ICAR AICE-JRF/SRF (PhD) · 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 AICE-JRF/SRF (PhD), core group 05 (Animal Genetics & Breeding)
Numerical problems on Hardy-Weinberg equilibrium, selection and mutation equilibria, effective population size, inbreeding coefficients by path counting, variance components, heritability from sire models, repeatability and correlated response; recall of cytogenetic abnormalities and named mutations.
How our questions for ICAR AICE-JRF/SRF spread across the units of Principles of Animal Genetics & Population Genetics. Difficulty: 0 easy, 9 medium, 22 hard.
| Heritability | 7 | |
| Forces changing gene frequency | 6 | |
| Hardy-Weinberg equilibrium | 4 | |
| Repeatability | 3 | |
| Genetic correlation | 3 | |
| Chromosomes and cytogenetics | 2 | |
| Genetic laboratory techniques | 2 | |
| Mendelian inheritance | 1 | |
| Gene action and epistasis | 1 | |
| Mutation | 1 | |
| Molecular genetics and the genetic code | 1 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
In a cross between two inbred lines differing at several additive loci of equal effect, 1 in 256 F2 animals is as extreme as one parent line. How many loci are segregating?
C. 4
The fraction of F2 as extreme as one parent is (1/4)^n. (1/4)^4 = 1/256, so n = 4. 8 comes from using (1/2)^n; 64 = 256/4 divides instead of taking the power; 2 uses the dihybrid fraction (1/16)^n.
Further reading: Griffiths AJF et al. Introduction to Genetic Analysis.
Three linked loci lie in the order A-B-C, with 12 cM between A and B and 8 cM between B and C. In a three-point test cross, double crossovers are observed at a frequency of 0.6%. What is the interference?
B. 0.375
Expected double crossovers = 0.12 x 0.08 = 0.0096 (0.96%). Coefficient of coincidence = 0.006/0.0096 = 0.625. Interference = 1 - 0.625 = 0.375. 0.625 is the coincidence, not the interference.
Further reading: Griffiths AJF et al. Introduction to Genetic Analysis.
A herd of 1,000 animals typed at a codominant locus contains 500 AA, 200 Aa and 300 aa. Test for Hardy-Weinberg equilibrium. Which statement is correct?
D. Chi-square is about 340 on 1 df; the herd is not in equilibrium, with a heterozygote deficit
p = (1,000 + 200)/2,000 = 0.6; expected AA 360, Aa 480, aa 160. Chi-square = 140^2/360 + 280^2/480 + 140^2/160 = 340.3. With 3 classes, df = 3 - 1 - 1 (p estimated) = 1; far above 3.84. Observed Aa (200) is far below expected (480): a deficit, suggesting inbreeding, subdivision or selection.
Further reading: Falconer DS, Mackay TFC. Introduction to Quantitative Genetics, 4th ed.
In a randomly mating cat population at equilibrium, 20% of males are orange (X^O Y). What proportion of females is expected to be tortoiseshell (X^O X^o)?
C. 0.32
Male frequency of an X-linked phenotype equals the allele frequency: q = 0.2, p = 0.8. Tortoiseshell females = 2pq = 2 x 0.8 x 0.2 = 0.32. 0.04 is the orange female frequency (q^2); 0.16 is pq, forgetting the 2.
Further reading: Falconer DS, Mackay TFC. Introduction to Quantitative Genetics, 4th ed.
Genotypes AA, Aa and aa have relative fitnesses 1, 1 and 0.8. The frequency of a is 0.4. What is the change in the frequency of a after one generation of selection?
D. -0.0198
delta q = -sq^2(1 - q)/(1 - sq^2) = -0.2 x 0.16 x 0.6/(1 - 0.2 x 0.16) = -0.0192/0.968 = -0.0198. -0.0192 forgets the denominator (mean fitness); -0.0320 is sq^2; -0.0800 is -sq, forgetting to square q.
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.
A cow has three lactation records averaging 600 kg above her herd-mates. Repeatability of lactation yield is 0.4. What is her most probable producing ability (as a deviation from the herd mean)?
D. +400 kg
MPPA = nr/(1 + (n - 1)r) x deviation = 1.2/1.8 x 600 = +400 kg. +240 kg uses r x 600 (as for a single record); +720 kg uses nr without the denominator.
Further reading: Bourdon RM. Understanding Animal Breeding, 2nd ed.
Heritability of a trait is 0.3 and repeatability 0.5. What is the heritability of the mean of four records per animal?
A. 0.48
h^2(mean of n) = nh^2/(1 + (n - 1)r) = 1.2/(1 + 1.5) = 0.48. 1.20 omits the denominator; 0.60 uses h^2/r.
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.
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