Metabolism and Survival

This chapter covers Advanced Higher Biology content, extending beyond Higher level.

Metabolism

Enzymes

Enzymes are biological catalysts that speed up biochemical reactions by lowering the activation Energy. They are globular proteins with a specific active site complementary in shape to the Substrate.

Lock and Key model: The substrate fits exactly into the active site like a key in a lock. Rigid And specific.

Induced Fit model: The active site changes shape slightly to accommodate the substrate. More Widely accepted.

Factors affecting enzyme activity:

Temperature:

Rate increases with temperature (up to the optimum) due to increased kinetic energy
Beyond the optimum, the enzyme denatures (active site loses its shape)
Typical optimum for human enzymes: 37 $^\circ$ C

pH:

Each enzyme has an optimum pH
Extreme pH changes the charge of amino acids, disrupting hydrogen bonds and ionic bonds
Pepsin: optimum pH 2 (stomach); trypsin: optimum pH 8 (small intestine)

Substrate concentration:

As substrate concentration increases, rate increases
Eventually plateaus when all active sites are occupied (Vmax)

Enzyme concentration:

More enzyme molecules = more active sites = higher rate (provided substrate is not limiting)

Inhibitors:

Type	Mechanism	Reversibility
Competitive	Binds to active site; competes with substrate	Reversible
Non-competitive	Binds to allosteric site; changes active site shape	reversible

Michaelis-Menten Kinetics

The relationship between enzyme activity and substrate concentration is described by the Michaelis-Menten equation:

$v = \frac{V_{\max}[S]}{K_m + [S]}$

Where:

$v$ = reaction rate
$V_{\max}$ = maximum rate (when enzyme is saturated)
$[S]$ = substrate concentration
$K_m$ = Michaelis constant (substrate concentration at half $V_{\max}$ )

$K_m$ interpretation:

Low $K_m$ : High affinity between enzyme and substrate
High $K_m$ : Low affinity between enzyme and substrate

Lineweaver-Burk plot: A double reciprocal plot of $1/v$ vs. $1/[S]$ :

$\frac{1}{v} = \frac{K_m}{V_{\max}} \cdot \frac{1}{[S]} + \frac{1}{V_{\max}}$

Y-intercept: $1/V_{\max}$
X-intercept: $-1/K_m$
Gradient: $K_m/V_{\max}$

Effect of inhibitors on Lineweaver-Burk plots:

Competitive: Increases $K_m$ (apparent); $V_{\max}$ unchanged. Lines intersect on the y-axis.
Non-competitive: Decreases $V_{\max}$ ; $K_m$ unchanged. Lines intersect on the x-axis.

Worked Example: Determining $K_m$ and $V_{\max}$ from data.

An enzyme has the following rates at different substrate concentrations:

[S] (mM)	v ( $\mu$ Mol/min)
1	1.67
2	2.86
5	4.55
10	5.88
20	6.90
50	7.69
100	8.00

At very high [S], $v$ approaches $V_{\max} \approx 8$ $\mu$ Mol/min.

At $v = V_{\max}/2 = 4$ , $[S] = K_m \approx 5$ mM.

So $K_m = 5$ mM and $V_{\max} = 8$ $\mu$ Mol/min.

Cellular Respiration

Glycolysis (Cytoplasm)

Glucose ( $\mathrm{C_6\mathrm{H_{12}\mathrm{O_6$ ) is broken down to two molecules of pyruvate ( $\mathrm{C_3\mathrm{H_4\mathrm{O_3$ ).

