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We continue our glycolyis, we started of yesterday where we considered different metabolic processes
that actually go on in the body. And if we lo at the first slide here, the overall catabolic
processes lo at the break down of lipids polysaccharides and proteins. Now the break down of lipids
gets into components of fatty acids and glycerol. The polysaccharides break down to monosaccharide
and the proteins breakdown to amino acids.
Now in the anabolic processes we have these broken down amino acids and other factors
that actually get on into building up the other macro molecules that are required for
our bodily functions. Now what we are interested in is the break down of the monosaccharide
particularly the processes of glycolysis that takes glucose and breaks it down into pyruvate
and later on we will see how this pyruvate then gets on into the tricarboxylic acid cycle
or the kreb cycle to finally get to water and carbon dioxide and an off shoot of that
is the production of ATP where we studied oxidative phophorylation in the different
complex processes that require a number of electron transfer cofactors as well as certain
enzymes.
The overall equation for glycolysis is the break down of glucose into 2 pyruvate. Now
what we have here is we see how ATP is produced. As we continue with all the steps we will
see how ATP is produced in some of the steps but in the first set that we did, we found
ATP consumption.
This is the number of steps that we considered in our last class. We had glucose going to
glucose-6-phosphate, the enzyme being hexinase. We have to remember that a kinase is a transferase
that transfers a phosphate group, so in this process we had the break down of ATP to ADP
and the phosphate was transferred to glucose. This then went on to form the ketose from
aldose so we had fructose-6- phosphate that had the enzyme phosphoglucose isomerase acting
on it because this is an isomer the aldose and the ketose that we have here. The fructose-6-
phosphate then when on to form fructose1, 6-bisphosphate where we required another ATP
to be broken down and the enzyme used there was another kinase but in this time it was
phosphofructinase.
After this particular step we have aldolase come into the picture, aldolase is actually
what breaks up the 6 carbon membered ring into 2, 3 carbon membered rings where we have
glyceraldehyde-3-phosphates and dihydroxyacetone phosphate. Now these two are intertransferable
by the enzyme triose phosphate isomerase and we also learnt that the equilibrium of this
enzyme is such that the dihydroxy acetone phosphate is preferred to be formed in larger
quantities but to shift the equilibrium to the right where glyceraldehyde -3- phosphate
will be utilized because that is what continues the glycolysis process. So we have to have
glyceraldehyde-3- phosphate that is going to the glycolysis or continue with the breakdown
and we have to remember here that we now have 2, 3 carbon units instead of the 6 carbon
unit that we started off with.
So this is our step number 6, if you remember we reached glyceraldehyde-3- phosphate after
the action of aldolase on fructose 1, 6 bisphosphate. So our next step is the formation of 1, 3
bisphosphoglycerate from glyceraldehyde-3-phosphate. So we are now adding a phosphate but we have
not used ATP in this step, we are using Pi. So the reaction here is glyceraldehyde-3-
phosphate plus *** plus Pi in a reversible step giving you 1, 3 bisphosphoglycerate plus
NADH plus H plus.
So this is our sixth step where what has been done is we have added a phosphate using a
*** plus forming NADH, in a reversible step. Now in this case what happens this is the
only step in glycolysis where *** plus is actually reduced to NADH. In this step that
uses glyceraldehyde-3- phosphate dehydrogenase because you have this hydrogen being removed
from the aldehyde group and you have a phosphate introduced by the Pi that comes in as a factor
to form 1, 3 bisphosphoglycerate. And what we have here now is if you notice we are going
to actually create an acid from this because this now as a form where we have C double
bond OPO3 2 minus. This was initially an aldehyde and we are gradually getting to an acid.
Now what is going to happen is in this case there is a cysteine thiol, all of us know
that in the cysteine residues we have the SH thiol group. Now what happens is the cysteine
thiol is present at the active site of the glyceraldehyde-3- phosphate dehydrogenase
and that actually forms an intermediate with the acetaldehyde of the glyceraldehyde-3-
phosphate, the aldehyde that reacts with the cysteine thiol to form what is called thioemiacetal.
So what happens here like we learnt yesterday in one of the reactions we had lysine come
into the picture, here we have cysteine we will also seen one of the enzymes has hystedene.
So these are the particular side chain of the enzymes that are going to take part in
the overall reactions for the transformation to occur.
