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#1




Albinism Marriage and Genetic Probability
A 38 y man with tyrosinase negative oculocutanous albinism (AR disease, crs 11) marries a woman with tyrosinase positive oculocutanous albinism (AR disease, crs 15). What is the probability that their first child will be affected by albinism?
A. 0%; B. 25%; C. 50%; D. 75%; E. 100% 
#2




A. 0%......

#3




A. no relation
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#4




What am I missing answer could be 0 or 50% ? Depending upon the carrier state of woman .
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#5




can any1 explain in a bit detailed manner...:sorry::sorry:

#6




Quote:
if a man is a carrier of recessive allele on chromosome 15 then the probability of albinism in a child would be 50%. so summing it up it would give us 100%. I don't think this is the case here. I might be wrong though. I'll go with the assumption that none of them is a carrier of the other one's diseasecausing allele so the risk will be 0%. 
#7




if woman has AR disease on chr 15( two defective copies) and father has AR disease on chr 11( again two defective copies)the child will inherit one copy of chr 15 from mother (defective copy) and another copy of 15 from father (normal)
the child will also inherit one copy of chr 11 from mother(normal) and another copy from father(defective copy) so the child has one bad copy of 11 and another bad copy of 15, which means NO DISEASE....0% risk.... 
The above post was thanked by:  
Kabutar111 (07262012) 
#8




Quote:
There is of course a chance that in between the lines we should also consider what Kabutar111 wrote, that one of the parents is a carrier of the other mutated allele, but as I explained in th eprevious post I don't think this is the case here let's see what Teaona says 
#9




Casandra and anomali are right. The answer is A, O%, because the mutations causing tyrosinase positive albinism and tyrosinase negative albinism are at 2 different loci. That means that all of the offspring will be heterozygous carriers of both mutations but none will be affected.

#10




Great post

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