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Photosynthesis in Higher Plants - C3 and C4 pathways

Grade 11CBSEBiology

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

The biosynthetic phase of photosynthesis (Dark Reaction) occurs in the stroma of chloroplasts and does not directly depend on light but relies on the products of the light reaction, ATPATP and NADPHNADPH.

In the C3C_3 pathway (Calvin Cycle), the primary acceptor of CO2CO_2 is a 5-carbon sugar called RuBPRuBP (Ribulose-1,5-bisphosphate), and the first stable product is a 3-carbon compound, 3PGA3-PGA (3-phosphoglyceric acid).

The Calvin cycle consists of three stages: Carboxylation (fixation of CO2CO_2 into 3PGA3-PGA), Reduction (formation of glucose using ATPATP and NADPHNADPH), and Regeneration (renewal of the CO2CO_2 acceptor RuBPRuBP).

In the C4C_4 pathway (Hatch-Slack pathway), plants exhibit 'Kranz anatomy'. The primary CO2CO_2 acceptor is PEPPEP (Phosphoenolpyruvate) located in the mesophyll cells, and the first stable product is OAAOAA (Oxaloacetic acid), a 4-carbon compound.

Photorespiration is a wasteful process occurring in C3C_3 plants when RuBisCORuBisCO binds with O2O_2 instead of CO2CO_2 at high temperatures and high O2O_2 concentration, leading to the formation of phosphoglycolate and no synthesis of sugar or ATPATP.

C4C_4 plants are more efficient than C3C_3 plants in high temperatures and low CO2CO_2 concentrations because they have a mechanism to increase the concentration of CO2CO_2 at the enzyme site, thus minimizing photorespiration.

📐Formulae

6CO2+12H2OLight/ChlorophyllC6H12O6+6H2O+6O26CO_2 + 12H_2O \xrightarrow{Light/Chlorophyll} C_6H_{12}O_6 + 6H_2O + 6O_2

For 1 molecule of glucose in C3 plants: 18ATP+12NADPH\text{For } 1 \text{ molecule of glucose in } C_3 \text{ plants: } 18 ATP + 12 NADPH

For 1 molecule of glucose in C4 plants: 30ATP+12NADPH\text{For } 1 \text{ molecule of glucose in } C_4 \text{ plants: } 30 ATP + 12 NADPH

Net reaction of Carboxylation in C3:RuBP+CO2+H2ORuBisCO2(3PGA)\text{Net reaction of Carboxylation in } C_3: RuBP + CO_2 + H_2O \xrightarrow{RuBisCO} 2(3-PGA)

💡Examples

Problem 1:

Calculate the number of ATPATP and NADPHNADPH molecules required to fix 66 molecules of CO2CO_2 in a Maize plant.

Solution:

30ATP30 ATP and 12NADPH12 NADPH.

Explanation:

Maize is a C4C_4 plant. In the C4C_4 pathway, 22 extra ATPATP are required per CO2CO_2 molecule fixed (to pump C4C_4 acids) in addition to the 3ATP3 ATP used in the Calvin cycle. Thus, 5ATP5 ATP per CO2CO_2 fixed. For 6CO26 CO_2, total ATP=6×5=30ATP = 6 \times 5 = 30. The NADPHNADPH requirement remains 22 per CO2CO_2, so 6×2=126 \times 2 = 12.

Problem 2:

Why does RuBisCORuBisCO function as an oxygenase in C3C_3 plants during high temperatures?

Solution:

Due to the competitive binding at the active site of RuBisCORuBisCO where the affinity for O2O_2 increases as temperature rises.

Explanation:

RuBisCORuBisCO has a dual affinity for CO2CO_2 and O2O_2. When the concentration of O2O_2 increases or temperature rises, the enzyme's active site binds with O2O_2 instead of CO2CO_2, initiating the C2C_2 cycle (Photorespiration) where RuBPRuBP is oxidized to one molecule of 3PGA3-PGA and one molecule of phosphoglycolate.

C3 and C4 pathways - Revision Notes & Key Diagrams | CBSE Class 11 Biology