Now in this case
what we have, here is our enzyme with the cysteine group attached to it. Anyway what
we have here, you can see the enzyme has the SH group attached to this. Now what is happening
here, this SH creates a thioemiacetal intermediate by reacting with the aldehyde of the glyceraldehyde-3-
phosphate. So this is our glyceraldehyde. So this is carbon number 1, carbon number
2, carbon number 3 because now remember we have broken down our glucose into a 3 carbon
system.
So the enzyme, the cysteine group that is present in the active site of glyceraldehyde-3-phosphate
dehydrogenase is forming a thioemiacetal intermediate with the aldehyde of the glyceraldehyde-3-
phosphate. So now we have this linked to the enzyme, now this part is linked to the enzyme
then *** plus is reduced to NADH in the process we get an acyl thioester intermediate where
is this thioester? You see this CH OH is now C double bond O and the S is part of the cysteine.
So it becomes a thioester an acyl thioester intermediate.
Now what happens is then the Pi that is the phosphate comes and cleaves the S C bond and
your enzyme is regenerated. And in the event what happens, the phosphate adds on to the
first carbon atom. So now you have 1, 3 bisphosphoglycerate. Is that clear? So what happens is we have
this acyl thioester that is attacked by the Pi which actually cleaves the substrate or
rather in this case it is now become the product and regenerates the enzyme. So again we have
enzyme cysteine with SH that now is ready to act upon another glyceraldehyde-3- phosphate.
So basically what we have here is from glyceraldehyde-3- phosphate with the help of this cysteine of
the enzyme we have the formation of thioemiacetal intermediate that then forms an acyl thioester
intermediate which then is attacked by Pi which yields 1, 3 bisphosphoglycerate and
regenerates the enzyme. So that is our step. Now we have landed up with 1, 3 bisphosphoglycerate,
now what happens is this phosphate is taken up by ADP to produce ATP. So we have now the
first reaction where we have produced ATP in our glycolysis step. So ADP, one of the
phosphates from 1, 3 bisphosphoglycerate is the first one is lost to ADP which produces
ATP, in the event it forms 3 phosphoglycerate.
So we now have an acid here not an aldehyde at carbon number 1 and all of this remember
is enzymatic. So initially we had, what was the first thing that we had here? We had glyceraldehyde-3-
phosphate. So this was our aldehyde, I have to create pyretic acid so I have to create
an acidic group at position number 1. The first step in doing that is the formation
of 1, 3 bisphosphoglycerate. In the formation of 1, 3 bisphosphoglycerate the enzyme glyceraldehyde-3-
phosphate dehydrogenase utilizes a cysteine thiol to bring about the reaction. And in
the next step we lose the phosphate to ADP which produces ATP and we have 1, 3 bisphosphoglycerate
plus ADP go to 3 phosphoglycerate plus ATP.
The next step is a mutase, if you remember we considered in our last class all the enzymes
that are actually going to take part in the glycolysis steps. The mutase shifts the phosphate
from one position to the other. So the mutase enzyme shifts a myt from one carbon atom to
another, in this case the 3 phosphor is shifted to the second carbon, so the product is 2-
phosphoglycerate.
So instead of 3 phosphoglycerate after the action of this enzyme phosphoglycerate mutase
which shifts the phosphate myt from the third carbon atom to the second carbon atom. So
the phosphate is shifted from the OH on C3 to the OH on C2, so we have a shift here.
Now in this case from 3- phosphoglycerate to 2-phosphoglycerate the enzyme which is
phosphoglycerate mutase has a histidine. And this histidine is present again in the active
side of the enzyme and what it does is, it helps in the phosphate transfer by accepting
the phosphate and then donating it again. So the mechanism is such that the histidine
accepts the phosphate from 3-phosphoglycerate and then donates it back but to the second
carbon atom so this would obviously depend upon the position of where the carbon atoms
are actually located in the active site of phosphoglycerate mutase.
And in the intermediate that we have here, we have an additional actually 2, 3 bisphosphoglycerate
and this finally looses the phosphate that is the third carbon atom into becoming 2-phosphoglycerate.
So now we have already formed the acid we are gradually getting to the step where we
are going to form an enol and then a pyruvate.
