Rank the following from the most ATP that could be made to the...

Question

Rank the following from the most ATP that could be made to the least ATP that could be made:

a. 1 glucose molecule processed via a fermentation pathway (consider that glycolysis is the first stage of the process)

b. A lipid made of glycerol and three 10-carbon fatty acid chains entering cellular respiration

c. 1 glucose molecule entering the Entner–Doudoroff pathway

d. 1 glucose molecule entering cellular respiration

Accepted Answer

Hello, everybody. And welcome back. Let's take a look at our next problem when completely oxidized in aerobic respiration, which of the following molecules will generate the most at P A. Two pyruvic molecules produced as a result of glycolysis B A maltose molecule after hydrolysis by the maltase enzyme C A molecule of A 16 carbon ba acid or D A sucrose molecule after complete hydrolysis into glucose and fructose. So let's think about each of these molecules. Um We have our problem telling us they're all going through aerobic respiration. That's easy to see with. We've got um two sugars here as well as prut being the process of glycolysis. So this, we're maybe a little more familiar with but maybe a little more surprising looking at choice C A 16 carbon fatty acid. So we need to think about the metabolism process of fatty acids, knowing that it will be going through aerobic respiration. So when we think about the the metabolism of fatty acids, fatty acids get broken down two carbons at a time by beta oxidation into acetyl coa. And that should sound familiar to us because as we recall, acetyl coa is the molecule that enters the Krebs cycle in aerobic respiration. So we have a commonality with our other molecules here and that we're coming to the point of they will be generating acetyl coa. So we need to compare what the yield is here. So if it's broken down two carbons at a time, we have a 16 carbon fatty acid. So we're going to end up with eight acetyl coa as a result of the metabolism of the 16 carbon fatty acid. So now let's look at our other molecules. Well, let's note that with maltose and sucrose, both are disaccharide maltose. When hydroly hydrolyzed, excuse me, is going to generate two glucose molecules and then sucrose by a more complicated process will end up yielding the same thing. So it yields one molecule glucose. One of fructose that fructose will get broken down in sort of a slightly more complicated process. Fructose metabolism involves the liver. But essentially at the end of all this, the essential yield will be two glucose molecules. So I'm going to put like three arrows here indicating a multi step process ending up in two glucose molecules. So note that B and D give us the same starting point of two glucose molecules. So we only have to look at the yield for one of them because they're going to be the same. So we have two glucose molecules. And let's remember the steps of our process of starting with glucose going through a aerobic respiration to complete oxidation, we have. If we start with one glucose molecule, we go through glycolysis and generate two pyruvate molecules and then go through pyruvate oxidation to end up with two acetyl coa molecules. And then that of course will enter the creb cycle. So right away without doing any math at all, looking at choice A that starts with two pyruvate molecules. That would be the result of starting with one glucose molecule. So that must yield less A TP than B or D which essentially start with two glucose molecules or they're going to yield twice as much A TP. So we can roll out choice A right away because we're starting with a half the amount of glucose. So we just need to compare our disaccharide with our 16 carbon fatty acid. Well, let's think about from the equivalent of two glucose molecules. How much acetyl coa do we generate? Well, two glucose molecules will result in four pyruvate molecules which will resort in result in four acetyl coa molecules. Well, just looking at this, if I'm in a hurry on my test and I don't want to have to do a lot of math to figure out my A TP yield, I would pick choice C and just go on, go with that because we have literally twice the amount of acetyl coa um that will enter that creb cycle. However, we want to be a little careful here because the processes of glycolysis and pyruvate oxidation themselves produce some A TP. The creb cycle is the largest producer for sure. But we may want to be thorough and double check and make sure these molecules going through these earlier processes aren't going to end up adding up to more A TP than our fatty acid that just starts from ace coa and has none of that A TP yield from earlier. So let's think about our steps and their A TP yield. So the process of glycolysis is going to generate two A TP molecules directly and then it will generate two N A DH molecules. And when N A DH