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?

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Ashly Harmon

Biology

Today, 14:38

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?

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Everett Stephens · 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%.