Category: Class 12

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 7 Applications of Differential Calculus Ex 7.10 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.10

Choose the most suitable answer from the given four alternatives:

Question 1.
The volume of a sphere is increasing in volume at the rate of 3π cm³/ sec. The rate of change of its radius when radius is \(\frac { 1 }{ 2 }\) cm
(a) 3 cm/s
(b) 2 cm/s
(c) 1 cm/s
(d) \(\frac { 1 }{ 2 }\) cm/s
Solution:
(a) 3 cm/s
Hint:
Volume V = \(\frac { 4 }{ 3 }\) πr³
Given \(\frac { dV }{ dt }\) = 3π cm³/sec
Differentiating w.r.t. ‘t’ r = \(\frac { 1 }{ 2 }\) cm

Rate of change of radius is 3 cm/sec.

Question 2.
A balloon rises straight up at 10 m/s. An observer is 40 m away from the spot where the balloon left the ground. Find the rate of change of the balloon’s angle of elevation in radian per second when the balloon is 30 metres above the ground.
(a) \(\frac { 3 }{ 25 }\) radan/sec
(b) \(\frac { 4 }{ 25 }\) radian/sec
(c) \(\frac { 1 }{ 5 }\) radian/sec
(d) \(\frac { 1 }{ 3 }\) radian/sec
Solution:
(b) \(\frac { 4 }{ 25 }\) radian/sec
Hint:

Question 3.
The position of a particle moving along a horizontal line of any time t is given by s(t) = 3t² – 2t – 8. The time at which the particle is at rest is
(a) t = 0
(b) t = \(\frac { 1 }{ 3 }\)
(c) t = 1
(d) t = 3
Solution:
(b) t = \(\frac { 1 }{ 3 }\)
Hint:
s(t) = 3t² – 2t – 8
Velocity V = \(\frac { ds }{ dt }\) = 6t – 2
When the particle comes to rest,
velocity V = 0
6t – 2 = 0
t = \(\frac { 1 }{ 3 }\)

Question 4.
A stone is thrown, up vertically. The height reaches at time t seconds is given by x = 80t – 16t². The stone reaches the maximum! height in time t seconds is given by
(a) 2
(b) 2.5
(c) 3
(d) 3.5
Solution:
(b) 2.5
Hint:
x = 80t – 16t²
Velocity V = \(\frac { ds }{ dt }\) = 80 – 32t
When it reaches the maximum height
V = 0 ⇒ 80 – 32t = 0
t = \(\frac { 80 }{ 32 }\) = 2.5

Question 5.
Find the point on the curve 6y = x³ + 2 at which y-coordinate changes 8 times as fast as x-coordinate is
(a) (4, 11)
(b) (4, -11)
(c) (-4, 11)
(d) (-4, -11)
Solution:
(a) (4, 11)
Hint:

x = ±4
When x = 4, y = 11
∴ Point on the curve is (4, 11).

Question 6.
The abscissa of the point on the curve f(x) = \(\sqrt { 8-2x }\) at which the slope of the tangent is -0.25?
(a) -8
(b) -4
(c) -2
(d) 0
Solution:
(b) -4
Hint:
f(x) = \(\sqrt { 8-2x }\)
f'(x) = –\(\frac { 2 }{ 2\sqrt { 8-2x } }\) = –\(\frac { 1 }{ \sqrt { 8-2x } }\)
Slope of the tangent is – 0.25
ie„ f'(x) = -0.25
–\(\frac { 1 }{ \sqrt { 8-2x } }\) = -0.25 = \(\frac { -1 }{ 4 }\)
\(\sqrt { 8-2x }\) = 4
8 – 2x = 16
x = –\(\frac { 8 }{ 2 }\) = -4
Abscissa x = -4

Question 7.
The slope of the line normal to the curve f(x) = 2 cos 4x at x = \(\frac { π }{ 12 }\) is
(a) -4√3
(b) -4
(c) –\(\frac { √3 }{ 12 }\)
(d) 4√3
Solution:
(c) –\(\frac { √3 }{ 12 }\)
Hint:
f(x) = 2 cos 4x
f'(x) = -8 sin 4x
Slope of the normal at x = \(\frac { π }{ 12 }\) is

Question 8.
The tangent to the curve y² – xy + 9 = 0 is vertical when
(a) y = 0
(b) y = ±√3
(c) –\(\frac { 1 }{ 2 }\)
(d) y = ±3
Solution:
(d) y = ±3
Hint:
y² – xy + 9 = 0 ……… (1)
2y\(\frac { dy }{ dx }\) – (x\(\frac { dy }{ dx }\) + y) = 0
\(\frac { dy }{ dx }\) (2y – x) = y
\(\frac { dy }{ dx }\) = \(\frac { y }{ 2y-x }\)
When the tangent is vertical \(\frac { dy }{ dx }\) = ∞
i.e., \(\frac { y }{ 2y-x }\) = \(\frac { 0 }{ 1 }\)
⇒ 2y – x = 0
2y = x
sub in (1)
y² – 2y² + 9 = 0
⇒ y² = 9
y = ±3

Question 9.
The angle between y² = x and x² = y at the origin is
(a) tan^-1 \(\frac { 3 }{ 4 }\)
(b) tan^-1 (\(\frac { 4 }{ 3 }\))
(c) \(\frac { π }{ 2 }\)
(d) \(\frac { π }{ 4 }\)
Solution:
(c) \(\frac { π }{ 2 }\)
Hint:
y² = x and x² = y are the standard forms of parabolas for which y-axis and x-axis are the two tangents respectively.
Angle between x-axis and y-axis is \(\frac { π }{ 2 }\)

Question 10.
The value of the limit \(\lim _{x \rightarrow 0}\) (cot x – \(\frac { 1 }{ x }\)) is
(a) 0
(b) 1
(c) 2
(d) ∞
Solution:
(d) ∞
Hint:

Question 11.
The function sin⁴ x + cos⁴ x is increasing in the interval
(a) [ \(\frac { 5π }{ 8 }\), \(\frac { 3π }{ 4 }\) ]
(b) [ \(\frac { π }{ 2 }\), \(\frac { 5π }{ 8 }\) ]
(c) [ \(\frac { π }{ 4 }\), \(\frac { π }{ 2 }\) ]
(d) [ 0, \(\frac { π }{ 4 }\) ]
Solution:
(c) [ \(\frac { π }{ 4 }\), \(\frac { π }{ 2 }\) ]
Hint:
f(x) = sin⁴x + cos⁴x
f'(x) = 4 sin³ x cos x – 4 cos³ x sin x
f'(x) = 0 ⇒ 4 sin x cos x (sin²x – cos²x) = 0
sin x = 0; cos x = 0; sin² – cos² x = 0
x = 0; x = \(\frac { π }{ 2 }\); sin² x = cos² x
x = \(\frac { π }{ 4 }\)

In [0, \(\frac { π }{ 2 }\) ], f'(x) = -ve ⇒ f(x) is decreasing
In [ \(\frac { π }{ 2 }\), \(\frac { π }{ 2 }\) ], f'(x) = +ve ⇒ f(x) is increasing

Question 12.
The number is given by Rolle’s theorem for the function x³ – 3x², x ∈ [0, 3] is
(a) 1
(b) √2
(c) \(\frac { 3 }{ 2 }\)
(d) 2
Solution:
(d) 2
Hint:
f(x) = x³ – 3x²
f'(x) = 3x² – 6x
f'(x) = 0
⇒ 3x (x – 2) = 0
x = 0, 2

Question 13.
The number given by the Mean value theorem for the function \(\frac { 1 }{ x }\), x ∈ [1, 9] is
(a) 2
(b) 2.5
(c) 3
(d) 3.5
Solution:
(c) 3
Hint:

Question 14.
The minimum value of the function |3 – x| + 9 is
(a) 0
(b) 3
(c) 6
(d) 9
Solution:
(d) 9
Hint:
f(x) = |3 – x | + 9
Minimum value of |3 – x | = 0
Minimum value of |3 – x| + 9 = 0 + 9 = 9 and No Maximum value.

Question 15.
The maximum slope of the tangent to the curve y = e^x sin x, x ∈ [0, 2π] is at
(a) x = \(\frac { π }{ 4 }\)
(b) x = \(\frac { π }{ 2 }\)
(c) x = π
(d) x = \(\frac { 3π }{ 2 }\)
Solution:
(b) x = \(\frac { π }{ 2 }\)
Hint:
y = e^x sin x, x ∈ [0, 2π] dy
Slope ‘S’ = \(\frac { dy }{ dx }\) = e^x cos x + e^x sin x
S = e^x (cos x + sin x)
\(\frac { dS }{ dx }\) = e^x (-sin x + cos x) + (cos x + sin x)e^x
= e^x (2 cos x)
For maximum or minimum,
\(\frac { dS }{ dx }\) = 0 ⇒ 2e^x cos x = 0
e^x = 0 is not possible
∴ cos x = 0
x = \(\frac { π }{ 2 }\)

Question 16.
The maximum value of the function x² e^-2x, x > 0 is
(a) \(\frac { 1 }{ e }\)
(b) \(\frac { 1 }{ 2e }\)
(c) \(\frac { 1 }{ e^2 }\)
(d) \(\frac { 4 }{ e^4 }\)
Solution:
(c) \(\frac { 1 }{ e^2 }\)
Hint:
Let f(x) = x²e^-2x, x > 0
f'(x) = -2x² e^-2x + e^-2x (2x)
f'(x) = 0 ⇒ -2xe^-2x(x – 1) = 0
x = 0 and x = 1
f(x) attains maximum at x = 1 as f”(x) < 0
when x = 1
∴ Maximum value f(1) = (1)² e^-2 = \(\frac { 1 }{ e^2 }\)

Question 17.
One of the closest points on the curve x² – y² = 4 to the point (6, 0) is
(a) (2, 0)
(b) (√5, 1)
(c) (3, √5)
(d) (\(\sqrt { 13 }\), -√3)
Solution:
(c) (3, √5)
Hint:
x² – y² = 4
y² = x² – 4
y = ±\(\sqrt { x^2-4 }\)
Any point on the curve is (x, ± \(\sqrt { x^2-4 }\))
Distance between (6, 0) and (x, ± \(\sqrt { x^2-4 }\)) is \(\sqrt { (x-6)^2+x^2-4 }\)
Substituting all’ the given options, we get minimum distance.
∴ Required point is (3, √5)

Question 18.
The maximum value of the product of two positive numbers’, when their sum of the squares is 200, is
(a) 100
(b) 25√7
(c) 28
(d) 24\(\sqrt { 14 }\)
Solution:
(a) 100
Hint:
Given x² + y² = 200
y² = 200 – x²
y = \(\sqrt { 200-x^2 }\)
Product P = xy = x\(\sqrt { 200-x^2 }\)
\(\frac { dP }{ dx }\) = \(\frac { x(-2x) }{ 2\sqrt{200-x^2} }\) + \(\sqrt{200-x^2}\) …. (1)
\(\frac { dP }{ dx }\) = 0 ⇒ -2x² + 200 = 0
x² = 100
x = 10
∴ y = \(\sqrt{200-100}\) = 10
∴ Maximum product is P = xy
= (10) (10)
= 100

Question 19.
The curve y = ax⁴ + bx² with ab > 0
(a) has no horizontal tangent
(b) is concave up
(c) is concave down
(d) has no points of inflection
Solution:
(d) has no points of inflection
Hint:
y = ax⁴ + bx²
\(\frac { dy }{ dx }\) = 4ax³ + 2bx
\(\frac { d^2y }{ dx^2 }\) = 12ax² + 2b
\(\frac { d^2y }{ dx^2 }\) = 0 ⇒ 12ax² + 2b = 0
x² = –\(\frac { b }{ 6a }\)
x is unreal.
Hence no points of inflection.

Question 20.
The point of inflection of the curve y = (x – 1)³ is
(a) (0, 0)
(b) (0, 1)
(c) (1, 0)
(d) (1, 1)
Solution:
(c) (1, 0)
Hint:
y = (x – 1)³
\(\frac { dy }{ dx }\) = 3(x – 1)²
\(\frac { d^2y }{ dx^2 }\) = 6(x – 1)
\(\frac { d^2y }{ dx^2 }\) = 0 ⇒ 6(x – 1) = 0
x = 1
y = f(x)
\(\frac { dy }{ dx }\) = f'(x)
\(\frac { d^2y }{ dx^2 }\) = f”(x)
In (-∞, 1), f”(x) < 0, curve is concave down In (1, ∞), f”(x) > 0, curve is concave up
f”(x) changes its sign when passing through x = 1
when x = 1, y = 0
∴ (1, 0) is the point of inflection.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Bio Zoology Guide Pdf Chapter 6 Evolution Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Bio Zoology Solutions Chapter 6 Evolution

12th Bio Zoology Guide Evolution Text Book Back Questions and Answers

Question 1.
The first life on Earth originated …………………..
(a) in air
(b) on land
(c) in water
(d) on mountain
Answer:
(c) in water

Question 2.
Who published the book “Origin of species by Natural Selection” in 1859?
(a) Charles Darwin
(b) Lamarck
(c) Weismann
(d) Hugo de Vries
Answer:
(a) Charles Darwin

Question 3.
Which of the following was the contribution of Hugo de Vries?
(a) Theory of mutation
(b) Theory of natural Selection
(c) Theory of inheritance of acquired characters
(d) Germplasm theory
Answer:
(a) Theory of mutation

Question 4.
The wings of birds and butterflies is an example of………………………………..
(a) Adaptive radiation
(b) convergent evolution
(c) divergent evolution
(d) variation
Answer:
(b) Convergent evolutions

Question 5.
The phenomenon of “Industrial Melanism” demonstrates……………………….
(a) Natural selection
(b) induced mutation
(c) reproductive isolation
(d) geographical isolation
Answer:
(a) Natural selections

Question 6.
Darwin’s finches are an excellent example of……………………….
(a) connecting links
(b) seasonal migration
(c) adaptive radiation
(d) parasitism
Answer:
(c) Adaptive radiations

Question 7.
Who proposed the Germplasm theory?
(a) Darwin
(b) August Weismann
(c) Lamarck
(d) Alfred Wallace
Answer:
(b) August Weismann

Question 8.
The age of fossils can be determined by
(a) electron microscope
(b) weighing the fossils
(c) carbon dating
(d) analysis of bones
Answer:
(c) Carbon dating

Question 9.
Fossils are generally found in
(a) igneous rocks
(b) metamorphic rocks
(c) volcanic rocks
(d) sedimentary rocks
Answer:
(d) Sedimentary rocks

Question 10.
Evolutionary history of an organism is called……………………….
(a) ancestry
(b) ontogeny
(c) phylogeny
Answer:
(c) Phylogeny

Question 11.
The golden age of reptiles was……………………….
(a) Mesozoic era
(b) Cenozoic era
(c) Paleozoic era
(d) Proterozoic era
Answer:
(a) Mesozoic era

Question 12.
Which period was called “Age of fishes”?
(a) Permian
(b) Triassic
(c) Devonian
(d) Ordovician
Answer:
(c) Devonian

Question 13.
Modem man belongs to which period?
(a) Quaternary
(b) Cretaceous
(c) Silurian
(d) Cambrian
Answer:
(a) Quaternary

Question 14.
The Neanderthal man had the brain capacity of………………………..
(a) 650 – 800cc
(b) 1200cc
(c) 900cc
(d) 1400cc
Answer:
(d) 1400cc

Question 15.
List out the major gases seems to fie found in the primitive Earth.
Answer:
CO₂, NH₃, UV and Water vapour

Question 16.
Explain the three major categories in which fossilization occur.
Answer:
(i) Actual remains is the most common method of fossilization. When marine animals die, their hard parts such as bones and shells, etc. are covered with sediments and are protected from further deterioration. They get preserved as such as they are preserved in vast ocean the salinity in them prevents decay. The sediments become hardened to form definite layers or strata. For example, Woolly Mammoth that lived 22 thousand years ago were preserved in the frozen coast of Siberia as such. Several human beings and animals living in die ancient city of Pompeii were preserved intact by volcanic ash which gushed out from Mount Vesuvius.

(ii) Petrifaction – When animals die the original portion of their body may be replaced molecule for a molecule by minerals and the original substance being lost through disintegration. This method of fossilization is called petrifaction. The principal minerals involved in this type of fossilization are iron pyrites, silica, calcium carbonate, and bicarbonates of calcium and magnesium.

(iii) Natural moulds and casts – Even after disintegration, the body of an animal might leave indelible impression on the soft mud which later becomes hardened into stones. Such impressions are called moulds. The cavities of the moulds may get filled up by hard minerals and get fossilized, which are called casts. Hardened faecal matter termed as coprolites, occur as tiny pellets. Analysis of the coprolites enables us to understand the nature of diet, the prehistoric animals thrived.

Question 17.
Differentiate between divergent evolution and convergent evolution with one example for each.
Answer:
Divergent Evolution :
Divergent evolution is a result of homology. Erg: The wings of bird and the forelimbs of human both are homologous structures modified according to functions. In birds, it is used for flight and in humans used for writing and other purposes.

Convergent Evolution :
Convergent evolution is a result of analogy, E.g: Root modification in sweet potato, and stem modification in potato are analogous structures both performing same function i.e., storage,

Question 18.
How does Hardy-Weinberg’s expression (p² + 2pq + q² = 1) explain that genetic equilibrium is maintained in a population? List any four factors that can disturb the genetic equilibrium.
Answer:
The allele frequencies in a population are stable and are constant from generation to generation in the absence of gene flow, genetic drift, mutation, recombination and natural selection. If a population is in a state of Hardy Weinberg equilibrium, the frequencies of alleles and genotypes or sets of alleles in that population will remain same over generations. Evolution is a change in the allele frequencies in a population over time. Hence population in Hardy Weinberg is not evolving.

Suppose we have a large population of beetles, (infinitely large) and appear in two colours ’ dark grey (black) and light grey, and their colour is determined by ‘A’ gene. ‘AA’ and ‘Aa’ beetles are dark grey and ‘aa’ beetles are light grey. In a population let’s say that ‘ A’ allele has frequency (p) of 0.3 and ‘a’ allele has a frequency (q ) of 0.7. Then p+q= 1.

If a population is in Hardy Weinberg equilibrium, the genotype’frequencycan be estimated by Hardy Weinberg equation.

(p + q)² = p² + 2pq + q²
p² = frequency of AA
2pq= frequency of Aa
q² = frequency of aa
p = 0.3, q = 0.7 then,
p² = (0.3)2 = 0.09 = 9 %AA
2pq = 2(0.3) (0.7) = 0.42 = 42 % Aa
q² = (0.7)2 0.49 = 49 % aa’

Hence the beetle population appears to be in Hardy- Weinberg equilibrium. When the beetles in Hardy- Weinberg equilibrium reproduce the allele and genotype frequency in the next generation would be: Let’s assume that the frequency of ‘A’ and ‘a’ allele in the pool of gametes that make the next generation would be the same, then there would be no variation in the progeny. The genotype frequencies of the parent appears in the next generation.
(i.e. 9% AA, 42% Aa and 49% aa).

If we assume that the beetles mate randomly (selection of male gamete and female gamete in the pool of gametes), the probability of getting the offspring genotype depends on the genotype of the combining parental gametes.

Question 19.
Explain how mutations, natural selection and genetic drift affect Hardy Weinberg equilibrium.
Answer:
Natural selection occurs when one allele (or combination of alleles of differences) makes an organism more or less fit to survive and reproduce in a given environment. If an allele reduces fitness, its frequencies tend to drop from one generation to the next.

The evolutionary path of a given gene (i.e) how its allele’s change in frequency in the population across generation, may result from several evolutionary mechanisms acting at once. For example, one gene’s allele frequencies might be modified by both gene flow and genetic drift, for another gene, mutation may produce a new allele, that is favoured by natural selection.

Genetic drift / Sewall Wright Effect is a mechanism of evolution in which allele frequencies of a population change over generation due to chance (sampling error). Genetic drift occurs in all population sizes, but its effects are strong in a small population. It may result in a loss of some alleles (including beneficial ones) and fixation of other alleles. Genetic drift can have major effects, when the population is reduced in size by natural disaster due to bottle neck effect or when a small group of population splits from the main population to form a new colony due to founder’s effect.

Although mutation is the original source of all genetic variation, mutation rate for most organisms is low. Hence new mutations on allele frequencies from one generation to the next is usually not large.

Question 20.
How did Darwin explain fitness of organisms?
Answer:
Organisms struggle for food, space and mate. As these become a limiting factor, competition exists among the members of the population. Darwin denoted struggle for existence in three ways Intra specific struggle between the same species for food, space and mate.
Interspecific struggle with different species for food and space.

Struggle with the environment to cope with the climatic variations, flood, earthquakes and drought, etc.
According to Darwin, nature is the most powerful selective force. He compared origin of species by natural selection to a small isolated group. Darwin believed that the struggle for existence resulted in the survival of the fittest. Such organisms become better adapted to the changing environment.

Question 21.
Mention the main objections to Darwinism.
Answer:
Some objections raised against Darwinism were Darwin failed to explain the mechanism of variation.

1. Darwinism explains the survival of the fittest but not the arrival of the fittest.
2. He focused on small fluctuating variations that are mostly non-heritable.
3. He did not distinguish between somatic and germinal variations.
4. He could not explain the occurrence of vestigial organs, overspecialization of some organs like large tusks in extinct mammoths and oversized antlers in the extinct Irish deer, etc.

Question 22.
Taking the example of Peppered moth, explain the action of natural selection. What do you call the above phenomenon?
Answer:
Natural selection can be explained clearly through industrial melanism. Industrial melanism is a classical case of Natural selection exhibited by the peppered moth, Bistort betularia. These were available in two colours, white and black. Before industrialization peppered moth both white and black coloured were common in England. Pre-industrialization witnessed white colpured background of the wall of the buildings hence the white coloured moths escaped from their predators.

Post industrialization, the tree trunks became dark due to smoke and soot let out from the industries. The black moths camouflaged on the dark bark of the trees and the white moths were easily identified by their predators. Hence the dark coloured moth population was selected and their number increased when compared to the white moths. Nature offered positive selection pressure to the black coloured moths. The above proof shows that in a population, organisms that can adapt will survive and produce more progenies resulting in increase in population through natural selection.

Question 23.
Darwin’s finches and Australian marsupials are suitable examples of adaptive radiation – Justify the statement.
Answer:
Darwin’s finches are the birds whose common ancestor arrived on the Galapagos about 2 million years ago. During that time, Darwin’s finches have evolved into 14 recognized species differing in body size, beak shape and feeding behavior. Changes in the size and form of the beak have enabled different species to utilize different food resources such as insects, seeds, nectar from cactus flowers and blood from iguanas, all driven by Natural selection. Genetic variation in the ALX1 gene in the DNA of Darwin finches is associated with variation in the beak shape. Mild mutation in the ALX1 gene leads to phenotypic change in the shape of the beak of the Darwin finches.

Marsupials in Australia and placental mammals in North America are two subclasses of mammals they have adapted in similar way to a particular food resource, locomotory skill or climate. They were separated from the common ancestor more than 100 million years ago and each lineage continued to evolve independently.

Despite temporal and geographical separation, marsupials in Australia and placental mammals in North America have produced varieties of species living in similar habitats with similar ways of life. Their overall resemblance in shape, locomotory mode, feeding and foraging are superimposed upon different modes of reproduction. This feature reflects their distinctive evolutionary relationships.

Over 200 species of marsupials live in Australia along with many fewer species of placental mammals. The marsupials have undergone adaptive radiation to occupy the diverse habitats in Australia, just as the placental mammals have radiated across North America.

Question 24.
Who disproved Lamarck’s Theory of acquired characters? How?
Answer:
Lamarck’s “Theory of Acquired characters” was disproved by August Weismann who conducted experiments on mice for twenty generations by cutting their tails and breeding them. All mice bom were with tail. Weismann proved that change in the somatoplasm will not be transferred to the next generation but changes in the germplasm will be inherited.

Question 25.
How does the Mutation Theory of De Vries differ from Lamarck and Darwin’s view on the origin of new species?
Answer:
According to de Vries, sudden and large variations were responsible for the origin of new species, whereas Lamarck and Darwin believed in gradual accumulation of all variations as the causative factors in the origin of new species.

Question 26.
Explain stabilizing, directional, and disruptive selection with examples.
Answer:
i. Stabilising selection (centripetal selection): This type of selection operates in a stable environment as shown in fig. The organisms with average phenotypes survive whereas the extreme individuals from both ends are eliminated. There is no speciation but the phenotypic stability is maintained within the population over a generation. For example, measurements of sparrows that survived the storm clustered around the mean, and the sparrows that failed to survive the storm clustered around the extremes of the variation showing stabilizing selection.

ii. Directional Selection: The environment” which undergoes gradual change is subjected to directional selection, as shown in fig. This type of selection removes the individuals from one end towards the other end of phenotypic distribution. For example, size differences between male and female sparrows. Both male and female look alike externally but differ in body weight. Females show directional selection in relation to body weight.

iii. Disruptive selection: (centrifugal selection) When homogenous environment changes into heterogenous environment this type of selection is operational as shown in fig. The organisms of both the extreme phenotypes are selected, whereas individuals with average phenotype are eliminated. This results in splitting of the population into subpopulation/species. This _ is a rare form of selection but leads to formation of two or more different species. It is also ( called adaptive radiation. (E.g:) Darwin’s finches beak size in relation to seed size inhabiting Galapagos islands. Group selection and sexual selection are other types of selection. The two major group selections are Altrusim and Kin selection.

Question 27.
Rearrange the descent in human evolution.
Answer:
Australopithecus → Homo erectus → Homo sapiens → Ramapithecus → Homo habilis
Ramapithecus → Australopithecus → Homo habilis → Homo erectus → Homo sapiens

Question 28.
Differentiate between the eating habit and brain size of Australopithecus and Ramapithecus.
Answer:

Question 29.
How does the Neanderthal man differ from the modern man in appearance?
Answer:
Neanderthal man differ from the modem human in having semierect posture, flat cranium, sloping forehead, thin large orbits, heavy brow ridges, protruding jaws and no chin.

Question 30.
Mention any three similarities found common in Neanderthal man and Homo sapiens.
Answer:
Common characters showed by Neanderthal man and Homo sapiens are:

• Usage of Fire
• Burying of dead bodies
• Protecting themselves from predators

Question 31.
According to Darwin, organic evolution is due to………………….
(а) Intraspecific competition
(b) Interspecific competition
(c) Competition within closely related species. ‘
(d) Reduced feeding efficiency in one species due to the presence of interfering species.
Answer:
(d) Reduced feeding efficiency in one species due to the presence of interfering species.

Question 32.
A population will not exist in Hardy – Weinberg equilibrium if……………….
(a) Individuals mate selectively
(b) There are no mutations
(c) There is no migration
(d) The population is large
Answer:
(a) Individuals mate selectively

12th Bio Zoology Guide Evolution Additional Important Questions and Answers

12th Bio Zoology Guide Evolution One Mark Questions and Answers

Question 1.
Identify the incorrect statement in concern with Neanderthals.
(a) Neanderthal human were found in Germany.
(b) They possessed flat cranium.
(c) They used to bury their dead.
(d) Their brain size is of 650 – 800 cc.
Answer:
(d) Their brain size is of 650 – 800 cc

Question 2.
Which of the following statement does not satisfy Hardy Weinberg principle?
(a) A population undergoing random mating
(b) Small-sized population
(c) Population where there is no mutation or gene flow
(d) Absence of natural selection
Answer:
(b) Small-sized population

Question 3.
Match column I with column II

(a) a – iii b – ii c – iv d – i
(b) a – iv b – iii c – i d – ii .
(c) a – iii b – iv c – i d – ii
(d) a – ii b – iii c – i d – iv
Answer:
(a) a – iii b-ii c-iv d-i

Question 4.
Placental mammals develop during……..
(a) Eocene
(b) Oligocene
(c) Pliocene
(d) Paleocene
Answer:
(d) Paleocene

Question 5.
Identify the correct sequence from oldest to youngest
(а) Cambrian → Permian → Devonian → Silurian → Ordovician
(б) Permian → Silurian → Devonian → Ordovician → Cambrian
(c) Permian → Devonian → Silurian → Cambrian → Ordovician
(d) Cambrian → Ordovician → Silurian → Devonian → Permian
Answer:
(d) Cambrian → Ordovician → Silurian Devonian → Permian

Question 6.
Match the scientists with their terminologies used

(a) a – iii b-iv c-ii d- i
(b) a – ii b-iv, c -i d- Hi
(c) a – Hi b – i c- iv d- ii
(d) a-i b-iv c-iii d-ii
Answer:
(b) a – ii b-iv c-i d – iii

Question 7.
Anatomical structures that have similar functions but not similar structures are called
(a) Homologous structures
(b) Vestigial structures
(c) Analogous structures
(d) Generalized structures
Answer:
(c) Analogous structures

Question 8.
Who propounded the theory of recapitulation?
(a) Ernst Von Haeckel
(b) Charles Darwin
(c) Thomas Huxley
(d) Oparin
Answer:
(c) Ernst Von Haeckel

Question 9.
Mammal in human male is…………….
(a) Atavistic organ
(b) Rudimentary Organ
(c) Vestigial organ
(d) Homologous structure
Answer:
(c) Vestigial organ

Question 10.
Which of the following is/are not examples of analogous structure
(a) Wings of Birds and Bats
(b) Wings of Birds and Insects
(c) Thom of Bougainvillea and Tendril of cururbita
(d) Flippers of Penguins and Dolphins
(i) a, b, c (ii) a and c (iii) b and d (iv) All the above
Answer:
(ii) a and c

Question 11.
identify the mismatched pairs
(a) Thom of Bougainvillea and Tenrdril of cucurbita – Analogy
(b) Forelimbs of whale and cat – Analogy
(c) Octopus eye & Mammalian eye – Homology
(d) Root of sweet potato & stem of potato – Homology
Answer:
(a) Thorn of Bougainvillea & Terdril of crucurbita – Analog

Question 12.
Witnesses for evolution are found in ……………….
(a) Rocks
(b) Ocean beds
(c) Fossils
(d) Desert
Answer:
(c) Fossils

Question 13.
Assertion (A): Oparin used the term coacervates
Reason (R): Coacervates are colloidal particles in an aqueous environment
(a) Both A and Rare incorrect
(b) Both A and R are correct
(c) Both A and R are correct. R explains A.
(d) A is correct R is incorrect
Answer:
(c) Both A and R are correct. R explains A.

Question 14.
According to the theory of spontaneous generation, life originated from…………….
(a) Cosmic particles
(b) Non-living materials
(c) Coacervates
(d) Sea
Answer:
(b) Non-living materials

Question 15.
Assertion (A): Hardy – Weinberg principle states that allelic frequency of a population remains constant
Reason (R) : Constancy is maintained through natural selection and mutation
(a) A is true R is false
(b) A is false R is true
(c) Both A and R are true
(d) R explains A
Answer:
(a) A is true R is false

Question 16.
Calculate the allelic frequency of Aa. frequency of 0.7
(a) 0.67
(b) 0.42
(c) 0.36
(d) 0.59
Answer:
(b) 0.42

Question 17.
Match the following Evolutionary Human

(d) a – iv b-i c – ii d – iii
ib) a – ii b – iv c – iii d – i
(c) a – ii b – iii c – iv d – i
{d) a – iii b-i c – ii d – iv
Answer:
(a) a-iv b-i c-ii d-iii

Question 18.
Genetic drift leads to………………
(a) Mutation
(b) Bottleneck effect
(c) Immigration
(d) Isolation
Answer:
(b) Bottleneck effect

Question 19.
Atavism refers to…………….
(a) Inheritance of trial by mother
(b) Inheritance of trial by father
(c) Criss-cross inheritance
(d) Inheritance of characters not shown by parents
Answer:
(d) Inheritance of characters not shown by parents

12th Bio Zoology Guide Evolution Two Marks Questions and Answers

Question 1.
State the theory of spontaneous generation.
Answer:
According to the theory of spontaneous generation or Abiogenesis, living organisms originated from non-living materials and occurred through stepwise chemical and molecular evolution over millions of years. Thomas Huxley coined the term abiogenesis.

Question 2.
List the four eras of geological time scale.
Answer:

1. Precambrian era
2. Paleozoic era
3. Mesozoic era
4. Cenozoic era

Question 3.
Which periods of the Paleozoic era are referred to as

• Age of fishes
• Invertebrates

Answer:

• Age of fishes – Devonian period
• Age of invertebrates – Cambrian period

Question 4.
Point out the epochs of Carboniferous period.
Answer:

•  Pennsylvanian
• Mississippian

Question 5.
Compare relative dating with absolute dating.
Answer:
Relative dating is used to determine a fossil by comparing it to similar rocks and fossils of known age. Absolute dating is used to determine the precise age of a fossil by using radiometric dating to measure the decay of isotopes.

Question 6.
Wing of a cockroach and the wing of parrot. What do you infer from this statement with reference to evolution?
Answer:
Both the wings of cockroach and bird are different in structure but similar in their function. Thus, they are the analogous structure that brings about convergent evolution.

Question 7.
Name the scientists who propounded the following theories.

1. Mutation theory
2. Chemical theory of evolution

Answer:

1. Mutation theory was propounded by Hugo de Vries.
2. The chemical theory of evolution was propounded by Oparin and Haldane

Question 8.
Define fossilization and mention its types.
Answer:
‘ Fossilization is the process by which plant and animal remains are preserved in sedimentary rocks. It is of three major types,

• Actual remains
• Petrifaction
• Natural moulds and casts.

Question 9.
Name the principle minerals involved in petrifaction.
Answer:
Iron pyrites, silica, calcium carbonate and bicarbonates of calcium and magnesium.

Question 10.
What is meant by petrifaction?
Answer:
When animals die the original portion of their body may be replaced molecule for molecule by minerals and the original substance being lost through disintegration. This method of fossilization is called petrifaction. The principle minerals involved in this type fossilization are iron pyrites, silica, calcium carbonate and bicarbonates of calcium and magnesium.

Question 11.
Define analogous organ with an example.
Answer:
Organisms having different structural patterns but similar function are termed as analogous structures. For example, the wings of birds and insects are different structurally but perform the same function of flight that brings about convergent evolution.

Question 12.
Mention any four organs homologous to human hand.
Answer:
Flippers of whale, wings of bat, wings of bird and forelimb of horse.

