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fruit fly of genotype a a b h (parent I)is crossed to another fruit lty of genotype ala b/b (parent 2),2 pts) 7. A fruit fly of ssed to another fruit fly of genotype a/a b/b (parent 2). (2 The progeny of this cross were: Genotype a la b/b ala b/b a la b/b ala b/b Number of individuals 38 37 12 13 a) What gametes were produced by parent 1 and in what proportions? b) Do these proportions demonstrate independent assortment of the two genes? c) What can be concluded from these proportions? d) Draw the chromosomes that contain these genes of parents 1 and 2 using the appropriate symbols.
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Answer #1

It is given that the genotype of parent 1 is a+/a- b+/b

a) Four types of gametes will be produced. These are: a+b+, a-b, a+b, a-b+

Parent 2 has genotype a/a- b/b

Two types of gametes will be produced by this parent. These will be ab and a-b

The results will be 8 types of genotypes:   a+/a b+/b, a-/a b/b, a+/a b/b, a-/a b+/b

a+/a- b+/b, a-/a- b/b, a+/a- b/b, a-/a-b+/b

b)

The results are not showing independent assortment. Had it been independent assortment, then all the above 8 genotypes would have been there, and in equal proportion.

c)

Total genotypes = 38+37+12+13=100

Parental types = 38+37 = 75

Recombinants = 12+13= 25

Since parental types >>>recombinants it indicates that genes are linked. The linked genes are a+/b+ and a-/b

d)

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