Photosynthesis & Cellular Respiration

How to Use This Guide

• Photosynthesis — all light reactions, Calvin cycle, HL detail
• Cellular Respiration — glycolysis, Krebs, ETC, chemiosmosis
• HL / AHL Only — extra depth required at Higher Level
• MCQ Practice — styled like real IB Paper 1 questions
• Exam Alerts — the traps and mistakes that cost marks

Aligned to IB Biology 2025 syllabus — C1.2 Cell Respiration — C1.3 Photosynthesis

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Section 1: The Big Picture

Before any detail, you must understand where each process happens inside the cell. The single most common source of lost marks in MCQs is confusing locations. Fix this first and many questions become easy.

Cell Map — Where Every Reaction Occurs

Overall Equations

The two processes are the INVERSE of each other.

Photosynthesis:

$6\text{CO}_2 + 6\text{H}_2\text{O} + \text{light energy} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2$

• $\text{O}_2$ comes from splitting $\text{H}_2\text{O}$ (photolysis) — NOT from $\text{CO}_2$
• $\text{CO}_2$ is FIXED into glucose — it is a REACTANT

Aerobic Cellular Respiration:

$\text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2 \rightarrow 6\text{CO}_2 + 6\text{H}_2\text{O} + \sim38\text{ ATP}$

Anaerobic — animals:

$\text{Glucose} \rightarrow 2\text{ Lactate} + 2\text{ ATP}$

Anaerobic — yeast:

$\text{Glucose} \rightarrow 2\text{ Ethanol} + 2\text{CO}_2 + 2\text{ ATP}$

MCQ Practice

Which statement is correct about the oxygen released during photosynthesis?

A. It is produced when $\text{CO}_2$ is broken down in the stroma

B. It is a by-product of the Calvin cycle

C. It comes from the photolysis of water in the thylakoid ← CORRECT

D. It is produced when $\text{NADPH}$ is oxidised

Why: $\text{O}_2$ is released during PHOTOLYSIS — the splitting of water molecules on the thylakoid membrane using light energy. $\text{CO}_2$ is fixed (incorporated) during the Calvin cycle; it does not release $\text{O}_2$. This is one of the most frequently tested facts in photosynthesis MCQs.

Watch: Photosynthesis and Respiration Overview

Amoeba Sisters · 8 min · Quick visual summary of photosynthesis vs respiration

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Section 2: Photosynthesis

Photosynthesis converts light energy into chemical energy (stored in glucose). Two main stages: Light-dependent reactions on the thylakoid membrane, and the Calvin cycle in the stroma. Both stages must be understood in detail for HL.

2.1 Chloroplast Structure

Chloroplast Adaptations HL

AHL — B2.2.5

Thylakoid membrane — large surface area:

• Gives maximum space for photosystems, electron carriers, and $\text{ATP}$ synthase
• Grana (stacked) greatly multiply the surface area per chloroplast

Thylakoid lumen — small volume:

• $\text{H}^+$ ions pumped in by ETC quickly build up a steep concentration gradient
• A steep gradient = strong proton motive force = more efficient $\text{ATP}$ synthesis
• Small volume means the same number of $\text{H}^+$ ions creates a higher concentration

Stroma — compartmentalises Calvin cycle:

• Keeps RuBisCO and substrates ($\text{RuBP}$, $\text{CO}_2$) in high concentrations together
• Physically separate from cytoplasm — unique chemical environment maintained
• $\text{CO}_2$ diffuses directly from stomata into the stroma

Double membrane (envelope):

• Maintains a controlled internal environment
• Acts as a selective barrier — regulates what enters/exits the chloroplast

2.2 Photosynthetic Pigments and Light Absorption

Pigments absorb specific wavelengths of light. When a photon of the right wavelength hits a pigment, it excites an electron to a higher energy level. This energy drives the light reactions. Multiple pigments capture a broader range of wavelengths, increasing efficiency.

Absorption and Action Spectra

Pigment	Wavelengths Absorbed / Role
Chlorophyll a	Red (680 nm) + blue-violet (430 nm). Main reaction-centre pigment. P680 in PS II, P700 in PS I
Chlorophyll b	Blue (450 nm) + orange-red. Accessory pigment — absorbs and transfers energy to Chl a
Carotenoids	Blue-violet (400-500 nm). Accessory pigments. Reflect yellow-orange wavelengths. Also protect chlorophyll from excess light damage
Phycoerythrin / phycocyanin	Green and yellow (in algae/cyanobacteria). Fill the absorption gaps of chlorophylls

Photosystems and Pigment Arrays HL

AHL — C1.3.5

Antenna complex:

• Hundreds of accessory pigment molecules arranged around each reaction centre
• Absorb photons of various wavelengths and pass energy by resonance to the reaction centre
• Acts like a funnel — greatly increases the effective area for light capture
• Accessory pigments can NOT pass electrons directly to the ETC

Reaction centre (one special pair of chlorophyll a):

• PS II reaction centre = P680 (absorbs 680 nm red light)
  • P680 donates an excited electron to the electron transport chain
  • $\text{P680}^+$ is the strongest biological oxidising agent (strong enough to split water)
• PS I reaction centre = P700 (absorbs 700 nm far-red light)
  • Re-energises electrons received from PS II
  • Passes electrons to ferredoxin, which ultimately reduces $\text{NADP}^+$

MCQ Practice

Which of the following correctly describes a photosystem’s antenna complex?

A. A single chlorophyll a molecule that absorbs red light only

B. An array of accessory pigments that absorb light and pass energy to the reaction centre ← CORRECT

C. The site where water is split to release oxygen

D. A protein complex that pumps $\text{H}^+$ across the thylakoid membrane

Why: The antenna complex is an array of many accessory pigment molecules that absorb light of various wavelengths and transfer the energy by resonance to the reaction centre. Water is split at the reaction centre of PS II (not the antenna). $\text{H}^+$ pumping is done by the cytochrome b6f complex in the ETC.

2.3 Light-Dependent Reactions

The light-dependent reactions occur on the thylakoid membrane. They produce $\text{ATP}$, $\text{NADPH}$, and $\text{O}_2$ — the $\text{O}_2$ is a waste product. Two types: non-cyclic photophosphorylation (both PS I and PS II) and cyclic photophosphorylation (PS I only).

