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#1




Simple Genetics; Chances of fetus being affected!
28 yr old pregnant lady in 1st trimester comes for her 1st Antenatal check up. Her mother suffered from Huntigtons chorea who died few months back. She has no symptoms suggestive of the same disease. Her previous two pregnancies were uneventful and she has a 4yr old boy and 11yr old girl both of whom are doing well. Her husband is healthy and no such family history is present on his side. She is now worried about her the 3rd child and wants to find out if it will be affecte in the future.
What are the chances of the 3rd child going to be affected with the disease in future? a. 20% b. 33% c. 50% d. 25% e. 22% ===================== Q may sound silly, like procrastinating about the unborn.. Lol.. Prepared it myself.. let me know if m wrong somewhere. 
#2




my answer
Huntington's disease is an autosomal dominant disease. ok, so we must assume here that the patient (mom) is a (so far asymptomatic) heterozygous (otherwise it wouldn't make much sense, would it?). If so, and her partner is healthy and is not a carrier of a Huntingtin gene, the risk for ANY of her kids to suffer from the disease is 50%.

#3




Add on:
Quote:
She could be homo or heterozygous.. We should take into consideration both situations.. Assume grand mother is heterozygous then ? 
The above post was thanked by:  
Casandra (05122012) 


#4




.
if she's a healthy homozygous (no Huntingtin allele present in her genome) than the risk of the disease for her baby is 0%. that's why I wrote in my post before that to me only assumption of her being a heterozygous makes sense.

#5




.
Quote:
if a grandmother is a homozygous (HH) the probability of the mother being affected (Hh) is 1, and the probability of the child getting H allele from the mother is 1/2, so the risk for the child to be affected is 1 x 1/2 = 50%. if a grandmother is a heterozygous (Hh) then the probability of the mother being affected (Hh as well) is 1/2 and the probability of her child getting the H allele (=being Hh) is 1/2. So the risk for the child to be affected in this case is 1/2 x 1/2 = 1/4. So if we take into consideration those two possibilities the overall risk for this child will be: 1/2 + 1/4 = 3/4 or 75%. But I can't see it amongst the given possible answers... 
#6




Quote:
Ya, you are right.. I realised it late.. I wanted to edit it but ran out of time. Was sure someone would come up with the correct ans anyways .. 
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