Now in the ninth step we have enolase, enolase actually we have now reached 2-phosphoglycerate,
we had 3-phosphoglycerate we have now formed 2-phosphoglycerate by an enzyme called mutase.
After 2- phosphoglycerate we have now an enolate intermediate. What is that enolate intermediate?
It has formed with the loss of H plus here which has formed double bond at this position
between the carbon atom 1 and 2 we have a double bond formation. Then with the loss
of OH we have a double bond between carbons 2 and 3.
So this negative charge of the oxygen will come back here, this forms a double bond here
and the OH is lost. Is the mechanism clear? So we have the O come back here to form the
double bond then this double bond shifts here, the OH is lost. So we get phosphoenolpyruvate,
so we are gradually getting to our final step which is going to be the formation of pyruvate
from glucose.
So we have in our ninth step enolase that catalyses the formation of phosphoenolpyruvate
from 2-phosphoglycerate which is actually nothing but the dehydration reaction. So essentially
you are losing this hydrogen and this OH into forming a double bond between carbon atoms
2 and 3, that is essentially what is happening. So you have CH2 double bond C with a phosphate
attached and C O minus a carboxylate group. So essentially you are having a dehydration
where you are loosing this hydrogen and this OH and you are forming a double bond. Now
what else do you have to loose to form the pyruvate? The phosphate, so that will be our
next step but there is some other notes here where we have the dehydration reaction is
magnesium depended.
The 2 magnesium ions interact with oxygen atoms of the substrate carboxyl group at the
active site. So what happens is we have a magnesium interacting with the oxygen atoms
of the carboxylic group of 2- phosphoglycerate that helps stabilize. We have two O minus
so what happens is the magnesium ions stabilize this enolate ion and then we have an extraction
of the hydrogen from carbon number 2 when a Lysine actually helps in doing that. So
the active site in this case would be the presence of Lysine and the presence of 2 magnesium
ions that would help stabilize the enolate intermediate.
So our next step now is the formation of pyruvate where we have to loose the phosphate. This
is the last step where we have formed pyruvate, pyruvate is CH3 C double bond O C OH that
is pyruvic acid. So we have phosphoenolpyruvate that was formed from 2 phosphoglycerate, from
2- phosphoglycerate we formed phosphoenolpyruvate and we now know that to form pyruvic acid
we have to have CH3 C double bond O COOH or other pyruvate would be CH3 C double bond
O COO minus. So we have to loose this phosphate. And who takes up the phosphate? ADP.
So now we form another ATP here in this step of glycolysis. And the enzyme here is pyruvate
kinase that is shifting the phosphate, remember the kinase shifts the phosphate. It is taking
the phosphate from phosphoenolpyruvate and giving it to ADP and forming ATP. The intermediate
is enol pyruvate that actually then forms the pyruvate. So our last step is phosphoenolpyruvate
plus ADP going to pyruvate plus ATP. So eventually we broken down the glucose and we have finally
formed pyruvate. This pyruvate is then later on going into through acetylCoA it is going
to go into the tricarboxylic acid cycle and from there we will have the production of
carbondioxide and water which you will see in a later class.
Now the formation or the phosphate transfer from PEP. What is PEP? Phosphoenolpyruvate
to ADP is a spontaneous reaction. PEP has a larger delta G of phosphate hydrolysis than
ATP. What is that mean? It is going to loose the phosphate very easily and ADP will take
up this phosphate to form ATP and because you have these anionic residues there are
certain cations that bind to the anionic residues of the active site of pyruvate kinase to bring
about this particular reaction. So remember that apart from the electron transfer cofactors,
apart from the side change that are present since we have carboxylate anions or inolate
ions. We have some cations specifically that are used to stabilize the intermediates. So
if we lo at the overall glycolysis steps now. We have 10 steps.
So if we just reiterate we have glucose in the first step going to glucose-6-phosphate.
You have to remember that you have to form the pyruvate from the glucose. So the first
step is glucose with the help of hexinase and ATP forming glucose-6- phosphate. The
next step is an isomerase where we form fructose-6- phosphate; we then need to form fructose 1,
6 bisphosphate which means that we need another kinase and another ATP. This you should remember
in this form, so you have glucose, glucose-6- phosphate and since you have to form fructose
1, 6 bisphosphate you have to form fructose- 6- phosphate.
