ICAR AIEEA (PG): Master's entrance, Veterinary Science and Animal Sciences · India · NTA for ICAR
Original practice questions written to the published syllabus · not past papersInheritance from Mendel to molecules, and the population and quantitative genetics that animal breeding is built on.
In the exams: ICAR AICE-JRF core group 05; ICAR AIEEA-PG 14 (AGB); TNPSC AGB 20; ASRB NET (numerical problems, per coaching sources)
ICAR AIEEA (PG), Code 14 Animal Sciences (AGB)
Mendelian and epistatic ratios, sex linkage, chromosome numbers of domestic species, genetic code features, Hardy-Weinberg calculations, heritability and response to selection.
How our questions for ICAR AIEEA spread across the units of Principles of Animal Genetics & Population Genetics. Difficulty: 25 easy, 33 medium, 5 hard.
| Molecular genetics and the genetic code | 8 | |
| Gene action and epistasis | 7 | |
| Chromosomes and cytogenetics | 7 | |
| Mendelian inheritance | 6 | |
| Hardy-Weinberg equilibrium | 6 | |
| Forces changing gene frequency | 6 | |
| Sex-linked inheritance | 5 | |
| Mutation | 5 | |
| Genetic laboratory techniques | 5 | |
| Heritability | 4 | |
| Repeatability | 2 | |
| Genetic correlation | 2 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
A heterozygous polled bull (Pp) is mated to four heterozygous polled cows (Pp), producing one calf each. Polled is completely dominant. What is the probability that exactly two of the four calves are horned?
D. 0.211
Binomial with p(horned) = 1/4: C(4,2) x (1/4)^2 x (3/4)^2 = 6 x 0.0625 x 0.5625 = 0.211. 0.035 forgets the 6 possible birth orders; 0.375 wrongly uses p = 1/2 (6/16); 0.063 is just (1/4)^2.
Further reading: Griffiths AJF et al. Introduction to Genetic Analysis.
What is the diploid chromosome number of a mule (horse mare x donkey jack)?
D. 63
The horse gamete carries 32 chromosomes and the donkey gamete 31, giving 63 (64 is the horse's diploid number, 62 the donkey's, and 126 adds both diploid sets). The unpaired set disrupts meiosis, so mules are almost always sterile.
Further reading: Nicholas FW. Introduction to Veterinary Genetics, 3rd ed.
A test cross of an AaBb bull to aabb cows (pooled over many matings) gives 420 AB, 410 ab, 90 Ab and 80 aB offspring. What is the map distance between the loci, and what is the bull's linkage phase?
B. 17 cM; coupling (AB/ab)
The largest classes (AB, ab) are parental, so the bull is AB/ab (coupling). RF = (90 + 80)/1000 = 0.17, i.e. 17 cM. 8.5 cM counts only one recombinant class; 83 is the parental fraction.
Further reading: Griffiths AJF et al. Introduction to Genetic Analysis.
Which mutagen characteristically causes frameshift mutations by inserting between stacked base pairs?
B. Acridines
Acridines intercalate and cause insertions or deletions. 5-bromouracil is a base analogue causing transitions; UV makes thymine dimers; nitrous acid deaminates bases, causing transitions.
Further reading: Griffiths AJF et al. Introduction to Genetic Analysis.
Why do dideoxynucleotides terminate chain growth in Sanger sequencing?
C. They lack the 3'-hydroxyl group needed to form the next phosphodiester bond
DNA polymerase adds each nucleotide to a 3'-OH. A dideoxynucleotide is incorporated normally but has no 3'-OH, so extension stops at that base.
Further reading: Griffiths AJF et al. Introduction to Genetic Analysis.
In a randomly mating cattle population, 91% of animals are polled (P, dominant) and 9% horned. What proportion of the polled animals is homozygous PP?
A. 0.538
q^2 = 0.09, q = 0.3, p = 0.7. PP = p^2 = 0.49, Pp = 2pq = 0.42. Among polled: 0.49/0.91 = 0.538. 0.490 is PP as a share of the whole population; 0.462 is the Pp share among polled.
Further reading: Falconer DS, Mackay TFC. Introduction to Quantitative Genetics, 4th ed.
A herd has an allele frequency of 0.2. Each generation 10% of breeding animals are replaced by imported animals with an allele frequency of 0.6. What is the allele frequency in the next generation?
B. 0.24
q1 = q0 + m(qm - q0) = 0.2 + 0.1(0.6 - 0.2) = 0.24. 0.26 adds 0.1 x 0.6 without removing the replaced natives; 0.40 simply averages 0.2 and 0.6, ignoring m.
Further reading: Falconer DS, Mackay TFC. Introduction to Quantitative Genetics, 4th ed.
For litter size in sows, genetic variance = 30, permanent environmental variance = 20 and temporary environmental variance = 50. What is the repeatability?
B. 0.50
r = (VG + VEp)/VP = 50/100 = 0.50. 0.30 is the broad heritability (VG/VP), which repeatability always equals or exceeds; 0.20 omits VG, and 0.80 adds the temporary instead of the permanent environmental variance.
Further reading: Falconer DS, Mackay TFC. Introduction to Quantitative Genetics, 4th ed.
Which statement about repeatability (r) and heritability (h^2) of the same trait is correct?
C. h^2 can never exceed r
r = (VG + VEp)/VP includes VA and more, so r is the upper limit of both broad and narrow heritability.
Further reading: Falconer DS, Mackay TFC. Introduction to Quantitative Genetics, 4th ed.
For two traits, the additive genetic covariance is 6, and the additive genetic variances are 16 and 9. What is the genetic correlation?
A. 0.50
rA = covA/(sigma_Ax sigma_Ay) = 6/(4 x 3) = 0.50. 0.04 = 6/144 uses variances instead of SDs; 0.38 = 6/16 divides by one variance only, and 0.67 = 6/9 by the other.
Further reading: Falconer DS, Mackay TFC. Introduction to Quantitative Genetics, 4th ed.
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