Question

Transketolase: \(2A + 2G\text{-}3P \rightarrow 2\) xylulose-5-phosphate \(+ 2B\). In the above reaction A and B are respectively

• A. Sedoheptulose-7-phosphate and erythrose-4-phosphate
• B. Fructose-6-phosphate and ribose-5-phosphate
• C. Fructose-6-phosphate and erythrose-4-phosphate
• D. DHAP and fructose-6-phosphate

Hint:

Remember the important PPP reaction: \[ 2F6P + 2G3P \leftrightarrow 2Xu5P + 2R5P \] This reaction is catalyzed by transketolase and frequently appears in competitive examinations.

Answer 1

The Correct Option is B

Concept: Transketolase is a key enzyme of the Pentose Phosphate Pathway (PPP). It transfers a two-carbon unit from a ketose donor to a ketose or aldose acceptor. The reactions of the non-oxidative phase of PPP help interconvert sugars having different carbon numbers. A well-known transketolase reaction is: \[ 2 \text{Fructose-6-phosphate} + 2 \text{Glyceraldehyde-3-phosphate} \rightarrow 2 \text{Xylulose-5-phosphate} + 2 \text{Ribose-5-phosphate} \]

Step 1: Compare the given reaction with the standard PPP reaction. Given: \[ 2A + 2G\text{-}3P \rightarrow 2 \text{Xylulose-5-phosphate} + 2B \] This exactly corresponds to the reversible transketolase reaction: \[ 2 \text{Fructose-6-phosphate} + 2 \text{Glyceraldehyde-3-phosphate} \rightarrow 2 \text{Xylulose-5-phosphate} + 2 \text{Ribose-5-phosphate} \]

Step 2: Identify A. Comparing both equations: \[ A = \text{Fructose-6-phosphate} \]

Step 3: Identify B. Similarly: \[ B = \text{Ribose-5-phosphate} \]

Step 4: Verify carbon balance. Left side: \[ 2(6) + 2(3)=18 \] Right side: \[ 2(5)+2(4)=18 \] Thus the reaction is balanced.

Step 5: Final conclusion. \[ \boxed{ A=\text{Fructose-6-phosphate}, \quad B=\text{Ribose-5-phosphate} } \] Hence option (B) is correct.