Question

Two genes, A and B, are located 30 map units apart. A mating between an individual homozygous dominant for both traits and one homozygous recessive for both traits is conducted. The F1 progeny is then test-crossed to a homozygous recessive individual. What proportion of their offspring is expected to be dominant for both traits?

Answer 1

Answer:

35%

Explanation:

If two genes are 30 map units apart, 30% of the produced gametes will be recombinant.

A mating between an individual homozygous dominant for both traits (AB/AB) and one homozygous recessive for both traits (ab/ab) is conducted.

The F1 will be heterozygous for both genes: AB/ab.

The F1 progeny is then test-crossed to a homozygous recessive individual:

AB/ab X ab/ab

The possible offspring will be:

• Parental (70%): AB/ab and ab/ab
• Recombinant (30%): Ab/ab and aB/ab

Since 30% of all the gametes produced by the F1 individual will be recombinant, 70% will be parental. As there are two types of parental gametes, each of them will have a frequency of 35%.

The offspring that will have a dominant phenotype for both traits has the genotype AB/ab with a proportion of 35%.