How do "dormant" yeast interact with the beer?

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massdestruction

massdestruction

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I keep reading over and over again that it's good to let the beer sit on the yeast cake after fermentation because it "cleans up flavors" and so forth. But some people claim that the flavors go away with time, independent of whether the beer was racked or left in primary (on the cake)? Can anyone explain how this is thought to work? Is it known that specific reactions take place between the "dormant" yeast and the beer? Or is this just a matter of many people claiming to have successful brews using both methods?
 
I'll make bold and try a brief explanation. Much of what goes on in fermentation and after can be described in terms of redox reactions. Thus, acetaldehyde, CH3CHO is reduced to ethanol, CH3CH2OH and diacetyl, CH3CHOCHOCH3, to acetoin, CH3CHOHCHOCH3, and then 2,3 butane diol, CH3CHOHCHOHCH3. "Dormant" yeast maintain the reduced state which is necessary for these reactions (and there are many, many more) to take place favoring reduction. If yeast are removed then exposure to the slightest amount of oxygen results in an oxidized state and not only is acetaldehyde not reduced to ethanol nor diacetyl to butane diol but other alcohols can be oxidized to aldehydes - the compounds most often associated with staling.

Take beer off yeast and the flavors will continue to change but generally in the more oxidized direction - not the one usually wants to go. For commercial breweries, where the product is packaged with all yeast removed, it is absolutely critical to have the "airs" in the package as low as possible to give a reasonable shelf life.
 
@ajdelange thanks for your reply. So, is it that the presence of the yeast in the cake, though generally dormant, are available to "activate" and scrub any transient levels of oxygen that may arise from other reactions taking place, thus preventing oxidation of other compounds? The exposure to oxygen (e.g. during racking) as a threat to the beer makes perfect sense to me, based on the types of reactions you've described above. But I'm wondering if this is the real reason that one should "leave it on the cake". In other words, is it not really the presence of dormant yeast that matters as much as not exposing the beer to oxygen during transfer? Perhaps a little of both?
 
The yeast isn't necessarily dormant just because it appears that fermentation is complete, or it's flocculated out of suspension. Especially if there is still "work" to be done, in terms of cleaning up it's byproducts. If the yeast is above it's dormancy temp, which is usually is unless we're cold crashing. It is still very active in the conditioning process, which is the true "secondary fermentation" that new brewers get confused with "secondary vessels."

Here's John Palmer's explanation of the Secondary fermentation Phase

The fermentation of malt sugars into beer is a complicated biochemical process. It is more than just the conversion of sugar to alcohol, which can be regarded as the primary activity. Total fermentation is better defined as three phases, the Adaptation or Lagtime phase, the Primary or Attenuative phase and a Secondary or Conditioning phase. The yeast do not end Phase 2 before beginning Phase 3, the processes occur in parallel, but the conditioning processes occur more slowly. As the majority of simple sugars are consumed, more and more of the yeast will transition to eating the larger, more complex sugars and early yeast by-products. This is why beer (and wine) improves with age to a degree, as long as they are on the yeast. Beer that has been filtered or pasteurized will not benefit from aging.

The reactions that take place during the conditioning phase are primarily a function of the yeast. The vigorous primary stage is over, the majority of the wort sugars have been converted to alcohol, and a lot of the yeast cells are going dormant - but some are still active.

The Secondary Phase allows for the slow reduction of the remaining fermentables. The yeast have eaten most all of the easily fermentable sugars and now start to turn their attention elsewhere. The yeast start to work on the heavier sugars like maltotriose. Also, the yeast clean up some of the byproducts they produced during the fast-paced primary phase. ...
Click to expand...

And this;

Leaving an ale beer in the primary fermentor for a total of 2-3 weeks (instead of just the one week most kits recommend), will provide time for the conditioning reactions and improve the beer. This extra time will also let more sediment settle out before bottling, resulting in a clearer beer and easier pouring. And, three weeks in the primary fermentor is usually not enough time for off-flavors to occur.
Click to expand...

That's the conditioning reactions he is talking about in the first section I quoted.

There will be plenty of non dormant yeast floating around in your beer for a long time. That's why you can drop sugar into beer in secondary (or after a week in primary) and have a krauzen pretty rapidly, the yeast isn't simply just coming off the bottom. The stuff in suspension is not dormant.

If you haven't read the discusion in here, I suggest you do. This topic has pretty much been done to death there. https://www.homebrewtalk.com/f163/secondary-not-john-palmer-jamil-zainasheff-weigh-176837/
 
massdestruction said:
@ajdelange thanks for your reply. So, is it that the presence of the yeast in the cake, though generally dormant, are available to "activate" and scrub any transient levels of oxygen that may arise from other reactions taking place, thus preventing oxidation of other compounds?
Click to expand...

Yeast do scrub oxygen, of course, but it is not enough to "neutralize" the oxygen to achieve the reduction of diacetyl and acetaldehde. Hydrogen (and in particular, the electrons it carries) must be supplied to diacetyl and acetaldehyde to reduce them.

massdestruction said:
The exposure to oxygen (e.g. during racking) as a threat to the beer makes perfect sense to me, based on the types of reactions you've described above. But I'm wondering if this is the real reason that one should "leave it on the cake". In other words, is it not really the presence of dormant yeast that matters as much as not exposing the beer to oxygen during transfer? Perhaps a little of both?
Click to expand...

Yes, it is both. Obviously, the more yeast, the more active they are and the less oxygen there is the more "reduced" conditions will be in the beer. In my brewing (which is mostly lagers) the beer goes straight from the fermenter to the keg under counter pressure. The kegs are prepared in such a way that there is no oxygen in them. The fermenter's chill bands keep the yeast in suspension. I never get diacetyl and I never get acetaldehyde. And I don't get wet cardboard flavors after more than a year. The beer is stable. Other guys on this forum who take similar steps have reported the same results.
 
The stuff in the cake is important.You may want to consider that most of my experience is with lagers and ales that are almost lagers (like Kölsch) but it is settled yeast that do the job during lagering. I do transfer with lots of yeast in suspension but within a couple of weeks they are stuck to the bottom of the keg (at least that's what we hope for i.e. clear beer shortly after kegging). At this point lagering is far from over and the beer will continue to improve for a couple of months.

