A woman with normal vision whose father was color-blind marries a man with normal vision. … — JAMB UTME

A woman with normal vision whose father was color-blind marries a man with normal vision. What is the probability that any male child born to this couple will be color-blind?

A
$25\%$
$50\%$ Answer
C
$0\%$
D
$100\%$

Answer

The probability that any male child born to this couple will be color-blind is

50\%

The mother is a carrier (

X^C X^c

) because she inherited the color-blindness allele (

X^c

) from her affected father. When crossed with a normal male (

X^C Y

), sons inherit their Y chromosome from the father and one of the mother's two X chromosomes. Therefore, there is a

50\%

chance that a son receives the

X^c

allele and manifests color blindness.

Step-by-Step Solution

Determine the parental genotypes.

The woman's father was color-blind (

X^c Y

), so she inherited the

X^c

allele. Being phenotypically normal, her genotype is

X^C X^c

(heterozygous carrier). The man has normal vision, so his genotype is

X^C Y

Red-green color blindness is an X-linked recessive trait.

Construct a Punnett square for the cross

X^C X^c \times X^C Y

The possible offspring genotypes are

X^C X^C

(normal female),

X^C X^c

(carrier female),

X^C Y

(normal male), and

X^c Y

(color-blind male).

Each parent contributes one sex chromosome to offspring.

Calculate the probability specific to male offspring.

The male offspring genotypes are

X^C Y

and

X^c Y

. Out of these

2

possible male outcomes,

1

is color-blind (

X^c Y

). The probability among male children is

\frac{1}{2} = 50\%

The question asks specifically for the probability among sons, requiring evaluation within the male subset only.

Key Concept

Sex-Linked Inheritance and X-Linked Recessive Traits

Estimated Time:1m 30s

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