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Biology

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Feb 20, 2024

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Uploaded by SuperHumanCrowMaster93

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 hypotheses (X-linked dominant, X-linked recessive, autosomal dominant, 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)

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