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So far, from various studies we have seen that (anaerobic) glucose metabolism leads to the production of 2 mol of lactate and 2 mol of ATP per mol of glucose consumed; the process also requires ATP and inorganic phosphate. The amount of glucose consumed, and of lactate formed, depends on the amount of ADP provided.
For the following studies, chicken breast muscle was homogenised in phosphate buffer at pH 7.4, then subjected to high speed centrifugation to remove cell debris, nuclei and mitochondria. The supernatant fraction (essentially cytosol, but also containing ribosomes and microsomes, was then dialysed against four changes of buffer to remove any endogenous substrates and other low molecular weight compounds. Each incubation in the following sets of experiments contained the dialysed homogenate from 1 g of fresh muscle tissue.
Experiments were conducted incubating this preparation with one of:
- [14C-U]glucose at a specific activity of 1 µCi /mmol
- [32P]ATP at a specific activity of 1 µCi /mmol
- [32P]inorganic phosphate at a specific activity of 1 µCi /mmol
In the first set of experiments, incubations contained:
- 200 µmol glucose
- 200 µmol ATP
- 200 µmol ADP
- 200 µmol sodium phosphate
After incubation at 30C for 30 min, incubations were stopped by adding trichloroacetic acid to denature proteins, centrifuged and the supernatant was neutralised with sodium hydroxide. Aliquots of the supernatant were then subjected to high pressure liquid chromatography to measure both the amounts of various metabolites present and and the radioactivity in each. The results are shown as total amount of each metabolite and the specific radioactivity (µCi / mmol) - mean values ± sd from 3 x replicate incubations.
The following results were obtained:
metabolite |
amount |
specific radioactivity
(µCi /mmol) |
|||
µmol |
from [14C-U]glucose |
from [32P]ATP |
from [32P] inorganic phosphate |
||
glucose |
100 ± 5 |
0.99 ± 0.02 |
- |
- |
|
glucose 6-phosphate |
3.2 ± 0.4 |
0.98 ± 0.03 |
0.99 ± 0.03 |
0 |
|
fructose 6-phosphate |
2.5 ± 0.3 |
0.99 ± 0.03 |
0.98 ± 0.03 |
0 |
|
fructose 1,6-bisphosphate |
3.3 ±
0.7 |
0.99 ± 0.02 |
1.99 ± 0.04 |
0 |
|
glyceraldehyde 3-phosphate |
94 ± 1.2 |
0.48 ± 0.03 |
0.99 ± 0.02 |
0 |
|
dihydroxyacetone phosphate |
98 ± 1.4 |
0.49 ± 0.02 |
1 |
0 |
|
ATP |
0 |
- |
- |
- |
|
ADP |
248 ± 5 |
- |
0 |
0 |
There was no detectable formation of lactate.
What conclusions can you draw from these results?
It is difficult to explain at this stage why there was no formation of lactate or ATP. Indeed, all of the ATP seems to have been converted to ADP, whereas in previous studies we have seen phosphorylation of ADP to ATP associated with glucose metabolism.
It is obvious from these results that glucose has been phosphorylated to glucose 6-phosphate; it then seems likely that glucose 6-phosphate has been isomerised to fructose 6-phosphate. Fructose 6-phosphate has then been phosphorylated to fructose 1,6-bisphosphate, and it looks as is fructose 1,6-bisphosphate has been cleaved to yield glyceraldehyde 3-phosphate and dihydroxyacetone phosphate. Neither of these three carbon compounds has undergone any further metabolism.
A further experiment was set up in which the amount of ATP per incubation was varied, and only glucose disappearance and the appearance of glyceraldehyde 3-phosphate and dihydroxyacetone phosphate were measured.
The results were as follows:
What conclusions can you draw from these results?
These graphs show that there is a requirement for 2 mol of ATP for each mol of glucose metabolised, and that 1 mol each of glyceraldehyde 3-phpsphate and dihydroxyacetone phosphate is formed per mol of glucose. These results thus confirm the conclusions drawn from the first experiment with dialysed muscle supernatant.
Like red blood cells, muscle preparations that are incubated anaerobically normally produce lactate from glucose.
Why do you think it may be that the dialysed muscle preparation did not form lactate?
The fact that the dialysed muscle preparation did not produce any lactate, although muscle incubated under anaerobic conditions forms lactate from glucose, suggests that there may be a low molecular weight compound that is required for the full pathway of glucose metabolism, and this has been lost in dialysis. A classical way of testing this hypothesis is to boil a sample of the tissue preparation before dialysis, centrifuge it to remove denatured protein, then add some of the supernatant to the incubation mixture. This preparation is commonly known as a kochsaft (from the German, meaning literally cooking or cooked juice).
