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