ANS3384

Problem Set 1

Review of Mendelian Genetics as Applied to Livestock

1. Terms with which you should be familiar:

dominant recessive incomplete dominant sex-linked sex-influenced sex-limited

multiple alleles epistasis co-dominant homozygous (homozygote) heterozygous (heterozygote) linkage groups

alleles mutants (mutations) phenotype genotype partial penetrance

2. Explain why a female which has produced one male progeny with a sex-linked recessive defect can be expected to have ½ of all future male progeny with the defect regardless of the male she is bred to. Also, will normal sons of such a female be expected to produce affect progeny?

3. In cattle the following alleles have all been found to be located at the S locus S^H, Hereford pattern; S^P, Pinzgauer pattern; S⁺, solid nonspotted; and s, recessive Holstein-type spotting. S^H and S^P are co-dominant to each other, and both are incompletely dominant over S⁺ and completely dominant over s. Determine the phenotypes that will be produced from the following crosses.

a. Hereford (S^HS^H) × Holstein (ss)

b. Hereford (S^HS^H) × Pinzgauer (S^PS^P)

c. (Hereford × Pinzgauer) (S^HS^P) × Angus (S⁺S⁺)

d. Hereford-Holstein (S^Hs) × Angus-Holstein (S⁺s)

e. Pinzgauer-Angus (S^PS⁺) × Holstein (ss)

f. Hereford-Angus (S^HS⁺) × Hereford (S^HS^H)

4. The roan coloration of cattle is produced by the action of an incompletely dominant gene, R in the heterozygous state. When homozygous, R removes pigment from all of the coat producing a white animal. What types of progeny would be expected from the following matings? (The roan mutant isindependent of other color producing factors. Shorthorn cattle are homozygous red. Red is produced by the recessive mutant, e; black is controlled by its dominant allele, E^d.

a. Red roan × Red roan d. Red × Red

b. White × Red roan e. Black × Red roan

c. Blue roan × Red f. Black × White

Extra Credit

5. Explain a procedure that could be used to incorporate the mutant responsible for the production of blue eggs into a strain of Leghorns and then maintain it in the population while backcrossing to leghorns to restore productivity. The blue egg mutant is dominant and closely linked (<5 m.u.) to the mutant responsible for pea comb (also dominant) Araucana chickens lay blue eggs and have pea combs. Leghorns lay white eggs and have simple combs.

ASG 3313

Answers to Problem Set 1

2. If a normal female produces a male progeny with a sex-links recessive defect, one of her X chromosomes carries the defect and, therefore, half of all her future male progeny will also express the defect. The sire of her progeny does not influence this since he passes only the Y chromosome to his male progeny. Normal sons of such a female must have received a normal X chromosome (one not carrying the defective gene) from her and thus, should not produce the defect.

3. a. all S^Hs - Hereford pattern

b. all S^HS^P - Hereford and Pinzgauer patterns on the same animal

c. ½ S^HS⁺ - white face, perhaps with some white on the underline

½ S^PS⁺ - white on rump and tail only

d. ¼ S^HS⁺ - white face - perhaps with some white on the underline

¼ S^Hs - Hereford pattern - perhaps with some white on the underline

¼ S⁺s - no white spotting (solid)

¼ ss - Holstein spotting (irregular)

e. ½ S^Ps - Pinzgauer pattern

½ S⁺s - no white spotting (solid)

f. ½ S^HS^H - Hereford pattern

½ S^HS⁺ - white face - perhaps with some white on the underline

4. a. ¼ red (r⁺r⁺ ee)

½ red roan (Rr⁺ ee)

¼ white (RR ee)

b. ½ white (RR ee)

½ red roan (Rr⁺ ee) - This answer assumes the white was ee.

c. Assume the blue roan is E^de Rr⁺, then the cross of E^de Rr⁺ × ee r⁺r⁺ yields:

¼ E^de Rr⁺ - blue roan ¼ E^de r⁺r⁺ - black

¼ ee Rr⁺ - red roan ¼ee r⁺r⁺ - red

d. all red (ee r⁺r⁺)

e. Assuming the black animal is homozygous the cross of E^dE^d r⁺r⁺ × ee Rr⁺ yield:

½ E^de Rr - blue roan

½ E^de rr - black

If the black animal is heterozygous for red, E^der⁺r⁺, its cross with ee Rr⁺ animals will yield:

¼ E^de Rr⁺ - blue roan ¼ E^de r⁺r⁺ - black

¼ ee Rr⁺ - red roan ¼ ee r⁺r⁺ - red

f. Assuming the black is E^de r⁺r⁺ in genotype and the white is E^de RR is genotype, the expected progency ratios would be:

¾ E^d_ Rr⁺ - blue roan

¼ ee Rr⁺ - red roan

If the white parent is ee RR and the black parent is E^de r⁺r⁺, the cross will yield:

½ E^de Rr⁺ - blue roan

½ ee Rr⁺ - red roan

5. First, cross leghorn hens, (O represents the blue mutant, o⁺ signifies normal white), P - pea comb, p⁺ - single comb) with Araucana roosters. All F₁ hens should lay blue eggs and have pea combs,. Second, mate the F₁ hens to Leghorn roosters. From the progeny select those pullets with pea combs as they should also lay blue eggs. When these pullets (‹ Leghorn) begin to lay, cull any that lay white eggs (these will be few as they are crossovers). Third, continue this procedure for several generations, then mate hens with pea combs and that lay blue eggs to roosters with pea combs (nearly all of these will also carry the blue egg mutant). Through progeny testing of pea comb males that result from this mating, homozygous males can be located. They are progeny tested by matings to Leghorn hens. Homozygous males will produce only daughters laying blue eggs - test a set of 10 or more daughters per male. These males can then be used in the population to produce a homozygous stock of basically Leghorn ancestry and performance but laying blue eggs.