Catabolism of Carbohydrates: Glycolysis: Study with Video Lessons, Practice Problems & Examples

Glycolysis is a metabolic pathway consisting of 10 biochemical reactions that convert glucose into pyruvate, producing ATP and NADH. The process begins with phosphorylation of glucose to form glucose-6-phosphate, followed by isomerization to fructose-6-phosphate. Subsequent reactions involve further phosphorylation, cleavage, and isomerization, ultimately yielding glyceraldehyde-3-phosphate (G3P) and dihydroxyacetone phosphate (DHAP). Key transformations include oxidation to 1,3-bisphosphoglycerate and substrate-level phosphorylation to generate ATP. The pathway concludes with the conversion of phosphoenolpyruvate (PEP) to pyruvate, highlighting the importance of energy transfer in cellular metabolism.

Glycolysis Reactions

Hey, everyone. In this video, let's give a quick overview of glycolysis. Now, here we're going to say that glycolysis is a sequence of 10 biochemical reactions. And, we're going to say reactions 1 to 5. So the first half split 1 glucose molecule into 2 Glyceraldehyde 3 Phosphate Molecules.

Now, 2 Glyceraldehyde 3 Phosphate, we're going to say abbreviated as G3P. And then the second half, we're going to say that's reactions 6 to 10. Well, they convert this Glyceraldehyde 3-Phosphate to Pyruvate and produce high-energy molecules. These high energy molecules are ATP and NADH. Alright.

So just remember, glycolysis, 10 biochemical reactions. The first half is reactions 1 to 5. The second half is reactions 6 to 10.

Glycolysis Reaction 1

Reaction 1 represents a phosphorylation. In it, we have the Carbon 6 hydroxyl group of the Alpha Glucose; it attacks the gamma phosphorus of ATP. So here, we're going to have a basic residue of some type; it's going to deprotonate this OH. The bond falls here to oxygen, but then we can have one of these lone pairs come out, attack this phosphorus, the end result is for this phosphorus to break the connection to this oxygen. Now we could talk about it doing a type of substitution where the pi bond of oxygen breaks up, and it comes back down later, but again, the end result is us just kicking away this oxygen here. So in that way, we have this phosphate group that has been attached to this oxygen of the alpha glucose.

So, in essence, we're going to say this is an SN2 reaction, more accurately. And as a result of this, we transform alpha glucose to alpha glucose 6-phosphate, we've lost a phosphate group on ATP, and now it is just ADP. So this represents reaction 1 of glycolysis.

Glycolysis Reaction 2

Now, reaction 2 is an isomerization reaction. Here, Alpha Glucose 6 Phosphate isomerizes to Alpha Fructose 6 Phosphate. Here we're going to say, isomerization takes place through an enediol intermediate. Here we have a cyclic form of Alpha Glucose 6 Phosphate, and remember that it can also exist in its chained form. So here we have it as just Glucose 6 Phosphate, and what we can do here is we're going to have a basic residue come and deprotonate.

This bond breaks and falls here to make a double bond. At the same time, this pi bond would break and grab this H⁺, which is how we end up with this structure here, our enediol. Now, here another basic residue comes in, deprotonates, causing this bond to break and fall here to make a double bond. And this pi bond would break and grab an H⁺. So at the end, we've created a CH₂OH on the top, and this becomes our carbonyl to give us fructose 6 phosphate.

This can close in on itself to make the cyclic form which is alpha fructose 6 phosphate. So, this is the mechanism of reaction 2 of glycolysis.

Glycolysis Reaction 3

Now in reaction 3, we have phosphorylation. Here, alpha fructose 6 phosphate undergoes phosphorylation to yield beta fructose 1,6-bisphosphate. Now here we separate this into reactions 3a and 3b. In reaction 3a, mutarotation occurs before phosphorylation. So here we have mutarotation, where we're going from Alpha Fructose 6 Phosphate.

We open up the chain, and then it closes into giving us Beta Fructose 6 Phosphate. Here we'd say we have a basic residue, which would deprotonate here. This bond would break, go to the oxygen. This oxygen would attack this, so this is alpha, beta, gamma. It would attack the gamma, phosphorus by sn2 , causing this bond here to break.

