Category: Class 11

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Chemistry Guide Pdf Chapter 9 Solutions Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions

11th Chemistry Guide Solutions Text Book Back Questions and Answers

Textbook Evaluation:

I. Choose the best answer:

Question 1.
The molality of a solution containing 1.8 g of glucose dissolved in 250 g of water is
a) 0.2 M
b) 0.01 M
c) 0.02 M
d) 0.04 M
Answer:
d) 0.04 M

Question 2.
Which of the following concentration terms is / are independent of temperature
a) molality
b) molarity
c) mole fraction
d) a and b
Answer:
d) a and b

Question 3.
Stomach acid, a dilute solution of HCl can be neutralized by reaction with aluminium hydroxide Al(OH)₃ + 3HCl (aq) -> AlCl₃ + 3H₂O. How many milliliters of 0.1 M Al(OH)₃ solution is needed to neutralize 21 ml of 0.1 M HCl?
a) 14 mL
b) 7 mL
c) 21 mL
d) none of these
Answer:
b) 7 mL

Question 4.
The partial pressure of nitrogen in air is 0.76 atm and its Henry’s law constant is 7.6 × 10⁴ atm at 300 K. What is the mole fraction of nitrogen gas in the solution obtained when air is bubbled through water at 300 K ?
a) 1 × 10^-4
b) 1 × 10⁴
c) 2 × 10^-5
d) 1 × 10^-5
Answer:
d) 1 × 10^-5

Question 5.
The Henry’s law constant for the solubility of Nitrogen gas in water at 350 K is 8 × 10⁴ atm. The mole fraction of nitrogen in air is 0.5. The number of moles of Nitrogen from air dissolved in 10 moles of water at 350 K and 4 atm pressure is
a) 4 × 10^-4
b) 4 × 10⁴
c) 2 × 10^-2
d) 2.5 × 10^-4
Answer:
d) 2.5 × 10^-4

Question 6.
Which one of the following is incorrect for an ideal solution?
a) ∆H_mix = 0
b) ∆U_mix =0
c) ∆P = P_observed – P_{calculated by Raoults law} = 0
d) ∆G_mix = 0
Answer:
d) ∆G_mix = 0

Question 7.
Which one of the following gases has the lowest value of Henry’s law constant?
a) N₂
b) He
c) CO₂
d) H₂
Answer:
c) CO₂

Question 8.
P₁ and P₂ are the vapour pressures of pure liquid components, 1 and 2 respectively of an ideal binary solution If x₁ represents the mole fraction of component 1, the total pressure of the solution formed by 1 and 2 will be
a) P₁ + x₁ (P₂ – P₁)
b) P₂ – x₁ (P₂ + P₁)
c) P₁ – x₂(P₁ – P₂)
d) P₁ + x₂(P₁ – P₂)
Answer:
c) P₁ – x₂(P₁ – P₂)

Question 9.
Osomotic pressure (π) of a solution is given by the relation
a) π = nRT
b) πV = nRT
c) πRT = n
d) none of these
Answer:
b) πV = nRT

Question 10.
Which one of the following binary liquid mixtures exhibits positive deviation from Raoults law?
a) acetone + chloroform
b) water + nitric acid
c) HCl + water
d) ethanol + water
Answer:
d) ethanol + water

Question 11.
The Henry’s law constants for two gases A and B are x and y respectively. The ratio of mole fractions of A to B 0.2. The ratio of mole fraction of B and A dissolved in water will be
a) \(\frac{2 x}{y}\)

b) \(\frac{y}{0.2 x}\)

c) \(\frac{0.2 x}{y}\)

d) \(\frac{5 x}{y}\)
Answer:
d) \(\frac{5 x}{y}\)

Question 12.
At 100°C the vapour pressure of a solution containing 6.5g a solute in 100g water is 732 mm. If K_b = 0.52, the boiling point of this solution will be
a) 102°C
b) 100°C
c) 101°C
d) 100.52°C
Answer:
c) 101°C

Question 13.
According to Raoults law, the relative lowering of vapour pressure for a solution is equal to
a) mole fraction of solvent
b) mole fraction of solute
c) number of moles of solute
d) number of moles of solvent
Answer:
b) mole fraction of solute

Question 14.
At same temperature, which pair of the following solutions are isotonic?
a) 0.2 M BaCl₂ and 0.2 M urea
b) 0.1 M glucose and 0.2 M urea
c) 0.1 M NaCl and 0.1 M K₂SO₄
d) 0.1 M Ba(NO₃)₂ and 0.1 M Na₂SO₄
Answer:
d) 0.1 M Ba(NO₃)₂ and 0.1 M Na₂SO₄

Question 15.
The empirical formula of a non – electrolyte (X) is CH₂O. A solution containing six grams of X exerts the same osmotic pressure as that of 0.025 M glucose solution at the same temperature. The molecular formula of X is
a) C₂H₄O₂
b) C₈H₁₆O₈
c) C₄H₈O₄
d) CH₂O
Answer:
b) C₈H₁₆O₈

Question 16.
The K_H for the solution of oxygen dissolved in water is 4 × 10⁴ atm at a given temperature. If the partial pressure of oxygen in air is 0.4 atm, the mole fraction of oxygen in solution is
a) 4.6 × 10³
b) 1.6 × 10⁴
c) 1 × 10^-5
d) 1 × 10⁵
Answer:
c) 1 × 10^-5

Question 17.
Normality of 1.25 M sulphuric acid is
a) 1.25 N
b) 3.75 N
c) 2.5 N
d) 2.25 N
Answer:
c) 2.5 N

Question 18.
Two liquids X and Y on mixing gives a warm solution. The solution is
a) ideal
b) non-ideal and shows positive deviation from Raoults law
c) ideal and shows negative deviation from Raoults Law
d) non-ideal and shows negative deviation from Raoults Law
Answer:
d) non-ideal and shows negative deviation from Raoults Law

Question 19.
The relative lowering of vapour pressure of a sugar solution in water is 2.5 × 10^-3. The mole fraction of water in that solution is
a) 0.0035
b) 0.35
c) 0.0035/18
d) 0.9965
Answer:
d) 0.9965

Question 20.
The mass of a non – volatile solute (molar mass 80 g mol^-1) which should be dissolved in 92g of toluene to reduce its vapour pressure to 90%
a) 10 g
b) 20 g
c) 9.2 g
d) 8 g
Answer:
d) 8 g

Question 21.
For a solution, the plot of osmotic pressure (π) versus the concentration (c in mol L^-1) gives a straight line with slope 310 R where ‘R’ is the gas constant. The temperature at which osmotic pressure measured is
a) 310 × 0.082 K
b) 310° C
c) 37°C
d) \(\frac{310}{0.082}\) K
Answer:
c) 37°C

Question 22.
200 ml of an aqueous solution of a protein contains 1.26 g of protein. At 300 K, the osmotic pressure of this solution is found to be 2.52 × 10^-3 bar. The molar mass of protein will be (R = 0.083 L bar mol^-1 K^-1}
a) 62.22 kg mol^-1
b) 12444 g mol^-1
c) 300 g mol^-1
d) None of these
Answer:
a) 62.22 kg mol^-1

Question 23.
The Van’t Hoff factor (i) for a dilute aqueous solution of the strong electrolyte barium hydroxide is
a) 0
b) 1
c) 2
d) 3
Answer:
d) 3

Question 24.
Which is the molality of a 10% w/w aqueous sodium hydroxide solution?
a) 2.778
b) 2.5
c) 10
d) 0.4
Answer:
b) 2.5

Question 25.
The correct equation for the degree of an associating solute, ‘n’ molecules of which undergoes association in solution, is
a) α = \(\frac{\mathrm{n}(\mathrm{i}-1)}{\mathrm{n}-1}\)

b) α² = \(\frac{n(1-i)}{(n-1)}\)

c) α = \(\frac{\mathrm{n}(\mathrm{i}-1)}{1-\mathrm{n}}\)

d) α = \(\frac{\mathrm{n}(1-\mathrm{i})}{\mathrm{n}(1-\mathrm{i})}\)
Answer:
c) α = \(\frac{\mathrm{n}(\mathrm{i}-1)}{1-\mathrm{n}}\)

Question 26.
Which of the following aqueous solutions has the highest boiling point?
a) 0.1 M KNO₃
b) 0.1 M Na₃PO₄
c) 0.1 M BaCl₂
d) 0.1 M K₂SO₄
Answer:
b) 0.1 M Na₃PO₄

Question 27.
The freezing point depression constant for water is 1.86° K Kg mol^-1. If 5 g Na₂SO₄ is dissolved in 45 g water, the depression in freezing point is 3.64°C. The Vant Hoff factor for Na₂SO₄ is
a) 2.57
b) 2.63
c) 3.64
d) 5.50
Answer:
a) 2.57

Question 28.
Equimolal aqueous solutions of NaCl and KCl are prepared,. If the freezing point of NaCl is -2°C, the freezing point of KCl solution is expected to be
a) -2°C
b) -4°C
c) -1°C
d) 0°C
Answer:
a) -2°C

Question 29.
Phenol dimerises in benzene having van’t Hoff factor 0.54. What is the degree of association?
a) 0.46
b) 92
c) 46
d) 0.92
Answer:
d) 0.92

Question 30.
Assertion:
An ideal solution obeys Raoults Law.
Reason:
In an ideal solution, solvent – solvent as well as solute – solute interactions are similar to solute-solvent interactions.
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

II. Write brief answer to the following questions:

Question 31.
Define:
(i) Molality
(ii) Normality
Answer:
(i)Molality :
Molality (m) is defined as the number of moles of the solute dissolved in one kilogram (Kg) of the solvent. The units of molality are moles per kilogram, i.e., mole kg^-1. The molality is preferred over molarity if volume of the solution is either expanding or contracting with temperature.
molality (m) = \(\frac{\text { Number of mole of solute }}{\text { mess of solvent in } \mathrm{kg}}\)

ii) Normality:
Normality (N) of a solution is defined as the number of gram equivalents of the solute present in one liter of the solution. Normality is used in acid-based redox titrations.
Normality (N) = \(\frac{\text { Number of gram equivalents of solute }}{\text { Volume of solution in litre }}\)

Question 32.
a) What is a vapour pressure of liquid?
Answer:
“The pressure exerted by the vapors above the liquid surface which is in equilibrium with the liquid at a given temperature is called vapor pressure”.

b) What is a relative lowering of vapour pressure?
Answer:
The relative lowering of vapour pressure is defined as the ratio of lowering of vapour pressure to the vapour pressure
of pure solvent (P₀) RLVP = \(\frac{p^{0}-P}{P^{0}}\)

Question 33.
State and explain Henry’s law.
Answer:
“The partial pressure of the gas in vapor phase (vapour pressure of the solute) is directly proportional to the mole fraction (x) of the gaseous solute in the solution at low concentrations”. This statement is known as Henry’s law.
Henry’s law can be expressed as,
P_solute α x_{solute in solution}
P_solute = K_H x_{solute solution}
Here, P_solute represents the partial pressure of the gas in vapour state which is commonly called as vapour pressure. X_solute in solution represents the mole fraction of solute in the solution. K_H is a empirical constant with the dimensions of pressure.

Question 34.
State Raoult law and obtain expression for lowering of vapour pressure when nonvolatile solute is dissolved In solvent.
Answer:
In an ideal solution, the vapour pressure of the solution is decreased when a non-volatile solute is dissolved in a solvent. The magnitude of decrease in the vapour pressure of the solution depends on the amount of solute added.
Let us consider the solution with the following features.
Mole fraction of the solvent = x_A
Mole fraction of the solute = x_B
Vapour pressure of the pure solvent = P°_A
Vapour pressure of solution = P
As the solute is nonvolatile, the vapour pressure of the solution is only due to the solvent. Therefore, the vapour pressure of the solution (P) will be equal to the vapour pressure of the solvent (P_A) over the solution.
i.e., P = P_A
According to Raoult’s law, the vapour pressure of solvent over the solution is equal to the product or its vapour pressure in a pure state and its mole fraction.
P_A = P°_A x_A or
P = P°_A x_A

Question 35.
What is molal depression constant? Does it depend on nature of the solute?
Answer:
If m = 1 then ∆T_f = K_f
“Then K_f is equal to the depression in freezing point for 1 molal solution”. No, it does not depends on nature of the solute.

Question 36.
What is osmosis?
Answer:
“The phenomenon of the flow of solvent through a semipermeable membrane from pure solvent to the solution is called osmosis”. Osmosis can also be defined as “the excess pressure which must be applied to a solution to prevent the passage of solvent into it through the semipermeable membrane”. Osmotic pressure is the pressure applied to the solution to prevent osmosis.

Question 37.
Define the term ‘isotonic solution’.
Answer:
Two solutions having same osmotic pressure at a given temperature are called isotonic solutions.

Question 38.
You are provided with a solid. ‘A’ and three solutions of A dissolved in water – one saturated, one unsaturated, and one supersaturated. How would you determine which solution is which?
Answer:
(A) Unsaturated solution:
It can dissolve salt an additional to it.
(B) Saturated solution:
Further solubility of salt does not takes place but solubility can takes place on heating.
(c) Supersaturated solution:
Solubility of salt do not takes place on even an further heating.

Question 39.
Explain the effect of pressure on the solubility.
Answer:
Generally, the change in pressure does not have any significant effect in the solubility of solids and liquids as they are not compressible. However, the solubility of gases generally increases with increase of pressure.

Consider a saturated solution of a gaseous solute dissolved in a liquid solvent in a closed container. In such a system, the following equilibrium exists.
Gas                      ⇌     Gas
(in gaseous state) (in solution)

According to Le-Chatelier principle, the increase in pressure will shift the equilibrium in the direction which will reduce the pressure. Therefore, more number of gaseous molecules dissolves in the solvent and the solubility increases.

Question 40.
A sample of 12 M Concentrated hydrochloric acid has a density 1.2 M gL^-1 calculate the molality.
Solution:
Given:
Molarity = 12 M HCl
density of solution = 1.2 g L^-1
In 12 M HCl solution, there are 12 moles of HCl in 1 litre of the solution.
Molality = \(\frac{\text { no of moles of solute }}{\text { mass of solvent (in } \mathrm{kg} \text { ) }}\)
Calculate mass of water(solvent)
mass of 1 litre HCl solution = density × volume
= 1.2 × gmL^-1 × 1000 mL = 1200 g
mass of HCl = no. of moles of HCl × molar mass of HCl
= 12 mol × 36.5 g mol^-1
= 438 g.
mass of water = mass of HCl solution – mass of HCl
mass of water = 1200 – 438 = 762 g
molality(m) = \(\frac{12}{0.762}\) = 15.75 m

Question 41.
A 0.25 M glucose solution, at 370.28 K has approximately the pressure as blood does what is the osmotic pressure of blood?
Solution:
C = 0.25 M
T = 370.28 K
(π)_glucose = CRT
(π) = 0.25 mol L^-1 × 0.082L atm K^-1mol^-1 × 370.28K
= 7.59 atm

Question 42.
Calculate the molality of a solution containing 7.5 g glycine(NH₂-CH₂-COOH) dissolved in 500g of water.
Solution:

Question 43.
Which solution has the lower freezing point? 10 g of methanol (CH₃OH) in 100g g of water (or) 20 g of ethanol (C₂H₅OH) in 200 g of water.
Solution:
∆T_f = K_f i.e
∆T_f α m
m_CH3-OH = \(\frac{\left(\frac{10}{32}\right)}{0.1}\)
= 3.125 m

m_C2H5-OH = \(\frac{\left(\frac{20}{46}\right)}{0.2}\)
= 2.174 m

∴ Depression in freezing point is more in methanol solution and it will have lower freezing point.

Question 44.
How many moles of solute particles are present in one liter of 10^-4 M potassium sulphate?
Answer:
In 10^-4 M K₂SO₄ solution, there are 10^-4 moles of potassium sulphate.
K₂SO₄ molecule contains 3 ions (2K+ and 1 SO₄^2-)
1 mole of K₂SO₄ molecule contains 3 × 6.023 × 10²³ ions
10^-4 mole of K₂SO₄ contains 3 × 6.023 × 10²³ × 10^-4 ions
= 18. 069 × 10¹⁹

Question 45.
Henry’s law constant for solubility of methane in benzene is 4.2 × 10^-5 mm Hg at a particular constant temperature. At this temperature calculate the solubility of methane at
i) 750 mm Hg
ii) 840 mm Hg.
Solution:
(K_H)_benzene = 4.2 × 10^-5 mm
Solubility of methane =?
P = 750 mm Hg P = 840 mm Hg
According to Henrys Law,
P = K_H X_{in solution}
750 mm Hg = 4.2 × 10^-5 mm Hg. X_{in solution}
⇒ X_{in solution} = \(\frac{750}{4.2 \times 10^{-5}}\)

i. e solubility = 178. 5 × 10⁵
similarly at P = 840 mm Hg
solubility = \(\frac{840}{4.2 \times 10^{-5}}\) = 200 × 10^-5

Question 46.
The observed depression in freezing point of water for a particular solution is 0.093°C calculate the concentration of the solution in molality. Given that molal depression constant for water is 1.86 K Kg mol^-1.
Solution:
∆T_f = 0.093°C = 0.093 K, m = ?
K_f = 1.86 K Kg mol^-1
∆T_f = K_f.m
∴ m = \(\frac{\Delta \mathrm{T}_{\mathrm{f}}}{\mathrm{K}_{\mathrm{f}}}=\frac{0.093 \mathrm{~K}}{1.86 \mathrm{~K} \mathrm{Kg} \mathrm{mol}^{-1}}\)
= 0.05 mol Kg^-1 = 0.05 m

Question 47.
The vapour pressure of pure benzene (C₆H₆) at a given temperature is 640 mm Hg. 2.2 g of non – volatile solute is added to 40 g of benzene. The vapour pressure of the solution is 600 mm Hg. Calculate the molar mass of the solute?
Solution:
P°C₆H₆ = 640 mm Hg
W₂ = 2.2 g (non volatile solute)
W₁ = 40 g (benzene)
P_solution = 600 mm Hg
M₂ =?

11th Chemistry Guide Solutions Additional Questions and Answers

I. Choose the best answer:

Question 1.
6.02 × 10²⁰ molecules of urea ate present in 200 ml of its solution, The concentration of urea solution is (N₀ = 6.02 × 10²³ mol^-1)
a) 0.001 M
b) 0.01M
c) 0.02 M
d) 0.10 M
Answer:
c) 0.02 M

Question 2.
Calculate the molarity and normality of a solution containing 0.5 g of NaOH dissolved in 500 ml solution
a) 0.0025 M, 0.025 N
b) 0.025 M, 0.025 N
c) 0.25 M, 0.25 N
d) 0.025M, 0.0025 N
Answer:
b) 0.025 M, 0.025 N

Question 3.
5 ml of N HCl, 20 ml of N/2 H₂SO₄ and 30 ml of N/3 HNO₃ are mixed together and volume made to one liter. The normality of the resulting solution is
a) \(\frac{\mathrm{N}}{40}\)

b) \(\frac{\mathrm{N}}{10}\)

c) \(\frac{\mathrm{N}}{20}\)

d) \(\frac{\mathrm{N}}{5}\)
Answer:
a) \(\frac{\mathrm{N}}{40}\)

Question 4.
At 25°C, the density of 15 M H₂SO₄ is 1.8 g cm^-3. Thus, mass percentage of H₂SO₄ in aqueous solution is
a) 2%
b) 81.6%
c) 18%
d) 1.8%
Answer:
b) 81.6%

Question 5.
Mole fraction of C₃H₅(OH)₃ in a solution of 36 g of water and 46 g of glycerine is :
a) 0.46
b) 0.36
c) 0.20
d) 0.40
Answer:
c) 0.20

Question 6.
The molality of a urea solution in which 0.0100 g of urea, [(NH₂)₂CO] is added to 0.3000 dm³ of water at STP is
a) 0.555 m
b) 5.55 × 10^-4
c) 33.3 m
d) 3.33 × 10^-2 m
Answer:
b) 5.55 × 10^-4

Question 7.
15 grams of methyl alcohol is dissolved in 35 grams of water. What is the mass percentage of methyl alcohol in solution?
a) 30%
b) 50%
c) 70%
d) 75%
Answer:
a) 30%

Question 8.
A 3.5 molal aqueous solution of methyl alcohol (CH₃OH) is supplied. What is the mole fraction of methyl alcohol in the solution?
a) 0.100
b) 0.059
c) 0.086
d) 0.050
Answer:
b) 0.059

Question 9.
In which mode of expression of concentration of a solution remains independent of temperature?
a) Molarity
b) Normality
c) Formality
d) Molality
Answer:
d) Molality

Question 10.
Calculate the molarity of pure water (d = 1 g/L)
a) 555 M
b) 5.55 M
c) 55. 5 M
d) None
Answer:
c) 55. 5 M

Question 11.
Calculate the quantity of sodium carbonate (anhydrous) required to prepare 250 ml solution
a) 2.65 grams
b) 4.95 grams
c) 6.25 grams
d) None of these
Answer:
a) 2.65 grams

Question 12.
Find the molality of H₂SO₄ solution whose specific gravity is 1.98 g ml^-1 and 95 % by volume H₂SO₄
a) 7.412
b) 8.412
c) 9.412
d) 10.412
Answer:
c) 9.412

Question 13.
Calculate molality of 1 liter solution of 93 % H₂SO₄ by volume. The density of solution is 1.84 g ml^-1
a) 9.42
b) 10.42
c) 11.42
d) 12.42
Answer:
b) 10.42

Question 14.
Calculate the molality and mole fraction of the solute in aqueous solution containing 3.0 g of urea per 250 gm of water (Mol.wt. of urea = 60).
a) 0.2 m, 0.00357
b) 0.4 m, 0.00357
c) 0.5 m, 0.00357
d) 0.7 m, 0.00357
Answer:
a) 0.2 m, 0.00357

Question 15.
Calculate normality of the mixture obtained by mixing 100 ml of 0.1 N HCl and 50 ml of 0.25 N NaOH solution.
a) 0.0467 N
b) 0.0367 N
c) 0.0267 N
d) 0.0167 N
Answer:
d) 0.0167 N

Question 16.
300 ml 0.1 M HCl and 200 ml of 0.03 M H₂SO₄ are mixed. Calculate the normality of the resulting mixture
a) 0.084 N
b) 0.84 N
c) 2.04 N
d) 2.84 N
Answer:
a) 0.084 N

Question 17.
What weight of oxalic acid (H₂C₂O₄.2H₂O) is required to prepare, 1000mL of N/10 solution?
a) 9.0 g
b) 12.6 g
c) 6.3 g
d) 4.5 g
Answer:
c) 6.3 g

Question 18.
Which of the following units is useful in relating concentration of solution with its vapour pressure?
a) Mole fraction
b) Parts per million
c) Mass percentage
d) Molality
Answer:
a) Mole fraction

Question 19.
The pressure under which liquid and vapour can co-exist at equilibrium is called the
a) Limiting vapour pressure
b) Real vapour pressure
c) Normal vapour pressure
d) Saturated vapour pressure
Answer:
b) Real vapour pressure

Question 20.
CO(g) is dissolved in H₂O at 30°C and 0.020 atm. Henry’s law constant for this system is 6.20 × 10⁴ atm. Thus, mole fraction of CO(g) is
a) 1.72 × 10^-7
b) 3.22 × 10^-7
c) 0.99
d) 0.01
Answer:
b) 3.22 × 10^-7

Question 21.
H₂S gas is used in qualitative analysis of inorganic cations. Its solubility in water at STP is 0.195 mol kg^-1. Thus, Henry’s law constant ( in atm raolaT1) for H₂S is
a) 2.628 × 10^-4
b) 5.128
c) 0.185
d) 3.826 × 10³
Answer:
b) 5.128

Question 22.
Which of the following is correct for a solution showing positive deviations from Raoult’s law?
a) ∆V = +ve, ∆H = + ve
b) ∆V = -ve, ∆H = – ve
c) ∆V = + ve, ∆H = -ve
d) ∆V = – ve, ∆H = +ve
Answer:
a) ∆V = +ve, ∆H = + ve

Question 23.
If liquids A and B form an ideal solution
a) The entropy of mixing is zero
b) The Gibbs free energy is zero
c) The Gibbs free energy as well as the entropy of mixing are each zero
d) The enthalpy of mixing is zero
Answer:
d) The enthalpy of mixing is zero

Question 24.
Water and ethanol form non – ideal solution with positive deviation from Raoult’s law. This solution, will have vapour pressure
a) equal to vapour pressure of pure water
b) less than vapour pressure of pure water
c) more than vapour pressure of pure water
d) less than vapour pressure of pure ethanol
Answer:
c) more than vapour pressure of pure water

Question 25.
Which of the following is less than zero for ideal solutions?
a) ∆H_mix
b) ∆V
c) ∆G_mix
d) ∆S_mix
Answer:
c) ∆G_mix

Question 26.
Which of the following shows negative deviation from Raoult’s law?
a) CHCl₃ and CH₃COCH₃
b) CHCl₃ and C₂H₅OH
c) C₆H₅CH₃ and C₆H₆
d) C₆H₆ and CCl₄
Answer:
a) CHCl₃ and CH₃COCH₃

Question 27.
Given at 350 K, P°_A = 300 torr and P°_B = 800 torr, the composition of the mixture having a normal boiling point of 350 K is :
a) X_A = 0.08
b) X_A = 0.06
c) X_A = 0.04
d) X_A = 0.02
Answer:
a) X_A = 0.08

Question 28.
In mixture A and B, components show – ve deviation as :
a) ∆V_mix is + ve
b) A – B interaction is weaker than A – A and B – B interaction
c) ∆H_mix is + ve
d) A – B interaction is stronger than A – A and B – B interaction
Answer:
d) A – B interaction is stronger than A – A and B – B interaction

Question 29.
If liquid A and B form ideal solution, then:
a) ∆V_mix is = 0
b) ∆V_mix = 0
c) ∆G_mix =0, ∆S_mix = 0
d) ∆S_mix = 0
Answer:
b) ∆V_mix = 0

Question 30.
Which liquid pair shows a positive deviation from Raoult’s law ?
a) Acetone – chloroform
b) Benzene – methanol
c) Water – nitric acid
d) Water – hydrochloric acid
Answer:
b) Benzene – methanol

Question 31.
For A and B to form an ideal solution which of the following conditions should be satisfied ?
a) ∆H_mixing =0
b) ∆V_mixing =0
c) ∆S_mixing =0
d) All three conditions mentioned above
Answer:
d) All three conditions mentioned above

Question 32.
Two liquids are mixed together to form a mixture which boils at same temperature, and their boiling point is higher than the boiling point of either of them so they shows.
a) no deviation from Raoult’s law
b) positive, deviation from Raoult’s law
c) negative-deviation from Raoult’s law
d) positive or negative deviation from Raoult’s law depending upon the composition
Answer:
c) negative-deviation from Raoult’s law

Question 33.
Molal elevation constant of liquid is:
a) the elevation in b.p. which would be produced by dissolving one mole of solute in 1oo g of solvent
b) the elevation of b.p. which would be produced by dissolving 1 mole solute in 10 g of solvent
c) elevation in b.p. which would be produced by dissolving 1 mole of solute in 1000g of solvent
d) none of the above
Answer:
c) elevation in b.p. which would be produced by dissolving 1 mole of solute in 1000g of solvent

Question 34.
The vapour pressure of pure liquid solvent is 0.50 atm. When a non – volatile solute B is added to the solvent, its vapour pressure drops to 0.30 atm. Thus, mole fraction of the component B is
a) 0.6
b) 0.25
c) 0.45
d) 0.75
Answer:
a) 0.6

Question 35.
The mass of a non – volatile solute (molecular mass = 40) which should be dissolved in 114 g octane to reduce its vapour pressure to 80 % will be
a) 20 g
b) 30 g
c) 10 g
d) 40 g
Answer:
c) 10 g

