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8. (2 pts) In Drosophila dorsoventral patterning, cactus encodes the protein that holds the Dorsal (DI) transcription factor in the cytoplasm thus preventing its nu...

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8. (2 pts) In Drosophila dorsoventral patterning, cactus encodes the protein that holds the Dorsal (DI) transcription factor in the cytoplasm thus preventing its nuclear localization: Remember that Dl is localized in a nuclear gradient with highest levels in the ventral region, and is required for ventral cell fates. cactus is degraded upon phosphorylation by the pelle kinase after Toll receptor signaling; spatzle is the Toll ligand. What would be the phenotype of embryos derived from a female h

8. (2 pts) In Drosophila dorsoventral patterning, cactus encodes the protein that holds the Dorsal (DI) transcription factor in the cytoplasm thus preventing its nuclear localization: Remember that Dl is localized in a nuclear gradient with highest levels in the ventral region, and is required for ventral cell fates. cactus is degraded upon phosphorylation by the pelle kinase after Toll receptor signaling; spatzle is the Toll ligand. What would be the phenotype of embryos derived from a female heterozygous for a Toll dominant allele that makes a ligand- independent receptor (active everywhere along the dorsoventral axis) and homozygous for a spatzle loss of function allele with respect to: the cuticular pattern dorsalized or ventralized (circle the correct answer) the subcellular localization of the dorsal (DI) protein in the ventral region nuclear or cytoplasmic



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In the tiny model plant Arabidopsis, the recessive allele $h y g$ confers seed resistance to the drug hygromycin, and her, a recessive allele of a different gene, confers seed resistance to herbicide. A plant that was homozygous hyg/hyg$\cdot$her/her was crossed with wild type, and the $\mathrm{F}_{1}$ was selfed. Seeds resulting from the $\mathrm{F}_{1}$ self were placed on petri dishes containing hygromycin and herbicide. a. If the two genes are unlinked, what percentage of seeds are expected to grow? b. In fact, 13 percent of the seeds grew. Does this percentage support the hypothesis of no linkage? Explain. If not, calculate the number of map units between the loci. c. Under your hypothesis, if the $\mathrm{F}_{1}$ is testcrossed, what proportion of seeds will grow on the medium containing hygromycin and herbicide?

The drawing shown is a sample list of Gina types at Locus A for 10 individuals to calculate the observed frequency of recess civil eels or, in this case, the lower case a weenie dicked solve. For Q and use the following equation. I am going to use the highlighting tool to show you where I'm getting these numbers from your home. Ozai Ghous Recess Civil eels need to be counted first and then multiplied by two highlighted in yellow. There are four homos, I guess. Recessive alleles. You're hetero Ziegesar Leal's need to be counted next. In this case, it will be your lower and uppercase A's or one dominant Alil and one recess of Alil. I have highlighted them in blue, and we have four. We then divide this by N, which is the total number of individuals. They're 10, and I have highlighted them in red. We know that two times four is eight eight plus four is 12, two times 10 equals 20 and 12, divided by 20 equals 200.6. So, using the equation, we have determined that the frequency of the recess of Lille is also equal to Q, which is 0.6 now we will solve for the frequency of the dominant Alil, also known as P. This is solved by using the following equation. I will highlight the information again so you know where all of the numbers air coming from. First, we need to count the Homo Zika's dominant A Leal's and insert them into the equation. This is going to be highlighted in red. There are two of them total. Then we're going to count the hetero Ziegesar Leal's, which are going to be highlighted in yellow. There are four and lastly, the total number of individuals is also needed to solve the equation. This is going to be highlighted in blue. There are 10 now. We saw the equation two times two is four. Four plus four is eight. Two times 10 is 20 on finally eight, divided by 20 is 200.4. This tells us the frequency of the Homo dominant trait, or P, is point. For now, we're going to solve for the observed frequency of Homo zegas recessive genes. Using the same information, we will count the total number of Homo zegas recess of the Leal's. This is going to be highlighted in yellow. There are four, then we're going to divide it by the total number of individuals. This will be highlighted in blue. There are 10 four. Divided by 10 is 100.4, so the frequency of Homo Zika's recessive is 0.4. Next, we will calculate the frequency for hetero Ziegesar, Leo's. This is solved by counting the total number of heterocyclic individuals. This is highlighted in green and there are four. Then we divide that by the total number of individuals, which is highlighted in blue, there are 10 four. Divided by 10 is 100.4, So the hetero Zika's frequency is 0.4. Now we will solve for the Homo Zegas dominant frequency. We need to count the total number of Homo Zegas dominant genes there, too, and they're highlighted in yellow. We then needed to divide that by the total number of individuals, which is 10 again. It's highlighted in blue two. Divided by 10 is 100.2, so the Homo Zika's dominant frequency is point to. If we use the hardy Weinberg expectation, weaken solve for the frequency of the next generation. Using the following equations, we will start with the Homo Zika's dominant genotype. I will highlight the numbers that we need in yellow. This could be solved by P. Squared P is 0.4. So 0.4 squared is 0.16 This means 0.16 is the Ho Mose, I guess. Dominant frequency. Now we will solve for the hedgerows, I guess. One. This will be highlighted in green. The equation needed is two times p. Times Q. This means we need to multiply two times 20.6 times 0.4. This equals 0.48 So 0.48 is your hetero Zika's frequency. Lastly, we will calculate the Ho Mose, I guess. Recessive frequency. This is highlighted in red, thick way Asian needed is Q squared. So 0.6 squared equals 0.36 This is your home owes, I guess recessive frequency for the Hardy Weinberg expectations toe hold True, certain conditions must be followed if they are not met. Some of the observed Gina types may differ. The expectations are as followed. There could be no mutations. There could be no selection among Gino types. There can be no gene flow. The population size is infinite and the mating is random. If these conditions air held, we should expect to see the frequencies we calculated

