Baldness (b) is inherited as x-linked dominant disease. an affected male marries a homozygous recessive female. what is the probability that their male offspring will be affected by this disease. (hint: determine the genotype of the parents and then perform the punnett square) 0% 25% 50% 100% i know its not 50%

In this case, (B) is dominant and (b) is recessive.

The genotype of the male is XB Yb. Diseases can only be linked to X chromosomes. In males, they cannot be carriers of a disease since they only have one X chromosome. Since the male is affected by the disease, he will have the dominant trait (B) on his only X chromosome and the recessive trait (b) on his Y chromosome because Y chromosomes are not dominant. Even if the trait is recessive in heterozygous females, it will always be expressed in males.

The genotype of the female will be Xb Xb. Since she is homozygous recessive, both traits will be the same. She will have the recessive trait (b) on each X chromosome, this means she does not have the disease or is a carrier.

So here’s the Punnett square equation:
Bx by • bx bx

The probability of the couple having a male is of course always going to be 50%.

For the male offspring genotypes, I got bx by and bx by. Both of the male offspring’s X chromosomes are recessive, which means they will not be affected by the disease.

So the probability of the male offspring having the disease is 0%.

I hope this helped. (:

Cand d are correct because the rest are just showing your mistakes and you don’t want people who are reading a summary you wrote to see all the mistakes you made.