Glucose-6-phosphate dehydrogenase (G6PD) deficiency is an X-linked recessive metabolic con… — JAMB UTME

Glucose-6-phosphate dehydrogenase (G6PD) deficiency is an X-linked recessive metabolic condition in humans. If a man with normal enzyme activity ( $X^G Y$ ) marries a heterozygous carrier woman ( $X^G X^g$ ), what is the probability that any child born to them will be an affected male?

25%Answer
B
50%
C
75%
D
100%

Answer

The probability that any given child from this marriage will be an affected male is 25%.

The mother passes either

X^G

X^g

with equal probability (

0.5

). The father passes either

X^G

Y

with equal probability (

0.5

). An affected male must inherit the defective

X^g

allele from the mother (

p = 0.5

) and the

Y

chromosome from the father (

p = 0.5

). Multiplying these independent probabilities yields

0.5 \times 0.5 = 0.25

, or 25%.

Step-by-Step Solution

Identify parental genotypes and gametes

Father (

X^G Y

) produces sperm

X^G

and

Y

. Mother (

X^G X^g

) produces eggs

X^G

and

X^g

Determining gamete types is required to construct the genetic cross.

Construct the Punnett square to find offspring genotypes

Possible genotypes:

X^G X^G

(25% normal female),

X^G X^g

(25% carrier female),

X^G Y

(25% normal male),

X^g Y

(25% affected male).

Combining parental gametes shows all possible genetic combinations for their children.

Calculate the specific probability for an affected male among all children

The target genotype

X^g Y

occupies 1 out of 4 total squares =

1/4 = 25\%

The question asks for the probability relative to any child born, not restricted to sons only.

Key Concept

X-linked Recessive Inheritance Probability

Estimated Time:1m 30s

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