another probability problem

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another probability problem

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manu
Forum Newbie

Topics: 2
Posts: 3

01/29/06 - 08:41 PM

Problem: A couple are both heterozygous for the autosomal recessive allele for albinism. They have two children. What is the probability that both children will be phenotypically identical with regard to skin color?

My attempt to solve the problem:

Child1: p(albino) = ¼; p(normal) = ¾
Child2: p(albino) = ¼; p(normal) = ¾

P = ¼ ∙ ¼ = 1/16 = 0.0625 6.25% probability to be both albino

P = ¾ ∙ ¾ = 9/16 = 0.56 56% probability to be both phenotypically normal

My dilemma:

If I calculate the probability that the two are not identical, I obtain 18.75 (1/4 X 3/4) .... and still doesn't add up to 100% ... am I making sense? What is that I'm not considering?

Thanks,

Manu

anjushree
Forum Guru

Topics: 64
Posts: 386

02/04/06 - 10:58 AM

i think it will3/4=in75%of case that they will be phenotypically identical

robin082006
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02/21/06 - 10:04 PM

P = ¼ ∙ ¼ = 1/16 = 0.0625 6.25% probability to be both albino

P = ¾ ∙ ¾ = 9/16 = 0.56 56% probability to be both phenotypically normal

So the probability that both children will be phenotypically identical with regard to skin color is 56+6.25=62.25%

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manu
Forum Newbie

Topics: 2
Posts: 3

02/23/06 - 06:22 PM

Thank you.

robin082006 wrote:
P = ¼ ∙ ¼ = 1/16 = 0.0625 6.25% probability to be both albino

P = ¾ ∙ ¾ = 9/16 = 0.56 56% probability to be both phenotypically normal

So the probability that both children will be phenotypically identical with regard to skin color is 56+6.25=62.25%

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