Question

If 6 $CO_2$ molecules entered into the Calvin cycle, number of G-3-P formed, net gain of G-3-P and G-3-P participated in RUBP regeneration are:

• A. 12, 2 & 10
• B. 2, 12 & 10
• C. 10, 12 & 2
• D. 6, 12 & 10

Hint:

12 G3P produced; 10 go back to keep the cycle turning, 2 go out to make sugar.

Answer 1

The Correct Option is A

Step 1: Set up the Calvin cycle.
The question takes 6 molecules of $CO_2$ into the Calvin cycle. We track how many G-3-P are made, how many are the real gain, and how many go back to rebuild RuBP.

Step 2: Fixation step.
Each $CO_2$ joins one RuBP. So 6 $CO_2$ join 6 RuBP. This makes 12 molecules of 3-PGA, since each fixation gives two 3-PGA.

Step 3: Reduction step.
Each 3-PGA is changed into one G-3-P using ATP and NADPH. So 12 molecules of 3-PGA give $12$ molecules of G-3-P formed.

Step 4: Find the net gain.
To build one 6-carbon sugar the cycle keeps only 2 G-3-P. So the net gain is $2$ G-3-P.

Step 5: Find the part used for regeneration.
The remaining G-3-P go back to make RuBP so the cycle can run again. That is $12 - 2 = 10$ G-3-P used for regeneration.

Step 6: Write the answer in order.
Formed = 12, net gain = 2, used for regeneration = 10. This is the first option.
\[ \boxed{12,\ 2\ \&\ 10} \]