Many commercial (and home) brewers take steps to increase the area over which the yeast can settle as by putting cornies to be lagered on their sides or, in the case of commercial operators, using horizontal tanks with "beechwood chips" in them (the chips have lots of surface area).
 
ajdelange said:
Many commercial (and home) brewers take steps to increase the area over which the yeast can settle as by putting cornies to be lagered on their sides or, in the case of commercial operators, using horizontal tanks with "beechwood chips" in them (the chips have lots of surface area).
Click to expand...
I was gonna mention this but you beat me to it. The totally whacked out things some homebrewers will try. Not enough brews tasted to guesstimate the effectiveness. Also, I posted incorrect info in that thread, the beechwood chips are more like 18" long.
 
That depends on the type of conical. If it is equipped with chill bands the beer is colder next to the walls than in the center and sinks pushing warm beer up the axis so you get circulation that keeps the yeast in suspension. I used to think that was a con because you couldn't get clear beer into a keg. I've changed my thinking and now consider this a big plus.

Even if the fermenter is not equipped with chill bands yeast tend to settle out on the slope of the cone - I guess this would be increased the shallower the cone. Do people notice yeast on the sides? It tends to slip down as the cone is drained but you should be able to see it there if it is indeed accumulating there.
 
ajdelange said:
Yeast do scrub oxygen, of course, but it is not enough to "neutralize" the oxygen to achieve the reduction of diacetyl and acetaldehde. Hydrogen (and in particular, the electrons it carries) must be supplied to diacetyl and acetaldehyde to reduce them.
Click to expand...

This is a uneccessarily confusing. You make it sound as if someone has
to introduce hydrogen gas into the wort. The hydrogen is supplied by
the enzyme (for example, butanediol dehydrogenase) in an acid-base
reaction, many times the hydrogen is supplied by a histidine residue in
the enzyme. Overall the chemistry can be described by redox processes
but the individual reactions are almost always aqueous acid-base
chemistry.

Yeast while alive and in a liquid medium is a chemical reaction
factory, there are probably many thousands of reactions going on
continuously, oxygen is used for some of them and free oxygen in
the wort is absorbed through the cell membrane and used for
various purposes in the yeast cell. Inorganic reactions take place
outside the cell with oxygen, and these undesirable byproducts can then
be absorbed by the yeast and transformed by the cell metabolism
to other things.

Ray
 
I usually secondary after two weeks for pipline purposes and end up with enough yeast on the bottom of the secondary after another two weeks to pitch another beer. I used to drop a carbonation tab or two into the secondary to help with the head space but have decided it isn't worth the trouble.
 
rayg said:
This is a uneccessarily confusing. You make it sound as if someone has
to introduce hydrogen gas into the wort. The hydrogen is supplied by
the enzyme
Click to expand...
As yeast's main source if reducing power is NADH I'd guess (but I do not know - I don't know that much biochem in general) that NADH is the reducing agent for conversion of diacetyl to Acetoin:

2(CH3)(CHO)(CHO)(CH3) + H+ NADH --> 2(CH3)(CHOH)(CHO)(CH3) + NAD+

and for the conversion of acetoin to 2,3 butane diol:

2(CH3)(CHOH)(CHO)(CH3) + H+ NADH --> 2(CH3)(CHOH)(CHOH)(CH3) + NAD+


Seems to me that if the H were supplied by the enzyme, it would become oxidized and unable to catalyze further reduction.

rayg said:
(for example, butanediol dehydrogenase)
Click to expand...

The main enzyme involved is diacetyl reductase. While acetoin reductase is mentioned it is thought by some to be the same as diacetyl reductase. Sort of makes sense. First you do the carbonyl on one side of the molecule and then the one on the other.

But other reductases, in particular acetaldehyde reductase (alcohol dehydrogenase) apparently reduce it too.


rayg said:
in an acid-base
reaction, many times the hydrogen is supplied by a histidine residue in
the enzyme.
Click to expand...

Electrons are sometimes passed through the enzyme (oxidative phosphorylation) but I have never heard of the enzyme supplyng the electrons. Or the H+. Those usually come from acids produced by the yeast. But as I say, I'm pretty weak in biochem.

rayg said:
Overall the chemistry can be described by redox processes
but the individual reactions are almost always aqueous acid-base
chemistry.
Click to expand...

One of the "advantages" of the Lewis model is that as all bonding involves the exchange of electron pairs and an electron pair donor is an Lewis base and an electron pair acceptor a Lewis acid all chemical reactions are acid/base reactions. Sort of the unifying field theory of chemistry. If you think my post confused people, "clarifying" by taking them to this level is bound to send them from their terminals screaming.


rayg said:
Inorganic reactions take place
outside the cell with oxygen, and these undesirable byproducts can then
be absorbed by the yeast and transformed by the cell metabolism
to other things.
Click to expand...

This is exactly what happens with diacetyl. It is formed after packaging as acetolactate is oxidized to diacetyl non enzymatically (but it's still organic). This diacetyl is then be taken up by and reduced in the cell.
 
So it seems to me then, that the cake is not important in the stages immediately following primary fermentation, because there is so much active yeast still in suspension that the reaction surface area is many times greater than what's accumulated at the bottom. But further down the line when the beer clears, then sediment activity is all there is. Not to say the sediment isn't *at all* active, but just that it's more of an issue during lagering, for example. Does that jive?

BTW, lagers are primarily what I want to brew, so this is all great info. :)
 
rayg said:
The hydrogen is supplied by
the enzyme
Click to expand...

Just thought of something - did you mean the co-enzyme ( NADH)? It is the nicotinamide ring in NADH that provides the 2 electrons and H+ and the nicotinamide is quite similar to the histidine that you mention. But NADH isn't part of the enzyme - it get's tucked into the notch in the enzyme (assuming that NADH is the reducing agens). And it delivers a proton and 2 electrons i.e. a hydrogen atom and an electron. I don't see how you can call that an acid/base reaction (except in the sense that all reaction are under the Lewis approach). It looks like a classical biochem redox to me. Maybe the nomenclature has changed in the years since I've looked at this.
 
ajdelange said:
As yeast's main source if reducing power is NADH I'd guess (but I do not know - I don't know that much biochem in general) that NADH is the reducing agent for conversion of diacetyl to Acetoin:
Click to expand...

I have confirmed that it is indeed NADH which reduces diacetyl catalyzed by diacetyl reductase (EC 1.1.1.5).