And that's how we get this phosphate group getting attached. Phosphorylation occurs at carbon 1. This would be 1, 2, 3, 4, 5, and 6. So it occurs at carbon number 1. So here goes the phosphate group that we've added.

So one's on carbon 1, one's on carbon 6, and that's why it's beta Fructose 1,6-bisphosphate. We've lost a phosphate group from ATP and that's why it's ADP right now. Alright. So this will represent a phosphorylation of reaction 3 of glycolysis.

Glycolysis Reaction 4

Now, reaction 4 is a bond cleavage reaction, and it's separated into several parts here. So let's go a little bit slowly. We're going to say bond cleavage where fructose 1,6-bisphosphate undergoes what we call a retro-aldol reaction. Now, in 4a, we say an iminium intermediate is formed first. So here we have our fructose 1,6-bisphosphate; this will represent our iminium intermediate.

ENZ here means our enzyme with the amino group that's shown branching off of it because that's going to be the group that's going to engage in these mechanisms. And we're going to have our imine formation. Remember, an imine is just a carbon double bonded to a nitrogen. Alright. So what we're going to have here is we're going to have a retro-aldol reaction, which will give us 4b.

And here we're going to say that it's carbons 3 and 4. The bond between them of the intermediate is cleaved, and that's going to give us our enamine as well as our Glyceraldehyde 3-phosphate. Remember, an enamine is when you have two carbons double bonded and one of them single bonded to a nitrogen. Now the way this works is we're going to have our basic residue. It's going to come and deprotonate, which is going to cause this bond here to come and make a double bond, causing this red bond to break entirely, come here to make a double bond, and then kick this bond to the nitrogen.

Doing this is how we create the enamine structure here, and then this bottom portion. So if we number this, this would be 1, 2, 3, 4, 5, and 6. And this would be 1, 2, 3, which form the top part, and this would be 4, 5, and 6. So that one iminium intermediate here got split into these two here. Now, 4c and 4d, we're going to say, tautomerization and hydrolysis of the enamine produces our DHAP, which remember is dihydroxyacetone phosphate.

So here goes our enamine. It's going to undergo a keto-enol tautomerization type of mechanism. We're going to say here, this nitrogen decides to make a double bond here, causing this pi bond to break and grab an H₊. Doing that gives us this structure here, which is our Iminium ion. An Iminium ion is just a protonated imine.

Right? So this thing's making 4 bonds; that's why it's positively charged. And then we're going to say here through imine hydrolysis, we basically cleave this portion here and leave behind an oxygen. So we make a carbonyl group. And this represents our DHAP, which is our dihydroxyacetone phosphate group.

Alright. So again, these are parts 4a to 4d which represent the bond cleavage portion of glycolysis.

Glycolysis Reaction 5

Reaction 5 represents an isomerization reaction where DHAP is isomerized to G3P. Here, it occurs via an enzyme mechanism. So here we have our DHAP to start things off. We're going to use a basic residue, which is going to deprotonate, remove this hydrogen here, which causes this bond to break and move here to make a double bond. And this pi bond breaks so that we can grab this H+.

That gives us temporarily this cis-enediol, and we're going to say another basic residue comes in, deprotonates this now, falls here to make a double bond, causes this pi bond to break, and grab this H+. So through a keto-enol tautomerism type of mechanism, we create G3P here. Now, this could also be drawn this way, where the aldehyde is on the top portion here, and we have OH here, NH here, and CH₂O with our phosphate group. So, this would also be G3P. So here we have our isomerization or reaction 5 of glycolysis.

Glycolysis Example 1

Here in this question it says, "Identify the mechanism through which DHAP is converted into G3P." So, DHAP is our dihydroxyacetone phosphate group and here we have G3P which is Glyceraldehyde 3-phosphate. Remember that this occurs in reaction 5 of glycolysis, which represents an isomerization reaction. It occurs via an Enediol mechanism. So here, our answer would have to be option B.

Through an Enediol mechanism, we're able to undergo an isomerization reaction where Dihydroxyacetone is converted into Glyceraldehyde 3-phosphate or G3P.