Glucose is phosphorylated (2 ATP used)
6-carbon intermediate is split into two 3-carbon molecules
Each 3-carbon molecule is oxidised and phosphorylated
2 NADH and 4 ATP produced (net: 2 ATP, 2 NADH)

Link Reaction (Mitochondrial Matrix)

Pyruvate is decarboxylated and dehydrogenated:

$\mathrm{Pyruvate + \mathrm{NAD^+ + \mathrm{CoA \to \mathrm{Acetyl-CoA + \mathrm{CO_2 + \mathrm{NADH$

(Per glucose: 2 $\mathrm{CO_2$ 2 NADH)

Krebs Cycle (Mitochondrial Matrix)

Acetyl-CoA (2C) combines with oxaloacetate (4C) to form citrate (6C)
Citrate is decarboxylated and oxidised back to oxaloacetate
Per turn: 2 $\mathrm{CO_2$ 3 NADH, 1 FADH $\_2$ 1 ATP
Per glucose: 4 $\mathrm{CO_2$ 6 NADH, 2 FADH $\_2$ 2 ATP

Oxidative Phosphorylation (Inner Mitochondrial Membrane)

NADH and FADH $_2$ donate electrons to the electron transport chain
Electrons pass through a series of protein complexes (I, III, IV)
Energy released pumps protons from the matrix to the intermembrane space
Creates a proton gradient (proton motive force)
Protons flow back through ATP synthase, producing ATP
Oxygen is the final electron acceptor, forming water

ATP yield per glucose:

Glycolysis: 2 ATP (net) + 2 NADH $\to$ ~5 ATP
Link reaction: 2 NADH $\to$ ~5 ATP
Krebs cycle: 2 ATP + 6 NADH + 2 FADH $_2$ $\to$ ~22 ATP
Total: approximately 38 ATP (theoretical maximum; actual yield is about 30-32 ATP)

Photosynthesis

Light-Dependent Reactions (Thylakoid Membranes)

Light energy is absorbed by photosynthetic pigments (chlorophyll a, chlorophyll b, carotenoids) in Photosystem II and Photosystem I
Water is split by photolysis: $2\mathrm{H_2\mathrm{O \to 4\mathrm{H^+ + 4e^- + \mathrm{O_2$
Electrons pass through the electron transport chain, generating a proton gradient
ATP is produced by photophosphorylation
NADP is reduced to NADPH

Light-Independent Reactions (Calvin Cycle, Stroma)

$\mathrm{CO_2$ is fixed by ribulose bisphosphate carboxylase (RuBisCO) combining with ribulose bisphosphate (RuBP, 5C) to form two molecules of glycerate-3-phosphate (GP, 3C)
GP is reduced to triose phosphate (TP) using ATP and NADPH
Some TP is used to synthesise glucose and other organic compounds
Most TP is used to regenerate RuBP (using ATP)

Summary of the Calvin cycle:

3 turns of the cycle fix 3 $\mathrm{CO_2$ to produce 1 TP (G3P)
6 turns fix 6 $\mathrm{CO_2$ to produce 1 glucose molecule

Survival

Homeostasis

Homeostasis is the maintenance of a constant internal environment despite changes in the External environment.

Key principles:

Stimulus $\to$ Receptor $\to$ Coordinator $\to$ Effector $\to$ Response
Negative feedback loops maintain stability

Thermoregulation

Temperature regulation in mammals:

When too hot:

Vasodilation: Blood vessels near the skin surface dilate, increasing heat loss by radiation
Sweating: Evaporation of sweat removes heat
Piloerection (in some mammals): Fur lies flat to reduce insulation

When too cold:

Vasoconstriction: Blood vessels near the skin constrict, reducing heat loss
Shivering: Muscle contractions generate heat
Piloerection: Fur stands up to trap air (insulation)
Brown fat metabolism: Non-shivering thermogenesis

The thermoregulatory centre is in the hypothalamus. It receives input from thermoreceptors in The skin and in the hypothalamus itself.

Osmoregulation

The kidneys regulate water balance and blood composition.

