Download DUET Master 2018 DU MSc Genetics Question Paper With Answer Key

Download DUET (Delhi University Entrance Test conducted by the NTA) 2018 DU MSc Genetics Question Paper With Solution Key

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DUET Last 10 Years 2011-2021 Question Papers With Answer Key || Delhi University Entrance Test conducted by the NTA

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DU MSc Genetics

Topic:- DU_J18_MSC_GENETICS

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Which of the following statements about G proteins is FALSE? [Question ID = 52123]
1. They must be activated before the cell can make needed cAMP. [Option ID = 88487]
2. They become activated when bound to GDP. [Option ID = 88486]
3. They are involved in signal cascades. [Option ID = 88484]
4. They bind to and are regulated by guanine nucleotides. [Option ID = 88485]
Correct Answer :-
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They become activated when bound to GDP. [Option ID = 88486]
Which one of the following statement about nitrogen fixation is correct? [Question ID = 52072]
1. Plants convert atmospheric nitrogen to ammonia [Option ID = 88281]
2. Mutant strains of rhizobium are able to secrete excess protein into the soil [Option ID = 88283]
3. The enzyme nitrogenase reduces N2 to form ammonia [Option ID = 88280]
4. Ammonia is converted to N2, which is the form of nitrogen most easily absorbed by plants [Option ID = 88282]
Correct Answer :-

The enzyme nitrogenase reduces N2 to form ammonia [Option ID = 88280]
Which one of the following combination of scientist(s) and the experiment generated first conclusive evidence that DNA is the genetic material? [Question ID = 52061]
1. Watson and Crick who gave a model for the structure of DNA [Option ID = 88239]
2. Garrod, who postulated that Alkaptonuria, or black urine disease, is due to a mutation in the gene coding for important enzyme. [Option ID = 88237]
3. Beadle and Tatum, who used a mutational and biochemical analysis of the bread mold Neurospora to establish a direct link between genes and enzymes [Option ID = 88238]
4. Avery, MacLeod, and McCarty who repeated the transformation experiments and chemically characterized the transforming principle. [Option ID = 88236]
Correct Answer :-

Avery, MacLeod, and McCarty who repeated the transformation experiments and chemically characterized the transforming principle. [Option ID = 88236]
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Which one of the following is the main contributor of the ascent of sap in the xylem vessels ? [Question ID = 52073]
1. Root pressure [Option ID = 88284]
2. Transpiration and cohesive forces [Option ID = 88285]
3. Capillary action [Option ID = 88286]
4. Atmospheric pressure [Option ID = 88287]
Correct Answer :-

Transpiration and cohesive forces [Option ID = 88285]
Which one of the following, studies the transcripts and proteins expressed by a genome? [Question ID = 52071]
1. Structural genomics [Option ID = 88277]
2. Comparative genomics [Option ID = 88276]
3. Proteo genomics [Option ID = 88278]
4. Functional genomics [Option ID = 88279]
Correct Answer :-

Functional genomics [Option ID = 88279]

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Red hair is a recessive trait in humans. In a random mating population in Hardy-Weinberg equilibrium, approximately 9% of individuals are red-haired. What is the frequency of heterozygotes? [Question ID = 52140]
1. 49% [Option ID = 88553]
2. 18% [Option ID = 88555]
3. 42% [Option ID = 88554]
4. 81% [Option ID = 88552]
Correct Answer :-
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42% [Option ID = 88554]
GTPase domain made up of alpha-helix and Beta pleated sheets in a certain relative orientation is an example of [Question ID = 52110]
1. secondary structure [Option ID = 88433]
2. primary structure [Option ID = 88432]
3. quatenary structure [Option ID = 88435]
4. tertiary structure [Option ID = 88434]
Correct Answer :-

tertiary structure [Option ID = 88434]
The migration of a protein on an SDS polyacrylamide gel is best described as inversely proportional to the [Question ID = 52109]
1. isoelectric point [Option ID = 88429]
2. log of carbohydrate content [Option ID = 88430]
3. log of molecular weight [Option ID = 88431]
4. negative charge [Option ID = 88428]
Correct Answer :-

log of molecular weight [Option ID = 88431]
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Phenotypes such as beard in a woman is most likely a result of the malfunctioning of: [Question ID = 52080]
1. Thyroid [Option ID = 88315]
2. Pituitary [Option ID = 88312]
3. Adrenal cortex [Option ID = 88313]
4. Adrenal medulla [Option ID = 88314]
Correct Answer :-

