 7 Probability Distributions Ex 7.1 11

Question 9.
Defects in yarn manufactured by a local mill can be approximated by a distribution with a mean of 1.2 defects for every 6 metres of length. If lengths of 6 metres are to be inspected, find the probability of less than 2 defects.
Solution:
Given n = 6
mean np = 1.2
⇒ 6p = 1.2
p = \(\frac { 1.2 }{6}\) p = 0.2 (or) p = \(\frac { 1 }{5}\)
q = 1 – p = 1 – \(\frac { 1 }{5}\)
∴ q = \(\frac { 4 }{5}\)
The binomial distribution
p(x) = 6Cx (0.2)x(0.8)6-x
p(x < 2) = p(x = 0) + p(x = 1)
Samacheer Kalvi 12th Business Maths Guide Chapter 7 Probability Distributions Ex 7.1 12

Samacheer Kalvi 12th Business Maths Guide Chapter 7 Probability Distributions Ex 7.1

Question 10.
If 18% of the bolts produced by a machine are defective, determine the probability that out of the 4 bolts chosen at random
Solution:
n = 4
probability of defective bolts p = 18/100
9 = 1 – p = 1.018 = 0.82
The binoial distribution p(x) = 4Cx(0.18)x(0.82)4-x
= 4C1(0.18)1(0.82)4-1
= 4 × 0.18 × (0.82)³
= 0.72 × 0.551368
= 0.39698496
p(X = 1)= 0.3969 approximately

(ii) p(none will be defective)p(x = 0)
= 4C0(0.18)°(0.82)4-0
= (1)(1)(0.45212176)
p(x = 0) = 0.45212

(iii) p(almost 2 will be defective) = p(x ≤ 2)
= p(x = 0) + (p(x = 1) + p(x = 2)
= 4C0(0.18)°(0.82)4-0 + 4C1(0.18)1(0.82)4-1 + 4C2 (0.18)² (0.82)4-2
= (0.82)4 + 4 × (0.18) × (0.82)³ + \(\frac { 4×3 }{1×2}\) × (0.18)² (0.82)²
= 0.45212176 + (0.72 × 0.551368) + (6 × 0.0324 × 0.6724)
= 0.45212176 + 0.39698496 + 013071456
= 0.97982128
= 0.9798

Question 11.
If 18% of the bolts produced by a machine are defective, determine the probability that out of the 4 bolts chosen at random
(i) exactly one will be defective
(ii) none will be defective
(iii) atmost 2 will be defective
Solution:
p = 0.09 = \(\frac { 9 }{100}\)
q = 1 – p ⇒ q – 1 – \(\frac { 9 }{100}\)
q = \(\frac { 100-9 }{100}\)
q = \(\frac { 91 }{100}\)
In a binomial distribution p(x = x) = ncxpxqn-x
p(atleast one success) = p(x ≥ 1)
Samacheer Kalvi 12th Business Maths Guide Chapter 7 Probability Distributions Ex 7.1 13
Taking log on both sides log ≤ (0.66) ≤ n log(0.91)
∴ n ≥ log \(\frac { log(0.66) }{log(0.91)}\)
n ≥ 5
∴ 5 or more trials are needed

Samacheer Kalvi 12th Business Maths Guide Chapter 7 Probability Distributions Ex 7.1

Question 12.
Among 28 professors of a certain department, 18 drive foreign cars and 10 drive local made cars. If 5 of these professors are selected at random, what is the probability that atleast 3 of them drive foreign cars?
Solution:
Let p be the probability of foreign cars driven by the professors
p = \(\frac { 18 }{28}\) = \(\frac { 9 }{14}\)
Let q be the probability of local made cars driv¬en by the professors
q = \(\frac { 10 }{28}\) = \(\frac { 5 }{14}\)
and n = 5
The binomial distribution p(x = x) = ncxpxqn-x
Samacheer Kalvi 12th Business Maths Guide Chapter 7 Probability Distributions Ex 7.1 14

Question 13.
Out of 750 families with 4 children each, how many families would be expected to have (i) atleast one boy (ii) atmost 2 girls (iii) and children of both sexes? Assume equal probabilities for boys and girls?
Solution:
Assume equal probabilities for boys girls let p be the probability of having a boy Let x be the random variable for getting either a boy or a girl
∴ p = \(\frac { 1 }{2}\) and q \(\frac { 1 }{2}\) and n = 4
In binomial distribution p(x = 4) = ncxpxqn-x
Here the binomial distribution is p(X= x)
Samacheer Kalvi 12th Business Maths Guide Chapter 7 Probability Distributions Ex 7.1 15
= 1 – 0.0625
= 0.9375
for 750 families (p ≥ 1) = 750 × 0.9375
= 703.125
= 703(approximately)

(ii) p(almost 2 girls) = p(x ≤ 2)
= p(x = 0) = p(x = 1) + p(x = 2)
Samacheer Kalvi 12th Business Maths Guide Chapter 7 Probability Distributions Ex 7.1 16
For 750 families p(x ≤ 2) = 0.6875 × 750
= 515.625
= 516 (approximately)

(iii) p(children of both sexes) = p(x = 1) + p(x = 2)p(x = 3)
Samacheer Kalvi 12th Business Maths Guide Chapter 7 Probability Distributions Ex 7.1 17
= 0.875 x 750
For 750 families p(x = 2) = 0.875 × 750
= 656.25
= 656 (approximately)

Samacheer Kalvi 12th Business Maths Guide Chapter 7 Probability Distributions Ex 7.1

Question 14.
Forty percent of business travellers carry a laptop. In a sample of 15 business travelers,
Solution:
Given n = 5
p = \(\frac { 40 }{100}\) = \(\frac { 2 }{5}\)
q = 1 – p = 1 – \(\frac { 2 }{5}\) \(\frac { 5-2 }{5}\) = \(\frac { 3 }{5}\)
The binomial distributionp (X = x) = 15cx (\(\frac { 2 }{5}\))x (\(\frac { 3 }{5}\))15-x
(i) p(probability that 3 will have a laptop) = p(x = 3)
Samacheer Kalvi 12th Business Maths Guide Chapter 7 Probability Distributions Ex 7.1 18

(ii) p(12 of the traels will not have a laptop)
= 1 – p(x = 12)
= 1 – 15c12 (\(\frac { 2 }{5}\))12 (\(\frac { 3 }{5}\))15-12
Samacheer Kalvi 12th Business Maths Guide Chapter 7 Probability Distributions Ex 7.1 19

(iii) p(at least three of the travelers have a laptop)
= p(x ≥ 3)
p(x ≥ 3) = 1 – p (x < 3)
= 1 – [p(x = 0) + p(x = 1) + p(x = 2)]
p(x < 3) = p(x = 0) + p(x = 1) + p(x = 2)
Samacheer Kalvi 12th Business Maths Guide Chapter 7 Probability Distributions Ex 7.1 20

Question 15.
A pair 5f dice is thrown 4 times. If getting a doublet is considered a success, find the probability of 2 successes.
Solution:
In a throw of a pair dice the doublets are (1, 1),(1, 2) (3, 3),(4, 4),(5, 5),(6, 6)
probability of getting a doublet p = \(\frac { 6 }{36}\) = \(\frac { 1 }{6}\)
⇒ q = 1 – q = \(\frac { 5 }{6}\) and n = 4 is given
The probability of success = \(\left[\begin{array}{l}
4 \\
x
\end{array}\right]\) [latex]\frac { 1 }{6}[/latex]x [latex]\frac { 5 }{6}[/latex]4-x
Therefore the probability of 2 success are
Samacheer Kalvi 12th Business Maths Guide Chapter 7 Probability Distributions Ex 7.1 21

Question 16.
The mean of a binomial distribution is 5 and standard deviation is 2. Determine the distribution.
Solution:
In a binomial distribution
mean np = 5 → (1)
Standard deviation \(\sqrt { npq}\) = 2
squaring on body sides
npq = 4 → (2)
Samacheer Kalvi 12th Business Maths Guide Chapter 7 Probability Distributions Ex 7.1 22
n = 5 × 5 ⇒ n = 25
∴ the binomial distribution is
P(X = x) = ncxpxqn-x
(i.e) p(X = x) = \(\left[\begin{array}{l}
25 \\
x
\end{array}\right]\) (\(\frac { 1 }{5}\))x (\(\frac { 4 }{5}\))(25-x)

Samacheer Kalvi 12th Business Maths Guide Chapter 7 Probability Distributions Ex 7.1

Question 17.
Determine the binomial distribution for which the mean is 4 and variance 3. Also find P(X = 15)
Solution:
In a binomial distribution
mean np = 4 → (1)
variance npq = 3 → (2)
Samacheer Kalvi 12th Business Maths Guide Chapter 7 Probability Distributions Ex 7.1 23
n = 4 × 4 ⇒ n = 16
The binomial distribution is p(X = x)ncxpxqn-x
Samacheer Kalvi 12th Business Maths Guide Chapter 7 Probability Distributions Ex 7.1 24

Question 18.
Assume that a drug causes a serious side effect at a rate of three patients per one hundred. What is the probability that atleast one person will have side effects in a random sample of ten patients taking the drug?
Solution:
Here n = 10
Samacheer Kalvi 12th Business Maths Guide Chapter 7 Probability Distributions Ex 7.1 25
= 0.5344

Samacheer Kalvi 12th Business Maths Guide Chapter 7 Probability Distributions Ex 7.1

Question 19.
Consider five mice from the same litter, all suffering from Vitamin A deficiency. They are fed a certain dose of carrots. The positive reaction means recovery from the disease.
Assume that the probability of recovery is 0.73. What is the probability that atleast 3 of the 5 mice recover.
Solution:
n = 5
Let probability of recovery p = 0.73
q = 1 – p = 1 – 0.73
∴ q = 0.27
The binomial distribution is
p (x = x) = nCxpxqn-x
p(x = x) = 5Cx(0.73)x(0.27)5-x
p(atleast 3 of the 5 mice recover) = p(x ≥ 3)
= p(x = 3) + p(x = 4) + p(x = 5)
= 5C3 (0.73)³(0.27)5-3 + 5C4 (0.73)4 (0.27)5-4 + 5C5 (0.73)5 (0.27)5-5
5C2 (0.73)³ (0.27)² + 5c1 (0.73)4 (0.27)1 + 5c0(0.73)5(0.27)°
[\(\frac { 5×4 }{1×2}\) × 0.389017 × 0.0729] + [5 × 0.28398241 × 0.27] + (1 × 0.2073071593 × 1)
= 0.283593393 + 0.3833762535 + 0.2073071593
= 0.2836 + 0.3834 + 0.2073
= 0.8743

Question 20.
Consider five mice from the same litter, all suffering from Vitamin A deficiency. They are fed a certain dose of carrots. The positive reaction means recovery from the disease. Assume that the probability of recovery is 0.73. What is the probability that atleast 3 of the 5 mice recover.
Solution:
Success = 2 × fails
p = 2q ⇒ p = 2(1 – p)
p = 2 – 2p ⇒ p + 2p = 2
3p = 2 and p = 2/3
q = 1 – p = 1 – 2/3
q = 1/3 and n = 5
The binomial destribution is
p (X = x) = nCxpxqn-x
= 5C(2/3)x (1/3)
(i) p(three successes) = p(x = 3)
Samacheer Kalvi 12th Business Maths Guide Chapter 7 Probability Distributions Ex 7.1 26

Samacheer Kalvi 12th Business Maths Guide Chapter 7 Probability Distributions Ex 7.1

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Bio Zoology Guide Pdf Chapter 10 Neural Control and Coordination Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

11th Bio Zoology Guide Neural Control and Coordination Text Book Back Questions and Answers

 

Part -I.

I. Choose The Best Options

Question 1.
Which structure in the ear converts pressure waves to action potentials?
a) Tympanic membrane
b) Organ of Corti
c) Oval window
d) Semicircular canal
Answer:
b) Organ of Corti

Question 2.
Which of the following pairings is correct
a) Sensory nerve – afferent
b) Motor nerve – afferent
c) Sensory nerve – ventral
d) Motor nerve – dorsal
Answer:
a) Sensory nerve – afferent

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 3.
During synaptic transmission of nerve impulse, neurotransmitter (P) is released from synaptic vesicles by the action of ions
(Q). Choose the correct P and Q.
a) P = Acetylcholine, Q = Ca++
b) P = Acetylcholine, Q = Na+
c) P = GABA, Q = Na+
d) P = Cholinesterase, Q = Ca++
Answer:
a) P = Acetylcholine, Q = Ca++

Question 4.
Examine the diagram of the two cell types A and B given below and select the correct option.
a) Cell – A is the rod cell found evenly all over retina
b) Cell – A is the cone cell more concentrated in the fovea centralis
c) Cell – B is concerned with colour vision in bright light
d) Cell – A is sensitive to bright light intensities
Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination 1
Answer:
c) Cell – B is concerned with colour vision in bright light

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 5.
Assertion- The imbalance in concentration of Na+, K+ and proteins generates action potential.
Reason: To maintain the unequal distribution of Na+ and K+ , the neurons use electrical energy.
a) Both Assertion and Reason are true and Reason is the correct explanation of the Assertion.
b) Both Assertion and Reason are true but the Reason is not the correct explanations of Assertion.
c) Assertion is true, but Reason is false.
d) Both Assertion and Reason are false.
Answer:
a) Both Assertion and Reason are true and Reason is the correct explanation of the Assertion.

Question 6.
Which part of the human brain is concerned with the regulation of body temperature?
a) Cerebellum
b) Cerebrum
c) Medulla oblongata
d) Hypothalamus
Answer:
d) Hypothalamus

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 7.
The respiratory centre is present in the
a) Medulla oblongata
b) Hypothalamus
c) Cerebellum
d) Thalamus
Answer:
a) Medulla oblongata

Question 8.
Match the following human spinal nerves in column I with their respective number in column II and choose the correct option
Column I Column 11
(P) Cervical nerves
(Q) Thoracic nerve
(R) Lumbar nerve
(S) Coccygeal nerve
(i) 5 pairs
(ii) 1 pair
(iii) 12 pairs
(iv) 8 pairs
a) (P-iv), (Q-iii), (R-i), (S-ii)
b) (P-iii), (Q-i), (R-ii), (S-iv)
c) (P-iv), (Q-i), (R-ii), (S-iii)
d) (P-ii), (Q-iv), (R-i), (S-iii)
Answer:
a) (P-iv), (Q-iii), (R-i), (S-ii)

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 9.
Which of the following cranial nerve controls the movement of eye ball?
a) Trochlear nerve
b) Optic nerve
c) Olfactory nerve
d) Vagus nerve
Answer:
a) Trochlear nerve

Question 10.
The abundant intracellular cation is
a) H+
b) K+
c) Na+
d) Ca++
Answer:
b) K+

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 11.
Which of the following statements is wrong regarding conduction of nerve impulse.
a) In a resting neuron, the axonal membrane is more permeable to K+ ions and nearly impermeable to Na+ ions.
b) Fluid outside the axon has a high concentration of Na+ ions and low concentration of K+, in a resting neuron.
c) Ionic gradient s are maintained by Na+ K+ pumps across the resting membrane, which transport 3 Na ions outwards for 2K+ into the cell.
d) A neuron is polarized only when the outer surface of the axonal membrane possess a negative charge and its inner surface is positively charged.
Answer:
b) Fluid outside the axon has a high concentration of Na+ ions and low concentration of K+, in a resting neuron.

Question 12.
All of the following are associated with the myeline sheath except
a) Faster conduction of nerve impulses
b) Nodes of Ranvier forming gaps along the axon
c) Increased energy output for nerve impulse conduction
d) Saltatory conduction of action potential
Answer:
c) Increased energy output for nerve impulse conduction

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 13.
Several statements are given here in reference to cone cells which of the following option indicates all correct statements for cone cells? Statements
(i) Cone cells are less sensitive in bright light than Rod cells
(ii) They are responsible for colour vision
(iii) Erythropsin is a photo pigment which is sensitive to red colour light
(iv) They are present in fovea of retina
a) (iii), (ii) and (i)
b) (ii), (iii) and (iv)
c) (i), (iii) and (iv)
d) (i), (ii) and (iv)
Answer:
c) (i), (iii) and (iv)

Question 14.
Which of the following statement concerning the somatic division of the peripheral neural system is incorrect?
a) Its pathways innervate skeletal muscles
b) Its pathways are usually voluntary
c) Some of its pathways are referred to as reflex arcs
d) Its pathways always involve four neurons
Answer:
d) Its pathways always involve four neurons

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 15.
When the potential across the axon membrane is more negative than the normal resting potential, the neuron is said to be in a state of
a) Depolarization
b)FIyperpo1ariation
c) Repolarization
d) Hypopolaiization
Answer:
c) Repolarization

Question 16.
Why is the blind spot called so?
Answer:
Slightly below the posterior pole of the eye, the optic nerve and the retinal blood vessels enter the eye. This region is devoid of rods and cones. Hence, this region is called blind spot.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 17.
Sam’s optometrist tells him that his intraocular pressure is high. What is this condition called and which fluid does it involve?
Answer:

  • The increase in intraocular pressure leads to the disease called Glaucoma.
  • Any block in the canal of Schlemm increases the intraocular pressure of aqueous humor and leads to ‘Glaucoma’ where the optic nerve and the retina are compressed due to pressure.

Question 18.
The action potential occurs in response to a threshold stimulus: but not at sub-threshold stimuli. what is the name of the principle involved?
Answer:
When we cry, the tears come out of the tear glands under the eyelids and drain through the tear duct that empties into the nose. It mixes with mucus there and the nose runs.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 19.
Pleasant smell of food urged Ravi to rush into the Kitchen. Name the parts of the brain involved in the identification of food and emotional responses to odour
Answer:
The hypothalamus contains a pair of small rounded body called mamillary bodies that are involved in olfactory reflexes and emotional responses to odour
This is also act as centre for appetite thirst and heat regulation.

Question 20.
Cornea transplant in humans is almost never rejected State the reason.
Answer:
The cornea does not have blood vessels. Hence there is no possibility of rejection when the cornea is transplanted from one person to another person.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 21.
At the end of repolarisation, the nerve membrane gets hyperpolarized Why?
Answer:
If repolarization becomes more negative than the resting potential – 70mV to about – 90mV it is called hyperpolarization.

Question 22.
Label the parts of the neuron.
Answer:
Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination 2
A- Nucleus with nucleolus
B- Axolemma
C- Dendrites
D- Myelin sheath, Nucleus
E- Axon (Direction of signal transmission)
F- Node of Ranvier.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 23.
The choroid plexus secretes cerebral spinal fluid List the function it.
Answer:
Cerebrospinal fluid provides buoyancy to the central nervous system.

  1. It acts as a shock absorber for the brain and spinal cord.
  2. It nourishes the brain cells by transporting food and oxygen.
  3. It carries harmful metabolic wastes from the brain to the blood.
  4. It maintains constant pressure inside the cranial vessels.

Question 24.
What is the ANS controlling center? Name the parts that are supplied by the ANS?
Answer:
Controlling center – autonomic neural system. Controlling organs.

  1. Eyes
  2. Salivary Glands
  3. Heart
  4. Lungs
  5. Stomach
  6. Liver
  7. Kidney
  8. Intestines
  9. Bladder

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 25.
Why the limbic system is called the emotional brain. Name the parts of it?
Answer:
The limbic system is a set of components located on both sides of the thalamus present in the inner part of the cerebral hemisphere. It includes the olfactory bulbs, cingulate gyrus, mammillary body, amygdala, hippocampus and hypothalamus. The limbic system plays a primary role in the regulation of pleasure, pain, anger, fear, sexual feeling, affection and memory. Hence it is called the emotional brain.

Question 26.
Classify receptors based on the type of stimuli.
Answer:

ReceptorsStimulusEffector organs
Mechano receptorsPressure and vibrationMechano receptors are present in the cochlea of the inner ear and the semicircular canal and utriculus
ChemoreceptorsChemicalsTaste buds in the tongue and nasal epithelium
Thermo receptorsTemperatureSkin
PhotoreceptorsLightRod and cone cells of the retina in the eye.

Question 27.
Name the first five cranial nerves, their nature and their functions
Answer:

Cranial nervesNature of nerveFunction
I. Olfactory nerveSensorySense of smell
II.Optic nervesSensorySense of sight
III. Oculo motor nervesMotorMovement of the eye
IV. Trochlear nerveMotorRotation of the eyeball
V.Trigeminal nerveSensory and motor mixedThe functioning of face ball

Question 28.
The sense of taste is considered to be the most pleasurable of all senses. Describe the structure of the receptor involved with a diagram.
Answer:

  • The sense of taste is considered to be the most pleasurable of all senses.
  • The tongue is provided with many small projections called papillae.
  • Taste buds are located mainly on the papillae.
  • Taste buds are flask-shaped.

There are two major types.

  • Gustatory epithelial cells or taste cells.
  • Basal epithelial cells or repairing cells.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination 3

  • Long microvilli called gustatory hairs project from the tip of the gustatory cells and extend through a taste pore to the surface of the epithelium.
  • Gustatory hairs are the sensitive portion of the gustatory cells and they have sensory dendrites which send the signal to the brain.
  • The basal cells that act as stem cells divide and differentiate into new gustatory cells.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 29.
Describe the structures of olfactory receptors?
Answer:
The smell receptors are excited by air-borne chemicals that dissolve in fluids. The yellow coloured patches of olfactory epithelium form the olfactory organs that are located on the roof of the nasal cavity.

The olfactory epithelium is covered by a thin coat of mucus layer below and olfactory glands bounded connective tissues, above. It contains three types of cells: supporting cells, Basal cells and millions of pin-shaped olfactory receptor cells (which are unusual bipolar cells).

The olfactory glands and the supporting cells secrete the mucus. The unmyelinated axons of the olfactory receptor cells are gathered to form the filaments of olfactory nerve [cranial nerve-I] which synapse with cells of olfactory bulb.

The impulse, through the olfactory nerves, is transmitted to the frontal lobe of the brain for identification of smell and the limbic system for the emotional responses to odour.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination 4

Part-II.

11th Bio Zoology Guide Neural Control and Coordination Additional Important Questions and Answers

I. Choose The Correct Answer

Question 1.
What is the functional unit of the nervous system?
a) Neuroglial cells
b) Neuron
c) Nephron
d) Axon
Answer:
b) Neuron

Question 2.
Name the nutritive cells of neuron.
a) Axon
b) Neuroglia
c) Schwann’s cell
d) Epithelial cells.
Answer:
b) Neuroglia

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 3.
Name the cell organelle which is not seen in the nerve cell.
a) Mitochondria
b) Golgi apparatus
c) Centrioles
d) Nucleus
Answer:
c) Centrioles

Question 4.
Name the plasma membrane which surrounds the neurons?
a) Neurilemma
b) Axolemma
c) Myalin membrane
d) Sarcolemma
Answer:
a) Neurilemma

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 5.
Name the plasma membrane surrounds the axon?
a) Neurilemma
b) Myalin membrane
c) Sarcolemma
d) Axolemma
Answer:
d) Axolemma

Question 6.
Where is Nissl’s bodies seen in the nerve cell?
a) Cell body and dendrites
b) Cell body and axon
c) Cell body and myalin sheath
d) Cell body and end plate of axon.
Answer:
a) Cell body and dendrites

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 7.
Name the cell organell which is not seen in the axons.
a) Mitochondria
b) Golgi apparatus
c) Centriole
d) Endoplasmic reticulum
Answer:
b) Golgi apparatus

Question 8.
Cell forms the myalin sheath ……………..
a) Schwaan’s cell
b) Neuroglial cells
c) Nephron
d) Epithelial cells
Answer:
a) Schwaan’s cell

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 9.
Which part of the nerve cells do not contain myalin sheath.
a) Axon
b) Cell body
c) Dentrites
d) Axon end plate
Answer:
c) Dentrites

Question 10.
Which of the following does the work of phagocytosis of microbes during the infection of the brain
a) Schwann cells
b) Node of Ranvier
c) Neuroglial cells
d) Neurilemma
Answer:
c) Neuroglial cells

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 11.
Find out the correct statement
a) The short nerve fibres are called as dentrites.
b) The membrane surrounds the neuron is axolemma.
c) The longest sciatic nerve runs from the base of the spine to the big toe of each foot
d) Schwann cell’s do not synthesize myalin sheath
Answer:
c) The longest sciatic nerve runs from the base of the spine to the big toe of each foot

Question 12.
Find out the wrong pair
a) Multi polar neuron: One axon and two or more dentrites.
b) Bi polar neurons: One axon and more than two dentrites.
c) Uni polar neurons: Short process axon.
d) Synaptic Knob: Neurotransmitters
Answer:
b) Bi polar neurons: One axon and more than two dentrites.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 13.
This substance is more in the tissue fluid of cytoplasm of axolemma?
a) Sodium chloride and bicarbonates.
b) Nutritious substances and oxygen.
c) Potassium and magnesium phosphate.
d) All the above.
Answer:
c) Potassium and magnesium phosphate.

Question 14.
Match the following and find the answer.

1. Leakage channelsa. Opens through the sensation of touch and pressure.
2. Ligand-gated channelsb. They are stable.
3. Voltage gated channelsc. Opens through chemical stimulation.
4. Charged particlesd. Ionic channels are always open.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination 5
Answer:
c) i- B,ii – D,iii – C,iv – A

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 15.
Name the gaps in the myelin sheath between adjacent Schwann.
a) Nodes of Ranvier
b) Nodes of axon
c) Nodes of cyton
d) Nodes of dentrites
Answer:
a) Nodes of Ranvier

Question 16.
Where are bipolar neurons situated?
a) Cranial nerves
b) Olfactory lobes of brain
c) Spinal nerves
d) Skin
Answer:
b) Olfactory lobes of brain

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 17.
Where are Bipolar neurons situated?
a) Spinal cord
b) Retina
c) Inner ear
d) Brain.
Answer:
a) Spinal cord

Question 18.
Find out the odd one.
a) Schwann cell – Myalin sheath
b) Synaptic knob – Synaptic vesicles
c) Bipolar neuron – Cranial nerves
d) Multipolar neuron – Spinal nerves.
Answer:
c) Bipolar neuron – Cranial nerves

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 19.
Where is Bipolar neurons meet?
a) Synapses
b) Synaptic cleft
c) Synaptic vesicle
d) Synaptic Knob
Answer:
a) Synapses

Question 20.
Match and find the correct sequence.

I. Neuro junctiona. Neuron which receives a stimulus
II. Presynaptic neuronsb. Synaptic vesicles
III. Synaptic cleftc. Synapses
IV. Neurotransmitterd. Pre and post neuron junction

a) I-d II-a ¡II-c IV-d
b) I-a II-c III-d IV-b
c) I-a II-b III-c IV-d
d) I-c II-a III-d IV-b
Answer:
d) I-c II-a III-d IV-b

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 21.
Central nervous system forms from this layer during embryonic development.
a) Endoderm
b) Ectoderm
c) Mesoderm
d) Middle layer.
Answer:
b) Ectoderm

Question 22.
Find out whether the following statements are true or false. Find out the correct sequence.
i. The thick outer covering of the brain is dura matter
ii. The space between pia matter and dura matter is subarachnoid space.
iii. The membrane that surrounds the brain is piamalter
iv. The space between arachnoid and pia mater is subarachnoid space
a) I – False; II – False; III – True; IV – True
b) I-True; II-False; III-True; IV-True
c) I – True; II – False; III – False; IV – True
d) I-True; II-False; III-True; IV-False
Answer:
b) I-True; II-False; III-True; IV-True

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 23.
Which is considered as the seat of intelligence.
a) Cerebellum
b) Cerebrum
c) Medulla oblongata
d) Pons.
Answer:
b) Cerebrum

Question 24.
Find out whether the following statements are true or false.
a) At resting membrane potential the efflux of potassium is more than the influx of sodium from the outer surface of the neurilemma.
b) When the stimulus is sent from the cell body to the axon there is no creation of action potential.
c) The spike potential is +45mV.
d) The threshold potential is +55mV.
i) a) True b) True c) False d) False
ii) a) False b) True c) True d) True
iii) a) True b) False c) True d) True
iv) a) True b) False c) True d) False
Answer:
iv) a) True b) False c) True d) False

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 25.
Find out the wrong pair
a) Synapses – The junction of two neurons
b) Neurotransmitter – Postsynaptic neuron
c) Synaptic vesicles – A small bag filled with chemicals
d) Piamatter – Membrane which closely adheres to the brain.
Answer:
b) Neurotransmitter – Postsynaptic neuron

Question 26.
………………. is the part of the fore-brain.
a) Cerebrum – Mesencephalon
b) Cerebrum – Prosencephalon
c) Cerebrum – Diencephalon
d) Cerebrum – Parietal lobe.
Answer:
c) Cerebrum – Diencephalon

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 27.
…………………….. is useful in diagnosis of neurological and sleep disorders.
a) PET
b) CT
c) X – rays
d) EEG
Answer:
d) EEG

Question 28.
Name the structure which connects cerebral hemisphere
a) Choroid tissue
b) Corpus callosum
c) Neuroglia tissue
d) Pineal body
Answer:
b) Corpus callosum

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 29.
Match and find out the correct sequence.

I. Frontal lobea. Reading
II. Parietal lobeb. Hearing
III. Temporal lobec. Vision
IV. Occipital lobed. Memory

a) I-d II-a III-c IV-b
b) I-a II-c III-d IV-b
c) I-a II-b III-c IV-d
d) I-d II-a III-b IV-c
Answer:
d) I-d II-a III-b IV-c

Question 30.
Assertion: Statement A- The cortex of cerebrum consists of grey matter.
Statement B- The reason for the grey matter is due to the presence of non – myelinated sheath.
a) Statement A- True, Statement B – False
b) Statement A and B are true
c) Statement A- False the statement B- True.
d) Statement A and B are false.
Answer:
b) Statement A and B are true

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 31.
Match and find the correct answer.
1. Broca area a. Perception of taste.
2. The pre-central gyrus b. Voluntary movements
3. The postcentral gyrus c. Aggression
4. Amygdala d. Speech function
a) I-a II-b III-c IV-d
b) I-d II-b III-c IV-a
c) I-d II-b Ill-a IV-c
d) I-a II-b III-c IV-d
Answer:
c) I-d II-b III-a IV-c

Question 32.
Assertion: Statement S – Brain controls emotions and feelings.
Statement T – There are no sensory neurons in the brain. Hence it cannot sense the pain.
a) Statement – S – True. Statement T – False
b) Statement – S – True Statement T – True
c) Statement – S – False Statement T – False
d) Statement S – False Statement T – True
Answer:
b) Statement – S – True Statement T – True

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 33.
Which is considered as the relay centre for impulses
a) Medulla oblongata
b) Cerebrum
c) Hypothalamus
d) Thalamus
Answer:
d) Thalamus

Question 34.
Which indicates A, B, C and D in the following diagram
Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination 6
(a)
a) Temporal lobe
b) Frontal lobe
c) Parietal lobe
d) Cerebellum
(b)
a) Cerebellum
b) Parietal lobe
c) Temporal lobe
d) Frontal lobe
(c)
a) Cerebellum
b) Frontal lobe
c) Temporal lobe
d) Parietal lobe
(d)
a) Parietal lobe
b) Cerebellum
c) Frontal lobe
d) Temporal lobe
Answer:
(b)
a) Cerebellum
b) Parietal lobe
c) Temporal lobe
d) Frontal lobe

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 35.
Find out the wrong statement.
a) The pre-central gyrus is called a motor strip.
b) The post-central gyrus is called a sensory strip.
c) The medulla oblongata controls respiration.
d) Wernicke area of the brain involved in the comprehension of speech.
Answer:
d) Wernicke area of the brain involved in the comprehension of speech.

Question 36.
The melatonin hormone which controls the sleep-wake cycle is secreted by
a) Pineal body
b) Thalamus
c) Hypothalamus
d) Medulla oblongata
Answer:
a) Pineal body

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 37.
The lower portion of the midbrain consists of a pair of longitudinal nervous bands called
a) Emotional brain
b) Cerebral peduncles
c) Pons
d) Vermis
Answer:
b) Cerebral peduncles

Question 38.
The cauda equina presents in
a) Cerebrum
b) Thalamus
c) Hippo cambus
d) Spinal cord
Answer:
d) Spinal cord

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 39.
Match and find the correct answer.
i) Thalamus – A) Respiration Gastric Pits.
ii) Hypothalamus – B) Vision Hearing
iii) Mid brain – C) Satiety centre
iv) Medulla oblongata – D) Learning memory
Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination 7
Answer:
a) i- D,ii – C,iii – B,iv – A

Question 40.
The cluster of nerve tissue in the cell body of neurons are called as
a) Gyri
b) Sulci
c) Vermis
d) Nerve ganglion.
Answer:
d) Nerve ganglion.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 41.
Find out the wrong pair
a) Cervical nerve – 8 pairs
b) Thoracic nerve – 12 pairs
c) Sacrum nerve – 4 pairs
d) Coccyx nerve – 1 pair
Answer:
c) Sacrum nerve – 4 pairs

Question 42.
What is the amount of cerebrospinal fluid secreted in an adult?
a) 500 ml
b) 150 ml
c) 400 ml
d) 250 ml
Answer:
b) 150 ml

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 43.
The dissolved nicotene and alcohol in …………………… and ………………… minutes reach the brain.
a) 7 and 8 minutes
b) 6 and 7 minutes
c) 7 seconds and 6 minutes
d) 7 and 5 minutes.
Answer:
c) 7 seconds and 6 minutes

Question 44.
Name the structure that regulates homeostasis.
a) Mammillary body
b) Pineal body
c) Hypothalamus
d) Pituitary.
Answer:
a) Mammillary body

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 45.
What is brain stem.
a) The structure seen in between spinal cord and cerebellum.
b) The region between Diencephalon and spinal cord
c) The region between medulla oblongata and cerebrum.
d) The region between medulla oblongata and mid-brain.
Answer:
b) The region between Diencephalon and spinal cord

Question 46.
What is the function of corpora quadri gemina?
a) Vision and hearing
b) Vision and homeostasis
c) Hearing and sense of touch
d) Hearing and vomiting.
Answer:
a) Vision and hearing

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 47.
The region seen in the mid of cerebellum.
a) Corpus callosum
b) Pineal body
c) Pituitary
d) Vermis.
Answer:
d) Vermis.

Question 48.
What is the function of the interneuron?
a) Impulses from afferent and efferent neurons.
b) Impulses transmitted from efferent neurons to afferent neurons.
c) Impulses transmitted from central nervous system to receptor organs
d) From nerve transmitters to the spinal cord
Answer:
a) Impulses from afferent and efferent neurons.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 49.
Which part of the brain controls the unconditioned reflux?
a) The cortex of cerebrum
b) The medulla of brain
c) Cerebellum
d) pons.
Answer:
a) The cortex of cerebrum

Question 50.
……………………..pair of nerves from cranium and
…………………….. pair of nerves from spinal cord
a) 10 and 31
b) 12 and 31
c) 31 and 12
d) 31 and 10.
Answer:
b) 12 and 31

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 51.
How much cerebrospinal fluid is synthesized in adults?
a) 200ml
b) 300ml
c) 150ml
d) 100ml.
Answer:
c) 150ml

Question 52.
What is the function of vagus nerve?
a) It regulates the function of abdominal structure
b) It regulates the speech and swallowing
c) It regulates the homeostasis of the body
d) Taste perception
Answer:
a) It regulates the function of abdominal structure

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 53.
Name the neural system which is auto functioning and self-governed.
a) Somatic neural system
b) Automatic neural system
c) Peripheral neural system
d) Limbic system.
Answer:
b) Automatic neural system

Question 54.
Name the structure that regulates the autonomic neural system.
a) Cerebrum
b) Cerebellum
c) Hypothalamus
d) Pons.
Answer:
c) Hypothalamus

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 55.
Name the lubricating fluid which is secreted from the gland present at the base of the eyelashes.
a) Lacrymal gland
b) Cerebro-spinal fluid
c) Thymus gland
d) Sebaceous glands.
Answer:
d) Sebaceous glands.

Question 56.
Find the ABC and D in the diagram
Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination 8
a) a) Tear duct b) Sciera c) Iris d) Tear gland
b) a) Sciera b) Iris c) Tear gland d)Tear duct
c) a) Tear duct b) Iris c) Sclerad d)lTear duct
d) a) Tear gland b) Sclera c) Tear duct d) Iris.
Answer:
d) a) Tear gland b) Sclera c) Tear duct d) Iris.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 57.
Name the tissue which forms the non-vascular transparent coat cornea
a) Stratified squamous epithelium
b) Squamous epithelium
c) Canal of schlemm
d) All the above.
Answer:
a) Stratified squamous epithelium

Question 58.
Assertion: In bright light the circular muscle in the iris contract so that the size of the pupil decreases and the light enter is regulated Reason: In dim light, the radial muscle contract the pupil size increases and less light enters the eye.
a) The assertion true: but reason wrong
b) The assertion and reason are true
c) The assertion is true but reason does not explain the assertion
d) The assertion and reason are wrong.
Answer:
a) The assertion true: but reason wrong

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 59.
Name the yellow flat spot at the centre of the posterior region of the retina.
a) Blind spot
b) Scelera
c) Iris
d) Maculalutea
Answer:
d) Maculalutea

Question 60.
Which colour is perceived through chloropsin is sensitive to the medium wavelength of 530 mm
a) Red cones
b) Green cones
c) Blue cones
d) Red cells.
Answer:
b) Green cones

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 61.
Name the glands that secrete tear.
a) Lacrymal gland
b) Sebaceous gland
c) Mandibular gland
d) Choroid plexues
Answer:
a) Lacrymal gland

Question 62.
…………………… enzyme is present in the tear.
a) Ptyalin
b) Lysozyme
c) Lipase
d) Erypsin.
Answer:
b) Lysozyme

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 63.
Name the structure where more cones present.
a) Foveacenlralis
b) Maculalutea
c) Retina
d) Blind spot.
Answer:
a) Foveacenlralis

Question 64.
……………………………….. people can see the nearby object but cannot see the distant object.
a) Astigmatism
b) Presbiopia
c) Hypermetropia
d) Myopia
Answer:
d) Myopia

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 65.
The concave lens helps in correct ………………. defect
a) Myopia
b) Hyper metropia
c) Astigmatism
d) Presbiopia
Answer:
a) Myopia

Question 66.
Which of the defect is depicted in the following diagram:
Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination 9
a) Emmetropia
b) Myopia
c) Hypermetropia
d) Astigmatism.
Answer:
b) Myopia

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 67.
Match and find the correct sequence.

I. ConcaveSurgery
II. ConvexHypermetropia
III. CataractAstigmatism
IV. Cylindrical lensMyopia

a) 1 -a 2-b 3-c 4-c
b) 1 -d 2-b 3-a 4-c
c) 1-d 2-a 3-a 4-b
d) 1-d 2-c 3-b 4-c
Answer:
b) 1 -d 2-b 3-a 4-c

Question 68.
Name the organ which regulates vision perception and homeostasis.
a) Ear
b) Eye
c) Cerebrum
d) Cerebellum.
Answer:
a) Ear

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 69.
Find out the wrong pair.
a) Pinna – Eardrum
b) Oval window – Round window
c) Cochlea – Malleus
d) Perilymph – Endolymph
Answer:
c) Cochlea – Malleus

Question 70.
Indicate ABC and D in the Diagram.
Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination 10
a)
a) lncus
b) Eardrum
c) Organ of Corti
d) Tactoreal membrane.
b)
a) Organ of cortri
b) Incus
c) Tectoreal membrane
d)Eardrum
c)
a) Eardrum
b) Incus
c) Organ of Corti
d) Tectoreal membrane
d)
a) Eardrum
b) Incus
c) Tectoreal membrane
d) Organ of Corti
Answer:
d)
a) Eardrum
b) Incus
c) Tectoreal membrane
d) Organ of Corti

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 71.
Find out the wrong statement.
a) The equilibrium receptor regions called maculae are involved in detecting the linear movement of the head.
b) The otoliths are made up of calcium
c) The crista ampullaris is to detect rotational movement of the head
d) Severe hearing loss occurs with frequent exposure to sound with intensities greater than50db.
(The 90db will affect the hearing).
Answer:
d) Severe hearing loss occurs with frequent exposure to sound with intensities greater than50db.

Question 72.
A person on his way to a village in a car for weekend holidays after finishing the office work. As he is very tired he begins to feel drowsy. He turns up the car stereo volume opens the car window and has sips ice-cold water.
How do these actions keep him awake? The increase in the number of sensory stimuli he received is relayed to the cerebral cortex which gets activated and prevents sleeping,
a) Spinal cord
b) Cerebrum
c) Medulla oblongata
d) Cerebellum.
Answer:
c) Medulla oblongata

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 73.
Name the structure that connects the middle ear and pharynx.
a) Eustachian tube
b) Middle ear
c) Oval window
d) Organ of the cortii.
Answer:
a) Eustachian tube

Question 74.
Which part of the ear have bony labyrinth and membranous labyrinth?
a) External ear
b) Inner ear
c) Middle ear
d) None of the above.
Answer:
b) Inner ear

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 75.
Name the structure present in the bony labyrinth.
a) Cochlea, vestibule, maleus
b) Cochlea, vestibule Incus
c) Cochlea semicircular canals, stapes
d) Cochlea, vestibule semicircular canals.
Answer:
d) Cochlea, vestibule semicircular canals.

Question 76.
Where is the stereo cilia cells present in the ear?
a) Scala media
b) Scala vestibuli
c) Scala tymnani
d) Middle ear.
Answer:
a) Scala media

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 77.
Match and find the correct sequence.

I. Meissner’s corpusclesa) continuous pressure
II. Pacinian corpusclesb) hardness, pain
III. Ruffini endingsc) temperature
IV. Krause and bulbsd) light pressure

a) I-d, II-a, III-c, IV-d
b) I-d, II-b, III-c, V-a
c) I-d, II-b, Ill-a, IV-c
d) I-a, II-b, III-c, IV-d
Answer:
c) I-d, II-b, Ill-a, IV-c

Question 78.
When a cockroach tries to enter into the ear of a sleeping person, which one of the following process will start?
a) Unconditioned reflex
b) Neuro muscular fatigue
c) Stimulation of negative feedback mechanism
d) conditioned reflex
Answer:
a) Unconditioned reflex

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 79.
What is the number of nerve cells present in the brain
a) 90 million
b) 100 billion
c) 90 billion
d) 100 million
Answer:
b) 100 billion

(2 Marks)

II. Write Very Short Answer

Question 1.
What are the two branches of human nervous system?
Answer:

  1. Central nervous system
  2. Peripheral nervous system

Question 2.
What is meant by Neuroglia? What are its functions?
Answer:
The non-nervous special cells called neuroglia from the supporting cells of the nervous tissue.

  1. Providing nourishment to the neurons
  2. Repairing the injured tissues
  3. It helps in dividing and regeneration
  4. Acting as phagocyte cells to engulf the foreign particles at the time of any injury to the brain.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 3.
Differentiate the functions of myelinated and non-myelinated neurons.
Answer:

Myelinated neuronNon-myelinated neuron
They conduct impulses rapidlyConduct impulse slowly
Present in the medullaSeen in the cortex of the cerebrum.

Question 4.
Give notes on (a) Synaptic Knob (b) Neurotransmitters (c) Inter neural space.
Answer:

  • Synaptic Knob: Distant end of the axon terminates into a bulb
  • Synaptic vesicles: Vesicles filled with neurotransmitters
  • Inter neural space – The space between presynaptic and postsynaptic neurons.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 5.
What is meant by nerve impulses?
Answer:
A nerve impulse is a series of electrical impulses which travel along with the nerve fibre.

Question 6.
What is meant by resting potential?
Answer:
The electrical potential difference across the plasma membrane of a resting neuron is called the resting potential.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 7.
What is meant by threshold potential?
Answer:
During depolarization when enough Na+ ions enter the cell the action potential reaches a certain level called threshold potential. (-55 mV)

Question 8.
What is threshold stimulus?
Answer:
The particular stimulus which is able to bring the membrane potential to the threshold is called the threshold stimulus.

Question 9.
What is meant by all or none principle?
Answer:
The action potential occurs in response to a threshold stimulus but does not occur at sub-threshold stimuli. This is called the all or none principle.

Question 10.
What is the cause of brain tumours?
Answer:

  • Glial cells: Nerve cells do not divide but glial cells do not lose the ability to undergo cell division.
  • So most brain tumours of neural origin consist of glial cells.

Question 11.
What are meninges?
Answer:
The brain is covered by outer Duramater, the median Arachnoid mater and the inner Piamater. These membranes are called meninges.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 12.
What is meant by lazy gate?
Answer:
During hyper polarization the K+ion gates are more permeable to K+ even after reaching threshold level it closes slowly hence called lazy gates.

Question 13.
What is spike potential?
Answer:
Due to the rapid influx of Na+ ions the membrane potential shoots rapidly up +45 mV which is called the spike potential.

Question 14.
What is meant by repolarisation?
Answer:
When the membrane reaches the spike potential the potassium ions K+ efflux from the axolemma and hence form negative potential. This is called repolarisation.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 15.
What is Hyper polarisation?
Answer:
If repolarisation becomes more negative than the resting potential -70 mV to about -90 mV it is called hyper polarisation.

Question 16.
Which decides the speed of the conduction of a nerve impulse?
Answer:

  • The conduction speed of a nerve impulse depends on the diameter of axon.
  • The greater the axon’s diameter the faster is the conduction.

