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Review 3: Pyruvate & Fatty Acid Oxidation, Citric Acid Cycle, & Glycogen Metabolism

Fatty Acid Oxidation Practice 2


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so which compound here is an intermediate veda? Oxidation? This one's a little tricky. The answer is D And if you look a d notice how you have your co a on the end, which you need, right, You need your co A. They're so based on just that you can eliminate all of these. Right? So on Lee, these to have the CO a right. But now this one, this one has this oxygen here. I don't know what that's all about, right, So that's out also. Let's just take a closer look at D. De is basically our final our final round of beta oxidation, right? D is gonna be our last round of beta oxidation because look, if we cut it in half, there were gonna be left with two acetyl coa is you have one here because that's going to get a CO a attached to it, and then you have your other already formed right there. Now, let's take a look at this molecule. Why does this molecule require N a. D ph to reduce a carbon carbon double bond during beta oxidation? Well, the answer is because of this particular confirmation of double bonds that it has. So it's because of the confirmation of double bonds. And I don't I don't think you need thio really bother. Like knowing exactly what that confirmation is that's going to require an a D. PH. But just know that any D. P. H is used to reduce a carbon carbon double bond during beta oxidation due to specific confirmation of to double bonds, carbon, carbon, double bonds. Now, let's take a look at this last problem here. Consider the beta oxidation of the following fatty acid. Right. How many rounds of beta oxidation are we going to have to do here? Well, first, let's start by just looking at how maney carbons we have here. So if you count up your carbons, you hopefully we'll see that we actually have 11, right? 123456789 10, 11. Right. So we have an odd number of carbons, so you already know that this is gonna be a little trickier than normal. So let's put in our cut points, right? We're gonna cut there, there, there, there. And we do have these double bonds present, but they're not next to each other. So we're not gonna have a situation where we can't just use a nice SOM race to move the double bond around to get it into the right place. Right? That only comes up when we have double bonds that air next to each other. So that's not gonna happen here. Now, how many rounds of beta oxidation are necessary to convert it to acetyl coa? A. But only four, right? Four rounds. But have you got, um we're going to have to do something about this because that's going to come off this pro bono. He'll Coetzee, right? So let's think of the output of reduced electron carriers and ATP. Alright. So, uh, looking at this, this is going to generate. This is going to generate directly 1234 acetyl coa a right, Because this is going to go off thio second l CO A. And we don't know what's gonna happen to that. So I'm just going to rule that out, right? We don't know what's gonna happen in this took milk away. It could be used for other stuff. So who knows with that? But we know for sure that we're gonna be generating acetyl coa is So what's going to be the output here? All right, well, let's think about this. We're doing four rounds of beta oxidation, so that normally would mean for F A D. Right? But we have these two points oven saturation, so we're only gonna have to f a d h two. We are gonna have four n a d h. And of course, this is going to lead to three ATP in this is going thio, lead thio 10 http And now let's think about our acetyl coa is right. They're going through the citric acid cycle and that the four acetyl coa is So this is beta oxidation here. Here's citric acid cycle here. Those four asked Hilco is They're gonna make four f A T h two and those forest Yoko is there gonna make 12 en aims 12 n a d h. So again, f a d h two that is going to be 1.58 p. So we're going to make six ATP with this groups six ATP and then n a d h 25 Right, So this is gonna be 30 a. T p. And then don't forget that we're also gonna produce for GTP slash 80 p. I'm gonna count them in four of those. Now, there's one thing. One thing. We can't add this all up yet, because in producing the Sooke and Ilco way, we have to spend minus one ATP, right. You have to spend 18 p to make that. So let's total it up right now. So we are going Thio, um, add all of this together, and if we do, we get 52 ATP total. Right? Because again, uh, Theis issue here is that we're gonna lose 1 80 p uh, in making signal. Coetzee. So normally, this would actually all add up Thio 53. But we have to subtract that 18 p to make certain milk away. So 50 to 80 p. Total. Alright. That's all I have for this exam review. Good luck studying for your test. If you have any questions, please post them in the forum