Question 36.
The vapour pressure of pure liquid solvent A is 0.80 atm. When a non – volatile substance B is added to the solvent, its vapour pressure drops to 0.60 atm. Mole fraction of the component B in the solution is:
a) 0.50
b) 0.25
c) 0.75
d) 0.40
Answer:
b) 0.25

Question 37.
18 g of glucose (C₆H₁₂O₆) is added to 178.2 g of water. The vapour pressure of water for this aqueous solution at 100°C is :
a) 752.40 torr
b) 759.00 torr
c) 7.60 torr
d) 76.00 torr
Answer:
a) 752.40 torr

Question 38.
Calculate the vapour pressure of a solution at 100°C containing 3 g of cane sugar in 33 g of water, (at wt. C = 12, H = 1, O = 16)
a) 760 mm
b) 756.90 mm
c) 758.30 mm
d) None of these
Answer:
b) 756.90 mm

Question 39.
Lowering of vapour pressure due to a solute in 1 molal aqueous solution at 100°C is
a) 13.44 mm Hg
b) 14.12 mm Hg
c) 31.2 mm Hg
d) 35.2 mm Hg
Answer:
a) 13.44 mm Hg

Question 40.
The vapour pressure of a dilute aqueous solution of glucose is 750 mm Hg at 373 K. The mole fraction of the solute is
a) \(\frac{1}{76}\)

b) \(\frac{1}{7.6}\)

c) \(\frac{1}{38}\)

d) \(\frac{1}{10}\)
Answer:
a) \(\frac{1}{76}\)

Question 41.
When 3 g of a nonvolatile solute is dissolved in 50 g of water, the relative lowering of vapour pressure observed is 0.018 Nm^-2. Molecular weight of the substance is
a) 60
b) 30
c) 40
d) 120
Answer:
a) 60

Question 42.
Elevation in boiling point of a molar (1M) glucose solution (d = 1.2 gmL^-1) is
a) 1.34 K_b
b) 0.98 K_b
c) 2.40 K_b
d) K_b
Answer:
b) 0.98 K_b

Question 43.
Given, H₂O (l) ⇌ H₂O (g) at 373 K, ∆H° = 8.31 kcal mol^-1. Thus, boiling point of 0.1 molal sucrose solution is
a) 373. 52 K
b) 373.052 K
c) 373.06 K
d) 374.52 K
Answer:
c) 373.06 K

Question 44.
A solution of 0.450 g of urea (mol. Wt. 60) in 22.5 g of water showed 0.170°C of elevation in boiling point. Calculate the molal elevation constant of water.
a) 0.17°C
b) 0.45°C
c) 0.51°C
d) 0.30°C
Answer:
c) 0.51°C

Question 45.
At higher altitudes, water boils at temperature < 100°C because
a) temperature of higher altitudes is low
b) atmospheric pressure is low
c) the proportion of heavy water increases
d) atmospheric pressure becomes more
Answer:
b) atmospheric pressure is low

Question 46.
Which aqueous solution exhibits highest boiling point?
a) 0.015 M glucose
b) 0.01 M KNO₃
c) 0.015 M urea
d) 0.01 M Na₂SO₄
Answer:
d) 0.01 M Na₂SO₄

Question 47.
A solution of urea in water has boiling point of 100.15°C. Calculate the freezing point of the same solution if K_f and K_b for water are 1.87 K kg mol^-1 and 0.52 K kg mol^-1 respectively
a) – 0.54°C
b) – 0.44°C
c) – 0.64°C
d) – 0.34°C
Answer:
a) – 0.54°C

Question 48.
Which will have largest ∆T_b?
a) 180 g glucose in 1 kg water
b) 342 g sucrose in 1,000 g water
c) 18 g glucose in 100 g water
d) 65 g urea in 1kg water
Answer:
d) 65 g urea in 1kg water

Question 49.
An aqueous solution of glucose boils at 100.01°C. The molal elevation constant for water is 0.5 K mol^-1 kg. The number of molecules of glucose in the solution containing 100 g of water is
a) 6.023 × 10²³
b) 12.046 × 10²²
c) 12.046 × 10²⁰
d) 12.046 × 10²³
Answer:
c) 12.046 × 10²⁰

Question 50.
The latent heat of vaporization of water is 9700 cal/mole and if the b.p. is 100°C, ebullioscopic constant of water is
a) 0.513°C
b) 1.026°C
c.) 10.26°C
d) 1.832°C
Answer:
a) 0.513°C

Question 51.
If for a sucrose solution elevation in boiling point is 0.1 °C then what will be the boiling point of NaCl solution for same molal concentration
a) 0.1°
b) 0.2°C
c) 0.08°C
d) 0.01°C
Answer:
b) 0.2°C

Question 52.
The molal boiling point constant for water is 0.513°C kg mol^-1. When 0.1 mole of sugar is dissolved in 200 ml of water, the solution boils under a pressure of one atmosphere at
a) 100.513°C
b) 100.0513°C
c) 100.256°C
d) 101.025°C
Answer:
c) 100.256°C

Question 53.
The boiling point of 0.1 m K₄[Fe(CN)₆] is expected to be (K_b for water = 0.52 K kg mol’1)
a) 100.52°C
b) 100.10°C
c) 100.26°C
d) 102.6°C
Answer:
c) 100.26°C

Question 54.
The value of K_f for the water is 1.86K Kg mole^-1, calculated from glucose solution. The value of K_f for water calculated for NaCl solution will be :
a) = 1.86
b) < 1.86
c) > 1.86
d) zero
Answer:
a) = 1.86

Question 55.
The amount of urea to be dissolved in 500 cc of water (K_f = 1.86) to produce a depression of 0.186°C in the freezing point is :
a) 9 g
b) 6 g
c) 3 g
d) 0.3 g
Answer:
c) 3 g

Question 56.
Freezing point of an aqueous solution is – 0.186°C. Elevation of boiling point of the same solution is if K_b = 0.512 K molality^-1 and K_f= 1.86 K molality^-1
a) 0.186°C
b) 0.0512°C
c) 0.092°C
d) 0.237°C
Answer:
b) 0.0512°C

Question 57.
What should be the freezing point of aqueous solution containing 17 g of C₂H₅OH in 1000 g of water (K_f for water = 1.86 deg kg mol^-1)?
a) – 0.69°C
b) 0.34°C
c) 0.0°C
d) – 0.34°C
Answer:
a) – 0.69°C

Question 58.
The freezing point of equimolal aqueous solution will be highest for:
a) C₆H₅NH₃Cl
b) Ca(NO₃)₂
c) La(NO₃)₂
d) C₆H₁₂O₆
Answer:
d) C₆H₁₂O₆

Question 59.
Cryoscopic constant of a liquid
a) is the decrease in freezing point when 1 g of solute is dissolved per kg of the solvent
b) is the decrease in the freezing point when 1 mole of solute is dissolved per kg of the solvent
c) is the elevation for 1 molar solution
d) is a factor used for calculation of depression in freezing point
Answer:
b) is the decrease in the freezing point when 1 mole of solute is dissolved per kg of the solvent

Question 60.
Which of the following solution will have highest freezing point?
a) 2 M NaCl solution
b) 1.5 M AlCl₃ solution
c) 1 M Al₂(SO₄)₃ solution
d) 3 M Urea solution
Answer:
d) 3 M Urea solution

Question 61.
0.48 g of a substance is dissolved in 10.6 g of C₆H₆. The freezing point of benzene is lowered by 1.8°C. what will be the mol.wt. of the substance (K_f for benzene = 5)
a) 250.2
b) 90.8
c) 125.79
d) 102.5
Answer:
c) 125.79

Question 62.
Which of the following aqueous molal solution have highest freezing point?
a) Urea
b) Barium chloride
c) Potassium bromide
d) Aluminium sulphate
Answer:
a) Urea

Question 63.
What weight of NaCl is added to one liter of water so that ∆T_f/K_f = 1?
a) 5.85 g
b) 0.585 g
c) 0.0585 g
d) 0.0855 g
Answer:
c) 0.0585 g

Question 64.
A solution of glucose (C₆H₁₂O₆) is isotonic With 4 g of urea (NH₂ – CO – NH₂) per liter of solution. The concentration of glucose is :
a) 4 g/L
b) 8 g/L
c) 12 g/L
d) 14 g/L
Answer:
c) 12 g/L

Question 65.
A 5% solution of cane sugar (molar mass = 342) is isotonic with 1% of a solution of unknown solute. The molar mass of unknown solute in g/mol is
a) 136.2
b) 171.2
c) 68.4
d) 34.2
Answer:
c) 68.4

Question 66.
The weight of urea dissolved in 100 ml solution which produce an osmotic pressure of 20.4 atm, will be
a) 5 g
b) 4 g
c) 3 g
d) 6 g
Answer:
a) 5 g

Question 67.
In the phenomenon of osmosis, the membrane allow passage of _________.
a) Solute only
b) Solvent only
c) Both solute and solvent
d) None of these
Answer:
b) Solvent only

Question 68.
A 5.8% (wt./vol.) NaCl solution will exert an osmotic pressure closest to which one of the following:
a) 5.8% (wt./vol.) sucrose solution
b) 5.8% (wt./vol.) glucose solution
c) 2 molal sucrose solution
d) 1 molal glucose solution
Answer:
c) 2 molal sucrose solution

Question 69.
Osmotic pressure of a sugar solution at 24°C is 2.5 atmospheres. Determine the concentration of the solution in gram mole per liter.
a) 0.0821 moles/liter
b) 1.082 moles/liter
c) 0.1025 moles/liter
d) 0.0827moles/liter
Answer:
c) 0.1025 moles/liter

Question 70.
What is the freezing point of a solution that contains 10.0g of glucose C₆H₁₂O6 in 100 g of H₂O? K_f = 1.86° C/m.
a) – 0.186°C
b) + 0.186°C
c) – 0.10°C
d) – 1.03°C
Answer:
d) – 1.03°C

Question 71.
The order of osmotic pressure of equimolar solutions of BaCl2, NaCl and glucose will be:
a) BaCl₂ > NaCl > glucose
b) NaCl > BaCl₂ > glucose
c) glucose > BaCl₂ > NaCl
d) glucose > NaCl > BaCl₂
Answer:
a) BaCl₂ > NaCl > glucose

Question 72.
The wt. of urea dissolved in 100 ml solution which produce an osmotic pressure of 20.4 atm, will be
a) 5 g
b) 4 g
c) 3 g
d) 6 g
Answer:
a) 5 g

Question 73.
A compound MX₂ has observed and normal molar masses 65.6 and 164 respectively. Calculate the apparent degree of ionization of MX₂:
a) 75%
b) 85%
c) 65%
d) 25%
Answer:
a) 75%

Question 74.
The freezing point of 0.2 molal K₂SO₄ is – 1.1°C. Calculate van’t Hoff facor and percentage degree of dissociation of K₂SO₄. K_f for water is 1.86°
a) 97.5
b) 90.75
c) 105.5
d) 85.75
Answer:
a) 97.5

Question 75.
For 0.1M solution, the colligative property will follow the order
a) NaCl > Na₂SO₄ > Na₃PO₄
b) NaCl > Na₂SO₄ ≈ Na₃PO₄
c) NaCl < Na₂SO₄ < Na₃PO₄
d) NaCl < Na₂SO₄ = Na₃PO₄
Answer:
c) NaCl < Na₂SO₄ < Na₃PO₄

Question 76.
P_H of a 0.1M monobasic acid is found to be 2. Hence its osmotic pressure at a given temp. T K is
a) 0.1 RT
b) 0.11 RT
c) 1.1 RT
d) 0.01 RT
Answer:
b) 0.11 RT

Question 77.
Which has the highest boiling point?
a) 0.1 m Na₂SO₄
b) 0.1 m Al(NO₃)₃
c) 0.1 m MgCl₂
d) 0.1 m C₆H₁₂O₆ (glucose)
Answer:
b) 0.1 m Al(NO₃)₃

Question 78.
Aluminium phosphate is 100% ionized in 0.01 molal aqueous solution. Hence ∆T_b/ K_b is:
a) 0.01
b) 0.015
c) 0.0175
d) 0.02
Answer:
d) 0.02

Question 79.
1.0 molal aqueous solution of an electrolyte X₃Y₂ is 25% ionized. The boiling point of the solution is (K_b for H₂O = 0.52 K kg/mol)
a) 373.5 K
b) 374.04 K
c) 377.12 K
d) 373.25 K
Answer:
b) 374.04 K

Question 80.
The freezing point of 0,05 m solutions of a non – electrolyte in water is
a) -1.86 °C
b) -0.93°C
c)-0.093°C
d) 0.93°C
Answer:
c)-0.093°C

Question 81.
For an ideal solution containing a non – volatile solute, which of the following expression is correctly represented?
a) ∆T_b = K_b × m
b) ∆T_b = K_b × M
c) ∆T_b = K_b × 2m
d) ∆T_b = K_b × 2M
Where m is the molality of the solution and K_b is molal elevation constant.
Answer:
a) ∆T_b = K_b × m

Question 82.
If 5.85 g of NaCl are dissolved in 90 g of water, the mole fraction of NaCl is
a) 0.1
b) 0.2
c) 0.3
d) 0.0196
Answer:
d) 0.0196

Question 83.
What will be the molarity of a solution containing 5g of sodium hydroxide in 250 ml solution?
a) 0.5
b) 1.0
c) 2.0
d) 0.1
Answer:
a) 0.5

Question 84.
If 5.85 g of NaCl (molecular weight 58.5) is dissolved in water and the solution is made up to 0.5 liter, the molarity of the solution will be
a) 0.2
b) 0.4
c) 1.0
d) 0.1
Answer:
a) 0.2

Question 85.
To prepare a solution of concentration of 0.03 g/ml of AgNO₃, what amount of AgNO₃ should be added in 60ml of solution
a) 1.8
b) 0.8
c) 0.18
d) None of these
Answer:
a) 1.8

Question 86.
How many g of dibasic acid (mol.wt. 200) should be present in 100ml of its aqueous solution to give decinormal strength?
a) 1 g
b) 2 g
c) 10 g
d) 20 g
Answer:
a) 1 g

Question 87.
The molarity of a solution of Na₂CO₃ having 10.6 g/500 ml of solution is
a) 0.2 M
b) 2 M
c) 20 M
d) 0.02 M
Answer:
a) 0.2 M

Question 88.
Molecular weight of glucose is 180, A solution of glucose which contains 18 g per liter is
a) 2 molal
b) 1 molal
c) 0.1 molal
d) 18 molal
Answer:
c) 0.1 molal

Question 89.
0.5 M of H₂SO₄ is diluted from lliter to 10 liters, normality of resulting solution is
a) 1 N
b) 0.1 N
c) 10 N
d) 11 N
Answer:
b) 0.1 N

Question 90.
An aqueous solution of glucose is 10% in strength. The volume in which 1 g mole of it is dissolved will be
a) 18 liters
b) 9 liters
c) 0.9 liters
d) 1.8 liters
Answer:
d) 1.8 liters

Question 91.
When 1.80 g glucose dissolved in 90 g of H₂O, the mole fraction of glucose is
a) 0.00399
b) 0.00199
c) 0.0199
d) 0.998
Answer:
b) 0.00199

Question 92.
A 5 molar solution of H₂SO₄ is diluted from 1 liter to 10 liters. What is the normality of the solution?
a) 0.25 N
b) 1 N
c) 2N
d) 7 N
Answer:
b) 1 N

Question 93.
Normality of 2 M sulphuric acid is
a) 2 N
b) 4 N
c) N/2
d) N/4
Answer:
b) 4 N

Question 94.
What is the molarity of H₂SO₄ solution, that has a density 1.84 g/cc at 35°C and Contains solute 98% by weight
a) 4.18 M
b) 8.14 M
c) 18.4 M
d) 18 M
Answer:
c) 18.4 M

Question 95.
Which of the following is a colligative property?
a) Osmotic pressure
b) Boiling point
c) Vapour pressure
d) Freezing point
Answer:
a) Osmotic pressure

Question 96.
The vapour pressure of benzene at a certain temperature is 640 mm of Eg. A non – volatile and non – electrolyte solid weighing 2.175 g is added to 39.08 g of benzene. The vapour pressure of the solution is 600 mm of Hg. What is the molecular weight of solid substance?
a) 49.50
b) 59.6
c) 69.5
d) 79.8
Answer:
c) 69.5

Question 97.
The average osmotic pressure of human blood is 7.8 bar at 37°C. What is the concentration of an aqueous NaCl solution that could be used in the Mood stream?
a) 0.16 mol/L
b) 0.32 mol/L
c) 0.60 mol/L
d) 0.45 mol/L
Answer:
b) 0.32 mol/L

Question 98.
The osmotic pressure in atmospheres of 10% solution of cane sugar at 69°C is
a) 724
b) 824
c) 8.21
d) 7.21
Answer:
c) 8.21

Question 99.
The molal boiling point constant for water is 0.513°C kg mol^-1. When 0.1 mole of sugar is dissolved in 200 ml of water, the solution boils under a pressure of one atmosphere at
a) 100.513°C
b) 100.0513°C
c) 100.256°C
d) 101.025°C
Answer:
c) 100.256°C

Question 100.
The freezing point of a solution prepared from 1.25 g of a non – electrolyte and 20 g of water is 271.9 K. If molar depression constant is 1.8 K mole^-1 then molar mass of the solute will be
a) 105.7
b) 106.7
c) 115.3
d) 93.9
Answer:
a) 105.7

Question 101.
Osmotic pressure of 0.1 M solution of NaCl and Na₂SO₄ will be
a) same
b) osmotic pressure of NaCl solution will be more than Na₂SO₄ solution
c) osmotic pressure of Na₂SO₄ solution will be more than NaCl
d) osmotic pressure of NaSO₄ will be less than that of NaCl solution
Answer:
c) osmotic pressure of Na₂SO₄ solution will be more than NaCl

Question 102.
At 25 °C the highest osmotic pressure is exhibited by 0.1 M solution of
a) CaCl₂
b) KCl
c) Glucose
d) Urea
Answer:
a) CaCl₂

Question 103.
Azeotropic mixture of HCl and water has
a) 84% HCl
b) 22.2% HCl
c) 63 % HCl
d) 20.2 HCl
Answer:
d) 20.2 HCl

Question 104.
The boiling point of water (100°C) becomes 100.25°C, if 3 grams of a nonvolatile solute is dissolved in 200 ml of water. The molecular weight of solute is (K_b for water is 0.6 K Kg mol^-1)
a) 12.2 g mol^-1
b) 15.4 g mol
c) 17.3 g mol^-1
d) 20.4 g mol
Answer:
c) 17.3 g mol^-1

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

Question 1.
Define solution.
Answer:
A solution is homogeneous mixture of two or more substances, consisting of atom, ions or molecules. The constituent of the homogeneous mixture present in a lower amount is called the solute, and the one present in larger amount is called the solvent. For example, when a small amount of NaCl dissolved in water.

Question 2.
What is saturated solution?
Answer:
A saturated solution is one that contains the maximum amount of a solute that can dissolve in a solvent at a specific temperature.
For example, the solubility of NaCl in 100 g of water at 20°C is 36 g but at other temperatures, or in other solvents, is different.

Question 3.
What is unsaturated solution?
Answer:
An unsaturated solution is the one that contains less amount of solute than its capacity to dissolve.

Question 4.
What is Supersaturated solution?
Answer:
A super saturated solution is one that contains more dissolved solute than the saturated solution. It is generally not stable and eventually the dissolved solute will separate as crystals.

Question 5.
What is mass percentage?
Answer:
The mass percentage of a component in a solution is the mass of the component present in 100 g of the solution.
Mass percentage of component = \(\frac{\text { Mass of the component in the solution } \times 100}{\text { Total mass of the solution }}\)

Question 6.
What is parts per million (ppm)?
Answer:
If the amount of solute in solution is very much less, then the concentration is expressed as parts per million (ppm).
Parts per million (ppm) = \(\frac{\text { Mass of the solute }(\mathrm{mg})}{\text { Mass of the solvent }} \times 10^{6}\)

Question 7.
What is Molarity?
Answer:
Molarity (symbol M) is defined as the number of moles of solute present in a liter of solution. The units of molarity are moles per liter (mol L^-1) or moles per cubic decimeter (mol dm^-3)
Molarity(m) = \(\frac{\text { Number of ntoles of solute }}{\text { Volume of solution in liter }}\)

Question 8.
What is Non – ideal solution?
Answer:
The solutions which do not obey Raoult’s law over the entire range of concentration, are called non – ideal solutions. For a non – ideal solution, there is a change in the volume and enthalpy upon mixing, i.e.
∆H_mixing ≠ 0 & ∆V_mixing ≠ 0.

Question 9.
State Raoult’s law.
Answer:
According to Raoult’s law, the vapor pressure of solvent over the solution is equal to the product of its vapor pressure in pure state and its mole fraction.
P_A = P°_A X_A or
P = P°_AX_A

Question 10.
Define boiling point.
Answer:
The boiling point of a liquid is the temperature at which its vapour pressure becomes equal to the atmospheric pressure (1 atm).

Question 11.
Define freezing point.
Answer:
Freezing point is defined as “the temperature at which the solid and the liquid states of the substance have the same vapour pressure”.

Question 12.
What is Osmotic pressure?
Answer:
Osmotic pressure can be defined as “the pressure that must be applied to the solution to stop the influx of the solvent (to stop osmosis) through the semi permeable membrane”.

Question 13.
State Dalton’s law.
Answer:
According to Dalton’s law of partial pressure the total pressure in a closed vessel will be equal to the sum of the partial pressures of the individual components.

Question 14.
What is elevation of boiling point? Give it.
Answer:
The temperature difference between the solution and pure solvent is called elevation of boiling point,
∆T_b = T – T°
Unit of ∆T_b is K Kg mole^-1.

III. Short Question and answers(3 Marks):

Question 1.
What is mole fraction?
Answer:
Mole fraction, x, of solute is defined as the ratio of the number of moles of the components divided by the total number of the moles of all the component present in the solution.
Mole fraction of the solute x_solvent = \(\frac{\text { Moles of the component }}{\text { total number of moles of all the component in solution }}\)

Question 2.
What is volume percentage?
Answer:
It is defined as the volume of the component present in 100mL of the solution. If V_a is a volume of solute and V_b is volume of solvent, then
Vollume percentage of solute = \(\frac{V_{a} \times 100}{V_{a}+V_{b}}\)

Question 3.
What are the advantages of using standard solution?
Answer:

1. The error due to weighing the solute can be minimized by using concentrated stock solution that requires large quantity of solute.
2. We can prepare working standards of different concentrations by diluting the stock solution, which is more efficient since consistency is maintained.
3. Some of the concentrated solutions are more stable and are less likely to support microbial growth than working standards used in the experiments.

Question 4.
Limitations of Henry’s law.
Answer:
Henry’s law is applicable at moderate temperature and pressure only. Only the less soluble gases obeys Henry’s law. The gases reacting with the solvent do not obey Henry’s law.
Example:
Ammonia or HCl reacts with water and hence does not obey this law.
NH₃ + H₂O ⇌ NH₄⁺ + OH^–

Question 5.
What is Van’t Hoff factor?
Answer:
It is defined as the ratio of the actual molar mass to the abnormal (calculated) molar mass of the solute. Here, the abnormal molar mass is the molar mass calculated using the experimentally determined colligative property.
i = \(\frac{\text { Normal (actual) molar mass }}{\text { obseved (abnormal) molar mass }}\)

Question 6.
What is ideal solution?
Answer:
An ideal solution is a solution in which each component i.e. the solute as well as the solvent obeys the Raoult’s law over the entire range of concentration.
For an ideal solution,

1. There is no change in the volume on mixing the two components (solute & solvents) (∆V_mixing =0).
2. There is no exchange of heat when the solute is dissolved in solvent (∆H_mixing =0).
3. escaping tendency of the solute and the solvent present in it should be same as in pure liquids.
   Example:
   benzene & toluene; h-hexane & n-heptane; ethyl bromide & ethyl iodide; chlorobenzene & bromobenzene.

Question 7.
What are colligative properties?
Answer:
“ The properties of the solutions which depend only on the number of solute particles but not on the nature of the solute are called colligative properties”. The following four colligative properties are very important.

1. Relative lowering of vapor pressure (∆P)
2. Elevation of boiling point (∆T_b)
3. Depression of freezing point (∆T_b)
4. Osmotic pressure (π)

Question 8.
What are the significances of Osmotic pressure?
Answer:
Unlike elevation of boiling point (for 1 molal solution the elevation in boiling point is 0.512°C for water) and the depression in freezing point (for 1 molal solution the depression in freezing point is 1.86°C for water), the magnitude of osmotic pressure is large.

The osmotic pressure can be measured at room temperature enables to determine the molecular mass of bio molecules which are unstable at higher temperatures. Even for a very dilute solution the osmotic pressure is large.

Question 9.
0.24 g of a gas dissolves in 1L of water at 1.5 atm pressure. Calculate the amount of dissolved gas when the pressure is raised to 6.0 atm at constant temperature?
Solution:
P_solute = K_H X_{solute in solution}
At pressure 1.5 atm,
P₁ = K_HX₁ …………..(1)
At pressure 6.0 atm,
p₂ = K_HX₂ …………….(2)
Dividing equation (1) by (2)
From equation = \(\frac{P_{1}}{P_{2}}=\frac{X_{1}}{X_{2}}\)

There fore
\(\frac{0.24 \times 6.0}{1.5}\) = 0.96 g/L

Question 10.
Why the carbonated drinks are stored in a pressurized container?
Answer:
The carbonated beverages contain carbon dioxide dissolved in them. To dissolve the carbon dioxide in these drinks, the CO₂ gas is bubbled through them under high pressure. These containers are sealed to maintain the pressure. When we open these containers at atmospheric pressure, the pressure of the CO₂ drops to the atmospheric level and hence bubbles of CO₂ rapidly escape from the solution and show effervescence. The burst of bubbles is even more noticeable, if the soda bottle in warm condition.

Question 11.
An aqueous solution of 2% non volatile solute exerts a pressure of 1.004 bar at the boiling point of the solvent. What is the molar mass of the solute when P°_A is 1.013 bar?
Solution:
\(\frac{\Delta P}{P_{A}^{0}}=\frac{W_{B} \times M_{B}}{M_{B} \times W_{A}}\)

In a 2 % solution weight of the solute is 2 g and solvent is 98g
∆P = P_solution – P°_A = 1.013 – 1.004 bar
= 0.009 bar
M_B = \(\frac{\mathrm{P}_{\mathrm{A}}^{0} \times \mathrm{W}_{\mathrm{B}} \times \mathrm{M}_{\mathrm{A}}}{\Delta \mathrm{P} \times \mathrm{W}_{\mathrm{A}}}\)

M_B = 2 × 18 × 1.013/(98 × 0.009) = 41.3 g mol^-1

Question 12.
Ethylene glycol (C₂H₆O₂) can be at used as an antifreeze in the radiator of a car. Calculate the temperature when ice will begin to separate from a mixture with 20 mass percent of glycol in water used in the car radiator. K_f for water = 1.86K Kg mol^-1 and molar mass of ethylene glycol is 62 g mol^-1.
Solution:
Weight of solute (W₂) = 20 mass percent of solution means 20 g of ethylene glycol
Weight of solvent (water) W₁ = 100 – 20 = 80 g
∆T_f = K_f m
= \(\frac{\mathrm{K}_{\mathrm{f}} \times \mathrm{W}_{2} \times 1000}{\mathrm{M}_{2} \times \mathrm{W}_{1}}=\frac{1.86 \times 20 \times 1000}{62 \times 80}\)
= 7.5 K
The temperature at which the ice will begin to separate is the freezing of water after the addition of solute i.e 7.5 K lower than the normal freezing point of water (273 – 7.5 K) = 265.5 K

Question 13.
At 400 K 1.5 g of an unknown substance is dissolved in solvent and the solution is made to 1.5 L. Its osmotic pressure is found to be 0.3 bar. Calculated the molar mass of the unknown substance.
Solution :

Question 14.
The depression in freezing point is 0.24 K obtained by dissolving 1 g NaCl in 200g water. Calculate van’t – Hoff factor. The molar depression constant in 1.86 K Kg mol^-1
Solution :
Molar mass of solute = \(\frac{1000 \times K_{f} \times \text { mass of } N a C l}{\Delta T_{f} \times \text { mass of solvent }}\)

= \(\frac{1000 \times 1.86 \times 1}{0.24 \times 200}\)
= 38.75 g mol^-1

Theoretical molar mass of NaCl is = 58.5 g mol^-1

i = \(\frac{\text { Theoretical molar mass }}{\text { Experimental molar mass }}\)
= \(\frac{58.5}{38.75}\) = 1.50

IV. Long question and answers(5 Marks):

Question 1.
Explain the factors including the solubility of solute?
Answer:
Factors influencing the solute:
The solubility of a solute generally depends on the nature of the solute and the solvent in which it is dissolved. It also depends on the temperature and pressure of the solution.