Question 17.
What is septum pellucidum?
Answer:
A thin membrane which separates the lateral ventricles I and II is called the septum pellucidSamacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 18.
What is the difference between sub durai space and sub arachnoid space
Answer:

Sub durai spaceSub arachnoid space
A narrow sub dural space present in between arachnoid and durameterThe space between arachnoid membrane and piameter.

Question 19.
What is meant by sulcus and gyrus? What is its functions.
Answer:

  • Gyri – The convolution seen in the cerebrum.
  • Sulci – The shallow grooves increase the surface area of the cerebral cortex.

Question 20.
Name the lobes of cerebrum?
Answer:

  1. Pair of frontal
  2. Pair of parietal
  3. Pair of temporal
  4. Occipital

Question 21.
What is the function of proca?
Answer:
It involves in speech function.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 22.
What is meant by a blood brain barrier.
Answer:
It protects the brain by preventing many foreign substances in our vascular system from reaching the brain.

Question 23.
Which part of the cerebrum is called as sheet of emotions?
Answer:
The cortex of parietal occipital and temporal.

Question 24.
What is meant by motor area?
Answer:

  • The area which controls the voluntary muscular movements lies in the posterior part of the frontal lobes.
  • Functions They receive and interpret the sensory impulses.

Question 25.
What is the function of association area?
Answer:
They deal with memory communications learning and reasoning.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 26.
Where is pineal body situated? What is its functions?
Answer:

It is situated behind the choroid plexus the epithalamus forms short stalk
Functions
The melotoxin secreted by the pineal body regulates sleep wake cycle.

Question 27.
For a man to live all parts of the brain is important. How is brain divided into?
Answer:

  • Cerebrum
  • Thalamus
  • Hypothalamus
  • Pons
  • Cerebellum
  • Medulla oblongata

Question 28.
What is meant by corpus callosum?
Answer:
The two hemispheres of cerebrum is connected with the nerve band known as corpus callosum.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 29.
What is meant by brain stem? What are its parts?
Answer:
The region between spinal cord and diencephalon.

  1. Mid-brain
  2. Pons varoli
  3. Medulla oblongata

Question 30.
What is meant by corpora quadri gemina? What are its functions.
Answer:

  • The four papillae seen on the dorsal side of mid brain.
  • Perception of vision and hearing

Question 31.
What is meant by reflex arc?
Answer:

  • It is a fast involuntary unplanned sequence of actions that occurs in response to a particular stimulus.
  • The nervous elements that involved in carrying out reflex action constitute reflex arc.

Question 32.
What are the functions of cerebellum?
Answer:
The cerebellum controls and co-ordinates muscular movements and body equilibrium.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 33.
What is meant by receptors?
Answer:
The organ which made us aware of changes occur in our surrounding is called receptors.

Question 34.
What is meant by voluntary neural system?
Answer:
It is the part of the peripheral neural system associated with the voluntary control of body movements via skeletal muscles

Question 35.
Name the bacteriocidal substance present in the tears? Where is it seen? Name its secretion?
Answer:

  • Lysosome enzyme
  • Seen in tears.
  • Lacrymal glands. 1ml is secreted in a day.

Question 36.
What are mammillary bodies? What are its functions?
Answer:

  • The hypothalamus contains a pair of small rounded body called mammillary bodies.
  • Functions: It is involved in olfactory reflexes and emotional responses to odour.

Question 37.
What are cerebral peduncles?
Answer:
The region between diencephalon and the pons is called midbrain. It consists of a pair of longitudinal bands of nervous tissue called cerebral peduncles.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 38.
What is brain stem? What are its parts?
Answer:
It is the part of the brain between the spinal cord and the diencephalon.
Organs: Mid-brain, Pons, Medulla oblongata

Question 39.
What is caudo equina?
Answer:

  • After the 2nd lumbar vertebra the spinal nerves are greatly elongated
  • The thick bundle of elongated nerve roots appears as a horse’s tail called as cauda equina.

Question 40.
Where is enlargement region seen in the spinal cord?
Answer:

  • Cervical enlargement
  • Lumbar enlargement.

Question 41.
What is meant by reflex action?
Answer:

  • When a very quick response is needed the spinal cord can effect motor initiation as the brain and brings about an effect.
  • This rapid action by spinal cord is called reflex action.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 42.
What is meant by peripheral nervous system?
Answer:
All nervous tissue outside the CNS is the nervous system.

Question 43.
What is meant by mixed nerve?
Answer:
Each spinal nerve is a mixed nerve containing both afferent (sensory) and efferent (motor) fibers.

Question 44.
What is meant by pre ganglionic neuron?
Answer:
Its cell body is in the brain or spinal card. Its myelinated axon exits the CNS as part of cranial or spinal nerve and end in an autonomic ganglion.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 45.
What is autonomic ganglion?
Answer:
The axons of pre ganglionic neuron and cell bodies of post ganglionic neuron includes autonomic ganglion.

Question 46.
What is post ganglionic neuron?
Answer:
It conveys nerve impulses from autonomic ganglia to visceral effector organs.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 47.
What are the two branches of autonomous nervous system?
Answer:

  1. Sympathetic neural system
  2. Parasympathetic neural system.
  3. varying distances is called accommodation.

Question 48.
Give short notes on Lacrymal glands?
Answer:

  • The tear secreting glands are known as Lacrymal glands.
  • 1ml of tear is secreted in a day.
  • Tear salts, mucous and lysozyme enzyme to destroy bacteria.

Question 49.
What is conjunctiva?
Answer:
It is a thin protective mucous membrane found lining the outer surface of the eye ball.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 50.
Dilation and congestion of the blood vessels due to local irritation or infections. What is the disease? When will it occur?
Answer:

  • Conjunctivitis or madras eye.
  • Irritation or infection are the cause of blood shot eye.

Question 51.
What is sty?
Answer:
Infection of ciliary glands by bacteria causes a painful pus filled swelling called a sty.

Question 52.
What are the three types of cell present on the neural retina?
Answer:

  • Photo receptor cell
  • Cones
  • Rods
  • Bipolar and ganglion cells.

Question 53.
What is cataract?
Answer:
Due to the changes in nature of protein the lens become opaque called cataract.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 54.
What is aqueous humour and vitreous humour
Answer:

  • Aqueous humour – The fluid filled in between the cornea and iris.
  • Vitreous humour – Fluid filled in between lens and retina.

Question 55.
Give notes on lens fibres?
Answer:
Eye lens is transparent and biconvex made up of long columnar epithelial cells called lens fibre, these cells are formed of crystalline protein.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 56.
What are the three layers of eye ball?
Answer:

  1. Sclera – Outer coat
  2. Choroid – Pigmented middle layer
  3. Retina – Inner most layer

Question 58.
What is meant by maculae lutea?
Answer:
The yellow flat spot at the centre of the posterior region of the retina is called macula lutea. It is responsible for sharp detailed vision.

Question 59.
What is fovea centralis?
Answer:
A small depression present in the centre of the yellow spot is called fovea centralis which contains only cones.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 60.
What is blind spot?
Answer:
The optic nerves and the retinal blood vessels enter the eye slightly below the posterior pole whichis devoid of photoreceptors is called blind spot.

Question 61.
Why there is no cornea rejection during cornea transplantation?
Answer:

  • There is little or no possibility of cornea rejection.
  • This is because cornea does not have blood vessels.

Question 62.
What are the three ear ossicles?
Answer:

  1. Malleus – hammer bone
  2. Incus – anvil bone
  3. Stapes – stirrup bone

Question 63.
What are the three chambers of cochlea?
Answer:

  1. Scala vestibule
  2. Scala tympani
  3. Scala media

Question 64.
What is tectorial membrane?
Answer:
A roof like structure over hanging the organ of corti through out its length with the stiff gel membrane called tectorial membrane.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 65.
Name the receptors which are excited by airborne chemicals that dissolve in fluids?
Answer:
The receptors for taste and smell are the chemo receptors

Question 66.
What is tactile merkel disc?
Answer:
It is the light touch receptor lying in the deeper layer of epidermis.

Question 67.
What are the structures present in the membranous labyrinth?
Answer:

  • Cochlea
  • Vestibule
  • Semi circular canals

Question 68.
What are the three chambers of cochlea? Name the membrane which separates these chambers?
Answer:

  1. Scala vestibuli
  2. Scala tympanii
  3. Scala media

Separating membranes
Reisner’s membrane. It separates scala vestibuli from scala media
Basilar membrane. It separates scala media from scala tympani.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 69.
What is meant by stereo cilia?
Answer:

  • The organ of corti contains numerous hair cells on the basilar membrane.
  • Protruding from the apical part of each hair cell is hair like structures called stereo cilia.

Question 70.
What is proprioception?
Answer:
The ability to provide information about position and movements of the body is proprioception.

Question 71.
What is meant by vestibular system?
Answer:
This is the organ of balance this system is composed of fluid filled sacs and tubules.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 72.
Give notes on utriculus and sacculus?
Answer:
The utricle and saccule contain equilibrium receptor regions called maculae that are involved in detecting the linear movement of the head.

Question 73.
What is otolith?
Answer:

  • Maculae contains hair cells. These hair cells contain calcareous particles called otoliths.
  • It increases the inertia.

Question 74.
What are ampullae?
Answer:
At one end of each semicircular canal at its lower end has a swollen area called ampulla Each ampulla has sensory hair cells and supporting cells called crista ampullaris. The function is to detect rotational movement of the head.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 75.
Give the name of balancing structures which perform the following action.
a) Linear movement of the head
b) Changes occurring in body position.
c) Rotational movement of the head
Answer:

  • Maculae
  • Vestibular system
  • Crista ampularis

Question 76.
What is the unit of sound? Give notes on our ability of sound perception?
Answer:

  • The intensity of sound is measured in decibels. (dB)
  • 0-50 dB is the threshold of hearing for normal ear.
  • Prolonged exposure to sound with intensities greater than 90dB causes hearing loss.

Question 77.
What are chemo receptors?
Answer:

  • The receptors for taste and smell are the chemo receptors
  • The smell receptors are excited by air borne chemicals that dissolve in fluids.

Question 78.
What are olfactory organs? Where are they situated?
Answer:

  • The yellow coloured patches of olfactory epithelium form the olfactory organs.
  • They are located on the roof of the nasal cavity.

Question 79.
What are papillae?
Answer:
The tongue is provided with many small projection called papillae which is responsible for taste.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 80.
Give notes on taste buds.
Answer:

  • The taste bud cells are subjected to huge amounts of friction because of their location and are routinely burned by hot foods.
  • These are the most dynamic cells in the body and replaced every seven to ten days.

 

(3 Marks)

III. Short Answers

Question 1.
What are the three main functions of nervous system.
Answer:

  • Sensory functions: It receives sensory input from internal and external environment.
    Motor functions: It transmits motor commands from the brain to the skeletal and muscular system.
  • Autonomic functions: Reflex actions.

Question 2.
How do we classify neurons on the basis of its functions.
Answer:

  • Afferent neurons – That take sensory impulses to the central nervous system from the sensory organs.
  • Efferent neurons – That carry motor impulses from CNS to the effector organ.
  • Inter neurons – That lie entirely within the CNS between the afferent and efferent neurons.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 3.
What is meant by depolarisation?
Answer:

  • The inside of the axolemma becomes positively charged and negatively charged outside.
    This reversal of electrical charge is called depolarisation.

Question 4.
How is nerve impulses transmitted?
Answer:
Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination 11

  • Inner to the axolemma the cytoplasm contains the infra cellular fluid with large amounts of potassium and magnesium phosphate with negatively charged proteins and other organic molecules.
  • Outside the axolemma contains large amounts of sodium chloride bicarbonates CO2 andmetabolic wastes
  • Due to the charged particles present in the inner and outside of the axolemma are responsible for the conduction of nervous impulses.

Question 5.
What is meant by repolarisation?
Answer:

  • When the axolemma reaches the spike potential the sodium voltage – gate closes and potassium – voltage gate opens.
  • It checks influx of Na+ ions and initiation the efflux of K ions which lowers the number of positive ions within the cell. Thus the potential falls back towards the resting potential.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination 12

  • The reversal of membrane potential inside the axolemma to negative occurs clue to efflux of K+ ions. This is called repolarisation.

Question 6.
What is meant by hyper polarisation.
Answer:

  • If repolarisation becomes more negative than the resting potential -70 mV to about -90 mV.
    • It is called hyper polarisation. During this K+ ion gate are more permeable to K+ even after reaching the threshold level as it closes slowly hence called lazy gates.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination 13

  • The membrane potential return to its original resting state when K+ ion channel close completely.
  • During hyper polarization the Na+ voltage gate remains closed.

Question 7.
Give an account of the conduction speed of a nerve impulse.
Answer:

  • If the axon’s diameter is greater the conduction will be faster.
  • The myelinated axon conducts the impulse faster than the non – myelinated axon.

Question 8.
What is meant by saltatory conduction?
Answer:
The myelinated axon conduction the impulse faster than the non – myelinated axon.
The voltage gated Na+ and K+ channels are concentrated at the nodes of Ranvier as a result the impulse jumps node to node. This is called saltatory conduction.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 9.
Give an account of the membranes (meninges) of brain?
Answer:
Brain is covered by three cranial meninges.

  1. The outer thicker layer is duramater which lines the inner surface of the cranial cavity.
  2. The inner most layer which is adhered to the brain is piameter.
  3. The median thin layer is arachnoid mater by a narrow subdural space.
  4. The piameter is separated from the arachnoid mater by the subarachnoid space.

Question 10.
List out the functions of brain lobes?
Answer:

StructureFunction
FrontalBehaviour intelligence memory movement
ParietalLanguage reading sensation
TemporalSpeech hearing memory
OccipitalVisual processing

Question 11.
Why is thalamus considered as a important relay centre for impulse?
Answer:

  • It serves as a relay centre for impulses between the spinal and brain and cerebrum.
  • Information is sorted and edited.
  • It plays a key role in learning and memory.
  • It is a co-ordinating centre for sensory and motor signaling.

Question 12.
Why there may be a death due to the affectation of medulla oblongata?
Answer:

  • Medulla controls cardio-vascular reflexes respiration and gastric secretions.
  • If medulla is getting affected the circulation and respiration is affected.
  • That may cause a death of a person.

Question 13.
Give notes on mid brain?
Answer:

  • The midbrain is located between the diencephalon and the pons.
  • The lower portion of the mid-brain consists of a pair of longitudinal bands of nervous tissue called cerebral peduncles.
  • This relay impulses back and forth between cerebrum cerebellum pons and medulla.
  • The dorsal position of the mid Brain consist of four rounded bodies called corpora quadrigemina which acts as a reflex center for vision and hearing.

Question 14.
What are the features we develop when there is a functional deficiency of serotonin and nor epinephrine?
Answer:
Features:

  • negative mood
  • loss of interest
  • an in ability to experience pleasure
  • suicidal tendencies.

Drugs:
Anti depressant drug can treat all this features.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 15.
Give notes on cerebro spinal fluid?
Answer:

  • 150ml of cerebro spinal fluid is secreted in an adult.
  • 500ml of cerebro spinal fluid is secreted in a day.
  • Every 8 hours this fluid is rejuvenated.
  • Choroid plexux carries harmful metabolic wastes from the brain to the blood.

Question 16.
Based on their position how the receptors are classified?
Answer:
1. Exteroceptors – They are located at the surface of the body.
These senses the hearing vision touch taste and smell.

2. Introceptors: They are located in the visceral organs and blood vessels.
They are sensitive to internal stimuli.

3. Prerprioceptors:
They provide information about position and movements of the body.

Question 17.
Your friend is returning home after his visit to USA. All at home are waiting for his arrival.
How would you feel? State the division of ANS that predominates and mention few changes take place in your body?
Answer:

  • The sympathetic and parasympathetic neural system are is mainly involved.
  • On seeing his friend out of the happiness his autonomic nervous system is stimulated and he gets tears in his eyes.
  • After sometime the parasympathetic stops the secretion of tears.

Question 18.
What is meant by depression? What are its symptoms?
Answer:

  • Depression is a functional deficiency of serotonin and nor epinephrine.
  • This disorder is characterized by a pervasive negative mood, loss of interest an inability to experience pleasure and suicidal tendencies.

Question 19.
When dust falls on our eyes like eyelids close immediately not waiting for our willingness, on touching a hot pan the hand is with drawn rapidly. Do you know how this happens?
Answer:
Reflex axon.
When a very quick response is needed the spinal cord can effect motor initiation as the brain and brings about an effect it is due to fast action of spinal cord.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 20.
What are the differences between conditioned and unconditioned reflex.
Answer:

Unconditioned reflexConditioned reflex
In born reflex for an unconditioned stimulus.It’s response to a stimulus acquired by learning.
It does not need any past experience or trainingDoes not naturally exists in animals.
Eg: Blinking of an eye when a dust particle about to fall in itEg: Excitement of salivary gland on seeing a food

Question 21.
What is the difference between exteroceptors and interoceptors.
Answer:

ExteroceptorsInteroceptors
Located at or near the surface of the bodyThey are located in the visceral organs and blood vessels.
These receive impulses from hearing vision touch taste and smell.They are sensitive to internal stimuli.

Question 22.
Where are lens fiber seen? What is the substance that fills it?
Answer:

  • The lens is a transparent biconvex structure made up of slender columnar epithelial cells.
  • These cells are called as lens fibre.
  • These cells are formed of crystaline protein.

Question 23.
What is meant by accommodation of eye? List the structures that are involved in this process?
Answer:

  • The ability of the eyes to focus objects at varying distances is called accommodation.
  • This is achieved by suspensory ligament ciliary muscle and ciliary body

Question 24.
Name the structures that helps in fixing eyes in the eye ball?
Answer:
Eye is held in its position with the help of six extrinsic muscles.
They are
Superior
Interior
Lateral
Median rectus muscle
Superior oblique
Inferior oblique

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 25.
Name the accessory structure which protects eye?
Answer:

  • Eye lashes and the eye brows
  • They help to protect the eyeballs from foreign objects, perspiration and from direct sunrays.
  • Sebaceous glands or ciliary glands.
  • They secrete a lubricating fluid.
  • Lacrymal glands.
  • Secrete tears.
  • Tears contain salts mucus and lysozyme enzyme to destroy bacteria.

Question 26.
Give an account of aqueous humour?
Answer:

  • It supplies nutrients and oxygen to the lens cornea and retinal cells.
  • It is produced and drained at same rate.
  • It maintains a constant infra ocular pressure of about 16 mmHg.

Question 27.
Give short notes on glaucoma.
Answer:
Any block in the canal of Schlemm increases the intra ocular pressure of aqueous humour and leads to “glaucoma when the optic nerve and the retina are compressed due to pressure.

Question 28.
Name the detects of detraction.
Answer:

  • Myopia
  • Hyper metropia
  • Astigmatism
  • Cataract

Question 29.
Give notes on pigments of colour vision
Answer:

  • Redcones. It contains visual pigment erythropsin which is sensitive to long wavelength close to 560 nm.
  • Green cones – It has a pigment chloropsin which is sensitive to medium wavelength of 530 nm.
  • Blue cones – It has a pigment which is sensitive to short wavelength of 420 nm.

Question 30.
What is the difference between conducting hearing loss and neuro sensory hearing loss.
Answer:

Conducting Hearing lossNeuro sensory
This defect is due to the blockage of ear canal with ear waxThe defect may be in the organ of cortii
Rupture of earThe auditory nerve
Middle ear infection with fluid accumulation restriction of ossicular movementIn the ascending auditory pathways or auditory cortex

Question 31.
Give short notes on melanin.
Answer:

  • Melanocytes synthesize melanin.
  • It gives colour to skin and protects it from the sun.
  • Vitiligo is a condition in which the melanin pigment is lost from the areas of the skin causing white patches.
  • The leukoderma appears when melanocytes fails to synthesis melanin pigment.

Question 32.
What is meant by the proprioception?
Answer:

  • Balance is part of a sense called proprioception.
  • It is the ability to sense the position orientation and movement of the body.

Question 33.
Which is the fastest renewable cells in the body? How is it affected?
Answer:
Taste buds

  • Taste buds are subjected to huge amounts of friction because of their location and are routinely burned by hot foods.
  • These cells are replaced every 7-10 days.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 34.
Name the disease which can affect people of any age gender or ethnic group. This is not the contagious disease?
Answer:
Leucoderma is a condition in which the melanin pigment is lost from areas of the skin causing white patches.

Question 35.
What are meissner’s corpuscles?
Answer:

  • These are small light pressure receptors found just beneath the epidermis in the dermal papillae.
  • They are numerous in hairless skin areas such as finger tips and soles of the feat.

(5 Marks)

IV. Brief Answers

Question 1.
Describe structure of a neuron.
Answer:
Neuron is composed of three region.

  1. Cell body
  2. Dendrites
  3. axon.

Cell body:
The cell body is spherical. There is no centriole.
The plasma membrane covering the neuron is called neurilemma and axon is axolemma.

Dendrites:
The repeatedly branched short fibres coming out of the cell bod dendrites which transmit impulses towards the cell body in it’s cytoplasm Nissl’s granules are present.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination 14
Axon:
It is a long fibre that arises from a cone shaped area of the cell body. There is no golgi bodies and Nissis granules in its cytoplasm.
The axon of peripheral nerves is surrounded by Schwann’s cells to form myelin sheath. The myelin sheath is not continuous. There are gaps in the myelin sheath between an adjacent Schwann cells called nodes of Ranvier.
Each branch at the distal end of the axon terminates into a knob like structure called synaptic knob which possesses synaptic vesicles filled with transmitters. The axon transmits nerve impulses away from the body to neuro muscular junction. The myelinated nerve cell transmits impulses faster than non – myelinated nerve cells.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 2.
Describe about the different types of neuron with diagram?
Answer:
1. Multi polar neurons:
They have many processes with one axon and two or more dendrites.

2. Bipolar neurons:
They have two processes with one axon and one dendrites.
These are found in the retina of the eye, inner ear and the olfactory area of the brain.

3. Unipolar neurons:
They have a single short process and one axon.
Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination 15

Question 3.
Give an account of Ionic channels in the axolemma?
Answer:
Ionic channels:

  1. Leakage channels
  2.  Ligand – gated channels
  3. Voltage-gated channels.

1. Leakage channels:

    • These channels are always remain open.
    • K+ leakage channels are more in number than the Na+ leakage channels.
    • Sarco lemma has greater permeability to k+ icons than Na+ icons
    • These icons keep moving continuously to maintain the potential continuously to maintain the
    • potential difference across the axo lemma.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination 16

2. Ligand – gated channels:
These are chemic called gated channels which open or close in response to a chemical stimuli.
Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination 17

  • They are located between the pre synaptic membrane of the first axon and post synaptic membrane of the second.
  • The neurotransmitter acetylcholine opens ligand channels that allow Na+ and Ca++ ions diffuse inward and K+ icon diffuse outward.

Voltage-gated channels:
Answer:
This channels open in response to a physical stimulus in the form of vibration such as touch and pressure.

  • Sodium Voltage gated channels.
  • Potassium Voltage gated channels.

Question 4.
Give an account of resting potential?
Answer:

  • The electrical potential difference across the plasma membrane of a resting neuron is called the resting potential.
  • More potassium is getting out of the neurilemma rather than sodium which is getting into.
  • Hence the interior of the cell becomes negative.
  • In resting condition the axon membrane is more permeable to K+ and less permeable to Na+ cons whereas it remains impermeable to negatively charged protein icons.
  • In the outer side of axon there is low concentration of K+ and high concentration of Na+ ions.
  • This difference can be maintained by ATP – driven sodium – potassium pump. This exchange 3 Na+ outwards for 2K+ in the cells.
  • In neuron the resting membrane potential ranges from 40mv to 90mv.
  • And its normal value is 70mv.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 5.
a) What is meant by action – membrane potential?
b) What is Depolarisation?
Answer:
a) An action potential occurs when a neuron sends information down an axon away from the cell body.
b) When a nerve fibre is stimulated sodium voltage gate opens and makes the axo lemma permeable to Na++ and the potassium voltage gate closes and potassium is getting out of the axo lemma. The concentration reduces
Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination 18

  • The axo lemma becomes positively charged inside and negatively charged outside.
  • This reversal of electrical charge is called depolarisation potential reaches level called threshold potential (-55mV)
  • The stimulus which bring this threshold potential is called threshold stimulus.
  • The action potential occur in response to a threshold stimulus but does not occur at subthreshold stimuli. This is called all or none principle.
  • Due to the rapid influx of Na+ ions the membrane potential shoots rapidly up to +45mV which is called the spike potential.

Question 6.
What is synapsis?
Answer:
Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination 19

  • The junction between two neurons is called a synapses through which a nerve impulse is transmitted,
  • The first neuron involved in the synapse forms the pre synaptic neuron and the second neuron is the post – synaptic neuron
  • A small gap between the two neuron is synaptic cleft.
  • The axon terminals contain synaptic vesicles filled with neurotransmitters.
  • When an impulse arrives at the axon terminals it depolorizes the pre – synaptic membrane opening the voltage gated calcium channels.
  • Influx of calcium ions stimulates the synaptic vesicles toward the pre – synaptic membrane and fuse with it.
  • In the neurilemma the vesicles release their neurotransmitters into the synaptic cleft by exocytosis.
  • The released neurotransmitters bind to their specific receptors.
  • The entry of the ions can generate a new potential in the post synaptic neuron.
  • This excitatory post – synaptic potential causes depolarisation and in inhibitory post – synaptic potential causes hyperpolarisation.

Question 7.
In the following diagram indicate
Answer:
(A) – Frontal
(B) – parietal
(C) – occipital
(D) – temporal lobe
Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination 20

Question 8.
Describe the structure of human brain with a diagram?
Answer:

  • The brain is located in the cranial cavity and it is covered by three cranial membranes.
  • The outer layer — durameter the inner most layer piameter and the median thin layer arachnoid. The brain is divided into three major regions. Fore brain, Mid-brain, Hind

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination 21

  • It comprises of cerebrum and diencephalon.
  • Tire cerebral cortex is composed of grey and unmyelinated nerve cells.
  • The medulla is composed of white mater.
  • The surface of the cerebrum shows many convolutions and grooves. The folds are called gyri and the shallow groove is sulci.

There are eight lobes in cerebrum.
A pair of frontals, parietals temporals and occipital lobes.
The longitudinal fissure divides the cerebrum | longitudinally into two hemispheres The hemispheres are connected by a tract of nerve fibres called corpus callosum.
Cerebral cortex has three functional areas

Sensory areas :
It occurs in the parietal temporal and occipital lobes of the cortex.

Motor areas :
This controls voluntary muscular movement which lies in the posterior part of the frontal iobes.

Association area :
It lies in between cortex and diencephalan This involves in memory communication learning and reasoning.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 9.
Describe the structure of hind brain?
Answer:
Rhombencephalon forms the hind brain.
It comprises of cerebellum pons varolii and medulla oblongata.

Cerebellum:
It is the second largest part of the brain. It consists of two cerebellar hemisphere and central worm shaped part the vermis.

Function:
It controls and coordinates muscular movements and body equilibrium.
Any damage to cerebellum results in unco-ordinated voluntary muscle movements.

Pons varoli:
It lies in between the midbrain and the medulla oblongata.
This form a bridge between the two cerebellar hemisphere and connect the medulla oblongata with the other region of the brain.

Medulla oblongata:
This forms the posterior most part of the brain. It connects the spinal cord with various parts of the brain.
It receives and integrates signals from spinal cord and sends it to the cerebellum and thalamus.

Function:
It controls cardiovascular reflexes respiration and gastric secretions.

Question 10.
Describe the structures of diencephalon?
Answer:
Epithalamus

  • It is a non – nervous tissue.
  • The anterior part of epithalamus is vascular and folded to form the anterior choroid plexus.
  • The epithalamus forms a short stalk which ends in a rounded body called pineal body.
  • This secretes the hormone melotonin which regulates sleep and wake cycle.

Thalamus

  • It is formed of grey matter
  • It serves as a relay centre for impulses between the spinal cord brain stem and cerebrum.
  • It plays a key role in learning and memory.
  • It is a major co-ordinating centre for sensory and motor signalling.

Hypothalamus

  • It forms the floor of the diencephalon.
  • It has a pair of small rounded body called mammillary bodies
  • It involves in olfactory reflexes and emotional responses to odour.
  • It maintains homeostasis.
  • It controls the body temperature.
  • Urge for eating and drinking.
  • It also contains a group of neuro secretory cells which secrete the hypothalamic hormones.
  • It also acts as the satiety centre.

Question 11.
Give an account of ventricles of the brain.
Answer:

  • The brain has four hollow fluid filled spaces.
  • The c – shaped space found inside each cerebral hemisphere forms the lateral ventricles I and II which are separated by a thin membrane called the septum pellucidum. Each lateral ventricle communicates with the III ventricle through opening called foramen of mortro.
  • The III ventricle opens in to the IV th ventricle through a canal called aqueduct of sylvius. Choroid plexus is a network of blood capillaries found in the root of the ventricles and forms cerebro spinal fluid.

Functions

  • CSF provides buoyancy to the CNS.
  • It acts as a shock absorber.
  • It nourishes the brain by supplying food and oxygen.
  • It carries harmful metabolic wastes from the brain to the blood.
  • It maintains a constant pressure inside the cranial vessels

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 12.
Differentiate the conditioned stimulus from unconditioned stimulus.
Answer:

ConditionedUnconditioned
1. Not through learningAcquired by learning
2. In born reflexIt is not an in born reflex
3. It does not need any past experienceExperience makes it a part of the behaviour
4. Eg: Blinking of an eye when a dust fall in to it.Eg: Secretion of salivary gland on seeing the food

Question 13.
Tabulate the nerves and functions and its nature of 6-12th cranial nerves.
Answer:

Cranial nervesNature of nerveFunction
I Olfactory nerveSensorySense of smell
II Optic nerveSensorySense of sight
III Oculomotor nerveMotorMovement of the eye
IV Trochlear nerveMotorRotation of the eye ball
V Trigeminal nerveSensory and motor (mixed)Functioning of facial parts
VI Abducens nerveMotorRotation of the eye ball
VII Facial nerveMixedFunctioning of facial parts
VIII Auditory/ Vestibulocochlear nerveSensoryMaintains the equilibrium of the body/ Auditory function
IX Glossopharyngeal nerveMixedTaste and touch
X VagusMixedRegulation of the visceral organs
XI Spinal accessoryMotorMuscular movement of pharynx, larynx, neck and shoulder
XII HypoglossalMotorSpeech and swallowing

Question 14.
Tabulate the functions of sympathetic and para sympathetic nerve.
Answer:

Sympathetic nervous systemPara sympathetic nervous system
1. It dialates pupilIt constricts pupil of the eye
2. It inhibits the secretion of salivaIt stimulates saliva secretion
3. It increases the heart beatIt reduces the heart beat
4. It dialates bronchesIt constricts the bronchus
5. It inhibits digestionIt stimulates digestion
6. It increases the glucose releaseIt stimulates bile release
7. Stimulates epinephrine and nor epinephrine releaseIt reduces the epinephrine and nor epinephrine release
8. Inhibits peristalsis and secretionStimulates peristalsis and secretion
9. Reflexes bladderContracts bladder

Question 15.
Describe the structure of cross section of spinal cord.
Answer:

  • There are two indentations the posterior median sulcus and the anterior median fissure.
  • In the spinal cord the grey matter forms an inner butterfly shaped region surrounded by the outer white matter.
  • The grey matter consists of dendrites inter neurons and guai cells.
  • White matter consists of bundles of nerve fibres.
  • Each half of the grey matter is divided into a
  • Dorsal horn – Cell bodies of inter neurons.
  • Ventral horn – Efferent motor neurons supplying the skeletal muscle.
  • Lateral horn – Nerves supply to heart smooth muscles exocrine glands. Originate from the cell bodies.
  • Ascending tract – This carry sensory impulses to the brain.
  • Descending tract – This carry motor impulses to brain.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination 22

Question 16.
Give an account of the functional components of a reflex arc.
Answer:
Sensory receptor – It is a sensory structure that responds to specific stimulus.
Sensory neuron – This neuron takes the sensory impulse to the grey (afferent) matter of the
spinal cord through the dorsal root of the spinal cord.
Inter neuron – It may serve to transmit the impulses from the sensory neuron to the motor neuron.
Motor neuron – It transmits impulse from CNS to the effector organ
Effector neuron – It may be a muscle or gland which responds to the impulse received.
Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination 23

  • Lacrymal glands located in the upper lateral region of each orbit secrete tears.
  • Tears are secreted at the rate of 1 ml/day.
  • Tears contain salt mucus and lysozyme enzyme to destroy bacteria.
  • The protective mucus membrane present in the outer surface of the eye ball.
  • The eye has two compartments anterior and posterior compartments filled with aqueous humour and vitreous humour respectively.
  • The eye ball is consists of three layers sclera, vascular choroid and sensory retina.

Sclera

  • It consists of anterior cornea and the posterior sclera.
  • Cornea is composed of stratified squamous epithelium. Sclera forms the white of the eye and protects the eyeball.
  • At the junction of the sclera and the cornea is a channel called canal of schlemm which continuously drains out the excess of aqueous humour.

Choroid layer
It is highly vascularised pigmented layer.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 17.
Tabulate the differences between sympathetic and parasympathetic neural system.
Answer:

Sympathetic Neural System (SNS)Parasympathetic Neural System (PNS)
SNS originates in the thoracic and lumbar region of the spinal cord.PNS originates in the cranial region of the brain and the sacral region of the spinal cord.
Sympathetic ganglia are linked up to form a chain.Its ganglia remain isolated
Preganglionic fibres are short and the postganglionic fibres are long.Preganglionic fibres are long and the postganglionic fibres are short.
Noradrenaline is produced at the terminal ends of the postganglionic fibres at the effector organs. Hence the system is adrenergic.Acetylcholine is produced at the terminal ends of the postganglionic fibres at the effector organs. Hence the system is cholinergic.
Active during stressful conditions preparing the body to face them.Active during relaxing times restoring normal activity after a stress.
The overall effect is excitatory and stimulating.The overall effect is inhibitory.
It is considered as the flight or fight system.It is considered as “The rest and Digest System” or “The Feed and Breed System”.

Question 18.
Give an account of peripheral nervous system.
Answer:

  • All nervous tissue outside the central nervous system is the peripheral neural system.
  • It includes nerves ganglia enteric plexuses and sensory receptors.
  • Ganglia are sinai! masses of nervous tissue.
  • The neurons of these plexuses help in regulating the digestive system.
  • The specialised structure that helps to respond to changes in the environment are called sensory receptor.
  • This triggers nerve impulses along the afferent fibres to CNS.
  • PNS comprises of 12 pairs of cranial nerves and 31 pairs of spinal nerve.
  • The neural retina layer consists of cones and rods.
  • The yellow flat spot at the centre of the posterior region of the retina is called macula lutea.
  • A small depression present in the centre of the

Question 19.
Describe the structure of an eye.
Answer:

  • Eye is located in the orbit of the skull and held in its position with the help of six-extrinsic muscles.
  • Eye lashes and the eye brows help to protect the eyeballs.
  • Anteriorly the choroid thickens to form the ciliary body and coloured portion of the iris.
  • The aperture at the centre of the iris is the pupil through which the light enters.
  • The ability of the eyes to focus objects at varying distances is called accommodation which is achieved by suspensory ligament ciliary muscle and ciliary body.

Retina

  • It contains two region.
  • A sheet of non visual pigmented part neural visual region.
  • yellow spot is called fovea centralis which contains only cones.
  • The optic nerves and the retinal blood vessels enter the eye slightly below the posterior pole which is devoid of photo receptors called blind spot.

Question 20.
Draw the diagram of cross section of the eye.
Answer:
Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination 24

Question 21.
Describe about the mechanism of vision.
Answer:

  • When light enters the eye it gets refracted by the cornea, aqueous humor and lens and it is focused on the retina and excites the rod and cone cells.
  • The rods and cones contain the retinal a derivative of vitamin A and the photo pigment opsin.
  • Light induces dissociation of retinal from opsin and causes the structural changes in opsin.
  • This generates an action potential in the photo receptor cells.
  • It is transmitted by the optic nerves to the visual cortex of the brain via the optic nerves for the perception of vision.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination 25

Question 22.
List the refractive errors of eye.
Answer:
Myopia

  • The affected person cannot see distant objects.
  • As the eyeball is elongated or thickened lens the image of distant object is formed in front of the yellow spot.
  • This error can be corrected using concave lens.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination 26

Hyper metropia

  • Affected person cannot see the nearby objects.
  • This is due to a shortened eyeball and thin lens. So the image of closest object is converged behind the retina.
  • This defect can be corrected by using convex lens.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination 27

Presbyopia

  • Due to aging the lens loses elasticity and power of accommodation.
  • Convex lenses are used to correct this defect.

Astigmatism

  • It is due to the rough curvature of cornea or lens.
  • Cylindrical glasses are used to correct this error.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 23.
Describe the structure of an ear?
Answer:
Anatomically the ear is divided into three regions external ear middle ear and inner ear.
External ear:-

  • It consists of pinna external auditory meatus and ear drum.
  • The pinna collects the sound waves.
  • The external auditory meatus extends up to the tympanic membrane.
  • The tympanic membrane is covered with skin outside and with mucus membrane inside.

Middle ea:-

  • It is a air filled cavity in the temporal bone.
  • The middle ear contains three ossicles malleus incus and stapes.
  • The malleus is attached to the tympanic membrane and its head articulates with the incus.
  • The stapes is attached to the oval window in the inner ear.
  • The Eustachian tube connects the middle ear cavity with the pharynx

Inner ear:-

  • It is the fluid filled cavity consisting of two parts the bony labyrinth and the membranous labyrinths.
  • The bony labyrinth consists of three areas, cochlea, vestibule and semicircular canals.

a Cochlea is a coiled portion consist of 3 chambers namely.

  1. Scala vestibule
  2. Scala media
  3. Scala tympanii.

Scala vestibule and scala tympani are filled with peri lymph.
The scala media is filled with endo lymph.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination 28

Question 24.
Describe the structure and functions of organ of cortii.
Answer:

  • Organ of corti is a sensory ridge located on the top of the basilar membrane.
  • Basilar membrane contains numerous hair cells that are arranged in four rows.
  • Protruding from the apical part of each hair cell is hair like structures known as stereocilia.
  • On the organ of cortii a stiff gel membrane called tectorial membrane is situated.
  • During the conduction of sound wave sterocilia makes contact with tectorial membrane.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination 29

Question 25.
Give an account of mechanism of sound?
Answer:

  • Sound waves entering the external auditory meatus fall on the tympanic membrane and gets vibrated.
  • These vibrations are transmitted to the oval window through the three ear ossicles.
  • Since tympani membrane is 17-20 times larger than the oval windows. This pressure exerted on the oval window is about 20 times more than that on the tympanic membrane.
  • This pressure causes the round window to alternately bulge outward and inward mean while the basilar membrane along with the organ of corti move up and down.
  • These movements of hair alternately open and close the mechanically gated ion channel causes action potential.
  • This is propagated to the brain as sound sensation through cochlear nerve.

Question 26.
Differentiate the rods from cones.
Answer:

RodsCones
1. Helps in seeing in dim lightColour perception in bright light
2. Contains Rhodopsin pigmentContains photopsin
3. The protein retinol and vitamins aldehyde combine to form scotoptin called RhodopsinThe protein opsin and retinol combines to form photopsin.
4. There are 120 million rods are seen on the retina6-7 million cone cells are seen on the retina
5. Rods are seen richly over the surface of the foveal region.Cones are richly present on the foveal region.

Question 27.
A man is chating with his family in this time a house fly sits on his back at once he lifts the hand and strikes the fly. How can he know that? What are the structures involved in this perception?
Answer:
Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination 30

  • The skin senses this skin is the largest sense organ.
  • All over the skin sensory receptors of pressure heat cold and pain.
  • Following are the sensory receptors of skin.

Tactile merkel disc

  • It is a light touch receptor lying in the deeper layer of epdidermis.
  • Hair follicle receptors. These are light touch receptors lying around the hair follicles.

Meissner’s corpuscles

  • These are small light pressure receptors found just beneath the epidermis in the dermal papillae.
  • They are numerous in hair less skin areas such as finger tips and soles of the feet.

Pacinian corpuscles

  • These are the large egg shaped receptors found scattered in the dermis.
  • It detects different textures temperature hardness and pain.

Ruffini endings
This lie in the dermis responds to continuous pressure Krause end bulbs are thermo receptors that sense temperature.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 28.
Give notes on Defects of ear.
Answer:

  • Deafness may be temporary or permanent
  • Conductive deafness
  • Possible causes
  • The blockage of ear canal with ear wax.
  • Rupture of ear drum.
  • Middle ear infection with fluid accumulation,
  • Restriction of ossicular movement.
  • Sensory-neural deafness
  • The defect may be in the organ of cortii or the auditory nerve or in the ascending auditory pathways or auditory cortex.

Question 29.
a) What is meant by proprioception?
b) Give an account of organs of equilibrium?
Answer:
The ability to sense the position orientation and movement of the body is called proprioception.
1. Vestibular system: This is the organ of balance.
It is composed of a series of fluid filled sacs contain endolymph and are kept in the surrounding peri lymph.

Semicircular canals

  • The canals that lie posterior and lateral to the vestibule are semi circular canals.
  • At each end of semi circular canal has a swollen area called ampulla.
  • Each ampulla has a sensory area known as crista ampullaris.
  • It detects the rotational movement of the head.

Question 30.
Give notes on Gustatory receptors?
Answer:

  • The sense of taste is considered to be the most pleasurable of all senses.
  • The tongue is provided with many small projections called papillae which gives the tongue an abrasive feel.
  • Most taste buds are located on the tongue few are scattered on the soft palate inner surface of the cheeks.

Structure
Taste buds are flask shaped and consists of 50 -100 epithelial cells of two major types.

  1. ustatory epithelial cells: Taste cells.
  2. Basal epithelial cells

Gustatory epithelial cells:

  • Long gustatory hairs project from the tip of the gustatory cells and extends through a taste pore senses the taste.
  • The sensory dendrites send signal to the brain and senses the taste.
  • The basal cells differentiates in to new gustatory cells.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination

Question 31.
Describe the structure of the receptor involved with a diagram.
Answer:

  • Taste buds are flask shaped. There are two major types.
  • Gustatory epithelial cells or taste cells.
  • Basal epithelial cells or repairing cells.
  • Long micro villi called gustatory hairs project from the tip of the gustatory cells and extends through a taste pore to the surface of the epithelium.
  • Gustatory hairs are tire sensitive portion of the gustatory cells and they have sensory dendrites which send the signal to the brain.
  • The basal cells that act as stem cells divide and differentiate into new gustatory cells.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 10 Neural Control and Coordination 31

Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Chemistry Guide Pdf Chapter 12 Basic Concepts of Organic Reactions Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

11th Chemistry Guide Basic Concepts of Organic Reactions Text Book Back Questions and Answers

Textbook Evaluation:

I. Choose the best answer:

Question 1.
For the following reactions
(A) CH3CH2CH2Br + KOH → CH3 – CH = CH2 + KBr + H2O

(B) (CH3)3CBr + KOH → (CH3)3COH + KBr

(C) Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions 1
Which of the following statement is correct?
a) (A) is elimination, (B) and (C) are substitution
b) (A) is substitution, (B) and (C) are elimination
c) (A) and (B) are elimination and (C) is addition reaction
d) (A) is elimination, (B) is substitution and (C) is addition reaction
Answer:
d) (A) is elimination, (B) is substitution and (C) is addition reaction

Question 2.
What is the hybridisation state of benzyl carbonium ion?
a) sp2
b) spd2
c) sp3
d) sp2d
Answer:
a) sp2

Question 3.
Decreasing order of nucleophilicity is
a) OH > NH2 > -OCH3 > RNH2
b) NH2 > OH > -OCH3 > RNH2
c) NH2 > CH3O > OH > RNH2
d) CH3O > NH2 > OH > RNH2
Answer:
b) NH2 > OH > -OCH3 > RNH2

Question 4.
Which of the following species is not electrophilic in nature?
a) Cl+
b) BH3
c) H3O+
d) +NO2
Answer:
c) H3O+

Question 5.
Homolytic fission of covalent bond leads to the formation of
a) electrophile
b) nucleophile
c) carbocation
d) free radical
Answer:
d) free radical

Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions

Question 6.
Hyper conjucation is also known as
a) no bond resonance
b) Baker – nathan effect
c) both (a) and (b)
d) none of these
Answer:
c) both (a) and (b)

Question 7.
Which of the group has highest + I effect?
a) CH3
b) CH3 – CH2
c) (CH3)2 – CH-
d) (CH3)3 – C –
Answer:
d) (CH3)3 – C –

Question 8.
Which of the following species does not exert a resonance effect?
a) C6H5OH
b) C6H5Cl
c) C6H5NH2
d) C6H5NH3
Answer:
d) C6H5NH3

Question 9.
– I effect is shown by
a) – Cl
b) – Br
c) both (a) and (b)
d) – CH3
Answer:
c) both (a) and (b)

Question 10.
Which of the following carbocation will be most stable?
a) Ph3+C –
b) CH3+CH2
c) (CH3)2+CH
d) CH2 = CH – +CH2
Answer:
d) CH2 = CH – +CH2

Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions

Question 11.
Assertion:
Tertiary Carbocations are generally formed more easily than primary Carbocations ions.
Reason:
Hyper conjucation as well as inductive effect due to additional alkyl group stabilize tertiary carbonium ions.
a) both assertion and reason are true and reason is the correct explanation of assertion.
b) both assertion and reason are true but reason is not the correct explanation of assertion.
c) assertion is true but reason is false.
d) both assertion and reason are false.
Answer:
a) both assertion and reason are true and reason is the correct explanation of assertion.