Question 13.
Thorn of Bougainvillea and tendrils of Pisum sativum represent homology. How?
Answer:
The thorn of Bougainvillea and the tendrils of Curcurbita and Pisum sativum represent homology. The thorn in former is used as a defence mechanism from grazing animals and the tendrils of latter is used as a support for climbing.

Question 14.
Which type of evolution is brought out by homologous structures and analogous structures?
Answer:
Homologous structures brings about divergent evolution. Analogous structures brings about convergent evolution.

Question 15.
What are vestigial organs? Give example.
Answer:
Structures that are of no use to the possessor, and are not necessary for their existence are called vestigial organs. Vestigial organs may be considered as remnants of structures which were well developed and functional in the ancestors, but disappeared in course of evolution due to their non-utilization. E.g: Human appendix.

Question 16.
Human appendix is a vestige. Give reason.
Answer:
Human appendix is the remnant of caecum which is functional in the digestive tract of
herbivorous animals like rabbit. Cellulose digestion takes place in the caecum of these .
animals. Due to change in the diet containing less cellulose, caecum in human became functionless and is reduced to a vermiform appendix, which is vestigial.

Question 17.
What are connecting link? Give example.
Answer:
The organisms which possess the characters of two different groups (transitional stage) are called connecting links. Example Peripatus (connecting link between Annelida and Arthropoda) Archaeopteryx (connecting link between Reptiles and Aves).

Question 18.
Name one fossilised connecting link between reptiles and Aves also one living connecting link between Annelida and Arthropoda.
Answer:
Archaeopteryx – connecting link between Reptiles and Aves.
Peripatus – Connecting link between Annelida and Arthropoda.

Question 19.
Why it is considered as a connecting link?
Answer:
Peripatus is a worm that shown the characters of both Annelidia and Arthropoda. Hence it is a connecting link between Annelida and Arthropoda. ‘

Question 20.
Atavistic organs – comment.
Answer:
Sudden appearance of vestigial organs in highly evolved organisms is called atavistic organs. Example, presence of tail in human baby is an atavistic organ.

Question 21.
Define Ontogeny and Phytogeny.
Answer:
Ontogeny refers to the life history of an individual.
Phytogeny refers to the evolutionary history of a race. ‘

Question 22.
Who proposed the theory of recapitulation? State the theory.
Answer:
Ernst Von Haeckel proposed the theory of recapitulation, which states that life history of an individual briefly repeats the evolutionary history of the race. ,

Question 23.
Name few Neo – Lamarckists.
Answer:
Cope, Osborn, Packard and Spencer.

Question 24.
Who proposed the theory of acquired characters? Also mention the scientist who disproved it.
Answer:
The theory of acquired characters was proposed by Jean Baptise de Lamarck and it was disproved by August Weismann.

Question 25.
Point out the basic principles of Darwin’s theory of evolution.
Answer:
Over production, struggle for existence, Universal occurence of variation, Survival of fittest and Natural selection.

Question 26.
Name any four Neo – Darwinists.
Answer:
Gregor Mendel, August Weismann, Russel Wallace and Heinrich.

Question 27.
Enumerate the salient features of mutation theory.
Answer:

• Mutations or discontinuous variation are transmitted to other generations.
• In naturally breeding populations, mutations occur from time to time.
• There are no intermediate forms, as they are fully fledged.
• They are strictly subjected to natural selection.

Question 28.
Who proposed Mutation theory? Name the organism on which the experiment was carried out.
Answer:
Mutation theory was put forth by Hugo de Vries. Based on the experiments in Oenothera lamarckiana (The evening primrose plant).

Question 29.
What are the basic factors of modern synthetic theory that leads to evolution?
Answer:
Gene mutation, Chromosomal mutation, Genetic recombination, Natural selection and Reproductive isolation.

Question 30.
Name the scientists who supported modern synthetic theory.
Answer:
Sewell Wright, Dobzhansky, Huxley and Simpson.

Question 31.
Define point mutation.
Answer:
Gene mutation refers to the changes in the structure of the gene. It is also called gene / point mutation. It alters the phenotype of an organism and produces variations in their offsprings.

Question 32.
Point out the factors that alters allelic frequency of a population.
Answer:
Natural selection, Genetic drift, Mutation and Geneflow

Question 33.
Mention any two differences between Homo habilis and Homo erectus
Answer:
Homo habilis :

1. The brain capacity was between 650-800 cc.
2. They were probably vegetarians.

Homo erectus :

1. The brain capacity was around 900 cc.
2. They probably ate meat

Question 34.
Write a brief note on Homo sapiens with respect to evolution.
Answer:
Homo sapiens or modem human arose in Africa some 25,000 years ago and moved to other continents and developed into distinct races. They had a brain capacity of 1300 – 1600 cc. “fhey started cultivating crops and domesticating animals.

Question 35.
Define evolution.
Answer:
The term evolution describes heritable changes in one or more characteristics of a population of species from one generation to the other.

12th Bio Zoology Guide Evolution Three Marks Questions and Answers

Question 36.
Write a short note on Big Bang theory.
Answer:
Big bang theory explains the origin of universe as a singular huge explosion in physical terms. The primitive Earth had no proper atmosphere, but consisted of ammonia, methane, hydrogen and water vapour. The climate of the Earth was extremely high. UV rays from the Sun split up water molecules into hydrogen and oxygen. Gradually the temperature cooled and the water vapour condensed to form rain. Rain water filled all the depressions to form water bodies. Ammonia and methane in the atmosphere combined with oxygen to form carbon dioxide and other gases.

Question 37.
Theory of chemical evolution states that organisms have evolved from inorganic substances. If so, what was the atmospheric condition that favoured evolution?
Answer:
The atmosphere was devoid of O₂, and with high level of CO₂ NH₀3 and UV radiations.

Question 38.
Name the periods of Mesozoic era. Also mention the flora and fauna dominates during that periods.
Answer:
Mesozoic era is divided into three periods namely Triassic, Jurassic and Cretaceous.
Dominating Fauna : Reptiles and Dinosaurs Dominating Flora : Conifers, Ferns and Ginkgon.

Question 39.
Which era is referred as Age of Mammals? What are the periods of that era? And also mention the fauna during the periods.
Answer:
Cenozoic era is called as Age of Mammals.
Tertiary and Quaternary are the two periods of Cenozoic era.
Tertiary periods marks the abundance of mammalian fauna. Quaternary period marks the beginning of human social life.

Question 40.
Write a short note on Cenozoic era.
Answer:
Cenozoic era (Age of mammals) is subdivided into two periods namely Tertiary and Quaternary. Tertiary period is characterized by abundant mammalian fauna. This period is subdivided into five epochs namely, Paleocene (placental mammals, Eocene (Monotremes except duck billed Platypus and Echidna, hoofed mammals and carnivores), Oligocene (higher placental mammals appeared), Miocene (origin of first man like apes) and Pliocene (origin of man from man like apes). Quaternary period witnesses decline of mammals and beginning of human social life.

Question 41.
Name the gaseous mixture used in Urey – Miller’s experiment. Which type of physical force is applied to generate amino acids?
Answer:
Ammonia, Methane, Hydrogen, Water vapour are the gaseous mixture allowed to circulate over electric discharge from a tungsten electrode.

Question 42.
Which is the most common methods of fossilization? Explain how it occurs.
Answer:
Actual remains – The original hard parts such as bones, teeth or shells are preserved as such in the Earth’s atmosphere. This is the most commpn method of fossilization. When marine animals die, their hard parts such as bones and shells, etc., are covered with sediments and are protected from further deterioration. They get preserved as such as they are preserved in vast ocean; the salinity in them prevents decay. The sediments become hardened to form definite layers or strata. For example, Woolly Mammoth that lived 22 thousand years ago were preserved in the frozen coast of Siberia as such. Several human beings and animals living in the ancient city of Pompeii were preserved intact by volcanic ash which gushed out from Mount Vesuvius. ,

Question 43.
What are coprolites? Mention its role in phytogeny.
Answer:
Coprolites are the hardened faecal matters occurs as small pieces. Analysing the coprolites helps to understand the nature of diet of pre-historic animals.

Question 44.
What are moulds and casts?
Answer:
Even after disintegration, the body of an animal might leave indelible impression on the soft mud which later becomes hardened into stones. Such impressions are called moulds. The cavities of the moulds may get filled up by hard minerals and get fossilized, which are called casts.

Question 45.
How will you compute the age of fossil?
Answer:
The age of fossils can be determined using two methods namely, relative dating and absolute dating. Relative dating is used to determine a fossil by comparing it to similar rocks and fossils of known age. Absolute dating is used to determine the precise age of a fossil by using radiometric dating to measure the decay of isotopes.

Question 46.
“Ontogeny recapitulates phylogeny” – comment on the statement with example.
Answer:
The embryonic stages of a higher animal resemble the adult stage of its ancestors. Appearance of pharyngeal gill slits, yolk sac and the appearances of tail in human embryos are some of the examples.

Question 47.
Biogenetic law is not universal – justify.
Answer:
The biogenetic law is not universal and it is now thought that animals do not recapitulate the adult stage of any ancestors. The human embryo recapitulates the embryonic history and not the adult history of the organisms.

Question 48.
How macro molecules like DNA and RNA play their crucial role in evolutionary history?
Answer:
Molecular evolution is the process of change in the sequence composition of molecules such as DNA, RNA and proteins across generations. It uses principles of evolutionary biology and population genetics to explain patterns in the changes of molecules.

One of the most useful advancement in the development of molecular biology is proteins and other molecules that control life processes are conserved among species. A slight change that occurs over time in these conserved molecules (DNA, RNA and protein) are often called molecular clocks. Molecules that have been used to study evolution are cytochrome c (respiratory pathway) and rRNA (protein synthesis).

Question 49.
Explain the principles of Lamarckian theory.
Answer:

i. The theory of use and disuse – Organs that are used often will increase in size and those that are not used will degenerate. Neck in giraffe is an example of use and absence of limbs in snakes is an example for disuse theory.
ii. The theory of inheritance of acquired characters – Characters that are developed during ’ the life time of an organism are called acquired characters and these are then inherited.

Question 50.
Write a note on Mutation theory.
Answer:
Hugo de Vries put forth the Mutation theory. Mutations are sudden random changes that occur in an organism that is not heritable. De Vries carried out his experiments in the Evening Primrose plant {Oenothera lamarckiana) and observed variations in them due to mutation. According to de Vries, sudden and large variations were responsible for the origin of new species whereas Lamarck and Darwin believed in gradual accumulation of all variations as the causative factors in the origin of new species.

Question 51.
What do you mean by “adaptive radiation”? Give example.
Answer:
The evolutionary process which produces new species diverged from a single ancestral form becomes adapted to newly invaded habitats is called adaptive radiation. Adaptive radiations are best exemplified in closely related groups that have evolved in relatively short time. Darwin’s finches and Australian marsupials are best examples for adaptive radiation.

Question 52.
Darwins finches are the classical examples studied for adaptive radation. Explain.
Answer:
Darwin’s finches are the birds whose common ancestor arrived on the Galapagos about 2million year ago. During that time, Darwin’s finches have evolved into 14 recognized species differing in body size, beak shape and feeding behavior. Changes in the size and form of the beak have enabled different species to utilize different food resources such as insects, seeds and nectar from cactus flowers and blood from iguanas, all driven by Natural selection. Genetic variation in the ALX1 gene in the DNA of Darwin finches is associated with variation in the beak shape. Mild mutation in the ALX1 gene leads to phenotypic change in the shape of the beak of the Darwin finches.

Question 53.
What is micro evolution?
Answer:
Microevolution (evolution on a small scale) refers to the changes in allele frequencies within a population. Allele frequencies in a population may change due to four fundamental forces of evolution such as natural selection, genetic drift, mutation and gene flow.

Question 54.
Name the major types of Natural Selection.
Answer:

1. Stabilising Selection
2. Directional Selection
3. Disruptive Selection

Question 55.
What do you mean by gene flow?
Answer:
Movement of genes through gametes or movement of individuals in (immigration) and out (emigration) of a population is referred to as gene flow. Organisms and gametes that enter the population may have new alleles or may bring in existing alleles but in different proportions than those already in the population. Gene flow can be a strong agent of evolution.

Question 56.
Give an account on Genetic drift. Mention its impact over a population.
Answer:
Genetic drift is a mechanism of evolution in which allele frequencies of a population change over generation due to chance (sampling error). Genetic drift occurs in all population sizes, but its effects are strong in a small population. It may result in a loss of some alleles (including beneficial ones) and fixation of other alleles. Genetic drift can have major effects, when the population is reduced in size by natural disaster due to bottle neck effect or when a small group of population splits from the main population to form a new colony due to founder’s effect.

Question 57.
State Hardy – Weinberg equilibrium.
Answer:
The allele frequencies in a population are stable and are constant from generation to generation in the absence of gene flow, genetic drift, mutation, recombination and natural selection.

Question 58.
Write in brief about the characters of Australian ape man.
Answer:
Australopithecus lived in East African grasslands about 5 mya and was called the Australian ape man. He was about 1.5 meters tall with bipedal locomotion, omnivorous, semi erect, and lived in caves. Low forehead, brow ridges over the eyes, protruding face, lack of chin, low brain capacity of about 350 – 450 cc, human like dentition, lumbar curve in the vertebral column were his distinguishing features.

Question 59.
Who is Cro-Magnon?
Answer:
Cro-Magnon was one of the most talked forms of modem human found from the rocks of Cro-Magnon, France and is considered as the ancestor of modem Europeans. They were not only adapted to various environmental conditions, but were also known for their cave paintings, figures on floors and walls.

12th Bio Zoology Guide Evolution Five Marks Questions and Answers

Question 60.
Explain Oparin – Haldane hypothesis on evolution.
Answer:
According to the theory of chemical evolution primitive organisms in the primordial environment of the Earth evolved spontaneously from inorganic substances and physical forces such as lightning, UV radiations, volcanic activities, etc. Oparin (1924) suggested that the organic compounds could have undergone a series of reactions leading to more complex molecules.

He proposed that the molecules formed colloidal aggregates or ‘coacervates’ in an aqueous environment. The coacervates were able to absorb and assimilate organic compounds from the environment. Haldane (1929) proposed that the primordial sea served as a vast chemical laboratory powered by solar energy. The atmosphere was oxygen free and the combination of CO₂, NH₃ and UV radiations gave rise to organic compounds.

The sea became a ‘hot’ dilute soup containing large populations of organic monomers and polymers. They envisaged that groups of monomers and polymers acquired lipid membranes and further developed into the first living cell. Haldane coined the term prebiotic soup and this became the powerful symbol of the Oparin-Haldane view on the origin of life (1924-1929). Oparin and Haldane independently suggested that if the primitive atmosphere was reducing – and if there was appropriate supply of energy such as lightning or UV light then a wide range of organic compounds can be synthesized.

Question 61.
How Urey – Miller’s experiment supports the origin of life?
Answer:
Urey and Miller (1953), paved way for understanding the possible
synthesis of organic compounds that led to the appearance of living organisms is depicted in the Fjgure In their experiment, a mixture of gases was allowed to circulate over electric discharge from an tungsten electrode. A small flask was kept boiling and the steam emanating from it was made to mix with the mixture of

gases (ammonia, methane and hydrogen) in the large chamber that was connected to the boiling Fig. 6.1 Diagrammatic representation of Urey-Miller’s water. The steam condensed to form water which ran down the ‘U’ tube. Experiment was conducted continuously for a week and the liquid was analysed. Glycine, alanine, beta alanine and aspartic acid were identified. Thus Miller’s experiments had an insight as to the possibility of abiogenetic synthesis of large amount of variety of organic compounds in nature from a mixture of sample gases in which the only source of carbon was methane. Later in similar experiments, formation of all types of amino acids, and nitrogen bases were noticed.

Question 62.
Give a detailed account of Modern Synthetic Theory.
Answer:
Sewell Wright, Fisher, Mayer, Huxley, Dobzhansky, Simpson and Haeckel explained Natural Selection in the light of Post-Darwinian discoveries. According to this theory ‘ gene mutations, chromosomal mutations, genetic recombinations, natural selection and reproductive isolation are the five basic factors involved in the process of organic evolution.

• Gene mutation refers to the changes in the structure of the gene. It is also called gene/ point mutation. It alters the phenotype of an organism and produces variations in their offsprings.
• Chromosomal mutation refers to the changes in the structure of chromosomes due to deletion, addition, duplication, inversion or translocation. This too alters the phenotype of an organism and produces variations in their offspring.
• Genetic recombination is due to crossing over of genes during meiosis. This brings about genetic variations in the individuals of the same species and leads to heritable variations.
• Natural selection does not produce any genetic variations but once such variations occur » it favours some genetic changes while rejecting others (driving force of evolution).
• Reproductive isolation helps in preventing interbreeding between related organisms.

Higher Order Thinking Skills (HO’ts) Questions

Question 1.
Name the connecting link for the following groups of organisms.
(a) Annelida and Arthropoda
(b) Reptiles and Aves
(c) Pisces and Amphibians
(d) Reptiles and Mammals
Answer:
(a) Peripatus
(b) Archeopteryx
(c) Lung fish
(d) Platypus

Question 2.
Point out any four condition under which Hardy Weinberg’s equilibrium is not attained.
Answer:

•  Selected mating
• Flow of genes (either by immigration or emmigration)
• Occurance of mutation
• Definite population size

Question 3.
Why are analogous structures a result of convergent evolution?
Answer:
Analogous structures are not anatomically similar though they perform same function.

Question 4.
Organs which are of no use to the organism is called as vestige. Name any four vestigal organs that can be noticed in your body.
Answer:
Wisdom teeth, Mammae in male, Body hair and Coccyx.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Bio Zoology Guide Pdf Chapter 5 Molecular Genetics Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Bio Zoology Solutions Chapter 5 Molecular Genetics

12th Bio Zoology Guide Molecular Genetics Text Book Back Questions and Answers

Question 1.
Hershey and Chase experiment with bacteriophage showed that
(a) Protein gets into the bacterial cells
(b) DNA is the genetic material
(c) DNA contains radioactive sulphur
(d) Viruses undergo transformation
Answer:
(b) DNA is the genetic material

Question 2.
DNA and RNA are similar with respect to
(a) Thymine as a nitrogen base
(b) A single-stranded helix shape
(c) Nucleotide containing sugars, nitrogen bases and phosphates
(d) The same sequence of nucleotides for the amino acid phenyl alanine
Answer:
(c) Nucleotide containing sugars, nitrogen bases and phosphates

Question 3.
A mRNA molecule is produced by
(a) Replication
(b) Transcription
(c) Duplication
(d) Translation
Answer:
(b) Transcription

Question 4.
The total number of nitrogenous bases in human genome is estimated to be about
(a) 3.5 million
(b) 35000
(c) 35 million
(d) 3.1 billion
Answer:
(d) 3.1 billion

Question 5.
E. coli cell grown on 15N medium are transferred to 14N medium and allowed to grow for two generations. DNA extracted from these cells is ultracentrifuged in a cesium chloride density gradient. What density distribution of DNA would you expect in this experiment?
(a) One high and one low density band
(b) One intermediate density band
(c) One high and one intermediate density band
(d) One low and one intermediate density band
Answer:
(d) One low and one intermediate density band

Question 6.
What is the basis for the difference in the synthesis of the leading and lagging strand of DNA molecules?
(a) Origin of replication occurs only at the 5’ end of the molecules
(b) DNA ligase works only in the 3’ → 5’ direction
(c) DNA polymerase can join new nucleotides only to the 3′ ends of the growing stand
(d) Helicases and single-strand binding proteins that work at the 5’ end
Answer:
(d) DNA polymerase can join new nucleotides only to the 3’ end of the growing stand

Question 7.
Which of the following is the correct sequence of event with reference to the central dogma?
(a) Transcription, Translation, Replication
(b) Transcription, Replication, Translation
(c) Duplication, Translation, Transcription
(d) Replication, Transcription, Translation
Answer:
(d) Replication, Transcription, Translation

Question 8.
Which of the following statements about DNA replication is not correct?
(a) Unwinding of DNA molecule occurs as hydrogen bonds break
(b) Replication occurs as each base is paired with another exactly like it
(c) Process is known as semiconservative replication because one old strand is conserved in the new molecule
(d) Complementary base pairs are held together with hydrogen bonds
Answer:
(b) Replication occurs as each base is paired with another exactly like it

Question 9.
which of the following statements is not true about DNA replication in eukaryotes?
(a) Replication begins at a single origin of replication.
(b) Replication is bidirectional from the origins.
(c) Replication occurs at about 1 million base pairs per minute.
(d) There are numerous different bacterial chromosomes, with replication occurring in each at the same time.
Answer:
(d) There are numerous different bacterial chromosomes, with replication occurring in each at the same time

Question 10.
The first codon to be deciphered was ………………. which codes for ……………..
(a) AAA, proline
(b) GGG, alanine
(c) UUU, Phenylalanine
(d) TTT, arginine
Answer:
(c) UUU, Phenylalanine

Question 11.
Meselson and Stahl’s experiment proved
(a) Transduction
(b) Transformation
(c) DNA is the genetic material
(d) Semi-conservative nature of DNA replication
Answer:
(d) Semi-conservative nature of DNA replication

Question 12.
Ribosomes are composed of two subunits; the smaller subunit of a ribosome has a binding site for……………..and the larger subunit has two binding sites for two…………………
Answer:
mRNA, tRNA

Question 13.
Anoperonisa:
(a) Protein that suppresses gene expression
(b) Protein that accelerates gene expression
(c) Cluster of structural genes with related function
(d) Gene that switched other genes on or off
Answer:
(d) Cluster of structural genes with related function

Question 14.
When lactose is present in the culture medium:
(a) Transcription of lacy, lac z, lac a genes occurs
(b) Repressor is unable to bind to the operator
(c) Repressor is able to bind to the operator
(d) Both (a) and (b) are correct
Answer:
(d) Both (a) and (b) are correct

Question 15.
Give reasons: Genetic code is ‘universal’.
Answer:
The genetic code is universal. It means that all known living systems use nucleic acids and the same three base codons (triplet codon) direct the synthesis of protein from amino acids. For example, the mRNA (UUU) codon codes for phenylalanine in all cells of all organisms. Some exceptions are reported in prokaryotic, mitochondrial, and chloroplast genomes. However, similarities are more common than differences.

Question 16.
Name the parts marked ‘A’ and ‘B’ in the given transcription unit:

Answer:
A – Promoter site B – Structural gene

Question 17.
Differentiate – Leading strand and lagging strand
Answer:

Question 18.
Differentiate – Template strand and coding strand.
Answer:

1. Template Strand: During replication, DNA strands having the polarity 3’ → 5’ act as the template strand.
2. Coding Strand: During replication, a DNA strands having the polarity 5’ → 3’ act as a coding strand.

Question 19.
Mention any two ways in which single nucleotide polymorphism (SNPs) identified in human genome can bring revolutionary change in biological and medical science.
Answer:
Scientists have identified about 1.4 million locations, where single-base DNA differences (SNPs – Single nucleotide polymorphism – pronounced as ‘snips’) occur in humans. Identification of ‘SNIPS’ is helpful in finding chromosomal locations for disease-associated sequences and tracing human history.

Question 20.
State any three goals of the human genome project.
Answer:

• Identify all the genes (approximately 30000) in human DNA.
• Determine the sequence of the three billion chemical base pairs that make up the human DNA.
• To store this information in databases.

Question 21.
In E.coli, three enzymes 0- galactosidase, permease and transacetylase are produced in the presence of lactose. Explain why the enzymes are not synthesized in the absence of lactose.
Answer:
In the absence of lactose, the repressor protein binds to the operator and prevents the transcription of structural gene by RNA polymerase, hence the enzymes are not produced. However, there will always be a minimal level of lac operon expression even in absence of lactose.

Question 22.
Distinguish between structural gene, regulatory gene and operator gene.
Answer:
Structure of the operon: Each operon is a unit of gene expression and regulation and consists of one or more structural genes and an adjacent operator gene that controls the transcriptional, activity of the structural gene.

1. The structural gene codes for proteins, rRNA, and tRNA required by the cell.
2. Promoters are the signal sequences in DNA that initiate RNA synthesis. RNA polymerase binds to the promoter prior to the initiation of transcription.
3. The operators are present between the promoters and structural genes. The repressor protein binds to the operator region of the operon.

Question 23.
A low level of expression of lac operon occurs at all the time in E-coli. Justify the statement.
Answer:
One of the enzyme synthesized by lac operon is permease which is involved in the transport of lactose into the cells. If the lac operon gets inactivated, permease is not synthesized hence lactose cannot enter the cell. Lactose acts as a inducer, binding to the repressor protein and switch on the operator to initiate gene expression.

Question 24.
Why the human genome project is called a megaproject?
Answer:
The international human genome project was launched in the year 1990. It was a mega project and took 13 years to complete. The human genome is about 25 times larger than the genome of any organism sequenced to date and is the first vertebrate genome to be completed. Human genome is said to have approximately 3 >109 bp. HGP was closely associated with the rapid development of a new area in biology called bioinformatics.

Question 25.
From their examination of the structure of DNA, What did Watson and Crick infer about the probable mechanism of DNA replication, coding capability and mutation?
Answer:
Inference of Watson and Crick on DNA replication.
They concluded that each of the DNA strand in a helix act as template during DNA replication leading to formation of new daughter DNA molecules, which are complementary to parental strand, (i.e., Semi-conservative method of replication)

Inference on coding capability
During transcription, the genetic information in the DNA strand is coded to mRNA as complementary bases, (except for uracil in place of thymine in RNA)

Inference on mutation
Any changes in the nucleotide sequence of DNA leads to corresponding alteration in aminoacid sequence of specific protein thus confirming the validity of genetic code.

Question 26.
Why tRNA is called an adapter molecule?
Answer:
The transfer RNA, (tRNA) molecule of a cell acts as a vehicle that picks up the amino acids scattered through the cytoplasm and also reads specific codes of mRNA molecules. Hence it is called an adapter molecule. This term was postulated by Francis Crick.

Question 27.
What are the three structural differences between RNA and DNA?
Answer:
DNA :

1. Sugar is deoxyribose sugar. .
2. Double-stranded structure.
3. Nitrogen bases are Adenine, Guanine, Cytosine and Thymine.

RNA :

1. Sugar is ribose sugar
2. Single-stranded molecule.
3. Nitrogen bases are Adenine, Guanine, Cytosine and Uracil.

Question 28.
Name the anticodon required to recognize the following codons: AAU, CGA, UAU, and GCA.
Answer:
UUA, GCU, AUA and CGU.

Question 29.
a) Identify the figure given below
b) Redraw the structure as a replicating fork and label the parts
c) Write the source of energy for this replication and name the enzyme involved in this process.
d) Mention the differences in the synthesis of protein, based on the polarity of the two template strands.
Answer:
a) Replication fork
b)

c) Deoxy nucleotide, triphosphate acts as a energy source for replication. DNA polymerase is used for replication
d) mRNA contacting information for protein synthesis will developed from DNA strand having polariy 5 ’ → 3 ’

Question 30.
If the coding sequence in a transcription unit is written as follows:
5’ TGCATGCATGCATGCATGCATGCATGC 3’
Write down the sequence of mRNA.
Answer:
mRNA sequence is 3’ACGUACGUACGUUCGUACGUACGUACG5’

Question 31.
How is the two-stage process of protein synthesis advantageous?
Answer:
The split gene feature of eukaryotic genes is almost entirely absent in prokaryotes. Originally each exon may have coded for a single polypeptide chain with a specific function. Since exon arrangement and intron removal are flexible, the exon coding for these polypeptide subunits act as domains combining in various ways to form new genes. Single genes can produce different functional proteins by arranging their exons in several different ways through alternate splicing patterns, a mechanism known to play an important role in generating both protein and functional diversity in animals.

Introns would have arose before or after the evolution of eukaryotic gene. If introns arose late how did they enter eukaryotic gene? Introns are mobile DNA sequences that can splice themselves out of, as well as into, specific ‘target sites’ acting like mobile transposon-like elements (that mediate transfer of genes between organisms – Horizontal Gene Transfer – HGT). HGT occurs between lineages of prokaryotic cells, or from prokaryotic to eukaryotic cells and between eukaryotic cells. HGT is now hypothesized to have played a major role in the evolution of life on Earth.

Question 32.
Why did Hershey and Chase use radioactively labelled phosphorous and sulphur only? Would they have got the same result if they use radiolabelled carbon and nitrogen?
Answer:
Generally proteins contain sulphur but not phosphorous and nucleic acid (DNA) contains , phosphorous but not sulphur. Hence Hershey – Chase used radioactive isotopes of sulphur (35S) and phosphorus (32P) to keep separate track of viral protein and nucleic acid in culture medium. The expected result cannot be achieved, if radioactive carbon and nitrogen is used, since these molecules are present in both DNA and proteins.

Question 33.
Explain the formation of a nucleosome.
Answer:
Komberg proposed a model for the nucleosome, in which 2 molecules of the four histone proteins H2A, H2B, H3 and H4 are organized to form a unit of eight molecules called histone octamere.

The negatively charged DNA is wrapped around the positively charged histone octamere to form a structure called nucleosome. A typical nucleosome contains 200 bp of DNA helix. The histone octameres are in close contact and DNA is coiled on the outside of nucleosome.

Question 34.
It is established that RNA is the first genetic material. Justify giving reasons.
Answer:
Three molecular biologists in the early 1980’s (Leslie Orgel, Francis Brick and Carl Woese) independently proposed the ‘RNA world’ as the first stage in the evolution of life, a stage when RNA catalysed all molecules necessary for survival and replication. The term ‘RNA world’ first used by Walter Gilbert in 1986, hypothesizes RNA as the first genetic material on Earth. There is now enough evidence to suggest that essential life processes (such as metabolism, translation and splicing etc.,) evolved around RNA.

RNA has the ability to act as both genetic material and catalyst. There are several biochemical reactions in living systems that are catalysed by RNA. This catalytic RNA is known as ribozyme. But, RNA being a catalyst was reactive and hence unstable. This led to evolution of a more stable form of DNA, with certain chemical modifications. Since DNA is a double-stranded molecule having complementary strand, it has resisted changes by evolving a process of repair. Some RNA molecules function as gene regulators by binding to DNA and affect gene expression. Some viruses use RNA as the genetic material. Andrew Fire and Craig Mellow (recipients of Nobel Prize in 2006) were of the opinion that RNA is an active ingredient in the chemistry of life.