So you need an isomerase that is going to form fructose -6- phosphate from glucose -6-
phosphate, with the kinase you have fructose 1, 6 bisphosphate and aldolase acts upon the
open ketone form of fructose 1, 6 bisphosphate. That is what I showed you yesterday with the
Lysine acting there and we have now a break up into 2, 3 carbon myts. They are dihydroxy
acetone phosphate and the glyceraldehydes-3-phosphate. The glyceraldehydes-3- phosphate is going
to continue the glycolysis. So what is our glycolysis continued? We have glyceraldehydes-3-phosphate
with glyceraldehyde-3-phosphate dehydrogenase with the help of *** plus and Pi form 1, 3
bisphosphoglycerate. So you remember in this step when we have
to dehydrogenase, you are forming from the glyceraldehydes, you are forming the glycerate.
That is important because you have to form finally pyruvate which means you have to have
a carboxylic myt. So this is the step where you have that oxidation actually take place
from the aldehyde to the acid. So if you have a redox reaction, what is the enzyme that
you need? A dehydrogenase or an oxygenase some such enzyme that is going to bring about
the redox reaction and for that you need an electron transfer cofactor which in this case
is *** plus.
So this is what you would have to remember where you have glyceraldehyde -3- phosphate
that takes up *** plus and a Pi to form 1, 3 bisphosphoglycerate. Then it has to loose
the phosphate as simple as that. Now we need the bisphosphoglycerate to oxidized the aldehyde.
So for the oxidation of the aldehyde which in turn reduced *** plus to NADH because this
is a redox reaction. What is getting oxidized in this case? Glyceraldehyde is getting oxidized
to form glycerate. What is getting reduced? *** plus is getting reduced to NADH and the
enzyme is the dehydrogenase. So we have glyceraldehydes-3- phosphate going to glyceraldehydes-3-phosphate
dehydrogenase and in the event it forms 1, 3 bisphosphoglycerate.
Now that you have formed the glycerate, you have to loose the phosphate to form the pyruvate.
So the first step in the loss of the one of the phosphates is a phosphoglyceratekinase
that is going to loose the phosphate that is attached to the carboxylic first carbon
atom of the glycerate. And you have 3 phosphoglycerate formed. After you form 3 phosphoglycerate
there is a mutase reaction which shifts the phosphate myt from the third carbon to the
second carbon. So you have 3 phosphoglycerate form 2-phosphoglycerate then another enzyme
that helps in the dehydration is enolase that results in phosphoenolpyruvate. So after phosphoenolpyruvate
you have the enolic form of pyruvic acid which then forms pyruvate after the loss of the
phosphate. And who takes up this phosphate? ADP. So that comprises the whole series of
steps where glucose is broken down into pyruvate.
Now what we have to see is we have to do a balance of energy, we have to see how many
ATP’s are taken up, how many ATP’s are produced and to see whether the actual breakdown
of glucose is giving us any energy at all. There is one thing that we have to remember
that per glucose there are 2 glyceraldehyde 3 phosphates because what is the previous
step? We have two of these and in the triosephosphate isomerase we know that the equilibrium is
shifted to this side because this is being consumed in the further steps. So our balance
sheet for the phosphate bonds of ATP, how many ATP bonds are broken? How many are broken?
This is our whole series, we have one broken when we form glucose to glucose -6- phosphate
another broken when we form fructose 1, 6 bisphosphate from fructose -6- phosphate.
So our answer for the first question is 2. How many phosphate bonds of ATP are produced?
How many are produced? Is it 2 from every molecule of glucose 4, why? Because we have
two of these, 1 glyceraldehyde 3 phosphate is going to create 2 ATP’s but in our previous
step what do we have? We have two of these myts because we have broken down the 6 membered
ring in 6 carbons into 3 carbon molecules. So because of the 2, that is why I specifically
mentioned that there are 2 G3P that is glyceraldehyde 3 phosphates per glucose. So we have 4, so
how many bonds of ATP are produced? 4. And the net production per glucose is because
we are utilizing 2 and we are creating 4, so eventually we get some energy after we
break down a molecule of glucose.
