Question

As compared to \[{C_3}\] plants, how many additional molecules of ATP are needed for the net production of one molecule of hexose sugar by \[{C_4}\] plants?
A. Two
B. Six
C. Twelve
D. Zero

Answer 1

Hint: \[{C_3}\] plants (spinach, cotton, wheat and rice) uses Calvin or \[{C_3}\] cycle for the formation of one molecule of hexose sugar whereas \[{C_4}\] plants (maize and sugarcane) uses Hatch and Slack pathway for the production of hexose sugar.

Complete answer:
Calvin cycle:
The primary acceptor of \[C{O_2}\] is RuBP- a five carbon compound.
The first stable product is 3-phosphoglycerate (3C compound).
It occurs in the mesophyll cells of the leaves.
It is a slower process of carbon fixation.
3 ATP are consumed to fix one carbon dioxide molecule or to form one molecule of hexose sugar 18 (6*3) ATP is consumed.
Hatch-Slack pathway:
The primary acceptor of \[C{O_2}\] is PEP- a three carbon compound.
The first stable product is oxaloacetic acid (4C compound).
It occurs in the mesophyll (\[{C_4}\] cycle) and bundle sheath (\[{C_3}\] cycle) cells of the leaves.
It is a faster process of carbon fixation.
2 ATP is consumed to fix one carbon dioxide.
To form one molecule of hexose sugar or to fix 6 molecules of carbon dioxide 30 (6*5) ATP are used. 3 ATP from calvin cycle and 2 ATP from hatch-slack pathway.

So the correct answer is option C.

Note: The \[{C_4}\] plants show better yield and high productivity than \[{C_3}\] plants because they evolved a mechanism to avoid photorespiration (prevents loss of 25% of \[C{O_2}\]) and increases the concentration of \[C{O_2}\] at RuBisCo site and minimise the oxygenase activity of the enzyme.