Nephron structure:

Bowman’s capsule: Filtration of blood; forms glomerular filtrate
Proximal convoluted tubule: Selective reabsorption of all glucose, amino acids, and some water and ions
Loop of Henle: Countercurrent multiplier; creates a concentration gradient in the medulla for water reabsorption
Distal convoluted tubule: Fine-tuning of water and ion balance (under hormonal control)
Collecting duct: Water reabsorption under ADH control

ADH (Antidiuretic Hormone):

Produced by the hypothalamus, released by the posterior pituitary
Increases the permeability of the collecting duct to water
More ADH = more water reabsorbed = more concentrated urine
Triggered by increased blood osmolarity (detected by osmoreceptors in the hypothalamus)

Worked Example: ADH and water balance.

A person drinks 2 litres of water. What happens to their blood osmolarity and ADH levels?

The excess water is absorbed from the gut into the blood.
Blood osmolarity decreases (blood becomes more dilute).
Osmoreceptors in the hypothalamus detect the decrease in osmolarity.
The hypothalamus reduces ADH production and release.
The collecting duct becomes less permeable to water.
Less water is reabsorbed from the filtrate.
Large volumes of dilute urine are produced.
Blood osmolarity returns to normal.

Immune System

Innate immunity (non-specific):

Physical barriers: skin, mucous membranes, cilia, stomach acid
Phagocytosis: Phagocytes (macrophages, neutrophils) engulf and digest pathogens
Inflammation: Increased blood flow, increased permeability of capillaries, recruitment of phagocytes
Fever: Elevated temperature inhibits pathogen growth

Adaptive immunity (specific):

Cell-mediated response: T lymphocytes
Helper T cells: Release cytokines to stimulate B cells and cytotoxic T cells
Cytotoxic T cells: Kill infected body cells by releasing perforins
Memory T cells: Provide long-term immunity
Humoral response: B lymphocytes
Plasma cells: Produce and secrete antibodies
Memory B cells: Provide long-term immunity
Antibodies: Y-shaped proteins that bind to specific antigens on pathogens

Antibody structure:

Two heavy chains and two light chains
Variable region: Unique binding site for a specific antigen
Constant region: Determines the antibody class (IgG, IgM, IgA, IgE, IgD)

Primary vs. Secondary immune response:

Primary response: Initial exposure to antigen; slow response, produces IgM then IgG; memory cells formed
Secondary response: Subsequent exposure; rapid, stronger response; mainly IgG; due to memory cells

Vaccination:

Introduces a harmless form of the antigen (attenuated, dead, or subunit)
Stimulates primary immune response without causing disease
Memory cells provide protection against future infection
Herd immunity: When a high proportion of the population is immune, the spread of disease is reduced

Common Pitfalls

ATP yield: The theoretical maximum of 38 ATP per glucose is rarely achieved. Actual yield is about 30-32 due to proton leak and cost of transporting molecules.
$K_m$ interpretation: Low $K_m$ means HIGH affinity (less substrate needed to reach half $V_{\max}$ ), not low affinity.
Calvin cycle: 6 turns are needed to produce 1 glucose, not 1 turn.
ADH mechanism: ADH makes the collecting duct MORE permeable, not less. More ADH means more concentrated urine, not more dilute.
Innate vs. Adaptive immunity: Innate is non-specific and immediate; adaptive is specific and takes time to develop but provides memory.

