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At the end of the Krebs cycle, but before the electron transport chain, the oxidation of glucose has produced a net gain of
- a)3CO2, 5 NADH2, 1 FADH2, 2 ATP
- b)6CO2, 10 NADH2, 2 FADH2, 4 ATP
- c)6CO2, 10 NADH2, 2 FADH2, 38 ATP
- d)None of the above
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
At the end of the Krebs cycle, but before the electron transport chai...
At the end of the Krebs cycle, also known as the citric acid cycle or tricarboxylic acid cycle (TCA cycle), glucose has undergone a series of chemical reactions that result in the production of several molecules. These molecules include carbon dioxide (CO2), reduced forms of electron carriers such as NADH2 and FADH2, and a small amount of ATP.
The correct answer is option 'B': 6CO2, 10 NADH2, 2 FADH2, 4 ATP.
Here is a detailed explanation of each component:
1. Carbon Dioxide (CO2):
During the Krebs cycle, glucose is broken down into two molecules of pyruvate through a series of enzymatic reactions. Each pyruvate then enters the mitochondria and is further oxidized to produce one molecule of CO2. Since glucose produces two molecules of pyruvate, the total number of CO2 molecules produced is 2 x 3 = 6.
2. NADH2:
Throughout the Krebs cycle, each pyruvate molecule is completely oxidized, leading to the production of three molecules of NADH2. Since glucose produces two molecules of pyruvate, the total number of NADH2 molecules produced is 2 x 3 = 6.
3. FADH2:
During the Krebs cycle, one molecule of FADH2 is produced when succinate is converted to fumarate. Since glucose produces two molecules of succinate, the total number of FADH2 molecules produced is 2 x 1 = 2.
4. ATP:
During the Krebs cycle, one molecule of ATP is produced through substrate-level phosphorylation. Since glucose produces two molecules of ATP per cycle, the total number of ATP molecules produced is 2 x 2 = 4.
To summarize:
- CO2: 6 molecules
- NADH2: 10 molecules
- FADH2: 2 molecules
- ATP: 4 molecules
Thus, the net gain of glucose oxidation at the end of the Krebs cycle, but before the electron transport chain, is 6CO2, 10 NADH2, 2 FADH2, and 4 ATP, which matches option 'B'.
The correct answer is option 'B': 6CO2, 10 NADH2, 2 FADH2, 4 ATP.
Here is a detailed explanation of each component:
1. Carbon Dioxide (CO2):
During the Krebs cycle, glucose is broken down into two molecules of pyruvate through a series of enzymatic reactions. Each pyruvate then enters the mitochondria and is further oxidized to produce one molecule of CO2. Since glucose produces two molecules of pyruvate, the total number of CO2 molecules produced is 2 x 3 = 6.
2. NADH2:
Throughout the Krebs cycle, each pyruvate molecule is completely oxidized, leading to the production of three molecules of NADH2. Since glucose produces two molecules of pyruvate, the total number of NADH2 molecules produced is 2 x 3 = 6.
3. FADH2:
During the Krebs cycle, one molecule of FADH2 is produced when succinate is converted to fumarate. Since glucose produces two molecules of succinate, the total number of FADH2 molecules produced is 2 x 1 = 2.
4. ATP:
During the Krebs cycle, one molecule of ATP is produced through substrate-level phosphorylation. Since glucose produces two molecules of ATP per cycle, the total number of ATP molecules produced is 2 x 2 = 4.
To summarize:
- CO2: 6 molecules
- NADH2: 10 molecules
- FADH2: 2 molecules
- ATP: 4 molecules
Thus, the net gain of glucose oxidation at the end of the Krebs cycle, but before the electron transport chain, is 6CO2, 10 NADH2, 2 FADH2, and 4 ATP, which matches option 'B'.
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At the end of the Krebs cycle, but before the electron transport chai...
One molecule of the glucose that enters in the glycolysis produces 2 molecules pyruvate and 2 molecules of NADH2 and 2 ATP. 2 molecules of pyruvate will form 2 molecules of Acetyl Co-A and this will release 2 molecules of CO2 and 2 NADH2 . These 2 Acetyl Co-A will enter into Krebs cycle and will release 4CO2 , 6 NADH2 , 2 FADH2 and 2 ATP. so, the total gain before the ETC comes out to be 6 CO2 , 10NADH2 , 2FADH 2 and 4 ATP. Hence, the correct answer is '6CO2 , 10 NADH2 , 2 FADH2 , 4ATP'.
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At the end of the Krebs cycle, but before the electron transport chain, the oxidation of glucose has produced a net gain ofa)3CO2, 5 NADH2, 1 FADH2, 2 ATPb)6CO2, 10 NADH2, 2 FADH2, 4 ATPc)6CO2, 10 NADH2, 2 FADH2, 38 ATPd)None of the aboveCorrect answer is option 'B'. Can you explain this answer?
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