Non-Cyclic Photophosphorylation — The Z-Scheme

Non-cyclic products: $\text{ATP}$ + $\text{NADPH}$ + $\text{O}_2$

Electron source: $\text{H}_2\text{O}$ (split by photolysis at PS II)

Electron destination: $\text{NADPH}$ (electrons carried out of light reactions)

Cyclic Photophosphorylation (PS I Only)

Electron path: PS I → Fd → Cyt b6f → PC → PS I

Product: $\text{ATP}$ only (no $\text{NADPH}$, no $\text{O}_2$)

Used when: cell needs more $\text{ATP}$ relative to $\text{NADPH}$

Non-Cyclic vs Cyclic Photophosphorylation HL

AHL — C1.3.6 / C1.3.7

Non-cyclic — electrons flow in one direction (linear):

• $\text{H}_2\text{O} \rightarrow \text{PS II} \rightarrow \text{plastoquinone} \rightarrow \text{cyt b6f} \rightarrow \text{plastocyanin} \rightarrow \text{PS I} \rightarrow \text{Fd} \rightarrow \text{NADP}^+$
• Electrons are NOT recycled — they end up stored in $\text{NADPH}$
• Products: $\text{ATP}$ + $\text{NADPH}$ + $\text{O}_2$ (all three)

Cyclic — electrons loop back (no net products except ATP):

• $\text{PS I} \rightarrow \text{ferredoxin} \rightarrow \text{cyt b6f} \rightarrow \text{plastocyanin} \rightarrow \text{PS I}$ (loop)
• Products: $\text{ATP}$ ONLY
• No $\text{O}_2$ released, no $\text{NADPH}$ made, no water split
• Occurs when $\text{ATP}$/$\text{NADPH}$ ratio is low — supplements $\text{ATP}$ supply

Chemiosmosis in Thylakoids HL

Same principle as mitochondria

$\text{H}^+$ (protons) accumulate in thylakoid lumen from:

1. Photolysis: $2\text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4\text{e}^-$ (protons released into lumen)
2. PQ pumping: PQ carries $\text{H}^+$ from stroma to lumen as electrons pass through

• $\text{H}^+$ concentration in lumen >> stroma → steep proton gradient
• $\text{H}^+$ can ONLY move back into stroma through $\text{ATP}$ synthase channels
• Flow of $\text{H}^+$ down the gradient powers rotation of $\text{ATP}$ synthase → $\text{ATP}$

NADP Reduction HL

AHL — C1.3.7

• Ferredoxin passes 2 electrons to NADP reductase enzyme
• $\text{NADP}^+ + 2\text{e}^- + \text{H}^+\text{ (from stroma)} \rightarrow \text{NADPH}$
• $\text{H}^+$ comes from stroma, not from the lumen

Light-Dependent Reactions — Step by Step

1. Photon hits antenna complex of PS II. Energy passes by resonance to P680.
2. P680 absorbs energy — its electron is excited to a higher energy level.
3. Excited electron leaves P680, enters ETC via pheophytin then plastoquinone.
4. Photolysis: P680 is now electron-deficient ($\text{P680}^+$). It oxidises water: $2\text{H}_2\text{O} \rightarrow 4\text{H}^+ + 4\text{e}^- + \text{O}_2$. $\text{O}_2$ released as waste gas.
5. Electrons pass through cytochrome b6f complex. $\text{H}^+$ is actively pumped from stroma into the thylakoid lumen. This builds up the proton gradient.
6. Electrons carried by plastocyanin arrive at PS I (P700).
7. Second photon re-energises electrons at PS I. Electrons passed to ferredoxin.
8. Ferredoxin transfers electrons to NADP reductase. $\text{NADP}^+ + 2\text{e}^- + \text{H}^+ \rightarrow \text{NADPH}$.
9. $\text{H}^+$ gradient across thylakoid drives $\text{ATP}$ synthase. $\text{H}^+$ flows from lumen to stroma through $\text{ATP}$ synthase. $\text{ADP} + \text{P}_i \rightarrow \text{ATP}$ (photophosphorylation).

Chemiosmosis in the Thylakoid

MCQ Practice

Which combination of products is made by NON-CYCLIC photophosphorylation but NOT by cyclic photophosphorylation?

A. $\text{ATP}$ and $\text{CO}_2$

B. $\text{NADPH}$ and $\text{O}_2$ ← CORRECT

C. $\text{ATP}$ and $\text{RuBP}$

D. Glucose and $\text{NADPH}$

Why: Non-cyclic photophosphorylation produces $\text{ATP}$, $\text{NADPH}$, and $\text{O}_2$. Cyclic photophosphorylation produces $\text{ATP}$ ONLY — no $\text{NADPH}$ and no $\text{O}_2$ (because no photolysis occurs and electrons return to PS I). $\text{RuBP}$ and glucose are products of the Calvin cycle, not the light reactions.

2.4 The Calvin Cycle (Light-Independent Reactions)

The Calvin cycle occurs in the stroma. It uses $\text{ATP}$ and $\text{NADPH}$ from the light reactions to fix $\text{CO}_2$ into organic molecules. The key enzyme RuBisCO attaches $\text{CO}_2$ to $\text{RuBP}$. The cycle regenerates its own substrate — it is truly cyclic. Three turns = one net G3P; six turns = one glucose.

The Calvin Cycle — Complete Annotated Diagram

Per turn (1 $\text{CO}_2$ fixed):

$\text{Input: } 1\text{ CO}_2 + 3\text{ ATP} + 2\text{ NADPH} \quad \rightarrow \quad \text{Output: } \tfrac{1}{3}\text{ G3P net}$

To make 1 glucose (6C) from 6 $\text{CO}_2$: need 6 turns of the cycle.

Total: 18 $\text{ATP}$ + 12 $\text{NADPH}$ consumed.

Calvin Cycle — Full HL Mechanism HL

AHL — C1.3.8 / C1.3.9

Step 1: Carbon Fixation

• RuBisCO enzyme catalyses: $\text{CO}_2 + \text{RuBP (5C)} \rightarrow \text{unstable 6C intermediate}$
• 6C immediately splits into $2 \times \text{GP}$ (3-phosphoglycerate, 3C)
• $\text{GP}$ = the FIRST STABLE PRODUCT of $\text{CO}_2$ fixation

Step 2: Reduction — GP to G3P (triose phosphate)

• $\text{ATP}$ phosphorylates $\text{GP}$ → 1,3-bisphosphoglycerate (2 $\text{ATP}$ per $\text{CO}_2$)
• $\text{NADPH}$ reduces 1,3-BPG → $\text{G3P}$ (2 $\text{NADPH}$ per $\text{CO}_2$)
• $\text{G3P}$ is the organic product that exits the cycle (1 of every 6)

Step 3: Regeneration of RuBP

• 5 of every 6 $\text{G3P}$ molecules rearranged → ribulose-5-phosphate
• $\text{ATP}$ phosphorylates ribulose-5-P → $\text{RuBP}$ (3 $\text{ATP}$ per 3 $\text{CO}_2$ fixed)

