Every Genetics (Inheritance Biology) question CSIR-NET Life Sciences has set in its last 11 papers, sorted by topic, with 16 of them solved step by step. Answers follow NTA’s final answer keys.
Unit 8 of the CSIR-NET Life Sciences syllabus, Inheritance Biology, covers Mendelian genetics and its extensions, the concept of the gene and complementation, gene mapping, microbial genetics, human genetics and pedigrees, quantitative genetics, mutation, chromosomal alterations and extra-chromosomal inheritance. We went through all 11 papers from February 2022 to June 2025 and pulled out every question from this unit.
How much of the paper is Genetics?
Across the 11 papers, 108 of the 1,375 Part B and Part C questions (7.9%) came from Unit 8. That ranks it eighth of the 13 units. 68 of the 108 were 4-mark Part C questions.
| Paper | Part B | Part C | Total |
|---|---|---|---|
| June 2025 | 3 | 6 | 9 |
| December 2024, Shift 1 | 5 | 6 | 11 |
| December 2024, Shift 2 | 4 | 7 | 11 |
| June 2024, Shift 1 | 3 | 8 | 11 |
| June 2024, Shift 2 | 2 | 6 | 8 |
| December 2023, Shift 1 | 6 | 5 | 11 |
| December 2023, Shift 2 | 4 | 8 | 12 |
| June 2023, Shift 1 | 2 | 5 | 7 |
| June 2023, Shift 2 | 4 | 7 | 11 |
| February 2022, Shift 1 | 3 | 5 | 8 |
| February 2022, Shift 2 | 4 | 5 | 9 |
| All 11 papers | 40 | 68 | 108 |
Which Genetics topics come up most
Gene Mapping Methods and Extensions of Mendelian Principles together account for 42% of the unit’s questions.
16 solved Genetics PYQs
One or two from each major subtopic, from different papers. Try each one before you open the solution.
In a population with the ABO blood group system, if the frequency of the allele IA is 0.3, the frequency of the allele IB is 0.2, and the frequency of the allele i is 0.5, what would be the expected percentage of population with blood group A, considering that the population is under Hardy-Weinberg equilibrium?
- 19%
- 224%
- 339%
- 442%
Show answer and solution
Blood group A covers two genotypes, IAIA and IAi. Under Hardy–Weinberg their frequencies are p2 = 0.32 = 0.09 and 2pr = 2 × 0.3 × 0.5 = 0.30. Together: 0.39, or 39%.
UV mutagenesis was performed to isolate mutants of the lacZ gene. The mutation rate of this gene is 1 × 10−4 per cell division. Assuming that an E. coli culture was initiated from a culture density of 1 × 102 cells and grown to a density of 1 × 106 cells, how many lacZ mutants are expected in this population?
- 1~ 1
- 2~ 10
- 3~ 100
- 4~ 1000
Show answer and solution
Growing from 102 to 106 cells takes about 106 cell divisions (the new cells number 106 − 102). At 10−4 mutations per division, the expected number of mutational events is 106 × 10−4 = about 100 lacZ mutants.
Which one of the following relationships correctly represents the narrow sense heritability? (VA – additive variance, VD – dominance variance, VI – interaction variation, VGE – genotype × environmental variance, VT – total variance)
- 1VA / VT
- 2VA / VGE
- 3VA + VD + VI / VT
- 4VD / VT
Show answer and solution
Answer: (1) VA / VT
Narrow-sense heritability is the fraction of total phenotypic variance due to additive genetic variance: h2 = VA/VP (VT here). Only additive effects are reliably passed from parent to offspring, so it predicts the response to selection (R = h2S). Broad-sense heritability uses all genetic variance, (VA + VD + VI)/VT.
Drosophila melanogaster has 4 pairs of chromosomes. There is no recombination in males of this species. What is the number of possible combinations of chromosomes in gametes of males?
