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ASRB NET and ARS / SMS / STO Livestock & Poultry Breeding: practice questions

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 papers
10worked here, free
100written to this exam in this subject
12syllabus units

What this subject covers

Selection 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

Where the 100 questions fall

How our questions for ASRB NET and ARS / SMS / STO spread across the units of Livestock & Poultry Breeding. Difficulty: 26 easy, 45 medium, 29 hard.

Response to selection10
Inbreeding10
Outbreeding and crossbreeding10
Heterosis9
Genomic selection9
Selection index8
Selection experiments8
Sire evaluation and progeny testing8
Laboratory animal breeding8
Breeding strategies and national breeding policy7
Conservation of animal genetic resources7
Methods of selection6

10 practice questions, worked

Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.

1 · Response to selection · medium

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:

  1. 3.0 kg
  2. 25.5 kg
  3. 12.0 kg
  4. 22.5 kg
Answer and working

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.

2 · Response to selection · hard

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:

  1. Robertson effect
  2. Hill–Robertson effect
  3. Wahlund effect
  4. Bulmer effect
Answer and working

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.

3 · Selection index · easy

The selection index was first applied to animal breeding by:

  1. H. Fairfield Smith (1936)
  2. C.R. Henderson (1973)
  3. Sewall Wright (1921)
  4. L.N. Hazel (1943)
Answer and working

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.

4 · Selection experiments · medium

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:

  1. 0.30
  2. 3.33
  3. 0.050
  4. 0.23
Answer and working

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).

5 · Selection experiments · medium

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:

  1. 35 generations
  2. 50 generations
  3. 12.5 generations
  4. 70 generations
Answer and working

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.

6 · Inbreeding · hard

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:

  1. 0.25
  2. 0.375
  3. 0.50
  4. 0.3125
Answer and working

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.

7 · Sire evaluation and progeny testing · easy

The contemporary comparison method of sire evaluation was introduced in the United Kingdom by:

  1. Henderson (1973)
  2. Hazel (1943)
  3. Lush (1937)
  4. Robertson and Rendel (1954)
Answer and working

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.

8 · Sire evaluation and progeny testing · medium

Compared with lactation models, test-day models in dairy sire and cow evaluation:

  1. Use only the peak yield of each cow
  2. Ignore herd effects
  3. Use individual test-day milk records and account for the environment of each test day
  4. Can be used only for bulls with at least 100 daughters that have completed full lactations
Answer and working

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.

9 · Genomic selection · hard

The accuracy of genomic predictions declines over generations if the reference population is not updated, mainly because:

  1. SNP genotypes change as animals age
  2. Heritability rises each generation
  3. Candidates become less related to the reference animals and marker–gene associations decay through recombination
  4. The number of chromosomes changes
Answer and working

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.

10 · Conservation of animal genetic resources · easy

N'Dama cattle of West Africa are an important genetic resource mainly because of their:

  1. Very high milk yield
  2. Trypanotolerance
  3. Double muscling (myostatin)
  4. Natural polledness
Answer and working

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.

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Read the subject

Livestock & Poultry BreedingThe full study manual, free ASRB NET and ARS / SMS / STOFormat, sections and how it has changed

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Livestock & Poultry Breeding for other exams

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