Adrenal cortex [Option ID = 88313]
In bacteria, partial diploids for a specific gene can be generated by [Question ID = 52087]
1. Conjugation using a Hfr strain [Option ID = 88342]
2. Conjugation using a F' [Option ID = 88343]
3. Generalized transduction [Option ID = 88341]
4. Transformation of chromosomal DNA [Option ID = 88340]
Correct Answer :-

Conjugation using a F' [Option ID = 88343]

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In terms of lac operon regulation, what happens when E. coli is grown in medium containing both glucose and lactose? [Question ID = 52117]
1. Both CAP and lac repressor are bound to the DNA [Option ID = 88460]
2. CAP is bound to the DNA but the lac repressor is not [Option ID = 88461]
3. Lac repressor is bound to the DNA but CAP is not [Option ID = 88462]
4. Neither CAP nor the lac repressor are bound to the DNA [Option ID = 88463]
Correct Answer :-
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Neither CAP nor the lac repressor are bound to the DNA [Option ID = 88463]
The degree of genetic relatedness between offspring and their parents is [Question ID = 52141]
1. is one quarter [Option ID = 88559]
2. same as that between siblings [Option ID = 88558]
3. Lower than that between siblings [Option ID = 88557]
4. Higher than that between siblings [Option ID = 88556]
Correct Answer :-

same as that between siblings [Option ID = 88558]
1 map unit or centimorgam (cM) is equal to [Question ID = 52122]
1. 100% recombination [Option ID = 88483]
2. 1% recombination [Option ID = 88481]
3. 10% recombination [Option ID = 88482]
4. 0.1% recombination [Option ID = 88480]
Correct Answer :-

1% recombination [Option ID = 88481]
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Two genes X and Y are linked in cis. The genes are 20cM apart. If an individual with the genotype XxYy is test crossed what percentage of the progeny will have the genotype XxYy? [Question ID = 52150]
1. 40 [Option ID = 88594]
2. 20 [Option ID = 88593]
3. 80 [Option ID = 88595]
4. 10 [Option ID = 88592]
Correct Answer :-

40 [Option ID = 88594]
Two genes 'A' and 'B' are located on two different chromosomes of a diploid cell. If an individual heterozygous for the two genes is test-crossed what percentage of the progeny will be homozygous for at least one of the genes? [Question ID = 52132]
1. 100 [Option ID = 88523]
2. 75 [Option ID = 88522]
3. 50 [Option ID = 88521]
4. 25 [Option ID = 88520]
Correct Answer :-

75 [Option ID = 88522]

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Two genes are located 70cM apart in the same chromosome. The percentage of recombination between the two genes would be: [Question ID = 52136]
1. 35% [Option ID = 88537]
2. 70% [Option ID = 88536]
3. anywhere between 50% and 70% [Option ID = 88539]
4. =50% [Option ID = 88538]
Correct Answer :-
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=50% [Option ID = 88538]
A second mutation in the same gene restores the wild-type phenotype. This phenomenon is referred to as [Question ID = 52147]
1. intragenic suppression [Option ID = 88582]
2. reversion [Option ID = 88581]
3. intergenic complementation [Option ID = 88580]
4. epistasis [Option ID = 88583]
Correct Answer :-

intragenic suppression [Option ID = 88582]
During growth and division of E. coli, the daughter strand is recognized due to [Question ID = 52066]
1. Nicks in newly synthesized DNA [Option ID = 88257]
2. Hemi-methylation of newly synthesized DNA [Option ID = 88256]
3. Double stranded breaks in newly synthesized DNA [Option ID = 88258]
4. DNA damage in newly synthesized DNA [Option ID = 88259]
Correct Answer :-

