In humans, red-green color blindness is an X-linked recessive trait (X^c), whereas normal … — JAMB UTME

In humans, red-green color blindness is an X-linked recessive trait ( $X^c$ ), whereas normal vision is controlled by the dominant allele ( $X^C$ ). Albinism is an autosomal recessive disorder ( $a$ ), whereas normal skin pigmentation is controlled by the dominant allele ( $A$ ). A woman with normal vision and normal skin pigmentation, whose father was both color-blind and albino, marries a man with normal vision who is a carrier for albinism. If this couple produces a male child (son), what is the percentage probability that the son will be both color-blind and albino?

Answer: 12.5 %

Answer

The percentage probability that a son born to this couple will be both color-blind and albino is 12.5%.

The mother's father was albino (

aa

) and color-blind (

X^c Y

), meaning she inherited

a

and

X^c

from him. Given her normal phenotype, her genotype is

Aa X^C X^c

. The father is

Aa X^C Y

. When determining traits for a son, the son receives the

Y

chromosome from the father, so his vision phenotype depends entirely on which

X

chromosome he receives from his mother (50% chance of

X^c

). The probability of being albino from two carrier parents (

Aa \times Aa

) is 25% (

\frac{1}{4}

). Multiplying these independent probabilities yields

\frac{1}{2} \times \frac{1}{4} = \frac{1}{8}

, which equals 12.5%.

Step-by-Step Solution

Determine parental genotypes from the pedigree information provided.

Mother's genotype:

Aa X^C X^c

; Father's genotype:

Aa X^C Y

The mother received recessive alleles

a

and

X^c

from her affected father (

aa X^c Y

). The father is stated to have normal vision (

X^C Y

) and to be a carrier for albinism (

Aa

Calculate the probability of the male child inheriting the X-linked color blindness trait.

Probability of color-blind son =

\frac{1}{2}

(50%).

For male offspring, sex is fixed by inheriting the

Y

chromosome from the father. The mother has a 50% chance of passing her

X^c

allele.

Calculate the probability of the child inheriting autosomal albinism.

Probability of albino phenotype (

aa

) =

\frac{1}{4}

(25%).

Crossing two heterozygous carriers (

Aa \times Aa

) yields a 1 in 4 chance of an autosomal recessive

aa

offspring.

Apply the product rule for independent genetic events.

Combined probability =

\frac{1}{2} \times \frac{1}{4} = \frac{1}{8} = 12.5\%

Autosomal inheritance and X-linked inheritance are independent genetic events, so their probabilities are multiplied.

Key Concept

Independent assortment of an autosomal recessive trait and an X-linked recessive trait in human pedigree analysis.

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