12th Bio Zoology Guide Molecular Genetics Additional Important Questions and Answers

12th Bio Zoology Guide Molecular Genetics One Mark Questions and Answers

Question 1.
The term gene was coined by……………
Answer:
Wilhelm Johannsen

Question 2.
Whose experiment finally provided convincing evidence that DNA is the genetic material?
(a) Griffith experiment
(b) Avery, Macleod and McCarty’s experiment
(c) Hershey-Chase experiment
(d) Urey-Miller’s experiment
Answer:
(c) Hershey-Chase experiment

Question 3.
In Hershey – Chase experiment, the DNA of T₂ phase was made radioactive by using
(a) ³²P
(b) ³⁵S
(c) ³⁵P
(d) ³⁵S
Answer:
(a) ³²P

Question 4.
A nucleoside is composed of………………………………
(a) Sugar and Phosphate
(b) Nitrogen base and Phosphate
(c) Sugar and Nitrogen base
(d) Sugar, Phosphate and Nitrogenous base
Answer:
(c) Sugar and Nitrogen base

Question 5.
Identify the incorrect statement
(a) a base is a substance that accepts H+ ion
(b) Both DNA and RNA have four bases
(c) Purines have single carbon-nitrogen ring
(d) Thymine is unique for DNA
Answer:
(c) Purines have single carbon-nitrogen ring

Question 6.
Watson and Crick proposed their double helical DNA model based on the X-ray diffraction analysis of………………
(a) Erwin Chargaff
(b) Meselson and Stahl
(c) Wilkins and Franklin
(d) Griffith
Answer:
(c) Wilkins and Franklin

Question 7.
The term ‘RNA world’ was first used by………………
Answer:
Walter Gilbert

Question 8.
The distance between two consecutive base pairs in DNA is …………………
(a) 0.34 nm
(b) 3.4 nm
(c) 0.034 nm
(d) 34 nm
Answer:
(a) 0.34 nm

Question 9.
If the length of E. coli DNA is 1.36 mm, the number of base pairs is
(a) 0.36 x 10⁶m
(b) 4 x 10⁶m
(c) 0.34 x 10^-9nm
(d) 4 x 10^-9m
Answer:
(b) 4 x 10⁶m

Question 10.
Identify the proper sequence in the organisation of the eukaryotic chromosome.
(a) Nucleosome – Solenoid – Chromatid
(b) Chromatid – Nucleosome – Solenoid
(c) Solenoid – chromatin – DNA
(d) Nucleosome – solenoid – genophore
Answer:
(a) Nucleosome – Solenoid – Chromatid

Question 11.
Assertion (A) : Genophore is noticed in prokaryotes.
Reason (R) : Bacteria possess circular DNA without chromatin organisation.
(a) Both A and R are correct
(b) A is correct R is incorrect
(c) R explains A
(d) A is incorrect R is correct
Answer:
(c) R explains A

Question 12.
Assertion (A): Heterochromatin is transcriptionally active.
Reason (R): Tightly packed chromatin which stains dark.
(a) Both A and R are correct , (b) A is correct R is incorrect
(c) R explains A
(d) A is incorrect R is correct .
(d) A is incorrect R is correct
Answer:
(d) A is incorrect R is correct

Question 13.
Assertion (A): the semi-conservative model was proposed by Hershey and Chase.
Reason (R) : The daughter DNA contains only new strands.
(a) Both A and R are incorrect
(b) A is correct R is incorrect
(c) R explains A
(d) A is incorrect R is correct
Answer:
(a) Both A and R are incorrect

Question 14.
Komberg enzyme is called as ……………..
Answer:
DNA polymerase I

Question 15.
Replication of DNA occurs at phase of cell cycle.
(a) M
(b) S
(c) G₁
(d) G₂
Answer:
(b) S

Question 16.
Semi-conservative model of replication was proved by
(a) Hershey and Chase
(b) Griffith
(c) Meselson and Stahl
(d) Macleod and McCarty
Answer:
(c) Meselson and Stahl

Question 17.
How many types of DNA polymerases does a eukaryotic cell possess?
(a) two
(b) three
(c) four
(d) five
Answer:
(d) Five

Question 18.
Identify the incorrect statement
(a) Replication occurs at ori – site of DNA
(b) Deoxy nucleotide triphosphate acts as a substrate
(c) Unwinding of DNA strand is carried out by topoisomerase
(d) DNA polymerase catalyses the polymerization at 3-OH
Answer:
(c) Unwinding of DNA strand is carried out by topoisomerase

Question 19.
The discontinuously synthesized fragments of lagging strand are called………………
Answer:
Okazaki fragments

Question 20.
Retroviruses possess ………………..as genetic material.
Answer:
RNA

Question 21.
Which is NOT a part of transcription unit?
(a) Promoter
(b) Operator
(c) Structural gene
(d) Terminator
Answer:
(b) Operator

Question 22.
Goldberg – Hogness box of eukaryotes is equivalent to of prokaryotes.
Answer:
Pribnow box

Question 23.
Okazaki fragments are joined by the enzyme during DNA replication.
Answer:
DNA ligase

Question 24.
Match the following:

(a) A – iv, B – C – ii, D – iii
(b) A – i, B – ii, C – iii, D – iv
(c) A – ii, B – iii, C – i, D – ii
(d) A – iii, B – ii, C – ii, D – i
Answer:
(a) A – iv, B – C – ii, D – iii

Question 25.
The RNA polymerase of prokaryotes binds with ………………….factor to initiate polymerization.
(a) rho
(b) theta
(c) sigma
(d) psi
Answer:
(c) sigma A

Question 26.
Precursor
(a) Capung
(b) Tailing
(c) Splicing
(d) Transcribing
Answer:
(c) Splicing

Question 27.
Which of the following feature is absent in prokaryotes?
(a) Prokaryotes possess three major types of RNAs
(b) Structural genes are polycistronic
(c) Initiation process of transcription requires ‘P’ factor
(d) Split gene feature
Answer:
(d) Split gene feature

Question 28.
Which of the following sequence has completely translated?
(i) AGA, UUU, UGU, AGU, UAG
(ii) AUG, UUU, AGA, UAC, UAA
(iii) AAA, UUU, UUG, UGU, UGA
(iv) AUG,AAU,AAC,UAU,UAG
(a) i and ii
(b) ii only
(c) i and iii
(d) ii and iv
Answer:
(d) ii and iv

Question 29.
Capping of mRNA occurs using
(a) Poly A residues
(b) Methyl guanosine triphosphate
(c) Deoxy ribonucleotide triphosphate
(d) Ribonucleotide triphosphate
Answer:
(b) Methyl guanosine triphosphate

Question 30.
One of the aspect is not a feature of genetic code?
(a) Specific
(b) Degenerate
(c) Universal
(d) Ambiguous
Answer:
(d) Ambiguous

Question 31.
Which of the triplet codon is not a code of proline?
(i) CCU (ii) CAU (iii) CCG (zv) CAA
(a) i only
(b) ii and iv
(c) iii only
(d) all the above
Answer:
(b) ii and iv

Question 32.
Coding sequences found in split genes are called.
(a) Operons
(b) Introns
(c) Exons
(d) Cistron
Answer:
(c) Exons

Question 33.
Which of the following mRNA yields 6 aminoacids after translation?
(i) UCU UAU AGU CGA UGC AGU UGA AAA UUU
(ii) UGA AGA UAG GAG CAU CCC UAC UAU GAU
(iii) GUC UGC UGG GCU GAU UAA AGG AGC AUU
(iv) AUG UAC CAU UGC UGA UGC AGG AGC CCG
Answer:
(i) UCU UAU AGU CGA UGC AGU UGA AAA UUU

Question 34.
The transcription termination factor associated with RNA polymerase in prokaryotes is
(a) ∑
(b) σ
(c) ρ
(d) ∑
Answer:
(c) p

Question 35.
In a DNA double-strand, if guanine is 30%, what will be the percentage of thymine?
(a) 100%
(b) 20%
(c) 10%
(d) 70%
Answer:
(b) 20%

Question 36.
Identify the triplet pairs that code for Tyrosine
(a) UUU,UUC
(b) UAU, UAU
(c) UGC, UGU
(d) CAU, CAC
Answer:
(b) UAU, UAU

Question 37.
Match the following:

(a) A – ii B – i C – iv D – iii
(b) A – iii B – ii C – i D – iv
(c) A – iv B – i C – ii D – iii
(d) A – ii B – iii C – i D – iv
Answer:
(a) A-ii B-i C-iv D-iii

Question 38.
AUG code is for……………….
(a) Arginine
(b) Tyrosine
(c) Tryptophan
(d) Methionine
Answer:
(d) Methionine

Question 39.
The sequence of bases in coding strand of DNA is GAGTTAGCAGGC, then the sequence of codons in primary transcript is ……………
(d) CUCAUACGCCCG
(b) CUCAAUCGUCCG
(c) UCAGAUCUGCGC
(d) UUCAAUCGUGCG
Answer:
(b) CUCAAUCGUCCG

Question 40.
The promoter region of eukaryote is …………..
(a) TATAA
(b) AUGUT
(c) UUUGA
(d) AAAAU
Answer:
(a) TATAA

Question 41.
Match the following:

(a) A – iii B – i C – iv D – ii
(b) A – iii B – ii C – i D – iv
(c) A – iv B – i C – ii D – iii
(d) A – ii B – iii C – iv D – i
Answer:
(a) A – iii B-i C-iv D-ii

Question 42.
……………….number of codons, codes for cystine.
Answer:
Two

Question 43.
In sickle cell anaemia, the ……………….codon of p – globin gene is modified.
(a) Eighth
(b) Seventh
(c) Sixth
(d) Ninth
Answer:
(c) Sixth

Question 44.
Pick out the incorrect statement.
(a) tRNA acts as a adapter molecule
(b) Stop codons do not have tRNA’s
(c) Addition of amino acid leads to hydrolysis of tRNA {d) tRNA has four major loops
Answer:
(c) Addition of amino acid leads to hydrolysis of tRNA

Question 45.
Which of the following antibiotic inhibits the interaction between tRNA and mRNA?
(a) Neomycin
(b) Streptomycin
(c) Tetracycline
(d) Chloramphenicol
Answer:
(a) Neomycin

Question 47.
The cluster of genes with related function is called …………….
(a) Cistron
(b) Operon
(c) Muton
(d) Recon
Answer:
(b) Operon

Question 48.
Repressor protein of Lac operon binds to ……………of operon
(a) Promoter region
(b) Operator region
(c) terminator region
(d) inducer region
Answer:
(b) Operator region

Question 49.
Lac Z gene codes for ………………..
(a) Permease
(b) transacetylase
(c) P-galactosidase
(d) Aminoacyl transferase
Answer:
(c) p-galactosidase

Question 50.
Lac operon model was proposed by ………………
Answer:
Jacob and Monod

Question 51.
Approximate count of base pair in human genome is …………….
Answer:
3 x 10⁹ bp

Question 52.
Automated DNA sequences are developed by.
Answer:
Frederick Sanger

Question 53.
Which of the chromosome has a higher gene density?
(a) Chromosome 20
(b) Chromosome 19
(c) Chromosome 13
(d) Chromosome Y
Answer:
(b) Chromosome 19

Question 54.
Number of genes located in chromosome Y is
(a) 2968
(b) 213
(c) 2869
(d) 231
Answer:
(d) 231

Question 55.
How many structural genes are located in the lac operon of E.Coli?
(a) 4
(b) 3
(c) 2
(d) 1
Answer:
(b) 3

Question 56.
DNA fingerprinting technique was developed by
(d) Jacob and Monod
(b) Alec Jeffreys
(c) Frederick Sanger
Answer:
(b) Alec Jeffreys

Question 57.
In DNA fingerprinting, separation of DNA fragments is done by………
(a) Centrifugation
(b) Electrophoresis
(c) X-ray diffraction
(d) denaturation
Answer:
(b) Electrophoresis

Question 58.
SNP stands for
(a) Single nucleotide Polymorphism
(b) Single Nucleoside Polypeptide
(c) Single nucleotide Polymorphism
(d) Single nucleotide polymer
Answer:
(a) Single nucleotide Polymorphism

Question 59.
Specific sequences of mRNA that are not translated are……………..
Answer:
Untranslated Regions (UTR)

Question 60.
Non-coding or intervening DNA sequence is called ……………….
Answer:
Intron

Question 61.
………………is the monomer of DNA.
Answer:
Nucleotide

Question 62.
Which one of the following is wrongly matched?
(a) Transcription – Copying information from DNA to RNA
(b) Translation – Decoding information from mRNA to protein
(c) Replication – Making of DNA copies
(d) Splicing – Joining of exons with introns
Answer:
(d) ‘ Splicing – Joining of exons with introns

12th Bio Zoology Guide Molecular Genetics Two Marks Questions and Answers

Question 1.
Who proposed One gene – One enzyme hypothesis? Define it.
Answer:
George Beadle and Edward Tatum proposed the One gene – One enzyme hypothesis which states that one gene controls the production of one enzyme.

Question 2.
Differentiate nucleoside from nucleotide.
Answer:
Nucleoside :
Nucleoside subunit is composed of nitrogenous bases linked to a pentose sugar molecule.

Nucleotide :
Nucleotide subunit is composed of nitrogenous bases, a pentose sugar and a phosphate group.

Question 3.
State the key differences between DNA and RNA.
Answer:
DNA

1. DNA is made of deoxyribose sugar.
2. Nitrogenous bases of DNA are Adenine, Guanine, Cytosine and Thymine.

RNA:

1. RNA is made of ribose sugar.
2. Nitrogenous bases of RNA are Adenine,Guanine, Cytosine and Uracil.

Question 4.
Point out the nitrogenous bases of RNA.
Answer:
Adenine, Guanine, Cytosine and Uracil.

Question 5.
What makes the DNA and RNA acidic molecules?
Answer:
The phosphate functional group (P04) gives DNA and RNA the property of an acid at physiological pH, hence the name nucleic acid.

Question 6.
Which type of bond is formed

1. between a purine and pyrimidine base?
2. between the pentose sugar and adjacent nucleotide?

Answer:

1. Purine and pyrimidine bases are linked by hydrogen bonds.
2. Pentose sugar is linked to adjacent nucleotide by phosphodiester bonds.

Question 7.
DNA acts as genetic material for majority of living organisms and not the RNA. Give reasons to support the statement.
Answer:

1. RNA was reactive and hence highly unstable.
2. Some RNA molecules acts as gene regulators by binding to DNA and affect gene expression.
3. Uracil of RNA is less stable than thymine of DNA.

Question 8.
Name any two viruses whose genetic material is RNA.
Answer:

1. Tobacco Mosaic Virus (TMV)
2. Bacteriophage (B)

Question 9.
What are the properties that a molecule must possess to act as genetic material?
Answer:

• Self-replication
• Information storage
• Stability
• Variation through mutation

Question 10.
How many base pairs are present in one complete turn of DNA helix? What is the distance between two consecutive base pairs?
Answer:
There are ten base pairs in each turn with a distance of 0.34 x 10‘9 m between two adjacent base pairs.

Question 11.
What is a genophore?
Answer:
In prokaryotes such as E. coli though they do not have defined nucleus, the DNA is not scattered throughout the cell. DNA (being negatively charged) is held with some proteins (that have positive charges) in a region called the nucleoid. The DNA as a nucleoid is organized into large loops held by protein. DNA of prokaryotes is almost circular and lacks chromatin organization, hence termed genophore.

Question 12.
What is nucleosome? How many base pairs are there in a typical nucleosome?
Answer:
The negatively charged DNA is wrapped around the positively charged histone octamer to form a structure called nucleosome. A typical nucleosome contains 200 bp of DNA helix.

Question 13.
Expand and define NHC
Answer:
NHC : Non-histone Chromosomal protein.
In eukaryotes, apart from histone proteins, additional set of proteins are required for packing of chromatin at higher level and are referred as non – histone chromosomal proteins.

Question 14.
Differentiate between Heterochromatin and Euchromatin.
Answer:
Heterochromatin:

1. Region of nucleus where the chromatin are loosely packed and stains light are called Heterochromatin.
2. Transcriptionally inactive.

Euchromatin:

1. Region of the nucleus where the chromatin are tightly packed and stains dark are called Euchromatin.
2. Transcriptionally active.

Question 15.
Which is the widely accepted model of DNA replication? Who has proved it?
Answer:
Semi-conservative replication model. It was proved by Meselson and Stahl in 1958.

Question 16.
Name the chemical substance which is called by the name

1. Kornberg Enzyme
2. Ochoa’s enzyme

Answer:

1. DNA polymerase I is also known as Komberg enzyme.
2. Polynucleotide phosphorylase is also known as Ochoa’s enzyme.

Question 17.
Name the various types of prokaryotic DNA polymerase. State their role in replication process.
Answer:

Question 18.
What is the function of Deoxy nucleotide triphosphate in replication?
Answer:
Deoxy nucleotide triphosphate acts as substrate and also provides energy for polymerization reaction.

Question 19.
Given below are some events of eukaryotic replication. Name the enzymes involved in the process.

1. Unwinding of DNA
2. Joining of Okazaki fragments
3. Addition of nucleotides to a new strand
4. Correcting the repair

Answer:

1. Helicase
2. DNA ligase
3. DNA polymerase
4. Nuclease

Question 20.
Differentiate leading strand from lagging strand
Answer:
Leading strand :

1. Leading strand has the polarity 3’ →‘ 5’.
2. Replication is continuous.

Lagging strand :

1. Lagging strand has the polarity 5’ →‘ 3’
2. Replication is discontinuous.

Question 21.
What are Okazaki fragments?
Answer:
The discontinuously synthesized fragments of the lagging strand are called the Okazaki fragments are joined by the enzyme DNA ligase.

Question 22.
What is a replication fork?
Answer:
At the point of origin of replication, the helicases and topoisomerases (DNA gyrase) unwind and pull apart the strands, forming a Y-Shaped structure called the replication fork. There are two replication forks at each origin.

Question 23.
Apart from DNA polymerase, name any other four enzymes which were involved in DNA replication of eukaryotic cell.
Answer:
DNA ligase, Topoisomerase (DNA gyrase), Helicase and Nuclease.

Question 24.
Who proposed the central dogma? Write its concept.
Answer:

Question 25.
Define transcription and name the enzyme involved in this process.
Answer:
The process of copying genetic information from one strand of DNA into RNA is termed transcription. This process takes place in presence of DNA dependent RNA polymerase.

Question 26.
What is TATA box? State its function.
In eukaryotes, the promoter has AT-rich regions called TATA box or Goldberg-Hogness box. It acts as a binding site for RNA polymerase.

Question 27.
Structural gene of eukaryotes differ from prokaryotes. How?
In eukaryotes, the structural gene is monocistronic coding for only one protein whereas in prokaryotes the structural gene is polycistronic coding for many proteins.

Question 28.
What are the two major components of prokaryotic RNA polymerase? How do they act?
Answer:
Bacterial (prokaryotic) RNA polymerase consists of two major components, the core enzyme and the sigma subunit. The core enzyme (P1, p, and a) is responsible for RNA synthesis ” whereas a sigma subunit is responsible for recognition of the promoter.

Question 29.
Distinguish between exons and introns.
Answer:
Exons:
Expressed sequences (Coding sequences) of a eukaryotic gene

Introns:
Intervening sequences (non-coding sequences) of an eukaryotic gene

Question 30.
Define splicing.
Answer:
The process of removing introns from hnRNA is called splicing.

Question 31.
What is capping and tailing?
Answer:
In capping an unusual nucleotide, methyl guanosine triphosphate is added at the 5’ end of hnRNA, whereas adenylate residues (200-300) (Poly A) are added at the 3’ end in tailing.

Question 32.
If a double-stranded DNA has 20% of cytosine, calculate the percentage of adenine in DNA.
Answer:
Cytdsine = 20, hence Guanine = 20
As per ChargafFs rule (A+T) = (G+C) =100
Percent of Thymine + Adenine = 20 + 20 = 100
(T + A) = (20 + 20) = 100
(T + A) = 100 – (20 + 20)
T + A = 100 – 40
T + A = 60
Therefore the percent of Adenine will be 60/2 = 30%.

Question 33.
Mention the dual functions of AUG.
Answer:
AUG has dual functions. It acts as a initiator codon and also codes for the amino acid methionine.

Question 34.
How many codons are involved in termination of translation. Name them.
Answer:
Three codons terminate translation process. They are UAA, UAG and UGA.

Question 35.
Degeneracy of codon – comment.
Answer:
A degenerate code means that more than one triplet codon could code for a specific amino acid. For example, codons GUU, GUC, GUA and GUG code for valine.

Question 36.
Point out the exceptional categories to universality of genetic code.
Answer:
Exceptions to the universal nature of genetic code are noticed in prokaryotic mitochondrial and chloroplast genomes.

Question 37.
What are non-sense codons?
Answer:
UGA, UAA, and UAG are the non-sense codons, which terminate translation.

Question 38.
Name the triplet codons that code for

1. Tyrosine
2. Histidine

Answer:

1. Tyrosine – UAU, UAC
2. Histidine – CAU, CAC

Question 39.
Why hnRNA has to undergo splicing?
Answer:
Since hnRNA contains both coding sequences (exons) and non-coding sequences (introns) it has to undergo splicing to remove introns.

Question 40.
State the role of following codons in the translation process

1. AUG
2. UAA

Answer:

1. AUG is the initiator codon and also codes for methionine.
2. UAA is a terminator codon.

Question 41.
Given below is mRNA sequence. Mention the aminoacids sequence that is formed after its translation.
3’AUGAAAGAUGGGUAA5’
Answer:
Methionine – Lysine – Aspartic acid – Glycine

Question 42.
Name the four codons that codes valine.
Answer:
GUU, GUC, GUA and GUG.

Question 43.
The base sequence in one of the DNA strand is TAGC ATGAT. Mention the base sequence in its complementary strand.
Answer:
The complementary strand has ATCGTACTA.

Question 44.
Why t-RNA is called as adapter molecule?
Answer:
The transfer RNA (tRNA) molecule of a cell acts as a vehicle that picks up the amino acids scattered through the cytoplasm and also reads specific codes of mRNA molecules. Hence it is called as adapter molecule.

Question 45.
What do you mean by charging of tRNA? Name the enzyme involved in this process.
Answer:
The process of addition of amino acid to tRNA is known as aminoacylation or charging and the resultant product is called aminoacyl- tRNA (charged tRNA). Aminoacylation is catalyzed by an enzyme aminoacyl – tRNA synthetase.

Question 46.
What is UTR’s?
Answer:
mRNA also have some additional sequences that are not translated and are referred to as Untranslated Regions (UTR). UTRs are present at both 5’ end (before start codon) and at 3’ end (after stop codon).

Question 47.
What is the S – D sequence?
Answer:
The 5’ end of the mRNA of prokaryotes has a special sequence which precedes the initial AUG start codon of mRNA. This ribosome binding site is called the Shine – Dalgarno sequence or S-D sequence. This sequences base-pairs with a region of the 16Sr RNA of the small ribosomal subunit facilitating initiation.

Question 48.
Define translation unit.
Answer:
A translation unit in mRNA is the sequence of RNA that is flanked by the start codon on 5’ end and stop codon on 3’ end and codes of polypeptide.

Question 49.
Mention the inhibitory role of tetracycline and streptomycin in bacterial translation.
Answer:
Tetracycline inhibits binding between aminoacyl tRNA and mRNA.
Streptomycin inhibits initiation of translation and causes misreading.

Question 50.
At what stage, does the gene expression is regulated?
Answer:
Gene expression can be controlled or regulated at transcriptional or translational levels.

Question 51.
What is a operon? Give example.
Answer:
The cluster of genes with related functions is called operon.
E.g: lac operon in E.coli.

Question 52.
Considering the lac operon of E.coli, name the products of the following genes.

1. a) i gene
2. (b) lac Z gene
3. (c) lac Y gene
4. (d) lac a gene

Answer:

1. i gene – repressor protein
2. lac Z gene – fS-galactosidase
3. Lac Y gene – Permease
4. lac a gene – transacetylase

Question 53.
Expand

1. ETS
2. YAC.

Answer:

1. ETS: Expressed Sequence Tags
2. YAC: Yeast Artificial Chromosomes

Question 54.
Name the human chromosome that has

1. most number of genes
2. least number of genes

Answer:

1. Chromosome 1 has the maximum number of genes (2968 genes)
2. Chromosome Y has least genes (231 genes)

Question 55.
What are SNPs? Mention its uses.
Answer:
SNPs : Single nucleotide polymorphism. It helps to find chromosomal locations for disease-associated sequences and tracing human history.

Question 56.
Mention any four areas where DNA fingerprinting can be used.
Answer:

• Forensic analysis
• Pedigree analysis
• Conservation of wildlife
• Anthropological studies

12th Bio Zoology Guide Molecular Genetics Three Marks Questions and Answers

Question 57.
Classify nucleic acid-based on sugar molecules.
Answer:
There are two types of nucleic acids depending on the type of pentose sugar. Those containing deoxyribose sugar are called Deoxyribo Nucleic Acid (DNA) and those with ribose sugar are known as Ribonucleic Acid (RNA). The only difference between these two sugars is that there is one oxygen atom less in deoxyribose.

Question 58.
Both purines and pyrimidines are nitrogen bases yet they differ. How?
Answer:
Both purines and pyrimidines are nitrogen bases. The purine bases Adenine and Guanine have double carbon-nitrogen ring, whereas cytosine and thymine bases have single carbon-nitrogen ring.

Question 59.
How 5’ of DNA differ from its 3’?
Answer:
The 5’ of DNA refers to the carbon in the sugar to which phosphate (P04V) functional group is attached. The 3’ of DNA refers to the carbon in the sugar to which a hydroxyl (OH) group is attached.

Question 60.
State Chargaff’s rule.
Answer:
According to Erwin Chargaff,
a) Adenine pairs with Thymine with two hydrogen bonds.
b) Guanine pairs with Cytosine with three hydrogen bonds.

Question 61.
Chemically DNA is more stable than RNA – Justify.
Answer:
In DNA, the two strands being complementary, if separated (denatured) by heating can come together (renaturation) when appropriate condition is provided. Further 2’ OH group present at every nucleotide in RNA is a reactive group that makes RNA liable and easily degradable. RNA is also known to be catalytic and reactive. Hence, DNA is chemically more stable and chemically less reactive when compared to RNA. Presence of thymine instead of uracil in DNA confers additional stability to DNA.

Question 62.
Write in simple about semi-conservative mode of DNA replication.
Answer:
Semi-conservative replication was proposed by Watson and Crick in 1953. This mechanism of replication is based on the DNA model. They suggested that the two polynucleotide strands of DNA molecule unwind and start separating at one end. During this process, covalent hydrogen bonds are broken. The separated single strand then acts as template for the synthesis of a new strand. Subsequently, each daughter double helix carries one polynucleotide strand from the parent molecule that acts as a template and the other strand is newly synthesised and complementary to the parent strand.

Question 63.
Draw a simplified diagram of nucleosome and label it.
Answer:

Question 64.
What is a primer?
Answer:
A primer is a short stretch of RNA. It initiates the formation of new strand. The primer produces 3’-OH end on the sequence of ribonucleotides, to which deoxyribonucleotides are added to form a new strand.

Question 65.
Both strands of DNA are not copied during transcription. Give reason.
Answer:
Both the strands of DNA are not copied during transcription for two reasons.

1. If both the strands act as a template, they would code for RNA with different sequences. This in turn would code for proteins with different amino acid sequences. This would result in one segment of DNA coding for two different proteins, hence complicate the genetic information transfer machinery.

2. If two RNA molecules were produced simultaneously, double-stranded RNA complementary to each other would be formed. This would prevent RNA from being translated into proteins.

Question 66.
What do you mean by a template strand and coding strand?
Answer:
DNA dependent RNA polymerase catalyses the polymerization in only one direction, the strand that has the polarity 3’→5’ acts as a template, and is called the template strand. The other strand which has the polarity 5’→3’ has a sequence same as RNA (except thymine instead of uracil) and is displaced during transcription. This strand is called coding strand.

Question 67.
Name the factors that are responsible for initiation and termination of transcription in prokaryotes.
Answer:
Sigma factor is responsible for initiation of transcription.
Rho factor is responsible for termination of transcription.

Question 68.
Name the major RNA types of prokaryotes and mention their role.
Answer:
In prokaryotes, there are three major types of RNAs: mRNA, tRNA, and rRNA. All three \ RNAs are needed to synthesize a protein in a cell. The mRNA provides the template, tRNA
brings amino acids and reads the genetic code, and rRNAs play structural and catalytic role during translation.

Question 69.
Define genetic code.
Answer:
P The order of base pairs along DNA molecule controls the kind and order of amino acids found in the proteins of an organism. This specific order of base pairs is called genetic code.

Question 70.
Explain Wobble hypothesis.
Answer:
Wobble Hypothesis is proposed by Crick (1966) which states that tRNA anticodon has the t. ability to wobble at its 5’ end by pairing with even non-complementary baSe of mRNA codon.’ According to this hypothesis, in codon-anticodon pairing the third base may not be complementary. The third base of the codon is called wobble base and this position is called wobble position. The actual base pairing occurs at first two positions only. The importance of Wobbling hypothesis is that it reduces the number of tRNAs required for polypeptide synthesis and it overcomes the effect of code degeneracy.

Question 71.
Explain the nature of eukaryotic ribosomes.
Answer:
The ribosomes of eukaryotes (80 S) are larger, consisting of 60 S and 40 S subunits. ‘S’ denotes the sedimentation efficient which is expressed as Svedberg unit (S). The larger subunit in eukaryotes consist of a 23 S RNA and 5Sr RNA molecule and 31 ribosomal proteins. The smaller eukaryotic subunit consist of 18Sr RNA component and about 33 proteins.

Question 72.
Expand and define ORF.
Answer:
Any sequence of DNA or RNA, beginning with a start codon and which can be translated into a protein is known as an Open Reading Frame (ORF).

Question 73.
What are the components of the initiation complex of prokaryotic translation?
Answer:
Initiation of translation in E. coli begins with the formation of an initiation complex, consisting of the 30S subunits of the ribosome, a messenger RNA and the charged N-formyl methionine tRNA (f^met -t RNA f^met), three proteinaceous initiation factors (IF 1, IF2, IF3), GTP (Guanine Tri Phosphate) and Mg ²⁺.

Question 74.
Explain the components of operon.
Answer:
Structure of the operon: Each operon is a unit of gene expression and regulation and consists of one or more structural genes and an adjacent operator gene that controls transcriptional activity of the structural gene.

• The structural gene codes for proteins, rRNA and tRNA required by the cell.
• Promoters are the signal sequences in DNA that initiate RNA synthesis. RNA polymerase binds to the promoter prior to the initiation of transcription.
• The operators are present between the promoters and structural genes. The repressor protein binds to the operator region of the operon.

12th Bio Zoology Guide Molecular Genetics Five Marks Questions and Answers

Question 75.
Describe the Hershey and chase experiment. What is concluded by their experiment?
Answer:
Alfred Hershey and Martha Chase (1952) conducted experiments on bacteriophages that infect bacteria. Phage T₂ is a virus that infects the bacterium Escherichia coli. When phages (virus) are added to bacteria, they adsorb to the outer surface, some material enters the bacterium, and then later each bacterium lyses to release a large number of progeny phage. Hershey and Chase wanted to observe whether it was DNA or protein that entered the bacteria. All nucleic acids contain phosphorus, and contain sulphur (in the amino acid cysteine and methionine).

Hershey and Chase designed an experiment using radioactive _ isotopes of Sulphur (³⁵S) and phosphorus (³²P) to keep separate track of the viral protein and nucleic acids during the infection process. The phages were allowed to infect bacteria in culture medium which containing the radioactive isotopes ³⁵S or ³²P. The bacteriophage that grew in the presence of ³⁵S had labelled proteins and bacteriophages grown in the presence of ³²P had labelled DNA. The differential labelling thus enabled them to identity DNA and proteins of the phage. Hershey and Chase mixed the labelled phages with unlabeled E. coli and allowed bacteriophages to attack and inject their genetic material.

Soon after infection (before lysis of bacteria), the bacterial cells were gently agitated in a blender to loosen the adhering phase particles. It was observed that only ³²P was found associated with bacterial cells and ³⁵S was in the surrounding medium and not in the bacterial cells. When phage progeny was studied for radioactivity, it was found that it carried only ³²P and not ³⁵S. These results clearly indicate that only DNA and not protein coat entered the bacterial cells. Hershey and Chase thus conclusively proved that it was DNA, not protein, which carries the hereditary information from virus to bacteria.

Question 76.
Explain the properties of DNA that makes it an ideal genetic material.
Answer:
Self Replication: It should be able to replicate. According to the rule of base pairing and complementarity, both nucleic acids (DNA and RNA) have the ability to direct duplications. Proteins fail to fulfill this criteria.

Stability: It should he stable structurally and chemically. The genetic material should be stable enough not to change with different stages of life cycle, age or with change in physiology of the organism. Stability as one of property of genetic material was clearly evident in Griffith’s transforming principle. Heat which killed the bacteria did not destroy some of the properties of genetic material. In DNA the two strands being complementary, if separated (denatured) by heating can come together (renaturation) when appropriate condition is provided. Further 2’ OH group present at every nucleotide in RNA is a reactive group that makes RNA liable and easily degradable. RNA is also known to be catalytic and reactive. Hence, DNA is chemically more stable and chemically less reactive when compared to RNA. Presence of thymine instead of uracil in DNA confers additional stability to DNA.

Information storage: It should be able to express itself in the form of ‘Mendelian characters’. RNA can directly code for protein synthesis and can easily express the characters. DNA, however depends on RNA for synthesis of proteins. Both DNA and RNA can act as a genetic material, but DNA being more stable stores the genetic information and RNA transfers the genetic information.

Variation through mutation: It should be able to mutate. Both DNA and RNA are able to mutate. RNA being unstable, mutates at a faster rate. Thus viruses having RNA genome with shorter life span can mutate and evolve faster. The above discussion indicates that both RNA and DNA can function as a genetic material. DNA is more stable, and is preferred for storage of genetic information.

Question 77.
How the DNA is packed in a eukaryotic cell? ft
Answer:
In eukaryotes, organization is more complex. Chromatin is formed by a series of repeating units called nucleosomes. Komberg proposed a model for the nucleosome, in which 2 molecules of the four histone proteins H2A, H2B, H3 and H4 are organized to form a unit of eight molecules called histone octamere. The negatively charged DNA is wrapped around the positively charged histone octamere to form a structure called nucleosome. A typical * nucleosome contains 200 bp of DNA helix.

The histone octameres are in close contact and DNA is coiled on the outside of nucleosome. Neighbouring nucleosomes are connected by linker DNA (HI) that is exposed to enzymes. The DNA makes two complete turns around the histone octameres and the two turns are sealed off by an HI molecule. Chromatin lacking HI has a beads-on-a-string appearance in which DNA inters and leaves the nucleosomes at random places.

HI of one nucleosome can interact with 33l of the neighbouring nucleosomes resulting in the further folding of the fibre. The chrof&atin fiber in interphase nuclei and mitotic chromosomes have a diameter that vary between 200-300 nm and represents inactive chromatin. 30 nm fibre arises from the folding of nucfeosbme, chains into a solenoid structure having six nucleosomes per turn. This structure is stabilized by interaction between different HI molecules. DNA is a solenoid and packed about,%)_folds.

The hierarchical nature of chromosome structure is illustrated. Additional set of^pteins are required for packing of chromatin at higher level and are referred to as non-histone chromosomal proteins (NHC). In*,a typical nucleus, some regions of chromatin are Ibosely packed (lightly stained) and are referred to as euchromatin. The chromatin that is,-tightly packed (stained darkly) is called heterochromatin. Euchromatin is transcriptionally active and heterochromatin is transcriptionally inactive.

Question 78.
Meselson and Stahl’s experiment proved the semi-coflBptervation mode of DNA replication. Explain.
Answer:
The mode of DNA replication was determined in 1958 by Meselson and Stahl. They designed an experiment to distinguish between semi-conservative, conservative and dispersive replications. In their experiment, they grew two cultures of E.coli for many generations in separate media. The ‘heavy’ culture was grown in a medium in which the nitrogen source (NH₄Cl) contained the heavy isotope ₁₅N and the ‘ light’ culture was grown in a medium in which the nitrogen source contained light isotope ₁₄N for many generations.

At the end of growth, they observed that the bacterial DNA in the heavy culture contained only ₁₅N and in the light culture only ₁₅N. The heavy DNA could be distinguished from light DNA (₁₅N from ₁₄N) with a technique called Cesium Chloride (CsCl) density gradient centrifugation. In this process, heavy and light DNA extracted from cells in thtytwo cultures settled into two distinct and separate bands (hybrid DNA). ;

The heavy culture (₁₅N) was then transferred into a medium that had only NH₄Cl, and took samples at various definite time intervals (20 minutes duration). After the first replication, they extracted DNA and subjected it to density gradient centrifugation. The DNA settled into a band that was intermediate in position between the previously determined heavy and light bands. After the second replication (40 minutes duration), they again extracted DNA samples, and this time found the DNA settling into two bands, one at the light band position and one at intermediate position. These results confirm Watson and Crick’s semi-conservative replication hypothesis.

Question 79.
Give a detailed account of a transcription unit.
Answer:
A transcriptional unit in DNA is defined by three regions, a promoter, the structural gene and a terminator. The promoter is located towards the 5 ’ end. It is a DNA sequence that provides binding site for RNA polymerase. The presence of promoter in a transcription unit, defines the template and coding strands. The terminator region located towards the 3’ end of the coding strand contains a DNA sequence that causes the RNA polymerase to stop transcribing.

In eukaryotes the promoter has AT-rich regions called TATA box (Goldberg- Hogness box) ‘ and in prokaryotes this region is called Pribnow box. Besides promoter, eukaryotes also require an enhancer.

The two strands of the DNA in the structural gene of a transcription unit have opposite polarity. DNA dependent RNA polymerase catalyses the polymerization in only one direction, the strand that has the polarity 3’→5’ acts as a template, and is called the template strand. The other strand which has the polarity 5 ’→3’ has a sequence same as RNA (except thymine instead of uracil) and is displaced during transcription. This strand is called the coding strand (Fig. 5.7).

