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In cats, the genotype $A A$ produces tabby fur color; $A a$ is also a tabby, and $a a$ is black. Another gene at a different locus is epistatic to the gene for fur ...

Question

In cats, the genotype $A A$ produces tabby fur color; $A a$ is also a tabby, and $a a$ is black. Another gene at a different locus is epistatic to the gene for fur color. When present in its dominant $W$ form $(W W$ or $W w),$ this gene blocks the formation of fur color and all the offspring are white; $w w$ individuals develop normal fur color. What fur colors, and in what proportions, would you expect from the cross Aa $W w imes$ Aa Ww?

In cats, the genotype $A A$ produces tabby fur color; $A a$ is also a tabby, and $a a$ is black. Another gene at a different locus is epistatic to the gene for fur color. When present in its dominant $W$ form $(W W$ or $W w),$ this gene blocks the formation of fur color and all the offspring are white; $w w$ individuals develop normal fur color. What fur colors, and in what proportions, would you expect from the cross Aa $W w \times$ Aa Ww?



Answers

In cats, the genotype $A A$ produces tabby fur color; $A a$ is also a tabby, and $a a$ is black. Another gene at a different locus is epistatic to the gene for fur color. When present in its dominant $W$ form $(W W$ or $W w),$ this gene blocks the formation of fur color and all the offspring are white; $w w$ individuals develop normal fur color. What fur colors, and in what proportions, would you expect from the cross Aa $W w \times$ Aa Ww?

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.

Hello, everyone. Today I will be helping you with the 38th problem. Are the chapter 12 problems up? This question is asking one of the proportion Ah Fino types of the offspring and in what proportions. So it's basically asking like the cats that are blocked yellow calico and, um, it says coat color is carried on the X chromosome. A yellow cat is crossed with a black cat. Um, assume that the offspring are both male and female, so we know that. But we don't know which if the yellow cat is the female or if the black cats, the female and we don't know how calico shows up, whether it's, um, it's an X and X. Like if the one X is black and one ex is yellow, we don't know. So there's not enough information to do so or know what to deal with this. So have you found the sellable and I hope you have a great day. Thank you

So here we have upon it square that's representing the coloring in the Tigers. And so it says that in tigers the normal color jeans see, that's our capital. C is the dominant is the dominant gene, while the white color gene the lower case, he is going to be recessive. So what this is going to mean is that for any tiger who has a dominant color Gina Capital C, they're gonna have normal coloring. So capital C, Capital C or even Capital C lower case C because the capital C is always going to dominate. But if a tiger has to lower case sees their two recessive genes, that means that the tiger will be white. So this Planet square is showing the possibilities of crossing to taggers, each of whom have the recess of white jeans. So first know that these two parents are both going to be normal coloring because they have the dominant gene, but they have the ability to pass on the recessive gene. So we're gonna find a model that's going to be used to represent the Pundits Square and write it as a polynomial. So what this means is that the offspring are going to be the product of parent one and parent, too, their each of their genetic makeup. Now for a parent one, their genetic makeup is 50% dominant in 50% process it, and in this case, it happens to be the same for parents to 50% dominant and 50% processes. So the offspring will be the product of the two parents now, because we have the same binomial multiplying each other. What we could do is we could write this as five dominant see plus five recess of C squared because that's multiplying itself. Then I could apply my special products formula, which says that if I have the some the square of a sum to a binomial some that's being squared, I can write this as a squared plus to a B plus B squared. So what I can do is I can square a square B, multiply the two together and then double it, so I'm going to kind of go in that order. I find that to be helpful. So 0.5 c squared, so 0.5 times five is going to be 0.25 c squared and then 0.5 lower case C squared would be the same, but with the recessive gene. So 0.25 c squared. Now I'm going to dio a tense B So I know that C dominant times recessive is going to give me capital C Lower Casey 0.5 times 0.51 half times one half is 10.25 But then I still have to double it. And so that's gonna come out as 0.5. So my polynomial that is used to represent the Planet Square is going to be 0.25 dominant C squared plus 0.5 dominant recessive plus 0.25 recessive squared.

So this next question were given, Um, that Big C, um is full color C H equals chin shilla color and C H equals the him a land which is a white color, um, with black extremities and then a c, which is just all white. And then we are given, um, we're supposed to give the jeong types of the parents for each cross. So if there is a full color and albino and it yields have color and half a Bayno, the appearance must be must have this Gino type because they want to One scientific ratio in the progeny result from a cross of a head rose. I go with a homeless, I guess recessive because I'll buy knows assess recesses. All other Leo's a full color parent must have an albino Leo and the albino parent must be homeless. I guess for the albino olio, um, B is a 1,000,000 within albino and you get half and half, so this would yield a CHC times and for the same reason as a B 1 to 1 ratio, the progeny indicates appearance must be ahead. Arose. I go any homeless, I guess recessive for seeing a full color and albino and you get half full color and half chin Chila. So this must be this time we get and 1 to 1 ratio. We have a chance Chila progeny instead of an albino. Therefore, the head Rosie goes full color Parent must have a chin chill olio as well as a dominant full color Leo, the albino parent has to be home was I guess because, um, albino is recessive toe all other Leal's Krusty is full color and a Himalayan. You get half full color, quarter him Milan and quarter all by now. So this must be a combination of these Gino types. So the 1 to 2 to one ratio in the progeny indicates that both parents airhead ozai goats. Both must have another final deal because he all buying a progeny must have inherited the albino olio from each parent and our last crosses a full color and a full color and you get 3/4 full color and one court albino so they must have this genotype. The 3 to 1 ratio indicates that both parents are hetero zygotes. Both parents must have an albino olio for the albino project to result


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