CSIR NET Life Sciences Exam June 2016 Model Questions with Answer Key Part 2 (CSIR NET MCQ 04)

net practice test june december 2016

CSIR JRF NET Life Sciences June 2016
CSIR NET Model Questions Part 4 (CSIR NET MCQ – 04)
Biology MCQ (Multiple Choice Questions in Life Science)
(Sample/Model/Practice Questions for JRF/NET Life Science Examination, ICMR JRF, DBT JRF, GATE, ICAR NET, PG Entrance)

1. Bergmann’s rule refers to a general tendency of mammals to be

a. Large in size in colder areas of their distribution
b. Smaller in size in areas of their distribution
c. Darker-pigmented in warmer areas of their distribution
d. Lighter-pigmented in warmer areas of their distribution

2. EDTA is used as a :

a. Oxidizing agent
b. Reducing agent
c. Chelating agent
d. Both (b) and (c)

3. Which of the following element plays an important role in nitrogen fixation

a. Manganese
b. Molybdenum
c. Zinc
d. Copper

4. Pribnow box is the component of _______

a. Operators
b. Promoters
c. Enhancers
d. Activators

5. Presence of an extra chromosome in a eukaryotic cell is most likely due to

a. Linkage
b. Fertilization
c. Transposition
d. Non-disjunction

6. A hollow ball of cells best describes a :

a. Blastula
b. Morula
c. Gastrula
d. Gamete

7. Which of the following vectors can carry the longest piece of foreign DNA?

a. Plasmids
b. Bacteriophage
c. Cosmids
d. Yeast artificial chromosome (YAC)

8. Conversion of nitrite to nitrate in soil is done by the bacteria:

a. Azotobacter
b. Nitrosomonas
c. Nitrobacter
d. Pseudomonas

9. Which of the following pair of relatives will have the highest genetic correlation?

a. First double cousins
b. Half siblings
c. Brothers
d. Broter-sister

10. UGA and UAA are nonsense codons. What kind of single point mutation would cause reversion of UAG to a meaningful codon?

a. Transition
b. Transversion
c. Frameshift
d. Inversion

11. Which process reverses the greenhouse gas effect in the atmosphere?

a. Photosynthesis
b. Transpiration
c. Enteric fermentation in livestock
d. Burning fossil fuels

12. Wild animals also occurs outside forest because of:

a. Habitat loss and fragmentation
b. They are not exterminated outside forest
c. There is abundant food resources associated with humans outside forests
d. All of the above

13. Polyandry is a mating system in which:

a. One male mates with several females in a mating season
b. One female mates with only one male in her life time
c. One female mates with several males in a mating season
d. One male mates with several females in his life time

14. A competitive inhibitor of an enzyme

a. Decrease the Km
b. Decrease the Vmax
c. Decrease both Km and Vmax
d. Increase the Km

15. A scientist studying courtship behavior hypothesizes that while courting female beetles, males display their horns as a signal of their health. If he is correct which of the following must be true

a. Females can estimate male horn length
b. Female appendages are not affected by their health
c. Females always mate with large males
d. Both a and c

16. Among the extant reptiles, which group is phylogenetically closely related to aves?

a. Turtles
b. Lizards
c. Snakes
d. Crocodiles

17. An allele has different effects depending on whether it was inherited from the father or mother. This is most likely due to:

a. Sex linkage
b. Imprinting
c. Penetrance
d. Epistasis

18. Which of the following sequence correctly describes the cell cycle?

a. G1 → G2 → M → Cytokinesis
b. M → G1 → G2 → Cytokinesis
c. Cytokinesis → Mitosis → G1 → G2
d. S → G2 → M → Cytokinesis → G1

19. The genome of virus has a composition of 25% G, 25% C, 25% A and 25%T. The genome is a

a. ssDNA
b. dsDNA
c. ssRNA
d. dsRNA

20. In the Meselson-Stahl DNA replication experiment, what percent of the DNA was composed of one light strand and one heavy strand after one generation of growth in 14N containing growth media?

a. 25
b. 50
c. 75
d. 100.

Answer key and Explanations:

1. Ans. (a). Large in size in colder areas of their distribution

Bergmann’s rule: species of larger size are found in colder environments, and species of smaller size are found in warmer regions.

Allen’s rule: the limbs, ears, and other appendages of the animals living in cold climates tend to be shorter than in animals of the same species living in warm climates. Shorter and more compact body parts have less surface area than elongated ones and thus radiate less body heat.

