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copyright © 2008 - 2015 David A Bender

 

Breathless after sprinting

WS is a sprinter; he weighs 75 kg. At the end of a 100 m race he is breathing rapidly and deeply, and continues to do so for several minutes. His plasma lactate and pyruvate were measured before and immediately after the race, and again 30 min later, when his breathing had returned to normal.

	lactate (mmol /L)	pyruvate (mmol /L)
before the race	0.5	0.11
immediately after	11.5	0.09
30 minutes later	1.0	0.12

 

Can you explain why he was hyperventilating at the end of the race?

For reasons that we will be exploring in the exercise, he is acidotic, with a high plasma concentration of lactate. Note, however, that unlike the cases of PC and BD, who were considered in the exercise on Life-threatening acidosis in an alcoholic - and in a hunger striker given intravenous glucose, WS's plasma pyruvate is within the normal range at all times.

The usual response to acidosis is to increase the rate of breathing, so as to expel carbon dioxide, and so shift the equilibrium below to the left, lowering the hydrogen ion concentration in the bloodstream, and so raising plasma pH.

His oxygen consumption was also measured for 10 minutes before the race, and over the 30 minutes after the race. At rest he was consuming 0.5 L oxygen per minute. Over the 30 minutes after the race his average oxygen consumption was 14% higher, with a total consumption over 30 minutes of 17.12 litres of oxygen.

What we have to consider is where the lactate has come from, how it disappears during the 30 minutes after the race and why he consumes 14% more oxygen over the thirty minutes after he has completed the race.

In a series of experiments with anaesthetised fasting dogs cannulae were placed in the femoral artery and popliteal vein to permit measurement of metabolites taken up, or put out, by the gastrocnemius-plantaris muscle group after gentle electrical stimulation (1 twitch per second) and after vigorous electrical stimulation (5 twitches per second). Cannulae were also placed in the hepatic artery and hepatic vein, to permit measurement of metabolites taken up and put out by the liver. The result are shown as the arterio-venous difference as nmol /g tissue /minute; positive values indicate output from the tissue, negative values indicate uptake by the tissue. Figures show mean ± sd for 3 x replicate experiments.

	muscle		liver
	gentle stimulation	vigorous stimulation	gentle stimulation	vigorous stimulation
glucose	-215 ± 12	-885 ± 15	+100 ± 12	+660 ± 20
oxygen	-4515 ± 50	-6912 ± 50	-1150 ± 50	-1800 ± 50
lactate	+20 ± 5	+1112 ± 50	-50 ± 10	-1100 ± 40

 

What conclusions can you draw from these results?

In response to gentle stimulation muscle is mainly aerobic - less than 10% of the glucose taken up is released as lactate, with the remainder being metabolised to carbon dioxide and water.

In response vigorous stimulation there is the expected large increase in uptake of both glucose and oxygen, but there is also a large increase in lactate output, so that now more than 60% of the glucose taken up is released as lactate.

When muscle is only gently stimulated, the liver is putting out a small amount of glucose - presumably this is arising partly from its glycogen reserves, and partly as a result of gluconeogenesis from various substrates.

When muscle is stimulated vigorously, the liver takes up a large amount of lactate, and puts out a large amount of glucose. Oxygen consumption also increases at this time.

We can therefore conclude that in vigorous exercise muscle metabolises as much glucose aerobically as it can, but is presumably limited by the availability of oxygen, and in order to produce enough ATP to maintain a high rate of work output, it metabolises much glucose anaerobically, forming lactate in order to re-oxidise the NADH formed in glycolysis.

We can also conclude that the lactate is taken up by the liver, and is used for resynthesis of glucose, which is then exported, and presumably taken up by the muscle, either for immediate use or for synthesis of glycogen reserves in the muscle.

We can summarise this as an inter-organ metabolic cycle of anaerobic glycolysis in the muscle and gluconeogenesis in the liver.

We now need to consider what is involved in the conversion of lactate back to glucose.

There are three steps in glycolysis that are irreversible under physiological conditions:

The phosphorylation of glucose to glucose 6-phosphate, catalysed by hexokinase (and in the liver also by glucokinase)

The phosphorylation of fructose 6-phosphate to fructose 1,6-bisphosphate catalysed by phosphofructokinase

The reaction catalysed by pyruvate kinase, in which ADP is phosphorylated to ATP by transfer of phosphate from phosphoenolpyruvate.

The first two reactions are irreversible under physiological conditions because the ratio of ATP : ADP is ~ 500 : 1, and ATP binds very much better to these enzymes than does ADP.

For the first two reactions, there are separate enzymes involved in gluconeogenesis, glucose 6-phosphatase and fructose 1,6-bisphosphatase, which catalyse simple hydrolysis of the phosphate group, releasing inorganic phosphate. Fructose 1,6-bisphosphatase and glucose 6-phosphatase are only found in tissues that catalyse gluconeogenesis (liver, kidney and possibly the small intestine)

What would you expect to see if phosphofructokinase and fructose 1,6-bisphosphatase were equally active at the same time?

If these opposing enzymes were equally active, the net result would be futile cycling between fructose 6-phosphate and fructose 1,6-bisphosphate and hydrolysis of ATP to ADP and inorganic phosphate to no good purpose apart from generation of heat.

In later exercises you will investigate the regulation of glycolysis and gluconeogenesis, and you will see that these two enzymes are regulated in opposite directions in response to hormones and other controlling factors:

Factors that increase the activity of phosphofructokinase (and so increase the rate of glycolysis) also decrease the activity of fructose 1,6-bisphosphatase.

