Question

Х

Test

Patterned baldness is a sex influenced trait more expressed in a male. If b+ is normal allele making non-bald phenotype and b is mutant leading to bald phenotype, for a cross between a bald mother and normal father, what is the probability of a bald son?

Zero

25%

50%

100%​

Answer 1

Answer:

50 per cent is right answer

Answer 2

Probability is 50%.

Explanation:

• b+ is normal allele responsible for non bald phenotype. b+b genotype denote non baldness in female but baldness in male because trait is more expressed in male than female. b+b+ genotype denote non baldness in both male and female.
• b is a mutant allele responsible for bald phenotype so presence of bb denote bald phenotype in both male and female.
• so bald mother genotype will be bb and normal father genotype will be b+b+. If we cross bb×b+b+ all the children would have the genotype b+b.
• b+b represent non baldness in female and baldness in male.

• probability of being a bald son= probability of being a son× probability of being bald in case of male= $\frac{1}{2}$×1=$\frac{1}{2}$. so the probability of a bald son will be 50%