Hemi-methylation of newly synthesized DNA [Option ID = 88256]
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In a typical gene cloning experiment, by mistake a researcher introduced the DNA of interest within ampicilin resistant gene instead of lac z gene. The competent cells were allowed to take up the plasmid and then plated in the media containing ampicilin, X-gal and IPTG and subjected to blue-white screening. Considering all plasmids were recombinant which one of the following statements correctly describes the outcome of the experiment? [Question ID = 52096]
1. All of the bacteria would grow and give white colonies. [Option ID = 88379]
2. The bacteria which took up the plasmids would grow and give blue colonies. [Option ID = 88376]
3. The bacteria which took up the plasmids would not grow. [Option ID = 88377]
4. The bacteria which took up the plasmids would form white colonies. [Option ID = 88378]
Correct Answer :-

The bacteria which took up the plasmids would not grow. [Option ID = 88377]
In a four-point (ABCD) cross between Hfr and F strains of E. coli, the pair-wise frequencies of recombination fell in the following order :
AB > AC > AD

The most probable order of these genes on the bacterial chromosome would be: [Question ID = 52088]
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1. ABDC [Option ID = 88347]
2. ABCD [Option ID = 88344]
3. ADCB [Option ID = 88346]
4. ACDB [Option ID = 88345]
Correct Answer :-
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ADCB [Option ID = 88346]
The RNA components of ribosomes are synthesized in the ________ [Question ID = 52116]
1. nucleolus [Option ID = 88458]
2. endoplasmic reticulum [Option ID = 88459]
3. nucleus [Option ID = 88457]
4. Cytoplasm [Option ID = 88456]
Correct Answer :-

nucleolus [Option ID = 88458]
Within the aqueous environment of an animal cell, sugars are stored as polymers rather than as monomers. If the sugars were stored as monomers instead of polymers, which of the following properties would be LEAST affected? [Question ID = 52099]
1. pH [Option ID = 88391]
2. Freezing point [Option ID = 88388]
3. Boiling point [Option ID = 88389]
4. Viscosity [Option ID = 88390]
Correct Answer :-

pH [Option ID = 88391]
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The key difference between dominance and epistasis is: [Question ID = 52144]
1. dominance deals with two alleles; epistasis deals with two genes [Option ID = 88568]
2. dominance fits with Mendel's laws; epistasis is an exception to independent assortment [Option ID = 88571]
3. epistasis is a case of incomplete dominance [Option ID = 88569]
4. dominance expresses a relationship between two alleles; epistasis involves three or more [Option ID = 88570]
Correct Answer :-

dominance deals with two alleles; epistasis deals with two genes [Option ID = 88568]
The end product of glycolysis of a glucose molecule is: [Question ID = 52093]
1. 2 Pyruvate, NADH2 and 2 ATP [Option ID = 88364]
2. Pyruvate, NADH2 and ADP [Option ID = 88366]
3. 2 Pyruvate, 2NADH2 and ATP [Option ID = 88365]
4. Pyruvate, 2H+, 2e- and 4 ATP [Option ID = 88367]
Correct Answer :-
During eukaryotic cell division, metaphase to anaphase transition is regulated by degradation of [Question ID = 52095]
1. Cyclin B1 [Option ID = 88372]
2. Aurora A kinase [Option ID = 88374]
3. CDK1 [Option ID = 88373]
4. Polo-like kinase [Option ID = 88375]
Correct Answer :-

Cyclin B1 [Option ID = 88372]
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Fluorescence recovery after photobleaching in live cells is used to determine [Question ID = 52105]
1. Co-localization of proteins [Option ID = 88412]
2. Diffusion of proteins [Option ID = 88414]
3. Distance between two organelles [Option ID = 88413]
4. Nucleic acid compactness [Option ID = 88415]
Correct Answer :-

Diffusion of proteins [Option ID = 88414]
Monozygotic twin studies in humans are useful because: [Question ID = 52155]
1. twins have a greater likelihood of being heterozygous for a given trait [Option ID = 88613]
2. more refined estimates can be made regarding location of the genes on chromosomes [Option ID = 88612]
3. they allow a true estimate of the environmental influences on phenotypic variation [Option ID = 88614]
4. they allow a true estimate of the genetic influence on phenotypic variation [Option ID = 88615]
Correct Answer :-

they allow a true estimate of the environmental influences on phenotypic variation [Option ID = 88614]