Nature of solute and solvent:
Sodium-chloride, an ionic compound, dissolves readily iff a polar solvent such as water, but it does not dissolve in non polar-organic solvents such as benzene or toluene. Many organic compounds dissolve -readily in organic solvents and do not dissolve in water. Different gases dissolve in water to different extents: for example, ammonia is more soluble than oxygen in water.

Effect of temperature:

Solid solute in liquid solvent:
Generally, the solubility of a solid solute in a liquid solvent increases with increase in temperature. When the temperature is increased, the average kinetic energy of the molecules of the solute and the solvent increases. The increase in kinetic energy facilitates the solvent molecules to break the intermolecular attractive forces that keep the solute molecules together and hence the solubility increases.

When a solid is added to a solvent, it begins to dissolve. i.e. the solute leaves from the solid state (dissolution). After some time, some of the dissolved solute returns back to the solid state (recrystallisation). If there is excess of solid present, the rate of both these processes becomes equal at a particular stage. At this stage an equilibrium is established between the solid solute molecules and dissolved solute molecules.

Solute(solid) ⇌ Solute(dissolved)

According to Le-Chatelier principle, if the dissolution process is endothermic, the increase in temperature will shift the equilibrium towards left i.e solubility increases, for an exothermic reaction, the increase in temperature decreases the solubility. The solubilities, of ammonium nitrate, calcium Chloride, ceric sulphate nano-hydrate, and sodium chloride in water at different temperatures are given in the following graph.

Plot of solubility versus temperature for selective compounds

The following conclusions are drawn from the above graph:

1. The solubility of sodium chloride does not vary appreciably as the maximum solubility is achieved at normal temperature. In fact, there is only 10 % increase in solubility between 0° to 100 °C.
2. the dissolution process of ammonium nitrate is endothermic, the solubility increases steeply with increase in temperature.
3. In the case of ceric sulphate, the dissolution is exothermic and the solubility decreases with increase in temperature.
4. Even though the dissolution of calcium chloride is exothermic, the solubility increases moderately with increase in temperature. Here, the entropy factor also plays a significant role in deciding the position of the equilibrium.

Gaseous solute in liquid solvent:
In the case of gaseous solute in liquid solvent, the solubility decreases with increase in temperature. When a gaseous solute dissolves in a liquid solvent, its molecules interact with solvent molecules with weak intermolecular forces. When the temperature increases, the average kinetic energy of the molecules present in the solution also increases.

The increase in kinetic energy breaks the weak intermolecular forces between the gaseous solute and liquid solvent which results in the release of the dissolved gas molecules to the gaseous state. Moreover, the dissolution of most of the gases in liquid solvents is an exothermic process, and in such processes, the increase in temperature decreases the dissolution of gaseous molecules.

Question 2.
Explain vapour pressure of liquid in liquids binary solution?
Answer:
Now, let us consider a binary liquid solution formed by dissolving a liquid solute ‘A’ in a pure solvent B in a closed vessel. Both the components A and B present in the solution would evaporate and an equilibrium will be established between the liquid and vapour phases of the components A and B.

The French chemist Raoult, proposed a quantitative relationship between the partial pressures and the mole fractions of two components A & B, which is known as Raoult’s Law. This law states that “in the case of a solution of volatile liquids, the partial vapour pressure of each component (A & B) of the solution is directly proportional to its mole fraction”.
According to Raoult’s law,
P_A ∝ x_A
P_A = k x_A
when x_A = 1, k = P°_A
where p°A is the vapour pressure of pure component A’ at the same temperature. Therefore,
P_A = P°_A x_A
Similarly, for component ‘B’
P_B =P°_B x_B
x_A and x_B are the mole fraction of the components A and B respectively.

According to Dalton’s law of partial pressure the total pressure in a closed vessel will be equal to the sum of the partial pressures of the individual components.
Hence,
P_total = P_A + P_B
Substituting the values of P_A and P_B from equations in the above equation,
P_total = X_A P°_A + X_BP°_B
We know that X_A + X_B = 1 or X_A = 1 – X_B
Therefore,
P_total = (1 – X_B)P°_A + X_BP°_B
P_total = P°_A + X_B(P°_B – P°_A)
The above equation is of the straight¬line equation form y = mx + c. The plot of P_totalversus x_B will give a straight line with (P°_B – P_A) as slope and P°_A as the y intercept.

Let us consider the liquid solution containing toluene (solute) in benzene (solvent).

The variation of vapour pressure of pure benzene and toluene with its mole fraction is given in the graph.

Solution of benzene in toluene obeying Raoult’s law

The vapour pressures of pure toluene and pure benzene are 22.3 and 74.7 mmHg, respectively. The above graph shows, the partial vapour pressure of the pure components increases linearly with the increase in the mole fraction of the respective components. The total pressure at any composition of the solute and solvent is given by the following straight line (represented as red line) equation.
P_Solution = P°_toluene + X_benzene(P_benzene – P_toluene)

Question 3.
Explain Raoult’s law for the binary solution of Non-volatile solutes in liquids?
Answer:
When a nonvolatile salute js dissolved in a pure solvent, the vapour pressure of the pure solvent will decrease. In such solutions, the vapour pressure of the -Solution will depend only on the solvent ‘molecules as the solute is nonvolatile.

For example, when sodium chloride is added to the water, the vapor pressure of the salt solution is lowered. The vapour pressure of the solution is determined by the number of molecules of the solvent present in the surface at any time and is proportional to the mole fraction of the solvent.

Rate of vapourization reduced by presence of nonvolatile solute.

P_solution ∝ X_A
Where X_A is the mole fraction of the solvent
P_solution = kX_A
When X_A = 1, k = P°_solvent
(P°_solvent is the partial pressure of pure solvent)
P_Solution = P°_Solvent
\(\frac{P_{\text {solution }}}{P_{\text {solvent}}^{a}}\) = X_A

1 – \(\frac{p_{\text {Solution }}}{p_{\text {Solvent }}^{0}}\) = 1 – X_A

\(\frac{p_{\text {Solvent }}^{o}-p_{\text {Solurion }}}{P_{\text {Solvent }}^{0}}\) = X_B
Where X_B is the fraction of the solute
(∴ x_A + x_B = 1, X_B = 1 – X_A)

The above expression gives the relative lowering of vapour pressure. Based on this expression, Raoult’s Law can also be stated as “the relative lowering of vapour pressure of an ideal solution containing the nonvolatile solute is equal to the mole fraction of the solute at a given temperature”.

Question 4.
What is ideal solution? Write special features and characters of ideal solution.
Answer:
An ideal solution is a solution in which each component i.e. the solute as well as the solvent obeys the Raoult’s law over the entire range of concentration. Practically no solution is ideal over the entire range of concentration. However, when the concentration of solute is very low, the dilute solution behaves ideally.

If the two components present in the solution (A and B) are identical in size, structure, and having almost similar intermolecular attractive forces between them (i.e. between A-A, B-B, and B-A) and then the solution tends to behave like an ideal solution.

For an ideal solution:
(i) there is no change in the volume on mixing the two components, (solute & solvents). (∆V_mixing = 0)
(ii) there is no exchange of heat when the solute is dissolved in solvent (∆H_mixing = 0).
(iii) escaping tendency of the solute arid the solvent present in it should be same as in pure liquids.
Example:
benzene & toulene; n-hexane & n-heptane; ethyl bromide & ethyl iodide; chlorobenzene & bromobenzene.

Question 5.
Explain Non – ideal solution with strong positive deviation.
Answer:
The solutions which do not Raoult’s law over the entire range, of concentration, are called non-ideal solutions. For a non-ideal solution, there is a change in the volume and enthalpy upon mixing, i.e.
∆H_mixing ≠ 0 &, ∆V_mixing ≠ 0. The deviation of the non-ideal solutions from the Raoult’s law can either be positive or negative.

Non-ideal solutions – positive deviation from Rauolt’s Law:
The nature of the deviation from the Rauolt’s law can be explained in terms ©f the intermolecular interactions between solute and solvent (B). Consider a case in which the intermolecular attractive forces between A and B are weaker than those between the molecules of A (A-A) and molecules of B (B-B).

The molecules present in such a solution have a greater tendency to escape from the solution when compared to the ideal solution formed by A and B, in which the intermolecular attractive forces (A-A, B-B, A-B) are almost similar. Consequently, the vapour pressure of such non-ideal solution increases and it is greater than the sum of the vapour pressure of A and B as predicted by the Raoult’s law. This type of deviation is called positive deviation.

Here, P_A > p°_A X_A and p_B > P°_B X_B
Hence P_total > p°_A X_A + P_B X_B

Let us understand the positive deviation by considering a solution of ethyl alcohol and water. In this solution the hydrogen bonding interaction between ethanol and water is weaker than those hydrogen bonding interactions amongst themselves (ethyl alcohol-ethyl alcohol and water-water interactions).

This results in the increased evaporation of both components from the aqueous solution of ethanol. Consequently, the vapour pressure of the solution is greater than the vapour pressure predicted by Raoults law. Here, the mixing process is endothermic i.e. ∆H_mixing > 0 and there will be slight increase in volume(∆V_mixing > 0).

Example:
Ethyl alcohol cyclohexane, Benzene & acetone, Carbon tetrachloride & chloroform, Acetone & ethyl alcohol, Ethyl alcohol & water.

Question 6.
Explain Non – ideal solution with strong negative deviation.
Answer:
Let us consider a case where the attractive forces between solute (A) and solvent t (B) are stronger – thtSSar intermolecular attractive forces between the individual components (A – A & B – B). Here, the escaping tendency of A and B will be lower when compared with an ideal solution formed by A and B. Hence, the vapour pressure of such solutions will be lower than the sum of the vapour pressure of A and B. This type of deviation is called negative deviation. For the negative deviation,
P_A < P°_A X_A and P_B < P°_B X_B.

Let us consider a solution of phenol and aniline. Both phenol and aniline form hydrogen bonding interactions amongst themselves. However, when mixed with aniline, the phenol molecule forms hydrogen bonding interactions with aniline, which are stronger than the hydrogen bonds formed amongst themselves.

Formation of new hydrogen bonds considerably reduce the escaping tendency of phenol and aniline from the solution. As a result, the vapour pressure of the solution is less and there is a slight decrease in volume (∆V_mixing < 0) on mixing. During this process evolution of heat takes place i.e. ∆H_mixing < 0 (exothermic)

Example:
Acetone + chloroform, Chloroform + diethyl ether, Acetone + aniline, Chloroform + Benzene.

Question 7.
Explain the factors responsible for deviation from Raoult’s law.
Answer:
Factors responsible for deviation from Raoult’s law:
The deviation of solution from ideal behavior is attributed to the following factors.

i) Solute-solvent interactions:
For an ideal solution, the interaction between the solvent molecules (A-A), the solute molecules (B-B) and between the solvent & solute molecules (A-B) are expected to be similar. If these interactions are dissimilar, then there will be a deviation from ideal behavior.

ii) Dissociation of solute:
When a solute present in a solution dissociates to give its constituent ions, the resultant ions interact strongly with the
solvent and cause deviation from Raoult’s law.
For example, a solution of potassium chloride in water deviates from ideal behavior because the solute dissociates to give K and Cl ion which form strong ion-dipole interaction with water molecules.
KCl(s) + H₂O (l) → K⁺(aq)+ Cl^–(aq)

iii) Association of solute:
Association of solute molecules can also cause deviation from ideal behaviour. For example, in solution, acetic acid exists as a dimer by forming intermolecular hydrogen bonds, and hence deviates from Raoult’s law.

iv) Temperature:
An increase in temperature of the solution increases the average kinetic energy of the molecules present in the solution which causes decrease in the attractive force between them. As result, the solution deviates from ideal behaviour.

v) Pressure:
At high pressure the molecules tend to stay close to each other and therefore there will be an increase in their intermolecular attraction. Thus, a solution deviates from Raoult’s law at high pressure.

vi) Concentration:
If a solution is sufficiently dilute there is no pronounced solvent-solute interaction because the number of solute molecules are very low compared to the solvent. When the concentration is increased by adding solute, the solvent-solute interaction becomes significant. This causes deviation from the Raoult’s law.

Question 8.
How would you determine the molar mass of solute from Tb?
Answer:
The elevation of boiling point is directly proportional to the concentration of the solute particles.
∆T_b ∝ m
m is the concentration of solution expressed in molality.
∆T_b = K_bm
Where
K_b = molal boiling point elevation constant or Ebullioscopic constant.
∆T_b = \(\frac{K_{b} \times W_{B} \times 1000}{M_{B} \times W_{A}}\)

M_b = \(\frac{\mathrm{K}_{\mathrm{b}}}{\Delta \mathrm{Tb}} \times \frac{\mathrm{W}_{\mathrm{B}} \times 1000}{\mathrm{~W}_{\mathrm{A}}}\)

Question 9.
How would you determine the molar mass of solute from T?
Answer:
1f the solution is prepared by dissolving
W_B g of solute in W_B g of solvent, then the molality is,

m = \(\frac{\text { Number of moles of solute } \times 1000}{\text { weight of solvent in grams }}\)

Number of moles of solute = \(\frac{W_{B}}{M_{B}}\)
Where, M_B = molar mass of the solute
Therefore,

m = \(\frac{\mathrm{W}_{\mathrm{B}} \times 1000}{\mathrm{M}_{\mathrm{B}} \times \mathrm{W}_{\mathrm{A}}}\)

∆T_f = \(\frac{K_{f} \times W_{B} \times 1000}{M_{B} \times W_{A}}\)

Molar mass can be calculated using

M_B = \(\frac{\mathrm{Kb} \times \mathrm{W}_{\mathrm{B}} \times 1000}{\Delta \mathrm{T}_{\mathrm{b}} \times \mathrm{W}_{\mathrm{A}}}\)

Question 10.
How would you determine the molar mass of solute form A?
Answer:
According to van’t Hoff equation π = cRT
c = \(\frac{n}{V}\)
Here, n = number of moles of solute dissolved in ‘V’ liter of the solution.
Therefore,
π = \(\frac{n}{V}\)RT
πV = nRT
If the solution is prepared by dissolving W_Bg of nonvolatile solute in W_A g of solvent,
then the number of moles ‘n’ is,
n = \(\frac{\mathrm{W}_{\mathrm{B}}}{\mathrm{M}_{\mathrm{B}}}\)
since, M_B = molar mass of the solute
Substituting the ‘n’ value, we get,
π = \(\frac{\mathrm{W}_{\mathrm{B}}}{\mathrm{V}} \frac{\mathrm{RT}}{\mathrm{M}_{\mathrm{B}}}\)

M_B = \(\frac{W_{B}}{V} \frac{R T}{\pi}\)
From the above equation molar mass of the solute can be calculated.

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Bio Zoology Guide Pdf Chapter 2 Kingdom Animalia Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Bio Zoology Guide Chapter 2 Kingdom Animalia

11th Bio Zoology Guide Kingdom Animalia Text Book Back Questions and Answers

Part I

I. Choose The Best Options

Question 1.
The symmetry exhibited in cnidarians is
a. Radial
b. Bilateral
c. Pentamerous radial
d. Asymmetrical
Answer:
a. Radial

Question 2.
Sea anemone belongs to phylum
a. Protozoa
b. Porifera
c. Coelenterata
d. Echinodermata
Answer:
c. Coelenterata

Question 3.
The excretory cells that are found in platyhelminthes are
a. Protonephridia
b. Flame cells
c. Solenocytes
d. All of these
Answer:
b. Flame cells

Question 4.
In which of the following organisms, self fertilization is seen.
a. Fish
b. Round worm
c. Earthworm
d. Liver fluke
Answer:
d. Liver fluke

Question 5.
Nephridia of Earthworms are performing the same functions as
a. Gills of prawn
b. Flame cells of Planaria
c. The trachea of insects
d. Nematoblasts of Hydra
Answer:
b. Flame cells of Planaria

Question 6.
Which of the following animals has a true coelom?
a. Ascaris
b. Pheretima
c. Sycon
d. Taenia solium
Answer:
b. Pheretima

Question 7.
Metameric segmentation is the main feature of
a. Annelida
b. Echinodermata
c. Arthropoda
d. Coelenterata
Answer:
a. Annelida

Question 8.
In Pheretima locomotion occurs with the help of
a. circular muscles
b. longitudinal muscles and setae
c. circular, longitudinal muscles and setae
d. parapodia
Answer:
c. circular, longitudinal muscles and setae

Question 9.
Which of the following have the highest number of species in nature?
a. Insects
b. Birds
c. Angiosperms
d. Fungi
Answer:
a. Insects

Question 10.
Which of the following is a crustacean?
a. Prawn
b. Snail
c. Sea anemone
d. Hydra
Answer:
a. Prawn

Question 11.
The respiratory pigment in cockroach is
a. Haemoglobin
b. Haemocyanin
c. Heamoerythrin
d. None of the above
Answer:
d. None of the above

Question 12.
Exoskeleton of which phylum consists of chitinous cuticle?
a. Annelida
b. Porifera
c. Arthropoda
d. Echinodermata
Answer:
c. Arthropoda

Question 13.
Lateral line sense organs occur in
a. Salamander
b. Frog
c. Water snake
d. Fish
Answer:
d. Fish

Question 14.
The limbless amphibian is
a. Icthyophis
b. Hyla
c. Rana
d. Salamander
Answer:
a. Icthyophis

Question 15.
Four chambered heart is present in
a. Lizard
b. Snake
c. Scorpion
d. Crocodile
Answer:
d. Crocodile

Question 16.
Which of the following is not correctly paired?
a. Humans – Ureotelic
b. Birds – Uricotelic
c. Lizards – Uricotelic
d. Whale – Ammonotelic
Answer:
d. Whale – Ammonotelic

Question 17.
Which of the following is an egg laying mammal?
a. Delphinus
b. Macropus
c. Ornithorhynchus
d. Equus
Answer:
c. Omithorhynchus

Question 18.
Pneumatic bones are seen in
a. Mammalia
b. Aves
c. Reptilia
d. Sponges
Answer:
b. Aves

Question 19.
Match the following columns and select the correct option.

a. p – (ii), q – (i), r – (iii), s – (iv)
b. p – (iii), q – (iv), r – (ii), s – (i)
c. p – (ii), q – (iv), r – (i), s – (iii)
d. p – (i), q – (ii), r – (iii), s – (iv)
Answer:
b. p – (iii), q – (iv), r – (ii), s – (i)

Question 20.
In which of the following phyla, the adult shows radial symmetry but the larva shows bilateral symmetry?
a. Mollusca
b. Echinodermata
c. Arthropoda
d. Annelida
Answer:
b. Echinodermata

Question 21.
Which of the following is correctly matched?
a. Physalia – Portuguese man of war
b. Pennatula – Sea fan
c. Adamsia – Sea pen
d. Gorgonia-Sea anemone
Answer:
a. Physalia – Portuguese man of war

Question 22.
Why are spongin and spicules important to a sponge?
Answer:
Spongin and spicules provide support and support the soft body parts of the sponges. The spicules give the sponges rigidity and form to the sponges.

Question 23.
What are the four characteristics common to most animals?
Answer:

1. Cellular structure
2. The nature of coelom ;
3. Notochord
4. Segmentation or absence of segmentation.

Question 24.
List the features that all vertebrates show at some point in their development.
Answer:
All vertebrates possess notochord during the embryonic stay. li is repLaced by vertebra) column. All vertebrates possess pained appendages such as fins or lunits. Skin is covered by a protective skeleton comprising of scales. fiathcrs hairs, claws, nails, etc. Respiration is aerobic through gills, skin. buccopharyngeal cavity’ and lungs. All vertebrates have a muscular heart with two, three, or four chambers and kidneys for excretion and osmoregulation.

Question 25.
Compare closed and opened circulatory system
Answer:

Question 26.
Compare Schizocoelom with enterocoelom
Answer:

Question 27.
Identify the structure that the archenteron becomes in a developing animal.
Answer:
The archenteron becomes the cavity of the digestive tract.

Question 28.
Observe the animal below and answer the following questions
a. Identify the animal
b. What type of symmetry does this animal exhibit?
c. Is this animal Cephalized?
d. How many germ layers does this animal have?
e. How many openings does this animal’s digestive system have?
f. Does this animal have neurons?
Answer:
a) Sea anemone
b) Bilateral symmetry
c) It is not a cephalized animal
d) Diploblastic animal
e) One
f) Yes.

Question 29.
Choose the term that does not belong in the following group and explain why it does not belong?
Answer:

• The notochord, cephalization, dorsal nerve cord, and radial symmetry.
• Notochord, cephalization, and dorsal nerve cord are the characteristic features of chordates.
• The radial symmetry is not a characteristic feature of chordate.
• It is the feature of cnidarian and adult echinoderms. Hence it does not belong to the group.

Question 30.
Why flatworms are called acoelomates?
Answer:
The body cavity is formed from mesoderm but in flatworms, there is nobody cavity their body is solid with a perivisceral cavity.

Question 31.
What are flame cells?
Answer:
Flame cells are the specialized excretory cells in flatworms. They help in excretion and osmoregulation.

Question 32.
Concept Mapping – Use the following terms to create a concept map that shows the major characteristic features of the phylum Nematoda: Roundworms, pseudocoelomates, digestive tract, cuticle, parasite, sexual dimorphism
Answer:

Question 33.
In which phyla is the larva trochophore found?
Answer:
Trochopore larva is seen in the Phylum – Annelida.

Question 34.
Which of the chordate characteristics do tunicates retain as adults?
Answer:
Ventral and tabular heart. Respiration is through gill slits.

Question 35.
List the characteristic features that distinguish cartilaginous fishes from living jawless fishes.
Answer:

Question 36.
List three features that characterise bony fishes.
Answer:

1. These fishes have a bony endoskeleton.
2. The skin is covered by ganoid, cycloid or ctenoid scales.
3. Gills are covered by an operculum.
4. They are ammonotelic.
5. They have mesonephric kidneys.
6. External fertilization is seen.

Question 37.
List the functions of air bladder in fishes.
Answer:

• Air bladder may be connected to the gut or not.
• They help in gaseous exchange.
• In ray-finned fishes, they help in buoyancy.

Question 38.
Write the characteristics that contribute to the success of reptiles on land.
Answer:

• The characteristics that contribute to the success of reptiles on land are as follows:
• The presence of dry and cornified skin with epidermal scales or scutes prevents the loss of water.
• The presence of metanephric kidney.
• They are uricotelic (they excrete uric acid to prevent the loss of water).

Question 39.
List the unique features of a bird’s endoskeleton.
Answer:

• The endoskeleton is fully ossified.
• The long bones are hollow with air cavities. So that they can easily fly with lesser weight.

Question 40.
Could the number of eggs or young ones produced by an oviparous and viviparous female be equal? Why?
Answer:
No. The number of eggs or young ones produced by an oviparous and viviparous female cannot be equal. When the oviparous animals lay eggs in the external environment or in the medium, the chance of survival and successful development into the adults are not certain. But in the case of viviparous animals, young ones are nurtured by the adult animals. Hence, oviparous animals lay more eggs if they are fertilized in the medium or in water.

Part II

11th Bio Zoology Guide Kingdom Animalia Additional Important Questions and Answers

I. Choose The Best Options

Question 1.
…………………. is the first group of animals to exhibit tissue-level organization.
a. Cnidaria
b. Porifera
c. Mollusca
d. Echinodermata
Answer:
a. Cnidaria

Question 2.
Name the organs formed from ectoderm.
a. Heart
b. Hair
c. Muscle
d. Intestine
Answer:
b. Hair

Question 3.
The mesoglea seen in between the ectoderm and endoderm is present in …………………. phylum.
a. Platyhelminthes
b. Arthropoda
c. Annelida
d. Coelenterates
Answer:
d. Coelenterates

Question 4.
Inporiferans through ………………. pores water enters into the body and goes out through.
a. Osculum Ostia
b. Ostia Osculum
c. Mouth Ostia
d. Mouth Osculum
Answer:
b. Ostia Osculum

Question 5.
Choose the correct option.
a. Segmentation – Annelida
b. Archenteron – Heart Formation
c. Ostia – Sea anemone
d. Polyp Medusa – Phylum Ptenopora
Answer:
a. Segmentation – Annelida

Question 6.
Find out the correct and wrong statement and find out the correct sequence
I. In the phylum cnidaria on the tentacles nematocysts are present.
II. In evolutionary history the annelid is the first segmented animal.
III. The roundworms are diploblastic bilateral animals.
IV. The arthropods excrete through flame cells.
a. I – False, II – False, III – True, IV – True
b. I – True, II – False, III – False, IV – True
c. I – True, II – True, III – False, IV – Flase
d. I – False, II – True, III – True, IV – False
Answer:
c. I – True, II – True, III – False, IV – False

Question 7.
The organism that shows the regeneration character
a. Planaria
b. Liver fluke
c. Tapeworm
d. Leech
Answer:
a. Planaria

Question 8.
What is the excretory organ of roundworm?
a. Flame cells
b. Rennet glands
c. Green glands
d. Malphigeal tubules
Answer:
b. Rennet glands

Question 9.
The coelom of phylum Arthropoda is
a. Pseudo coelom
b. Eucoelom
c. Schizo coelom
d. Enter coelom
Answer:
c. Schizo coelom

Question 10.
Name the respiratory organ of Mollusca.
a. Ctenidia
b. Gills
c. Book lungs
d. Trachea
Answer:
a. Ctenidia

Question 11.
Name the excretory organ of Cephalo Chordata.
a. Mesonephridia
b. Metanephridia
c. Protonephridia
d. Flame cells
Answer:
d. Flame cells

Question 12.
Name the organism which has both features of chordate and non-chordates?
a. Balanoglossus
b. Ascidia
c. Amphioxces
d. Salpa
Answer:
a. Balanoglossus

Question 13.
…………………….. are called as tunicates
a. Urochordates
b. Cephalo chordates
c. Vertebrata
d. Hemi Chordata
Answer:
a. Urochordates

Question 14.
The eggs of birds are ………………..
a. Megalecithal
b. Mesolecithal
c. Telolocithal
d. Alecithal
Answer:
a. Megalecithal

Question 15.
Find the correct answer by matching.
A. Sponges – I. Mesoglea
B. Open circulation – II. Asymmetrical
C. Diploblastic animal – III. Echinodermata
D. Snails – IV. Coanocytes
a. A – IV, B -I, C – II, D – III
b. A -I, B – II, C – III, D – IV
c. A – IV, B – III, C -I, D – II
d. A – IV, B – II, C – III, D -I
Answer:
C. Diploblastic animal – III. Echinodermata

Question 16.
Find out the correct pair.
a. Planula – Planeria
b. Regeneration – Annelida
c. Trochopore larva – Cnidaria
d. Veliger larva – Mollusca
Answer:
d. Veliger larva – Mollusca

Question 17.
Which one of the following is not correctly paired?
a) Ctenophora – Veliger
b) Annelida – Trochophore
c) Cnidaria – Planula
d) Porifera – Parenchymula
Answer:
a) Ctenophora – Veliger

Question 18.
Find out the wrong statement.
a) In most animals the coelom lies between the body wall and the alimentary canal
b) In acoelomate organisms the free movement of the interval organ is restricted.
c) The cavity formed from the mesoderm is pseudo coelom.
d) If in a body cavity scattered pouches are seen then the coelom is pseudo coelom.
Answer:
c) The cavity formed from the mesoderm is pseudo coelom.

