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What is the energy yield in ATP associated with each of the following?a. $mathrm{NADH} longrightarrow mathrm{NAD}^{+}$b. glucose $longrightarrow 2$ pyruvatec. 2 pyr...

Question

What is the energy yield in ATP associated with each of the following?a. $mathrm{NADH} longrightarrow mathrm{NAD}^{+}$b. glucose $longrightarrow 2$ pyruvatec. 2 pyruvate $longrightarrow 2$ acetyl $mathrm{CoA}+2 mathrm{CO}_{2}$

What is the energy yield in ATP associated with each of the following? a. $mathrm{NADH} longrightarrow mathrm{NAD}^{+}$ b. glucose $longrightarrow 2$ pyruvate c. 2 pyruvate $longrightarrow 2$ acetyl $mathrm{CoA}+2 mathrm{CO}_{2}$



Answers

Which molecules are produced in glycolysis and used
in fermentation?
a. acetyl-CoA and NADH
b. lactate, ATP, and $\mathrm{CO}_{2}$
c. glucose, ATP, and $\mathrm{NAD}^{+}$
d. pyruvate and $\mathrm{NADH}$

All right, so in this question, we're being asked to find out how much a tee pee harmony net molecules of 80 p do you make during like Collis is And like Collis is is the process of going from glucose, which is a six carbon molecule to Piru of eight, which is a three carbon molecule. So for every glucose you are making two Piru of eight and pyro of eight, then goes on either into fermentation or the citric acid cycle, and Luca and Black Collis ISS deals with this first part off the central metabolic pathway. The first stepping, like Hollis is, is right here It is breaking down glucose into 23 carbon molecules called glycerol Height three phosphate. And this step uses to a T P. Now, over here, we also have 23 carbon molecules. Ah, that glucose gets broken down into. And then from here to here would just basically have different processing steps off the three carbon molecules. Those processing steps there two occasions where we are actually producing a teepee one here and one here, and just to spell them out in the first instance, you are going Ah, from 13 This fossil glitz a rate 23 p g, which is three forceful, glittery. And in the second stuff in the second step, where you are getting a teepee is the last step Where is basically producing Piru of eight. So sorry, I forgot the two here. So basically overall England Collis is you're getting minus two plus two plus two. And if you add all of this together you get two nets, 80 p that are produced during glycol assis.

So what kind of molecules are made doing like Hollis is in Glade? Collis is you are taking glucose, and the ultimate product that you are making is Piru Bait and pirouette can then go into fermentation or the electron transport chain. So glucose gets broken down 2 to 3 carbon molecules which then undergo some processing to make Piru bait. So there are a bunch of intermediate three carbon molecules in here and this process, however, you are making some in a D. H. And you are making some a T peak. And in the diagram that I have here, a D. H is right here 80 piece right here, right? And because is a six carbon molecule that gets broken up into 23 carpet molecules, you essentially have, ah, one of each happening for each pathway of processing these three carbon molecules. So in glucose, you get final products of are of eight in a D. H and a teepee. So what are our answer? Choices are answer choices. Are any D h Piru bait 80 p and F 80 eights, too. Out of all of those, if a d. H two is the only one that is not here, so f a D h two is not a byproduct of like Hollis is. So this is our answer choice d.

All right. So for part A, we are given the equation for the hydraulics is of 80 p. You're given that it's Delta G is 30.5 killer jewels. So, as always, we're gonna use our equation here. Delta G equals r T lnk, and we're just gonna plug in values again. Always know that our, um is 8.34 uh, or 8.314 Excuse me, Jules, perk on mole. And because this is in jewels, we need to change our delta G into jewels. So whoops. By multiplying by 1000 to get Jules, you could get 30,000 500 she waas So now let's just plug that into our equation. So 30,000 untraded. Joerres equals negative. 8.314 times 200 groups. Oh, my mouse is not working today Times 298 Kelvin Um, times l N K. So if you multiply these together and then divide this by that value and then you could take e to whatever number you get, you will find K. So if you want to take a minute and do those calculations, you should get a K of 1.38 times tend to the fifth. Perfect. All right, So the next part asks us to find the Delta G of reaction for the, um, metabolism of glucose. So we're given the reaction and we're going to use this formula products minus reactions. I think it's always helpful to write out kind of a little skeleton, and then you can just plug in values where you need to go. So the products we have we're gonna need the Delta G of CO two plus the Delta G of H 20 minus the Delta G of 02 plus the delta G of glucose. Remember, whenever their coefficients in the reaction, you need to put them in this equation as well. So you're gonna have six times the Delta G of C 02 plus six times the delta G of H 20 and six times the Delta G of 02 It's very important to remember that so you can find the Delta Gee of reactions, most likely in your textbook. You can also google them, but I have provided them right here. So from here it's just a plug and truck. So you're going to put this value here will substitute these values there, that value there and this value there. And that should come out to be a delta G of reaction. Um, who? 1000 879. Kill Ege loves Great. Moving on to see, um, it's asking you for how maney molecules of 80 p could be produced by metabolism of one mole of glucose. So the best way to think about this is you want to get, um, one mole of 80 p, or I should just say we want to have let me go back here. Sorry. On moles of 80 p come over moles of glucose. And we can use this Do this using our delta trees. So we know our Delta G for 80 p given in the problem is naked 30.5. So if we multiply that by our Delta G um, uh, what the Delta G We just calculated for the metabolism of glucose right up here knowing that we need to cancel out this killer jewels on the bottom. So we're gonna put 280 earth. Excuse me. 2879. Kill it. Jules. Up here, huh? One mole of glucose. This will come out to give you 94.4 moors, 80 p for every one glucose. So problems like these, it's always best to think about what you already know. So we knew we knew the two Delta G's and then look at the units to see what you can cancel out in order to get the units that you want your final answer to be in. All right. So d the percent yield again, this is super easy. So we're going to use or formula here actual over theoretical times 100%. So the actual is what we actually get in real life. So they give this to us. They say that it is 36 so we can say 36 let me see. They put, yeah, 36 months. So 36 malls in the theoretical is what we just calculated using our Delta G's. So we're gonna plug that number in over 94 point four moles times 100%. That is going to give you 38.1%

Which of these is produced by both cellular respiration on by the light. Independent reactions of the two synthesis a glucose be at Deep Sea Co two e Peru pick acids on e extreme. Okay, a glucose. This is produced by the Calvin Cycle. It is not produced by respiration. Be a tip off his Our answer. 80 p is produced by respiration on also the light dependent reactions of photosynthesis. Let's go through the of us for completion. Co two is produced by aerobic respiration, but not photosynthesis. Aerobic assets is a product of a glycol, Asus Um, Arabic phase of respiration, not photosynthesis. Oxygen is a waste product of photosynthesis, but it is not produced by respiration. It could be used for respiration for every bit. Respiration definitely not always product.


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