Interdependence of Light and Dark Reactions HL

AHL — C1.3.9

• If light stops → $\text{ATP}$ and $\text{NADPH}$ drop → GP ACCUMULATES (cannot be reduced)
• $\text{G3P}$ falls, $\text{RuBP}$ falls (cannot be regenerated)
• If $\text{CO}_2$ drops → RuBisCO slows → RuBP accumulates, $\text{GP}$ falls
• Both stages are tightly coupled — neither can operate without the other

Calvin’s Experiment

• Fed $^{14}\text{C}$-labelled $\text{CO}_2$ to Chlorella algae, used paper chromatography
• $\text{GP}$ (3-PGA) was the FIRST labelled product detected
• Proved the C3 pathway — 3-carbon first product

Calvin Cycle — Step by Step

1. $\text{CO}_2$ diffuses from atmosphere into stroma. RuBisCO binds $\text{CO}_2$ to $\text{RuBP}$ (5C).
2. Unstable 6C intermediate immediately splits into $2 \times \text{GP}$ (3-phosphoglycerate, 3C). This is the first stable carbon fixation product.
3. $\text{ATP}$ phosphorylates each $\text{GP}$ molecule → 1,3-bisphosphoglycerate. (2 $\text{ATP}$ used per $\text{CO}_2$)
4. $\text{NADPH}$ reduces 1,3-BPG → $\text{G3P}$ (triose phosphate). (2 $\text{NADPH}$ used per $\text{CO}_2$)
5. 1 out of every 6 $\text{G3P}$ molecules exits the cycle to make glucose, fatty acids, or amino acids.
6. The remaining 5 $\text{G3P}$ molecules enter the $\text{RuBP}$ regeneration pathway.
7. $\text{ATP}$ is used to phosphorylate ribulose-5-phosphate → $\text{RuBP}$. (3 $\text{ATP}$ per 3 $\text{CO}_2$ fixed)
8. $\text{RuBP}$ is ready to accept another $\text{CO}_2$. The cycle repeats.

MCQ Practice

If light intensity suddenly drops to zero during active photosynthesis, what immediate change is observed in the Calvin cycle?

A. $\text{RuBP}$ levels increase because carbon fixation continues

B. $\text{GP}$ levels decrease because it can no longer be made

C. $\text{GP}$ accumulates because it cannot be reduced to $\text{G3P}$ without $\text{NADPH}$ ← CORRECT

D. $\text{CO}_2$ levels inside the chloroplast increase

Why: With no light: $\text{ATP}$ and $\text{NADPH}$ production stops. RuBisCO can still make $\text{GP}$ (as long as $\text{RuBP}$ and $\text{CO}_2$ are present), but $\text{GP}$ CANNOT be reduced to $\text{G3P}$ because $\text{NADPH}$ is needed for that step. So $\text{GP}$ builds up. $\text{G3P}$ falls. $\text{RuBP}$ is gradually used up and not regenerated. This interdependence question is extremely common in HL exams.

MCQ Practice

Which molecule is the DIRECT product of carbon fixation by RuBisCO?

A. $\text{G3P}$ (triose phosphate)

B. Glucose

C. $\text{RuBP}$

D. $\text{GP}$ (3-phosphoglycerate) ← CORRECT

Why: RuBisCO catalyses $\text{CO}_2 + \text{RuBP} \rightarrow \text{unstable 6C} \rightarrow 2 \times \text{GP}$ (3-phosphoglycerate). $\text{GP}$ is the FIRST stable product of carbon fixation. $\text{G3P}$ is made AFTER $\text{GP}$ is reduced using $\text{ATP}$ and $\text{NADPH}$. Glucose is made after multiple turns of the cycle.

Watch: Photosynthesis — Light Reactions and Calvin Cycle

Khan Academy · 13 min · Light-dependent reactions, photosystems I and II, and the Z-scheme

Khan Academy · 11 min · Calvin cycle step by step — carbon fixation, reduction, regeneration of RuBP

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Section 3: Cellular Respiration

Cellular respiration releases energy from organic molecules and stores it as $\text{ATP}$. Four stages: Glycolysis (cytoplasm) → Link Reaction (matrix) → Krebs Cycle (matrix) → Electron Transport Chain and Chemiosmosis (inner membrane). Understanding each stage’s location, inputs, outputs, and mechanism is essential for HL.

3.1 ATP and NAD — Energy Carriers

ATP Structure and the Energy Cycle

• $\text{ATP}$ drives: active transport, movement, anabolism (synthesis reactions)
• $\text{ATP}$ is NOT stored — it is rapidly recycled (~40 kg $\text{ATP}$ cycled per day in humans)

NAD (Nicotinamide Adenine Dinucleotide)

• Oxidised form: $\text{NAD}^+$ (empty — can accept electrons)
• Reduced form: $\text{NADH}$ (loaded — carrying electrons + H)

$\text{NAD}^+ + 2\text{H} \rightarrow \text{NADH} \;(+ \text{H}^+)$

• Substrate is oxidised (loses H)
• $\text{NAD}^+$ is reduced (gains H)

$\text{FADH}_2$: similar to $\text{NADH}$ but carries slightly less energy. Used at one specific step in the Krebs cycle (succinate → fumarate).

Role of NAD as Hydrogen Carrier HL

AHL — C1.2.7

• $\text{NAD}^+$ acts as the hydrogen (electron + proton) ACCEPTOR during substrate oxidation
• Each $\text{NADH}$ carries 2 electrons + 1 $\text{H}^+$ to the electron transport chain
• At the ETC, $\text{NADH}$ is OXIDISED → $\text{NAD}^+$ is regenerated
• $\text{NAD}^+$ must be regenerated continuously — without it, glycolysis and Krebs stop
• In aerobic conditions: $\text{NAD}^+$ regenerated at the ETC ($\text{O}_2$ accepts electrons at end)
• In anaerobic conditions: $\text{NAD}^+$ regenerated by fermentation (pyruvate or acetaldehyde accepts electrons)
• $\text{FADH}_2$ also donates electrons to the ETC but enters at a later complex → yields slightly less $\text{ATP}$ than $\text{NADH}$ (~2 vs ~3 $\text{ATP}$ per molecule)

3.2 Glycolysis

Location: CYTOPLASM (cytosol). Glycolysis is the ONLY stage shared by aerobic and anaerobic respiration. One glucose (6C) → two pyruvate (3C), with a net yield of 2 $\text{ATP}$ and 2 $\text{NADH}$. No oxygen required.