- 1Four
- 2Eight
- 3Sixteen
- 4Thirty-two
Show answer and solution
Answer: (3) Sixteen
With no crossing over in Drosophila males, each chromosome goes into the gamete intact. Each of the 4 pairs sends one of two homologues into a gamete, independently of the others, so there are 24 = 16 possible chromosome combinations.
Which one of the following statements is correct for dosage compensation in human?
- 1X-chromosome inactivation in female occurs in a zygote immediately after fertilization.
- 2X-chromosome inactivation is non-random, in some individuals maternal X is inactivated while in others paternal X-chromosome is inactivated.
- 3Y-chromosome of males is seen as Barr body.
- 4The body of female is a mosaic of cells, some having paternal X- and others having maternal X-inactivated.
Show answer and solution
Answer: (4) The body of female is a mosaic of cells, some having paternal X- and others having maternal X-inactivated.
X-inactivation in human females happens in the early embryo (around implantation), not in the zygote. It is random in each cell, and the choice is then inherited by that cell’s descendants. So a woman is a mosaic of patches with either the paternal or the maternal X inactive, as calico cats show for coat colour. The Barr body is the inactive X, never the Y.
Which one of the following options contains only those types of mapping populations that are characterized by ‘true-breeding’ individuals?
- 1RILs and F2 populations
- 2F2 populations and BC1F2
- 3Doubled haploid populations and BC1F2
- 4Doubled haploid populations and RILs
Show answer and solution
Answer: (4) Doubled haploid populations and RILs
Recombinant inbred lines (RILs) are selfed for many generations from F2 plants until nearly homozygous. Doubled haploids are completely homozygous in one step. Both are true-breeding and “immortal”: they can be grown again and phenotyped in many environments. F2 and BC1F2 plants still segregate.
While developing genetic maps, Alfred Sturtevant proposed that genetic distances are additive. From test crosses involving two genes, if the genetic distance between genes A and B was observed to be 15 cM and between B and C was 10 cM, then the percentage of recombinants observed between A and C would be 25, given that the arrangement is A—B—C. This will be observed only when there is:
- 1No interference
- 2Complete interference
- 3Negative interference
- 4Variable interference
Show answer and solution
Answer: (2) Complete interference
Observed recombination between the outer genes A and C equals the sum of the two intervals (15 + 10 = 25%) only if there are no double crossovers, since a double crossover restores the parental arrangement of A and C. No double crossovers at all means a coefficient of coincidence of 0, which is complete interference.
A plant heterozygous for a dominant trait was selfed. The progeny had 140 plants showing the dominant trait and 20 plants showing the recessive trait. A researcher hypothesised that there are two genes with identical functions that control the dominant trait. The researcher also proposed that the two genes are not linked. The researcher carried out a chi-square test to test the hypothesis. Which one of the following options is the correct chi-square value (rounded to second decimal) obtained by the researcher?
- 122.86
- 213.33
- 310.67
- 45.71
Show answer and solution
Answer: (3) 10.67
Two unlinked genes with identical (duplicate) functions give a 15:1 ratio, since only aabb shows the recessive trait. Of 160 progeny, expect 150 : 10. χ2 = (140 − 150)2/150 + (20 − 10)2/10 = 0.67 + 10.00 = 10.67. That is above 3.84 at 1 df, so the hypothesis is rejected.
A cruciform structure of chromosomes during meiosis is a characteristic feature of:
- 1Translocation
- 2Inversion
- 3Deletion
- 4Duplication
Show answer and solution
Answer: (1) Translocation
In a reciprocal translocation heterozygote, each of the four chromosomes involved has homology with two others. To pair every segment at pachytene they form a cross-shaped (cruciform) quadrivalent. Its alternate or adjacent segregation produces balanced or unbalanced gametes. An inversion heterozygote forms a loop instead.
A colour blind father has a daughter who is also colour blind and has Turner’s syndrome. The genotype of the daughter is due to:
- 1Translocation event in the father.
