Question 1

The table below gives the populations (in thousands) of ten species (A-J) in four areas consisting of the number of habitats given within brackets against each. Study the table and answer the questions which follow –Which area shows maximum species diversity?

Accepted Answer

Species diversity is a measurement of an ecosystem's species richness and species evenness. Species richness is the number of species per unit area. As the area increases, the number of species also increases. Here species diversity is maximum in 's'. From A to J all have different species richness and maximum in number.

Question 2

Which of the following figures shows correct relative contribution of greenhouse gases to global warming?

Accepted Answer

The relative contribution of green house gases to global warming shows 60% carbon dioxide, 20% methane, 14% CFCs, and 6% Nitrous oxide.

Question 3

A fertilized egg is blue print of future development. Explain?

Accepted Answer

A fertilized egg divides mitotically to form a number of cells. These cells show morphogenetic movements to arrange themselves in three germinal layers. This arrangement initiates the phases of differentiation and specialization. Different groups of cells of a particular germinal layer change their shape and get differentiated into structures or organs of the individual. All the cells with different developmental potentialities form the different organs (brain, eye, heart, skin, etc.).

Question 4

Give reasons why:  (i) Most zygotes in angiosperms divide only after certain amount of endosperm is formed.  (ii) Groundnut seeds are exalbuminous and castor seeds are albuminous.  (iii) Micropyle remains as a small pore in the seed coat of a seed.  (iv) Integuments of an ovule harden and the water content is highly reduced, as the seed matures.  (v) Apple and cashew are not called true fruits.

Accepted Answer

(i) Formation of endospemis ensures supply of nutrients to the embryo.(ii) In groundnut food is stored in cotyledon and seeds have no residual endosperm as it is completely used up during embryo development. The castor seeds retain part of endosperm as it is not completely used up during embryo development.(iii) This facilitates entry of oxygen and water during seed germination.(iv) The seed coat hardens and moisture content of seeds is reduced considerably to enable the seed to remain viable for a long time during storage.(vi) Because in these fruits, the thalamus also contributes to fruit formation.

Question 5

The gene I that controls the ABO blood grouping in human beings has three alleles IA, IB and i(a) How many different genotypes are likely to be present in human population?(b) Also, how many phenotypes are possibly present?

Accepted Answer

(a) Six genotypes are likely to be present in human population: IAIA, IAI, IBIB, IBI, IAIB, ii(b) Four phenotypes are possible: A, B, AB and O.

Question 6

A pedigree chart given here, presents a particular generation which shows a trait irrespective of sexes (i.e., present in both male and female). Neither of the parents of the particular generation shows that trait. Draw your conclusion on the basis of the pedigree.

Accepted Answer

The trait is autosome linked and recessive in nature. Both the parents are carrier (i.e., heterozygous) hence among offspring few show the trait irrespective of sex. The other offspring are either normal or carrier.

Question 7

What would happen if histones were to be mutated and made rich in amino acids, aspartic acid, and glutamic acid in place of basic amino acids such as lysine and arginine?

Accepted Answer

If histone proteins were rich in acidic amino acids instead of basic amino acids then they may not have any role in DNA packaging in eukaryotes as DNA is also negatively charged molecule. The packaging of DNA around the nucleosome would not happen. Consequently, the chromatin fibre would not be formed.

Question 8

Recall the experiment done by Frederick Griffith. If RNA, instead of DNA was the genetic material, would the heat-killed strain of strepts have transformed the R-strain into a virulent strain? Explain your answer.

Accepted Answer

RNA is more labile and prone to degradation (owing to the presence of 2' OH group in its ribose). Hence heat-killed S-strain may not have retained its ability to transform the R-strain into virulent form if RNA was its genetic material.

Question 9

You are repeating the Hershey-Chase experiment and are provided with two isotopes; 32P and 15N (in place of 35S in the original experiment). How do you expect your results to be different?

Accepted Answer

The use of 15N will be inappropriate because the method of detection of 35S and 15N is different (32P being a radioactive isotope while 15N is not radioactive but is the heavier isotope of Nitrogen). Even if 15N was radioactive then its presence would have been detected both inside the cell (15N incorporated as a nitrogenous base in DNA) as well as in the supernatant because 15N would also get incorporated in the amino group of amino acids in proteins. Hence the use of 15N would not give any conclusive results.

Question 10

There is only one possible sequence of amino acids when deduced from given nucleotides. But multiple nucleotide sequences can be deduced from a single amino acid sequence. Explain this phenomenon.

Accepted Answer

Some amino acids are coded by more than one codon (known as the degeneracy of codons), hence on deducing a nucleotide sequence from an amino acid sequence, multiple nucleotide sequences will be obtained. For example, Ile has three codons: AUU, AUC, and AUA hence a dipeptide MetIle can have the following nucleotide sequence:(i) AUG - AUU(ii) AUG - AUC (iii) AUG - AUA and if, we deduce amino acid sequence the above nucleotide sequences, all the three, will code for Met-Ile.

Question 11

Would it be appropriate to use DNA probes such as VNTR in DNA fingerprinting of a bacteriophage?

Accepted Answer

Bacteriophage does not have repetitive sequences such as VNTRs in its genome as its genome is very small and have all the coding sequence. DNA finger printing is not done for phages.