Question 12.
Heterolytic fission of C – Br bond results in the formation of
a) free radical
b) Carbanion
c) Carbocation
d) Carbanion and Carbocation
Answer:
d) Carbanion and Carbocation

Question 13.
Which of the following represent a set of nucleophiles?
a) BF3, H2O, NH2-
b) AlCl3, BF3, NH3
c) CN, RCH2, ROH
d) H+, RNH3+, :CCl2
Answer:
c) CN, RCH2, ROH

Question 14.
Which of the following species does not acts as a nucleophile?
a) ROH
b) ROR
c) PCl3
d) BF3
Answer:
d) BF3

Question 15.
The geometrical shape of carbocation is
a) Linear
b) tetrahedral
c) Planar
d) Pyramidal
Answer:
c) Planar

Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions

II. Write brief answer to the following questions:

Question 16.
Write short notes on
a) Resonance
b) Hyper Conjucation
Answer:
a) Resonance (or) Mesomeric effect:
The resonance is a chemical phenomenon which is observed in certain organic compounds possessing double bonds at a suitable position. Certain organic compounds can be represented by more than one structure and they differ only in the position of bonding and lone pair of electrons. Such structures are called resonance structures (canonical structures) and this phenomenon is called resonance. This phenomenon is also called mesomerism or mesomeric effect.

For example, the structure of aromatic compounds such as benzene and conjugated systems like 1,3 – butadiene cannot be represented by a single structure, and their, observed properties can be explained on the basis of a resonance hybrid.

In 1,3 butadiene, it is expected that the bond between C1 – C2 and C3 – C4 should be shorter than that of C2 – C3, but the observed bond lengths are of same. This property cannot be explained by a simple structure in which two n bonds localised between C1 – C2 and C3 – C4. Actually the n electrons are delocalised as shown below.

Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions 2

The resonating structures are called canonical forms and the actual structure lies between these three resonating structures, and is called a resonance hybrid. The resonance hybrid is represented as below.

Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions 3

Similar to the other electron displacement effect, mesomeric effect is also classified into positive mesomeric effect (+M or +R) and negative mesomeric effect (-M or -R) based on the nature of the functional group present adjacent to the multiple bond.

b) Hyper Conjucation:
The delocalisation of electrons of a bond is called as hyper conjugation. It is a special stabilising effect that results due to the interaction of electrons of a σ -bond (usually C – H or C – C) with the adjacent, empty non-bonding p-orbital or an anti¬bonding σ* or π*-orbitals resulting in an extended molecular orbital. Unlike electromeric effect, hyper conjugation is a permanent effect.

It requires an α-CH group or a lone pair on atom like N, O adjacent to a π bond (sp2 hybrid carbon). It occurs by the overlapping of the σ-bonding orbitai or the orbital containing a lone pair with the adjacent π-orbital or p-orbital.

Example:
In propene, the σ -electrons of C-H bond of methyl group can be delocalised into the π -orbital of doubly bonded carbonas represented below.

Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions 4

In the above structure the sigma bond is involved in resonance and breaks in order to supply electrons for delocalization giving rise to 3 new canonical forms. In the contributing canonical structures: (Il), (III) & (IV) of propene, there is no bond between an α-carbon and one of the hydrogen atoms. Hence the hyper conjugation is also known as “no bond resonance” or “Baker Nathan effect”. The structures (II), (III) & (IV) are polar in nature.

Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions

Question 17.
What are electrophiles and nucleophiles? Give suitable examples for each.
Answer:
Nucleophiles are reagents that has high affinity for electron positive centers. They possess an atom has an unshared
pair of electrons, and hence it is in search for an electro positive centre where it can have an opportunity to share its electrons to form a covalent bond, and gets stabilised. They are usually negatively charged ions or electron rich neutral molecules (contains one or more lone pair of electrons). All Lewis bases act as nucleophiles.
Example :
Ammonia(NH3) and amines (RNH2), water (H2O), alcohols (ROH) and ethers ((R-O-R)

Electrophiles:
Electrophiles are reagents that are attracted towards negative charge or electron rich center. They are either positively charged ions or electron deficient neutral molecules. All Lewis acids acts as electrophiles. Neutral molecules like SnCl4 can also act as an electrophile, as it has vacant d – orbitals which can accommodate the electrons from others.
Example:
Carbon dioxide (CO2), dichlorocarbene (: CCl2), Aluminium chloride (AlCl3), boron trifluoride (BF3) and Ferric Chloride (FeCl3)

Question 18.
Show the heterolysis of covalent bond by using curved arrow notation and complete the following equations. Identify the nucelophile in each case.
i) CH3 – Br + KOH →
ii) CH3 – O – CH3 + HI →
Answer:
(i) CH3 – Br + KOH →
CH3 – Br + KQH → CH3OH + KBr
Nucleophile is : OH

ii) CH3 – O – CH3 + HI →
H            I
CH3 – O – CH3 + HI → CH3OH + CH3 I
Nucleophile is : I

Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions

Question 19.
Explain inductive effect with suitable example.
Answer:
Inductive effect is defined as the change in the polarisation of a covalent bond due to the presence of adjacent bonds, atoms or groups in the molecule. This is a permanent phenomenon.

Let us explain the inductive effect by considering ethane and ethylchloride as examples. The C-C bond in ethane is non polar while the C-C bond in ethyl chloride is polar. We know that chlorine is more electronegative than carbon, and hence it attracts the shared pair of electron between C-Cl in ethyl chloride towards itself. This develops a slight negative charge on chlorine and a slight positive charge on carbon to which chlorine is attached.

To compensate it, the C1 draws the shared pair of electron between itself and C2 This polarisation effect is called inductive effect. This effect is greatest for the adjacent bonds, but they also be felt farther away. However, the magnitude of the charge separation decreases rapidly, as we move away from C2 and is observed maximum for 2 carbons and almost insignificant after 4 bonds from the active group.
Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions 5

It is important to note that the inductive effect does not transfer electrons from one atom to another but the displacement effect is permanent. The inductive effect represents the ability of a particular atom or a group to either withdraw or donate electron density to the attached carbon. Based on this ability the substituents are classified as +I groups and -I groups. Their ability to release or withdraw the electron through sigma covalent bond is called +I effect and -I effect respectively.

Highly electronegative atoms and atoms of groups which are cany a positive charge are electron withdrawing or-I group.
Example :
-F , -Cl, -COOH, -NO2 NH2

Higher the electronegativity of the substituent, greater is the -I effect. The order of the -I effect of some groups are given below.
NH3 > NO2 > CN > SO3H > CHO > CO > COOH > COCl > CONH2 > F > Cl > Br > I > OH > OR > NH2 > C6H5 > H

Highly electropositive atoms and atoms are groups which carry a negative charge are electron donating or +1 groups.
Example:
Alkali metals, alkyl groups such as methyl, ethyl, negatively charged groups such as CH3O, C2H5O, COO etc.

Lesser the electronegativity of the elements, greater is the +1 effect. The relative order of +1 effect of some alkyl groups is given below.
-C(CH3)3 > -CH(CH3)2 > -CH2CH3 > -CH3

Question 20.
Explain electromeric effect with suitable example.
Answer:
Electromeric is a temporary effect which operates in unsaturated compounds (containing > C = C <, > C = 0, etc…) in the presence of an attacking reagent.
Let us consider two different compounds.
(i) compounds containing carbonyl group ( > C = O)
(ii) unsaturated compounds such as alkenes ( > C = C < )

When a nucleophile approaches the carbonyl compound, the n electrons between C and O is instantaneously shifted to the more electronegative oxygen. This makes the carbon electron deficient and thus facilitating the formation of a new bond between the incoming nucleophile and the carbonyl carbon atom.
Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions 6

On the other hand when an electrophile such as H+ approaches an alkene molecule, the π electrons are instantaneously shifted to the electrophile and a new bond is formed between carbon and hydrogen. This makes the other carbon electron deficient and hence it acquires a positive charge.

Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions 7

The electrometric effect, is denoted as E effect. Like the inductive effect, the electrometric effect is also classified as + E and – E based on the direction in which the pair of electron is transferred to form a new bond with the attacking agent.

Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions 8

When the π electron is transferred towards the attacking reagent, it is called + E (positive electrometric) effect.

The addition of H+ to alkene as shown above is an example of +E effect.
When the n electron is transferred away from the attacking reagent, it is called, -E (negative electrometric) effect.
Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions 9
The attack of CN on a carbonyl carbon, as shown above, is an example of – E effect.

Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions

Question 21.
Give examples for the following types of organic reactions
(i) β – elimination
(ii) Electrophilic substitution
Answer:
(i) β – elimination:
In this reaction two substituents are eliminated from the molecule, and a new C – C double bond is formed between the carbon atoms to which the eliminated atoms/groups are previously attached. Elimination reaction is always accompanied with change in hybridisation.
Example:
n – Propyl bromide on reaction with alcoholic KOH gives propene. In this reaction hydrogen and Br are eliminated.
Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions 10

(ii) Electrophilic substitution:

Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions 11
Here, Y+ is an electrophile.
Example :
Nitration of Benzene

Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions 12

11th Chemistry Guide Basic Concepts of Organic Reactions Additional Questions and Answers

I. Choose the best answer:

Question 1.
In organic reactions the reactant is called as ______.
a) intermediate
b) product
c) substrate
d) by product
Answer:
c) substrate

Question 2.
Heterolysis of C – Cl bond produces
a) Two free radicals
b) Two carbonium ions
c) Two carbanions
d) One cation and one anion
Answer:
d) One cation and one anion

Question 3.
Heterolysis of propane gives
a) Methyl and ethyl free radicals
b) Methyl carbocation and ethyl free radicals
c) Methyl anion and ethyl carbocation
d) Methyl free radical and ethyl carbocation
Answer:
c) Methyl anion and ethyl carbocation

Question 4.
Removal of hydride ion from a methane molecule will give a
a) Methyl radical
b) Carbonium ion
c) Carbanion
d) Methyl group
Answer:
b) Carbonium ion

Question 5.
The shape of carbonium ion is
a) Planar
b) Linear
c) Pyramidal
d) Tetrahedral
Answer:
a) Planar

Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions

Question 6.
The shape of methyl free radicals is
a) Planar
b) Pyramidal
c) Tetrahedral
d) Linear
Answer:
a) Planar

Question 7.
The reaction, (CH3)3C – Br -» (CH3)3 C+ + Br is an example of
a) Homolytic fission
b) Heterolytic fission
c) Cracking
d) All the above
Answer:
b) Heterolytic fission

Question 8.
Which of the following contains only three pairs of electrons
a) Carbocation
b) Carbanion
c) Free radical
d) All of these
Answer:
a) Carbocation

Question 9.
Which of the following species is paramagnetic
a) A carbaonium ion
b) A free radical
c) A carbanion
d) All of these
Answer:
b) A free radical

Question 10.
In carbonium ion the carbon bearing the positive charge is
a) sp hybridized
b) sp2 hybridized
c) sp3 hybridized
d) un hybridized
Answer:
b) sp2 hybridized

Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions

Question 11.
Among the following the true property about
Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions 13 is
a) non – planar
b) C+ is sp2 – hybridized
c) Electrophile can attack on C+
d) Does not undergo hydrolysis
Answer:
b) C+ is sp2 – hybridized

Question 12.
The geometry of a methyl carbanion is likely to be
a) Pyramidal
b) Tetrahedral
c) Planar
d) Linear
Answer:
a) Pyramidal

Question 13.
Which of the following statements are correct for nucleophile?
a) All negatively charged species are nucleophiles
b) Nucleophiles are Lewis bases
c) Alkenes, alkynes, benzene and pyrrole are nucleophiles
d) All are correct
Answer:
d) All are correct

Question 14.
Which one of the following statement is not correct for electrophile?
a) Electron deficient species are electrophiles
b) Electrophiles are Lewis acids
c) All positively charged species are electrophiles
d) AlCl3 and SO3 are electrophiles
Answer:
c) All positively charged species are electrophiles

Question 15.
Electrophiles are
a) Lewis bases
b) Lewis acids
c) Amphoteric
d) All of these
Answer:
b) Lewis acids

Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions

Question 16.
Electrophiles are
a) Electron loving species
b) Electron hating species
c) Nucleus loving reagents
d) Nucleus hating reagents
Answer:
a) Electron loving species

Question 17.
Nucleophiles are
a) Electron loving
b) Electron hating
c) Nucleus loving
d) Nucleus hating
Answer:
c) Nucleus loving

Question 18.
Which of the following statement is false about an electrophile?
a) Electron – deficient species
b) An acidic reagent
c) A reagent which attacks an electron – deficient site in a molecule
d) A species which seeks a pair of electrons
Answer:
c) A reagent which attacks an electron – deficient site in a molecule

Question 19.
Methyl carbanium ion is
a) Electrophile
b) Lewis acid
c) Nucleophile
d) Both (a) and (b)
Answer:
c) Nucleophile

Question 20.
Ammonia molecule is
a) A nucleophile
b) An electron deficient
c) A electrophile
d) An acid
Answer:
a) A nucleophile

Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions

Question 21.
Methyl carbanion is
a) Electrophile
b) Lewis acid
c) Both (a) and (b)
d) Nucelophile
Answer:
d) Nucelophile

Question 22.
Which of the following statements is correct about inductive effect?
a) Implies the transfer of lone pair of electrons from more electronegative atom to lesser electronegative atom in a molecule.
b) Implies the transfer to lone pair of electrons from lesser electronegative atom to the more electronegative atom in a molecule
c) Increases with increase in distance
d) Implies the atoms ability to cause bond polarization
Answer:
d) Implies the atoms ability to cause bond polarization

Question 23.
The displacement of electrons in a multiple bond in the presence of attacking reagent is called
a) inductive effect
b) electromeric effect
c) resonance
d) Hyper-conjugation
Answer:
b) electromeric effect

Question 24.
The electromeric effect in organic compounds is a
a) Temporary effect
b) Permanent effect
c) Temporary or permanent effect
d) All of the above
Answer:
a) Temporary effect

Question 25.
Which of the following is not the correct condition for resonance?
a) The positions of all the atomic nuclei in the resonating structures may be differ
b) The resonating structures must have the same number of unpaired or paired electrons
c) The molecules exhibiting resonance must be planar in nature
d) The resonating structures must have nearly the same energies
Answer:
a) The positions of all the atomic nuclei in the resonating structures may be differ

Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions

Question 26.
Resonance is due to ______.
a) delocalization of sigma electrons
b) migration of H atoms
c) migration of proton
d) delocalization of pi electrons
Answer:
d) delocalization of pi electrons

Question 27.
Resonance in benzene is accompanied by delocalization of n – electrons. Each n electron is attached with
a) 4 carbons
b) 2 carbons
c) 3 carbons
d) 6 carbons
Answer:
d) 6 carbons

Question 28.
Reaction mechanism describes
a) Sequential account of each step describing the details of electron movement
b) energy changes during bond breaking and bond formation
c) kinetics of the reaction
d) all of these
Answer:
d) all of these

Question 29.
Substitution reaction may be
a) Free radical substitution
b) Nucleophilic substitution
c) Electrophilic substitution
d) All are correct
Answer:
d) All are correct

Question 30.
CH3 – CH2 – X + KOH(alc) → CH2 = CH2 + KX + H2O is
a) addition reaction
b) substitution reaction
c) elimination reaction
d) molecular rearrangement
Answer:
c) elimination reaction

Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions

Question 31.
CH4 + Cl2 Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions 14 CH3Cl + HCl is an example
for
a) free radical substitution reaction
b) electrophilic substitution reaction
c) nucleophile substitution reaction
d) nueclophilic addition reaction
Answer:
a) free radical substitution reaction

Question 32.
C6H6 + 3 Cl2 Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions 14 C6H6Cl
a) free radical substitution reaction
b) electrophilic addition reaction
c) free radical addition reaction
d) nueclophilic addition reaction
Answer:
c) free radical addition reaction

Question 33.
Which one of the following orders is correct regarding the – I effect of the substituents?
a) – NR2 < – OR > – F
b) -NR2 > – OR > – F
c) -NR2 < – OR< – F
d) -NR2 > – OR < – F
Answer:
c) -NR2 < – OR< – F

Question 34.
Decreasing – I power of given groups is
(1) -CN
(2) -NO2
(3) – NH3+
(4) -F
a) 2 > 1 > 4 > 3
b) 2 > 3 > 4 > 1
c) 3 > 2 > 4 > 1
d) 3 > 2 > 1 > 4
Answer:
d) 3 > 2 > 1 > 4

Question 35.
Which of the following belongs to – I group?
a) – C6H5
b) – CH3
c) – CH2CH3
d) – C (CH3)3
Answer:
a) – C6H5

Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions

Question 36.
Which of the following functional group shows + R effect?
a) – CHO
b) – NO2
c) -CN
d) – NR2
Answer:
d) – NR2

Question 37.
Hyper conjugation is
a) σ – π conjugation
b) Due to delocalization of σ and π bonds
c) No bond resonance
d) All of these
Answer:
d) All of these

Question 38.
+ R power of the given groups
(1) -O
(2) – NH2
(3) – OH
(4) – NHCOCH3 in decreasing order is
a) 1 > 2 > 3 > 4
b) 4 > 3 > 2 > 1
c) 1 > 3 > 2 > 4
d) 1 > 4 > 3 > 2
Answer:
a) 1 > 2 > 3 > 4

Question 39.
In pyridine:
Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions 15
Number of conjugated electrons are:
a) 6
b) 8
c) Zero
d) 5
Answer:
a) 6

Question 40.
In hyper conjugation, the atom involved is
a) β – H atom
b) α – H atom
c) γ – H atom
d) All
Answer:
b) α – H atom

Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions

Question 41.
Hyper conjugation involves overlap of the following orbtials
a) σ – σ
b) σ – p
c) p – p
d) π – π
Answer:
b) σ – p

Question 42.
+ I effect is shown by
a) – NO2
b) – Cl
c) – Br
d) – CH3
Answer:
d) – CH3

Question 43.
Zero inductive effect is exerted by
a) C6H5
b) – H
c) CH3
d) – Cl
Answer:
b) – H

Question 44.
Hyper conjugation is most useful for stabilizing which of the following carbocations?
a) neo – Pentyl
b) tert – Butyl
c) iso – Propyl
d) Ethyl
Answer:
b) tert – Butyl

Question 45.
Hyper conjugation phenomenon is possible in
a) H2C = CH2
b) CH3CH2 – CH = CH2
c) C6H5CH = CH2
d) (CH3)3C – CH = CH2
Answer:
b) CH3CH2 – CH = CH2

Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions

Question 46.
Among the following carbocations which is more stable
a) (C6H5)3C+
b) Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions 16
c) Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions 17
d) Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions 18
Answer:
a) (C6H5)3C+

Question 47.
The compound which gives the most stable carbonium ion on dehydration
a) Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions 19
b) Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions 20
c) CH3 – CH2 – CH2 – CH2OH
d) Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions 21
Answer:
b) Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions 20

Question 48.
Which carbocation is more stable?
Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions 22
Answer:
a) Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions 23

Question 49.
Arrange the following groups in order of decreasing – R (Or – M) power
(1) NO2
(2) SO3H
(3) CF3
(4) CHO
a) 1 > 3 > 2 > 4
b) 1 > 2 > 3 > 4
c) 1 > 4 > 3 > 2
d)4 > 3 > 2 > 1
Answer:
a) 1 > 3 > 2 > 4

Question 50.
The reaction intermediate produced by homolytic cleavage of bond is called
a) carbocations
b) carbanions
c) free radicals
d) carbenes
Answer:
c) free radicals

Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions

Question 51.
Most stable carbocation is
a) CH3+CH2
b) +CH2CHCl2
c) +CH2CH2Cl
d) +CH2 – CH2NO2
Answer:
a) CH3+CH2

Question 52.
Carbocation is a reaction intermediate in which of the following reactions
a) E1 reactions
b) Electrophilic addition reactions of alkenes and alkynes
c) SN1 reactions
d) All of the above
Answer:
d) All of the above

Question 53.
Halogenation of an alkane takes place through the intermediate
a) Carbocation
b) carbanion
c) carbon free radical
d) carbene
Answer:
c) carbon free radical

Question 54.
Which free radical is the most stable?
Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions 24
Answer:
a) Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions 25

Question 55.
The most stable carbonium ion is
a) Methyl carbonium ion
b) Primary carbonium ion
c) Secondary carbonium ion
d) Tertiary carbonium ion
Answer:
d) Tertiary carbonium ion

Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions

Question 56.
Stability of which intermediate is not govern by hyperconjugation?
a) Carbon cation
b) carbon anion
c) carbon free radical
d) carboniumion
Answer:
b) carbon anion

Question 57.
Which of the following is an electrophilic reagent?
a) H2O
b) OH
c) NO2+
d) All of these
Answer:
c) NO2+

Question 58.
CH2 = CH2 + Br2 → CH2Br – CH2Br is an example for
a) addition reaction
b) substitution reaction
c) elimination reaction
d) molecular rearrangement
Answer:
a) addition reaction

Question 59.
R – X + OH(aq) → R – OH + X+(aq) is an example for
a) addition reaction
b) substitution reaction
c) elimination reaction
d) molecular rearrangement
Answer:
b) substitution reaction

Question 60.
C6H6 + Cl2 Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions 26 C6H5Cl + HC1 is an example for
a) Free radical substitution reaction
b) Electrophilic substitution reaction
c) Nucleophilic addition reaction
d) Nucelophilic substitution reaction
Answer:
b) Electrophilic substitution reaction

Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions

Question 61.
Alkenes readily undergo
a) Substitution reactions
b) Addition reactions
c) Elimination reactions
d) Rearrangement reactions
Answer:
b) Addition reactions

Question 62.
Nitration of benzene is
a) nucleophilic substitution
b) nucleophilic addition
c) electrophilic substitution
d) free radical substitution
Answer:
c) electrophilic substitution

Question 63.
The reaction (CH3)3C – Br Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions 27 (CH3)3 C – OH is
a) elimination reaction
b) substitution reaction
c) free radical reaction
d) displacement reaction
Answer:
b) substitution reaction

Question 64.
Inductive effect involves
a) displacement of σ – electrons
b) delocalization of π – electrons
c) delocalization of σ – electrons
d) displacement of π – electrons
Answer:
b) delocalization of π – electrons

Question 65.
Compound which shows positive mesomeric effect
a) H2C = CH – Cl
b) C6H5 – NO2
c) H2C = CH – CH2Cl
d) both b & c
Answer:
a) H2C = CH – Cl

Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions

Question 66.
The stability order, in the following carbocations,
CH3CH2+ (I), (CH3)2 +CH (II), (CH3)3 C+ (III), +CH3 (IV)
a) I > IV > III > II
b) I > II > III > IV
c) III > IV > I > II
d) III > II > I > IV
Answer:
d) III > II > I > IV

Question 67.
Which is most stable carbocation?
a) n – propyl cation
b) iso – propyl cation
c) Ethyl cation
d) Triphenylmethyl cation
Answer:
d) Triphenylmethyl cation

Question 68.
Which one of the following carbanions is least stable?
a) CH3CH2
b) HC ≡ C
c) (C6H5)3C
d) (CH3)3C
Answer:
d) (CH3)3C

Question 69.
Among the following, the strongest nucleophile is
a) C2H5SH
b) CH3COO
c) CH3NH2
d) NCCH2
Answer:
a) C2H5SH

Question 70.
Which of the following is least reactive in a nucleophile substitution reaction?
a) (CH3)3C – Cl
b) CH2 = CHCl
c) CH3CH2Cl
d) CH2 = CHCH2Cl
Answer:
b) CH2 = CHCl

Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions

II. Very short question and answers (2 Marks):

Question 1.
What is mechanism of reaction?
Answer:
Many chemical reactions are depicted in one or more simple steps. Each step passes through an energy barrier, leading to the formation of short lived intermediates or transition states. The series of simple steps which collectively represent the chemical change, from substrate to product is called as the mechanism of the reaction. The slowest step in the mechanism determines the overall rate of the reaction.

Question 2.
What is Positive Mesomeric effect? Give example.
Answer:
Positive resonance effect occurs, when the electrons move away from substituent attached to the conjugated system. It occurs, if the electron releasing substituents are attached to the conjugated system. In such cases, the attached group has a tendency to release electrons through resonance. These electron releasing groups . are usually denoted as +R or +M groups.
Example:
– OH , – SH , – OR , – SR, – NH2, – O – etc…

Question 3.
What is Negative Mesomeric effect? Give example.
Answer:
Negative resonance effect occurs, when the electrons move towards the substituent attached to the conjugated system. It occurs if the electron withdrawing substituents are attached to the conjugated system. In such cases, the attached group has a tendency to withdraw electrons through resonance. These electron withdrawing groups are usually denoted as -R or -M groups.
Example:
NO2, > C = O, – COOH, – C ≡ N etc..

Question 4.
Explain hyper conjugate effect of vinyl chloride molecule.
Answer:
Hyper conjugation effect is also observed when atoms/groups having lone pair of electrons are attached by a single bond, and in conjugation with a π bond. The lone pair of electrons enters into resonance and displaces π electrons resulting in more than one structure.
Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions 28

Question 5.
What are the types of substitution reaction?
Answer:
In this reaction an atom or a group of atoms attached to a carbon atom is replaced by a new atom or a group of atoms. Based on the nature of the attacking reagent, these reactions can be classified as

  • Nucleophilic substitution
  • Electrophilic substitution
  • Free radical substitution

Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions

Question 6.
Explain free radical substitution reaction with the suitable example.
Answer:
Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions 29
Aliphatic electrophilic substitution
A general aliphatic electrophilic substitution is represented as
Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions 30

Question 7.
Explain Nucleophilic Addition reaction with suitable example.
Answer:
Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions 31
Example:
Addition of HCN to acetaldehyde
Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions 32

Question 8.
Explain free radical addition reaction with suitable example.
Answer:
A general free radical addition reaction can be represented as below.
Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions 33
Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions 34
In the above reaction, Benzoyl perioxide acts as an radial initiator. The mechanism involves free radicals.

Question 9.
Write a note on Functional Group inter conversion.
Answer:
Organic synthesis involves functional group inter conversions. A particular functional group can be converted into other functional group by reacting it with suitable reagents. For example: The carboxylic acid group (-COOH) presents in organic acids can be transformed to a variety of other functional group such as – CH2-OH, – CONH2, – COCl by treating the acid with LiAlH4, NH3, and SOCl2 respectively.

Question 10.
After cutting an apple it turn brown. Why?
Answer:
Apples contain an enzyme called polyphenol oxidase (PPO), also known as tyrosinase. Cutting an apple exposes its cells to the atmospheric oxygen and oxidizes the phenolic compounds present in apples. This is called the enzymatic browning that turns a cut apple brown. In addition to apples, enzymatic browning is also evident in bananas, pears, avocados and even potatoes.

Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions

III. Short answer questions (3 Marks):

Question 1.
Explain hybridization of carbon in carbocation.
Answer:
In a carbocation, the carbon bearing positive charge is sp2 hybridised and hence it has a planar structure. In the reaction involving such a carbocation, the attack of a negatively charged species (nucleophiles) take place on either side of the carbocation as shown below.
Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions 35
Shape of carbocation

The carbanions are generally pyramidal in shape and the lone pair occupies one of the sp3 hybridized orbitais. An alkyl free radical may be either pyramidal or planar.
Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions 36
Shape of carbanion

Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions 37
Shape of Carbon radical

Question 2.
Write three types of electron movement in organic reaction.
Answer:

  • lone pair becomes a bonding pair.
  • bonding pair becomes a lone pair.
  • a bond breaks and becomes another bond.

Type: 1
A lone pair to a bonding pair
Example:
Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions 38

Type: 2
A bonding pair to a lone pair
Example:
Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions 39

Type: 3
A bonding pair to an another bonding pair
Example:
Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions 40

Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions

Question 3.
Explain hyper conjugate effect of acrylonitrile.
Answer:
When electronegative atoms or group of atoms are in conjugation with a π – bond,they pull π – electrons from the multiple bond. In case of carbocations, greater the number of alky groups attached to the carbon bearing positive charge, greater is number of the hyper conjucate structure. thus the stability of various carbocations decreases in the order :
3° Carbocation > 2° Carbocation > 1° Carbocation.

Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions 41

Question 4.
What are the different types of organic reactions?
Answer:
Organic compounds undergo many number of reactions, however in actual sense we can fit all those reactions into the below mentiohed six categories.

  1. Substitution reactions
  2. Addition reactions
  3. Elimination reactions
  4. Oxidation and reduction reactions
  5. Rearrangement reactions
  6. Combination of the above

Question 5.
Explain nucleophilc substituion reaction with suitable example.
Answer:
This reaction can be represented as
Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions 42
Here Y is the incoming nuclophile or and attacking species and x is the leaving group.

Example:
Hydrolysis of alkyl halides by aqueous KOH.
CH3Br  Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions 43 CH3OH + Br
Aliphatic nucleophilic substitution reactions take place either by SN1 or SN2 mechanism.

Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions

Question 6.
Explain Electrophilic Addition Reaction with suitable example.
Answer:
A general electrophilic addition reaction can be represented as below.
Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions 44
Example:
Bromination of alkene to give bromo alkane
Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions 45

Question 7.
What are oxidation and reduction reactions? Give suitable example.
Answer:
Many oxidation and reduction reactions of organic compounds fall into one of the four types of reaction that we already discussed but others do not. Most of the oxidation reaction of organic compounds involves gain of oxygen or loss of hydrogen Reduction involves gain of hydrogen and loss of oxygen.
Example:
Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions 46

Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions 47

Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions

IV. Long answer questions (5 Marks):

Question 1.
Explain different types of Fission of a covalent bond.
Answer:
All organic molecules contain covalent bonds which are formed by the mutual sharing of electrons between atoms. These covalent bonds break in two different ways, namely homolytic cleavage (symmetrical splitting) and heterolytic cleavage (unsymmetrical splitting). The cleavage of a bond in the substrate is influenced by the nature of the reagent (attacking agent).

Homolytic Cleavage:
Homolytic cleavage is the process in which a covalent bond breaks symmetrically in such way that each of the bonded atoms retains one electron. It is denoted by a half headed arrow (fish hook arrow). This type of cleavage occurs under high temperature or in the presence of UV light in a compound containing non polar covalent bond formed between atoms of similar electronegativity.

In such molecules, the cleavage of bonds results into free radicals. They are short lived and are highly reactive. The type of reagents that promote holmolytic cleavage in substrate are called as free radical initiators. For example Azobisisobutyronitile (AIBN) and peroxides such as benzoyl peroxide are used as free radical initiators in polymerisation reactions.

Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions 48

As a free radical with an unpaired electron is neutral and unstable, it has a tendency to gain an electron to attain stability. Organic reactions involve homolytic fission of C – C bonds to form alkyl free radicals. The stability of alkyl free radicals is in the following order:
C(CH3)3 > CH(CH3)2 > CH2CH3 > CH3

Heterolytic Cleavage:
Heterolytic cleavage is the process in which a covalent bond breaks unsymmetrically such that one of the bonded atoms retains the bond pair of electrons. It results in the formation of a cation and an anion. Of the two bonded atoms, the most electronegative atom becomes the anion and the other atom becomes the cation. The cleavage is denoted by a curved arrow pointing towards the more electronegative atom.

For example, in tert-butyl bromide, the C – Br bond is polar as bromine is more electronegative than carbon. The bonding electrons of the C – Br bond are attracted more by bromine than carbon. Hence, the C – Br undergoes heterolytic cleavage to form a tert-butyl cation during hydrolysis.

Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions 49

Let us consider the cleavage in a carbon-hydrogen (C-H) bond of aldehydes or ketones We know that the carbon is more electronegative than hydrogen and hence the heterolytic cleavage of C-H bonds results in the formation of carbanion (carbon bears a negative charge). For example in aldol condensation the OH- ion abstracts a a-hydrogen from the aldehyde, which leads to the formation of the below mentioned carbanion.

Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions 50

Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions

Question 2.
Explain the acidic nature of carboxylic acid.
Answer:
When a halogen atom is attached to the carbon which is nearer to the carboxylic acid group, its -I effect withdraws the bonded electrons towards itself and makes the ionisation of H easy. The acidity of various chioro acetic acid is in the following order. The strength of the acid increases with increase in the -J effect of the group attached to the carboxyl group.

Trichioro acetic acid > Dichloro acetic acid > Chloro acetic acid > acetic acid.
Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions 51

Similarly, the following order of acidity in the carboxylic acids is due to the +I effect of alkyl group.
Samacheer Kalvi 11th Chemistry Guide Chapter 12 Basic Concepts of Organic Reactions 52

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Miscellaneous Problems

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 6 Random Variable and Mathematical Expectation Miscellaneous Problems Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Miscellaneous Problems

Question 1.
The probability function of a random variable X is given by
Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Miscellaneous Problems 1
Evaluate the following probabilities.
(i) P(X ≤ 0)
(ii) P(X ≤ 0)
(iii) P( X ≤ 2) and
(iv) P(0 ≤ X ≤ 10)
Solution:
(i) From the data

x-2010
P(x = x)1/41/41/2

(i) P(x ≤ 0) = P( x = -2) + P(x = 0)
= \(\frac { 1 }{4}\) + \(\frac { 1 }{4}\) = \(\frac { 2 }{4}\) = \(\frac { 1 }{2}\)

(ii) P(x < 0) = P(x = -2) = \(\frac { 1 }{4}\)

(iii) P(|x| ≤ 2) = P(-2 ≤ X ≤ 2)
= P(X = -2) + P(X = 0) = \(\frac { 1 }{4}\) + \(\frac { 1 }{4}\)
= \(\frac { 2 }{4}\) = \(\frac { 1 }{2}\)

(iv) p(0 ≤ X ≤ 10) = P (X = 0) + P(X = 10)
= \(\frac { 1 }{4}\) + \(\frac { 1 }{2}\) = \(\frac { 1+2 }{4}\) = \(\frac { 3 }{4}\)

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Miscellaneous Problems

Question 2.
The probability function of a random variable X is given by
Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Miscellaneous Problems 2
(a) Compute: (i) P(1 ≤ X ≤ 2) and (ii) P(X = 3).
(b) Is X a discrete random variable? Justify your answer.
Solution:
W.K.T Probability density Function
f(x) = \(\frac { d[F(x)] }{dx}\)
Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Miscellaneous Problems 3
(ii) P(X = 3) = 0
(b) X is not discrete since f is not a step function.

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Miscellaneous Problems

Question 3.
The p.d.f. of X is defined as
f(x) = \(\left\{\begin{array}{l}
\mathrm{k}, \text { for } 0<x \leq 4 \\
0, \text { otherwise }
\end{array}\right.\)
Solution:
Let X and a random variable if a Probability density function
Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Miscellaneous Problems 4

Question 4.
The probability distribution function of a discrete random variable X is
Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Miscellaneous Problems 5
where k is some constant. Find (a) k and (b) P(X > 2).
Solution:
(a) Let X be the random variable of a probability distribution function
W.K.T Σpi = 1
P(x = 1) + P(x = 3) + P(x = 5) = 1
2k + 3k + 4k = 1
9k – 1 ⇒ k = 1/9

(b) P(x > 1) = P(x = 3) + P(x = 5)
= 3k + 4k = 7k
= 7(1/9) = 7/9

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Miscellaneous Problems

Question 5.
The probability distribution function of a discrete random variable X is
f(x) = \(\left\{\begin{array}{l}
\mathrm{a}+\mathrm{b} x^{2}, 0 \leq x \leq 1 \\
0, \text { otherwise }
\end{array}\right.\)
where a and b are some constants. Find (i) a and b if E(X) = 3/5 (ii) Var(X).
Solution:
Let X be due continuous variable of density function
Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Miscellaneous Problems 6
Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Miscellaneous Problems 7
Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Miscellaneous Problems 8

Question 6.
Prove that if E(X) = 0, then V(X) = E(X²)
Solution:
V(X) = E(X²) – [E(X)]²
= E(X²) – 0 {Given that E(X) = 0}
Var(X) = E(X²)
Hence proved.

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Miscellaneous Problems

Question 7.
What is the expected value of a game that works as follows: I flip a coin and, if tails pay you ₹ 2; if heads pay you ₹ 1. In either case I also pay you ₹ 50.
Solution:
Let x be the remain variable denoting the amount paying for a game of flip coin then x and takes 2 and 1
P(X = 2) = \(\frac { 1 }{2}\) (getting a head)
p(X = 1) = \(\frac { 1 }{2}\) (getting a tail)
Hence the probability of X is

x21
P(x = x)1/21/2

Expected value E(X) = \(\sum_{ x }\)x P(x)
= (2)(\(\frac { 1 }{2}\)) + 1 (\(\frac { 1 }{2}\))
= 1 + \(\frac { 1 }{2}\) = \(\frac { 3 }{2}\)
Since I pay you ₹ 50 in either case
E(X) = 50 × 3/2 = ₹ 75

Question 8.
Prove that, (i) V(aX) = a²V(X) , and (ii) V(X + b) = V(X).
Solution:
LHS = V(ax)
= E(ax)² – [E(ax)]²
= a² E(x²) – [aE(x)]²
= a²E(x²) – a²E(x)]²
= a²E(x²) – E(x)²]
= a²v(x)
= RHS
Hence proved

(ii) LHS = V(x + b)
= E (x + b)² – [E(x + b)]²
= E(x² + 2bx + b²) – [E(x) + b]² –
= E(x²) + 2bE(x) + b² – [E(x)]² + b² + 2bE(x)]
= E(x²) + 2bE(x) + b² -[E(x)]² + b² – 2bE(x)]
= E(x²) – [E(x)]²
= V(x)
= RHS
LHS = RHS Hence proved.

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Miscellaneous Problems

Question 9.
Consider a random variable X with p.d.f.
f(x) = \(\left\{\begin{array}{l}
3 x^{2}, \text { if } 00 \\
0, \text { otherwise }
\end{array}\right.\)
Find the expected life of this piece of equipment.
Solution:
Let X be the random variable
Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Miscellaneous Problems 9

Question 10.
The time to failure in thousands of hours of an important piece of electronic equipment used in a manufactured DVD player has the density function
f(x) = \(\left\{\begin{array}{l}
2 e^{-2 x}, x>0 \\
0, \text { otherwise }
\end{array}\right.\)
Find the expected life of this piece of equipment.
Solution:
Let X be the random variable
Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Miscellaneous Problems 10

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Miscellaneous Problems

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.3

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 6 Random Variable and Mathematical Expectation Ex 6.3 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.3

Question 1.
Value which is obtained by multiplying possible values of random variable with probability of occurrence and is equal to weighted average is called
(a) Discrete value
(b) Weighted value
(c) Expected value
(d) Cumulative value
Solution:
(c) Expected value

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.3

Question 2.
Demand of products per day for three days are 21, 19, 22 units and their respective probabilities are 0.29, 0.40, 0.35. Profit per unit is 0.50 paisa then expected profits for three days are
(a) 21, 19, 22
(b) 21.5, 19.5, 22.5
(c) 0.29, 0.40, 0.35
(d) 3.045, 3.8, 3.85
Solution:
(d) 3.045, 3.8, 3,85
Hint:

x211922
P(x)0.290.400.35

E(X) = \(\sum_{ x }\) xP(x)
For Day 1
E(X) = 21 × 0.29
= 6.09
Expected profit = 6.09 × 0,50 = 3.054
For Day 2
E(X) = 19 × 0.40
= 7.6
Expected Profit = 7.6 × 0.50 = 3.8
For Day 3
E(X) = 22 × 0.35
= 7.7
Expected Profit = 7.7 × 0.50 = 3.85

Question 3.
Probability which explains x is equal to or less than particular value is classified as
(a) discrete probability
(b) cumulative probability
(c) marginal probability
(d) continuous probability
Solution:
(b) Cumulative Probability

Question 4.
Given E(X) = 5 and E(Y) = -2, then E(X – Y) is
(a) 3
(b) 5
(c) 7
(d) -2
Solution:
(c) 7
Hint:
E(x – 7) = E(x) – E(y) = 5 – (-2) = 7

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.3

Question 5.
A variable that can assume any possible value between two points is called
(a) discrete random variable
(b) continuous random variable
(c) discrete sample space
(d) random variable
Solution:
(b) Continuous random Variable

Question 6.
A formula or equation used to represent the probability distribution of a continuous random variable is called
(a) probability distribution
(b) distribution function
(c) probability density function
(d) mathematical expectation
Solution:
(c) Probability density function

Question 7.
If X is a discrete random variable and p(x) is the probability of X, then the expected value of this random variable is equal to
(a) Σf(x)
(b) Σ[x + f(x)]
(c) Σf(x) + x
(d) Σxp(x)
Solution:
(d) Σxp(x)

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.3

Question 8.
Which of the following is not possible in probability distribution?
(a) Σp(x) > 0
(b) Σp(x) = 1
(c) Σxp(x) = 2
(d) p(x) = -0.5
Solution:
(d) P(x) = -0.5

Question 9.
If c is a constant, then E(c) is
(a) 0
(b) 1
(c) c f(c)
(d) c
Solution:
(d) c

Question 10.
A discrete probability distribution may be represented by
(a) table
(b) graph
(c) mathematical equation
(d) all of these
Solution:
(d) all of these

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.3

Question 11.
A probability density function may be represented by:
(a) table
(b) graph
(c) mathematical equation
(d) both (b) and (c)
Solution:
(d) both (b) and(c)

Question 12.
If c is a constant in a continuous probability distribution, then p(x = c) is always equal to
(a) zero
(b) one
(c) negative
(d) does not exist
Solution:
(a) Zero

Question 13.
E[X – E(X)] is equal to
(a) E(X)
(b) V(X)
(c) 0
(d) E(X) – X
Solution:
(c) 0

Question 14.
E[X – E(X)]² is
(a) E(X)
(b) E(X)²
(c) V(X)
(d) S.D(X)
Solution:
(c) V(X)

Question 15.
If the random variable takes negative values, then the negative values will have
(a) positive probabilities
(b) negative probabilities
(c) constant probabilities
(d) difficult to tell
Solution:
(a) Positive probabilities

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.3

Question 16.
If we have f(x) = 2x, 0 ≤ x ≤ 1, then f(x) is a
(a) probability distribution
(b) probability density function
(c) distribution function
(d) continuous random variable
Solution:
(b) Probability density function

Question 17.
\(\int_{ -∞ }^{∞}\) f(x) dx is always equal to
(a) zero
(b) one
(c) E(X)
(d) f(x) + 1
Solution:
(b) one

Question 18.
A listing of all the outcomes of an experiment and the probability associated with each outcome is called
(a) probability distribution
(b) probability density function
(c) attributes
(d) distribution function
Solution:
(a) Probability distribution

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.3

Question 19.
Which one is not an example of random experiment ?
(a) A coin is tossed and the outcome is either a head or a tail
(b) A six-sided die is rolled
(c) Some number of persons will be admitted to a hospital emergency room during any hour.
(d) All medical insurance claims received by a company in a given year.
Solution:
(d) All medical insurance claims received by a company in a given year

Question 20.
A set of numerical values assigned to a sample space is called
(a) random sample
(b) random variable
(c) random numbers
(d) random experiment
Solution:
(b) random variable

Question 21.
A variable which can assume finite or countably infinite number of values is known as
(a) continuous
(b) discrete
(c) qualitative
(d) none of them
Solution:
(b) Discrete

Question 22.
The probability function of a random variable is defined as
Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.3 1
Then k is equal to
(a) zero
(b) \(\frac { 1 }{4}\)
(c) \(\frac { 1 }{15}\)
(d) one
Solution:
(c) k = 1/15
Hint:
W.K.T Σi=1 P(xi) = 1
p(x= -1)+ p(x= -2) + p(x = 0) + p(x = 1)
+ p(x = 2) = V
k + 2k + 3k + 4k + 5k = 1
5k = 1 ⇒ k = 1/5

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.3

Question 23.
If p(x) = \(\frac { 1 }{10}\), x = 10, then E(X) is
(a) zero
(b) \(\frac { 6 }{8}\)
(c) 1
(d) -1
Solution:
(c) 1
Hint:
P(x) + 1/10 and x = 10
E(x) = Xp(x) = 10(1/10) = 1

Question 24.
A discrete probability function p(x) is always
(a) non-negative
(b) negative
(c) one
(d) zero
Solution:
(a) non-negative

Question 25.
In a discrete probability distribution the sum of all the probabilities is always equal to
(a) zero
(b) one
(c) minimum
(d) maximum
Solution:
(b) one

Question 26.
An expected value of a random variable is equal to it’s
(a) variance
(b) standard deviation
(c) mean
(d) con variance
Solution:
(c) mean

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.3

Question 27.
A discrete probability function p(x) is always non-negative and always lies between
(a) 0 and ∞
(b) 0 and 1
(c) -1 and +1
(d) -∞ and ∞
Solution:
(b) 0 and 1

Question 28.
The probability density function p(x) cannot exceed
(a) zero
(b) one
(c) mean
(d) infinity
Solution:
(b) One

Question 29.
The height of persons in a country is a random variable of the type
(a) discrete random variable
(b) continuous random variable
(c) both (a) and (b)
(d) neither (a) not (b)
Solution:
(b) Continuous random variable

Question 30.
The distribution function F(x) is equal to
(a) P(X = x)
(b) P(X ≤ x)
(c) P (X ≥ x)
(d) all of these
Solution:
(b) p(x ≤ x)

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.3

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.2

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 6 Random Variable and Mathematical Expectation Ex 6.2 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.2

Question 1.
Find the expected value for the random variable of an unbiased die
Solution:
When a un based die is thrown , any one of the number 1, 2, 3, 4, 5, 6, can turn up that x denote the random variable taking the values from 1 to 6
Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.2 1

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.2

Question 2.
Let X be a random variable defining number of students getting A grade. Find the expected value of X from the given table
Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.2 2
Solution:
Let X be a random variable taking values 0, 1, 2, 3,
Expected value E(x) = Σp1x1
= (0.2 × 0) + (0.1 × 1) + (0.4 × 2) + (0.3 × 3)
= 0 + 0.1 + 0.8 + 0.9
∴ E(x) = 0.18

Question 3.
The following table is describing about the probability mass function of the random variable X
Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.2 3
Find the standard deviation of x.
Solution:
Let x be the random variable taking the values 3, 4, 5
E(x) = Σpixi
= (0.1 × 3) + (0.1 × 4) + (0.2 × 5)
0.3 + 0.4 + 1.0
E(x) = 1.7
E(x²) = Σpixi²
= (0.1 × 3²)+ (0.1 × 4²) + (0.2 × 5²)
= (0.1 × 9) + (0.1 × 16) + (0.2 × 25)
E(x²) = 7.5
Var(x) = E(x²) – (E(x)]²
= 7.5 – (1.7)²
= 7.5 – 2.89
Var(x) = 4.61
Standard deviation(S.D) = \(\sqrt { var (x)}\)
= \(\sqrt { 4.61}\)
σ = 2.15

Question 4.
Let X be a continuous random variable with probability density function
fx (x) = \(\left\{\begin{array}{l}
2 x, 0 \leq x \leq 1 \\
0, \text { otherwise }
\end{array}\right.\)
Find the expected value of X.
Solution:
Let x be a continuous random variable. In the probability density function, Expected
Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.2 4

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.2

Question 5.
Let X be a continuous random variable with probability density function
fx (x) = \(\left\{\begin{array}{l}
2 x, 0 \leq x \leq 1 \\
0, \text { otherwise }
\end{array}\right.\)
Find the mean and variance of X.
Solution:
Let x be a continuous random variable. In the probability density function,
Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.2 5
Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.2 6

Question 6.
In an investment, a man can make a profit I of ₹ 5,000 with a probability of 0.62 or a loss of ₹ 8,000 with a probability of 0.38 . Find the expected gain.
Solution:

X5000-8000
P(x = x)0.620.38

Let x be the random variable of getting gain in an Investment
E(x) be the random variable of getting gain in an Investment
E(x) = ΣPixi
= (0.62 × 5000) + [0.38 × (-8000)]
= 3100 – 3040
E(x) = 60
∴ Expected gain = ₹ 60

Question 7.
What are the properties of mathematical expectation?
Properties of Mathematical expectation
Solution:
i. E(a) = a, where ‘a’ is a constant.
ii. E(aX) = aE(X)
iii. E(aX + b) = aE(X) + b, where ‘a’ and ‘b’ are constants.
iv. If X ≥ 0, then E(X) ≥ 0
v. V(a) = 0
vi. If X is random variable, then V(aX + b) = a²V(X)

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.2

Question 8.
What do you understand by mathematical expectation?
Solution:
The average value of a random Phenomenon is termed as mathematical expectation or expected value.
The expected value is weighted average of the values of a random variable may assume

Question 9.
How do you define variance in terms of Mathematical expectation?
Solution:
The variance of X is defined by
Var(X) = Σ[x – E(X)]² p(x)
If X is discrete random variable with probability mass function p(x).
Var(X) = \(\int_{ -∞ }^{∞}\) [x- E(X)]² fx (x) dx
If X is continuous random variable with probability density function fx (x).