you know, eventually down the line is used in the process of chem osmosis, it's going to generate some more A TP. It's going to generate approximately three A TP molecules per N A DH molecule. This is not exact, it's not always the same, it depends on the conditions, but we'll go with this and we'll just assume that the conditions will be the same for all of our molecules. So that would generate about two times three A TP molecules or approximately six A TP molecules, meaning that, that the whole process of glycolysis will generate eight A TP molecules. So we'll put a little circle around that. So let's go ahead and highlight it, you know, a little Tidier, little blue highlight there for eight A TP yielded by one molecule of glucose. Now, it gets a little more complicated because starting with one molecule glucose, that general its two molecules of pyruvate. So when we look at pur oxidation per molecule of pyruvate, it generates one molecule of N A DH, but we have two pyros. So for this 2 to 2, we would say two multiplied by, we'll say one N A DH P pyruvate. And this is important because we're not necessarily starting with two prut in some of our sugars up here, but one A DH per piru generates about three A P, as we said, and there's no a TP degenerated directly from this step. So we just need to think about the um that the N A DH molecules. So about three A TP per molecule at that step. And then finally, when a molecule of acetyl coa goes through the creb cycle, and again, we need to look at multiplying by two, just the way I wrote it because I had two acetyl coa coming from my one molecule glucose, two multiplied by. While in the creb cycle, our yield is one A TP that's produced directly three and A DH molecules. So that will end up leading to about three multiplied by three or nine ATP molecules. And then it also generates one molecule of fa DH two oops forgot either which and again, this is an approximate yield, not always exact will yield approximately two A TP molecules. So all of that in there will be a yield of 12 a TP molecules per molecule of acetyl coate. Again, remember if I'm looking at the whole process, I started with one glucose and therefore have two acetylate. But again, I'm covering different molecules. So I need to make sure I'm looking at my yield per molecule and I'm just going to go back and note to my py pyro oxidation step that yield of three A TP is per pyruvate and want to keep really straight what molecules I'm talking about. So now let's go back and compare our disaccharide and our um fatty acid. So for our disaccharide, I'm just gonna scroll up here because I'm running out of space, scroll up a little bit and make some more room at the bottom. So for B and D maltose and sucrose which each generate about two molecules of glucose. What is our yield of A TP? Well, we're going to have two molecules of glucose. All right, two glucose. So that will generate again, we have eight molecules of A TP per molecule glucose going through glycolysis. So that will be a yield of 16 A TP molecules plus. Now, my two glucose are going through glycolysis and becoming four pyruvate molecules. So my poor, when I look at my yield of A TP per pyruvate going through pyruvate oxidation, it's three A TPS per pyruvate. So that will yield about 12 ATP molecules plus. Well, now my four pates are getting oxidized to four acetyl coa molecules that are going to go through the crep cycle. And we know that for every turn of the crept cycle, for every molecule of the coa I generate about 12 A TP. So now let's look at my yield from two molecules of glucose as a result of either maltose or sucrose metabolism. So that will be 16 plus 12 plus 12, that's going to equal 76 A TP molecules. So you can see we're because we, how many molecules we have in these earlier steps, we generate a bit more A TP than we might expect. So it may come down to being closer than we think just by looking at the number of acetyl coa molecules. So now let's look at choice C where we know we just started in directly with our eight acetyl coa molecules. So we know that we, again, for every acetyl coa, we generate 12 A TP, approximately 12 A TP. So that's just going to be multiplied by 12 A TP PERL ka molecule. So that's going to equal approximately 96 A TP molecules. So as we suspected our fatty acid that generates eight acetyl coa molecules does yield more A TP than R two disaccharide, but it came a little bit closer than you would think. You don't want to ignore those earlier steps. So our answer is choice C A molecule of A 16 carbon fatty acid is going to be what yields the most A TP. After being completely oxidized in aerobic respiration. See you in the next video.

Key Concepts

ATP Yield in Cellular Respiration

Fermentation and ATP Production

Metabolism of Lipids and the Entner–Doudoroff Pathway

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