Practice Questions

Explain how competitive and non-competitive inhibitors affect enzyme activity, with reference to the Michaelis-Menten equation.
Calculate the number of ATP molecules produced from the complete oxidation of one molecule of glucose, showing the contribution from each stage.
Explain how the structure of the mitochondrion is adapted for its function in aerobic respiration.
Describe the role of ADH in osmoregulation and explain what happens when a person drinks a large volume of water.
Compare the primary and secondary immune responses, explaining why vaccination provides protection.
Explain the principle of negative feedback using thermoregulation as an example.
Draw and label a Lineweaver-Burk plot showing the effect of a competitive inhibitor on enzyme kinetics.
Explain the role of the loop of Henle in producing concentrated urine in desert animals.
Describe the process of oxidative phosphorylation, explaining the role of the electron transport chain and ATP synthase.
Explain why the Calvin cycle stops in the dark, even though it does not directly require light.
Describe the structure of an antibody and explain how antibodies provide specific immunity against pathogens.
Explain how vaccination leads to herd immunity and why this is important for protecting immunocompromised individuals.
A student measures the effect of pH on the activity of pepsin. Sketch the expected graph and explain its shape.
Describe the process of phagocytosis and explain how it differs from the action of antibodies.
Explain why fever can be beneficial during an infection, with reference to enzyme activity.
Calculate the $K_m$ and $V_{\max}$ of an enzyme from the following data: [S] = 2 mM, v = 3.33 $\mu$ Mol/min; [S] = 10 mM, v = 6.67 $\mu$ Mol/min; [S] = 50 mM, v = 8.33 $\mu$ Mol/min.
Explain how the countercurrent multiplier in the loop of Henle creates a concentration gradient in the kidney medulla.
Compare and contrast the light-dependent and light-independent reactions of photosynthesis.
Describe the role of cytokines in the immune response and explain how they coordinate the activities of different immune cells.
Explain why type 1 diabetes is an autoimmune disease and describe how it affects blood glucose regulation.

Review: Aerobic Respiration — Detailed ATP Accounting

The precise ATP yield from aerobic respiration depends on the shuttle system used to transport NADH From glycolysis into the mitochondria and on the actual number of protons pumped per complex.

Using modern estimates (2.5 ATP per NADH, 1.5 ATP per FADH $_2$ ):

Stage	NADH	FADH $_2$	Direct ATP	Total ATP
Glycolysis	2	0	2	7
Pyruvate oxidation	2	0	0	5
Krebs cycle	6	2	2	20
Total	10	2	4	32

Note: The actual yield may be lower (approximately 30 ATP) due to proton leak across the inner Mitochondrial membrane and the cost of transporting ATP out of the mitochondria.

Glycolysis NADH shuttle systems: The cytoplasmic NADH produced in glycolysis cannot cross the Inner mitochondrial membrane. Two shuttle systems transport the electrons:

Malate-aspartate shuttle: Transfers electrons to mitochondrial NAD $^+$ Producing NADH inside the mitochondrion (yields 2.5 ATP per NADH). This is the more efficient shuttle.
Glycerol-3-phosphate shuttle: Transfers electrons to mitochondrial FAD, producing $\mathrm{FADH_2$ (yields 1.5 ATP per $\mathrm{FADH_2$ ). This is less efficient.

The choice of shuttle system affects the total ATP yield. Using the malate-aspartate shuttle gives a Total of 32 ATP; using the glycerol-3-phosphate shuttle gives approximately 30 ATP.

Review: Photosynthesis — Light-Dependent Reactions in Detail

Photosystem II (P680):

Light energy is absorbed by antenna pigments and transferred to the reaction centre chlorophyll P680.
An electron in P680 is excited to a higher energy level and is captured by the primary electron acceptor.
P680 $^+$ is a very strong oxidant and extracts electrons from water (photolysis): $2\mathrm{H_2\mathrm{O \to 4\mathrm{H^+ + 4e^- + \mathrm{O_2$ .
The electrons pass through the electron transport chain (plastoquinone, cytochrome b6f complex, plastocyanin), pumping protons from the stroma into the thylakoid lumen.

Photosystem I (P700):

Light energy excites electrons in P700.
Electrons are passed to ferredoxin, then to $\mathrm{NADP^+$ reductase.
$\mathrm{NADP^+$ reductase reduces $\mathrm{NADP^+$ to NADPH using the electrons and a proton from the stroma.

Chemiosmosis in chloroplasts:

The proton gradient across the thylakoid membrane (higher $\mathrm{H^+$ concentration in the lumen) Drives ATP synthesis by ATP synthase. This is called photophosphorylation. The proton gradient is Generated by: (1) splitting of water (releases $\mathrm{H^+$ into the lumen), (2) pumping of $\mathrm{H^+$ by the cytochrome b6f complex, and (3) removal of $\mathrm{H^+$ from the stroma by $\mathrm{NADP^+$ reductase.