Glycolysis — Step-by-Step Flow

	Per glucose
$\text{ATP}$ used	2 (phosphorylation)
$\text{ATP}$ made	4 (substrate-level phosphorylation)
Net $\text{ATP}$	+2
$\text{NADH}$ produced	2
Pyruvate made	2 x 3C

Glycolysis — HL Mechanism HL

AHL — C1.2.8

Stage 1 — Phosphorylation:

• 2 $\text{ATP}$ hydrolysed → glucose receives 2 phosphate groups
• Forms fructose-1,6-bisphosphate (6C, 2 phosphates)
• Why: Adding phosphates destabilises the molecule, making it easy to split AND traps glucose inside the cell (phosphorylated glucose cannot cross membranes)

Stage 2 — Lysis:

• Fructose-1,6-bisphosphate → $2 \times$ triose phosphate ($\text{G3P}$, 3C each)
• Each $\text{G3P}$ retains 1 phosphate group from the phosphorylation step

Stage 3 — Oxidation:

• Each $\text{G3P}$ oxidised — removes 2H (electrons + protons)
• $\text{NAD}^+$ accepts the hydrogen → $\text{NADH}$ formed (2 total)
• A second inorganic phosphate ($\text{P}_i$) is added to each $\text{G3P}$
• Molecule now has 2 phosphate groups and is highly energised

Stage 4 — ATP Formation (substrate-level phosphorylation):

• Each of the 2 phosphate groups is transferred directly to $\text{ADP}$
• 4 $\text{ATP}$ produced in total (2 phosphates transferred per pyruvate x 2 $\text{G3P}$)
• Net: 4 made - 2 used = 2 $\text{ATP}$ net

MCQ Practice

A cell is fed glucose labelled with $^{14}\text{C}$ at every carbon atom. After glycolysis only (no further reactions), in which molecule(s) would ALL the $^{14}\text{C}$ be found?

A. $\text{CO}_2$ only

B. $\text{ATP}$ and $\text{NADH}$

C. Pyruvate ← CORRECT

D. $\text{CO}_2$ and pyruvate equally

Why: Glycolysis converts glucose to pyruvate — NO $\text{CO}_2$ is released in glycolysis. All 6 carbons end up in the 2 pyruvate molecules (3C each). $\text{CO}_2$ is released in the LINK REACTION (1 $\text{CO}_2$ per pyruvate) and KREBS CYCLE. $\text{ATP}$ and $\text{NADH}$ carry no carbon atoms from glucose.

3.3 Anaerobic Respiration

When $\text{O}_2$ is absent, cells must regenerate $\text{NAD}^+$ without using the ETC. The ONLY purpose of fermentation is to recycle $\text{NAD}^+$ so glycolysis can continue. Two pathways exist — one in animals, one in yeast.

Lactate Fermentation (Animals / Humans)

$\text{Pyruvate (3C)} + \text{NADH} \xrightarrow{\text{lactate dehydrogenase}} \text{Lactate (3C)} + \text{NAD}^+$

• Pyruvate is REDUCED to lactate
• $\text{NADH}$ is OXIDISED back to $\text{NAD}^+$ — this is the WHOLE POINT
• $\text{NAD}^+$ recycled → glycolysis can continue producing $\text{ATP}$
• Lactate accumulates in muscle → transported to liver (Cori cycle)
• When $\text{O}_2$ returns: lactate → pyruvate → aerobic pathway (oxygen debt)
• Net $\text{ATP}$: 2 per glucose (glycolysis only)
• REVERSIBLE reaction

Ethanol Fermentation (Yeast)

$\text{Pyruvate (3C)} \xrightarrow{\text{pyruvate decarboxylase}} \text{Acetaldehyde (2C)} + \text{CO}_2$

$\text{Acetaldehyde} + \text{NADH} \xrightarrow{\text{alcohol dehydrogenase}} \text{Ethanol (2C)} + \text{NAD}^+$

• $\text{CO}_2$ is released (used in baking = bread rises; fermentation = carbonation)
• Ethanol accumulates — toxic to yeast at concentrations > ~15%
• IRREVERSIBLE — yeast cannot convert ethanol back to pyruvate
• Net $\text{ATP}$: 2 per glucose (glycolysis only)

Comparison

	Both pathways	Animals	Yeast
Purpose	Regenerate $\text{NAD}^+$	Lactate (3C)	Ethanol (2C) + $\text{CO}_2$
$\text{ATP}$ yield	2 only	2 only	2 only
Location	Cytoplasm	Cytoplasm	Cytoplasm
Reversibility	—	Reversible	Irreversible

Anaerobic Respiration — HL Points HL

AHL — C1.2.9 / C1.2.10 / C1.2.17

Lipids vs Carbohydrates as Respiratory Substrates HL

AHL — C1.2.17

• Lipids yield ~2x more energy per gram than carbohydrates
• Why: Lipids are more reduced — they contain more C-H bonds per carbon (more hydrogen to be oxidised → more $\text{NADH}$ → more $\text{ATP}$ from ETC)
• Lipids contain very little oxygen already bound → more oxidation can occur
• Glycerol component → enters glycolysis
• Fatty acids → beta-oxidation in mitochondrial matrix → Acetyl-CoA → Krebs

MCQ Practice

Why must $\text{NAD}^+$ be regenerated during anaerobic respiration?

A. $\text{NAD}^+$ is needed to drive the electron transport chain

B. $\text{NAD}^+$ is required for oxidation of glucose in the Krebs cycle

C. Without $\text{NAD}^+$, glycolysis cannot oxidise triose phosphate, so $\text{ATP}$ production stops ← CORRECT

D. $\text{NAD}^+$ is needed to split water during photolysis

Why: Glycolysis REQUIRES $\text{NAD}^+$ in the oxidation step to accept hydrogen from triose phosphate. Without $\text{NAD}^+$, this step stalls, glycolysis cannot proceed, and no $\text{ATP}$ is produced. Fermentation regenerates $\text{NAD}^+$ by using pyruvate (or acetaldehyde) as an electron acceptor instead. The ETC is not available in anaerobic conditions.

3.4 The Link Reaction

Location: mitochondrial matrix. Pyruvate (3C) from glycolysis is converted to Acetyl-CoA (2C), releasing $\text{CO}_2$. This connects glycolysis to the Krebs cycle. Two link reactions occur per glucose (two pyruvate molecules).

