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A man who has alkaptonuria marries a woman who has hereditary sucrose intolerance. Both are autosomal recessive diseases and both map to 3q with a distance of 10 centimorgan separating the two loci. What is the chance they will have a child with alkaptonuria and sucrose intolerance?
A. 0%
B. 12.5%
C. 25%
D. 50%
E. 100%

thank you


hum, my calculations point to 12,5 % ...but hm, not my special field, indeed.


If I'm correct the chance of having a child with both disorders is 0%.

So the father has alkaptonuria. Both of his chromosomes are mutated (he has the disease), so all he can transfer through the germline for this particulalr locus is mutated alkaptonuria genes (both are mutated although we don't know if it's identical mutation).

Same thing with the mother. She has sucrose intolerance, a recessive disease too. It means that both of her chromosome 3 (one inherited from her mom, another from her dad) are mutated, so all her ova will contain a mutated version of the sucrose intolerance gene.

To simplify matters, we will assume that neither of the parents are carriers of the others' disease (in other words, they carry the wildtype gene for the disease they don't suffer).

So any offspring from this couple, will inherit one alkaptonuria gene (coming from the diseased father) and one sucrose intolerance gene (coming from the diseased mother). So all of the offspring will have one copy of each gene but they'll also have one copy of the wildtype gene. So the chance the child will be a carrier of both mutated genes will be 100%, but because for a recessive disease you require having both copies of the same gene mutated, then the chance of getting a child with both diseases is zero.


Edited by xenopus on Jul 12, 2012 - 1:41 PM. : clarity


Graphically would be something like this:

A = wildtype normal alkaptonuria gene

a = mutated abnormal alkaptonuria gene

Su = wildtype normal Sucrose intolerance gene

  su = mutated abnormal Sucrose intolerance gene


Father: Chr 3q (he has 2 chromosomes #3).

a/Su + a/Su

Gametes (spermatozoa):  a/Su or a/Su


Mother: Chr 3q (she has 2 chromosomes #3)

A/su + A/su

Gametes (ova):  A/su or A/su


a/Su + A/su (one boy, let’s say)


a/Su + A/su (one girl, let’s say)


So for each locus, an offspring has one mutated gene and a good gene too (aA for alkaptonuria locus or Su/su for sucrose intolerance locus), so he/she is heterozygous for that locus. But in order to get the disease he/she needs both copies/locus to be abnormal. That’s not the case so the chance is zero.



Xenopus, thank you for your cordial and detailed answer. It's easily understood for me. Sorry for replying this late. But here is where I got lost from your answer. What if both the parents are also carriers of the other disease, in other words, their genotypes are a/Su + a/su and A/su + a/su , respectively? If one of their offsprings gets the copy of a/su from both its parents coincidently, it might become a patient of both diseases. Should we also consider such a circumstance?




Sorry, I hadn't seen the reply to my post.

@callmecrazyjoe, the scenario you depict in which the parents are carriers of their spouse's mutated gene is likely, but it's too complicated for a USMLE question. If that's the case, we would need to take into consideration the 10 centimorgans of distance between these two genes to accomodate for any recombination occurring in parent's gonads due to synapsis at the meiotic stage. Then, we would need to compute probabilities with diverging outcomes (in the event of recombination or rather linkage disequilibrium - no recombination). In the answers there are no options with diverging outcomes, so I believe we don't need to calculate such complicated outcome. Tough problem to explain, although it's a bit easier to imagine (but much easier to get mixed up and fail the answer).

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