DO NOT UNDERSTANT THIS PPP LOGIC(BIOCHEM)

[The8]

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You have 30 moles of glucose 6-phosphate. How many moles of ribulose 5-phosphate would be produced via the non-oxidative pathway?

"Glucose 6-phosphate has 6 carbons, while ribulose 5-phosphate has 5. To answer this question correctly, we should know that the total number of carbons should be conserved in the non-oxidative pathway. 30 moles of glucose 6-phosphate contains 180 moles (30*6) of carbon. Therefore, the total number of carbon atoms from all of the ribulose 5-phosphate produced should also total 180 moles. Since each mole of ribulose 5-phosphate contains 5 moles of carbon, there should be 180/5 = 36 moles of ribulose 5-phosphate produced."

This answer makes no sense to me.... you need other carbon sources to make any other compound.

 

[theonlytycrane]

g6p -> ribulose-5-p occurs either through the oxidative pathway in which we lose 1 Co2 / g6p or through the non-oxidative pathway in which carbons are conserved. For the non-oxidative path, g6p would be converted to f6p + g3p -> ribulose-5-p with the total number of carbons being the same.

 

[laczlacylaci]

g6p -> ribulose-5-p occurs either through the oxidative pathway in which we lose 1 Co2 / g6p or through the non-oxidative pathway in which carbons are conserved. For the non-oxidative path, g6p would be converted to f6p + g3p -> ribulose-5-p with the total number of carbons being the same.

So would this mean that 30 moles of g6p going through the non-oxidative phase would still be 30? Since g6p->f6p+g3p->ru5p
Wouldn't it be the same if g6p goes through the oxidative phase? g6p->ru5p->non-oxidative phase

 

[theonlytycrane]

@laczlacylaci

For the non-oxidative phase, the # of carbons is conserved. 30 mol * 6 carbons (g6p) = 180 carbons. 180 carbons / 5 (r5p) = 36 mol of r5p (we get a bit more than 30 mol which we started with).

For the oxidative phase, even though we start with 180 carbons, we lose 1 carbon as CO2 / mol of g6p so we only have 150 carbons for r5p. r5p is only 5 carbons long so we get 30 mol of r5p in this path (a bit less than through the non-oxidative phase, but the same # of mol as which we started with).

 

[The8]

@laczlacylaci

For the non-oxidative phase, the # of carbons is conserved. 30 mol * 6 carbons (g6p) = 180 carbons. 180 carbons / 5 (r5p) = 36 mol of r5p (we get a bit more than 30 mol which we started with).

For the oxidative phase, even though we start with 180 carbons, we lose 1 carbon as CO2 / mol of g6p so we only have 150 carbons for r5p. r5p is only 5 carbons long so we get 30 mol of r5p in this path (a bit less than through the non-oxidative phase, but the same # of mol as which we started with).

i dont understand though..... you still have the G3P as well so you have more like 8 carbons total....

 

[theonlytycrane]

i dont understand though..... you still have the G3P as well so you have more like 8 carbons total....

Each 12 carbons (2 molecules of g6p) go through as 1 f6p and 2 g3p.

 

[The8]

Well then that limits the amount of mols of ribulose you can attain. 1 f6p and 2 g3ps will end up (through transketolase) to obtain 1 erythrose and 1 xylulose(equivalent to ribulose/ribose through epimerization/isomerization). But the erythrose is useless.... it can't be transformed into anything. I hope you understand what I am confused on; your explanations have been good but I am caught up on how on earth you actually get 36 mols of the 5 carbon compound from the 6 carbon compound glucose like what enzymes let you get from point A(all glucose) to point B(all ribose)?

 

[theonlytycrane]

There's 3 steps that take ribose-5-p, and 2 xylulose-5-p (15 carbons total) to 2 f-6-p and 1 g-3-p (15 carbons total). Below is the "forward" direction of the non-oxidative phase. So the question is referring to going backwards through it.

Step1: 5C + 5C -> 7C + 3C

Step2: 7C + 3C -> 6C + 4C

Step3: 4C + 5C -> 6C + 3C

 

[wizzed101]

You have 30 moles of glucose 6-phosphate. How many moles of ribulose 5-phosphate would be produced via the non-oxidative pathway?

"Glucose 6-phosphate has 6 carbons, while ribulose 5-phosphate has 5. To answer this question correctly, we should know that the total number of carbons should be conserved in the non-oxidative pathway. 30 moles of glucose 6-phosphate contains 180 moles (30*6) of carbon. Therefore, the total number of carbon atoms from all of the ribulose 5-phosphate produced should also total 180 moles. Since each mole of ribulose 5-phosphate contains 5 moles of carbon, there should be 180/5 = 36 moles of ribulose 5-phosphate produced."

This answer makes no sense to me.... you need other carbon sources to make any other compound.

That answer is wrong. If all G-6-P are converted to ribulose-5-phosphate, the ratio is always 1 to 1 because CO2 is lost along the pathway. There is no way to recover it. This is biochem not general chem. The answer is 30 moles.

 

[theonlytycrane]

@wizzed101

Backwards through non-oxidative?

 

[wizzed101]

Impossible.

Animals are not autotroph and plants use atmospheric CO2. There is no pathway that can recover a CO2 once lost.

