In Calvin cycle 1.3 - Bisphosphoglyceric acid is reductive dephosphorylated to form

Options

A. 3 - PGA
B. G - 3 - P
C. PEP
D. RuBP

Answer

To solve the question regarding the product formed from the reductive dephosphorylation of 1,3-bisphosphoglyceric acid in the Calvin cycle, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Starting Compound**:
- The starting compound in the question is 1,3-bisphosphoglyceric acid (1,3-BPG), which is also referred to as 1,3-bisphosphoglycerate.

2. **Understand the Process**:
- The question states that 1,3-BPG undergoes reductive dephosphorylation. This means that the compound will lose a phosphate group (dephosphorylation) and will also gain electrons (reduction).

3. **Reduction Process**:
- The reduction process in the Calvin cycle involves the use of NADPH, which donates electrons (and protons) to the 1,3-BPG. This results in the formation of a new compound.

4. **Dephosphorylation**:
- During dephosphorylation, one phosphate group is removed from 1,3-BPG. This step is crucial as it transforms the compound into a simpler form.

5. **Resulting Compound**:
- After the reduction and dephosphorylation of 1,3-BPG, the resulting compound is glyceraldehyde 3-phosphate (G3P).

6. **Conclusion**:
- Therefore, the final answer to the question is that 1,3-bisphosphoglyceric acid is reductively dephosphorylated to form glyceraldehyde 3-phosphate (G3P).

### Final Answer:
1,3-bisphosphoglyceric acid is reductively dephosphorylated to form glyceraldehyde 3-phosphate (G3P).

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