ASRB NET and ARS / SMS / STO (Agricultural Scientists Recruitment Board) · India · Agricultural Scientists Recruitment Board
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)
How our questions for ASRB NET and ARS / SMS / STO spread across the units of Principles of Animal Genetics & Population Genetics. Difficulty: 15 easy, 41 medium, 23 hard.
| Forces changing gene frequency | 11 | |
| Heritability | 10 | |
| Chromosomes and cytogenetics | 9 | |
| Hardy-Weinberg equilibrium | 9 | |
| Gene action and epistasis | 7 | |
| Mutation | 6 | |
| Molecular genetics and the genetic code | 6 | |
| Genetic laboratory techniques | 5 | |
| Genetic correlation | 5 | |
| Mendelian inheritance | 4 | |
| Repeatability | 4 | |
| Sex-linked inheritance | 3 |
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.
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.
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.
Genotypes AA, Aa and aa have relative fitnesses 0.9, 1.0 and 0.7. What is the equilibrium frequency of allele a?
D. 0.25
Heterozygote advantage with s = 0.1 against AA and t = 0.3 against aa: q = s/(s + t) = 0.1/0.4 = 0.25. 0.75 = t/(s + t) is the frequency of A.
Further reading: Falconer DS, Mackay TFC. Introduction to Quantitative Genetics, 4th ed.
At a single locus, genotypic values are AA = +10, Aa = +4 and aa = -10 (so a = 10, d = 4). The frequency of A is 0.6. What is the additive genetic variance contributed by this locus?
D. 40.63
alpha = a + d(q - p) = 10 + 4(0.4 - 0.6) = 9.2. VA = 2pq alpha^2 = 2 x 0.6 x 0.4 x 84.64 = 40.63. 48.00 ignores dominance (uses a instead of alpha); 3.69 is VD = (2pqd)^2; 9.20 is alpha itself.
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.
An analysis of repeated fleece weights gives a between-animal variance component of 12 and a within-animal variance component of 18. What is the repeatability?
D. 0.40
r = sigma^2_between/(sigma^2_between + sigma^2_within) = 12/30 = 0.40. 0.67 = 12/18 ignores the within-animal part of the total; 0.60 = 18/30 is its complement; 1.50 = 18/12 inverts the ratio.
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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