Question

A man having the genotype EEFfGgHH can produce P number of genetically different sperms, and a woman of genotype liLLMnNn can generate Q number of genetically different eggs. Determine the values P and Q

Answer 1

$\huge\mathbb\orange{Heya\:Mate..!}$

The number of types of gametes by an organism=2n

where n is number of heterozygous pairs of genes. Since man with genotype EEFfGgHH is heterozygous for 2 pairs of genes (F/f and G/g); total types of gametes formed by him will be

2n= 22=4.

=8 types of gametes.

Answer 2

Explanation:

P=4,Q=4

P=4.Q=8

P=8,Q=4

P=8,Q=8

Answer :

B

Solution :

NA