Question

A colourblind father has a daughter with normal vision and the daughter marries a man with normal vision. What is the probability of her son being a carrier for colourblindness? Explain with pedigree chart.

Hint:

Males cannot be carriers for X-linked traits; only females can. Sons are either normal or affected.

Answer 1

Step 1: Background.
Colourblindness is an X-linked recessive disorder. \[\begin{array}{rl} \bullet & \text{Females have two X chromosomes (XX).} \\ \bullet & \text{Males have one X and one Y chromosome (XY).} \\ \bullet & \text{Since males have only one X chromosome, if it carries the defective allele, they are colourblind.} \\ \bullet & \text{Females become carriers when they have one defective allele and one normal allele.} \\ \end{array}\]

Step 2: Cross explanation.
\[\begin{array}{rl} \bullet & \text{Father = X\textsuperscript{c}Y (colourblind).} \\ \bullet & \text{Mother (normal, not a carrier) = X\textsuperscript{N}X\textsuperscript{N}.} \\ \bullet & \text{Daughter inherits X\textsuperscript{c} from father and X\textsuperscript{N} from mother = X\textsuperscript{N}X\textsuperscript{c} (carrier, normal vision).} \\ \bullet & \text{Daughter marries a normal man (X\textsuperscript{N}Y).} \\ \end{array}\]

Step 3: Punnett square.
\[ \begin{array}{|c|c|c|} \hline & X^N & Y
\hline X^N & X^N X^N & X^N Y
\hline X^c & X^N X^c & X^c Y
\hline \end{array} \]

Step 4: Results.
\[\begin{array}{rl} \bullet & \text{Daughters: 50% normal (X\textsuperscript{N}X\textsuperscript{N}), 50% carriers (X\textsuperscript{N}X\textsuperscript{c}).} \\ \bullet & \text{Sons: 50% normal (X\textsuperscript{N}Y), 50% colourblind (X\textsuperscript{c}Y).} \\ \end{array}\]

Step 5: Answer.
Probability of her son being a carrier = 0, because males cannot be carriers (they have only one X chromosome). A son will either be normal (50%) or colourblind (50%).