Question

From the following lists, choose the correct combinations.

\[ \begin{array}{|c|l|l|l|} \hline & \text{List-I (Substrate)} & \text{List-II (Enzyme)} & \text{List-III (Product)} \\ \hline \text{I} & \text{PGA + ATP} & \text{Phosphoglycerokinase} & \text{Bis PGA} \\ \hline \text{II} & \text{Bis PGA + NADPH} & \text{G-3-P dehydrogenase} & \text{G-3-P} \\ \hline \text{III} & \text{Xylulose} & \text{Epimerase} & \text{Ribulose} \\ \hline \text{IV} & \text{Fructose + G-3-P} & \text{Transketolase} & \text{Xylulose} \\ \hline \end{array} \]
The correct answer is (and = all, only = only)

• A. I and II only
• B. II, III and IV only
• C. I, II and IV only
• D. I, II, III and IV

Hint:

Important Calvin cycle enzymes: \[ \boxed{ \begin{aligned} &\text{Phosphoglycerokinase} &\text{G-3-P dehydrogenase} &\text{Epimerase} &\text{Transketolase} \end{aligned} } \] These enzymes participate in reduction and regeneration phases of the Calvin cycle.

Answer 1

The Correct Option is D

Step 1: Recall the Calvin cycle reactions. \[ \mathrm{PGA + ATP} \xrightarrow{\text{Phosphoglycerokinase}} \mathrm{Bis\ PGA} \] Hence, Statement I is correct. \[ \mathrm{Bis\ PGA + NADPH} \xrightarrow{\text{G-3-P dehydrogenase}} \mathrm{G\!-\!3\!-\!P} \] Hence, Statement II is correct. Xylulose is converted into ribulose by \[ \boxed{\text{Epimerase}.} \] Hence, Statement III is correct. Fructose-6-phosphate and glyceraldehyde-3-phosphate react in the presence of \[ \boxed{\text{Transketolase}} \] to produce xylulose phosphate during regeneration. Hence, Statement IV is correct.

Step 2: Choose the correct combinations. Therefore, \[ \boxed{\text{I, II, III and IV}} \] are correct. Hence, the correct option is \[ \boxed{(D).} \]