How many ATP are produced when glucose is oxidized to pyruvate compared to when glucose is oxidized to CO₂ and H₂O?

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Hello everybody. And welcome back. Our next problem is calculate the number of A TP molecules generated when 7.00 molecules of glucose are fully oxidized to carbon dioxide and water. So let's recall that this oxidation process involves several different steps. And we need to look not just at the direct yield, um A TP but also the yield of N A DH and fa DH two. Because these are electron carriers that bring their electrons over to the electron transport chain where in the process of chemi osmosis, they're used to generate more A TP. So we need to think of the yield of both or all three of those things. So let's think about this process of glucose oxidation. We're going to walk through it with the example of one mole of glucose just to make it simpler. And then we go back and calculate for seven molecules. So one mole of glucose is going to go through the process of glycolysis. This splits glucose in half and generates two moles of pyruvate. So we always need to take into account that too there and thinking of the yield of all our different molecules glycolysis has a yield of two net A TP molecules. It actually generates four, but then uses up two and then it has a yield of two N A DH molecules. So let's highlight that as our first yield there, then we have our two moles of pyrite. And our next step is pyruvate oxidation. And in pyruvate oxidation, each mole of pyruvate oxidized is converted to acetyl Coase. So we're converted to two moles of acetyl coa and each mole of py rebate is going to generate one mole of N A DH. So, so have two moles of pr we're getting two moles of N A DH. So there's our yield there. And now a silk away, of course, is going to enter the crib cycle and generating carbon dioxide and water at the end. But we need to account for a TP production. So for every cycle on the creb cycle, so don't forget we have two moles of acetyl coa. So we need two multiplied by one mole of A TP. So directly yield just one mole per cycle. So that's two A TP. And then we generate two multiplied by three moles of N A DH, three moles of N A DH per cycle. So we're going to generate six moles of N AD and then two multiplied by one mole of FA H two, generating two moles of fa DH two. So now we just need to go and calculate how much AD P that's going to be, so are six molecules of N A DH do that calculation over here, six molecules, six moles N A DH. And when we look at that yields, yields can vary just a little bit depending on the conditions of our oxidative phosphorylation. And the yield is 2.5 moles of a TP per mole of N A DH because it can be between two and three, 2.5 moles of A TP per mole of N A DH. So that is going to yield 15 moles of A TP. And then we have our two moles of fa DH two. And again, yield varies a bit. But in general, 1.5 moles of A TP, it generated per mole of fa DH two. So we do that multiplication neck is this yield of three moles of ATP. So we've taken into account that our glucose is split into two multiply R. We have this multiplier of two everywhere. So we just need to add everything all up. Now, we still have some N A DH that we haven't accounted for in terms of how much AC P will generate. So our two N A DH from glycolysis will be do calculation over here two moles N A DH multiplied by 2.5 moles A TP promote than ADH. So that will yield five moles of A TP. There are five moles of a TP. And that's the case for my other two N a DH generated from pate oxidation that in there. And then we know later on we have two moles of fa DH two. So now let's just go through and highlight the various eels we're getting here. So we have from glycolysis two A TP plus five moles of A P from RN A DH. So I'm going to circle those amounts from pyruvate oxidation are two moles of N A DH yield five moles of A TP. So we will circle that as well. And then from our crib cycle, we've got two moles of A TP, six moles of N A DH which yielded 515 moles of A TP. And we had two moles of fa th too. And from our calculation, we did on the other side, we see that yields three moles of A TP. So we will highlight and circle those and we'll just circle those. A TP yields from creb cycle, direct A TPATP from N A DH A TP from FA DH two. So add that all together and we have two plus five moles of ATP from glycolysis which is seven, seven moles at TP from my glyco glycolysis process. So circle that and then PVI oxidation five more and creb cycle gives us two plus 15 plus three. So 20 altogether. Now we've managed to get a bit confusing with all our numbers here. But if we look at, we'll put asterisks by our yield of a TP per step So from glycolysis, our total yield of A TP was seven moles from pyruvate oxidation. Our yield was five moles. And from the crab cycle, our yield was 20 moles. So now I'm gonna roll this up a little just so I have more space down at the bottom. So for every one mole of glucose, we generate five plus seven oops, we should say seven plus five. That's first there to keep in order seven plus five plus 20 equals 32 malls of at P. But we're talking about seven molecules of glucose. So we would need to multiply by seven. And that is going to give us a total of 224 molecules of ATP as our yield. I'll highlight that in pink to make it stand out from all the yellow hair. And that is going to be our final answer to this question. So we've calculated the number of A TP molecules generated when 7.00 molecules of glucose are fully oxidized to carbon dioxide and water. And that is 224 molecules of A TP. See you in the next video.

How many ATP are produced when glucose is oxidized to pyruvate compared to when glucose is oxidized to CO₂ and H₂O?

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Key Concepts

Glycolysis

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How many ATP are produced when glucose is oxidized to pyruvate compared to when glucose is oxidized to CO2 and H2O?

Verified video answer for a similar problem:

Key Concepts

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How many ATP are produced when glucose is oxidized to pyruvate compared to when glucose is oxidized to CO₂ and H₂O?