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Biochemistry · Final exam review

Biochemistry final exam review: 8 units, worked examples and a graded practice test

This review groups common first-semester biochemistry topics into eight connected units: pH sets the charge on amino acids, charge and shape affect what enzymes do, and free energy helps explain metabolism. Use the checklist to find your gaps, read the short recaps and four worked problems, then take the 18-question practice test to see which units need more work.

Short on time? Take the 18-question practice test first, then review only the units it flags.

Topic checklist: what can you already do?

Tick each unit you could explain to a friend without notes. Anything left unticked is where your study time goes. Your ticks stay in this browser only.

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Your syllabus may also include nucleic acids, gluconeogenesis, glycogen metabolism or fatty acid breakdown. Check it before you plan your week.

How the units connect

Some questions cross units, like how a pH change alters an enzyme or why a poison stops ATP production. Two chains tie much of the course together:

  1. pH
  2. charge on amino acids
  3. protein shape
  4. enzyme activity
  1. free energy
  2. glycolysis
  3. citric acid cycle
  4. electron transport
  5. ATP

If a question seems to belong to two units, follow the chain. A drop in pH can change the charge on side chains, which can change the fold or how well the active site binds substrate.

Recap of the 8 units

Each unit has core ideas to review, one number worth knowing, and a common mistake to avoid.

Unit 1 pH and buffers

pH is a log scale, so each pH unit is a 10-fold change in [H⁺]. A buffer is a weak acid (HA) mixed with its conjugate base (A⁻); it soaks up added acid or base with only a small pH change.

  • Henderson-Hasselbalch: pH = pKa + log([A⁻]/[HA]). When pH = pKa, the acid is exactly half deprotonated.
  • Every 1 pH unit above the pKa multiplies the base-to-acid ratio by 10. One unit below divides it by 10.
  • A buffer works best within about 1 pH unit of its pKa. To pick a buffer, pick the pKa closest to the target pH.
  • Blood is held near pH 7.4 mainly by the bicarbonate system, at about 20 parts HCO₃⁻ to 1 part dissolved CO₂/H₂CO₃.

Number to know: pH = pKa means 50% deprotonated; pKa + 1 means about 91%; pKa + 2 means about 99%.

Classic trap: Flipping the ratio. The base (A⁻) goes on top. If pH is above the pKa, the base form wins.

Go deeper: Henderson-Hasselbalch practice problems.

Unit 2 Amino acids, charge and pI

Every amino acid has an α-carboxyl group (pKa about 2) and an α-amino group (pKa about 9 to 10). Seven also have an ionizable side chain. The pI is the pH where the average net charge is zero.

  • Charge rule: a group is mostly protonated when pH is below its pKa and mostly deprotonated when pH is above it.
  • pI rule: write the forms from fully protonated to fully deprotonated, find the neutral form, and average the two pKa values on either side of it.
  • No ionizable side chain: average the two backbone pKa values. Acidic (Asp, Glu): average the two lowest. Basic (His, Lys, Arg): average the two highest.
  • At pH 7, Asp and Glu side chains are negative, Lys and Arg are positive, and His is mostly neutral (its pKa is about 6).
  • In electrophoresis, a molecule at a pH below its pI is positive and moves toward the cathode (−).

Number to know: Side-chain pKa values from OpenStax: Asp 3.65, Glu 4.25, His 6.00, Cys 8.18, Tyr 10.07, Lys 10.53, Arg 12.48.

Classic trap: Averaging all three pKa values for an amino acid with a charged side chain. Only the two around the neutral form count.

Go deeper: Amino acid quiz and Isoelectric point practice.

Unit 3 Protein structure and hemoglobin

Protein structure is a stack of four levels, and each level is held by different interactions.

  • Primary: the amino acid sequence, joined by covalent peptide bonds.
  • Secondary: α-helices and β-sheets, held by hydrogen bonds between backbone C=O and N–H groups.
  • Tertiary: the overall 3D fold, driven mostly by the hydrophobic effect (nonpolar side chains bury inside), plus side-chain hydrogen bonds, salt bridges and disulfide bonds.
  • Quaternary: how separate chains pack together, as in hemoglobin's four subunits.
  • Denaturing (heat, urea, extreme pH) disrupts a protein's fold but usually leaves peptide bonds intact. Some proteins can refold when the denaturing conditions are removed.
  • Hemoglobin vs myoglobin: myoglobin's oxygen curve is hyperbolic; hemoglobin's is sigmoidal because its subunits bind O₂ cooperatively. Lower pH, more CO₂ (the Bohr effect), more 2,3-BPG and higher temperature all shift hemoglobin's curve to the right, so it lets go of O₂ more easily in working tissue.

Number to know: Fetal hemoglobin binds 2,3-BPG weakly, so its curve sits to the left of adult hemoglobin: higher O₂ affinity.

