Question

A man with genotype EEFfGgHH produces ‘P’ number of different sperms. A woman with genotype liLLMmNn can form ‘Q’ types of ova. What is correct about ‘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

Hint: Among the 23 pairs of chromosomes, the 23rd pair are sex chromosomes (which determine whether the offspring will be male or female) , XX for females, or XY for males. They participate in sex determination. The female produces ova while the male produces the sperm.

Complete answer:
In the question, it is asked that the number of types of gametes that can be formed in the male and female. The formula for the number of types of gamete in an organism is given by 2^{n}, where ‘n’ is the number of heterozygous (two different alleles) pairs of genes.
The man has the genotypes of EEFfGgHH, out of this ‘fF’ and ‘Gg’ is heterozygous. The value of ‘n’ will be 2. Hence, the total number of gametes formed in man will be 2^{2}, which is equal to 4. Similarly, the woman has the genotype of liLLMmNn, out of this ‘li’, ‘Mm’, and ‘Nn’ are heterozygous pairs. The value of ‘n’ will be 3. Hence, the total number of gametes formed in the woman will be 2^{3}, which is equal to 8.
Thus, a man with a genotype of EEFfGgHH will produce 4 types of sperms. A woman with genotype IiLLMmNm will produce 8 types of ova.
So the correct answer is, ‘P-4, Q-8.’

Note:
- At the time of spermatogenesis, there will be two types of sperm cells produced in equal numbers, one containing an X-chromosome and the other containing a Y chromosome.
- Since in males, one sex chromosome is X and the other is Y, they are called heterogametic while the females are homogametic because of two X chromosomes.