Factors that increase the activity of fructose 1,6-bisphosphatase (and so increase the rate of gluconeogenesis) also decrease the activity of phosphofructokinase.

This regulation is not absolute - both enzymes are active to some extent at all times, but the activity of one is considerably greater than the activity of the other, so that there is net flux through the pathway (towards glycolysis when the activity of phosphofructokinase is greater) or towards glycolysis when the activity of fructose 1,6-bisphosphatase is greater).

What is the possible benefit of both enzymes being active (to some extent) at the same time?

There are two possible benefits:

In response to stimulation, and in preparation for exercise, glycolysis in muscle may increase 1000-fold. If both enzymes are active to some extent then a relatively small change in the activity of both, in opposite directions, permits a rapid change in the rate of glycolysis. This allows both rapid and very sensitive regulation of the rates of glycolysis and gluconeogenesis.

Futile cycling generates heat, and this may be important in maintaining body temperature - a mechanism for non-shivering thermogenesis that is distinct from the uncoupling of electron transport and oxidative phosphorylation in response to activation of uncoupling proteins in muscle and brown adipose tissue. As with uncoupling proteins, this may also be important in maintenance of body weight and increasing energy expenditure to match food intake in excess of requirements.

The diagram shows the reaction of pyruvate kinase.

 

 

 

Why do you think the reaction is irreversible under physiological conditions?

The immediate product of the reaction is enolpyruvate, which undergoes a non-enzymic reaction to form pyruvate. This means that there is no enolpyruvate available to undergo phosphorylation to yield phosphoenolpyruvate.

Pyruvate is converted to phosphoenolpyruvate in a two step pathway, with the intermediate formation of oxaloacetate.

As shown on the right, 2 mol of ATP equivalent (1 mol each of ATP and GTP, (which is equivalent to ATP) are required to form phosphoenolpyruvate from pyruvate.

 

 

 

 

 

 

 

How many mol of ATP equivalents are required to convert 2 mol of lactate to 1 mol of glucose?

Click here to download a printable copy of the pathway of glycolysis

2 x Pyruvate to oxaloacetate requires 2 x ATP

2 x oxaloacetate to phosphoenolpyruvate requires 2 x GTP

2 x 3-phosphoglycerate to bisphosphoglycerate requires 2 x ATP

So there is a need for a total of 6 x ATP equivalents per 2 mol of lactate converted to glucose.

(Note that if the glucose formed is to be converted to glycogen in the liver there will be a need for an additional 1 mol of ATP equivalent. If glucose is exported and used for synthesis of glycogen in muscle there will be a need for an additional 2 mol of ATP. The reasons for this will become apparent in a later exercise. For now you can assume that the lactate is simply converted to glucose).

Although there is a requirement for 2 x NADH to reduce 2 mol of bisphosphoglycerate to glyceraldehyde 3-phosphate, the oxidation of 2 mol of lactate to pyruvate provide this.

The oxidation of acetyl CoA to carbon dioxide and water in the citric acid cycle yields 10 x ATP.

What is the ATP yield from the total oxidation of lactate to carbon dioxide and water?

Oxidation of lactate to pyruvate yields NADH = ~2.5 ATP

Oxidation of pyruvate to acetyl CoA yields NADH =~2.5 ATP

Oxidation of acetyl CoA to carbon dioxide and water yields 10 ATP

So, total yield from oxidation of lactate to carbon dioxide and water = 15 ATP.

WS weighs 75 kg.

Before the race he was consuming 0.5 L oxygen per minute = 15 L per 30 min
During the 30 minutes after the race he consumed 17.12 L of oxygen.
This additional 2.12 L of oxygen = (2.12 / 15) x 100 = a 14% increase in oxygen consumption.

Before the race his plasma lactate was 0.5 mmol /L
immediately after the race it was 11.5 mmol /L
30 minutes later is was 1.0 mmol/L

Assuming that extra-cellular fluid is 20% of body weight, what is the total amount of lactate that he metabolises in the 30 minutes after the race?

0.2 x 75 kg = 15 L extracellular fluid
over 30 min his plasma lactate falls by 10.5 mmol /L = 10.5 x 15 = 157.5 mmol total lactate

How much of lactate will have to be oxidised to carbon dioxide and water to provide the ATP required to synthesise glucose from 157.5 mmol lactate?

metabolism of 2 mol of lactate to glucose requires 6 x ATP

metabolism of 157.5 mol of lactate to glucose requires (157.5 / 2) x 6 = 472.5 x ATP

metabolism of 1 mol of lactate to carbon dioxide and water yields 15 x ATP

amount of lactate to be metabolised = 472.5 / 15 = 31.5 mol

The oxidation of lactate to carbon dioxide and water, like that of glucose, consumes 0.746 L oxygen per gram.

Why does the oxidation of lactate to carbon dioxide and water require oxygen?

The oxygen is needed to oxidise reduced cytochrome a/a3 in the mitochondrial electron transport chain. Remember that the final step in the electron transport chain is the reduction of oxygen to water.

What is his additional oxygen requirement (in L oxygen) over the 30 minutes after the end of the race?

The molecular mass of lactate is 90.1.

31.5 mmol lactate = 31.55 x 90.1 = 2838 mg = 2.838 g

2.838 x 0.746 = 2.12 L additional oxygen consumed.

This is what is known as oxygen debt - the increased consumption of oxygen to permit the metabolism of sufficient lactate (or other metabolic fuel) to provide the ATP required to convert the remainder of the lactate to glucose.

 

Summarise the key points of this exercise