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

Question

A man having the genotype EEFfGgHH can produce P number of genetically different sperms, and a woman of genotype liLLMmNn can generate Q number of genetically different eggs. Determine the values of P and Q. (A) P = 4, Q = 4 (B) P = 4, Q = 8 (C) P = 8, Q = 4 (D) P = 8, Q = 8

Tardigrade Admin · Accepted Answer

Types of gametes = 2n, where n is number of heterozygous loci. Thus, man will produce 22 = 4 types of gametes and woman will produce 23 = 8 types of gametes.

Q. A man having the genotype $EEF f G g HH$ can produce $P$ number of genetically different sperms, and a woman of genotype $l i LL M m N n$ can generate $Q$ number of genetically different eggs. Determine the values of $P$ and $Q$ .

$P = 4$ , $Q = 4$

$P = 4$ , $Q = 8$

$P = 8$ , $Q = 4$

$P = 8$ , $Q = 8$

Types of gametes $= 2^{n}$ , where $n$ is number of heterozygous loci. Thus, man will produce $2^{2} = 4$ types of gametes and woman will produce $2^{3} = 8$ types of gametes.

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

P=4, Q=4

P=4, Q=8

P=8, Q=4

P=8, Q=8