- 2Translocation event in the mother.
- 3Non-disjunction event in the mother.
- 4Non-disjunction event in the father.
Show answer and solution
Answer: (3) Non-disjunction event in the mother.
A colour-blind girl with Turner syndrome (45,X) has one X, and it carries the colour-blindness allele, so it came from her colour-blind father. She received no sex chromosome from her mother, so the egg must have come from non-disjunction in the mother.
Which one of the following conditions associated with chromosome 15 may cause Prader-Willi syndrome?
- 1Paternal uniparental disomy
- 2Maternal uniparental disomy
- 3Imprinting of 15q11-q13 locus in maternal copy
- 4Imprinting of 15q23-q25 locus in paternal copy
Show answer and solution
Answer: (2) Maternal uniparental disomy
Prader–Willi syndrome results from loss of the paternally expressed genes at 15q11–q13 (such as SNRPN), which are silenced on the maternal copy. A paternal deletion causes it, and so does maternal uniparental disomy: both chromosome 15s from the mother, so no active paternal copy. Paternal UPD causes Angelman syndrome.
Two genes a and b are located at a distance of 10cM. Individuals of the genotype AaBb are sib-mated. The two genes are linked in trans. What percentage of the progeny is expected to have the genotype aabb?
- 10.25
- 20.01
- 36.25
- 425
Show answer and solution
Answer: (1) 0.25
In trans (repulsion), the parental gametes are Ab and aB, and the recombinant ab and AB make up 10% together, 5% each. An aabb offspring needs an ab gamete from both parents: 0.05 × 0.05 = 0.0025, or 0.25%.
A fly with apricot coloured eye was crossed with a sepia eyed fly of opposite sex. In F1 all flies were wild type. The genes responsible for the two phenotypes were
- 1allelic
- 2non-allelic
- 3pseudo-allelic
- 4paralogous genes
Show answer and solution
Answer: (2) non-allelic
The F1 is wild type, so the two recessive mutations complement each other. Each parent supplies a working copy of the gene the other lacks. They are therefore in different genes (non-allelic). Mutations in the same gene would give mutant F1 flies.
The dominant and recessive alleles of a gene are ‘A’ and ‘a’, respectively. In 1000 offspring, if 500 are ‘aa’ and 500 are of the other genotypes, which one of the following is the most likely combination of parental genotypes?
- 1Aa and Aa
- 2AA and Aa
- 3Aa and aa
- 4AA and aa
Show answer and solution
Answer: (3) Aa and aa
A 1:1 ratio of aa to everything else is the signature of a test cross, Aa × aa. Half the offspring get a from the heterozygote (aa) and half get A (Aa). Aa × Aa would give only 25% aa, and any cross involving AA gives none.
Which one of the following statements best illustrates a dominant negative mutation?
- 1A mutation in a growth factor receptor gene leads to an overactive receptor that signals constantly, even in the absence of a signal.
- 2A mutation in an enzyme-coding gene decreases its activity, but enough enzyme activity remains in heterozygotes to maintain normal metabolic function.
- 3A mutation in a transcription factor gene reduces its ability to bind DNA, but there is no effect on gene expression unless both alleles are mutated.
- 4A mutation in a structural protein gene produces an altered protein that can interact with its wild type counterpart and disrupt its function.
Show answer and solution
Answer: (4) A mutation in a structural protein gene produces an altered protein that can interact with its wild type counterpart and disrupt its function.
A dominant negative mutant protein still interacts with the normal one but spoils its function, typically by joining a multimer or complex and inactivating it. One mutant copy can therefore knock out most activity in a heterozygote. Option 1 is a gain of function, and options 2 and 3 describe recessive loss of function.
