In humans, hemophilia is an X-linked recessive disorder. If a carrier woman (X^H X^h) marr… — JAMB UTME

In humans, hemophilia is an X-linked recessive disorder. If a carrier woman ( $X^H X^h$ ) marries a hemophilic man ( $X^h Y$ ), what is the probability that any daughter born to this couple will be a carrier of hemophilia?

A
25%
50%Cevap
C
75%
D
100%

Cevap

50%

For female offspring (

XX

), the father always contributes an

X^h

chromosome. The mother contributes either an

X^H

chromosome (50% chance) or an

X^h

chromosome (50% chance). Thus, 50% of the daughters will have the heterozygous genotype

X^H X^h

, making them carriers.

Adım Adım Çözüm

Determine parental genotypes and gamete types

Mother is

X^H X^h

(gametes:

X^H, X^h

). Father is

X^h Y

(gametes:

X^h, Y

Identifying parental gametes is necessary to cross all potential allele combinations.

Determine genotypes of female offspring (

XX

)

Female offspring inherit

X^h

from the father and either

X^H

X^h

from the mother, resulting in

X^H X^h

(carrier female) and

X^h X^h

(affected female).

Female children always receive one

X

chromosome from each parent.

Calculate the probability specific to female offspring

Out of 2 possible female genotypes (

X^H X^h

and

X^h X^h

), exactly 1 is a carrier (

X^H X^h

), giving a probability of

\frac{1}{2}

or 50%.

The question restricts the probability domain specifically to daughters.

Anahtar Kavram

Sex-linked inheritance and gender-specific probability calculations

Tahmini Süre:1m 30s

Soru

Cevap

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