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1. Three (3) recessive genes of Felis Cats and their inheritancepattern was observed in a lab. On the X chromosome, the (b) blackgene exists. The gene results in wi...

Question

1. Three (3) recessive genes of Felis Cats and their inheritancepattern was observed in a lab. On the X chromosome, the (b) blackgene exists. The gene results in wild type cats to have white fur,so the cats with mutation have black fur. The (r) round gene is onthe second chromosome. The gene results in wild type cats to havestraight paws, so the cats with mutation have round paws. The (s)short gene is on the third chromosome. The gene results in wildtype cats to have long tail, so the cats with

1. Three (3) recessive genes of Felis Cats and their inheritance pattern was observed in a lab. On the X chromosome, the (b) black gene exists. The gene results in wild type cats to have white fur, so the cats with mutation have black fur. The (r) round gene is on the second chromosome. The gene results in wild type cats to have straight paws, so the cats with mutation have round paws. The (s) short gene is on the third chromosome. The gene results in wild type cats to have long tail, so the cats with mutation have short tail. A cross was performed with two (2) parent cats: Parent (1): b/b+ ; r/r+ ; s/s Parent (2): b/Y ; r/r+ ; s/s+ a) What would be the appropriate phenotypes of these parents 1 & 2? b) Of the progenies from crossing parent 1 & 2, what percentage of male progenies would have the looks of complete wild type? c) Of the progenies from crossing parent 1 & 2, what percentage of the progenies would have the looks of neither parent 1 nor parent 2? d) Of the progenies from crossing parent 1 & 2, what percentage of progenies would be pure breeding for traits on the autosomes?



Answers

In cats, curled ears result from an allele $(C u)$ that is dominant over an allele $(c u)$ for normal ears. Black color results from an independently assorting allele $(G)$ that is dominant over an allele for gray $(g) .$ A gray cat homozygous for curled ears is mated with a homozygous black cat with normal ears. All the $\mathrm{F}_{1}$ cats are black and have curled ears. a. If two of the $\mathrm{F}_{1}$ cats mate, what phenotypes and proportions are expected in the $\mathrm{F}_{2} ?$ b. An $\mathrm{F}_{1}$ cat mates with a stray cat that is gray and possesses normal ears. What phenotypes and proportions of progeny are expected from this cross? PICTURE CANT COPY

So what we have here is across a die hybrid. Cross looking at two traits that are independently a sorting, they're not linked and um you have a Hamas august dominant cat that has the curled ears and the black fur with a Hamas I guess recessive cat that has normal ears and gray for. And so this is the parent generation. And so when we do that cross, we know for sure that all of those offspring are going to be hetero zegas, they're going to be um curly and black, but their genes are going to be heterosexuals, meaning they have one alil for the dominant trait and one alil for the recessive trait. Now the question asks us to then determine the genotype of the F. Two generation, who are at least the fino types. And so the F two generation is when we cross two of these together and what ends up happening is a phenotype ratio of nine 23 23 to 1. And you could work that out on a punnett square. Or you could just make your life easier and have that memorized. There are going to be nine of the offspring who exhibit both dominant traits. Nine of the offspring are going to be curled, have curled ears with black fur. Only one of the offspring is going to be gray with normal ears. And then the other two sets of groups. One will three of them will have curled ears with gray for or normal ears with black fur. And so um memorizing that phenotype ratio for this common um test cross is definitely going to help you in the next part of the question. It tells us that one of the F one individuals who's hetero zegas meets a stray cat in an alley that has the recessive traits of normal ears and gray for. And it says what are the expected outcomes of this? And so whenever a dyke hybrid cross happens where there's an Hatteras Egas individual, doubleheader, rosie angus crossed with a Hamas I guess recessive individual. The ratios are always going to be 1 to 1 to 1 to 1. There's going to be one individual that has those dominant um traits for both traits. So curled ears and black for there's going to be one individual that are the probability I guess not one individual but one probability um of this group of four that's going to have curled with the gray, one that will have normal ears with the black and one that will have um the normal ears with the gray. Now remember these? These are ratios right. And so each of these probabilities is one out of 41 out of 41 out of 41 out of four. If you were to do a die hybrid cross, that's because this is four out of 16. Four of these, um, four of the 16 gina types would code for the scene, a type of curled ears and black for.