Worked Example: Products of the light-dependent reactions.

For every 2 water molecules split:

1 $\mathrm{O_2$ released
4 electrons enter the ETC
4 $\mathrm{H^+$ released into the lumen
Approximately 3 ATP produced (by chemiosmosis)
2 NADPH produced

To fix 6 $\mathrm{CO_2$ molecules (to make one glucose), the light reactions must produce 18 ATP and 12 NADPH.

Review: Enzyme Inhibition in Metabolic Regulation

Enzyme inhibition is a key mechanism for regulating metabolic pathways.

Competitive inhibition: The inhibitor has a similar structure to the substrate and competes for The active site. The effect can be overcome by increasing substrate concentration. Examples: Malonate inhibits succinate dehydrogenase in the Krebs cycle; methotrexate inhibits dihydrofolate Reductase (used in cancer chemotherapy).

Non-competitive inhibition: The inhibitor binds to an allosteric site (different from the active Site), changing the enzyme’s conformation and reducing its activity. Increasing substrate Concentration does not overcome the inhibition. Examples: heavy metal ions (Pb $^{2+}$ Hg $^{2+}$ ) Bind to -SH groups and denature enzymes; cyanide inhibits cytochrome c oxidase.

End-product inhibition (feedback inhibition): The final product of a metabolic pathway inhibits An enzyme earlier in the pathway, preventing overproduction. Example: ATP inhibits Phosphofructokinase (PFK) in glycolysis, and citrate (from the Krebs cycle) also inhibits PFK.

Worked Example: Feedback inhibition of the Krebs cycle.

When ATP levels are high, the cell does not need to produce more energy through the Krebs cycle. High ATP inhibits several enzymes in the pathway:

Phosphofructokinase (glycolysis) is inhibited by ATP and citrate.
Pyruvate dehydrogenase (link reaction) is inhibited by ATP, NADH, and acetyl-CoA.
Isocitrate dehydrogenase (Krebs cycle) is inhibited by ATP and NADH.

This coordinated inhibition prevents unnecessary energy production when the cell already has Sufficient ATP.

Review: The Immune Response — Vaccination

How vaccines work:

A vaccine introduces an antigen (or a weakened/inactivated pathogen) into the body without causing Disease. This triggers the primary immune response, producing memory B cells and memory T cells. When the actual pathogen enters the body, the memory cells mount a rapid secondary response, Producing large quantities of antibodies before the pathogen can cause illness.

Herd immunity: When a high proportion of a population is immune to a disease (through Vaccination or previous infection), the spread of the disease is limited because there are too few Susceptible individuals for the pathogen to infect. This protects even those who cannot be Vaccinated (e.g., newborns, immunocompromised individuals). The threshold for herd immunity varies By disease: measles requires approximately 95% immunity; polio requires approximately 80%.

Types of vaccines:

Type	Example	Mechanism
Live attenuated	MMR vaccine	Weakened form of the pathogen
Inactivated	Influenza vaccine	Killed pathogen, cannot reproduce
Subunit	Hepatitis B	Specific antigens from the pathogen
Toxoid	Tetanus vaccine	Inactivated toxin from the pathogen
mRNA	COVID-19 vaccines	mRNA encoding a viral antigen (e.g., spike protein)

Review: The Kidney and Osmoregulation in Detail

Structure of the nephron in detail:

The nephron is the functional unit of the kidney. Each kidney contains approximately 1 million Nephrons.

