Genetics

Phenylketonuria, a Metabolic disease in humans, is caused by a recessive allele, k. If two heterozygotes marry and plan a family of six children: a)what is the probability that all will be unaffected? b)what is the probability that three will be unaffected and three will be affected? c)what is the probability that at least two will be unaffected? d)what is the probability that the first child will be an unaffected girl?
I don't even know how to approach this problem. Much help is needed. Thanks

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  1. Prob each is affected= 1/2*1/2= 1/4

    prob all unaffected= (3/4)^6

    prob three affected= (1/4)^3 * (3/4)^3 * ways.

    now ways = 6!/3!3!

    at least two will be unaffected
    combination ways
    uuaaaa 6!/(2!4!)= 15
    uuuaaa 6!/(3!3!)=40
    uuuuaa 6!/(2!4!) =15
    uuuuua 6!/5!1! =6
    uuuuuu 6!/6! =1

    check those, I did them in my head
    Pr(at least two)= sum prob*combinations
    = 15((3/4)^2(1/4)^4)+40((3/4)^3(1/4)^3)+ ....

    pr first child ug?
    Pr= (3/4)*1/2

  2. I don't get b or c at alll. What are the u's and a's? What are the exclamation points?

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