An albino woman marries a phenotypically normal man with no family history of albinism. What percentage of their children are likely to be albino?

0%

Explanation:

This involves a single gene coding for melanin pigment in the skin. The recessive form of this gene (allele) is albinism while the dominant form is a normal skin. The question highlights that the normal male parent (dominant) has no family history of albinism, this means that nobody in their family has ever possessed albinism or has been a carrier/heterozygous for the trait. This is because a carrier/heterozygote will always produce the recessive trait (albinism) if crossed with another carrier or albino individual.

In a nutshell, the normal male parent is homozygous for the dominant trait. Hence, if a homozygous dominant male is crossed with an albino woman (homozygous recessive), all their offsprings will possess a heterozygous genotype and dominant phenotype. Meaning that none of their children will ba an albino.

Hence, the percentage of their child being an albino is 0.

she forgot to record a transaction

the answer would be 2, if crossing over doesn’t occur 2 out of 4 of the daughter cells will have that allele.

hope this : )