In humans, hypertrichosis of the ear pinna is a Y-linked (holandric) trait, whereas red-gr… — JAMB UTME

In humans, hypertrichosis of the ear pinna is a Y-linked (holandric) trait, whereas red-green color blindness is an X-linked recessive disorder ( $X^c$ ). A man with hypertrichosis and normal vision marries a phenotypically normal woman whose father was color-blind. What is the probability that their first child will be a male displaying both hypertrichosis and red-green color blindness?

25%Cevap
B
50%
C
12.5%
D
6.25%

Cevap

25%

The correct answer is 25% because the mother is a carrier (

X^C X^c

) and the father carries hypertrichosis on his Y chromosome (

X^C Y^H

). For a child to be a male with both conditions, he must receive the

Y^H

chromosome from his father (probability 0.5) and the

X^c

allele from his mother (probability 0.5). Multiplying these independent events gives

0.5 \times 0.5 = 0.25

or 25%.

Adım Adım Çözüm

Determine parental genotypes

Father =

X^C Y^H

; Mother =

X^C X^c

The father has normal vision (

X^C

) and hypertrichosis (

Y^H

). The mother is phenotypically normal but inherited

X^c

from her color-blind father (

X^c Y

Determine offspring gamete combinations via a Punnett square

Female offspring:

X^C X^C

(25%),

X^C X^c

(25%); Male offspring:

X^C Y^H

(25%),

X^c Y^H

(25%)

Sons receive the

Y^H

chromosome from the father and either

X^C

X^c

from the mother.

Calculate the total probability for a male child with both traits

P(\text{Male with both traits}) = P(\text{Inheriting } Y^H) \times P(\text{Inheriting } X^c) = 0.50 \times 0.50 = 0.25 = 25\%

The question asks for the probability among all potential offspring, requiring the product of receiving the

Y^H

chromosome (50%) and the

X^c

allele (50%).

Anahtar Kavram

Simultaneous X-linked and Y-linked trait inheritance

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