Classic trap: Calling an α-helix a tertiary feature. Backbone hydrogen bonds are secondary structure; side-chain interactions are tertiary.

Unit 4 Enzyme kinetics and inhibition

Enzymes speed reactions by lowering the activation energy. They do not change ΔG or where the equilibrium lies.

  • Michaelis-Menten: v = Vmax[S] / (Km + [S]). Km is the [S] that gives half of Vmax; a lower Km means the enzyme reaches half speed at a lower substrate concentration.
  • kcat = Vmax / [E]total, the turnovers per enzyme per second. kcat/Km measures catalytic efficiency.
  • Lineweaver-Burk: 1/v = (Km/Vmax)(1/[S]) + 1/Vmax. The y-intercept is 1/Vmax and the x-intercept is −1/Km.
  • Competitive: Km goes up, Vmax unchanged (lines cross on the y-axis). Uncompetitive: Km and Vmax both drop by the same factor (parallel lines). Noncompetitive (pure): Vmax drops, Km unchanged (lines cross on the x-axis). Mixed: Vmax drops and Km changes.
  • Allosteric enzymes give sigmoidal v vs [S] curves and do not follow Michaelis-Menten.

Number to know: At [S] = Km, v = ½Vmax. At [S] = 3Km, v = ¾Vmax. At [S] = 9Km, v = 0.9Vmax.

Classic trap: Saying a competitive inhibitor lowers Vmax. Enough substrate still wins, so Vmax is unchanged.

Go deeper: Michaelis-Menten practice problems and Competitive vs noncompetitive inhibition.

Unit 5 Carbohydrates, lipids and membranes

Practice reading structures here: small changes in bonds or fatty acid tails can change the answer.

  • Glucose is an aldohexose. In its ring form, C1 is the anomeric carbon: α has the C1 OH on the opposite side of the ring from C6 (drawn down in a Haworth projection), β on the same side (drawn up).
  • Glycosidic bonds: maltose is Glc α(1→4) Glc, lactose is Gal β(1→4) Glc, sucrose is Glc α1→β2 Fru. Glycogen and amylopectin are α(1→4) chains with α(1→6) branches; cellulose is β(1→4).
  • A reducing sugar has a free anomeric carbon. Sucrose ties up both anomeric carbons, so it is nonreducing.
  • Fatty acids: cis double bonds put kinks in the tail, so unsaturated fats pack loosely and stay fluid. Longer, saturated tails pack tightly and raise the melting point.
  • Membranes are phospholipid bilayers. Cholesterol buffers fluidity: it stiffens the membrane at high temperature and keeps it from packing solid at low temperature.

Number to know: Humans digest α(1→4) and α(1→6) bonds but cannot digest the β(1→4) bonds of cellulose.

Classic trap: Calling sucrose reducing because it contains glucose. Check whether an anomeric carbon is free.

Unit 6 Free energy and ATP

ΔG tells you direction, not speed. A negative ΔG means the reaction can go forward on its own; a positive ΔG means it needs an energy input.

  • ΔG = ΔG°′ + RT ln Q. ΔG°′ is the standard value at pH 7; Q is products over reactants as they actually are in the cell. RT is about 2.58 kJ/mol at 37 °C (310 K).
  • At equilibrium, ΔG = 0 and ΔG°′ = −RT ln K′eq.
  • Coupling: when two reactions share an intermediate, their ΔG°′ values add. An unfavorable step goes forward when it is paired with a more favorable one, usually ATP hydrolysis.
  • ATP hydrolysis has ΔG°′ ≈ −30.5 kJ/mol (−7.3 kcal/mol). Inside a cell the real ΔG is more negative, about −57 kJ/mol.

Number to know: A positive ΔG°′ reaction can still run in a cell if the products are kept low (Q much less than 1).

Classic trap: Thinking a negative ΔG means a fast reaction. Speed depends on activation energy, which is the enzyme’s job.

Unit 7 Glycolysis and the citric acid cycle

Glycolysis splits glucose into 2 pyruvate in the cytoplasm. Pyruvate dehydrogenase then turns each pyruvate into acetyl-CoA in the mitochondrion, and the citric acid cycle oxidizes the acetyl group.

  • Glycolysis, net per glucose: 2 ATP, 2 NADH, 2 pyruvate.
  • Irreversible steps: hexokinase, phosphofructokinase-1 (PFK-1) and pyruvate kinase. PFK-1 is the committed, main control point: ATP and citrate slow it; AMP and fructose-2,6-bisphosphate speed it up.
  • Without oxygen, cells turn pyruvate into lactate to regenerate NAD⁺, so glycolysis can keep going.
  • Pyruvate dehydrogenase: pyruvate → acetyl-CoA + CO₂ + NADH.
  • One citric acid cycle turn: 3 NADH, 1 FADH₂, 1 GTP (or ATP) and 2 CO₂. One glucose runs two turns.

