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  #1  
Old 07-25-2012
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Genetics 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%
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Old 07-25-2012
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A. 0%......
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Old 07-25-2012
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A. no relation
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Old 07-25-2012
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What am I missing answer could be 0 or 50% ? Depending upon the carrier state of woman .
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Old 07-25-2012
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can any1 explain in a bit detailed manner...:sorry::sorry:
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Old 07-25-2012
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Quote:
Originally Posted by Kabutar111 View Post
What am I missing answer could be 0 or 50% ? Depending upon the carrier state of woman .
if a woman is a carrier of recessive allele on chromosome 11 then the probability of albinism in a child would be 50%
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 disease-causing allele so the risk will be 0%.
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Old 07-26-2012
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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....
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Old 07-26-2012
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Quote:
Originally Posted by anomali View Post
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....
yeah, exactly that's why I answered A) 0%

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
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Old 07-26-2012
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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.
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Great post
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