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18) A cell arrested at G2. Which of the following most likely caused this arrest? a) The tumor suppressor protein p53 has been activated to promote synthesis of the...

Question

18) A cell arrested at G2. Which of the following most likely caused this arrest? a) The tumor suppressor protein p53 has been activated to promote synthesis of the p21 Cdk inhibitor protein b) The APC ubiquitin ligase has added a polyubiquitin tag to (ubiquitylation of) the protein securin c) The kinetochore microtubules have been unable to capture all chromosomes d) The cell has a mutation in the Cdc25 phosphatase , where this protein has lost all its functions e) The cohesin proteins have bee

18) A cell arrested at G2. Which of the following most likely caused this arrest? a) The tumor suppressor protein p53 has been activated to promote synthesis of the p21 Cdk inhibitor protein b) The APC ubiquitin ligase has added a polyubiquitin tag to (ubiquitylation of) the protein securin c) The kinetochore microtubules have been unable to capture all chromosomes d) The cell has a mutation in the Cdc25 phosphatase , where this protein has lost all its functions e) The cohesin proteins have been activated19) A cell arrested at the mitotic checkpoint (also known as the spindle assembly checkpoint). Which of the following most likely caused this arrest? a) The tumor suppressor protein p53 has been activated to promote synthesis of the p21 Cdk inhibitor protein b) The cell has a mutation in the Cdc25 phosphatase. where this protein has lost all its functions c) The cohesin proteins have been activated d) The kinetochore microtubules have been unable to capture all chromosomes e) The APC ubiquitin ligase has added a polyubiquitin tag to (ubiquitylation of) the protein securin



Answers

Which of the following statements are correct? Explain your answers. A. Ribosomes are cytoplasmic structures that, during protein synthesis, become linked by an mRNA molecule to form polyribosomes. B. The amino acid sequence Leu-His-Arg-Leu-Asp-Ala-GlnSer-Lys-Leu-Ser-Ser is a signal sequence that directs proteins to the ER. C. All transport vesicles in the cell must have a v-SNARE protein in their membrane. D. Transport vesicles deliver proteins and lipids to the cell surface. E. If the delivery of prospective lysosomal proteins from the trans Golgi network to the late endosomes were blocked, lysosomal proteins would be secreted by the constitutive secretion pathways shown in Figure $15-30$ F. Lysosomes digest only substances that have been taken up by cells by endocytosis. G. $N$ -linked sugar chains are found on glycoproteins that face the cell surface, as well as on glycoproteins that face the lumen of the ER, trans Golgi network, and mitochondria.

Already one from 30 of chapter 10. And this was all about the p 53 right? And it's asking us Well, if it is, uh, damaged. What? What would be lost? And so it really answers. We you know, if it's not damaged, what happens? Right? And so the p 53 what happens is it will sort of stop, stop our if there's, um, damage in the DNA, right. So we have damaged or incorrect DNA it was gonna do is it's going to stop thes cell cycle. Rice was putting it on pause. It's gonna attempt to then repair its repair, the DNA, and then one of two things will happen. Either the repair was good. Okay. And then it will restart thes cell cycle or thea repair. It was unable to repair it, right? Eso were unable to repair it, and then it's gonna do a toast. Is this gonna sort of it's just gonna is this gonna kill, kill everything that's happening? So end and the process and the cell cell duplication. And and so if our p 53 is damaged, Okay, so if the p 53 is taken out, then instead of the damaged DNA being able to to go into the stop cycle, etcetera. Instead, it's just going to, uh the cell cycle will just continue, And it will keep trying to duplicate, which can cause cancers and such, because it's not ideal, right? S So we were looking at was lost. Well, was lost. His is this sort of three step process here. Okay, It's the being able to check the quality of the D N A. Recruiting the repair and signs or, uh, and ah, trickling the apoptosis if necessary, So the correct answer is going to be.

