[gonzo]

Hi everyone,

I have some confusion about some steps in glycolysis where I can't account for the loss and addition of a hydrogen atom. Hopefully you can help.

The first is from the step between glyceraldehyde-3-phosphate to 1,3-bisphosphoglycerate (not sure how to type formulas here, so bear with me please, and P is the whole phosphate group)

H
C=O
CHOH
CH2-O-P

->

P-O-C=O
       CHOH
       CH2-O-P

with the payoff of NAD+ -> NADH +H+

Now I count 5 hydrogen atoms in glyceraldehyde-3-phosphate and if 2 of them are transfered to NADH +H+ through oxidization how come there are still 4 hydrogen atoms in 1,3-biphosphoglycerate?

My second issue is in the last step - phosphoenolpyruvate to pyruvate. This time a hydrogen atom is added, but I don't understand how that can happen while producing 2 ATP, where is all this energy coming from in this reaction? I read that PEP is very unstable, but guess I'd like some more information to understand better.

O-
C=O
C-O-P
CH2 (double bonded carbon)

->

O-
C=O
C=O
CH3

payoff 2 ADP -> 2 ATP

[Arkcon]

Dangit.  Where did that mono-atomic hydrogen go, it was right here a min ... *farts fire*  Ah, there it went. 

OK.  Look at this Wikipedia page here: http://en.wikipedia.org/wiki/Glycolysis#Pay-off_phase (scroll down a bit)  The extra hydrogen comes as a cation for the phosphate group.  You will encounter more problems like these, with unaccounted for reactants, and I'm sorry the biology texts aren't more explicit.  Maybe some are better.  But the environment with in the mitochondria contains all the H+ the reactions will need.

[Babcock_Hall]

There are a couple of questions here.  The thermodynamic favorability of transferring a phosphoryl group from PEP to ADP is partly a result of the favorable free energy of enolpyruvate becoming pyruvate (a keto-enol equilibrum issue).  I found a citation and posted it here about six months ago.  There are two PEPs produced for every one glucose; therefore, a total of two ATPs are made.

The other thing I would suggest is to make sure that you clearly understand the differences among proton transfer, hydrogen atom transfer, and hydride ion transfer.  IMO following the electrons is often more enlightening about some biochemical processes than following the protons is.

[Dan]

(not sure how to type formulas here, so bear with me please, and P is the whole phosphate group)

See Formatting Guidelines.

[gonzo]

Thank you both.

Yes, I guess my problem is with discerning which transfer it is. I'm looking for electron transfers through this whole process as I'm basically trying to keep score of the energy going towards the electron transport chain. I have the same problem at a couple of points in the citric acid cycle, not being able to explain a couple of hydrogens.

If we take the first example I posted, the wiki article says that the Pi is an HPO4-- that dissociates to give the extra H+ (although I'm also looking for an electron here.. what am I missing?)

I'm using Campbell/Reece's Biology intro book and I like it, but it doesn't go into a lot of detail about how to better follow the electron transfer and it's not the easiest area for me. There's useful info on how energy is transferred in redox reactions and then it proceeds with cellular respiration, but I feel I'm lacking some info to analyze each step better.

To take another example: In the citric acid cycle isocitrate drops a CO2 and then is oxidized to give another NAD+ -> NADH +  H+ payoff. But again I can't account for both electrons:

isocitrate:

     COO-
     CH2
    HC-COO-
HO-CH
     COO-

->

alpha-ketoglutarate:

COO-
CH2
CH2
C=O
COO-

so here the middle COO- being dropped means that as CO2 is lost we are left with the electron from the anion and can pick up an H+ from the enviroment to bind to ketoglutarate I assume? That accounts for the ´3 carbon and that way the other 2 hydrogen are accounted for when oxidized. Is this correct? I'm trying to piece it all together.

[Babcock_Hall]

Assigning and Using Oxidation Numbers in Biochemistry Lecture Courses,  Journal of Chemical Education 2000;77(11):1428-1432.
http://pubs.acs.org/doi/pdf/10.1021/ed077p1428

Bentley et al., Oxidation Numbers in the Study of Metabolism, BIOCHEMISTRY AND MOLECULAR BIOLOGY EDUCATION, Vol. 30, No. 5, pp. 288–292, 2002
https://facultystaff.richmond.edu/~jbell2/288.pdf

One way you can account for electrons is to assign oxidation numbers to all carbon atoms.  When the oxidation number becomes more positive, there was a loss of one or more electrons.  One way to assign oxidation numbers is:

ON (C) = (# of bonds to oxygen or nitrogen) - # of bonds to hydrogen).

This is the method suggested in the first of the two references.  The second reference above uses a different formalism, but the two methods provide the same answer for NAD:  NADH is two electrons reduced, relative to NAD (from the transfer of a hydride ion).  I am not sure I understand your question with respect to isocitrate, but why don't you try assigning oxidation numbers and see what happens.  Please let us know what you find.

[Babcock_Hall]

It looks to be in balance when I assign oxidation numbers to all of the carbon atoms.

[Yggdrasil]

Thank you both.

