Download Chem 32 Solutions to Section 15.4 – 15.6 Homework Problems

Chem 32
Solutions to Section 15.4 – 15.6 Homework Problems
15.78 Only pathway “c” produces energy that can be harnessed to make ATP. Pathway “a”
does not produce or consume a significant amount of energy. Pathway “b” is an activation step,
and consumes energy (the body breaks down ATP in this step).
15.86 The starting materials of the citric acid cycle are acetyl-CoA, ADP, phosphate, FAD,
three molecules of NAD+, and three water molecules. The products are coenzyme A, ATP,
FADH2, three molecules of NADH, three hydrogen ions, and two molecules of CO2.
(Oxaloacetic acid is consumed in the first step, but is made in the last step.)
15.88 Milk spoils when certain bacteria carry out lactic acid fermentation. The product that
gives the milk its “spoiled” aroma and flavor is lactic acid.
15.90 a) The starting materials of alcoholic fermentation are glucose, ADP, and phosphate ions.
The products are ethanol, CO2, and ATP.
b) Alcoholic fermentation produces two molecules of ATP for every molecule of glucose.
15.92 The total ATP yield from one pyruvate ion is 12.5 molecules of ATP. Here is how you
can calculate this number.
Oxidative decarboxylation (pyruvate → acetyl-CoA): 1 NADH = 2.5 ATP
Citric acid cycle (acetyl-CoA → CO2):
1 ATP
3 NADH = 7.5 ATP
1 FADH2 = 1.5 ATP
15.96 The total ATP yield from one molecule of ethanol is 15 molecules of ATP. Here is how
you can calculate this number.
Ethanol → acetaldehyde:
1 NADH = 2.5 ATP
Acetaldehyde → acetyl-CoA:
1 NADH = 2.5 ATP
Citric acid cycle (acetyl-CoA → CO2):
1 ATP
3 NADH = 7.5 ATP
1 FADH2 = 1.5 ATP
15.98 a) The total ATP yield from three molecules of glucose is 80 molecules of ATP.
3 glucose → 5 pyruvate:
5 ATP
5 NADH = 12.5 ATP
5 pyruvate → 5 acetyl-CoA:
(Each pyruvate gives us 1 NADH)
5 NADH = 12.5 ATP
5 acetyl-CoA → 10 CO2
(Each acetyl-CoA gives us
1 ATP, 3 NADH, 1 FADH2)
5 ATP
15 NADH = 37.5 ATP
5 FADH2 = 7.5 ATP
b) For one molecule of glucose, the ATP yield is 80 ÷ 3 = 26.7 molecules of ATP.
15.102 a) The ATP yield from a molecule of capric acid is 64 molecules of ATP.
The first step in the oxidation of any fatty acid is the activation step, which breaks down
two molecules of ATP. Next, the fatty acid is broken down into acetyl-CoA. For a ten-carbon
fatty acid, this requires four cycles of beta-oxidation, and the result is five molecules of acetylCoA. Finally, the acetyl-CoA is broken down to CO2 by the citric acid cycle.
Activation step:
–2 ATP
Beta oxidation (4 cycles)
4 NADH = 10 ATP
4 FADH2 = 6 ATP
Citric acid cycle (5 cycles)
5 ATP
15 NADH = 37.5 ATP
5 FADH2 = 7.5 ATP
The overall total is 64 molecules of ATP. (Don’t forget to account for the two molecules that
were broken down during the activation step.)
b) Beta oxidation forms 16 molecules of ATP, as shown above.
15.118 The total ATP yield from one molecule of isoleucine is 33.5 molecules of ATP.
isoleucine → acetyl-CoA
(the reaction in the problem)
1 ATP
8 NADH = 20 ATP
3 FADH2 = 4.5 ATP
acetyl-CoA → CO2 (citric acid cycle)
1 ATP
3 NADH = 7.5 ATP
1 FADH2 = 1.5 ATP
converting NH4+ into urea
–2 ATP
Adding these up gives 33.5 molecules of ATP.
15.122 a) The yield is 8.3 moles of ATP per 100 g of asparagine (the calculator answer is
8.3255124 moles). The molecular formula of asparagine is C4H8N2O3, so the formula weight of
asparagine is 132.124. The ATP yield per 100 g of asparagine is:
11 moles ATP
100 g ×
= 8.3 moles ATP
132.124 g asparagine
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