Question 10.
Define mathematical expectation in tears of discrete random variable?
Solution:
Let X be a discrete random variable with probability mass function (p.m.f.) p(x). Then, its expected value is defined by
E(X) = \(\sum_{ x }\) x p(x) ……(1)

Question 11.
State the definition of mathematical expectation using continous random variable.
Solution:
If X is a continuous random variable and fix) is the value of its probability density function at x, the expected value of X is
E(X) = \(\int_{ -∞ }^{∞}\) x f(x) dx …….(2)

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.2

Question 12.
In a business venture a man can make a profit of ₹ 2,000 with a probability of 0.4 or have a loss of ₹ 1,000 with a probability of 0.6. What is his expected, variance and standard deviation of profit?
Solution:
Let X be the random variable of getting profit in a business

X2000-1000
P(x = x)0.40.6

E(x) = Σxxpx(x)
= (0.4 × 2000) +[0.6 × (-1000)]
= 800 – 600
E(X) = 200
∴ Expected value of profit = ₹ 200
E(X²) = Σx² Px(x)
= [(2000)² × 0.4] + [(-1000)² × 0.6]
= (4000000 × 0.4) + (1000000 × 0.6)
E(X²) = 2200000
Var(X) = E(X²) – [E(X)]²
= 22000000 – (200)²
= 2200000 – 40000
Var(X) = 21,60,000
Variance of his profit = ₹ 21,60,000
Standard deviation(S.D) = \(\sqrt { var (x)}\)
σ = \(\sqrt { 2160000}\)
σ = 1469.69
Standard deviation of his profit is ₹ 1,469.69

Question 13.
The number of miles an automobile tire lasts before it reaches a critical point in tread wear can be represented by a p.d.f.
Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.2 7
Find the expected number of miles (in thousands) a tire would last until it reaches the critical tread wear point.
Solution:
We know that,
Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.2 8

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.2

Question 14.
A person tosses a coin and is to receive ₹ 4 for a head and is to pay ₹ 2 for a tail. Find the expectation and variance of his gains.
Solution:
Let X denote the amount the person receives in a game
Then X takes values 4,-2 and
So P(X = 4) = P (of getting a head)
= \(\frac { 1 }{2}\)
P(X = – 2) = P (of getting a tail)
= \(\frac { 1 }{2}\)
Hence the Probability distribution is

X4-2
P(x = x)1/21/2

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.2 9
E(x²) = 10
Var(x) = E(x²) – E(x²)
= 10 – (1)²
Var(x) = 9
∴ His expected gain = ₹ 1
His variance of gain = ₹ 9

Question 15.
Let X be a random variable and Y = 2X + 1. What is the variance of Y if variance of X is 5?
Solution:
Var(X) = 5
Var(Y) = var (2x + 1) {∴ v(ax + b) = a²v(x)}
= (2)² var(X)
= 4(5)
Var() = 20

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.2

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Bio Zoology Guide Pdf Chapter 9 Locomotion and Movement Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

11th Bio Zoology Guide Locomotion and Movement Text Book Back Questions and Answers

Part I

Question 1.
Muscles are derived from
a) ectoderm
b) mesoderm
c) endoderm
d) neuro ectoderm
Answer:
b) mesoderm

Question 2.
Muscles are formed by
a) myocytes
b) leucocytes
c) osteocytes
d) lymphocytes
Answer:
a) myocytes

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 3.
The muscles attached to the bones are called
a) skeletal muscle
b) cardiac muscle
c) involuntary muscle
d) smooth muscles
Answer:
a) skeletal muscle

Question 4.
Skeletal muscles are attached to the bones by
a) tendon
b) ligament
c) pectin
d) fibrin
Answer:
a) tendon

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 5.
The bundle of muscle fibres is called
a) Myofibrils
b) fascicle
c) sarcomere
d) sarcoplasm
Answer:
b) fascicle

Question 6.
The pigment present in the muscle fibre to store oxygen is
a) myoglobin
b) troponin
c) myosin
d) actin
Answer:
a) myoglobin

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 7.
The functional unit of a muscle fibre is
a) sarcomere
b) sarcoplasm
c) myosin
d) actin
Answer:
a) sarcomere

Question 8.
The protein present in the thick filament is
a) myosin
b) actin
c) pectin
d) leucin
Answer:
a) myosin

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 9.
The protein present in the thin filament is
a) myosin
b) actin
c) pectin
d) leucin
Answer:
b) actin

Question 10.
The region between two successive Z-discs is called a
a) sarcomere
b) microtubule
c) myoglobin
d) actin
Answer:
a) sarcomere

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 11.
Each skeletal muscle is covered by
a) epimysium
b) perimysium
c) endomysium
d) hypomysium
Answer:
a) epimysium

Question 12.
Knee joint is an example of
a) saddle joint
b) hinge joint
c) pivot joint
d) gliding joint
Answer:
b) hinge joint

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 13.
Name of the joint present between the atlas and axis is
a) synovial joint
b) pivot joint
c) saddle joint
d) hinge joint
Answer:
b) pivot joint

Question 14.
ATPase enzyme needed for muscle contraction is located in
a) actinin
b) troponin
c) myosin
d) actin
Answer:
c) myosin

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 15.
Synovial fluid is found in
a) Ventricles of the brain
b) Spinal cord
c) immovable joint
d) freely movable joints
Answer:
d) freely movable joints

Question 16.
Inflammation of joints due to accumulation of uric acid crystals is called as
a) Gout
b) myasthenia gravis
c) osteoporosis
d) osteomalacia
Answer:
a) Gout

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 17.
Acetabulum is located in
a) collar bone
b) hip bone
c) shoulder bone
d) thigh bone
Answer:
b) hip bone

Question 18.
Appendicular skeleton is
a) girdles and their limbs
b) vertebrae
c) skull and vertebral column
d) ribs and sternum
Answer:
a) girdles and their limbs

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 19.
The type of movement exhibits by the macrophages are
a) flagellar
b) ciliary
c) muscular
d) amoeboid
Answer:
d) amoeboid

Question 20.
The pointed portion of the elbow is
a) acromion process
b) glenoid cavity
c) olecranon process
d) symphysis
Answer:
c) olecranon process

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 21.
Name the different types of movement
Answer:

  • Amoeboid movement
  • Ciliary movement
  • Flagellar movement
  • Muscular movement

Question 22.
Name the filaments present in the Sarcomere.
Answer:
Thick and thin filaments are the two types of filaments present inside the sarcomere.

Question 23.
Name the contractile proteins present in the skeletal muscle.
Answer:

  • Myosin – Thick filament
  • Actin – Thin filament

Question 24.
When describing a skeletal muscle. What does “Striated mean?
Answer:
Each skeletal muscle fibre has a repeated series of dark and light bands. The dark A-bands and light I-bands give a striated appearance to the muscle.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 25.
How does an isotonic contraction takes place?
Answer:
In isotonic contraction, the length of the muscle changes but the tension remains constant.
(eg) lifting dumbbells and weight lifting.

Question 26.
How does an isometric contraction take place?
Answer:
In isotonic contraction, the length of the muscle changes but the tension remains constant. The force produced is unchanged, e.g., lifting dumbbells and weight lifting.

Question 27.
Name the bones of the skull.
Answer:
The cranial bones are 8 in number.

  1. Paired parietal
  2. Paired temporal
  3. frontal
  4. Sphenoid
  5. Occipital
  6. Ethmoid

Question 28.
Which is the only jointless bone in the human body?
Answer:
The jointless bone is the hyoid bone in our throat. The hyoid bone (lingual bone) is a horseshoe.

Question 29.
List the three main parts of the axial skeleton.
Answer:

  1. Cranium
  2. Hyoid (Lingual)
  3. Vertebral column
  4. Thoracic cavity.

Question 30.
How is tetany caused?
Answer:
Due to the deficiency of parathyroid hormone the level of calcium decreases in the blood that leads to rapid muscle spasm called tetany.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 31.
How does rigor mortis happen?
Answer:
After the death of an individual, the membrane of muscle cells becomes more permeable to calcium ions. This happens due to the partial contraction of skeletal muscles. The contracted muscles are unable to relax. This condition is known as rigor mortis.

Question 32.
What are the different types of rib bones that form the rib cage?
Answer:

  • True rib bones (First 7 pairs)
  • False rib bones (8,9,10th pairs)
  • Floating rib bones (11 and 12th pair)

Question 33.
What are the bones that make the pelvic girdle?
Answer:

  • Ilium
  • Ischium
  • Pubis

Question 34.
List the disorders of the muscular system.
Answer:

  • Muscle fatigue
  • Atrophy
  • Muscle pull
  • Muscular dystrophy

Arthritis

  • Osteoarthritis
  • Rheumatoid arthritis
  • Gout

Question 35.
Explain the sliding-filament theory of muscle contraction.
Answer:
Sliding filament theory is an active process. It is proposed by Andrw F. Huxley in 1954 and Rolf Niedergerke.
Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement 1

  • Muscle contraction is initiated by a nerve impulse sents by the central nervous system through a motor neuron
  • When the nerve impulse reaches neuromuscular junction acetylcholine is released and created action potential.
  • This action potential triggers the release of calcium from the sarcoplasmic reticulum
  • The released calcium ions bind to troponin on thin filaments.
  • The active sites are exposed to the heads of myosin to form a cross bridge. Hence actin and myosin form a protein complex called actomyosin.
  • Utilizing the energy released from the hydrolysis of ATP the myosin head rotates until it forms a 90° angle with a long axis of the filament.
  • The power stroke begins after the myosin head and hinge region tilt from a 90° angle to a 45° angle.
  • The cross-bridge transforms into strong high force bond which allows the myosin head to swells it.
  • When the myosin head swells it pulls the attached actin filament towards the centre of the A – band.
  • The myosin returns back to its relaxed state and releases ADP and phosphate ions. A Newer ATP molecule binds to the head of myosin and the cross-bridge is broken.
  • At the end of each power stroke each myosin head detaches from actin then swivels back and binds to a new actin molecule to start another contraction cycle.
  • The power stroke repeats many times and the thin filaments move toward the centre of the sarcomere.
  • In this process, there is no change in the lengths of thick or thin filaments.
  • The Z – discs attached to the actin filaments are also pulled inwards from both sides causing the shortening of the sarcomere. This process continues.
  • When motor impulse stops the calcium ions are purnbed back into the sarcoplasmic reticulum results in the masking of the active sites of the actin filament and the myosin head fails to bind with the actin and causes Z – discs back to their original relaxed position.

Question 36.
What are the Benefits of regular exercise?
Answer:

  1. The benefits of regular exercise are:
  2. The muscles used in exercise grow larger and stronger.
  3. The resting heart rate goes down.
  4. More enzymes are synthesized in the muscle fibre.
  5. Ligaments and tendons become stronger.
  6. Joints become more flexible.
  7. Protection from a heart attack.
  8. Influences hormonal activity.
  9. Improves cognitive functions.
  10. Prevents obesity.
  11. Promotes confidence, esteem.
  12. Aesthetically better with a good physique.
  13. Overall well-being with good quality of life.
  14. Prevents depression, stress, and anxiety.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Part II.

11th Bio Zoology Guide Locomotion and Movement Additional Important Questions and Answers

I. Choose The Best Options

Question 1.
The pseudopodia of amoeba is formed from
a) Cytoplasm
b) Nucleoplasm
c) Sarcoplasm
d) all the above
Answer:
a) Cytoplasm

Question 2.
Where is ciliary movement taking place?
Answer:
a) Respiratory tract
b) Stomach
c) oesophagus
d) Reproductive tract
Answer:
d) Reproductive tract

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 3.
Name the movement of spermatozoa
a) amoeboid movement
b) flagellated movement
c) Ciliary movement
d) None of the above
Answer:
b) flagellated movement

Question 4.
Muscles are made up of ………………….
a) Myocytes
b) Lymphocytes
c) adenocytes
d) leucocytes
Answer:
a) Myocytes

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 5.
The cytoplasm of the muscle fibre is called:
a) Myofibril
b) Sarcomere
c) Sarcoplasm
d) Sarcolemma
Answer:
c) Sarcoplasm

Question 6.
Name the connective tissue which covers the entire muscle?
a) Epimycium
b) Perimycium
c) Endomyciurn
d) Mesornyscium
Answer:
a) Epimycium

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 7.
Name the membrane which covers each fasiculi
a) Epimycium
b) Perirnycium
c) Endomyciurn
d) Mesomvscium
Answer:
b) Perirnycium

Question 8.
Name the membrane which covers each muscle fibre?
a) Epimycium
b) Perimycium
c) Endomycium
d) Mesomysciurn
Answer:
c) Endomycium

Question 9.
Match and Find the Correct Pair
1. Sarcoplasm – a) Respiratory pigment
2. Myoglobin – b) Glucose giver
3. Glycosome – c) Unit of skeletal muscle
4. Sarcomere – d) cytoplasm
Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement 2
Answer:
a) I- d, II – a, III – b,IV- c

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 10.
Find the correct and wrong statement and arrange the following statement.
1. The contraction of muscle fibres depends on the actin and myosin protein.
2. The thick muscle fibres depends on Myosin.
3. Each meromyosin molecule will have a globular head with a long arm.
4. The head of the meromyosin bears an actin-binding site and an ATP binding site.
Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement 3
Answer:
a) True True False True

Question 11.
Name the protein that regulates muscle contraction.
a) Tropomyosin and actin
b) Troponin and myosin
c) tropomyosin and troponin
d) None of the above
Answer:
c) tropomyosin and troponin

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 12.
Who has proposed sliding – filaments hypothesis?
a) Andrew F. Huxley and Rolf Nieder gerke
b) Andrew F. Huxley and Nelson
c) Andrew F. Pluxley and Darwin
d) Andrew F. Huxley and Mendal
Answer:
a) Andrew F. Huxley and Rolf Nieder gerke

Question 13.
Which chemical initiates the opening of multiple gated channels in sarcolemma?
a) Epinephrine
b) Norepinephrine
c) acetylcholine
d) erythromycin
Answer:
c) acetylcholine

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 14.
Find out the wrong pair
a) Fast – oxidative fibres – have high ATP ase activity
b) Slow – oxidative fibres – low rates of ATP ase activity
c) Oxidative fibres – less number of mitochondria
d) Red muscle fibres – oxidative fibres
Answer:
c) Oxidative fibres – less number of mitochondria

Question 15.
The skeletal system is formed from this layer!
a) Ectoderm
b) Mesoderm
c) Endoderm
d) Neuroderm
Answer:
b) Mesoderm

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 16.
………………………… a number of bones forms the endoskeleton of man?
a) 210
b) 220
c) 206
d) 209
Answer:
c) 206

Question 17.
Hove many bones are there in the axial and appendicular skeleton?
a) 80 and 126
b) 126 and 80
c) 80 and 120
d) 80 and 118
Answer:
a) 80 and 126

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 18.
How many bones are there in the facial and cranial bones?
a) 14 and 9
b) 14 and 8
c) 14 and 10
d) 14 and 12
Answer:
b) 14 and 8

Question 19.
Name the opening of the temporal bone
a) External auditory meatus
b) Nasal opening
c) Optic opening
d) Mouth
Answer:
a) External auditory meatus

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 20.
Name the U-shaped single bone present at the base of the buccal cavity.
a) Palantine bone
b) Hyoid bone
c) Ethmoid bone
d) Sphenoid bone
Answer:
b) Hyoid bone

Question 21.
How many bones from the vertebral column?
a) 33
b) 32
c) 30
d) 36
Answer:
a) 33

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 22.
Match and find the correct pair
1) Cervical vertebrae – a) – 5
2) Thoraic – b)-l
3) Pelvic bone – 0-7
4) Cocyx bone – d) -12
Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement 4
Answer:
a) I -c, II – d, III – a,IV – b

Question 23.
In which bone ………….. situated?
a) Cervical vertebra
b) thoracic vertebra
c) Pelvic vertebra
d) all the above
Answer:
d) all the above

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 24.
Name the first vertebra?
a) Atlas
b) Maleus
c) Incus
d) Stapes
Answer:
a) Atlas

Question 25.
Name the second vertebra?
a) Atlas
b) Axis
c) Maleus
d) Stapes
Answer:
b) Axis

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 26.
How many are the number of ribs?
a) 14 pair
b) 13 pair
c) 12 pair
d) 15 pair
Answer:
c) 12 pair

Question 27.
Match and find the correct answer
1) ribs – a) – 1, 7
2) True ribs – b) -11,12
3) False ribs – c) – Bicephalic
4) Floating ribs – d) – 8,10
Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement 5
Answer:
b) I – c, II – a, III – d,IV -b

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 28.
How many bones are there appendicular skeleton?
a) 130
b) 140
c) 126
d) 122
Answer:
c) 126

Question 29.
Which forms the appendicular skeleton?
a) Upper and hindlimbs
b) Upper limbs and thoracic bones
c) Hind and vertebral column
d) Hind and cranial bones
Answer:
a) Upper and hindlimbs

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 30.
Name the bones which joint the axial and appendicular skeleton.
a) Clavicle bone
b) Scapula
c) Acromian process
d) None of the above
Answer:
a) Clavicle bone

Question 31.
Flow many bones are there in the upper arm.
a) 30
b) 32
c) 34
d) 36
Answer:
a) 30

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 32.
Find the wrong pair.
a) Wrist bone – 8
b) Fore arm bones – 30
c) Facial bone -16
d) Cranial bones – 8
Answer:
c) Facial bone -16

Question 33.
Find the wrong pair
a) Palm bones – 5
b) Phalanges -14
c) Thoracic bone – 12
d) Vertebral column – 33
Answer:
d) Vertebral column – 33

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 34.
Match and find the correct pair.
1) Cervical vertebra – a) – 12
2) Thoracic vertebra – b) – 1
3) Pelvic vertebra – c) – 7
4) Coccyx – d)-5
Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement 6
Answer:
a) I- c, II – a, III – d, IV – b

Question 35.
The oxidative skeletal muscle fibres are termed as
a) Fatty muscle fibres
b) White muscle fibres
c) red muscle fibres
d) yellow muscle
Answer:
c) red muscle fibres

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 36.
Whether the following statement is correct or wrong find out the correct sequence.
1) Pelvic bone Ilium, Ischium, and pubis
2) Acetabulam cavity is present in the sacrum
3) The head of the thigh bone fits in the acetabulum cavity
4) The pubic bones articulate anteriorly at the pubic symphysis
Sequence:
a) 1 – True, 2 – False, 3 – True, 4 – True
b) 1 – False, 2 – False, 3 – True, 4 – True
c) 1 – False, 2 – True, 3 – True, 4 – True
d) 1 – True, 2 – True, 3 – False, 4 – True
Answer:
a) 1 – True, 2 – False, 3 – True, 4 – True

Question 37.
Which is the prominent bone of the pelvic bone?
a) Ilium
b) Ischium
c) Pubis
d) all the above
Answer:
a) Ilium

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 38.
What is the number of hind limbs?
a) 32
b) 37
c) 30
d) 34
Answer:
c) 30

Question 39.
Name the longest bone
a) Femur
b) Humerus bone
c) Tibia
d) all the above
Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement 7
Answer:
a) I -a, II – d, III – c, IV -b

Question 40.
Match and find the correct pair.
1)Patella – a)-14
2) Tarsus – b) -5
3) Metatarsus – c) – 7 bones
4) Phalanges – d) – kneecap
Answer:
3) Metatarsus – c) – 7 bones

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 41.
Name the membrane which covers the femur bone.
a) Periosteum
b) Endosteum
c) Osteoclast
d) Osteoblast
Answer:
a) Periosteum

Question 42.
Name the joints seen in the cranial region
a) Fibrous joints
b) Cartilagenous joints
c) Diarthroses joints
d) Synovial joints
Answer:
a) Fibrous joints

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 43.
The Synovial fluid is seen in joints.
a) Cartilagenous joints
b) Diarthroses joints
c) Fibrous joints
d) all the above
Answer:
b) Diarthroses joints

Question 44.
Name the diseases due to the deficiency of acetylcholine.
a) Myasthenia gravis
b) Tetany
c) Duchene muscular dystrophy
d) Muscle pull
Answer:
a) Myasthenia gravis

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 45.
Name the disease due to the deficiency of ATP
a) Muscle fatigue
b) Muscle pull
c) Muscular dystrophy
d) Tetany
Answer:
a) Muscle fatigue

Question 46.
A tear in the muscle
a) Muscle fatigue
b) Muscle pull
c) Atrophy
d) Muscular dystrophy
Answer:
b) Muscle pull

Question 47.
Name the arthritis due to aging.
a) Osteoarthritis
b) Rheumatoid arthritis
c) Gout
d) Osteoporosis
Answer:
a) Osteoarthritis

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 48.
Osteoporosis is due to
a) Calcium
b) Sodium
c) Magnesium
d) Potassium
Answer:
a) Calcium

Question 49.
The deposition of urate crystals on the joints is called as
a) Gout
b) Rheumatoid
c) arthritis
d) Osteoporosis
Answer:
a) Gout

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 50.
Assertion A: Acetylcholine is secreted in the Neuromuscular junction.
Reason B: If acetylcholine is not secreted there won’t be any multiple gated channels in the sarcolemma.
a) A true B wrong
b) A true B this explains the action of A
c) A wrong B true
d) A wrong B wrong
Answer:
b) A true B this explains the action of A

Question 51.
Assertion A: The upper limbs are attached to the pectoral girdles.
Reason B: The pectoral girdles are very light and allow the mobility of the hand
a) A wrong B wrong
b) A True B does not explains the A
c) A True B explains the functions of A
d) A True B wrong
Answer:
c) A True B explains the functions of A

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 52.
Assertion A: In the pelvis bone a deep socket is present called acetabulum.
Reason B: The head of the femur bone fits in the acetabulum.
a) A True B explains the functions of A
b) A wrong B explains the functions of A
c) A True B wrong
d) B does not explains the structure of B
Answer:
a) A True B explains the functions of A

Question 53.
Assertion A: The pelvis of female is shallow wide and flexible in nature
Cause B: This helps during pregnancy.
a) A True B explains the functions of A
b) A wrong B explains the functions of A
c) A True B wrong
d) B does not explains the structure of A
Answer:
a) A True B explains the functions of A

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 54.
Assertion A: The lower arm carries the entire weight of the body and is subjected to exceptional forces when we jump or run.
Reason B: To bear the weight of the body it has 46 bones
a) A True B True
b) A wrong B wrong
c) A True B does not explains the functions of A
d) A True B Wrong
Answer:
c) A True B does not explains the functions of A

Question 55.
Assertion A: The cranium belongs to immovable fixed joints.
Reason B: Structures of the flat skull bones are fibrous joints.
a) A True B Wrong
b) A True B explains the structure of A
c) A True B True
d) B does not explains the structure of A
Answer:
b) A True B explains the structure of A

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 56.
Assertion A: The decreased synthesis of acetylcholine in the neuromuscular junction, causes myasthenia gravis.
Reason B: This leads to muscle fatigue weakness, paralysis.
a) A and B are True
b) A and B are wrong
c) A True B wrong
d) A wrong B True
Answer:
a) A and B are True

Question 57.
Match and arrange the sequence :
Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement 8
Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement 9
Answer:
a) I – D, II – C, III-A,IV – B

Question 58.
Match and arrange the sequence:
Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement 10
Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement 11
Answer:
b) I – C, II – D, III – B, IV -A

2 marks

II. Very Short answers

Question 1.
What is amoeboid movement?
Answer:
The movement of cells by streaming movements of the cytoplasm forming pseudo-podia is known as amoeboid movement, e.g., macrophages.

Question 2.
What are the types of muscles?
Answer:

  • Skeletal muscle
  • Visceral muscles
  • Cardiac muscles

Question 3.
Name the muscle protein?
Answer:

  • Actin
  • Myosin

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 4.
Name the regulatory proteins in the thin filaments.
Answer:

  • Tropomyosin
  • Troponin

Question 5.
Classify the muscles on the basis of their rate of shortening?
Answer:

  • Fast contraction fibre
  • Slow contraction fibre

Question 6.
On the basis of ATP formation, how are muscles classified?
Answer:

  • Oxidative fibres
  • Glycolytic fibres

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 7.
How are muscles classified on the basis of ATP are activity?
Answer:

  • Fast oxidative fibres
  • Slow oxidative fibres

Question 8.
What is perimysium?
Answer:
The connective tissue covering around each fascicle is the perimysium.

Question 9.
Differentiate the oxidative fibre from the glycolytic fibre.
Answer:

Oxidative fibreGlycolytic fibre
1. Numerous mitochondriaThere are few mitochondria
2. Depends on blood flowNot depend on blood flow
3. Myoglobin is presentNo myoglobin
4. These are known as red muscle fibresThese are called muscle fibres as white muscle fibres

Question 10.
What is sarcoplasm?
Answer:
The cytoplasm of the muscle fibre is called the sarcoplasm.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 11.
What is meant by exo skeleton?
Answer:
It is a rigid hard case present outside the body of animals (eg) Cockroach.

Question 12.
What is endo skeleton?
Answer:
It is found inside the body of vertebrates. It is composed of bones and cartilages, (eg) Man.

Question 13.
What are the two types of endo skeletons?
Answer:

  1. Axial skeleton.
  2. Appendicular skeleton.

Question 14.
What are Glycosomes?
Answer:
Glycosomes are the granules of stored glycogen that provide glucose during the period of muscle fiber activity.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 15.
What is a brainbox?
Answer:
The cranial bones form the hard protective outer covering of the brain and called the brain box.

Question 16.
Name the ear ossicles?
Answer:

  • Malleus
  • incus
  • stapes

Question 17.
Give notes on Jaw bones?
Answer:

  • It is composed of the upper jaw which is formed of the maxilla and the lower jaw is formed of the mandible.
  • The upper jaw is fused with the cranium and is immovable.
  • The lower jaw is connected to the cranium and is movable.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 18.
Name the openings of the skull?
Answer:

  • The orbits
  • Nasal cavity
  • Foramen magnum

Question 19.
What is meant by foramen magnum?
Answer:

  • It is a large opening found at the posterior base of the skull.
  • Through this opening, the medulla oblongata of the brain descends down as the spinal cord.

Question 20.
What are oxidative fibres?
Answer:
The muscle fibres that contain numerous mitochondria and have a high capacity for oxidative phosphorylation are classified as oxidative fibres. They are also called red muscle fibres.

Question 21.
Name the first two bones of the vertebral column.
Answer:

  1. Atlas
  2. Axis

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 22.
What are the functions of the vertebral column?
Answer:

  • It protects the spinal cord.
  • Supports the head
  • Serves as the point of attachment for the ribs and musculature of the back.

Question 23.
Give short notes on sternum?
Answer:

  • The sternum is a flat bone on the mid ventral line of j the thorax.
  • It provides space for the attachment of the thoracic ribs and abdominal muscles.

Question 24.
What is a hydrostatic skeleton?
Answer:
The skeleton found in soft-bodied invertebrates is called a hydrostatic skeleton. It is a fluid-filled cavity encircled by muscles, e.g., the earthworm.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 25.
What are true ribs?
Answer:
The first seven pairs of ribs are called true ribs. Dorsaily they are attached to the thoracic vertebrae and ventrally connected to the sternum.

Question 26.
What are false ribs?
Answer:
The 8th, 9th, and 10th pairs of ribs do not articulate directly with the sternum but joined with the seventh rib.

Question 27.
What is the sternum?
Answer:
The sternum is a flat bone on the midventral line of the thorax. It provides space for the attachment of the thoracic ribs and abdominal muscles.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 28.
What is meant by appendicular skeleton?
Answer:
The bones of the upper and lower limbs along with their girdles constitute the appendicular skeleton. It is composed by 126 bones.

Question 29.
White the 3 segments of lower limb.
Answer:

  1. The thigh
  2. the leg or the shank and
  3. the foot.

Question 30.
What is meant by acromian process?
Answer:
The scapula has a slightly elevated ridge called the spine which projects as flat expanded process called the acromion.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 31.
What is meant by glenoid cavity?
Answer:
Below the acromian is a depression called the glenoid cavity which articulates with the head of the humerus to form the shoulder joint.

Question 32.
What is meant by olecranon process?
Answer:
The radius and ultra bones present in the fore arm that form the pointed portion of the elbow called olecranon process.

Question 33.
What is meant by carpal tunnel?
Answer:

  • There are 8 bones in the wrist arranged in two rows of four each.
  • The anterior surface of the wrist has tunnel like appearance. This tunnel is termed as carpal tunnel.

Question 34.
Name the bones which forms the coxal bones.
Answer:

  • Ilium
  • Ischium
  • Pubis

Question 35.
What is meant by pubic symphysis?
Answer:
Ventrally the two halves of the pelvic girdle meet and form the pubic symphysis containing fibrous cartilage.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 36.
Where calcium ion binds with the muscle fibre? Name the molecules which binds with calcium?
Answer:

  • The calcium released from sarcoplasma binds with the thin fibre of the muscle.
  • The released calcium binds to troponin a thin filaments.

III. Fill Up The Blanks With Suitable Options

1. Scapula – Acromian process
……………. – Bones of upper arm
2. First 7 pair of rib bones – True ribs
11 and 12th pair of ribs – …………….
3. Cervical vertebrae – 7
……………. – Lumbar bones
4. Skull bones – 22
……………. – Skull bones
5. Thick fibres – Myosin
……………. – Thin fibres
6. Amoeboid movement – Macrophage cells
……………. – Sperm cells
Answer:

  1. Olecronan process
  2. Floating ribs
  3. 5
  4. 8
  5. Actin
  6. Flagellated movement

3 marks

IV. Short answers

Question 1.
Give short notes on skeletal muscle and their covering membranes.
Answer:

  • Each muscle is made up of bundles of muscle fibres called fascicle. Each muscle fibre contains rod like structures called myofibrils.
  • The connective tissue covering the muscle is the epimysium.
  • The covering around each fascicle is the perimysium.
  • The muscle fibre is surrounded by endomysium.

Question 2.
Give the structure of a skeletel muscle fibre.
Answer:

  • Each muscle fibre is thin and elongated.
  • Most of them taper at one or both ends.
  • Muscle fibres are surrounded by sarcolemma the cytoplasm of the muscle fibre is called the sarcoplasm.
  • It contains glycosomes myoglobin and sarcoplasmic reticulum.
  • Myoglobin is a red coloured respiratory pigment and glycosomes are reserved glycogen.
  • Muscle fibres contain muscle protein actin and myosin.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 3.
Give notes on slow oxidative fibres.
Answer:

  • These fibres have low rates of myosin ATP hydrolysis but have the ability to make large amounts of ATP.
  • This type of fibres seen in long distance swimmers and long distance runners.

Question 4.
Give notes on fast – oxidative fibres.
Answer:

  • These fibres have high myosin ATP ase activity and can make large amounts of ATP.
  • They are suited for rapid action.

Question 5.
Give notes on fast glycolytic fibres.
Answer:

  • These fibres have myosin ATP ase activity but cannot make as much ATP as oxidative fibres because their source of ATP is glycolysis.
  • These fibres are best suited for rapid intense actions such as short sprint at maximum speed.

Question 6.
Name the facial bones.
Answer:
There are 14 facial bones.

  1. Pair of maxilla
  2. Pair of Zygomatic
  3. Pair of Palantine
  4. Pair of lacrimal
  5.  Pair of Nasal
  6. Mandible or lower jaw
  7. Vomer

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 7.
Give notes on fibrous joints.
Answer:

  • They are immovable fixed joints in which no movement between the bones is possible.
  • Sutures of the flat skull bones are fibrous joints.

Question 8.
Give notes on cartilaginous joints.
Answer:
They are slightly movable joints in which the joint surface are separated by a cartilage and slight movement is only possible.

Question 9.
Give notes on synovial joints.
Answer:
They are freely movable joints the articulating bones are separated by a cavity which is filled with synovial fluid.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 10.
Give notes on myasthenia gravis.
Answer:

  • It is an auto immune disorder affecting the action of acetylcholine at neuro muscular junction leading to fatique.
  • Weakening and paralysis of skeletal muscles.
  • Acetylcholine receptors on the sarcolemma are blocked by antibodies leading to weakness of muscles.
  • When the disease progresses it can make chewing swallowing talking and even breathing difficult.

Question 11.
Give notes on muscle fatique.
Answer:

  • It is the inability of a muscle to contract after repeated muscle contraction.
  • This is due to lack of ATP and accumulation of lactic acid by anaerobic break down of glucose.

Question 12.
Give notes on Atrophy of muscles.
Answer:

  • A decrease in the activity of muscles results in the atrophy of muscles.
  • There is a reduction in the size of the muscle and makes the muscle to become weak which occurs with lack of usage as in chronic bed ridden patients.

Question 13.
What is meant by muscle pull?
Answer:

  • Muscle pull is actually a muscle tear.
    Atraumatic pulling of the fibres produces a tear known as sprain.
  • This can occur due to sudden stretching of muscle beyond the point of elasticity.
  • Back pain is a common problem caused by muscle pull due to improper posture with static sitting for long hours.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 14.
What is meant by muscular dystrophy?
Answer:

  • The group of diseases collectively called the muscular dystrophy are associated with the progressive degeneration of skeletel muscle fibres weakening the muscles and leading to death from lung or heart failure.
  • (eg) Duchene muscular dystrophy.

Question 15.
What is meant by skeletel muscle glycogen analysis?
Answer:

  • This is used to measure an athlete’s muscle glycogen.
  • Muscle glycogen provides the main source of energy during anaerobic exercise.
  • A single glycogen molecule may contain 5000 glucose molecules.

Question 16.
Give notes on osteoporosis?
Answer:

  • It occurs due to deficiency of vitamin D and hormonal imbalance.
  • It causes rickets in children and osteomalasia in adult females.
  • The bones become soft and fragile.
  • It can be minimized with adequate calcium intake vitamin D intake and regular physical activities.

Question 17.
What is carpal tunnel syndrome?
Answer:

  • The narrow passage bounded by bones and ligaments in the wrist get narrowed and pinches the median nerve.
  • This syndrome is mostly seen among the clerks, software professional and people who constantly play or text in mobile phones.

5 Marks

V. Give Detailed Answers 

Question 1.
Describe about the different types of movements with examples.
Answer:
Types of Movements :

  1. Amoeboid movements
  2. Ciliary movements
  3. Flagellar movements
  4. Muscular movements

1. Amoeboid movements:
Cells such as macrophages exhibit amoeboid movement for engulfing pathogens by pseudopodia formed by the streaming movement of the cytoplasm.

2. Ciliary movements :
This type of movement occurs in the respiratory and genital passages.

3. Flagellar movement:
This type of movement occurs in the cells which are having flagella or whiplike motile organels. (eg) Sperm cells.

4. Muscular movement:
The movement of hands legs jaws are caused by the contraction and relaxation of the muscle.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 2.
Describe the structure of sarcomere.
Answer:

  • The unit of the skeletal muscle is sarcomere A sarcomere is the region of a myofibril between two successive z – discs.
  • It contains an ‘A’ band with a half I band which are perfectly aligned with one another.
  • This type of arrangement gives the cell a striated appearance.
  • Each dark band has a lighter region in its middle called the M – zone.
  • Each H – zone is bisected vertically by a dark line called the M – line.
  • The I bands have a darker mid line area called the z – disc.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement 12

  • Inside the sarcomere two types of filaments are present namely the thick filaments and thin filaments.
  • The thick filaments extend the entire length of the A band, the thin filaments extend across the I band and partly into the A – band.
  • The invagination of the sarcolemma forms transverse (T- tubules) tubules and they penetrate into the junction between the A and I bands.

Question 3.
Describe the structure of muscle protein.
Answer:
Contraction of the muscle depends on the presence of contractile proteins such as actin and myosin.
Myosin fibre:

  • The thick filaments are composed of the protein myosin.
  • Each myosin molecule is made up of monomer called meromyosin.
  • The meromyosin have a globular head with a j short arm and a tail.
  • The short arm have heavy meromyosin and the tail portion have the light meromyosin.
  • The head bears an actin – binding site and an ATP binding site
  • It also contains ATP ase enzyme that split ATP to generate energy for the contraction of muscle.

Actin filament:

  • Actin has polypeptide subunits called globular actin or G – actin and filamentous form F – actin.
  • Each thin filament is made of two F – actins helically wound to each other.
  • Each F – actin is a polymer of monomeric G – actins, It also contains a binding site for myosin.
  • The thin filament contain several regulatory protein like tropomyosin, troponin, which help in regulating the contraction of muscles along with actin and myosin.

Thick filament:
Each thick filament consists of many myosin molecules whose heads produce at opposite ends of the filament Portion of a thick filament

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement 13

Thin filament:
A thin filament consists of two strands of actin subunits twisted into a helix plus two types of regulatory proteins (troponin and tropomyosin) Portion of a thin filament.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement 14

Question 4.
Give the schematic representation of muscle contraction.
Answer:
Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement 15

Question 5.
Give the four important features of skeletal muscles.
Answer:

  1. Excitability :
    Based on the chemical and electrical excitation the muscles contracts.
  2. Contractility:
    It is the ability of the muscle which gives movement to the attached organs.
  3. Conductivity:
    The excitation at one part of the muscle is connected to the other part of the muscle.
  4. Elasticity :
    The ability of the muscle to returns to its original position after the extension of the muscles.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 6.
Explain the bones that form the skull?
Answer:
The skull is composed of two sets of bones – cranial and facial bones. It consists of 22 bones of which 8 are cranial bones and 14 are facial, bones. The cranial bones form the hard protective outer covering of the brain and called the brain box. The capacity of the cranium is 1500 cm3.

These bones are joined by sutures which are immovable. They are paired parietal, paired temporal and individual bones such as the frontal, sphenoid, occipital and ethmoid. The large hole in the temporal bone is the external auditory meatus. In the facial bones maxilla, zygomatic, palatine, lacrimal, nasal are paired bones whereas mandible or lower jaw and vomer are unpaired bones. They form the front part of the skull.

A single U-shaped hyoid bone is present at the base of the buccal cavity. It is the only bone without any joint. Each middle ear contains three tiny bones- malleus, incus, and stapes collectively are called ear ossicles. The upper jaw is formed of the maxilla and the lower jaw is formed of the mandible.

The upper jaw is fused with the cranium and is immovable. The lower jaw is connected to the cranium by muscles and is movable. The most prominent openings in the skull are the orbits and the nasal cavity. The foramen magnum is a large opening found at the posterior base of the skull. Through this opening, the medulla oblongata of the brain descends down as the spinal cord.

Question 7.
Give an account of vertebral column.
Answer:

  • It consists of 33 serially arranged vertebrae which are interconnected by a cartilage known as inter vertebral disc.
  • The vertebral column extends from the base of the skull to the pelvis and forms the main framework of the trunk.

It has five major regions, they are

  1. Cervical – 7
  2. Thoracic vertebrae -12
  3. Lumbar vertebrae – 5
  4. Sacrum – 5 sacral vertebrae found in the infant which are fused to form one bone in the adult.
  5. Coccyx – 1 – 4 Coccy geal vertebrae found in the infant which are fused to form one bone in the adult.

Question 8.
Give an account of the ribcage.
Answer:

  • There are 12 pairs of ribs.
  • Each rib bone is connected dorsally to the vertebral column and ventrally to the sternum.
  • It has two articulation surfaces on its dorsal end called bicephalic.
  • The first 7 pairs of ribs are called true ribsorvertebro – sternal ribs.
  • Dorsally they are attached to the thoracic vertebrae and ventrally connected to the sternum with the help of hyaline cartilages.
  • The 8th, 9th and 10th pairs of ribs do not articulate directly with the sternum but joined with the cartilaginous part of the seventh rib.
  • These are called false ribs or vertebro – chondral ribs.
  • The last 11th and 12th pairs of ribs are not connected ventrally.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement 16

  • They are called as floating ribs or vertebral ribs.
  • Thoracic vertebrae ribs and sternum form the rib cage.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 9.
Give an account of the pectoral girdle?
Answer:

  • The upper limbs are attached to the pectoral girdles.
  • These are very light and allow the upper limbs a degree of mobility not seen any where else in the body.
  • The girdle is formed of two halves.
  • Each pectoral girdle consists of a clavicle or collar bone and a scapula.
  • The scapula is a large triangular bone situated in the dorsal surface of the ribcage between the second and seventh ribs.
  • It has a elevated expanded process called the acromion.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement 17

  • The clavicle articulates this process.
  • Below the acromion is a depression called the glenoid cavity which articulates with the head of the humerus to form the shoulder joint.
  • Each clavicle is a long slender bone with two curvatures which lie horizontally and connect the axial skeleton with the appendicular skeleton.