(2 marks)

II. Very Short Questions

Question 1.
What are pinococytes?
Answer:
In sponges, the outer surface is formed of plate-like cells that maintain the size and structure of the sponges are called pinococytes.

Question 2.
What are choanocytes?
Answer:
The inner layer of sponges is formed of flagellated collar cells called coanocytes. They maintain water flow through the sponges thus facilitating respiratory and digestive functions.

Question 3.
Define tissue.
Answer:
Cells that perform similar functions are aggregated to form tissues.

Question 4.
Define organ? Which was the first animal to have organ system?
Answer:

• Different kinds of tissues aggregate to form an organ to perform a specific function.
• In phylum Platyhelminthes, the organ level of organisation is first formed.

Question 5.
Differentiate between a complete digestive system from an incomplete digestive system.
Answer:

Question 6.
What are diploblastic animals?
Answer:

• Animals in which the cells are arranged in two embryonic layers the ectoderm and endoderm are diploblastic animals.
• The ectoderm gives rise to the epidermis.
• The endoderm gives rise to the tissue lining the gut.

Question 7.
What is radial symmetry?
Answer:
When any plane passing through the central axis of the body divides an organism into two identical parts, it is called radial symmetry, e.g. Cnidarian.

Question 8.
What is protostomia?
Answer:
In Eumetazoans, the embryonic blastopore that develops into the mouth are known as protostomia.

Question 9.
What are deutrostomia ?
Answer:
Eumetazoans in which the anus is formed from or near the blastopore and the mouth is formed away from the blastopore are deuterostomes.

Question 10.
List the excretory organs of phylum Arthropoda?
Answer:

• Malphigean tubules
• Green glands
• Coxal glands

Question 11.
Differentiate the respiratory pigment haemoglobin from haemocyanin.
Answer:

Question 12.
What are the advantages of bilaterally symmetrical animals?
Answer:
The bilaterally symmetrical animals can seek food, locate mates, escape from predators and move more efficiently. These animals have dorsal-ventral sides and anterior, posterior ends, right and left sides. They exhibit cephalization with sense organs and brain at the anterior end of the animal.

Question 13.
What is cleidoic egg?
Answer:
If the female organisms lay cleidoic eggs or shelled egg then it is known as cleidoic eggs.

Question 14.
What are the extraembryonic membranes present in reptiles?
Answer:

1. Amnion
2. Allantois
3. Chorion
4. Yolk sac

Question 15.
Name the muscles that help pigeons to fly. Write the kingdom, phyllum, and class for pigeon.
Answer:
a. Pectoralis major b. Pectoralis minor
(i) kingdom – Animalia
(ii) phylum – Chordata
(iii) class – Aves

( 3 marks)

III. Short Questions

Question 1.
What are the structures formed from ectoderm endoderm and mesoderm?
Answer:

1. Edoderm – Skin, Hair, Nerves, Nail, Teeth
2. Mesoderm – Muscles, Bones, Heart
3. Endoderm – Intestine, Lungs, Liver.

Question 2.
Name the parts A, B, and C in the diagram?
Answer:

A) Ectoderm
B) Pseudo coelom
c) Mesodorm

Question 3.
Differentiate parazoa from eumetazoa?
Answer:

Question 4.
Match


a) I – b, II – d, III – a, IV – c
b) I – a, II – b, III – d, IV – c
c) I – b, II – a, III – d, IV – c
Answer:
a) I – b, II – d, III – a, IV – c

Question 5.
Distinguish between hibernation and aestivation.
Answer:
Hibernation:

• The dormancy period for animals during winter is called hibernation.
• It is known as winter sleep.

Aestivation:

• The dormancy period for animals during summer is called Aestivation.
• It is known as summer sleep.

Question 6.
In the given diagram Balanoglossus mark A, B, and C.

Answer:
A) Proboscis
B) Collarette
c) Genital wings

Question 7.
Give any five characteristic features of Urochordata?
Answer:

1. They are exclusively marine.
2. They are mostly sessile some pelagic or free-swimming.
3. The body is covered by a tunic.
4. The coelom is absent.
5. The notochord is present only in the tail region of the larval stage.
6. The circulatory system is open type.

Question 8.
Look at the picture given below and answer questions.
a) What is the name of this fish?
b) What is the name of the larva of this fish?
c) What is the shape of the mouth?

Answer:
a) iprey
b) Ammocete
c) Circular

Question 9.
What are the characteristic features of amphibia?
Answer:

• Amphibians live both in aquatic as well as terrestrial habitats.
• They are poikilothermic.
• They have two pairs of limbs.
• They may have a tail or may not be present.
• Their skin is smooth or rough.
• The heart is three-chambered.
• They excrete urea as an excretory product.
• The kidneys are mesonephric.
• They are oviparous and development is indirect.

(5 Marks)

IV. Essay Questions

Question 1.
Explain various patterns of organisation in animals.
Answer:
Animals exhibit different patterns of organisation:
The cellular level of organisation:

• Cells are loosely arranged without the formation of tissues.
• There is a division of labour among the cells, e.g., sponges.

Tissue level of organisation:

• Cells which perform a similar function are grouped into tissues.
• The tissues perform a common function, e.g., cnidarians.

Organ level of organisation:
Different kinds of tissues aggregate to form an organ to perform a specific function e.g., flatworms and other hyper phyla.

Organ system level of organisation:

• The tissues are organised to form organs and organ systems.
• All the organ system function in a coordinated manner.

Question 2.
What is coelom? Describe its types?
Answer:
Body cavity lined with mesoderm is meant as a coelom. This lies between the body wall and the alimentary canal.

1.Pseudo coelom:
The body cavity is not lined by the mesodermal epithelium and the mesoderm is formed as scattered pouches between the ectoderm and endoderm. (Eg.) Roundworm

2.Eucoelom:
The coelom is a fluid-filled cavity that develops within the mesoderm and is lined by mesodermal epithelium called the peritoneum.

3.Schizocoelomates:
In these animals, the body cavity is formed by splitting mesoderm. (Eg.) Annelids.

4.Entero coelomate:
The body cavity is formed from the mesodermal pouches of the archenteron. (Eg.) Echinodermata

Question 3.
Compare Platyhelminthes with Aschelminthes?
Answer:

Question 4.
Classify animals based on coelom.
Answer:
The cavity between the body wall and the gut wall is called coelom. If the animals do not have a coelom, they are called acoelomates. e.g., flatworms. In some animals, the body cavity is not fully lined by the mesodermal epithelium. The mesoderm is formed as scattered pouches between the ectoderm and endoderm. Such a body cavity is called a pseudocoel. The animals which have pseudocoel e.g. roundworms.

If the coelom develops within the mesoderm and is lined by mesodermal epithelium it is called eucoelom. The animals which have true coelom are called eucoelomates. If the body cavity is formed by splitting mesoderm, the animals are called schizocoelomates e.g., Annelids, arthropods, and mollusks. If the body cavity is formed from the mesodermal ‘ pouches of archenteron, the animals are called enterocoelomate animals, e.g., echinoderms, hemichordates, and chordates.

Question 5.
What are the characteristic features of Hemichordata?

• They possess the characters of invertebrates and chordates.
• This phylum consists of soft worm-like organisms.
• They are triploblastic coelomate animals.
• They are bilaterally symmetrical.
• Their circulatory system is simple and open type.
• They are ciliary feeders.
• Respiration is through paired gill silts opening into the pharynx.
• Excretion is through the glomerulus.
• The nervous system is primitive sexes are separate.
• In its development, there is a free-swimming to maria larva.

Question 6.
What are the general characters of the phylum Vertebrata?
Answer:

1. They possess notochord during the embryonic stage only.
2. The notochord is replaced by a cartilaginous or bony vertebral column in the adult.
3. They possess paired appendages such as fins or limbs.
4. Skin is covered by skeleton consists of scales, feathers, hairs, claws, nails.
5. Respiration is through the gills skin buccopharyngeal cavity and lungs.
6. The heart is with two or three or four chambers.
7. Kidneys are for excretion and osmoregulation.

Question 7.
Write the general characters of the phylum cnidaria.
Answer:

• The cnidaria is aquatic, radially symmetrical, and diploblastic.
• The tentacles have stinging cells called cnidocytes or cnidoblasts or nematocysts.
• They exhibit a tissue level of organisation.
• They have a central gastrovascular cavity called coelenteron.
• Digestion is both extracellular and intracellular.
• Alternation of generation is seen in cnidarians which have polyp and medusa forms.
• Development is indirect with planula larva e.g. Physalia.

Question 8.
What are the general characters of mammals?
Answer:

• The body is covered by hairs.
• They are found in a variety of habitats.
• The presence of the mammary gland is the most unique feature of mammals.
• They have two pairs of limbs.
• The skin consists of sweat glands and sebaceous glands.
• Exo skeleton includes horns spines, scales claws, etc.
• Teeth are thecodont heterodont and diphyodont.
• The heart is four-chambered and posses a left systematic arch.
• Mammals have a large brain when compared to other animals.
• Their kidneys are metanephric and are ureotelic.
• All are homeothermic.

Question 9.
Give three distinct features of all chordates that are seen at some stage of their life cycle? What is the fate of two characters out of three in the matured adults?
Answer:

1. Presence of notochord below the nerve chord and above the alimentary canal.
2. The presence of the nerve cord lies above the notochord and below the dorsal body wall.
3. Presence of pharyngeal gill slits in all chordates at some stage of their life cycle.

Features saw in the matured adult animals

Question 10.
Compare the chordates with non-chordates?
Answer:

Question 11.
What are the parts of ABCD in the model diagram of Chordata?

Answer:
A) Dorsal Nerve Cord
B) Notochord
C) Mouth
D) Pharyngeal gill clefts
E) Muscle segment

Question 12.
Look into the given diagram and answer the question.
a) What is the name of the organism.
b) What is the respiratory organ of this animal.
c) What type of metamorphosis is seen?
d) Whether this organism contain nerve card?
e) What is the outer covering of it’s body?

Answer:
a) Ascidian
b) Gill clefts
c) Retrogressive metamorphosis
d) The larva consists of a nerve cord.
e) Tunic

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Bio Zoology Guide Pdf Chapter 1 The Living World Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Bio Zoology Guide Chapter 1 The Living World

11th Bio Zoology Guide Living World Text Book Back Questions and Answers

Part I

Question 1.
A living organism is differentiated from a non-living structure based on
a. Reproduction
b. Growth
c. Metabolism
d. All the above
Answer:
d. All the above

Question 2.
A group of organisms having similar traits of a rank is
a. Species
b. Taxon
c. Genus
d. Family
Answer:
b. Taxon

Question 3.
Every unit of classification regardless of its rank is
a. Taxon
b. Variety
c. Species
d. Strain
Answer:
a. Taxon

Question 4.
Which of the following is not present in the same rank?
a. Primata
b. Orthoptera
c. Diptera
d. Insecta
Answer:
a. Primata

Question 5.
What taxonomic aid gives comprehensive information about a taxon?
a. Taxonomic Key
b. Herbarium
c. Flora
d. Monograph
Answer:
a. Taxonomic Key

Question 6.
Who coined the term biodiversity?
a. Walter Rosen
b. AG Tansley
c. Aristotle
d. AP de Candole
Answer:
a. Walter Rosen

Question 7.
Cladogram considers the following characters
a. Physiological and Biochemical
b. Evolutionary and Phylogenetic
c. Taxonomic and systematic
d. None of the above
Answer:
b. Evolutionary and Phylogenetic

Question 8.
The molecular taxonomic tool consists of
a. DNA and RNA
b. Mitochondria and Endoplasmic reticulum
c. Cell wall and Membrane proteins
d. All the above
Answer:
a. DNA and RNA

Question 10.
Differentiate between probiotics and pathogenic bacteria
Answer:

Question 11.
Why mule is sterile?
Answer:
Mule gets one set of chromosomes (32) from the male parent, horse and one set of chromosomes (31) from the female parent, donkey. These two sets of chromosomes do not match with each other and cannot produce gametes by meiosis. Hence mule is sterile in nature.

Question 12.
What is the role of Charles Darwin in relation to the concept of species?
Answer:
Charles Darwin’s book on Origin of Species explains the evolutionary connections of species by the process of natural selection.

Question 13.
Why elephants and other wild animals are entering the human living areas?

• For the construction of houses, dams, and factories forests are destroyed. The area surface of forests is also getting reduced.
• As the bull elephant is hunted for their tusks the cow elephant during breeding season enters in to the dwelling area of people.

Question 14.
What is the difference between a Zoo and a wildlife sanctuary?

Question 15.
Can we use recent molecular tools to identify and classify organisms?
Answer:
The recent molecular taxonomical tools can be used to identify and classify the organism. The following molecular techniques and approaches are used in molecular tools.

1. DNA barcoding – Short genetic marker in an organism’s DNA to identify whether it belongs to a particular species.
2. DNA hybridization – Measures the degree of genetic similarity between pools of DNA sequences.
3. DNA fingerprinting – to identify an individual from a sample of DNA by looking at unique patterns in their DNA.
4. Restriction Fragment Length Polymorphism (RFLP) Analysis – the difference in homologous DNA sequences can be detected by the presence of fragments of different lengths after digestion of DNA samples.
5. Polymerase chain reaction (PCR) sequencing- to amplify a specific gene or portion of the gene.

Question 16.
Explain the role of Latin and Greek names in Biology.
Answer:
Aristotle (384 to 322 BC) was the first to classify all animals in his Historia Animalium in Latin. He classified the living organisms into plants and animals. Animals were classified as walking (terrestrial), flying (birds), and swimming (aquatic) based on their locomotion.

He classified the animals with red blood cells as Enaima and those without red blood cells as Anima. Though his method of classification had limitations, his contribution to biology was remarkable. Theophrastus did his research on the classification of plants. He was known as the Father of Botany.

Part II 

11th Bio Zoology Guide The Living World Additional Important Questions and Answers

Question 1.
Biodiversity is
a. A species live in a particular ecosystem.
b. Presence of a large number of species in a particular ecosystem.
c. A species live in a different ecosystem.
d. Many species live in more than one ecosystem.
Answer:
b. Presence of a large number of species in a particular ecosystem.

Question 2.
Aristotle has classified organisms based on the following category of locomotion.
a. Walking & bore dwellers
b. Flying & arboreal
c. Swimmers & aquatic
d. All the above.
Answer:
c. Swimmers & aquatic
d. All the above.

Question 3.
Who is “Father of Botany”?
a. Theophrastus
b. John Ray
c. Carolus Linnaeus
d. Aristotle
Answer:
a. Theophrastus

Question 4.
Whose researchers confirm that species is a fundamental unit of classification.
a. John Ray
b. R.H. Whittaker
c. CarlWoese
d. Cavalier-Smith
Answer:
a. John Ray

Question 5.
Find the correct pair.
1. Domestic Cat – Felis silvestris
2. Wildcat – Felis margarita
3. Wildcat – Felis Domestica
4. Tiger – Panthera tigers
Answer:
4. Tiger – Panthera tigers

Question 6.
Who has developed binomial nomenclature.
a. Carolous Linnaeus
b. Augustin
c. Aristotle
d. Ernst Haeckel
Answer:
a. Carolous Linnaeus

Question 7.
Find the unrelated pair.
a. Carl Woese – Trinominal hypothesis
b. Cavalier-Smith – Seven kingdom system
c. Male Lion and female Tiger results in – Hinny
d. Male Tiger and female Lion results in – Tigon
Answer:
c. Male Lion and female Tiger results in – Hinny

Question 8.
The three domains classification is based on the difference in the gene.
a. 60s rRNA
b. 70s rRNA
c. l6s rRNA
d. m RNA
Answer:
c. l6s rRNA

Question 9.
The prokaryotes that produce methane gas belongs to …………………… kingdom.
a. Monera
b. Eukarya
c. Bacteria
d. Archaea
Answer:
d. Archaea
Question 10.
Find out the correct sequence by matching.
A. Augustin Pyramus de Candole – Father of Botany
B. Aristotle – Father of Modern Taxonomy
C. Carolous Linnaeus – Father of Taxonomy
D. Theophrastus – Introduces Taxonomy
Answer:
D. Theophrastus – Introduces Taxonomy

Question 11.
Crosses between animals – Match.
A. Male Horse + Female Donkey – Tigon
B. Male Donkey + Female Horse – Tiger
C. Male Lion + Female Tiger – Mule
D. Male Tiger + Female Lion – Hinny
a) A-II, B -1, C – IV, D – III
b) A-IV, B -1, C – II, D – III
c) A-I, B-II, C-III, D-IV
d) A-IV, B-I, C-II, D-III
Answer:
A-II, B -I, C – IV, D – III

Question 12.
Three domain classification was proposed by:
a. Cavalier-Smith
b. R.H. Whittaker
c. Carolus Linnaeus
d. Carl Woese
Answer:
d. Carlwoese

Question 13.
Find out the wrong pair
a. Peacock – Pavocristatus
b. Tiger – Pantheratigeris
c. Man -Homosapiens
d. Domestic crow – Salcopopsindica
Answer:
d. Domestic crow – Salcopopsindica

Question 14.
Find the correct match.
1. John ray -a. Five kingdom concept
2. Linnaeus -b. Cladogram
3. Ernest Haeckel -c. Binomial nomenclature
4. R.H. Whittaker – d. Methodus Plantarum
a. 1 -d,2-c,3-b,4-a
b. 1-a,2-b,3-c,4-d
c. 1 – c, 2 – a, 3 – b, 4 – d
d. 1 – d, 2 – c, 3 – a, 4 – b
Answer:
a. 1 -d,2-c,3-b,4-a

Question 15.
Where are the 80s and 70s ribosomes seen in Eukaryotic cells?
a. Cytoplasm – Chloroplast
b. Mitochondrial – Golgi apparatus
c. Chloroplast – Endo plasm reticulum
d. Nucleus – Lysosomes
Answer:
a. Cytoplasm – Chloroplast

( 2 marks)

II. Very Short Questions

Question 1.
Classification of organisms is necessary.
Answer:
Classification of organisms is necessary to recognize, identify them, and differentiate closely related species.

Question 2.
What are the unique characteristic features of living organisms?
Answer:

• Cellular organization
• Nutrition
• Respiration
• Metabolism
• Movement
• Reproduction
• Excretion
• Homeostasis

Question 3.
The mating between different species produces sterile offsprings.
Answer:
The maternal and paternal chromosomes of the offsprings produced by the mating between different species are not identical and hence gametes are not produced by meiotic division.

Question 4.
What are the scientific stages of taxonomy?
Answer:

• Characterization
• Identification
• Nomenclature
• Classification

Question 5.
Why are molecular tools used now to study taxonomy?
Answer:
Molecular tools are accurate and authentic. Hence they are used to study taxonomy.

Question 6.
What is phylogenetic or cladistics classification?
Answer:
It is a classification based on evolution and genetic relationship.

Question 7.
What is the phylogenetic tree?
Answer:
It’s a method of representing evolutionary relationships with the help of a tree diagram known as a cladogram.

Question 8.
What is a cladogram?
Answer:
Arranging organisms on the basis of their similar or derived characters produced a phylogenetic tree or cladogram.

Question 9.
What are the three domains of life indicate?
Answer:
This system emphasizes the separation of prokaryotes into two domains.

Question 10.
How Archaea differ from bacteria?
Answer:
If differs in cell wall composition and in membrane composition and rRNA type.

Question 11.
What is the seven taxonomic hierarchy?
Answer:

1. Kingdom
2. Phyla
3. Class
4. Order
5. Family
6. Genus
7. Species

Question 12.
Define species?
Answer:
It is a group of animals having similar morphological features and is reproductively isolated to produce fertile offspring.

Question 13.
Define ‘Family’?
Answer:
It is a taxonomic category which includes a group of related genera with less similarity as compared to genus and species.

Question 14.
Define order?
Answer:
Order is an assemblage of one or more related families which show few common features. (Eg) Family Candiae and Felidae are placed in the order Carnivora.

Question 15.
Define class.
Answer:
Class includes one or more related orders with some common characters.

Question 16.
Define Phylum.
Answer:
The group of classes with similar distinctive characteristics constitute phylum.

Question 17.
Define animal kingdom.
Answer:
All living animals belonging to various phyla are included in the kingdom.

Question 18.
What are the features that we have to keep in mind in naming them scientifically?
Answer:

• Morphology
• Genetic information
• Habitat
• Feeding pattern
• Adaptations
• Evolutions

Question 19.
On whose guidelines naming animals in a scientific way is done?
Answer:
The naming of the organism is based on the guidelines of the international code of Zoological nomenclature.

Question 20.
What are taxonomical keys?
Answer:
Keys are based on a comparative analysis of the similarities and dissimilarities of organisms. There are separate keys for different taxonomic categories.

Question 21.
What is a museum?
Answer:
Biological museums have a collection of preserved plants and animals for study and ready reference.

Question 22.
Define Zoological parks.
Answer:
These are places where wild animals are kept in protected environments under human care.
It enables us to study their food habits and behaviour.

Question 23.
What are marine parks?
Answer:
Marine organisms are maintained in protected environments.

Question 24.
What are printed taxonomical tools?
Answer:

• Identification cards
• Description
• Field guides
• Manuals

Question 25.
What is the phylogenetic tree?
Answer:
It is the inferred evolutionary relationships upon similarities and differences in their physical or genetic characters.

Question 26.
Define phylogeny.
Answer:
Relationships among various biological species based upon similarities and differences in their physical or genetic characteristics.

Question 27.
What are shared characters?
Answer:
A shared character is one that two lineages have in common.

Question 28.
What are derived characters?
Answer:
Derived character is one that evolved in the lineage leading up to a clade.

Question 29.
Vandaloor Zoological park.
Answer:

• It is situated in the South-Western Part of Chennai.
• It spreads over an area of 1500 acres.
• It is one of the largest zoological parks in India.
• The Zoo houses 2553 species of both flora and fauna.

( 3 marks)

III. Short Questions

Question 1.
Define ecosystem.
Answer:
The ecosystem is defined as a community of living organisms (plants and animals), non-living things (minerals, climate, soil, sunlight, and water), and their interrelationships, e.g. Forest and grassland.

Question 2.
On which criteria the systematic classification is done?
Answer:

• Evolutionary history.
• Environmental adaptations.
• Environmental relationship.
• The interrelationship between species.

Question 3.
Give an account of Aristotle’s classification?
Answer:

•  In his book ‘History of Animals,’ he classifies plants and animals into two categories.
• Based on locomotion walking, flying, swimming,
• He classifies the organisms on the basis of blood.
• He classifies the animals into two as ‘Enaima’ with blood and those without blood as’ Anaima’

Question 4.
Who has developed the five kingdom classification?
Answer:

1. R.H. Whittaker proposed the five-kingdom classification.
2. It is based on cell structure.
3. Mode of nutrition.
4. Mode of reproduction.
5. Phylogenetic relationships.

The kingdoms are

• Monera
• Protista
• Fungi
• Plantae
• Animalia

Question 5.
What are the special features of frogs that are identified in Western Gauts?
Answer:

• This frog has shiny purple skin.
• There is a light blue ring around the eyes.
• It has a pointy big nose.
• It’s Zoological name Nasikabatrachus Bhupathi.

Question 6.
What is trinomial nomenclature
Answer:
Giving three names to the species is meant as trinomial nomenclature.
When members of any species have large variations then a trinomial system is used.
The species is classified into subspecies and this is an extension of binominal nomenclature system which has an addition of subspecies. Followed by Genus name species subspecies name is also added.

Question 7.
What are the limitations of Aristotle’s classification?
Answer:
Many organisms were not fitting into his classification. Frogs have lungs and they are amphibians while their larva, the tadpole is aquatic and respires through gills. It is difficult to classify frogs according to his method. All flying organisms such as birds, bats, flying insects were grouped together. Ostrich, emu and a penguin are flightless birds and hence they cannot be classified by his method.

 (5 marks)

V. Essay Questions

Question 1.
List the defects of Aristotle’s classification.
Answer:

• Aristotle’s classification system had limitations and many organisms were not fitting into his classification.
• The tadpoles of frog are born in water and have gills but when they metamorphosed into adult frogs they have lungs and can live both in water and on land. There is no answer to this question.
• Based on locomotion birds bats and flying insects were grouped either just by observing one single characteristic feature the flying ability.
• On the contrary to the above-said example, the ostrich emu and penguin are all birds but cannot fly. He did not classify them as birds.

Question 2.
What is special about the Domain Archaea?
Answer:

• This domain includes single-celled organisms the prokaryotes.
• They have the ability to grow in extreme conditions like volcano vents hot springs and polar ice caps hence are called extremophiles.
• They are capable of synthesizing their food without sunlight and oxygen by utilizing hydrogen sulphide and other chemicals from the volcanic vents.
• Some of them produced methane.
• Few live in salty environments and called Halophiles.
• Some thrive in acidic environments and are called thormoacidophiles.

Question 3.
What is special about the domain bacteria?
Answer:

1. Bacterias are prokaryotic.
2. They do not have a definite nucleus and do not have histones.
3. They have circular DNA.
4. They do not possess membrane-bound organelles except for 70s ribosomes.
5. Their cell wall contains peptidoglycans.
6. Many are decomposers. Some are photo-synthesizers and few cause diseases.
7. There are beneficial probiotic bacteria. (Eg.) Cyanobacteria produce oxygen.

Question 4.
What is the significance of cladistic classification?
Answer:
Cladistic classification takes into account ancestral characters (traits commons for the entire group) and derived characters (traits whose structure and function differ from the ancestral characters). The accumulation of derived characters resulted in the formation of new subspecies.

Question 5.
What are the basic roles to be followed in naming the animals?
Answer:

• The scientific name should be italicized in printed form and if handwritten it should be underlined separately.
• The generic name’s first alphabet should be in uppercase.
• The specific name should be in lower case.
• The scientific names of any two organisms are not similar.
• The name of the scientist who first publishes the scientific name may be written after the species name along with the year of publication.
• (Eg.) Lion – Felis Leo Linn . 1758 (or) Felis Leo L. 1758

Question 6.
What are the rules to be followed in the nomenclature of organisms?
Answer:
The scientific name should be italicized in printed form and the generic name and specific name should be underlined separately if it is handwritten.

• The first alphabet of the generic name should be of uppercase.
• The specific name (species) should be in lower case letters.
• The name or abbreviated name of the scientist who first published the scientific name may be written after the specific (species) name along with the year of publication, e.g. Felis Leo Linn., 1958.
• If the specific (species) name is framed after any person’s name, the name of the species shall end with i, ii, or ae. e.g. Ground – dwelling lizard Cyrtodactylus varadgirii.

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Bio Botany Guide Pdf Chapter 11 Transport in Plants Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 11 Transport in Plants

11th Bio Botany Guide Transport in Plants Text Book Back Questions and Answers

Part – I

Question 1.
In a fully turgid cell:
(a) DPD = 10 atm; OP = 5 atm; TP = 10 atm
(b) DPD = 0 atm; OP =10 atm; TP = 10 atm
(c) DPD = 0 atm; OP = 5 atm; TP = 10 atm
(d) DPD = 20 atm; OP = 20 atm; TP = 10 atm
Answer:
(b) DPD = 0 atm; OP =10 atm; TP = 10 atm

Question 2.
Which among the following is correct?
i) apoplast is fastest and operate in nonliving part
ii) Transmembrane route includes vacuole
in) Symplast interconnect the nearby cell through plasma desmata
iv) Symplast and the transmembrane route is in the living part of the cell
a) i and ii
b) ii and iii
c) iii and iv
d) i, ii, iii, iv
Answer:
d) i, ii, iii, iv

Question 3.
What type of transpiration is possible in the xerophyte Opuntia?
(a) Stomatal
(b) Lenticular
(c) Cuticular
(d) All the above
Answer:
(b) Lenticular

Question 4.
Stomata of a plant open due to
a) Influx of K⁺
b) Effrilx of K⁺
c) Influx of Cl^–
d) Influx of OH^–
Answer:
a) Influx of K⁺

Question 5.
Munch hypothesis is based on:
(a) translocation of food due to TP gradient and imbibition force
(b) ranslocation of food due to TP
(c) translocation of food due to imbibition force
(d) None of the above
Answer:
(b) ranslocation of food due to TP

Question 6.
If the concentration of salt in the soil is too high and the plants may wilt even if the field is thoroughy irrigated. Explain
Answer:
High salt concentration results in high be osmotic potential of the soil solution, so the plant has to use more energy to absorb water. Under extreme salinity conditions, plants may be unable to absorb water and will wilt even if the surrounding soil is thoroughly irrigated. This is also referred to as the osmotic or water deficit effect of salinity.

