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 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 AIEEA (PG), Code 14 Animal Sciences (AGB)
Definitions and formulae: selection differential and intensity, generation interval, methods of selection, crossbreeding systems, heterosis, sire evaluation methods, Indian composite breeds and breeding policy.
How our questions for ICAR AIEEA spread across the units of Livestock & Poultry Breeding. Difficulty: 26 easy, 30 medium, 3 hard.
| Outbreeding and crossbreeding | 8 | |
| Response to selection | 6 | |
| Inbreeding | 6 | |
| Heterosis | 6 | |
| Laboratory animal breeding | 6 | |
| Breeding strategies and national breeding policy | 5 | |
| Sire evaluation and progeny testing | 5 | |
| Conservation of animal genetic resources | 5 | |
| Selection index | 4 | |
| Methods of selection | 3 | |
| Genomic selection | 3 | |
| Selection experiments | 2 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
In a pig herd, mean body weight at 180 days is 90 kg. The boars and gilts selected as parents (mass selection in both sexes) average 102 kg at the same age. Heritability is 0.25. The expected response in the next generation is:
A. 3.0 kg
S = 102 − 90 = 12 kg. R = h²S = 0.25 × 12 = 3.0 kg. Multiplying h² by the selected mean (25.5) or by the herd mean (22.5) uses the wrong basis; 12 kg is the selection differential, not the response.
The selection index was first applied to animal breeding by:
D. L.N. Hazel (1943)
Smith (1936) introduced the discriminant function for plant selection; Hazel (1943) developed the index for animals using genetic and phenotypic parameters. Henderson developed BLUP; Wright developed path coefficients and the inbreeding coefficient.
Two half sibs are mated. Their common parent has an inbreeding coefficient of 0.2. The inbreeding coefficient of their offspring is:
B. 0.150
F = (½)^(1+1+1) × (1 + FA) = 0.125 × 1.2 = 0.150. 0.125 ignores the ancestor's own inbreeding; adding 0.2 directly (0.325) is not how the formula works; 0.25 is the full-sib value.
The mule is produced by mating:
A. A male donkey (jack) with a female horse (mare)
Jack × mare gives the mule; stallion × jenny gives the hinny. Both are almost always sterile because horse and donkey chromosome numbers differ.
The contemporary comparison method of sire evaluation was introduced in the United Kingdom by:
D. Robertson and Rendel (1954)
Robertson and Rendel compared each daughter with contemporary heifers by other sires in the same herd and season. Henderson developed BLUP; Hazel the selection index; Lush wrote the foundational text Animal Breeding Plans.
Compared with lactation models, test-day models in dairy sire and cow evaluation:
C. Use individual test-day milk records and account for the environment of each test day
Test-day models analyse each recording separately, allowing for the specific herd-test-date environment and making use of incomplete lactations, which improves accuracy. They include herd effects.
N'Dama cattle of West Africa are an important genetic resource mainly because of their:
B. Trypanotolerance
N'Dama and West African Shorthorn cattle tolerate trypanosome infection in tsetse areas, an adaptation that is hard to recreate and a key reason for conserving them.
A mouse strain is considered inbred after at least how many consecutive generations of brother × sister mating?
C. 20
By convention, an inbred strain requires at least 20 consecutive generations of brother × sister (or parent × offspring) mating, giving F of about 0.986.
Starting from a non-inbred base, three consecutive generations of full-sib mating give an inbreeding coefficient of:
D. 0.500
Using Ft = ¼(1 + 2Ft−1 + Ft−2): 0.25, 0.375, then ¼(1 + 0.75 + 0.25) = 0.500. 0.375 is the second generation; adding 0.25 per generation (0.75) overstates it; 0.125 is half-sib mating.
F1 hybrid mice produced by crossing two inbred strains are:
C. Genetically uniform and vigorous, but they do not breed true
All F1 animals receive one set of chromosomes from each inbred parent, so they are identical to each other and heterozygous wherever the strains differ, showing heterosis. Mating F1 × F1 produces a segregating F2.
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