Respuesta :
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%