Question 7.
How phosphorylase enzyme open the stomata in starch sugar interconversion theory?
Answer:

• The discovery of enzyme phosphorylase in guard cells by Hanes (1940) greatly supports the starch-sugar interconversion theory.
• The enzyme phosphorylase hydrolyses starch into sugar and high PH followed and the opening takes place during the night.

Day:

1.
2. Photosynthesis occur
3. pH – increased
4. Movement of water from
5. subsidiary cells to guard cells
6. Guard cells become turgid
7. Opening of stomata

Night:

1.

2. No Photosynthesis
3. pH – lowered
4. Movement of water from guard cells
5. Guard cells become flaccid
6. Closure of stomata

Question 8.
List out the non-photosynthetic parts of a plant that need a supply of sucrose?
Answer:
The non-photosynthetic parts of a plant that need a supply of sucrose:

1. Roots
2. Tubers
3. Developing fruits and
4. Immature leaves.

Question 9.
What are the parameters which control water potential?
Answer:
1. Slatyer and Taylor (1960) introduced the concept of water potential.
Definition – water potential is the potential energy of water in a system – compared to pure water when temperature and pressure are kept constant.

2. It is also a measure of how freely water molecules can move in a particular environment or system. Water potential is denoted by the Greek symbol  Ψ (psi) and measured in Pascal (Pa). At standard temperature, the water potential of pure water is zero

3. Addition of solute to pure water decreases the kinetic energy thereby decreasing the water potential, from zero to negative.

4. So, Comparatively a solution always has low water potential than pure water. In a group of cells with different water potential, a water potential gradient is generated.

5. Water will move from higher water potential to lower water potential.
When potential ( Ψ) can be determined by. Solute concentration or Solute potential ( Ψ_s) Pressure potential ( Ψ_p)
By correlating two factors, water potential is written as (Ψw=Ψ_s)+Ψ_p

a) Solute potential (Ψ_s) or Osmotic potential

• Denotes the effect of dissolved solute on water potential.
• In pure water, the addition of solute reduces its free energy and lowers the water potential value from zero to negative.
• Thus the value of solute potential is always negative. In a solution at standard atmospheric pressure, water potential is always equal to solute potential (Ψw = Ψ_s ).

b) Pressure Potential (Ψ_p)

• Pressure potential is a mechanical force working against the effect of solute potential.
• Increased pressure potential will increase water potential and water enters cells and cells become turgid.
• This positive hydrostatic pressure within the cell is called Turgor, pressure likewise, withdrawal of water from the cell decreases the water potential and the cell becomes flaccid.

Question 10.
An artificial cell made of selectively permeable membrane immersed in a beaker (in the figure) Read the values and answer the following questions?

a) Draw an arrow to indicate the direction of water movement
b) Is the solution outside the cell isotonic, hypotonic or hypertonic?
c) Is the cell isotonic, hypotonic, or hypertonic?
d) Will the cell become more flaccid, more turgid or stay in original size?
e) With reference to artifical cell state, the process is endomosis or exomosis? Give reasons
Answer:
a)

(b) Outside solution in hypotonic.
(c) The cell is hypertonic.
(d) The cell becomes more turgid.
(e) The process is endo – osmosis because the solvent (water) moves inside the cell.

Reason: Endomosis is defined as the osmotic entry of solvent into a cell when it is placed in pure water/Hypotonic solution. The solution in the beaker outside the cell is pure water. ( Ψw = 0), and water enters into the artificial cell which is placed inside the beaker of pure water, (i.e) from hypotonic to hypertonic solution.

Part II 

11th Bio Botany Guide Transport in Plants Additional Important Questions and Answers

I – Choose The Correct Answers

Question 1.
In plants, cell to cell transport is aided by:
(a) diffusion alone
(b) osmosis alone
(c) imbibition alone
(d) all the three above
Answer:
(d) all the three above

Question 2.
The smell from a lightened incense stick or mosquito coil or open perfume bottle in a closed room is due to
a) Osmosis
b) Facilitated diffusion
c) Simple diffusion
d) imbibition
Answer:
c. Simple diffusion

Question 3.
Which of the following statements are correct?
(i) Cell membranes allow water and non-polar molecules to permeate by simple diffusion.
(ii) Polar molecules like amino acids can also diffuse through the membrane.
(iii) Smaller molecules diffuse faster than larger molecules.
(iv) Larger molecules diffuse faster than smaller molecules.

(a) (i) and (iv) only
(b) (i) and (iii) only
(c) (i) and (ii) only
(d) (ii) and (iv) only
Answer:
(b) (i) and (iii) only

Question 4.
Solute potential is also known as
a) Water potential
b) Pressure potential
c) Osmotic potential
d) Maic potential
Answer:
c. Osmotic potential

Question 5.
The swelling of dry seeds is due to a phenomenon called:
(a) osmosis
(b) transpiration
(c) imbibition
(d) none of the above
Answer:
(c) imbibition

Question 6.
Cell A has an osmotic potential of -20 bars and a pressure potential of +6 bars. What will be its water potential?
a) -14 bars
b) +14 bars
c) -20 bars
d) +20 bars
Answer:
a. -14 bars

Question 7.
The OP and TP of two pairs of cells A – B, and X-Y are under
a) Cell A: OP=-I0atm, TP=4atm
b) Cell B : OP = l0atm, TP = 6atm
c) Cell X: Op =-l0atm, TP = 4atm
d) CeIlY: OP = -Katm, TP = 4atm
The net movement of water shall be from
a) A toB and X to Y
b) A to B and Y toX
c) B to A and X to Y
d) B to A and Y to X
Answer:
d. B to A and Y to X

Question 8.
Water potential is influenced by which of the two factors among the given four
I) Concentration
II) Pressure
III) Temperature
IV) gravity
a) I & II
b) II & III
c) III & IV
d) I & IV
Answer:
a) I & II

Question 9.
………………………. is equal to TP and is positive except plasmolysed cell and in xylem vessel where it is negative
a) Water potential
b) Pressure potential
c) Solute potential
d) Hydrostatic potential
Answer:
b. Pressure potential

Question 10.
Diffusion Pressure Deficit (DPD) was termed by Meyer in:
(a) 1928
(b) 1828
(c) 1936
(d) 1938
Answer:
(d) 1938

Question 11.
Imbibants present in plants are generally
a) Hydrothermic
b) Hydrostatic
c) Hydrophilic
d) Hydrophobic
Answer:
c. Hydrophilic

Question 12.
Kramer (1949) recognised two distinct mechanisms, which independently operate in the absorption of water in plants are:
(a) osmosis and diffusion
(b) imbibition and diffusion
(c) diffusion and absorption
(d) active absorption and passive absorption
Answer:
(d) active absorption and passive absorption

Question 13.
Root pressure is totally apsent in Gymnosperms because
a) Trachea absent
b) Tracheids absent
c) Trees are tall
d) Trees are comparatively short
Answer:
a. Trachea absent

Question 14.
When respiratory inhibitors like KCN, chloroform are applied:
(a) there is a decrease in the rate of respiration and an increase in the rate of absorption of water.
(b) there is an increase in the rate of respiration and a decrease in the rate of absorption of water.
(c) there is a decrease in the rate of respiration and also a decrease in the rate of absorption of water.
(d) there is an increase in the rate of respiration and also in the rate of absorption of water.
Answer:
(c) there is a decrease in the rate of respiration and also a decrease in the rate of absorption of water.

Question 15.
Find the DPD in a flaccid cell if its OP is 10
a) 20
b) 30
c) 10
d) 40
Answer:
c.10

Question 16.
Pulsation theory was proposed by:
(a) Strasburger
(b) Godsey
(c) J.C. Bose
(d) C.V. Raman
Answer:
(c) J.C. Bose

Question 17.
When a cell is kept in 0.5m solution of sucrose it’s volume does not alter. If the same cell is placed in 0.5M solution of sodium chloride, the volume of the cell
a) Increase
b) Decrease
c) cell will be pIasrnoysed
d) Will does not show any change
Answer:
d. Will does not show any change

Question 18.
Indicate the correct statements:
(i) Root pressure is absent in gymnosperms.
(ii) Root pressure is totally absent in angiosperms.
(iii) There is a relationship between the ascent of sap and root pressure.
(iv) There is no relationship between the ascent of sap and root pressure.

(a) (i) and (ii)
(b) (ii) and (iii)
(c) (ii) and (iv)
(d) (i) and (iv)
Answer:
(d) (i) and (iv)

Match The Following & Find Out The Correct Order

Question 19.
I) Water potential – A) Turgor pressure
II) Solute potential – B) Osmotic potential + Pressure potential
III) Matric potential – C) Osmotic potential
IV) Pressure potential – D) Imbibition pressure

Answer:
b) B C D A

Question 20.
I) Leaves – A) Antitransport
II) Seed – B) Transpiration
III) Roots – C) Negative osmotic potential
IV) Aspirin – D) Imbibition
V) Plasmolyced cell – E. Absorption

Answer:
a) C B D E A

Question 21.
I) Transport of substance from a region of lower concentration to a region of higher concentration is with the expenditure of energy – A. Antiport
II) The movement of two types of molecules across the membrane in opposite direction – B. Symport The movement of a molecule across III) a membrane independent of other molecules – C. Active port
IV) The movement of two types of molecules across the membrane in the same direction – D. Uniport

Answer:
c) C D A B

Question 22.
I) Passive transport – A) Uphill transport
II) Active transport – B) Short distance transport
HI) Cell to cell transport – C) Long-distance transport
IV) Ascent of sap – D) Downhill transport

Answer:
b) D A B C

Question 23.
The length and breadth of stomata is:
(a) about 10 – 30μ and 2 – 10μ respectively
(b) about 10 – 14μ and 3 – 10μ respectively
(c) about 10 – 40μ and 3 – 10μ respectively
(d) about 5 – 30μ and 5 – 10μ respectively
Answer:
(c) about 10 – 40μ and 3 – 10μ respectively

Question 24.
A membrane that permits the solvent and not the solute to pass through it is termed is
a) Permeable,
b) impermeable
c) semipermeable
d) differentially permeable
Answer:
c. Semi permeable

Question 25.
Who did observe that stomata open in light and close in the night:
(a) Unger
(b) Sachs
(c) Boehm
(d) Von Mohl
Answer:
(d) Von Mohl

Question 26.
The phosphorylase enzyme in guard cells supports the starch-sugar interconversion theory. The above reaction is:
(a) oxidation reaction
(b) hydrolyses reaction
(c) reduction reaction
(d) none of the above
Answer:
(b) hydrolyses reaction

Question 27.
If a cell kept in a solution of unknown concentration gets deplasmolysed the solution is
a) hypotonic
b) hypertonic
c) isotonic
d) detonic
Answer:
a. hypotonic

Question 28.
A cell placed in a strong salt solution will shrink because
a) the cytoplasm will decompose
b) mineral salts will break the cell wall
c) salt will leave the cell
d) water will leave by exosmosis
Answer:
d. water will leave by exosmosis

Question 29.
Phenyl Mercuric Acetate (PMA), when applied as a foliar spray to plants:
(a) induces partial stomatal closure for two weeks.
(b) induces partial stomatal opening for two weeks.
(c) induces partial stomatal closure for four weeks.
(d) induces stomatal closure permanently
Answer:
(a) induces partial stomatal closure for two weeks.

Question 30.
The osmotic pressure of cell sap is maximum in
a) Hydrophytes
b) Halophytes
c) Xerophytes
d) Mesophytes
Answer:
b. Halophytes

Question 31.
Say true or false and on that basis choose the right answer.
I) In facilitated diffusion, molecules move across the cell membrane with the help of special proteins, with the expenditure of energy
II) Porin is a larger transport protein, facilitates smaller molecules to pass through.
III) Aquaporins are recognized to transport urea, CO₂, NH3 metalloid & ROS
IV) The carrier proteins structure does not get modified due to its association with the molecules

Answer:
d. False True True False

Question 32.
I) Hypertonic is a strong solution (low solvent/high solute/ low Ψ )
II) Hypotonic is a weak solution (high solvent/low or zero solutes/ high Ψ)
III) Hypertonic is the weak solution (high solvent/low or zero solutes/high Ψ)
IV) Hypotonic is a strong solution (low solvent / high solute/low Ψ)

Answer:
b. True True False False

Question 33.
From sieve elements sucrose is translocated into sink organs such as root, tubers etc and this process is termed as:
(a) Xylem unloading
(b) Xylem uploading
(c) Phloem unloading
(d) Phloem uploading
Answer:
(c) Phloem unloading

Question 34.
The value of pure water is zero in which three aspects of the given options
I) Osmotic pressure
II) Osmotic potential
III) Water potential
IV) Pressure potential
a) I, II, & III
b) II, III & IV
c) I, Ill & IV
d) I, II & IV
Answer:
a. I, II & III

Question 35.
Gases such as oxygen and carbon dioxide cross the cell membrane by
a) Passive diffusion through the lipid bilayer
b) Primary active transport
c) Specific gas transport proteins
d) Secondary active transport
Answer:

Question 36.
Hydathodes are generally present in plants that grow in:
(a) dry places
(b) moist and shady places
(c) sunny places
(d) deserts
Answer:
(b) moist and shady places

Question 37.
Why sugars are transported in the form of su-crose in phloem?
a) It is inactive and highly soluble
b) It is active
c) It yields high ATP
d) It is lighter in weight.
Answer:
a. It is inactive and highly soluble

Question 38.
Unloading of pholem at sink includes
a) Passive transport
b) diffusio
c) Osmosis
d) Active transport
Answer:
d. Active transport

Question 39.
The liquid coming out of the hydathode of grasses is:
(a) pure water
(b) not pure water
(c) a solution containing a number of dissolved substances
(d) saltwater
Answer:
(c) a solution containing a number of dissolved substances

Question 40.
In a flaccid cell
a) DPD = OP
b) DPD = TP
c) TP = OP
d) OP = O
Answer:
a. DPD = OP

Question 41.
The pathway of water movement involving living part of a cell is
a) Apoplast pathway
b) symplast pathway
c) Transmembrane pathway
d) Lateral conduction
Answer:
b. Symplast pathway

Question 42.
The ascent of sap is
a) Upward movement of water in plants
b) downward movement of water in plants
c) upward and downward movement of water plants
d) None of the above
Answer:
a. upward movement of the water plants

Question 43.
High tensile strength of water is due to
a) Adhesion only
b) cohesion only
c) Both (a) and (b)
d) None of these
Answer:
c. Both (a) and (b)

Question 44.
Maximum transpiration occur in
a) Mesophytes
b) Xerophytes
c) Hydrophytes
d) Epiphytes
Answer:
a. Mesophytes

Question 45.
Supply ends in transport of solutes are
a) green leaves
b) root and stem
c) xylem and phloem
d) Hormones and enzymes
Answer:
c. Xylem and phloem

Question 46.
For guttation in plants, the process responsible is
a) Root pressure
b) Atmospheric pressure
c) Imbibition
d) None of these
Answer:
a. Root pressure

Question 47.
Which of the following theories for Ascent of sap was proposed by famous Indian scientist. J.C. Bose.
a) Transpiration pull theory
b) Pulsation theory
c) Root pressure theory
d) Atmospheric pressure theory
Answer:
b. Pulsation theory

Question 48.
Which of the following plant material is an efficient water imbibant?
a) Lignin
b) Pectin
c) Cellulose
d) Agar
Answer:
d. Agar

Question 49.
Which of the following helps in the Ascent of sap?
a) Root pressure
b) Transpiration
c) Capillarity
d) All the above
Answer:
d. All the above

Question 50.
In a girdled plant which of the following dies first?
a) Shoot
b) root
c) Both die simultaneously
d) None – the plant survives
Answer:
b. root

Question 51.
Assertion:-A Imbibition is also diffusion
Reason -R The movement of water in the above process is along a concentration gradient.
a) Both A and Rare true and R is correct explanation of A
b) Both A and R are true but R is not the correct explanation of A
c) A true but R false
d) Both A and Rare false
Answer:
a) Both A and R are True and R is correct explanation of A

Question 52.
Assertion: – A In rooted plant, the transport of water and minerals in xylem is essentially multi-directional
Reason – R Organic compound and nuitrient undergoes undirectional transport only
Answer:
d) Both A and R are false

Question 53.
Assertion: – A The adsorption of water by solid particles of an adsorbant with out forming a solution is known as imbibition
Reason: – R The liquid which is imbided is known as imbibate
Answer:
b) Both A and R are true but R is not the correct explanation of A

Question 54.
Assertion: – A In phloem loading, food is transported to the sink
Reason – R Food is transported from source to sink ‘
Answer:
d) Both Assertion ‘A’ and Reason ‘R’ are false

Question 55.
Assertion – A: Xylem a principal water conducting ’
Reason -R: It has been recognised by girdling or ringing experiments
Answer:
a) Both A and R are True R is the correct explanation of A

Question 56.
Assertion: – A In phloem, sugar are translocated in non reducing form
Reason – R Non reducing sugars are most reactive sugars
Answer:
c) Assertion is true but Reason is false

Question 57.
Assertion: AIn ringing experiment a narrow continuous band of tissues external to the phloem is removed
Reason: R Ringing experiment proves that phloem is involved in water transport ’
Answer:
d) Both A and R are false

II. Two Mark Questions

Question 1.
What is the need for the transport of materials in plants?
Answer:
Water absorbed from roots must travel up to leaves by xylem for food preparation by photosynthesis. Likewise, food prepared from leaves has to travel to all parts of the plant including roots.

Question 2.
What is osmosis
Answer:
It is a special type of diffusion almost same like simple diffusion but has a selectively permeable membrane is here, through which osmosis occur.
(OR)
It is the movement of water molecules from a place of its higher concentration, to the place of its lower concentration through a semipermeable membrane.

Question 3.
Define the term diffusion.
Answer:
The net movement of molecules from a region of their higher concentration to a region of their lower concentration along a concentration gradient until an equilibrium is attained.

Question 4.
The touch me plant closes its leaves at the touch – Explain.
Answer:

• In the ‘Touch me not’ plant the touching act as stimulus, and it closes the leaves.
• When we touch the plant, at that time the stem releases some chemicals, which force water to move out of the cell leading to the loss of Turgor pressure and the leaves droop down However after sometime they become normal.

Question 5.
What is meant by Porin?
Answer:
Porin is a large transporter protein found in the outer membrane of plastids, mitochondria and bacteria which facilitates smaller molecules to pass through the membrane.

Question 6.
Define water potential
Answer:

• The potential energy of water in a system compared to pure water when both temperature and pressure are ketp same.
• It is a measure of how freely water molecules can move in a given environment
• Water potential of pure water is = 0

Question 7.
Define Diffusion Pressure Deficit.
Answer:

• Termed by Meyer (1938)
• The difference between the Diffusion pressure of the solution and its solvent at a particular temperature and atmospheric pressure of the solution and its solvent at a particular temperature and atmospheric pressure is called DPD.

Question 8.
Differentiate between short distance and Long Distance Transport.
Answer:

Question 9.
Differentiate between Passive & Active Transport
Answer:

Question 10.
Give two examples of the phenomenon of Imbibition.
Answer:
two examples for the phenomenon of Imbibition:

1. The swelling of dry seeds.
2. The swelling of wooden windows, tables, doors due to high humidity during the rainy season.

Question 11.
Explain carbonic Acid Exchange theory.
Answer:

• Soil solution act as a medium of ion-exchange
• The CO₂ released by roots combine with water to form carbonic acid (H₂CO₃)
• Carbonic acid dissociates into H⁺ + HCO₃ in the soil solution.
• H⁺ ions exchange with cations adsorbed on clay particles and cations from micelles get released int c.

Question 12.
Give Answer in a sentence or two Distinguish between (i) Exomosis & Endomosis (ii) Apoplast & Symplast (iii) Cohesion & Adhesion (v) Influx & Efflux
Answer:

Question 13.
What is meant by osmotic pressure?
Answer:
When a solution and its solvent (pure water) are separated by a semipermeable membrane, the pressure is developed in the solution, due to the presence of dissolved solutes. This is called osmotic pressure (OP).

Question 14.
Define Root Pressure.
Answer:

• Stephen Hales – coined the term
• Stoking (1956) Defined the term.
• A pressure developing in the tracheary elements of the xylem as a result of metabolic activities of the root.

Question 15.
Define the term osmosis.
Answer:
Osmosis (Latin: Osmos – impulse, urge) is a special type of diffusion. It represents the movement of water or solvent molecules through a selectively permeable membrane from the place of its higher concentration (high water potential) to the place of its lower concentration (low water potential).

Question 16.
Why plants transport sugars as sucrose and not as starch or Monosaccharide (Glucose & Fructose)
Answer:

Question 17.
What are the three types of plasmolysis?
Answer:
Three types of plasmolysis occur in plants:

1. Incipient plasmolysis
2. Evident plasmolysis
3. Final plasmolysis.

Question 18.
Identify the diagram and Neatly label the parts
Answer:

The given diagram is the structure of Hydathode
A-Guard cell
B-Epithem
C-Tracheids

Question 19.
Identify the Diagram & Label the parts.
Answer:

The given diagram explain Reverse osmosis
A – Pressure
B – Pure water
C – Saltwater
D – Membrane

Question 20.
Differentiate between Ascent of sap and Translocation of solute.
Answer:

Question 21.
Give any two objections to starch-sugar interconversion theory.
Answer:
Two objections to starch – sugar interconversion theory:

1. In monocots, the guard cell does not have starch.
2. There is no evidence to show the presence of sugar at a time when starch disappears and stomata open.

Question 22.
Differentiate between cuticular and Lenticular Transpiration.
Answer:

Question 23.
Mention any two uses of anti – transpirants.
Answer:
Two uses of anti – transpirants:

1. Anti – transpirants reduce the enormous loss of water by transpiration in crop plants.
2. Useful for seedling transplantations in nurseries.

Question 24.
Give notes an Aquaporin.
Answer:

• Water pore – Aquaporin in KBC was discovered by Peter Agre (Nobel Prize for chemistry – 2003)
• Water channel protein is present in PM.
• Regulate the massive amount of water transport across PM
• 30 types of Aquaporins are known from maize

They also transporter

• glycerol
• urea
• CO2
• NH
• metalloids & Reactive oxygen species (ROS)

Function:

• They increase the permeability of the membrane of water
• They confer drought and salt, stress tolerance.

Question 25.
Define the term Ion – Exchange.
Answer:
Ions of external soil solution are exchanged with the same charged (anion for anion or cation for cation) ions of the root cells.

Question 26.
A. Differentiate between Cohesion and Adhesion and
B. Add a note on their significance.
Answer:
A.

B. The cohesive and Adhesive forces work together to form an unbroken continuous water column in xylem.
The magnitude of cohesive force is much high (350 atm) and is more than enough to ascent sap in the tallest
trees.

III. 3 Mark Questions

Question 1.
Compare and Contrast Diffusion & Osmosis.
Answer:

Question 2.
Differentiate between osmotic pressure it and osmotic potential
Answer:

Question 3.
Do you have an R.O. Purifier ¡n your house? Explain the principle behind it.
Answer:

• Yes / No – R.O. is working on the principle of osmosis. but in the reverse direction.
• In regular osmosis water moves from its higher concentration to its lower concentration through the selectively permeable membrane but here water moves from lower concentration to higher concentration through selectively permeable membrane.
• Since against concentration gradient, there is the expenditure of energy, to apply pressure, to force water in a reverse direction.
• Eg- Desalination plants to purify seawater also work like R-O-Purifiers Movement of Water in house hold usage.

Question 4.
Difference between various plasmolysis types
Answer:

Question 5.
Define Antitranspirant.
Answer:
Antitranspirant is any material applied to plants to retard or reduce the rate of transpiration – without disturbing the process of gaseous exchange, for respiration and photosynthesis.
Eg. Colourless plastics silicone oil and low viscosity waxes.

Question 6.
What are the inducers of stomatal closure.
Answer:

• Natural antitranspirants usually induce stomatal closure
  Eg. CO2 – inhibits photorespiration – thereby induces stomata! closure
• Some chemicals, when applied as a foliar spray can induce stomatal closure for 2 – 3 weeks.
  Eg. (PMA) Phenyl Mercuric Acetate & (ABA) Abscisic Acid.

Question 7.
Fill in the blanks in the tabulations given below

Answer:
1) Slatyer & Taylor 2) Kramer 3) J.C. Bose

Question 8.

Answer:
1) Inhibit the movement of both solvent and solute molecules
2) Semipermeable
3) Tonoplast & plasmalemma

Question 9.
Give the flow chart to cell to cell transport in plants.
Answer:

Question 10.
Explain the capillary theory of Boehm (1809).
Answer:
Capillary theory: Boehm (1809) suggested that the xylem vessels work like a capillary tube. This capillarity of the vessels under normal atmospheric pressure is responsible for the ascent of sap. This theory was rejected because the magnitude of the capillary force can raise water level only up to a certain height. Further, the xylem vessels are broader than the tracheid which actually conducts more water and against the capillary theory.

Question 11.
Explain Phloem loading?
Answer:
Definition:
The products of Photosynthesis from the Mesophyll of leaves to sieve elements of phloem is known as phloem loading. (Just like the cement sack manufactured in a factory being loaded in a vehicle to be transported the respective site)
It involves 3 steps.
Step I:

1. The chloroplast has photosynthate in the form of starch or Trlose phosphate
2. It is transported to the cytoplasm, where it is converted into Sucrose.

Question 12.
Explain the theory of photosynthesis in guard cells observed by Von Mohl with its demerits.
Answer:
Von Mohl (1856) observed that stomata open in light and close in the night. According to him, chloroplasts present in the guard cells photosynthesize in the presence of light resulting in the production of carbohydrate (Sugar) which increases osmotic pressure in guard cells. It leads to the entry of water from other cells and the stomatal aperture opens. The above process vice versa in the night leads to the closure of stomata.

Demerits:

1. The chloroplast of guard cells is poorly developed and incapable of performing photosynthesis.
2. The guard cells already possess much amount of stored sugars.

IV. 5 Mark Questions

Question 1.
Explain ‘routes’ of Water Absorption in the roots.
Answer:

• Introduction
• Root hair & other epidermal cells – By imbibition absorb water from soil –
• By osmosis moves radically & centripetally – across
  • cortex
  • Endodermis
  • Pencycle & Xylem

There are 3 Routes

• Apoplast
• Symplast
• Transmembrane route

I. Apoplast ( GK – Apo – Away) Everything external to PM
1. Cell walls
2. Extra Cellular Space
3. Interior of dead cells (vessel elements Tracheids)
Movement is continuous exclusively through the cell wall or nonliving part of the plant without crossing any membrane.

II. Symplast (GK – Sym = within)
Entire mass of cytosol of all the living cells in a plant + plasmo desmata + inter connecting cytoplasmic channel.
In the movement water has to cross PM, to enter cytoplasm of outer root cell; then move within adjoining
cytoplasm through plasmodesmata around the vacuoles without the necessity to cross more membrane it reaches xylem.

III. Trans – Membrane Route

• Water enters a cell on one side and exits from the other side.
• It crusses 2 membranes for each cell (also through to no plast).