The structural gene may be monocistronic (eukaryotes) or polycistronic (prokaryotes). In eukaryotes, each mRNA carries only a single gene and encodes information for only a single protein and is called monocistronic mRNA. In prokaryotes, clusters of related genes, known as operon, often found next to each other on the chromosome are transcribed together to give a single mRNA and hence are polycistronic.

Question 80.
Explain the transcription process in prokaryotes with the needed diagram.
Answer:
In prokaryotes, there are three major types of RNAs: mRNA, tRNA, and rRNA.
All three RNAs are needed to synthesize a protein in a cell. The mRNA provides the template, tRNA brings amino acids and reads the genetic code, and rRNAs play a structural and catalytic role during translation. There is a single DNA-dependent RNA polymerase that catalyses the transcription of all types of RNA. It binds to the promoter and initiates transcription (Initiation).

The polymerases binding sites are called promoters. It uses nucleoside triphosphate as substrate and polymerases in a template depended fashion following the rule of complementarity. After the initiation of transcription, the polymerase continues to elongate the RNA, adding one nucleotide after another to the growing RNA chain. Only a short stretch of RNA remains bound to the enzyme, when the polymerase reaches a terminator at the end of a gene, the nascent RNA falls off, so also the RNA polymerase.

The RNA polymerase is only capable of catalyzing the process of elongation. The RNA polymerase associates transiently with initiation factor sigma (a) and termination factor rho (p) to initiate and terminate the transcription, respectively. Association of RNA with these factors instructs the RNA polymerase either to initiate or terminate the process of transcription.

In bacteria, since the mRNA does not require any processing to become active and also since transcription and translation take place simultaneously in the same compartment (since there is no separation of cytosol and nucleus in bacteria), many times the translation can begin much before the mRNA is fully transcribed. This is because the genetic material is not ‘ separated from other cell organelles by a nuclear membrane consequently; transcription and translation can be coupled in bacteria.

Question 81.
Write the salient features of genetic code.
Answer:
The salient features of genetic code are as follows:

• The genetic codon is a triplet code and 61 codons code for amino acids and 3 codons do not code for any amino acid and function as stop codon (Termination).
• The genetic code is universal. It means that all known living systems use nucleic acids and the same three base codons (triplet codon) direct the synthesis of protein from amino
• acids. For example, the mRNA (UUU) codon codes for phenylalanine in all cells of all organisms. Some exceptions are reported in prokaryotic, mitochondrial, and chloroplast genomes. However, similarities are more common than differences.
• A non-overlapping codon means that the same letter is not used for two different codons. For instance, the nucleotide sequence GUTJ and GUC represents only two codons.
• It is comma less, which means that the message would be read directly from one end to the other i.e., no punctuation are needed between two codes.
• A degenerate code means that more than one triplet codon could code for a specific amino acid. For example, codons GUU, GUC, GUA and GUG code for valine.
• Non-ambiguous code means that one codon will code for one amino acid.
• The code is always read in a fixed direction i.e. from 5’ → 3’ direction called polarity.
• AUG has dual functions. It acts as an initiator codon and also codes for the amino acid methionine.
• UAA, UAG (tyrosine), and UGA (tryptophan) codons are designated as termination (stop) codons and also are known as “non-sense” codons.

Question 82.
Mutations on genetic code affect the phenotype. Describe with an example.
Answer:
The simplest type of mutation at the molecular level is a change in nucleotide that substitutes one base for another. Such changes are known as base substitutions which may occur spontaneously or due to the action of mutagens. A well studied example is sickle cell anaemia in humans which results from a point mutation of an allele of (3-haemoglobin gene (PHb).

A haemoglobin molecule consists of four polypeptide chains of two types, two a chains and two β-chains. Each chain has a heme group on its surface. The heme groups are involved in the binding of oxygen. The jruman blood disease, sickle cell anaemia is due to abnormal haemoglobin. This abnormality in haemoglobin is due to a single base substitution at the sixth codon of the beta globingene from GAG to GTG in β -chain of haemoglobin. It results in a change of amino acid gluconic acid to valine at the 6th position of the β -chain. This is the classical example of point mutation that results in the change of amino acids residue glutamic acid to valine. The mutant haemoglobin undergoes polymerisation under oxygen tension causing the change in the shape of the RBC from biconcave to a sickle shaped structure.

Question 83.
Explain the mechanism of AteArperon of the E-coli.
Answer:
The Lac (Lactose) operon: The metabolism of lactose in E.coli requires three enzymes – permease, β-galactosidase (P-gat) and transacetylase. The enzyme permease is needed for entry of lactose into the cell, Pjgglactosidase brings about hydrolysis of lactose to glucose and galactose, while transacetylqSSgtransfers acetyl group from acetyl Co A to P-galactosidase. The lac operon consists of one-regulator gene (T gene refers to inhibitor) promoter sites (p), and operator site (o). Besides these, it has three structural genes namely lac z, y and lac a. The lac ‘z’ gene codes for P-gaiaqtttsidase, lac ‘y’ gene codes for permease and ‘a’ gene codes for transacetylase.

Jacob and Monod proposed the classical model of Lac operon to explain gene expression and regulation in E.coli. In lac a polycistronic structural gene is regulated by a common promoter and regulatory genfc When the cell is using its normal energy source as glucose, the ‘i’ gene transcribes a repressor mRNA and after its translation, a repressor protein is produced. It binds to the operator region of the operon and prevents translation, as a result, β-galactosidase is not produced. In the absence of preferred carbon source such as glucose, if lactose is available as an energy source for the bacteria then lactose enters the cell as a result of permease enzyme. Lactose acts as an inducer and interacts with the repressor to inactivate it. The repressor protein binds to the operator of the operon and prevents RNA polymerase from transcribing the operon. In the presence of inducer, such as lactose or allolactose, the repressor is inactivated by interaction with the inducer. This allows RNA polymerase to bind to the promotor site and transcribe the operon to produce lac mRNA which enables formation of all the required enzymes needed for lactose metabolism. This regulation of lac operon by the repressor is an example of negative control of transcription initiation.

Question 84.
What are the objectives of the Human Genome project?
Answer:

• The main goals of the Human Genome Project are as follows:
•  Identify all the genes (approximately 30000) in human DNA.
• Determine the sequence of the three billion chemical base pairs that make up the human DNA.
• To store this information in databases.
• Improve tools for data analysis.
• Transfer related technologies to other sectors, such as industries.
• Address the ethical, legal and social issues (ELSI) that may arise from the project.

Question 85.
Write the salient features of the Human Genome Project.
Answer:

• Although human genome contains 3 billion nucleotide bases, the DNA sequences that encode proteins make up only about 5% of the genome.
• An average gene consists of 3000 bases, the largest known human gene being dystrophin with 2.4 million bases.
• The function of 50% of the genome is derived from transposable elements such as LINE Molecular Genetics 315 and ALU sequence.
• Genes are distributed over 24 chromosomes. Chromosome 19 has the highest gene density. Chromosomes 13 and the Y chromosomes have the lowest gene densities.
• The chromosomal organization of human genes shows diversity.
• There may be 35000-40000 genes in the genome and almost 99.9 nucleotide bases are exactly the same in all people.
• Functions for over 50 percent of the discovered genes are unknown.
• Less than 2 percent of the genome codes for proteins.
• Repeated sequences make up very large portion of the human genome. Repetitive sequences have no direct coding functions but they shed light on chromosome structure, dynamics and evolution (genetic diversity).
• Chromosome 1 has 2968 genes, whereas chromosome ’Y’ has 231 genes.
• Scientists have identified about 1.4 million locations, where single-base DNA differences (SNPs – Single nucleotide polymorphism – pronounce as ‘snips’) occur in humans.
• Identification of ‘SNIPS’ is helpful in finding chromosomal locations for disease-associated sequences and tracing human history.

Question 86.
Describe the principle involved in the DNA fingerprinting technique.
Answer:
The DNA fingerprinting technique was first developed by Alec Jeffreys in 1985. The DNA of a person and fingerprints are unique. There are 23 pairs of human chromosomes with 1.5 million pairs of genes. It is a well known fact that genes are segments of DNA which differ in the sequence of their nucleotides. Not all segments of DNA code for proteins, some DNA .segments have a regulatory function, while others are intervening sequences (introns) and still others are repeated DNA sequences.

In DNA fingerprinting, short repetitive nucleotide sequences are specific for a person. These nucleotide sequences are called variable number tandem repeats (VNTR). The VNTRs of two persons generally show variations and are useful as genetic markers. DNA fingerprinting involves identifying differences in some specific regions in DNA sequence called repetitive DNA, because in these sequences, a small stretch of DNA is repeated many times. These repetitive DNA are separated from bulk genomic DNA as different peaks during density gradient centrifugation. The bulk DNA forms a major peak and the other small peaks are referred to as satellite DNA.

Depending on base composition (A: T rich or G : C rich), length of segment and number of repetitive units, the satellite DNA is classified into many subcategories such as micro-satellites and mini¬satellites, etc. These sequences do not code for any proteins, but they form a large portion of the human genome. These sequences show a high degree of polymorphism and form the basis of DNA fingerprinting. DNA isolated from blood, hair, skin cells, or other genetic evidences left at the scene of a crime can be compared through VNTR patterns, with the DNA of a criminal suspect to determine guilt or innocence. VNTR patterns are also useful in establishing the identity of a homicide victim, either from DNA found as evidence or from the body itself.

Question 87.
Draw a flow chart depicting the steps of the DNA fingerprinting technique.
Answer:

Higher Order Thinking Skills (HOTs) Questions

Question 1.
An mRNA strand has a series of triplet codons of which the first three codons are given below
(a) AUG (b) UUU (c) UGC
(i) Name the amino acid encoded by these triplet codons.
(ii) Mention the DNA sequence from which these triplet codons would have transcribed?
Answer:
(i) AUG codes for Methionine UUU codes for Phenylalanine UGC codes for Cysteine
(ii) TAC sequence of DNA is transcribed to AUG
AAA sequence of DNA is transcribed to UUU
ACG sequence of DNA is transcribed to UGC

Question 2.
Given below are the structures of tRNA molecules which are involved in translation process. In one tRNA, codon is mentioned but not the amino acid. In another tRNA molecule, amino acid is named and not the codon. Complete the figure by mentioning the respective amino acids and codons.
Answer:

Question 3.
A DNA fragment possesses 32 adenine bases and 32 cytosine bases. How many total number of nucleotides does that DNA fragment contains? Explain.
Answer:
128 nucleotides. Adenine always pair Thymine base.
If there are 32 adenine bases then there must be 32 Thymine bases. Similarly cytosine pairs with guanine. If cytosine bases are 32 in number the guanine bases will be equal to cytosine. So it makes a total of 128 nucleotides.

Question 4.
Following is a DNA sequence representing a part of the gene TAC TCG CCC TAT UAA CCC AAA ACC TCT using this derive A.

1. The RNA transcript
2. The spliced mRNA (consider all the codons with two Aderine bases are introns)
3. The total number of amino acids coded by the mRNA

Answer:

1. RNA transcript: AUG UGC GGG AUA GGG UUU UGG AGA
2. Spliced mRNA: AUG UGC GGG GGG UUU UGG
3. 6 amino acids are coded by mRNA

Question 5.
Complete the molecular processes by naming them

1. DNA → DNA
2. mRNA → Protein
3. RNA transcript → mRNA

Answer:

1. Replication
2. Translation
3. Splicing

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Chemistry Guide Pdf Chapter 3 p-Block Elements – II Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Chemistry Solutions Chapter 3 p-Block Elements – II

12th Chemistry Guide p-Block Elements – II Text Book Questions and Answers

Part – I – Text Book Evaluation

I. Choose the correct answer

1. In which of the following, NH₃ is not used?
a) Nessler’s reagent
b) Reagent for the analysis of IV group basic radical
c) Reagent for the analysis of III group basic radical
d) Tollen’s reagent
Answer:
a) Nessler’s reagent

2. Which is true regarding nitrogen?
a) least electronegative element
b) has low ionisation enthalpy than oxygen
c) d – orbitals available
d) ability to form pπ -pπ bonds with itself
Answer:
d) ability to form pπ -pπ bonds with itself

3. An element belongs to group 15 and 3rd period of the periodic table, its electronic configuration would be
a) Is² 2s² 2p⁴
b) Is² 2s² 2p³
c) Is² 2s² 2p⁶ 3s2² 3p²
d) Is² 2s² 2p⁶ 3s² 3p³
Answer:
d) Is² 2s² 2p⁶ 3s² 3p³

4. Solid (A) reacts with strong aqueous NaOH liberating a foul smelling gas(B) which spontaneously burn in air giving smoky rings. A and B are respectively
a) P₄(red) and PH₃
b) P₄ (white) and PH₃
c) S₈ and H₂S
d) P₄(white) and H₂S
Answer:
b) P₄ (white) and Ph₃

5. On hydrolysis, PCl₃ gives
a) H₃PO₃
b) PH₃
c) H₃PO₄
d) POCl₃
Answer:
a) H₃PO₃

6. P₄O₆ reacts with cold water to give
a) H₃PO₃
b) H₄P₂O₇
c) HPO₃
d) H₃PO₄
Answer:
a) H₃PO₃

7. The basicity of pyrophosphorous acid (H₄P₂O₃) is
a) 4
b) 2
c) 3
d) 5
Answer:
b) 2

8. The molarity of given orthophosphoric acid solution is 2M. Its normality is
a) 6N
b) 4N
c) 2N
d) none of these
Answer:
a) 6N
Normality = M x basicity =2×3 = 6

9. Assertion : bond dissociation energy of fluorine is greater than chlorine gas
Reason : chlorine has more electronic repulsion than flourine
a) Both assertion and reason are true and reason is the correct explanation of assertion.
b) Both assertion and reason are true bu t reason is not the correct explanation of assertion
c) Assertion is true bu t reason is false
d) Both assertion and reason are false
Answer:
d) Both assertion and reason are false The converse is true

10. Among the following, which is the strongest oxidizing agent?
a) Cl₂
b) F₂
C) Br₂
d) l₂
Answer:
b) F₂

11. The correct order of the thermal stability of hydrogen halide is (PTA – 4)
a) Hl > HBr > HCl > HF
b) HF > HCl > HBr > HI
c) HCl > HF > HBr > HI
d) HI > HCl > HF > HBr
Answer:
b) HF > HCl > HBr > HI

12. Which one of the following compounds is not formed?
a) XeOF₄
b) XeO₃
c) XeF₂
d) NeF₂
Answer:
d) NeF₂

13. Most easily liquefiable gas is
a) Ar
b) Ne
c) He
d) Kr
Answer:
c) He

14. XeFg on complete hydrolysis produces
a) XeOF₄
b) XeO₂F₂
c) XeO₃
d) XeO₂
Answer:
c) XeO₃

15. Which of the following is strongest acid among all?
a) HI
b) HF
c) HBr
d) HCl
Answer:
a) HI
Reason : H-I bond is weakest.

16. Which one of the following orders is correct for the bond dissociation enthalpy of halogen molecules? (NEET)
a) Br₂ > I₂ > F₂ > Cl₂
b) F₂ > Cl₂ > Br₂ > I₂
c) I₂ > Br₂ > Cl₂ > F₂
d) Cl₂ > Br₂ > F₂ > I₂
Answer:
d) Cl₂ > Br₂ > F₂ > I₂

17. Among the following the correct order of acidity is (NEET)
a) HClO₂ < HClO < HClO₃ < HClO₄
b) HClO₄ < HClO₂ < HClO < HClO₃
c) HClO₃ < HClO₄ < HClO₂ < HClO
d) HClO < HClO₂ < HClO₃ < HClO₄
Answer:
d) HClO < HClO₂ < HClO₃ < HClO₄

18. When copper is heated with cone HNOs it produces
a) Cu (NO₃)₂, NO and NO₂
b) Cu (NO₃)₂ and N₂O
c) Cu (NO₃)₂ and NO₂
d) Cu (NO₃)₂ and NO
Answer:
c) Cu (NO₃)₂ and NO₂

II. Answer the following questions.

Question 1.
What is the inert pair effect?
Answer:
In p-block elements, as we go down the group, two electrons present in the valence s-orbital become inert and are not available for bonding (only p-orbital involves chemical bonding). This is called inert pair effect.

Question 2.
Chalcogens belong to p-block.
Answer:
Give reason.

• Chalcogens belong to p-block elements.
• Because their outer electronic configuration is ns² np⁴.
• In these elements the last electron enters np orbital.
• Hence they belong to p-block elements.

Question 3.
Explain why fluorine always exhibits an oxidation state of -1?
Answer:
Fluorine the most electronegative element than other halogens and cannot exhibit any positive oxidation state. Fluorine does not have a d-orbital while other halogens have d-orbitals. Therefore fluorine always exhibits an oxidation state of-1 and others in the halogen family shows +1, +3, +5 and +7 oxidation states.

Question 4.
Give the oxidation state of halogen in the following
Answer:
a) OF₂
b) O₂F₂
c) Cl₂O₃
d) I₂O₄

Question 5.
What are interhalogen compounds? Give examples
Answer:
Each halogen combines with other halogens to form a series of compounds called interhalogen compounds. For example, Fluorine reacts readily with oxygen and forms difluorine oxide (F₂O) and difluorine dioxide (F₂O₂).

Question 6.
Why fluorine is more reactive than other halogens? (PTA – 1, 3)
Answer:
Fluorine is the most reactive element among halogens. This is due to the low value of F – F bond dissociation energy.

Question 7.
Give the uses of helium. (PTA – 2)
Answer:

1. Helium and oxygen mixture is used by divers in place of air oxygen mixture. This prevents the painful dangerous condition called bends.
2. Helium is used to provide an inert atmosphere in the electric arc welding of metals
3. Helium has the lowest boiling point hence used in cryogenics (low-temperature science).
4. It is much less denser than air and hence used for filling air balloons.

8. What is the hybridisation of iodine in IF₇? Give its structure. (PTA – 5)
Answer:

9. Give the balanced equation for the reaction between chlorine with cold NaOH and hot NaOH
Answer:
Chlorine reacts with cold NaOH to give sodium hypochlorite

Chlorine reacts with hot NaOH to give sodium chlorate

10. How will you prepare chlorine in the laboratory? (PTA – 2)
Answer:
1. Chlorine is prepared by the action of cone, sulphuric acid on chlorides in presence of manganese dioxide.
4NaCl + MnO₂ + 4H₂SO₄ → Cl₂ + MnCl₂ + 4NaHSO₄ + 2H₂O

2. It can also be prepared by oxidising hydrochloric acid using various oxidising agents such as manganese dioxide, lead dioxide, potassium permanganate or dichromate.
PbO₂ + 4HCl → PbCl₂ + 2H₂O + Cl₂
MnO₂ + 4HCl → MnCl₂ + 2H₂O + Cl₂
2KMnO₄ + 16HCl → 2KCl + 2MnCl + 8H₂O + 5Cl₂
K₂Cr₂O₇ + 14HCl → 2KCl + 2CrCl₃ + 7H₂O + 3Cl₂

3. When bleaching powder is treated with mineral acids chlorine is liberated
CaOCl₂ + 2HCl → CaCl₂ + H₂O + Cl₂
CaOCl₂ + H₂SO₄ → CaSO₄ + H₂O + Cl₂

11. Give the uses of sulphuric acid.
Answer:
Sulphuric acid is used

• In the manufacture of fertilisers, ammonium sulphate and superphosphates.
• In the manufacture of other chemicals such as hydrochloric acid, nitric acid etc.,
• as a drying agent.
• in the preparation of pigments, explosives, etc.,

12. Give a reason to support that sulphuric acid is a dehydrating agent. (PTA – 1)
Answer:

• Sulphuric acid is highly soluble in water.
• It has a strong affinity towards water.
• Hence it can be used as a dehydrating agent.
• When dissolved in water, it forms mono (H₂SO₄.H₂O) and dihydrates (H₂SO₄.2H₂O) and the reaction is exothermic.
  ex
  C₁₂H₂₂O₁₁ + H₂SO₄ → 12C + H₂SO₄.11H₂O
  HCOOH + H₂SO₄ → CO + H₂SO₄.H₂O

13. Write the reason for the anomalous behaviour of Nitrogen.
Answer:
1. Due to its small size, high electronegativity, high ionisation enthalpy and absence of d-orbitals.

2. N, has a unique ability to form pπ – pπ multiple bond whereas the heavier members of this group (15) do not form pπ – pπ bond, because their atomic orbitals are so large and diffused that they cannot have effective overlapping.

3. Nitrogen exists a diatomic molecule with triple bond between the two atoms whereas other elements form single bond in the elemental state.

4. N cannot form dπ – pπ bond due to the absence of d – orbitals whereas other elements can.

14. Write the molecular formula and structural formula for the following molecules
Answer:
a) Nitric add
b) dinitrogen pentoxide
c) phosphoric acid
d) phosphine

15. Give the uses of argon.
Answer:
Argon prevents the oxidation of hot filament and prolongs the life in filament bulbs.

16. Write the valence shell electronic configuration of group -15 elements.
Answer:
Valence shell electronic configuration of group 15 elements is ns²np³

17. Give two equations to illustrate the chemical behaviour of phosphine.
Answer:
Basic Nature :
Phosphine is weakly basic and forms phosphonium salts.
PH₃ +HI → PH₄ I
PH₄I + H₂O \(\underrightarrow { \triangle } \) PH₃ + H₃O⁺ + I^–
It react with halogen to give phosphorous penta halidae
PH₃ +4Cl₂ → PCl₅ + 3HCl

Combustion:
When phosphine is heated with air or oxygen it burns to give metaphosphoric acid

18. Give a reaction between nitric acid and a basic oxide.
Answer:
Nitric acid rects with a basic oxide to form salt and water.
3 FeO + 10HNO₃ → Fe (NO₃)₃ + NO + 5H₂O

19. What happens when PCl₅ is heated?
Answer:
On heating Phoshorous pentachloride decomposes into phosphorus trichloride and chlorine

20. Suggest a reason why HF is a weak acid, whereas binary acids of all other halogens are strong acids.
Answer:

• HF is only slightly ionised, hence it is a weak acid
• Other halogen acids are almost completely ionised, hence they are strong acids.
• Among halogen acids, the electronegativity difference is maximum (1.9) in HF acid.
• Hence the bond between H and F is stronger and the acid HF is weaker.

21. Deduce the oxidation number of oxygen in hypofluorous acid – HOF.
Answer:
Oxidation number of F = -1
Oxidation number of H = +1
Oxidation number of O in HOF = x
(+1) + x + (-1) = 0
x = 0
Oxidation number of O in HOF = 0

22. What type of hybridisation occur in (PTA – 5)
a) BrF₅
b) BrF₃

23. Complete the following reactions
1. NaCl+ MnO₂ + H₂SO₄ →
2. NaNO₂ + HCl →
3. P₄ + Na0H + H₂O →
4. AgNO₃ + PH₃ →
5. Mg + HNO₃ →

8. Sb + Cl₂ →
9. HBr + H₂SO₄ →
10. XeF₆ + H₂O →
11. XeO₆^4- + Mn²⁺ + H⁺ →
12. XeOF₄ + SiO₂ →

Answer:

III. Evaluate yourself

1. Write the products formed in the reaction of nitric acid (both dilute and concentrated) with zinc.

12th Chemistry Guide p-Block Elements – II Additional Questions and Answers

Part II – Additional Questions

I. Match the following

1.

Answer:

2.

Answer:

3.

Answer:

4.

Answer:

II. Assertion and Reason

i) Both A and R are correct, R explains A.
ii) A is correct, R is wrong
iii) A is wrong, R is correct
iv) Both A and R are correct, but R does not explain A.

1. Assertion (A) : Aqueous solution of potash Alum is acidic
Reason (R) : Aluminium sulphate undergo hydrolysis. (PTA – 2)
Answer:
i) Both A and R are correct, R is explanation of A

2. Assertion (A) : Elements belonging to group 16 are called chalcogens
Reason (R) : Group 16 elements are saltforming elements
Answer:
ii) A is correct, R is wrong
Correct Reason : Group 16 elements are ore forming elements

3. Assertion (A) : Among halogen acids, HF has low melting and boiling points
Reason (R) : In HF hydrogen bond is present.
Answer:
iii) A is wrong, R is correct.
Correct A : Among halogen acids HF has high melting and boiling points.

4. Assertion (A) : A small piece of Zinc dissolved in dilute nitric acid but hydrogen gas is not evolved. (PTA – 3)
Reason (R) : HNO₃ is an oxidising agent and this oxidizes hydrogen.
Answer:
ii) A is correct but R is wrong.

III. Pick out the Correct statement

1. i) Oxygen is diamagnetic
ii) Oxygen forms hydrogen bonds
iii) Oxygen exists in two allotropic forms
iv) Oxygen exists as a triatomic gas
a) (i) &(ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
b) (ii) & (iii)
Correct statement: (i) Oxygen is paramagnetic (iv) Oxygen exists as a diatomic gas

2. i) Sulphur exists in crystalline as well as an amorphous form
ii) Rhombic sulphur has a characteristic yellow colour and composed of Sg molecules.
iii) When heated slowly above % C monoclinic sulphur is converted into Rhombic sulphur
iv) At around 140°C Rhombic sulphur melts to form mobile pale yellow liquid called X sulphur.
a) (i) &(ii)
b) (i) & (iii)
c) (ii) & (iii)
d) (iii) & (iv)
Answer:
a) (i) & (ii)
Correct statement : (iii) When heated slowly above 96°C, Rhombic sulphur is converted into monoclinic sulphur
(iv) At around 140°C the monoclinic sulphur melts to form mobile pale yellow liquid called X sulphur

3. i) H₂SO₄ is a dibasic acid
ii) H₃PO₃ is a tribasic acid
iii) H₃PO₄ is a dibasic acid
iv) H₃PO₂ is a monobasic acid
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
d) (i) & (iv)
Correct statement:
(ii) H₃PO₃ is a dibasic acid
(iii) H₃PO₄ is a tribasic acid

4. i) Krypton is used in cryogenics.
ii) Neon is used in high-speed electronic flashbulbs used by photographers.
iii) Helium is used to provide an inert atmosphere in electric arc welding of metals.
iv) Radon is used as a source of gamma rays.
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
c) (iii) & (iv)
Correct statement: (i) Helium is used in cryogenics
(ii) Xenon is used in high speed electronic flash bulbs used by photographers.

IV. Pick out the incorrect statement

1. i) In inter halogen compounds the central atom will be the smaller halogen
ii) Interhalogen compounds can be formed only between two halogen atoms.
iii) Flourine can act as a central atom.
iv) Interhalogens are strong oxidising agents
a) (i) & (ii)
b) (i) & (iii)
c) (ii) & (iii)
d) (i) & (iv)
Answer:
b) (i) & (iii)
Correct statements:
(i) In interhalogen compounds the central atom will be the larger halogen
(iii) Flourine cannot act as a central atom.

2. i) Nitrogen reacts with group 2 metals to form ionic nitrides.
ii) Ammonia is less soluble in water.
iii) Liquid nitrogen is used in biological preservation.
iv) In the conversion of metal oxides to metal ammonia acts as an oxidising agent.
a) (i) & (ii)
b) (ii) & (iii)
c) (i) & (iii)
d) (ii) & (iv)
Answer:
d) (ii) & (iv)
Correct statements:
(ii) Ammonia is extremely soluble in water.
(iv) In the conversion of metal oxides to metal ammonia acts as a reducing agent

3. i) When reacted with metals, nitric acid liberates hydrogen
ii) Chromium when reacted with nitric acid becomes passive due to the formation of nitrate on its surface.
iii) In most of the reactions nitric acid acts as an oxidising agent
iv) Fuming nitric acid contains oxides of nitrogen
a) (i) & (ii)
b) (ii) & (iii)
c) (ii) & (iv)
d) (i) & (iv)
Answer:
a) (i) & (ii)
Correct statement : (i) When reacted with metals, nitric acid does not liberate hydrogen
(ii) Chromium when reacted with nitric acid becomes passive due to the formation of oxide on its surface

4. The rate of decomposition of ozone increases sharply in alkaline solution.
ii) In acidic solution ozone exceeds the oxidising power of fluorine and atomic oxygen
iii) Considerable amount of ozone is formed in the upper atmosphere by the action of UV light
iv) The shape of the Ozone molecule is linear
a) (i) &(ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
d) (i) & (iv)
Correct statement : (i) The rate of decomposition of ozone drops sharply in alkaline solution.
(iv) The shape of the ozone molecule is bent.

V. Pick out the odd man out

1. w.r.t oxidation number pick the odd man out.
a) HPO₃
b) H₃PO₃
c) H₃PO₄
d) H₄P₂O₇
Answer:
b) H₃PO₃ – O.N is +3 while in others the O.N of phosphorous is +5

2. w.r.t the reaction with sulphuric acid pick the odd man out
a) Gold
b) Silver
c) Platinum
d) Copper
Answer:
d) Copper – copper reacts with sulphuric acid while others do not

3. w.r.t reactivity pick the odd man out
a) F₂
b) Cl₂
c) Br₂
d) I₂
Answer:
a) F₂ – F₂ is more reactive than other halogens

4. w.r.t the ability to form oxoacids pick the odd man out
a) fluorine
b) chlorine
c) bromine
d) iodine
Answer:
a) Flour – Fluorine forms only one oxoacid where as other halogens form more than one oxoacid.

