Question

A man having the genotype EEFfGgHH can produce P number of genetically different sperms, and a woman of genotype IiLLMmNn 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

Answer 1

The correct option is B P = 4, Q = 8

The number of types of gametes by an organism =$2^n$ 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 $2^n$= $2^2$=4.

The women with genotype IiLLMmNn is heterozygous for 3 pairs of genes (I/I, M/m and N/n), she will produce total $2^n$=$2^3$=8 types of gametes. This makes option B correct answer.