A mutation in a Drosophila gene causes a reduction in eye size, if animals are grown at 29°C but not at 18°C. This happens even if the animals bear a single copy of this mutation. Based on this information, this mutation can be described as:
- 1Temperature insensitive, dominant
- 2Temperature sensitive, recessive
- 3Temperature insensitive, recessive
- 4Temperature sensitive, dominant
Show answer and solution
Answer: (4) Temperature sensitive, dominant
The phenotype appears at 29 °C but not at 18 °C, so the mutation is temperature-sensitive, typically a protein that misfolds at high temperature. It shows with a single copy, so it is dominant.
All 108 Genetics questions, by subtopic
Each link opens the question on its full paper page. Press Show answer and solution there to see NTA’s final key and a worked solution.
Gene Mapping Methods (27 questions)
| Paper and question | Part | Question |
|---|---|---|
| Jun 2025, Q111 | C | Given below are a few statements related to genetics and linkage mapping in plants. Which one of the following options correctly describes… |
| Jun 2025, Q137 | C | Following are a few statements regarding mapping populations and principles of genetic mapping in crop species. Which one of the following… |
| Dec 2024 S1, Q78 | C | Quantitative Trait Loci (QTLs) can be identified using two main approaches: bi-parental matings (BPM) and Genome-Wide Association Studies… |
| Dec 2024 S2, Q77 | C | Sturtevant introduced the concept of recombination frequency-based genetic maps. Since genetic distances between genes may be… |
| Dec 2024 S2, Q118 | C | Given below are a few statements. Which one of the following options represents the correct sequence of terms to fill in the blanks in the… |
| Jun 2024 S1, Q60 | B | Which one of the following statements regarding principles of linkage mapping in plants is correct? |
| Jun 2024 S1, Q139 | C | A plant breeder plans to introgress a gene for pathogen resistance (R) from a wild species (B) into a cultivated variety (A). Panel I in… |
| Jun 2024 S1, Q140 | C | The following gel patterns are that of DNA markers observed in parents (P1 and P2), F1 from a cross between them and doubled haploid… |
| Jun 2024 S2, Q107 | C | The figure summarizes the observation following a cross between two haploid strains of Neurospora crassa having alleles A and a… |
| Jun 2024 S2, Q116 | C | A cross was made between wild type female Drosophila melanogaster and mutant males which are yellow bodied (y) and crossveinless (cv). The… |
| Dec 2023 S1, Q49 | B | Which one of the following is true for Genome-wide association study (GWAS)? |
| Dec 2023 S1, Q50 | B | Which one of the following options contains only those types of mapping populations that are characterized by ‘true-breeding’ individuals? |
| Dec 2023 S1, Q115 | C | This is a hypothetical example. In a plant, a single gene governs flower color. The wild type color is red, while the mutant is white. It… |
| Dec 2023 S2, Q49 | B | While developing genetic maps, Alfred Sturtevant proposed that genetic distances are additive. From test crosses involving two genes, if… |
| Dec 2023 S2, Q113 | C | A cross was carried out between two strains of Neurospora carrying alleles ‘A’ and ‘a’, respectively. The cross led to the following octad… |
| Dec 2023 S2, Q114 | C | Given below are two figures (X and Y) representing molecular markers and their profiles in parental lines (P1 and P2) and F1 progeny. The… |
| Dec 2023 S2, Q115 | C | The loci for three mutations on X-chromosome, yellow body colour (y), cross-vein less (cv), and forked bristles (f) are shown below in the… |
| Dec 2023 S2, Q143 | C | Given below are two figures (X and Y) representing the segregation of molecular markers in 12 individuals of a mapping population (1 to… |
| Jun 2023 S1, Q49 | B | Segregation of alleles can occur either at anaphase I or anaphase II of meiosis. Which one of the following is an ideal model system for… |