So this what we have? We have 2 ATP expended, 4 ATP produced that is 2 from each of the
two 3 carbon fragments from glucose and the net production there is 2 phosphate bonds
of ATP or 2 ATP rather per glucose. So our overall step is going to be glucose plus 2
*** plus, where did we use this *** plus? We use just one here, two here because we
have again 2 of G3P. So this is where we are using two of them. Then 2 ADP and 2 Pi? Where
was Pi used again? In the same step where you had *** plus, we used the Pi to form the
phospo and what is finally formed? 2 pyruvate from 1 glucose two 3 carbon myts 2 NADH plus
2 ATP. So that is our total pathway.
Now in aerobic organisms you have pyruvate that is produced in glycolysis oxidized to
carbon dioxide that is what we are going to see when we do the kerbs cycle. This is also
known as the TCA cycle and the NADH, you have NADH produced here, this NADH that is produced
in glycolysis and also in the kerbs cycle is reoxidized. What is it means to be reoxidizing
NADH? Means you have to get back *** plus because you need *** plus for the glycolysis,
now it is already NADH. The only way you can get back is by the respiratory chain that
is the oxidative phophorylation which we studied with production of a lot of additional ATP
which we studied with the proton pump. There was mainly FADH and NADH that was used there.
Now another thing that we have to look at is anaerobic organisms, when we have aerobic
organisms we have the formation of pyruvate but in this case what happens there is the
formation of lactate instead of pyruvate where for example even in our muscles where there
is a lack of oxygen at times pyruvate is formed in to lactate. Now in anaerobic organisms,
what is an anaerobic organism? They function without the presence of oxygen.
They lack a respiratory chain in the absense but in the formation of glyceraldehyde 3 phosphate
two 1, 3 bis phosphoglycerate where we have the *** plus and the Pi. This is the one of
the reactions in glycolysis. Now the NADH produced in glycolysis must be reoxidized
through some other reaction because it is needed for this reaction and the NADH is reoxidized
as pyruvate is converted to a more reduced compound later on that may excreted. This
is one of the cases and the complete pathway including glycolysis and the reoxidation is
actually called fermentation, we will see that. What we have here is now we created
pyruvate from glucose, that was our final step in our glycolytic cycle. Now we can reform
*** plus, remember what did we take in the bis phosphoglycerate formation. We took ***
plus and made NADH out of it. Now lactate dehydrogenase, it actually reduces the keto
in pyruvate and in the event it oxidizes NADH to *** plus, again it is a redox reaction.
So the enzyme is going to be a dehydrogenase. Now for example this is what happens in skeletal
muscles that ferment glucose to lactate during exercise when aerobic metabolism cannot keep
up with energy requirement. So what happens is this pyruvate is further, what happens?
There is a reduction of the ketone in the pyruvate and this reduction will eventually
do what? Oxidize NADH to *** plus, so the lactate released to the blood may be taken
up by other tissues or by muscle after exercise and converted via the reversible lactate dehydrogenase
back to pyruvate to get into krebs cycle.
So this is sort of a stop gap situation where we would have in the lack of aerobic metabolism
that would lead to the lactate formation where there is not enough oxygen present. Now some
anaerobic organisms metabolize pyruvate to ethanol. This process you have learnt fermentation
like yeast this is a reaction that yeast often do where you have industrial chemistry where
you learn about a lot of this, where we have NADH go to *** plus by alcohol dehydrogenase.
In the first step the pyruvate looses carbon dioxide to form acetaldehyde that is one step
of the reaction. This acetaldehyde then in the presence of alcohol dehydrogenase forms
ethanol and in the event NADH is oxidized to *** plus. So they metabolize pyruvate to
ethanol which is actually a waste product. This is a process in fermentation, so animal
cells also contain alcohol dehydrogenase which oxidizes ethanol to acetaldehyde.
So we have ethanol then we have *** plus that is getting reduced to NADH and what is it
doing? It is oxidizing ethanol to acetaldehyde and then further oxidation is also possible
to acetate where you have *** plus going to NADH plus H plus. So what are you doing in
this case? You are reducing *** plus in the event oxidizing acetaldehyde to acetate. In
fact the animal cells that contain alcohol dehydrogenase in the forming of acetaldehyde,
this is what results in the hang over that you get from alcohol consumption.
You hear of hangovers from alcohol consumption, what happens is this ethanol gets oxidized
to acetaldehyde in the body by this particular enzyme alcohol dehydrogenase and this acetaldehyde
actually reacts with certain proteins to give you that numb feeling. So it is all bio chemistry
that goes on. So this our final step of glycolysis where we have our glucose 2 ADP 2 *** plus
going to 2 phosphate plus 2 phosphate that forms 2 pyruvate 2 ATP 2 NADH plus 2 H plus.
