Question

A woman who is a carrier for hemophilia marries a man who does not have hemophilia. What are the chances of them having a son or daughter with hemophilia? Show your work.

I need help with this short-answer/essay/written problem. I think the answer to this problem is 50% due to the fact that a child inherits EXACTLY 50% of the chromosomes from each of both parents, but I'm not 100% sure. Many thanks to whomever lends me a helping hand with this problem!

Answer 1

If only the mother is a carrier, then none of the children will show the trait, but all offspring would have a 50% chance of being a carrier.

Hh x HH

1/2 HH

1/2 Hh

If the father is a carrier, than there is a 50% chance of the children being a carrier and a 25% chance of the children expressing the trait.

Hh x Hh

1/4 HH

1/2 Hh

1/2 hh

The question doesn’t say whether or not the father is a carrier — it just says that he doesn’t express the trait.

Answer 2

"a man who does not have hemophilia." it says very clearly that he is not a carrier.

Answer 3

Oh wait never mind, I didn't catch the difference between being a carrier and actually expressing the trait/ not expressing it.

Answer 4

Thank you so much for your response, Anonymous! I really appreciate it!

Answer 5

If only the mother is a carrier, then none of the children will show the trait, but all offspring would have a 50% chance of being a carrier.

Hh x HH

1/2 HH

1/2 Hh

If the father is a carrier, than there is a 50% chance of the children being a carrier and a 25% chance of the children expressing the trait.

Hh x Hh

1/4 HH

1/2 Hh

1/4 hh <== typo correction

The question doesn’t say whether or not the father is a carrier — it just says that he doesn’t express the trait.

Answer 6

To answre this question, we need to know a few facts about hemophilia.

It is an X-linked recessive defect, which means that it is present only on the x-chromosome, and in females, which has 2 x-chromosomes, will require both copies to be defective to have the disorder. If only one x-chromosome is affected, she will be a carrier.

On the other hand, males have only one x-chromosome (the other one is a y-chromosome). Therefore males will either exhibit the disorder, or be just normal. Males cannot be carriers.

For the given problem, the mother is a carrier, and the father does not have the disorder.

Using
X=chromosome with disorder, and
x=normal,
we have the following Punnett square
x y
X Xx Xy

x xx xy

This means that there will be four outcomes, each with probability 25%:

Answer 7

continued:
(where Xx is the carrier mother, xy is the normal father)
Outcomes (each 25%)
Xx = daughter is a carrier
xx = daughter is free from hemophilia
Xy = son had hemophelia
xy = son is free from hemophelia

Answer 8

Yeah, I thought about answering it that way, but the author of the question did not state that it was a sex linked trait. Usually, it is stated with these types of questions.

Answer 9

I guess it is part of the exercise to test knowledge of the well-known sex-linked traits.

Answer 10

Probably so, but the question is a simple punett square question, and isn't something that is generally discussed in a general genetics course, but it has been a long time for me. It's usually discussed in a Human Genetics course, and the chromosome number or type of chromosome is given. But the question does say male or female offspring, so I suppose. I should have just went with my first thought and included punett square results for sex-linked traits, but didn't feel like typing that much work.

Answer 11

Regarding : "but didn't feel like typing that much work."
I consider a proper answer related to Punnett squares should indicate the genotypes and interpret them in terms of phenotypes, whether asked for explicitly or not. Again, I consider that as a part of the exercise.

For future reference, whenever a specific disorder is named, you would be expected to either know (as in a test) or look up and learn the characteristics of the disorder/disease. This way, we accumulate knowledge as we go.

Wish you great success in your course!

Answer 12

The kid would have a 25% chance of showing the trait. If its asking for carrier too, that would be a different story but for my explanation I will focus on the 25% expressing.

Letters in parenthesis are supposed to be in superscript. The mom is a carrier so her genotype would be:
X(H)X(h)
The dad is not a carrier so he would be:
X(H)Y
Now after putting those genotypes into a punnet square, you will get:
25% X(H)X(H) which is healthy female (non-carrier)
25% X(H)X(h) which is carrier female
25% X(H)Y which is healthy male
25% X(h)Y which is effected male w/disease

Now if the question was asking how much of a chance does the child have of being a carrier OR directly affected, then the answer would be 50% but only showing would be 25%