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
Answers
Answered by
2
Answer: P= 4 and Q = 8
Explanation:
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.
THEREFORE, P=5 and Q=8
Similar questions