Link Reaction — Full Diagram

	Per pyruvate	Per glucose
$\text{CO}_2$ released	1	2
$\text{NADH}$ produced	1	2
Acetyl-CoA formed	1	2

• CoA = Coenzyme A (acts as a carrier/handle for the acetyl group)
• CoA is released when Acetyl-CoA enters Krebs, and is recycled

3.5 The Krebs Cycle

Location: mitochondrial matrix. The Krebs cycle fully oxidises Acetyl-CoA, releasing $\text{CO}_2$ and producing large amounts of $\text{NADH}$ and $\text{FADH}_2$. These electron carriers deliver energy to the ETC. The 4C oxaloacetate starting molecule is regenerated each turn — making it truly cyclic.

The Krebs Cycle — Complete Annotated Diagram

Per turn (per Acetyl-CoA = per 1/2 glucose):

Product	Amount
$\text{CO}_2$ released	2
$\text{NADH}$ produced	3
$\text{FADH}_2$ produced	1
$\text{ATP}$ (substrate-level phosphorylation)	1

Per glucose (2 turns):

$4\text{ CO}_2 + 6\text{ NADH} + 2\text{ FADH}_2 + 2\text{ ATP}$

Krebs Cycle — HL Detail HL

AHL — C1.2.12

Condensation: Acetyl-CoA (2C) + Oxaloacetate (4C) → Citrate (6C) + CoA released. CoA is recycled to pick up another acetyl group from the link reaction.

1st Decarboxylation: Isocitrate → alpha-ketoglutarate. $\text{CO}_2$ released + $\text{NAD}^+$ reduced to $\text{NADH}$.

2nd Decarboxylation: alpha-ketoglutarate → Succinyl-CoA. $\text{CO}_2$ released + $\text{NAD}^+$ reduced to $\text{NADH}$.

Substrate-level Phosphorylation: Succinyl-CoA → Succinate. 1 $\text{ATP}$ produced directly (not via ETC). CoA released.

FAD Reduction: Succinate → Fumarate. FAD reduced to $\text{FADH}_2$ (not $\text{NAD}^+$).

• Why FAD here? This oxidation step has insufficient energy to reduce $\text{NAD}^+$. Succinate dehydrogenase enzyme is embedded in the inner mitochondrial membrane.

Malate Oxidation: Malate → Oxaloacetate. $\text{NAD}^+$ reduced to $\text{NADH}$. Oxaloacetate regenerated → cycle continues.

Products summary (per turn): 3 $\text{NADH}$ + 1 $\text{FADH}_2$ + 1 $\text{ATP}$ + 2 $\text{CO}_2$ (+ regenerated oxaloacetate)

Krebs Cycle — Step by Step

1. Acetyl-CoA (2C) enters the matrix. CoA is released and recycled to the link reaction.
2. Condensation: acetyl group (2C) combines with oxaloacetate (4C) → citrate (6C).
3. Citrate rearranged → isocitrate (same formula, different structure).
4. 1st decarboxylation: isocitrate → alpha-ketoglutarate (5C) + $\text{CO}_2$. $\text{NAD}^+ \rightarrow \text{NADH}$.
5. 2nd decarboxylation: alpha-ketoglutarate → succinyl-CoA (4C) + $\text{CO}_2$. $\text{NAD}^+ \rightarrow \text{NADH}$.
6. Substrate-level phosphorylation: succinyl-CoA → succinate + 1 $\text{ATP}$. CoA released.
7. Oxidation: succinate → fumarate. $\text{FAD} \rightarrow \text{FADH}_2$.
8. Hydration: fumarate + $\text{H}_2\text{O}$ → malate.
9. Oxidation: malate → oxaloacetate. $\text{NAD}^+ \rightarrow \text{NADH}$. Oxaloacetate ready for next turn.

MCQ Practice

How many molecules of $\text{CO}_2$ are produced per turn of the Krebs cycle (per acetyl-CoA)?

A. 1

B. 2 ← CORRECT

C. 3

D. 4

Why: Two $\text{CO}_2$ molecules are released per turn of the Krebs cycle — one at each of the two decarboxylation steps (isocitrate → alpha-KG and alpha-KG → succinyl-CoA). Per glucose (2 turns): 4 $\text{CO}_2$ from Krebs + 2 $\text{CO}_2$ from the link reaction = 6 $\text{CO}_2$ total from Krebs + link (matching the 6 $\text{CO}_2$ in the overall equation).

MCQ Practice

Which is the only Krebs cycle step that produces $\text{FADH}_2$ rather than $\text{NADH}$?

A. Isocitrate → alpha-ketoglutarate

B. alpha-ketoglutarate → succinyl-CoA

C. Succinate → fumarate ← CORRECT

D. Malate → oxaloacetate

Why: FAD (rather than $\text{NAD}^+$) is the hydrogen acceptor at the succinate → fumarate step. This is because the oxidation of succinate releases insufficient energy to reduce $\text{NAD}^+$. Succinate dehydrogenase is the enzyme, and it is embedded in the inner mitochondrial membrane. All other oxidative steps use $\text{NAD}^+$.

3.6 Electron Transport Chain and Chemiosmosis

Location: INNER MITOCHONDRIAL MEMBRANE. $\text{NADH}$ and $\text{FADH}_2$ from glycolysis, the link reaction, and the Krebs cycle deliver electrons to protein complexes embedded in the inner membrane. As electrons flow down the chain, $\text{H}^+$ is pumped into the intermembrane space. The proton gradient drives $\text{ATP}$ synthesis through chemiosmosis. $\text{O}_2$ is the FINAL electron acceptor.

Electron Transport Chain and Chemiosmosis — Diagram

• $\text{NADH}$ → ~3 $\text{ATP}$
• $\text{FADH}_2$ → ~2 $\text{ATP}$ (enters at ubiquinone, skips Complex I)
• $\text{O}_2$ is ESSENTIAL — without it, ETC stops, $\text{H}^+$ not pumped, no $\text{ATP}$
• Water is produced at Complex IV (matrix side of inner membrane)
• $\text{ATP}$ yield from ETC per glucose: ~34 $\text{ATP}$

ETC and Chemiosmosis — HL Mechanisms HL

AHL — C1.2.13 to C1.2.16

Energy Transfer to ETC (AHL C1.2.13):

• $\text{NADH}$ delivers electrons to Complex I (NADH dehydrogenase)
• $\text{FADH}_2$ delivers electrons to Complex II (succinate dehydrogenase) via ubiquinone
• As electrons pass through each complex, energy released pumps $\text{H}^+$ into IMS
• $\text{FADH}_2$ yields LESS $\text{ATP}$ than $\text{NADH}$ because it bypasses Complex I (less $\text{H}^+$ pumped)

Proton Gradient Generation (AHL C1.2.14):

