CSIR JRF NET Life Sciences June 2016
Model Questions Part 03 (CSIR NET MCQ – 03)
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). Which of the following does not contribute to protein diversity?
a. RNA editing
b. RNA splicing
c. RNA interference
d. Alternative initiation of translation
(2). Enzymes act by:
a. Increasing the activation energy for a reaction
b. Lowering the activation energy for a reaction
c. Increasing the free energy of the system
d. Decreasing the free energy of the system
(3). A carbon footprint is
a. A greenhouse gas emission
b. A way to make accurate tracing of animal track
c. The amount of carbon stored within a living thing
d. The total amount of circulating carbon in the biosphere
(4). A messenger RNA is 336 bases long including the initiation and termination codon. The number of amino acids in the polypeptide translated from this is:
(5). A mutation occurs in the lac repressor protein such that it can no longer bind to lactose. What effect will this mutation have on expression of proteins of the lac operon in the presence of lactose?
a. It will have no effect since lactose shuts off the operon
b. It will decrease expression of the lac operon proteins
c. It will increase the expression of lac operon proteins
d. It will have no effect since the operon is modulated by another protein
(6). Thomas was purifying an enzyme from a homogenate of muscle cells. He went through seven steps of purification and fond that the enzyme activity was the same as the homogenate value. On the eight step, when the protein was highly pure, the enzyme activity rose to five times that of the homogenate value. Can you suggest a possible reason?
a. The homogenate assays a wrong
b. Step eight has co-purified an activator of the benzene
c. The enzyme has an inhibitor present in the muscle cell homogenate
d. The temperature at step eight was just right for the enzyme activity
(7). Fertilizers washed into lakes are a major cause of
d. Bio rectification
(8). Charles Darwin discussed all of the following except:
a. Natural selection removes organisms that are poorly adapted to their environment
b. Individual within a species exhibit variability in form and function
c. Organisms produce more offspring than can survive
d. Gene mutations are the sources of variation for evolution
(9). Where would you expect to find chemosynthetic organisms?
a. Deep sea thermal vents
b. Hypersaline lakes, like the dead sea
c. Streams polluted with domestic sewage
d. Polar ice caps
(10). The group of organisms that is now separated from the other groups of fungi based on their motile spores and cellulose rich cell wall is
(11). The copy number of a transgene in plants can be deciphered by:
a. Southern blotting|
b. Northern blotting
c. South western blotting
d. Far western blotting
(12). Which of the following method is the most appropriate for estimating the population density of burrowing animals?
a. Quadrat sampling
b. Line transect sampling
c. Tag-recapture method
d. Nearest neighbor distance method
(13). For the flower induction, the vernalization signal in plants is perceived mainly by:
a. Young leaves subtending the apical meristem
b. Mature leaves near the root shoot junction
c. All vegetative parts
d. Shoot apical meristem
(14). Alkaloid production in plants is regulated by change in the endogenous pool of:
d. Abscisic acid
(15). The different beak morphologies of Darwin’s finches on the Galapagos Islands is best explained by:
a. Genetic variation
b. Dietary differences
c. Different habitats
d. All of these
e. None of these
(16). Let us consider three independently assorting autosomal genes in Drosophila: A, B and C. What is the probability that a female of genotype AaBBCc when corssed to a male of AAbbCc will produce an offspring of genotype AABbcc?
(17). Insectivorous bats hunt at night by using echolocation (biosonar). Which of the following aspects of the target insect does not affect the loudness of the echo?
c. Position of wings
d. Rate of approach
(18). Two students independently isolated lactate dehydrogenase from chicken heart and measured its activity as a function of substrate concentration, each determining Vmax and Km of their own preparation. They both got the same Km but different Vmax. One student said they had isolated different forms of the same enzyme, the other said they had isolated the same form. To resolve their argument they must:
a. Measure the molecular weight of the enzyme
b. Measure enzyme activity as a function of time after addition of substrate
c. Determine the turnover number of each preparation
d. All the above
(19). As you increase the n (number of measurements) in an experiment, which quantity do you expect to decrease?
c. Standard error of the mean
d. Both (b) and (c)
(20). A drosophila fruit fly female weights about 2 milligrams. The approximate blood volume of this fly will be:
a. One milliliter
b. One microliter
c. One nanolitre
d. One picolitre
Answer key and Explanations:
1. Ans. (c). RNA interference
RNA interference inhibits the gene expression by destructing specific mRNA molecules. RNA interference can act as a gene regulation method at the translation level; however, it never will increase the diversity of a gene product. RNA editing and RNA splicing are the post transcriptional modifications of RNA. Usually some viruses and very rarely some eukaryotes use these methods to increase the protein diversity by changing the site of editing in the primary transcript. Alternate initiation of translation is another strategy adopted by many viruses for the production of many protein form a limited amount of DNA sequence.
