Weight loss in a patient with advanced cancer

Objectives

By the end of this exercise you should be able to:

  • Calculate a person's energy deficit from his / her weight loss
  • Calculate a person's BMR from resting oxygen consumption
  • Explain why BMR is elevated in cancer cachexia and explain what is meant by the term hypermetabolism
  • Describe the energy-expensive metabolic cycling between anaerobic glycolysis in the tumour and gluconeogenesis in the liver

Start

AB is an 80 year old man who has metastatic cancer of the pancreas. He is 170 cm tall, and weighs 49 kg. He has lost 2 kg in body weight over the last 4 weeks - the condition of cancer cachexia.

What is his body mass index, and how would you classify it?

height ^2 = 1.7 x 1.7 = 2.89

BMI = weight / (height m^2) = 49 / 2.89 = 16.96

He is undernourished

His mean skinfold thickness is 1.9 mm, suggesting that he has negligible reserves of adipose tissue, and he shows considerable wasting of muscle, so we can assume that most of his weight loss is muscle.

The composition of muscle is:

  • 79% water
  • 17% protein (at 17 kJ /gram)
  • 3% fat (at 37 kJ /gram)

What was his overall energy deficit over the last 4 weeks?

1 kg muscle =

  • 170 g protein at 17 kJ /g = 2890 kJ
  • 30 g fat at 37 kJ /g = 1110 kJ

Total = 4000 kJ = 4 MJ / kg muscle loss

He has lost 2 kg over 4 weeks = 8 MJ total deficit = 8000/28 = 286 kJ /day

For his age and weight we would expect his basal metabolic rate (BMR) to be between 4.7 - 4.9 MJ /day. His oxygen consumption at rest was measured and was found to be 11.2 L /hour.

Regardless of the fuel being consumed, 1 L of oxygen is equivalent to energy expenditure of 20 kJ.

What is his BMR now?

11.2 L oxygen x 20 kJ = 224 kJ /hour x 24 = 5.376 kJ = 5.4 MJ /day

This means that his BMR is 0.6 MJ /day or 12.5% higher than the average for a man of his age and weight.

We need to try to explain why his BMR is so much higher than would be expected.

A blood sample was taken one morning before breakfast, and the following results were obtained:

 
AB
reference range
glucose, mmol /L
3.1
3.0 - 5.5
lactate, mmol /L
2.5
0.4 - 0.6
pyruvate, mmol /L
0.05
0.04 - 0.06
standard liver function tests
normal
-

What conclusions can you draw from these results?

His plasma glucose is low, but within the reference range.

He is slightly acidotic, with an elevated plasma lactate, and normal pyruvate.

Under what conditions might you expect to see a moderately elevated plasma lactate concentration?

As we saw in the exercise breathless after sprinting, an elevated plasma lactate is associated with vigorous physical activity. Although 2.5 mmol /L is not as high as seen in WS after a 100 m sprint, it is within the range you would expect for someone who had just climbed 2 or 3 flights of stairs. However, AB has not climbed any stairs, and indeed the blood sample was taken while he was sitting up in bed - and therefore more or less at rest.

Apart from muscle, which other tissue normally produces lactate?

Red blood cells. They have no mitochondria and are therefore only capable of anaerobic glycolysis. They take up glucose from the bloodstream and put out lactate, which is used in the liver as a substrate for gluconeogenesis.

Since the lactate has obviously not come from muscle as a result of physical activity, and there is no reason to believe that AB's red blood cells are producing significantly more lactate than normal, why might his plasma lactate be so high?

One possibility is that his liver is not clearing the lactate produced by red blood cells, so that the plasma concentration is elevated. However, we are told that the results of "standard liver function tests" were normal, so this is unlikely.

This leaves the most likely source of lactate as the tumour and its metastases. There is a great deal of evidence that many tumours have impaired oxidative metabolism and are relatively anaerobic, producing a significant amount of lactate, which is then used for gluconeogenesis in the liver.

Such cycling between anaerobic glycolysis in the tumour and gluconeogenesis in the liver would explain a considerable part of his increased basal metabolic rate. As we saw in the exercise breathless after sprinting, there is a cost of 6 x ATP for each 2 mol of lactate converted to glucose in the liver, and this has to be provided by additional oxidative metabolism in the liver. (We will come back to AB in later exercises, when we we will consider other metabolic problems associated with advanced cancer).

In order to test this hypothesis that AB's problem is increased cycling between anaerobic glycolysis in the tumour and gluconeogenesis in the liver, AB was given an intravenous infusion of [13C-1]glucose, and a blood sample was taken an hour later. 13C is the stable isotope of carbon, which can be detected by mass spectrometry or its nuclear magnetic resonance (NMR) signal. Using 13C-NMR it is possible not only to determine the amount of 13C present, but also to determine which carbon atoms are labelled.

Which carbon of lactate will contain the label from [13C-1]glucose?

Click here to download the pathway of glycolysis.

If you look at the pathway of glycolysis and follow through the fate of carbon-1 of glucose you will see that it becomes carbon-3, the methyl group of lactate. (Remember that carbon atoms are numbered from the functional group for which the compound is named)

Which carbon atom(s) of glucose would you expect to be labelled if there is gluconeogenesis from [13C-3]lactate?

Since carbon-3 of lactate becomes carbon-3 of glyceraldehyde 3-phosphate and carbon-3 of dihydroxyacetone phosphate, you would expect to see the label in carbon-1 and carbon-4 of fructose 1,6-bisphosphate, and hence also in carbon-1 and carbon-4 of glucose.

For reasons that will become apparent in a later exercise, there is some degree of randomisation of the label, because there are alternative pathways for the metabolism of the single mitochondrial pool of oxaloacetate (mainly the citric acid cycle), so that label will be found in other carbon atoms of glucose as well.

When a blood sample from AB was subjected to 13C-NMR after he had received the infusion of [13C-1] glucose, there was 2 - 3 times more 13C label in carbon 4 of glucose than in glucose from cancer patients whose weight was stable who received a similar infusion of [13C-1]glucose.

What conclusions can you draw from this?

This suggests that AB, who is hypermetabolic and losing weight, is indeed recycling more glucose through anaerobic glycolysis (presumably in the tumour) and gluconeogenesis in the liver than cancer patients whose weight is stable.

The ATP cost of resynthesising glucose from lactate will explain his high BMR.

Key points from this exercise:

  • Many people with advanced cancer suffer from cancer cachexia - significant weight loss with an elevated basal metabolic rate - so-called hypermetabolism.
  • A major contributor to hypermetabolism in cancer cachexia is cycling between anaerobic glycolysis in the tumour and (ATP-expensive) gluconeogenesis in the liver.
  • The extent of glucose cycling in this way can be estimated by measuring the appearance of label from [13C-1]glucose into other carbon atoms of glucose as a result of resynthesis of glucose from labelled lactate.