Question

Meiotic drive is a phenomenon observed occasionally in which a heterozygous genotype does not produce a 1: 1 proportion of functional gametes, usually because one of the gametic types is not formed or fails to function. Suppose that an allele D shows meiotic drive such that heterozygous Dd genotypes form 3/4 D-bearing and 1/4 d-bearing functional gametes. What is the expected ratio of genotypes in the F2 generation of a monohybrid cross under the assumptions stipulated below? (Hint: Use Punnett squares.)

(a) The meiotic drive occurs equally in both sexes.

(b) The meiotic drive occurs only in females.

Answer 1

Step-by-step explanation:

To find the expected ratio of genotypes in the F2 generation of a monohybrid cross, you can use Punnett squares.

(a) If the meiotic drive occurs equally in both sexes, you can represent the cross as follows:

F1 generation:

DD x Dd

F2 generation:

DD Dd dd

DD Dd dd

In this case, the expected ratio of genotypes in the F2 generation is 1:2:1, which means that there is a 1/4 chance of producing a DD genotype, a 1/2 chance of producing a Dd genotype, and a 1/4 chance of producing a dd genotype.

(b) If the meiotic drive occurs only in females, you can represent the cross as follows:

F1 generation:

DD x dd

F2 generation:

DD Dd dd

DD Dd dd

In this case, the expected ratio of genotypes in the F2 generation is 3:1, which means that there is a 3/4 chance of producing a DD or Dd genotype and a 1/4 chance of producing a dd genotype.