 

[The8]

@wizzed101 nah, I think you are wrong because non-oxidative does not use CO2, like that is its purpose. I don't know what you are talking about regarding CO2. @theonlytycrane I understand the logic, but the exact mechanism is a bit too complicated, I guess the assumption that carbons are always conserved is a given..... essentially there is always SOME way to go from one sugar to the other using transketolase and transaldolase. Regarding you're logic about the oxidative pathway however, I think that CO2 is not really a matter stoichiometrically, basically because the EQUATION states 1 mol of a 6 carbon molecule becomes a 5 carbon molecule, while there is a loss of carbon you get 1:1 ratio regardless.

 

[wizzed101]

@wizzed101

Backwards through non-oxidative?

@The8 Oh I see now. My bad. You mean the reverse of the non-oxidative pathway. I didn't consider the reverse part of the PPP. Still, that doesn't make any sense because Ribulose 5 phosphate is just an intermediate to make either ribose-5-phosphate or be fed into glycolysis. The reverse is upregulated in order to make more ribose-5-phosphate, and as you can see ribose-5-phosphate can be made without being converted back into ribulose 5 phosphate. Only xylulose 5 phosphate needs converting.

So for every 15 C, only 10 C (2 xylulose 5 phosphate) is actually converted into ribulose 5 phosphate at some point, the other 5 C is already in ribose 5 phosphate form.

Starting from glucose, without going through the PPP, 2/3 of the carbon is converted to ribulose 5 phosphate. 30 moles of glucose should make 20 moles of ribulose 5 phosphate in total!

 

[theonlytycrane]

For every 5 molecules of glucose (30C) we can convert this to 4 f-6-p and 2 g-3-p (30C).

Going backwards through the non-oxidative phase, we can do 4 f-6-p and 2 g-3-p (30C) to 4 xylulose-5-p and 2 ribose-5-p (30C).

Finally, we can convert 4 xylulose-5-p and 2 ribose-5-p (30C) to 6 molecules of ribulose-5-p (30C).

 

[wizzed101]

But it will not convert the 2 ribose-5-phosphate at the same time the condition for the reverse pathway exists!

The point of the reverse pathway is to make ribose-5-phosphate for consumption!

 

[The8]

@theonlytycrane Can you draw out the transketolase/transaldolase reactions you are reffering to from start to finish. I don't see how you are getting 2 of something and four of something when transketolase and transaldolase always transfer carbons around so you go from 2 MOLES sugar to another 2 MOLES sugar. You can't gain and lose moles. I think you may be correct but I'd just like to see your logic if you don't mind its a little complicated at this point.

 

[theonlytycrane]

@The8 Yeah I can, is the mechanism for the above steps unclear? Just wanting to know if there's anything I can specifically clarify.

The 4 f-6-p and 2 g-3-p total to 30 carbons, which are the carbons that come from 6 molecules of glucose (36 carbons, but 6 carbons are lost as CO2).

 

[The8]

@theonlytycrane There is no CO2 lost in the nonoxidative stretch. Also, my main problem is that with 30 glucose molecules, and ONLY using the nonoxidative pathway, I don't see a mechanism consistent with the CONSERVATION OF MATTER in terms of making ribulose. Basically I don't get how you can take each of the 30 glucose molecules you start with and turn them into ribulose using TRANSALDOLASE and TRANSKETOLASE where you are KEEPING TRACK of each molecule.... the way you describe the process I have no track of the molecules it seems like you are adding matter even if your carbons are kept constant.... basically how do you take exactly 30 molecules of glucose and rearrange them into the ribulose molecules.... as each reaction REQUIRES 2 sugars as input and outputs two sugars of either n-2, n+2 or n-3, n+3. The reactions are also very specific.... if you could please describe exactly how you get from the START n = 6, 30 of these sugars to n =5, X of these sugars.... find X which should end up being 36 according to NextStep. Where are you getting all these sugars from on a STOICHOMETRIC basis; I don't see your answer STOICHOMETRICALLY making sense at all with regards to the starting 30 moles when each reaction requires input of 2 sugars. (Sorry for spelling stoich wrong)

 

[theonlytycrane]

let's find a starting place that we can agree on:

30 mol of g-6-p = 180 carbons. fair?

target: ribulose-5-p (5 carbons).

 

[The8]

Sure i agree with you there.... there are indeed 180 carbons. Now the question is stoichometrically, what is your next step, NOT looking at the carbons but in terms of the physical molecules of glucose. WE can both agree hopefully that you cannot just look at a molecule in terms of its atoms, as this is too simplistic of an approach, we must look at it as it is, a grouping of atoms that can only be changed in this case by Transaldolase and Transketolase.

 

[theonlytycrane]

I'm not simplifying anything, but I believe that the mechanistic steps might be confusing to you. As outlined above:

5 g-6-p -> 4 f-6-p + 2 g-3-p. Now let's take 2 of those f-6-p and 1 of the g-3-p and go backwards:

f-6-p + g-3-p -> erythrose-4-p + xylulose-5-p via transketolase.

erythrose-4-p + f-6-p -> g-3-p + sedo-heptulose-7-p via transaldolase.

g-3-p + sedo-heptulose-7-p -> xylulose-5-p + ribose-5-p via transketolase.

Net: 2 f-6-p + g-3-p -> 2 xylulose-5-p + ribose-5-p. xylulose-5-p and ribose-5-p can both be converted to ribulose-5-p.

 

[The8]

Oh wow, perfectly done man. Was totally confused before. I am assuming normally we don't need to know that much but just assume that any carbon rearrangements are essentially plausible? Can we make any sugar from another is the question through this.

 

[theonlytycrane]

it's a confusing pathway, but I'm glad it helped 🙂

I'm not sure about any sugar -> any sugar, but sugars are broken down into glycolytic intermediates to jump into the common metabolic pathways.