So the question here gives us a scenario where we have an unusual cat shown is crossed with a cat with non curled ears here. So all of the kittens born from the thing this cross here had non curled ears. So later, when he's Austrian across with each other, the FEMA traffic ratio was three toe, one non curled thio curled ears. So one wants us to determine what conclusions could be made about the inheritance off, um, curled ears here. So assuming that the unusual cat waas homeless, obvious, dominant and the non curled ears is a homeless isis recessive trait, we can assume that the kittens here are going to be hetero. Zayed's XIII gets so for use are as non curled and lower case r as curled. Essentially, what we can see is that if we cross, our are are But we're going to get in this particular scenario is gonna be a mix of the two like that. So what weaken Therefore see is that as a result of this, are are here, which is gonna be curled in this particular case must therefore be a recess if traits. So let's get through the options that we have eso for eight. It says you are results of crossing over so of crossing over occurred then. Therefore, essentially, it's a very rare trait. So in this case, the ratio would not be three once. That cannot be a correct answer for me here. It states that it is a dominant trait. So if curled ears were dominant here, it would therefore make it such that you would see it in the higher ratio. So in this case is in smaller ratio. So that cannot be a for see here as we way we saw in the diagram that I showed it is a recessive traits. So that is gonna be correct. Answer. And for D here say it's that more crosses you need to be done to determine how the trait is inherited. So in this case, as I walked you through, there really isn't any need for more tests to be done as become essentially deduce what type of trade it's gonna be. So that's not a correct answer. So the correct answer here is gonna be see where its processes

Everybody, This is Ricky. Today we're walking through Problem 37 from chapter 12. We're told that calico cats can be black, yellow or calico, which, uh is a combination of both black and yellow and that also, the genes that determine this are located on the X chromosome. So what we can tell about this? These types of genes are there x linked and that they are co dominant? Because we see this, um, calico expression, which would only occur if I've, um, both the black and yellow A wheels had the same levels of dominance. So x linked plus co dominant Ohio's of this video is helpful, and I will see you in the next one.