Bowman’s capsule and glomerulus (ultrafiltration): Blood enters the glomerulus (a knot of capillaries) at high pressure. The afferent arteriole is wider than the efferent arteriole, which creates high hydrostatic pressure. Water, glucose, amino acids, urea, and ions are forced out of the blood through the basement membrane into the Bowman’s capsule. Large proteins and blood cells are retained. The basement membrane acts as a filter, allowing only small molecules to pass through.
Proximal convoluted tubule (selective reabsorption): All glucose, all amino acids, most water and ions, and some urea are reabsorbed. Glucose and amino acids are reabsorbed by active transport (secondary active transport coupled to $\mathrm{Na^+$ reabsorption). Water follows by osmosis. The PCT has many mitochondria to provide ATP for active transport and microvilli to increase the surface area for reabsorption.
Loop of Henle (countercurrent multiplication):

Descending limb: Permeable to water but not to ions. Water leaves the filtrate by osmosis into the increasingly concentrated medulla. The filtrate becomes more concentrated.
Thin ascending limb: Impermeable to water. $\mathrm{Na^+$ and $\mathrm{Cl^-$ diffuse out passively.
Thick ascending limb: Actively pumps out $\mathrm{Na^+$ , $\mathrm{K^+$ And $\mathrm{Cl^-$ . Impermeable to water. The medulla becomes increasingly concentrated from cortex to papilla (approximately 300 mOsm at the cortex to 1200 mOsm at the papilla).

Distal convoluted tubule: Fine-tuning of water and ion balance under hormonal control:

ADH: Makes the collecting duct more permeable to water.
Aldosterone: Increases $\mathrm{Na^+$ reabsorption and $\mathrm{K^+$ secretion in the DCT and collecting duct.

Collecting duct: Water reabsorption is controlled by ADH. In the presence of ADH, aquaporin channels are inserted into the membrane, and water is reabsorbed into the concentrated medulla. The maximum urine concentration in humans is approximately 1200 mOsm (four times more concentrated than blood).

Worked Example: Comparing urine production in different conditions.

Condition	ADH Level	Collecting Duct Permeability	Urine Concentration	Urine Volume
Well hydrated	Low	Low	Low (dilute)	High
Normal	Normal	Moderate	Moderate	Moderate
Dehydrated	High	High	High (concentrated)	Low
Diabetes insipidus	Very low	Very low	Very low (dilute)	Very high

Review: Thermoregulation in Detail

The hypothalamus is the body’s thermostat. It receives input from two types of thermoreceptors:

Peripheral thermoreceptors: Located in the skin. Detect changes in external temperature.
Central thermoreceptors: Located in the hypothalamus. Detect changes in blood temperature.

Heat loss mechanisms (when core temperature rises above 37 $\degree$ C):

Vasodilation: Arterioles near the skin surface dilate, increasing blood flow to the skin. More heat is lost by radiation and convection.
Sweating: Sweat glands secrete sweat onto the skin surface. Evaporation of water absorbs latent heat, cooling the skin. Each gram of water evaporated removes approximately 2.4 kJ of heat.
Piloerection: In some mammals, fur lies flat to reduce insulation. In humans, this mechanism is largely vestigial (goose bumps).

Heat conservation mechanisms (when core temperature falls below 37 $\degree$ C):

Vasoconstriction: Arterioles near the skin surface constrict, reducing blood flow to the skin. Less heat is lost.
Shivering: Rapid, involuntary contractions of skeletal muscles generate heat as a by-product of respiration.
Piloerection: Hair stands up, trapping a layer of insulating air.
Brown fat metabolism: Brown adipose tissue (found in newborns and between the shoulder blades in adults) contains many mitochondria with uncoupling protein (UCP1). The ETC operates without ATP synthesis, releasing energy as heat (non-shivering thermogenesis).

Worked Example: Why shivering generates heat.

Shivering involves rapid, involuntary contractions of skeletal muscles. Muscle contraction requires ATP, which is produced by cellular respiration. Respiration is only about 40% efficient, meaning That 60% of the energy from glucose is released as heat. During shivering, the increased rate of Respiration in muscles generates significantly more heat than normal, raising the core body Temperature.

Review: Enzyme Kinetics — Worked Examples

Worked Example: Effect of a competitive inhibitor.