So the equations are indeed

ajdelange said:
2(CH3)(CHO)(CHO)(CH3) + H+ + NADH --> 2(CH3)(CHOH)(CHO)(CH3) + NAD+

and for the conversion of acetoin to 2,3 butane diol:

2(CH3)(CHOH)(CHO)(CH3) + H+ + NADH --> 2(CH3)(CHOH)(CHOH)(CH3) + NAD+
Click to expand...

A proton and an electron (i.e. a hydrogen atom) are passed to each of the 2 diacetyls in the first reaction which shows 2 protons and 2 electrons because we get 2 electrons and 1 proton from the NADH and need an extra from the solvent (beer). The 2 electrons and 1 proton (a hydride ion if you prefer) come from the pyridine ring in the NADH: NADH --> 2e- + H+ + NAD+ = H- + NAD+. The hydrogen ion from the solvent is needed to pair with the extra electron. This is classical bio redox (including the fact that the redox potential depends on pH).

So we know the most of the hydrogen comes from the NADH (with the rest from the solvent)but the remaining question is "Where does the NADH come from". The answer is NAD+. It is constantly being reduced by yeast in the energy generation phase of glycolysis. In fact one of the reasons alcohol is produced in fermentation is because it is a byproduct of the process which transforms NADH into the NAD+ required in that phase. IOW as long as the yeast are not totally dormant i.e. there is something for them to munch on they will produce NADH by recycling NAD+ and thus remain able to reduce diacetyl. There should be no question in anyone's mind that it is NADH which reduces acetaldehyde to ethanol: CH3CHO + NADH + H+ --> CH3CH2(OH) + NAD+. So again, as long as the basic metabolism is running, if slowly, acetaldehyde should be reduced too.
 
ajdelange said:
As yeast's main source if reducing power is NADH I'd guess (but I do not know - I don't know that much biochem in general) that NADH is the reducing agent for conversion of diacetyl to Acetoin:

2(CH3)(CHO)(CHO)(CH3) + H+ NADH --> 2(CH3)(CHOH)(CHO)(CH3) + NAD+

and for the conversion of acetoin to 2,3 butane diol:

2(CH3)(CHOH)(CHO)(CH3) + H+ NADH --> 2(CH3)(CHOH)(CHOH)(CH3) + NAD+


Seems to me that if the H were supplied by the enzyme, it would become oxidized and unable to catalyze further reduction.
Click to expand...

The reaction you wrote above is an acid/base reaction. The ultimate
source of the H+ (the acid) is some redox reaction far earlier in
the sequence, but ultimately it is used in an aqueous acid base reaction.
For a mechanism involving histidine as an acid/base, see the diagram
labeled "Serine Protease Mechanism" on this page:
http://employees.csbsju.edu/hjakubowski/classes/ch331/catalysis/olcatenzmech.html

Your molecular structure and stoichiometry above is incorrect. You've
written:
2(CH3)(CHO)(CHO)(CH3) + H+ NADH --> 2(CH3)(CHO)(CHO)(CH3) + H+ NADH --> 2(CH3)(CHOH)(CHO)(CH3) + NAD+ + NAD+

But the diketone on the left should be

(CH3)(CO)(CO)(CH3) + 2H+ + 2NADH - - > (CH3)(CHOH)(CHOH)(CH3)

Each carbonyl requires 2 moles of H, one formally H+ and one
formally H- (minus), the NADH supplies the H-, the H+ is
supplied by the enzyme usually through some acidic source such
as a protonated histidine or a protonated serine or protonated
tryptophan. A reverse reaction would be the conversion of
ethanol to ethanal:
CH3CH2OH + NAD+ ---> CH3CH=O + NADH + H+
Here the NAD+ picks up an H- and the H+ is picked
up by some part of the enzyme (a histidine or serine residue
or other etc).

But all of this is overkill for most of the members of this
list. All the op needs to know is that the yeast are chemical
factories that absorb things from solution and convert them
to something else, one thing in particular being O2.

I know you are trying to help by explaining, but if you are
going to be very technical, you have to at least be correct,
otherwise you are just creating more confusion. Your comment
about Lewis structures involving nothing but acid/base
chemistry is also not correct, but I'm not going to get into
that here because I don't think too many here would care.

Ray
 
I wish I could follow you guys...it sounds interesting...Ill prolly eventually pick up my chem book and go over it and a biochem book sooner or later...
 
rayg said:
Your molecular structure and stoichiometry above is incorrect...
But the diketone on the left should be

(CH3)(CO)(CO)(CH3) + 2H+ + 2NADH - - > (CH3)(CHOH)(CHOH)(CH3)
Click to expand...

Yes, that was a stupid mistake on my part.

rayg said:
Each carbonyl requires 2 moles of H, one formally H+ and one
formally H- (minus),
Click to expand...

Yes though you've shown it all in one step and it usually goes diacetyl -> acetoin -> butane dione.

rayg said:
the NADH supplies the H-,
Click to expand...

So at least we agree that part of the hydrogen comes from the coenzyme - that's progress.

rayg said:
the H+ is
supplied by the enzyme usually through some acidic source such
as a protonated histidine or a protonated serine or protonated
tryptophan.
Click to expand...

Why would it take it from some protonated residue on the enzyme if it were readily available from the solvent. Indeed with respect to...

rayg said:
A reverse reaction would be the conversion of
ethanol to ethanal:
CH3CH2OH + NAD+ ---> CH3CH=O + NADH + H+
Here the NAD+ picks up an H- and the H+ is picked
up by some part of the enzyme (a histidine or serine residue
or other etc).
Click to expand...

...Stryer, in Biochemistry p 320 describes exactly this reaction with the phrase "one hydrogen atom of the substrate is directly transferred to NAD+, whereas the other appears in the solvent". In the reverse reaction, especially in fermentation where the pH is low, why wouldn't it come from the solvent by way of the multiple acids produced by the yeast in order to establish the low pH levels at which they compete best? I'll agree that any acid residue with a pK < pH of the solvent would give up it's protons but why would the acetaldehyde (in the reverse reaction) prefer those? Are you perhaps so used to thinking in terms of physiologic pH that you've forgotten that fermenting beer pH is a couple of units lower?

