Small changes in temperature can have large physiological effects

The story opening this chapter described a world-class marathon runner failing to finish an important race because of heat stress. As many of us have experienced, our ability to do hard work is reduced when we get overheated. Because high tissue temperatures can damage cells, shutting off a muscle’s ability to do work may serve a protective function.

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investigating life

Can the Work Capacity of Muscle Be Increased by Extracting Heat from the Palms of the Hands?

experiment

Original Paper: Grahn, D. A., V. H. Cao, C. M. Nguyen, M. T. Liu and H. C. Heller. 2012. Work volume and strength training responses to resistive exercise improve with periodic heat extraction from the palm. Journal of Strength and Conditioning Research 26(9): 2558–2569.

At the beginning of the chapter you learned that mammals dissipate excess heat through their non-hairy skin areas. Those skin areas are small, but they have special high-volume blood vessels that transfer heat from the body core to the skin. Humans have these same blood vessel adaptations. In this experiment, heat was extracted from the palm of a hand, and muscle work capacity was measured.

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investigatinglife work with the data follows below.

work with the data

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This experiment involves a simple outcome measure: number of bench presses in 6 sets done to muscle failure. The experimental manipulation is heat extraction from a palm between sets. The control condition is an equivalent period of rest between sets without palmar cooling. The main question is whether there is a difference in the work volume when cooled versus when not cooled. However, there are some complicating factors: (1) the subjects varied considerably in their initial work capacity, and (2) since the treatments were sequential there could be an order effect. To minimize the order effect, the treatments were randomized as to palmar cooling or no cooling. Your analysis must first standardize the data, then test for an order effect, and finally test for a treatment effect.

How can you standardize the data? Because the subjects began the experiment at different ability levels, the absolute increase in repetitions cannot be used as a comparative outcome measure. An increase of 5 repetitions is more impressive for a subject who began at 25 than for one who began at 50—a 20 percent increase in contrast to a 10 percent increase. For each individual and each treatment it is necessary to convert the raw data to percent increase.

How can you test for a treatment effect? Each subject was cooled in one treatment and not cooled in the other treatment. To see if there is a difference in the percent increase due to palmar cooling, you can average the percent increases for all subjects in their cooling treatments and for all subjects in their control treatments. Using a paired t-test, you can assess whether palmar cooling had a significant effect on the outcome.

QUESTIONS

Question 1

Because this experiment involved repetitions of a physical activity, you would expect there to be a conditioning effect whether the subjects were cooled or not, and you might expect the improvements to be greater in the second treatment than in the first. To test for an order effect, compare the percent increases for all subjects in treatment 1 to the percent increases for all subjects in treatment 2. For statistical analysis use a paired t-test (see Appendix B). What do you conclude about a possible order effect in your collection of data?

Standardizing results to percent increases in treatments 1 and 2, averaging and calculating standard deviations for those two groups, comparing the two groups to determine if there is a possible order effect through use of a paired t-test.

Percent increase, experiment 1 Percent increase, treatment 2
Subject Control group (no cooling) Treatment group (palmar cooling)
1 35.29 11.11
2 24.00 27.59
3 4.55 5.80
4 28.30 12.50
5 25.86 35.44
6 3.28 14.93
7 2.70 76.74
8 6.00 28.57
9 –5.26 28.95
Treatment group Control group
10 39.39 7.87
11 11.49 5.62
12 30.00 8.75
13 32.31 39.44
14 57.89 26.19
15 42.62 21.59
16 53.33 18.67
17 48.98 1.35
Mean 25.93 21.83
Standard deviation 18.80 17.53

The P-value (0.058) shows there is no significant difference between the two data sets and therefore there cannot be an order effect.

Question 2

Each subject was cooled in one treatment and not cooled in the other treatment. To see if there is a difference in the percent increase due to palmar cooling, average the percent increases for all subjects in their cooling treatments and for all subjects in their control treatments. Using a paired t-test, you can assess whether palmar cooling had a significant effect on the outcome. What do you conclude about a treatment effect in your experimental results?

The standardized values (percent increases) are sorted according to treatment—palmar cooling or no cooling. Means and standard deviations are calculated and a paired t-test is applied to the data.

Subject Palmar cooling No cooling
1 11.11 35.29
2 27.59 24.00
3 5.80 4.55
4 12.50 28.30
5 35.44 25.86
6 14.93 3.28
7 76.74 2.70
8 28.57 6.00
9 28.95 –5.26
10 39.39 7.87
11 11.49 5.62
12 30.00 8.75
13 32.31 39.44
14 57.89 26.19
15 42.62 21.59
16 53.33 18.67
17 48.98 1.35
Mean 32.80 14.95
Standard deviation 18.54 12.88

The P value of < 0.008 is highly significant, indicating that the increases during the palmar cooling treatments were greater than the increases during the no cooling treatments.

Question 3

Is the hypothesis supported or disproven? Explain.

The hypothesis that extracting excess heat from the body will increase the capacity of an individual’s muscles to do work is supported by these data. Normalizing the data to percent increases and then testing for order effects showed that there was no effect from whether the cooling or the control treatment came first. Testing for treatment effect, however, yielded a highly significant benefit to the cooling treatment.

A similar work with the data exercise may be assigned in LaunchPad.

Pyruvate kinase is an enzyme critical for production of ATP in muscles; it catalyzes the last step in glycolysis, producing pyruvate that can enter the mitochondria and be metabolized to produce ATP (see Figure 9.12). Without ATP, muscles fatigue. Muscle pyruvate kinase inactivates around 40°C, which shuts off muscle function and thereby prevents thermal damage to the muscle.The thermal inactivation of muscle pyruvate kinase is probably a protective adaptation to prevent thermal damage to the muscle. You may thus wonder, if we could facilitate the extraction of heat from muscle, could we increase its capacity to do work? We consider this question in Investigating Life: Can the Work Capacity of Muscle Be Increased by Extracting Heat from the Palms of the Hands?