In the fermentation instead of pyruvate we are going to get lactate, so we have the glucose
that with 2 ADP plus 2 Pi will give us 2 lactate plus 2 ATP because remember when we are forming
the pyruvate to the lactate, you just go back a step this NADH, *** plus does not come into
the picture because if we look at this reaction, we have the *** plus and the Pi here forming
NADH but that is not the NADH is utilized in the pyruvate reaction to lactate. So we
do not have *** plus NADH feature in the fermentation where we are creating lactate from glucose
because that pyruvate to lactate conversion requires *** plus NADH.
So if we now look at all the enzymes that were involved there are 10 enzymes here for
the 10 steps in glycolysis. Hexinase, isomerase, so now you should be able to tell me exactly
what each of these steps are doing. Hexinase creates glucose 6 phosphate from glucose,
isomerase creates fructose 6 phosphate from glucose 6 phosphate, phophofructinase creates
the fructose 1, 6 bis phosphate from the fructose 6 phosphate, aldolase breaks it up into dihydroxy
acetone phosphate and glyceraldehyde 3 phosphate.
The isomerase inter converts dihydroxy acetone phosphate and glyceraldehyde 3 phosphate then
we have glyceraldehyde 3 phosphate dehydrogenase then we have phosphoglycerate kinase that
is going to create the best phosphoglycerate. The best phosphoglycerate is then has a mutase
acting on it where it creates from three phosphoglycerate it goes to 2 phosphoglycerate then we have
the enolase that creates the phosphoenol pyruvate that then looses because when you have the
enolase you have to have an enol present there, so you have phopho enol pyruvate that phosphoenol
pyruvate looses the phosphorus by pyruvate kinase to from pyruvate. So these are all
the steps of glycolysis and if we now look at the specific free energy changes involved
in the steps we have a delta G zero prime.
A prime means a biological standard where the temperature is taken as 37 degree centigrade
instead of 25 degree centigrade because you are looking at the free energy changes that
are happening in the body when the reactions are taking place. In each of the kinase reaction
this is just the delta G with the concentrations that we have, this is delta G zero. So how
do we calculate delta G? Using delta G zero and you need the concentrations of the reactants
and the products, that is all you need. How do you do? You calculate an equilibrium constant
for a delta G zero process and then you have the reaction quotient that is going to depend
upon the concentration of the reactants and the products that you have and you can calculate
the delta G associated with the process provided you know the delta G zero.
So considering that the concentrations of products and the reactants can be calculated
or can be obtained, we get a delta G associated with these enzymes, 10 of these enzymes and
if you notice the kinase enzymes are the ones that are most spontaneous. They have high
kilo joule per mole value and the reason why they have this is because this is a highly
spontaneous reaction where it is going to loose the phosphate. These are the ones that
use couple reactions with the ATP that provides the energy actually to give you the product.
So what we have is if we look now at the energetics of phophorylation, we are looking at the first
step hexokinase, the glucose actually needs the phosphate to become glucose 6 phosphate,
the delta G zero associated with that is actually plus 14 kilo joules per mole. So there is
no way it is going to happen by itself but you have a couple reaction that is ATP, the
hydrolysis of ATP the high energy phosphate bond that is actually going to break ATP into
ADP plus Pi. This Pi is going to be supplied to glucose, this free energy change is minus
30 kilojoules per mole. So we have this couple reaction so we use the half reaction to determine
the delta G zero for the whole reactions. And we have therefore an over an overall reaction
that is going to be glucose plus ATP giving us glucose 6 phosphate plus ADP with a favorable
free energy standard free energy change that we have. Now depending on the concentrations
of glucose 6 phosphate that you have and the concentrations of glucose and the relative
amounts of ATP and ADP that you have, the free energy change of the reaction is going
to change. Because what is that reaction that we have? If we look at a delta G zero value,
now what you can have is now for each of these you can have an E zero also associated with
it because remember you have a redox reaction taking place.