In the reaction Quinone + 2H+ + 2e- <--> Hydroquinone no enzyme is required but the protons have to come from somewhere and somewhere is the solvent. That's why the redox potential for this reaction depends on pH as do the redox potentials computed (in, for example Mathhew and van Holde (Biochemistry - p526) for the acetaldehyde/ethanol couple. Stryer bears a copyright of 1988 and Matthew and van Holde 1996. Has something new come to light? Is there a different way of looking at things today? The literature abounds with references to this as a redox reaction - in fact it (the ethanol/acetaldehyde reaction) is often cited as an example of a biochemical redox, as I expect you must know. If I want to compute the change in Gibbs energy for this reaction I can do it using the published redox potential. If that's no longer valid, how would I do it?

rayg said:
But all of this is overkill for most of the members of this
list. All the op needs to know is that the yeast are chemical
factories that absorb things from solution and convert them
to something else, one thing in particular being O2.
Click to expand...
Agreed, but this is the science topic.

rayg said:
I know you are trying to help by explaining, but if you are
going to be very technical, you have to at least be correct
Click to expand...

I'll stand by what I've posted (with the exception of the errors I've acknowledged).



rayg said:
otherwise you are just creating more confusion.
Click to expand...

I thought my original explanation was fine as it is. Untill convinced otherwise I have to believe that these are redox reactions - I've just seen that stated i n too many places over the years. Telling readers that they are acid base, given that so many sources call them redox, seems to be more likely to confuse whether that model is valid or not. The only way I can, at this point see that as working would be the Lewis definitions. I almost thought I had it figured out because the nitrogen in the pryridine ring in the nicatinamide part of NADH is clearly a Lewis base but I'm not knowledgeable enough to take it any further than that.

rayg said:
Your comment
about Lewis structures involving nothing but acid/base
chemistry is also not correct, but I'm not going to get into
that here because I don't think too many here would care.
Click to expand...

That comment, and I'm amazed I was able to find the reference, comes from the description of the Lewis definitions in "Concise Inorganic Chemistry" by J.D. Lee. He says on p 265 "Almost all reactions become acid-base reactions under this system". I forgot the "almost". Are we quibbling over the "almost" or is this statement just flat out wrong? Again, it's an older book - 1991 copyright.

Do I need to buy a whole new set of textbooks?
 
I figured I resolve the redox-acid question by drawing a pe - pH diagram. Guess what. I can't becaue pe cancels out. While the individual half reactions are definitely redox the overall isn't.
 
ajdelange said:
Why would it take it from some protonated residue on the enzyme if it were readily available from the solvent. Indeed with respect to...
Click to expand...

You don't seem to understand why enzymes exist at all. Why does
an enzyme accelerate the rate of a reaction? Part of the answer
is that the substrate doesn't have to randomly bump into another
molecule, it is supplied by the enzyme at the active site. Another
reason is that these reactions are almost all stereoselective. The
acid has to be added to one particular face of the carbonyl.
Butanediol dehydrogenase converts butanediol to (R)-2-acetoin,
while diacetyl reductase converts butanediol to (S)-2-acetoin. You
can see that here:
http://www.genome.jp/kegg-bin/show_...apno=00650&mapscale=1.0&show_description=show



ajdelange said:
...Stryer, in Biochemistry p 320 describes exactly this reaction with the phrase "one hydrogen atom of the substrate is directly transferred to NAD+, whereas the other appears in the solvent".
Click to expand...

I would go back and reread it, since almost all of these proton transfers
involve a specific location in the active site of the enzyme. I can't
tell what Stryer really means from that brief sentence.

ajdelange said:
That comment, and I'm amazed I was able to find the reference, comes from the description of the Lewis definitions in "Concise Inorganic Chemistry" by J.D. Lee. He says on p 265 "Almost all reactions become acid-base reactions under this system". I forgot the "almost". Are we quibbling over the "almost" or is this statement just flat out wrong? Again, it's an older book - 1991 copyright.
Do I need to buy a whole new set of textbooks?
Click to expand...

Not every textbook contains completely reliable or completely
precise explanations. I can't tell what he really means because
all I have is that sentence to go by, but just based on that,
it is wrong. There is a whole class of reactions, free-radical
reactions, that are drawn with Lewis structures:
http://chemistry2.csudh.edu/rpendarvis/Radicals.html
and there are plenty of inorganic redox reactions that can
be drawn with Lewis structures.

I think part of the reason for your confusion is that you are
having some trouble understanding "oxidation/reduction" in
the context of organic molecules such as alcohols/ketones.
An organic chemist may say "I've oxidized the alcohol to
a ketone", but that is just a "formal" oxidation, meaning
no real oxidation has taken place with the alcohol, the
bonds have just been rearranged and hydrogen lost. The
real redox reaction takes place on one of the hydrogens that
is lost and the reagent. For example, the Jones reagent
"oxidizes" secondary alcohols to ketones:

http://www.adichemadi.com/engine.html#organic/organicreagents/jones/jonesreagent1.html

Nice video:
http://chemed.chem.wisc.edu/~tsts/stahl/stahl9-12/Croxidant.html

but what really happens is, if you follow the mechanism, one
hydrogen is oxidized to H+, and the CrVI is reduced to CrIV
(which then disproportionates to CrIII and CrVI.)

The enzyme reactions are ultimately very complex because
the source of the substrates and reactants is most often
some long chain of reactions taking place somewhere else
other than the spot where the final reaction takes place.
You might be able to figure out the source of everything
if you want to by examining closely the pathways in
the KEGG database link above.

Ray
 
rayg said:
You don't seem to understand why enzymes exist at all. Why does
an enzyme accelerate the rate of a reaction? Part of the answer
is that the substrate doesn't have to randomly bump into another
molecule, it is supplied by the enzyme at the active site.
Click to expand...

My understanding has always been that the enzyme binds the substrate (stereospecifically) and coenzyme in such an orientation that the concentrations of the bits that need to react are locally high enough to drive the reaction in the desired direction by Le Chatelier's principal. I can see how there might be additional "pockets" for protons in a case where a proton is required, especially where pH is high and protons rare but I guess I don't know why nature would bother to evolve the enzyme in that way if the protons are so readily available in the solvent as they are at the low pH of fermenting wort.

rayg said:
I would go back and reread it, since almost all of these proton transfers
involve a specific location in the active site of the enzyme. I can't
tell what Stryer really means from that brief sentence.
Click to expand...

I guess it is, or was, a fairly common misconception because Lehninger makes this same mistake in 2 different places in his classic. Now if you had said "almost all" or "many" in earlier posts we probably wouldn't be discussing this particular aspect because I have no doubt that this is probably often the case. I just have this lingering doubt about low pH situations.



rayg said:
Not every textbook contains completely reliable or completely
precise explanations. I can't tell what he really means because
all I have is that sentence to go by, but just based on that,
it is wrong.
Click to expand...