Yes, I guess my problem is with discerning which transfer it is. I'm looking for electron transfers through this whole process as I'm basically trying to keep score of the energy going towards the electron transport chain. I have the same problem at a couple of points in the citric acid cycle, not being able to explain a couple of hydrogens.

If we take the first example I posted, the wiki article says that the Pi is an HPO4-- that dissociates to give the extra H+ (although I'm also looking for an electron here.. what am I missing?)

I'm using Campbell/Reece's Biology intro book and I like it, but it doesn't go into a lot of detail about how to better follow the electron transfer and it's not the easiest area for me. There's useful info on how energy is transferred in redox reactions and then it proceeds with cellular respiration, but I feel I'm lacking some info to analyze each step better.

To take another example: In the citric acid cycle isocitrate drops a CO2 and then is oxidized to give another NAD+ -> NADH +  H+ payoff. But again I can't account for both electrons:

isocitrate:

     COO-
     CH2
    HC-COO-
HO-CH
     COO-

->

alpha-ketoglutarate:

COO-
CH2
CH2
C=O
COO-

so here the middle COO- being dropped means that as CO2 is lost we are left with the electron from the anion and can pick up an H+ from the enviroment to bind to ketoglutarate I assume? That accounts for the ´3 carbon and that way the other 2 hydrogen are accounted for when oxidized. Is this correct? I'm trying to piece it all together.

If you've taken organic chemistry, it might be useful to look at the arrow pushing mechanism shown on the wikipedia page for isocitrate dehydrogenase (https://en.wikipedia.org/wiki/Isocitrate_dehydrogenase#Catalytic_mechanism).  Some biochemistry text books (e.g. Voet & Voet) may also have a more detailed discussion of the enzyme's mechanism than a general introductory biology text.

[Babcock_Hall]

@OP, Note that the hydride transfer takes place before decarboxylation, not after decarboxylation.  How does the timing facilitate the reaction?

[gonzo]

Thank you everyone for your replies, I've been away a bit with illness, but tried to figure out this oxidation number thing now from the linked pdf. Unfortunately it seems each time I try to count I get different numbers..

I was using the isocitrate -> alpha-ketoglutarate example and my numbers just keep turning out all wrong.. counting oxidation numbers on the carbon I go from +9 to +6

[Babcock_Hall]

For the five carbon atoms of alpha-ketoglutarate, I get +3, +2, -2, -2, and +3.  We need to also remember that CO₂ is produced, and its carbon has an oxidation number of +4.  Why don't you provide the six oxidation numbers for the six carbons of isocitrate?

[gonzo]

I did forget the CO2, but I still seem to manage new numbers, here's what I get:

Isocitrate:

     COO-
     CH2
    HC-COO-
HO-CH
     COO-

Using electrons gained/lost +n = charge

COO- means 3 electrons lost and the charge is -1 so:
-3 +n = -1
meaning n = +2
and for the subsequent carbons I get -2; -1; +2; 0; +2; in total +3

alpha-ketoglutarate:

COO-
CH2
CH2
C=O
COO-

I get +2; -2; -2; +2; +2; in total +2

So my numbers don't match up with yours and they don't seem to match up with one another either..

[Babcock_Hall]

I am not sure, but you may be confusing formal charge with oxidation number, or perhaps the issue revolves around which atom has the charge.  None of the carbon atoms is charged in alpha ketoglutarate (two oxygen atoms are).  One way to calculate the oxidation number of carbon is ON (C) = (# of bonds to O, N, etc.) - (# of bonds to H).  This is the formalism suggested by Halkides (2000).  Another way to calculate the oxidation number of carbon is Σ electrons gained (lost) + n = 0.  The letter n represents the oxidation number in the formalism given by Bentley et al, 2002.  For carbon number 2 (the ketone carbon) -2 is the number of electrons lost to oxygen.  -2 + n = 0, and n = 2.  When an atom is charged, the charge replaces the zero in the formula above.

[gonzo]

The latter approach was the one I was using, I'm not sure what you say I'm doing wrong? Like you said, with the ketone it is -2 + n = 0 so n = 2 which I also wrote it as..

I'm a little confused about how to do the charges, I thought I had to do it for each group, so that in COO- carbon has three bonds to oxygen (-3) and the oxygen has a negative charge, thus I thought it would be -3 +n = -1 so that n = 2.

How do I incorporate the charges of oxygen if not in that way?

[Babcock_Hall]

With respect to carboxylate groups, according to the formalism in the first paper
ON (C) = 3 - 0 = +3
This paper does not treat oxidation numbers on atoms other than carbons.

I am more familiar with the formalism in the first paper than I am with the formalism of the second, but here goes:
For the carbon atom I would calculate -3 + n = 0; therefore n = +III
For the oxygen atom with a negative charge, I would calculate (+1) + n = -1 ; therefore n = -II
For the oxygen atom w/o a negative charge, I would calculate (+2) + n = 0 ; therefore n = -II

In the three bonds to oxygen, the oxygen owns the electrons.  Also the oxygen atom, not the carbon, has a formal charge of -1.  The Bentley paper gives +III as the oxidation number for Carbon-1 of the acetate ion in Table III.

What do you calculate for isocitrate?