A woman who is a carrier for red-green colour blindness (X^C X^c) marries a man with norma… — JAMB UTME

A woman who is a carrier for red-green colour blindness ( $X^C X^c$ ) marries a man with normal colour vision ( $X^C Y$ ). What is the probability that any son born to this couple will be colour-blind?

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

Answer

The probability that any son born to this couple will be colour-blind is

50\%

A carrier mother has the genotype

X^C X^c

, meaning half of her eggs carry the normal allele (

X^C

) and half carry the recessive colour-blindness allele (

X^c

). All sons inherit a Y chromosome from their father and an X chromosome from their mother. Therefore, each son has a

50\%

chance of inheriting the

X^c

chromosome and being colour-blind (

X^c Y

Step-by-Step Solution

Determine parental genotypes and gametes

Mother (

X^C X^c

) produces gametes

X^C

and

X^c

in equal proportions (

50\%

each). Father (

X^C Y

) produces gametes

X^C

and

Y

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

Determine genotypes of male offspring

Sons inherit the Y chromosome from their father and an X chromosome from their mother. Possible male genotypes are

X^C Y

(normal vision) and

X^c Y

(colour-blind).

Male offspring inherit their sex-defining Y chromosome strictly from the father.

Calculate the probability among sons

Out of 2 possible male genotypes (

X^C Y

and

X^c Y

), 1 represents a colour-blind son, giving a probability of

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

The question specifically asks for the probability among sons, not total offspring.

Key Concept

X-linked recessive inheritance and gender-restricted offspring probabilities

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

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