before we move on to the fate of D H A. P. I want to talk a little bit about carbon numbering because your professor likes to ask questions about it and because there's a problem with the image we have here now, from the previous step, we were left with D H A P, which we see here, and the carbon numbering is all fine on this. Right? Um, thes these numbers are, uh, reflecting the numbers from the original glucose molecules. So this was initially carbon one this initially carbon to in this initially carbon three of glucose and the G three p that we generated from the last step wound up with carbons 45 and six from our initial glucose molecule. This step that is carried out by tryouts phosphate by summaries and has a Delta G that's close to zero, making it readily reversible converts D h a. P into g three p. But notice here that are carbon numbering on the G three p is actually reflecting the carbon numbering from the G three p of the previous step. And this is probably because the person who made these images, um or made these diagrams was getting a little lazy, and they just wanted to reuse the molecule. But I want to make sure that I correct this here for you guys. Because I know that your professor, um, your professor cares a lot about knowing the carbon numbers. So this is actually carbon number and three from this d h a p molecule or rather, carbons 12 and three coming from the D. H a P. But those numbers are actually corresponding to the carbons of the original glucose. Um, now bear in mind that the carbon numbering of the naming scheme and I know this is just getting really confusing, so I'm tryingto make this is clear, it's possible. But the carbon numbering off the naming scheme here is actually different than the carbon numbers that I'm adding to these images. The carbon numbers I'm adding into the images and will be continuing to add into the images are reflecting the carbons from the initial glucose molecule. You know, this is glycerol to hide three phosphate. So in the context of this molecule, the phosphate is actually bound to you carbon three of the molecule. But that's actually carbon one from our glucose molecule. Right? So these numbers that I'm adding in are the numbers from our original glucose molecule. All right, moving on to step number six. It's important to note that from this point forward, everything needs to be mentally doubled on. That's because with one molecule of glucose, at this point, we'd actually have two molecules of G three p. But I'm only gonna show what happens to one molecule of GDP because there's no point just copying and pasting the same image, right? So just bear in mind that all of the subsequent reactions from this point forth, uh, are actually happening twice for every one molecule of glucose. So Step six, we have blister Aldo hide three phosphate di hydrogen ace Delta G close to zero. And you can see that in this reaction, um, were being shown the numbers from our initial G three t, But if we were showing, uh, the numbers from the G three p that used to be D H a P that this would be carbon one thio end three. So right, you can think of this a 65 or four or 12 or three anyhow, this reaction is gonna take G three p, and it's going to take an 80 plus and an inorganic phosphate. And it's going thio. Add a phosphate group onto that G three p, making it +13 Biss Foster obliterate. And in the process, it's actually going to reduce n a d plus tau n a t h. And we're basically going to say, See later to that any D H For now, it's gonna come back and be important when we talk about, um the well, cellular respiration and specifically electron transport. For now, we can just say bye bye, an a D h. It's nice knowing you. There are two of you produced for everyone. Glucose. See you later. Anyhow, Looking at our molecules, let's put in those carbon numbers. So we actually are adding a phosphate group on here. So this is from art again from our original glucose. This is carbon three thio and one or 65 and four. Right, So either or now moving on to step seven, we have three phosphate blistery kindness and you see that the Delta G is negative here. This is a favorable reaction. It's actually a little closer um, it's more readily reversible than it appears. Mhm. It's actually a little closer to zero at cellular conditions. And in this reaction, we're gonna take 313 spots, obliterate and ADP. And we're going Thio remove one of those phosphate groups to make three foster blistery. And in the process, we're going to create an ATP through substrate level phosphor elation. So taking Ah, look at our previous molecule. You can see that this would be one or six two or five and three or for and we're actually just going to remove the phosphate group that we had just put on. I'm sorry. I did my numbers backwards. This should be three or one. This should be two or five. That was That was right. And this should be, um I'm sorry. Six. I mean, four. And this should be one or six. Sorry about that. Um, anyhow, so we're going to remove the phosphate group that we just added, Um and that's going to leave us with three foster obliterate. And this is again get our carbon numbers. Right. That's three or four. This is two or five. And this here is one or six. Okay, now, Step eight. We have three foster obliterate mu tastes a delta G close to zero. That's when we've got our reversible arrows there. And we're gonna take three foster eggless A rate, and we are going to form to Foster illustrates. So we're basically just going to be moving that phosphate group over, um, by, like, one position. So not a huge change here and again. Just put our numbers in. We have three, 43 or four rights, not three and three and four. This is two or five, and this is one or six. And just to be super thorough again, here's 34 two or five and one or six. Okay, let's flip the page.