Duchenne muscular dystrophy is an X-linked recessive genetic condition in humans. If a fem… — JAMB UTME

Duchenne muscular dystrophy is an X-linked recessive genetic condition in humans. If a female carrier ( $X^D X^d$ ) marries a male affected by the condition ( $X^d Y$ ), what is the probability that their first child will be an affected female?

$25\%$ Answer
B
$50\%$
C
$75\%$
D
$0\%$

Answer

The probability that their first child will be an affected female is 25% (or 1/4).

The correct answer is 25%. A cross between a carrier mother (

X^D X^d

) and an affected father (

X^d Y

) yields four equally likely offspring genotypes:

X^D X^d

(carrier female),

X^d X^d

(affected female),

X^D Y

(unaffected male), and

X^d Y

(affected male). Since only 1 of the 4 possible outcomes represents an affected female (

X^d X^d

), the probability is 1 out of 4, or 25%.

Step-by-Step Solution

Determine the parental genotypes

Mother =

X^D X^d

(carrier female); Father =

X^d Y

(affected male).

Duchenne muscular dystrophy is an X-linked recessive disorder, so the father must carry the recessive allele on his single X chromosome.

Construct a Punnett square for the cross

Gametes: Mother produces

X^D

and

X^d

; Father produces

X^d

and

Y

.
Offspring Genotypes:

X^D X^d

(carrier female),

X^d X^d

(affected female),

X^D Y

(unaffected male),

X^d Y

(affected male).

Each of the 4 combinations is equally likely (25% chance each).

Identify the target phenotype among all possible offspring

The target genotype for an affected female is

X^d X^d

, which accounts for 1 out of 4 possible total outcomes.

Probability =

\frac{1}{4} = 25\%

Key Concept

X-linked recessive inheritance patterns and probability calculations across total offspring versus sex-specific subsets.

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