a woman whose father had hemophilia and whose mother was normal married a man whose father did not have hemophilia but whose mother did . if the couple's first child was a son what are the chances that he would show the trait of hemophilia?

Question

a woman whose father had hemophilia and whose mother was normal married a man whose father did not have hemophilia but whose mother did . if the couple's first child was a son what are the chances that he would show the trait of hemophilia?

Answer 1

recall: human chromosome is represented as XX for females and XY for males. for hemophilic people, h is assigned next to the X chromosome (note, hemophilia is carried or present only on the X chromosome).

Ok..the father of the woman (the mother of the boy) has hemophilia so has genotype; XhY

mother of the woman has no hemophilia so she is XX.

the grandparent from the father is not hemophilic and so XY, whereas the grandmother is hemophilic and so XhXh.

Ok, now use cross each grandparents and see if the couples are infected;

The mother's;

father x mother = XhY x XX = XhX, XhX, Xy and Xy. So the mother of the boy is infected because 100% of the females are infected (XhX) and so she has XhX genotype.

The father's;

father x mother = XY x XhXh = XhXh, XX, XhY, XhY. So yeb, the father is also infected with a genotype XhY.

Finally, we cross the couples together

father is XhY and mother is XhX;

XhY x XhX = XhXh, XhX (carrier), XhY and XY. So, the chances are;

2/4 infected female i.e. 50%
1/4 infected male i.e. 25%
1/4 no infected male i.e. 25%

so a son would have 25% chance of getting the trait.

...see if that helps you..

Answer 2

To determine the chances that the couple's first son would show the trait of hemophilia, we need to consider the inheritance pattern of hemophilia.

Hemophilia is an X-linked recessive disorder, meaning it is carried on the X chromosome. Males have one X and one Y chromosome, while females have two X chromosomes.

Based on the information given, we can create a Punnett square to analyze the possibilities:

Mother's genotype: XhX (Xh represents the hemophilia gene)
Father's genotype: XH Xh (XH represents the normal gene, Xh represents the hemophilia gene)

Using the Punnett square, we can determine the potential genotypes and phenotypes of their offspring:

XH Xh
------------------
Xh | XHXh | XhXh
------------------
Xh | XhXH | XhXh

From the Punnett square, we can see that there are four possible combinations of alleles for the child's genotype: XHXh, XhXh, XhXH, and XHXH. The first combination (XHXh) represents a carrier female who does not show symptoms of hemophilia. The second combination (XhXh) represents a son with hemophilia. The third and fourth combinations (XhXH and XHXH) represent carrier females like the first combination.

So, out of the four possible combinations, only one (XhXh) represents a son with hemophilia. Therefore, there is a 1 in 4 chance (25% chance) that the couple's first son would show the trait of hemophilia.