Number to know: Per glucose, the cycle alone makes 6 NADH, 2 FADH₂ and 2 GTP; pyruvate dehydrogenase adds 2 NADH.

Classic trap: Forgetting the factor of 2. One glucose gives 2 pyruvate, so everything after glycolysis happens twice.

Go deeper: Glycolysis quiz and Citric acid cycle practice.

Unit 8 Electron transport and ATP yield

NADH and FADH₂ hand their electrons to the electron transport chain in the inner mitochondrial membrane. The energy released pumps protons out of the matrix, and ATP synthase uses the proton flow back in to make ATP.

  • Complexes I, III and IV pump protons. Complex II (succinate dehydrogenase) does not, which is why FADH₂ is worth less ATP than NADH.
  • O₂ is the final electron acceptor and becomes water at Complex IV.
  • Current values: about 2.5 ATP per NADH and 1.5 per FADH₂. That gives about 32 ATP per glucose with the malate-aspartate shuttle, or 30 with the glycerol-3-phosphate shuttle. Older books say 36 to 38.
  • Inhibitors: rotenone blocks electron entry through Complex I, but electrons can still enter through Complex II. Cyanide and carbon monoxide block Complex IV. Oligomycin blocks ATP synthase and slows electron flow as the proton gradient builds. Each reduces ATP synthesis, but their effects on electron flow differ.
  • Uncouplers such as 2,4-dinitrophenol (DNP) let protons leak back without ATP synthase. Oxygen use goes up, ATP synthesis goes down, and the energy is lost as heat.

Number to know: 10 NADH × 2.5 + 2 FADH₂ × 1.5 + 4 ATP/GTP made directly = 32 ATP per glucose.

Classic trap: Saying an uncoupler slows oxygen use. Electron flow speeds up because nothing holds back the pumps.

Go deeper: Citric acid cycle practice.

Four worked examples

These four examples cover common calculation types. Try each one on paper before reading the steps.

Example 1: Phosphate buffer at blood pH pH and buffers

Problem: The phosphate buffer pair H₂PO₄⁻ / HPO₄²⁻ has pKa 6.86. At pH 7.40, what is the ratio [HPO₄²⁻]/[H₂PO₄⁻], and what percent of the phosphate is in the HPO₄²⁻ form?

  1. Set up Henderson-Hasselbalch with the base form on top: 7.40 = 6.86 + log([HPO₄²⁻]/[H₂PO₄⁻]).
  2. Subtract: log(ratio) = 0.54.
  3. Undo the log: ratio = 100.54 ≈ 3.47.
  4. Turn the ratio into a percent: 3.47 / (3.47 + 1) = 3.47 / 4.47 ≈ 0.776.

Answer: ratio ≈ 3.5 to 1, so about 78% HPO₄²⁻.

Use the pKa supplied in a problem. Phosphate pKa values in tables vary with conditions; some sources list a value closer to 7.2.

Sanity check: pH is about half a unit above the pKa, so the base form should win, but by less than 10 to 1. It does.

Example 2: The pI of histidine, and which way it moves Amino acids

Problem: Histidine has pKa values of 1.82 (α-COOH), 6.00 (side chain) and 9.17 (α-NH₃⁺). Find its pI. In electrophoresis at pH 7.40, which way does it move?

  1. Start fully protonated at very low pH: COOH, side chain +1, NH₃⁺ +1. Net charge +2.
  2. Remove protons in order of pKa. After 1.82 the carboxyl is COO⁻: net +1. After 6.00 the side chain is neutral: net 0. After 9.17 the amino group is NH₂: net −1.
  3. The neutral form sits between pKa 6.00 and pKa 9.17, so average those two: (6.00 + 9.17) / 2 = 7.585.
  4. At pH 7.40 the pH is just below the pI, so histidine carries a tiny net positive charge (about +0.02 on average).

Answer: pI ≈ 7.59. At pH 7.40 it drifts slowly toward the cathode (the negative electrode).

Note: this is the one amino acid where "basic, so average the two highest" gives a pI near 7. That is because histidine's side chain pKa is close to neutral.

Example 3: Rate with and without a competitive inhibitor Enzymes

Problem: An enzyme has Km = 2.0 mM and Vmax = 120 µmol/min. (a) What is v at [S] = 6.0 mM? (b) A competitive inhibitor is added at [I] = 10 µM with Ki = 5.0 µM. What is the new v at the same [S]? (c) How do the Lineweaver-Burk intercepts change?

