Category: IIT JAM Biotechnology Previous Year Question paper Topic Genetics

Category: IIT JAM Biotechnology Previous Year Question paper Topic Genetics

IIT JAM Biotechnology Previous Year Question paper Topic Genetics

35. A Variable Number of Tandem Repeats (VNTR) locus has 15 different alleles. The number of genotypes possible in a population for this VNTR is ________.      

VNTR Genotype Calculation

34. Which one of the following processes causes the Founder effect in a population?  (A) Genetic drift (B) Mutations (C) Natural selection (D) Genetic recombination

Founder Effect Explained

33. Assume that the gene UNC is essential for the coordinated movement of a nematode and that the wild-type allele U is dominant over the mutant allele u. Similarly, the wild-type allele D of another gene DPY, which is responsible for the normal body length, is dominant over the mutant allele d. Assume the UNC and DPY are on two different chromosomes. If a female of genotype UUDD mates with a male of the genotype uudd, the percentage of the F2 progeny that will display uncoordinated movement but will have normal body length is         

Dihybrid Cross Probability Calculation

32. In a large wild flower population, assume that no new mutations occur and that no natural selection operates. What factor(s) will affect the frequency of a genotype in this population?  (A) Non-random mating (B) Gene flow (C) Out-breeding within the population (D) Invasion of a new pathogen that kills a large number of individuals in the population

Factors Affecting Genotype Frequency

31. In maize, the genes for colored seed and round seed are dominant over the genes for colorless seed and shrunken seed. Pure breeding strains of the double dominant variety were crossed with the double recessive variety and a test cross of the F1 generation produced the following:  Phenotypes                            Number of seeds Colored, round seed                   380 Colorless, shrunken seed           396 Colored, shrunken seed              14 Colorless, round seed                  10 For the above, the distance between the genes for seed color and seed shape on the chromosomes would be centimorgan units.

Gene Mapping in Maize Using Test Cross

30. A population is in Hardy-Weinberg equilibrium for a gene with only two alleles (“A” and “a”). If the gene frequency of the allele “A” is 0.7, genotype frequency of heterozygous “Aa” is ______.    

Hardy-Weinberg Equilibrium

29. If a colour-blind woman marries a normal man, the chance that their boy child will be colour-blind is %.       

Colour Blindness Inheritance Probability

28. Both strands of a DNA molecule are labeled with radioactive thymidine and are allowed to duplicate in an environment containing non-radioactive thymidine. The number of DNA molecules that will contain radioactive thymidine after three duplications is      .

Semiconservative DNA Replication Numerical

27. The total number of genetically different types of gametes that will be produced by a heterozygous plant carrying the genotypes AABbCc is.         

Number of Different Gametes Produced by AABbCc

26. Heterozygous female fruit flies with gray body and purple eyes were mated with homozygous males with black body and red eyes. The number of offspring obtained and their phenotypes are shown below: Calculate the recombination frequency.   

Recombination Frequency Calculation in Fruit Fly Test Cross

25. A rare genetic disorder resulting from homozygosity for a recessive allele (r) occurs in 2 out of every 10,000 individuals in a population. Assuming that (i)the disorder is not lethal, (ii)the disorder does not impact reproductive success, (iii)no new mutations are introduced in the population, and (iv)the population follows Hardy-Weinberg equilibrium, the percentage (%) of the carriers in the population that pass the r allele to offspring is_________. (rounded off to 1 decimal)

Hardy-Weinberg Carrier Frequency Calculation

24. In a farm animal breeding programme, the animal with the dominant A phenotype, the recessive b phenotype, the dominant D phenotype, and the recessive e phenotype are commercially important. The inheritance of these traits follows Mendelian laws. From the tetra-hybrid cross of two genotypes AaBbDdEe and AaBbDdEe, the expected frequency of offspring that will show all the above-mentioned desired phenotypes is ________________. (rounded off to 3 decimals)

Tetrahybrid Cross Probability Calculation

23. In a plant species, the genotype DDEE is crossed with the genotype ddee; and the F₁ is test crossed. Considering that the two genes are linked and 20 map unit (cM) apart in the chromosome, the percentage (%) of the test cross progeny with the genotype ddee is ___________. (answer in integer)

Linked Genes Test Cross Numerical

22. In humans, Down syndrome is caused by  (A) Trisomy 16  (B) Trisomy 18  (C) Trisomy 21  (D) Trisomy 22

Down Syndrome Explained

21. A population of 1000 plants in Hardy-Weinberg equilibrium has genotypes: RR = 640, Rr = 320, rr = 40. The frequency of r allele (in percentage) is __________ (rounded off to the nearest integer).

Hardy-Weinberg Allele Frequency Calculation

20. Three genes x, y, and z are located on a chromosome in a linear order. If the recombination frequencies between x and y is 0.15, and between y and z is 0.10, then the expected frequency of double crossovers is __________ (rounded off to three decimal places).      