Classify each one of the following reactions as phosphorylation (P), isomerization (I), or neither (N).

a) ____ Conversion of glucose into glucose-6-phosphate.

b) ____ Conversion of glucose-6-phosphate into fructose-6-phosphate.

c) ____ Conversion of DHAP into G3P.

d) ____ Cleavage of fructose-1,6-bisphosphate into DHAP and G3P.

• A. a) P; b) N; c) I; d) N
• B. a) I; b) P; c) N; d) P
• C. a) P; b) I; c) I; d) N
• D. a) I; b) I; c) N; d) P

Which one of the following compound pairs is produced by cleavage of fructose-1,6-bisphosphate?

• A. Dihydroxyacetone phosphate and glyceraldehyde-3-phosphate
• B. glyceraldehyde-3-phosphate and glyceraldehyde-2-phosphate
• C. Glyceraldehyde-3-phosphate and CO₂
• D. Dihydroxyacetone phosphate and 3-phosphoglycerate

Identify which carbon atoms of fructose 1,6-bisphosphate become carbonyl groups in dihydroxyacetone phosphate and glyceraldehyde 3-phosphate when it undergoes a retro aldol reaction in step 4 of glycolysis.

• A. C3 and C4
• B. C2 and C5
• C. C3 and C5
• D. C2 and C4

Glycolysis Reaction 6

Now, reaction 6 represents an oxidation twice. Here, G3P undergoes oxidation to produce 1,3-bisphosphoglycerate, which we abbreviate as 1,3BPG. Here, in reaction 6A, a hemithioacetyl is formed first. This happens by a basic residue coming in, deprotonating the thiol portion. This bond breaks and hits this carbonyl carbon by nucleophilic addition.

That causes this pi bond here to break and grab NH⁺. This gives us initially this hemithioacetyl group. Now, here another basic residue will come in and deprotonate yet again, causing this bond to break and form a carbonyl group here. This hydrogen leaves as a hydride to attack this aldehyde in this position here, causing this pi bond to resonate here, and this one to resonate onto this nitrogen. Now remember, NAD⁺ is our coenzyme which facilitates oxidation.

So, 6B is NAD⁺ oxidation, producing a thioester. So here what we have is this thioester that's created as a result. Now going to 6C, we're going to say that the thioester undergoes an SN2 reaction with a phosphate ion. So here, this phosphate ion comes in and attacks this carbonyl carbon causing a pi bond to break to grab an H⁺. So what we have here now is this tetrahedral intermediate.

We're going to say next, a basic residue was present which will deprotonate, fall here to make a double bond, and then we're going to say that this bond here, this polar bond, is going to break, grab this H⁺, creating the structure that's leaving. At the end, what we form is 1,3-bisphosphoglycerate as our product. Right? This represents reaction 6 of glycolysis.

Glycolysis Reaction 7

Reaction 7 represents phosphate transfer twice because we're dealing with 2 molecules. Here, 1,3-bisphosphoglycerate (13BPG) produces 3-phosphoglycerate (3PG) by losing a phosphate group. And this occurs via an NAS mechanism. Alright. So here, for simplicity, we're going to start out with our 13BPG, and we're going to say here that this oxygen is going to attack this phosphorus, causing this pi bond to resonate up, giving us this pentacovalent intermediate initially.

This oxygen then decides to reform its pi bond because remember it's an NAS mechanism, coming here to make a pi bond yet again, causing this bond to break and go to the oxygen. So now that's how we're left with a carboxyl group here. And we've attached a phosphate group to ADP, so we've created ATP as a result. So at the end, we now have 3PG in reaction 7 of glycolysis.

Glycolysis Reaction 8

In reaction 8, we have isomerization happening twice. So 3-phosphoglycerate (3PG) undergoes isomerization to yield 2-phosphoglycerate (2PG), and this occurs via two nucleophilic acyl substitution (NAS) reactions. We're dealing with nucleophilic acyl substitution. Here, a basic residue will come in and deprotonate, causing this bond to break and go to the oxygen, and at the same time, this oxygen will come in, hit the phosphorus causing the pi bond to break up to the oxygen. But it is an NAS reaction, so a lone pair will decide to come back down to reform our pi bond.