Question 2.
Draw & Explain the structure of Stomata.
Answer:
1. Definition:
The epidermis of leaves and green stems possess many small pores called – Stomata

2. Length & Breadth
The length – 10- 40μ The Breadth – 3 – 10μ
Number Mature leaves contain 50- 500 stomata / mm²

3. Structure
a. Guard Cells – A pair of Kidney shaped cells (semilunar) surrounded a small opening called stoma
b. Subsidiary Cells – Guard cells attached to surrounding epidermal cells known as subsidiary cells or accessory cells.

• The inner wall of guard cell is thicker
• The stoma open into an interior substomatal cavity.

Question 3.
Explain osmosis by Potato osmoscope Experiment.
Answer:
Aim : To demonstrate osmosis by Potato osmoscope
Apparatus used: Potato tuber, beaker containing water, sugar solution and pin.

Definition:
Diffusion of water or solvent from the region of higher water potential to a region of lower water potential
is known as osmosis.

Procedure:
Take a peeled potato tuber and make a cavity inside with the help of a knife fill the cavity with concentrated sugar solution and mark the initial level.
Place this set up in a beaker containing pure water After 10 minutes observe the sugar solution level and record your observation.

Observation:
There is rise in the level of the solution. in the cavity of the tuber due to osmosis
Inference: Osmosis has occured, through the potato osmoscope

Question 4.
Measure Transpiration with Ganong’s Photometer
Answer:
Aim: To measure the rate of Transpiration with Ganong’s Potometer
Apparatus needed: Ganong’s Potometer, a twig, beaker, water, split rubber cork, and vaseline.

Procedure:

• Ganong’s Potometer is a horizontal graduated tube which is bent in opposite directions at the ends.
• A reservoir is fixed to the horizontal tube hear the wider end Reservoir has stop cock to regulate water flow.
• A twig is fixed to the wider arm through the split cork. The apparatus is filled with water with water from reservoir.
• The apparatus is made air tight by applying vaseline.
• The other bent end of the horizontal tube is dipped into a beaker containing coloured water.

An air bubble is introduced into the graduated tube at the narrow end. Keep the apparatus in bright sunIght
and observe

Observation:
As the twig transpires, the air bubble move towards the twig.
This loss is compensated by water ohsorption from the beaker.

inference:
By the experiment we can study the rate of Transpiration and rate of transpiration is equal to the rate of water absorption.

Question 5.
Explain Mechanism of Translocation by Munch Mass flow Hypothesis
Answer:
Munch – Proposed it in 1930 Crafts – elaborated it in 1938
Definition: Organic substances (solute) move from a region of high osmotic pressure (mesophyll) to
region of low OP along TP gradient.

Example – Physical system :
Chamber ‘A’ & chamber ‘B’ made up of semi permeable membrane connected by a tube ‘T’
A – Contain highly concentrated sugar solution (hypertonic)
B – Contain dilute sugar solution (hypotonic)
A – draws water from the reservoir by Endosmosis – TP of chamber ‘A’ increased

• Continuous entry of water in to A – TP increased
• Flow of solute from chamber A to B thro TP gradient.
• The movement continues till both Aand B attain isotonic condition (equilibrium)
  (However if new sugar solution added to A system will start to run again)
  Example (Biological system)
• Chamber A (Source) – (Equivalent to) – Mesophyll cells of leaves (High concentration of soluble food)
• Chamber B (Sink) – (Equivalent to) – Cells of stem & Roots (Consumption end)
• TubeT – (Analogous to) – Sieve tube to phloem

Steps :
1. Xylem (Reservoir) – Movement of water (Endomosis) – Mesophyll cells (TP increase)
2. Mesophyll cells (High TP) Source – enmass movement of  organic solutes through Phloem by TP Gradient  – Cells of stem & Root (low TP) (Sink)

Question 6.
Explain the theory of K⁺ transport – or Explain the mechanism of stomatal movement
Answer:
Introduction:
Levit (1974) – Proposed it
Raschke (1975) – Elaborated it
Steps:

This process of exchange of ions is called Actie ion exchange ( consume ATP) or Energy

• Increased K⁺ ions in the Guard cells – balanced by CP ions
• Increase in solute concentration (Hypertonic) Decrease in water potential
• Water enters into Guard cells from subsidiary cells
• Wall pressure increase Turgor pressure, Turgid guard cells – fall apart & opens the stoma

• Exit of H⁺
• Intake of K⁺
• Exit of K⁺
• Loss of H2O
• Uptake of H2O⁺
• Turgidity of Guard Cells
• Accumulation of CO2 – Lowering of pH
• Opening of Stoma.
• Activation of ABA
• Closure of Stoma.

Question 7.
Explain Cytochrome Pump Theory (or) Explain Carrier concept of Active Absorption, through cytochrome Pump theory.
Answer:
Lunde gardth & Burstom (1933)- Proposed the Cytochrome

Pump theory:

• There is correlation between Respiration & Anion absorption.
• when a plant is transferred from water to salt solution, the rate of respiration increases – known as Anion respiration – or salt respiration

The Assumptions of Cytochrome pump theory:

• The mechanism of anion and cation absorption is different.
• Anion – absorption – through cytochrome pump or chain by Active process
• An oxygen gradient is responsible for oxidation at outer surface of the membrane and reduction at the inner surface.

Explanation:

• On the inner surface, the enzyme dehydrogenase Produces protons (W) and electrons (e)

• Anions are picked up by oxidized cytochrome oxidase and transferred to the other members of the chain.
• Theory assumes the passive movement of cations (C⁺) along the electrical gradient created by the accumulation of anions (A^–) at the inner surface of the membrane.

Defects :

• • Cations also induce respiration
  • to fail to explain the selective uptake of ions
  • It explains absorption of anions only.

Question 8.
Explain the opening and closing of stomata by a starch – sugar – Interconversion theory.
Answer:
i) Lloyd (1908)
According to him, turgidity of Guard cell is due to interconversion of starch → sugar

• Day time:
  Guard cells have sugar → so turgid → opening of stomata
• Nighttime:
  Guard cells have starch → so loose turgidity (become flaccid) → closure of stomata

ii) Sayre (1920)
According to him, the pH of Guard cell determine opening and closing of stomata

• Day time: Guard cells have high pH →so turgid → opening of stomata
• Nighttime: Guard cells have low pH → become flaccid → closure of stomata to be elaborate
• Day time: Utilisation of CO₂. in photosynthesis → Starch into sugar → high pH → high Turgor pressure→Opening of Stomata
• Night Time: No photosynthesis, so the accumulation of CO₂ → sugar to starch → low pH → decrease in TP → closure of stomata

iii) Hanes (1940)
According to Hanes – Enzyme phosphorylase is responsible for starch sugar conversion in the guard cells.

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Bio Botany Guide Pdf Chapter 15 Plant Growth and Development Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 15 Plant Growth and Development

11th Bio Botany Guide Plant Growth and Development Text Book Back Questions and Answers

Part – I.

Question 1.
Select the wrong statement from the following:
(a) Formative phase of the cells retain the capability of cell division.
(b) In elongation phase development of central vacuole takes place.
(c) In maturation phase thickening and differentiation takes place.
(d) In maturation phase, the cells grow further.
Answer:
(d) In maturation phase, the cells grow further.

Question 2.
If the diameter of the pulley is 6 inches, length of pointer is 10 inches and distance travelled by pointer is 5 inches. Calculate the actual growth in length of plant.
a) 3 inches
b) 6 inches
c) 12 inches
d) 30 inches
Answer:
options are wrong, (correct Ans = 1.5 inches)
Solution:
Step I:
Diameter of the Pulley=6 inches
Radius of the pulley \(=\frac{6}{2}\)= 3 inches
Actual growth in length= Distance travailed by pointer x Radius of the pulley Length of the pointer
=\(\frac{5 \times 3}{10}\) =1.5=1.5.
Answer:
1.5 inches

Question 3.
In unisexual plants, sex can changed by the application of
a) Ethanol
b) Cytokinins
c) ABA
d) Auxin
Answer:
c) ABA

Question 4.
Select the correctly matched one
A) Humanurine i) Auxin-B
B) Corn gram oil ii) GA₃
C) Fungs iii) Abscisic acid II
D) Herring fish sperm iv) Kinetin
E) Unripcrnaizegrains v) AuxinA
F) Young cotton boils vi) Zeatin

Answer:
b) A – v, B – i, C – ii, D – iv, E – vi, F – iii

Question 5.
Seed dormancy allows the plants to:
(a) overcome unfavorable climatic conditions
(b) develop healthy seeds
(c) reduce viability
(d) prevent deterioration of seeds
Answer:
(a) overcome unfavorable climatic conditions

Question 6.
What are the parameters used to measure growth of plants?
Answer:
Growth in plants can be measured in terms, of

• Increase in length or girth (roots and stems)
• Increase in fresh or dry weight
• Increase in area or volume (fruits and leaves)
• Increase in a number of cells produced.

Question 7.
What is plasticity?
Answer:
Plasticity refers to the environmental heterophylly seen in Butter cup plant (Ranunculus). In this aquatic plant, the leaves in the air is normal, where as the leaves submerged underwater are highly thin and hairy highly adapted to do carbon assimilation Developmental heterophlly seen in the juvenile plant leaves of cotton and corianter. Where the young leaves have a different shape from the mature leaves is not considered as plasticity.

Question 8.
Write the physiological effects of Cytokinins.
Answer:

1. Cytokinin promotes cell division in the presence of auxin (IAA).
2. Induces cell enlargement associated with IAA and gibberellins
3. Cytokinin can break the dormancy of certain light-sensitive seeds like tobacco and induces seed germination.
4. Cytokinin promotes the growth of lateral bud in the presence of apical bud.
5. Application of cytokinin delays the process of aging by nutrient mobilization. It is known as the Richmond Lang effect.
6. Cytokinin:
   • increases rate protein synthesis
   • induces the formation of inter-fascicular cambium
   • overcomes apical dominance
   • induces the formation of new leaves, chloroplast and lateral shoots.
7. Plants accumulate solutes very actively with the help of cytokinins.

Question 9.
Describe the mechanism of photoperiodic induction of flowering.
Answer:
Mechanism of photoperiodic induction of flowering.

• The physiological change on flowering due to the relative length of light and darkness (photoperiod) is called Photoperiodism.
• The photoperiod required to induce flowering is called critical day length. Eg. 12 hours in Maryland Mammoth’s Tobacco Xanthium 15.05 hours.

Photoperiodic induction:

• An appropriate photoperiod in 24 hours cycle constitutes one inductive cycle. Plants may require one or more inductive cycles for flowering.
• The phenomenon of conversion of leaf primordia into flower primordia under the influence of suitable inductive cycles is called photoperiodic induction. Example: Xanthium (SDP) -1 inductive cycle and Plantago (LDP) -25 inductive cycles.

Site of photoconductive perception:

• Leaves are the parts that receive photoperiodic stimulus (PPS), again it is only leaves that synthesize floral hormones and translocate them to the apical tip to promote flowering.
• This can be demonstrated by experiments conducted in the Cocklebur plant. Which is an SD plant. The nature of flower-producing stimulus has been elusive so far. It is believed by physiologists that a hormone is responsible for it, Chailakyan (1936) named it as Florigen It is not possible to isolate it.

Question 10.
Give a brief account of programmed cell death (PCD).
Answer:
Senescence is controlled by plants’ own genetic program and the death of the plant or plants part consequent to senescence is called Programmed Cell Death. In short senescence of an individual cell is called PCD. The proteolytic enzymes involving PCD in plants are phytases and in animals are caspases. The nutrients and other substrates from senescing cells and tissues are remobilized and reallocated to other parts of the plant that survives.

The protoplasts of developing xylem vessels and tracheids die and disappear at maturity to make them functionally efficient to conduct water for transport. In aquatic plants, aerenchyma is normally formed in different parts of the plant such as roots and stems which enclose large air spaces that are created through PCD. In the development of unisexual flowers, male and female flowers are present in earlier stages, but only one of these two completes its development while the other aborts through PCD.

Part-II.

11th Bio Botany Guide Plant Growth and Development Additional Important Questions and Answers

I. Choose The Correct Answer

Question 1.
The open form of the growth occurs in:
(a) leaves and flowers
(b) stem and root
(c) leaves and stem
(d) stem and flowers
Answer:
(b) stem and root

Question 2.
An example of a De-Differentiating cell is ………………
a) Tracheary element
b) shoot apex
c) Cork cambium
d) root apex
Answer:
c) Cork cambium

Question 3.
Primary growth of the plant is due to the activity of:
(a) phloem parenchyma
(b) phloem meristem
(c) vascular cambium
(d) apical meristem
Answer:
(d) apical meristem

Question 4
Choose the natural Auxin of the following
a) Anti Auxin
b) NAA
c) 2.4.D
d) IndoleAcetic Acid (IAA)
Answer:
d) Indole Acetic Acid (IAA)

Question 5.
Thickening and differentiation of cells take place during:
(a) elongation phase
(b) formative phase
(c) maturation phase
(d) flowering phase
Answer:
(c) maturation phase

Question 6.
The hormone present in Coconut milk is
a) Gibberellins
b) Ethylene
c) Cytokinin
d) Auxin
Answer:
c) Cytokinin

Question 7.
The total growth of the plant consists of four phases in the following order.
(a) Log phase, lag phase, decelerating phase and maturation phase
(b) Log phase, lag phase, maturation phase and decelerating phase
(c) Lag phase, log phase, maturation phase and decelerating phase
(d) Lag phase, log phase, decelerating phase and maturation phase
Answer:
(d) Lag phase, log phase, decelerating phase and maturation phase

Question 8.
Which of the following Phytóhormone does not occur naturally in plants?
a) 2. 4. D
b) GibberellicAcid
c) 6. Furfuryl amino purine
d) IAA
Answer:
a) 2.4.D

Question 9.
Absence of light may lead to the yellowish color in plants and this is called:
(a) venation
(b) etiolation
(c) estivation
(d) vernation
Answer:
(b) etiolation

Question 10.
Apical dominance is caused when Auxin
a) Concentration is more than Cytokinins
b) Concentration is less than Cytokinins
c) and Cytokinin concentration are equal
d) and Cytokinin concentration are fluctuating
Answer:
a) Concentration is more than Cytokinins

Question 11.
Indicate a plant growth regulator from the following:
(a) cytocin
(b) cytokinins
(c) acetic acid
(d) methylene
Answer:
(b) cytokinins

Question 12.
Which prevents premature fall of fruit?
a) NAA
b) Ethylene
c) GA₃
d) Zeatin
Answer:
a) NAA

Question 13.
The activity of synergistic effect involves the activity of:
(a) auxin and gibberellins
(b) auxin and ethylene
(c) ABA and gibberellins
(d) none of the above
Answer:
(a) auxin and gibberellins

Question 14.
The term Auxin was coined by
a) Went
b) Darwin
c) Smith
d) Garner
Answer:
a) Went

Question 15.
The term auxin was first coined by:
(a) Charles Darwin
(b) Kogl
(c) F.W. Went
(d) Smith
Answer:
(c) F.W. Went

Question 16.
The term Gibberellin was coined by
a) Went
b) Kurosawa
c) Skoog
d) Yabuta
Answer:
d) Yabuta

Question 17.
Indicate a synthetic auxin.
(a) Indole Acetic Acid
(b) Phenyl Acetic Acid
(c) Indole Butyric Acid
(d) Naphthalene Acetic Acid
Answer:
(d) Naphthalene Acetic Acid

Question 18.
The mineral required for the synthesis of IAA is
a) Copper
b) Magnesium
c) Zinc
d) Boron
Answer:
c) Zinc

Question 19.
Auxin stimulates:
(a) transpiration
(b) respiration
(c) flowering
(d) none of the above
Answer:
(b) respiration

Question 20.
The most widely occurring Cytokinin in plants is
a) ABA
b) NAA
c) TNT
d) IPA
Answer:
d) IPA

Question 21.
Who established the structure of gibberellic acid?
(a) Brain etal
(b) Kurosawa
(c) Cross et al
(d) Yabuta and Sumiki
Answer:
(c) Cross etal

Question 22.
The term Florigen was coined by
a) Maheswari
b) Chailakyan
c) R Gane
d) Richmond Lang
Answer:
b) Chailakyan

Question 23.
Cytokinins inducing cell division was first demonstrated by:
(a) Haberlandt
(b) Charles Darwin
(c) Clarke
(d) Hubert
Answer:
(a) Haberlandt

Question 24.
Which of the following is a bioassay for Cytokinins?
a) Chlorophyll preservation test
b) Dwarf maize Assay
c) Seed germination Assay test
d) Neem cotyledon Assay
Answer:
d) Neem cotyledon Assay

Question 25.
Indicate correct statements.
(i) Genes are intracellular factors for growth.
(ii) Temperature has no role in the growth of plant.
(iii) Oxygen has a vital role in the growth of plants.
(iv) CIN ratio of soil does not affect the growth of plant.
(a) (i) and (iv)
(b) (ii) and (iv)
(c) (i) and (iii)
(d) (ii) and (iii)
Answer:
(c) (i) and (iii)

Question 26.
Avena curvature test as a BioAssay for
a) Auxins
b) GA₃
c) Cytokinin
d) Ethylene
Answer:
a) Auxins

Question 27.
The stress phytohormones (Abscisic acid) was first isolated by:
(a) Linn et al
(b) Addicott et al
(c) Edward et al
(d) Stone and Black
Answer:
(b) Addicott et al

Question 28.
The Gibberellins have been commercially exploited for
a) increasing the size of grapefruits
b) inducing rooting in stem cuttings
c) breaking the dormancy in seeds
d) production of disease-resistant varieties
Answer:
c) breaking the dormancy in seeds

Question 29.
Pick out the correct statement from the following:
(i) Abscisic acid is found abundantly inside the chloroplast of green cells.
(ii) ABA is a powerful growth promotor.
(iii) ABA is formed from the pentose phosphate pathway.
(iv) ABA has anti-auxin and anti-gibberellin properties.
(a) (i) and (iv)
(b) (i) and (ii)
(c) (ii) and (iii)
(d) (ii) and (iv)
Answer:
(a) (i) and (iv)

Question 30.
Biennials can be induced to flower in the first season itself by treatment with
a) Auxin
b) Kinetin
c) GA
d) ABA
Answer:
c) GA

Question 31.
Pea and barley are classified under:
(a) short-day plants
(b) short long day plants
(c) long day plants
(d) long short day plants
Answer:
(c) long day plants

Question 32.
Auxin a was isolated from human urine by
a) F.W. went
b) Charles Darwin
c) Kogl and Haugen Smith
d) Denny
Answer:
c) Kogl and Haugen smith

Question 33.
Usually, Xanthiumpensylvanicum will flower under:
(a) long day condition
(b) short long day condition
(c) photo neutral condition
(d) short-day condition
Answer:
(d) short-day condition

Question 34.
The most widely occurring Cytokinin in plants is
a) Indole Acetic Acid (LAA)
b) Indole Butyric Acid (IBA)
c) Pentenyl Adenine (IPA)
d) Naphthalene  Acetic Acid (NAA)
Answer:
c) Pentenyl Adenine (IPA)

Question 35.
Who found out the phytochrome in plants?
(a) Butler et al
(b) Michell et al
(c) Boumick et al
(d) Gamers and Allard
Answer:
(a) Butler et al

Question 36.
Scientists, those who are connected with Ethylene
(I) Denny
(II) R. Gane
(III) Kurosawa
(IV) Cocken
Options:
a) (I) (II) & (III)
b) (II) (III) & (IV)
c) (I) (II) & (IV)
d) (I) (III) & (IV)
Answer:
c) (I)(II)& (IV)

Question 37.
Pick out the wrong statement from the following:
(a) Vernalization increases the cold resistance of plants
(b) It increases the resistance of plants to fungal disease
(c) Vemalizatiqn increase the vegetative period of the plant
(d) It accelerates the plant breeding
Answer:
(c) Vemalizatiqn increase the vegetative period of the plant

Question 38.
Day-neutral plants are
a) Sugarcane & Coleus
b) Bryophyllum& Night Jasmine
c) Wheat, rice & Oats.
d) Potato, Tomato & Cotton
Answer:
d) Potato, Tomato & Cotton

Question 39.
In apple and plum, the method of breaking seed dormancy involves the process of:
(a) impaction
(b) Scarification
(c) exposing to red light
(d) Stratification
Answer:
(d) Stratification

Question 40.
Xanthium (Cocklebur) requires …………….. hours of light to induce flowering,
a) 12
b) 9
c) 15.05
d) 13.05
Answer:
c) 15.05

Question 41.
The hormone that cannot be isolated
a) IAA
b) ABA
c) NAA
d) Florigen
Answer:
d) Florigen

Question 42.
The term Photoperiodism was coined by
a) Went
b) Butler
c) Gamer
d) Skoog
Answer:
c) Garner

Question 43.
ABA acts as antagonistic to
a) Ethylene
b) Cytokinin
c) Gibberellic acid
d) IAA
Answer:
c) Gibberellic acid

Question 44.
If a short-day plant, flowering is induced by
a) Long nights
b) Photo periods less than 12 hrs
c) Photoperiods shorter than critical value and uninterrupted long night
d) Short photoperiods and interrupted long nights
Answer:
c) Photoperiods shorter than critical value and uninterrupted long night.

Question 45.
Phytochrome is
a) Reddish phytohormone
b) Bluish biliprotein pigment
c) Photoreceptor of apical bud
d) Unstable pigment molecule
Answer:
b) Bluish biliprotein pigment

Question 46.
The growth & ripening is induced by Ethylene in
a) Tropical fruits
b) Temperate fruits
c) Climacteric fruits
d) Nonclimacteric fruits
Answer:
c) Climacteric fruits

Question 47.
The bioassay of ABA was done with
a) Rice
b) Wheat
c) Maize
d) Barley
Answer:
a) Rice

Question 48.
Four types of senescence were recognized by
a) Leopold
b) Gamer
c) Addicott
d) Cocken et al
Answer:
a) Leopold

Question 49.
The final stage of senescence is
a) PCD
b) Scarification
c) Yellowing
d) Abscission
Answer:
d) Abscission

Question 50.
Match & Find out the Correct Answer

Answer:
b) D C A B

Question 51.
Match the following and Find the Correct Answer

Answer:
b) D E A B C

II. Assertion (A) & Reason (R)

Question 52.
a. Both A & R are true and ‘R’ is the correct explanation of A
b. Both A & R are true but ‘R’ is not the correct explanation of A
c. A is true but R is False
d. Both A and ‘R’ are False
Assertion (A): The shoot Apical meristems are the only source of Auxin synthesis
Reason (R): Dormancy of lateral buds over Apical buds is due to Auxin
Answer:
C. A is true but R is False

Question 53.
Assertion (A): Hormones are also called Growth regulator
Reason (R): Hormones promote or inhibit plant growth
Answer:
A. Both Assertion (A) and Reason (R) are true and Reason is the correct explanation of Assertion.

Question 54.
Assertion (A): In many land Mammoth flowering occurred at different times at different latitude
Reason (R): Many land Mammoth is a tobacco variety
Answer:
b. Both Assertion (A) and, Reason (R) are true and Reason is not the correct explanation of Assertion.

III. 2 Mark Questions

Question 1.
Define closed form of growth in plants.
Answer:
Leaves, flowers, and fruits are limited in growth or of determinate or closed-form growth.

Question 2.
Compare between Absolute and Relative growth rates
Answer:

Question 3.
Name the phases of growth in ‘S’ shaped growth curve.
Answer:

• Lag phase
• Log phase
• Decelerating phase
• Maturation phase

Question 4.
Mention the phase of growth in plants
Answer:
I. Formative phase
II. Elongation phase
III. Maturation phase

Question 5.
Distinguish between absolute growth rate and relative growth, rate.
Answer:
Absolute growth rate:
An increase in total growth of two organs measured and compared per unit time is called absolute growth rate.

Relative growth rate:
The growth of the given system per unit time expressed per unit initial parameter is called relative growth rate.

Question 6.
What is the Grand period of growth
Answer:
The total period from initial to the final stage of growth is called Grand period of growth.
When plotted against time the growth curve is ‘S’ shaped, (sigma curve) it is also known as Grand Period curie consists of 4 phases

1. Lag,
2. Log,
3. Decelerating,
4. Maturation.

Question 7.
What is meant by the dedifferentiation of plant cells?
Answer:
Differentiated cells, after multiplication again lose the ability to divide and mature to perform specific functions. This is called redifferentiation, eg: Secondary xylem and Secondary phloem.

Question 8.
Define Phytohormone.
Answer:
The chemical substances synthesized by plants and thus naturally occuring are known as Phytohormones. Eg. Auxin, Gibberellins.
Recently 2 groups – Brassinosteroids, Polyamines were also known to behave like hormones.

Question 9.
Mention any two synthetic auxins.
Answer:

• 2, 4 – Dichloro Phenoxy Acetic Acid (2, 4 – D)
• 2, 4, 5 – Trichloro Phenoxy Acetic Acid (2, 4, 5 – T)

Question 10.
State 3 characteristic features of phytohormones.
Answer:

1. They are produced in root tips and stem tips and leaves (do not have specialized cells or organs for secretion)
2. The transfer of hormones takes place through the conducting system (xylem and phloem)
3. They are required in trace quantities
4. They either promote, inhibit or modify growth.

Question 11.
Name the natural auxins present in plants.
Answer:

• Indole Acetic Acid (IAA)
• Indole Propionic Acid (IPA)
• Indole Butyric Acid (IBA)
• Phenyl Acetic Acid (PAA)

Question 12.
Give the historial significance of Agent Orange
Answer:

• Mixture of two phenoxy herbicides – 2.4. D and 2.4.5 T together known as Agent orange.
• This Agent orange, was used by USA in Vietnam war as chemical warfare weapon to defoliate forests in Vietnam.

Question 13.
Does the trimming of plants in gardens have any scientific explanation?
Answer:

• Yes, trimming of plants removes apical buds and hence apical dominance is prevented the lateral buds sprout and give a beautiful bushy appearance and aesthetic value.
• Also in tea estates, this trimming develops more lateral branches and more tea leaves thus it has commercial significance.

Question 14.
Where do you find cytokinin hormone in plants?
Answer:
The distribution of cytokinin in plants is not as wide as those of auxin and gibberellins but found mostly in roots. Cytokinins appear to be translocated through xylem.

Question 15.
What is bolting?
Answer:

• When treated with Gibberellins the rose the plants (genetic dwarf) exhibit excessive internodal growth.
• This sudden elongation of a stem followed by flowering is called bolting.

Question 16.
What is Richmond Lang effect?
Answer:
Application of Cytokinin delays the process of aging by nuitrient mobilization.

Question 17.
Why do you call Abscisic acid (ABA) as stress hormone?
Answer:
It inhibits the shoot growth and promotes growth of root system. This character protect the plants from water stress. Hence, ABA is called as stress hormone.

Question 18.
Define photoperiodism & Critical day length.
Answer:

• The physiological change on flowering due to relative length of light and darkness (Photoperiod) is called
  Photoperiodism.
• The photoperiod required to induce flowering is called critical day length Eg.
  • Mary land mammoth (tobacco variety) requires 12 hours of light.
  • Cocklebur required 15.05 hours of light.

Question 19.
Write down the importance of photoperiodism in plants.
Answer:

• The knowledge of photoperiodism plays an important role in hybridisation experiments.
• Photoperiodism is an excellent example of physiological pre-conditioning that is using an external factor to induce physiological changes in the plant.

Question 20.
What is the importance of photoperiodism?
Answer:

• The knowledge of photoperiodism an important role in hybridization experiments.
• It is an excellent example of physiological preconditioning that is using an external factor to induce physiological changes in the plant.

Question 21.
What is meant by Epigeal germination?
Answer:
During epigeal germination, cotyledons are pushed out of the soil. This happens due to the elongation of the hypocotyl.
Eg: Castor and Bean.

Question 22.
Define Vernalization.
Answer:

• It is a process by which many annuals and biennials are induced to flower when subjected to low-temperature exposure.
• T.d. Lysenko first used the term.

Question 23.
Define the term phytogerontology.
Answer:
The branch of botany which deals with ageing, abscission and senescence is called Phytogerontology.

