Question

a.) There are genes C and D that are 16 map units apart. The cross parent genotypes are: CC:DD x cc:dd. What would be the proportion of gametes C:D in the F1 generation?

b.) There are genes C and D that are 20 map units apart. The parental cross was CC:dd x cc:DD and F1 individuals were crossed, Cc:Dd x Cc:Dd. What is the proportion of the F2 offspring that have the CC:DD genotype?

Answer 1

Answer a) :

Given that, genes C and D are 16 map units (mu) apart. Thus, there would be 16% recombinant and the remaining 84% would be parental type.

Given cross is CCDD × ccdd.

CD and cd are parental type gametes, whereas cD and Cd are recombinant ones.

Parental types account for 84%, thus CD and cd would be present in equal proportion of 84, i.e.

= (1/2) × 84

= 42%

Thus, 42% of F1 gametes will be CD.

Answer b)

The given cross parental cross is CCdd × ccDD. These will be the parental genotype. The recombinants would be CCDD and ccdd.

C and D genes are 20 mu apart. Thus there would be a total of 20% recombinants.

CCDD = (1/2) × 20

= 10%