An enzyme has $K_m = 5$ mM and $V_{\max} = 10$ $\mu$ Mol/min. A competitive inhibitor is added at a Concentration that doubles the apparent $K_m$ .

With inhibitor: apparent $K_m = 10$ mM, $V_{\max}$ unchanged at 10 $\mu$ Mol/min.

At $[S] = 5$ mM (original $K_m$ ):

Without inhibitor: $v = (10 \times 5) / (5 + 5) = 5$ $\mu$ Mol/min (50% of $V_{\max}$ ).
With inhibitor: $v = (10 \times 5) / (10 + 5) = 3.33$ $\mu$ Mol/min (33.3% of $V_{\max}$ ).

At very high $[S]$ (e.g., 100 mM):

Without inhibitor: $v \approx 10$ $\mu$ Mol/min.
With inhibitor: $v \approx 10$ $\mu$ Mol/min.

The competitive inhibitor reduces the rate at low $[S]$ but has no effect at very high $[S]$ Because the substrate outcompetes the inhibitor.

Worked Example: Effect of a non-competitive inhibitor.

The same enzyme with $K_m = 5$ mM and $V_{\max} = 10$ $\mu$ Mol/min. A non-competitive inhibitor Reduces $V_{\max}$ to 6 $\mu$ Mol/min but does not change $K_m$ .

At $[S] = 5$ mM:

Without inhibitor: $v = 5$ $\mu$ Mol/min.
With inhibitor: $v = (6 \times 5) / (5 + 5) = 3$ $\mu$ Mol/min.

At very high $[S]$ (e.g., 100 mM):

Without inhibitor: $v \approx 10$ $\mu$ Mol/min.
With inhibitor: $v \approx 6$ $\mu$ Mol/min.

Unlike the competitive inhibitor, the non-competitive inhibitor reduces the maximum rate even at Very high substrate concentrations, because it reduces the number of functional enzyme molecules Regardless of substrate concentration.

Review: C4 and CAM Photosynthesis (Advanced Higher)

C4 photosynthesis: In C4 plants (e.g., maize, sugarcane), $\mathrm{CO_2$ is initially fixed by PEP carboxylase in mesophyll cells to form oxaloacetate (4C), which is converted to malate. Malate Is transported to bundle-sheath cells, where $\mathrm{CO_2$ is released and enters the Calvin cycle. PEP carboxylase has a much higher affinity for $\mathrm{CO_2$ than RuBisCO and does not bind $\mathrm{O_2$ Minimising photorespiration.

CAM photosynthesis: In CAM plants (e.g., cacti, pineapples), stomata open at night to fix $\mathrm{CO_2$ into malic acid, which is stored in vacuoles. During the day, stomata close, and malic Acid is decarboxylated to release $\mathrm{CO_2$ for the Calvin cycle. This temporal separation Minimises water loss while maintaining carbon fixation.

Worked Example: Comparing water use efficiency.

A C3 plant loses approximately 500 g of water per gram of $\mathrm{CO_2$ fixed. A C4 plant loses Approximately 250 g of water per gram of $\mathrm{CO_2$ fixed. A CAM plant loses approximately 50 g Of water per gram of $\mathrm{CO_2$ fixed.

CAM plants are the most water-efficient because they close their stomata during the day, minimising Transpirational water loss. This is why CAM plants dominate in arid environments.

Review: Summary Table of Respiration vs. Photosynthesis

Feature	Respiration	Photosynthesis
Overall equation	$\mathrm{C_6\mathrm{H_{12}\mathrm{O_6 + 6\mathrm{O_2 \to 6\mathrm{CO_2 + 6\mathrm{H_2\mathrm{O_2$	$6\mathrm{CO_2 + 6\mathrm{H_2\mathrm{O_2 \to \mathrm{C_6\mathrm{H_{12}\mathrm{O_6 + 6\mathrm{O_2$
Energy change	Exergonic ( $\Delta G \lt 0$ )	Endergonic ( $\Delta G \gt 0$ )
Electron donor	Glucose	Water
Electron acceptor	$\mathrm{O_2$	$\mathrm{NADP^+$
ATP produced	~30-32 per glucose	~18 consumed per glucose
Location	Cytoplasm, mitochondria	Chloroplasts
Organisms	All living organisms	Plants, algae, some bacteria
ETC location	Inner mitochondrial membrane	Thylakoid membrane
Proton gradient site	Intermembrane space	Thylakoid lumen

