The mouse autosomal genes B and S are linked and 38 map units apart.Genotypes B S/B S and b s/b s are intercrossed,and the F₁ is testcrossed to b s/b s.The proportion of B-S- progeny will be: A)0.19. B)0.31. C)0.50. D)0.38. E)0.76.

Question

Quizplus · Accepted Answer

Since the genes are 38 map units apart, the recombination frequency between them is 38/100 = 0.38. For the B-S- progeny, both the B and S alleles must have been lost through recombination. Therefore, the probability of producing a B-S- progeny is equal to the recombination frequency (0.38). In the testcross, the genotype of the F₁ parent is not relevant as it will always produce gametes containing either B or S. Thus, the proportion of B-S- progeny will be 0.38 multiplied by the proportion of b s gametes produced by the testcross parent. Since the testcross parent is homozygous for b s, it can only produce b s gametes. Therefore, the proportion of B-S- progeny will be 0.38 multiplied by 0.5 (since half the gametes produced by the testcross parent will be b s). Thus, the proportion of B-S- progeny will be 0.19, or 19%.

The Mouse Autosomal Genes B and S Are Linked and 38