Question

A phenotypically wilptype F1 female fruit fly that was heterozygous for genes controlling body color and wing length was crossed to a homozygous mutant male tester with black body (mutant allele b) and vestigial wings (mutant allele vg).

a. Symbolize the genotype of the dihybrid female and tester male using appropriate Drosophila gene symbolism using b+ and vg+ for the wildtype dominant alleles; b and vg for the recessive mutant alleles.

b. What is the genotype of the F1 dihybrid female (above)?

c. Assuming the F1 dihybrid female has alleles linked in coupling phase, draw her chromosomes showing the correct position of the b and vg loci.

d. The F1 dihybrid female was originally formed by crossing two different homozygous parents. Show the genetic configuration of these parents showing the position and conditions of the alleles on their chromosomes.

e. The cross against the tester produced the following progeny:

gray body, normal wings 126

gray body, vestigial wings 24

black body, normal wings 26

black body, vestigial wings 124

Given the data, what would initially lead you to suspect that the genes are linked?

f. Determine the frequency of recombination.

g. Draw a genetic map for the two genes.

Answer 1

a & b).

b+ b vg+ vg (female) x (male) bb vg vg

c).

b+ vg+ / b vg (F1 dihybrid female)

d).

b+ vg+ / b+ vg+   x b vg / b vg ----Parents

b+ vg+ / b vg (F1 dihybrid female)

e).

If the genes are not linked, the progeny would have 1:1:1:1 ratio. But the number of different phenotypes deviates the ratio, so that we can conclude that the genes are linked.

	b vg
b+vg+	b+b vg+vg (wild-type)
b vg+	bb vg+vg (black boy and normal wings)
b+ vg	b+b vg vg (normal body, vestigial wings)
b vg	bb vg vg (black body, vestigial wings)

f).

Recombination frequency = (number of recombinants / Total progeny) 100

RF = (24 + 26 / 300)100

RF = 16.67%

RF = Distance between the genes

g).

The distance between the two genes = 16.67 map units

b-----16.67 mu------vg