Question

BACKGROUND INFORMATION:

Vinnie is a professional road cyclist participating in the 2020 Giro d’Italia, a three-week race nearly 3,600 km in distance with some stages exceeding altitudes of 2,700 m above sea level (i.e., Stelvio Pass, Italy). As the newly hired head of the high-performance team, it is your responsibility to manage his preparation for the event. You must demonstrate to your team a fundamental understanding of physiology, the chronic adaptations expected from his training, and methods that could be implemented to improve the likelihood of success.

QUESTION:

The longest leg of the race is 228 km, during which energy intake is critical. Glycolysis and β-oxidation are processes that break down carbohydrates (i.e., glucose, sucrose, fructose) and fatty acids, respectively.

A. Individually, what is the total ATP yield from one molecule of glucose and one molecule of palmitate?

B. How did you reach these numbers (i.e., substrate yield and use)?

C. Based on what you know about the yield and speed of these two pathways, do you recommend Vinnie ingest glucose or fatty acids during the race? (Hint, the cross-over concept)

Answer 1

A. 1 molecule of glucose yields 34 ATP molecules

1 molecule of palmitate yields 108 ATP molecules

B. Oxidation of Glucose:

Glycolysis (Breakdown of gluccose to 2 pyruvic acid molecules): 4 ATP + 2 NADH+H

Oxidative decarboxylation of 2 Pyruvic acid to produce 2 Acetyl CoA: 2 NADH+H

Kreb's Cycle (oxidation of acetyl CoA to CO2 and H2O): 3 NADH+H + 1 FADH2 + 1 ATP (from 1 Acetyl CoA molecule)

Therefore 2 molecules of Acetyl CoA yields: 6 NADH+H + 2 FADH2 + 2 ATP

Therefore,

Total =10 NADH+H + 2 FADH2 + 6 ATP

1 NADH+H = 2.5 ATP molecules, 1 FADH2 = 1.5 ATP molecules

Therefore total number of ATP molecules = 10 x 2.5 + 2 x 1.5 + 6

= 25 + 3 + 6

= 34

Initially, 2 ATP molecules are used up during glycolysis

Therefore net production = 34 - 2 = 32 ATP molecules from 1 molecule of glucose.

Oxidation of palmitic acid:

Plamitic acid is the C16 satrurated fatty acid.

It undergoes 7 rounds of beta oxidation to yield 8 molecules of Acetyl CoA

During 1 round of acetyl CoA - 1 NADH+H and 1 FADH2 are produced

Therefore during 7 rounds-----> 7 NADH+H and 7 FADH2 are produced

1 molecule of Acetyl CoA ------> 3 NADH+H + 1 FADH2 + 1 ATP
8 molecules of Acetyl CoA -------> 24 NADH+H + 8 FADH2 + 8 ATP

therefore total ATP equivalents ---> 31 NADH+H + 15 FADH2 + 8 ATP

= 31 x 2.5 ATP + 15 x 1.5 ATP + 8

= 77.5 + 22.5 + 8

= 108 ATP

C. For part III i am not sure.