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Each of two parents has the genotype green divided by brown​, which consists of the pair of alleles that determine eye color​, and each parent contributes one of those alleles to a child. Assume that if the child has at least one green ​allele, that color will dominate and the​ child's eye color will be green. a. List the different possible outcomes. Assume that these outcomes are equally likely. b. What is the probability that a child of these parents will have the brown divided by brown ​genotype? c. What is the probability that the child will have green eye color​?

Respuesta :

david8644

Answer:

A)BB-Bg-Bg-GG

B)The probability is 25%

C)The probability is 75%

Explanation:

In order to solve this you just have to create a punnet square, where the parents are shown to both have brown and green coloured eyes genes, so it would be like this:

               Mother:      Brown                                 Green

Father↓  

Brown                      Brown-Brown                 Brown-Green

Green                      Brown-Green                  Green-Green

As the instructions say that if any kid had one green allele that would be the dominant, only the Brown-Brown will show the fenotype, so there´s 1/4 chance to happen, or 25%, the other probability of having green eyes is 3/4 or 75%

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