• Complexes I, III, and IV all pump $\text{H}^+$ from matrix into intermembrane space (IMS)
• IMS has small volume → $\text{H}^+$ accumulates quickly → steep concentration gradient
• Electrochemical gradient = proton motive force (concentration + charge difference)

Chemiosmosis and ATP Synthesis (AHL C1.2.15):

• $\text{H}^+$ can ONLY re-enter the matrix through $\text{ATP}$ synthase (Complex V)
• Flow of $\text{H}^+$ down the gradient causes mechanical rotation of $\text{ATP}$ synthase
• Rotation drives conformational changes in $\text{ATP}$ synthase → $\text{ADP} + \text{P}_i \rightarrow \text{ATP}$
• ~3 $\text{H}^+$ needed per $\text{ATP}$ molecule (approximate, accepted by IB)

Role of $\text{O}_2$ (AHL C1.2.16):

• $\text{O}_2$ is the TERMINAL (final) electron acceptor at Complex IV
• $\text{O}_2 + 4\text{e}^- + 4\text{H}^+ \rightarrow 2\text{H}_2\text{O}$ (water produced in the matrix)
• WITHOUT $\text{O}_2$: electrons cannot leave Complex IV → ETC completely blocks
• Blocked ETC → $\text{H}^+$ not pumped → no gradient → no $\text{ATP}$ synthesis
• $\text{NADH}$/$\text{FADH}_2$ cannot be oxidised → $\text{NAD}^+$ runs out → Krebs and link reaction stop
• ONLY glycolysis (with fermentation for $\text{NAD}^+$ regeneration) continues without $\text{O}_2$

ETC and Chemiosmosis — Step by Step

1. $\text{NADH}$ (from glycolysis, link reaction, Krebs) passes electrons to Complex I. $\text{FADH}_2$ passes electrons to ubiquinone (bypassing Complex I).
2. Electrons flow from Complex I → ubiquinone → Complex III → cytochrome c → Complex IV.
3. At each complex (I, III, IV), energy from electron flow pumps $\text{H}^+$ from matrix into the intermembrane space.
4. $\text{H}^+$ accumulates in the intermembrane space — builds up a steep concentration gradient (proton motive force).
5. $\text{H}^+$ can ONLY flow back into the matrix through $\text{ATP}$ synthase channels.
6. $\text{H}^+$ flowing through $\text{ATP}$ synthase causes it to rotate — this mechanical energy drives synthesis of $\text{ATP}$ from $\text{ADP} + \text{P}_i$ (oxidative phosphorylation).
7. At Complex IV: electrons are transferred to $\text{O}_2$. $\text{O}_2$ + electrons + $\text{H}^+$ → $\text{H}_2\text{O}$. Water is the final product of the ETC.
8. $\text{NAD}^+$ and FAD are regenerated and return to the Krebs cycle and link reaction to pick up more electrons.

MCQ Practice

Why does aerobic respiration stop if oxygen is removed, even though glycolysis does not require oxygen?

A. Oxygen is needed to activate RuBisCO in the matrix

B. Without oxygen the ETC stops, $\text{NADH}$ cannot be oxidised, $\text{NAD}^+$ runs out, and the Krebs cycle stops ← CORRECT

C. Oxygen is required for the condensation reaction in the Krebs cycle

D. Without oxygen, pyruvate cannot enter the mitochondrial matrix

Why: Oxygen is the terminal electron acceptor at the end of the ETC. Without it, electrons cannot leave Complex IV, the ETC backs up, $\text{NADH}$ cannot be oxidised to $\text{NAD}^+$, and $\text{NAD}^+$ runs out. Without $\text{NAD}^+$, the Krebs cycle and link reaction cannot proceed. Only glycolysis continues, using fermentation to regenerate $\text{NAD}^+$. This is a classic chain-of-consequence MCQ.

MCQ Practice

A cell is treated with a drug that blocks $\text{ATP}$ synthase. Which prediction is correct?

A. The ETC stops immediately because $\text{ATP}$ synthase is needed to power it

B. The proton gradient across the inner membrane breaks down

C. The proton gradient increases because $\text{H}^+$ is still pumped but cannot return ← CORRECT

D. $\text{NADH}$ production increases because more substrate is oxidised

Why: If $\text{ATP}$ synthase is blocked, $\text{H}^+$ can no longer flow back into the matrix. However, the ETC continues pumping $\text{H}^+$ into the IMS (as long as $\text{O}_2$ and $\text{NADH}$ are available). The gradient therefore INCREASES but no $\text{ATP}$ is made. Eventually the gradient becomes so steep that ETC pumping slows (back-pressure), reducing $\text{NADH}$ oxidation too.

Watch: Cellular Respiration — Glycolysis, Krebs Cycle, and ETC

Amoeba Sisters · 10 min · Clear overview of all four stages of cellular respiration

Khan Academy · 15 min · Oxidative phosphorylation — electron transport chain and chemiosmosis in detail

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Section 4: ATP Yield — The Full Accounting

Understanding the $\text{ATP}$ yield at each stage and why the ETC produces so much more $\text{ATP}$ than glycolysis is crucial for both MCQ and extended-response questions.

Complete ATP Yield Table — Per Glucose

Stage	Location	Inputs	Outputs	$\text{ATP}$ per glucose
Glycolysis	Cytoplasm	Glucose	2 Pyruvate, 2 $\text{NADH}$	2 net
Link Reaction	Mito. matrix	2 Pyruvate	2 Acetyl-CoA, 2 $\text{CO}_2$, 2 $\text{NADH}$	0
Krebs Cycle	Mito. matrix	2 Acetyl-CoA	4 $\text{CO}_2$, 6 $\text{NADH}$, 2 $\text{FADH}_2$	2
ETC + $\text{ATP}$ synthase	Inner mito. membrane	10 $\text{NADH}$, 2 $\text{FADH}_2$	$\text{H}_2\text{O}$	~34
TOTAL	Both	Glucose + $\text{O}_2$	$\text{CO}_2$ + $\text{H}_2\text{O}$	~38

Why Does the ETC Produce ~34 ATP While Glycolysis Makes Only 2?

Glycolysis and Krebs use substrate-level phosphorylation — $\text{ATP}$ is made DIRECTLY when a phosphate group is transferred to $\text{ADP}$. This only happens at a few specific steps → low yield.