Question 10.
Describe the structure of the upper limb.
Answer:

  • The upper limb consists of 30 separate bones and is specialized for mobility.
  • The region between the shoulder and elbow is the humerus.
  • The head of humerus articulates with the glenoid cavity of the scapula and forms the shoulder joint.
  • The distel end of humerus articulates with the two fore arm bones the radius and ulna
  • Olecranon process is situated at the upper end of the ulna which forms the pointed portion of the elbow.
  • The hand consists of carpals metacarpals and phalanges.
  • Carpals the wrist bones 8 in number are arranged in two rows of four each and form a tunnel termed as carpal tunnel.
  • Meta carpals the palm bones are 5 in number and phalanges the digit bones are 14 in number.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement 18

Question 11.
Give an account of pelvic girdle.
Answer:

  • The pelvic girdle is a heavy structure specialised for weight-bearing.
  • It is composed of two hib bones called coxal bones that secure the lower limbs to the axial skeleton.
  • Together with the sacrum and coccyx the hib bones form the basin like bonypelvis.
  • Each coxal bone consists of three fused bones ilium, ischium, and pubis.
  • At the point of fusion of these three bones forms a deep hemispherical socket called the acetabulum present on the lateral surface of the pelvis.
  • It receives the head of the femur at hip joint and helps in the articulation of the femur.
  • Ventrally the two halves of the pelvic girdle meet and form the pubic symphysis containing fibrous cartilage.
  • The ilium is the superior flaring portion of the hip bone. Each ilium forms a secure joint with the sacrum posteriorly.
  • The ischium is a curved bar of bone. The ‘V’ shaped pubic bones articulate anteriorly at the pubic symphysis.
  • The pelvis of male is deep and narrow with larger heavier bones and the female is shallow wide and flexible in nature and this helps during pregnancy which is influenced by female hormones.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement 19

Question 12.
Give an account of the lower limb.
Answer:

  • The lower limb consists of 30 bones which carries the entire weight of the erect body and is subjected to exceptional forces when we jump or run.
  • The bones of the lower limbs are thicker and stronger than the upper limbs.
  • Each lower limb consists of the thigh, the leg or the shank and the foot.
  • The femur is the strongest and longest bone of the body.
  • The head of femur articulates with the acetabulum of the pelvis to form the hip joint.
  • The tibia and fibula form the skeleton of the shank.
  • A thick triangular patella forms the knee cap which protects the knee joint arteriorly and improves the leverages of thigh muscles acting across the knee.
  • The foot includes the bones of ankle the tarsus (7) the metatarsus (5) and the phalanges or toe ebones. (14)
  • The foot supports our body weight and acts as a lever to propel the body forward while walking and running.
  • The phalanges of the foot are smaller than those of the fingers.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement 20

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 13.
Give an account of a structure of a typical long bone.
Answer:

  • The typical long bone has a diaphysis, epiphyses and membranes.
  • A tubular diaphysis or shaft forms the long axis of the bone and has a central medullary cavity.
  • The epiphyses are the bone ends.
  • Compact bone forms the exterior of epiphyses and their interior contains spongy bone with red marrow.
  • The region where the diaphysis and epiphysis meet is called metaphysis.
  • The external surface of the entire bone except the joint surface is covered by a double-layered membrane called the periosteum.
  • The outer fibrous layer is dense irregular connective tissue.
  • The inner osteogenic layer consists of osteoblasts cell. ( bone-forming cells) and osteoclasts cells (E bone – destroying cells)
  • There are primitive stem cells osteogenic cells that give rise to the osteoblasts.
  • The periosteum is richly supplied with nerve fibres lymphatic vessels and blood vessels.
  • Internal bone surfaces are covered with a delicate connective tissue membrane called the endosteum It also contains osteoblasts and osteoclasts cells.
  • Between the epiphysis and diaphysis growthplate or epiphyseal plate is present.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement 21

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 14.
Write a short note on Rib cage?
Answer:
There are 12 pairs of ribs. Each rib is a thin flat bone connected dorsally to the vertebral column and ventrally to the sternum. It has two articulation surfaces on its dorsal end, hence called bicephalic.

The first seven pairs of ribs are called ‘true ribs or vertebro-stemal ribs. Dorsally they are attached to the thoracic vertebrae and ventrally connected to the sternum with the help of hyaline cartilages.

The 8th, 9th and 10th pairs of ribs do not articulate directly with the sternum but joined with the cartilaginous (hyaline cartilage) part of the seventh rib. These are called ‘false ribs’ or vertebro-chondral ribs.

The last 11th and 12th pairs of ribs are not connected ventrally. Therefore, they are called as ‘floating ribs’ or vertebral ribs. Thoracic vertebrae, ribs and sternum together form the ribcage. Rib cage protects the lungs, heart, liver and also plays a role in breathing.

Question 15.
Tabulate the differentiate of joints man
Answer:

Pivot jointbetween atlas and axis
Gliding jointbetween the carpals
Saddle jointbetween the carpal and meta carpal
Ball and socket jointbetween humerus and pectoral gridle
Hinge jointKnee joint
Condyloid or Angular or EllipsoidJoint between radius joint and carpal

Question 16.
Draw the diagram of different types of fracture and arrange them.
Answer:
Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement 22

Question 17.
Bones of the skeletal system. Table: 1 Bones of skeletal system
Answer:
Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement 23
Notes:

  1. The strongest muscle in the human: Massetter in cheeks
  2. The smallest muscle in the human: Middle ear in stapedius
  3. Well moving muscle: Tongue
  4. The largest muscle in the human: Buttock in Glutens Maximus
  5. The longest muscle in the human : Hip to knee (sartorius)
  6. Total number of bones is adults = 206

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 18.
Explain the basic categories of exercise and physical activity?
Answer:
Exercise and physical activity fall into four basic categories. Endurance, strength, balance, and flexibility. Endurance or aerobic activities increase the breathing and heart rate. They keep the circulatory system healthy and improve overall fitness.

Strength exercises make the muscles stronger. They help to stay independent and cany out everyday activities such as climbing stairs and carrying bags.

Balance exercises help to prevent falls which is a common problem in older adults. Many strengthening exercises also improve balance.

Flexibility exercises help to stretch body muscles for more freedom of joint movements.

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.1

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 6 Random Variable and Mathematical Expectation Ex 6.1 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.1

Question 1.
Construct cumulative distribution function for the given probability distribution.
Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.1 1
Solution:
F(0) = P(x ≤ 0) = p(0) = 0.3
F(1) = P(x ≤ 1) = p(0) + p(1)
= 0.3 + 0.2 = 0.5
F(2) = P(x ≤ 2) = P(0) + P(1) + P(2)
= 0.3 + 0.2 + 0.4 = 0.9
F(3) = P(x ≤ 3) = P(0) + P(2) + P(3) + P(4)
= 0.3 + 0.2 + 0.4 + 0.1 = 1

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.1

Question 2.
Let X be a discrete random variable with the following p.m.f
Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.1 2
Find and plot the c.d.f. of X.
Solution:
F(3) = P(x ≤ 3) = P(3) = 0.3
F(5) = P(x ≤ 5) = P(x = 3) + (x = 5)
= 0.3 + 0.2 = 0.5
F(8) = P(x ≤ 8) = P(3) + P(5) + P(8)
= 0.3 + 0.2 + 0.3
= 0.8
F(10) = P(x ≤ 10)
= P(3) + P(5) + P(8) + P(10)
= 0.3 + 0.2 + 0.3 + 0.2
= 1
Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.1 3

Question 3.
The discrete random variable X has the following probability function.
Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.1 4
where k is a constant. Show that k = \(\frac { 1 }{18}\)
Solution:
From the data
P(x = 2) = kx
= 2k
P(x = 4) = kx
= 4k
P(x = 6) = kx
= 6k
P(x = 8) = k(x – 2)
= k(8 – 2) = 6k
Since P(X = x) is a probability mass function
\(\sum_{x=2}^{8}\) P(X = x) = 1
\(\sum_{i=2}^{∞}\) P(xi) = 1
(i.e) P(x = 2) + P(x = 4) + P(x = 6) + P(x = 8) = 1
2k + 4k + 6k + 6k = 1
18k = 1
∴ k = \(\frac { 1 }{18}\)

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.1

Question 4.
The discrete random variable X has the probability function
Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.1 5
Solution:
From the data
P(x = 1) = k
p(x = 2) = 2k
p(x = 3) = 3k
P(x = 4) = 4k
Since P(X = x) is a probability Mass function
\(\sum_{x=1}^{4}\) P(X = x) = 1
\(\sum_{i=1}^{∞}\) P(xi) = 1
p(x = 1) + P(x = 2) + P(x = 3) + P(x = 4) = 1
P(x = 1) + P(x = = 2) + P(x = 3) + P(x = 4) = 1
k + 2k + 3k + 4k = 1
10k = 1
k = \(\frac { 1 }{10}\)
∴ k = 0.1

Question 5.
Two coins are tossed simultaneously. Getting a head is termed as success. Find the t probability distribution of the number of successes.
Solution:
Let X is the random variable which counts the Number of Heads when the coins are tossed the outcomes are stated below
Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.1 6

Question 6.
The discrete random variable X has the probability function.
Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.1 7
(i) Find k
(ii) Evaluate p( x < 6), p(x ≥ 6)and p(0 < x < 5) (iii) If P(X ≤ x) > 1, 2 then find the minimum value of x.
Solution:
(i) Since the condition of Probability Mass function
\(\sum_{i=2}^{∞}\) P(xi) = 1
\(\sum_{i=0}^{7}\) P(xi) = 1
P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) + P(x = 4) + P P(x = 6) + p(x = 7) = 1
0 + k + 2k + 2k + 3k + k² + 2k² + 7k² + k = 1
10k² + 9k – 1 = 0
10k² + 10k – k – 1 = 0
10k(k + 1) -1(k + 1) = 0
(k + 1)(10k – 1) = 0
k + 1 = 0; 10k – 1 = 0
k = -1 10k = 1
k = \(\frac { 1 }{10}\)
k = -1 is not possible

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.1

(ii) P(x < 6)
P(x < 6) = P(x = 0) + P(x = 1) + P(x = 2)+ P(x = 3) + p(x = 4) + P(x = 5)
= 0 + k + 2k +2k + 3k + k²
= 8k + k²
Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.1 8
P(x ≥ 6) = P(x = 6) + p(x = 7)
= 2k² + 7k² + k
= 9k² + k
Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.1 9
P(0 < x < 5)
= P(x = 1) + P(x = 2) + p(x = 3) + P(x = 4)
= k + 2k + 2k + 3k
= 8k = 8(\(\frac { 1 }{10}\))
∴ P(0 < x < 5) = \(\frac { 8 }{10}\)

(iii) We want the minimum value of x for which P(X ≥ x) > \(\frac { 1 }{2}\)
Now P(X ≤ 0) = 0 < \(\frac { 1 }{2}\)
P(X ≤ 1) = P(x = 0) + P(X = 1) = 0 + k = k
\(\frac { 1 }{10}\) < \(\frac { 1 }{2}\)
P( X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)
= 0 + k + 2k = 3k
\(\frac { 3 }{10}\) \(\frac { 1 }{2}\)
P(X ≤ 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 2) + P(X = 3)
= 0 + k + 2k + 2k = 5k
= \(\frac { 5 }{10}\) = \(\frac { 1 }{2}\)
P(X ≤ 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 2) + P(X = 3) + P(X = 4)
= 0 + k + 2k + 2k + 3k = 8k
= \(\frac { 8 }{10}\) > \(\frac { 1 }{2}\)
This Shows that the minimum value of X for which P(X ≤ x) > \(\frac { 1 }{2}\) is 4

Question 7.
The distribution of a continuous random variable X in range (-3, 3) is given by p.d.f.
Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.1 10
Verify that the area under the curve is unity.
Solution:
Since(-3, 3) in the range of given p.d.f
Area under the curve A = \(\int_{-∞}^{∞}\) f(x)dx = \(\int_{-3}^{3}\) f(x)dx
Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.1 11
Hence the Area under the curve is unity

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.1

Question 8.
A continuous random variable X has the following distribution function:
Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.1 12
Find (i) k and (ii) the probability density function.
Solution:
(i) Here F(3) – F(1) = 1
K(3 – 1)4 – O = 1
K(2)4 = 1
K(16) = 1
k = \(\frac { 1 }{16}\)

(ii) The Probability density function
f(x) = \(\frac { d(F(x)) }{dx}\) = \(\left\{\begin{array}{l}
4 k(x-1)^{3}, 1<x \leq 3 \\
0 \text { else where }
\end{array}\right.\)

Question 9.
The length of time (in minutes) that a certain person speaks on the telephone is found to be random phenomenon, with a probability function specified by the probability density function f(x) as
Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.1 13
(a) Find the value of A that makes f (x) a p.d.f.
(b) What is the probability that the number of minutes that person will talk over the phone is (i) more than 10 minutes, (ii) less than 5 minutes and (iii) between 5 and 10 minutes.
Solution:
(a) Since f(x) is a probability density Function
Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.1 14
b (i) more than 10 minutes
Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.1 15
Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.1 16

Question 10.
Suppose that the time in minutes that a person has to wait at a certain station for a train is found to be a random phenomenon with a probability function specified by the distribution function
Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.1 17
(a) Is the distribution function continuous? If so, give its probability density function?
(b) What is the probability that a person will have to wait (i) more than 3 minutes, (ii) less than 3 minutes and (iii) between 1 and 3 minutes?
Solution:
(i) Yes, the distribution function is continuous on [0, 4]
The probability density function
Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.1 18
(b) The probability that a person will have to wait
(i) more than 3 minutes
Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.1 19
(iii) between 1 and 3 minutes
Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.1 20

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.1

Question 11.
Define random variable.
Solution:
A random variable (r.v.) is a real valued function defined on a sample space S and taking values in (-∞, ∞) or whose possible values are numerical outcomes of a random experiment.

Question 12.
Explain What are the types of random variable.
Solution:
Random variables are classified into two types namely discrete and continuous random variables these are important for practical applications in the field of Mathematics and Statistics.

Question 13.
Define dicrete random Variable
Solution:
A variable which can assume finite number of possible values or an infinite sequence of countable real numbers is called a discrete random variable.

Question 14.
What do you understand by continuous random variable?
Solution:
A random variable X which can take on any value (integral as well as fraction) in the interval is called continuous random variable.

Question 15.
Describe what is meant by a random variable
Solution:
If S is a sample space with a probability measure and x is a real valued function defined over the elements of S then x is called a random variable A random variable is also called a change variable.

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.1

Question 16.
Distinguish between discrete and continuous random variable
Solution:
Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.1 21

Question 17.
Explain the distribution function of random variable
Solution:
The discrete cumulative distribution function or distribution function of a real valued discrete random variable X takes the countable number of points x1, x2, …. with corresponding probabilities p(x1), p(x2),… and then the cumulative distribution function is defined by
Fx(x) = P(X ≤ x), for all x ∈ R
i.e. Fx (x) = \(\sum_{x \leq x}\) p(xi)

Question 18.
Explain the terms (i) probability Mass function (ii) probability density function and (iii) probability
Solution:
(i) If X is a discrete random variable with distinct values x1, x2, …. xn, …, then the function, denoted by Px(x) and defined by
Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.1 22
This is defined to be the probability mass function or discrete probability function of X.

(ii) The probability that a random variable X takes a value in the interval [t1, t2] (open or closed) is given by the integral of a function called the probability density function fx(x):
P(t1 ≤ X ≤ t2) = \(\int_{t_{1}}^{t_{2}}\) fx(x)dx.

(iii)The probability distribution of a random variable X is defined only when we have the various values oft the various values of the random variable e.g. x1, x2 …… xn togather with respective probabilities p1, p2, p3 …… p4 satisfying
\(\sum_{i=1}^{n}\) Pi = 1

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.1

Question 19.
What are the properties of (i) discrete random variable and (ii) Continuous random variable
Solution:
(i) The probability mass function p(x) must satisfy the following conditions
(i) p(xi) ≥ 0 ∀ i,
(ii) \(\sum_{i=1}^{∞}\) p(xi) = 1

(ii) The probability density functions/)(x) or simply ; by/(x) must satisfy the following conditions.
(i) f(x) ≥ 0 ∀ x and
(ii) \(\int_{-∞}^{∞}\) f(x) dx = 1

Question 20.
State the Properties of distribution function.
Solution:
The function Fx(x) or simply F(x) has the following properties.
(i) 0 ≤ F(x) ≤ 1, -∞ < x < ∞
(ii) F(-∞) = \(\lim _{x \rightarrow-\infty}\) F(x) = 0 and F(+∞) = \(\lim _{x \rightarrow ∞}\) F(x) = 1.
(iii) F(.) is a monotone, non-decreasing function; ; that is, F(a) < F(b) for a < b.
(iv)F(.) is continuous from the right; that is, \(\lim _{h \rightarrow 0}\) F(x + h) = F(x).
(v) F(x) = \(\frac { d }{dx}\) F(x) = f(x) ≥ 0;
(vi) F'(x) = \(\frac { d }{dx}\) F(x) = f(x) ⇒ dF(x) = f(x)dx
dF(x) is known as probability differential of X.
(vii) P(a ≤ x ≤ b) = \(\int_{a}^{b}\) f(x)dx = \(\int_{-∞}^{b}\) f(x)dx – \(\int_{-∞}^{a}\) f(x)dx
= P(X ≤ b) – P(X ≤ a)
= F(b) – F(a)

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.1

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Bio Zoology Guide Pdf Chapter 8 Excretion Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

11th Bio Zoology Guide Excretion Text Book Back Questions and Answers

Part – I

Question 1.
Arrange the following structures in the order that a drop of water entering the nephron would encounter them.
a) Afferent arteriole
b) Bowman’s Capsule
c) Collecting duct
d) Distal tubule
e) Glomerulus
f) Loop of Henle
g) Proximal tubule
h) Renal Pelvis
Answer:
(a), (e), (b), (g), (f), (d), (c), (h)

Question 2.
Name the three filtration barriers that solutes must come across as they move from plasma to the lumen of Bowman’s capsule. What components of the blood are usually excluded by these layers.
Answer:
The three filtration barriers that solutes must come across as they move from plasma to the lumen of Bowman’s capsule are,

  1. Glomerular capillary endothelium
  2. Basal lamina or basement membrane
  3. Epithelium of Bowman’s capsule.
  4. Blood corpuscles and plasma protein are excluded by these layers.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 3.
What forces promote glomerular filtration? What forces oppose them? What is meant by net filtration pressure?
Answer:

  • Glomerulus hydrostatic pressure
  • Glomerulus pressure
  • Opposing pressure: Colloidal osmotic pressure, Capsular hydrostatic pressure
  • Net filtration pressure = Glomerular hydrostatic pressure – (Colloidal osmotic pressure + capsular hydrostatic pressure.

Question 4.
Identify the following structures and explain their significance in renal physiology?
Answer:
a) Juxtaglomerular apparatus It is a specialized tissue in the afferent arteriole of the nephron that consists of macula densa and granular cells.

  • Action: The macula densa cells sense distal tubular flow and aspect afferent arteriole diameter.
  • Actions of Glandular cells: When there is a fall in glomerular blood flow it can activate Juxta glomerulus cells to release renin from the liver which converts angiotensinogen to angiotensin I.
  • Angiotensin converting enzyme converts angiotensin I to angiotensin II.
  • It stimulates Na+ reabsorption in the proximal convoluted tubule by vasoconstriction of the blood vessel and increases the glomerular blood pressure.

b) Podocytes:

  • The visceral layer of Bowman’s capsule is made of epithelial cells called podocytes.
  • The podocytes and the foot processes and the basement membrane of glomerulus act as a filtering membrane.
  • Podocytes act in glomerulus filtration.

c) Sphincters in the bladder:

  • The sphincters of the bladder regulate urination.
  • When the bladder is full the stretching of the urinary bladder stimulates the CNS through the sensory neurons of the parasympathetic nervous system and brings about contraction of the bladder.
  • Somatic motor neurons induce the sphincters to close.
  • Smooth muscles contract results in the opening of the internal sphincters and relaxing the external sphincter.
  • The sphincter opens and the urine is expelled out.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 5.
In which segment of the nephron most of the re-absorption of substances takes place?
Answer:
In the proximal convoluted tubule.

Question 6.
When a molecule or ion is reabsorbed from the lumen of the nephron, Where does it go?
Answer:
When a molecule or ion is reabsorbed from the lumen of the nephron, it goes to the bloodstream through an efferent arteriole which carries blood away from the glomerulus. If a solute is filtered and not reabsorbed from the tubule, it goes along with urine.

Question 7.
Which segment is the site of secretion and regulated reabsorption of ions and pH homeostasis?
Answer:

  • 1. Distal tubule; 2. Collecting duct
  • In the renal tubules H+ and NH4 are secreted and liberated into the tubules and excreted through
    urine thus maintaining acid base balance.
  • For each H+ ions from the liberated filtrate one Na+ is reabsorbed in the tubules.
  • The secreted HCO3, PO3 and NH3 combines to form carbonic acid and phosphoric acids.
  • Thus the H+ are maintained and their reabsorption is prevented.

Question 8.
What solute is normally present in the body to estimate GFR in humans?
Answer:

  • The glomerulus filtrate consists of water glucose amino acids creatinine protein salts and urea.
  • These solutes decide the glomerular filtration rate.

Question 9.
Which part of the autonomic nervous system is involved in the micturition process?
Answer:
Parasympathetic nervous system.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 10.
If the afferent arteriole of the nephron constricts, what happens to the GFR in that nephron? If the efferent arteriole constricts what happens to the GFR in that nephron? Assume that no autoregulation takes place.
Answer:

  • Blood enters the glomerulus faster with greater force through the afferent arteriole and leaves the glomerulus through the efferent arterioles much slower.
  • Because afferent arteriole is wider than efferent arteriole.
  • The constriction of this does not affect the rate of filtration.

Question 11.
The concentration of urine depends upon which part of the nephron.
Answer:
a) Bowman’s capsule
b) Length of Henle’s loop
c) PCT
d) network of capillaries arising from the glomerulus.
Answer:
b) Length of Henle’s loop

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 12.
If Henle’s loop were absent from mammalian nephron, which one of the following is to be expected?
a) There will be no urine formation
b) There will be hardly any change in the quality and quantity of urine formed.
c) The urine will be more concentrated
d) The urine will be more dilute
Answer:
d) The urine will be more dilute

Question 13.
What will happen if the stretch receptors of the urinary bladder wall are totally removed?
a) Micturition will continue
b) Urine will continue to collect normally in the bladder
c) There will be micturition
d) Urine will not collect in the bladder
Answer:
a) Micturition will continue

Question 14.
The end product of the Ornithine cycle is
a) Carbon dioxide
b) Uric acid
c) Urea
d) Ammonia
Answer:
c) Urea

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 15.
Identify the wrong match.

a) Bowman’s capsuleGlomerular filtration
b) DCTAbsorption of glucose
c) Henle’s loopConcentration of urine
d) PCTAbsorption of Na+ and K+ ions

Answer:
b) DCT – Absorption of glucose

Question 16.
Podocytes are the cells present on the
a) Outer wall of Bowman’s capsule
b) Inner wall of Bowman’s capsule
c) Neck of the nephron
d) Wall glomerular capillaries
Answer:
b) Inner wall of Bowman’s capsule

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 17.
Glomerular filtrate contains
a) Blood without blood cells and proteins
b) Plasma without sugar
c) Blood with proteins but without cells
d) Blood without urea
Answer:
c) Blood with proteins but without cells

Question 18.
Kidney stones are produced due to the deposition of uric acid and Silicates
b) Minerals
c) Calcium Carbonate
d) Calcium oxalate
Answer:
d) Calcium oxalate

Question 19.
Animal requiring the minimum amount of water to produce urine is
a) Ureotelic
b) Ammonotelic
c) Uricotelic
d) Chemotelic
Answer:
c) Uricotelic

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 20.
Aldosterone acts at the distal convoluted tubule and collecting duct resulting in the absorption of water through
a) Aquaporins
b) Spectrins
c) GLUT
d) Chloride channels
Answer:
a) Aquaporins

Question 21.
The hormone which helps in the reabsorption of water in kidney tubules is
a) Cholecystokinin
b) Angiotensin Il
c) Antidiuretic hormone
d) Pancreozymin
Answer:
c) Antidiuretic hormone

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 22.
Malpighian tubules remove excretory products from
a) Mouth
b) Oesophagus
c) Haemolymph
d) Alimentary canal
Answer:
c) Haemolymph

Question 23.
Identify the biological term excretion, glomerulus, urinary bladder, glomerular filtrate, ureters, urine, Bowman’s capsule, urinary system, reabsorption, micturition, osmosis, proteins.
Answer:

  • A liquid which gathers in the bladder – Urine
  • Produced when blood is filtered in a Bowman’s capsule – Glomerular filtrate
  • The temporary storage of urine – Urinary bladder
  • A ball of intertwined capillaries – Glomerulus
  • Removal of unwanted substances from the body – Excretion
  • Each contains a glomerulus – Bowman’s Capsule
  • Carry urine from the kidneys to the bladder – Ureter
  • The scientific term for urination – Micturition
  • Regulation of water and dissolved substances in the blood and tissue fluid – Osmoregulation
  • Consists of the kidneys ureters and bladder Excretory system
  • Removal of useful substances from glomerular filtrate reabsorption
  • What solute the do blood contain that is not present in the glomerular filtrate? – Plasma Protein

Question 24.
With regards to toxicity and the need for dilution in water how different are ureotelic and ureocotelic excretions? Give examples of animals that use these types of excretions?
Answer:

  • The type of nitrogenous end product (Urea or Uric acid or ammonia) of an animal excretion depends upon the habitat of the animal.
  • For the Excretion of Ammonia, more water is needed.
  • Animals that excrete most of their nitrogen in the form of ammonia are called ammonites.
  • Fishes Amphibians. Uric acid can be eliminated with a minimum loss of water and is called Uricoteles.
  • Reptiles, birds insects, landrails. Mammals and terrestrial amphibians excrete urea and are called ureoteles.
    Nitrogenous

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion 1

Question 25.
Differentiate protonephridia from metanephridia.
Answer:

ProtonephridiaMetanephridia
1. Proto – first it is a network of dead-end tubules lacking internal opening. (E.g) PlatyhelminthesMeta – after They have a tubular internal opening called nephrostome. (E.g) Earthworm
2. Flame cells are excretory structures.The excretory products are filters and selectively reabsorbed.
3. Excretory product excretes through nephridioporeThe excretory product excretes through the nephridiopore.
4. Excretory structure are usually osmo regulatorsExcretory structures are osmo regulators and helps in excretion.

Question 26.
What is the nitrogenous waste produced by amphibian larvae and by adult animals?
Answer:
The adult amphibian’s excretory product is urea. The larval form of amphibian’s excretory product is ammonia.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 27.
How is urea formed in the human body?
Answer:
Urine formation involves three main processes namely Glomerular Alteration, tubular reabsorption and tubular secretion.

I. Glomerular filtration:

  • Blood enters the kidney from the renal artery into the glomerulus.
  • Blood is composed of water colloidal proteins, sugars and nitrogenous end product.
  • The filteration is started in the glomerulus. The fluid that leaves the glomerular capillaries and enters the Bowman’s capsule is called the glomerular filtrate. It formed 170-180 litre within 24 hours.
  • Glomerular membrane has large surface area. Blood enters the glomerulus faster with greater force through the afferent arteriole and leaves the glomerulus through the efferent arterioles much slower.
  • This is because afferent arteriole is wider than efferent arteriole and the glomerular hydrostatic pressure is around 55 mm Hg.
  • Molecules larger than 5mm are barred from entering the tubule.

The two opposing forces are contributed by the plasma proteins in the capillaries.

  1. Colloidal osmotic pressure – 30 mm Hg
  2. Hydrostatic pressure -15 mm Hg

Both pressures combine. 30 mm Hg + 15mm Hg = 45mm Hg

  • The net filtration pressure of 10mm Hg is responsible for the renal filtration.
  • Net filtration pressure – Glomerular
  • Hydrostatic Pressure – Colloidal Osmotic pressure + Capsular hydrostatic pressure.

Net filteration pressure = 55 mm Hg – (30mm Hg +15mm Hg) = 10mm Hg

SubstanceConcentration in blood Plasm / g dm-3Concentration in glomerular filtrate / g dm-3
Water900900
Proteins80.0.05
Aminoacids0.50.5
Glucose1.01.0
SubstanceConcentration in blood PIasm /gdm-3Concentration glomerular filtrate/gdm-3
Urea0.30.3
Uric Acid0.040.04
Creatinine0.010.01
Inorganic ions (mainly (Na+, K+and Cl)7.2

 

7.2

II. Tubular reabsorption:

  • The volume of filtrate formed per day is around 170 – 180l.
  • Urine released is around 1.51 per day. 99% of the glomerular filtrate is reabsorbed by the renal tubules.

Reabsorption in proximal convoluted tubule:

  • Glucose lactate amino acids Na in the filtrate are reabsorbed in the PCT.
  • Sodium is reabsorbed by active transport through a sodium-potassium pump in the PCT.
  • Small amounts of urea and uric acid are reabsorbed.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion 2

Reabsorption in Henle Loop:

  • Descending limb of Henle’s loop is permeable to water due to the presence of aqua porlns but not permeable to salts.
  • Water is lost in the descending limb. Hence Na and Cl get concentrated in the filtrate.
  • Ascending limb of Henle’s loop is impermeable to water but permeable to Na+, Cl andK+.

Distal Convoluted Tubule:

  • It recovers water and secretes potassium into tubule.
  • Na+, Cl-, and water remain in the DCT.
  • Reabsorption of HCO3 takes place to regulate the blood pH.
  • Collecting the tubule is permeable to water potassium ions are actively transported into the tubule and Na+ to produce concentrated urine.

Tubular Secretion:

  • Once it enters the collecting duct water is absorbed and concentrated hypertonic urine is formed.
  • For every H secreted into the tubular filtrate, a Na+ is absorbed by the tubular cell.
  • The H+ secreted combines with HCO+3, HPO3, and NH+ and gets fixed as H2CO3, H2PO3, and NH4+.
  • Since H+ gets fixed in the fluid reabsorption of H+is prevented.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 28.
Differentiate cortical from medullary nephrons.
Answer:
In the renal tubules proximal convoluted tubule and distal convoluted tubule are situated in the cortical region of the kidney and the loop of Henle is in the medullary region.

Cortical NephronMedullary Nephron
1. The loop of Henle is too short and extends only very little into the medulla.Nephrons have a long loop of Henle that run deep in to the medulla.
2. There is no vasa recta.Vasa recta is present.

Question 29.
What vessels carry blood to the kidneys? Is this blood arterial or venous?
Answer:
Renal artery right and left artery. The blood is arterial.

Question 30.
Which vessels drain filtered blood from the kidneys?
Answer:
Renal vein. The filtered blood is taken from the kidney to inferior vena cava through renal vein.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 31.
What is tubular secretion? Name the substances secreted through the renal tubules?
Answer:

  • In the distal convoluted tubule the urine becomes hypotonic.
  • It contains urea and salts pass from peri tubular blood in to the cells of DCT.
  • For every H+ secreted a Na+ is absorbed by the tubular cell.
    The H+ secreted combines HCO3, HPO3 and NH3 and becomes carbonic acid phosphoric acid and ammonium.
  • Since H+ gets fixed in the fluid reabsorption of H+ is prevented.

Question 32.
How are the kidneys involved in controlling blood volume? How is the volume of blood in the body related to arterial pressure?
Answer:

  • When the volume of blood decreases the flow pressure, decreases.
  • This can be sensed by hypothalamus and osmo receptors are stimulated and anti diuretic hormone is secreted from the neuro hypophysis.
  • The aquaporins in the proximal convoluted tubules and collecting tubule reabsorb water. Hence the blood volume increases and the blood pressure increases.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 33.
Name the three main hormones that are involved in the regulation of the renal function?
Answer:

  • Renin
  • Angio Tensin I
  • AngioTensin II

Question 34.
What is the function of anti diuretic hormone? Where is it produced and What stimuli increases or decreases its secretion?
Answer:

  • ADH which is also called as vasopressin or Anti diuretic hormone is secreted from neuro hypophysis.
  • Fluid loss or if blood pressure increases the osmoreceptors of hypothalamus is stimulated and hence neuro hypophysis is stimulated and secretes ADH.
  • When the fluid level and pressure is maintained due to the negative feed back mechanism ADH secretion stops.

Question 35.
What is the effect of aldosterone on kidneys and where is it produced?
Answer:

  • Due to the stimulation of Angiotensin II the adrenal cortex secretes aldosterone. That causes reabsorption of Na+, K+ excretion and absorption of water from distal convoluted tubule and collecting tubule.
  • This mechanism is known as Renin – Angio tensin Aldosterone system.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 36.
Explain the heart’s role in secreting a hormone that regulates renal function? What hormone is this?
Answer:

  • Excessive stretch of cardiac atrial cells cause an increase in blood flow to the atria of the heart and release Atrial Natriuretic Peptide or factor (ANF) travels to the kidney where H increases Na+ excretion and increases the blood flow to the glomerulus,
  • Acting on the afferent glomerular arterioles as a vasodilator or an efferent arterioles as a vasoconstrictor.
  • The first identical natri uretic hormone is atrial natriuretic horome of heart.

Part – II.

11th Bio Zoology Guide Excretion Additional Important Questions and Answers

I. Choose The Correct Answer.

Question 1.
Name the organ that regulates the acid base and water regulation?
a) Kidney
b) Liver
c) Skin
d) Lungs
Answer:
a) Kidney

Question 2.
Whether the following statements are true or false find the correct sequence. Find the correct sequence.
I) According to the environmental changes organism change their osmotic concentration and are called osmo confirmers.
II) Osmo regulators maintain their internal osmotic concentration irrespective of their external osmotic environment.
III) The Euryhaline animals are able to tolerate only narrow fluctuations in the salt concentration?
IV) The steno haline animals can tolerate wide fluctuations in the salt concentration.
Series:
a) I – False, II – True, III – False, IV – True
b) I – True, II – True, III – False, IV- False
c) I – False, II – True, III – False, IV – True
d) I – False, II – False, III – True, IV – True
Answer:
b) I – True, II – True, III – False, IV- False

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 3.
Name the toxic formed due to the degeneration of amino acid.
a) Urea
b) Uric Acid
c) Ammonia
d) Carbonic Acid
Answer:
c) Ammonia

Question 4.
Find out the odd one out.
a) Ammonoteles – Aquatic Amphibians
b) Urico teles – Birds
c) Ureoteles – Land amphibians
d) Ureoteles – Earthworm (When it is in water)
Answer:
d) Ureoteles – Earthworm (When it is in water)

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 5.
What is the functional unit of kidney?
a) Nephron
b) Neuron
c) Synapsis
d) Glomerulus
Answer:
a) Nephron

Question 6.
What type of Urine is formed in organism with long Henle’s loop?
a) Isotonic
b) less concentrated
c) Concentrated
d) None of the above
Answer:
c) Concentrated

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 7.
In the marine organism the kidney with glomerulus produce ………………. type of concentrated urine.
a) Concentrated than body fluid
b) Equal to the concentration of body
c) Less than the concentration of body fluids
d) None of the above
Answer:
b) Equal to the concentration of body

Question 8.
The average weight of human kidney,
a) 100-120g
b)150-200g
c) 120-170g
d) 120-150g
Answer:
c) 120-170g

Question 9.
What is the outer covering of the kidney?
a) Renal fascia perirenal fat capsule pleura
b) Renal fascia perirenal fat capsule peri cardial membrane
c) Renal fascia, Peri renal fat capsule meninges
d) Renal fascia perirenal fat capsule fibrous capsule
Answer:
d) Renal fascia perirenal fat capsule fibrous capsule

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 10.
What is meant by renal corpuscle?
a) Glomerulus and Bowman’s Capsule
b) Glomerulus and Malpighian capsule
c) Glomerulus and nephron
d) Glomerulus and Henle’s loop
Answer:
d) Glomerulus and Henle’s loop

Question 11.
Regarding renal tubules find the correct sequences.
a) Glomerulus Bowman’s capsule Malpighian capsule uriniferous tubules.
b) Proximal convoluted tubules thick descending loop thin ascending limb distal convoluted tubule.
c) Proximal convoluted tubule, Henle’s loop. Distal convoluted tubule.
d) Proximal convoluted tubule, thin descending limb thick ascending limb distal convoluted tubule.
Answer:
d) Proximal convoluted tubule, thin descending limb thick ascending limb distal convoluted tubule.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 12.
Urea is synthesisted through this cycle.
a) Citric cycle
b) Carboxylic Acid cycle
c) Ornithinecycle
d) Arginine cycle
Answer:
c) Ornithinecycle

Question 13.
What is the pressure in the afferent arthery of Glomerulus?
a) 55 mm Hg
b) 50 mm Hg
c) 57 mm Hg
d) 58 mm Hg
Answer:
a) 55 mm Hg

Question 14.
What is the amount of filteration of the glomerulus in 24 hours?
a) 200l
b) 180l
c)190l
d) 170l
Answer:
b) 180l

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 15.
What should be the size of the molecule that can pass through the tubules from the plasma filtrate?
a) 7 nm
b) 4 nm
c) 5 nm
d) 6 nm
Answer:
c) 5 nm

Question 16.
What is the pressure in glomerulus?
a) 50 mm Hg
b) 60 mm Hg
c) 55 mm Hg
d) 62 mm Hg
Answer:
c) 55 mm Hg

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 17.
The glomerular pressure encounter ………….. of colloidal osmotic pressure and …………… of hydrostatic pressure.
a) 30mmHg;15mmHg
b) 15mmHg;30mmHg
c) 40 mm Hg; 30 mm Hg
d) 30mmHg;40mmHg
Answer:
a) 30mmHg;15mmHg

Question 18.
What is the net filteration pressure?
a) 55mmHg – (15mmHg + 30mmHg) = 10mm Hg
b) 55 mm Hg – (30 mm Hg +15 mm Hg) =10 mm Hg
c) 55 mm Hg – (35 mm Hg + 5 mm Hg) = 15 mm Hg
d) 55 mm Hg – (5 mm Hg + 35 mm Hg) = 15 mm Hg
Answer:
b) 55 mm Hg – (30 mm Hg + 15 mm Hg) =10 mm Hg

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 19.
What is the amount of filtrate formed in one minute?
a) 100 ml -125 ml
b) 100 ml -150 ml
c) 120 ml -125 ml
d) 130 ml -140 ml
Answer:
c) 120 ml -125 ml

Question 20.
What is the amount of filtrate formed in a day?
a) 170 ml-200 ml
b) 170 ml-190 ml
c) 170 ml-180 ml
d) 170 ml-185 ml
Answer:
c) 170 ml-180 ml

Question 21.
What is the amount of urine excreted in a day?
a) 1.5l
b) 1l
c) 2l
d) 2.5l
Answer:
a) 1.5l

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 22.
Which of the following hormones are secreted by the kidney?
a) Renin
b) Gastrin
c) Calcitriol
d) secretin
a) (ii) and (iii)
b) (i) and (iii)
c) (ii) and (iv)
d) (i) and (ii)
Answer:
b) (i) and (iii)

Question 23.
Where is aquaporins present?
a) Proximal convoluted tubule
b) Distal convoluted tubule and ascending limb of Henle
c) The descending limb of Henle and distal convoluted tubule
d) All the above
Answer:
b) Distal convoluted tubule and ascending limb of Henle

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 24.
Name the hormone that helpsin the reabsorption of water in distal convoluted tubule.
a) Vasopressin
b) Serotonin
c) Oxytocin
d) All the above
Answer:
a) Vasopressin

Question 25.
Where is Renin synthesized in Kidney?
a) Afferent arteriole
b) Efferent arteriole
c) Vasarecta
d) Collecting duct
Answer:
a) Afferent arteriole

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 26.
Name the hormone that helpsin the conversion of plasma protein angioten- sinogen into angiotensin I
a) Renin
b) Vasopressin
c) ADH
d) STH
Answer:
a) Renin

Question 27.
What is micturition?
a) Excretion of urine from the urinary bladder
b) The formation of urine in the glomerulus
c) Urine formation in distal convoluted tubule
d) the absorption of urine in Bowman’s capsule
Answer:
a) Excretion of urine from the urinary bladder

Question 28.
According to the food that maneats the pH of urine ……………… to ……………. can be altered.
a) pH 4.8-7.5
b) PH 4.9-7.9
c) pH 4.5-8.0
d) pH 4.4 -7.4
Answer:
c) pH 4.5-8.0

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 29.
What is the reason for the yellow colour of the urine?
a) Urochrome
b) Haemoerythrin
c) Haemocyanin
d) None
Answer:
a) Urochrome

Question 30.
What is the normal pH of urine?
a) pH 6.0
b) pH 5.3
c) pH 6.4
d) pH 5.9
Answer:
a) pH 6.0

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 31.
What is the amount of CO2 released from lungs in a day?
a) 20l
b) 19l
c)18l
d)119l
Answer:
c)18l

Question 32.
Confirm
Statement A: The important function of sweat gland is to cool the body.
Statement B: The sweat glands excrete sodi¬um chloride urea and lactic acid.
a) Statement A-True B-True
b) Statement A-True B-False
c) Statement A-FalseB-True
d) Statement A – False B – False
Answer:
a) Statement A-True B-True

Question 33.
Confirmation:
Statement S: When kidney fails suddenly there is more chance of recovery.
Statement T: In the chronic kidney failure there may not be any chance of recovery.
a) Statement S-True T-True
b) Statement S-True T-False
c) Statement S-True T-True
d) Statement S – True T – False
Answer:
c) Statement S-True T-True

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 34.
Confirmation:
Statement A: The kidney infection leads to inflammation of bladder and kidney
Statement B: Urination with pain, urinary urgency bloodytinged urine.
a) Statement A – True Statement B explain the symptom of A.
b) A – True – The statement B does not explains the statement A
c) A and B are false
d) A False B – True
Answer:
a) Statement A – True Statement B explain the symptom of A.

Question 35.
What is the normal urea level in the blood
a) 17-30mg/100ml
b) 30-35mg/ 100ml
c) 10-15 mg/100 ml
d) 5-10 mg/100 ml
Answer:
a) 17-30mg/100ml

Question 36.
Find the wrong pair.
a) Renal stone – nephrolithiasis
b) Urine – Urochrome
c) Deficiency of ADH – Urine out put decreases
d) Skin- Lactic acid excretion
Answer:
c) Deficiency of ADH – Urine out put decreases

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 37.
Confirmation:
Statement A: Bright’s disease is due to the infection of streptococcus in children
statement B: There is inflammation of glomerulus.
a) Statement A and Bare false
b) Statement A True B explain the A
c) A True B False
d) A False B – True
Answer:
b) Statement A True B explain the A

Question 38.
Find the wrong pair.
a) Heporin – Anticoagulating factor
b) Glomerulo nephritis – Accumulation of water in the body
c) Primary kidney – Meso nephridia
d) Uremia – Increase in the blood urea level
Answer:
c) Primary kidney – Meso nephridia

Question 39.
How much urine can be stored up in the bladder?
a) 300-600 ml
b) 200-300 ml
c) 400-700 ml
d) 500-800 ml
Answer:
a) 300-600 ml

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 40.
For how much time urine can be held in the bladder?
a) 6 hours
b) 2 hours
c) 5 hours
d) 3 hours
Answer:
c) 5 hours

Question 41.
The urinary bladder is made up this muscle ……………………….
a) Detrusor muscle
b) Striated muscle
c) Sphincter muscle
d) None of the above
Answer:
a) Detrusor muscle

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 42.
Match the following:

1. Steno halineI Shark
2. EurihalineII Otter
3. Osmo regulatorsIII Goldfish
4. Osrno confirmersIV Salmon

a) 1-IV 2-III 3-II 4-1
b) 1 -II 2-II 3-IV 4-1
c) 1 -III 2-IV 3-II 4-1
d) 1-1 2-II 3-IV 4-III
Answer:
c) 1 -III 2-IV 3-II 4-1

Question 43.
Find out the less toxic waste among the following:
a) Urea
b) Uric acid
c) Ammonia
d) Creatinine
Answer:
a) Urea

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 44.
Find out the ammonotelic organisms?
a) Reptiles
b) Birds
c) Aquatic amphibians
d) Frog
Answer:
c) Aquatic amphibians

Question 45.
Find out the wrong pair.
a) Rennette cells – Annelida
b) Molluscs – Metanephridia
c) Amphioxus – Mesonephridia
d) Tapeworm – Flame cells
Answer:
c) Amphioxus – Mesonephridia

Question 46.
Find the excretory structure of prawn?
a) Malpighian tubules
b) Green glands
c) Rennette cells
d) Peyer gland
Answer:
b) Green glands

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 47.
Kidney with noglomerulus form this types of urine?
a) Hypertonic
b) Isotonic
c) Hypotonic
d) Neutrotonic
Answer:
c) Hypotonic

Question 48.
Match the following:

1. Conical tissue massesa) Columns of Bertini
2. Extension in renalb) Renal pelvis
3. Broad part of Hilumc) Calyces
4. Projection of pelvisd) Medullary pyramids

a) 1-d 2-a 3-b 4-c
b) 1-a 2-b 3-c 4-d
c) 1-b 2-a 3-c 4-d
d) 1-a 2-c 3-b 4-d
Answer:
a) 1-d 2-a 3-b 4-c

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 49.
How much urine can be held in the urinary bladder?
a) 300-800ml
b)300-600ml
c) 100-400ml
d)200-300ml
Answer:
b)300-600ml

Question 50.
The muscles of urinary bladder is called by this name?
a) sphincter muscle
b) Detrusor muscle
c) constricting muscle
d) Peristaltic muscle
Answer:
b) Detrusor muscle

Question 51.
Name the nephron where the Henle’s loop is short?
a) Medullary nephron
b) Cortical nephron
c) Juxta medullary nephron
d) Medulla nephron
Answer:
b) Cortical nephron

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 52.
The vessel which runs parallel to the loop of Henle is called by this name.
a) Efferent artery
b) Afferent artery
c) Vasarecta
d) Vasanervosa
Answer:
c) Vasarecta

Question 53.
What is the process of urea formation?
a) Ross cycle
b) Ornithine cycle
c) Urea cycle
d) b and c
Answer:
b) Ornithine cycle

Question 54.
What is the pressure in the afferent arteriole?
a) 35 mm Hg
b) 55 mm Hg
c) 20 mm Hg
d) 40 mm Hg
Answer:
b) 55 mm Hg

Question 55.
What is the filtrate entered into the Bowman’s capsule from glomerulus?
a) Secondary urine
b) Primary urine
c) Tertiary urine
d) Quarternary urine
Answer:
b) Primary urine

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 56.
Find the correct option
Assertion: The glomerular filtrate resembles the blood
Reason: The glomerulus filtered the blood received from efferent artery
a) Assertion True Reason True
b) Assertion False Reason False
c) Assertion True The Reason does not explains the statement
d) Assertion True Reason explains A
Answer:
c) Assertion True The Reason does not explains the statement

Question 57.
What is the fate of water in the descending limb of Henle?
a) The concentrations of Na reduces
b) The concentration of Cl ions reduces
c) Na and Cl gets concentrated in the filtrate.
d) Formation of Hypotonic solution.
Answer:
c) Na and Cl gets concentrated in the filtrate.

