c Reade orocsudhedu bbcswebdavpid36792 32dtconten d7666156 1

c Reade oro.csudh.edu bbcswebdav/pid-36792 32-dt-conten d-7666156 1/cou 2154 20604 /Midterm l-Extra9620credit.pdf pid-3679232-d I I Chegg Study I Guided Solutions and Study Help I Chegg.com dh.edu/bbcs. bd Assignments BIO340-01 2154 20604 22. Joan is phenotypically normal, but had a child with the autosomal recessive disease cystic fibrosis (CF from a previous marriage. Joan\'s father has hemophila A, a sex- linked recessive condition where the blood fails to clot properly. Joan now intends to marry Bill, who is also phenotypically normal but who has a sister, Jill, with CF. Bill\'s parents are phenotypically normal, and there is no history of hemophilia A in his family Assume that Joan and Bill do marry and have a child. What is the probability that this child will have CF, but will not have hemophilia A?

Solution

Hemophilia is a sex linked disease .

For the child to have hemophilia, both parents should have h allele(mutant allele responsible for hemophilia).

But since bills parent have no history of hemophilia Bill doesn\'t have h allele.

Hence the probability of Joan\'s and Bill\'s child to not have hemophilia is 1.

X^h

Y

X^H

X^hX^H

X^HY

X^H

X^HX^H

X^HY

Now for CF

Since Joan is normal but his child from previous marriage has CF then he must heterozygous .

Since Bill\'s sister has CF but she is normal, she can heterozygous or homozygous(without the disease carrying allele), both with equal probability.

Hence the probability of Bill being heterozygous is 0.5.

The child cannot have CF if either both parents are not heterozygous or one of the parents is homozygous carrying the disease allele.

Probability of both parents being heterozygous=1*0.5=0.5

If parents are heterozygous using Punnet\'s square

C

c

C

CC

Cc

c

Cc

cc

The probability of child having having CF disease will be 0.25.

Probability of child having CF =(Probability of Child having CF when parents are heterozygous)*(Probability of parents being heterozygous)=0.25*0.5=0.125

Probability of child having CF but not hemophilia=(Probability of child having CF)*(Probability of child not having hemophilia)=0.125*1=0.125

Hence the probability of child having CF but not hemophilia will be 0.125

	Xh	Y
XH	XhXH	XHY
XH	XHXH	XHY