The ETC uses oxidative phosphorylation — a completely different mechanism:

• 10 $\text{NADH}$ + 2 $\text{FADH}_2$ carry electrons to the ETC
• As electrons pass through protein complexes, their energy pumps $\text{H}^+$
• A large proton gradient drives $\text{ATP}$ synthase to make $\text{ATP}$
• ~2.5 $\text{ATP}$ per $\text{NADH}$ and ~1.5 $\text{ATP}$ per $\text{FADH}_2$ (approximately 3 and 2 for IB)
• 10 $\text{NADH}$ x 3 = 30 $\text{ATP}$
• 2 $\text{FADH}_2$ x 2 = 4 $\text{ATP}$
• Total from ETC: ~34 $\text{ATP}$

MCQ Practice

Which statement explains why $\text{FADH}_2$ yields fewer $\text{ATP}$ molecules than $\text{NADH}$ during oxidative phosphorylation?

A. $\text{FADH}_2$ carries fewer electrons than $\text{NADH}$

B. $\text{FADH}_2$ cannot enter the mitochondrial matrix

C. $\text{FADH}_2$ donates electrons at a later point in the ETC, bypassing Complex I, so fewer $\text{H}^+$ are pumped ← CORRECT

D. $\text{FADH}_2$ is oxidised in the cytoplasm, not the mitochondria

Why: $\text{FADH}_2$ delivers electrons directly to ubiquinone (between Complexes I and III), bypassing Complex I entirely. Since Complex I pumps $\text{H}^+$ as electrons pass through it, skipping it means fewer $\text{H}^+$ are pumped into the IMS, creating a smaller contribution to the proton gradient, and therefore fewer $\text{ATP}$ molecules are synthesised per $\text{FADH}_2$.

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Section 5: MCQ Strategy and Common Traps

IB Biology Paper 1 MCQs test whether you know exact locations, exact sequences, exact products, and can follow chains of consequence. The most effective strategy: read the question, predict your answer before looking at options, then check. This prevents attractive wrong answers from misleading you.

The Most Common MCQ Traps — Memorise These

Five MCQ Strategy Rules for IB Biology

1. Predict before reading options: Cover the answers. Decide what you think the answer is. Then check. This avoids being misled by plausible-sounding distractors.
2. Location questions — go to your mental map: Ask: is this a membrane process (ETC, light reactions) or a fluid/matrix process (Krebs, Calvin, glycolysis)? This eliminates 2-3 wrong options instantly.
3. “Cannot proceed because…” questions — trace the chain: Work step by step: what stops first? What runs out next? For example: no $\text{O}_2$ → ETC stops → $\text{NADH}$ not oxidised → $\text{NAD}^+$ runs out → Krebs stops.
4. Product counting questions — add up from your diagrams: Know exact yields: Glycolysis = 2 $\text{ATP}$; Link = 0 $\text{ATP}$, 2 $\text{NADH}$; Krebs = 2 $\text{ATP}$, 6 $\text{NADH}$, 2 $\text{FADH}_2$, 4 $\text{CO}_2$; ETC = ~34 $\text{ATP}$.
5. “Increases/decreases” questions — think about the feedback: e.g. “light stops → what happens to GP?” → $\text{ATP}$ and $\text{NADPH}$ drop → GP CANNOT be reduced → GP ACCUMULATES. Always follow the logic, not the memory.

MCQ Practice

During aerobic respiration, where are the protons ($\text{H}^+$) that flow through $\text{ATP}$ synthase coming from?

A. From the hydrolysis of $\text{ATP}$ in the matrix

B. From water produced at Complex IV

C. From the oxidation of $\text{NADH}$ and $\text{FADH}_2$, pumped into the intermembrane space by the ETC ← CORRECT

D. From the breakdown of glucose during glycolysis

Why: When $\text{NADH}$ and $\text{FADH}_2$ deliver electrons to the ETC, the energy released pumps $\text{H}^+$ from the matrix into the intermembrane space (via Complexes I, III, and IV). These accumulated $\text{H}^+$ then flow back through $\text{ATP}$ synthase into the matrix, driving $\text{ATP}$ synthesis. Water is produced AT Complex IV (where $\text{O}_2$ accepts electrons), not as a source of $\text{H}^+$ for $\text{ATP}$ synthase.

MCQ Practice

A plant is moved from red light to green light only. What happens to the rate of photosynthesis?

A. It increases because plants can use all visible light equally

B. It stays the same because the Calvin cycle does not require light

C. It decreases significantly because chlorophyll absorbs very little green light ← CORRECT

D. It stops completely because green light has too little energy to excite electrons

Why: Chlorophyll a and b absorb mainly blue and red light — they REFLECT green light (which is why leaves look green). Very little green light is absorbed, so photosynthesis rate drops sharply. It does not stop completely because carotenoids absorb some green-adjacent wavelengths. The action spectrum confirms photosynthesis is lowest in the green region.

MCQ Practice

Which statement about the relationship between photosynthesis and aerobic respiration is correct?

A. Both processes occur only in plant cells

B. The $\text{ATP}$ produced in photosynthesis is directly used in respiration

C. Oxygen produced in photosynthesis is the same oxygen consumed in aerobic respiration ← CORRECT

D. Both processes produce carbon dioxide as a final waste product

Why: $\text{O}_2$ produced by photolysis in the light reactions is the same $\text{O}_2$ that acts as the terminal electron acceptor in the mitochondrial ETC. This links the two processes: the oxygen cycle. $\text{ATP}$ from photosynthesis is NOT directly transferred to respiration — it is used to make glucose in the Calvin cycle, and glucose is then respired. Respiration occurs in ALL living cells, not just plants. Respiration releases $\text{CO}_2$; photosynthesis CONSUMES $\text{CO}_2$.

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Section 6: HL Content — Complete Checklist HL

This checklist covers every AHL (Additional Higher Level) point for photosynthesis and respiration in the 2025 IB Biology syllabus. If you are HL, every item on this list is assessable.