Question 58.
How can the pH of blood be regulated?
a) Due to glucose reabsorption
b) Due to the reabsorption of HCO3
c) Due to the reabsorption of Na+
d) Due to the reabsorption of Cl
a) 1-2 II-4 III-3 IV-1
b) 1-1 II-2 III-3 IV-4
c) 1-4 II-3 III-2 IV-1
d) 1-2 II-l III-3 IV-4
Answer:
b) 1-1 II-2 III-3 IV-4

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 59.
Match the following
Answer:

I. Potassium1. Descending limb of Renie
II. Water2. Active transport
III. Glucose3. Proximal convoluted tubule
IV. Na, cl, lc4. Aqua porin

Question 60.
How is osma regulation in medulla maintained?
a) Juxtamedullary
b) Counter current
c) Positive current exchanger
d) Negative control
Answer:
b) Counter current

Question 61.
Find the wrong pair.
a) Vasa recta – Proximal convoluted tubule
b) ADH – Neuro hypophysis
c) Specialized nephron tissue – Juxtaglomerular apparatus
d) Renin – Glanularcell
Answer:
a) Vasa recta – Proximal convoluted tubule

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 62.
This synthesizes angio tenginogen.
a) Kidney
b) Liver
c) Malpighian tubule
d) Glomerular
Answer:
b) Liver

Question 63.
This increases the absorption of sodium ions.
a) Renin
b) Angiotensin I
c) Angio tensin II
d) Angio tensinogen
Answer:
c) Angio tensin II

Question 64.
Name the hormone that decreases the blood pressure
a) Angiotensin I
b) Atrial natriuretic peptide
c) Vasopressin
d) ADH
Answer:
b) Atrial natriuretic peptide

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 65.
This decreases the excretion of Renin?
a) Aldosterone
b) vasopressin
c) Atrial natriuretic peptide
d) Ventrical natriuretic peptide
Answer:
c) Atrial natriuretic peptide

Question 66.
What is the pH of Urine?
a) 6.5
b) 7.0
c) 6.0
d) 8
Answer:
c) 6.0

Question 67.
What is the reason for the yellow colour of urine?
a) Uro chrome
b) Cytochrome
c) Chlorochrome
d) Phyto chrome
Answer:
a) Uro chrome

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 68.
The increase in the level of urea
a) Uremia
b) ureomia
c) Uricmia
d) Ketosis
Answer:
a) Uremia

Question 69.
The accumulation of salt in the blood.
a) Poly urea
b) Oligo urea
c) Polydipsia
d) Polyphagia
Answer:
b) Oligo urea

Two marks

II. Very short answer 

Question 1.
What is meant by stenohaline organism?
Answer:
The stenohaline animals can tolerate only narrow fluctuations in the salts concentration. (eg) Goldfish.

Question 2.
What is meant by Eury haline animals.
Answer:
Eury haline animals are able to tolerate wide fluctuations in the salt concentrations. (eg) Salmons Tilapia.

Question 3.
What is meant by Osmoregulation?
Answer:
It is the control of tissue osmotic pressure which acts as a driving force for movement of water across biological membrances.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 4.
What is meant by Ionic regulation?
Answer:
It is the control of ionic composition of body fluids.

Question 5.
What are the nitrogenous waste formed due to the degen ration of aminoacid?
Answer:

  • Ammonia
  • Urea
  • Uric acid

Question 6.
Write is kidney situated in the body?
Answer:

  • The kidney lie in the superior lymbar region between the levels of the last thoracic and third lumber vertebra close to the dorsal inner wall of the abdominal cavity.
  • The right kidney is placed lower than the left kidney.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 7.
What is meant by renal hilum?
Answer:
The centre of the inner concave surface of the kidney has a notch called the renal hilum through which ureter blood vessels and nerves innervate.

Question 8.
What is meant by malpighian capsule or renal corpuscle?
Answer:
The Bowman’s capsule and the glomerulus together constitute the renal corpuscle.

Question 9.
What are podocytes?
Answer:

  • The visceral wall of gromerulus is made of epithelial cells called podocytes.
  • The podocytes end in foot processes which cling to the basement membrance of the glomerulus.

Question 10.
What are cortical nephrons?
Answer:
The loop of Henle is too short and extends only very little into the medulla and are called cortical nephrons.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 11.
What is meant by Juxta medullary nephrons?
Answer:
Some nephrons have very long loop of Henle that run deep in to the medulla and are called Juxta medullary nephrons.

Question 12.
What is vasa recta?
Answer:
The efferent arteriole serving the juxta medullary nephron forms bundles of long straight vessel called vasa recta and runs parallel to the loop of Henle.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 13.
List the three important process of urine formation?
Answer:

  1. Glomerulus filtration
  2. Tubular reabsorption
  3. Tubular secretion

Question 14.
What is meant by Glomerulus filtrate? What is its composition?
Answer:
The blood comes to the glomerulus are filtered and enters the Bowman’s capsule is called glomerular filtrate
Composition Water glucose aminoacids and nitrogenous wastes.

Question 15.
What is meant by Net filtration pressure?
Answer:
The two opposing forces against the glomerular blood pressure

  1. Collodial osmotic pressure is 30 mm Hg
  2. Capsular hydrostatic pressure is 15mm Hg

Net filtration pressure- Glomerular
Hydrostatic pressure – Colloidal
Osmotic pressure + Capsular hydrostatic pressure = 55 mm Hg – 30 mm Hg + 15 mm Hg = 10 mm Hg

Question 16.
What is meant by Glomerular filtration rate?
Answer:
It is the volume of filtrate formed in a minute in all nephrons of both the kidneys.

Question 17.
What is the amount of glomerular filtrate?
Answer:
In adults the GFR is 120 -125 ml per minute.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 18.
What is meant by primary filtrate?
Answer:
The filtrate enters from glomerulus to the Bowman’s capsule is called as primary filtrate.

Question 19.
Why glomercular filtrate resembles blood plasma?
Answer:
In the glomerular filtrate all the contents that present in the blood except the plasma protein is present.

Question 20.
How much filtrate is formed in one day?
Answer:
The amount of filtrate formed in a day is 170 to 180l.

Question 21.
What is meant by selective permeability?
Answer:

  • Some substances present in the glomerular filtrate is essential for our body.
  • Hence these molecules are reabsorbed in a tubules. This process is called as selective reabsorption.

Question 22.
Name the process in which selective reabsorption is taking place?
Answer:

  • Passive transport
  • Active transport
  • Diffusion
  • Osmosis

Question 23.
What are aqua porins?
Answer:
Aquaporins are membrane transport proteins that allow water to move across the epithelial cells.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 24.
What are the molecules that move into the filtrate of nephron?
Answer:
Nitrogen potassium ammonia creatinine and organic acids.

Question 25.
What is meant by Isotonic solution?
Answer:
Isotonic condition of a solution indicates no passage of water across the membrane separating two such solution.

Question 26.
What is meant by Hypotonic solution?
Answer:
The solution in which there is a loss of water then that solution is a hypotonic solution.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 27.
What is meant by hypertonic solution?
Answer:
When two solutions A and B are separated by a semi permeable membrane when water move from solution A to B across the membrane then the B solution is hypertonic and the solution A where the water loses is known as hypotonic solution.

Question 28.
Name the organ that the angiotensin II acton?
Answer:

  • Heart
  • Kidney
  • Brain
  • Adrenal cortex

Question 29.
What are the symptoms of diabetes mellitus?
Answer:

  • Excess glucose and ketone bodies in the urine
  • Poly dipsia – Excessive drinking of water
  • Polyurea – Excretion of large quantities of urea
  • Polyphagia – Excessive appetite

Question 30.
Name the organ that excrete nitrogen other than kidney?
Answer:

  • Lungs
  • Liver
  • Skin

Question 31.
What are the significance of sweat glands?
Answer:

  • Sweat produced by the sweat glands helps to cool the body.
  • It excretes Na+ and Cl small quantities of urea and lactate.

Question 32.
What is meant by nephrolithiasis?
Answer:
It is the formation of hard stone like masses in the renal tubules of renal pelvis.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 33.
What is meant by phleothotomy or lithotripsy?
Answer:
Renal stones can be removed by the technique pyleothotomy or lithotripsy.

Question 34.
What is meant by Bright’s disease?
Answer:
The inflammation of the glomeruli of both kidney of children due to the streptococcal infection is called as Bright’s disease.

Question 35.
What is meant by haemodialysis?
Answer:
The process of removing toxic urea from the blood of renal failure patients is known as haemodialysis.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 36.
Why females are prone to urinary tract infection than men?
Answer:
Females are prone to recurring urinary tract infection as they have shorter urethra.

Question 37.
Why men are finding difficult to urinate in their old age?
Answer:
With age prostate in males may enlarge which forces urethra to tighten restricting a normal urinary flow.

Question 38.
What is the change in Urine formation when there is a deficiency of ADH?
Answer:
When there is a deficiency of ADG the reabsorption of water from the proximal convoluted tubule decreases leads to dilute urine formation.

Question 39.
Why there is a increase in the body fluid when we drink large volume of water with out eating anything salty?
Answer:

  • When we drink or eat salty products the Na+ enters into the body fluids.
  • The sodium ions helps in the reabsorption of water
  • But when we drink only water as there is no sodium ions the tubules cannot reabsorb water.
  • Hence there is a increase in the urine output.

Question 40.
Consider how different foods affect water and salt balance and how the excretory system must respond to maintain homeostasis.
Answer:

  • When we eat different foods the salt present in them goes to renal tubules.
  • If sufficient salts are present in the body fluids this salt may not be reabsorbed when it goes through nephron and excreted through urine.
  • If there is a decrease in the volume of water in body fluids sodium ions are reabsorbed. Hence water is reabsorbed from filtrate and water volume is increased.

Question 41.
Name the different process that maintains water level? When there is a severe loss of water in the body?
Answer:

  • The blood vessels supplies to skin constricts and thus there is a decrease in the secretion of sweat prevents loss of water.
  • There is a reduction in the glomerular blood pressure and the rate of filtration decreases.
  • The reaborption of water in the proximal distal convoluted tubules increase.
  • There is absorption of water from the small intestine and large intestine and thus increases the water content in the blood.

Question 42.
What is meant by Osmolarity?
Answer:

  • The solute concentration of a solution of water is known as osmolarity,
  • The unit is millosmoles / litre (mOsm /l)

Question 43.
What is meant by aquaporins? What are its functions?
Answer:
Aqua porins are water permeable channels.
Functions
It helps in allowing water to move across the epithelial cells in relation to the osmotic difference from the lumen to the interstitial fluid.

Question 44.
How can we measure that there is a efficient glomerular filtration?
Answer:
If the renal clearance is equal to the glomerular filtration rate with little reabsoption and secretion. Then we know the kidney is functioning efficiently.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 45.
What is the Significance of having long and short Henle’s loop of Nephrons?
Answer:

  • The main function of Henle’s loop is to reabsorb water from filtrate.
  • If the length of the loop is longer then there is more reabsorption of water and if the lengh of Henle’s loop is shorter then the reabsorption of water is less.

Question 46.
Give notes on Capillary to capsule.
Answer:
Bloodcells and most blood proteins are too big to cross the capsular membrane into the capsule space. But the membrane’s slits and pores allow through water mineral salts polypeptides and other small molecules including waste such as urea ammonia and creatinine.

Question 47.
Give short notes on Blood enters the glomerulus.
Answer:
Blood flows from renal arteriole into the knot of capillaries. It enters at pressure which will force water and other out of the capillaries into the capsular space.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 48.
Give notes on filteration in proximal convoluted tubule?
Answer:
Proximal tubule is nearer to the Bowman’s capsule. This region allows much water to be reabsorbed into the capillaries and surronding fluids as well as glucose mineral salts and other useful substances.

Question 49.
Give notes on filteration in peritubular capillaries.
Answer:
It is also called the vasarecta this network reabsorbs upto 99 percent of the water in the tubule as well as various other substance using active pumps it also moves sodium from the blood to the tubule.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 50.
Give notes on filteration in Henle’s loop (Ascending)
Answer:

  • As the loop of the Henle dipsin to the renal medcula more water moves from the tubule into the blood as well as small amounts of salts and some urea and creatinine.
  • Some acids and amines may move into the tubule in which ammonia cango in both the direction.

Question 51.
Give notes on filteration in distal tubule?
Answer:

  • Distal tubule is far from capsule. This region may see water go in or out of the tubule depending on the concentration of water already in the tubule/ while hydrogen and potassium ions move to regulate both blood and urine pH.
  • Acids amines and ammonia compounds may also transported into the tubule.

Question 52.
Give notes on filteration in collecing duct.
Answer:

  • Fine adjustment of urine composition continues into the collecting duct system.
  • About 5 percent of all the water and sodium being reabsorbed into the blood is recovered here.

Question 53.
Give notes on venousflow?
Answer:

  • Blood flowing away from the nephrons carries 99 % of its orginal water.
  • 98% of its sodium calcium and cholrides and about 40% of its urea.

Question 54.
We are not consuming urea. But in our body area is produced. Why?
Answer:
Through Arnithine cycle in the lives the nitrogenous waste formed due to the breakdown of amino acid creates urea.

Question 55.
What is meant by Ionic regulation?
Answer:
It is the control of the ionic composition of body fluids.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 56.
What are stenohaline animals?
Answer:
They can tolerate only narrow fluctuations in the salt concentration. Ex: Goldfish

Question 57.
What are Euryhaline animals?
Answer:

  • They are able to tolerate wide fluctuations in the salt concentrations. Ex: Artemia Salmons.
  • Acids amines and ammonia compounds may also transported into the tubule.

Question 58.
List the nitrogenous wastes
Answer:
Ammonia urea uricacid.

Question 59.
What are the other nitrogenous wastes of protein metabolism?
Answer:
Alantonin Alantoic acid, Ornithuriacid creatinine creatine purines pyramidines and pterines.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 60.
Which are called ammoneteles?
Answer:
Animals that excreate most of its nitrogen inthe form of ammonia are called ammonoteles.

Question 61.
What are called as uricoteles?
Answer:
Animals that excrete uric acid crystals with minimum loss of water are called uricoteles.

Question 62.
What are ureoteles?
Answer:
Mammals and terrestrial amphibians mainly excrete urea are called ureoteles.

Question 63.
How is Reptiles produced lesshypotonic urine?
Answer:
Reptiles have reduced glomerulus or lack glomerulus and Henles loop and hence produce very little hypotonic urine.

Question 64.
How is mammals produced concentrated urine?
Answer:
Mammalian kidneys produce concentrated urine due to the presence of Henle’s loop.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 65.
What are the three coverings of kidney?
Answer:

  • Renal facia
  • Perirenal fat capsule
  • Fibrous capsule

Question 66.
What are medullary pyramids?
Answer:
The medulla is divided into a few conical tissue masses called medullary pyramids.

Question 67.
What is meant by renal columns of Bertini?
Answer:
The part of cortex that extends in between the medullary pyramids is the renal columns of Bertini.

Question 68.
What is meant by renal pelvis?
Answer:
A broad funnel shaped space inner to the hilum is called renal pelvis.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 69.
What is calyces?
Answer:
The projection in the pelvis is called calyces.

Question 70.
What are cortical nephrons?
Answer:
The loop of Henle is too short and extends only very little in to the medulla and are called cortical nephrons.

Question 71.
What is meant by juxta medullary nephron?
Answer:
Some nephrons have very long loop of Henle that run deep into the medulla and are called Juxta medullary nephrons.

Question 72.
What is the functions of aquaporins?
Answer:
This helps in allow water to move across the epithelial cells in relation to the osmotic difference from the human to the interstitial fluid.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 73.
What is meant by juxta glomerular apparatus?
Answer:
It is a specialized tissue in the afferent arteriole of the nephron that consists of maculadensa and granular cells.

Question 74.
What is meant by micturition?
Answer:
The process of release of urine from the bladder is called micturition.

Question 75.
What are the symptoms of diabetes mellitus?
Answer:
Presence of glucose and ketone bodies in the urine.

Question 76.
How is lung acting as a excretory organ?
Answer:
Lungs remove large quantities of carbondixide 181 / day and significant quantities of water every day

Question 77.
What is meant by renal clearnace?
Answer:
The amount of solute passing from the urine in a given period of time is renal clearance.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 78.
How can we estimate the efficiency of kidney?
Answer:

  • The renal clearance is equal to the glomerular filtrate then there is efficient filtration with little reabsorption and secretion.
  • Thus we can estimate the clearance is equal.

Three Marks

Short Answer –

Question 1.
a) List the ma j or nitrogenous wastes.
b) What are other wastes formed during protein metabolism?
Answer:
a) 1. Major nitrogenous wastes.

  • ammonia
  • Urea
  • Uricacid

b) Other Wastes:

  • Trimethy lamine oxide TMO
  • Quanine
  • Allantonin
  • Creatinine
  • Creatine
  • Purines

Question 2.
Give notes on ammonoteles uricoteles and ureoteles.
Answer:
Ammonoteles:

  • Animals that excrete most of its nitrogen in the form of ammonia are called ammonoteles.
    (e.g) fishes Amphibians aquatic insects.
  • In bony fishes ammonia diffuses out acrossthe body surface.

Uricoteles:

  • Animals which excrete uricacid crystals with a minimum loss of water is called.
  • Uricoteles (e.g) Reptiles Birds land snails and insects.

Ureoteles:

  • Animals which excrete urea as a nitrogenous wastes are called ureoteles
  • (e.g) Mammals, terrestrial amphibians.

Question 3.
Nephrons are the functional and structural unit of kidney’s. What is the relationship between glomerulus, Henle’s loop and urine formation?
Answer:

  • Reptiles have reduced glomerulus or lack glomerulus and Henle’s loop and produce hypotonic urine (dilute)
  • Mammalian kidneys produce concentrated urine due to the presence of long Henle’s loop.
  • Aglomerular kidneys of marine fishes produce little urine that is iso osmotic to the body fluid.
  • Amphibians and fresh water fishlack Henle’s loop hence produce dilute urine.

Question 4.
Differentiate the cortical nephron fron juxta medullary nephron
Answer:
Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion 3

Question 5.
Give Short notes on capillary bed of the nephron:
Answer:
The first capillary bed of the nephron is the glomerulus.
The other is peritubular capillaries.
1. Glomerulus:
Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion 4
Blood enters into the glomerulus through afferent arteriole and drained by the efferent arteriole.

2. Peritubular capillaries:

  • The efferent arteriole forms a fine cappillary network around the renal tubule called the peritubular capillaries.
  • The efferent arteriole serving the juxta medullary nephrons forms bundles of long straight vessel called vasa recta.
  • Vasa recta is absent in cortical nephrons.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion 5

Question 6.
What happens to the filtrate that comes to the proximal convoluted tubule? (or) Explain about reabsorption?
Answer:
Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion 6

  • In the proximal convoluted tubule glucose lacticacid aminoacid sodium ions are reabsorbed.
  • Sodium is reabsorbed – potassium pump in the proximal convoluted tubule.
  • Small amounts of urea and uric acid are also reabsorbed.

Question 7.
What happen to the filtrate that comes to the Henle’s loop? (or) Explain the reabsorptionin the Henle’s loop?
Answer:
Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion 7

Descending Loop:
The aquaporin present in the descending limb of Henle permeable to water but not permeable to salts.
Hence Na+ and cl gets concentrated in the filtrate.
Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion 8

Ascending Limb:
It is impermeable to water but permeable to solutes like Na+cl and K+.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 8.
Give an account of tubular reabsorption?
Answer:
The volume of filterate formed perday is 170-180 is and the urine released in a day is 1.5l

  • Nearly 99 % of the glomerular filtrate is reabsorbed by the renal tubules. It is called selective reabsorption.
  • Reabsorption is taken place by the tubular epithelial cells in different segments of the nephron by active transport or passive transport diffusion and osmosis.

Question 9.
What is happenning to the filtrate in distal convoluted tubule (or) Reabsorption taking place here?
Answer:

  • Depending on the body’s need the reabsorption taking place here and is regulated by hormones.
  • Reabsorption of bicarbonate HCO3 takes place to regulate the blood pH.
  • Homestasis of K+ and Na+ in the blood is also regulated in this region.

Question 10.
Name the structures that regulate the functioning of kidney?
Answer:

  • Hypothalamus
  • Juxta glomerular apparatus
  • Heart

Question 11.
What is meant by diabetes incipidus?
Answer:
If there is deficiency or absence of ADH that leads to dilute urine called diabetes incipidus.
Symptoms.

  • Excessive thirst
  • Excretion of large quantities of dilute urine.
  • FaIl in blood pressure.

Question 12.
Give notes on juxta glomerular apparatus?
Answer:

  • Specialized tissue in the afferent arteriole of ncphron is the juxta glomerular apparatus.
  • It consists of macula densa and granular cells.
  • The macula densa cells sense distal tubular flow and affect afferent alteriole diameter.
  • The granular cells secrete renin.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 13.
Why female are prone to urinary tract infections? (Urethritis)
Answer:

  • Female’s urethra is very short and its external opening is close to the analopening.
  • Hence improper toilet habits can easily carry faecal bacteria into the urethra.
  • The urethral mucusa is continuous with the urianary tract and the inflammation of the urethra is called urethriti’s.

Question 14.
What is cystitis?
Answer:
The urinary tract infection leads to inflammation of bladder called cystiti’s.
Symptoms:

  • Painful urination
  • Urinary Urgency
  • Cloudy or bloodtingedurine
  • Back pain head ache offen occurs

Question 15.
What is meant by renal failure? What are its types?
Answer:
When the kidney fails to excrete wastes may lead to accumulation of urea with marked reduction in the out put called renal failure.
Types
Acute renal failure
Chronic renal failure

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 16.
Why the chronic renal failure is dangerous than acute renal failure?
Answer:

  • Though the kidney stops its function abruptly there are chances for recovery of kidney function in acute renal failure.
  • But in chronic failure there is a progressive loss of function of the nephron which gradually decreases the function of kidney.

Question 17.
What is meant by glomerulo nephritis or Bright’s disease? What are its symptoms?
Answer:
Inflammation of the glomerulus of both the kidneys due to the strepto coccal infection in children is called as Bright’s disease Symptoms

  • Haematuria
  • Proteinuria
  • Salt and water retention – Oligouria (Low urine out put)
  • Hypertension and Pulmonaryoedema

Question 18.
a) What is meant by kidney transplantations.
b) Where is graft kidney received from?
c) What are the steps to be taken to avoid graft rejection?
Answer:

  • Transfer of healthy kidney from one person (donor) to another person with kidney failure is called kidney transplantation.
  • The donated kidney may be taken from a healthy person who is declared brain death or from sibling or close relatives.
  • Immuno supressive drugs are administered to the patient to avoid tissue rejection.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 19.
a) Name the hormone that the dipsticks contain which tests urine?
b) Which colour indicates the presence of glucose in the urine?
Answer:

  • Glucose oxidase and peroxidase.
  • Brown coloured compound is produced.

Question 20.
What are Osmo confirmers?
Answer:
Osmo confirmers are able to change their internal osmotic concentration with change in external environments asin marine and sharks. molluscs

Question 21.
What are Osmo regulators?
Answer:
They maintain their internal osmotic concentration irrespective of their external osmotic environment, (eg) Otters.

Question 22.
List the excretory structures of different organisms.
Answer:

  • Protonephridia
  • Meta nephridia
  • Flame cells – Platy helminthes
  • Rennette cells – Nematodes
  • Malpighian tubules – Insects
  • Greenglands – Prawns

Question 23.
What is meant by filtration slits?
Answer:

  • The visceral layer of glomerulus is made of epithelial cells called podocytes and ends in foot processes which cling to the basement membrane of the glomerulus.
  • The openings between the foot processes are called filtration slits.

Question 24.
Draw the diagram of ornithine cycle?
Answer:
Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion 9

Question 25.
Why there is a pressure reduction when the blood goes through efferent arteriole?
Answer:

  • Blood enters the glomerulus faster with greater force through afferent arteriole.
  • Because the afferent arteriole is broader than efferent arteriole that is why the pressure reduces when it goes through the efferent arteriole.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 26.
What are the changes taking place in our body when there is a fluid loss?
Answer:

  • The osmo receptors in the hypothalamus is stimulated.
  • The neurohypophysis is stimulated and anti diuretic hormone is liberated.
  • The aquaporins in the tubuler are increased and water is reabsorbed and enters into the interstitial cell and the water loss is rectified.

Question 27.
How is skin acted as a excretory organ?
Answer:

  • Skin excretes Na+ and Cl small quantities of urea and lactate.
  • Sebaceous glands eliminated certain substances like steroids, hydrocarbons and waxes.

Question 28.
What is meant by urethritis?
Answer:
The urethral mucosa is continuous with the urinary tract. The infection in the urethra is called urethritis.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 29.
What is meant by Cystitis?
Answer:
The infection in the urethra can ascend the tract to cause bladder inflammation called cystitis.

Question 30.
What is meant by Phelitis or Pyelone phritis?
Answer:
The bladder infection ascend to the renal inflammation called pyelitis or pyelonephritis.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 31.
What are the two types of renal failure?
Answer:

  1. Acute renal failure
  2. Chronic renal failure

1. Acute renal failure

  • In acute renal failure the kidney stops its function abruptly.
  • There are chances for recovery of kidney function.

2. Chronic renal failure
In chronic renal failure there is a progressive loss of function of the nephron which gradually decreases the function of kidneys.

Question 32.
What is meant by Uremia?
Answer:
Uremia is characterized by increase in Urea uric acid and creatinine in blood.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 33.
What is meant by Nephrolithiasis?
Answer:
It is a formation of hard stone like masses in the renal tubules of renal pelvis.

Question 34.
How is water excess taken through drinking too much fruit juice regulated?
Answer:

  • When we drink much fruit juice the osmo receptors in hypothalamus is not stimulated and hence the secretion of vaso pressin from neuro hypophysis is reduced.
  • The aquaporin escapes from collecting duct to cytoplasm and hence water reabsorption is prevented and formed dilute urine.

Question 35.
What is uremia?
Answer:
Uremia is a condition in which there is a increase level of urea uric acid and creatinine in blood.

Question 36.
What is the amount of Urea present in the blood?
Answer:

  • The level of urea in the blood is 17 -30 mg /100ml.
  • In chronic kidney failure there is 10 times increase in urea level.

Five marks

IV. Detailed Answers –

Question 1.
Name the different excretory structure and different organisms.
Answer:
1. Invertebrate – Protonephridia / Meta nephridia
2. Platvhelminthes – Flamecells
3.  Amphioxues – Solenocytes
4 Nematodes – Rennette cells
5. Annelida – Metanephridia
6. Insects – Malpighian tubules
7. Prawn / Crustaceans – Green glands / Antenna! glands

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 2.
The concentration of urine depends on the structure of nephron? Explain?
Answer:

  1. In the reptiles the glomerulus is reduced or there may be no glomerulus and 1-lenle’s loop and hence produces dilute urine (chypotonic).
  2. In the mammals the long Henle’s loop produces concentrated urine (hypertonic)
  3. A glomerular kidneys of marine fishes produce little urine that is isoosmotic to the body fluid.
  4. Amphibians and fresh water fish lack Henle’s loop hence produce dilute urine.

Question 3.
Draw the following diagram and mark the following parts.
Answer:
A) Aorta
B) Renal vein
C) Ureter
D) Urinary bladder
Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion 10

Question 4.
a. What is the weight of kidneys? What are its outer coverings?
b. Draw the L.S of kidney and name the parts.
c. Explain the internal structure of kidney
Answer:
a. Each kidney weighs an average of 120 – 170 gms.
The outer layer of the kidney is covered by three layers of supportive tissue namely renal fascia perirenal fat capsule fibrous capsule.

b. Draw the LS of kidney and name the parts.
Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion 18

c. Internal Structure of kidney

  • The longitudinal section of kidney shows an outer cortex inner medulla and pelvis.
  • The inner concave surface of the kidney is renal hilum through which ureter blood vessels and nerves enter.
  • Inner to the hium is a funnel shaped renal pelvis with projection called calyces.
  • The calyces collect the urine and empties in to the ureter.
  • The medulla consists of conical tissues called medullary pyramids or renal pyramids.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 5.
a) What is the structural and functional unit of kidney.
b)Draw the diagram of nephron and name the parts.
c) Give short notes on Malpighian body or Renal Corpuscle.
Answer:
a) Hie structural and functional unit of kidney is nephron. It is composed of Malpighian body or renal corpuscle and Urine ferous tubule.
b) Structure of nephron
Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion 11
c. Malpighian body/Renal Corpuscle.

  • The Bowman’s capsule and the glomerulus together constitutes Malpighian corpuscle.
  • Bowman’s capsule is made up of two layers. It contains blood vessels called glomerules.
  • The endothelial of glomerulus has many pores. The viscral layers of glomerulus is made of epithelial cells called podocytes.
  • Tire podocytes end in foot processes which cling to the basement membrance of the glomerulus.
  • The openings between the foot processes are called filtration slits.

Question 6.
a) Give the different regions of renal tubule.
b) Where is renal tubule present in the kidney?
c) How is renal tubules differentiated depending on the Henle’s loop?
Answer:
a)
1. proximal convoluted tubule.
2. Henle’s loop
a. Thindescending limb of Henle’s loop.
b. Thick ascending limb.

2. Distal convoluted tubules
The distal convoluced tubules opens in to acollecting duct.
Several collecting ducts fuse to form papillary duct that delivers urine in to the calvces which opens into the renal pelvis.
b) The PCT and DCT are situated in the cortical region of the kidney.
The loop of Henle is in the medulla region.
c) The loop of Henle is too short and extends only very little into the medulla and are called cortical nephron.
Some nephrons have very long loops of Henle that run deep into the medulla and are called Juxta medullary nephrons.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 7.
a) Give notes on capillaries of nephron
b) Give an account of blood vessels of glomerulus.
c) What is vasa recta? Where is it seen?
Answer:
Capillaries of nephron
1. Glomerulus capillary bed
2. Peritubular capillaries

1. Glomerular capillarybed

  • It consists of afferent and efferent arteriole.
  • The afferent arteriole is broader than efferent arteriole.
  • The efferent arteriole that comes out of the glomerulus forms a fine capillary network around the renal tubule called the peritubular capillaries.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion 12
Vasa recta
The efferent arteriole serving the juxta medullary nephrons form bundles of long straight vessel called vasa recta and runs parallel to the loop of Henle.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 8.
a) Describe the three important process in the urine formation?
b) Give an account on glomerular filtration
Answer:
a. Main Processes

  • Clornerular Filtration
  • Tubular reabsorption
  • Tubular secretion

b. Glomerular filtration

  • Blood enters the kidney from the renal artery.
  • Blood enters the glomerulus faster with greater force through the afferent arterioles because the afferent arteriole is wider than efferent arteriole.
  • The glomerular hydrostatic pressure is around 55 mm Hg.
  • The colloidal osmotic pressure is 30 mm Hg. Capsular hydrostatic pressure is 15 mm Hg. The net pressure of 10 mm Hg is responsible for the renal filtration.
  • Next filtration pressure 55 mm Hg – (30 mm Hg +15 mm Hg) 10 mm Hg.
  • The kidneys produce 1801 of glomerular filtrate 24 hours and 120 -125 ml/min.
  • This filtrate will enter into the Bowman’s capsule and called primary urine. It contains more water colloidal protein, glucose salts and nitrogenous waste.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion 13

Question 9.
a) Why glomerular filtrate resembles blood plasma?
b) Tabulate the concentration of substances in the blood plasma and in the glomerular filtrate.
Answer:
As the glomerular filtrate forms it contain all the substances except plasma protein of blood. Hence it resembles blood.

SubstanceConcentration in blood Plasma / g dm-3Concentration in glomerular filtrate / g dm-3
Water900900
Proteins80.00.05
Aminoacids0.50.5
Glucose1.01.0
Urea0.30.3
Uric Acid0.040.04
Creatinine0.010.01
Inorganic ions (mainly (Na+, K+ and Cl )7.27.2

Question 10.
a) What is meant by Tubular Secretion?
b) Give an account of tubular secretion of nephron.
Answer:
Tubular Secretion:

  • The collecting tubule of nephron secrete H+ NH4 , Creatinine and Organic acid and liberated into the tubules and excreted through urine.
  • Most of the water is absorbed in the proximal convoluted tubule.
  • Na is exchanged for water in the loop of Henle.
  • The hypotonic fluid enters the distal convol uted tubule.
  • Substances such as urea and salts pass from peritubular blood into the cells as distal convoluted tubule and then to collecting duct.
  • Water is absorbed and concentrated hypertonic urine is form ed.
  • For every H+ secreted into the tubular filtrate a Na+ is absorbed by the tubular cell.
  • The H secreted combines with HCO3, HPO3 and NH3 and gets fixed as carbonic acid CH2CO3 and Phosphoric acid CH2PO4
  • Since H+ gets fixed in the fluid reabsorption of H+ is prevented.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 11.
a) In which process concentrated urine is formed?
b) How is concentrated urine formed in the Hen le’s loop.
Answer:
a) The major function of Henle’s loop is to concentrate Na+ and Cl.

  • There is low osmolarity near the cortex and high osmolarity towards the medulla.
  • This osmolarity in the medulla is due to the presence of the solutes transporters and is maintained by
    the arrangement of the loop of Henle collecting duct and vasa recta.
  • The osmolarity of interstitial fluid is 300 m Osm.
  • The Henle’s loop create a countercurrent multiplier.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion 14

  • As the fluid enters the descending limb water moves from the lumen into the inter stitial fluid the osmolarity reduces.
  • To counteract this dilution the region of the ascending limb actively pumps solutes from the lumen into the interstitial fluid and the osmolarity increases to about 1200 m OSM in medula.

b) The vasa recta maintains the medullary osmotic gradient via counter current exchanger.

  • The counter current exchanger of vasa recta preserves the medullary gradient while removing reabsorbed water and solutes.
  • The vasa recta leaves the kidney at the junction between the cortex and medulla.
  • When the blood leaves the efferent arteriole and enters vasa recta the osmolarity in the medulla increases (1200 rnOsm) and result in passive up take of solutes and loss of water.
  • As the blood enters the cortex the osmolarity in the blood decreases and the blood loses solutes and gain water to form concentrated urine.

Question 12.
How is vasa recta helps in producing concentrated urine?
Answer:
Vasa recta maintains the medullary osmotic gradient via counter current exchanger.
Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion 15

  • Vasa recta preserves the medullary gradient while removing reabsorbed water and solutes through counter current exchanges.
  • The vasa recta leave the kidney at the junction between the cortex and medulla.
  • The interstitial fluid at this point is iso – osmotic to blood.
  • When the blood leaves the efferent arteriole and enters the vasa recta the osmalarity in the medulla increases (1200 mOsm) and results in passive up take of solutes and loss of water.
  • As the blood enters the cortex the osmolarity in the blood decreases (300mOsm) and the blood loses solutes and gains water to form concentrated urine.
  • Human kidneys can produce urine nearly four times concentrated than the initial filtrate formed.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 13.
a) What are the structures that regulate kidney functioning?
b) What is the role of ADH in Urine formation.
c) What are the symptoms of diabetes insipidus.
Answer:
a) The structures that regulate kidney functioning

  • Hypothalamus
  • Juxta glomerular apparatus
  • Heart

b) The functions of ADH

  • When there is excessive loss of fluid from the body or when there is an increase in the blood pressure the osmo receptors of the hypothalamus is stimulated.
  • The osmo receptors stimulate the neurohypophysis to secrete antidiuretic hormone (AOH) or vasopressin.
  • ADH facilitates reabsorption of water by increasing the number of aquaporins on the cell surface of the distal convoluted tubule and collecting duct and prevents excessloss of water.

Question 14.
a) Name the cell that secretes the enzyme renin.
b) Where is granular cell present?
c) What is the role of renin in Osmoregulation.
Answer:
a) Renin is secreted by granular cells.
b) Granular cells are present in the afferent arteriole.
c) The role of renin

  • A fall in glomerular blood flow blood pressure and filtration rate can activate granular cells of juxta glomerular cells to release renin.
  • Renin converts the plasma protein angiotensinogen into angiotensis I and angiotensin II.
  • Angiotensis II stimulates Na+ reabsorption in the proximal convoluted tubule by vaso constriction of the blood vessels and increases the glomerular blood pressure.
    Angiotensis II stimulates adrenal cortex to secrete aldo sterone that causes reabsorption of Na+ ,K+ excretion and absorption of water.
  • This increases the glomerular blood pressure and glomerular filtration rate.
  • Hence renin regulates the osmoregulation.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 15.
a) Where is atrial natriuretic peptide liberated from?
b) Write its significance in short.
Answer:
a) This is liberated from atrium of heart.
b) Use of Atrial natriuretic peptide

  • It increases Na excretion and increases the blood flow to the glomerulus.
  • It acts on the afferent glomerular arteriole as a vaso dilator or an efferent arteriole as a vaso constrictor.
  • It reduces aldosterone from adrenal cortex and renin secretion.
  • Thus decreases the angiotensin II.
  • The atrial natri uretic factor acts antagonistically to renin angiotensin system aldosterone and vasopressin.

Question 16.
a) What is micturition?
b) How is central nervous system regulates urination?
Answer:
a) The process of release of Urine from the bladder is called micturition or urination.
b) Urine formed by the nephrons is ultimately carried to the urinary bladder where it is stored till it receives a signal from the central nervous system.

  • The stretch receptors present inthe urinary bladder are stimulated when it gets filled with urine.
  • At the same time the internal sphincters opens and relaxing the external sphincter.
  • The sphincter opens and the urine is expelled out.

Question 17.
Answer for the following questions.
a) What is the average excretion of an adult human?
b) What is the pH of Urine
c) How much pH is differed due to the food we eat?
d) What is the reason for the yellow colour of urine?
e) How much urea is excreted in a day?
f) If there is more glucose, and ketone what does it indicates?
Answer:

  • 1.5l
  • pH = 6
  • pH = 4.5-8
  • Uro chrome
  • 25 – 30 g
  • Diabetes melitus

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 18.
Answer for the following question in excretion by other organs.
a) What are the other structures that excrete nitrogenous wastes rather than kidney?
b) How much CO2 is excreted through lungs?
c) What are the wastes excreted by digestive systems? ’
d) What are the glands that excrete waste through skin?
e) What is the main function of sweat glands?
f) What is the second important function of sweat gland?
g) Name the substance excreted by sebaceous glands.
h) Name the substances excreted by sebaceous glands.
i) Name the waste excreted through saliva.
Answer:

  • Lungs, Liver and skin
  • 18 Litre, much Water
  • Bilirubin, Biliverdin, Cholesterol, Vitamins and drugs.
  • Sweat glands and sebaceous glands
  • To cool the body
  • Na+ and Cl, small quantities of Urea and Lactate excretion.
  • Sebum
  • Steroid Hydrocarbon and wax.
  • Nitrogenous wastes

Question 19.
Draw the schematic representations of renin hormone in the regulation of body fluid concentration.
Answer:

Question 20.
a) What is meant by Haemodialysis?
b) Why is it called as artifical kidney?
c) Give an account of Haemodialysis.
Answer:
a) in the patients of kidney failure toxic urea can be removed from the blood by a process called haemodialysis.
b) The dialyzing machine is a artificial kidney.
c) Hameodialysis
Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion 17

  • A dialyzing machine consists of a long cellulose tube surrounded by the dialysing fluid in a water bath.
  • A patient’s blood is drawn from a convenient artery and pumped into the dialysing unit after adding an anti coagulant like heparin.
  • The tiny poresin the dialysis tube allows small molecules such as glucose salts and urea to enter into the water bath.
  • Where as blood cells and protein molecules do not enter these pores the cleared blood is then pumped back tothe body through a vein.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 21.
Give notes on kidney transplantation?
Answer:

  • This involves transfer of healthy kidney from one person (donor) to another person who is with kidney failure.
  • The donated kidney may be taken from a healthy person who is declared brain dead or from sibling or close relatives to minimize the chances of rejection by the immune system of the host.
  • Immuno suppressive drugs are usually administered to the patient to avoid tissue rejection.

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Chemistry Guide Pdf Chapter 11 Fundamentals of Organic Chemistry Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

11th Chemistry Guide Fundamentals of Organic Chemistry Text Book Back Questions and Answers

Textbook Evaluation:

I. Choose the best answer:

Question 1.
Select the molecule which has only one π bond.
a) CH3 – CH = CH – CH3
b) CH3 – CH = CH – CHO
c) CH3 – CH = CH – COOH
d) All of these
Answer:
a) CH3 – CH = CH – CH3

Question 2.
In the hydrocarbon Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 1 the state of hybridization of carbon 1, 2, 3, 4 and 7 are in the following sequence.
a) sp, sp, sp3, sp2, sp3
b) sp2, sp, sp3, sp2, sp3
c) sp, sp, sp2, sp, sp3
d) none of these
Answer:
a) sp, sp, sp3, sp2, sp3

Question 3.
The general formula for alkadiene is
a) CnH2n
b) CnH2n – 1
c) CnH2n – 2
d) CnHn – 2
Answer:
c) CnH2n – 2

Question 4.
Structure of the compound whose IUPAC name is 5, 6 – dimethylhept – 2 – ene is
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 2
Answer:
a) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 14

Question 5.
The IUPAC name of the compound is
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 3
a) 2, 3 – Dimethylheptane
b) 3 – methyl – 4 – ethyloctane
c) 5 – ethyl – 6- methyloctane
d) 4 – Ethyl – 3 methyloctane
Answer:
d) 4 – Ethyl – 3 methyloctane

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 6.
Which one of the following names does not fit a real name?
a) 3 – Methyl – 3- hexanone
b) 4 – Methyl – 3 – hexanone
c) 3 – Methyl – 3 hexanol
d) 2 – Methyl cyclo hexanone
Answer:
a) 3 – Methyl – 3- hexanone

Question 7.
The IUPAC name of the compound CH3 – CH = CH – C ≡ CH is
a) Pent – 4- yn – 2 – ene
b) Pent – 3- en – 1- yne
c) Pent – 2 – en – 4 – yne
d) Pent – 1 yn – 3 – ene
Answer:
b) Pent – 3- en – 1- yne

Question 8.
IUPAC name of  Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 4 is
a) 3, 4, 4 – Trimethylheptane
b) 2 – Ethyl – 3, 3, – dimethyl heptane
c) 3, 4, 4 – Trimethyloctane
d) 2 – Butyl – 2 – methyl – 3 ethyl – butane
Answer:
c) 3, 4, 4 – Trimethyloctane

Question 9.
The IUPAC name of
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 5 is
a) 2,4,4 – Trimethylpent – 2 – ene
b) 2,4,4 – Trimethylpent – 3 – ene
c) 2,2,4 – Trimethylpent – 3 – ene
d) 2,2,4 – Trirnethylpent – 2 – ene
Answer:
c) 2,2,4 – Trimethylpent – 3 – ene

Question 10.
The IUPAC name of the compound Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 6 is
a) 3 – Ethyl – 2 – hexene
b) 3 – Propyl – 3 – hexene
c) 4 – Ethyl – 4 – hexene
d) 3 – Propyl – 2 – hexene
Answer:
a) 3 – Ethyl – 2 – hexene

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 11.
The IUPAC name of the compound
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 7 is
a) 2 – Hydroxypropionic acid
b) 2 – Hydroxy Propanoic acid
c) Propan – 2 – ol – 1 – oic acid
d) 1 – Carboxyethanol
Answer:
b) 2 – Hydroxy Propanoic acid

Question 12.
The IUPAC name of Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 8 is
a) 2 – Bromo – 3- methyl butanoic acid
b) 2 – methyl – 3 bromo butanoic acid
c) 3 – Bromo – 2 – methylbutanoic acid
d) 3 – Bromo – 2, 3 – dimethyl propanoic acid
Answer:
c) 3 – Bromo – 2 – methylbutanoic acid

Question 13.
The structure of isobutyl group in an organic compound is
a) CH3 – CH2 – CH2 – CH2

b) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 9

c) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 10

d) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 11
Answer:
c) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 10

Question 14.
The number of stereoisomers of 1, 2 – dihydroxy cyclopentane
a) 1
b) 2
c) 3
d) 4
Answer:
c) 3

Question 15.
Which of the following is optically active?
a) 3 – Chloropentane
b) 2- Chloro propane
c) Meso – tartaric acid
d) Glucose
Answer:
d) Glucose

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 16.
The isomer of ethanol is
a) acetaldehyde
b) dimethyl ether
c) acetone
d) methyl carbinol
Answer:
b) dimethyl ether

Question 17.
How many cyclic and acyclic isomers are possible for the molecular formula C3H6O?
a) 4
b) 5
c) 9
d) 10
Answer:
c) 9

Question 18.
Which one of the following shows functional isomerism?
a) ethylene
b) Propane
c) ethanol
d) CH2Cl2
Answer:
c) ethanol

Question 19.
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 12 are
a) resonating structure
b) tautomers
c) Optical isomers
d) Conformers
Answer:
b) tautomers

Question 20.
Nitrogen detection in an organic compound is carried out by Lassaigne’s test. The blue colour formed is due to the formation of
a) Fe3[Fe(CN)6]2
b) Fe4 [ Fe(CN)6]3
c) Fe4[Fe(CN)6]2
d) Fe3[Fe(CN)6]3
Answer:
b) Fe4 [ Fe(CN)6]3

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 21.
Lassaigne’s test for the detection of nitrogen fails in
a) H2N – CO – NH.NH2.HCl
b) NH2 – NH2.HCl
c) C6H5 – NH – NH2.HCl
d) C6H5CONH2
Answer:
c) C6H5 – NH – NH2.HCl

Question 22.
Connect pair of compounds which give blue colouration / precipitate and white precipitate respectively, when their Lassaigne’s test is separately done.
a) NH2 NH2 HCl and ClCH2 – CHO
b) NH2 CS NH2 and CH3 – CH2Cl
c) NH2 CH2 COOH and NH2 CONH2
d) C6H5NH2 and ClCH2 – CHO
Answer:
d) C6H5NH2 and ClCH2 – CHO

Question 23.
Sodium nitropruside reacts with sulphide ion to give a purple colour due to the formation of
a) [Fe (CN)5 NO]3-
b) [Fe (NO)5 CN]+
c) [Fe (CN)5 NOS]4-
d) [Fe (CN)5 NOS]3-
Answer:
c) [Fe (CN)5 NOS]4-

Question 24.
An organic Compound weighing 0.15 g gave on carius estimation, 0.12 g of silver bromide. The percentage of bromine in the Compound will be close to
a) 46 %
b) 34 %
c) 3.4 %
d) 4.6 %
Answer:
b) 34 %

Question 25.
A sample of 0.5 g of an organic compound was treated according to Kjeldahl’s method. The ammonia evolved was absorbed in 50mL of 0.5 M H2SO4 The remaining acid after neutralization by ammonia consumed 80 mL of 0.5 M NaOH. The percentage of nitrogen in the organic compound is.
a) 14 %
b) 28 %
c) 42 %
d) 56 %
Answer:
b) 28 %

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 26.
In an organic compound, phosphorus is estimated as
a) Mg2P2O7
b) Mg3(PO4)2
c) H3PO4
d) P2O 5
Answer:
a) Mg2P2O7

Question 27.
Ortho and para – nitro phenol can be separated by
a) azeotropic distillation
b) destructive distillation
c) steam distillation
d) cannot be separated
Answer:
c) steam distillation

Question 28.
The purity of an organic – compound is determined by
a) Chromatography
b) Crystallization
c) melting or boiling point
d) both (a) and (c)
Answer:
d) both (a) and (c)

Question 29.
A liquid which decomposes at its boiling point can be purified by
a) distillation at atmospheric pressure
b) distillation under reduced pressure
c) fractional distillation
d) steam distillation
Answer:
b) distillation under reduced pressure

Question 30.
Assertion:
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 13
Reason:
The principal functional group gets lowest number followed by double bond (or) triple bond.
a) both the assertion and reason are true and the reason is the correct explanation of assertion.
b) both assertion and reason are true and the reason is not the correct explanation of assertion.
c) assertion is true but reason is false.
d) both the assertion and reason are false
Answer:
a) both the assertion and reason are true and the reason is the correct explanation of assertion.