Question 24.
Distinguish between Epigeal and Hypogeal germination.
Answer:

Question 25.
Define seed dormancy and what are its types.
Answer:
The condition of a seed when it fails to germinate even in suitable environmental condition is called seed dormancy.
There are two types
(I) Innate dormancy (II) Imposed dormancy.

Question 26.
What is Scarification?
Answer:

• By mechanical and chemical treatments like cutting or chipping of hard tough sed coat and use of organic solvents to remove waxy or fatty compounds are called scarification.
• It is a method of breaking dormancy of the seeds.

Question 27.
Distinguish between Re differentiation and Devernalization.
Answer:

Question 28.
Define Senescence.
Answer:

• Ageing or getting old is called senescence.
• It refers to all collective, progressive and deteriorative processes which ultimately lead to complete loss of organization and function (Eg. leaves turn yellow and fall off from plant).

Question 29.
What is impaction in seed dormancy.
Answer:

• In some seeds water and oxygen are unable to penetrate micropyle due to blockage by cork cells.
• These seeds are shaken vigorously to remove the plug
• The process of removing the plug or block is called impactation.

Question 30.
What is called stratification in seed dormancy?
Answer:

• The break dormancy, some plant seeds have to be exposed to well aerated, moist conditions under low temperature (0°c to 10°c) for weeks to months.
• This kind of seed dormancy breaking treatment is known as stratification.
• The stratified soil layers should be given a low-temperature treatment for a certain period so as to induce germination.
• Eg. the Seeds of Rosaceae plants Apple, Plum, Peach, etc.

Question 31.
What are the 4 types of Senescence?
Answer:
Leopold (1961) explained 4 types they are

1. Overall senescence
2. Top senescence
3. Deciduous senescence
4. Progressive senescence.

Question 32.
What is the Abscission layer or Abscission Zone?
Answer:
Abscission is marked internally at the place of petiole by a distance zone of few layers of thin-walled cells arranged transversely. This zone is called Abscission Zone, which leads to Abscission of the leaf.

Question 33.
The photoperiodic response will not be possible in a defoliated plant. Give scientific reasons.
Answer:
Yes, a defoliated plant will not respond to photoperiodic change because the hormonal substance responsible for flowering is present in the leaves of the plant.

Question 34.
What is gas chromatography?
Answer:

• It is a bioassay technique by which Ethylene can be measured.
• It helps in the detection of the exact amount of ethylene from different plant tissues like lemon and orange.

Question 35.
Give the occurrence and precursors of Gibberellins and Cytokinins.
Answer:

IV. 3 Mark Questions

Question 1.
Explain Arithmetic growth rate and Geometric growth rate by diagrams.
Answer:

Question 2.
Explain stages in growth by drawing the sigmoid curve.
Answer:

• The total period from the initial to the final stage of growth is called the Grand period of growth.
• The graph that is drawn by taking time and rate of growth is ‘S’ shaped. It is known as a sigmoid curve.

It has 4 stages:

1. Lag phase
2. Log phase
3. Decelerating phase
4. Maturation phase

Question 3.
Mention the internal factors, that affect the growth of plants.
Answer:

• Genes are intracellular factors for growth.
• Phytohormones are intracellular factors for growth, eg: auxin, gibberellin, cytokinin.
• C/N ratio.

Question 4.
Mention the Agricultural role of Auxin.
Answer:

• Eradicate weeds: Eg. 2.4 D and 2.4.5.7
• Formation of seedless fruits: (Parthenocarpic fruits) Eg. Synthetic Auxin.
• Break dormancy.
• Induction of flowering: In pineapple NAA induce flowering
• Increase the number of female flowers: Eg. Cucurbita.

Question 5.
List out the agricultural applications of auxins.
Answer:

• It is used to eradicate weeds, eg: 2,4 – D and 2,4,5 – T.
• Synthetic auxins are used in the formation of seedless fruits (Parthenocarpic fruit).
• It is used to break the dormancy in seeds.
• Induce flowering in Pineapple by NAA & 2,4 – D.
• Increase the number of female flowers and fruits in cucurbits.

Question 6.
What are the Precursors of Gibberellins?
Answer:

• Gibberellins are chemically related to terpenoids (natural rubber, Carotenoids, and steroids) formed by 5-C precursors and an Isoprenoid unit called Iso Pentenyl Pyrophosphate (IPP) through a number of intermediates.
• The primary precursors are Acetate.

Question 7.
What are the uses of ethylene in agriculture?
Answer:

• Ethylene normally reduces flowering in plants except in Pineapple and Mango.
• It increases the number of female flowers and decreases the number of male flowers.
• Ethylene spray in the cucumber crop produces female flowers and increases the yield.

Question 8.
Explain the mechanism of Vernalization by Hypothesis of hormonal involvement.
Answer:
I. Vernalization: According to Purvis

II. Devernalization

Question 9.
What are the practical applications of Vernalization?
Answer:

• It shortens the vegetative period and induces the plant to flower earlier.
• It increases the cold resistance of the plants.
• It increases the resistance of plants to fungal disease.
• Plant breeding can be accelerated.

Question 10.
What is meant by the viability of seeds?
Answer:

• Viable means the living condition of the seed
• The shelf life of the seed after which it cannot germinate is known as the viable period
• It varies from plant to plant

Question 11.
Differentiate between climacteric and Non-climacteric fruits.
Answer:

Question 12.
Differentiate between scarification & Stratification in breaking seed dormacy
Answer:

Question 13.
Mention the factors causing dormancy of seeds.
Answer:

• Hard, tough seed coat causes barrier effect as impermeability of water, gas and restriction of the expansion of embryo prevents seed germination.
• Many species of seeds produce imperfectly developed embryos called rudimentary embryos which promotes dormancy.
• Lack of specific light requirement leads to seed dormancy.
• A range of temperatures either higher or lower cause dormancy.
• The presence of inhibitors like phenolic compounds which inhibits seed germination cause dormancy.

Question 14.
What are the factors that affect senescence?
Answer:

Question 15.
What are the morphological and Anatomical changes due to Abscission?
Answer:

• Abscission Zone: formed at the base of petiole
• Greenish grey in colour by rows of 2 to 15 cells thick primary wall and middle lamella
• The dissolution of by pectinase & Cellulase
• Formation tyloses – that block conduction of vessels
• Degradation of chlorophyll – Colour of leaves changes and leaves fall off.
• After Abscission – Suberization of outer layer of cells by the development of periderm.

Question 16.
Write down the significance of Abscission.
Answer:
1. Abscission separates dead parts of the plant like old leaves and ripe fruits.
2. Helps in dispersal of fruits and continuing the life cycle.
3. Abscission of leaves (in deciduous plants) helps in water conservation during summer.
4. Helps in vegetative propagation (Shedding of gemmae or plantlets) Eg. Bryophyta.

V. 5 Mark Questions

Question 1.
Describe an experiment to measure the growth of a plant or By lever Auxanometer measure the rate of growth in stem tip.
Answer:
Experiment:
1. Arc auxanometer:
The increase in the length of the stem tip can easily by measured by an arc auxanometer. If consists of a small pulley to the axis of which is attached a long pointer sliding over a graduated arc. A thread one end of which is tied to the stem tip and another and to a weight passes over the pulley tightly. As soon as the stem tip increases in length, the pulley moves and the pointer slide over the graduated arc (Refer Figure) The reading is taken. The acutal increase in the lengthwm stem is then calculated by knowing the length of the pointer and the radius of the pulley. If the radius of the pulley is 4 inches and the length of pointer 20 inches the actual growth is measured as follows:

Actual growth in length\(=\frac{\text { Distance travelled by the pointer radius of the }}{\text { Length of the pointer }}\)
For example, actual growth in length \(=\frac{10 \times 4 \text { inches }}{20 \text { inches }}\) = 2 inches

Question 2.
Explain the physiological effect of Auxin? Add a note on its agricultural applications.
Answer:
Cell elongation:
Promotes cell elongation in stem & Coleoptile

Root growth:
At extremely low concentration – promote root growth, at high concentrations it inhibits elongation of roots, but induce more lateral roots

Apical dominance:
Suppression of growth of lateral buds – by apical bud is known as Apical dominance
Prevents Abscission

Secondary growth:
Promotion of cell division in the cambium, responsible for secondary growth this property is exploited in tissue culture. (Callus foundation)

Respiration Stimulates respiration Induces Vascular differentiation.

Question 3.
Give the Agricultural application of Auxin.
Answer:
Weedicide
2.4.D & 2.4.5. T – Weedicides to remove weeds

Induce parthenocarpy
Synthetic auxins used to induce parthenocarpy (formation of seedless fruits).

Break dormancy
Used to break seed dormancy.

Induce flowering in Pineapple Eg. NAA & 2.4.D

Induce female flowers (numbers)
Eg. Cucumber.

Question 4.
Explain physiological effects of Gibberellins
Answer:

• Induction of cell division & cell elongation – Extraordinary stem elongation.
• Reversal of dwarfism & Bolting – Rosette (genetic dwarfism) plants when treated with Gibberellins exhibit excessive enter nodal growth – This sudden elongation of a stem followed by flowering is called Bolting.
• Breaks dormancy – in Potato tubers.
• Biennials flower in the 1 st year – Instead of cold exposure, if biennials treated with Gibberellins flower in the 1st year itself.

Question 5.
Write an essay on the role of ethylene on plant physiology and agriculture.
Answer:
Almost all plant tissues produce ethylene gas in minute quantities.
1. Discovery:
In 1924, Denny found that ethylene stimulates the ripening of lemons. In 1934, R. Gane found that ripe bananas contain abundant ethylene. In 1935, Cocken et al., identified ethylene as a natural plant hormone.

2. Occurrence:
Maximum synthesis occurs during climacteric ripening of fruits and tissues undergoing senescence. It is formed in almost all plant parts like roots, leaves, flowers, fruits and seeds.

3. Transport in plants:
Ethylene can easily diffuse inside the plant through intercellular spaces.

4. Precursor:
It is a derivative of amino acid methionine, linolenic acid and fumaric acid.

5. Bioassay (Gas Chromatography):
Ethylene can be measured by gas chromatography. This technique helps in the detection of exact amount of ethylene from different plant tissues like lemon and orange.

6. Physiological Effects:

• Ethylene stimulates respiration and ripening in fruits.
• It stimulates radial growth in sterft and roof and inhibits linear growth.
• It breaks the dormancy of buds, seeds and storage organs.
• It stimulates the formation of an abscission zone in leaves, flowers and fruits. This makes the leaves to shed prematurely.
• Inhibition of stem elongation (shortening the internode).
• In low concentration, ethylene helps in root initiation.
• Growth of lateral roots and root hairs. This increases the absorption surface of the plant roots.
• The growth of fruits is stimulated by ethylene in some plants. It is more marked in climacteric fruits.
• Ethylene causes epinasty.

7. Agricultural role:

• Ethylene normally reduces flowering in plants except in Pineapple and Mango.
• It increases the number of female flowers and decreases the number of male flowers.
• Ethylene spray in cucumber crops produces female flowers and increases the yield.

Question 6.
Explain the physiological Effects of Cytokinins.
Answer:

• With IAA – Promotes cell division With IAA & GA – Induces cell enlargement
• Breaks dormancy of light-sensitive seeds (tobacco) induces seed germination.
• Promotes growth of lateral
• buds even in the presence of apical bud.
• Delays the process of aging by nutrient mobilization known as Richmond Lang effect.
• Induces rate of protein synthesis.
• Induces the formation of interfascicular cambium Overcomes apical dominance
• Induces the formation of new leaves chloroplast and lateral shoots.
• Induces Accumulation of solutes.

Question 7.
Write down the physiological effects of Ethylene
Answer:

• Stimulates respiration and thereby ripening of fruits
• Stimulates radial growth in stem and root and inhibits linear growth.
• breaks dormancy of
  1. buds
  2. Seeds
  3. Storage Organs
• Stimulates abscission 2 one formation in
  1. leaves
  2. flowers
  3. fruits (so leaves shed prematurely)
• Prevents stem elongation by preventing internodal growth
• Root growth in low concentration
• Stimulates growth of lateral roots and root hairs and increase the absorptive surface
• Ripening of fruits – Increases ripening in climacteric fruits (Mango, banana) etc.
• It causes epinasty

Question 8.
Describe the methods of breaking the dormancy of seeds in plants.
Answer:
The dormancy of seeds can be broken by different methods. These are:
1. Scarification:
Mechanical and chemical treatments like cutting or chipping of hard tough seed coat and use of organic solvents to remove waxy or fatty compounds are called Scarification.

2. impaction:
in some seeds, water and oxygen are unable to penetrate micropyle due to blockage by cork cells. These seeds are shaken vigorously to remove the plug which is called Impaction.

3. Stratification:
Seeds of rosaceous plants (Apple, Plum, Peach, and Cherry) will not germinate until they have been exposed to well aerated, moist conditions under low temperature (0°C to 10°C) for weeks to months. Such treatment is called Stratification.

4. Alternating temperatures: Germination of some seeds is strongly promoted by alternating daily temperatures. An alternation of low and high temperature improves the germination of seeds.

5. Light:
The dormancy of photoelastic seeds can be broken by exposing them to red light.

Question 9.
Define photoperiodism – Classify plants based on photoperiodism
Answer:
a. The physiological change on flowering due to the relative length of light and darkness is called photoperiodism.

• Gamer and Allard (1920) coined the term
• They studied photoperiodism in Biloxi variety of soybean (Glycine max) and Many land mammoth varieties of tobacco.

b. Depending on photoperiodic responses plants are classified into several types.
1. L.D. Plants (Long Day) The photoperiod required to induce flowering is called critical day length depending on critical day length if it is long (more than 12 hours) and with short nights. Eg. Pea Barley and Oats
Short LD Plants: These are Long day plants but need short day length during the early period of growth for flowering Eg. Wheat, Rye

2. SD Plants: Plants requiring short critical day length for flowering or a long night.
Eg. Tobacco, Cocklebur, Soya, Rice, and Chrysanthemum.
Long SD Plants: Actually SD plants but need long days during the early period of growth for flowering Eg. Some SPS of Bryophyllum & Night Jasmine.

3. Intermediate day plants:
These require a photoperiod between a long day and a short day for flowering Eg. Sugarcane and coleus.

4. Day Neutral plants:
There are a number of plants which can flower in all possible photoperiods, known as photo neutral or hiterterminate plants. Eg. Potato, Rhododendron, Tomato & Cotton.

Question 10.
Describe the role of phytochrome in inducing Flowering
Answer:
Definition:
It is a bluish biliprotein responsible for the perception of light in the photophysiological process, existing in two different forms is mainly involved in flower induction, (i.e) Pr and PFr.

• Butler et al(1959) named the pigment.
• It exists in two interconvertible forms

Mechanism:

Other functions:
Play a role in seed germination and changes in membrane conformation.

Question 11.
Write an Essay on Vernalization
Answer:
Definition:

• Many biennials and perennials are induced to flower by low-temperature exposure (O°c to 5°c) This process is called Vernalization.
• T.D. Lysenko – Coined the term.

Mechanism of Vernalization:
2 theories explain the mechanism of vernalization.

1. Hypothesis of Phasic development (T.D. Lysenko),
The development of the annual plant has 2 phases.

1. Thermostate-Vegetatine stage requiring low temperature and suitable moisture.
2. Photo stage -high temperature needs to synthesize florigen.

2. Hypothesis of hormonal involvement (Purvis 1961)

Vernalization has several steps

The technique of Vernalization:

• Seeds soaked in water
• Allowed to germinate at 10°C to 12°C
• Transferred to low temperature for few days to 30 days (3°C to 5°C).
• Germinated seeds after the low temp, treatment are allowed to dry & then sown.
• Quickened flowering than untreated (control seedling)

Devernalization:
The reversal of the effect of vernalization is called Devemalization.

• Practical Applications:
• Vernalization shortens the vegetative period and induces the plant to flower earlier
• It increases cold resistance
• It increases fungal resistance
• It accelerates Plant Breeding.

Question 12.
Define Senescence and give its types
Answer:
Definition:
Getting old or Ageing is call d senescence in plants.
It refers to all collective, progressive, and deteriorative processes which ultimately lead to complete loss of organization and function.
Types – 4 types (Leopold -1961)

1. Overall senescence: When the entire plant gets affected and dies – Eg. Annuals – Wheat & Soybeans, Perennials – Agave & Bamboo
2. Top senescence: Occur in aerial parts only Eg. Parrennials – Banana and Gladiolus
3. Deciduous senescence: Occur only in leaves Eg. Decidual plants – Elm and Maple
4. Progressive Senescence: Occur in Annuals occur in old leaves first followed by new leaves than stem and finally root system.

Question 13.
Explain the physiology of senescence and Factors affecting senescence Physiology of Senescence :
Answer:

• Change in the structure of cells
• Vacuoles act like lysosome-secrete hydrolytic enzymes.
• Reducation in photosynthetic rate (due to loss of chlorophyll & accumulation of anthocyanin)
• The decrease in Starch content,  Protein content
• Decrease in …………. r RNA level due to increased activity of enzyme RNA ase
• Degeneration of DNA – by increased activity of enzyme DNA ase

Factors affecting senescence :

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Bio Botany Guide Pdf Chapter 14 Respiration Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 14 Respiration

11th Bio Botany Guide Respiration Text Book Back Questions and Answers

Part-I.

Question 1.
The number of ATP molecules formed by complete oxidation of one molecule of pyruvic acid is:
(a) 12
(b) 13
(c) 14
(d) 15
Answer:
(a) 12

Question 2.
During oxidation of two molecules of cytosolic NADH + H⁺, number of ATP molecules produced in plants are
a) 3
b) 4
c) 6
d) 8
Answer:
b) 4

Question 3.
The compound which links glycolysis and Krebs cycle is:
(a) succinic acid
(b) pyruvic acid
(c) acetyl CoA
(d) citric acid
Answer:
(c) acetyl CoA

Question 4.
Assertion (A): Oxidative phosphorylation takes place during the electron transport chain in mitochondria.
Reason (R): Succinyl Co A is phosphorylated into succinic acid by substrate phosphorylation.
a) A and R is correct. R is correct explanation of A
b) A and R is correct but R is not the correct explanation of A,
c) A is correct but R is wrong
d) A and R is wrong
Answer:
c) A is correct but R is wrong

Question 5.
Which of the following reaction is not involved in Krebs cycle.
(a) Shifting of phosphate from 3C to 2C
(b) Splitting of Fructose 1,6 bisphosphate of into two molecules 3C compounds.
(c) Dephosphorylation from the substrates
(d) All of these
Answer:
(d) All of these

Question 6.
What are enzymes involved in phosphorylation and dephosphorylation reactions in EMP pathway?
Answer:
(i) Enzymes involved in phosphorylation are
a) Hexokinase and phospnofructio kinase.
(ii) Enzymes involved in dephosphorylation are
a) Phosphoglycerate Kinase
b) Pyruvate Kinase

Question 7.
The respiratory quotient is zero in succulent plants. Why?
Answer:
In some succulent plants like Opuntia, Bryophyllum carbohydrates are partially oxidised to organic acid, particularly malic acid without the corresponding release of CO₂ but O₂ is consumed hence the RQ value will be zero.

Question 8.
Explain the reactions taking place in mitochondrial inner membrane.
Answer:
Electron and hydrogen (proton) transport takes place across four multiprotein complexes (I-IV). They are.
1. Complex-I (NADH dehydrogenase).
It contains a flavoprotein (FMN) and associated with non-heme iron Sulphur protein (Fe-S). This complex is responsible for passing electrons and protons from mitochondrial NADFI (Internal) to Ubiquinone (UQ)
NADH+H⁺UQ ⇌ NAD^–+UQH₂

2. In plants, an additional NADH dehydrogenase (External) complex is present on the outer surface of inner membrane of mitochondria which can oxidise cytosolic NADH + H⁺.
Ubiquinone (UQ) or Coenzyme Quinone (CoQ) is a small, lipid-soluble electron, proton carrier located within the inner membrane of mitochondria).

3. Complex-II (succinic dehydrogenase) It contains FAD flavoprotein is associated with non-heme iron Sulphur (Fe-S) protein. This complex receives electrons and protons from succinate in Kerbs cycle and is converted into fumarate and passes to ubiquinone.
Succinate + UQ Fumaraic LQH₂

4. Complex-III (Cytochrome bcj complex) This complex oxidises reduced ubiquinone (ubiquinol) and transfers the electrons through Cytochrome bc1 Complex (Iron Sulphur centci bcl complex) to cytochrome c.

5. Complex IV (Cytochrome c oxidase) Complex IV is the terminal oxidase and brings about the reduction of 1/2 O₂ to H₂O. TWO protons are needed to form a molecule of H₂O (terminal oxidation).

Question 9.
What is the name of alternate way to glucose breakdown? Explain the process in involved in it?
Answer:

• Pentose phosphate pathway is the alternate pathway for breakdown of glucose.
• Pentose phosphate pathway was described by Warburg, Dickens and Lipmami (1938).
• It is also known as Hexose monophosphate shunt (HMP shunt) or Direct oxidative phase and non – oxidative phase.
• The oxidative phase convert six molecules of six carbon Glucose 6 phosphate to 6 molecules of five-carbon sugar Ribulose – 5 Phosphate with loss of 6CO₂ and generation of 12 NADPH + H⁺

Non oxidative pathway convert Ribulose – 5 – phosphate molecules to various intermediates such as
Ribose – 5 – phosphate (5C)
Xylulose – 5 – phosphate (5C)
Glyceraldehyde – 3 – phosphate (3C)
Sedoheptulose – 7 – phosphate (7C) and
Erythrose – 4 – phosphate (4C)

Finally five molecules of glucose 6 – phosphate is regenerated
6 x Glucose – 6 – phosphate + 12NADP⁺ + 6H₂O
↓
5 x glucose – 6 – phosphate + 6CO₂ + Pi + 12 NADPH + 2H⁺
The net result of complete oxidation of one glucose – 6 – phosphate yield 6CO₂ and 12 NADPH + H⁺. The oxidative pentose phosphate pathway is controlled by glucose – 6 – phosphate dehydrogenase enzyme which is inhibited by high ratio of NADPH to NADP^+.

Question 10.
How will you calculate net products of one sucrose molecule upon complete oxidation during aerobic respiration as per recent view?
Answer:
When the cost of transport of ATPs from the matrix into the cytosol is considered, the number will be 2.5 ATPs for each NADH + H⁺ and 1.5 ATPs for each FADH₂ oxidized during the electron transport system. Therefore, in plant cells net yield of 30 ATP molecules for complete aerobic oxidation of one molecule of glucose. But in those animal cells (showing malate shuttle mechanism) net yield will be 32 ATP molecules. Since the sucrose molecule gives, two molecules of glucose and net ATP in plant cell will be 30 × 2 = 60.
In an animal cell, it will be 32 × 2 = 64.

Part-II.

11th Bio Botany Guide Respiration Additional Important Questions and Answers

I. Choose The Correct Answer

Question 1.
The term respiration was coined by:
(a) Lamark
(b) Kerb
(c) Pepys
(d) Blackman
Answer:
(c) Pepys

Question 2.
Black man divided respiration into floating respiration and protoplasmic respiration based on respiratory ………..
a) Quotient
b) respiratory reaction
c) respiratory pathway
d) substrate
Answer:
d) substrate

Question 3.
The discovery of ATP was made by:
(a) Lipman
(b) Hans Adolt
(c) Warburg
(d) Karl Lohman
Answer:
(d) Karl Lohman

Question 4.
The type of respiration which is rare and liberates toxic ammonia
a) Protoplasmic respiration
b) floating respiration
c) Aerobic respiration
d) Anaerobic respiration
Answer:
a) Protoplasmic respiration

Question 5.
On hydrolysis, one molecule of ATP releases energy of:
(a) 8.2 Kcal
(b) 32.3 kJ
(c) 7.3 Kcal
(d) 7.8 Kcal
Answer:
(c) 7.3 Kcal

Question 6.
To convert Kcal to KJ multiply by 4.18(100 Kcal=418 KJ) calculate the amount KJ energy for 7.3 Kcal
a) 30.6 KJ
b) 32.06 KJ
c) 29.03 KJ
d) 5.01 KJ
Answer:
a) 30.6 KJ

Question 7.
Identify the link reaction:
(a) conversion of glucose into pyruvic acid
(b) conversion of glucose into ethanol
(c) conversion of acetyl CoA into CO₂ and water
(d) conversion of pyruvic acid into acetyl coenzyme – A
Answer:
(d) conversion of pyruvic acid into acetyl coenzyme – A

Question 8.
Food materials like carbohydrate, fat and proteins are completely oxidised into CO₂, H₂O and energy in ………………. respiration
a) anaerobic
b) aerobic
c) Bacterial respiration
d) Facultative
Answer:
b) aerobic

Question 9.
Kreb’s cycle is a:
(a) catabolic pathway
(b) anabolic pathway
(c) amphibolic pathway
(d) hydrolytic pathway
Answer:
(c) amphibolic pathway

Question 10.
Which process is occur in yeast and some bacteria and 2 ATP molecules are produced during this process.
a) Anaerobic respiration
b) aerobic respiration
c) mixed fermentation
d) CAC cycle
Answer:
a) Anaerobic respiration

Question 11.
The oxidation of one molecule of NADH + H⁺ gives rise to:
(a) 2 ATP
(b) 3 ATP
(c) 4 ATP
(d) 2.5 ATP
Answer:
(b) 3 ATP

Question 12.
Net product in Glycolysis are
a) 4ATP and 2NADH + H⁺
b) 2 ATP and 2NADH + H⁺
c) 6ATP
d) 24 ATP
Answer:
b) 2ATP and 2NADH + H⁺

Question 13.
Cyanide acts as electron transport chain inhibitor by preventing:
(a) synthesis of ATP from ADP
(b) flow of electrons from NADH + H⁺
(c) flow of electrons from cytochrome a₃ to O₂
(d) oxidative phosphorylation
Answer:
(c) flow of electrons from cytochrome a₃ to O₂

Question 14.
Which is the raw material for the formation of chlorophylls, cytochrome, phytochrome and pyrrole substance
a) acetyl COA
b) Pyruvic acid
c) Malic acid
d) Succinyl COA
Answer:
d) Succinyl COA

Question 15.
End products of fermentation in yeast is:
(a) pyruvic acid and CO₂
(b) lactic acid and CO₂
(c) ethyl alcohol and CO₂
(d) mixed acid and CO₂
Answer:
(c) ethyl alcohol and CO₂

Question 16.
Which cycle is considered as amphibolic pathway.
a) Calvin cycle
b) Glycolysic
c) ETS chain
d) Kreb cycle
Answer:
d) Kreb cycle

Question 17.
The external factors that affect respiration are:
(a) temperature, insufficient O₂ and amount of protoplasm
(b) temperature, insufficient O₂ and high concentration of CO₂
(c) temperature, high concentration of CO₂ and respiratory substrate
(d) temperature, high concentration of CO₂ and amount of protoplasm
Answer:
(b) temperature, insufficient O_2, and high concentration of CO₂

Question 18.
The complex system responsible for passing electrons and protons from mitochondria to ubiquinone is ………………..
a) Complex I
b) Complex II
c) Complex III
d) Complex IV
Answer:
a) Complex I

Question 19.
The oxidative pentose phosphate pathway is controlled by the enzyme:
(a) glucose, 1, 6 diphosphate dehydrogenase
(b) glucose 6 phosphate dehydrogenase
(c) fructose – 6 – phosphate dehydrogenase
(d) none of the above
Answer:
(b) glucose 6 phosphate dehydrogenase

Question 20.
Which are the high energy phosphate groups in ATP
a) adenine
b) Pentose sugar
c) Last two phosphate group
d) First two phosphate group
Answer:
c) Last two phosphate group

Question 21.
In-plant tissue erythrose is used for the synthesis of:
(a) Erythromycin
(b) Xanthophyll
(c) Erythrocin
(d) Anthocyanin
Answer:
(d) Anthocyanin

Question 22.
How many ATP molecules are produced when a molecule of glucose undergo fermentation?
a) TwoATPs
b) Six ATPs
c) Eight ATPs
d) one ATP
Answer:
a) Two ATPs

Question 23.
Identify the electron transport inhibitor:
(a) phosphoenol
(b) dinitrophenol
(c) xylene
(d) indol acetic acid
Answer:
(b) dinitrophenol

Question 24.
Enzymatic reaction for partial oxidation of glucose in the absence of oxygen is present in
a) Some Bacteria
b) Yeast fungus
c) A and B
d) Bryophytes
Answer:
c) A and B

Question 25.
Cyanide resistant respiration is known to generate heat in thermogenic tissues as high as:
(a) 35° C
(b) 38° C
(c) 40° C
(d) 51° C
Answer:
(d) 51° C

Question 26.
Match the Column I with the enzyme responsible for its production in column II
Answer:

Answer:
a) A-4,B -1,C-2,D-3.