Review: Anaerobic Respiration in Detail

Lactate fermentation in muscle cells:

During intense exercise, oxygen supply to muscle cells may be insufficient for aerobic respiration. Muscle cells switch to anaerobic respiration:

$\mathrm{Glucose \to 2 \mathrm{pyruvate \to 2 \mathrm{lactate + 2 \mathrm{ATP$

Pyruvate is reduced to lactate by the enzyme lactate dehydrogenase (LDH), which regenerates $\mathrm{NAD^+$ from NADH. This allows glycolysis to continue producing ATP even in the absence of Oxygen. However, the net yield is only 2 ATP per glucose (compared to approximately 30-32 ATP for Aerobic respiration).

The oxygen debt: After exercise, the accumulated lactate must be metabolised. Lactate is Transported to the liver, where it is converted back to pyruvate. Some pyruvate enters the Krebs Cycle, and some is converted back to glucose via gluconeogenesis. This process requires oxygen, Which is why breathing rate remains elevated after exercise (to repay the oxygen debt).

Alcoholic fermentation in yeast:

$\mathrm{Glucose \to 2 \mathrm{pyruvate \to 2 \mathrm{ethanol + 2 \mathrm{CO_2 + 2 \mathrm{ATP$

Pyruvate is decarboxylated to acetaldehyde (by pyruvate decarboxylase) and then reduced to ethanol (by alcohol dehydrogenase). This regenerates $\mathrm{NAD^+$ for glycolysis. Alcoholic fermentation Is exploited in brewing (beer) and baking (bread — $\mathrm{CO_2$ causes the dough to rise).

Comparing fermentation products:

Feature	Lactic acid fermentation	Alcoholic fermentation
Organisms	Animals, some bacteria	Yeast, some plants
End products	Lactate	Ethanol and $\mathrm{CO_2$
Enzyme (step 1)	Lactate dehydrogenase	Pyruvate decarboxylase
Enzyme (step 2)	—	Alcohol dehydrogenase
ATP yield	2 ATP per glucose	2 ATP per glucose
Reversible?	Yes (Cori cycle in liver)	No (ethanol cannot be reconverted)

Commercial applications of fermentation:

Brewing: Yeast ferment sugars in malted barley to produce ethanol. The $\mathrm{CO_2$ produced carbonates the beer. Different yeast strains produce different flavour compounds (esters, phenols).
Baking: Yeast ferment sugars in dough, producing $\mathrm{CO_2$ which causes the dough to rise. The ethanol evaporates during baking.
Cheese production: Bacteria ferment lactose to lactic acid, lowering the pH and causing milk proteins (casein) to coagulate, forming the cheese curd.
Yoghurt production: Lactobacillus bacteria ferment lactose to lactic acid, thickening the milk and producing the characteristic tart flavour.

Key comparison: aerobic vs anaerobic respiration:

Aerobic respiration yields approximately 30-32 ATP per glucose; anaerobic respiration yields only 2 ATP per glucose.
Aerobic respiration completely oxidises glucose to $\mathrm{CO_2$ and $\mathrm{H_2\mathrm{O$ ; anaerobic respiration produces partially oxidised products (lactate or ethanol).
Aerobic respiration requires oxygen as the final electron acceptor in the electron transport chain; anaerobic respiration uses organic molecules as the final electron acceptor.
Aerobic respiration occurs in the mitochondria (in eukaryotes); anaerobic respiration occurs in the cytoplasm.

Mark this page as reviewed