CSIR JRF NET Life Sciences December 2015 Examination Model Question Paper with Answer Key and Explanations Part 02 (Biology MCQ 017)


Biology MCQ 017


NET December 2015 Model Question

(1). Arabidopsis and rice have diploid chromosome number of 10 and 24, respectively. Assuming no crossing over taking place, genetic variation among F2 individuals in a genetic cross is likely to be:

a. Same in both species but not zero
b. More in Arabidopsis
c. More in rice
d. Zero in both the species

(2). In tryptophan operon, tryptophan acts as:

a. Repressor
b. Activator
c. Co-repressor
d. Co-activator

(3). Which of the following statement is CORRECT?

a. Plants adapted to cold environment have higher ratio of ‘unsaturated to saturated’ fatty acids in their membrane compared to those adapted to hot environment
b. Plants adapted to cold environment have lower ratio of ‘unsaturated to saturated’ fatty acids in their membrane compared to those adapted to hot environment
c. Plants adapted to cold environment have same ratio of ‘unsaturated to saturated fatty acids in their membrane compared to those adapted to hot environment
d. Plants do not have any unsaturated fatty acids in the membrane

(4). The recombination frequencies between three genes x, y, and z are as follows:

x – y = 2.6%, y – z = 1.4% and x – z = 1.2%. Then the gene order is:

a. x – z – y
b. x – y – z
c. y – x – z
d. z – x – y

(5). A mutant phenotype due to a nonsense mutation can be rescued by a mutation in tRNA gene. This rescue is an example of:

a. Induced mutation
b. Suppressor mutation
c. Spontaneous mutation
d. Deletion mutation

(6). Ames test is performed to detect:

a. Mutagen
b. pH
c. nutrient stress
d. salinity

(7). Positive selection of T cells ensures:

a. MHC restriction
b. Self-tolerance

c. TCR engagements
d. Activation by co-stimulatory signals

(8). A DNA binding motif is:

a. Helix-loop-helix
b. Helix-turn-helix
c. Helical wheel
d. Loop-helix-loop

(9). Bacterial cell lysis by lysozyme is due to the:

a. Hydrolysis of α-1,4-glycosidic bonds between the N-acetylglucosamine and N-acetylmuramic acid
b. Inhibition of cell wall synthesis
c. Hydrolysis of pentapeptide bridges
d. Hydrolysis of β-1,4-glcosidic bonds between the N-acetylglucosamine and N-acetylmuramic acid

(10). Amino acids responsible for N-linked and O-linked glycosylation of proteins are:

a. Asparagine and Aspartic acid
b. Glutamine and Serine
c. Glutamic acid and Serine
d. Asparagine and Threonine


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Answer key and Explanations:

1. Ans. (c). More in rice
In the question, it is mentioned that crossing over is not taking place, thus the possible chance of variation is the independent assortment of chromosomes during the first meiotic metaphase. This is based on the orientation of maternal and paternal chromosome in the metaphase plate and this alignment decides the chromosome diversity in the gametes. Here in the question, both the parents are diploid (2n). The probable number of different combination chromosomes in the games is equal to the value of ploidy number to the power of haploid set of chromosome.
i.e., in Arabidopsis it is 25 = 32

In rice it is 212 = 4096
Thus the variation we can predict after sexual reproduction will be more in individuals with more chromosome numbers. In the question, rice has more chromosome, the variation in the F1 or F2 will be more in rice than Arabidopsis.

2. Ans. (c). Co-repressor
trp-operon is an example of negative repressible operon. trp-operon is switched on in the absence of tryptophan (the operator will be blocked by the repressor protein in the presence of tryptophan). The repressor of the trp-operon is a monomer and it is constitutively expressed in low level in the cell. When tryptophan is present in the cells, the trp-repressor monomers associated to tetramer and they bind to tryptophan to form a functional repressor, which can bind to the operator region of trp-operon. Once operator region is accompanied by the functional trp-repressor protein, the RNA polymerase cannot bind to the promoter and hence transcription of mRNA is inhibited. In the absence of tryptophan, the repressor is unable to binds to the operator and thus the operon is switched on. Thus in trp-operon, the tryptophan is a co-repressor which binds to the non-functional repressor to form a functional repressor-co-repressor complex.

3. Ans. (a). Plants adapted to cold environment have higher ratio of ‘unsaturated to saturated’ fatty acids in their membrane compared to those adapted to hot environment.
A cell, for its survival, has to maintain the fluid of the membrane all the time. The fluidity of the membrane is depended on the body temperature and also the temperature of the surrounding. Cells can maintain the fluidity of its membrane by changing the fatty acid composition of its lipid bilayer. Typically plants or animals adapted to cold environment contain large number of long chain unsaturated fatty acids in the membrane and they have a higher unsaturated to saturated fatty acid ratio. Increased unsaturation and increased chain length of fatty acids increase the fluidity of the membrane and this helps the cells to prevent the solidification of membrane in lower temperature and thereby the cell can maintain the fluidity of the membrane.

4. Ans. (a). x – z – y
The percentage of recombination (crossing over) between two genes is directly related to the distance between them. In the question maximum recombination percentage is given between x and y genes (2.6%) and thus these genes should be placed in maximum distance. Furthermore, the sum of recombination percentage between y and z (1.4%) and x and z (1.2%) is equal to the recombination percentage of x and y (1.4 + 1.2 = 2.6%).

5. Ans. (b). Suppressor mutation
A suppressor mutation is a second mutation that alleviates or revert the phenotypic effects of an already existing mutation.

6. Ans. (a). Mutagen

7. Ans. (a). MHC Restriction
In MHC restriction, the T cells recognize a peptide antigen only when it is bound to a host body’s own MHC antigen.

8. Ans. (b). Helix-turn-helix
Option (a) is also correct
Common DNA binding domains/motifs are Helix-turn-helix, Zinc finger domain, Leucine zipper, Winged helix, Winged helix turn helix, Helix-loop-helix, HMG-box and Wor3 domain.

9. Ans. (d). Hydrolysis of β-1,4-glcosidic bonds between the N-acetylglucosamine and N-acetylmuramic acid

10. Ans. (d). Asparagine and Threonine

Glycosylation is the covalent attachment (through glycosidic bond)of carbohydrates (glycan) parts by to other macro molecules in the cells such as proteins and lipids. Such complex molecules are called glycol-conjugates.

There are two types of glycosylation based on the type of glycosidic bond formed in the glycosylation reaction.

(1). O-Linked glycosylation

(2). N-Linked glycosylation

JRF NET Life Sciences December 2015 Model Question Paper Set 2 by Easybiologyclass

O-linked glycosylation: here the gycan part forms a glycosidic bond between the anomeric carbon atom of carbohydrate and the – OH group of Serine (Ser) or Threonine (Thr) amino acid residue of the protein (O-linked).

N-linked glycosylation: here the carbohydrate forms a glycosidic bond between the anomeric carbon atom and the amide nitrogen of Asparagine (Asn) residue of the protein (N-linked)

There are two categories of glyco-conjugates of proteins

(1). Glycoproteins

(2). Proteoglycans.

Glycoprotein: A glyco-conjugate of protein and carbohydrates where the protein part is the major part by weight than the glycan part. Even though in glycoproteins, the glycan part is small, the structural complexity of the carbohydrate part will be more here.

Proteoglycan: here usually the glycan part (carbohydrate) part will be bulkier than the protein part. Example: Glycosaminoglycans


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