Question 27.
Which one is wrongly matched

Answer:
B. Glycolysis – Twenty four ATP

II. 2 Marks Questions

Question 1.
Define respiration?
Answer:
Respiration is a biological process in which oxidation of various food substances like carbohydrates, proteins, and fats take place and as a result of this, energy is produced where O₂ is taken in and CO₂ is liberated.

Question 2.
Name some High energy compounds present in a cell
Answer:

• ATP → Adenosine Tri Phosphate
• GTP → Guanosine Tri Phosphate
• UTP → Uridine Tri Phosphate

Question 3.
What do you understand by compensation of point?
Answer:
The point at which CO₂ released in respiration is exactly compensated by CO₂ fixed in photosynthesis that means no net gaseous exchange takes place, it is called the compensation point.

Question 4.
What is Anaerobic respiration? What are its steps?
Answer:

• In the absence of molecular oxygen-glucose is incompletely degraded into either ethyl alcohol (or) Lactic acid.
• It includes two steps (i) Glycolysis (ii) Fermentation

Question 5.
What is anaerobic respiration?
Answer:
In the absence of molecular oxygen-glucose is incompletely degraded into either ethyl alcohol or lactic acid. It includes two steps:

1. Glycolysis
2. Fermentation

Question 6.
What is Link reaction?
Answer:
In aerobic respiration, Conversion of Pyruvic acid into acetyl coenzyme – A in the mitochondrial matrix with two molecules of NADH + H⁺ and 2 CO₂. This is called Link reaction (or) transition reaction.

Question 7.
Who is Sir Hans Adolf Krebs?
Answer:
Sir Hans Adolf Krebs was born in Germany on 25th August 1900. He was awarded Nobel Prize for his discovery of Citric acid cycle in Physiology in 1953.

Question 8.
Why Kreb cycle is called as citric acid cycle (or) Tri Carboxylic acid Cycle?
Answer:

• TCA cycle starts with condensation of acetyl COA with oxaloacetate in the presence of water to yield Citri acid (or) Citrate.
• So it is also known as citric acid cycle (or) Tricarboxylic acid cycle.

Question 9.
Mention the role of NADH dehydrogenase enzyme in the electron transport system.
Answer:
NADH dehydrogenase contains a flavoprotein (FMN) and associated with non – heme iron Sulphur protein (Fe – S). This complex is responsible for passing electrons and protons from mitochondrial NADH (Internal) to Ubiquinone (UQ).

Question 10.
Which cycle is amphibolic pathway? Why? The Krebs cycle is called an amphibolic pathway.
Answer:

• Kreb cycle is primarily a catabolic pathway Later it is an anabolic pathway too.
• Hence it is called amphibolic pathway.

Question 11.
Mention any two electron transport chain inhibitors.
Answer:
Two electron transport chain inhibitors:

1. 2, 4 DNP (Dinitrophenol) – It prevents synthesis of ATP from ADP, as it directs electrons from Co Q to O₂.
2. Cyanide – It prevents flow of electrons from Cytochrome a₃ to O₂.

Question 12.
How many ATP molecules are produced in aerobic respiration present in plants?
Answer:
In aerobic respiration net gain of 36 ATP molecules produced in complete oxidation of glucose.

Question 13.
What are the significances of Respiratory Quotient?
Answer:
The significances of Respiratory Quotient:

1. RQ value indicates which type of respiration occurs in living cells, either aerobic or anaerobic.
2. It also helps to know which type of respiratory substrate is involved.

Question 14.
Who was awarded the Nobel Prize for coupling of oxidation and phosphorylation in mitochondria?
Answer:
Peter Mitchell, a British Biochemist received Nobel Prize for chemistry in 1978.

Question 15.
Mention any two industrial uses of alcoholic fermentation.
Answer:
Two industrial uses of alcoholic fermentation:

1. In bakeries, it is used for preparing bread, cakes, biscuits.
2. In beverage industries for preparing wine and alcoholic drinks.

Question 16.
Define mixed acid fermentation.
Answer:

• Formation of Lactic acid, ethanol, formic acid and gases like CO₂ and H₂ from pyruvic acid.
• eg. Enterobacteriaceae.

Question 17.
Mention any two internal factors, that affect the rate of respiration in plants.
Answer:
Two internal factors, that affect the rate of respiration in plants:

1. The amount of protoplasm and its state of activity influence the rate of respiration.
2. The concentration of respiratory substrate is proportional to the rate of respiration.

Question 18.
Why microorganisms respire an anaerobically?
Answer:

• Bacteria are prokaryotes and they are devoid of membrane-bound organelle mitochondria.
• So they are respire anaerobically.

Question 19.
Write down any two significance of the pentose phosphate pathway.
Answer:
Two significance of pentose phosphate pathway:

1. HMP shunt is associated with the generation of two important products.
2. Coenzyme NADPH generated is used for reductive biosynthesis and counter damaging the effects of oxygen-free radicals.

Question 20.
Complete the following Picture.

Answer:
A. Compensation point
B. Rate of Respiration

Question 21.
Write the missing A and B.

Answer:
A. Ribose
B. Adenine

III. 3 Mark Questions

Question 1.
In the biosphere how do plants and animals are complementary systems, which are integrated to sustain life?
Answer:
In plants, oxygen enters through the stomata and it is transported to cells, where oxygen is utilized for energy production. Plants require carbon dioxide to survive, to produce carbohydrates, and to release oxygen through photosynthesis, these oxygen molecules are inhaled by humans through the nose, which reaches the lungs where oxygen is transported through the blood and it reaches cells. Cellular respiration takes place inside or the cell for obtaining energy.

Question 2.
What is Respiration?
Answer:

• Breaking of C-C bonds of complex organic compounds through oxidation within the cells.
• The energy released during respiration is stored in the form of ATP and heat is liberated.
• It occurs in all the living cells of organisms.

Question 3.
What are the factors associated with the compensation point in respiration?
Answer:
The two common factors associated with compensation points are CO₂ and light. Based on this there are two types of compensation points. They are the CO₂ compensation point and light compensation point. C₃ plants have compensation points ranging from 40 – 60 ppm (parts per million) CO₂ while those of C₄ plants range from 1 – 5 ppm CO₂.

Question 4.
Differentiate floating respiration and protoplasmic respiration.
Answer:

Question 5.
What is a redox reaction?
Answer:
NAD⁺ + 2e^– + 2H⁺ → NADH + H⁺
FAD + 2e^– + 2H⁺ → FADH₂
When NAD⁺ (Nicotinamide Adenine Dinucleotide – oxidized form) and FAD (Flavin Adenine Dinucleotide) pick up electrons and one or two hydrogen ions (protons), they get reduced to NADH + H⁺ and FADH₂ respectively. When they drop electrons and hydrogen off they go back to their original form. The reaction in which NAD⁺ and FAD gain (reduction) or lose (oxidation) electrons are called redox reactions (Oxidation-reduction reactions). These reactions are important in cellular respiration.

Question 6.
Define ETS (or) Electron transport chain (or) What is the importance of ETS and oxidative Phosphorylation in respiration.
Answer:

• Electron transport chain and oxidative phosphorylation remove hydrogen atoms from the products of glycolysis, link reaction, and Kreb cycle.
• It releases water molecule with energy in the form of ATP molecules in the mitochondrial inner membrane.

Question 7.
Mention the significance of Kreb’s cycle.
Answer:
The significance of Kreb’s cycle:

1. TCA cycle is to provide energy in the form of ATP for metabolism in plants.
2. It provides carbon skeleton or raw material for various anabolic processes.
3. Many intermediates of the TCA cycle are further metabolised to produce amino acids, proteins, and nucleic acids.
4. Succinyl CoA is the raw material for the formation of chlorophylls, cytochrome, phytochrome, and other pyrrole substances.
5. α – ketoglutarate and oxaloacetate undergo reductive amination and produce amino acids.
6. It acts as a metabolic sink which plays a central role in intermediary metabolism.

Question 8.
Write the differences between ubiquinone and Cytochrome C.
Answer:

Question 9.
Write down the characteristic of Anaerobic respiration.
Answer:
The characteristic of Anaerobic respiration:

1. Anaerobic respiration is less efficient than aerobic respiration.
2. A limited number of ATP molecules is generated per glucose molecule.
3. It is characterized by the production of CO₂ and is used for Carbon fixation in photosynthesis.

Question 10.
RQ will be less than one in Red colour Parts Present in Plants? Why?
Answer:

• Red colour parts present in plants is due to the presence of anthocyanin
• Synthesis of anthocyanin require more O₂ than CO₂ evolved.
• So RQ will be less than one.

Question 11.
Write down any three external factors, that affect respiration in plants.
Answer:
Three external factors, that affect respiration in plants:

1. The optimum temperature for respiration is 30°C. At low temperatures and very high temperatures rate of respiration decreases.
2. When a sufficient amount of O₂ is available the rate of aerobic respiration will be optimum and anaerobic respiration is completely stopped. This is called the Extinction point.
3. The high concentration of CO₂ reduces the rate of respiration.

Question 12.
Define Lactic acid fermentation.
Answer:
Formation of Lactic acid from pyruvic acid is Lactic acid fermentation.
Eg. Bacillus bacteria, fungi, muscles of vertebrates.

Question 13.
What is the Pentose phosphate pathway?
Answer:

• It is an alternate pathway for break down of glucose.
• It takes place in the cytoplasm of mature plant cells.
• In this pathway glucose 6 phosphate molecule is converted to Ribulose 5 phosphate with CO₂ and NADPH^– + H⁺.

Question 14.
How alcoholic beverages like beer and wine is made?
Answer:

• The conversion of pyruvate to ethanol takes place in malted barley and grapes through fermentation.
• Yeast Carryout this process under anaerobic conditions and this conversion increases ethanol concentration.
• If the concentration increases It’s toxic effect kills yeast cells and the left out is called beer and wine respectively.

IV. 5 Mark Questions

Question 1.
Ambulate the differences between aerobic and anaerobic respiration.
Answer:

Question 2.
Draw the flow chart diagram For Glycolysis (or) EMP pathway.
Answer:

Question 3.
Explain the Pay – off phase of EMP Pathway of Glycolysis (or) Explain the oxidative phase of Glycolysis (or) Triose phase of Glycolysis.
Answer:

• Two molecules of glyceraldehyde 3 – phosphate oxidatively phosphorylated into two molecules of 1-3
  bisphospho glycerate.
• During this reaction 2 NAD⁺ is reduced to 2NADH+ H⁺ by glyceraldehyde 3-phosphate dehydrogenase.
• Further reactions are carried out by different enzymes at the end two molecules of pyruvate are produced.
• In this phase 4 ATPS are produced (at step 7 and step 10)
• Through Direct transfer of phosphate from substrate molecule to ADP and is converted into ATP is called substrate Phosphotylation. (or) Direct Phosphorylation (or) transphosphorylation.
• During the reaction (at step 9)2 phospo glycerate dehydrated into phosphoenol pyurvate, a water molecule is removed by the enzyme enolase.
• As a result enol group is formed within the molecule. This process is called Enolation.

Energy Budge of pay off phase:

• In the payoff phase totally 4 ATP and 2NADH + H⁺ molecules are produced.
• Since 2 ATP molecules are already consumed in the preparatory phase the net products in glycolysis are 2ATP and 2NADH + H⁺

Question 4.
Explain the preparatory phase of Glycolysis (or) EMP pathway (or) Describe the energonic phase phase of Glycolysis (or) EMP pathway. Describe the hexose phase of Glycolysis (or) EMP pathway.
Answer:
Glycolysis is a linear series of reactions in which 6- carbon glucose split into two molecules of 3 carbon pyruvic acid.
Preparatory phase:

• Glucose enters glycolysis which is the end product of photosynthesis.
• Glucose is phosphorylated into glucose 6 phosphate by the enzyme hexokinase and subsequent reactions are carried out by different enzymes.
• At the end of this phase fructose 1,6 – bisphote is cleaved into glyceraldehyde 3- phosphate and dihydroxyacetone phosphate by the enzyme aldolase.
• These two are Isomers.
• Dihydroxyacetone phosphate is isomerised into glyceraldehyde 3- phosphate by the enzyme triose phosphate isomerase.
• Now two molecules of glyceraldehyde 3 phosphate enter into pay off phase.

During the preparatory phase, two ATP molecules are àonsumed.

Question 5.
Explain pyruvate oxidation (or) Link reaction of Glycolysis.
Answer:

• Two molecules of pyruvate formed by glycolysis in the cytosol enter into mitochondnalrnatrxi.
• In aerobic respiration this pyruvate with coenzyme A is oxidatively decarboxylated into acetyl CoA by pyruvate dehydrogenase complex. .
• It produces two molecules of NADH + H⁺ and 2CO₂

It is also called transition reaction (or) Link reaction.

The pyruvate dehydrogenase complex consists of three distinct enzymes.

1. Pyruvate dehydrogenase
2. Dihydroiipoyil transacetylase
3. Dihydrolipoyil dehydrogenase and 5 coenzymes TPP (thymine pyrophosphate)
   • NAD⁺
   • FAD
   • COA and lipoate.

Question 6.
Draw the flow chart diagram for the Kreb cycle (or) Citric acid cycle.
Answer:

Question 7.
Explain Kreb cycle (or) Citric acid cycle (or) TCA cycle.
Answer:

• Two molecules of acetyl CoA formed from link reaction now enter into Kreb Cycle.
• It is named after its discoverer German Biochemist Sir Hans Adolf Kreb (1937).
• It is takes place in the mitochondrial matrix and inner membrane of mitochondria.
• The enzymes needed for TCA cycle are found in the mitochondrial matrix except for succinate dehydrogenase which is found in the mitochondrial inner membrane.
• First step starts with condensation of acetyl CoA with oxaloacetate in the presence of water to yield citric acid (or) citrate.
• It is followed by the action of different enzymes in cyclic manner.
• During the conversion of succinyl CoA to succinate by the enzyme succinyl CoA synthetase a molecule of ATP Synthesis from Substrate without entering the electron transport chain is called substrate-level phosphorylation.
• Kreb Cycle is repeated twice for every glucose molecule.
• Where two molecules of pyruvic acid produces six molecules of CO₂, eight molecules of NADH+H⁺ two molecules of FADH₂ and two molecules of ATP.

Question 8.
Significance of Kreb Cycle.
Answer:

• TCA cycle is to provide energy in the form of ATP for metabolism in plants.
• It provides carbon skeleton or raw material for various anabolic process.
• many intermediates of TCA cycle are further metabolised to produce amino acids, proteins and nucleic acids.
• Succinyl CoA is raw material for formation of chlorophyll, cytochrome, phytochrome and other pyrroles
  substances.
• α – ketoglutarate and oxaloacetate undergo reductive amination and produce amino acids.
• it acts as metabolic sink which plays a central role in intermediary metabolism.

Question 9.
Write four Electron transport chain in hibitors.
Answer:

• 2,4 DNP (Dinitrophenol) – It prevents the synthesis of ATP from ADP, as it directs electrons from CoQ to O₂
• Cyanide – It prevents the flow of electrons from Cytochrome a3 to O₂
• Rotenone – It prevents flow of electrons from NADH + H⁺ / FADH₂ to Co Q
• Oligomycin – It inhibits oxidative phosphorylation

Question 10.
Tabulate Net Products of ATP gained during aerobic respiration per glucose molecule.
Answer:

Question 11.
Experiment to demonstrate the production of CO₂ in aerobic respiration.
Answer:

• Take small quantity of any seed (groundnut or bean seeds) and allow them to germinate by imbibing them.
• While they are germinating place them in a conical flask.
• A small glass tube containing 4 ml of freshly prepared Potassium hydroxide (KOH) solution is hing into the conical flask with the help of a thread and tightly close the one holed cork.
• Take a bent glass tube, the shorted end of which is inserted into the conical flask through the hole in the cork.
• The longer end is dipped in a beaker containing water.
• Observe the position of initial water level in bent glass tube.
• This experimental setup is kept for two hours.
• After two hours, the level of water rises in the glass tube. It is because the CO₂ evolved during aerobic
  respiration by germinating seeds will be absorbed by KOH solution and the level of water will rise in the glass tube.
• CO₂ + 2KOH → K₂CO3 + H₂O

Question 12.
Compare Alcoholic fermentation.
Answer:

Question 13.
Write the Industrial uses of alcoholic fermentation.
Answer:

• In bakeries, it is used for preparing bread, cakes, biscuits.
• In beverage industries for preparing wine and alcoholic drinks.
• In producing vinegar and in tanning, curing of leather.
• Ethanol is used to make gasohol (a fuel that is used for cars in Brazil).

Question 14.
Tabulate the comparison between glycolysis and fermentation.
Answer:

Question 15.
Explain the demonstration of alcoholic fermentation.
Answer:

• Take a Kuhne’s fermentation tube which consists of an upright glass tube with a side bulb
• Pour 10% sugar solution mixed with baker’s yeast into the fermentation tube the side tube is filled plug the mouth with lid.
  After some time, the glucose solution will be fermented. The solution will give out an alcoholic smell.
• The level of the solution in the glass column will fall due to the accumulation of CO₂ gas.
• It is due to the presence of zymase enzyme  yeast which converts the glucose solution into alcohol and CO₂
• Now introduce a pellet of KOH into the tube, the KOH will absorb CO₂ and the level of solution will rise in the upright tube.
• This experiment proves during fermentation CO₂ gas is evolved.

Question 16.
Write the Factors (Internal and External) which affect the process of respiration.
Answer:
External Factors:

• The optimum temperature for respiration is 30°C. At low temperatures and very high temperatures rate, respiration decreases.
• When sufficient amount of O₂ is available the rate of aerobic respiration will be optimum and anaerobic respiration  is completely stopped. This is called Extinction point.
• The high concentration of CO₂ reduces the rate of respiration.
• A plant or tissue transferred from water to salt solution wi li increase the rate of respiration. It is called silt respiration.
• Light is an indirect factor affecting the rate of respiration.
• Wounding of plant organs stimulates the rate of respiration in that region.

Internal Factors:

• The concentration of respiratory substrate is proportional to the rate of respiration
• The amount of protoplasm and its state of activity influence the rate of respiration.

Question 17.
Write about the alternate pathway for glucose break down (or) Write about pentose phosphate pathway. (or) Phosphogluconate pathway (or) War burg – Dickens Lipmann pathway (or)Hexose Monophosphate pathway (or) HMP Shunt pathway.
Answer:

• The pentose phosphate pathway was described by Warburg, Dickens, and Lipmann (1938). Hence, it is also called Warburg – Dickens Lipmann pathway.
• It takes place in the cytoplasm of mature plant cells. It is an alternate way for break4own of glucose.
• It consists of two phases, oxidative phase, and non-oxidative phase.
• The oxidative events concert six molecules of six carbon Glucose 6 phosphate to 6 molecules of five-carbon sugar Ribulose -5 phosphate with loss of 12 NADPH + H⁺ (not NADH).
• The remaining reactions known as non oxidative pathway, covert Rihulose 5phosphate molecules to various intermediates such as Ribose – 5 – phosphate (5C), Xylulose – 5 – phosphate (5C), Glyceraldehyde – 7 – Phosphate (7C), and Eiythrose -4- phosphate (4C).
• Finally, five molecules of glucose -6- phosphate is regenerated. The overall reaction is:
  6 x Glucose – 6 – Phosphate + 12NADP⁺ + 6H₂O
  ↓
  5 x Glucose-6- Phosphate + 6CO₂ + Pi + 12NADPH + 12H⁺
  
• The net result of complete oxidation of one glucose-6-phosphate yield 6CO₂ and12NADPH+H⁺

Question 18.
Draw the cycle for pentsoe phosphate pathway (or) Draw the flow chart for HMP Shunt.
Answer:

Question 19.
Write about the Significance of Pentose Phosphate pathway.
Answer:

• HMP shunt is associated with the generation of two important products. NADPH and pentsoe sugars, which play a vital role in anaholic reactions.
• Coenzyme NADPH generated is used by reductive bisynthesìs and counter damaging the effects of oxygen-free radicals.
• Ribose – 5 – phosphate and its derivatives are used in the synthesis of DNA, RNA, ATP, NAD, FAD and
  Coenzynie A. .
• Erythrose is used for the synthesis of anthocyanin Jignin and other aromatic compounds.

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Computer Science Guide Pdf Chapter 14 Classes and Objects Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Computer Science Solutions Chapter 14 Classes and Objects

11th Computer Science Guide Classes and Objects Text Book Questions and Answers

Book Evaluation

Part I

Choose The Correct Answer
Question 1.
The variables declared inside the class are known as data members and the functions are known as
a) data functions
b) inline functions
c) member functions
d) attributes
Answer:
c) member functions

Question 2.
Which of the following statements about member functions are True or False?
i) A member function can call another member function directly with using the dot operator.
ii) Member function can access the private data of the class.
a) i-True, ii-True
b) i-False, ii-True
c) i-True, ii-False
d) i-False, ii-False
Answer:
b) i-False, ii-True

Question 3.
A member function can call another member function directly, without using the dot operator called as
a) sub function
b) sub member
c) nesting of member function
d) sibling of member function
Answer:
c) nesting of member function

Question 4.
The member function defined within the class behave like
a) inline functions
b) Non inline function
c) Outline function
d) Data function
Answer:
a) inline functions

Question 5.
Which of the following access specifier protects data from inadvertent modifications?
a) Private
b) Protected
c) Public
d) Global
Answer:
a) Private

Question 6.
class x
{
inty;
public:
x(int z)
{
y=z;
}
} x1[4];
intmain( )
{
x x2(10);
return 0;
}
How many objects are created for the above program?
a) 10
b) 14
c) 5
d) 2
Answer:
c) 5

Question 7.
State whether the following statements about the constructor are True or False.
i) constructors should be declared in the private section.
ii) constructors are invoked automatically when the objects are created.
a) True, True
b) True, False
c) False, True
d) False, False
Answer:
c) False, True

Question 8.
Which of the following constructor is executed for the following prototype ?
add display (add &); // add is a class name
a) Default constructor
b) Parameterized constructor
c) Copy constructor
d) Non Parameterized constructor
Answer:
c) Copy constructor

Question 9.
What happens when a class with parameterized constructors and having no default constructor is used in a program and we create an object that needs a zero- argument constructor?
a) Compile-time error
b) Domain error
c) Runtime error
d) Runtime exception
Answer:
a) Compile-time error

Question 10.
Which of the following create a temporary instance?
a) Implicit call to the constructor
b) Explicit call to the constructor
c) Implicit call to the destructor
d) Explicit call to the destructor
Answer:
b) Explicit call to the constructor

Part – II

Very Short Answers

Question 1.
What are called members?
Answer:
The class comprises members. Members are classified as Data Members and Member functions. Data members are the data variables that represent the features or properties of a class. Member functions are the functions that perform specific tasks in a class.

Question 2.
Differentiate structure and class though both are user-defined data types.
Answer:
The only difference between structure and class is the members of the structure are by default public whereas it is private in class.

Question 3.
What is the difference between the class and object in terms of oop?
Answer:
Object:

• Object is an instance of a class.
• Object is a real-world entity such as pen, laptop, mobile, chair, etc.
• Object allocates memory when it is created.

Class:

• Class is a blueprint or template from which objects are created.
• Class is a group of similar objects.
• Class doesn’t allocate memory when it is created.

Question 4.
Why it is considered a good practice to define a constructor though a compiler can automatically generate a constructor?
Answer:
A user-defined constructor is the best method of initialise array of objects and normal objects.

Question 5.
Write down the importance of the destructor.
Answer:
The purpose of the destructor is to free the resources that the object may have acquired during its lifetime. A destructor function removes the memory of an object which was allocated by the constructor at the time of creating an object.

Part – III

Short Answers

Question 1.
Rewrite the following program after removing the syntax errors if any and underline the errors:
#include<iostream>
#include<stdio.h>
classmystud
{ intstudid =1001;
char name[20];
public
mystud( )
{ }
void register ( ) {cin>>stdid;gets(name);
}
void display ( )
{ cout<<studid<<“: “<<name<<endl;}
}
int main( )
{ mystud MS;
register.MS( );
MS.display( );
}
Answer:
MODIFIED PROGRAM:
#include<iostream>
#include<stdio.h>
class mystud
{
int studid;
char name[20];
public:
mystud( )
{
studid=1001;
}
void register ( )
{
cin>>stdid;
gets(name);
}
void display ( )
{
cout<<studid<<“: “<<name<<endl;
}
};
int main( )
{
mystud MS;
MS.reqister( );
MS.display( );
}

Question 2.
Write with an example of how will you dynamically initialize objects?
Answer:
Dynamic initialization of Objects:
When the initial values are provided during runtime then it is called dynamic initialization.
Program to illustrate dynamic initialization
#include
using namespace std;
class X
{
int n;
float avg;
public:
X(int p,float q)
{
n=p;
avg=q;
}
void disp( )
{
cout<<“\n Roll numbe:-” <<n;
cout<<“\nAverage :-“<<avg;
}
};
int main( )
{
int a ; float b;
cout<<“\nEnter the Roll Number”;
cin>>a; .
cout<<“\nEnter the Average”;
cin>>b;
X x(a,b); // dynamic initialization
x.disp( );
return 0;
}
Output
Enter the Roll Number 1201
Enter the Average 98.6
Roll number:- 1201
Average :- 98.6

Question 3.
What are advantages of declaring constructors and destructor under public access ability?
Answer:

When constructor and destructor are declared under public:

1. we can initialize the object while declaring it.
2. we can explicitly call the constructor.
3. we can overload constructors and therefore use multiple constructors to initialize objects automatically.
4. we can destroy the objects at the end of class scope automatically (free unused memory).

However, some C++ compiler and Dev C++ do not allow to declare constructor and destructor under private section. So it is better to declare constructor and destructor under public section only.

Question 4.
Given the following C++ code, answer the questions (i) & (ii).
Answer:
class TestMeOut
{
public:
~TestMeOut( ) //Function 1
{
cout<<“Leaving the examination hall”<<endl;
}
TestMeOut( ) //Function 2
{
cout<<“Appearing for examination'<<endl;
}
void MyWork( ) //Function 3
{
cout<<“Attempting Questions//<<endl;
}
};
i) In Object-Oriented Programming, what is Function 1 referred to as and when does it get invoked/called?
Function 1 is called a destructor. It will be automatically invoked when the object goes out of scope (ie. at the end of a program).

ii) In Object-Oriented Programming, what is Function 2 referred to as, and when does it get invoked/called?
Function2 is called a constructor. It will be automatically invoked when an object comes into scope.(ie. at the time of object creation).

Question 5.
Write the output of the following C++ program code:
Answer:
#include<iostream>
using namespace std;
class Calci
{
char Grade; .
int Bonus;
public:
Calci( )
{
Grade=’E’;
Bonus=0;
}//ascii value of A=65
void Down(int G)
{
Grade-=G;
}
void Up(int G)
{
Grade+=G;
Bonus++;
}
void Show( )
{
cout<<Grade<<“#”<<Bonus<<endl;
}
};
int main( )
{
Calci c;
c.Down(3);
c.Show( );
c.Up(7);
c.Show( );
c.Down(2);
c.Show( );
return 0;
}
Output