So, we're going to have a lot of arrows moving here. We're remaking our pi bond here, which is going to cause this bond to break. Initially, what we're going to have is this 2,3-bisphosphoglycerate molecule, and then we have our freed enzyme. Now, here it can attack this phosphorus, also by NAS. So we've already done one NAS, and now we're going to do it again.

We're going to hit this phosphorus here causing this pi bond to break and go to the oxygen. It is an NAS reaction, so it's going to reform its pi bond. Doing that, moving it back like that, will cause this bond to break to the oxygen, and we use a lone pair on the oxygen to grab an H⁺. So, it's a lot of arrow movements, but the end result is our 2PG here. This structure becomes CH₂OH, which is our 2PG at the end of this section of glycolysis.

Glycolysis Reaction 10

The last reaction of glycolysis is reaction 10 and this is a phosphate transfer that happens twice. Here phosphoenolpyruvate (PEP) yields pyruvate by losing a phosphate group. We're going to say 10a is an NAS reaction with ADP, which produces enolpyruvate. Now, here we have phosphoenolpyruvate. We're going to have here, the oxygen of this phosphate group is going to come out and attack this phosphorus, causing this bond to move up.

Well, we know this is an NAS reaction, so that oxygen with its lone pair will come back down to reform the pi bond, causing this bond here to break. So this bond will go to the oxygen. At the same time or following this, we'd have this oxygen using its lone pairs to grab this H⁺. So if we look, that's how we get this OH here. The pi bond is not affected.

We've added a phosphate group to ADP, so that's why ATP is lost or created here. So we have our enolpyruvate which undergoes a keto-enol tautomerization type of mechanism, and we'd say that would give us this pyruvate at the end. Right. So the whole process at the end, we get pyruvate being formed.

Glycolysis Example 2

Which one of the following glycolysis reactions will produce an ATP molecule? The key to getting this question correct is understanding that an ADP molecule is involved, and through a phosphate transfer, a phosphate group is moved towards it. So ADP would change into ATP. Looking at our choices, we need to see if there is a reduction in the number of phosphate groups from our reactant to the next molecule. If we take a look here, we have 3-phosphoglycerate to 2-phosphoglycerate.

There is not a reduction in the number of phosphate groups; all that's happening is the phosphate group is moving from position 3 to position 2. So this wouldn't work. In this one, we have Glyceraldehyde 3-phosphate to 1,3-bisphosphoglycerate. We haven't increased the number of phosphate groups. So, there isn't a phosphate group that we're giving over to ADP to change it into ATP.

Same thing with C, glucose to glucose 6-phosphate. The answer here is D because we go from 1,3-bisphosphoglycerate to 3-phosphoglycerate. We've lost one of our phosphate groups. That's because that phosphate group got transferred over to ADP, changing it into ATP, producing our ATP molecule. So here, the answer would have to be option D.

Identify the mechanism for each of the following reactions as NAS, E1cB, or S_N2.

a) ____ Dehydration of 2-phosphoglycerate to produce phosphoenolpyruvate.

b) ____ Conversion of 1,3-bisphosphoglycerate into 3-phosphoglycerate.

c) ____ Conversion of glucose into glucose-6-phosphate.

d) ____ Conversion of 3-phosphoglycerate into 2-phosphoglycerate.

• A. a) NAS; b) E1cB; c) S_N2; d) NAS
• B. a) E1cB; b) NAS; c) S_N2; d) NAS
• C. a) NAS; b) E1cB; c) S_N2; d) E1cB
• D. a) E1cB; b) S_N2; c) NAS; d) S_N2

If the P atom in 3-phosphoglycerate is labelled with phosphorous-32, where does the label end up after it is isomerized to 2-phosphoglycerate in step 8 of glycolysis?

• A. Ends up as the phosphate group in 2-phosphoglycerate.
• B. Ends up as free phosphate group because it is exchanged with a free phosphate from the solution outside the enzyme.
• C. On the histidine residue of the enzyme that catalyzes the reaction.
• D. Ends up as the γ-phosphate group of ATP that’s produced in the reaction.