Sponsor Area

Principles Of Inheritance And Variation

Question
CBSEENBI12047315

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.

  • P - 4; Q - 4

  • P - 4; Q - 8

  • P - 8; Q - 4

  • P - 8; Q - 8

Solution

B.

P - 4; Q - 8

The given genotype of man = EEFfGgHH

 P = 2n = 22 = 4 (where, n = number of heterozygotes)

i.e., EFGH, EfGH, EFgH, EfgH.

The given genotype of woman = IiLL Mm Nn

 Q = 2n = 23 = 8 (where, n = number of heterozygosity)

i.e., ILMN, ILmN, ILMn, ILmn, iLMN, iLmN, iLMn, iLmn.