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Plasmid vectors used in cloning experiments often contain a fragment encoding the N-terminal 146 amino acids of ß-galactosidase gene because: [Question ID = 52067]
1. It enables the plasmid vector to replicate in E. coli host cells [Option ID = 88263]
2. It facilitates the ligation of the insert into the vector [Option ID = 88262]
3. It allows selection of E. coli host cells that contain plasmid in which the insert has been ligated [Option ID = 88261]
4. It allows selection of E. coli host cells that contain the plasmid [Option ID = 88260]
Correct Answer :-
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It allows selection of E. coli host cells that contain plasmid in which the insert has been ligated [Option ID = 88261]
Calvin cycle represents one of the following phenomenon: [Question ID = 52074]
1. Oxidative carboxylation [Option ID = 88290]
2. Dark respiration [Option ID = 88289]
3. Reductive carboxylation [Option ID = 88291]
4. Dark phosphorylation [Option ID = 88288]
Correct Answer :-

Reductive carboxylation [Option ID = 88291]
When water is sprinkled on a red-hot iron plate, the drops become spherical and do not vaporize at once because: [Question ID = 52063]
1. A layer of water vapour is formed between the plate and the drops which prevent heat conduction [Option ID = 88246]
2. At this place the temperature of the hot plate falls [Option ID = 88244]
3. Water molecules aggregate into drops [Option ID = 88247]
4. Boiling point of water rises [Option ID = 88245]
Correct Answer :-

• A layer of water vapour is formed between the plate and the drops which prevent heat conduction [Option ID = 88246]
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The initial mechanism for repairing nucleotide errors in DNA during replication is ________ [Question ID = 52113]
1. thymine dimers [Option ID = 88447]
2. nucleotide excision repair [Option ID = 88446]
3. DNA polymerase proofreading [Option ID = 88445]
4. mismatch repair [Option ID = 88444]
Correct Answer :-

DNA polymerase proofreading [Option ID = 88445]
Why is it that inhaling nitric oxide reduces blood pressure only in lung tissue and not elsewhere in the body? Because [Question ID = 52077]
1. nitric oxide cannot cross plasma membranes [Option ID = 88301]
2. nitric oxide cannot enter the bloodstream [Option ID = 88303]
3. other body tissues use a different signalling molecule [Option ID = 88300]
4. nitric oxide breaks down quickly and thus cannot travel far [Option ID = 88302]
Correct Answer :-

nitric oxide breaks down quickly and thus cannot travel far [Option ID = 88302]

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A non-competitive inhibitor of an enzyme-catalyzed reaction: [Question ID = 52062]
1. reduces KM and increases Vmax [Option ID = 88242]
2. no effect on KM and reduces Vmax [Option ID = 88241]
3. reduces KM and reduces Vmax [Option ID = 88243]
4. increases KM and increases Vmax [Option ID = 88240]
Correct Answer :-
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no effect on KM and reduces Vmax [Option ID = 88241]
Removal of gene activity of A from a linear pathway results in higher than normal levels of transcripts from gene B. A reasonable hypothesis would be that: [Question ID = 52075]
1. Gene B must act upstream to gene A [Option ID = 88292]
2. Increase in transcript B abundance is an experimental error [Option ID = 88295]
3. Gene A has no relation to transcript of gene B [Option ID = 88293]
4. Gene B acts downstream to gene A and is regulated by A directly or indirectly [Option ID = 88294]
Correct Answer :-

Gene B acts downstream to gene A and is regulated by A directly or indirectly [Option ID = 88294]
A condition where the genotypic ratio obeys Mendelian laws while the phenotypic ratio does not is referred as: [Question ID = 52149]
1. test cross [Option ID = 88590]
2. back cross [Option ID = 88591]
3. epistasis [Option ID = 88589]
4. incomplete dominance [Option ID = 88588]
Correct Answer :-

