In humans, the allele for brown eyes (B) is dominant to the allele for blue eyes (b). What is the probability that two heterozygous brown-eyed parents will have a child with blue eyes?

but what is the percentage

bobpursely, Y U so stupid?

To determine the probability of two heterozygous brown-eyed parents having a child with blue eyes, we need to understand the inheritance pattern of eye color in humans.

In this case, the allele for brown eyes (B) is dominant over the allele for blue eyes (b). Dominant alleles are represented by capital letters and recessive alleles by lowercase letters.

Given that both parents are heterozygous (i.e., they have one dominant allele, B, and one recessive allele, b) for the brown eye trait, their genotypes would be Bb.

When these two parents reproduce, there are four potential combinations of alleles that their child can inherit: BB, Bb, Bb, or bb. The first three combinations would result in the child having brown eyes because they have at least one dominant allele for brown eyes. On the other hand, the last combination (bb) would result in the child having blue eyes.

Based on these possibilities, there is a 25% chance (1 out of 4) that two heterozygous brown-eyed parents will have a child with blue eyes.

Bb x Bb= BB, Bb, Bb, bb Looks like only one bb kid out of four.

Are you serious? 1/4 is 25 percent. If you are above 7th grade, you need to review what percent means.