Question

A woman's father has ornithine transcarbamylase deficiency (OTD), an X-linked recessive disorder producing mental deterioration if not properly treated. The woman's mother is homozygous for the wild-type allele.
Part A: What is the woman's genotype?
Express your answer using D to represent the dominant allele and d to represent the recessive allele.

Part B: If the woman has a son with a normal man, what is the chance the son will have OTD?

Part C: If the woman has a daughter with a man who does not have OTD, what is the chance the daughter will be a heterozygous carrier of OTD? What is the chance the daughter will have OTD?
A) 50% chance to be heterozygous carrier of OTD, 50% chance to have OTD.
B) 100% chance to be heterozygous carrier of OTD, 0% chance to have OTD.
C) 50% chance to be heterozygous carrier of OTD, 25% chance to have OTD.
D) 100% chance to be heterozygous carrier of OTD, 50% chance to have OTD.
E) 50% chance to be heterozygous carrier of OTD, 0% chance to have OTD.

Part D: Identify a male with whom the woman could produce a daughter with OTD.
A) dY
B) dd
C) Dd
D) DY

Part E) For the instance you identified in part (d), what proportion of daughters produced by the woman and the man are expected to have OTD?
A)75%
B)50%
C)100%
D)25%

Part F
For the instance you identified in part (d), what proportion of sons of the woman and the man are expected to have OTD?
A)100%
B)75%
C)50%
D)25%

Answer 1

Answer:

Part A - Dd

Part B - 50%

Part C - Option E

Part D - dY

Part E - 50%

Part F - 50%

Explanation:

Given,

Let the allele for dominant trait (wild type) be "D" and the allele for recessive ornithine transcarbamylase deficiency (OTD) be "d"

A X-linked recessive gene affects male offspring in a family tree though a female remains a carrier until it mates with a male with the a recessive diseased allele.

Part A

Father's genotype would be Dd. Hence it will pass d gene  (X diseased gene ) to her daughter.

Thus , Woman's genotype is Dd

Part B

Genotype of Normal man - DD

DD * Dd

DD, Dd, Dd, DD

One of the two males will have  ornithine transcarbamylase deficiency (OTD), an X-linked recessive disorder producing mental deterioration hence, the probability is 50%

Part C

Option E is correct

Only one of the female offspring will have ornithine transcarbamylase deficiency (OTD), an X-linked recessive disorder which will be expressed if other allele is also X linked recessive.

Hence, The chance of being a carrier will be one out of two female offspring i.e 50% and and chance of having OTD is 0%

Part D

dY

Part E

dY * Dd

DY, dY, dD, dd

one out of the two female is diseased. Hence, the probability is 50%

Part F

50%