Question

In three Calvin cycle, the gross number of PGAL molecules produces at the cost of ATP and $NADPH_2$ is
A. 3PGAL, 3ATP, $3NADPH_2$
B. 6PGAL, 6ATP, $6NADPH_2$
C. 18PGAL, 18ATP, $18NADPH_2$
D. 9PGAL, 9ATP, $9NADPH_2$

Answer 1

Hint: Calvin cycle is the process of fixation of Carbon dioxide into three-carbon sugars. It takes 6 turns of the Calvin cycle to produce one molecule of glucose. It consumes 18 ATP and 12NADPH to form one glucose molecule.

Complete answer:
Option A: Once the Calvin cycle uses up 2 molecules of ATP in the carbon fixation step.
Therefore, this is the incorrect option.
Option B: Calvin cycle has 3 steps. The first stage is carbon fixation. Here one molecule of carbon dioxide binds with 5 carbon RuBP sugar and forms 2 molecules of a 3 carbon compound, 3PGA. This process uses 2 ATP. Three Carbon dioxide use up 6 ATP similarly and produce 6 molecules of 3PGA.
The next step is a reduction. Here 6 molecules of 3PGA from the previous step are reduced into 6 molecules of G3P or PGAL. This is happening at an energy expenditure of 6 NADPH2. One G3P or PGAL is used to form glucose. The other five are used to regenerate the RuBP used up in the first step.
 Therefore in three turns of the Calvin cycle, 6 PGAL is formed at the expense of 6 ATP and 6 NADPH. This is the correct answer.

Option C: In 3 repetition of the Calvin cycle, only 6 molecules of PGAL are produced. Therefore, this is the incorrect option.

Option D: In three Calvin cycles 9 molecules of PGAL are not produced. Therefore, this is the incorrect option.

Hence, the correct answer is option (B)

Note: In the last step of the Calvin cycle, regeneration of RuBP happens from five PGAL molecules. This process uses 3 ATP per three cycles. The question is referring to the part before this. Therefore this is not included in the calculation.