There are certainly many errors in textbooks but usually they are a misplaced sign or printing a C where the author meant a Cl or something of that magnitude. Here, the "error" is of a much greater magnitude. He is listing, point by point, the ways in which the Lewis definitions are distinguished from the Lowry- Brønsted, Arrhenius and other definitions. If it were dead wrong I'd think one of the reviewers/collaborators would have caught it.

rayg said:
There is a whole class of reactions, free-radical
reactions, that are drawn with Lewis structures:
Click to expand...
No question. Remember, it didn't say all, it said "almost all" and perhaps that's a stretch. I have never seen a claim of that breadth made elsewhere but I have seen statements to the effect that, for example, all complexation reactions become acid/base netralizations under the Lewis system.

rayg said:
I think part of the reason for your confusion is that you are
having some trouble understanding "oxidation/reduction" in
the context of organic molecules such as alcohols/ketones.
An organic chemist may say "I've oxidized the alcohol to
a ketone"
Click to expand...

No. I figured out a long time ago that in biological systems "oxidize" means take away a hydrogen or add an oxygen and that "reduce" means the opposite. When one says he is reducing an aldehyde he simply means that he's adding 2 hydrogens. It's probably well to remember that this is just a model in which oxidation and reduction formally refer to the changes in oxidation numbers of individual atoms within the molecule, not the molecule itself. Diactyl and acetoin both have oxidation number 0. However, according to the established rules for calculating oxidation number, each of the carbonyl carbons in diacetyl has oxidation number +II whereas in acetoin that same carbon has oxidation number 0. So when we speak of reduction of diacetyl we really mean the reduction of the carbonyl carbon. At the same time the H- ion has oxidation state -I whereas the H+ ion has state +I and the 2 hydrogens in the acetoin have oxidation number +II. Thus the carbon has been reduced by 2 and the hydride oxidized by 2 which we may think of as the transfer of 2 electrons. But it's just a model. The reaction is simply 2H + diacetyl --> acetoin. As everyone has been taught that reduction involves the transfer of electrons, and as in this case 2 electrons are transferred per carbonyl to the diacetyl (along with 2 protons) I often say it is hydrogen, and in particular its electrons) that are responsible for the reduction. Which is, I think, a pretty simple and reasonably accurate way to look at it.

Thanks taking the trouble to dig out the references. I have looked at a couple of them thus far and will check out the rest.
 
ajdelange said:
My understanding has always been that the enzyme binds the substrate (stereospecifically) and coenzyme in such an orientation that the concentrations of the bits that need to react are locally high enough to drive the reaction in the desired direction by Le Chatelier's principal. I can see how there might be additional "pockets" for protons in a case where a proton is required, especially where pH is high and protons rare but I guess I don't know why nature would bother to evolve the enzyme in that way if the protons are so readily available in the solvent as they are at the low pH of fermenting wort.
Click to expand...

As I explained, because proximity helps the rate of the reaction, and
the H+ has to be added stereoselectively. This isn't debatable, you
are talking about this as if it's 1890 and we don't know how these things
work. I showed you an example of protonated histidine earlier and
another proton transfer from a tyrosine residue in my last post.


ajdelange said:
I guess it is, or was, a fairly common misconception because Lehninger makes this same mistake in 2 different places in his classic. Now if you had said "almost all" or "many" in earlier posts we probably wouldn't be discussing this particular aspect because I have no doubt that this is probably often the case. I just have this lingering doubt about low pH situations.
Click to expand...

I doubt there is a mistake on his part, but I'm pretty sure you are
misunderstanding what's being said. I don't remember any Leninger
quote from your previous post, you were quoting Stryer and some other
guy.

ajdelange said:
There are certainly many errors in textbooks but usually they are a misplaced sign or printing a C where the author meant a Cl or something of that magnitude. Here, the "error" is of a much greater magnitude. He is listing, point by point, the ways in which the Lewis definitions are distinguished from the Lowry- Brønsted, Arrhenius and other definitions. If it were dead wrong I'd think one of the reviewers/collaborators would have caught it.
Click to expand...

Now you are confusing "Lewis Acid-Base Theory" with "Lewis Structures."

ajdelange said:
No question. Remember, it didn't say all, it said "almost all" and perhaps that's a stretch. I have never seen a claim of that breadth made elsewhere but I have seen statements to the effect that, for example, all complexation reactions become acid/base netralizations under the Lewis system.
Click to expand...

I was not talking about Lewis acids/bases. That system is usually
used in conjunction with inorganic reagents, like aluminum trichloride,
which is a Lewis acid because it can accept electrons. We were
talking about Lewis structures, which are the diagrams used to explain
organic chemical reaction mechanisms, which is what this discussion
is about.


ajdelange said:
So when we speak of reduction of diacetyl we really mean the reduction of the carbonyl carbon. At the same time the H- ion has oxidation state -I whereas the H+ ion has state +I and the 2 hydrogens in the acetoin have oxidation number +II.
Click to expand...

Nope, of the two hydrogens added to the carbonyl, once they are
added they are both neutral (one electron of the bond goes to carbon,
one to hydrogen). *Before* the addition, one H- is added and one
H+ is added, again the total is neutral.

ajdelange said:
Thus the carbon has been reduced by 2 and the hydride oxidized by 2 which we may think of as the transfer of 2 electrons. But it's just a model.
Click to expand...

But your understanding of the model is incorrect, see my previous comment.

ajdelange said:
The reaction is simply 2H + diacetyl --> acetoin. As everyone has been taught that reduction involves the transfer of electrons, and as in this case 2 electrons are transferred per carbonyl to the diacetyl (along with 2 protons) I often say it is hydrogen, and in particular its electrons) that are responsible for the reduction. Which is, I think, a pretty simple and reasonably accurate way to look at it.
Click to expand...

No, it's not accurate or event partly correct. No electrons have
been lost or gained by the carbon in going from carbonyl to alcohol.
The actual redox chemistry takes place elsewhere.

Ray
 
rayg said:
I doubt there is a mistake on his part, but I'm pretty sure you are
misunderstanding what's being said. I don't remember any Leninger
quote from your previous post, you were quoting Stryer and some other
guy.
Click to expand...