A series of incubations was set up containing both the dialysed muscle preparation and either buffer or the kochsaft from boiled (but undialysed) muscle preparation, together with:
- 200 µmol glucose
- 200 µmol ATP
- 200 µmol ADP
- 200 µmol sodium phosphate
After incubation at 30C for 30 min, incubations were stopped by adding trichloroacetic acid to denature proteins, centrifuged and the supernatant was neutralised with sodium hydroxide. Aliquots of the supernatant were then subjected to high pressure liquid chromatography to measure glucose, lactate, ATP and ADP. The following results were obtained:
control (+ buffer) | + kochsaft | |
glucose | 100 |
100 |
lactate | 0 |
80 |
ATP | 0 |
400 |
ADP | 300 |
0 |
What conclusions can you draw from these results?
The control incubations show the same results as before - no lactate is formed and 2 mol of ATP are consumed per mol of glucose metabolised. However, with the kochsaft added there is now formation of lactate, and instead of net consumption of ATP there is phosphorylation of ADP to ATP.
The obvious next step is to discover what it is in the kochsaft that is required for the complete metabolism of glucose to lactate. After a number of experiments it was discovered that the compound concerned is nicotinamide adenine dinucleotide (NAD), a coenzyme that undergoes oxidation and reduction reactions.
Using uv absorption spectrometry it is easy to follow redox reactions using NAD or NADH - both absorb uv light at 259 nm, but the reduced coenzyme (NADH) also absorbs at 340 nm. This means that an oxidation reaction can be followed by measuring the increase in absorbance at 340 nm, and a reduction reaction by measuring the disappearance of absorbance at 340 nm.
A series of incubations was set up containing the dialysed muscle preparation and varying amounts of NAD, together with:
- 200 µmol glucose
- 200 µmol ATP
- 200 µmol ADP
- 200 µmol sodium phosphate
After incubation at 30C for 30 min, incubations were stopped by adding trichloroacetic acid to denature proteins, centrifuged and the supernatant was neutralised with sodium hydroxide. Aliquots of the supernatant were then subjected to high pressure liquid chromatography to measure glucose, lactate, NAD and NADH. An initially unidentified compound was also found in the incubation mixtures, which was later identified as pyruvate.
The following results were obtained:
What conclusions can you draw from these results?
Even a very small amount of NAD seems to be adequate to permit full consumption of glucose (limited presumably by the amount of either ADP or ATP added). This suggests that NAD has a catalytic role on the pathway of glucose metabolism. However, at the end of the incubations almost al of the NAD had been reduced to NADH.
When only a small amount of NAD was added, all of the glucose consumed was converted to lactate (2 mol of lactate per mol of glucose consumed). However, as more NAD was added, an increasing proportion of the glucose was metabolised to pyruvate rather than lactate. Note the the sum of lactate + pyruvate formed is more or less constant, at 2 mol of (lactate or pyruvate) per mol of glucose consumed.
What is the relationship between lactate and pyruvate, and how might they be interconverted?
You can either say that lactate is the product of reduction of pyruvate, or that pyruvate is the product of oxidation of lactate - they form a redox pair.
What would you expect to observe if you incubated the dialysed muscle supernatant preparation with lactate and NAD?
What would you expect to observe if you incubated the dialysed muscle supernatant preparation with pyruvate and NADH?
If you incubated with lactate and NAD, you would expect to see stoichiometric loss of lactate, appearance of pyruvate and appearance of NADH (detectable by its absorbance at 240 nm)
If you incubated with pyruvate and NADH, you would expect to see stoichiometric appearance of lactate, disappearance of NADH and disappearance of pyruvate.
There is an enzyme, lactate dehydrogenase, that catalyses the interconversion of lactate and pyruvate, linked to utilisation of NAD and NADH:
This suggests that NAD is reduced to NADH in the metabolism of glucose to form pyruvate, then lactate is formed to re-oxidise the NADH back to NAD.
The same set of incubations was performed using only 20 µmol of ATP, rather than 200 µmol. The results were the same. What conclusion can you draw from this?
We already know that when glucose is metabolised to lactate there is a net yield of 2 ATP per glucose metabolised. EWe also know that there is a cost of 2 x ATP to convert glucose to fructose 1,2-bisphosphate. Obviously, since only a small amount of ATP (20 µmol in this experiment) is required to permit the metabolism of much more glucose (100 µmol in this experiment), then the total amount of ATP formed in the metabolism of 1 mol of glucose must be more than 2 mol of ATP formed. Not only the ADP added initially but also that resulting from phosphorylation of glucose to glucose 6-phosphate and fructose 6-phosphate to fructose 1,6-bisphosphate, is also phosphorylated to ATP.
This suggests that altogether there is probably formation of 4 mol of ATP per mol of glucose metabolised to lactate.
