Genes A and Bare located 20 map units apart on the X chromosome in humans. Thus

Question

Genes A and Bare located 20 map units apart on the X chromosome in humans. Thus they are X-linked. The two genes show simple dominance to their recessive alleles, a and b, respectively. Given the following genotypes for the parents: X^AB//X^ab times X^Ab//Y What will be the frequency of offspring that are male and show both dominant traits (A and B)? What will be the frequency of offspring that are female showing both dominant traits (A and B? In the fruit fly, Drosophila melanogaster, crossing over is absent in the males. Suppose you are interested in the relationship between two linked genes on chromosome 2. The genes are for black body (b) and for curved wings (c, each of which is recessive to the normal (B) body color and normal (C) wing shape. In your laboratory, females that are heterozygous at the body color and wing shape loci are mated with a black-bodied male with curved wings. The offspring of these matings were counted, with the following results: 367 normal body, curved wing 131 normal body, normal wing 139 black body, curved wing 363 black body, normal wing What is the cross-over frequency between these two loci? How many map units separate these two loci? Show the pattern of linkage (coupling or repulsion) in the female flies of this mating?

Explanation / Answer

40a. ANS: The frequency of offspring that are male and show both dominant traits (A and B) is 0.20.

40b. ANS: The frequency of offspring that are female showing both dominant traits (A and B) is 0.25.

41a. ANS: The cross-over frequency between these two loci is 27%.

41b. ANS: 27 many map units separate these two loci.

41c. ANS: The pattern of linkage (coupling or repulsion) in the female flies of this mating is Cb//cB.

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