| Jun 2023 S1, Q113 | C | Different types of mapping populations that can be created using a variety of methods are presented as I to IV in the figure below: A list… |
| Jun 2023 S1, Q115 | C | Assume that the genes w+ and cv+ are located 20 cM apart on the X chromosome of Drosophila melanogaster. Mutations in w+ and cv+ give rise… |
| Jun 2023 S1, Q118 | C | Alleles A, a, B and b can be distinguished on the basis of their mobility on an agarose gel. These genes are present on the same… |
| Jun 2023 S2, Q113 | C | The following represents selected AFLP bands (I to V) observed in parents (P1 and P2), F1 progeny and 20 doubled haploid (DH) progeny… |
| Jun 2023 S2, Q117 | C | To obtain recombinant products during meiosis, a double-strand break in the DNA yields crossovers needed for chiasmata formation. The… |
| Feb 2022 S2, Q50 | B | Two genes a and b are located at a distance of 10cM. Individuals of the genotype AaBb are sib-mated. The two genes are linked in trans… |
| Feb 2022 S2, Q114 | C | The following figure represents a physical map and a genetic map for 5 different genes (a to e). Which one of the following statements… |
| Feb 2022 S2, Q117 | C | Doubled haploids (DH) are plants derived from single immature pollen and doubled artificially to form diploids. A DH population was created… |
Extensions of Mendelian Principles (18 questions)
| Paper and question | Part | Question |
|---|---|---|
| Jun 2025, Q71 | C | An embryo lethal recessive mutation in a plant gene “R” located on chromosome 1 was rescued by an ectopically integrated GFP-tagged wild… |
| Jun 2025, Q85 | C | In a dioecious plant species, a single gene is responsible for petal colour. A cross between a true breeding red female and a true breeding… |
| Dec 2024 S1, Q109 | C | In Drosophila melanogaster, a cross was performed, and the resulting progeny are indicated below. The F1 progeny were sib-mated and the F2… |
| Dec 2024 S2, Q63 | B | Which one of the following statements is correct? |
| Dec 2024 S2, Q97 | C | A mutation in a plant gene is female gametophyte lethal. This mutant allele can be normally transmitted through pollen. However, when… |
| Jun 2024 S1, Q117 | C | Given below are figures representing four different situations/examples of genetic and environmental (temperature) effects on plant height… |
| Jun 2024 S1, Q132 | C | Given below are statements on concepts of genetics. Which one of the following options represents the most appropriate sequence of terms to… |
| Jun 2024 S2, Q79 | C | In a plant species, the following pathways contribute to seed color. The wild type phenotype of seed color is red. • A recessive mutation… |
| Dec 2023 S1, Q39 | B | Which one of the following statements is correct for dosage compensation in human? |
| Dec 2023 S1, Q51 | B | In a family, the father has an X-linked mutation causing a late-onset lethal disorder and the mother is not a carrier. Based on the above… |
| Dec 2023 S1, Q113 | C | The following represents a Southern hybridization of restricted genomic DNA, probed with a DNA fragment corresponding to gene ‘Z’. ‘Z’ is a… |
| Dec 2023 S1, Q114 | C | Epistasis is observed between different genes in flower color of sweet pea. A red variety on selfing yields red:white in ratio 9:7. The… |
| Dec 2023 S1, Q116 | C | Shown below is the inversion product of the X chromosome of wild type Drosophila. The w+ gene coding for the red eye colour is near the… |
| Jun 2023 S2, Q49 | B | A plant heterozygous for a dominant trait was selfed. The progeny had 140 plants showing the dominant trait and 20 plants showing the… |
| Jun 2023 S2, Q78 | C | The following statements were made about X-chromosome inactivation in humans: Which one of the following options represents the combination… |
| Feb 2022 S1, Q50 | B | Human polydactyly traits having extra fingers or toes are caused by a dominant allele. In a screening it was found that out of 42… |