VI. Choose the best answer.

1. The principal gas present in atmosphere is
a) O₂
b) N₂
C) H₂
d) CO₂
Answer:
b) N₂

2. The basicity of hypophosphorous acid is (PTA – 2)
a) 1
b) 2
c) 3
d) 4
Answer:
a) 1

3. Chile salt petre is
a) NaNO₂
b) NaNO₃
c) KNO₂
d) KNO₃
Answer:
b) NaNO₃

4. Indian salt petre is
a) NaNO₂
b) NaNO₃
c) KNO₂
d) K.NO₃
Answer:
d) KNO₃

5. Inert character of nitrogen is due to its
a) high electronegativity
b) low electro negativity
c) high bonding energy
d) low bonding energv
Answer:
c) high bonding energy

6. Which of the following is caro’s acid?
a) H₂S₂O₈
b) H₂S₂O₇
c) H₂SO₃
d) H₂SO₃
Answer:
c H₂SO₅

7. Haber’s process is used for the synthesis of
a) NO₂
b) HNO₃
c) NH₃
d) N₂O
Answer:
a) NH₃

8. The substance used in cryosurgery for producing low temperature is
a) liquid oxygen
b) liquid nitrogen
c) liquid hydrogen
d) liquid ammonia
Answer:
b) liquid nitrogen

9. Urea on hydrolysis gives
a) NO₂
b) HNO₃
c) NH₃
d) N₂O
Answer:
a) NH₃

10. The catalyst used in Haber’s process is
a) Ni
b) Fe
c) Co
d) Pt
Answer:
b) Fe

11. The smell of ammonia is
a) rotten egg
b) rotten fish
c) pungent
d) garlic
Answer:
c) pungent

12. Like water, ammonia is a fairly good ionising solvent, because its dielectric constant is
a) considerably low
b) considerably high
c) equal to zero
d) equal to one
Answer:
b) considerably high

13. The process used for the manufacture of nitric acid is known as
a) Haber’s process
b) Deacon’s process
c) contact process
d) Ostwald’s process
Answer:
d) Ostwald’s process

14. With excess of chlorine, ammonia reacts to give an explosive substance
a) N₂
b) NH₄NO₃
c) NH₄Cl
d) NCl₃
Answer:
d) NCl₃

15. The deep blue colour compound formed when excess of ammonia is added to aqueous solution of copper sulphate is
a) [Cu(NO₃)₂]
b) [Cu(NH₃)₂]²⁺
c) [Cu(NH₃)₄]²⁺
d) [Cu(NH₃)₂]⁺
Answer:
c) [Cu(NH₃)₄]²⁺

16. The shape of ammonia molecule is
a) tetrahedral
b) pyramidal
c) square planar
d) octahedral
Answer:
b) pyramidal

17. The bond angle in ammonia is
a) 104°
b) 104°28′
c) 107°
d) 180°
Answer:
c) 107°

18. The colour of Pure nitric acid is
a) colourless
b) brown
c) pale green
d) green
Answer:
a) colourless

19. Fuming nitric acid contains oxides of
a) sulphur
b) hydrogen
c) nitrogen
d) carbon
Answer:
c) nitrogen

20. Nitric acid can act as
a) an acid
b) an oxidising agent
c) nitrating agent
d) all of the above
Answer:
d) all the above

21. The formula of hyponitrous acid is (MARCH 2020)
a) H₂N₂O₂
b) H₄N₂O₄
c) HOONO
d) HNO₂
Answer:
a) H₂N₂O₂

22. The oxidising power of oxo acids follows the order
a) HOX > HXO₂ > HXO₃ > HXO₄
b) HXO₄ > HXO₃ > HXO₂ > HOX
c) HXO₃ > HXO₄ > HXO₂ > HOX
d) HOX > HXO₄ > HXO₃ > HXO₂
Answer:
a) HOX > HXO₂ > HXO₃ > HXO₄

23. White phosphorous is kept under
a) kerosene
b) water
c) alcohol
d) ether
Answer:
b) water

24. White phosphorous becomes yellow phosphorous due to
a) hydrolysis
b) reduction
c) oxidation
d) displacement
Answer:
c) oxidation

25. In the conversion of yellow phosphorous into phosphine, phosphorous acts as
a) oxidising agent
b) reducing agent
c) catalyst
d) hydrolysing agent
Answer:
b) reducing agent

26. In the conversion of phosphorous into orthophosphoric acid, the catalyst used is
a) Cl₂
b) Br₂
c) I₂
d) F₂
Answer:
c) I₂

27. Which is used in match boxes?
a) White phosphorous
b) Red phosphorous
c) Black phosphorous
d) Scarlet phosphorous
Answer:
b) Red Phosphorous

28. The acid having O-O bond in its structure (PTA – 6)
a) H₂SO₃
b) H₂S₂O₆
c) H₂S₂O₈
d) H₂S₄O₆
Answer:
c) H₂S₂O₈

29. The smell of phosphine is
a) rotten egg
b) rotten fish
c) pungent
d) garlic
Answer:
b) rotten fish

30. The compounds used in Holme’s signal are
a) CaC₂ & Ca₃P₂
b) AlP & Ca₃P₂
c) CaC₂ & P₄
d) AlP & P₄
Answer:
a) CaC₂ & Ca₃P₂

31. The gases liberated in Holme’s signal are
a) C₂H₂ & CH₄
b) C₂H₂ & Ph₃
c) C₂H₄ & PH₃
d) CH₄ & Ph₃
Answer:
b) C₂H₂ & Ph₃

32. The formula of pyrophosphoric acid is
a) H₄P₂O₆
b) H₄P₂O₇
c) H₃PO₂
d) H₃PO₃
Answer:
b) H₄P₂O₇

33. Thermodynamically stable allotrophic form of sulphur is
a) Rhombic sulphur
b) Monoclinic sulphur
c) Plastic sulphur
d) Colloidal sulphur
Answer:
a) Rhombic sulphur

34. The gas found in volcanic eruptions is
a) NO₂
b) NO
c) SO₂
d) SO₃
Answer:
c) SO₂

35. The hybridisation of sulphur in SO₂ is
a) sp
b) sp²
c) sp³
d) dsp²
Answer:
b) sp²

36. The gas liberated when dilute sulphuric acid reacts with metals is
a) SO₂
b) SO₃
c) H₂
d) O₂
Answer:
c) H₂

37. The gas liberated when cone, sulphuric acid reacts with metals is
a) SO₂
b) SO₃
c) H₂
d) O₂
Answer:
a) SO₂

38. When sulphuric acid reacts with barium chloride solution, the white precipitate formed is
a) PbSO₄
b) BaSO₄
c) (CH₃COO)₂SO₄
d) PbCl₂
Answer:
b) BaSO₄

39. The halogen which exists as a liquid is
a) flourine
b) chlorine
c) bromine
d) iodine
Answer:
c) bromine

40. The halogen which exists as a solid is
a) flourine
b) chlorine
c) bromine
d) iodine
Answer:
d) iodine

41. Chlorine is manufactured by
a) Haber’s process
b) Deacon’s process
c) Contact process
d) Ostwald’s process
Answer:
b) Deacon’s process

42. The colour of chlorine gas is
a) colourless
b) brown
c) greenish yellow
d) pale green
Answer:
c) greenish yellow

43. Aqua regia is a mixture of cone. HCl and cone. HNO₃ in the ratio
a) 1 : 3
b) 3 : 1
c) 2 : 3
d) 3 : 2
Answer:
b) 3 :1

44. The halogen acid which forms hydrogen bond is
a) HF
b) HCl
c) HBr
d) HI
Answer:
a) HF

45. Among halogen acids, the strongest bond is present in
a) HF
b) HCl
c) HBr
d) HI
Answer:
a) HF

46. Among halogen acids, the weakest bond is present in
a) HF
b) HCl
c) HBr
d) HI
Answer:
d) HI

47. Among halogen acids, the strongest acid is
a) HF
b) HCl
c) HBr
d) HI
Answer:
d) HI

48. Among halogen acids, the weakest acid is
a) HF
b) HCl
c) HBr
d) HI
Answer:
a) HF

49. The correct order of acid strength is
a) HF > HCl > HBr > HI
b) HF < HCl < HBr < HI
c) HF > HCl < HBr > HI
d) HF < HCl > HBr < HI
Answer:
b) HF < HCl < HBr < HI

50. Which is more reactive towards hydrogen?
a) flourine
b) chlorine
c) bromine
d) iodine
Answer:
a) flourine

51. The number of bond pair and lone pair of electrons present in the interhalogen compound BrF₃ is
a) 1 & 3
b) 3 & 2
C) 5 & 1
d) 7 & 0
Answer:
b) 3 & 2

52. The oxidation number of oxygen in F₂O is
a) -2
b) -1
c) +2
d) +1
Answer:
c) +2

53. The oxidation number of chlorine in Cl₂O₇ is
a) +1
b) +4
c) +6
d) +7
Answer:
d) +7

54. The strongest oxidising agent among the following is
a) chlorous acid
b) chloricacid
c) hypochlorous acid
d) perchloric acid
Answer:
c) hypochlorous acid

55. The first ionisation energy of noble gases is in the order
a) He < Ne < Ar < Kr
b) He > Ne > Ar > Kr
c) He < Ne > Ar < Kr
d) He > Ne < Ar > Kr
Answer:
b) He > Ne > Ar > Kr

56. Among noble gases, chemical reactivity is shown by
a) He & Ne
b) Ar & Kr
c) Kr & Xe
d) Xe & Rn
Answer:
c) Kr & Xe

57. Which among the following is used in cryogenics?
a) He
b) Ne
c) Ar
d) Kr
Answer:
a) He

58. Which is used for filling air balloons?
a) He
b) Ne
c) Ar
d) Kr
Answer:
a) He

59. Which is used in advertisement sign boards?
a) He
b) Ne
c) Ar
d) Kr
Answer:
b) Ne

60. Lamps used in airports as approaching lights is filled with
a) He
b) Ne
c) Ar
d) Kr
Answer:
d) Kr

VII. Two Mark Questions

Question 1.
How is pure nitrogen gas prepared?
Answer:
Pure nitrogen gas is prepared by the thermal decomposition of sodium azide at about 575 K
2NaN₃ \(\underrightarrow { 573K } \) 2Na + 3N₂

Question 2.
Nitrogen does not form any penta halides like phosphorus, why?
Answer:
Nitrogen does not form pentahalide although it exhibit +5 oxidation state. Due to the absence of d-orbitals.
It cannot undergo sp3d hybridization and hence cannot form pentahalides.

Question 3.
What is Haber’s process?
Answer:
The synthesis of ammonia from nitrogen and hydrogen at high pressure and optimum temperature in presence of iron catalyst is known as Haber’s process.

∆H_f = -46.2 Kjmol^-1

Question 4.
Write the uses of nitrogen
Answer:
Nitrogen is used
In the manufacture of ammonia, nitric acid, and calcium cyanamide etc.
Liquid nitrogen is used for producing low temperature required in cryosurgery and so used in biological preservation.

Question 5.
How is ammonia prepared in the laboratory?
Answer:
Ammonia is prepared in the laboratory by heating an ammonium salt with a base
2NH₄Cl + CaO → CaCl₂ + 2NH₃ + H₂O

Question 6.
Write about the reducing property of ammonia.
Answer:
When passed over heated metallic oxides Ammonia reduces metal oxides into metal
3PbO + 2NH₃ → 3Pb + N₂ + 3H₂O

Question 7.
What happens when ammonia reacts with excess of chlorine?
Answer:
With excess of chlorine ammonia reacts to give an explosive substance nitrogen trichloride
2NH₃ + 6Cl₂ → 2NO₃ + 6HO

Question 8.
On standing nitric acid becomes yellow in colour why?
Answer:

• Pure nitric acid is colourless
• Fuming nitric acid contains oxides of nitrogen
• It decomposes on exposure to sunlight or on being heated into nitrogendioxide, water and oxygen.
  4HNO₃ → 4NO₂ + 2H₂O + O₂
• Due to this reaction, pure nitric acid or its concentrated solution becomes yellow on standing

Question 9.
Prove that nitric acid is an oxidising agent.
Answer:
Non metals like carbon, sulphur are oxidised by nitric acid.
C + 4HNO₃ → CO₂ + 4NO₂ + 2H₂O
S + 2HNO₃ → H₂SO₄ + 2NO

Question 10.
Prove that nitric acid is a nitrating agent.
Answer:
Nitric acid replaces hydrogen atom from organic compounds with nitronium ion NO₂⁺. This is called nitration.

Question 11.
Write the uses of nitric acid is used
Answer:

• as an oxidising agent.
• in the preparation of aqua regia.
• Salts of nitric acid are used in photography (AgNO₃) and gunpowder for fire arms (NaNO₃)

Question 12.
How is nitrous oxide prepared?
Answer:
By Heating ammonium nitrate nitrous oxide is prepared
NH₄NO₃ → N₂O + 2H₂O

Question 13.
How is nitrous acid prepared?
Answer:
By treating nitrite salt with acids, nitrous acid is prepared
Ba(NO₂)₂ + H₂SO₄ → 2HNO₂ + BaSO₄

Question 14.
What is phosphorescence?
Answer:
White phosphorous undergoes spontaneous slow oxidation in air giving a greenish yellow glow which is visible in the dark. This is known as phosphorescence. The main product of this slow oxidation is P₂O₃.

Question 15.
How is phosphine prepared?
Answer:
Phosphine is prepared by the action of sodium hydroxide with white phosphorous in an inert atmosphere of carbon dioxide

Question 16.
How is orthophosphoric acid prepared in ‘ the laboratory?
When phosphorous is treated with cone, nitric acid in the presence of iodine catalyst, it is oxidised to orthophosphoric acid.

Question 17.
Write the uses of phosphorous Phosphorous is used
Answer:

• in match boxes
• For the production of certain alloys such as phosphor bronze.

Question 18.
What happens when phosphine is heated in the absence of air?
Answer:
Phosphine decomposes into its elements when heated in the absence of air at 317 K
4PH₃ \(\underrightarrow { 317K } \) P₄ + 6H₂

Question 19.
Write about the reducing property of phosphine?
Answer:
Phosphine reduces silver nitrate into silver
PH₃ + 6AgNO₃ + 3H₂O → 6Ag + 6HNO₃ + H₃PO₃

Question 20.
How is phosphorous trichloride prepared?
Answer:

• When a slow stream of chlorine is passed over white phosphorous, PCl₃ is obtained.
• It is also prepared by treating white phosphorous with thionyl chloride.
  P₄ + 8SOCl₂ → 4PCl₃ + 4SO₂ + 2S₂Cl₂

Question 21.
Ozone (O₃) acts as a powerful oxidizing agent why? (PTA – 5)
Answer:
Ozone is not a very stable compound under normal conditions and decomposes readily on heating to give a molecule of oxygen and nascent oxygen.
Nascent oxygen, being a free radical, is very reactive
O₃ \(\underrightarrow { \triangle } \) O₂ + [O]

Question 22.
Write the uses of oxygen
Answer:

• Oxygen is one of the essential components for the survival of living organisms.
• Oxygen is used in oxyacetylene welding.
• Liquid oxygen is used as a rocket fuel.

Question 23.
How is sulphur dioxide prepared in the laboratory?
Answer:
SO₂ is prepared in the laboratory by treating a metal or metal sulphite with sulphuric acid
Cu + 2H₂SO₄ → CuSO₄ + SO₂ + 2H₂O
SO₃^2- + 2H⁺ → SO₂ + H₂O

Question 24.
Illustrate the oxidising property of SO₂.
Answer:
SO₂ oxidises hydrogen sulphide to sulphur and magnesium to magnesium oxide.
2H₂S + SO₂ → 3S + 2H₂O
2Mg + SO₂ → 2MgO + S

Question 25.
Write about contact process.
Answer:

• In contact process SO₂ is oxidised to SO₃
• It is used in the manufacture of sulphuric acid.
  

Question 26.
Write the uses of sulphurdioxide.
Answer:

• SO₂ is used in bleaching hair, silk, wool etc.
• SO₂ is used for disinfecting crops and plants in agriculture

Question 27.
Write about the structure of sulphr dioxide.
Answer:

• Sulphur undergoes sp² hybridisation.
• A double bond arises between S and O due to pπ – dπ overlapping

Question 28.
Illustrate the dehydrating property of sulphuric acid.
Answer:

Question 29.
Show that sulphuric acid is a dibasic acid.
Answer:
H₂SO₄ forms two types of salts with NaOH Hence it is dibasic.

Question 30.
How is chlorine is manufactured by Deacon’s process?
Answer:

• A mixture of air and hydrochloric acid is passed up a chamber containing a number of shelves, containing pumice stones soaked in cuprous chloride.
• Hot gases at about 723 K are passed through a jacket that surrounds the chamber.
  
• Chlorine obtained is dilute and used for the manufacture of bleaching powder

Question 31.
Write about the bleaching action of chlorine.
Answer:
Chlorine is a strong oxidising and bleaching agent since it produces nascent oxygen.
H₂O + Cl₂ → HCl + HOCl (Hypochlorous acid)
HOCl → HCl +[0]
Colouring matter + Nascent oxygen → Colourless oxidation product.

The bleaching of chlorine is permanent.

Question 32.
How is bleaching powder prepared? (MARCH 2020)
Answer:
Bleaching powder is prepared by passing chlorine gas through dry slaked lime (calcium hydroxide)
Ca(OH)₂ + Cl₂ → CaOCl₂ + H₂O

Question 33.
Write the uses of chlorine Chlorine is used
Answer:

• In the purification of drinking water.
• In the bleaching of cotton textiles, paper, and rayon.
• In the extraction of gold and platinum.

Question 34.
How is hydrochloric and prepared in the laboratory?
Answer:
Hydrochloric add is prepared by the action of sodium chloride and cone, sulphuric acid
NaCl + H₂SO₄ → NaHSO₄. + HCl
NaHSO₄ + NaCl → Na₂SO₄. + HCl
Dry hydrochloric acid is obtained by passing the gas through cone, sulphuric acid

Question 35.
How is xenon trioxide prepared?
Answer:
2XeF₆ + SiO₂ \(\underrightarrow { 50°C } \) 2XeOF₄ + SiF₄
2XeOF₄ + SiO₂ → 2XeO₂F₂ + SiF₄
2XeO₂F₂ + SiO₂ → 2XeO₃ + SiF₄
(or)
XeF₆ + 3H₂ → XeO₃ + 6HF

Question 36.
How is sodium per xenate obtained?
Answer:
When XeF₆ reacts with 2.5 M NaOH, sodium per xenate is obtained.
2XeF₆ + 16NaOH → Na₄XeO₆ + Xe + O₂ + 12NaF + 8H₂O

Question 37.
Show that sodium per xenate is a strong oxidising agent
Answer:
Sodium per xenate oxidises manganese (II) ion into permanganate ion even in the
absence of a catalyst
5XeO^4-₆ + 2Mn²⁺ + 14H⁺ → 2MnO^–₄ + 5XeO₃ + 7H₂O

Question 38.
Give reason: ICl is more reactive than l₂ (PTA – 3)
Answer:

• This is because inter-halogen compounds are in general more reactive than halogens due to w eaker inter-halogen X-X bond than X-X bond.
• So, I₂ is more stable and less reactive than ICl.

VIII. Three Mark Questions

Question 1.
Write about the structure of ammonia.
Answer:

• Ammonia molecule is pyramidal in shape.
• Hybridisation of nitrogen is sp³.
• The shape must be tetrahedral but in one of the tetrahedral positions a lone pair of electrons from the nitrogen atom is present, hence it is pyramidal.
• The N – H bond distance is 1.016 A0.
• The H – H bond distance is 1.645 A°.
• The bond angle is 107°
  

Question 2.
How does red phosphorous react with oxygen?
Answer:
Red phosphorous reacts with oxygen on heating to give phosphorous trioxide and phosphorous pentoxide.
P₄ + 3O₂ → P₄O₆
P₄ + 5O₂ → P₄O₁₀

Question 3.
How is pure phosphine prepared?
Answer:
Pure phosphine is prepared by heating phosphorous acid

A pure sample of phosphine is prepared by heating phosphonium iodide with caustic soda solution.

Question 4.
What happens when phosphine is heated with air?
Answer:
When phosphine is heated with air or oxygen, it undergoes combustion to give meta phosphoric acid.

Question 5.
Write about Holmes signal
Answer:

• In a ship during distress, a container with calcium carbide and calcium phosphide mixture is pierced and thrown into the sea.
• The mixture reacts with seawater liberating acetylene and phosphine gases.
• The liberated phosphine catches fire and ignites acetylene.
• These burning gases with lot of smoke serves as a signal to the approaching ships.
• This is known as Holme’s signal.

Question 6.
Write about the structure of phosphine
Answer:

• Phosphorous shows sp³ hybridisation.
• Three orbitals are occupied by bond pair electrons.
• Fourth orbital is occupied by lone pair of electrons.
• Hence instead of tetrahedral, PH₃ has a pyramidal shape.
• Bond angle is reduced to 94°
  

Question 7.
How is oxygen prepared in the laboratory?
Answer:
Oxygen is prepared in the laboratory by the decomposition of hydrogen peroxide in presence of Mn02 catalyst or by the oxidation of potassium permanganate.

5H2O₂ + 2MnO₄^– + 6H⁺ → 5O₂ + 8H₂O + 2Mn²⁺

Question 8.
Write about ozone
Answer:

• Ozone is an allotropic form of oxygen
• Ozone is triatomic gas
• Although negligible amounts of ozone occurs at sea level, it is formed in the upper atmosphere by the action of UV light.
• In the laboratory ozone is prepared by passing electrical discharge through oxygen.
• At a potential of 20,000 V about 10% of oxygen is converted into ozone, it gives a mixture known as ozonised oxygen.
• Pure ozone is obtained as a pale blue gas by the fractional distillation of liquefied ozonised oxygen.
  
• Ozone molecule has a bent shape and symmetrical with delocalised bonding between the oxygen atoms.

Question 9.
Write about the reducing property of sulphur dioxide
Answer:

SO₂ reduces chlorine into hydrochloric acid
SO₂ + 2H₂O + Cl₂ → H₂SO₄ + 2HCl

SO₂ reduces potassium permanganate into manganese sulphate (Mn²⁺).
2KMnO₄ + 5SO₂ + 2H₂O → K₂SO₄ + 2MnSO₄ + 2H₂SO₄

SO₂ reduces potassium dichromate into chromic sulphate (Cr³⁺)
K₂Cr₂O₇ + 3SO₂ + H₂SO₄ → K₂SO₄ + Cr₂(SO₄)₃ + H₂O

Question 10.
Write about the bleaching action of sulphur dioxide.
Answer:
In presence of water, sulphur dioxide bleaches coloured wool, silk, sponges and straw into colourless due to its reducing property

When the bleached product (Colourless) is allowed to stand in air, it is reoxidised by atmospheric oxygen to its original colour.

Hence bleaching action of sulphurdioxide is temporary

Question 11.
Explain the manufacture of sulphuric acid
Answer:

• Sulphur dioxide is produced by burning sulphur or iron pyrites in oxygen / air
  S + O₂ → SO₂
  4FeS₂ + 11O₂ → 2Fe₂O₃ + 8SO₂
• SO₂ is oxidised to SO₃ by air in presence of V₂O₅ or platinised absestos
  
• SO₃ is absorbed in cone H₂SO₄ and oleum is produced
  SO₃ + H₂SO₄ H₂S₂O₇
• Oleum is converted into sulphuric acid by diluting it with water.
  H₂S₂O₇ + H₂O → 2H₂SO₄
• To maximize the yield the plant is operated at 2 bar pressure and 720 K.
  96% pure H₂SO₄ is obtained.

Question 12.
Show that sulphuric acid is an oxidising agent
Answer:
Sulphuric acid is an oxidising agent as it produces nascent oxygen
H₂SO₄ → H₂O + SO₂ + [O] (Nascert oxygen)

Sulphuric acid oxidises carbon into carbon dioxide
C + 2H₂SO₄ → 2SO₂ + 2H₂O + CO₂

Sulphuric acid oxidises phosphorous into orthophosphoric acid
P₄ + 10H₂SO₄ → 4H₃PO₄ + 10SO₂ + 4H₂0

Sulphuric acid oxidises iodide into iodine.
H₂SO₄ + 2HI → 2SO₂ + 2H₂O + I₂

Question 13.
What is the action of sulphuric acid on metals?
Answer:
Dilute sulphuric acid reacts with metals liberating hydrogen gas.
Zn + H₂SO₄ ZnSO₄ + H₂ ↑

Hot cone. Sulphuric acid reacts with metals to give sulphates and sulphur dioxide
Cu + 2H₂SO₄ → CuSO₄ + 2H₂O + SO₂ ↑

Sulphuric acid does not react with noble metals like gold, silver and platinum.

Question 14.
Give the test for sulphate / sulphuric acid
Dilute solution of sulphuric acid / Sulphates react with barium chloride or lead acetate solution to give a white precipitate

Question 15.
How is chlorine manufactured by the electrolytic process?
Answer:
When brine solution (NaCl) is electrolyzed, Na⁺ and Cl^– ions are formed.
Na⁺ ions react with OH^– ions of water forming sodium hydroxide.
Hydrogen and chlorine are liberated as gases.
NaCl → Na+ +Cl^–
H₂O → H+ +OH^–
Na⁺ + OH^– → NaOH
At cathode : H⁺ + e^– → H
H + H → H₂
At anode : Cl^– → Cl + e^–
Cl + Cl → Cl₂

Question 16.
What is aqua regia? What is its action on gold?
Answer:
Aqua regia is a mixture of three parts of cone, hydrochloric acid and one part of cone, nitric acid.
This is used for dissolving gold, platinum etc.
AU + 4H⁺ + NO₃^– + 4Cl^– → AuCl₄^– +NO + 2H₂O

Question 17.
HF acid is a weaker acid at low concentration, but becomes stronger as the concentration increases why?
Answer:
0.1 M Solution HF is 10% ionised, hence it is a weak acid.
But 5 M, 15 M solution of HF is stronger due to the equilibrium.

At high concentration, the equilibrium involves the removal of flouride ions and increases the hydrogen ion concentration, HF becomes stronger acid as the concentration increases.

Question 18.
HF acid is not stored in glass bottles why? (MARCH 2020)
Answer:
HF attacks silica and silicates present in glass bottles.

Hence HF is not stored in glass bottles. But HF is stored in Teflon bottles.
SiO₂ + 4HF → SiF₄ + 2H₂O
Na₂SiO₃ + 6HF → Na₂SiF₆ + 3H₂O

Question 19.
Mention the characteristic of interhalogen compounds (PTA – 2)
Answer:

• The central atom will be the larger halogen.
• It can be formed only between two halogens and not more than two halogens.
• Fluorine can’t act as a central atom because it is the smallest among halogens and highly electronegative.
• They are strong oxidising agents and undergo auto ionization.

Question 20.
Give the preparation of xenon fluorides
Answer:

Question 21.
What is the hybridisation in XeOF₂? Give its structure. (PTA – 1)
Answer:

IX. Five Mark Questions

Question 1.
How is nitric acid manufactured using Ostwald’s process?

• Ammonia prepared by Haber’s process is mixed about 10 times of air.
• This mixture is preheated and passed into the catalyst chamber where they come in contact with platinum gauze.
• The temperature rises about 1275 K.
• The metallic gauze brings about the rapid catalytic oxidation of ammonia resulting in the formation of NO.
• NO is oxidized to NO₂
  4NH₃ +5O₂ → 4NO + 6H₂O +120 KJ
  2NO + O₂ → 2NO₂
• NO₂ produced is passed through a series of adsorption towers.
• NO₂ reacts with water to give nitric acid.
• Nitric acid formed is bleached by blowing air.
  6NO₂ + 3H₂O → 4HNO₃ + 2NO + H₂O

Question 2.
Explain the action of nitric acid on metals with one example.
Primary reaction:
Metal nitrate is formed with the release of nascent hydrogen.
3Cu + 6HNO₃ → 3CU(NO₃)₂ + 6(H)

Secondary reaction:
Nascent hydrogen produces the reduction products of nitric acid
6(H) + 3HNO₃ → 3HNO₂ + 3H₂O

Tertiary reaction:
With dilute acid, the secondary products decompose to give final products.
3HNO₂ → HNO₃ + 2NO + H₂O
Hence overall reaction is
3Cu + 8HNO₃ → 3CU(NO₃)₂ + 2NO + 4H₂O

With concentrated acid the secondary products react to give the final products.
HNO₂ + HNO₃ → 2NO₂ + H₂O
Hence overall reaction is
Cu + 4HNO₃ → Cu(NO₃)₂ + 2NO₂ + 2H₂O

Question 3.
Write the preparation of nitrogen oxides.
Answer:

Question 4.
Write the preparation of oxoacids of nitrogen
Answer:

Question 5.
Explain the structure of oxides of phosphorus
Answer:
Phosphorus trioxide :

• In P₄O₆, four phosphorous atoms lie at the corners of a tetrahedron and six oxygen atoms along the edges.
  The P – O bond distance is 165.6 pm which is shorter than the single bond distance of the P-O bond (184 pm)
• This is due to Pπ – dπ bonding
• This results in the considerable double bond character
  

Phosphorous Pentoxide:

• In P₄O₁₀ each P atom forms a single bond with three oxygen atoms and a coordinate bond with one oxygen atom.
• Terminal coordinate P-O bond length is 143 pm
• This is less than the expected single bond distance
• This may be due to lateral overlap of filled
  p – Orbitals of an oxygen atom with empty
  d – Orbital on phosphorous.

Question 6.
Write the structure of and basicity oxoacids of phosphorous. (PTA – 3)

Answer:

Question 7.
Write the preparation of oxoacids of phosphorous
Answer:

Question 8.
Write the structure of oxo acids of sulphur.
Answer:

Question 9.
List any five compounds of Xenon and mention the type of hybridization and structure of the compounds (PTA – 6)
Answer:

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 7 Applications of Differential Calculus Ex 7.9 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.9

Question 1.
Find the asymptotes of the following curves:
(i) f(x) = \(\frac { x^2 }{ x^2-1 }\)
(ii) f(x) = \(\frac { x^2 }{ x+1 }\)
(iii) f(x) = \(\frac { 3x }{ \sqrt{x^2+2} }\)
(iv) f(x) = \(\frac { x^2-6x-1 }{ x+3 }\)
(v) f(x) = \(\frac { x^2+6x-4 }{ 3x-6 }\)
Solution:
(i) f(x) = \(\frac { x^2 }{ x^2-1 }\)
The function becomes undefined when x = 1 and x = -1
∴ x = 1 and x = -1 are the vertical asymptotes

As ‘x’ gets larger (Positive or negative) the function, the function attaining the value 1.
∴ y = 1 is horizontal asymptote

(ii) f(x) = \(\frac { x^2 }{ x+1 }\)

The function becomes undefined when x = -1
∴ Vertical asymptote is x = -1 and there is no horizontal asymptote.
No horizontal asymptote exists for the curve. Oblique asymptote can be obtained by polynomial long division method.

Oblique (or) slant asymptote is y = x – 1

(iii) f(x) = \(\frac { 3x }{ \sqrt{x^2+2} }\)
No vertical asymptotes
Horizontal asymptotes RHL (Right Hand Limit)


y = 3 and y = -3 are the Horizontal asymptotes
Slant asymptotes’. No such slant asymptotes exist for the given curve.

(iv) f(x) = \(\frac { x^2-6x-1 }{ x+3 }\)

When x = -3, the function becomes undefined.
∴ x = -3 is the vertical asymptote.
No Horizontal asymptote exist for the curve.
Oblique asymptote can be obtained by polynomial long division method

∴ y = x – 9 is the slant (or) oblique asymptote.

(v) f(x) = \(\frac { x^2+6x-4 }{ 3x-6 }\)

The function becomes undefined when x = 2.
∴ x = 2 is the vertical asymptote.
No Horizontal asymptote exist for the given curve.
Oblique asymptote can be obtained by polynomial long division method.
∴ y = \(\frac { x }{ 3 }\) + \(\frac { 8 }{ 3 }\) (or) 3y = x + 8 is the slant asymptote.

Question 2.
Sketch the graphs of the following functions
(i) y = –\(\frac { 1 }{ 3 }\) (x³ – 3x + 2)
(ii) y = x \(\sqrt { 4-x }\)
(iii) y = \(\frac { x^2+1 }{ x^2-4 }\)
(iv) y = \(\frac { 1 }{ 1+e^{-x} }\)
(v) y = \(\frac { x^3 }{ 24 }\) – log x
Solution:
(i) y = –\(\frac { 1 }{ 3 }\) (x³ – 3x + 2)

Factorizing we get
y = –\(\frac { 1 }{ 3 }\) (x – 1)² (x + 2) = f(x)
The domain and the range of the given function f(x) are the entire real line.
Putting y = 0, we get x = 1, 1, – 2. Hence the x-intercepts are (1, 0) and (- 2, 0) and by putting x = 0. We get y = –\(\frac { 2 }{ 3 }\). Therefore, the y-intercept is (0, –\(\frac { 2 }{ 3 }\))
f'(x) = \(\frac { (3x^2-3) }{ 3 }\) = -(x² – 1) = 1 – x²
f'(x) = 0 ⇒ 1 – x² = 0 ⇒ x = ±1
The critical points of the curve occur at x = ± 1 .
f”(x) = -2x
f”(1) = – 2 < 0, ∴ f(x) is maximum at x = 1 and the local maximum is f(1) = o
f”(-1) = 2 > 0, ∴ f(x) is minimum at x = -1 and the local minimum is
f(-1) = –\(\frac { 4 }{ 3 }\)
f”(x) = – 2x < 0 ∀ x > 0, ∴ The function is concave downward in the positive real line.
f”(x) = 2x > 0 ∀ x < 0, ∴ The function is concave upward in the negative real line.
Since f”(x) = 0 at x = 0 and f”(x) changes its sign when passing through x = 0.
Hence the point of inflection is (0, –\(\frac { 2 }{ 3 }\))
The curve has no asymptotes.

(ii) y = x\(\sqrt { 4-x }\)

y = x\(\sqrt { 4-x }\) = f(x)
where x > 4 the curve does not exist and it exists for x ≤ 4
∴ The domain is (-∞, 4] and the Range is (-∞, \(\frac { 16 }{ 3√3 }\) ]
The curve passes through the origin. The curve intersects x-axis at (4, 0).
f'(x) = –\(\frac { x }{ 2 \sqrt{4-x} }\) + \(\sqrt { 4-x }\) = \(\frac { 8-3x }{ 2 \sqrt{4-x} }\)
f'(x) = 0 ⇒ 8 – 3x = 0 ⇒ x = \(\frac { 8 }{ 3 }\)
∴ Critical point of the curve occur at x = \(\frac { 8 }{ 3 }\)
f”(x) = \(\frac { 3x-16 }{ 4(4-x)^{\frac{3}{2}} }\)
f”(\(\frac { 8 }{ 3 }\)) = –\(\frac { 3√3 }{ 4 }\) < 0
∴ f(x) is maximum at x = \(\frac { 8 }{ 3 }\) and the local maximum f(\(\frac { 8 }{ 3 }\)) = \(\frac { 16 }{ 3√3 }\) and local minimum is 0 at x = 4 (from the graph)
f”(x) = \(\frac { 3x-16 }{ 4(4-x)^{\frac{3}{2}} }\) < 0 ∀ x < 4
∴ The curve is concave downward in the negative real line.
No point of inflection exists.
As x → ∞, y → ±∞ , and hence the curve does not have any asymptotes.

(iii) y = \(\frac { x^2+1 }{ x^2-4 }\)

The domain of the given function f(x) is (-∞, -2) ∪ (-2, 2) ∪ (2, ∞)
ie. x < -2 (or) -2 < x < 2 (or) x > 2.
Range of f(x) is (-∞, –\(\frac { 1 }{ 4 }\)) ∪ (1, ∞)
i.,e. f(x) ≤ –\(\frac { 1 }{ 4 }\) (or) f(x) > 1.
Putting y = 0, x is unreal. Hence, there is no ‘x’ intercept.
By putting x = 0, we get y = –\(\frac { 1 }{ 4 }\).
∴ y intercept is (0, –\(\frac { 1 }{ 4 }\))
f'(x) = –\(\frac { 10x }{ (x^2-4)^2 }\)
f'(x) = 0 ⇒ x = 0,
∴ The critical point is at x = 0
f'(x) = \(\frac { 10(x^2-4)(3x^+4) }{ (x^2-4)^4 }\)
f'(0) = –\(\frac { 5 }{ 8 }\) < 0,
∴ f(x) is maximum at
x = 0. Hence the local maximum is f(0) = –\(\frac { 1 }{ 4 }\)
No points of inflection exist for the curve.
When x = ± 2, y = ∞
∴ Vertical asymptotes are x = 2 and x = -2 and Horizontal asymptote is y = 1.

(iv) y = \(\frac { 1 }{ 1+e^{-x} }\)

The Domain of the function f(x) is the entire real line.
ie., (-∞, ∞) ⇒ -∞ < x < ∞ and the range is (0, 1) ie., 0 < f(x) < 1
No ‘x’ intercept for f(x) and when x = 0
y = \(\frac { 1 }{ 2 }\)
∴ The ‘y’ intercept is (0, \(\frac { 1 }{ 2 }\))
f'(x) = \(\frac { e^{-x} }{ (1+e^{-x})^2 }\)
f'(x) = 0 ⇒ which is absurd. Hence there is no extremum.
No vertical asymptote for the curve exist and the Horizontal asymptotes are y = 1 and y = 0.

(v) y = \(\frac { x^3 }{ 24 }\) – log x

The curve exists only for positive values of ‘x’ (x > 0) ie., domain is (0, ∞) and
The range is (\(\frac { 1 }{ 3 }\) – log e², ∞)
No intersection points are possible
f'(x) = \(\frac { x^2 }{ 8 }\) – \(\frac { 1 }{ x }\)
f'(x) = 0 ⇒ x³ – 8 = 0 ⇒ x = 2
∴ Critical point occur at x = 2
f'(x) = \(\frac { x }{ 4 }\) + \(\frac { 1 }{ x^2 }\)
f”(2) = \(\frac { 3 }{ 4 }\) > 0,
∴ f(x) is mimmum at x = 2 and the local minimum is f(2) = \(\frac { 1 }{ 3 }\) – log e²
No point of inflection exists.
No Horizontal asymptotes are possible, but the vertical asymptote is x = 0 (y-axis).

