Albinism in humans is controlled by a recessive gene (c).Suppose two heterozygou

Question

Albinism in humans is controlled by a recessive gene (c).Suppose two heterozygous individuals (genotype Cc) marry. Assumethey have six children.

A. What is the probability that at least one child will bealbino?

B. What is the probability that three will be albino andthree will have pigmentation?

C. A challenge: What is the probability that three of thechildren will be boys AND all three boys will be albino? Hintassume gender is independent of whether or not a child turns out tobe albino. Not too tricky if you think carefully!

Explanation / Answer

Ok first off when doing probability you must determine the odds ofeach event separately then multiply them together. for examples what is the proability that I can flip two coins andget head both time. the probability of getting heads on a coin is1/2. so 1/2*1/2= 1/4 when you have heterozygous parents they can produce up to fourpossible outcomes: CC, Cc, Cc, cc(this is the only one who isalbino) so 1/4 for an albino kid A) here only one kid has to be albino so the rest technically dontmatter 1/4*4/4*4/4*4/4*4/4*4/4=1/4=.25=25% B) here they tell us that 3 will be albino(1/4) and that 3 will benormal(3/4) (1/4*1/4*1/4*3/4*3/4*3/4)*100%= 0.659179688% C)here three albino boys gender probability boys(1/2), albino probability (1/4) (1/4*1/4*1/4*1/2*1/2*1/2)*100%=0.1953125 %

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