| Feb 2022 S1, Q83 | C | Following statements were made about imprinting in the human genome. Select the option with all the correct statements. |
| Feb 2022 S2, Q112 | C | A Drosophila male carrying an X-linked temperature sensitive recessive mutation that is lethal at 29°C but viable at 18°C is mated to: If… |
Human Genetics (17 questions)
| Paper and question | Part | Question |
|---|---|---|
| Jun 2025, Q120 | C | The below pedigree shows the inheritance pattern of gene X. A loss of function mutation of the gene leads to a disorder in adults. Further… |
| Dec 2024 S1, Q22 | B | In a population with the ABO blood group system, if the frequency of the allele IA is 0.3, the frequency of the allele IB is 0.2, and the… |
| Dec 2024 S1, Q133 | C | The pedigree (Fig A) represents the inheritance of a monogenic disorder, caused by a defective enzyme encoded by a mutant allele. The… |
| Dec 2024 S2, Q80 | C | Two closely related individuals, suffering from a congenital disease, have several children. A genetic counselor constructs the following… |
| Jun 2024 S1, Q72 | C | Cystic fibrosis is caused by a recessive allele. Roughly one out of every 500 individuals (0.20%) have this disease. Using the… |
| Jun 2024 S1, Q141 | C | Recessive mutations in the human dysferlin gene lead to Limb Girdle type II muscular dystrophy. The gene is located on the second… |
| Jun 2024 S2, Q41 | B | The pedigree in Panel (i) represents the inheritance pattern of a given trait. The trait is NOT 100% penetrant. Panel (ii) represents PCR… |
| Dec 2023 S1, Q52 | B | In a mutagenesis experiment, the following pedigree was obtained. All progeny had the same phenotype. The mutation is most likely to have occurred |
| Dec 2023 S2, Q50 | B | Which one of the following statements about determination of ABO blood types is INCORRECT? |
| Dec 2023 S2, Q52 | B | Monozygotic and dizygotic twins are used to study the onset of mental illness. The influence of genetic factors on these diseases can be… |
| Dec 2023 S2, Q82 | C | Following statements were made for fragile X syndrome: Which one of the following options provides combination of all correct statements? |
| Dec 2023 S2, Q117 | C | This is a hypothetical example. The pedigree for a monogenic trait is given below. The shaded individuals show spotted skin color while the… |
| Jun 2023 S1, Q114 | C | Red-blue colour blindness is a human X-linked recessive disorder. The two parents with normal colour vision have two sons. Son 1 has 47… |
| Jun 2023 S2, Q28 | B | Which one of the following conditions associated with chromosome 15 may cause Prader-Willi syndrome? |
| Jun 2023 S2, Q52 | B | A colour blind father has a daughter who is also colour blind and has Turner’s syndrome. The genotype of the daughter is due to: |
| Feb 2022 S1, Q111 | C | The pedigree given below represents the genotype at four different loci for the children in generation III. Which one of the given… |
| Feb 2022 S2, Q49 | B | A newborn baby got mixed up with other babies in a hospital. If the mother is of O blood group and is Rh +ve and the father is of AB blood… |
Microbial Genetics (10 questions)
| Paper and question | Part | Question |
|---|---|---|
| Jun 2025, Q40 | B | Two temperature-sensitive mutants of Escherichia coli were mated with each other and none of the resulting exconjugants showed the… |
| Jun 2025, Q119 | C | A transformation experiment is performed with two Pneumococcal strains. One strain is resistant to four drugs – A, B, C, and D, whereas the… |
| Jun 2024 S1, Q113 | C | The plaque morphology of wild type and rII mutants of T4 bacteriophage following infection of different E. coli strains is summarized… |
| Jun 2024 S2, Q111 | C | Five different strains of Salmonella (1, 2, 3, 4, 5) which can utilize lactose (Lac+) as the sole carbon source but cannot synthesize… |
| Jun 2024 S2, Q118 | C | Mutants of bacteriophage that carry deletions can be used to rapidly locate mutational sites of newly obtained mutants. The mapping is… |
