Question

The first stable product identified in Calvin cycle of ${C_3}$ and ${C_4}$ plants are respectively
A. 3-PGA and 3-PGA
B. 3-PGA and OAA
C. 1,3-di PGA and malate
D. 3-PGA and malate

Answer 1

Hint: The Calvin cycle is the chemical reaction cycle that plants undergo to "fix" carbon from $C{O_2}$ into three-carbon sugars. Later, these three-carbon compounds can be converted by plants and animals into amino acids, nucleotides, and more complex sugars, such as starches. These reactions occur outside the thylakoid membranes in the stroma, the fluid-filled area of a chloroplast.

Complete answer:
1, 3-di PGA is a transitional stage during $C{O_2}$ fixation/reduction between glycerate 3-phosphate and glyceraldehyde 3-phosphate. Malate is a ${C_4}$ - dicarboxylate that results from the malic acid deprotonation of both carboxyl groups. As a fundamental metabolite, it has a function.
The 3 carbon compound, that is, 3 phosphoglyceric acid (PGA), is the first stable compound formed in the Calvin cycle. Because of this the Calvin cycle is called the ${C_3}$ cycle as well. In the ${C_4}$ plant, Oxaloacetic Acid (OAA) is a 4 carbon compound that is the first stable product. This oxaloacetic acid is then converted into a 4 C compound called malate. This 4 C compound then reaches the sheath cell of the bundle where the Calvin cycle in a chloroplast takes place. Here as well, 3 phosphoglyceric acid (PGA) is the first stable product formed.

Hence, the correct answer is option (A).

Note: In both glycolysis and the Calvin cycle, 3-PGA is a biochemically important metabolic intermediate. When referring to the Calvin cycle, this anion is also represented as PGA. 3-phosphoglycerate is the result of the spontaneous split of an unstable 6-carbon intermediate formed upon $C{O_2}$ fixation in the Calvin cycle. For each molecule of $C{O_2}$ that is set, two equivalents of 3-phosphoglycerate are thus formed.