
Oxidation of one molecule of glucose in aerobic respiration produces
A. 2 ATP
B. 30 ATP
C. 36 ATP
D. 38 ATP
Answer
558.6k+ views
Hint: Aerobic respiration requires oxygen to produce energy or ATP. It involves three phases i.e., glycolysis, oxidative decarboxylation, and citric acid cycle.
Complete answer:
Aerobic respiration involves three phases to carry out the oxidation of one molecule of glucose.
Phase 1: Glycolysis
It takes place in the cytosol of cells in all living organisms and can occur either in the presence or absence of oxygen. In the presence of oxygen, glycolysis converts one molecule of glucose into two molecules of pyruvate. The reaction for glycolysis has been stated below:
Glucose$ + $2 $NA{D^ + }$ $+$2 ${P_i}$ $ + $2 ADP$\to$2 Pyruvate$ + $2 ${H^ + }$ $ + $2 NADH $ + $2 ATP$ + $2 ${H_2}O$
Total ATP produced during glycolysis:
$ \Rightarrow $2 ATP$ + $2 NADH (Since, 1 NADH = 3 ATP)
$ \Rightarrow $2 ATP$ + $ 2$ \times $3 ATP = 8 ATP
Phase 2: Oxidative decarboxylation
In this phase, the pyruvate produced in glycolysis is oxidized to acetyl-CoA and$C{O_2}$by the pyruvate dehydrogenase complex. In this conversion, 1 NADH and 1 molecule of$C{O_2}$are formed.
Since two pyruvate molecules are produced during glycolysis therefore the oxidative decarboxylation of 2 pyruvate molecules will yield 2 NADH molecules. So, total ATP yield in this phase can be calculated: As, 1 NADH = 3 ATP; therefore 2 NADH = 2$ \times $3 ATP = 6 ATP.
Phase 3: Citric acid cycle
Also known as Kreb’s cycle or tricarboxylic acid cycle, takes place in the mitochondria under aerobic conditions. The net gain from a single cycle is 3 NADH, 1$FAD{H_2}$, and 1 GTP. As, from one glucose molecule, two pyruvate molecules are formed therefore two cycles will be formed for the complete breakdown. So, the total yield will be 6 NADH, 2$FAD{H_2}$, and 2 GTP.
Total ATP = 6 NADH$ + $2$FAD{H_2} + $2 GTP. (1 GTP = 1 ATP; 1$FAD{H_2}$= 2 ATP)
$ \Rightarrow $Total ATP = 6$ \times $3 ATP$+$2$\times$2 ATP$ + $2$ \times $1 ATP = 18 ATP$ + $4 ATP$ + $2 ATP = 24 ATP.
So, total ATP yield in the oxidation of one molecule of glucose during aerobic respiration can be summarized below:
Total ATP = ATP yield in glycolysis$ + $ATP yield in oxidative decarboxylation$ + $ATP yield in a citric acid cycle.
$ \Rightarrow $Total ATP = 8 ATP$ + $6 ATP$ + $24 ATP = 38 ATP.
So, after looking at the whole derivation of ATP molecules during the three phases of aerobic respiration, it can be concluded that only option (D) i.e., 38 ATP is correct while the other options are incorrect.
Hence, the correct answer is option (D).
Note: The aerobic respiration process is a far more energy-efficient process as it generates 38 molecules of ATP leading to the complete breakdown of glucose into carbon dioxide while the anaerobic respiration generates only 2 ATP molecules per glucose molecule.
Complete answer:
Aerobic respiration involves three phases to carry out the oxidation of one molecule of glucose.
Phase 1: Glycolysis
It takes place in the cytosol of cells in all living organisms and can occur either in the presence or absence of oxygen. In the presence of oxygen, glycolysis converts one molecule of glucose into two molecules of pyruvate. The reaction for glycolysis has been stated below:
Glucose$ + $2 $NA{D^ + }$ $+$2 ${P_i}$ $ + $2 ADP$\to$2 Pyruvate$ + $2 ${H^ + }$ $ + $2 NADH $ + $2 ATP$ + $2 ${H_2}O$
Total ATP produced during glycolysis:
$ \Rightarrow $2 ATP$ + $2 NADH (Since, 1 NADH = 3 ATP)
$ \Rightarrow $2 ATP$ + $ 2$ \times $3 ATP = 8 ATP
Phase 2: Oxidative decarboxylation
In this phase, the pyruvate produced in glycolysis is oxidized to acetyl-CoA and$C{O_2}$by the pyruvate dehydrogenase complex. In this conversion, 1 NADH and 1 molecule of$C{O_2}$are formed.
Since two pyruvate molecules are produced during glycolysis therefore the oxidative decarboxylation of 2 pyruvate molecules will yield 2 NADH molecules. So, total ATP yield in this phase can be calculated: As, 1 NADH = 3 ATP; therefore 2 NADH = 2$ \times $3 ATP = 6 ATP.
Phase 3: Citric acid cycle
Also known as Kreb’s cycle or tricarboxylic acid cycle, takes place in the mitochondria under aerobic conditions. The net gain from a single cycle is 3 NADH, 1$FAD{H_2}$, and 1 GTP. As, from one glucose molecule, two pyruvate molecules are formed therefore two cycles will be formed for the complete breakdown. So, the total yield will be 6 NADH, 2$FAD{H_2}$, and 2 GTP.
Total ATP = 6 NADH$ + $2$FAD{H_2} + $2 GTP. (1 GTP = 1 ATP; 1$FAD{H_2}$= 2 ATP)
$ \Rightarrow $Total ATP = 6$ \times $3 ATP$+$2$\times$2 ATP$ + $2$ \times $1 ATP = 18 ATP$ + $4 ATP$ + $2 ATP = 24 ATP.
So, total ATP yield in the oxidation of one molecule of glucose during aerobic respiration can be summarized below:
Total ATP = ATP yield in glycolysis$ + $ATP yield in oxidative decarboxylation$ + $ATP yield in a citric acid cycle.
$ \Rightarrow $Total ATP = 8 ATP$ + $6 ATP$ + $24 ATP = 38 ATP.
So, after looking at the whole derivation of ATP molecules during the three phases of aerobic respiration, it can be concluded that only option (D) i.e., 38 ATP is correct while the other options are incorrect.
Hence, the correct answer is option (D).
Note: The aerobic respiration process is a far more energy-efficient process as it generates 38 molecules of ATP leading to the complete breakdown of glucose into carbon dioxide while the anaerobic respiration generates only 2 ATP molecules per glucose molecule.
Recently Updated Pages
The number of solutions in x in 02pi for which sqrt class 12 maths CBSE

Write any two methods of preparation of phenol Give class 12 chemistry CBSE

Differentiate between action potential and resting class 12 biology CBSE

Two plane mirrors arranged at right angles to each class 12 physics CBSE

Which of the following molecules is are chiral A I class 12 chemistry CBSE

Name different types of neurons and give one function class 12 biology CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

What is 1s 2s 2p 3s 3p class 11 chemistry CBSE

Discuss the various forms of bacteria class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

An example of chemosynthetic bacteria is A E coli B class 11 biology CBSE