For this problem. We are taking a look at the simple cross of cats and their resulting blood types. And we're being asked to work backwards from the Fino types were given to the Gino types of the parents and the information weren't given is that the alil for type A blood is dominant over the illegal for tight beat blood. What this means is that head Rosie Vegas organisms will be a type home is. I guess dominant organisms will also have type a blood. And home was, I guess, recess ITV cats will be tight, be blood. This is the only way this becomes a really important thing when we go to actually answer our problems. Um, because this is the only way we can get this Fino type. And the second thing we know, obviously, because this is what the question is asking us to work with is the FINA types of the parents and the offspring. And so we're given and these things and were asked to figure out what which of these they're most likely to be. And this means that we can actually use our pundits squares in order to work backwards, and I'll show you how we can do that with the first problem. So in this question we're told that a type A cat and a type B cat were made it and as a result, they had four type A kittens and three tight beak. It's we know that the male parrot he had to have at least one dominant olio. Because if we look at our different Gino types, but both of the type a Gino types have in common is at least one dominant illegal. And so I will go ahead and draw that in here. Now the female parrot, she is tight beat, which means that she has toe have those two recess Ivo Leal's So her Gino type is homo zegas. Process it here. And if we fill out this row weaken, See, we have, uh, individual offspring. That would be Taipei because their head Rosie Vegas and they have the dominant trait. But that's only some of the offspring here. The rest of them, as we can see from our actual resulting litter, I have type B blood. This means that they had to receive a recess of Lille from both parents and the only way that could work is if the father also had a recess of Leo. So if he was hetero Xigris, then we would see these results. And that matches what happened in reality. So in this case, the female catch waas homocide goes recessive because she was tight be and the male cat he was header rows. August dominant because half of the offspring had the dominant trait and the other half showed The recessive trait is their fino type, and we'll repeat this process with all of our other parts of this question. So let's move on to part B. In this case, both parents were told had type B blood. This means that much like the female parent in this example, that they were both homes, I guess recessive because again, that is the only way they could show that trade as their quino type. And if we fill out the pundits square that results from across like this just to make sure we're not making a mistake and that our pundits square results match up with the actual offspring that were born, we can see that all of the offspring from across of these homes August recess of parents are also home is, I guess, recessed. So all of these organisms would have type B blood, which is the case in our problem. So we know that both of the parents had toe have this home, was, I guess recessive gene o type now on to part C. Now, in this case, we have an interesting scenario because, like in part A, we have made it a type A cat with a tight be cat. But unlike part a, all of the offspring or Taipei And so in this case, it's the father who is ah tightened Be So he is home a Xigris recessive. So we'll draw that. And we know that the mother, because she was type A, she had to have at least one dominant olio. And if we draw out these squares and are upon its square, we have organisms with head rose, I guess Dominant Fianna View. No types. And they would have the dominant fino tech. They would have type a blood. In this case, all of the offspring have this trait. So that means all of them received at least one dominant alil from one of their parents. And we know that the father didn't have any dominant wheels. So the Onley place that could come from would be from the mother. So that means in this case, uh, the father was Hamas, I guess. Recessive And the mother was homos, I guess. Dominant. So she had two copies of that dominant type A Leo. And that's the difference between that problem. And this problem is that in this case, both parents were helping Vegas. A movie on department D. Here we have a case where we have crossed to ah, dominant type A parents and our resulting cross has mostly dominant results. But some recess ITV offspring, some type B kittens, Which means that all we know is that the parents had tohave at least one dominance polio. And if we were to cross these out, we would see some offspring would be Hamas, I guess, dominant just as the mother and this question waas. But what about the rest of them? Well, we know that the majority of them had, ah this type a ah fina type, So most of them received a dominant alil. But for the two type B individuals, that means that these kittens had to receive to recess civil eels from their parents. Which means that the only way this could happen is if both parents had a recessive alil for Type B blood. And if we finish out this pundit square, this actually makes sense with what happened in our problem. Because, as you can see, most of the organisms are type A because they have the one dominant a little at least, but some of them are type B. They have their homes, I guess, recessive. And so in this case, these resulting kittens came from a pairing of two hetero, zegas dominant organisms. And that explains how we could get some recess ITV kittens from to type a parents moving on to her next section. We have another question. Kind of like D, in which both parents or Taipei so we know that they have to have at least one dominant olio. But in this case, all of the offspring are typing. This means that all of the offspring have to have at least one the dominant olio and the Onley way. This could work out as we can see here, where they're sort of the links in the parent genomes that are filled in by the offspring genomes. We know that all of the parents had to have enough quote unquote dominant A Leal's in order so that all of their offspring would also have that. If that wasn't the case, then some of them might be recessive. Now there is actually a another option here, and you might already be thinking that. So in this example, both parents are home was, I guess, dominant We could have a case where, instead of both, being home was like a stall minute. One might be header rows, I guess, and I'll show you how that would look as well. In that case, all offspring would receive at least one type a olio, and so they would all have that dominant trait. They were all test for type A blood, but and the same thing for the parents as well, but they wouldn't be carrying a recess. ITV khalil. So we can't really determine whether or not both parents are homo zegas or one is Hamas, I guess dominant and the other is head rose, I guess, dominant without further testing or another cross with this generation. So for this question, it might be best to save both homes, I guess. Just because that is sort of the easiest, uh, solution. But know that it is an option for one to be head rose itis. And finally, we have our last example where we have one parent with type B blood and we know what to do with them because we know they have to be home was, I guess, recesses and another parent with type A flood. So we know that parent has to have at least one dominant Ulliel. And if we fill out the pundits Square that results from this matchup from this information that we've been able to jot down, we can see offspring that would be type A because they have the one dominant alil we are told, similar to our first problem where some had type B blood. We know that at least one of these offspring is going to have to recess itv a Leal's. We're own colors. That's the parental color, not the kitten color. We know that, uh, just reiterate both parents have to contribute. At least, you know, they all contribute one of you And in order to have an offspring with type B blood. Both parents have tohave that recess ITV olio, which, and if we continue to draw out our results, yes, this would make it more likely. So let's back up for a second. This is essentially the same pairing as part A. But in this case, you see it's the offspring or split roughly half and half, not so much in this case, where most of the offspring or type A So what's happening here? Well, this is still be viable answer because while do the probability, if you were to mate these parents an infinite number of times and counter LaRosa ring, then we would expect half of them to have type a blood and half of them to have to be blood. But obviously we don't do that. And a wheels are matched up, usually randomly on what that means is that just due to random chance, we can have more offspring from this hetero zegas category than we do from the homes, I guess. Recessive category. And that explains how we got this Ah resulting offspring batch. But again, the only way that we could even get one type B kitten is if both of the parents have a recess civil eel, which would mean that parents that one parent is home was August recessive because she was checked beef and the other was header rows, I guess dominant because of being type A. And so that's how you solve these problems. Where, um, you might see this a lot in other test prep or ah, testing scenarios where you're told the FINA types as we were here and asked to work backwards to find the Gino types. And you would employ the same sorts of tactics where we'd see what the offspring Gina types are and what traits they are, what Gina has they would have tohave based on that information.


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