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Problem 66 Hard Difficulty

Refer back to the scenario of the previous exercise. In the figure below, the genotypes of both members of Generation I are known, as is the genotype of the male member of Generation II. We know that hamster II2 must be black-colored thanks to her father, but suppose that we don't know her genotype exactly (as indicated by $B$ - in the figure).
(a) What are the possible genotypes of hamster $\mathrm{II} 2,$ and what are the corresponding probabilities?
(b) If we observe that hamster III has a black coat (and hence at least one $B$ gene), what is the probability her genotype is $B b ?$
(c) If we later discover (through DNA testing on poor little hamster III) that her genotype in $B B,$ what is the posterior probability that her mom is also $B B ?$

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Video Transcript

okay for this problem, We have a little bit of set up. We need to dio. So we are talking about hamster genetics where a capital B is a dominant black gene. A lower case B is a recessive brown gene. So that means that if we have B B or capital, be lower B than the hamster is black. If you have lower, lower, be lower be than the hamsters brown. We have current family tree where we have Gen one has a big B Lobi, I'll say up B and Lobi actually, as 11 and up beat up be to they come together and they create gen two. Hmm. This ones labeled number two for some reason where we know that they have well, they have toe have a dominant black gene. We don't know what their other gene is. They mate with an up be Lobi. There's number one there and we are considered We consider the third generation that comes from that. So for the first problem, for the first part, we want to figure out what the possible Gina types for hamster to to so to to what the possible Gina types are so if we consider it they, Either way, they will get a dominant gene from number one to a dominant black gene. So really, it's up to whether they get a dominant black gene from number one or a process of brown gene from number one. So really, there's just a 50% chance it just depends on what they get from number one. So there's a 50% chance B B or UPI UPI a 50% chance of up Be low now for part B. Oh, if we observe. So we observe that 32 has a black coat, so they are be blank. We want to figure out what is the probability? Yeah, that the Gina type is, uh, up be Lobi. So what we can do here is we can work through this using a tree diagram. So we have our starting point. We have Where to, To is up the up be That has a 50% chance. Then it branches off where 31 is up. Be happy. That has a 50% chance mhm or 31 can have up below. B has a 50% chance. That comes from the fact that if this one's capital B, or if number 22 is be upbeat up be, then it's the exact same situation repeated. The probabilities are insane. So then we have. The other possibility. Is that to To is up the Lobi that has a 50% chance, which means that there's actually going to be three possibilities here. That 31 is, um, B A P that has a 25% chance. It should be pretty obvious why that's the case. We have total of four different possible combinations, one as getting a dominant from both. It will also be that one is getting recessive from both at 25% chance, and then there's a 50% chance of getting a dominant from both. So we know that hamster 31 has a black coat, which means that either this case, this case, this case or this case, the probability of each of these cases following the branches. So we have 0.5 times 0.5 at 0.25 Then for this guy, he 0.5 times 0.5 another 0.25 Here you have 0.5 times 0.25 0.125 Then lastly, we have for the, uh, up the Lobi case, we have 0.25 again. Uh huh. So the possibility that the if we consider everything that we have here the probability of a black coat. So label that as probability of e blank is equal to that 0.25 Or actually, we can approach this as the last note. Is that the probability of getting a brown for generation 30.125 So the total probability of getting a black coat going to be 0.25 plus 0.250 point 25 plus 0.125 uh, which is going to be 0.75 plus 0.125 Which is going to be, um yeah. One moment. Here. Yeah. 0.875 That's the probability that we get in black coat for a generation three. Then out of this, if you look what out of those possible cases, what are the chances that we have the feet or the genotype up be Lobi? So we'd have a 0.25% chance from this guy in this one, and we would have a 0.225% chance this one. So that means that we're going to have ability of up below be given be something. It's going to be 0.5 divided by 0.875 which is going to give us 0.57 Lastly, um, for part C. If we later discover that the genotype for hamster 31 is be, then we need to find out what is the possibility. So we need to find out what the possibility is that her mom was also BB, so that number 22 was also BB. So basically, it's probability that to to is capital B Capital B, given that 31 is capital B Capital B. So we could go through this with a more formal way. But I'm working through it with the tree. So if we consider the different possible ways that the mother is B B or ah, upbeat, upbeat, rather so we have out of all the different possibilities. There's the 0.25 chance that we have Mother BB and 31 is B B 0.25 is on the top for us there. And then we look at what are the different possible ways that we would end up with the upbeat, upbeat case. So it's either the mother's BB the child is our up PFP or the mother is up below V and the child is up. The up be So the total, um, possible ways that we could have the upbeat, upbeat case. Yeah, adds up to 0.25 plus 0.125 that it's going to be about zero 0.67 And if we think about it, makes a little bit of sense. Essentially, in that there's yeah, out of all the different end possibilities, there are three in which we have up the up be for number 31 and one for Yeah, there are three possibilities where we have uppy uppy for 31 and on Lee, one of them would come from when the mother is up below B. It was more likely that's ah if number 31 is, uh, PFP that the mother was up. Ian up. Be up. Be as well