In some cases you would have a redox reaction taking place where you would have this minus
n FE. From this you can determine the equilibrium constant. What is the equation? Equilibrium,
then you can have a delta G. What is that? That is delta G zero plus RT ln Q. What is
this Q? This Q is your reaction quotient. What is the reaction quotient? It is the ratio
of the concentration of the products to the reactants. So based on that you can find the
delta G associated with this. Now if we look at this first reaction that we have here the
fact that glucose 6 phosphate concentration is low. In this case what is the Q going to
be? It is going to be glucose 6 phosphate ADP divided by glucose ATP concentration that
is what Q is going to be.
Now if you have a high concentration of products that could compensate more than the delta
G zero value negative that you have here and make an overall delta G of the reaction positive.
What am I talking about? I am talking about this factor, this is negative but the ratio
of the product to the reactants is important here. You cannot have too higher concentration
of the products because that will over compensate for the negative delta G zero that you have
here making this non spontaneous. This is what happens in most biological reactions,
the product concentrations are pretty low making the reaction spontaneous because your
delta G zero of the coupled reaction is negative.So the concentration of glucose 6 phosphate in
this particular reaction is actually low and the overall Gibbs free energy that you get
is minus 27 kilo joules per mole. So what you can do with this value is actually determined
what the ratio of the products to the reactants are, you know what delta G zero is, you know
what delta G is. So you can find the ratio of the products to the reactants and when
you are considering, if it is not mentioned a delta G zero prime value means you use a
temperature of 37 degree centigrade. So that would be how you would actually calculate
the energetics of every step of glycolysis.
We can calculate the energetics associated with every step of glycolysis and find an
overall energetics associated with this. Now the spontaneous reaction that we did see that
was with this particular case this enzyme is hexokinase and the other ones that I showed
you were phophofructinase and pyruvate kinase. These are the 3 enzymes that actually catalyze
spontaneous reactions and the control of these enzymes actually determines the rate of glycolysis
because each of these particular reactions if you go back and look at the glycolysis
pathway, it is these three reactions that are irreversible.
So once glucose forms glucose 6 phosphate it is trapped in the cell. So the control
of these enzymes actually determines the rate of the glycolysis pathway. So there are two
types of control, we have local control and global control. Local control means that it
involves dependence of enzyme catalyzed reactions on concentrations of the pathway substrates
or intermediates within a cell. So what is this local control? This is exactly what I
was speaking about in the last slide where we have the delta G values associated with
the amount of product and reactant that you have.
So if my product concentration gets too high, what is going to happen? My reaction will
be non spontaneous, so that is where you would have local control. If glucose 6 phosphate
has too higher concentration, the delta G associated with the expression that we have
written here will make this non spontaneous. So this is regulated by the amount glucose
6 phosphate that is formed because remember I mentioned in the last class that glucose
6 phosphate actually inhibits hexokinase. So what is going to happen? If it inhibits
hexokinase then this enzyme cannot act on another glucose molecule which means that
the concentration of glucose 6 phosphate is getting too high, making this Q too large
which is offsetting the delta G zero preventing a reaction from taking place. You do not want
that to happen. If you want the glycolysis reaction to go forward then you have to have
this regulation. You do not want too much of glucose to be broken down, that is where
the regulation takes place.
So we have local control which involves the dependence of enzyme catalyzed reactions on
concentrations of pathway substrates or intermediates within a cell. You now realize what this means
that the concentrations of the intermediates, the concentrations of the products and the
substrates are extremely important in determining the dependence of these catalyzed reactions.
Then you have global control that actually involves hormone activated production of second
messengers, these are certain messengers that are there in the body that actually regulate
the cellular reactions for the benefit of the organism as a whole. So this is like a
more global approach where it could work on the cell itself preventing the cell from acting
or preventing certain reactions taking place all together. But at the local level the concentration
dependence are due to the variations in the product and reactant. Now let us look at one
such example, this is one step of the glycolytic pathway.
If you look at this, you see it has 2 phosphates attached carbon number 1 and carbon number
6 you have a ketose. So this is fructose 1, 6 bisphosphate, it has broken down into dihydroxy
acetone phosphate and glyceraldehyde -3- phosphate. So the enzyme that has acted here is aldolase
in a reverse aldol reaction. So this is what you should be able to recognize, this is the
key step in the glycolysis. Why? Because it is actually breaking down the glucose, previously
all the steps that led to fructose 1, 6 bisphosphate still have the 6 carbon atoms in it. This
step, the aldolase step is the one that you are breaking the 6 carbon unit into 3 carbon
units. Now we have a delta G0 of this particular reaction having a value of plus 23.8 kilo
joules per mole.