Inheritance and in particular flies or dress. Ophelia. And here we have a capital s for a straight wing, a little less for a curved wing. Big are for red eyes, and little are for brown eyes. Okay, so part A just wants us to very simple use these letters to represent the parental gina types. So in the first cross, we're doing too true. Breeding tells us that in the first sense, So that would be true. Beauty means Hamas, I guess, for either the dominant trait or messiahs for the recessive traits, this would be the parent cross. And then, of course, this gives us the standard F one generation that is hetero, I guess. And so that we make a test cross with this generation in another set of true breeding. Recess it. This? Yeah. Okay, so it was just thio construct thes, and l b is we're going thio, um, do our standard Planet Square. So if we have this heros, I guess that could break down into for possible game. It's, um, sort of one of every flavor, hoping this a little tick on the S helps separate the capital. The dominant versus the recess civil lower case. I hate using S s because they look relatively the same big or small. Okay, and then, But the game, it's for the recessive. We only have one option e Technically, it would make for game. It's with the exact same, um, recessive in each. But we don't need to make for columns. And so we'll do our square here and we get a head rose, I guess for both, Then the hello Zegas in one but recessive in the other and then homeless Vegas recessive. Okay, so this would be b and then see is just to sort of say, Well, what, um, are these gene attempts and fuel types, right? So the distribution here is 25% of each for both the geno types and FINA types. A few Typically, though, this first road would be the straight and, uh, straight winged and red eyed. Then we have the straight wing because it has the dominant s. But this time the brown eyes let me get into the courage. Wings were recessive, double recessive for us to the next to both the curved wings. And then we see the dominant I color. So for this third row will be red eyes. And in the last row, recess in both, we will have curved wings and brown eyes. Okay? And that's hopefully get some practice, doing things in the standard way there. But then we get to D, and we want to mix it up a little bit. So these telling us, as it happens, these air on the same chromosome, which means they're linked. So then we do a different cross on. I tend to write my linked crosses a little bit differently. Like to just group them together. So that would mean that our first parent, the very first cross with the true breeding would I'm sorry about the ticks here, Um, would look like this. And so then I group these linked traits like this. So then the only kind of offspring ends up being this linked dominant with a linked recesses set. Of course. This problem then wanted us to cross this with another true breeding recess. If I've perhaps they put a tick on that one where it doesn't belong. So this is the recessive okay? And so But when they're linked, then we only really get two options, right? so because this is the same. No matter which way it goes, we either get a combination with the first pair or with the second pair. So the only offspring here end up being either. How does I guess? In both or Home was, I guess, in both. And so that would mean we would maybe expect without crossover 50% of each and the FINA types would of course be 50% um, straight winged and red eyed for this one. That's headers, I guess. And then, um, curdling and brown eyed for the home side is recessive okay, but E, as they move on, wants us to suppose that 500 flies are produced in a test cross and to apply, uh, math medic math methods to calculate the expected number. So if we look at the map that were given, we can see that the wings the gene different wings occurs at 75.5 map units aan den. The one for eyes occurs up 104.5 map units s The difference here is 29 map units now a recombination frequency. So one map unit is going to be equal to a 0.1 recombination frequency. That means these are having a recombination occur 29% of the time. Okay, so we initially predicted that we would have 50%. Would be this Hera's, I guess in that 50% would be homeless, I guess. Successive. But it's 29% are gonna recall. I recommend it. Then we No, that 71% will be of this kind. Okay, which means that's gonna be 35.5% each. So if we take 500 figure out what 35.5% of that is, that means we'll have approximately. I'm getting 177.5, so you can say 170. Can't may have half a fly more on down 177 of both of these gino types. Mhm. Okay. And so that leaves. 29% will be the other two. They become been, we re accommodated ones. So that's when we have hello zegas in one, but not the other. Or, um, it's also 14 and a half percent each. So if we take 500 multiply that by 14 and a half percent, that means approximately 72 flies. Each would be of the hetero zegas in one and not the other. Okay, if you have these altogether, do surrounding where two flies short. So, um, maybe we could add to thio the dominant where the recombination ones, Um it doesn't really matter, but that concludes this slightly lengthy, multi part problem.

