A woman’s father has ornithine transcarvamylase deficiency (OTD), an X-linked re

Question

A woman’s father has ornithine transcarvamylase 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.

What is the woman’s genotype?

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

If the woman has a daughter with a man who does not have OTD, what is the chance the daughter will be heterozygous carrier of OTD? What is the chance the daughter will have OTD?

Identify a male with whom the woman could produce a daughter with OTD.

For the instance you just identified in part (d), what proportion of daughters produced by the woman and the man are expected to have OTD? What proportion of sons of the woman and the man are expected to have OTD?

Explanation / Answer

Let the allele for condition OTD be Xo

The genotype of father will be XoY and genotype of mother will be XX. Thus, the genotype of woman (offspring) will be  XoX.

If the woman has a son with a normal man then the cross will be between XoX and XY. Thus, the male progeny can have genotype XoY or XY. This means that there will be 50% chance that the son will have OTD.

If the woman has a daughter with a man who does not have OTD then the cross will be between XoX and XY.  Thus, the female progeny can have genotype XoX or XX. This means there is 50% chance the daughter will be heterozygous carrier of OTD. There is 0% chance that the daughter will have OTD.

Male with whom the woman can produce a daughter with OTD will have genotype XoY.

The cross will be between XoX and XoY. This means that daughters will have XoX or XoXo as genotype. Thus, 50% of daughters produced by the woman and man are expected to have OTD. From the cross the  sons will have XoY or XYas genotype. This means that 50% of sons produced by the woman and man are expected to have OTD.

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