Double Crossover Frequency Calculation

19. Which one of the following genetic crosses and their results indicates cytoplasmic inheritance? (A) Wild-type male × mutant female → 100% progeny are mutant (B) Wild-type male × mutant female → 25% progeny are wild-type (C) Mutant male × wild-type female → 50% progeny are mutant (D) Mutant male × wild-type female → 75% progeny are wild-type

Cytoplasmic Inheritance Explained

18. Which one of the following statements about polyploidy is correct?  (A) Autopolyploids are derived from a single species (B) Autopolyploids are derived from two different species (C) Allopolyploids are derived from a single species (D) Allopolyploids are not fertile when mated with each other

Polyploidy Explained

17. A genetic linkage map represents the  (A) relative locations of genes on a chromosome (B) distribution of the mutational hotspots (C) phylogenetic linkage among organisms (D) accurate physical distances among loci

Genetic Linkage Map Explained

16. The inability in humans to taste capsaicin resides in a single gene difference between two alleles P and p. The allele P for tasting is dominant over the nontasting allele. In a population of 400 individuals in Hardy-Weinberg equilibrium, 64 are nontasters. How many individuals are heterozygous for the gene? (A) 64 (B) 128 (C) 144 (D) 192

Hardy-Weinberg Equilibrium Numerical

15. The pedigree given below shows individuals affected (shaded circles/rectangles) by chronic hypertension. Assuming 100% penetrance, the inheritance of this trait is  (A) autosomal dominant (B) autosomal recessive (C) sex-linked dominant (D) sex-linked recessive

Pedigree Analysis of Autosomal Dominant Inheritance

14. In mice, a trait is determined by a dominant allele Y and recessive allele y. What proportion of the offspring from a YY × yy cross is expected to be homozygous recessive in F1 generation? (A) 0 (B) 0.25 (C) 0.5 (D) 1

YY × yy Cross Explained

13. In a randomly mating population, the frequency of ‘A’ allele is 0.7. What is the frequency of ‘Aa’ genotype in the next generation according to Hardy-Weinberg’s law? (up to two decimal places)   

Hardy-Weinberg Equilibrium Calculation

12. How many pairs of autosomal chromosomes are there in normal humans?          

Number of Autosomal Chromosome Pairs in Humans

11. A single copy of an allele in sickle-cell heterozygous individuals reduces the frequency and severity of malaria. The reason for this is  (A) Low oxygen binding capacity of hemoglobin (B) Single amino acid substitution in hemoglobin deforms the red blood cells (C) Abnormal hemoglobin is toxic for malaria parasite (D) Malaria parasite escapes the deformed red blood cells

Sickle Cell Trait and Malaria Resistance Explained

10. Which of the following enzymes is absent in a person suffering from Alkaptonuria?  (A) Tyrosinase (B) Homogentisic acid oxidase (C) Catechol dioxygenase (D) Phenylalanine hydroxylase

Alkaptonuria Explained

9. In bacteria, the σ factor that plays a major role in transcription during the stationary phase is (A) σ⁷⁰ (B) σ⁵⁴ (C) σ²⁸ (D) σ³²

PTC Tasting Inheritance Explained

8. DNA gyrase can (A) cut single-stranded DNA (B) relax supercoiled DNA (C) introduce negative supercoiling in DNA (D) not utilize ATP

Frameshift Mutation Explained

7. The number of “Barr Bodies” present in a somatic cell of a woman suffering from Turner syndrome is ____.  (A) 0 (B) 1 (C) 2 (D) 3

Barr Bodies in Turner Syndrome Explained

6. In a population at Hardy-Weinberg equilibrium, for gene-X only two alleles, namely A and a, are found. If frequency of allele A is 0.2 and the frequency of allele a is 0.8, the frequency of the heterozygote genotype Aa in that population will be _____ (correct to 2 decimal places).

Hardy-Weinberg Equilibrium Heterozygote Frequency Calculation

5. The maximum number of genotypes possible for gametes formed from a diploid cell of the genotype AaBBCcDd is _____.  

Maximum Number of Gametes from AaBBCcDd

4. For a gene present on human chromosome 4, the maximum number of alleles that may be detected by sequencing the genome of 5 males and 10 females is _____.      

Maximum Number of Alleles in 5 Males and 10 Females

3. The difference between mitosis and meiosis I is                                                    (A) Sister chromatids separate in mitosis, whereas homologous chromosomes separate in meiosis I (B) The nuclear membrane is absent during mitotic metaphase, but not in meiotic metaphase (C) The DNA is double helical in meiosis I but not in mitosis (D) Unlike in mitotic metaphase, chromosomes do not align at the equatorial plate in meiosis I

Mitosis vs Meiosis I Explained

2. In a genetic cross between plants bearing violet flowers and green seeds (VVGG), and white flowers and yellow seeds (vvgg), the following phenotypic distribution was obtained in the F₂ progeny: i) 2340 plants with violet flowers and green seeds ii) 47 plants with violet flowers and yellow seeds iii) 43 plants with white flowers and green seeds iv) 770 plants with white flowers and yellow seeds Which one of the following interpretations explains the above phenotypic distribution?              (A) Same genes control both flower and seed colors (B) Genes for flower and seed colors are genetically interacting (C) Genes for flower and seed colors are present on the same chromosome (D) Flower color in this plant species is a polygenic trait

Genetic Linkage Explained

1. If the blood groups of mother and father are AB and O, respectively, what are the blood groups possible for their child? (2021) (A) AB or A (B) AB (C) A or B (D) AB, A, B or O

ABO Blood Group Inheritance Explained

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