Question

1. Color blindness is a recessive, X-linked trait. A normal sighted man marries a female

who is a carrier for colorblindness (but can see color fine herself. ) What are the chances

their female children will be colorblind? What about their male children?

Cross (parents):

х

Genotype ratio:

Phenotype ratio:

1. 2.

1.

2.

Answer 1

Step-by-step explanation:

When crossing sex-linked traits, we can use a Punnet square, with some modifications:

`X^nY*X^bX^n`

with n being normal and b being color-blindness. We get the following results:

`X^nX^b, X^nX^n, X^bY, X^nY`

Out of 2 possible male children, 1 will be colorblind, and out of 2 possible female children, 0 will be colorblind.

So there's a 1/2 chance the male children will be blind and a 0/2 chance the female children will be blind.

Genotype ratio - 1:1:1:1

Phenotype ratio - 3:1 (normal:colorblindness)