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In human, cell cycle is regulated by all of the following EXCEPT: [Question ID = 52126]
1. Ubiquitinylation of proteins [Option ID = 88496]
2. Synthesis of cyclin proteins [Option ID = 88498]
3. Proteolysis of Cdks [Option ID = 88497]
4. Proteolysis of cyclin proteins [Option ID = 88499]
Correct Answer :-
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Proteolysis of Cdks [Option ID = 88497]
If a cell makes both a signalling molecule and the receptor for that signalling molecule, what is this mode of signalling termed? [Question ID = 52081]
1. Endocrine [Option ID = 88317]
2. Juxtacrine [Option ID = 88319]
3. Autocrine [Option ID = 88318]
4. Paracrine [Option ID = 88316]
Correct Answer :-

• Autocrine [Option ID = 88318]
In meiosis, an inversion in one member of a pair of homologous chromosomes will most likely lead to: [Question ID = 52138]
1. Chromosomes with duplications and deficiencies [Option ID = 88545]
2. Increased recombination frequency in the inverted region [Option ID = 88546]
3. Nondisjunction of the affected chromosome [Option ID = 88544]
4. Mispairing of the affected chromosome with a non-homologous chromosome [Option ID = 88547]
Correct Answer :-

Chromosomes with duplications and deficiencies [Option ID = 88545]
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In which of the following regions of a eukaryotic gene will a point mutation most likely have a major negative impact on the function of the encoded protein? [Question ID = 52108]
1. The third nucleotide of a codon in the first exon [Option ID = 88426]
2. The first nucleotide of a codon in the first exon [Option ID = 88427]
3. The TATA box in the promoter [Option ID = 88424]
4. The 5'UTR [Option ID = 88425]
Correct Answer :-

The first nucleotide of a codon in the first exon [Option ID = 88427]
In an experiment, clones of a plant is grown in a field. The plants were observed to be of different heights. When a graph is plotted for frequency of plants (Y-axis) against different heights(X-axis). A bell shaped curve was obtained. From the above it can be concluded that the observed variation in height is due to [Question ID = 52143]
1. It being a polygenic trait [Option ID = 88564]
2. environment influencing different genotypes differently [Option ID = 88567]
3. Variation in genotype [Option ID = 88566]
4. Environmental effect [Option ID = 88565]
Correct Answer :-

Environmental effect [Option ID = 88565]

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A cross between two independent mutants of Drosophila with vestigial wings results in all the F1 progeny being wild type. This is because of: [Question ID = 52154]
1. Dominance [Option ID = 88608]
2. Suppression [Option ID = 88611]
3. Complementation [Option ID = 88610]
4. Epistasis [Option ID = 88609]
Correct Answer :-
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Complementation [Option ID = 88610]
In E. coli different subsets of genes are transcribed under different stress conditions such as heat shock or nitrogen starvation. RNA polymerase achieves this by employing different sets of [Question ID = 52115]
1. beta subunit [Option ID = 88454]
2. alpha subunit [Option ID = 88455]
3. omega subunit [Option ID = 88453]
4. sigma subunit [Option ID = 88452]
Correct Answer :-

sigma subunit [Option ID = 88452]
Somatic cell hybridisation between man and mouse cells results in [Question ID = 52069]
1. Loss of mouse chromosomes [Option ID = 88268]
2. Loss of human chromosomes [Option ID = 88269]
3. Chromosome fusions [Option ID = 88270]
4. Chromosomal aberrations [Option ID = 88271]
Correct Answer :-

Loss of human chromosomes [Option ID = 88269]
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Identify the correct match between the animal (flatworm, earthworm, roundworm) and its body cavity type (acoelomate, coelomate, pseudocoelomate): [Question ID = 52084]
1. Roundworm - pseudocoelomate; Earthworm - coelomate; Flatworm - acoelomate [Option ID = 88330]
2. Roundworm - pseudocoelomate; Earthworm - acoelomate; Flatworm -
  This download link is referred from the post: DUET Last 10 Years 2011-2021 Question Papers With Answer Key || Delhi University Entrance Test conducted by the NTA

This post was last modified on 29 January 2020

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