Question 12

In a particular plant species majority of the plants bear purple flowers. Very few plants bear white flowers. No intermediate colours are observed. If you are given a plant bearing purple flowers, how would you ascertain that it is a pure breed for that trait? Explain.

Accepted Answer

To ascertain whether the plant is of pure breed for purple coloured flowers a test cross will be made. In a test cross the given purple coloured plant will be crossed with homozygous plant with white flowers. If all the flowers of the progeny are purple, the plant is homozygous dominant.

Question 13

A cross between a red flower bearing plant and a white flower bearing plant of Antirrhinum produced all plants having pink flowers. Work out a cross to explain how this is possible.

Accepted Answer

This type of inheritance is called incomplete dominance as the red color allele (R) is not completely dominant over the recessive (r) allele. So partial expression of both alleles in heterozygotes results in a phenotype intermediate between those of two homozygotes, i.e., the pink color of flowers.

Question 14

In a phenotype intermediate between those of two homozygotes, i.e., the pink color of flowers. In a typical monohybrid cross the F,- population ratio is written as 3:1 for phenotype but expressed as 1:2:1 for genotype. Explain with the help of an example.

Accepted Answer

This particular monohybrid cross is between a homozygous tall plant and a homozygous dwarf plant.

Question 15

AB and CD represent two strands of a DNA molecule. When this molecule undergoes replication, forming a replication fork between A and C in the above.(i) Name the template strands for replication.(ii) Using which strand as the template, will there be a continuous synthesis of a complementary DNA strand? (iii) Complementary to which strand will Okazaki segments get synthesized and discontinuous synthesis will occur.(iv) What are template strands and Okazaki pieces?(v) In which direction is a new strand synthesized?

Accepted Answer

(i) AB, CD (ii) AB (iii) CD (iv) Template strands: Parental DNA strands complementary to which new strands of DNA are synthesized; Okazaki fragments are short segments of DNA synthesized on lagging strand. (v) 5'-3'.

Question 16

Haemophilia is a sex-linked recessive disorder of humans. The pedigree chart given below shows the inheritance of haemophilia in one family. Study the pattern of inheritance and answer the questions given.(a) Give all the possible genotypes of the members 4, 5 and 6 in the pedigree chart. (b) A blood test shows that individual 14 is a carrier of haemophilia. The member numbered 15 has recently married member number 14. What is the probability that their first child will be a haemophilic male?

Accepted Answer

(a) No. 4 Carrier female, No.-5 Haemophilic boy and No. 6 Normal boy(b) Hence, the chances of the first male child to be haemophilic are 25% or 1/4th.

Question 17

Study the pedigree chart given, showing the inheritance pattern of blood groups in a family and answer the following questions(a) Give the possible genotypes of the individuals 1 and 2. (b) Which antigen or antigens will be present on the plasma membranes of the RBC's of individuals 5 and 9.(c) Give the genotypes of the individuals 3 and 4.

Accepted Answer

(a) Individual 1 - IBIO; Individual 2 - IAIO(b) 5- Antigens for both A and B are present.      9 - Lacks any of the antigen A or B.(c) The genotypes of 3 and are IBIO and IAIOrespectively.

Question 18

Study the given pedigree chart showing the pattern of blood group inheritance in a family (a) Given the genotype of the following:(i) Parents (ii) The individual 'X' in the second generation (b) State the possible blood groups of the individual 'Y' in third generation.(c) How does the inheritance of this blood group explain codominance?

Accepted Answer

(a) (i) Parents IAIO- A, IOIO- B (ii) X individual IBIO or IOIA genotype(b) Individual 'Y' blood groups IO, IO, IO, IA genotype (c)IA and IB when stay together, show the phenomena of codominance and express themselves in pressure of each other. In heterozygous hybrid, when both alternative alleles coexist, both the alleles show codominance.

Question 19

(a) Construct a complete transcription unit with promoter and terminator on the basis of hypothetical template strand given below(b) Write the RNA strand transcribed from the above transcription unit along with its polarity

Accepted Answer

(a) Transcription unit

Question 20

Study the following carefully and explain why mutation (A) did not cause any sickle cell anemia inspite of change in the molecular structure of the gene which codes for Haemoglobin, where as a similar mutation (B) did. (The question is based on properties of the genetic code. c= codon, a= amino acid, Hb= Haemoglobin)Codons for Hb:           C1​- C2​- C3​- C4​- C5​- GAA- GAA- C8​ .......Amino acids in Hb:     a1​- a2​- a3​- a4​- a5​- Glutamic acid- Glutamic acid- a8​ ........ (Normal haemoglobin)Mutation (A):              C1​- C2​- C3​- C4​- C5​- GAG- GAA- C8​ .......                                      a1​- a2​- a3​- a4​- a5​- Glutamic acid- Glutamic acid- a8​ ........ (Normal haemoglobin)Mutation (B):               C1​- C2​- C3​- C4​- C5​- GUG- GAA- C8​ .......                                      a1​- a2​- a3​- a4​- a5​- Valine- Glutamic acid- a8​ ........ (Sickle-cell haemoglobin)

Accepted Answer

- Genetic code degenerate i.e., more than one code for one amino acid.- Both GAG GAA code for glutamic acid -Mutation of third base/nucleotide - no change in phenotype in mutation  A/Wobble hypothesis - In case of change on 2nd codon of triplet code as in Mutation B, codon stands for a different amino acid valine.- Hb becomes different / normal Hb becomes Hbs / Structure of protein changed.

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