AP BIOLOGY

PHOTOSYNTHESIS

THE CALVIN CYCLE

Question [CLICK ON ANY CHOICE TO KNOW THE RIGHT ANSWER]
In the process of efficiently creating 3 glucose molecules, how many PGA molecules would be produced?
A
18
B
6
C
36
D
12
Explanation: 

Detailed explanation-1: -During carboxylation, CO2 combines with RuBP (primary acceptor molecule of CO2) and forms an unstable 6 carbon compound. This reaction is catalysed by the enzyme RuBisCO. The unstable 6 carbon compound splits into two molecules of 3-PGA. Hence, for every one CO2 molecule fixed, two molecules of 3-PGA are produced.

Detailed explanation-2: -3-Phosphoglyceric acid (3PG, 3-PGA, or PGA) is the conjugate acid of 3-phosphoglycerate or glycerate 3-phosphate (GP or G3P). This glycerate is a biochemically significant metabolic intermediate in both glycolysis and the Calvin-Benson cycle. The anion is often termed as PGA when referring to the Calvin-Benson cycle.

Detailed explanation-3: -When three CO2start text, C, O, end text, start subscript, 2, end subscript molecules enter the cycle, six G3P molecules are made. One exits the cycle and is used to make glucose, while the other five must be recycled to regenerate three molecules of the RuBP acceptor.

Detailed explanation-4: -Each turn of the cycle involves only one RuBP and one carbon dioxide and forms two molecules of 3-PGA.

There is 1 question to complete.