2. Ans. (b). Lowering the activation energy for a reaction
Activation energy: energy required to start a chemical reaction. This is the minimum energy input required to a system to start the reaction. By lowering the activation energy, enzymes facilitate an alternate way to initiate the reaction very quickly. Different enzymes adopt different strategies to reduce the activation energy.
3. Ans. (a). A green-house gas emission
Carbon footprint is the measure of greenhouse gases (particularly carbon dioxide and methane) produced by any industry or organization and it is expressed in CO2e (Carbon dioxide equivalent). Carbon footprint cannot be measured accurately because large amount of greenhouse gasses are also produce by natural activities like volcanic eruption, forest fire etc.
4. Ans. (a). 110
Length of messenger RNA = 336
Each codon consists of 3 nucleotides including start and stop codons
Start and stop codon do not code for amino acid in the protein
Thus we can deduct 6 (3 + 3 = 6) nucleotides from total nucleotides
i.e. 336 – 6 = 330
Each amino acid is coded by 3 nucleotides, hence total number of amino acids in the protein will be 330/3 = 110
5. Ans. (b). It will decrease the expression of lac operon protein
Lac operon is a negative inducible operon. This means that the gene is turned off by the regulatory factor called lac repressor unless the inducer binds to it. In the absence of lactose, the lac repressor will binds to the operator and prevent the expression of lac genes. In the presence of lactose or allolactose (the inducer), the inducer binds to the repressor and forms a repressor inducer complex. Once the repressor inducer complex is formed, it will not be able to bins to the operator and thus the lac genes will be transcribed by the RNA polymerase enzyme. In the question, the mutation in lac repressor is in such a way that it is not allowing the binding of inducer (lactose) to the repressor, the repressor will continue to stay on the operator and this will inhibit the expression of lac genes.
6. Ans. (c). The enzyme has an inhibitor present in the muscle cell homogenate
The best option is option (c). The inhibitor in very minute concentration can also inhibit the enzymatic activity.
7. Ans. (a). Eutrophication
Eutrophication: an ecological process by the excessive growth of algal population in the aquatic body due to the addition of nutrients such as phosphates and nitrates through pollution by detergents, fertilizers and sewages. The algal bloom ultimately causes the increase of BOD and it kills the entire flora and fauna of the aquatic body and the water become unfit to support any life forms.
8. Ans. (d). Gene mutations are the source of variation for evolution
Darwin in his theory of natural selection described the process of evolution based on his observations in Galapagos Islands. In Darwinism, Darwin highlighted the importance of variation in speciation. He suggested that the desirable variation occurring in the progeny will be inherited to the next generation, and the accumulation of desirable variation ultimately leads to the formation of new species. Even though the ‘variation’ is one of the significant factors in the process evolution by Darwin, he failed to explain the cause of variation. Now we know that the cause of variation in a population is mutation, recombination, and genetic drift.
9. Ans. (a). Deep sea thermal vents
Chemosynthesis is a biological process of synthesizing carbohydrates by energy obtained from the oxidation of inorganic substances such as hydrogen sulfide, or methane in the absence of light. Organisms which are able to synthesize carbohydrates by chemosynthesis are called chemoautotrophs (chemotrops). Very few microbes are able to do chemosynthesis. These microbes usually live in the deep hydrothermal vents where there is no light for photosynthesis.
10. Ans. (d). Oomycetes
Oomycetes are traditionally included in the Kingdom fungi. They include some of the plant pathogenic fungi like Pythium, Phytopthora etc. Recent phylogenetic analysis shows that Oomycetes are phylogenetically more related to protists and some brown algae. The most important difference of Oomycetes from other fungi is that, Oomycetes possess cellulose in their cell wall instead of chitin.
11. Ans. (a). Southern blotting
12. Ans. (c). Tag-recapture method
In tag-capture method, a portion of the population is captured, marked, and released. Later, another portion is captured and the number of marked individuals within the sample is counted. Since the number of marked individuals within the second sample should be proportional to the number of marked individuals in the whole population, an estimate of the total population size can be obtained by dividing the number of marked individuals by the proportion of marked individuals in the second sample. The method is most useful when it is not practical to count all the individuals in the population.
13. Ans. (a). Young leaves subtending the apical meristem
Vernalization: induction of flowering in plants by clod treatment. The hypothetical hormone involved in floral induction during vernalization is vernalin.
14. Ans. (b). Jasmonates
Jasmonate is a lipid based hormone signaling molecule in plants. It influence a wide range of response in plants starting from growth, photosynthesis, metabolism of biochemical pathways and development of reproductive parts.
15. Ans. (d). all of these
16. Ans. (c). 1/8
17. Ans. (c). Position of wings
18. Ans. (c). Determine the turnover number of each preparation
19. Ans. (c). Standard error of the mean
20. Ans. (b). One microliter
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