HL Photosynthesis — What You Must Know

AHL Topic	Key Points to Know
Chloroplast adaptations (B2.2.5)	Large thylakoid membrane surface area; small lumen volume builds $\text{H}^+$ gradient fast; stroma compartmentalises Calvin cycle enzymes
Photosystem structure (C1.3.5)	Antenna complex = array of accessory pigments; reaction centres P680 (PS II) and P700 (PS I); $\text{P680}^+$ is strongest biological oxidising agent
Non-cyclic vs cyclic photophosphorylation (C1.3.6)	Non-cyclic: PS I + PS II, products $\text{ATP}$ + $\text{NADPH}$ + $\text{O}_2$. Cyclic: PS I only, product $\text{ATP}$ only. Electron flow directions.
NADP reduction (C1.3.7)	Ferredoxin → NADP reductase → $\text{NADP}^+ + 2\text{e}^- + \text{H}^+\text{(stroma)} \rightarrow \text{NADPH}$
Chemiosmosis in thylakoid (C1.3.12)	$\text{H}^+$ pumped into lumen by PQ + cyt b6f; small lumen volume → steep gradient; $\text{H}^+$ returns via $\text{ATP}$ synthase → $\text{ATP}$
Calvin cycle mechanism (C1.3.8)	3 stages: carbon fixation (RuBisCO, $\text{CO}_2$ + $\text{RuBP}$ → $\text{GP}$), reduction ($\text{ATP}$ + $\text{NADPH}$, $\text{GP}$ → $\text{G3P}$), regeneration ($\text{ATP}$, $\text{G3P}$ → $\text{RuBP}$). Per glucose: 18 $\text{ATP}$ + 12 $\text{NADPH}$
Interdependence of stages (C1.3.9)	Light stops → $\text{ATP}$/$\text{NADPH}$ fall → $\text{GP}$ accumulates → $\text{G3P}$ falls → $\text{RuBP}$ falls. $\text{CO}_2$ falls → $\text{RuBP}$ accumulates, $\text{GP}$ falls.
Calvin’s experiment	$^{14}\text{CO}_2$ + paper chromatography → $\text{GP}$ was first stable product; proved C3 pathway

HL Respiration — What You Must Know

AHL Topic	Key Points to Know
NAD as hydrogen carrier (C1.2.7)	$\text{NAD}^+$ accepts H during oxidation of substrate → $\text{NADH}$; $\text{NADH}$ delivers electrons to ETC; $\text{NAD}^+$ regenerated by ETC (aerobic) or fermentation (anaerobic)
Glycolysis stages (C1.2.8)	4 stages: phosphorylation (-2 $\text{ATP}$), lysis, oxidation (2 $\text{NADH}$), $\text{ATP}$ formation (+4 $\text{ATP}$). Net: 2 $\text{ATP}$ + 2 $\text{NADH}$. Cytoplasm. Both aerobic and anaerobic.
Lactate fermentation (C1.2.9)	Pyruvate → lactate + $\text{NAD}^+$. Purpose: regenerate $\text{NAD}^+$. Reversible. No extra $\text{ATP}$.
Yeast fermentation (C1.2.10)	Pyruvate → acetaldehyde + $\text{CO}_2$, then acetaldehyde + $\text{NADH}$ → ethanol + $\text{NAD}^+$. Irreversible. Used in baking/brewing.
Link reaction (C1.2.11)	Pyruvate → Acetyl-CoA. Decarboxylation + oxidation. Per pyruvate: 1 $\text{CO}_2$ + 1 $\text{NADH}$. Matrix.
Krebs cycle (C1.2.12)	Per turn: 2 $\text{CO}_2$, 3 $\text{NADH}$, 1 $\text{FADH}_2$, 1 $\text{ATP}$. Key intermediates: citrate (6C), oxaloacetate (4C). Matrix.
ETC and proton gradient (C1.2.13/14)	$\text{NADH}$ → Complex I, $\text{FADH}_2$ → ubiquinone (skips I). $\text{H}^+$ pumped at I, III, IV into IMS. Gradient = proton motive force.
Chemiosmosis (C1.2.15)	$\text{H}^+$ flows through $\text{ATP}$ synthase → $\text{ADP}$ + $\text{P}_i$ → $\text{ATP}$. ~3 $\text{H}^+$ per $\text{ATP}$. $\text{NADH}$ → ~3 $\text{ATP}$, $\text{FADH}_2$ → ~2 $\text{ATP}$.
Role of $\text{O}_2$ (C1.2.16)	$\text{O}_2$ = terminal electron acceptor at Complex IV. $\text{O}_2 + 4\text{e}^- + 4\text{H}^+ \rightarrow 2\text{H}_2\text{O}$. Without $\text{O}_2$: ETC stops, $\text{NAD}^+$ runs out, Krebs stops.
Lipids vs carbohydrates (C1.2.17)	Lipids: higher energy per gram, more C-H bonds, less O already bound. Fatty acids → Acetyl-CoA via beta-oxidation. Glycolysis and anaerobic respiration use carbohydrates ONLY.

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Section 7: Photosynthesis vs Cellular Respiration — Side-by-Side Comparison

These two processes are the inverse of each other. Understanding how they connect is one of the most tested concepts in IB Biology.

Overview Comparison

Detailed Comparison Table

Feature	Photosynthesis	Cellular Respiration
Overall equation	$6\text{CO}_2 + 6\text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2$	$\text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2 \rightarrow 6\text{CO}_2 + 6\text{H}_2\text{O} + \sim38\text{ ATP}$
Energy conversion	Light energy → chemical energy	Chemical energy → ATP
Organelle	Chloroplast	Mitochondrion (+ cytoplasm)
Organisms	Plants, algae, cyanobacteria	ALL living cells
When	Day only (requires light)	24/7 (day and night)
Gas exchange	Takes in $\text{CO}_2$, releases $\text{O}_2$	Takes in $\text{O}_2$, releases $\text{CO}_2$
Water	Consumed (photolysis)	Produced (at ETC)
Glucose	Produced (Calvin cycle)	Consumed (glycolysis)
Electron carriers	$\text{NADPH}$ (carries electrons)	$\text{NADH}$ and $\text{FADH}_2$ (carry electrons)
ATP production method	Chemiosmosis (thylakoid)	Chemiosmosis (inner membrane) + substrate-level
$\text{H}^+$ gradient location	Thylakoid lumen (inside)	Intermembrane space
Key enzyme	RuBisCO (carbon fixation)	ATP synthase (ATP production)
Stages	Light reactions → Calvin cycle	Glycolysis → Link → Krebs → ETC

What They Share (Similarities)

Shared Feature	In Photosynthesis	In Respiration
Chemiosmosis	$\text{H}^+$ gradient across thylakoid membrane drives ATP synthase	$\text{H}^+$ gradient across inner mitochondrial membrane drives ATP synthase
Electron transport chain	Thylakoid membrane (PS II → PS I)	Inner mitochondrial membrane (Complexes I-IV)
ATP synthase	In thylakoid membrane	In inner mitochondrial membrane
Electron carriers	$\text{NADPH}$	$\text{NADH}$, $\text{FADH}_2$
Compartmentalisation	Reactions separated between thylakoid and stroma	Reactions separated between matrix and inner membrane
$\text{CO}_2$ involvement	Fixed by RuBisCO	Released by decarboxylation

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IB Biology HL — Photosynthesis & Cellular Respiration — Complete Study Guide — 2025 Syllabus — Good luck!