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A colourblind man marries a woman with b...

A colourblind man marries a woman with bormal sight who has no histroyh of colour blindess in her family. What is the probability of their grandson becoming blind

A

0.5

B

1

C

nil

D

0.25

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem of determining the probability of the grandson becoming colorblind, we will follow these steps: ### Step 1: Understand the Genetics of Color Blindness Color blindness is an X-linked recessive trait. This means that the gene responsible for color blindness is located on the X chromosome, and a male (XY) needs only one copy of the recessive allele to express the trait, while a female (XX) needs two copies. ### Step 2: Determine the Genotypes of the Parents - The colorblind man has the genotype **X^cY** (where X^c represents the X chromosome with the color blindness allele). - The woman with normal sight and no family history of color blindness has the genotype **XX** (both X chromosomes are normal). ### Step 3: Determine the Possible Genotypes of Their Children When these two individuals have children, we can set up a Punnett square to determine the possible genotypes of their offspring. - The man can pass on either **X^c** or **Y**. - The woman can pass on either **X**. The possible combinations are: - From the man: X^c (colorblind) or Y (normal male) - From the woman: X (normal) The potential offspring genotypes will be: 1. **X^cX** (female carrier of color blindness) 2. **XY** (normal male) ### Step 4: Determine the Probability of the Grandson Being Colorblind The grandson will be the child of one of the daughters (X^cX) and a normal male (XY). The possible combinations for the grandson would be: - If the daughter (X^cX) passes on X^c and the normal male passes on Y, the grandson will be **X^cY** (colorblind). - If the daughter passes on X and the normal male passes on Y, the grandson will be **XY** (normal). Thus, the probabilities are: - 50% chance of being colorblind (X^cY) - 50% chance of being normal (XY) ### Conclusion The probability of their grandson becoming colorblind is **50%**. ---
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