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Q11. The pedigree below is for & rare (and mild) heredity disorder of the skin How is the disorder inherited (justify Your answer) Give the genotypes for as man...

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Q11. The pedigree below is for & rare (and mild) heredity disorder of the skin How is the disorder inherited (justify Your answer) Give the genotypes for as many individual as possible Consider the 4 unaffected children of parents III.4 and IILS. What is the probability of having 4 unaffected child?

Q11. The pedigree below is for & rare (and mild) heredity disorder of the skin How is the disorder inherited (justify Your answer) Give the genotypes for as many individual as possible Consider the 4 unaffected children of parents III.4 and IILS. What is the probability of having 4 unaffected child?



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The accompanying pedigree is for a rare, but relatively mild, hereditary disorder of the skin. 1 II II IV a. How is the disorder inherited? State reasons for your answer. b. Give genotypes for as many individuals in the pedigree as possible. (Invent your own defined allele symbols.) c. Consider the four unaffected children of parents III- 4 and III- $5 .$ In all four-child progenies from parents of these genotypes, what proportion is expected to contain all unaffected children?

This trait of cataracts is our decimal dominant? The way you can tell is look at the first generation in the first generation. Nearly, oh, kids are affected and their kids are affected in subsequent generations. This is different. If you look at X linked traits, we don't have to get into great detail here. Excellent traits are divided by biological sex. Excellent traits are usually unless they're dominant in males and occasionally in females, but you won't necessarily see them in the first generation. So use this information and eliminate some relevant, irrelevant choices like X linkage, because we can see that males and females are affected just about equally. And we know that trade is dominant because it's visible in the first generation. If the trait was not visible in the first generation, it would most likely be experience because it would most likely be because it's an honest, normal, recessive trait. Now that's a part. The part asks. What's the probability of of a couple person five in the pedigree having a child that's a girl with the disease? The easiest way to do this is we know that person Number five has cataracts and most likely she is a carrier for the recessive trait. So that's person five. She marries and has this child with an uninfected person so normal and so we can easily do a quick pennant square. Put person fives a Leal's on top of the windows, put the other parent on the side windows and then bring the illegals together and we can see that their Children have a 50% chance of having cataracts. So 0.5 to have cataracts. Well, that takes care of this part of the equation or the problem. But we need to take into account that there's only a 50% chance that they will have a girl the other 50% chance they will have a boy. And so the answer is 1/4. It is a 25% chance or 1/4 chance that person number five, who has a kid with a normal cap division. We'll have a girl that has the disease of cataracts

In this question, we're told that a man who has hemophilia has a daughter who is unaffected by hemophilia and then were asked a series of questions about the probability that this daughter will have Children who are affected by hemophilia. So first of all, we have to remember that hemophilia is a sex linked trait. Um, and it's carried on the X chromosome. So the father of this woman that we're going to be speaking of has the genotype of an affected X and a why and hemophilia is actually recessive. But because men on Lee have one X, um, if they haven't affected Alil on that x chromosome, then that even if it's a recessive alil, it will show up. The phenotype will be expressed. But, um, in this case, he has a daughter, and the daughter is unaffected. Well, we know then that she must have at least one normal alil for hemophilia because that is, um, that is the dominant trait. And because she receives one x from her father and the Onley X her father has to give is an affected X. She actually has one affected. Next. So she's a carrier for hemophilia. She's a carrier. Okay, so in the question, we're told that, um, this woman marries a man who is unaffected by hemophilia. So here's Here's this woman. And then she married a man who is unaffected by hemophilia. So he has the dominant trait, and he has one. And so we're asked the likelihood that their Children will be affected. So let's look to see through penn and square here, um, to see what the possible Gammy combinations are so going across the top, we have two dominant axis. So this would be a daughter who is unaffected by hemophilia and is not a carrier. We have a dominant X and A y. So this would be an unaffected male child. Then we have one daughter who would be unaffected but is a carrier for hemophilia and a possibility that there would be a son who is affected by hemophilia. So now we're asked, what's the likelihood that, uh, this couple will have a daughter? Um, that is affected is the The husband is the father. And then this is the mother. Okay? More asked. What's the likelihood that the two of these individuals will have a daughter who is affected who has hemophilia? What? Well, in order to be affected, a girl has tohave two X is that are both affected Leal's or that have illegal for hemophilia and in this case, neither daughter. Neither daughter here has, um, has this genotype where you'd have to have to Lille. That's because the man in this union, the father has a dominant alil here, right? So that dominant Alil, um, will be passed on to both of his daughters. And so the daughters, even if they receive the the recess of Alil from their mother, will still express a normal phenotype. So there is 0% chance that they haven't affected daughter 0% chance. All right, The next question is, what is the likelihood that they have a son who is affected? Uh huh. So this just doesn't happen. Okay, so what's the likelihood that they have a son who is affected? Well, we can see here that there's one son that is affected so we can say that one out of two. There's one out of two chance because you could have the phenotype where there's a dominant Lille. Or you could have the phenotype from the mother where there is a no effective Lille s Oh, there is a one and two chance that the sun will be affected. Okay. And then the next question is, uh what if they had four sons and what's the probability that they would all be affected? So the easiest way to do this is to use the law of Probability, which states that, um if you to look for, to find the probability of multiple simultaneous events or independent events, but that they would occur simultaneously. You can multiply, multiply the probability that each individual event occurs on its own. So if you were to have four sons, that it would be a one and two chance for all four sons, so each time you had a son would be a one in two chance that they have hemophilia, and we're wondering what happened. What what if this occurred all simultaneously? Well, what we can do is multiply these probabilities together, and that will tell us the probability that all of these events will occur simultaneously, and that is a one in 16 chance. So there's a one in 16 chance that if this couple had four sons, that all of them would be affected. So let's go ahead and circle our answers in green. So it's a 0% chance that they have a daughter who was affected because, um, because the father will always tough on his normal X chromosome, there is a one in 2% or one in two chance that the, um, that they will have a son who is affected because the mother will pass on either her dominant, um, ex or her recessive X. There's a one in two chance, and there is a one in 16 chance that this that if they had four sons that all of them will simultaneously will be affected by, um, hemophilia.

