ICAR AICE-JRF/SRF (PhD) · India · NTA for ICAR
Original practice questions written to the published syllabus · not past papersSelection and mating systems that improve herds and flocks, sire evaluation, genomic selection and the conservation of animal genetic resources.
In the exams: ICAR AICE-JRF core group 05; ICAR AIEEA-PG 14 (AGB); EGEL Plus reproduction and genetics 15
ICAR AICE-JRF/SRF (PhD), core group 05 (Animal Genetics & Breeding)
Numerical problems on selection response per year, four-pathway gain, selection index weights, inbreeding coefficients and effective population size, heterosis and retained heterosis, progeny-test accuracy; theory of selection limits, BLUP, genomic selection and open nucleus schemes.
How our questions for ICAR AICE-JRF/SRF spread across the units of Livestock & Poultry Breeding. Difficulty: 0 easy, 15 medium, 26 hard.
| Selection experiments | 6 | |
| Genomic selection | 6 | |
| Response to selection | 4 | |
| Selection index | 4 | |
| Inbreeding | 4 | |
| Methods of selection | 3 | |
| Heterosis | 3 | |
| Sire evaluation and progeny testing | 3 | |
| Outbreeding and crossbreeding | 2 | |
| Breeding strategies and national breeding policy | 2 | |
| Conservation of animal genetic resources | 2 | |
| Laboratory animal breeding | 2 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
A dairy population has these selection intensities (i), accuracies (r) and generation intervals (L, years) on the four pathways: sires of sons i = 2.5, r = 0.80, L = 7; sires of daughters i = 1.8, r = 0.75, L = 6; dams of sons i = 1.4, r = 0.60, L = 5; dams of daughters i = 0.3, r = 0.50, L = 4.5. The additive genetic standard deviation of milk yield is 500 kg. Expected annual genetic gain is:
D. 96.4 kg/year
ΔG = (Σ i r) σA / Σ L. Σ i r = 2.0 + 1.35 + 0.84 + 0.15 = 4.34; Σ L = 7 + 6 + 5 + 4.5 = 22.5. ΔG = 4.34 × 500 / 22.5 = 96.4 kg/year. Dividing by the mean interval instead of the sum (385.8) counts each pathway's gain four times; ignoring accuracy gives 133.3; dividing by 4 instead of Σ L gives gain per generation-equivalent, 542.5.
Trait X (heritability 0.40) is selected with intensity 1.4. Trait Y has heritability 0.10 and phenotypic standard deviation 5 units, and the additive genetic correlation between X and Y is 0.6. What is the correlated response in Y per generation?
B. 0.84 units
CRY = i hX hY rA σPY = 1.4 × 0.632 × 0.316 × 0.6 × 5 = 0.84. Direct selection on Y with the same intensity would give i h²Y σPY = 0.70, so here indirect selection is slightly better because X is more heritable. Omitting hY gives 2.66; using h² instead of h for both traits gives 0.168.
The reduction in additive genetic variance caused by selection itself, through negative covariances (gametic phase disequilibrium) between loci among selected parents, is known as the:
D. Bulmer effect
Bulmer (1971) showed that selection creates negative disequilibrium between loci, lowering additive variance in the next generation until equilibrium. The Hill–Robertson effect concerns interference between linked loci under selection in finite populations; the Wahlund effect is the deficit of heterozygotes when subpopulations are pooled.
Over six generations of selection for litter weight, the cumulative selection differential was 150 g and the cumulative response (measured against a control line) was 45 g. The realised heritability is:
A. 0.30
Realised h² = cumulative response / cumulative selection differential = 45/150 = 0.30. Inverting the ratio gives 3.33, impossible for a heritability; dividing again by the number of generations (0.050) double-counts time, since both totals already span six generations; 0.23 wrongly divides by S + R (45/195).
In a population with effective size 25, Robertson's theory gives the maximum half-life of response to selection (the number of generations to reach half the limit) as about:
A. 35 generations
Half-life is at most about 1.4Ne generations: 1.4 × 25 = 35. 50 is 2Ne, the multiplier for total response; 12.5 is Ne/2; 70 multiplies 1.4 by 2Ne instead of Ne.
A sire whose own inbreeding coefficient is 0.25 is mated to his own daughter (out of an unrelated dam). The inbreeding coefficient of the offspring is:
D. 0.3125
The sire is the common ancestor; n1 = 0 (he is the sire) and n2 = 1 (his daughter is the dam). F = (½)^(0+1+1) × (1 + 0.25) = 0.25 × 1.25 = 0.3125. 0.25 ignores the sire's own inbreeding; 0.375 would be the second generation of full-sib mating; 0.50 confuses F with the parent–offspring relationship.
Long runs of homozygosity (ROH) in an animal's genome indicate:
D. Inbreeding from a recent common ancestor
Recombination breaks identical-by-descent segments each generation, so long ROH come from recent common ancestors and short ROH from distant ones. Crossbreeding produces heterozygosity, not long ROH.
Single-step genomic BLUP (ssGBLUP) is used because it:
A. Combines genotyped and non-genotyped animals in one evaluation, blending pedigree and genomic relationships
In most populations only a minority of animals is genotyped. The single-step method merges the pedigree matrix A and genomic matrix G into one relationship matrix so that all records and all animals contribute.
The accuracy of genomic predictions declines over generations if the reference population is not updated, mainly because:
C. Candidates become less related to the reference animals and marker–gene associations decay through recombination
Genomic prediction depends on linkage disequilibrium and relationships between candidates and the reference; both erode with recombination and distance of relationship. Adding new phenotyped animals to the reference each generation keeps accuracy up.
Recombinant inbred strains are produced by:
A. Inbreeding many separate lines from the F2 of a cross between two inbred strains
Each recombinant inbred line fixes a different mosaic of the two parental genomes, making a permanent mapping panel. Backcrossing a gene gives a congenic strain; a mutation within a strain gives a coisogenic strain; substituting a chromosome gives a consomic strain.
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