Okay, So from 20 is all about the, uh, my Trojan activated protein kinase cascade. Right? And so these are triggered by r r t k receptors. So those those with an intra and like, extra sailor component, Okay. And it's what's gonna happen. Is the the M a p K pathway. Basically, that sort of has this r e S r a f This is a whole cascade, right? Going threes. Proteins here, uh, the any K protein on dhe, the e r k protein. And this this cascade this phosphor relation cascade. Uh, this, You know, this is also sort of called this, uh, this It is a parent to see this map. Okay, pathway. This is what will trigger cell division. Right and subdivision equals equals growth. So this is gonna because growth or division however you want to. I think about it, okay? And so we have a list of over what would cause uncontrolled growth. Right? And so we want to see what sort of mutations would cause more of this to happen. That's just looking through, uh, all these options. I think sometimes it's easier to pick out the things that are in incorrect right. So, for example, one of the items listed under a is a loss of function in the, uh, map ke pathway. Right? So if we have a loss of function of our map ke Okay, well, so that would be this not functioning. Okay, but that doesn't function that causes growth. So there's no way that that mutation would lead to uncontrolled growth rate s. So then we know that a a can't be one of our answers, right? And so if we look at, then if you look it see, be here one of the ones listed and b is to have the eye cap of the Greek happe here, uh, be permanently bound to the and F Kappa. Be no. Remember the eye Kappa B is an inhibitor of the nf Capa be. And this this guy will, uh, lied to sort of starts. You're transcription of your DNA, right? So this this is gonna transcribe. And so if this is currently inhibited, it could never enter the nucleus and start that process. You certainly need more deanna to duplicate, right? So this, uh, the eye Kappa be permanently bound would would would certainly not allow for for, uh, this additional and extra growth. Okay. And then if we want to take a look at our choices in D Okay, so this one has a loss of function. Mutation is well in both the r s and R f. Right, But we need those. We need those to get to this growth and division. That's part of our our map. Cape halfway. Okay, so this this r E s slash r a f protein lost would not lead to more growth. Okay, so it sort of just leaves us with, see, right. But if our RS protein was unable to head drawl size, then it's not gonna sort of do it needs in the cascade. And if we have loss of function for the loss of function of of the inhibitor, right, then our NF capa be will always be able to enter, uh, the nucleus. So we would we would sort of get that function back. So that's that checks out. That's good. Uh, if we gain a mutation in the genes for the pathway so there's gain of function somewhere in here, that means it's more functional and creating more growth division that checks out that would certainly help. And then, lastly, an unregulated phosphor relation. Cascade writes that this cascade is sort of happening in here. And if that's unregulated, that means that it has the potential to sort of grew out of control and not check itself. Um, so So all the all the options here and see do seem to check out, and that is our answer is letter C.

We are given a list of eight statements and are asked to stay whether or not they are true or false. Our first statement is kind of sing moves Endo, Plas, Mick Ridiculous membranes along micro tubules so that the network of E R. Tubules becomes stretched throughout this cell. This is true. A continual outward movement of E. R is required in the absence of micro tubules. The ER collapses toward the center. Our second statement is without acting. Cells conform a functional my tonic spindle and pulled their chromosomes apart, but cannot divide. This is true. Acton is needed to make the contract tile ring that causes the physical cleavage between the two daughter cells, whereas the my tonic spindle that partitions the chromosomes is composed of micro tubules. Our third statement is Lamma Lemelle, a podia and Philip Odiah are feelers that a cell in extends to find anchor points on the sub stratum that it will then crawl over. This is true. Both extensions are associated with transmit they proteins that protrude from the plasma membrane and able and enable the cell to form new anchor points on the sub stratum. Our fourth statement is GTP is hide realized by two villain Because the bending or Phleger Fla Gela This is false to cause bending 80 p is hide realized is hide realized by the dining motor proteins motor proteins? Our statement for E is cells having on intermediate filament network that cannot be deep. Prelim arised would die. This is false. Cells cannot divide without rearranging their intermediate you intermediate filaments. But many terminally differentiated and long loose cells such as a nerve cells have stable intermediate filaments that are not known to deep prelim arised. So this is false. Some cells, some intermediate filaments in some cells. So intermediate filaments in some cells are not required to prelim arise to deep polarized. Uh, our statement is the plus ends of micro tubules go faster because they have a larger GTP cap. This is false. The rate of growth is independent of the size of the GTP Cup. The plus and minus ends have different growth rates because they have physically distinct binding sites for the to the incoming tubular in sub units. The rate of addition of two billion sub units differs at the two ends G, the trans verse two bills in muscle cells are an extension of the plasma membrane with which they are continuous. Similarly, the SOCKEM is the Sarko plasma critic. You'll, um, is a new extension of the end of clothes. Make particular This is true. Both are examples of how the same membrane can have. Regions that are highly specialized for a particular function in the same membrane can have highly specialized regions. A. I activation of myson movement on act in filaments is triggered by the false relation of troponin in some situations and by calcium ion binding to troponin in others. This is false mice and movement is activated by the phosphor relation of my sin or by the calcium binding to troponin.

Hey everybody today, we're talking about this question revealing, we're trying to reveal which cell cycles the earliest one to be affected and that's going to be by being stopped by the way. You need to come in there. Um p 21 which is a tumor suppressor protein inhibit C. D. AK one and then we have these and we're going to go through each of them. So G think of growth s synthesis during that late phase, we're going to have cycling, binds with C. D. K. two, cycling dependent penis. Again, s phase synthesis, It's also dealing with c. d. K. two. Um G two is going to have to deal with C. d. K. one but it's not going to be stopped. That's actually going to encourage that. So not stop. And the micah tick pro phase is mps is really going to be involved here. Which is maturation promoting factor. So if we just know what the phases are involved with and that's it, we can figure out that it has to be A. Because and A. That's how we're getting that tumor to stop Because p. 21 is inhibiting C. D. K. one as opposed to the rest of it being promoted like an D. So we know that the answer has to be A and during this we're going to have contact inhibition. So if you think about tumors, their uncontrolled growth, that's malignant. So if you can stop that early on you can stop the tumor. Which is what I think of as b. The answer has to be a because that's where it's actually being stopped as opposed to promote it. Have a great day. I hope you learned something.


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