Total Energy from Glucose - Video Tutorials & Practice Problems
On a tight schedule?
Get a 10 bullets summary of the topic
1
concept
Total Energy From Glucose Concept 1
Video duration:
2m
Play a video:
Video transcript
Hey, everyone. So in this video, we're gonna talk about the total energy from glucose. Now, here we're gonna say complete oxidation of one glucose molecule produces the following. And here we're looking at the metabolism of glucose molecule through glycolysis, pyruvate oxidation and the Krebs cycle also known as the citric acid cycle. Now here with glycolysis, remember there are 10 reactions involved. The premise is that we have one glucose molecule that's split in half. Here, it will create two pyruvate molecules. We'd also create two N A DH high energy molecules as well as two A TP molecules. For pyruvate oxidation. We now have our two pyruvate molecules. We'd say that they each would make one acetyl coa two acetyl coa. In total, we have two and a DH molecules being formed and two carbon dioxides. Now, finally, we have Krebs cycle. Now, creb cycle, we're dealing with our two acetyl coa and with creb cycle, we're also talking about the citric acid cycle. Same thing. We have our two acetyl coa which means we've gone from the cytosol to the mitochondrial matrix. And because of that, we're going to say acetyl coa we would make six N A DH S, we'd make two FA DH twos, we'd make two A TP and then four carbon dioxides. Remember there's two aceto coa. So we go through the crept cycle twice in order to create this total number of products at the end. Now, this is just talking about stage three of our food metabolism. We know beyond stage three, we're dealing with the etc and oxidated phosphorylation. This would create even more A TP molecules than we've seen at this point. For these three portions, glycolysis, pyd oxidation and the crep cycle slash the citric acid cycle. These are the totals we get for each one of the products formed.
2
example
Total Energy From Glucose Example 1
Video duration:
2m
Play a video:
Video transcript
In this example question, it says glycolysis yields how many A TP and how many N A DH molecules? Now remember with glycolysis, we have our phase A which is our energy consuming phase there, we have our glucose and we're gonna say in reactions one and three, we consume an A TP molecule. So that means we're gonna consume two A TP in phase A. But then for phase B, so this is A and then for phase B, we're gonna say this is the energy producing stage in phase B its reactions seven plus 10 where we create a TP. So here we're going to create two A TP. But remember where we have two pyruvate molecules involved here from the two glyceraldehyde three phosphates. So we'd multiply this number times two. So it's four A TPS that are being created within phase B. But here we're talking about really the net when we're talking about yields here, we're talking about what's the net value of A TP created? We consumed two and made four. So really it's two A PS are net value here. We'd say for N A DH molecules, we say that happens under phase B as well. That happens in reaction six. And we're gonna say reaction six, we make one N A DH molecule. But then we have to remember to multiply it by two. Since we're dealing with two glyceraldehyde three phosphates, so the net would be two N A each molecules then pyruvate oxidation yields what? Well, here we have two pyruvate, each pyruvate creates one N A DH. And since there's two of them, that means overall, we yield two N A DH molecules, right? So this is what we'd say. Glycolysis yields two net A TPS two N A DH molecules and pyruvate oxidation yields two N A DH molecules. These would be our answers for the following example question.
3
concept
Total Energy From Glucose Concept 2
Video duration:
2m
Play a video:
Video transcript
So here we're going to say that the total energy yield from one glucose molecule is summarized in the table below. So here we're dealing with glycolysis, pyruvate oxidation, the creb cycle A K A, the citric acid cycle and then oxidated phosphorylation, remember etc and oxidated phosphorylation are combined in terms of stage four. Now, here we're gonna say to calculate the total that we have in this box. Remember A TP yield through oxidated phosphorylation. We have 10 N A DH molecules. Each one contributes on average 2.5 A TP. So that'd be 25. And then we have two fa DH S fa DH twos, each one at 1.5 A TP. On average, that's three. So we have 28 A TP molecules. Theoretically, we've already figured out under the crept cycle. We have four carbon dioxides, two A TPS, uh two FA DH twos and six N A DH S. We know here that we have our oxy acetate as the end molecule within our crept cycle or citric acid cycle. Now, let's go to glycolysis with glycolysis. We'd have no carbon dioxide uh being formed here, we have two A TP net A TP being formed zero fa DH two and then a net of two N A DH S. Our end product here would be two pyruvate. Remember we're slicing glucose in half. So to get to the two pyruvate at the end of reaction, tending glycolysis with pyruvate oxidation, we have our two pyruvate and we're gonna say here through its oxidation, we're gonna create two carbon dioxides, we're gonna create zero A TP, zero FA DH two. And then we're gonna create two N A DH s. Here. We're going to say we end with two acetyl coa as we go from the cytosol to the matrix or mitochondrial matrix because of the presence of oxygen. Now here, crypt cycle acetyl coa goes to oxy acetate and then that gets shuttled over to our etc and oxidated phosphorylation. If we look at all these totals, we can add them up. We'd have six carbon dioxides. Here, we'd have 32 A TP two fa DH twos. And then we'd have here 10 N A DH twos. So this is how we come up with these theoretical values based on each of these stages, all happening from one glucose molecule.
4
Problem
Problem
Which stage of glucose metabolism produces the majority of ATP?
A
Citric acid cycle
B
Digestion
C
ETC + Oxidative Phosphorylation
D
Glycolysis
5
Problem
Problem
How many total equivalent ATP molecules would be produced from 3 moles of glucose through glycolysis in aerobic environment?
A
2 ATP
B
6 ATP
C
7 ATP
D
21 ATP
6
Problem
Problem
The total equivalent ATP yield from: 2 pyruvate → 2 Acetyl CoA + 2 CO2.
A
0 ATP
B
2 ATP
C
5 ATP
D
10 ATP
Do you want more practice?
We have more practice problems on Total Energy from Glucose