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Bio Zoology Guide Pdf Chapter 11 Biodiversity and its Conservation Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Bio Zoology Solutions Chapter 11 Biodiversity and its Conservation

12th Bio Zoology Guide Biodiversity and its Conservation Text Book Back Questions and Answers

Question 1.
Which of the following region has maximum biodiversity?
(a) Taiga
(b) Tropical forest
(c) Temperate rain forest
(d) Mangroves
Answer:
(b) Tropical forest

Question 2.
Conservation of biodiversity within their natural habitat is ……………….
(a) In-situ conservation
(b) Ex-situ conservation
(c) In vivo conservation
(d) In vitro conservation
Answer:
(a) In-situ conservation

Question 3.
Which one of the following is not coming under in-situ conservation?
(a) Sanctuaries
(b) Natural parks
(c) Zoological park
(d) Biosphere reserve
Answer:
(c) Zoological park

Question 4.
Which of the following is considered a hotspot of biodiversity in India?
(a) Western ghats
(b) Indo-gangetic plain
(c) Eastern Himalayas
(d) A and C
Answer:
(d) A and C

Question 5.
The organization which published the red list of species is ………………
(a) WWF
(b) IUCN
(c) ZSI
(d) UNEP
Answer:
(b) IUCN

Question 6.
Who introduced the term biodiversity?
(a) Edward Wilson
(b) Walter Rosen
(c) Norman Myers
(d) Alice Norman
Answer:
(b) Walter Rosen

Question 7.
Which of the following forests is known as the lungs of the planet Earth?
(a) Tundra forest
(b) Rain forest of north east India
(c) Taiga forest
(d) Amazon rain forest
Answer:
(d) Amazon rain forest

Question 8.
Which one of the following are at high risk extinction due to habitat destruction?
(a) Mammals
(b) Birds
(c) Amphibians
(d) Echinoderms
Answer:
(c) Amphibians

Question 9.
Assertion: The Environmental conditions of the tropics are favourable for speciation and diversity of organisms.
Reason: The climate seasons, temperature, humidity and photoperiod are more or less stable and congenial.
(a) Both Assertion and Reason are true and Reason explains Assertion correctly.
(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.
Answers:
(a) Both Assertion and Reason are true and Reason explains Assertion correctly.

Question 10.
Define endemism.
Answer:
Endemism: A species or a taxon which is confined to a specific area
E..g: Nilgiri Thar

Question 11.
How many hotspots are there in India? Name them.
Answer:
India encloses 4 biodiversity hotspots. They are

• Himalayan
• Indo-Burma
• Western Ghats
• Sunderland

Question 12.
What are the three levels of biodiversity?
Answer:

• Genetic Diversity
• Species Diversity
• Community / Ecosystem Diversity

Question 13.
Name the active chemical found in the medicinal plant Rauwolfia vomitoria. What type of diversity does it belong to?
Answer:
Rauwolfia vomitoria can be cited as an example of genetic diversity. Reserpine is an active chemical extracted from Rauwolfia vomitoria.

Question 14.
“Amazon forest is considered to be the lungs of the planet”-Justify this statement.
Answer:
Amazon rain forest is known as the “lungs of the planet” since a great proportion of CO₂ released due to anthropogenic activities are uptaken by their dense tropical forests, which is _ homologous to the functioning of human beings except for the difference in gases.

Question 15.
‘Red data book’-What do you know about it?
Answer:
Red Data Book or Red list is a catalogue of taxa facing the risk of extinction. IUCN – International Union of Conservation of Nature and Natural Resources, which is renamed as WCU – World Conservation Union (Morges Switzerland) maintains the Red Databook. The concept of Red list was mooted in 1963. The purpose of preparation of Red List are:

• To create awareness on the degree of threat to biodiversity
• Identification and documentation of species at high risk of extinction
• Provide global index on declining biodiversity
• Preparing conservation priorities and help in conservation of action

Information on international agreements on conservation of biological diversity Red list has eight categories of species.

1. • Extinct
   • Extinct in wild
   • Critically Endangered
   • Endangered
   • Vulnerable
   • Lower risk
   • Data deficiency
   • Not evaluated

Question 16.
Extinction of a keystone species led to loss of biodiversity – Justify.
Answer:
A keystone species is an organism that helps define an entire ecosystem. Without the keystone species, a particular ecosystem would be dramatically disturbed or even ceased. Keystone species either directly or indirectly affects every species in a particular ecosystem. If a keystone species is lost or removed no other organism would compensate its ecological niche.

E.g. Jaguar is a keystone species. As a top predator, it plays a crucial role in the ecosystem. Without jaguar, there is an exponential increase in herbivoral population that would decimate the plants of the ecosystem. At one point even the herbivore populations also get declined due to the lack of vegetation. Thus jaguar acts a keystone species.

Question 17.
Compare and Contrast the insitu and exsitu conservation.
Answer:

Question 18.
What are called endangered species? Explain with examples.
Answer:
A species which has been categorized as very likely to become extinct. E.g., Bengal tiger, Polar bears.

Question 19.
Why do we find a decrease in biodiversity distribution, if we move from the tropics towards the poles?
Answer:
There is a decrease in biodiversity as we move from tropics towards poles due to a drop in temperature which makes the condition unfavourable for the majority of organisms to survive.

Question 20.
What are the factors that drive habitat loss?
Answer:
Natural habitats are destroyed for the purpose of settlement, agriculture, mining, industries, and construction of highways. As a result, species are forced to adapt to the changes in the environment or move to other places. If not, they become victim to predation, starvation, disease and eventually die or results in human animal conflict.

Question 21.
Where are biodiversity hotspots normally located? Why?
Answer:
Hotspots are areas characterized with high concentration of endemic species experiencing unusual rapid rate of habitat modification loss. Norman Myers defined hot spots as “regions that harbour a great diversity of endemic species and at the same time, have been significantly impacted and altered by human activities.”

Question 22.
Why is biodiversity so important and worthy of protection?
Answer:
Biodiversity reflects the number of different organisms and their relative frequencies in an ecological system and constitutes the most important functional component of a natural ecosystem. It helps to maintain ecological processes, create soil, recycle nutrients, influence climate, degrade waste, and control diseases. It provides an index of the health of an ecosystem. The survival of the human race depends on the existence and well-being of all life forms (plants and animals) in the biosphere.

Question 23.
Why do animals have greater diversification than plant diversity?
Answer:
The distribution of plants and animals is not uniform around the world. Organisms require different sets of conditions for their optimum metabolism and growth. Plants in general have the ability to withstand and overcome environmental fluctuations. Moreover, majority of plants are autotrophs so they can synthesize their own food, hence they show minimal modifications.
In case of animals, they have to tolerate climatic fluctuations, migrate to other areas in search of food, or adapt themselves or their body parts according food they consume thus varying from ancestors leading to evolution of new species. Hence it is understood that climatic condition food source, predation, competition and other natural forces lead to much diversification among animals than in plants.

Question 24.
Alien species invasion is a threat to endemic species – substantiate this statement.
Answer:
Exotic species are organisms often introduced unintentionally or deliberately for commercial purpose, as biological control agents and other uses. They often become invasive and drive away the local species and is considered as the second major cause for extinction of species. Tilapia fish (Jilabi kendai) (Oreochromis mosambicus) introduced from east coast of South Africa in 1952 for its high productivity into Kerala’s inland waters, became invasive, due to which the native species such as Puntius dubius and Labeo kontius face local extinction.
Amazon sailfin catfish is responsible for destroying the fish population in the wetlands of Kolkata. The introduction of the Nile Perch, a predatory fish into Lake Victoria in East Africa led to the extinction of an ecologically unique assemblage of more than 200 nature species of cichlid fish in the lake.

Question 25.
Mention the major threats to biodiversity caused by human activities. Explain.
Answer:
Human activities, both directly and indirectly are today’s main reason for habitat loss and biodiversity loss. Fragmentation and degradation due to agricultural practices, extraction (mining, fishing, logging and harvesting) and development (settlements, industrial and associated infrastructures) leads to habitat loss and fragmentation leads to the formation of isolated, small and scattered populations and as endangered species.

Some of the other threats include specialised diet, specialized habitat requirement, large size, small population size, limited geographic distribution and high economic or commercial value. Large mammals by virtue of their size require larger areas to obtain the necessities of life – food, cover and mates than do smaller mammals.

The individual home range of Lion can be . “ about 100 square Km. Mammals have specialized dietary needs such as carnivores, frugivores and the need to forage over much larger areas than general dietary herbivores and omnivores. Mammals also have low reproductive output other than small rodents.

Question 26.
What is mass extinction? Will you encounter one such extinction in the near future. Enumerate the steps to be taken to prevent it.
Answer:
The Earth has experienced quite a few mass extinctions due to environmental catastrophes. A mass extinction occurred about 225 million years ago during the Permian, where 90% of shallow-water marine invertebrates disappeared.

Question 27.
In northeastern states, the jhum culture is a major threat to biodiversity – substantiate.
Answer:
In shifting cultivation, plots of natural tree vegetation are burnt away and the cleared patches 1 are farmed for 2-3 seasons, after which their fertility reduces to a point where crop production is no longer profitable.

The farmer then abandons this patch and cuts down a new patch of forest trees elsewhere for crop production. This system is practiced in the north-eastern regions of India. When vast areas are cleared and burnt, it results in loss of forest cover, pollution, and discharge of CO₂, which in turn attributes to a loss of habitat and climate change which has an impact on the faunal diversity of that region.

Question 28.
List out the various causes for biodiversity losses.
Answer:
The major causes of biodiversity decline are:

• Habitat loss, fragmentation, and destruction (affects about 73% of all species),
• Pollution and pollutants (smog, pesticides, herbicides, oil slicks, and GHGs).
• Climate change.
• Introduction of alien/exotic species.
• Overexploitation of resources (poaching, indiscriminate cutting of trees, overfishing, hunting, and mining).
• Intensive agriculture and aquacultural practices.
• Hybridization between native and non-native species and loss of native species
• Natural disasters (Tsunami, forest fire, earthquake,s, and volcanoes).
• Industrialization, Urbanization, infrastructure development, Transport – Road and Shipping activity, communication towers, dam construction, unregulated tourism, and monoculture are a common area of specific threats.
• Co-extinction

Question 29.
How can we contribute to promoting biodiversity conservation?
Answer:

• identify and protect all threatened species
• identify and conserve in protected areas the wild relatives of all the economically important organisms
• identify and protect critical habitats for feeding, breeding, nursing, resting of each species
• resting, feeding, and breeding places of the organisms should be identified and protected.
• Air, water, and soil should be conserved on a priority basis
• Wildlife Protection Act should be implemented

Question 30.
‘Stability of a community depends upon its species diversity’ – Justify the statement.
Answer:
Species diversity leads to a stable community because an area with more species diversity always leads to higher productivity thus maintains a stable community.

Question 31.
Write a note on
(i) Protected areas
(ii) Wildlife sanctuaries
(iii) WWF.
Answer:
(i) Protected areas are biogeographical areas, where biological diversity along with natural and cultural resources is protected, maintained, and managed through legal measures. Protected areas include national parks, wildlife sanctuaries, community reserves, and biosphere reserves.

(ii) Any area other than the area comprised of any reserve forest or the territorial waters can be notified by the State Government to constitute as a sanctuary if such area is of adequate ecological, faunal, floral, geomorphological, natural or zoological significance. This is for the purpose of protecting, endangered factual species. Some restricted human activities are allowed inside the sanctuary area. Ecotourism is permitted, as long as animal life is undisturbed.

(iii) WWF stands for World Wide Fund for nature is an international NGO working in the field of wildlife conservation.

12th Bio Zoology Guide Biodiversity and its Conservation Additional Important Questions and Answers

12th Bio Zoology Guide Biodiversity and its Conservation One Mark Questions and Answers

Question 1.
Who coined the term Bio-diversity?
Answer:
Walter Rosen.

Question 2.
Which is not an indices of species diversity?
(a) Alpha diversity
(b) Beta diversity
(c) Delta diversity
(d) Gamma diversity
Answer:
(c) Delta diversity

Question 3.
The total number of mega biodiversity countries in the world is …………….
(a) Twelve
(b) Fifteen
(c) Seventeen
(d) Nineteen
Answer:
(c) Seventeen

Question 4.
How many biogeographic zones are there in India?
(a) Twelve
(b) Seventeen
(c) Ten
(d) Fifteen
Answer:
(c) Ten

Question 5.
The most important pattern of biodiversity is ……………………..
(a) Longitudinal gradient in diversity
(b) Latitudinal gradient in diversity
(c) Polar gradient diversity
(d) Equatorial gradient in diversity
Answer:
(b) Latitudinal gradient in diversity

Question 6.
Which of the following denotation is correct regarding increasing diversity?
(a) Poles < Equator
(b) Equator < Pole
(c) Pole = Equator
(d) Latitude = Longitude
Answer:
(a) Poles < Equator

Question 7.
Select the proper sequence indicating the increasing order of biodiversity.
(a) Polar, Temperate and Polar
(b) Tropics, Temperate and Polar
(c) Temperate, Tropic and Polar
(d) Polar, Tropic and Temperate
Answer:
(a) Polar, Temperate and Polar

Question 8.
Select the correct linear equation describing the species-area relationship.
(a) loc C = log S + Z log A
(b) Z log A = log S + log C
(c) log S = log C + Z log A
(d) log C = log S ± Z log A
Answer:
(c) log S = log C + Z log A

Question 9.
If meat-eating animals are called carnivores, how do you call the animals that thrive mostly on fruits?
Answer:
Frugivore

Question 10.
The wild ass is endemic to ………………..
(a) the Western Ghats
(b) Deccan Peninsula
(c) the Himalayas
(d) Indian desert
Answer:
(d) Indian desert

Question 11.
Which is considered as the Biogeographical Gateway of India?
(a) the Himalayas
(b) Andaman & Nicobar
(c) North – East India
(d) Mumbai
Answer:
(c) North – East India

Question 12.
Species introduced deliberately in an area are referred to as …………………
(a) Endemic species
(b) Vulnerable species
(c) Exotic species
(d) Extinct species
Answer:
(c) Exotic species

Question 13.
Tilapia fish (Oreochromis Mozambique) is an exotic breed from ……………………
(a) Mexico
(b) South Africa
(c) Canada
(d) Central America
Answer:
(b) South Africa

Question 14.
Mention the correct number of biodiversity hotspots identified throughout the world.
(a) 29
(b) 16
(c) 34
(d) 46
Answer:
(c) 34

Question 15.
Which is not an accepted biodiversity hotspot of India?
(a) Indian Himalayas
(b) Western Ghats
(c) Indo-Burma
(d) Deccan Plateau
Answer:
(d) Deccan Plateau

Question 16.
A species is considered extinct
(a) When its member is confined to a particular area
(b) When its member is maintained in a non-native area
(c) When none of its members is alive in the native area
(d) When none of its members alive anywhere in the world
Answer:
(d) When none of its members are alive anywhere in the world.

Question 17.
The concept of Red list was noted in …………………
(a) 1953
(b) 1963
(c) 1973
(d) 2003
Answer:
(6) 1963

Question 18.
Match the following
(a) Tiger reserves in India – (i) 4
(b) Hotspots in India – (ii) 104
(c) Biosphere reserves in India – (iii) 27
(d) National parks in India – (iv) 18
Answer:
a – iii, b – i, c – iv, d – ii

Question 19.
Statement 1: Biodiversity is the assemblage of different life form.
Statement 2: The term biodiversity was introduced by Edward Wilson.
(a) Statement 1 is correct, statement 2 in incorrect
(b) Statement 1 is incorrect, statement 2 in correct
(c) Both the statements are correct
(d) Both the statements are incorrect
Answer:
(fl) Statement 1 is correct, statement 2 is incorrect

Question 20.
Statement 1: India is the seventh-largest country in the world in terms of area.
Statement 2: It includes ten biogeographic areas.
(a) Statement 1 is correct, statement 2 is incorrect
(b) Statement 1 is incorrect, statement 2 is correct
(c) Both the statements are correct
(d) Both the statements are incorrect
Answer:
(c) Both the statements are correct

Question 21.
Statement 1: the Western Ghats extend from South Gujarat to Karnataka.
Statement 2: Wild ass is an endemic species of Western Ghats
(a) Statement 1 is correct, statement 2 is incorrect
(b) Statement 1 is incorrect, statement 2 is correct
(c) Both the statements are correct
(d) Both the statements are incorrect
Answer:
(d) Both the statements are incorrect

Question 22.
Statement 1: Exotic species are a non-native organisms.
Statement 2: Sailfin catfish is an exotic species in India.
(a) Statement 1 is correct, statement 2 in incorrect
(b) Statement 1 is incorrect, statement 2 in correct
(c) Both the statements are correct
(d) Both the statements are incorrect
Answer:
(c) Both the statements are correct

12th Bio Zoology Guide Biodiversity and its Conservation Two Marks Questions and Answers

Question 1.
Define biodiversity.
Answer:
Biodiversity is the variability among living organisms from all sources, including terrestrial, marine, and other aquatic ecosystems and the ecological complexes of which they are part. This includes diversity within species, between species and ecosystems of a region.

Question 2.
Who coined the term biodiversity? Name the levels of biodiversity.
Answer:
The term biodiversity was coined by Walter Rosen (1986). The levels of biodiversity are – Genetic diversity, Species diversity, and ecosystem diversity.

Question 3.
What is species richness? Give example.
Answer:
The number of species per unit area at a specific time is called species richness, which denotes the measure of species diversity.
E.g., Western ghats have more amphibian species than Eastern ghats.

Question 4.
Enlist the factors that determine the distribution of biodiversity.
Answer:
Temperature, precipitation, distance from the equator (latitudinal gradient), altitude from sea level (altitudinal gradient) are some of the factors that determine biodiversity distribution patterns.

Question 5.
What are the most important causes of biodiversity loss?
Answer:
Habitat loss, fragmentation, and destruction.

Question 6.
Name any two alien animal species introduced in India.
Answer:
(a) Tilapia fish (Oreochromis Mozambique)
(b) African Apple snail (Achatina Fulica)

Question 7.
Name any four biogeographic zones in India.
Answer:
(a) the Himalayas
(b) Indian deserts
(c) Trans-Himalayan region
(d) Western ghats.

Question 8.
What do you mean by the term co-extinction?
Answer:
Coextinction of a species is the loss of a species as a consequence of the extinction of another.
E.g., orchid bees and forest trees by cross-pollination.

Question 9.
What are the natural causes of biodiversity loss?
Answer:
Natural threats include spontaneous jungle fires, tree falls, landslides, defoliation by insects or locust attacks.

Question 10.
Define hotspots.
Answer:
Hotspots are areas characterized with high concentration of endemic species experiencing unusual rapid rate of habitat modification loss.

Question 11.
What will be the consequences of the loss of biodiversity?
Answer:
Loss of biodiversity has a immense impact on plant and animal life. The loss of diversity leads to,

• Extinction of species
• Dramatic influence on the food chain and food web in.
• It will lead to immediate danger for food necessity

Question 12.
Name the four biodiversity hotspots in India.
Answer:
(a) the Himalayas
(b) Western Ghats
(c) Indo-Burma region
(d) Sundaland

Question 13.
What does IUCN stand for?
Answer:
IUCN – International Union for Conservation of Nature and Natural resources.

Question 14.
Define species extinction. What makes a species to become extinct?
Answer:
A species is considered extinct when none of its members are alive anywhere in the world. Environmental changes and population characteristics are the two major reasons for species extinction.

Question 15.
When a species is considered as locally extinct?
Answer:
A species is considered to be locally extinct when it is no longer found in an area it once inhabited but is still found elsewhere in the wild.

Question 16.
State the mission of IUCN.
Answer:
IUCN’s mission is to influence, encourage, and assist societies throughout the world to conserve nature and to ensure that any use of natural resources is equitable and ecologically sustainable.

Question 17.
What is the Red list? How many categories of species are mentioned in the Red List?
Answer:
Red List or Red databook is a catalogue of taxa facing a risk of extinction. It has 8 categories of species.

Question 18.
Mention any four categories of species mentioned in the Red data book.
Answer:

• Extinct
• Endangered
• Extinct in wild
• Vulnerable

Question 19.
How a national park can be defined?
Answer:
The national park is a natural habitat that is notified by the state government to be constituted as a National Park due to its ecological, faunal, floral, geomorphological, or zoological association of importance. No human activity is permitted inside the national park except the activities permitted by the Chief Wildlife Warden of the state.

Question 20.
Name any two species that are extinct due to human activities.
Answer:

• Dodo of Mauritius
• Steller’s cow of Russia

Question 21.
Define in-situ conservation.
Answer:
Conservation of animals in their natural habitat is called in-site conservation.
E.g., National parks.

Question 22.
What is the goal of “Project Tiger”?
Answer:
The project ensures a viable population of Bengal tigers in their natural habitats, protecting them from extinction and preserving areas of biological importance as a natural heritage.

Question 23.
Give the number of national parks in India. Name any two of them in Tamil Nadu.
Answer:
India has 104 National Parks. Guindy National Park (Chennai) and Mudumalai National Park (Nilgiris) are located in Tamil Nadu.

Question 24.
State the role of Biosphere Reserve.
Answer:
Biosphere Reserves are designated to deal with the conservation of biodiversity, economic and social development, and maintenance of associated cultural values.

Question 25.
Name few endangered species protected in Arignar Anna Zoological Park.
Answer:
Royal Bengal Tiger, Lion Tailed Macaque, Nilgiri Langur, and Gray Wolf.

Question 26.
Give the names of two methods of in-situ conservation.
Answer:

• Wildlife Sanctuaries
• Biosphere reserve

12th Bio Zoology Guide Biodiversity and its Conservation Three Marks Questions and Answers

Question 27.
Point out the biosphere reserves in Tamil Nadu.
Answer:

• Nilgiris (Tamil Nadu – Kerala)
• Agasthya malai (Tamil Nadu – Kerala – Karnataka)
• Gulf of Mannar (Tamil Nadu)

Question 28.
Write a note on Sacred Groves.
Answer:
A sacred grove or sacred woods are any groves of trees that are of special religious importance to a particular culture. Sacred groves feature in various cultures throughout the world.

Question 29.
What is ex-situ conservation?
Answer:
Ex-situ conservation of selected rare plants/ animals in places outside their natural homes. It includes offsite collections and gene banks.

Question 30.
Why Red list is prepared periodically?
Answer:
The purpose of preparation of Red List are:

• To create awareness on the degree of threat to biodiversity
• Identification and documentation of species at high risk of extinction
• Provide global index on declining biodiversity
• Preparing conservation priorities and help in conservation of action
• Information on international agreements on the conservation of biological diversity

Question 31.
Name the types of extinctions.
Answer:

• Natural Extinction
• Mass Extinction
• Anthropogenic Extinction

Question 32.
Point out the human activities that threaten biodiversity.
Answer:
Direct and indirect human activities have a detrimental effect on biodiversity. Direct human, activities like change in local land use, species introduction or removal, harvesting, pollution, and climate change contribute a greater pressure on the loss of biodiversity. Indirect human drivers include demographic, economic, technological, cultural, and religious factors.

Question 33.
Extinction of Dodo bird led to the danger of Calvaria tree – Justify,
Answer:
Another example for co-extinction is the connection between Calvaria tree and the extinct bird of Mauritius Island, the Dodo. The Calvaria tree is dependent on the Dodo bird for completion of its life cycle. The mutualistic association is that the tough horny endocarp of the seeds of Calvaria tree are made permeable by the actions of the large stones in birds gizzard and digestive juices thereby facilitating easier germination. The extinction of the Dodo bird led to the imminent danger of the Calvaria tree coextinction.

Question 34.
Give an account on slash and burn agriculture.
Answer:
In shifting cultivation, plots of natural tree vegetation are burnt away and the cleared patches are farmed for 2-3 seasons, after which their fertility reduces to a point where crop production is no longer profitable. The farmer then abandons this patch and cuts down a new patch of forest trees elsewhere for crop production.

This system is practiced in the north-eastern regions of India. When vast areas are cleared and burnt, it results in loss of forest cover, pollution and discharge of CO₂ which in turn attributes to a loss of habitat and climate change which has an impact on the faunal diversity of that regions.

Question 35.
Impact of Industrialization on Biodiversity – Comment.
Answer:
Industrialization is a major contributor to climate change and a major threat to biodiversity. Energy drives our industries, which is provided by burning of fossil fuels. This increases the emission of CO₂, a GHG, leading to climate change. Due to large scale deforestation, the emitted CO₂ cannot be absorbed fully, and its concentration in the air increases.

Climate change increases land and ocean temperature changes precipitation patterns and raises the sea level. This, in turn, results in melting of glaciers, water inundation, less predictability of weather patterns, extreme weather conditions, outbreak of squalor diseases, migration of animals and loss of trees in forest. Thus, climate change is an imminent danger to the existing biodiversity.

Question 36.
What are exotic species? Explain with example.
Answer:
Exotic species are organisms often introduced unintentionally or deliberately for commercial purpose, as biological control agents and other uses. They often become invasive and drive away the local species and is considered as the second major cause for extinction of species. Exotic species have proved harmful to both aquatic and terrestrial ecosystems.
Tilapia fish (Jilabi kendai) (Oreochromis mosambicus) introduced from east coast of South Africa in 1952 for its high productivity into Kerala’s inland waters, became invasive, due to which the native species such as Puntius dubius and Labeo kontius face local extinction.

Question 37.
Write a brief note on Habitat fragmentation.
Answer:
Habitat fragmentation is the process where a large, continuous area of habitat is both, reduced in area and divided into two or more fragments. Fragmentation of habitats like forest land into croplands, orchard lands, plantations, urban areas, industrial estates, transport, and transit systems has resulted in the destruction of complex interactions amongst species, (food chain and webs) destruction of species in the cleared regions, annihilation of species restricted to these habitats (endemic) and decreased biodiversity in the habitat fragments.

Animals requiring large territories such as mammals and birds are severely affected. The elephant corridors and migratory routes are highly vulnerable. The dwindling of many well-known birds (sparrows) and animals can be attributed to this.

Question 38.
Write a note on the biogeographic area – the Gangetic plains.
Answer:
Gangetic Plains: These plains are relatively homogeneously defined by the Ganges river system and occupy about 11% of the country’s landmass. This region is very fertile and extends up to the Himalayan foothills. Fauna includes rhinoceros, elephant, buffalo, swamp deer, hog-deer.

Question 39.
Compare Alpha diversity with Beta diversity.
Answer:
(i) Alpha diversity: It is measured by counting the number of taxa (usually species) within a particular area, community or ecosystem.
(ii) Beta diversity: It is species diversity between two adjacent ecosystems and is obtaining by comparing the number of species unique to each of the ecosystem.

Question 40.
What is species diversity?
Answer:
Species diversity refers to the variety in number and richness of the species in any habitat. The number of species per unit area at a specific time is called species richness, which denotes the measure of species diversity. The Western Ghats have greater amphibian species diversity than the Eastern Ghats. The more the number of species in an area the more is the species richness. The three indices of diversity are – Alpha, Beta and Gamma diversity.

Question 41.
State the principle of Stockholm declaration – 1972.
Answer:
The natural resources of the Earth, including air, water, land, flora, and fauna of natural ecosystems must be safeguarded for the benefit of the present and future generations through careful planning and management, as appropriate – Principle of the Stockholm Declaration. 1972.

12th Bio Zoology Guide Biodiversity and its Conservation Five Marks Questions and Answers

Question 42.
Give an account on genetic diversity and community diversity.
Answer:
Genetic diversity refers to the differences in genetic make-up (number and types of genes) between distinct species and to the genetic variation within a single species; also covers genetic variation between distinct populations of the same species. Genetic diversity can be measured using a variety of molecular techniques. India has more than 50,000 genetic variants of Paddy and 1000 variants of Mango. Variation of genes of a species increases with diversity in size and habitat. It results in the formation of different races, varieties and subspecies. Rauwolfia vomitaria, a medicinal plant growing in different ranges of the Himalayas shows differences in the potency and concentration of the active ingredient reserpine due to genetic diversity. Genetic diversity helps in developing adaptations to changing environmental conditions.

Community/Ecosystem diversity is the variety of habitats, biotic communities, and ecological processes in the biosphere. It is the diversity at ecosystem level due to diversity of niches, trophic levels and ecological processes like nutrient cycles, food webs, energy flow, and several biotic interactions. India with its alpine meadows, rain forests, mangroves, coral reefs, grasslands, and deserts has one of the greatest ecosystem diversity on Earth.

Question 43.
Why Tropical regions are rich in biodiversity?
Answer:
The reasons for the richness of biodiversity in the Tropics are:

• Warm tropical regions between the tropic of Cancer and Capricorn on either side of the equator possess congenial habitats for living organisms.
• Environmental conditions of the tropics are favourable not only for speciation but also for supporting both variety and number of organisms.
• The temperatures vary between 25°C to 35°C, a range in which most metabolic activities of living organisms occur with ease and efficiency.
• The average rainfall is often more than 200 mm per year.
• Climate, seasons, temperature, humidity, photoperiods are more or less stable and encourage both variety and numbers.
• Rich resource and nutrient availability.

Question 44.
What is the significance of the slope of regression in a species-area relationship?
Answer:
German Naturalist and Geographer Alexander von Humboldt explored the wilderness of South American jungles and found that within a region the species richness increased with the increasing area but upto a certain limit. The relationship between species richness and area for a wide variety of taxa (angiosperm plants, birds, bats and freshwater fishes) turned out to be the rectangular hyperbola. On a logarithmic scale, the relationship is a straight line described by the equation.
log S = log C + Z log A
where S = Species richness
A = Area
Z = Slope of the line
(regression coefficient)
C = Y-intercept

Regression coefficient Z generally has a value of 0.1-0.2 regardless of taxonomic group or region. However, in case of the species-area relationship in very large areas like entire continents, the slope of the line appears to be much steeper (Z-value in the range of 0.6-1.2). For example, in the case of the fruit eating (frugivorous) birds and mammals in the tropical forests of different continents, the slope is found to be a steeper line.

Question 45.
Point out any 5 functional attributions of biodiversity.
Answer:
The major functional attributes are:

• continuity of nutrient cycles or biogeochemical cycles (N₂, C, H₂O, P, S cycles)
• soil formation, conditioning or maintenance of soil health (fertility) by soil microbial diversity along with the different trophic members
• increases ecosystem productivity and provide food resources
• act as water traps, filters, water flow regulators, and water purifiers (forest cover and vegetation)
• climate stability (forests are essential for rainfall, temperature regulation, C02 absorption, which in turn regulate the density and type of vegetation)
• forest resource management and sustainable development

Question 46.
Explain in detail about various types of extinctions.
Answer:
There are three types of Extinctions
i. Natural extinction: It is a slow process of replacement of existing species with better adapted species due to changes in environmental conditions, evolutionary changes, predators and diseases. A small population can get extinct sooner than a large population due to inbreeding depression (less adaptivity and variation)

ii. Mass extinction: The Earth has experienced quite a few mass extinctions due to environmental catastrophes. Amass extinction occurred about 225 million years ago during the Permian, where 90% of shallow-water marine invertebrates disappeared.

iii. Anthropogenic extinctions: These are abetted by human activities like hunting, habitat destruction, overexploitation, urbanization and industrialization. Some examples of extinctions are Dodo of Mauritius and Steller’s sea cow of Russia. Amphibians seem to be at higher risk of extinction because of habitat destruction. The most serious aspect of the loss of biodiversity is the extinction of species. The unique information contained in its genetic material (DNA) and the niche it possesses are lost forever.

Question 47.
Give a comparative account on ex-situ conservation.
Answer:
Ex-Situ Conservation: It is conservation of selected rare plants/ animals in places outside their natural homes. It includes offsite collections and gene banks.

Offsite Collections: They are live collections of wild and domesticated species in Botanical gardens, Zoological parks, Wildlife safari parks, Arborata (gardens with trees and shrubs). The organisms are well maintained for captive breeding programmes. As a result, many animals which have become extinct in the world continue to be maintained in Zoological Parks. As the number increases in captive breeding, the individuals are selectively released in the wild. In this way, the Indian crocodile and Gangetic dolphin have been saved from extinction.

Gene Banks: Gene banks are a type of biorepository which preserves genetic materials. Seeds of different genetic strains of commercially important plants can be stored for long periods in seed banks, gametes of threatened species can be preserved in viable and fertile condition for long periods using cryopreservation techniques. However, it is not economically feasible to conserve all biological wealth and all the ecosystems. The number of species required to be saved from extinction far exceeds the conservation efforts.

Higher Order Thinking Skills (HOTs) Questions

Question 1.
What will be the result, if the relationship between species richness and area for wide variety of taxa are plotted on a long scale?
Answer:
A rectangular hyperbola.

Question 2.
What may be the reasons for the entry of wild .lives into the agriculatural lands or towns?
Answer:

• Habitat lose / Habitat fragmentation
• Lack of food or water source

Question 3.
When does a species is categorized as endangered?
Answer:
A species that has been categorized as very likely to become extinct is an endangered species.

