Question

If a colour blind man marries a carrier woman for colour blindness, chances of their children to become colour blind is

• A. 100%
• B. 75%
• C. 50%
• D. 25%

Hint:

Read the question carefully: - "Chance of {sons} being colorblind" = 50% (1 out of 2 sons). - "Chance of {children} being colorblind" = 50% (2 out of 4 children). In this cross, both values happen to be the same, but they can differ.

Answer 1

The Correct Option is C

Step 1: Determine Genotypes:
Colour blindness is an X-linked recessive disorder.

• Colour blind man: $X^c Y$
• Carrier woman: $X^+ X^c$ (where $X^+$ is normal, $X^c$ is the colorblind allele)

Step 2: Perform the Cross:
Parents: $X^+ X^c$ (Female) $\times$ $X^c Y$ (Male) \begin{center} \begin{tabular}{|c|c|c|} \hline Gametes & $X^c$ (Male) & $Y$ (Male)
\hline $X^+$ (Female) & $X^+ X^c$ (Carrier Daughter) & $X^+ Y$ (Normal Son)
\hline $X^c$ (Female) & $\mathbf{X^c X^c}$ (Colour blind Daughter) & $\mathbf{X^c Y}$ (Colour blind Son)
\hline \end{tabular} \end{center}
Step 3: Calculate Probability:
Total possible children genotypes: 4 ($X^+X^c, X^+Y, X^cX^c, X^cY$).

• Colour blind children: $X^cX^c$ (Daughter) and $X^cY$ (Son).
• Number of colour blind children = 2.
• Total children = 4.
• Percentage = $\frac{2}{4} \times 100 = 50%$.

Step 4: Final Answer:
The chance of children being colour blind is 50%.