Question

A colour blind man marries a woman with normal sight who has no history of colour blindness in her family. What is the probability of their grandson being colour blind ?

• A. Nil
• B. 0.25
• C. 0.5
• D. 1

Answer 1

The Correct Option is B

 To solve the question of determining the probability of a grandson being colour blind, we first need to understand the genetics of color blindness and the inheritance pattern involved.

Step 1: Understanding Genetics of Color Blindness

Color blindness is a sex-linked recessive trait located on the X chromosome. The following genotypes are relevant:

• X^N - Normal vision gene
• X^c - Color blindness gene

For males (XY), having one X^c will result in color blindness since they only have one X chromosome.

For females (XX), two X^c genes are needed for color blindness; hence, they are carriers if they possess one X^c and one X^N.

Step 2: Analyzing the Parents' Genotypes

• The man is color blind, so his genotype is X^cY.
• The woman has normal sight with no color blindness history, so her genotype is X^NX^N.

Step 3: Determining the Children's Genotypes

Now, we calculate the possible genotypes of their children:

• Male Children: Genotypes will be X^NY (normal vision) and they inherit X^N from the mother and Y from the father.
• Female Children: They can only be X^NX^c (carriers but not color blind), inheriting one X^c from the father and X^N from the mother.

Step 4: Considering the Next Generation (Grandchildren)

Assuming the carrier daughter (X^NX^c) marries a man with normal vision (X^NY), their children might be:

• Sons: 50% X^NY (normal sight), 50% X^cY (color blind)
• Daughters: 50% X^NX (normal), 50% X^NX^c (carrier)

Therefore, each male child has a 50% chance of being color blind.

Conclusion

The probability that the grandson is color blind, accounting for the possibility of the carrier daughter having a son (which is a 50% chance), is:

\(\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}\) or 0.25