Question 4.
Give any two examples of anthropogenic extinction.
Answer:
Dodo of Mauritius Steller’s cow of Russia

Question 5.
Mention any two species that had become extinct very recently.
Answer:

• George, the tree snail (Achatinella apexfulva)
• Sudan – Northern white rhinoceros (Ceratotherium simum)

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Bio Zoology Guide Pdf Chapter 4 Principles of Inheritance and Variation Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Bio Zoology Solutions Chapter 4 Principles of Inheritance and Variation

12th Bio Zoology Guide Principles of Inheritance and Variation Text Book Back Questions and Answers

Question 1.
Haemophilia is more common in males because it is a………………
(a) Recessive character carried by Y-chromosome
(b) Dominant character carried by Y-chromosome
(c) Dominant trait carried by X-chromosome
(d) Recessive trait carried by X-chromosome
Answer:
(d) Recessive trait carried by X-chromosome

Question 2.
ABO blood group in man is controlled by …………………….
(a) Multiple alleles
(b) Lethal genes
(c) Sex linked genes
(d) Y-linked genes
Answer:
(a) Multiple alleles

Question 3.
Three children of a family have blood groups A, AB and B. What could be the genotypes of their parents?
(a) I^AI^B and ii
(b) I^A1^O and I^BI^O
(c) I^B I^B and I^AI^A
(d) I^AI^A and ii
Answer:
(b) I^A1^O and I^BI^O

Question 4.
Which of the following is not correct?
(a) Three or more alleles of a trait in the population are called multiple alleles.
(b) A normal gene undergoes mutations to form many alleles
(c) Multiple alleles map at different loci of a chromosome
(d) A diploid organism has only two alleles out of many in the population
Answer:
(c) Multiple alleles map at different loci of a chromosome

Question 5.
Which of the following phenotypes in the progeny are possible from the parental combination
(a)AandB only
(b) A,B andAB only
(c) AB only
(d) A,B,AB and O
Answer:
(d) A,B,AB and O

Question 6.
Which of the following phenotypes is not possible in the progeny of the parental genotypic combination I^AI^O x l^Al^B ?
(a) AB
(b) O
(c) A
(d) B
Answer:
(b) O

Question 7.
Which of the following is true about Rh factor in the offspring of a parental combination DdXDd (both Rh positive)?
(a) All will be Rh positive
(b) Half will be Rh positive
(c) About 3/4 will be Rh negative
(d) About one fourth will be Rh negative
Answer:
(d) About one fourth will be Rh negative

Question 8.
What can be the blood group of offspring when both parents have AB blood group?
(a) AB only
(b) A, B and AB
(c) A, B, AB and O
(d) A and B only
Answer:
(b) A, B and AB

Question 9.
If the childs blood group is ‘O’ and fathers blood group is ‘A’ and mother’s blood group is ‘B’ the genotype of the parents will be …………….
(a) I^AI^A and I^BI^O
(b) I^AI^O and I^BI^O
(c) I^AI^O and I^OI^O
(d) I^OI^O and I^BI^B
Answer:
(b) I^AI^O and I^BI^O

Question 10.
XO type of sex determination and XY type of sex determination are examples of …………………
(a) Male heterogamety
(b) Female heterogamety
(c) Male homogamety
(d) Both (b) and (c)
Answer:
(a) Male heterogamety

Question 11.
In an accident there is great loss of blood and there is no time to analyse the blood group Question which blood can be safely transferred?
(a) ‘O’ and Rh negative
(b) ‘O’ and Rh positive
(c) ‘B’ and Rh negative
(d) ‘AB’ and Rh positive
Answer:
(a) ‘O’ and Rh negative

Question 12.
Father of a child is colourblind and mother is carrier for colourblindness, the probability of the child being colourblind is ………………
(a) 25%
(b) 50%
(c) 100%
(d) 75%
Answer:
(b) 50%

Question 13.
A marriage between a colourblind man and a normal woman produces
(a) All carrier daughters and normal sons
(b) 50% carrier daughters, 50% normal daughters
(c) 50% colourblind sons, 50% normal sons
(d) All carrier offsprings
Answer:
(a) All carrier daughters and normal sons

Question 14.
Mangolism is a genetic disorder which is caused by the presence of an extra chromosome number.
(a) 20
(b) 21
(c) 4
(d) 23
Answer:
(b) 21

Question 15.
Klinefelters’ syndrome is characterized by a karyotype of………………
(a) XYY
(b) XO
(c) XXX
(d) XXY
Answer:
(d) XXY

Question 16.
Females with Turners’syndrome have………………
(a) Small uterus
(b) Rudimentary ovaries
(c) Underdeveloped breasts
(d) All of these
Answer:
(d) All of these

Question 17.
Pataus’ syndrome is also referred to as………………
(a) 13-Trisomy
(b) 18-Trisomy
(c) 21-Trisomy
(d) None of these
Answer:
(a) 13-Trisomy

Question 18.
Who is the founder of Modem Eugenics movement?
(a) Mendel
(b) Darwin
(c) Fransis Galton
(d) Karl pearson
Answer:
(c) Fransis Galton

Question 19.
Improvement of human race by encouraging the healthy persons to marry early and produce large number of children is called………………
(a) Positive eugenics
(b) Negative eugenics
(c) Positive euthenics
(d) Positive euphenics
Answer:
(a) Positive eugenics

Question 20.
The ……………… deals with the control of several inherited human diseases especially inborn errors of metabolism.
(a) Euphenics
(b) Eugenics
(c) Euthenics
(d) All of these
Answer:
(a) Euphenics

Question 21.
“Universal Donor” and “Universal Recipients” blood group are ……………… and ………………  respectively.
(a) AB, O
(b) O, AB
(c) A, B
(d) B, A
Answer:
(b) O, AB

Question 22.
ZW-ZZ system of sex determination occurs in………………
(a) Fishes
(b) Reptiles
(c) Birds
(d) All of these
Answer:
(d) All of these

Question 23.
A co-dominant blood group is
(a) A
(b) AB
(c) B
(d) O
Answer:
(b) AB

Question 24.
Which of the following is incorrect regarding ZW-ZZ type of sex determination?
(a) It occurs in birds and some reptiles
(b) Females are homogametic and males are heterogametic
(e) Male produce two types of gametes
(d) It occurs in gypsy moth
Answer:
(b) Females are homogametic and males are heterogametic

Question 25.
What is haplodiploidy?
Answer:
In haplodiploidy, the sex of the offspring is determined by the number of sets of chromosomes it receives. Fertilized eggs develop into females (Queen or Worker) and unfertilized eggs develop into males (drones) by parthenogenesis. It means that the males have half the number of chromosomes (haploid) and the females have double the number (diploid).

Question 26.
Distinguish between heterogametic and homogametic sex determination systems.
Answer:
Heterogametic Sex :

1. Organisms producing two different types of gametes.
2. Example: Human male.
   Sperm with X chromosome
   Sperm with Y chromosome

Homogametic Sex :

1. Organisms producing only one type of gametes.
2. Example: Human female.
   Every egg produced contain X chromosomes.

Question 27.
What is Lyonisation?
Answer:
Lyonisation is a process of inactivation of one of the X chromosomes in some females.

Question 28.
What is criss-cross inheritance?
Answer:
Inheritance of genes from a male parent to female child and then to male grandchild or female parent to male child and then to female grandchild. E.g., X-linked gene inheritance.

Question 29.
Why are sex-linked recessive characters more common in male human beings?
Answer:
Sex linked inherited traits are more common in males than females because, males are hemizygous and therefore express the trait when they inherit one mutant allele.

Question 30.
What are holandric genes?
Answer:
The genes present in the differential region of the Y chromosome are called Y- linked or holandric genes. The Y linked genes have no corresponding allele in X chromosome.

Question 31.
Mention the symptoms of Phenylketonuria.
Answer:
Severe mental retardation, light pigmentation of skin and hair. Phenylpyruvic acid is excreted in the urine.

Question 32.
Mention the symptoms of Down’s syndrome.
Answer:
Severe mental retardation, defective development of the central nervous system, increased separation between the eyes, flattened nose, ears are malformed, mouth is constantly open and the tongue protrudes.

Question 33.
Differentiate Intersexes from Supersexes.
Answer:Intersexes:
Intersexes refers to the individuals having the characteristics of both female and male sexes and their sexual anatomy does not seem to fit the typical definition of male or female.

Supersexes:
Supersexes ar formed as a result of an abnormal combination of sex chromosomes.
Example: Super males in humans human beings have 44+XYY chromosomes.

Question 34.
Explain the genetic basis of ABO blood grouping in man.
Answer:
Multiple allele inheritance of ABO blood groups
Blood differs chemically from person to person. When two different incompatible blood types are mixed, agglutination (clumping together) of erythrocytes (RBC) occurs. The basis of these chemical differences is due to the presence of antigens (surface antigens) on the membrane of RBC and epithelial cells. Karl Landsteiner discovered two kinds of antigens called antigen ‘A’ and antigen ‘B’ on the surface of RBC’s of human blood. Based on the presence or absence of these antigens three kinds of blood groups, type ‘A’, type ‘B’, and type ‘O’ (universal donor) were recognized. The fourth and the rarest blood group ‘AB’ (universal recipient) was discovered in 1902 by two of Landsteiner’s students Von De Castelle and Sturli.

Bernstein in 1925 discovered that the inheritance of different blood groups in human beings is determined by a number of multiple allelic series. The three autosomal alleles located on chromosome 9 are concerned with the determination of blood group in any person. The gene controlling blood type has been labeled as ‘L’ (after the name of the discoverer, Landsteiner) or I (from isoagglutination). The I gene exists in three allelic forms, I^A, I^B and I^O. I^A specifies A antigen. I^B allele determines B antigen and I^O allele specifies no antigen. Individuals who possess these antigens in their fluids such as the saliva are called secretors.

Each allele (I^A and I^B) produces a transferase enzyme. I^A allele produces N-acetyl galactose transferase and can add N-acetyl galactosamine (NAG) and I^B allele encodes for the enzyme galactose transferase that adds galactose to the precursor (i.e. H substances). In the case of I^O/I^O allele no terminal transferase enzyme is produced and therefore called “null” allele and hence cannot add NAG or galactose to the precursor.

From the phenotypic combinations it is evident that the alleles I^A and I^B are dominant to 1°, but co-dominant to each other (I^A = I^B). Their dominance hierarchy can be given as (I^A=I^B> 1^O). A child receives one of three alleles from each parent, giving rise to six possible genotypes and four possible blood types (phenotypes). The genotypes are I^AI^A , I^AI^O, I^BI^B, I^BI^O, I^AI^B and I^OI^O.

Question 35.
How is sex determined in human
Answer:
Genes determining sex in human beings are located on two sex chromosomes, called allosomes. In mammals, sex determination is associated with chromosomal differences between the two sexes, typically XX females and XY males. 23 pairs of human chromosomes include 22 pairs of autosomes (44A) and one pair of sex chromosomes (XX or XY). Females are homogametic producing only one type of gametes (egg), each containing one X chromosome while the males are heterogametic producing two types of sperms with X and Y chromosomes. An independently evolved XX: XY system of sex chromosomes also exist in Drosophila.

Question 36.
Explain male heterogamety.
Answer:
Male heterogamety (XY males) is a type of sex determination in which males produce two different types of gametes. For example, human males produce two kinds of sperms that is sperm with X-chromosome and sperms with Y-chromosome.

Question 37.
Brief about female heterogamety.
Answer:
Female heterogamety (ZO females) refers to the condition, where female produces two types of egg cells. Some with Z chromosome and some without Z chromosome.

Question 38.
Give an account of genetic control of Rh factor?
Answer:
Genetic control of Rh factor
Fisher and Race hypothesis: Rh factor involves three different pairs of alleles located on three different closely linked loci on the chromosome pair. This system is more commonly in use today, and uses the ‘Cde’ nomenclature.
In the given figure, three pairs of Rh alleles (Cc, Dd and Ee) occur at 3 different loci on homologous chromosome pair-1. The possible genotypes will be one C or c, one D or d, one E or e from each chromosome. For e.g. CDE/cde; CdE/cDe; cde/cde; CDe/CdE etc. All genotypes carrying a dominant ‘D’ allele will produce Rh+positive phenotype and double recessive genotype ‘dd’ will give rise to Rh negative phenotype.

Wiener Hypothesis
Wiener proposed the existence of eight alleles (R¹, R², R⁰, R^z, r, r¹, r¹¹, r^y) at a single Rh locus. All genotypes carrying a dominant ‘R allele’ (R¹, R² ,R⁰ ,R^z) will produce ‘Rh-positive’ ^phenotype and double recessive genotypes (rr, rr¹, rr¹¹, rr^y) will give rise to Rh-negative phenotype.

Question 39.
Explain the mode of sex determination in honeybees.
Answer:
In hymenopteran insects such as honeybees, ants and wasps, a mechanism of sex determination called haplodiploidy mechanism of sex determination is common. In this system, the sex of the offspring is determined by the number of sets of chromosomes it receives. Fertilized eggs develop into females (Queen or Worker) and unfertilized eggs develop into males (drones) by parthenogenesis. It means that the males have half the number of chromosomes (haploid) and the females have double the number (diploid), hence the name haplodiploid for this system of sex determination.

This mode of sex determination facilitates the evolution of sociality in which only one diploid female becomes a queen and lays the eggs for the colony. All other females which are diploid having developed from fertilized eggs help to raise the queen’s eggs and so contribute to the queen’s reproductive success and indirectly to their own, a phenomenon known as Kin Selection. The queen constructs their social environment by releasing a hormone that suppresses fertility of the workers.

Question 40.
Discuss the genic balance mechanism of sex determination with reference to Drosophila?
Answer:
XX-XY type (Lygaeus Type) sex determination is seen in Drosophila. The females are homogametic with XX chromosomes, while the males are heterogametic with X and Y chromosomes. Homogametic females produce only one kind of egg, each with one X chromosome, while the heterogametic males produce two kinds of sperms some with X chromosome and some with Y chromosome.

The sex of the embryo depends on the fertilizing sperm. An egg fertilized by an ‘X’ bearing sperm produces a female, if fertilized by a ‘Y’ bearing sperm, a male is produced.

Question 41.
What are the applications of Karyotyping?
Answer:

• Karyotyping helps in gender identification.
• It is used to detect the chromosomal aberrations like deletion, duplication, translocation, non-disjunction of chromosomes.
• It helps to identify the abnormalities of chromosomes like aneuploidy.
• It is also used in predicting the evolutionary relationships between species.
• Genetic diseases in human beings can be detected by this technique.

Question 42.
Explain the inheritance of sex linked characters in human being.
Answer:
Haemophilia is commonly known as bleeder’s disease, which is more common in men than women. This hereditary disease was first reported by John Cotto in 1803. Haemophilia is caused by a recessive X-linked gene. A person with a recessive gene for haemophilia lacks a normal clotting substance (thromboplastin) in blood, hence minor injuries cause continuous ’bleeding, leading to death. The females are carriers of the disease and would transmit the disease to 50% of their sons even if the male parent is normal. Haemophilia follows the characteristic criss-cross pattern of inheritaitce.

Question 43.
What is extra chromosomal inheritance? Explain with an example.
Answer:
The cytoplasmic extra nuclear genes have a characteristic pattern of inheritance which does not resemble genes of nuclear chromosomes and are known as Extrachromosomal/ Cytoplasmic inheritance.

Question 44.
Comment on the methods of Eugenics.
Answer:
Eugenics refers to the study of the possibility of improving the qualities of human population.

Methods of Eugenics:

• Sex-education in school and public forums.
• Promoting the uses of contraception.
• Compulsory sterilization for mentally retarded and criminals.
• Egg donation.
• Artificial insemination by donors.
• Prenatal diagnosis of genetic disorders and performing MTP
• Gene therapy
• Cloning
• Egg/sperm donation of healthy individuals.

12th Bio Zoology Guide Principles of Inheritance and Variation Additional Important Questions and Answers

12th Bio Zoology Guide Principles of Inheritance and Variation One Mark Questions and Answers

Question 1
If a colorblind female marries a normal male, their sons will be ………………
(a) All normal visioned
(b) All color blinded
(c) One half normal visioned other half colorblind
(d) Three fourth colorblind one fourth normal
Answer:
(c) One half normal visioned other half colorblind

Question 2
Excess hair growth on pinna is a feature noticed only in males because ……………
(a) Males produce more testosterone
(b) gene responsible for the character is located in Y-chromosome
(c) Estrogen suppresses the character in females
(d) females act only as a carriers for this character
Answer:
(b) gene responsible for the character is located in Y-chromosome

Question 3.
ABO blood group is a classical example for ………………..
(a) Multiple allelism
(b) Pleotropism
(c) Incomplete dominance
(d) Polygenic mechanism
Answer:
(a) Multiple allelism

Question 4
Unit of heredity is ……………….
(a) allele
(b) allelomorph
(c) trait
(d) gene
Answer:
(d) gene

Question 5.
Identify the proper dominance hierarchy.
(a) I^A = I^O > I^B
(b) I^A = I^B > O
(b) I^A = I^B > O
(d) I^B = I^A > O
Answer:

Question 6
Haemophilia is more common in human males than human females. The reason is due to
(а) X-linked dominant gene
(b) X-linked recessive gene
(c) Y-linked recessive gene
(d) Allosomal abnormality
Answer:
(b) X-linked recessive gene

Question 7.
Identify the correct statement.
(a) Homozygous sex chromosome (XX) produce males in Drosophila
(b) Homozygous sex chromosome (ZZ) determine female sex in birds
(c) Heterozygous sex chromosome (XO) determine male sex in grasshopper
(d) Heterozygous sex chromosome (ZW) determine male sex in gypsy moth
Answer:
(c) Heterozygous sex chromosome (XO) determine male sex in grasshopper

Question
Which blood group doesnot possess antibodies?
(a) I^AI^B
(b) I^OI^O
(c) I^AO
(d) I^BI^B
Answer:
(a) I^AI^B

Question 9.
Assertion (A): On diagnosis, Ramu is reported to have underdeveloped testis and gynecomastia.
Reason (R): His karyotype reveals XXY condition.
(а) A is right but R is wrong
(b) R explains A
(c) Both A and R are wrong
(d) Both and R are right but R is not the correct explanation of A
Answer:
(b) R explains A

Question 10.
Pick out the odd man.
(a) Klinefelter’s syndrome
(b) Turner’s syndrome
(c) Huntington’s chorea
(d) 13-Trisomy
Answer:
(c) Huntington’s chorea

Question 11.
Pick the odd one out regarding Mendelian disorder.
(a) Thalassemia
(b) phenylketonuria
(c) Albinism
(d) Huntington’s chorea
Answer:
(d) Huntington’s chorea

Question 12.
Match the following:

(a) A – iv, B – ii, C – i, D – iii
(b)A – ii, B – iv, C – iii, D – i
(a)A – iii, B – i, C – ii, D – iv
(c) A – i, B – iv, C – iii, D – iii
Answer:
(a) A – iv, B – ii, C – i, D – iii

Question 13.
Identify the proper ratio of normal visioned individuals against colorblind individuals, if colorblind carrier female marries a normal male.
(a) 1 : 1
(b) 3:1
(c) 1 : 3
(d) All four are normal visioned
Answer:
(c) 1 : 3

Question 14.
Pick out the correct statement.
(i) Karyotyping helps in gender identification
(ii) Holandric genes are located on X-chromosome
(iii) Trisomy-21 is an allosomal abnormality
(iv) Cooley’s anemia is an autosomal recessive disorder
(a) i, iii, iv are correct
(c) i and iv are correct
Answer:
(c) i and iv are correct

Question 15.
DOPA stands for ……………….
(a) 3,4- dihydroxy phenylacetate
(b) 3, 4 – dihydroxy phenylalanine
(c) 3,4- dihydroxy phenyl aspartate
(d) 3, 4- dihydroxy phenol aldehyde
Answer:
(b) 3,4 – dihydroxy phenylalanine

Question 16.
The type of antibody generated against Rh antigen is ….
(a)IgE
(b) IgG
(c) IgA
(d) IgB
Answer:
(b) IgG

Question 17.
Which of the following symbol is used in the pedigree analysis to represent unspecified sex?

Answer:

Question 18:
A colorblind man marries a woman with normal sight who has no history of color blindness in her family. What is the probability of their grandson being colorblind?
(a) 1/4
(b) 3/4
(c) 2/4
(d) 4/4
Answer:
(a) 1/4

Question 19.
Multiple alleles are located……………………
(a) at different loci on homologous chromosome
(b) at same locus on homologous chromosome
(c) at different loci on non-homologous chromosome
(d) at different chromosomes
Answer:
(b) at same locus on homologous chromosome

Question 20.
Identify the incorrect statement regarding haplodiploidy.
(g) Haplodiploidy is noticed in honeybees and drosophila
(b) Unfertilized eggs develop into drones
(c) Fertilized eggs develop into queen and worker bees
(d) Males have half the total chromosomal number
Answer:
(a) Haplodiploidy is noticed in honeybees and drosophila

Question 21.
I^A and I^B genes of ABO blood group are
(a) Co-dominant
(b) Pleotropic
(c) Dominant and recessive
(d) Epistatic
Answer:
(a) Co-dominant

Question 22.
Which one of the following crosses show 3 : 1 ratio of normal visioned versus carrier blind?
(a) X^CX^C x X⁺Y
(b) X⁺ X^C x X^C Y^–
(c) X⁺X^C x X⁺Y^–
(d) X⁺X⁺ x X^CY^–
Answer:
(c) X⁺X^C x X⁺Y^–

12th Bio Zoology Principles of Inheritance and Variation Two Marks Questions and Answers

Question 1.
Define multiple allelism.
Answer:
When three or more alleles of a gene that control a particular trait occupy the same locus on the homologous chromosome of an organism, they are called multiple alleles and their inheritance is called multiple allelism.

Question 2.
Name the discoverers of antigen A, B and AB.
Answer:
Antigens A and Antigen B was discovered by Karl Landsteiner. Antigen AB was discovered by Von De Castelle and Sturli.

Question 3.
What happens if type A blood is injected to a person having B blood group? Explain the reason.
Answer:
When two different incompatible blood types are mixed, agglutination (clumping together) of erythrocytes (RBC) occurs. The basis of these chemical differences is due to the presence of antigens (surface antigens) on the membrane of RBC and epithelial cells.

Question 4.
State the allelic forms of I gene and mention its chromosomal location.
Answer:
The I gene exists in three forms: I^A, I^B and I^O. The alleles are located on chromosome 9.

Question 5.
Write the possible genotypes for a person having a B-blood group.
Answer:
The possible genotypes of a B-blood group person are I^BI^B or I^BI^O.

Question 6.
State Wiener Hypothesis on Rh-factor.
Wiener proposed the existence of eight alleles (R¹, R², R⁰, R^z, r, r¹, r¹¹, r^y) at a single Rh locus. All genotypes carrying a dominant ‘R allele’ (R¹, R² ,R⁰ ,R^z) will produce ‘Rh-positive’ phenotype and double recessive genotypes (rr, rr¹, rr¹¹, rr^y) will give rise to Rh-negative phenotype.

Question 7.
Distinguish between homogametic and heterogametic condition with example.
Answer:
Homogametic organism:

1. Organism producing only one type of gametes.
2. e.g. Human female (Only X)

Heterogametic organism :

1. Organism producing two different types of gametes.
2. e.g. Human Male (X and Y)

Question 8.
Name any four organism expressing ZW-ZZ type of sex determination.
Answer:
Gypsy moth, fishes, reptiles and birds.

Question 9.
Expand (a) SRY (b) TDF
Answer:
SRY – Sex Determining region Y
TDF – Testes Determining Factor

Question 10.
Define Barr body.
Answer:
In 1949, Barr and Bertram first observed a condensed body in the nerve cells of female cat which was absent in the male. This condensed body was called sex chromatin by them and was later referred as Barr body.

Question 11.
Based on Lyon’s hypothesis, mention the number of Barr bodies in XXY males, XO females.
Answer:
XXY males – One Barr body.
XO females – No Barr body.

Question 12.
State Lyon’s hypothesis.
Answer:
Lyon’s hypothesis states that in mammals the necessary dosage compensation is accomplished by the inactivation of one of the X chromosomes in females so that both males and females have only one functional X chromosome per cell.
Mary Lyon suggested that Barr bodies represented an inactive chromosome, which in females becomes tightly coiled into a heterochromatin, a condensed and visible form of chromatin (Lyon’s hypothesis). The number of Barr bodies observed in cell was one less than the number of X-Chromosome. XO females have no Barr body, whereas XXY males have one Barr body.

Question 13.
Mention few X-linked inherited diseases.
Answer:
Red-green colour blindness or daltonism, haemophilia and Duchenne’s muscular dystrophy.

Question 14.
Define Karyotyping.
Answer:
Karyotyping is a technique through which a complete set of chromosomes is separated from a cell and the chromosomes are arranged in pairs. An idiogram refers to a diagrammatic representation of chromosomes.

Question 15.
Explain the inheritance pattern of Y-linked genes for example.
Answer:
Genes in the non-homologous region of the Y-chromosome are inherited directly from male to male. In humans, the Y-linked or holandric genes for hypertrichosis (excessive development of hairs on pinna of the ear) are transmitted directly from father to son, because males inherit the Y chromosome from the father. Female inherits only X chromosome from the father and are not affected.

Question 16.
Observe the symbol used in pedigree analysis and give the proper terms they represent.
Answer:

Question 17.
Write a brief note on pedigree analysis.
Answer:
Pedigree is a “family tree”, drawn with standard genetic symbols, showing the inheritance pathway for specific phenotypic characters. Pedigree analysis is the study of traits as they have appeared in a given family line for several past generations.

Question 18.
What do you mean by ‘Mendelian disorder’.
Answer:
Alteration or mutation in a single gene causes Mendelian disorders. These disorders are transmitted to the offsprings on the same line as the Mendelian pattern of inheritance. E.g., Thalassemia.

Question 19.
Name any four Mendelian disorders.
Answer:
(a) Thalassemia (b) Albinism (c) sickle cell anaemia (d) Huntington’s chorea

Question 20.
What is the phenotype of (a) I^AI^O (b) I^OI^O
Answer:
(a) I^AI^O – A blood group person
(b) I^OI^O – O blood group person

Question 21.
On which chromosomes does HBA1 gene and HBB genes are located?
Answer:
HBA1 gene is located on chromosome 16.
HBB gene is located on chromosome 11.

Question 22.
Complete the equation.

Answer:
(a) A = Phenylalanine hydroxylase
(b) B = Tyrosinase

Question 23.
Write a note on Huntington’s chorea.
Answer:
Huntington’s chorea is inherited as an autosomal dominant lethal gene in man. It is characterized by involuntary jerking of the body and progressive degeneration of the nervous system, accompanied by gradual mental and physical deterioration. The patients with this disease usually die between the age of 35 and 40.

Question 24.
Comment on Trisomy-21.
Answer:
Trisomic condition of chromosome – 21 results in Down’s syndrome. It is characterized by severe mental retardation, defective development of the central nervous system, increased separation between the eyes, flattened nose, ears are malformed, mouth is constantly open and the tongue protrudes.

Question 25.
Mention the genetic makeup of Turner’s syndrome person and Klinefelter’s syndrome , person.
Answer:
Klinefelter’s syndrome – 44AA+XXY
Turner’s syndrome – 44AA+XO

Question 26.
List out any four clinical symptoms of Klinefelter’s syndrome.
Answer:
Gynaecomastia, high pitched voice, under developed genetalia and tall with long limbs.

12th Bio Zoology Principles of Inheritance and Variation Three Marks Questions and Answers

Question 27.
Write the types of sex-determination mechanisms does the following crosses as shown. Give an example for each.
(a) Female XX with Male XO (6) Female ZW with Male ZZ
Answer:
(a) Male heterogamety. e.g., Human beings.
(b) Female heterogamety. e.g., Birds.

Question 28.
What are the enzymes encoded by the alleles I^A, I^B and I^O?
Answer:
IA allele produces N-acetyl galactose transferase and can add N-acetyl galactosamine (NAG) and I^B allele encodes for the enzyme galactose transferase that adds galactose to the precursor (i.e. H substances). In the case of I^O/I^O allele no terminal transferase enzyme is produced and therefore called “null” allele and hence cannot add NAG or galactose to the precursor.

Question 29.
Draw a tabular column representing various types of blood group in human beings, their phenotypes, genotypes, antigens and respective antibodies.
Answer:
Genetic basis of the human ABO blood groups:

Question 30.
Give an account on Rhesus factor.
Answer:
Rhesus or Rh – Factor: The Rh factor or Rh antigen is found on the surface of erythrocytes. It was discovered in 1940 by Karl Landsteiner and Alexander Wiener in the blood of rhesus monkey, Macaca rhesus and later in human beings. The term ‘Rh factor’ refers to “immunogenic D antigen of the Rh blood group system. An individual having D antigen are Rh D positive (Rh+) and those without D antigen are Rh D negative (Rh”)”. Rhesus factor in the blood is inherited as a dominant trait.

Naturally occurring Anti D antibodies are absent in the plasma of any normal individual. However if an Rh” (Rh negative) person is exposed to Rh+ (Rh positive) blood cells (erythrocytes) for the first time, anti D antibodies are formed in the blood of that individual. On the other hand, when an Rh positive person receives Rh-negative blood no effect is seen.

Question 31.
How Erythroblastosis foetalis can be prevented?
Answer:
If thefmother is Rh negative and foetus is Rh positive, anti D antibodies should be administered to the mother at 28th and 34th week of gestation as a prophylactic measure. If the Rh-negative mother delivers Rh positive child then anti D antibodies should be administered to the mother soon after delivery. This develops passive immunity and prevents the formation of anti D antibodies in the mothers blood by destroying the Rh foetal RBC before the mother’s immune system is sensitized. This has to be done whenever the woman attains pregnancy.

Question 32.
Explain XX-XO type of sex determination.
Answer:
XX-XO method of sex determination is seen in bugs, some insects such as cockroaches and grasshoppers. Pi The female with two X chromosomes are homogametic Gametes (XX) while the males with only one X chromosome are heterogametic (XO). The presence of unpaired X chromosomes determines the male sex. The males PI Generation with unpaired ‘X’ chromosome produce two types of sperms, one half with X chromosome and other half without X chromosome. The sex of the offspring depends upon the sperm that fertilizes the egg.

Question 33.
Name the type of sex-determination mechanism of the following organisms.
(a) Gypsy moth (A) Human beings (c) Butterflies
Answer:
(a) Gypsy moth -ZW – ZZ type (ZW-females, ZZ – males)
(b) Human beings – XX – XY type (XX-females, XY – males)
(c) Butterflies – ZO – ZZ type (ZO-females, ZZ – males)

Question 34.
Complete the following cross.

Answer:

Question 35.
Role of Y- chromosome is crucial for maleness – Justify.
Answer:
Current analysis of Y chromosomes has revealed numerous genes and regions with potential genetic function; some genes with or without homologous counterparts are seen on the X. Present at both ends of the Y chromosome are the pseudoautosomal regions (PARs) that are similar with regions on the X chromosome which synapse and recombine during meiosis.

The remaining 95% of the Y chromosome is referred as the Non-combining Region of the Y (NRY). The NRY is divided equally into functional genes (euchromatic) and non-functional genes (heterochromatic). Within the euchromatin regions, is a gene called Sex determining region Y (SRY). In humans, absence of Y chromosome inevitably leads to female development and this SRY gene is absent in X chromosome. The gene product of SRY is the testes determining factor (TDF) present in the adult male testis.

Question 36.
Color blindness is a perfect example for criss-cross of inheritance – Justify the statement.
Answer:
A marriage between a colour blind man and a normal visioned woman will produce normal visioned male and female individuals in F1 generation but the females are carriers. The marriage between a F1 normal visioned carrier woman and a normal visioned male will produce one normal visioned female, one carrier female, one normal visioned male and one colour blind male. Colour blind trait is inherited from the male parent to his grandson through carrier daughter, which is an example of criss-cross pattern of inheritance.

Question 37.
How the Karyotype of lymphocytes was prepared by Tjio and Levan?
Answer:
Preparation of Karyotype Tjio and Levan (1960) described a simple method of culturing lymphocytes from the human blood. Mitosis is induced followed by addition of colchicine to arrest cell division at metaphase stage and the suitable spread of metaphase chromosomes is photographed. The individual chromosomes are cut from the photograph and are arranged in an orderly fashion in homologous pairs. This arrangement is called a karyotype. Chromosome banding permits structural definitions and differentiation of chromosomes.

Question 38.
What is a genetic disorder? Mention its types?
Answer:
A genetic disorder is a disease or syndrome that is caused by an abnormality in an individual -DNA. Abnormalities can range from a small mutation in a single gene to the addition or subtraction of an entire chromosome or even a set of chromosomes. Genetic disorders are of two types namely, Mendelian disorders and chromosomal disorders.

Question 39.
Explain the genetic basis of Phenylketonuria.
Answer:
Phenylketonuria is an inborn error of Phenylalanine metabolism caused due to a pair of autosomal recessive genes. It is caused due to mutation in the gene PAH (phenylalanine hydroxylase gene) located on chromosome 12 for the hepatic enzyme “phenylalanine hydroxylase”.

This enzyme is essential for the conversion of phenylalanine to tyrosine. Affected individual lacks this enzyme, so phenylalanine accumulates and gets converted to phenylpyruvic acid and other derivatives. It is characterized by severe mental retardation, light pigmentation of skin and hair. Phenylpyruvic acid is excreted in the urine.

Question 40.
Give an account of Patau’s syndrome.
Answer:
Trisomic condition of chromosome 13 results in Patau’s syndrome. Meiotic non-disjunction is thought to be the cause for this chromosomal abnormality. It is characterized by multiple and severe body malformations as well as profound mental deficiency. Small head with small eyes, cleft palate, malformation of the brain and internal organs are some of the symptoms of this syndrome.

Question 41.
Define aneuploidy.
Answer:
Failure of chromatids to segregate during cell division resulting in the gain or loss of one or more chromosomes is called aneuploidy. It is caused by the non-disjunction of chromosomes.

12th Bio Zoology Principles of Inheritance and Variation Five Marks Questions and Answers

Question 42.
What do you mean by “syndrome”? Give two examples.
Answer:
Group of signs and symptoms that occur together and characterize a particular abnormality is called a syndrome, e.g., Down’s syndrome and Turner’s syndrome.

Question 42.
Explain in detail about Erythroblastosis foetalis.
Answer:
Rh incompatability has great significance in childbirth. If a woman is Rh-negative and the man is Rh positive, the foetus may be Rh positive having inherited the factor from its father. The Rh negative mother becomes sensitized by carrying Rh positive foetus within her body. Due to damage of blood vessels, during child birth, the mother’s immune system recognizes the Rh antigens and gets sensitized. The sensitized mother produces Rh antibodies. The antibodies are IgG type which are small and can cross placenta and enter the foetal circulation. By the time the mother gets sensitized and produce anti ‘D’ antibodies, the child is delivered.

Usually no effects are associated with exposure of the mother to Rh positive antigen during the first child birth, subsequent Rh positive children carried by the same mother, may be exposed to antibodies produced by the mother against Rh antigen, which are carried across the placenta into the foetal blood circulation. This causes haemolysis of foetal RBCs resulting in haemolytic jaundice and anaemia. This condition is known as Erythroblastosis foetalis or Haemolytic disease of the new bom (HDN).

Question 43.
Decribe female heterogamy and its types.
Answer:
Heterogametic Females:
In this method of sex determination, the homogametic male possesses two ‘X’ chromosomes as in certain insects and certain vertebrates like fishes, reptiles and birds producing a single type of gamete; while females produce dissimilar gametes. The female sex consists of a single ‘X’ chromosome or one ‘X’ and one ‘Y’ chromosome. Thus the females are heterogametic and produce two types of eggs. Heterogametic females are of two types, ZO-ZZ type and ZW-ZZ type. ,

ZO-ZZ Type
This method of sex determination is seen in certain moths, butterflies and domestic chickens. In this type, the female possesses single ‘Z’ chromosome in its body cells and is heterogametic (ZO) producing two kinds of eggs some with ‘Z’ chromosome and some without ‘Z’ chromosome, while the male possesses two ‘Z’ chromosomes and is homogametic (ZZ).