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

II. Write brief answers to the following questions:

Question 31.
Give the general characteristics of organic compounds.
Answer:
All organic compounds have the following characteristic properties.

  1. They are covalent compounds of carbon and generally insoluble in water and readily soluble in organic solvent such as benzene, toluene, ether, chloroform etc…
  2. Many of the organic compounds are inflammable (except CCl4). They possess low boiling and melting points due to their covalent nature.
  3. Organic compounds are characterized by functional groups. A functional group is an atom or a specific combination of bonded atoms that react in a characteristic way, irrespective of the organic molecule in which it is present. In almost all the cases, the reaction of an organic compound takes place at the functional group. They exhibit isomerism which is a unique phenomenon.

Question 32.
Describe the classification of organic compounds based on their structure.
Answer:
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 15

Question 33.
Write a note on homologous series.
Answer:
Homologous series:
A series of organic compounds each containing a characteric functional group and the successive members differ from each other in molecular formula by a CH2 group is called homologous series.
Example:
Alkanes:
Methane (CH4), Ethane (C2H6), Propane (C3H8) etc..
Alcohols:
Methanol (CH3OH), Ethanol (C2H5OH) Propanol (C3H7OH) etc…

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 34.
What is meant by a functional group?
Identify the functional group in the following compounds.
a) acetaldehyde
b) oxalic acid
e) dimethyl ether
d) methylamine
Answer:
Functional group:
An atom or group of atoms within a molecule that shows characteristics set of physical and chemical properties.
a) acetaldehyde → – CHO
b) oxalic acid → – COOH
c) di methyl ether → – O –
d) methyiamine → – NH2

Question 35.
Give the general formula for the following classes of organic compounds
a) Aliphatic monohydric alcohol
b) Aliphatic ketones
c) Aliphatic amines
Answer:
a) Aliphatic monohydric alcohol → CnH2n + 2O
b) Aliphatic ketones → CnH2nO
c) Aliphatic amines → CnH3n + 2N

Question 36.
Write the molecular formula of the first six members of homologous series of nitre alkanes.
Answer:

  • CH3NO2
  • C2H5NO2
  • C3H7NO2
  • C4H9NO2
  • C5H11NO2
  • C6H13NO2

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 37.
Write the molecular formula and possible structural formula of the first four members of homologous series of carboxylic acids.
Answer:
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 16

Question 38.
Give the IUPAC names of the following compounds.
i) (CH3)2CH – CH2 – CH(CH3) – CH(CH3)2

ii) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 17

iii) CH3 – O – CH3

iv) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 18

v) CH2 = CH – CH – CH2

vi) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 19

vii) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 20

viii) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 21

ix) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 22

x) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 23

xi) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 24

xii) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 25

xiii) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 26

xiv) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 27

xv) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 28

xvi) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 29
Answer:
(i) 2, 3, 5 – trimethyihexane
(ii) 2 – bromo – 3 – methylbutane
(iii) methoxymethane
(iv) 2 – hydroxybutanal
(y) buta – 1 ,3 – diene
(vi) 4 – chloropent – 2 – yne
(vii) 1 – bromobut – 2 – ene
(viii) 5 – oxohexanoic acid
(ix) 3 – ethyl – 4 – ethenylheptane
(x) 2, 4, 4 – trimethylpent – 2 – ene
(xi) 2 – methyl -1 – phenyipropan – 1 -amine
(xii) 2, 2 – dimethyl – 4oxopentanenitrile
(xiii) 2 – ethoxypropane
(xiv) 1 – fluoro – 4 – methyl – 2 -nitrobenzene
(xv) 3 – bromo – 2 – methylpentanal
(xvi) Acetophenone

Question 39.
Give the structure for the following compound
(i) 3 – ethyl – 2 methyl – 1 – pentene
(ii) 1, 3, 5 – Trimethyl cyclohex – 1 – ene
(iii) tetry butyl iodide
(iv) 3 – Chlorobutanal
(V) 3 – Chlorobutanol
(vi) 2 – Chloro – 2 – methyl propane
(vii) 2, 2 – dimethyl – 1 – chloropropane
(viii) 3 – methylbut -1- ene
(ix) Butan – 2, 2 – diol
(x) Octane – 1, 3 – diene
(xi) 1, 3 – Dimethylcyclohexane
(xii) 3 – Chlorobut – 1 – ene
(xiii) 2 – methylbutan – 3 – ol
(xiv) acetaldehyde
Answer:
(i) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 30

(ii) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 31

(iii) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 32

(iv) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 33

(v) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 34

(vi) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 35

(vii) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 36

(viii) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 37

(ix) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 38

(x) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 39

(xi) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 40

(xii) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 41

(xiii) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 42

(xiv) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 43

Question 40.
Describe the reactions involved in the detection of nitrogen in an organic compound by Lassaigne method.
Answer:
This method involves the conversion of covalently bonded N, S or halogen present in the organic compounds to corresponding water soluble ions in the form of sodium salts For this purpose a freshly cut piece of Na of the size of a paper, dried it by pressing between the folds of a filter paper is taken in a fusion tube in an iron tand clamping it just near the upper end and it is gently heated.

When it melts to a shiñing globule, put a pinch of the organic compound on it. Heat the tube with the tip of the flame till all reaction ceases and it becomes red hot. Now plunges it in about 50 mL of distilled water taken in a china dish and break the bottom of the tube by striding against the dish. Boil the contents of the dish for about 10 mts and filter. This filtrate is known as lassaignes extract or sodium fusion extract and it used for qualitative analysis of nitrogen, sulfur and halogens present in organic compounds.

Test for Nitrogen:
If nitrogen is present it gets converted to sodium cyanide which on reaction with freshly prepared ferrous sulphate and ferric ion followed by cone. HCl gives a Prussian blue color or green color or precipitate. It confirms the presence of nitrogen. HCl is added to dissolve the greenish precipitate of ferrous hydroxide produced by the excess of NaOH on FeSO4 which would otherwise mark the Prussian blue precipitate. The following reaction takes part in the formation of Prussian blue.
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 44

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 41.
Give the principle involved in the estimation of halogen in an organic compound by Carius method.
Answer:
Estimation of halogens (Carius method):
A known mass of the organic compound is heated with fuming HNO3 along with AgNO3. C, H & S gets oxidized to CO2, H2O, SO2 and halogen combines with AgNO3 to form a precipitate of silver halide.
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 45

The ppt of AgX is filtered, washed, dried and weighed. From the mass of AgX and the mass of the organic compound taken, percentage of halogens are calculated.

A known mass of the substance is taken along with fuming HNO3 and AgNO3 is taken in a clean carius tube. The open end of the Carius tube is sealed and placed in a iron tube for 5 hours in the range at 530 – 540 K Then the tube is allowed to cool and a small hole is made in the tube to allow gases produced to escape. The tube is broken and the ppt is filtered, washed, dried and weighed. From the mass of AgX obtained, calculations are made.

Calculation:
Weight of the organic compound: w g
Weight of AgCl precipitate = a g
143. 5 g of AgCl contains 35.5 g of Cl
∴ a g of AgCl contains \(\frac{35.5}{143.5}\) × a
w g Organic compound gives a g AgCl
Percentage of Cl in w g organic compound = \(\left(\frac{35.5}{143.5} \times \frac{\mathrm{a}}{\mathrm{w}} \times 100\right) \%\)
Let Weight of silver Bromide be ‘b’ g
188 g of AgBr contains 80 g of Br
∴ b g of AgBr contains \(\frac{80}{180} \times \frac{b}{w}\) of Br
w g Organic compound gives b g AgBr
Percentage of Br in w g organic compound = \(\left(\frac{80}{180} \times \frac{b}{w} \times 100\right) \%\)
Let Weight of silver Iodide be ‘c’ g
235 g of AgI contains 127 g of I
∴ C g of AgI contains \(\left(\frac{127}{235} \times \frac{c}{w}\right)\) of I
w g Organic compound gives c g AgI
percentage of I in w g organic compound = \(\left(\frac{80}{180} \times \frac{b}{w} \times 100\right) \%\)

Question 42.
Give a brief description of the principles of
(i) Fractional distillation
(ii) Column Chromatography
Answer:
(i) Fractional distillation:
This is one method to purify and separate liquids present in the mixture having their boiling point close to each other. In the fractional distillation, a fractionating column is fitted with distillation flask and a condenser. A thermometer is fitted in the fractionating column near the mouth of the condenser. This will enable to record the temperature of vapour passing over the condenser.

The process of separation of the components in a liquid mixture at their respective boiling points in the form of vapours and the subsequent condensation of those vapours is called fractional distillation. The process of fractional distillation is repeated, if necessary. This method finds a remarkable application in distillation of petroleum, coal-tar and crude oil.

(ii) Column Chromatography:
This is the simplest chromatographic method carried out in long glass column having a stop cock near the lower end. This method involves separation of a mixture over a column of adsorbent (Stationery phase) packed in a column. In the column a plug of cotton or glass wool is placed at the lower end of the column to support the adsorbent powder. The tube is uniformly packed with suitable adsorbent constitute the stationary phase. (Activated aluminum oxides (alumina), Magnesium oxide, starch are also used as adsorbents).

The mixture to be separated is placed on the top of the adsorbent column. Eluent which is a liquid or a mixture of liquids is allowed to flow down the column slowly. Different components depending upon the degree to which the components are adsorbed and complete separation takes place. The most readily adsorbed substances are retained near the top and others come down to various distances in the column.

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 43.
Explain paper chromatography.
Answer:
Paper chromatography (PC) is an example of partition chromatography. The same procedure is followed as in thin layer chromatography except that a strip of ‘ paper acts as an adsorbent. This method involves continues differential portioning of components of a mixture between stationary and mobile phase. In paper chromatography, a special quality paper known as chromatography paper is used. This paper act as a stationary phase.

A strip of chromatographic paper spotted at the base with the solution of the mixture is suspended in a suitable solvent which act as the mobile phase. The solvent rises up and flows over the spot. The paper selectively retains different components according to their different partition in the two phases where a chromatogram is developed. The spots of the separated colored compounds are visible at different heights from the position of initial spots . on the chromatogram. The spots of the separated colorless compounds may be observed either under ultraviolent light or by the use of an appropriate spray reagent.

Question 44.
Explain various types of constitutional isomerism (structural isomerism) in organic compounds.
Answer:
Structural isomerism:
This type of isomers have same molecular formula but differ in their bonding sequence.

(a) Chain or nuclear or skeletal isomerism:
These isomers differ in the way in which the carbon atoms are bonded to each other in a carbon chain or in other words isomers have similar molecular formula but differ in the nature of the carbon skeleton (ie. Straight or branched).
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 46

(b) Position isomerism:
If different compounds belonging to same homologous series with the same molecular formula and carbon skeleton, but differ in the position of substituent or functional group or an unsaturated linkage are said to exhibit position isomerism.
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 47

(c) Functional isomerism:
Different compounds having same molecular formula but different functional groups are said to exhibit functional isomerism.
Molecular formula C3H6O
CH3 – CH2 – CHO
propanal (aldehyde group)
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 48

Question 45.
Describe optical isomerism with suitable example.
Answer:
Compounds having same physical and chemical property but differ only in the rotation of plane of the polarized light are known as optical isomers and the phenomenon is known as optical isomerism.
Example:
Some organic compounds such as glucose have the ability to rotate the plane of the plane polarized light and there called are said to be optically active compounds and this property of a compound is called optical activity. The optical isomer, which rotates the plane of the plane polarised light to the right or in clockwise direction is said to be dextrorotary (dexter means right) denoted by the sign (+), whereas the compound which rotates to the left or anticlockwise is said to be leavorotatory (leavues means left) denoted by sign (-). Dextrorotatory compounds are represented as ‘d’ or by sign (+) and leavorotatory compounds are represented as ‘l’ or by sign (-).

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 46.
Briefly explain geametrical isomerism in alkene by considering 2-butene as an example.
Answer:
Geometrical isomers are the stereoisomers which have different arrangement of groups or atoms around a rigid frame work of double bonds. This type of isomerism occurs due to restricted rotation of double bonds, or about single bonds in cyclic compounds.

In alkenes, the carbon-carbon double bond is sp2 hybridized. The carbon-carbon double bond consists of a σ bond and a π bond. The π bond is formed by the head on overlap of sp2 hybrid orbitals. The n bond is formed by the side-wise overlap of ‘p’ orbitals. The presence of the π bond lock the molecule in one position. Hence, rotation around C = C bond is not possible. This restriction of rotation about C – C double bond is responsible for geometrical isomerism in alkenes.

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 49

These two compounds are termed as geometrical isomers and are distinguished from each other by the terms c is and trans. The c is isomer is one in which two similar groups are on the same side of the double bond. The trans isomers is that in which the two similar groups are on the opposite side of the double bond, hence this type of isomerism is often called cis- trans isomerism.

The cis-isomer can be converted to trans isomer or vice versa is only if either isomer is heated to a high temperature or absorbs light. The heat supplies the energy (about 62kcal/ mole) to break the n bond so that rotation about a bond becomes possible. Upon cooling, the reformation of the n bond can take place in two ways giving a mixture both cis and trans forms of trans-2-butene and cis-2-butene.
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 50

Question 47.
0.30 g of a substance gives 0.88 g of carbon dioxide and 0.54 g of water calculate the percentage of carbon and hydrogen in it.
Solution:
Weight of organic compound = 0.30 g
Weight of carbon dioxide = 0.88 g
Weight of water = 0.54 g

Percentage of carbon:
44 g of carbondioxide contains, carbon = 12 g
0.88 g of carbon dioxide contains, carbon = \(\frac{12 \times 0.88}{44}\) g
0.30 g substance contains,
carbon = \(\frac{12 \times 0.88}{44}\) g

100 g substance Contains \(\frac{12 \times 0.88}{44}\) × \(\frac{100}{0.30}\) = 80 g of carbon
Percentage of carbon = 80 %

Percentage of hydrogen:
18 g of water contains, hydrogen = 2 g
0.54 g of water contains, hydrogen = \(\frac{2 \times 0.54}{18}\) g
0.30 g of substance contains hydrogen = \(\frac{2 \times 0.54}{18 \times 0.30}\) g
100 g of substance contains = \(\frac{2 \times 0.54}{18 \times 0.30}\) × 100 g = 20 g of hydrogen
Percentage of hydrogen = 20 %

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 48.
The ammonia evolved form 0.20 g of an organic compound by Kjeldahl method neutralized 15 ml of N / 20 sulphuric acid solution. Calculate the percentage of Nitrogen.
Answer:
weight of organic compound = 0.20 g
Normality of acid = \(\frac{\mathrm{N}}{20}\)
Volume of standard acid neitralized by ammonia = 15 ml
1000 ml of N ammonia contains = 14 g of nitrogen
15 ml of ammonia of normality \(\frac{\mathrm{N}}{20}\) contains nitrogen = \(\frac{14 \times 15 \times 1}{1000 \times 20}\)

0.20 g of compound contains nitrogen = \(\frac{14 \times 15}{1000 \times 20}\)

100 g of compound contains nitrogen = \(\frac{14 \times 15 \times 100}{1000 \times 20 \times 0.20}\) = 5.25 g
Percentage of nitrogen = 5.25 %

Question 49.
0.32 g of an organic compound, after heating with fuming nitric acid and barium nitrate crystals is a sealed tube gave 0. 466 g of barium sulphate. Determine the percentage of sulphur in the compound.
Solution:
Mass of the substance taken = 0.32 g
Mass of BaSO4 formed = 0.466 g
Molecular mass of BaSO4 = 137 + 32 + 64 = 233
Then, mass of S in 0.466 g of BaSO4 = \(\frac{0.466 \times 32}{233}\)

Percentage of S in compound= \(\frac{0.466 \times 32 \times 100}{233 \times 0.32}\) = 20 %

Question 50.
0.24 g of an organic compound gave 0.287 g of silver chloride in the carius method. Calculate the percentage of chlorine in the compound.
Solution:
Mass of organic compound = 0.24 g
Mass of silver chloride = 0.287 g
143. 5 g AgCl contains = 35.5 g chlorine
0.287 g of AgCl contains = \(\frac{35.5}{143.5}\) × 0.287

Percentage of chlorine = \(\frac{35.5}{143.5} \times \frac{0.287}{0.24}\) × 100 = 29.58 %

Question 51.
In the estimation of nitrogen present in an organic compound by Dumas method 0.35 g yielded 20.7 mL of nitrogen at 15° C and 760 mm pressure. Calculate the percentage of nitrogen in the compound.
Solution:
Volume of N2 at NTP = \(\frac{V \times P}{t+273} \times \frac{273}{760}\)
= V0 ml
Substituting the various values in the above equation,
V0 = \(\frac{20.7 \times 760}{288} \times \frac{273}{760}\) = 19.62 ml

weight of 19.62 ml of Nitrogen = \(\frac{28}{22400}\) × 19.62 g

∴ Percentage of Nitrogen = \(\frac{28}{22400}\) × 19.62 × \(\frac{100}{0.35}\)
= 4.9 %

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

11th Chemistry Guide Fundamentals of Organic Chemistry Additional Questions and Answers

I. Choose the correct answer:

Question 1.
Organic compounds can be formed by
a) Plants only
b) Animals only
c) Plants and Animals
d) Plants, animals and can be synthesized in laboratory
Answer:
d) Plants, animals and can be synthesized in laboratory

Question 2.
Generally, organic compounds are
a) Amorphous
b) Complexes
c) Covalent
d) Electrovalent
Answer:
c) Covalent

Question 3.
The vital force theory was proposed by
a) Wohler
b) Berthlot
c) Berzelius
d) Kolbe
Answer:
c) Berzelius

Question 4.
The first carbon compound prepared from its elements is
a) Urea
b) Acetic acid
c) Methane
d) benzene
Answer:
b) Acetic acid

Question 5.
The first organic compound was synthesized in laboratory by
a) Wohler
b) Kolbe
c) Berzelius
d) Neil Barthlot
Answer:
a) Wohler

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 6.
The first organic compound synthesized in the laboratory from an inorganic compound is
a) NH4NCO
b) NH2 – CO – NH2
c) CH3COOH
d) CH4
Answer:
b) NH2 – CO – NH2

Question 7.
Marsh gas mainly contains
a) C2H2
b) C2H4
c) CH4
d) C2H6
Answer:
c) CH4

Question 8.
Hybridization at 2nd carbon in CH2 = CH – CH3 is
a) sp
b) sp2
c) sp3
d) sp3d
Answer:
b) sp2

Question 9.
Number of possible position isomers for Dichlorobenzene is
a) 2
b) 3
c) 4
d) 5
Answer:
b) 3

Question 10.
n – Butane and isobutane are a pair of
a) chain isomers
b) position isomers
c) metamers
d) functional isomers
Answer:
a) chain isomers

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 11.
Alkanols and Alkoxyalkanes are
a) Functional isomers
b) Keto – enol tautomers
c) Geometrical isomers
d) Not isomers at all
Answer:
a) Functional isomers

Question 12.
n – propyl alcohol and isopropyl alcohol are examples of
a) Position isomerism
b) Chain isomerism
c) Tautomerism
d) Geometrical isomerism
Answer:
a) Position isomerism

Question 13.
The number of structural alcoholic isomers for C4H10O is
a) 2
b) 3
c) 4
d) 5
Answer:
c) 4

Question 14.
Cycloalkanes are isomeric with
a) Alkadienes
b) Alkynes
c) Aromatic compounds
d) Olefins
Answer:
d) Olefins

Question 15.
Number of possible monochloro benzenes is
a) 1
b) 3
c) 5
d) 6
Answer:
a) 1

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 16.
Diethyl ether and n – propyl methyl ether are
a) Metamers
b) Chain isomers
c) Geometrical isomers
d) Position isomers
Answer:
a) Metamers

Question 17.
The total number of structural isomers for the compound of the formula C4H10O is
a) 7
b) 6
c) 4
d) 3
Answer:
a) 7

Question 18.
The number of primary alcoholic isomers with the formula C4H10O is
a) 1
b) 2
c) 3
d) 4
Answer:
b) 2

Question 19.
The compound which is not isomeric with diethyl ether is
a) n – propyl methyl ether
b) Butan – 1 – ol
c) 2 – Methylpropan – 2 – ol
d) Butanone
Answer:
d) Butanone

Question 20.
The number of isomeric amines possible for the formula C3H9N
a) 4
b) 3
c) 5
d) 6
Answer:
a) 4

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 21.
Which hybrid orbitals are involved in the CH3 – CH = CH – CH3 compound
a) sp and sp3
b) sp2 and sp3
c) sp and sp2
d) only sp3
Answer:
b) sp2 and sp3

Question 22.
Which of the following bonds is strongest?
a) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 51

b) > C = C <

c) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 52

d) – C – C –
Answer:
a) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 51

Question 23.
According to Huckel’s rule a compound, is said to be aromatic if’ it contains
a) 4n bonds
b) 4n atoms
c) (4n + 2) atoms
d) (4n + 2) π electrons
Answer:
d) (4n + 2) π electrons

Question 24.
Which of the following is an aromatic compound
a) Phenol
b) Naphthalene
c) Pyridine
d) All
Answer:
d) All

Question 25.
Which is a saturated compound?
a) alkanes
b) alkenes
c) alkynes
d) cyclo alkenes
Answer:
a) alkanes

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 26.
Which is an alicyclic compound?
a) benzene
b) cyclohexane
c) pyridine
d) pyrrole
Answer:
b) cyclohexane

Question 27.
Which of the following is not a cyclic compound?
a) Anthracene
b) Pyrrole
c) Phenol
d) Isobutylene
Answer:
d) Isobutylene

Question 28.
Functional group present in amides is
a) – COOH
b) – NH2
c) – CONH2
d) – COO –
Answer:
c) – CONH2

Question 29.
IUPAC name of ester is
a) Alkoxy alkane
b) Alkyl alkanoate
c) Alkanoyl halide
d) Alkanoic anhydride
Answer:
b) Alkyl alkanoate

Question 30.
IUPAC name of methyl cyanide is
a) Cyano methane
b) Ethanenitrile
c) Methane nitrile
d) Methyl – n – butyl amine
Answer:
b) Ethanenitrile

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 31.
The correct IUPAC name of Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 53 is
a) 1, 2 – diethyl butene
b) 2 – ethyl – 3- methyl pentene
c) 3, 4 – dimethyl hex – 3 – ene
d) 2, 3 – dimethyl pent – 2 – ene
Answer:
d) 2, 3 – dimethyl pent – 2 – ene

Question 32.
IUPAC name of CH2OH – CH2OH is
a) 1, 2 – dihydroxy ethane
b) ethylene glycol
c) ethane – 1, 2 – diol
d) ethane – 1, 2 – dial
Answer:
c) ethane – 1, 2 – diol

Question 33.
IUPAC name of CH ≡ C – CH = CH2 is
a) but – 3 – ene – 1 – yne
b) but – 1 – ene – 3 – yne
c) but – 1 – yne – 3 – ene
d) but – 3 – yne – 1 – ene
Answer:
b) but – 1 – ene – 3 – yne

Question 34.
The IUPAC name of Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 54 is
a) 4 – hydroxy – 2 – methyl pentanal
b) 2 – hydroxy – 4 – methyl pentanal
c) 4 – hydroxy – 2 – methyl pentanol
d) 2 – hydroxy – 4 – methyl pentanol
Answer:
a) 4 – hydroxy – 2 – methyl pentanal

Question 35.
3 – methyl penta -1, 3- diene is
a) CH2 = CH (CH2)2 CH3
b) CH2 = CHCH (CH3) CH2CH3
c) CH3CH = C(CH3)CH = CH2
d) CH3 = C = CH (CH3)2
Answer:
c) CH3CH = C(CH3)CH = CH2

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 36.
The structural formula of methyl amino methane is
a) (CH3)2 CH NH2
b) (CH3)3 N
c) (CH3)2 NH
d) CH3NH2
Answer:
c) (CH3)2 NH

Question 37.
Which of the following is the functional isomer of methyl acetate?
a) Ethyl acetate
b) Propanoic acid
c) Ethyl formate
d) Propanone
Answer:
b) Propanoic acid

Question 38.
The compound which is not isomeric with diethyl ether is
a) n – propyl methyl ether
b) 1 – Butanol
c) 2 – Methyl – 2 – propanol
d) Butanone
Answer:
d) Butanone

Question 39.
Which of the following pairs of compounds are tautomers?
a) Propanol & propanone
b) Ethanol & vinyl alcohol
c) Ethanol & allyl alcohol
d) Vinyl alcohol & ethanal
Answer:
d) Vinyl alcohol & ethanal

Question 40.
Which of the following compounds does not have any tertiary hydrogen atoms?
a) (CH3)3 CCH2 CH3
b) (CH3)2 CHCH2 CH3
c) (CH3)2 CHCH (CH3)2
d) (CH3)3 CH
Answer:
a) (CH3)3 CCH2 CH3

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 41.
The IUPAC name of Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 55 is
a) 2 – Methyl – 2 – butenoic acid
b) 3 – Methyl – 3 – butenoic acid
c) 3 – Methyl – 2 – butenoic acid
d) 2 – Methyl – 3 – butenoic acid
Answer:
c) 3 – Methyl – 2 – butenoic acid

Question 42.
The IUPAC name of Cinnamaldehyde is
a) 3 – Phenyl prop – 2 – enal
b) 1 – Phenyl – prop – 1 – enal
c) 1 – Phenyl – prop – 2 – enal
d) 3 – Phenyl – prop – 1 – enal
Answer:
a) 3 – Phenyl prop – 2 – enal

Question 43.
The IUPAC name of the Compound CH3 – CH(OH) – COOH is
a) Lactic acid
b) 2 – Hydroxy propanoic acid
c) 3 – Hydroxy propanoic acid
d) Carboxy propanol
Answer:
b) 2 – Hydroxy propanoic acid

Question 44.
The IUPAC name of the compound Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 56 is
a) 2 – Ethyl – ethyl acetate
b) Ethyl – 3 – methy1butanoate
c) Ethyl – 2 – methyl butanoate
d) 2- methyl butanoic acid
Answer:
c) Ethyl – 2 – methyl butanoate

Question 45.
The IUPAC name of the given compound Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 57 is
a) 2,2 — Dimethyl butane
b) lsohexane
c) 2, 3 – Dimethyl butane
d) Di isohexane
Answer:
c) 2, 3 – Dimethyl butane

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 46.
The IUPAC name of the given compound Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 58 is
a) Octyl cyclopentane
b) 3 – cyclopentyl octane
c) Cyclopentane octane
d) 6 – cyclopentyl octane
Answer:
b) 3 – cyclopentyl octane

Question 47.
The IUPAC name of Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 59 is
a) but – 2- ene – 2,3 – diol
b) pent – 2- ene – 2,3 – diol
c) 2 – methyl but – 2 – ene – 2,3 – diol
d) hex – 2- ene – 2,3 – diol
Answer:
b) pent – 2- ene – 2,3 – diol

Question 48.
The structure of 3-bromoprop-1-ene is
a) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 60

b) CH3 – CH = CH – Br

c) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 61

d) Br – CH2 – CH ≡ CH2
Answer:
d) Br – CH2 – CH ≡ CH2

Question 49.
Neo-heptyl alcohol is correctly represented as
a) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 62

b) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 63

c) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 64

d) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 65
Answer:
c) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 64

Question 50.
Number of dibromo derivatives possible for propane are
a) 2
b) 3
c) 1
d) 4
Answer:
d) 4

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 51.
The number of aromatic isomers possible for C7H8O is
a) 2
b) 3
c) 4
d) 5
Answer:
d) 5

Question 52.
Isomers of propanoic acid are
a) HCOOC2H5 and CH3COOCH3
b) H – COOC2H5 and C3H7COOH
c) CH3COOCH3 and C3H7OH
d) C3H7OH and CH3COCH3
Answer:
a) HCOOC2H5 and CH3COOCH3

Question 53.
IUPAC name of CH3 – CH (OCH3) – CH2 – NH2
a) 2-methoxy propanamine
b) 1-amino – 2-methoxy propane
c) 1-amino – 2-methyl – 2-methoxy ethane
d) 1 – methoxy- 2-amino propane
Answer:
a) 2-methoxy propanamine

Question 54.
IUPAC name of Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 66 is
a) 3 – cyanopentane – 1, 5 – dinitrile
b) Propane – 1, 2, 3-tri nitrile
c) 1, 2, 3-tri cyano propane
d) Propane 1, 2, 3-tricarbonitrile
Answer:
d) Propane 1, 2, 3-tricarbonitrile

Question 55.
The IUPAC name of the compound Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 67 is
a) 3 – Carboxylic pentane – 1,5 -dioic acid
b) Propane – 1, 2, 3 – trioic acid
c) 1, 2, 3- tricarboxylic propane
d) Propane – 1,2, 3 – tricarboxylic acid
Answer:
b) Propane – 1, 2, 3 – trioic acid

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 56.
The IUPAC name of the following compound
CH3 – C(CH3)2 – CH2 – CH = CH2 is
a) 2, 2 – Dimethyl – 4 – pentene
b) 4, 4 – Dimethyl – 1 – pentene
c) 1, 1, 1 – trimethyl – 3 – butene
d) 4, 4, 4 – trimethyl – 1 – butene
Answer:
b) 4, 4 – Dimethyl – 1 – pentene

Question 57.
The IUPAC name of Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 68 is
a) 2, 4 – Dimethyl pentan – 2 – ol
b) 2, 4 – Dimethyl pentan – 4 – ol
c) 2,2 – Dimethyl butan – 2- ol
d) Butan – 2 – ol
Answer:
a) 2, 4 – Dimethyl pentan – 2 – ol

Question 58.
The IUPAC name the compound Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 69 is
a) Butane – 2, 3, 4 – triol
b) Butane – 1,2, 3 – triol
c) Pentane – 1, 2, 3 – triol
d) 2, 3 dihydroxy butanol
Answer:
b) Butane – 1,2, 3 – triol

Question 59.
The IUPAC name the compound Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 70 is
a) 3, 3 – Dimethyl – 1 – cyclohexanol
b) 1, 1- Dimethyl – 3 – hydroxy cyclohexane
c) 3, 3 – Dimethyl – 1 – hydroxy cyclohexane
d) 1, 1 – Dimethyl – 3 – cyclohexanol
Answer:
a) 3, 3 – Dimethyl – 1 – cyclohexanol

Question 60.
The IUPAC name of the compound Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 71 is
a) 2 – Ethylprop – 2 – en – 1 – ol
b) 2 – Hydroxymethyl butan – 1- ol
c) 2 – Methylene butan – 1 – ol
d) 2 – Ethyl -3 hydroxyprop – 1 – ene
Answer:
a) 2 – Ethylprop – 2 – en – 1 – ol

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 61.
The number of possible alkynes with molecular formula C5H8 is
a) 2
b) 3
c) 4
d) 5
Answer:
b) 3

Question 62.
What is the IUPAC name of the following Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 72 is
a) 3 – chloro cyclo hexa – 1, 5 – diene
b) 5 – chloro cyclo hexa – 1, 3 – diene
c) 1 – chloro cyclo hexa – 2, 5 – diene
d) 2 – chloro cyclo hexa – 1, 4 – diene
Answer:
b) 5 – chloro cyclo hexa – 1, 3 – diene

Question 63.
What is the IUPAC name of the following Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 73 is
a) 6 – hydroxy cyclohex – 2 – ene – 1 – al
b) 4 – hydroxy cyclohex – 1 – ene – 3 – al
C) 2 – hydroxy cyclohex – 5 – ene – 1 – al
d) 1 – formyl cyclohex – 5 – ene – 2 – ol
Answer:
a) 6 – hydroxy cyclohex – 2 – ene – 1 – al

Question 64.
What is the IUPAC name of the following?
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 74
a) 1 – chloro – 2 – bromo – 4 – nitrobenzene
b) 1 – bromo – 2- chloro – 4 – mtrobenzene
c) 3 – bromo – 4 – chloro – nitrobenzene
d) 2 – bromo – 1 – chloro – 4- nitrobenzene
Answer:
d) 2 – bromo – 1 – chloro – 4- nitrobenzene

Question 65.
What is the IUPAC name of the following?
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 75
a) Ethenyl cyclo pentane
b) cyclopentyl ethene
c) cyclopentyl ethylene
d) vinyl cyclopentane
Answer:
b) cyclopentyl ethene

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 66.
The hybridization of carbon atoms in C – C single bond is HC ≡ C – CH = CH2 is
a) sp3 – sp3
b) sp2 – sp3
c) sp – sp2
d) sp3 – sp
Answer:
c) sp – sp2

Question 67.
The correct IUPAC name of the compound Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 76 is
a) 1, 4 – Butane dioicacid
b) Ethane – 1, 2 – dicarboxylic acid
c) Succinic acid
d) 1, 2 – Ethane dioic acid
Answer:
a) 1, 4 – Butane dioic acid

Question 68.
Correct statements about CH3 – CH2 – CN is
a) common name of the compound is ethylcyanide.
b) IUPAC name of the compound propane – 1 – nitrile.
c) secondary suffix of the compound is nitrile.
d) IUPAC name of the compound is ethane nitrile.
Answer:
a) common name of the compound is ethylcyanide.

Question 69.
Tautomerism is shown by
a) R – C ≡ N
b) R – NO2
c) R – OH
d) R – COOH
Answer:
b) R – NO2

Question 70.
Stereo isomers have different
a) Molecular mass
b) Molecular formula
c) Structural formula
d) Configuration
Answer:
d) Configuration

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 71.
Geometrical isomerism may be exhibited by compounds having atleast.
a) One double bond
b) One triple bond
c) One asymmetric carbon
d) One polar bond
Answer:
a) One double bond

Question 72.
The prefixes syn – and anti – are used to denote
a) structural isomers
b) conformational isomers
c) geometrical isomers
d) optical isomers
Answer:
c) geometrical isomers

Question 73.
d – tartaric acid and l – tartaric acid are
a) geometrical isomers
b) conformers
c) enantiomers
d) diastereomers
Answer:
c) enantiomers

Question 74.
The method of separation of enantiomers from racemic mixture is known as
a) inversion
b) recemisation
c) resolution
d) asymmetric synthesis
Answer:
c) resolution

Question 75.
Racemic mixture is optically inactive due to
a) internal compensation
b) external compensation
c) inversion
d) plane of symmetry
Answer:
c) inversion

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 76.
Meso isomers are possible when the organic compound contains
a) one asymmetric carbon
b) two or more dissimilar asymmetric carbons
c) similar asymmetric carbons
d) unsaturation
Answer:
c) similar asymmetric carbons

Question 77.
Optical inactivity of meso isomer is due to
a) element of symmetry and element of asymmetry
b) internal compensation
c) due to lack of asymmetric carbon
d) External compensation
Answer:
b) internal compensation

Question 78.
A racemic mixture is a mixture of
a) meso and its isomers
b) d and l isomers of same compound in equimolar proportions
c) d and l isomers of same compound in different proportions
d) mixture of d and meso isomers
Answer:
b) d and 1 isomers of same compound in equimolar proportions

Question 79.
Which of the following is optically active?
a) HOOC – CH2 – COOH
b) CH3 – CO – COOH
c) CH3 – CH(OH) – COOH
d) CH3 – CH2 – COOH
Answer:
c) CH3 – CH(OH) – COOH

Question 80.
Which of the following is optically active?
a) n – propanal
b) 2 – chlorobutane
c) n – butanal
d) 3 – pentanol
Answer:
b) 2 – chlorobutane

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 81.
The number of optical enantiomers of tartaric acid
a) 3
b) 2
c) 4
d) 1
Answer:
b) 2

Question 82.
Geometrical isomers differ in
a) position of substituents
b) Position of double bond
c) C – C bond length
d) Spatial arrangement of groups
Answer:
d) Spatial arrangement of groups

Question 83.
Which of the following exhibit cis – trans isomerism
a) propene
b) 1 – butene
c) 2 – butene
d) benzene
Answer:
c) 2 – butene

Question 84.
Which of the following show geometrical isomerism?
a) CH3CH = CHCH3
b) (CH3)2C = CH2
c) C2H5CH = CH2
d) CH3CH = CH2
Answer:
a) CH3CH = CHCH3

Question 85.
Which of the following does not show geometrical isomerism?
a) 1, 2 – dichloro – 1 – pentene
b) 1, 3 – dichloro – 2 – pentene
c) 1, 1 – dichloro – 1 – pentene
d) 1, 4 – dichloro – 2 – pentene
Answer:
c) 1, 1 – dichloro – 1 – pentene

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 86.
Geometrical isomerism is not shown by
a) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 77

b) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 78

c) CH2 = C(CI) CH3

d) CH3 – CH = CH – CH = CH2
Answer:
c) CH2 = C(CI) CH3

Question 87.
Which of the following is optically active?
a) Glycerine
b) Acetaldehyde
c) Glyceraldehyde
d) Acetone
Answer:
c) Glyceraldehyde

Question 88.
The minimum number of C atoms for a hydrocarbon to exhibit optical isomerism
a) 4
b) 5
c) 6
d) 7
Answer:
d) 7

Question 89.
Which of the following can form cis – trans isomers?
a) C2H5Br
b) (CH)2(COOH)2
c) CH3CHO
d) (CH2)2COOH
Answer:
b) (CH)2(COOH)2

Question 90.
No.of geometrical isomers possible for the compound CH3 – CH = CH – CH = CH – C2H5
a) 2
b) 3
c) 4
d) 5
Answer:
c) 4

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 91.
The number of geometrical isomers of CH3 – CH = CH – CH = CH – CH = CHCl is
a) 2
b) 4
c) 6
d) 8
Answer:
d) 8

Question 92.
Minimum number of C atoms for an alkene hydrocarbon, that shows geometrical & optical isomerism both
a) 5
b) 6
c) 7
d) 8
Answer:
c) 7

Question 93.
Among the following compounds which exhibits optical isomerism?
a) propanol
b) 2 – propanol
c) 1 – butanol
d) 2 – butanol
Answer:
d) 2 – butanol

Question 94.
Which of the mesoisomer?
a) CH2OHCHOHCHO
b) CH2OHCHOHCHOHCHO
c) HOOCCHOHCHOHCOOH
d) HOH2CCHOHCHOHCOOH
Answer:
c) HOOCCHOHCHOHCOOH

Question 95.
d – tartaric acid and l – tartaric acid can be separated by
a) Salt formation
b) Fractional distillation
c) Fractional crystallization
d) Chromatography
Answer:
a) Salt formation

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 96.
Paper chromatography is
a) Adsorption chromatography
b) partition chromatography
c) Ion exchange chromatography
d) all of these
Answer:
b) partition chromatography

Question 97.
Simple distillation can be used to separate liquids which differ in their boiling points at least by
a) 5°C
b) 10°C
c) 40 – 50°C
d) 100°C
Answer:
c) 40 – 50°C

Question 98.
In adsorption chromatography mobile phase will be
a) Only solid
b) Only liquid
c) Only gas
d) Liquid as well as gas
Answer:
d) Liquid as well as gas

Question 99.
Which of the following can be used as adsorbent in adsorption chromatography?
a) Silica gel
b) Alumina
c) Cellulose powder
d) All of these
Answer:
d) All of these

Question 100.
Two substances when separated out on the basis of their extent of adsorption by one material, the phenomenon is called
a) Chromatography
b) Crystallization
c) Sublimation
d) Steam distillation
Answer:
a) Chromatography

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 101.
Chromatographic technique is used for the separation of
a) Camphor
b) Alcohol & Water
c) Acetone and Methanol
d) Plant pigments
Answer:
d) Plant pigments

Question 102.
In column chromatography stationary phase is
a) only solid
b) only liquid
c) only gas
d) All of these
Answer:
a) only solid

Question 103.
Which of the following method is used for the purification of solids?
a) Distillation under reduced pressure
b) Distillation
c) Strain distillation
d) Sublimation
Answer:
d) Sublimation

Question 104.
Vacuum distillation is used to purify liquids which
a) are highly volatile
b) are explosive in nature
c) soluble in water
d) decomposes below their B.P’s
Answer:
c) soluble in water

Question 105.
Impure Napthalene is purified by
a) Fractional crystallization
b) Fractional distillation
C) solvent extraction
d) sublimation
Answer:
d) sublimation

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 106.
A very common adsorbent used in coloun^n chromatography is
a) Powdered charcoal
b) Alumina
c) Chalk
d) Sodium carbonate
Answer:
b) Alumina

Question 107.
Simple distillation of liquids involves simultaneously
a) Vapourisation and condensation
b) Condensation and vapourisation
c) Vapourisation and sublimation
d) Sublimation and condensation
Answer:
a) Vapourisation and condensation

Question 108.
The latest technique for the purification of organic compounds is
a) Fractional distillation
b) Chromatography
c) Vacuum distillation
d) Crystallization
Answer:
b) Chromatography

Question 109.
Fixed melting point of an organic compound informs
a) Purity of an organic compound
b) Conductivity of compound
c) Chemical nature of compound
d) Whether the compound is liquid or gas
Answer:
a) Purity of an organic compound

Question 110.
Lassaigne’s test is used in qualitative analysis to detect
a) Nitrogen
b) Sulphur
c) Chlorine
d) All of these
Answer:
d) All of these

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 111.
In Lassaigne’s method organic compound is fused with
a) Sodium metal
b) Zinc dust
c) Sodium carbonate and Zinc dust
d) Calcium metal
Answer:
a) Sodium metal

Question 112.
Presence of nitrogen in organic compound in Lassaigne’s extract as
a) Nitrogen gas
b) NH3
c) NO
d) CN
Answer:
d) CN

Question 113.
Medium of Sodium extract is
a) Neutral
b) Basic
c) Acidic
d) Depends on organic compound
Answer:
b) Basic

Question 114.
H2O vapours on passing through anhydrous CuSO4 turns it to
a) Green
b) Blue
c) Violet
d) White
Answer:
b) Blue

Question 115.
When a nitrogenous organic compound is fused with sodium, the nitrogen present in the compound is converted into
a) Sodium Nitrate
b) Sodium nitrite
c) Sodamide
d) Sodium cyanide
Answer:
d) Sodium cyanide

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 116.
In the Lassainge’s test the Sulphur present in the organic compound first changes into
a) Na2SO3
b) CS2
c) Na2SO4
d) Na2S
Answer:
d) Na2S

Question 117.
Which of the following elements in an organic compound cannot be detected by Lassaigne’s test?
a) N
b) S
c) Cl
d) H
Answer:
d) H

Question 118.
A compound which does not give a positive result in the Lassaigne’s test for nitrogen is
a) Urea
b) Hydroxyl amine
c) Glycine
d) Phenylhydrazine
Answer:
b) Hydroxyl amine

Question 119.
Lassaigne’s test gives a violet colouration with sodium nitroprusside, it indicates presence of
a) N
b) S
c) O
d) Cl
Answer:
b) S

Question 120.
The presence of halogen in an organic compound is detected by
a) Iodoform test
b) Silver nitrate test
c) Beilstein’s test
d) Million’s test
Answer:
c) Beilstein’s test

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 121.
The Beilstein’s test in a rapid test used for . organic compound^ to^ detect
a) Phosphorous
b) Sulphur
c) Halogens
d) Nitrogen
Answer:
c) Halogens

Question 122.
Liebig’s method is used for the estimation of
a) Nitrogen
b) Sulphur
c) Carbon and hydrogen
d) Halogens
Answer:
c) Carbon and hydrogen

Question 123.
In Kjeldahl’s method of estimation of nitrogen, copper sulphate act as
a) Oxidizing agent
b) reducing agent
c) Catalytic agent
d) Hydrolysing agent
Answer:
c) Catalytic agent

Question 124.
Percentage of carbon in an organic compound is determined by
a) Duma’s method
b) Kjeldahl’s method
c) Carius method
d) Liebig’s method
Answer:
d) Liebig’s method

Question 125.
Halogen can be estimated by
a) Duma’s method
b) Carius method
c) Leibig’s method
d) All of these
Answer:
b) Carius method

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 126.
In Garius method halogens are estimated
a) X2
b) BaX2
c) PbX2
d) AgX
Answer:
d) AgX

Question 127.
In Duma’s method nitrogen in organic compound is estimated in the form of
a) N2
b) NO
c) NH3
d) N2O5
Answer:
a) N2

Question 128.
In Kjeldahl’s method to estimate nitrogen, compound is heated with conc.H2SO4 in presence of
a) CaSO4
b) (NH4)2SO4
c) CuSO4
d) P2O5
Answer:
c) CuSO4

Question 129.
In organic compounds, Sulphur is estimated as
a) BaSO4
b) BaCl2
c) Ba3(PO4)2
d) H2SO4
Answer:
a) BaSO4

Question 130.
In the Liebig’s method, if ‘w’ is the mass of compound taken and ‘x’ is the amount of C0„ formed then
a) %C = \(\frac{12 \times x}{16 \times w}\)

b) %C = \(\frac{12}{44} \times \frac{\mathrm{w}}{\mathrm{x}} \times 100\)

c) %C = \(\frac{12}{44} \times \frac{x}{w} \times 100\)

d) %C = \(\frac{12}{44} \times \frac{x}{w}\)
Answer:
c) %C = \(\frac{12}{44} \times \frac{x}{w} \times 100\)

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 131.
In Dumas method for estimating nitrogen in organic compound, the gas finally collected is
a) N2
b) NO
c) NH3
d) H2
Answer:
a) N2

Question 132.
In Dumas method, the gas which is collected in Nitrometer is
a) N2
b) NO
c) NH3
d) H2
Answer:
a) N2

Question 133.
In Kjeldahl’s method, the nitrogen presence is estimated as
a) N2
b) NH3
c) NO2
d) N2O3
Answer:
b) NH3

Question 134.
In Kjeldahl’s method, nitrogen present in the organic compound is first converted into
a) NH3
b) (NH4)2SO4
c) N2
d) NO
Answer:
b) (NH4)2SO4

Question 135.
In Liebig’s method for the estimation of C and H, the combustion tube is passed over
a) CuO pellets
b) Copper turnings
c) Iron fillings
d) Zinc – copper couple
Answer:
a) CuO pellets

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 136.
Which gas is introduced into the combustion tube in Liebig’s method?
a) Pure and dry CO2
b) Pure and dry N2
c) Pure and dry O2
d) Pure and dry He
Answer:
c) Pure and dry O2

Question 137.
Chromatographic techniques of purification can be used for
a) Coloured compounds
b) Liquids
c) Solids
d) All of these
Answer:
d) All of these

Question 138.
Two substances when separated on the basis of partition co – efficient between two liquid phase, then the technique is known as
a) column chromatography
b) Paper chromatography
c) GLC
d) TLC
Answer:
b) Paper chromatography

Question 139.
Ortho and para nitro phenols can be separated by
a) crystallization
b) distillation
c) sublimation
d) solvent extraction
Answer:
b) distillation

Question 140.
In steam distillation, the sum of the vapour pressure of the volatile compound and that of water is
a) Equal to atmospheric pressure
b) Less than atmospheric pressure
c) More than atmospheric pressure
d) Exactly half of the atmospheric pressure
Answer:
a) Equal to atmospheric pressure

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 141.
Organic compound is fused with metallic sodium for testing nitrogen, sulphur, and halogens because
a) To make the solution alkaline
b) To convert into elemental state of nitrogen, sulphur, and halogens
c) To convert covalent compound into ionic compound
d) To decrease fusion temperature
Answer:
c) To convert covalent compound into ionic compound

Question 142.
Sodium extract gives blood red colour when treated with FeCl3, Formation of blood-red colour confirms the presence of
a) Only nitrogen
b) Only sulphur
c) Only halogens
d) Both Nitrogen and Sulphur
Answer:
d) Both Nitrogen and Sulphur

Question 143.
The compound not formed in the positive test for nitrogen with the Lussaigne’s solution of an organic compound is
a) Fe4[Fe(CN)6]3
b) Na3[Fe(CN)6]
c) Fe(CN)3
d) Na3[Fe(CN)5NOS]
a) b, c, d
b) a, b
c) a, b, c
d) a only
Answer:
a) b, c, d

Question 144.
The Lassaigne’s solution when heated with ferrous sulphate and acidified with sulphuric acid gave intense blue colour indicating the presence of nitrogen. The blue colour is due to the formation of
a) Na4[Fe(CN)6]
b) Fe3[Fe(CN)6]2
c) Fe2[Fe(CN)6]
d) Fe4[Fe(CN)6]3
Answer:
d) Fe4[Fe(CN)6]3

Question 145.
Which of the following compounds will answer Lassaigne’s test for nitrogen?
a) NH2NH2
b) NH2OH
c) NaCN
d) NaNO3
Answer:
c) NaCN

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 146.
In Dumas method 0.5 g of an organic compound containing nitrogen gave 112 ml of nitrogen at S.T.P. The percentage of nitrogen in the given compound is
a) 28
b) 38
c) 18
d) 48
Answer:
a) 28

Question 147.
0.73 g of organic compound on oxidation gave 1.32 g of carbon dioxide. The percentage of carbon in the given compound will be
a) 49.32
b) 59.32
c) 29.32
d) 98.64
Answer:
a) 49.32

Question 148.
In an estimation of S by Carius method 0.217 g of the compound gave 0.5825 g of BaSO4. Percentage of S is
a) 36.78 %
b) 35.50 %
c) 36.48 %
d) 35.69 %
Answer:
a) 36.78 %

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

II. Very short question and answers (2 Marks):

Question 1.
What are Acyclic compounds? Give suitable example.
Answer:
These are the compounds in which carbon atoms are linked to form open chain (straight or branched). These compounds may be saturated (all single bonds) or unsaturated (multiple bonds).
Example:
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 79

These acyclic compounds are known as open chain or aliphatic compounds.