2. Ans. (c). Chelating agent

Chelating agent: a multi-dentate ligand which can forms several bonds to a single metal ion. Chelating agents capture and hold metal ions form the medium. Example EDTA and porphyrin ring in hemoglobin and chlorophyll.

3. Ans. (b). Molybdenum

Atmospheric nitrogen fixation by microbes is facilitated by the enzyme nitrogenase. Nitrogenase enzyme activity requires Molybdenum as a co-factor.

4. Ans. (b). Promoters

Pribnow box is a six nucleotide consensus sequence (TATAAT) in the prokaryotic promoter. Pribnow box has a function similar to the TATA box in the promoter of eukaryotic genes.

5. Ans. (d). Non-disjunction

Correct segregation of chromosome during meiotic segregation is called disjunction. Disjunction ensures the correct distribution of chromosomes in the gametes. The abnormalities in the dis-junction are called non-dis-junction of chromosomes. Non dis-junction causes the formation of gametes with aneuploidy numerical aberrations of chromosomes such as presence of extra chromosomes or loss of chromosomes etc.

6. Ans. (a). Blastula

Blastula: a state in the embryological development of animals consists of a hollow sphere of cells called blastomeres. The inner fluid filled cavity of blastula is called blastocoele. Blastula is developed from morula, a solid mass of cells formed from zygote after many cleavages. The blastula precedes the formation of the gastrula in which the germ layers, ectoderm, mesoderm and endoderm, are differentiated.

7. Ans. (d), Yeast artificial chromosome

Yeast artificial chromosomes (YAC) can carry large DNA segments for cloning (up to 100–1000 kb segments of DNA). It can be used for the process of chromosome walking. Now yeast artificial chromosome technology is used very less. Today we use bacterial artificial chromosomes (BAC) for the cloning of long DNA stretches.

8. Ans. (c). Nitrobacter

Microbes involved in nitrogen cycle:

(1). Nitrogen fixation: process of conversion of atmospheric nitrogen to ammonia: Rhizobia, Frankia, Azospirillum, Klebsiella pneumaniae, Azotobacter vinelandii, cyanobacteria

(2). Nitrification: conversion of ammonim to nitrites: Nitrosomonas

(3). Nitrification: conversion of Nitrites to nitrates: Nitrobacter

(4). Denitrification: Conversion of nitrates to atmospheric nitrogen: Pseudomonas, Clostridium, Paracoccus denitrificans, thiobacillus denitrificans

icmr jrf exam model question paper by easybiologyclass

9. Ans. (c). Brothers

10. Ans. (b). Transversion

Transition: point mutation that changes a purine nucleotide to another purine (A ↔ G) or a pyrimidine nucleotide to another pyrimidine (C ↔ T).

Transversion: a point mutation that changes a purine to a pyrimidine or vice versa.

11. Ans. (a). Photosynthesis

Greenhouse effect is mainly caused by increased concentration of Carbon dioxide in the atmosphere. Photosynthesis can use CO2 from atmosphere to synthesize carbohydrates and thereby it can reduce the level of CO2 in the atmosphere.

12. Ans. (a). Habitat loss and fragmentation

Habitat loss: loss of original habitat of wild animals or plants.

Habitat fragmentation: large habitats were get fragmented into smaller bits primarily due to the activity of human such as construction of road, rail, dam or other buildings through a natural habitat. Both habitat loss and habitat fragmentation severely affect the biodiversity of the nature.

13. Ans. (c). One female with several males in a mating season

14. Ans. (d). Increase the Km

Km is Michaelis Menton constant, it is the substrate concentration required to reach half of Vmax. Km denotes the affinity of enzyme to its substrate. Increased Km value means decreased affinity and vice versa. Competitive inhibitors are structural analogues of substrates which compete for the active site of enzyme with the substrate. Thus in the presence of a competitive inhibitor the Km of the enzyme will be increased i.e., reduce the affinity of enzyme to its original substrate.

15. Ans. (a). Females can estimate male born length

16. Ans. (d). Crocodiles

17. Ans. (b). Imprinting

Genetic imprinting: certain genes are expressed in a parent-of-origin-specific manner

18. Ans. (d). S → G2 → M → Cytokinesis → G1

19. Ans. (b). dsDNA

20. Ans. (d). 100

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