ZW-ZZ type
This method of sex determination occurs in certain insects (gypsy moth) and in vertebrates such as fishes, reptiles and birds. In this method the female has one ‘Z’ and one ‘ W’ chromosome (ZW) producing two types of eggs, some carrying the Z chromosomes and some carry the W chromosome. The male sex has two ‘Z’ chromosomes and is homogametic (ZZ) producing a single type of sperm.

Question 44.
Write elaborately about the following Mendelian disorders.
(a) Thalassemia (b) Albinism
Answer:
(a) Thalassemia
Thalassemia is an autosomal recessive disorder. It is caused by gene mutation resulting in excessive destruction of RBC’s due to the formation of abnormal haemoglobin molecules. Normally haemoglobin is composed of four polypeptide chains, two alpha and two beta globin chains. Thalassemia patients have defects in either the alpha or beta globin chain causing the production of abnormal haemoglobin molecules resulting in anaemia.

Thalassemia is classified into alpha and beta based on which chain of haemoglobin molecule _ is affected. It is controlled by two closely linked genes HBA1 and HBA2 on chromosome 16. Mutation or deletion of one or more of the four alpha gene alleles causes Alpha Thalassemia. In Beta Thalassemia, production of beta globin chain is affected. It is controlled by a single gene (HBB) on chromosome 11. It is the most common type of Thalassemia and is also known as Cooley’s anaemia. In this disorder, the alpha chain production is increased and damages the membranes of RBC.

(b) Albinism
Albinism is an inborn error of metabolism, caused due to an autosomal recessive gene. Melanin pigment is responsible for skin colour. Absence of melanin results in a condition called albinism. A person with the recessive allele lacks the tyrosinase enzyme system, which is required for the conversion of dihydroxyphenylalanine (DOPA) into melanin pigment inside the melanocytes. In an albino, melanocytes are present in normal numbers in their skin, hair, iris, etc., but lack melanin pigment.

Question 45.
Discuss any two Allosomal anomalies in human.
Answer:
Allosomal abnormalities in human beings
Mitotic or meiotic non-disjunction of sex chromosomes causes allosomal abnormalities. Several sex chromosomal abnormalities have been detected. E.g. Klinefelter’s syndrome and Turner’s syndrome.

1. Klinefelter’s Syndrome (XXY Males)
This genetic disorder is due to the presence of an additional copy of the X chromosome resulting in a karyotype of 47,XXY. Persons with this syndrome have 47 chromosomes (44AA+XXY). They are usually sterile males, tall, obese, with long limbs, high pitched voice, under developed genetalia and have feeble breast (gynaecomastia) development.

2. Turner’s Syndrome (XO Females)
This genetic disorder is due to the loss of a X chromosome resulting in a karyotype of 45,X. Persons with this syndrome have 45 chromosomes (44 autosomes and one X chromosome) (44AA+XO) and are sterile females. Low stature, webbed neck, underdeveloped breast, rudimentary gonads lack of menstrual cycle during puberty, are the main symptoms of this syndrome.

Higher Order Thinking Skills (HOTs) Questions

Question 1.
On analysis, a person’s karyotype reveals an extra one chromosome of the twenty-first pair. What does this condition represent? which type of symptoms can be noticed in the person?
Answer:
Trisomy-21 or Down’s syndrome.
Symptoms – Mental retardation, malformed ears, protruded tongue, mouth is constantly open etc.

Question 2.
A female whose blood group is AB^– got conceived and later it is diagnoised that her – foetus possess B+. What measures would be taken to prevent the foetus from Haemolytic disease of Newborn (HDN)
Answer:
If the mother is Rh-negative and foetus is Rh-positive, anti D antibodies should be administered to the mother at 28th and 34th week of gestation as a prophylactic measure. If the Rh-negative mother delivers a Rh-positive child then anti D antibodies should be administered to the mother soon after delivery. This develops passive immunity and prevents the formation of anti D antibodies in the mother’s blood by destroying the Rh foetal RBC before the mother’s immune system is sensitized. This has to be done whenever the woman attains pregnancy.

Question 3.
The following table shows the genotypes for ABO blood grouping and these phenotypes. Complete the table by filling the gaps.

Answer:
2) I^AI^O
3) I^AI^B
4) O

Question 4.
Give one example for each of the following group of drugs, (a) Stimulants (b) Analgesic (c) Hallucinogens
Answer:
(a) Stimulants – Eg.: Nicotine
(b) Analgesic – Eg.: Opium
(c) Hallucinogens – Phencyclidine

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 7 Applications of Differential Calculus Ex 7.8 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.8

Question 1.
Find two positive numbers whose sum is 12 and their product is maximum.
Solution:
Let the two numbers be x, 12 – x.
Their product p = x (12 – x) = 12x – x²
To find the maximum product.
p'(x) = 12 – 2x
p”(x) = -2
p'(x) = 0 ⇒ 12 – 2x = 0 ⇒ 2x = 12
⇒ x = 6
at x = 6, p”(x) = -2 = -ve
⇒ p is maximum at x = 6
when x = 6, 12 – x = 12 – 6 = 6
So the two numbers are 6, 6

Question 2.
Find two positive numbers whose product is 20 and their sum is minimum.
Solution: Let the two positive numbers be ‘x’ and ‘y’
Given product is 20 ⇒ xy = 20 ⇒ y = \(\frac { 20 }{ x }\)
Sum S = x + y
S = x + \(\frac { 20 }{ x }\)
\(\frac { dS }{ dx }\) = 1 – \(\frac { 20 }{ x^2 }\)
For maximum or minimum, \(\frac { dS }{ dx }\) = 0
x² – 20 = 0 x² = 20
x = ±2√5
[x= -2√5 is not possible
\(\frac { d^2S }{ dx^2 }\) = \(\frac { 40 }{ x^3 }\)
at x = 2√5, \(\frac { d^2S }{ dx^2 }\) > 0
∴ Sum ‘S’ is minimum when x = 2√5
y = \(\frac { 20 }{ 2√5 }\) = 2√5
Minimum sum = 2√5 + 2√5 = 4√5

Question 3.
Find the smallest possible value of x² + y² given that x + y = 10.
Solution:
Given x + y = 10 ⇒ y = 10 – x
Let A = x² + y²
A = x² + (10 – x)²
\(\frac { dA }{ dx }\) = 2x + 2(10 – x)(-1)
For maximum or minimum,
\(\frac { dA }{ dx }\) = 0 ⇒ 2(2x – 10) = 0
x = 5
\(\frac { d^2A }{ dx^2 }\) = 4
at x = 5, \(\frac { d^2A }{ dx^2 }\) > 0
∴ A is minimum when x = 5
y = 10 – 5 = 5
∴ The smallest possible value of x² + y² is
(5)² + (5)² = 25 + 25 = 50

Question 4.
A garden is to be laid out in a rectangular area and protected by a wire fence. What is the largest possible area of the fenced garden with 40 meters of wire?
Solution:

Perimeter = 40 m
2(l + b) = 40 ⇒ l + b = 20
Let l = x m
b = (20 – x)m
Area = l × b = x(20 – x) = 20x – x²
To find the maximum area
A(x) = 20x – x²
A'(x) = 20 – 2x
A”(x) = -2
A'(x) = 0 ⇒ 20 – 2x = 0
⇒ x = 10
x = 10 is a maximum point
:. Maximum Area = x (20 – x)
= 10(20 – 10)
= 10 × 10 = 100 sq.m.

Question 5.
A rectangular page is to contain 24 cm² of print. The margins at the top and bottom of the page are 1.5 cm and the margins at the other sides of the page are 1 cm. What should be the dimensions’ of the page so that the area of the paper used is minimum?
Solution:
Let the width of the printed part be ‘x’ cm
Let the height Of the printed part be ‘y’ cm
Given, Area of the printed part = 24 cm²
i.e., xy = 24
y = \(\frac { 24 }{ x }\)

From the given data,
Width of the page
= x + 2(1) = x + 2 cm
Height of the page
= y + 2(1.5) = y + 3 cm
∴ Area of the paper
‘A’ = (x + 2) (y + 3)
= (x + 2) (\(\frac { 24 }{ x }\) + 3)
A = 24 + 3x + \(\frac { 48 }{ x }\) + 6
\(\frac { dA }{ dx }\) = 3 – \(\frac { 48 }{ x^2 }\)
For maximum or minimum,
\(\frac { dA }{ dx }\) = 0
3x² – 48 = 0
x² = 16
x = ±4 [∵ x cannot be negative
∴ x = 4
Now, \(\frac { d^2A }{ dx^2 }\) = \(\frac { 96 }{ x^3 }\)
at x = 4, \(\frac { d^2A }{ dx^2 }\) > 0
∴ Area is minimum when x = 4
y = \(\frac { 24 }{ 4 }\) = 6
∴ Dimensions of the page:
Width of the page = x + 2 = 4 + 2 = 6 cm
Height of the page = y + 3 = 6 + 3 = 9 cm

Question 6.
A farmer plans to fence a rectangular pasture adjacent to a river. The pasture must contain 1,80,000 sq. mtrs in order to provide enough grass for herds. No fencing is needed along the river. What is the length of the minimum needed fencing material?
Solution:

Let the length of the pasture be ‘x’ m
Let the breadth of the pasture be ‘y’ m
Given Area = 1,80,000
xy = 1,80,000
y = 1,80,000
For fencing, we need 2y + x
(one side is River)
Let P = 2y + x
P = 2(\(\frac { 180000 }{ 2 }\)) + x = \(\frac { 360000 }{ x }\) + x
\(\frac { dP }{ dx }\) = –\(\frac { 360000 }{ x^2 }\) + 1
For maximum or minimum,
\(\frac { dP }{ dx }\) = 0
⇒ – 360000 + x² = 0
x² = 360000
x = ±600
[x = -600 is not possible]
∴ x = 600
Now, \(\frac { d^2P }{ dx^2 }\) = \(\frac { 720000 }{ x^3 }\)
at x = 600, \(\frac { d^2P }{ dx^2 }\) > 0
∴ P is minimum when x = 600
y = \(\frac { 180000 }{ 600 }\) = 300
∴ Length of the minimum needed fencing material = 2y + x = 2(300) + 600 = 1200 m

Question 7.
Find the dimensions of the rectangle with the maximum area that can be inscribed in a circle of a radius of 10 cm.
Solution:
Let the length of the rectangle be ‘x’ cm
The breadth of the rectangle be ‘y’ cm

From the figure,
x² + y² = (20)² [Pythagoras Theorem
y² = 400 – x²
[∵ radius of the circle is 10 cm
y = \(\sqrt { 400-x^2 }\)
Now, Axea of the rectangle A = xy

For maximum or minimum,
\(\frac { dA }{ dx }\) = 0 ⇒ \(\frac { -2x^2+400 }{ \sqrt{400-x^2} }\)
x² = 200
x = ±10√2
x = -10√2 is not possible
∴ x = 10√2

Area of the rectangle is maximum When x = 10√2
y = \(\sqrt { 400-200 }\) = \(\sqrt { 200 }\) = 10√2
∴ x = y = 10√2
Length of the rectangle = 10√2 cm
Breadth of the rectangle = 10√2 cm
(Note: A largest rectangle that can be inscribed in a circle is a square)

Question 8.
Prove that among all the rectangles of the given perimeter, the square has the maximum area.
Solution:
Let x, y be the length and breadth of a rectangle and given perimeter is P (say)
ie. 2(x + y) = P
y = \(\frac { P }{ 2 }\) – x
Area of a rectangle ‘A’ = xy
A = x(\(\frac { P }{ 2 }\) – x) = \(\frac { P }{ 2 }\) x – x²
\(\frac { dA }{ dx }\) = \(\frac { P }{ 2 }\) – 2x
For maximum or minimum,
\(\frac { dA }{ dx }\) = 0 ⇒ \(\frac { P }{ 2 }\) – 2x = 0
x = \(\frac { P }{ 4 }\)
Now, \(\frac { d^2A }{ dx^2 }\) = -2
at x = \(\frac { P }{ 4 }\), \(\frac { d^2A }{ dx^2 }\) < 0
∴ Area of the rectangle is maximum when x = \(\frac { P }{ 4 }\)
Now, y = \(\frac { P }{ 2 }\) – x = \(\frac { P }{ 2 }\) – \(\frac { P }{ 4 }\) = \(\frac { P }{ 4 }\)
∴ Length of a rectangle = \(\frac { P }{ 4 }\)
Breadth of a rectangle = \(\frac { P }{ 4 }\)
Since Length = Breadth, the rectangle is a square.
Hence Proved.

Question 9.
Find the dimensions of the largest rectangle that can be inscribed in a semi-circle of radius r cm.
Solution:

Given radius of the semi-circle = ‘r’ cm
Let the length of the rectangle be ‘x’ cm
Let the breadth of the rectangle be ‘y’ cm
From the figure,

For maximum or minimum,
\(\frac { dA }{ dx }\) = 0
⇒ \(\frac { 1 }{ 2 }\) [ \(\frac { 4r^2-2x^2 }{ \sqrt{4x^2-x^2} }\) ] = 0
4r² – 2x² = 0
x² = 2r²
x = ± √2 r
x = -√2 r is not possible
∴ x = √2 r

∴ Length of the rectangle is √2 r cm
Breadth of the rectangle is \(\frac { r }{ √2 }\) cm

Question 10.
A manufacturer wants to design an open box having a square base and a surface area of 108 sq. cm. Determine the dimensions of the box for the maximum volume.
Solution:
Let ‘x’ be the length Of the box.
‘y’ be the height of the box.
Given, surface area = 108 sq.cm
i,e. 4(xy) + x² = 108
⇒ y = \(\frac { 108-x^2 }{ 4x }\)
Volume of the box V = x²y

maximum or minimum, \(\frac { dV }{ dx }\) = 0
⇒ 108 – 3x² = 0
x² = 36
x = ± 6 [x = – 6 is not possible
∴ x = 6
Now, \(\frac { d^2V }{ dx^2 }\) = –\(\frac { 6x }{ 4 }\) = –\(\frac { 3x }{ 2 }\)
at x = 6, \(\frac { d^2V }{ dx^2 }\) < 0
Volume of theboxis maximum when x = 6
y = \(\frac { 108-36 }{ 24 }\) = \(\frac { 72 }{ 24 }\) = 3
∴ Length of the box = 6 cm
Breadth of the box = 6 cm
Height of the box = 3 cm

Question 11.
The volume of a cylinder is given by the formula V = πr²h. Find the greatest and least values of V if r + h = 6.
Solution:
Given r + h = 6
⇒ r = 6 – h
Volume V = πr²h
V = π(6 – h)²h
\(\frac { dV }{ dh }\) = π [(6 – h)² (1) + 2h(6 – h) (-1)] = π(6 – h)[6 – 3h]
For maximum or minimum,
\(\frac { dV }{ dh }\) = 0
⇒ π (6 – h) (6 – 3h) = 0
⇒ h = 6, h = 2
h = 6 is not possible as r + h = 6
∴ h = 2
\(\frac { d^2V }{ dh^2 }\) = π [(6 – h)(-3) + (6 – 3h)(-1)] = π [6h – 24]
at h = 2, \(\frac { d^2V }{ dh^2 }\) < 0
∴ Volume of the cylinder is maximum when h = 2 and r = 6 – 2 = 4
greatest value of V = π(4)² (2) = 32 π
least value of V = 0

Question 12.
A hollow cone with a base radius of a cm and’ height of b cm is placed on a table. Show that) the volume of the largest cylinder that can be hidden underneath is \(\frac { 4 }{ 9 }\) times the volume of the cone.
Solution:

Cone
heigthof the cone = b cm
base radius = a cm
Cylinder
Let the base radius be ‘r’ cm
height be ‘h’ cm
From the figure, \(\frac { h }{ a-r }\) = \(\frac { b }{ a }\)
(using similar triangles property
⇒ h = \(\frac { b }{ a }\) (a – r)
= b – \(\frac { b }{ a }\) r
Volume of cylinder V = πr²h

For maximum or minimum,
\(\frac { dV }{ dr }\) = 0
⇒ br(2a – 3r) = 0
r = 0 and r = \(\frac { 2a }{ 3 }\)
r = 0 is not possible
∴ r = \(\frac { 2a }{ 3 }\)

Hence proved.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Bio Zoology Guide Pdf Chapter 3 Reproductive Health Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Bio Zoology Solutions Chapter 3 Reproductive Health

12th Bio Zoology Guide Reproductive Health Text Book Back Questions and Answers

Question 1.
Which of the following is correct regarding HIV, hepatitis B, gonorrhoea and trichomoniasis?
(a) Gonorrhoea is a STD whereas others are not.
(b) Trichomoniasis is a viral disease whereas others are bacterial.
(c) HIV is a pathogen whereas others are diseases.
(d) Hepatitis B is eradicated completely whereas others are not.
Answer:
(c) HIV is a pathogen whereas others are diseases.

Question 2.
Which one of the following groups includes sexually transmitted diseases caused by bacteria only?
(a) Syphilis, gonorrhoea and candidiasis
(b) Syphilis, chlamydiasis and gonorrhoea
(c) Syphilis, gonorrhoea and trichomoniasis
(d) Syphilis, trichomoniasis and pediculosis
Answer:
(b) Syphilis, chlamydiasis and gonorrhoea

Question 3.
Identify the correct statements from the following:
(a) Chlamydiasis is a viral disease.
(b) Gonorrhoea is caused by a spirochaete bacterium, Treponema palladium.
(c) The incubation period for syphilis is 2 to 14 days in males and 7 to 21 days in females.
(d) Both syphilis and gonorrhoea are easily cured with antibiotics.
Answer:
(d) Both syphilis and gonorrhoea are easily cured with antibiotics

Question 4.
A contraceptive pill prevents ovulation by …………….
(a) blocking fallopian tube
(b) inhibiting release of FSH and LH
(c) stimulating release of FSH and LH
(d) causing immediate degeneration of released ovum.
Answer:
(b) inhibiting release of FSH and LH

Question 5.
The approach which does not give the defined action of contraceptive is

Answer:
(b) Vasectomy – Prevents spermatogenesis

Question 6.
Read the given statements and select the correct option.
Statement 1: Diaphragms, cervical caps, and vaults are made of rubber and are inserted into the female reproductive tract to cover the cervix before coitus.
Statement 2: They are chemical barriers of conception and are reusable.
(a) Both statements 1 and 2 are correct and statement 2 is the correct explanation of statement 1.
(b) Both statements 1 and 2 are correct but statement 2 is not the correct explanation of statement 1.
(c) Statement 1 is correct but statement 2 is incorrect.
(d) Both statements 1 and 2 are incorrect.
Answer:
(c) Statement 1 is correct but statement 2 is incorrect.

Question 7.
Match column I with column II and select the correct option from the codes given below.

(a) A – (iv), B – (ii), C – (i), D-(iii)
(b) A – (iv), B – (i), C – (iii), D – (ii)
(c) A – (i), B – (iv), C – (ii), D – (iii)
(d) A – (iv), B – (i), C – (ii), D – (iii)
Answer:
(d) A-(iv), B-(i), C-(ii), D-(iii)

Question 8.
Select the incorrect action of hormonal contraceptive pills from the following
(a) Inhibition of spermatogenesis.
(b) Inhibition of ovulation.
(c) Changes in cervical mucus impairing its ability to allow passage and transport of sperms.
(d) Alteration in uterine endometrium to make it unsuitable for implantation.
Answer:
(a) Inhibition of spermatogenesis.

Question 9.
What is amniocentesis? Why a statutory ban is imposed on this technique?
Answer:
Amniocentesis is a prenatal technique used to detect any chromosomal abnormalities in the foetus and it is being often misused to determine the sex of the foetus. Once the sex of the foetus is known, there may be a chance of female foeticide. Hence, a statutory ban on amniocentesis is imposed.

Question 10.
Select the correct term from the bracket and complete the given branching tree
Answer:

Question 11.
Correct the following statements

1. Transfer of an ovum collected from donor into the fallopian tube is called ZIFT.
2. Transfering of an embryo with more than 8 blastomeres into uterus is called GIFT,
3. Multiload 375 is a hormone-releasing IUD.

Answer:

1. Transfer of 8 celled blastomeres collected from donor into the fallopian tube is called ZIFT.
2. Transfering of an embryo with more than 8 blastomeres into uterus is called IUT.
3. Multiload 375 is a copper-releasing IUD.

Question 12.
Which method do you suggest the couple to have a baby, if the male partner fails to inseminate the female or due to very low sperm count in the ejaculate?
Answer:
Micro-testicular sperm extraction (TESE):
Microsurgical sperm retrieval from the testicle involves a small midline incision in the scrotum, through which one or both testicles can be seen. Under the microscope, the seminiferous tubules are dilated and small amount of testicular tissue in areas of active sperm production are removed and improved for sperm yield compared to traditional biopsy techniques.

Question 13.
Expand the following:

1. ZIFT
2. ICSI

Answer:

1. Zygote intra-fallopian transfer
2. Intracytoplasmic sperm injection

Question 14.
What are the strategies to be implemented in India to attain total reproductive health?
Answer:

• Creating awareness and providing medical assistance to build a healthy society.
• Introducing sex education in schools to provide information about adolescence and adolescence related changes.
• Educating couples and those in the marriageable age groups about the available birth control methods and family planning norms.
• Creating awareness about care for pregnant women, post-natal care of mother and child, and the importance of breastfeeding.
• Encouraging and supporting governmental and non-governmental agencies to identify new methods and/or to improve upon the existing methods of birth control.

Question 15.
Differentiate foeticide and infanticide.
Answer:
Female foeticide refers to ‘aborting the female in the mother’s womb’.
Female infanticide is ‘killing the female child after her birth’.

Question 16.
Describe the major STDs and their symptoms.
Answer:

Cervical cancer:
Cervical cancer is caused by a sexually transmitted virus called Human Papillomavirus (HPV). HPV may cause abnormal growth of cervical cells or cervical dysplasia. The most common symptoms and signs of cervical cancer are pelvic pain, increased vaginal discharge and abnormal vaginal bleeding. The risk factors for cervical cancer include

1. Having multiple sexual partners
2. Prolonged use of contraceptive pills

Gervical cancer can be diagnosed by a Papanicolaou smear (PAP smear) combined with an HPV test. X-Ray, CT scan, MRI and a PET scan may also be used to determine the stage of cancer. The treatment options for cervical cancer include radiation therapy, surgery and chemotherapy.

Modem screening techniques can detect precancerous changes in the cervix. Therefore screening is recommended for women above 30 years once in a year. Cervical cancer can be prevented with vaccination. Primary prevention begins with HPV vaccination of girls aged 9-13 years before they become sexually active. Modification in lifestyle can also help in preventing cervical cancer. A healthy diet, avoiding tobacco usage, preventing early marriages, practicing monogamy and regular exercise minimize the risk of cervical cancer.

Question 17.
How are STDs transmitted?
Answer:
Sexually transmitted diseases (STD) are called as Sexually transmitted infections (STI). Normally STI are transmitted from person to person during intimate sexual contact with an infected partner. Infections like Hepatitis-B and HIV are transmitted sexually as well as by sharing of infusion needles, surgical instruments, etc with infected people, blood transfusion or from infected mother to baby.

Question 18.
Write the preventive measures of STDs.
Answer:
Prevention of STDs:

• Avoid sex with unknown partner/ multiple partners
• use condoms
• In case of doubt, consult a doctor for diagnosis and get complete treatment.

Question 19.
The procedure of GIFT involves the transfer of female gametes into the fallopain tube, can gametes be transferred to the uterus to achieve the same result? Explain.
Answer:
Gamete intra-fallopian transfer (GIFT): Transfer of an ovum collected from a donor into the fallopian tube. In this the eggs are collected from the ovaries and placed with the sperms in one of the fallopian tubes. The zygote travels toward the uterus and gets implanted in the inner lining of the uterus.

Question 20.
Amniocentesis, the foetal sex determination test, is banned in our country. Is it necessary? Comment.
Answer:
Amniocentesis is a prenatal technique used to detect any chromosomal abnormalities in the foetus and it is being often misused to determine the sex of the foetus. Once the sex of the foetus is known, there may be a chance of female foeticide. Hence, a statutory ban on amniocentesis is imposed.

Question 21.
Open Book Assessment ‘Healthy reproduction, legally checked birth control measures and proper family planning programmes are essential for the survival of mankind’ Justify.
Answer:
Reproductive health and proper family planning programmes are highly essential for the survival of mankind. Reproductive health refers to the state of physical, psychological and social well-being merely in absence of illness in all matters related to the reproductive system and its proper functioning. Family planning has the following health and social benefits.

1. Protecting the health of women by reducing frequent pregnancies.
2. Reducing abortions.
3. Providing stable population growth.
4. Assuming the infants well being by providing adequate time lapse between successive pregnancies.

12th Bio Zoology Guide Reproductive Health Additional Important Questions and Answers

12th Bio Zoology Guide Reproductive Health One Mark Questions and Answers

Question 1.
Which of the following is Not a natural contraceptive?
(a) Rhythm method
(b) Lactational amenorrhoea
(c) Progestasert
(d) Continuous abstinence
Answer:
(c) Progestasert

Question 2.
Identify the fungal STD(s) ………………
(i) Trichomoniasis
(ii) Genital herpes
(iii) Candidiasis
(a) Only (z)
(b) Only (z’zz)
(c) Only (zv)
Answer:
(A) Only (iii)

Question 3.
Match the following.

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

Question 4.
Pick out the incorrect statement regarding the character of a good contraceptive.
(a) It should be user friendly
(b) should not affect sexual drive
(c) side effects must be least
(d) should not be easily available
Answer:
(d) should not be easily available

Question 5.
Select the proper hormonal composition of oral contraceptive pills
(a) FSH & Prolactin
(b) Prolactin & TSH
(c) TSH & FSH
(d) FSH & LH
Answer:’
(d) FSH & LH

Question 6.
Assertion (A): IUD’s are inserted into the ovary.
Reason (R): IUD’s Increases phagocytosis of the sperm.
(a) Both A and R are correct
(b) Both A and R are incorrect
(c) A is correct R is incorrect
(d) A is incorrect R is correct
Answer:
(d) A is incorrect R is correct

Question 7.
Identify the mismatched pair.

Answer:
(c) Candidiasis (iii) Albugo Candida

Question 8.
In India, a family planning programme was initiated in …………………
(a) 1961
(b) 1981
(c) 1951
(d) 1971
Answer:
(c) 1951

Question 9.
Assertion (A): Amniocentesis helps to diagnose the chromosomal aberrations in foetus.
Reason (R): Amniocentesis is legalized is our country.
(a) Both A and R are wrong
(b) A is right and R is wrong
(c) R explains A
(d) A is the wrong R is right
Answer:
(b) A is right and R is wrong

Question 10.
The legalized marriageable age of female in India is …………….
(a) 19 years
(b) 20 years
(c) 18 years
(d) 21 years
Answer:
(c) 18 years

Question 11.
Identify the correct statement.
(a) Lactational amenorrhea is a permanent birth control method
(b) Condoms are made of polyethylene glycol and lambskin
(c) LNG -20 is a copper – releasing IUD
(d) Diaphragm covers the cervix thereby preventing sperm entry
Answer:
(d) Diaphragm covers the cervix thereby preventing sperm entry

Question 12.
According to WHO, India is the…………largest HIV affected country.
(a) first
(b) second
(c) third
(d) seventh
Answer:
(c) third

Question 13.
Identify the correct statement.
(a) MTP is the voluntary killing of an infant.
(b) MTP is legalized in India in 1974.
(c) Performing MTP during the second trimester is riskier.
(d) It is a surgical – based abortion.
Answer:
(c) Performing MTP during the second trimester is riskier.

Question 14.
Saheli contains a non-steroidal preparation called ………………
Answer:
Centchroman

Question 15.
The average foetal heartbeat rate is between ……………… beats per minute.
Answer:
120-160

Question 16.
World AIDS Day is observed on ………………
Answer:
11th July

Question 17.
Indian Government legalized MTP in ………………
Answer:
1971

Question 18.
In the chorionic villus sampling test, the tissue sample is taken from ………………
(a) amniotic fluid
(b) placental tissue
(c) Intestinal villi
(d) foetal liver
Answer:
(b) placental tissue

Question 19.
In ZIFT technique, the zygote is transferred at the stage of ………………
(a) 16blastomere
(b) morula
(c) 12 blastomere
(d) 8 blastomere
Answer:
(d) 8 blastomere

Question 20.
Given below are the basic steps in IVF treatment cycle. Select the proper sequence.
(i) Ovarian stimulation
(ii)Egg retrieval
(iii) fertilization
(iv) Embryo culture
(v) Embryo transfer
(a) (ii) → (iv) → (v) → (i) → (iii)
(b) (i) → (iii)) → (ii) → (v) → (iv)
(c) (i) → (ii) → (iii) → (iv) → (v)
(d) (ii) → (1) → (iii) → (v) → (iv)
Answer:
(c) (i) → (ii) → (iii) → (iv) → (v)

Question 21.
Enactment of ……………………..banned the identification of sex and to prevent prenatal abortion.
(a) POGSOAct
(b) POTAAct
(c) PCPNDTAct
(d) GOONDAAct
Answer:
(c) PCPNDT Act

Question 22.
Which is NOT a national health care programme?
(a) Pradhan Mantri Surakshit Matritva Abhiyan
(b) Pradhan Mantri Fiscal BhimaYojana
(c) RMNCH + A approach
(d) Janani Shishu Suraksha Karyakaram
Answer:
(b) Pradhan Mantri Fiscal Bhima Yojana

Question 23.
…………… is Known as an anti-sterility vitamin.
Answer:
Vitamin-E

12th Bio Zoology Guide Reproductive Health Two Marks Questions and Answers

Question 1.
Name any four national-level health care programmes run by the Indian Government.
Answer:

1. Janani Suraksha Yojana
2. Shishu Suraksha Karyakaram
3. RMNCH+A Programme
4. Pradhan Mantri Surakshit Matritva Abhiyan

Question 2.
Comment on ‘Saheli’
Answer:
Saheli is an oral contraceptive pill provided by Central Drug Research Institute in Lucknow, India. It contains a non-steroidal hormone preparation called centchroman.

Question 3.
What is Mayer – Rokitansky Syndrome?
Answer:
All women are bom with ovaries, but some do not have functional uterus. This condition is called Mayer-Rokitansky syndrome.

Question 4.
Point out any four STD’s caused by viruses.
Answer:

1. Genital herpes
2. Hepatitis-B
3. Genital warts
4. AIDS

Question 5.
Define Surrogacy.
Answer:
Surrogacy is a method of assisted reproduction or agreement whereby a woman agrees to carry a pregnancy for another person, who will become the newborn child’s parent after birth. Through In Vitro Fertilization (IVF), embryos are created in a lab and are transferred into the surrogate mother’s uterus.

Question 6.
Name the causative organism of (a) Syphilis (b) Genital warts
Answer:
(a) Syphilis is caused by Treponema palladium
(b) Genital warts caused by Human Papilloma vims.

Question 7.
Define Infertility.
Answer:
Inability to conceive or produce children even after unprotected sexual cohabitation is called infertility. That is, the inability of a man to produce sufficient numbers or quality of sperm to impregnate a woman or inability of a woman to become pregnant or maintain a pregnancy.

Question 8.
Expand the acronyms:

1. GIFT
2. ICSI

Answer:

1. GIFT – Gamete Intra – Fallopian Transfer.
2. ICSI – Intra Cytoplasmic Sperm Injection.

Question 9.
What is amniocentesis?
Answer:
Amniocentesis involves taking a small sample of the amniotic fluid that surrounds the foetus to diagnose chromosomal abnormalities.

Question 10.
How copper IUD’s provide contraception?
Answer:
Copper IUD’s release free copper and copper salts into the uterus and suppress the sperm motility. They remain in uterus for 5-10 years. Eg: NovaT, Cu T-380.

Question 11.
What do you mean by the term – coitus interruptus?
Answer:
Coitus interruptus is a oldest family planning method, where the male partner withdraws his penis before ejaculation, there by preventing the deposition of semen into vagina.

Question 12.
What is MTP?
Answer:
Medical Termination of Pregnancy (MTP) : Medical method of abortion is a voluntary or intentional termination of pregnancy in a non-surgical or non-invasive way. Early medical termination is extremely safe upto 12 weeks (the first trimester) of pregnancy and generally has no impact on a women’s fertility. Abortion during the second trimester is more risky as the foetus becomes intimately associated with the maternal tissue.

Question 13.
Which type of women are benefited by In Vitro Fertilization?
Answer:
IVF is used to treat women with blocked, damaged or absent fallopian tubes.

12th Bio Zoology Guide Reproductive Health Three Marks Questions and Answers

Question 1.
What are the steps taken by Government to overcome population explosion?
Answer:
To overcome the problem of population explosion, birth control is the only available solution.
People should be motivated to have smaller families by using various contraceptive devices. Advertisements by the Government in the media as well as posters/bills, etc., with a slogan Naam iruvar namakku iruvar (we two, ours two) and Naam iruvar namakku oruvar (we two, ours one) have also motivated to control population growth in Tamil Nadu. Statutory rising of marriageable age of the female to 18 years and that of males to 21 years and incentives given to pouples with small families are the other measures taken to control population growth in our country.

Question 2.
Lactational amenorrhoea – Comment.
Answer:
Menstrual cycles resume as early as 6 to 8 weeks from parturition. However, the reappearance of normal ovarian cycles may be delayed for six months during breastfeeding. This delay in ovarian cycles is called lactational amenorrhoea. It serves as a natural, but an unreliable form of birth control.

Question 3.
Write a note on sterilization procedure in the male.
Answer:
Vasectomy is the surgical procedure for male sterilisation. In this procedure, both vas deferens are cut and tied through a small incision on the scrotum to prevent the entry of sperm into the urethra. Vasectomy prevents sperm from heading off to penis as the discharge has no sperms in it.

Question 4.
Define Tubectomy.
Answer:
Tubectomy is the surgical sterilisation in women. In this procedure, a small portion of both fallopian tubes are cut and tied up through a small incision in the abdomen or through vagina. This prevents fertilization as well as the entry of the egg into the uterus.

Question 5.
Complete the table by filling the gaps.