Hemophilia is an X-linked recessive trait in humans. Huntington’s Disease is inherited with an autosomal dominant allele.

a. Mr. Y is unaffected by either condition. He marries Ms. X, who is unaffected by hemophilia but shows signs of Huntington’s Disease. Ms. X’s father has hemophilia but is unaffected by Huntington’s Disease. What is the probability that the first child born to Ms. X and Mr. Y will be a son with hemophilia and their second child will be a daughter with Huntington’s Disease and their third child will be a son who is unaffected by either condition?

b. For Ms. X and Mr. Y, what is the probability that the first child will be either a son with Huntington’s Disease but without hemophilia or a daughter that is unaffected by Huntington’s Disease but carries the allele for hemophilia?

c. Mr. U has hemophilia and Huntington’s Disease. Mr. U’s father is unaffected by either condition. Mr. U marries Ms. V, who is unaffected by either condition. Ms. V’s father has hemophilia. Out of three children, what is the probability that two will be girls with hemophilia and Huntington’s Disease?

d. What is the probability that Mr. U and Ms. V will have a son with hemophilia or a daughter with Huntington’s Disease as their first child?

e. What is the probability that Mr. U and Ms. V will have a son with hemophilia and Huntington’s Disease or a daughter with neither condition as their second child?

I have hard time setting up and understanding the problem. Here's what I did:

Huntington(Hnt) is autosomal dominant= H Hemophilia (Hem) is x linked recessive= h

since we're given that Miss X has Hnt then her genotype is H_ (underscore for an unknown 2nd allele). We're given that her dad suffers from Hem thus his genotype must be h(X^h,Y no allele on Y chromosome). Since, Miss X has Hnt (auto dom. does not skip generation usually) then her mom also had one thus her geno is H_. Thus her parental geno(P1) is H_ x h, by carrying out this punnet square we would get 0.25% chance for Hh female, h_ female,H male and a healthy male. Thus the probability of Miss X to be sick is 1/4

Now moving onto the cross between Mr Y and Mrs X

we have --(healthy male) x H_

thus doing the cross we get 0.25 chance to get H_ female,-- female,H male,-- male

thus there is basically 0% chance to get a hem son? this is where i get lost and my logic fails because the mother of Miss X could be Hh(have hemo recessive allele) but in that case:

--(healthy MR Y) x Hh(hetezygote Mrs X)

0.5% to get h-( X^h X) female and 0.5% to get h male(X^h Y). THus if the P to get healthy male is 0% the whole thing becomes zero??

How do I break down the problem more feasibly? as i find myself using the dirty math tricks to tailor my answer to the right one without understanding on the mechanics

correct answer is 1/128 which is (1/4)^3 x 1/2 but i just dont get where the 1/2 and third 1/4 came from (the possiblity to have a healthy male is 0% remember?)

I would appreciate if you could break it down to me

• $0.50\%$ is $\frac{5}{1000}$. Fifty percent is either $50\%$ or $0.5$. – Graham Kemp Sep 15 '15 at 2:06

Hemophilia is an X-linked recessive trait in humans. Huntington’s Disease is inherited with an autosomal dominant allele.

Let 'a' be the recessive allele for Hemophilia, 'A' the corresponding dominant allele. It is X-chromosome linked recessive gene, thus males are haploid for this gene.

Let 'B' be the dominant allele for Huntington's, and 'b' the corresponding recessive allele. It is autosomal dominant.

(Note: using different capitalisations of the same letter for the two diseases seems to have muddled your logic.)

a. Mr. Y is unaffected by either condition. He marries Ms. X, who is unaffected by hemophilia but shows signs of Huntington’s Disease. Ms. X’s father has hemophilia but is unaffected by Huntington’s Disease. What is the probability that the first child born to Ms. X and Mr. Y will be a son with hemophilia and their second child will be a daughter with Huntington’s Disease and their third child will be a son who is unaffected by either condition?

First a little deductive work.

Mister Y is 'A bb' - neither a dominant Huntington's gene, nor a recessive haemophilia gene.

Father X is genotype 'a bb' - no dominant Huntington's gene, but a recessive haemophilia on the X-chromosome.

Miz X is thus 'Aa Bb' - since she must receive the recessives from her father, and the dominants from her mother, thus showing Huntington's and not being a haemophiliac.

So the crossings are: $\rm 'A'\times'Aa' = \{\underbrace{'AA','Aa'}_{\text{female}},\underbrace{'A','a'}_{\text{male}}\}$ and $\rm 'bb'\times 'Bb' = \{'Bb', 'bb'\}$.

That is their children have a $50\%$ probability of showing Huntington's, their daughters will not be haemophiliac but have a $50\%$ probability of being carriers, but their sons have a $50\%$ probability of being haemophiliacs.

Notes:

You assumed that Father X and Mr. Y both had to be either homozygous recessive or heterozygous yet not showing with equal probability, but this does not follow.

Because we are given no measure of how likely someone with the gene is to develop the disease (or not), nor how prevalent the disease is in the general population, we need to assume Complete Dominance.   This is that that 'shows no sign of Huntington's', or 'unaffected' means 'does not have a dominant gene for it'.

Otherwise we can't make any meaningful assessment on the probabilities of Mr. Y or Father X having a dominant gene for the trait, let alone Mother X.

• Thanks for your help! But if we cross A x Aa we will have AA and Aa meaning no children get hemo, did you mean AAxAa? because then you would get the results you described – Jx1 Sep 15 '15 at 15:22
• $\rm A\times Aa$ is a male (XY) with one dominant on the X-chromosome, crossed with a heterozygous female (XX). Daughters will inherit the male's Dominant Allele and one of the females two; sons will only inherit one of their mother's X-linked Alelles. – Graham Kemp Sep 16 '15 at 1:11