1. From one molecule of glucose, the reactions of the citric acid cycle (a.k.a.
ID: 212167 • Letter: 1
Question
1. From one molecule of glucose, the reactions of the citric acid cycle (a.k.a. TCA, Krebs cycle) yield enough energy in the presence of oxygen to synthesize:
Select one:
a. 32 ATP
b. 20 ATP
c. 25 ATP
d. 30 ATP
e. 2 ATP
2. Which of the following components of the citric acid cycle (Krebs) contains the least hydrogen atoms?
Select one:
a. oxaloacetate
b. succinate
c. -ketoglutarate
d. citrate
3. Which of the following components of the citric acid cycle (Krebs) has the most hydrogen atoms?
Select one:
a. malate
b. -ketoglutarate
c. citrate
d. oxaloacetate
4. What is the maximum yield of ATP for the oxidation of one glucose molecule in a heart cell under aerobic conditions?
5. For each mole of glucose oxidized by a brain cell under aerobic conditions, how many moles of ATP are formed through oxidative phosphorylation only?
Explanation / Answer
Q.1 Answer isB. from one molecule of glucose under aerobic condition the kreb cycle yields energy sufficient to synthesis 20 ATPs.
Two molecules of pyruvate obtain from glycolysis get converted in 2 Acetyl CoA molecules in the linkage step. these 2 acetyl CoA molecule then undergo Kreb cycle and produces reducing equivalents NADH and FADH2. Each moleule of Acetyl CoA yields 3 NADH, 1 FADH2 , 1 ATP/GTP. so 2 Acetyl CoA molecules will give 6 NADH, 2FADH2, 2 ATP/GTP.
1 NADH YIELDS 2.5 ATP. SO 6 * 2.5= 15
1 FADH2 YIELDS 1.5 ATP SO 2*1.5 = 3
2 ATP/GTP = 2
TOTAL =20 ATP.
So answer is 20 ATP.
Q.2 Answer is A oxaloacetate. its formula is CH2CO(COOH)2
Q.3 Answer is C citrate. It has most hydrogen atoms.
Q.4 oxidation of one glucose molecule in a heart cell yields 38 or 36 ATP molecules. ( if you use 1 NADH GIVES 3 ATP AND 1 FADH2 GIVES 2 ATP FOR CALCULATION THEN ANSWER WILL BE 38. if you use 1 NADH = 2.5 ATP & 1 FADH2 = 2 ATP THEN ANSWER WILL BE 36 ATP)
Q.5 As NADH cannot enter mitochondria directly it follows glycerol-3-phosphate dehydrogenase pathway to enter mitochondria in Brain cell. By this pathway under aerobic condition oxidation of 1 molecule of glucose gives 30 ATP molecules.
1 MOLE of glucose = 6.023 * 1023 MOLECULES of glucose
1 molecule of glucose gives = 30 ATP so 6.023*1023 molecule of glucose will give
30 * 6.023*1023 = 1.806 * 1025 molecules of ATP.
1 mole of ATP = 6.023 *1023 ATP molecules
so 1.806 *1025 ATP molecules= 1.806 *1025 / 6.023*1023 = 29.98 moles of ATP.
so one mole glucose oxidized by brain cell under aerobic conditions gives 29.98 moles of ATP.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.