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Name LS7B Week 3 Lab Worksheet You are a genetic counselor and a couple has come to you for advice. They have a family history of cancer and are concerned that their unborn third child will also get cancer. They provide you with the following pedigrees for two different traits relating to defects in the enzymes separase and topoisomerase: A . Separase defect B. Topoisomerase 1 2) 3) 4 defect — [ These two pedigrees represent the same family. Genetic testing shows that individual 4 has only nonmutant alleles of both genes and individual 12 has only mutant alleles of both genes. Individuals 6, 8, 9, 12, and 14 have cancer. The couple just recently learned that their daughter (individual 14) has cancer and has both mutations. In this lab activity, you will use genetic hypothesis testing to determine the probability that the couple’s third unborn child will also inherit both mutations and be at risk for developing cancer. Part 1. Now that you’ve had some practice with using Punnett Squares in the week 2 lab and class, let’s examine the pedigrees presented above a bit further. 1. Let’s start by figuring out what kind of trait is caused by the separase mutation. Test each of the four possible hypoth (X-linked domi X-linked ive, | domi , autosomal recessive) and determine which hypothesis cannot be rejected. Write your final solution with a brief explanation. X-linked dominant is rejected because it is not possible for person 7 to be not affected X-linked recessive cannot be rejected because all females are carrier Autosomal dominant is rejected because it's not possible for both non-affected persons 3 and 4 to produce a child that is affected Autosomal recessive is rejected because person 4 has only non-mutant alleles of both which is RR; therefore, persons 3 and 4 cannot produce a child that is affected (rr) 2. Having determined which pattern of inheritance cannot be rejected, what is the chance that individuals 11 and 12 would have a child affected by the separase defect? XAr e 50% of child would be affected XAR XARXAr XARY MAAAAAAS XA XMXAr o XArY 3. Now let’s figure out what kind of trait is caused by the topoisomerase mutation. Test each of the four possible hypotheses and determine which hypothesis cannot be rejected. Write your final solution with a brief explanation. . X-linked dominant cannot be rejected because person 11 is X*rX”r and person 12 XARY, it works for the pedigree tree . X-linked recessive is rejected because it is not possible for person 8 to be affected since her mother is X*rX*r and father is X"RY . Autosomal dominant is rejected because person 12 has only mutant alleles of both which is RR; therefore, it is not possible for persons 11 and 12 to produce a child that is not affected . Autosomal recessive is rejected because person 8 and 9 are both affected which is rr, therefore, they cannot produce a child is not affected (person 13) 4, Having determined which pattern of inheritance cannot be rejected, what is the chance that individuals 11 and 12 would have a child affected by the topoisomerase defect? XOR e 50% of child would be affected, but XA XARXAr - XArY 100% of female would be affected. 0% XA XARXAr XATY of male would be affected. S. Based on your work so far, what is the overall probability that individuals 11 and 12 will have a child affected by both separase and topoisomerase defects? (72)(72)(1) + (2)(72)(0) = V4 Therefore, 25% of child would be affected

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