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You have created a mutant fly that lacks functional p53. What would be the result?All the cells in the fly would continuously proliferate, even inresponse to mutage...

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You have created a mutant fly that lacks functional p53. What would be the result?All the cells in the fly would continuously proliferate, even inresponse to mutagen exposureAll the cells in the fly would be arrested in the S phasebecause cyclin A would not be expressedAll the cells in the fly would be arrested in G1 because E2f1would never be released by RbAll the cells would be arrested in G2You have a group of ten friends, but it’s always up to you to bethe “primary organizer” of outing

You have created a mutant fly that lacks functional p53. What would be the result? All the cells in the fly would continuously proliferate, even in response to mutagen exposure All the cells in the fly would be arrested in the S phase because cyclin A would not be expressed All the cells in the fly would be arrested in G1 because E2f1 would never be released by Rb All the cells would be arrested in G2 You have a group of ten friends, but it’s always up to you to be the “primary organizer” of outings, parties and get-togethers. In science terms, you would be called the: Niewkoop Center Blastoderm Dorsal Lip of the Blastopore Ventral Lip of the Gastropod A chimeric mouse is an example of a complete transgenic organism. True False



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It is not an easy matter to assign particular functions to specific components of the basal lamina, since the overall structure is a complicated composite material with both mechanical and signaling properties. Nidogen, for example, cross-links two central components of the basal lamina by binding to the laminin $\gamma-1$ chain and to type IV collagen. Given such a key role, it was surprising that mice with a homozygous knockout of the gene for nidogen-1 were entirely healthy, with no abnormal phenotype. Similarly, mice homozygous for a knockout of the gene for nidogen-2 also appeared completely normal. By contrast, mice that were homozygous for a defined mutation in the gene for laminin $\gamma-1,$ which eliminated just the binding site for nidogen, died at birth with severe defects in lung and kidney formation. The mutant portion of the laminin $\gamma-1$ chain is thought to have no other function than to bind nidogen, and does not affect laminin structure or its ability to assemble into the basal lamina. How would you explain these genetic observations, which are summarized in Table $Q 19-1 ?$ What would you predict would be the phenotype of a mouse that was homozygous for knockouts of both nidogen genes?

Hey there Today we're talking about Dress Ophelia, which are flies, and they're used in a lot of different genetics problems. And yeah, you'll if you're studying genetics, you'll do way too much with these babies. So anyway, I know when you look at problems like this, it's very scary because it's a lot of words, but basically you wanna go through and first things first. I would read the question and look for a few things. You're gonna wanna look for the type of G. So if anything says, um that, for example, it talks about that this gene they're looking for causes the flies to shake and quiver. If that says, shake, quiver and die or they die all the time. We know, like at birth, we know that this is gonna be a lethal gene, and that completely changes things. So we go through, we see that it says it's an autism, a recessive gene, and you're trying to compete to figure out whether the quiver phenotype is linked with the recess of gene. For the vestigial wings, which are like reduce wings, the scientist crosses a fly that is Hamas, I guess, for quiver and vestigial traits with a fly Hamas, I guess, for the wild types and then uses the resulting F one females in a test grass. She obtains this information right here, which I got from your textbook. Okay, so I'm sure that sometimes the numbers air difference, but these are the ones that I'm using. And even if the numbers are different, what I'm doing will still be the same. Um, just the answer might be different. Okay, So and let's look at my chart right here. So we have the total, which is 6 50. We have the individuals involved, right for our crosses. And we're trying to use that to figure out whether or not the genes are linked and whether or not there it sorted independently. Okay, so we know a few things, right? We have to write down and let's right now. If you think so. You have this? I'd write it down. Just do you remember? It's important. These are what we're gonna be using for the genes from now on. Um so let's, uh, continue on. So here we go. So paws up here, if you need to write down that anyway, here we are, so first off, let's handle the linked um in across between a hetero zegas and Hamas egas tester. The percentage of re competence is less than 50. If that's the case, then the genes air said to be linked. Okay, her. And if you look at this right, you see that there is left, then 50. Um, you could figure that out. And it's not. It's not terribly difficult, right? We look at these numbers and we have appear going to write. Let's use black less and 50 percent. So if there's less than 50% than the genes are linked, so what was right equals linked. So since there's less then 50% right, look at that to 13 to 24 that are re competent, that means that these are like so the answer is yes, they are both late. So that's part A, I guess. Looks like with that Oh, now we're going to move on to be which is honestly not that difficult. It's just a chi square test. I will not be providing a chi square table that should be provided for you. Way too hard to copy, you know, like nine 1000 different columns or something. Not really. You'll never be expected ever to memorize the Chi square thing, but you will be expected. That's how I'm drinking Camel. I'm doing this to memorize the different ratios. So the ratio that we're looking for here is if it's an independent assortment, independent assortment equals, what's that ratio? Anybody remember? So the ratio not that I can actually talk to you equals one one. Do one the want. So that's what we're looking for for our ratio. So that's what we use to figure out are expected. So we know our total of 6. 50. Divide that by four equals 162.5. Okay, so we know that are expected. We also know how to get chi squares. Right? It's observed minus, expected squared. You should be reciting this in your sleep by this point over expected. We're going to use that to make a chart. So let's write out are observed. Okay. These are gonna be You're right. Observed. Expected. Usually I just write Owen e, but for you, I'm being nice. Is your special to me? Oh, minus E. For this part? Yeah, I'm reading it for one thing. You just take up too much screen. Screen room observed minus, expected squared. And Mom? Yeah, by the way. So all this right? That's what. Uh huh. Can I keep writing it in the angle, by the way? So let's use different color for our structure. So we're gonna need four things. Remember, we looked up there, so we want room for four. Stupid. Yeah. Okay. Sorry for my neighbor's dog, if you can hear it. All right. So let's let's transfer our information. Let's just do it the way they did it in the charts. We're gonna put V g plus s ps plus right here so that we haven't observed 2 30 for the V g s P s. So just the wild, we have 2 24 um, for V g S P s. Plus, we have 97 and B g plus SBS wild type equals 99. So by looking at that right now, we can guess that these are not independently sort, independently sorted. Remember, it's supposed to be a 1 to 1 to one ratio, which you can figure out based on the total is 162 0.5. All the way down. So all those were the same. Okay, 162.5. So we're gonna do some math. Absorbed. Minus expected 67.5. Next up, 61.5. Oh, we're getting negative numbers. Never fun. Negative. 65.5. That usually means you're not independently, sort of. You got really high, positive numbers, and I'm really high negative. Good sign. Sneak peek, Um, to the answers. We're going to get that. That is not gonna be independently assorted. So four or 556.25 You just see this up here? We're just doing what that says, and you can write X squared up here Chi Square. Or you can write that equation. I just wrote that for, like, screen space. Um, next up, 37 82.25 Next up member were squaring. And whenever you square something, that makes the negative cancel out. So for two 90.25 next up for zero 32.25 This is just I'm plugging it in my calculator while I'm doing this. But if I'm going too fast pause. Write down what I've done so far. You know the drugs. Okay, so now we have to do is divide that by e basically these numbers and we get 28 23.3, 26.4, 24 point eight. Gonna add all those up. Our total t equals one or 2.5. Okay. Um, next up, we have to figure out our degrees of freedom for degrees of freedom. It's just, uh, Rose like that. You're doing minus one. We have four minutes. One three. So we have three degrees of freedom. You know, our total is one of 2.5. Um, you look at your chi square table now, so you go to your degrees of freedom. You know your total for your chi square value. And you see that this is completely out of range. Okay? The hypothesis is rejected from from five. How exciting. And we know they use blue. It's exciting way. Know that these air Not independently. Sort of. Anyway, I hope