We can now return to our experiments with radioactive glucose, ATP and inorganic phosphate.
Experiments were conducted incubating an amount of the dialysed muscle supernatant preparation equivalent to 1 g of fresh muscle tissue with one of:
- [14C-U]glucose at a specific activity of 1 µCi /mmol
- [32P]ATP at a specific activity of 1 µCi /mmol
- [32P]inorganic phosphate at a specific activity of 1 µCi /mmol
In the first set of experiments, incubations contained:
- 200 µmol glucose
- 20 µmol ATP
- 20 µmol NAD
- 200 µmol ADP
- 200 µmol sodium phosphate
After incubation at 30C for 30 min, incubations were stopped by adding trichloroacetic acid to denature proteins, centrifuged and the supernatant was neutralised with sodium hydroxide. Aliquots of the supernatant were then subjected to high pressure liquid chromatography to measure both the amounts of various metabolites present and and the radioactivity in each. The results are shown as total amount of each metabolite and the specific radioactivity (µCi / mmol) - mean values ± sd from 3 x replicate incubations.
The following results were obtained:
metabolite |
amount |
specific radioactivity
(µCi /mmol) |
|||
µmol |
from [14C-U]glucose |
from [32P]ATP |
from [32P] inorganic phosphate |
||
glucose |
100 ± 5 |
0.99 ± 0.02 |
- |
- |
|
glucose 6-phosphate |
3.2 ± 0.5 |
0.98 ±
0.03 |
0.99 ±
0.03 |
0.98 ±
0.05 |
|
fructose 6-phosphate |
2.6 ± 0.6 |
0.99 ± 0.03 |
0.98 ± 0.03 |
0.99 ± 0.04 |
|
fructose 1,6-bisphosphate |
3.4 ± 0.7 |
0.99 ± 0.02 |
1.99 ± 0.04 |
1.98 ± 0.03 |
|
glyceraldehyde 3-phosphate |
2.9 ± 0.3 |
0.48 ± 0.03 |
0.98 ± 0.02 |
0.96 ± 0.02 |
|
dihydroxyacetone phosphate |
2.8 ± 0.2 |
0.47 ± 0.02 |
0.97 ± 0.03 |
0.96 ± 0.03 |
|
3-phosphoglycerate |
6.1 ± 0.3 |
0.49 ± 0.03 |
0.99 ± 0.02 |
0.96 ± 0.03 |
|
2-phosphoglycerate |
5.7 ± 0.2 |
0.51 ± 0.02 |
0.97 ± 0.03 |
0.98 ± 0.03 |
|
phosphoenolpyruvate |
6.2 ± 0.3 |
0.49 ± 0.03 |
0.99 ± 0.02 |
0.98 ± 0.03 |
|
1,3-bisphosphoglycerate |
5.9 ± 0.2 |
0.47 ± 0.03 |
0.98 ± 0.01 |
1.99 ± 0.04 |
|
pyruvate |
6.4 ± 0.3 |
0.49 ± 0.01 |
- |
- |
|
lactate |
160 ± 4 |
0.50 ± 0.02 |
- |
- |
|
ATP |
219 ± 3 |
- |
0 |
1.01 ± 0.02 |
|
ADP |
0 |
- |
- |
- |
The compounds shown in red are new intermediates that were not detected previously; they are shown in the order in which they elute from the high pressure liquid chromatography column.
The compounds shown in blue are those for which we are reasonably certain (from previous experiments) that this is the sequence in which they are formed.
What conclusions can you draw from these results?
As expected, under these conditions all of the ADP has been phosphorylated to ATP, and 2 mol of ATP have been formed for each mol of glucose consumed. Most of the glucose has been metabolised to lactate.
When [32P]ATP is used, both phosphate groups in fructose 1,6-bisphosphate arise from ATP, and this leads to one labelled phosphate in each of the phosphorylated intermediates. This suggests that the second phosphate incorporated into 1,2-bisphoglycerate has come from inorganic phosphate, not ATP.
This is confirmed by the experiments using [32P]inorganic phosphate - the specific activity of phosphate in 1,3-bisphosphoglycerate is double that of the other phosphorylated intermediates.
If the phosphate in fructose 1,6-biphosphate has come from ATP, as suggested by the experiments using [32P]ATP, why is it labelled in the experiments using [32P]inorganic phosphate?
There is net phosphorylation of ADP to ATP, and the experiments with [32P]inorganic phosphate show that the ATP that is formed incorporated the label from inorganic phosphate. This means that after a few reaction cycles most or all of the ATP present will be labelled with 32P.
We have seen previously that fructose 1,6-bisphosphate is cleaved to yield dihydroxyacetone phosphate and glyceraldehyde 3-phosphate, which accumulate in equal amounts when there is no NAD in the incubation medium. However, when NAD is added, neither accumulates to any significant extent.