Hey guys, this question address the basic principles of mentally in generics. The question presents a hair grammar and went to know the chance of the couple territory and to reform having sick Children knows that requires informs that inheritance in question has a recessive trait. There is that sick people have a genotype. The 21 human has the disease. There is she has a genotype multitude. Does not have the disease. No. So necessary. He has a a leo as he has a sick Darryl to reform necessary. He gave up a recess city. Aaliyah to her. Mhm. Does the U. two men can only be heros signals. The crossing of couple chill generates the falling probabilities of offspring. Therefore the individual tree to can only be heros signals since he does not have the disease and the human trick for is a closing between trajectory and to reform. We have the following Jenna types. Therefore the chance of having six Children is on health or 50 that for today released a didn't see you later.

I rarely start. Questions are starting answers with the pedigree already drawn are much work shown, but this is an elaborate pedigree, and I didn't want to waste a lot of time setting it up. So what I've done is re copied the pedigree from the book. And then I have shown you what the probable gina types of different individuals are. Um, and you can look at that in your leisure that what we're interested here is A and B. And you noticed that when we get here to the parents of a baby, things start to get interesting. So, um, we're interested in trying to figure out what the probability is that A and B were carrying the PK You, Leo, we don't know that much about some of these individuals. So that as much as we'd like what? For example, um, this individual A's father, right? Had parents who almost certainly the father was big p Big P and the mother would have been headers, I guess. Big three little p. So the question is, what's the probability that the that this person is big p little pea? Well, he certainly got a big P from his father. We don't know if he got a big here, a little b from his mother. So there's a 50% chance that he's big, he little p All right, So, um, when we come to a, then we can ask the same question. What's the probability that this that this person is big p little p Now he had to have gotten a big P from his mother? People coming from the outside will have the disorder and so most of probability that he will have the little P. Alil. Well, there's a 50% chance of Father has it. And if the filer has that, there's a 50% chance he would pick it up. So the probability that a is big the little P is 1/4. So a. The probability is 1/4 that he is Big d little piece for B. It's a somewhat similar situation. Um, there's a 2/3 chance they're bees. Mother is being p little Pino. We know she's not little. Be little because she has a disorder. There's a 1/3 chance. She's just Hamas, I guess. Dominant. There's a 2/3 chance that he is, um, Header is I guess so. For B, we had asked the probability. What's the probability? Um, the she will inherit the little P Alil. Well, B is going to have to get the big P Leo from her father. That's this fellow right here. Uh, and so the probably she's gonna get the little P Leo. It's simply going to be 2/3 times 1/2 which is going to be, ah, to thank you or 1/3. So there's a 1/3 chance that be he's being p little pee and there's 1/4 chance that a is big feel guilty. All right, so we're coming close to where we want to be. So we're asked, What's the probability then, um, that they would have an offspring whose little P. Little p who has the disease. I think this is a letter B in the work. All right. So for for their offspring, for an imaginary offspring from these two people, I mean town, What's the probability? Well, it's the probability that a is headers. Vegas. I'm probability that be etcetera zegas and assuming both of they are our headers, I guess the probability that the individual would be homes, I guess recessive that is little Pete Little P is 1/4. So think of the on its square that would just have to little peas in one corner. And so that means that the probability of that they have their first child, probably that that child would be little P little Pierre How the disorder is 1 48th So it's pretty small. Were then asked, What's the probability? Ah, a second child, Um, have bean the same home as I guess. Recessive. What's the? Probably that travel would also be that was, I guess, recessive. Assuming that the first child is normal if the first child is normal, we haven't really learned anything. So if the first child is normal, it could be big p Big P. It could be big P. Little p. We have not learned anything about the Gina types of the parents. And so even if the first child was normal, the probably the second child will get disease is still going to be 1 48 again. The reason is the first child is normal. We don't know any more than we do know. We know that they that the parents, um, contain each of the parents at least one of the parents contains. Ah, it's headers, I guess. But we already knew that going in because they don't have the disorder, right? So that's the same thing. 1 48 for D. Um, we are told that if the first house has the disease, what's it? Chance to the second child will be unaffected. So if the first child has the disease, the first has the disease in. But it tells us that both parents carry the Leo. So both parents are big P little Pete. So if the first child is little p little p, that means the parents are both big P. Little P. That's important news, right? We've learned something valuable there. And so the probability the the next child would be normal, of course, which weaken? Check with opponents Square, Um, is, ah again going to be 3/4. And the reason is 3/4. Is that his ears? That childhood in here in both dominantly ills? A dominantly A from one parent or the diamond Elio from the other parent, Um And so the problem, though, is that the third this next child would be normal would be 3/4. So learning that the first the couple has a child with the disorder tells us a lot about the makeup of the parents, and that gets a lot more information to work with.


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