I quoted Stryer and Mathews and Van Holde previously and subsequently found the same thing in Lehninger. Let me quote the label to fig 13-6 on p 342 of the 2nd edition from the latter:

"The malate-dehydrogenase reaction, showing the path of the hydrogen atoms removed by the enzyme. One is transferred as a hydride ion to the 4 position of the pyridine ring; the other appears as a hydrogen ion in the solution. (emphasis mine).

And then from Organic Chemistry, 6th Edition, Morrison and Boyd p1102 "Acetaldehyde gains a hydride ion from NADH, and a proton from the solvent.

If I'm misinterpreting those, I guess I need to take an English course. So I think it's time to stop telling me I'm reading it wrong. If you want to tell me that the results of new research have invalidated this model I would listen but to just say all the books are wrong or I'm reading them wrong doesn't seem supportable.



rayg said:
Now you are confusing "Lewis Acid-Base Theory" with "Lewis Structures."
Click to expand...

No, I am afraid it is you who are confused on this issue. I have been talking about Lewis acid/base theory from the first time I mentioned it. This should have been clear from the wording and context. I did think that maybe you were confusing the two but just couldn't believe that this would be the case.




rayg said:
I was not talking about Lewis acids/bases. That system is usually
used in conjunction with inorganic reagents, like aluminum trichloride,
which is a Lewis acid because it can accept electrons. We were
talking about Lewis structures, which are the diagrams used to explain
organic chemical reaction mechanisms, which is what this discussion
is about.
Click to expand...

I was (talking about Lewis acids). Perhaps based on understanding this you may change your opinion about the breadth of applicability of the Lewis model (or more to the point whether Lee still belongs on my burn list?) And they do have application in organic/biochemistry. Clearly a hydrogen ion is a Lewis acid whereas a hydride ion is a Lewis base. The pyridine ring in NAD is a Lewis base.



rayg said:
Nope, of the two hydrogens added to the carbonyl, once they are
added they are both neutral (one electron of the bond goes to carbon,
one to hydrogen). *Before* the addition, one H- is added and one
H+ is added, again the total is neutral.
Click to expand...

Yep. The hydride ion has 2 electrons and 1 proton so it's charge is negative. By the rules (IUPAC or Werner and Stumm, Aquatic Chemistry - you can add them to the list of books which are in error) for the definition of Oxidation number it's Oxidation number is, therefore -I. The Hydrogen ion has a unit positive charge so it's oxidation number is +I.

After the addition the rules define the oxidation state of each atom as the charge that would remain on it after removing all ligands leaving the bonding pair with the more electronegative of the 2 bonded atoms. As carbon is more electronegative than hydrogen the C-H bond pair are assigned to the carbon and the hydrogen's oxidation state is thus +I. For the other hydrogen, oxygen is assigned the pair. One of the carbon's bonding pairs will be assigned to the oxygen in the hydroxyl. The pair, from the other hydrogen, was assigned, as noted to the carbon. So +I for the pair that goes to the oxygen and -I for the pair from the hydrogen. Net 0. Of the 2 pairs in the C:C:C bond, half of each is assigned to the (originally) carbonyl carbon. So +I + -I + 0 + 0 = 0 and the carbon started at +II and got reduced to 0. Something had to be oxidized by the same amount. That's the hydride which goes from -I to +I, also a change of 2. So to explain the redox aspect of this you really need to use the H-, H+ model (remember, this is just a model) in order to have something to reduce the carbon.



rayg said:
But your understanding of the model is incorrect, see my previous comment.
Click to expand...

No it's not. You have to read what I said. I said the oxidation number of carbon changed. I did not say it took up electrons. Of course one of the pairs involved in the double bond splits with one electron going to the oxygen (to pair with the electron from one of the hydrogens) and the other stays with the carbon to pair with the electron from the other hydrogen so it looks as if that electron is gained by the carbon. And, of course, the other is very close to it in the oxygen. In fact, I guess I'd be pretty comfortable with a LCAO model in which both new electrons are shared by the carbon (but that isn't what I was thinking of). What I did say is that as people like to think of electron transfer when redox is mentioned and electrons are transferred to the vicinity of and perhaps right to the carbon, I feel comfortable explaining it that way.




rayg said:
No, it's not accurate or event partly correct. No electrons have
been lost or gained by the carbon in going from carbonyl to alcohol.
Click to expand...
Never said they were (they are gained by the molecule). I said the oxidation number of the carbon was changed. But as I noted above it looks as if one is and perhaps the other could be under certain bond formation models.

rayg said:
The actual redox chemistry takes place elsewhere.
Click to expand...

No, it takes place right here when the bonding electrons around the carbon in question rearrange themselves into 2 single bonds rather than one double bond. This reduces the oxidation number of the carbon. That's reduction. Also, as soon as the hydride gives its electrons to the more electronegative oxygen, it is oxidized.

Different set of rules for assigning oxidation numbers would lead to a different model and different conclusions. Perhaps that's where the disconnect is.
 
Before the carbonyl was converted to the alcohol, there were two
bonds to oxygen, one bond each to other carbons, or in the case of
the aldehyde, one to carbon and one to hydrogen, total 4 bonds. After
the conversion, there is one bond to oxygen, and an additional
bond to hydrogen, total 4 bonds. There is therefore no
change in the oxidation state of the carbon.

Here is a database of enzyme reaction mechanisms:

http://www.ebi.ac.uk/thornton-srv/databases/MACiE/

If you type in dehydrogenase in the top box of the search, you get
a list of enzymes. If you click on one of the enzymes in the list
you get a detailed description of the mechanism.

For malonate dehydrogenase:

http://www.ebi.ac.uk/thornton-srv/databases/cgi-bin/MACiE/getPage.pl?id=M0021

"Step 1 Asp278 deprotonates Lys183, which deprotonates the hydroxyl group of malate,
which eliminates a hydride ion that is added to NADP.
Step 2 The reduced intermediate decarboxylates with concomitant double bond rearrangement.
Step 3 The oxyanion collapses, reducing the C=C with concomitant deprotonation of Tyr112.
Step 4 Tyr112 deprotonates Lys183 which deprotonates Asp278 in an inferred return step."

Note the deprotonation (protonation going the other way):
"Asp278 deprotonates Lys183, which deprotonates the hydroxyl group of malate,
which eliminates a hydride ion that is added to NADP."

"Tyr112 deprotonates Lys183 which deprotonates Asp278 in an inferred return step."

No solvent is involved, protonation/deprotonation occurs via donation/acceptance
from the amino acid residues (aspartic acid, lysine and tyrosine in this case).