| Dec 2023 S2, Q51 | B | DNA from a strain of bacteria with genotype a+ b+ c+ d+ e+ was isolated and used to transform a strain of bacteria that was a− b− c− d− e−… |
| Jun 2023 S1, Q117 | C | T4 phages were plated on three E. coli bacterial plates labelled I, II and III. The phenotypes obtained are depicted in the picture below… |
| Jun 2023 S2, Q114 | C | Two mutations were isolated in bacteriophage, one causing clear plaque (c) and the other causing minute plaque (m). The genes responsible… |
| Feb 2022 S1, Q48 | B | DNA was isolated from a strain of bacterium with genotype a+ b+ c+ d+ e+ and transformed into a bacterial strain a− b− c− d− e−. The… |
| Feb 2022 S1, Q113 | C | P22 transduction was used to map the fadL gene. The result of two-factor crosses between fadL and two linked markers, purF and aroC are… |
Mutation (9 questions)
| Paper and question | Part | Question |
|---|---|---|
| Dec 2024 S1, Q33 | B | A mutation in a Drosophila gene causes a reduction in eye size, if animals are grown at 29°C but not at 18°C. This happens even if the… |
| Dec 2024 S1, Q45 | B | UV mutagenesis was performed to isolate mutants of the lacZ gene. The mutation rate of this gene is 1 × 10−4 per cell division. Assuming… |
| Dec 2024 S1, Q71 | C | The table below shows types of chemical mutagens and names of mutagens. Which one of the following options shows the correct match between… |
| Dec 2024 S2, Q29 | B | Which one of the following correctly describes how a mutagen induces a specific type of base change in DNA? |
| Dec 2024 S2, Q66 | B | Which one of the following statements best illustrates a dominant negative mutation? |
| Dec 2024 S2, Q111 | C | Normal yeast cells grow at 42°C. Five yeast haploid strains, with independent alleles of YFG1 having impaired cell growth at 42°C, were… |
| Dec 2024 S2, Q114 | C | Based on homology, a protein, CG2024, functions as a homotetramer. The function of each unit within the tetramer is essential for its… |
| Dec 2023 S1, Q25 | B | Fifteen spontaneous Ara− mutants of E. coli that were not able to utilize arabinose as a sole carbon source in minimal synthetic medium at… |
| Jun 2023 S2, Q118 | C | The following are some statements made regarding mutations: Which one of the following options represents the combination of all INCORRECT statements? |
Quantitative Genetics (9 questions)
| Paper and question | Part | Question |
|---|---|---|
| Jun 2025, Q38 | B | Which one of the following relationships correctly represents the narrow sense heritability? (VA – additive variance, VD – dominance… |
| Dec 2024 S1, Q93 | C | Given below are a few statements regarding gene actions observed in plants. Which one of the following options represents the combination… |
| Jun 2024 S1, Q59 | B | Select the correct combination of terms from plant breeding systems that represents selfing or promote selfing. |
| Jun 2024 S1, Q99 | C | Given below are a few statements on concepts of molecular breeding. Which one of the following options represents all correct statements? |
| Jun 2024 S2, Q49 | B | Which one of the following statements regarding genetics of quantitative traits in plants is INCORRECT? |
| Jun 2024 S2, Q98 | C | Given below are a few statements related to inheritance biology. Which one of the following options represents a combination of all correct… |
| Dec 2023 S2, Q138 | C | Following statements are made regarding heterosis breeding in plants. Which one of the following options represents the combination of all… |
| Feb 2022 S1, Q114 | C | Wheat plants can have kernels of white colour or in shades of red i.e. light red, red, dark red and very dark red (purple). A researcher… |
| Feb 2022 S2, Q51 | B | Which one of the following statements is true regarding heritability of a quantitative character? |
Concept of Gene – Alleles & Complementation (6 questions)
| Paper and question | Part | Question |
|---|---|---|