Question 2.
What arc Alicyclic Compounds? Give suitable example.
Answer:
These are saturated or unsaturated carbo-cyclic compounds which resemble the corresponding acylic compounds in their properties.
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 80

Question 3.
What are Aromatic heterocyclic Compounds? Give example.
Answer:
These are the heterocyclic compounds which possess aromaticity and resemble, the corresponding aromatic compounds in most of their properties. These are also called non-benzenoid aromatic compounds.
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 81

Question 4.
What is functional group? Give example.
Answer:
An atom or group of atoms within a molecule that shows a characteristics set of physical and chemical properties.
Example:
(i) – NH2 – amines
(ii) = NH – Imines
(iii) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 82

Question 5.
Explain the following terms in IUPAC system of nomenclature. of organic compounds.
(i) Root word
(ii) prefix
(iii) suffix
Answer:
(i) Root word:
Root word denotes the number of carbon atoms in the longest continuous chain in molecules.
(ii) prefix:
Prefix denotes the group(s) attached to the main chain which is placed before the root.
(iii) suffix:
Suffix denotes the funtional group and is paced after the root word.

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 6.
What is Metamerism? With suitable examples.
Answer:
Metamerism:
This type of isomerism is a special kind of structural isomerism arises due to the unequal distribution of carbon atoms on either side of the functional group or different alkyl groups attached to the either side of the same functional group and having same molecular formula. This isomerism is shown by compounds having functional group such as ethers, ketones, esters and secondary amines between two alkyl groups.
C4H10O
CH3 – O – C3H7 – Methyl propyl ether 1 – methoxypropane

C2H5 – O – C2H5 – diethyl ether ethoxyethane

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 83 – methyl iso – propyl ether 2 – methoxypropane.

Question 7.
What is meant by stereochemistry?
Answer:
The isomers which have same bond connectivity but different arrangement of groups or atoms in space are known as stereoisomers. This branch of chemistry dealing with the study of three-dimensional nature (spactial arrangement) of molecules is known as stereo chemistry. The metabolic activities in living organisms, natural synthesis and drug synthesis involve various stereoisomers.

Question 8.
What is enantiomerism?
Answer:
An optically active substance may exist in two or more isomeric forms which have same physical & chemical properties but differ in terms of direction of rotation of plane polarized light, such optical isomers which rotate the plane of polarized light with equal angle but in opposite direction are known as enantiomers and the phenom-enon is known as enantiomerism.
Example: d and l lactic acid.

Question 9.
What are the conditions for an organic compound is said to be optically active?
Answer:
(1) The molecule must contains at least one chiral or Asymmetric carbon atom.
(2) The object molecule should not be super impossable with its own mirror image.
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 84

Question 10.
How will you detect phosphorus present in the given organic compound?
Answer:
Test for phosphorous:
A solid compound is strongly heated with a mixture of Na2CO3 & KNO3. Phosphorous present in the compound is oxidized to sodium phosphate. The residue is extracted with water and boiled with Conc. HNO3.. A solution of ammonium molybdate is added to the above solution. A canary yellow coloration or precipitate shows the presence of phosphorous.

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 11.
Explain the sublimation process for the purification of organic compounds.
Answer:
Few substances like benzoic acid, naphthalene and camphor when heated pass directly from solid to vapor without melting (ie liquid). On cooling the vapours will give back solids. Such phenomenon is called sublimation. It is a useful technique to separate volatile and non-volatile solid. It has limited application because only a few substance will sublime.
Example:
naphthalene, benzoic acid.

Question 12.
What is the need for purifying an organic compound?
Answer:
In order to study the structure, physical properties, chemical properties and biological properties of organic compounds they must be in the pure state.

III. Short question and answers (3 Marks):

Question 1.
What is Homologous Series? Give suitable example.
Answer:
Homologous series:
It is a series of compounds in which the adjacent members differ by a – CH2 unit. Individual members of such series are called homologues, and the phenomenon is called as homology. All the members of such a series of alkane have general formula CnH2n + 2. Few members of this family are
CH4 – Methane
C2H6 – Ethane
C3H8 – Propane
C4H10 – Butane
C5H12 – Pentane

Question 2.
What are Alicycic heterocyclic Compounds? Give example.
Answer:
These heterocyclic compounds resemble the corresponding aliphatic compounds in most of their properties.
Example:
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 85

Question 3.
Write the IUPAC name of the following compounds.
a) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 86

b) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 87

c) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 88

d) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 89
Answer:
a) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 90

b) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 91

c) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 92

d) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 93

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 4.
Write the IUPAC name of the following compounds.
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 94
Answer:
a) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 95

b) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 96

c) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 97

d) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 98

Question 5.
Write the IUPAC name of the following compounds.
a) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 99
b) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 100

c) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 101

d) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 102
Answer:
a) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 103

b) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 104

c) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 105

d) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 106

Question 6.
Write the IUPAC name of the following compounds.
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 112
Answer:
a) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 113

b) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 114

c) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 115

d) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 116

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 7.
Write the IUPAC name of the following compounds.
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 107
Answer:
a) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 108

b) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 109

c) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 110

d) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 111

Question 8.
Classify the following compounds based on the structure.
i) CH ≡ C – CH2 – C ≡ CH

ii) CH3 – CH2 – CH2 – CH2 – CH3

iii) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 117

iv) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 118
Answer:
i) CH ≡ C – CH2 – C ≡ CH is unsaturated open chain compound

ii) CH3 – CH2 – CH2 – CH2 – CH3 is saturated open chain compound

iii) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 117 is aromatic benzenoid compound

iv) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 118 is alicyclic compound

Question 9.
Give two examples for each of the following type of organic compounds.
(i) non-benzenoid aromatic
(ii) aromatic heterocyclic
(iii) alicyclic
(iv) aliphatic open chain
Answer:
(i) non-benzenoid aromatic
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 118

(ii) aromatic heterocyclic
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 119

(iii) alicyclic
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 120

(iv) aliphatic open chain
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 121

Question 10.
Explain the copper oxide test for the detection of carbon and hydrogen present in the given organic compound.
Answer:
Copper oxide test:
The organic substance is mixed intimately with about three times its weight of dry copper oxide by grinding. The mixture is then placed in a hard glass test tube fitted with a bent delivery tube. The other end of which is dipping into lime water in an another test tube. The mixture is heated strongly and the following reaction take place.
C + 2CuO → CO2 + 2CuO
2H + CuO → H2O + Cu

Thus if carbon is present, it is oxidized to CO2 which turns lime water millcy. If hydrogen is also present, it will be oxidized to water which condenses in small droplets on the cooler wall of the test tube and inside the bulb. Water collected in the bulb is separated on anhydrous CuSO4 which turn blue. This confirms the presence of C and H in the compound. If however, H is not present water droplet is not obtained in the bulb.

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 11.
In an estimation of sulphur by carius method. 0.2175 g of the substance gave 0.5825 g of BaSO4 calculate the percentage composition of S in the compound.
Solution:
Weight of organic compound = 0.2175 g
Weight of BaSO4 = 0.5825 g
233 g of BaSO4 contains = \(\left(\frac{32}{233} \times \frac{0.5825}{0.2175}\right)\)
0.5825 g of BaSO4 contains
Percentage of S = \(\left(\frac{32}{233} \times \frac{0.5825}{0.2175} \times 100\right)\)
= 36.78 %

Question 12.
0.16 g of an organic compound was heated in a carius tube and H2SO4 acid formed
was precipitated with BaCl2. The mass of BaSO4 was 0.35 g. Find the percentage of
sulphur [30.04]
Answer:
Weight of organic substance (w) = 0.16 g
Weight of Barium sulphate (x) = 0.35 g
Percentage of sulphur = \(\left(\frac{32}{233} \times \frac{x}{w} \times 100\right)\)

= \(\left(\frac{32}{233} \times \frac{0.35}{0.16} \times 100\right)\)
= 30.04 %

Question 13.
0.284 g of an organic substance gave 0.287 g AgCl in a Carius method for the estimation of halogen. Find the percentage of Cl in the compound.
Solution:
Weight of the organic substance = 0.284 g
Weight of AgCl is = 0.287 g
143.5 g of AgCl contains = 35.5 g of chlorine
0.287 g of AgCl contains = \(\left(\frac{35.5}{143.5} \times \frac{0.287}{0.284}\right)\)

% of chlorine is = \(\left(\frac{35.5}{143.5} \times \frac{0.287}{0.284} \times 100\right)\)
= 24.56 %

Question 14.
0.185 g of an organic compound when treated with Conc. HNO3 and silver nitrate gave 0.320 g of silver bromide. Calculate the % of bromine in the compound.
(Ag = 108, Br = 80)
Answer:
Weight of organic substance (w) = 0.185 g
Weight of silver bromide (x) = 0.320 g
Percentage of bromine = \(\left(\frac{80}{188} \times \frac{x}{w} \times 100\right)\)

= \(\left(\frac{80}{188} \times \frac{0.32}{0.185} \times 100\right)\)
= 73.6%

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 15.
0.40 g of an lodo – substituted organic compound gave 0.235 g of AgI by carlus method. Calculate the percentage of iodine in the compound. (Ag = 108 I = 127).
Solution:
Weight of organic substance (w) = 0.40 g
Weight of silver iodide (x) = 0.235 g
Percentage of iodine = \(\left(\frac{127}{235} \times \frac{x}{w} \times 100\right)\)

= \(\left(\frac{127}{235} \times \frac{0.235}{0.40} \times 100\right)\)
= 31.75%

Question 16.
0.24 g of organic compound containing phosphorous gave 0.66 g of Mg2P2O7 by the usual analysis. Calculate the percentage of phosphorous in the compound.
Solution:
Weight of an organic compound = 0.24 g
Weight of Mg2P2O7 = 0.66 g
222 g of Mg2P2O7 contains = 62 g of P
0.66 g contains = \(\frac{62}{222} \times \frac{0.66}{0.24}\)

Percentage of p = \(\frac{80}{180} \times \frac{0.66}{0.24}\) = 76.80%

Question 17.
0.33 g of an organic compound containing phosphorous gave 0.397 g of Mg2P2O7 by the analysis. Calculate the percentage of P in the compound.
Solution:
Weight of organic substance (w) = 0.33 g
Weight of Mg2P2O7 (x) = 0.397 g
Percentage of phosphorous = \(\frac{62}{222} \times \frac{x}{w} \times 100\)

= \(\frac{62}{222} \times \frac{0.397}{0.33} \times 100\) = 33. 59 %

Question 18.
Explain simple distillation process with suitable example.
Answer:
The process of distillation involves the impure liquid when boiled gives out vapour and the vapour so formed is collected and condensed to give back the pure liquid in the receiver. This method is called simple distillation. Liquids with large difference in boiling point(about 40k) and do not decompose under ordinary pressure can be purified by simple distillation.
Example:
The mixture of C6H5NO2 (b.p 484K) &
C6H6 (354 K) and mixture of diethyl ether (b.p 308K) and ethyl alcohol (b.p 35 K).

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 19.
How will you purify an organic compound by differential extraction process?
Answer:
The process of removing a substance from its aqueous solution by shaking with a suitable organic solvent is termed extraction. When an organic substance present as solution in water can be recovered from the solution by means of a separating funnel. The aqueous solution is taken in a separating funnel with little quantity of ether or chloroform (CHCl3).

The organic solvent immiscible with water will form a separate layer and the contents are shaken gently. The solute being more soluble in the organic solvent is transfered to it. The solvent layer is then separated by opening the tap of the separating funnel, and the substance is recovered.

Question 20.
Explain the steam distillation process for purifying organic compound.
Answer:
This method is applicable for solids and liquids. If the compound to be steam distilled the compound should not decompose at the steam temperature, should have a fairly high vapour pressure at 373 K, it should be insoluble in water; the impurities present should be non-volatile.

The impure liquid along with little water is taken in a round-bottom flask which is connected to a boiler on one side and water condenser on the other side, the flask is kept in a slanting position so that no droplets of the mixture will enter into the condenser on the brisk boiling and bubbling of steam. The mixture in the flask is heated and then a current of steam passed in to it. The vapours of the compound mix up with steam and escape into the condenser.

The condensate obtained is a mixture of water and organic compound which can be separated. This method is used to recover essential oils from plants and flowers, also in the manufacture of aniline and turpentine oil.

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 21.
Write a note on azeotropic distillation with suitable example.
Answer:
These are the mixture of liquids that cannot be separated by fractional distillation. The mixtures that can be purified only by azeotropic distillation are called as azeotropes. These azeotropes are constant boiling mixtures, which distil as a single compound at a fixed temperature. Ethanol and water are the most common examples of azeotropic mixture in the ratio of 95.87 : 4.13.

In this method apart from azeotropic mixture a third component like C6H6, CCl4, ether, glycerol, glycol which act as a dehydrating agent depress the partial pressure of one component so that the boiling point of that component is raised sufficiently and thus other component will distill over.

Dehydrating agents like C6H6, CCl4 have low boiling points and reduce the partial vapour pressure of alcohol more than that of water whereas glycerol & glycol, etc. have high boiling point and reduce the partial vapour pressure of water more than that of alcohol.

Question 22.
What is Ring chain isomerism? Give an example.
Answer:
In this type of isomerism, compounds having same molecular formula but differ terms of bonding of carbon atom to form open-chain and cyclic structures.
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 123

Question 23.
Briefly explain geometrical isomerism in oximes with suitable example.
Ans:
Restricted rotation around C = N (oximes) gives rise to geometrical isomerism in oximes. Here ‘syn’ and ‘anti’ are used instead of cis and trans respectively. In the syn isomer the H atom of a doubly bonded carbon and -OH group of doubly bonded nitrogen lie on the same side of the double bond, while in the anti isomer, they lie on the opposite side of the double bond.
For eg:
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 124

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

IV. Long Question and answers (5 Marks):

Question 1.
Write the structure of the following compounds.
a) Cinnamic acid
b) Lactic acid
c) Phthalic acid
d) Tartaric acid
d) Benzoic acid
Answer:
a) Cinnamic acid
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 125

b) Lactic acid
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 126

c) Phthalic acid
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 127

d) Tartaric acid
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 128

d) Benzoic acid
C6H5COOH

Question 2.
Explain the methods for the representation of structure of organic compounds with suitable example.
Answer:
The structure of an organic compound can he represented using any one of the below mentioned methods.

  1. Lewis structure or dot structure,
  2. Dash structure or line bond structure,
  3. Condensed structure
  4. Bond line structure

We know how to draw the Lewis structure for a molecule. The line bond structure is obtained by representing the two electron covalent bond by a dash or line (-) in a Lewis structure. A single line or dash represents single a covalent bond, double line represents double bond (1 σ bond, 2 π bond) and a triple line represents triple bond (1 σ bond, 2 π bond). Lone pair of electrons on hetero atoms may or may not be shown. This represents the complete structural formula.

This structural formula can be further abbreviated by omitting some or all of the dashes representing covalent bonds and by indicating the number of identical groups attached to an atom by a subscript. The resulting expression of the compound is called a condensed structural formula.

For further simplification, organic chemists use another way of representing the structures in which only lines are used. In this type of representation of organic compounds, carbon and hydrogen atoms are not shown and the lines representing carbon-carbon bonds are shown in a zigzag fashion. The only atoms specifically written are oxygen, chlorine, nitrogen etc. These representations can be easily understood by the following illustration.

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 129

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 3.
Explain the molecular model method for the representation of structure of organic compounds.
Answer:
Molecular models:
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 130

Molecular models are physical devices that are used for a better visualisation and perception of three dimensional shapes of organic molecules. These are made of wood, plastic or metal and are commercially available.
(i) Frame work model
(ii) Ball and stick model &
(iii) space filling model.

In the frame work model only the bonds connecting the atoms themselves are shown. This model emphasizes the pattern of bonds of a molecule while ignoring the size of the atom.

In the ball and stick model, both the atoms and the bonds are shown. Ball represent atoms and the stick a bond. Compounds containing C = C can be best represented by using springs in place of sticks and this model is termed as ball and spring model.

The space filling model emphasizes the relative size of each atom based on its Vander Waals radius.

Question 4.
Explain briefly
(i) Fisher projection
(ii) sawhorse projection
(iii) Newman projection formula with neat example.
Answer:
(i) Fisher projection formula:
This is a method of representing three dimensional structures in two dimension. In this method, the chiral atom(s) lies in the plane of paper. The horizontal substituents are pointing towards the observer and the vertical substituents are away from the observer. Fisher projection formula for tartaric acid is given below.
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 131

Sawhorse projection formula:
Here the bond between two carbon atoms is drawn diagonally and slightly elongated. The lower left hand carbon is considered lying towards the front and the upper right hand carbon towards the back. The Fischer projection inadequately portrays the spatial relationship between ligands attached to adjacent atoms. The sawhorse projection attempts to clarify the relative location of the groups.
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 132

Newman projection formula:
In this method the molecules are viewed from the front along the carbon-carbon bond axis. The two carbon atom forming a bond is represented by two circles. One behind the other so that only the front carbon is seen. The front carbon atom is shown by a point whereas the carbon lying further from the eye is represented by the origin of the circle. Therefore, the C-H bonds of the front carbon are depicted from the circle while C-H bonds of the back carbon are drawn from the circumference of the circle with an angle of 120° to each other.
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 133

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 5.
What is Tautomerism? Explain different types of Tautomerism with suitable example.
Answer:
Tautomerism:
It is a special type of functional isomerism in which a single compound exists in two readily inter con-vertible structures that differ markedly in the relative position of atleast one atomic nucleus, generally hydrogen. The two dif¬ferent structures are known as tautomers. There are several types of tautomerism and the two important types are dyad and triad systems.

Dyad system:
In this system hydrogen atom oscillates between two directly linked polyvalent atoms.
Example:

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 134

In this example hydrogen atom oscillates between carbon & nitrogen atom.

Triad system:
In this system hydrogen atom oscillates between three polyvalent atoms. It involves 1, 3 migration of hydrogen atom from one polyvalent atom to other within the molecule. The most important type of triad system is keto-enol tautomerism and the two groups of tautomers are ketoform and enol-form. The polyvalent atoms involved are one oxygen and two carbon atoms. Enolisation is a process in which keto-form is converted to enol form. Both tautomeric forms are not equally stable. The less stable form is known as labile form.
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 135

Question 6.
How is sulphur detected by Lassaigne test?
Answer:
a) To a portion of the lassaigne’s extract, add freshly prepared sodium nitro prusside solution. A deep violet colouration is obtained. This test is also used to detect S2- in inorganic salt analysis.
Na2S + Na2 [Fe(CN5) NO] → Na4 [Fe (CN5) NOS]
sodium nitro prusside

b) Acidify another portion of lassaigne’s extract with acetic acid and add lead acetate solution. A black precipitate is obtained.
(CH3COO)2Pb + Na2S → PbS↓ + 2CH3COONa
(black ppt)

c) Oxidation test:
The organic substances are fused with a mixture of KNO3 and Na2CO3. The sulphur, if present is oxidized to sulphate.
Na2CO3 + S + 3O + Na2SO4 + CO2
The fused mass is extracted with water, acidified with HCl and then BaCl2 solution is added to it. A white precipitate indicates the pressure of sulphur.
BaCl2 + Na2SO4 -> BaSO4 + 2NaCl

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 7.
How is halogen detected by using Lassaigne test?
Answer:
To another portion of the Lassaigne’s filtrate add dil HNO3 warm gently and add AgNO3
Solution:
a) Appearance of curdy white precipitate soluble in ammonia solution indicates the presence of chlorine.
b) Appearance of pale yellow precipitate sparingly soluble in ammonia solution indicates the presence of bromine.
c) Appearance of a yellow precipitate insoluble in ammonia solution indicates the presence of iodine.

Na + Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 136 NaX ( where X = Cl, Br, I)
from organic compound
NaX + AgNO3 → AgX + NaNO3

If N or S is present in the compound along with the halogen, we might obtain NaCN and Na2S in the solution, which interfere with the detection of the halogen in the AgNO3 test Therefore we boil the lassaignes extract with HNO3 which decomposes NaCN and Na2S as

NaCN + HNO3 Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 137 NaNO3 + HCN T
Na2S + 2HNO3 Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 137 2NaNO3 + H2S
NaCN + AgNO3 Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 138 AgCN + NaNO3
white ppt confusing with AgCl
Na2S + AgNO3 → Ag2S ↓ + NaNO3
black ppt

Question 8.
How will you estimate carbon and hydrogen present in the given organic compound?
Answer:
Both carbon and hydrogen are estimated by the same method. A known weight of the organic substance is burnt in excess of oxygen and the carbon and hydrogen present in it are oxidized to carbon dioxide and water, respectively.
CXHY + O2 + XCO2 + \(\frac{\mathrm{y}}{2}\)H2O

The weight of carbon dioxide and water thus formed are determined and the amount of carbon and hydrogen in the original substance is calculated. The apparatus employed for the purpose consists of three units (1) oxygen supply (2) combustion tube (3) absorption apparatus.

Oxygen supply:
To remove the moisture from oxygen it is allowed to bubble through sulphuric acid contained in a drech gel bottle and then passed through a U-tube charged with soda lime to remove CO2. The oxygen gas free from moisture and carbon dioxide enters the combustion tube.

Combustion tube:
A hard glass tube open at both end is used for the combustion of the organic substance. It is filled with
(i) a role of oxidized copper gauze to prevent the backward diffusion of the product of combustion
(ii) a porcelain boat containing a known weight of the organic substance
(iii) coarse copper oxide packed in about 2/3 of the entire length of the tube, and kept in position by loose asbestos plugs on either side and
(iv) a roll of oxidized copper gauze placed towards the end of the combustion tube to prevent any vapours of the organic substance having the tube unoxidized. The combustion tube is enclosed in a furnace, and heated by a gas burner.

Absorption Apparatus:
The products of combustion containing moisture and carbon dioxide are then passed through the absorption apparatus which consists of
(i) a weighed U-tube packed with pumice soaked in Conc. H2SO4 to absorb water
(ii) a set of bulbs containing strong solution of KOH to absorb CO2 and finally
(iii) a guard tube filled with anhydrous CaCl2 to prevent the entry of moisture from atmosphere.

Procedure:
To start with, before loading it with the boat, the combustion tube is detached from the absorption unit. The tube is heated strongly to dry its content and CO2 present in it is removed by passing a current of pure, dry oxygen through it. It is then cooled slightly and connected to the absorption apparatus. The other end of the combustion tube is open for a while and the boat containing weighed organic substance is introduced.

The tube is again heated strongly till the substance in the boat is burnt away. This takes about 2 hours. Finally, a strong current of oxygen is passed through the combustion tube to sweep away any traces of carbon dioxide or moisture which may have been left in it. The U-tube and the potash bulbs are then detached and the increase in weight of each of them is determined.

Calculation:
Weight of the organic substance taken = w g
Increase in weight of H2O = x g
Increase in weight of CO2 = y g
18 g of H2O contain 2 g of hydrogen
∴ x g of H2O contain = \(\left(\frac{2}{18} \times \frac{x}{w}\right)\)
% of hydrogen = \(\left(\frac{2}{18} \times \frac{x}{w} \times 100\right)\) %

44 g of CO2 contains = 12 g of carbon
∴ y g of CO2 contain = \(\left(\frac{12}{44} \times \frac{y}{w}\right)\) g of carbon
Percentage of carbon = \(\left(\frac{12}{44} \times \frac{y}{w} \times 100\right)\) %

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 9.
Explain the Carius method for the estimation of sulphur in an organic compound.
Answer:
Estimation of sulphur Carius method:
A known mass of the organic substance is heated strongly with fuming HNO3, C & H get oxidized to CO2 & H2O while sulphur is oxidized to sulphuric acid as per the following reaction.

C Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 139 CO2
2H Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 139 H2O
S → SO2 Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 140 H2SO4

The resulting solution is treated with excess of BaCl2 solution H2SO4 present in the solution in quantitatively converted into BaSO4, from the mass of BaSO4, the mass of sulphur and hence the percentage of sulphur in the compound can be calculated.

Procedure:
A known mass of the organic compound is taken in clean carius tube and added a few mL of fuming HNO3. The tube is the sealed. It is then placed in an iron tube and heated for about 5 hours. The tube is allowed to cool to temperature and a small hole is made to allow gases produced inside to escape.

The carius tube is broken and the content collected in a beaker. Excess of BaCl2 is added to the beaker H2SO4 acid formed as a result of the reaction is converted to BaSO4. The precipitate of BaSO4 is filtered, washed, dried and weighed. From the mass of BaSO4, percentage of S is found.

Mass of the organic compound = w g
Mass of the BaSO4 formed = x g
233 g of BaSO4 contains = 32 g of sulphur

∴ x g of BaSO4 contain = \(\left(\frac{32}{233} \times \frac{x}{w}\right)\) g of S

Percentage of sulphur = (\(\left(\frac{32}{233} \times \frac{x}{w}\right)\) × 100) %

Question 10.
0.26 g of an organic compound gave 0.039 g of water and 0.245 g of carbon dioxide on combustion. Calculate the percentage of C & H.
Answer:
Weight of organic compound = 0.26 g
Weight of water = 0.039 g
Weight of CO2 = 0.245 g
Percentage of hydrogen
18 g of water contain 2 g of hydrogen
0.039 of water contain = \(\left(\frac{2}{18} \times \frac{0.039}{0.26}\right)\) of hydrogen

% of hydrogen = \(\left(\frac{0.039}{0.26} \times \frac{2}{18} \times 100\right)\) = 1.66%
Percentage of Carbon
44 g of CO2 contain 12 g of C
0.245 g of CO2 contains \(\left(\frac{12}{44} \times \frac{0.245}{0.26}\right)\) g of C

% of Carbon = \(\left(\frac{12}{44} \times \frac{0.245}{0.26} \times 100\right)\) = 25.69 %

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 11.
0.2346 of an organic compound yielded C, H & 0 0.2754 g of H2O and 0.4488 g CO2. Calculate the % composition.
Solution:
Weight of organic substance w = 0.2346 g
Weight of water (x) = 0.2754 g
Weight of CO2 (y), = 0.4488 g
Percentage of carbon = \(\left(\frac{12}{44} \times \frac{y}{w} \times 100\right)\)

= \(\left(\frac{12}{44} \times \frac{0.4488}{0.2346} \times 100\right)\) = 52.17 %

Percentage of hydrogen = \(\left(\frac{2}{18} \times \frac{x}{w} \times 100\right)\)
= \(\left(\frac{2}{18} \times \frac{0.2754}{0.2346} \times 100\right)\) = 13. 04%
Percentage of oxygen= [100 – (52.17 + 13.04)]
= 100 – 65.21 = 34.79 %

Question 12.
Explain the estimation of phosphours by Carius method.
Answer:
Carius method:
A known mass of the organic compound (w) containing phosphorous is heated with fuming HNO in a sealed tube where C is converted into CO2 and H to H2O. phosphorous present in organic compound is oxidized to phosphoric acid which is precipitated, as ammonium phosphomolybdate by heating with Cone. HNO3 and then adding ammonium molybdate.

H3PO4 + 12(NH4)2MoO4 + 21 HNO3 Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 141 (NH4)3PO4 + 21 NH4NO3 + 12HNO3

The precipitate of ammonium phosphomolybdate thus formed is filtered washed, dried, and weighed. In an alternative method, the phosphoric acid is precipitated as magnesium-ammonium phosphate by adding magnesia mixture (a mixture containing MgCl2, NH4Cl, and ammonia) This ppt is washed, dried, and ignited to get magnesium pyrophosphate which is washed, dried a weighed. The following are the reaction that takes place.

By knowing the mass of the organic compound and the, mass of ammonium phosphomolybdate or magnesium pyrophosphate formed, the percentage of P is calculated.

Mass of organic compound = w g
Weight of ammonium phosphomolybdate = x g
Weight of magnesium pyrophosphate = y g
Mole mass of (NH4)3PO4. 12MoO3 is = 1877 g
[3 × (14 + 4) + 31 + 4(16)]+ 12(96 + 3 × 16)
Molar mass of Mg3P2O7 is 222 g
(2 × 24) + (31 × 2) + (7 × 16)
1877 g of (NH4)3PO4. 12 MoO3 contains 31 g of P
X g of (NH4)3PO4. 12 MoO3 in w g of organic
compound contains \(\frac{31}{1877} \times \frac{x}{w}\) of phosphorous
Percentage of phosphorous = \(\frac{31}{1877} \times \frac{x}{w}\) (or)
227 of Mg2P2O7 contains 62 g of P
y g of Mg2P2O2 in w g of Organic compound contains = \(\frac{62}{227} \times \frac{y}{w}\) of P

Percentage of phosphorous = \(\frac{62}{227} \times \frac{y}{w}\)%

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 13.
Explain the estimation of nitrogen by Dumas method.
Answer:
1. Dumas method:
This method is based upon the fact that nitrogenous compound when heated with cupric oxide in an atmosphere of CO2 yields free nitrogen. Thus
Cx Hy Nz + (2x + y/2) CuO → x CO2 + y/2 H2O Z/2 N2 +(2X + y/2)Cu.

Traces of oxide of nitrogen, which may be formed in some cases, are reduced to elemental nitrogen by passing overheated copper spiral. The apparatus used in Dumas method consists of CO2 generator, combustion tube, Schiffs nitrometer.

CO2 generator:
CO2 needed in this process is prepared by heating magnetite or sodium bicarbonate contained in a hard glass tube or by the action of dil. HCl on marble in a kipps apparatus. The gas is passed through the combustion tube after being dried by bubbling through Cone. H2SO4 contained in a Drechel bottle.

Combustion Tube:
The combustion tube is heated in a furnace is charged with
a) A roll of oxidized copper gauze to prevent the back diffusion of the products of combustion and to heat the organic substance mixed with CuO by radiation
b) a weighed amount of the organic substance mixed with an excess of CuO,
c) a layer, of course, CuO packed in about 2/3 of the entire length of the tube and kept in position by loose asbestos plug on either side; this oxidizes the organic vapors passing through it, and
d) a reduced copper spiral which reduces any oxides of nitrogen formed during combustion to nitrogen.

Schiff’s nitro meter:
The nitrogen gas obtained by the decomposition of the substance in the combustion tube is mixed with considerable excess of CO2 It is estimated by passing nitrometer when CO2 is absorbed by KOH and the nitrogen gets collected in the upper part of graduated tube.

Procedure:
To start with the tap of nitro meter is left open. CO2 is passed through the combustion tube to expels the air in it. When the gas bubbles rising through, the potash solution fails to reach the top of it and is completely absorbed it shows that only CO2 is coming and that all air has been expelled from the combustion tube. The nitrometer is then filled with KOH solution by lowering the reservoir and the tap is closed. The combustion tube is now heated in the furnace and the temperature rises gradually.

The nitrogen set free from the compound collects in the nitro meter. When the combustion is complete a strong current of CO2 is sent through, the apparatus in order to sweep the last trace of nitrogen from it. The volume of the gas gets collected is noted after adjusting the reservoir so that the solution in it and the graduated tube is the same. The atmospheric pressure and the temperature are also recorded.

Calculation:
Weight of the substance taken = w g
Volume of nitrogen = V1 L
Room Temperature = T1 K
Atmospheric Pressure = P mm of Hg
Aqueous tension at room temperature = P’ mm of Hg
Pressure of diy nitrogen = (P – P’) = P’ mm of Hg.
Let P0V0 and T0 be the pressure, volume and temperature respectively of dry nitrogen at STP,
Then,

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 142

Calculation of percentage of nitrogen, 22.4 L of N2 at STP weight 28 g of N2
∴ V0 L of N2 at S.T.P weigh = \(\frac{28}{22.4}\) × V0
Wg Organic compound contain = \(\left(\frac{28}{22.4} \times \frac{\mathrm{V}_{\mathrm{o}}}{\mathrm{W}}\right)\) of nitrogen

∴ percentage of nitrogen = \(\left(\frac{28}{22.4} \times \frac{\mathrm{V}_{\mathrm{o}}}{\mathrm{W}}\right)\) × 100

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 14.
0.1688 g when analyzed by the Dumas method yield 31.7 ml of moist nitrogen measured at 14°C and 758 mm mercury pressure. Determine the % of N in the substance (Aqueous tension at 14°C = 12 mm)
Solution:
Weight of Organic compound = 0.168 g
Volume of moist nitrogen (V1) = 31.7 mL
= 31.7 × 10-3 L
Temperature (T1) = 14°C = 14 + 273 = 287 K
Pressure of Moist nitrogen (P) = 758 mm Hg
Aqueous tension at 14°C =12 mm of Hg
∴ Pressure of dry nitrogen = (P – P1)
= 758 – 12 = 746 mm of Hg
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 143

Question 15.
Explain the estimation of nitrogen by Kjeldahl’s method.
Answer:
This method is carried much more easily than the Dumas method. It is used largely in the analysis of foods and fertilizers. Kjeldahl’s method is based on the fact that when an organic compound containing nitrogen is heated with Conc. H2SO4, the nitrogen in it is quantitatively converted to ammonium sulphate. The resultant liquid is then treated with excess of alkali and then liberated ammonia gas absorbed in excess of standard acid. The amount of ammonia (and hence nitrogen) is determined by finding the amount of acid neutralized by back titration with same standard alkali.

Procedure:
A weighed quantity of the substance (0.3 to 0.5 g) is placed in a special long – necked Kjeldahl flask made of pyrex glass. About 25 mL of Cone. H2SO4 together with a little K2SO4 and CuSO 4 (catalyst) are added to it the flask is loosely stoppered by a glass bulb and heated gently in an inclined position. The heating is continued till the brown color of the liquid first produced, disappears leaving the contents clear as before. At this point all the nitrogen in the substance is converted to (NH4)2SO4.

The Kjeldahl flask is then cooled and its contents are diluted with same distilled water and then carefully transferred into a 1 lit round bottom flask. An excess NaOH solution is poured down the side of the flask and it is fitted with a Kjeldahl trap and a water condenser. The lower end of the condenser dips in a measured volume of excess the H2SO4 solution. The liquid in the round bottom flask is then heated and the liberated ammonia is distilled into sulphuric acid. The Kjeldahl trap serves to retain any alkali splashed up on vigorous boiling.

When no more ammonia passes over (test the distillate with red litmus) the receiver is removed. The excess of acid is then determined by titration with alkali, using phenolphthalein as the indicator.

Calculation:
Weight of the substance = w g
Volume of H2SO4 required for the complete neutralization of evolved NH3 = V ml.
Strength of H2S04 used to neutralize NH3 = N
Let the Volume and the strength of NH3 formed are V1 and N1 respectively.
We know that V1N1 = VN
The amount of nitrogen present in the = \(\frac{14 \times \mathrm{NV}}{1 \times 1000 \times \mathrm{w}}\)

Percentage of Nitrogen = \(\left(\frac{14 \times \mathrm{NV}}{1000 \times \mathrm{w}}\right)\) × 100 %

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 16.
0.6 g of an organic compound was kjeldalised and NH3 evolved was absorbed into 50 mL of semi – normal solution of H2SO4. The residual acid solution was diluted with distilled water and the volume made up to 150 mL. 20 mL of this diluted solution required 35 mL of \(\frac{\mathbf{N}}{20}\) NaOH solution for complete neutralization. Calculate the % of N in the compound.
Solution:
Weight of Organic compound = 0.6 g
Volume of sulphuric acid taken = 50 mL
Strength of sulphuric acid taken = 0.5 N
20ml of diluted solution of unreacted sulphuric acid was neutralized by 35 mL of 0.05 N Sodium hydroxide
Strength of the diluted sulphuric aicd = \(\frac{35 \times 0.05}{20}\) = 0.0875 N
Volume of the sulphuric acid remaining after reaction with Organic compound = V1 mL
Strength of the diluted H2SO4 = 0.5 N
Volume of the diluted H2SO4 = 150 mL
Strength of the diluted H2SO4 = 0.0875 N
V1 = \(\frac{150 \times 0.0875}{0.5}\) = 26.25 ml
Volume of H2SO4 consumed by ammonia = 50 – 26.25 = 23.75 ml
23.75 ml of 0.5 N H2SO4 = 23.75 ml of 0.5 N NH3
The amount of Nitrogen present in the 0.6 g of organic compound = \(\frac{14 g}{1000 m L \times 1 N}\) × 23.75 × 0.5 N = 0.166 g
Percentage of Nitrogen = \(\frac{0.166}{0.6}\) × 100 = 27.66%

Question 17.
Explain the various steps involved in the purification of organic compounds by crystallization process.
Answer:
(i) Selection of solvent:
Most of the organic substances being covalent do not dissolve in polar solvents like water, hence selection of solvent (suitable) becomes necessary. Hence the powdered organic substance is taken in a test tube and the solvent is added little by little with constant stirring and heating, till the amount added is just sufficient to dissolve the solute (ie) organic compound. If the solid dissolves upon heating and throws out maximum crystals on cooling, then the solvent is suitable. This process is repeated with other solvents like benzene, ether, acetone, and alcohol till the most suitably one is sorted out.

(ii) Preparation of solution:
The quantity to be purified is taken in a conical flash fitted with a reflux condenser. The solvent selected in first stage is also taken along with solute and the quantity of the solvent just enough to dissolve the whole solid on boiling. Small amount of are animal char-coal can be added before boiling to decolorize any colored substance. The heating may be done over a wire gauze or water bath depending upon the nature of liquid (ie) whether the solvent is low boiling or high boiling.

(iii) Filtration of hot solution:
The solution so obtained is filtered through a fluted filter paper placed in a hot water funnel.

(iv) Crystallization:
The hot filtrate is then allowed to cool. Most of the impurities are removed on the filter paper, the pure solid substance separate as crystal. When copious amount of crystal has been obtained, then the crystallization is complete. If the rate of crystallization is slow, it is induced either by scratching the walls of the beaker with a glass rod or by adding a few crystals of the pure compounds to the solution.

(v) Isolation and drying of crystals:
The crystals are separated from the mother liquor by filtration. Filtration is done under reduced pressure using a Bucher funnel. When the whole of the mother liquor has been drained into the filtration flask, the crystals are washed with small quantities of the pure cold solvent and then dried.

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 18.
How will you purify an organic compound by Thin layer chromatography?
Answer:
This method is another type of adsorption chromatography with this method it is possible to separate even minute quantities of mixtures. A sheet of a glass is coated with a thin layer of adsorbent (cellulose, silica gel or alumina). This sheet of glass is called chromoplate or thin layer chromatography plate. After drying the plate, a drop of the mixture is placed just above one edge and the plate is then placed in a closed jar containing eluent (solvent). The eluent is drawn up to the adsorbent layer by capillary action.

The components of the mixture move up along with the eluent to different distances depending upon their degree of adsorption of each component of the mixture. It is expressed in terms of its retardation factors (ie) Rf value
Rf = \(\frac{\text { Distance moved by the substance from baseline }(\mathrm{x})}{\text { Distance moved by the solvent from baseline }(\mathrm{y})}\)

The spots of colored compounds are visible on TLC plate due to their original color. The colorless compounds are viewed under UV light or in another method using iodine crystals or by using an appropriate reagent.