Let's assume that the ability to cross one's eyes is a mendelian trait, in that it is determined by only two alleles, which can be dominant, recessive, or a combination of the two. eye crossing is the dominant trait and the inability to cross one's eyes is the recessive trait. (this is a completely imaginary scenario.) a man marries a woman and they have a child together. the father is heterozygous dominant for the trait of eye-crossing, and the mother is homozygous recessive for the trait. what are the chances that their child will be born with the ability to cross his or her eyes?

Question

Biology

Jazlyn Soto

19 September, 12:29

Let's assume that the ability to cross one's eyes is a mendelian trait, in that it is determined by only two alleles, which can be dominant, recessive, or a combination of the two. eye crossing is the dominant trait and the inability to cross one's eyes is the recessive trait. (this is a completely imaginary scenario.) a man marries a woman and they have a child together. the father is heterozygous dominant for the trait of eye-crossing, and the mother is homozygous recessive for the trait. what are the chances that their child will be born with the ability to cross his or her eyes?

+5

Answers (1)

Know the Answer?

Frances Clark · Answer 1

Let 'C' denote the dominant trait of eye-crossing and 'c' denote the recessive trait of not being able to cross the eyes. The genotype of the father is heterozygous dominant for the trait of eye-crossing. This is denoted by 'Cc'. The mother is homozygous recessive for the trait of eye-crossing, denoted by 'cc'. The mating between the two will result in following genotypes: two of 'Cc' and two of 'cc'. Therefore the probability that the child will be able to cross his or her eyes is 0.5 or 50%.