iedu/webapos/lassessmenttake/launch.isp?course.assessment Jd 4 D which votsthese
ID: 190012 • Letter: I
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iedu/webapos/lassessmenttake/launch.isp?course.assessment Jd 4 D which votsthese+ist. D Web Sice Galery Aople YouTube Googe aps w wikipeda d from English to EnglishShow originai courseJd)02216,18conten_id",5766460.1&sten..; News lmorted From Safari Options w QUESTION 2 5 points Saved --O 1-2 5.0i Cystic fibrosis is an autosomal recessive genetic disorder that is inherited in a simple Mendelian manner (C - normal allele, c cystic fibrosis allele). The pedigree above shows a family containing individual's with cystic fibrosis. Note that affected individuals are shown are shaded, but carriers are not half-shaded Assume that individuals Iill-1 and II-2 decide to have more children. Please answer the following questions based on these individuals: 1.What is the genotype of IlI-1? 2. What is the genotype of II-2? 3. Which rule would you used to calculate the probablity that ll-1's and Iili-2's next child will be a boy AND have cystic fibrosis? 4. Calculate the probability that IlI-1's and Ill-2's next child will be a boy AND have cystic fibrosis 5, using the binomial expansion equation, calculate the probability that 2 out of the next 3 children for and II-2 will not have cystic fibrosis. To receive full redit, please indicate the answer (P) as well as x n-x P. and q. Binomial expansion equation For the toolbar, press ALT F10 (PQ or ALT-FN F10 (Mac). 1-CC words:3Explanation / Answer
1) Genotype of III-1 is Cc because one parent is affected having cc genotype and other parent is Cc. Other parent is heterozygous because his child is affected, so he has to have the affected allele.
2) Genotype of III-2 is cc because it is autosomal recessive disorder.
3) The rule of multiplication will be used to determine whether the next child is not and have cystic fibrosis.
4) Probablitiy of being boy is 1/2 and probability of being affected is also 1/2. So, the total probability is 1/2*1/2=1/4.
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