Question

1. a woman whose father was color blind marries a man who is color blind. what are the chances that their offspring will be color blind? (note: the woman is not color blind.)

Explanation:

Step1: Determine Genotypes

Color blindness is X - linked recessive. Let \( X^c \) be the allele for color blindness, \( X \) for normal. The woman's father was color blind (\( X^cY \)), so she gets \( X^c \) from him. Since she's not color blind, her genotype is \( X X^c \). The man is color blind, so his genotype is \( X^cY \).

Step2: Set Up Punnett Square

The woman's gametes: \( X \) and \( X^c \). The man's gametes: \( X^c \) and \( Y \).

	\( X \)	\( X^c \)
\( Y \)	\( X Y \) (normal male)	\( X^c Y \) (color - blind male)

Step3: Calculate Probabilities

Total offspring genotypes: 4. Color - blind offspring: \( X^c X^c \) and \( X^c Y \), which is 2 out of 4. So the probability is \( \frac{2}{4}=\frac{1}{2} \) or 50%.

Answer:

The chances that their offspring will be color blind is 50% (or \(\frac{1}{2}\)).