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Kipp_miniQ - recessive and extremely rare Frank’s father...

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Name: SID: GENETICS REVIEW 1) “Jazz Hands” is an extremely rare autosomal recessive disorder that makes one an incredible dancer. Frank and Jamie know they have a family history, and want to capitalize on this fact by creating a superstar child. Frank’s brother is affected, as is Jamie’s grandmother. What is the likelihood of their child having the “Jazz hands” phenotype? Of being a carrier? Having jazzhands: p(frank is Jj)*p(Jamie is Jj)*p(offspring is jj) (2/3)*(1/2)*(1/4)= 1/12 Carrier: p(frank is Jj)*p(transmission of j) + p(Jamie is Jj)*p(transmission of j) – p(both happen) =(2/3)*(1/2) + (1/2)*(1/2) – 1/12 (from above) = 6/12=1/2 2) Frank and Jamie also have a family history of “Crazy Legs”, another extremely rare, extremely profitable condition. Crazy legs is known to be sex linked
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Unformatted text preview: recessive, and extremely rare. Frank’s father and Jamie’s maternal grandfather were both affected. What is the likelihood of a crazy legged child? Of a crazy legs carrier? You know frank is not a carrier. Having crazylegs: p(frank is XY)*p(Jamie is XX * )*p(no wt X chromosomes in offspring) = (1)*(1/2)*(1/4) = 1/8 Carrier p(frank is XY)*p(Jamie is XX*)*p(offspring is XX*) = (1)*(1/2)*(1/4) = 1/8 Another way to look at it is that there are two possible 2x2 crosses, each with a 50% likelihood: XY x XX and XY x XX* Do both 2x2 crosses to see that only 1 of the 8 cells has a carrier, and only 1 cell has crazylegs. Bonus: What is the likelihood of a child with “Michael Jackson Syndrome” (both crazy legs and jazz hands)? p(crazylegs)*p(jazzhands)=(1/8)*(1/12)= 1/96...
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