Fatty Acid Oxidation Practice 2 - Video Tutorials & Practice Problems
On a tight schedule?
Get a 10 bullets summary of the topic
concept
Fatty Acid Oxidation Practice 2
Video duration:
7m
Play a video:
Video transcript
So, which compound here is an intermediate of beta oxidation? This one's a little tricky. The answer is d and if you look at d notice how you have your coA on the end which you need, right? You need your coA there. So based on just that, you can eliminate all of these, right? So only these 2 have the coA, 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. D is basically our final our final round of beta oxidation, right? D is going to be our our our last round of beta oxidation because look, if we cut it in half there, we're going to be left with 2 acetyl CoA's. You have one here because that's going to get a coA 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 NADPH 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 think you need to really bother like knowing exactly what that confirmation is that's going to require NADPH. But just know that NADPH is used to reduce a carbon carbon double bond during beta oxidation due to specific confirmation of 2 double bonds, carbon carbon double bonds. Now, let's take a look look at this last problem here. Consider the beta oxidation of the following fatty acid. Alright. How many rounds of beta oxidation are we going to have to do here? Well, first, let's start by, just looking at how many carbons we have here. So, if you count up your carbons, you hopefully will see that we actually have 11. Right? 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11. Right? So we have an odd number of carbons. So we already know that this is going to be little trickier than normal. So let's put in our cut points. We're going to 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 going to have a situation where, we can't just use an isomerase to move the double bond around to get it into the right place, right? That only comes up when we have double bonds that are next to each other. So, that's not going to happen here. Now, how many rounds of beta oxidation are necessary to convert it to Acetyl CoA? Well, only 4, right? Four rounds. But caveat, we are going to have to do something about this because that's going to come off as propanomal co A, right? So let's think of the output of reduced electron carriers and ATP. ATP. Alright. So, looking at this, this is going to generate this is going to generate directly 1, 2, 3, 4 acetyl CoA, right? Because this is going to go off to, succinyl CoA. And we don't know what's going to happen to that. So I'm just going to rule that out, right? We don't know what's going to happen to the succinyl CoA. It could be used for other stuff. So who knows with that? But we know for sure that we're going to be generating 1, 2, 3, 4 Acetyl CoA's. So, what's going to be the output here? Alright. Well, let's think about this. We are doing 4 rounds of beta oxidation. So that normally would mean 4 FAD, right? But we have these two points of unsaturation. So we're only going to have 2 FADH 2. We are going to have 4 NADH. And of course this is going to lead to 3 ATP and this is going to lead to 10 ATP. And now let's think about our acetyl coase, right? They're going through the citric acid cycle and that those 4 coase are going to make 4 FAD. So 4 acetyl CoA's are going to make 4 FADH2 and those 4 acetyl CoA's are going to make 12 n a whoops. 12 NADH. So, again, FADH 2 that is going to be 1.5 ATP. We're going to make 6 ATP with this. 6 ATP and then NADH 2.5. So this is going to be 30 ATP. And then don't forget that we're also going to produce 4 GTP gtpATP. I'm going to count them in. 4 of those. Now there's one thing. One thing. We can't add this all up yet because in producing the succinyl CoA, we have to spend minus 1 ATP, right? We have to spend 1 ATP to make that. So let's total it up right now. So we are going to add all of this together and if we do, we get 52 ATP total. Because again, the issue here is that we're going to lose 1 ATP, in making circonyl CoA. So normally, this would actually all add up to 53 but we have to subtract that 1 ATP to make circoneal CoA. So 52 ATP total. All right. 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.