mash
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A man who has Neurofibromatosis type 1 (autosomal dominant) marries a phenotypically normal woman. If they have five children, what is the probability that none of the children will be affected with this disorder? What is the probability that all of the children will be affected?
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| rida
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1/2 that none will be affected, and 0 for all being affected
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| uscan
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Is it .0315 for both cases?
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| mash
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ans is 1/32, 1/32
I think there is 50% chance of having d ds & 50% of not having it.
so, chances dat all 5 children have d ds r
(1/2)^5 =1/32
and chances of all children being normal r
(1/2)^5 = 1/32
Dont know whether this explanation is right or not.
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| rida
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oops, forgot about them having five children, i understand 50% of kids having it, but how is it 50% of all being normal?????
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| mash
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If u draw a punnet square with father being Aa(autosomal dominant) n mother being aa(homozygous normal), u get 50% normal (aa) n 50% Aa.
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| rida
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That is what i had done, but it asks, what is the probably that ALL the children are affected, shouldn't that be none cuz there will 50% affected and 50% not affected? Unless i am reading into this question wrong??
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| Idiopathic
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The probability that ALL FIVE will be affected is 0.5 x 0.5 x 0.5 x 0.5 x 0.5 (1/2 for each child, since they are isolated events). The same holds true for the likelihood of none of the children being affected, since it is 50/50 either way.
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| rida
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thanx idiopatihic!
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| uscan
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So my answer is right! 
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| Idiopathic
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Yes
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| mjl1717
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1) This is a good q because tests at least 2 or 3 steps.
2)Must know that an autosomal dominant and a normal phenotype
have a 50/50 chance of having afflicted or normal children.
( In A.D. the allele is expressed in a heterozygote) (5separate kids)
3) Now we must shift gears and use deductive reasoning (although some
are good Playing with numbers)
4)Probabilities are usually expressed as decimal fractions from 0 to 1
(absolute certainty)
5)* we use the multiplication rule from biostat- which states that the proba-
bility of two or more "statistically independent events' all occurring is
equal too the product of their individual probabilities.
6) So as Idiopathic says take that fraction 0.5 to the 5th power (the five
kids are considered independent events) :idea:
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