| Dec 2024 S1, Q61 | B | The disease phenylketonuria is caused by the null allele of the gene phenylalanine hydroxylase (PAH). This is an example of a recessive… |
| Dec 2024 S1, Q143 | C | Six mutant yeast haploids (His1-6) requiring histidine supplementation for viability were fused in pair-wise combinations to form diploids… |
| Dec 2023 S1, Q77 | C | P. aeruginosa makes a blue-green pigment called pyocyanin. To understand the pyocyanin biosynthetic pathway, mutants which cannot make… |
| Feb 2022 S1, Q49 | B | A fly with apricot coloured eye was crossed with a sepia eyed fly of opposite sex. In F1 all flies were wild type. The genes responsible… |
| Feb 2022 S1, Q112 | C | A Drosophila mutant (line A) with vestigial wings is isolated in a laboratory. The vestigial wing phenotype was observed to be recessive… |
| Feb 2022 S2, Q116 | C | A cross is made between Drosophila stocks, each with an independent mutant allele, resulting in white eye color. The mutant alleles (named… |
Structural & Numerical Chromosomal Alterations (5 questions)
| Paper and question | Part | Question |
|---|---|---|
| Dec 2023 S2, Q118 | C | The figure given below is of two homologous chromosomes paired during meiosis where one event of recombination occurred between two… |
| Jun 2023 S1, Q50 | B | Based on the image given below, select the option that describes it correctly: |
| Jun 2023 S2, Q51 | B | A cruciform structure of chromosomes during meiosis is a characteristic feature of: |
| Jun 2023 S2, Q77 | C | Columns X and Y of the following table list some treatment methods, reagents, and events that are related to human lymphocyte culture, and… |
| Feb 2022 S2, Q52 | B | Chromosomal inversions are balanced rearrangements and thus do not change the overall amount of genetic material. While inversions can… |
Mendelian Principles (4 questions)
| Paper and question | Part | Question |
|---|---|---|
| Jun 2025, Q42 | B | Drosophila melanogaster has 4 pairs of chromosomes. There is no recombination in males of this species. What is the number of possible… |
| Dec 2024 S1, Q64 | B | The dominant and recessive alleles of a gene are ‘A’ and ‘a’, respectively. In 1000 offspring, if 500 are ‘aa’ and 500 are of the other… |
| Dec 2024 S2, Q58 | B | In a population, the frequency of allele ‘a’ is 0.2 and that of allele ‘b’ is 0.1. Consider that there are two alleles for each of the… |
| Feb 2022 S2, Q115 | C | In 1990, Bhattacharya et al identified that the wrinkled seed character of pea as described by Mendel is caused by a transposon – like… |
Extra-Chromosomal Inheritance (3 questions)
| Paper and question | Part | Question |
|---|---|---|
| Dec 2024 S2, Q107 | C | In sesame, the seed coat color is of two types: white or brown. True breeding white- and brown-seeded plants were reciprocally crossed and… |
| Jun 2024 S1, Q67 | B | Brassica juncea has bisexual flowers. A mutation in the mitochondria leads to cytoplasmic male sterility (CMS). CMS can be restored by a… |
| Jun 2023 S2, Q115 | C | In Drosophila, a cross was set between a male homozygous for alleles s+/s+ (phenotype A) and a female homozygous of s/s (phenotype B) (‘s+’… |
PYQs for every CSIR-NET unit
How to revise Genetics with these questions
- Start with the two biggest subtopicsGene Mapping Methods and Extensions of Mendelian Principles are where most of the marks have been. Revise them first.
- Attempt before you readWork through a subtopic’s questions on paper, then check them against the answer keys.
- Find the repeatsThe same ideas keep coming back in new wording. When you get one wrong, look for its earlier version in the list above.
- Finish with a full paperTake a whole paper in live exam mode to practise choosing which Part C questions to attempt.
Keep going
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p class=”lbq-src”>Source: CSIR-NET Life Sciences question papers and final answer keys, National Testing Agency, February 2022 to June 2025. Subtopic tags, difficulty ratings and solutions: Lesson Basket.