This is the general way that all of these types of enzyme reactions work.

I don't know if you are misunderstanding, misquoting (deliberately or not)
or simply making up those textbook quotes, but they are wrong. Given that
you've misunderstood molality in another thread, written incorrect stoichiometry
and structures in this one, and don't understand the oxidation state of
carbon in organic molecules (see above), I'm afraid I can't trust what you are saying.
You seem to be trying to "win" this argument, but it's a waste of time
and therefore I'm bowing out. I've provided enough online references
from reputable sources for you to figure this out.

Ray
 
Okay then class. Did you take good notes? there WILL BE a test on this that will constitute 100% of your membership status.

1st Question: What is NADH?






Great stuff guys. Way too heady for me but was fun to read and insightful to some degree.
 
To leave the cake or not leave the cake that is the question..haha good stuff though I wish i could follow and give an opinion..

btw ray g what are your qualifications? just curious
 
CrystallineEntity said:
It's a scientific Showdown!!!!
Click to expand...

It's common for there to be disagreements between scientists (and I'm not one, BTW). There are ways to resolve these but egos sometimes get in the way. For example, on perhaps the simplest point in this discussion: the change in oxidation state of a carbonyl carbon when it is "reduced" to an alcohol, e.g. HCHO + 2H ---> CH3(OH), there are "rules" by which the oxidation state can be determined. The ones promulgated by the IUPAC (an international body that sees to such things for chemists) are quite simple to the point that I can paste them in here:

* * * * *
"Oxidation State

A measure of the degree of oxidation of an atom in a substance. It is defined as the charge an atom might be imagined to have when electrons are counted according to an agreed-upon set of rules: (l) the oxidation state of a free element (uncombined element) is zero; (2) for a simple (monatomic) ion, the oxidation state is equal to the net charge on the ion; (3) hydrogen has an oxidation state of 1 and oxygen has an oxidation state of -2 when they are present in most compounds. (Exceptions to this are that hydrogen has an oxidation state of -1 in hydrides of active metals, e.g. LiH, and oxygen has an oxidation state of -1 in peroxides, e.g. H2O2; (4) the algebraic sum of oxidation states of all atoms in a neutral molecule must be zero, while in ions the algebraic sum of the oxidation states of the constituent atoms must be equal to the charge on the ion. For example, the oxidation states of sulfur in H2S, S8 (elementary sulfur), SO2, SO3, and H2SO4 are, respectively: -2, 0, +4, +6 and +6. The higher the oxidation state of a given atom, the greater is its degree of oxidation; the lower the oxidation state, the greater is its degree of reduction."

* * * *


I would encourage each of you to apply these rules to the simple reaction above and see what you think the oxidation states are in the formaldehyde (left side) and methanol on the right side. Actually I'll do it for you as it's so simple and you can check my work if you like. In the formaldehyde (HCHO) there are two hydrogens (oxidation state +1 each according to the rules) and 1 oxygen (-2) according to the rules. The sum over all the atoms must be 0 so the oxidation state of carbon must be 0. In the methanol, CH3(OH), there are 4 hydrogens (+4) and one oxygen (-2) so the carbon must have an oxidation state of -2 because +4 -2 -2 = 0. The oxidation state of the carbon has been changed from +2 to 0, a reduction of 2. Refer again to the rules for hydrogen. The hydrogen atoms on the left each have an oxidation state of 0. On the right they have an oxidation state of +1 each. Thus the carbonyl alcohol has incurred reduction by 2 and the hydrogens oxidation by 1 each for a total of 2 and everyone should be happy. Matching oxidation and reduction to get chemical equations balanced is one of the main reasons for doing all this.

Now if you don't like that answer there are 2 approaches you can take. The more constructive one is to say "Gee, there's a discrepancy here between what the IUPAC rules give and what I compute. Why is that?" and that's what a good scientist does. The other is to simply declare the IUPAC rules and any textbook that uses them as wrong or, worse still, to accuse your faithful reporter of doctoring what he copied and pasted from the IUPAC website for the sake of winning an argument. This sort of thing, unfortunately, does happen, but is, fortunately, typically limited to those on opposite sides of the global climate change debate.

But the IUPAC rules are not wrong. The rules for assignment of oxidation number are, to quote one text "somewhat arbitrary" and the phrase "agreed-upon" in the IUPAC definition implies that. If my correspondent does not wish to subscribe to the IUPAC rules that's fine but he should, in that case, state whose rules he is following and what they are. It may be that in his industry a different set of rules are used. As an example of this, the "standard conditions" for redox potentials are different for biochemists than chemists of other stripe. In such a case a reasonable scientist says "There's the discrepancy! You are using IUPAC rules but in the YYY field we use the XXX rules." To say "IUPAC is wrong" is an unfortunate response because not only is it antiproductive but it tends to destroy the credibility of the one making such a statement.

For full disclosure: when I calculated the change in oxidation state earlier I used a slightly different, but consistent, set of rules (which you can find in chemistry textbooks or at http://www.newworldencyclopedia.org/entry/Oxidation_state where the IUPAC rules are also stated - look under "From a Lewis structure"). They give the same result but that doesn't necessarily mean that all sets of rules should.

I swore I wouldn't take any more flame bait here but I didn't realize others were following. I'm certainly happy that I was able to provide some entertainment and hope that anyone who has been following along will do his own research and come to his own conclusions.
 
ajdelange said:
........ As an example of this, the "standard conditions" for redox potentials are different for biochemists than chemists of other stripe. ............
Click to expand...

As I was reading the thread, this was my exact thought as to the source of the disagreement

ajdelange said:
I swore I wouldn't take any more flame bait here but I didn't realize others were following. I'm certainly happy that I was able to provide some entertainment and hope that anyone who has been following along will do his own research and come to his own conclusions.
Click to expand...

Thanks for the entertainment! For what it is worth, I'd have to say I found no flaws in your statements. I have taken quite a few advanced biochemistry classes (it has been a few years though). I've also conducted my fair share of enzyme assays but am mostly doing DNA/RNA stuff lately, so my hard core biochem. is a little rusty. My organic chemistry is much rustier.

:off:

Also, I have a pdf copy of Understanding Alkalinity and Hardness pt II and I love it! It got me into using lime to reduce my temporary hardness. This made a big change in smoothing out the bitterness of my beers. Thank you!!!!