If this four has on Lee the Alil TD President, it needs trip to fan to grow If sus present, it suppresses the l e o t d. And it behaves like a wild type, so it doesn't need trip to clean to grow. So that means that if t d s u is present, no trip to fan is required in eight parts. If you breathe a t d as view individual with a wild type individual, Theo F one will have the following distributions. You will have in equal ratios. So four individuals tds you, which would be the wild type. You would have a T d s U Plus and that would be that they need Trip van. There would be won t d plus s u plus and that would be a wild type. And finally there would be a TV plus s you and that would also be a wild time. And so the ratio for B would be three wild types to one individual that needs trip to plan

Let's start by outlining the parents in a the parents in a, um, and of normal or wild type eye color and wild type wing shape or weighing texture. And that fly is bread to a fly that has pink eyes and blistery wings. And so I'm gonna underline my peas because you can't tell the difference between a Capital P and a lower case P in the F one. What you have from this combination is 100% of the offspring with the genotype. Big teen, little and big. What being why little b. Why so the FINA type is 100% normalize and normal wings Now The F to the one key piece of information in this question was that these two genes are color and blistering wings are late, so we have to treat them like it's a mono hybrid Cross Nana Die Hybrid cross. And so, if you do that, the new uncover the idea that you have a 3 to 1 ratio where the wild type FINA type is three times more common, then the recessive pinkeye, blistering wing FINA type. If we look at B B, if you at first glance, looks like a, but it's a little different. So be you have big P Big P for wild type eyes with a fly that has blistery wings, and your across that fly with with a fly that has pink eyes and wild type wanes in the F one. 100% of your progeny is header rows. I go for eye color, and we have a little blistery, a Leo paired with a dominant Leo. And so these flies air all wild type, but that Onley produced two kinds of gammy knees, so you have a big P with blistery wing alil and then a little P with a wild type polio. And if you do a cross in the F two, you'll find a one to Ceuta, one FINA typical ratio where you have wild type eyes and blistery wind flies, and then you have to wild type for both traits. Um, any of two and then you have one. I think I, with normal wanes. And so because of the distribution of a Leal's and GAM eats and the fact that there's no crossing over, you get a 1 to 2 to one ratio in the Teoh Answer. C. what we have. If crossing over what happened and here in deep heart causing ever did not happen is you would have a pure re competent carrying a recessive trait. And so here you would have a fly that has peak eyes and blister wings. We don't have that in a or B, and so D asks us to draw what the chromosomes would look like if crossing over what happened. And so we can easily do that. I'm just gonna draw minus straight sticks. So if we look at B, here's the chromosomes that have been replicated in s face. And here's the other ones because we're undergoing the Asus, I'm gonna use blue, and we're gonna put the's Gino types here on our chromosomes and remember their replicated. So you're gonna have a pair for each game. Meat that's, you know, been fertilized and undergoing division to produce new fly babies. And so here, let's put Number one. I'm still put your little number one here and it number two here in the gaming. What's happening is you have a big P and its sister or it's replicated version. And then be, why? Because that you meet has the wild type of Leo for eyes and the blistery winged a Leo. And then it will be duplicated. Maybe I'll put this one in red. Then we're gonna do number two here. And so here this one will have the pink eye, a Leo and its replicas and normal wings, and it's replicate. If crossing over is to happen, we need to figure out a way to make that the chromosomes exchange their wheels so that pink little P and blistered little B Why are together. The easiest way to do that is to cross over between non sister crow motives. So these two can cross over and they will produce the following combinations of chromosomes. So let me get my black back and drove for more sticks, so this will be the result. So in blue, we have a parental. So I'm just copying this one p and B. Why? So that's a parental. The other parent is here, and I will switch to read and put the pink idol Leo here and the wild type for wing shape There. We can also go ahead and put the pink eyed Alil on this re confident and the wild type polio for eye color on this. Re confident. So these two are we confidence. And what we can do is we're gonna flip flop these two wing a Leal's. And so I put them in grace. You can see where they've been changed. Here. I'll just keep the color scheme. Okay. Here, you're gonna have a B. Why? And here you're going to have a little B y. This one will be what we're looking for with pink eyes and blistered wings. If crossing over didn't happen, this re competent would not be possible.


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