It is observed that the myself knock out for either of the genes need religion to no needle into have no finna topic. Editors. This is because the other type of negligence can replace the first one. Ah, the result here shows the for no types off mice with genetic defects and components off basil lemina, NATO than one. Have gene or garden minus minus. You know type none isn't to. 10. Gino Cortez minus minus fennel type. None lemon on comma one needed in binding side dilation, plus minus none on a lemon in gamma. One. NATO gin binding site minus minus. Direct birth Lemon gamma one is essential in need. Agent to bind. If Nadine could not bind, then there would be, uh, there would not be formation off basil lemina in the mice and the mice will die. That's the my sector Homo jazz. I goes mutants for lemon on gamma. One have severe fanatic. From this observation, it is found that the mice, which are Hamas egas for knockouts off both NATO agents will have more severe phenotype than obtained in the case off lemon in gamma one mutants here, um, the plus minus stands for Head Rodallega's and minus minus stands for home

So here we have a question about a mutant fly that becomes paralyzed when the temperature is raised. Um, the mutation effects the Dina man and so on and so forth. So we're gonna suggest why signal transmission as soon have some might require Dina hman. And now, on the basis of this hypothesis, what would be expected to see an electron micrografx of synapses of flies that were exposed to an elevated temperature. So synaptic transmission involves the release of neurotransmitter by exercise toeses. During this event, the membrane of the synaptic vesicles fuses with the mat plasma membrane of the nerve terminals. To make new synaptic vesicles membrane must be received. Um, from the plasma membrane by endo psychosis. This industry Tosa step is blocked of diamond is defective. Um, so does the block of dynasty, effective as the protein is required, A pinch off the classroom coated and aside eg best schools. The first clue deciphering the role of Dinah Hman came from electro micrografx of synapses of immune flies. There are many flask like imaginations of the plasma membrane, representing deeply in vaginal classroom coated pits that cannot pinch off. The collars visible on the next of these imaginations are made of meat and Dina Hman

Let's first review what a puff is. A puff chromosome is a chromosome that's experiencing high levels of transcription activity. So the researcher failed that there was a puff at three C, which is a gene region, or gene located on the X chromosome. What they found or he found, was that when the puff happened, but there was an increase in the production of protein factor. Four. So what this tells us is that as the transcription increases for protein factor four, we subsequently see puff at three C, and we can make the inference that the genes coding for protein factor. Four are found at three C on chromosome X. With the transcription activity of protein factor for decreases, the puff decreases. The puff disappears because there's no longer that, um, level of transformational activity in that region of the chromosome, so there would be no cuffs. So we can say that by the third marvel in start stage in Drosophila, the puff disappears because the level of protein factor for production has decreased. This information could be confirmed when you have flies with no salary glands, so they were unable to produce protein factor four. So in fact because they were unable to produce protein factor for you never would see a puff because the gene expression level for the area three C on the X chromosome would never be at a level to provide an opportunity for the chromosome to puff at that section as the chromosome.


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