The one thing I've always been curious to know, and keep meaning to test, is just how quickly will the temporary hardness come to equilibrium with CO2 in the air? I keep mulling this over in my head. If I bubbled air through a tank of water (say 10 gal) how quickly with it come to equilibrium (and remove a good chunk of the temporary hardness)? My ideal situation would be to have a large water tank with a spigot that I'd fill with water, bubble for a week or whatever, then drain off what I need for a batch, and then just top off and keep on bubbling. My understanding is that this is the same as removing hardness by boiling. All the boiling does is to speed up the reaction. So if one were patient enough, could one just put air bubbler on and wait?

ps. I 'd be tempted to first run the air through a water trap to help keep evaporation to a minimum.
 
pjj2ba said:
The one thing I've always been curious to know, and keep meaning to test, is just how quickly will the temporary hardness come to equilibrium with CO2 in the air? I keep mulling this over in my head. If I bubbled air through a tank of water (say 10 gal) how quickly with it come to equilibrium (and remove a good chunk of the temporary hardness)? My ideal situation would be to have a large water tank with a spigot that I'd fill with water, bubble for a week or whatever, then drain off what I need for a batch, and then just top off and keep on bubbling. My understanding is that this is the same as removing hardness by boiling. All the boiling does is to speed up the reaction. So if one were patient enough, could one just put air bubbler on and wait?
Click to expand...

Slowly! We (as respiring creatures) wouldn't be able to regulate blood pH by CO2 exchange were it not for an enzyme (so this isn't that far OT), carbonic anhydrase which speeds the dissolution thousand fold (or more - that's just a guess). If I do add chalk to water to create an authentic profile I bubble pure CO2 through and it often takes 24 hr or more to get all the chalk dissolved.

But this probably isn't the best way to remove alkalinity. As the air scrubs out CO2 lime will precipitate and ultimately the mix will come to the CO2/carbonic/bicarbonate/lime equilibrium which is determined by air CO2 content. That's about 8 something pH and there can be a fair amount of alkalinity in the water at that level. The way to do this is to sparge with air but at higher temperature where the CO2 is less soluble, more leaves, and more CaCO3 precipitates. This is of course the standard way of decarbonating. You don't even have to bring it to a boil as long as you are sparging with air. The reason you need to boil if you don't is so that steam bubbles can do the sparging job.
 
ajdelange said:
...there can be a fair amount of alkalinity in the water at that level.
Click to expand...

I should have gone on to say that this residual level is about what you'd get by boiling or lime treatment i.e. you will get decarbonated to about the same extent if you are willing to wait. Perhaps a twist on this is that water that is reasonably super saturated wrt CaCO3 may not form a precipitate at all or so slowly that it effectively stays super saturated. Adding some finely divided chalk to provide nucleation sites may help in this situation.
 
boss429 said:
To leave the cake or not leave the cake that is the question.
Click to expand...

In all the arcana I guess that question didn't really get answered. You want the beer to have low ORP (oxidation reduction potential) so that 1) existing carbonyl carbons (diacetyl, acetaldehyde) will be reduced and 2) no reduced (saturated) carbons get oxidized (to trans 2 nonenal or other manifestations of staling). The other thing you want is low pH.

The basic reduction reaction is

R(CO)R1 + NADH + H+ --> R(HCOH)R1 + NAD+

R and R1 stand for the rest of the molecules. You want CO converted to HCOH (i.e the carbon reduced). CO is what's responsible for the nasty taste/smell of acetaldehyde (and any other aldehyde) and diacetyl (and other VDK's). Even though diacetyl contains 2 carbonyl groups (the second one is part of R1), reducing either one reduces the strength of the objectionable flavor dramatically. So you want the reaction to procede to the right. LeChatelier's principal says that you do that by increasing the concentration of one or more of the reactants (on the left) and/or decreasing the concentration of one or more of the products (on the right). Active yeast do both. They:

1) produce acid (H+ ~ low pH)
2) produce NADH and
3) produce the NADH from NAD+.

Thus you want yeast present. The question is when and for how long. The following comments are based on lagers where diacetyl seems to be more of a bete noir than in ales.


Diacetyl not produced by yeast. The yeast form acetolactate and that gets non enzymatically oxidixed to diacetyl which must be reduced. So how long you keep the beer on the yeast depends on how you plan to manage diacetyl. In the Narziß method, the yeast are pitched cold and the temperature is allowed to rise slowly during the ferment. At the conclusion the temperature is raised, for a day or 2 accelerating the formation of diacetyl from acetolactate and the beer is then lagered, on the yeast, for a relatively brief period after which it can be racked or filtered off. Good for a commercial brewer.

In the traditional method the pitch is at a higher temperature (but still below 50 °F) and is not raised at the end so much less acetolactate is converted. Instead of that taking place over a day or two the conversion, and subsequent reduction, take place over the months of traditional lagering. At that time the beer is transferred off the yeast and packaged. It is also diacetyl (and acetaldehyde) free and relatively stable.

If you can keep the beer on the yeast, which as a home brewer, you can, beyond the lagering period you can keep the beer fresh for a year or more. Eventually, however, the yeast do give up the ghost and diacetyl begins to be noticeable again.

Last nite we were comparing my Boh Pils (done the traditional way) to the local GB guy's (obviously done the Narziß way because a commercial operation cannot afford the long lagering time). Very similar beers - equally free of diacetyl anyway.

An interesting third method has now emerged in the form of a product called Maturex. This is the enzyme acetolactate decarboxylase which takes acetolactate directly to acetoin thus bypassing the oxidation to diacetyl and subsequent reduction by the yeast. This means no raised temperature step (diacetyl rest) and very short lagering are possible (just long enough to establish low ORP and clean up acetaldehyde). Very good for commercial brewers.
 
ajdelange said:
Diacetyl not produced by yeast. The yeast form acetolactate and that gets non enzymatically oxidixed to diacetyl which must be reduced.
Click to expand...

Diacetyl is certainly produced by yeast. The reaction of acetoin to
diacetyl is reversible that's why the enzymes that do that have
two names in the KEGG database:

(R,R)-butanediol dehydrogenase;diacetyl reductase [EC:1.1.1.303]

and

diacetyl reductase [(S)-acetoin forming];(S)-acetoin dehydrogenase [EC: 1.1.1.304]
(see http://www.genome.jp/kegg/pathway/map/map00650.html
top center)

The pH inside the cell is not the same as the pH of the wort, so
it's not a good idea to relate wort chemistry to cell metabolism.

Ray
 

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