How can you begin to deduce the sequence of reactions involved in the metabolism of dihydroxyacetone phosphate and glyceraldehyde 3-phosphate?
One obvious experiment would be to add either dihydroxyacetone phosphate or glyceraldehyde 3-phosphate to the incubations with and without added NAD.
The following results were obtained:
No added NAD |
+ 20 µmol NAD |
|||
+ 20 µmol dihydroxyacetone phosphate | + 20 µmol glyceraldehyde 3-phosphate | + 20 µmol dihydroxyacetone phosphate | + 20 µmol glyceraldehyde 3-phosphate |
|
dihydroxyacetone phosphate | 9.8 ± 0.2 µmol |
9.9 ± 0.2 µmol |
0.5 ± 0.05 µmol |
0.1 ± 0.01 µmol |
glyceraldehyde 3-phosphate | 9.9 ± 0.3 µmol |
9.8 ± 0.2 µmol |
3.9 ± 0.2 µmol |
3.8 ± 0.3 µmol |
What conclusions can you draw from these results?
There is rapid equilibrium between dihydroxyacetone phosphate and glyceraldehyde 3-phosphate, and it seems likely that it is glyceraldehyde 3-phosphate that undergoes the next step in the pathway. Since this next reaction requires NAD, it is obviously an oxidation - again it is more likely that an aldehyde (glyceraldehyde 3-phosphate) will undergo oxidation than a ketone (dihydroxyacetone phosphate).
The product of oxidation of an aldehyde is a carboxylic acid, and it might seem obvious that the product of oxidation of glyceraldehyde 3-phosphate would be 3-phosphoglycerate. To test this a series of incubations was set up in which the muscle supernatant preparation was incubated with glyceraldehyde 3-phosphate and NAD, with or without added ADP.
Each incubation contained:
- 100 µmol glyceraldehyde 3-phosphate
- 100 µmol NAD
- zero or 200 µmol ADP
- 200 µmol sodium phosphate
The results were as follows (µmol of product formed or substrate remaining , mean ± sd for 3 x replicate incubations):
+ 100 µmol ADP | no added ADP | |
glyceraldehyde 3-phosphate | 25.1 ± 2.4 |
70.2 ± 0.2 |
3-phosphoglycerate | 5.8 ± 0.3 |
0 |
2-phosphoglycerate | 6.1 ± 0.4 |
0 |
1,3-bisphosphoglycerate | 5.9 ± 0.3 |
29.2 ± 1.2 |
pyruvate | 6.1 ± 0.4 |
0 |
lactate | 50.1 ± 2.3 |
0 |
When the experiment was repeated in the presence of [32P]inorganic phosphate, label was found in 1,3-bisphosphoglycerate.
What conclusions can you draw from these results?
As it is formed, the carboxylic acid group is esterified (presumably with inorganic phosphate, since no ATP was added, and label from inorganic phosphate is incorporated) to form 1,3-bisphosphoglycerate.
From the chemistry of the remaining intermediates between 1,3-bisphosphoglycerate and lactate, you should be able to deduce the sequence of reactions.
Key points from this exercise:
- The metabolism of glucose can occur anaerobically, leading to the formation of 2 mol of lactate per mol of glucose.
- Two mol of ATP are required to form fructose 1,6-bisphosphate from glucose.
- Fructose 1,6-bisphosphate is cleaved to dihydroxyacetone phosphate and glyceraldehyde 3-phosphate, which are interconvertible.
- Glyceraldehyde 3-phosphate undergoes oxidation (linked t the reduction of NAD to NADH) and phosphorylation (using inorganic phosphate) to form 1,3-bisphosphoglycerate.
- One phosphate esterified to 1,3-bisphosphoglycerate is transferred to ADP, forming ATP.
- Following a series of isomerisation reactions, phosphoenolpyruvate is formed, and its phosphate is also transferred to ADP forming ATP.
- Under anaerobic conditions the NADH formed by oxidation of glyceraldehyde 3-phosphate is re-oxidised by reducing pyruvate to lactate. (The fate of pyruvate under aerobic conditions will be considered in a later exercise.)
- Because there are two 3-carbon sugars formed from each mol of glucose, there is a yield of 4 x ATP per glucose metabolised. The 2 mol of ATP required to form fructose 1,6-bisphosphate from glucose must be subtracted from this, so there is a net yield of 2 x ATP per mol of glucose metabolised under anaerobic conditions.
- Under aerobic conditions the NADH formed can be re-oxidised in the mitochondrial electron transport chain, yielding an additional ~2.5 x ATP per NADH, and so an additional ~5 x ATP per glucose.