Why does photorespiration ($C_2$ cycle) occur in $C_3$ plants but is absent in $C_4$ plants?

RuBisCO has dual catalytic affinity for both $CO_2$ and $O_2$. When $O_2$ concentration is high or temperatures rise, RuBisCO binds $O_2$ to produce $1$ molecule of phosphoglycerate ($3C$) and $1$ molecule of phosphoglycolate ($2C$). In $C_4$ plants, Kranz anatomy physically separates primary $CO_2$ fixation in mesophyll cells (via PEP carboxylase, which has zero oxygenase activity) from the Calvin cycle in bundle sheath cells, maintaining a high internal $CO_2$ concentration around RuBisCO to completely suppress oxygenation.

How does the proton gradient in chloroplasts differ from that in mitochondria during ATP synthesis?

In chloroplasts, protons ($H^+$) are pumped into the internal thylakoid lumen from the stroma, making the lumen acidic ($pH \approx 4-5$). In mitochondria, protons are pumped from the matrix into the intermembrane space. In chloroplasts, the chemiosmotic force is dominated by a large $pH$ differential ($\Delta pH$) rather than an electrical membrane potential ($\Delta \psi$).

How do I quickly remember the total ATP and NADPH count for $C_3$ vs $C_4$ plants in NEET numerical problems?

Memorize the base formula per $CO_2$ fixed: $C_3 = 3\text{ ATP} + 2\text{ NADPH}$ and $C_4 = 5\text{ ATP} + 2\text{ NADPH}$. For a 6-carbon glucose ($C_6H_{12}O_6$), multiply by $6$: $C_3 = 18\text{ ATP} + 12\text{ NADPH}$ and $C_4 = 30\text{ ATP} + 12\text{ NADPH}$.

Photosynthesis in Higher Plants Notes, Formulas, Traps & PYQs | ExamiFly

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Key Formulas & High-Yield Equations

Overall Photosynthetic Equation

$6CO_2 + 12H_2O \xrightarrow{\text{Light, Chlorophyll}} C_6H_{12}O_6 + 6H_2O + 6O_2$

Strictly represents oxygenic photosynthesis where

H_2O

acts as the electron donor and is oxidized to release

O_2

Photolysis of Water Reaction

$2H_2O \xrightarrow{\text{Mn}^{2+}, \text{Cl}^-} 4H^+ + 4e^- + O_2$

Occurs strictly on the inner side of the thylakoid membrane, contributing

H^+

directly to the thylakoid lumen.

Stoichiometry of $C_3$ Calvin Cycle (Per Glucose)

$6 CO_2 + 18 ATP + 12 NADPH \rightarrow C_6H_{12}O_6 + 18 ADP + 12 NADP^+ + 18 P_i$

Requires 3

ATP

and 2

NADPH

per

CO_2

fixed under non-limiting light and normal atmospheric conditions.

Stoichiometry of $C_4$ Pathway (Per Glucose)

$6 CO_2 + 30 ATP + 12 NADPH \rightarrow C_6H_{12}O_6 + 30 ADP + 12 NADP^+ + 30 P_i$

Consumes an additional 2

ATP

per

CO_2

fixed compared to

C_3

plants to transport

PEP

back to mesophyll cells.

Proton Motive Force (Chemiosmosis)

$\Delta p = \Delta \psi - \frac{2.303 RT}{F} \Delta pH$

In chloroplasts, the proton motive force driving

ATP

synthesis is predominantly generated by a large

\Delta pH

across the thylakoid membrane rather than membrane potential

\Delta \psi

Blackman's Law of Limiting Factors

$\text{Photosynthetic Rate} = f(\text{Limiting Factor with Minimum Rate})$

Applies strictly when a biochemical process is conditioned by multiple variables; the overall rate is dictated by the factor nearest to its minimum value.

Solved High-Yield Sample Questions

Practice Mode

Question #1NEET 2024

To synthesize one molecule of glucose ( $C_6H_{12}O_6$ ), how many total $ATP$ and $NADPH$ molecules are consumed by a $C_4$ plant such as Maize?

Solution Locked

Step-by-Step Solution & Concept

Step 1

Calculate requirements for $1$ molecule of $CO_2$ in $C_4$ plants. Fixing $1\text{ }CO_2$ requires $3\text{ ATP}$ and $2\text{ NADPH}$ in the Calvin cycle, plus an additional $2\text{ ATP}$ in the $C_4$ dicarboxylic acid pathway to regenerate $PEP$ from pyruvate. Total per $CO_2 = 5\text{ ATP}$ and $2\text{ NADPH}$ .

Step 2

Multiply by $6$ to calculate for $1$ full molecule of glucose ( $C_6H_{12}O_6$ ), which contains $6$ carbon atoms: $\text{Total ATP} = 6 \times 5 = 30\text{ ATP}$ and $\text{Total NADPH} = 6 \times 2 = 12\text{ NADPH}$ .

Step 3

Match with the choices. A $C_4$ plant requires $30\text{ ATP}$ and $12\text{ NADPH}$ per glucose, making Option B correct.

Select an option & click Check Answer to reveal solution

Question #2NEET 2023

During non-cyclic photophosphorylation ( $Z$ -scheme), where does the photolysis of water occur and where do the released protons ( $H^+$ ) accumulate?

Solution Locked

Step-by-Step Solution & Concept

Step 1

Locate the Water Splitting Complex (Oxygen Evolving Complex), which is physically associated with Photosystem II ( $PS\text{ II}$ ) on the inner side of the thylakoid membrane.

Step 2

Understand proton transport direction. As $2H_2O \rightarrow 4H^+ + 4e^- + O_2$ , the $4H^+$ are released directly into the thylakoid lumen, increasing lumen proton concentration and building a $\Delta pH$ gradient across the membrane.

Step 3

Conclude that photolysis occurs on the inner surface of the thylakoid membrane, accumulating protons in the thylakoid lumen, confirming Option B as correct.

Select an option & click Check Answer to reveal solution

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Mastering Photosynthesis in Higher Plants: Formulas, Traps & Exam Strategy | ExamiFly

Chapter Overview

Core Subtopics & Syllabus Breakdown

Key Formulas & High-Yield Equations

Overall Photosynthetic Equation

Photolysis of Water Reaction

Stoichiometry of $C_3$ Calvin Cycle (Per Glucose)

Stoichiometry of $C_4$ Pathway (Per Glucose)

Proton Motive Force (Chemiosmosis)

Blackman's Law of Limiting Factors

Common Mistakes & Traps to Avoid

Past Year Question (PYQ) Trends

Solved High-Yield Sample Questions

Frequently Asked Questions

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