The Gibbs free energy is actually favorable. What is that mean I have a delta G lets just
write this down clearly. So I have a delta G and I have delta G zero plus RT ln Q, this
value is plus 23.8 kilo joules per mole. This value is minus 6 kilojoules per mole, so where
is the regulation coming from? It has to come from Q. So what we can do is we can actually
calculate assuming that fructose 1, 6 bisphosphate is 1 millimolar. I have these in equal concentration
because it is breaking down, it forming from this. So I can actually find out the concentrations
of glyceraldehyde 3 phosphate and dihydroxy acetone phosphate at equilibrium. How can
I do that at equilibrium? It will just be at the equilibrium constant that I get, from
the delta G0 value.
What do I get from the delta G zero value? I forgot to write a minus in the first one.
So I can get this because I know what my delta G zero. So I can get the ratio of the products
to the reactants based on considering 1 millimolar concentration of fructose 1, 6 bisphosphate
and I can also find the same ratio during glycolysis where my reaction is actually favorable.
So this whole idea gives you the energetics that are associated with the each of these
steps. Each of these steps has associated with this energy. So each of these energy
values can tell me what is actually happening and what delta G zero reaction actually have
the equilibrium. So this completes our discussion on glycolysis where we have actually broken
down glucose to form pyruvate. Our next step will now be to see how there is further degradation
of pyruvate in the tricarboxylic acid cycle where finally carbon dioxide and water will
be formed. We will do that in our next class. Thank you. In the last step of our metabolism
of carbohydrates we are going to consider today the tricarboxylic acid cycle or the
Krebs cycle.
Now, if we look at the three stages of cellular respiration, there are actually, the first
stage that we have gone through that is glycolysis for acetyl coenzyme A production. We have
already seen the breakdown of glucose to pyruvate and we will see now how pyruvate actually
gets to acetyl coenzyme A that then gets into the tricarboxylic acid cycle which eventually
leads to the production of carbon dioxide. And we have already considered the third step
where we have electron transport and oxidative phosphorylation that leads to the production
of ATP. So these are actually the three stages of cellular respiration and if we consider
the catabolic pathways of all the break down whether it is amino acids or fatty acids or
glucose they ultimately lead to acetyl CoA which is the key component of the TCA cycle
which we will see.
So what happens is what we have studied in glycolysis is glucose getting to pyruvate.
Now the reverse of this is also possible in a process that is called gluconeo genesis
where glucose is formed from pyruvate. It is not the exact reverse of the glycolysis
step but relatively since these involved some of the similar enzymes, we do have the pyruvate
getting back to the glucose. So we also saw how the lactate is formed from the pyruvate.
So we have lactate also getting to pyruvate and amino acids have actually entry points
at three points where it can get to pyruvate, you can get to acetyl CoA or you could get
directly to the tricarboxylic acid cycle. Fatty acids get into acetyl CoA. so far the
break down of glucose to pyruvate.
Now in general we have these three overall steps, all of them produce ATP. That is our
major concern in the production of energy. ………………So this is the lysine, you
see the lysine here and what are these chains? These are the lipoy acid chains, can you see
the lipoy acid chains, these are the lipoid acid chains that have the dythiol and the
dithyol one of them picks up the acetyl and has it linked to the sulphur of the lipoyamite
the lipoy acid part here. Now what happens is, coenzyme A now comes into the picture.
Coenzyme A now has to form acetyl OA so it picks up this acetyl forming acetyl OA. And
then you have the reduced SH SH. Now what has happened for this to act again? This has
to get back to the dichyol there so two of these Hs have to be removed. How can they
be removed? They can be removed by FAD. FAD then comes into the picture, picks up the
two hydrogen and this gets back to your oxidized lipoyllisine.
So this is just a general procedure where you would have these two,…… because you
realize that you have the acetate formed here directly already, but you have to have these
enzymatic steps to get the enzymes back to where they started with so that they can go
and act on another pyruvate. Usually what happens is that this gets to the oxidized
part so this can now take up another acetyl, another aceyl rather. So another aceyl can
now get attached to this. FAD has now been reduced to FADH2.