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30 Question (1 point)Istattempt Carla has Huntington $ disease (HD} which is genetic disorder caused by an Jutosomal dominant allele: Like most peopl wlth HD; Carla...

Question

30 Question (1 point)Istattempt Carla has Huntington $ disease (HD} which is genetic disorder caused by an Jutosomal dominant allele: Like most peopl wlth HD; Carla heterozygous. Her husband Ricardo doesnot have Huntington s discasc; What is tne chance they will have chlld with the disorder? Choose one; There is J 2598 chance B, Thereisa 509 chance Thereisa Oxchance There Isa 75%chancc

30 Question (1 point) Istattempt Carla has Huntington $ disease (HD} which is genetic disorder caused by an Jutosomal dominant allele: Like most peopl wlth HD; Carla heterozygous. Her husband Ricardo doesnot have Huntington s discasc; What is tne chance they will have chlld with the disorder? Choose one; There is J 2598 chance B, Thereisa 509 chance Thereisa Oxchance There Isa 75%chancc



Answers

More Genetics In Problem $33,$ we learned that for some diseases, such as sickle-cell anemia, an individual will get the disease only if he or she receives both recessive alleles. This is not always the case. For example, Huntington's disease only requires one dominant gene for an individual to contract the disease. Suppose that a husband and wife, who both have a dominant Huntington's disease allele $(S)$ and a normal recessive allele $(s)$ decide to have a child. (a) List the possible genotypes of their offspring. (b) What is the probability that the offspring will not have Huntington's disease? In other words, what is the probability that the offspring will have genotype $s s ?$ Interpret this probability. (c) What is the probability that the offspring will have Huntington's disease?

This problem is similar to problem 35 and that we are going to ask to find probabilities based on Gina types. We're going to use a pundit square to determine the genotype of the offspring. In part A. If the genotype of the parents are both capital as small S. Then the genotype of the offspring are capital S, capital S, capital S, small, S, capital S, Small S. And small S. Smallness. So the answer to a is the same as in number 35. Mhm. That is in part B. Since we want to know the probability of having to recess about wheels, we can see that only one out of four possibilities as small, as small as Therefore the probability is one divided by four or 25%. Yeah, part C. Is asking the probability of having at least one dominant lille And three out of four possibilities have at least one capital s. So we know that the probability is three divided by four Or 75%.

They're having four Children and probably that he shelled as a disease. 0.25 So then, for the probability that at least one child has it, it's one child Childrens. That disease has two years. It's gonna be equal. Teoh, when my eyes the probability that they have none, which is gonna be or zero times my success rate it's your 0.25 20 with power. I'm 0.75 to the fourth power, which that's gonna end up calling zero point 68 359 the B and the probability. At least three have the disease get the keys. And so then we'll have that equal to or choose three 0.25 cubed and your 0.75 or choose for 0.25 to the fourth. No end up equally your point zero five years of the nine

So the probability off getting a disease is point to fight. There is the probability off. Oh, hello Wicked is point 75 So therefore 30 wired for viability is one minus point 75 to the part off for which is 0.68 Do you fight on the require probability here is for support these four. C three into 0.25 to the power off three one replied with 0.75 plus 0.25 to the power off four, which is 0.0 fight 07

Asked about the genetic disease Huntington's sees where the probability that if both parents carrying the gene but do not have the actual disease to their carriers there is a 25 25% or 0.25 probability that the offspring will have it as well. And those newlywed House intends to have four Children were asked if I impart a probability that at least one child gets the disease so at least one child gets a di disease, which means we can subtract. It's, uh, at least one is equal to at least one. You probably have at least one. Getting is equal to one minus at most one minus, uh, the probably of No. One, no child getting it, no child positive. So the probability that there isn't gonna be no Children with Huntingdon disease is 0.75 you 75 to the fourth power and one minus their points in five to the fourth. Power gives a 0.68 0.68359 What is the probability that at least one child has the disease? Harpy were asked. What is the probability that at least three Children get the disease. So at least three Children means we're gonna add the probability of their being exactly three cases and four cases. So three cases exactly the Children game disease means 0.75 Is there a point 25 to the third power times 0.75 And because there could be many combinations of getting of three people testing positive between the four total Children, we're gonna multiplied by four factories of the total number of Children divided by and these exponents as factorial three factorial times, one factorial. And we can leave those Alford. And then we want to add the probability that at least three or the exactly four Children or all four Children are gonna have hunting this Huntington's disease. Which is there? A point to five race 1/5 power or fourth power? You therefore, Children, we employ dozens. Your calculator we get this answer is 0.508 078 This is the probability that at least three Children get Huntington's disease


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