  1. (a) v = Vmax[S] / (Km + [S]) = 120 × 6.0 / (2.0 + 6.0) = 720 / 8.0 = 90 µmol/min.
  2. (b) A competitive inhibitor multiplies Km by α = 1 + [I]/Ki = 1 + 10/5.0 = 3. Apparent Km = 3 × 2.0 = 6.0 mM. Vmax stays 120.
  3. v = 120 × 6.0 / (6.0 + 6.0) = 60 µmol/min.
  4. (c) The y-intercept, 1/Vmax = 1/120 min/µmol, does not move. The x-intercept moves from −1/2.0 = −0.50 mM⁻¹ to −1/6.0 ≈ −0.17 mM⁻¹, closer to zero.

Answer: 90 µmol/min without inhibitor, 60 µmol/min with it. Same y-intercept, x-intercept moves toward zero.

Example 4: Coupling and the real ΔG in a cell Free energy

Problem: (a) Glucose + Pi → glucose 6-phosphate has ΔG°′ = +13.8 kJ/mol. ATP hydrolysis has ΔG°′ = −30.5 kJ/mol. What is ΔG°′ for the hexokinase reaction, glucose + ATP → glucose 6-phosphate + ADP? (b) A different reaction has ΔG°′ = +5.0 kJ/mol, but in the cell [products]/[reactants] = 0.010. What is ΔG at 37 °C?

  1. (a) The two reactions share Pi, so they add: +13.8 + (−30.5) = −16.7 kJ/mol. The coupled reaction is favorable.
  2. (b) RT at 310 K = 8.314 J/(mol·K) × 310 K = 2577 J/mol ≈ 2.58 kJ/mol.
  3. ln(0.010) = −4.61, so RT ln Q = 2.58 × (−4.61) ≈ −11.9 kJ/mol.
  4. ΔG = +5.0 + (−11.9) = −6.9 kJ/mol.

Answer: (a) −16.7 kJ/mol. (b) About −6.9 kJ/mol, so the reaction runs forward in the cell even though ΔG°′ is positive.

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Practice test: 18 mixed questions

The questions are mixed across all 8 units, like a real final, with at least two per unit. Check each answer as you go, or answer them all and press Grade my test at the end for a score by unit. Wrong answers tell you what went wrong, and every question has a full solution.

Question 1 · pH and buffers

to 1

Show solution

log([HCO₃⁻]/[CO₂]) = 7.4 − 6.1 = 1.3. Ratio = 101.3 ≈ 20, the textbook 20 to 1.

Question 2 · pH and buffersYou need a buffer for pH 7.2. Which weak acid is the best choice?

Show solution

Pick the pKa closest to the target pH. H₂PO₄⁻ (pKa 6.86) is 0.34 units away; every other choice is farther.

Question 3 · Amino acids, charge and pI

pH units

Show solution

Fully protonated Asp is +1. Losing the α-COOH proton (1.88) gives 0; losing the side-chain proton (3.65) gives −1. The neutral form is between 1.88 and 3.65: pI = (1.88 + 3.65) / 2 ≈ 2.77.

Question 4 · Amino acids, charge and pIAt pH 7.0, which amino acid side chain carries a full positive charge most of the time?

Show solution

A group stays protonated when the pH is below its pKa. Lysine’s side chain (pKa ≈ 10.5) is protonated and positive at pH 7. Histidine’s (pKa 6.0) is mostly neutral.

Question 5 · Protein structure and hemoglobinWhich of these stabilize the tertiary structure of a protein? Select all that apply.

Show solution

Tertiary structure is held by side-chain interactions: the hydrophobic effect (the biggest driver), disulfide bonds, salt bridges and side-chain hydrogen bonds. Peptide bonds are primary structure, and glycosidic bonds belong to carbohydrates.

Question 6 · Protein structure and hemoglobinA protein is treated with urea and a reducing agent until it fully unfolds. Which level of structure is still intact?

Show solution

Only the primary structure, the sequence of peptide-bonded amino acids, survives. Anfinsen’s ribonuclease experiment showed the sequence alone can refold the protein once urea and the reducing agent are removed.

Question 7 · Protein structure and hemoglobinWhich change shifts hemoglobin’s oxygen-binding curve to the right?

Show solution

Right shift = lower O₂ affinity. More 2,3-BPG, lower pH, more CO₂ and higher temperature all shift right, which helps unload O₂ in active tissue.

Question 8 · Enzyme kinetics and inhibition

% of Vmax

Show solution

v / Vmax = [S] / (Km + [S]) = 12 / 16 = 0.75, so 75%. Shortcut: [S] = 3Km always gives ¾Vmax.

Question 9 · Enzyme kinetics and inhibitionAn inhibitor lowers both the apparent Km and the apparent Vmax by the same factor. On a Lineweaver-Burk plot, the lines with and without inhibitor are parallel. What kind of inhibitor is it?

Show solution

Parallel lines mean the slope Km/Vmax is the same. That happens when Km and Vmax drop by the same factor, which is uncompetitive inhibition: the inhibitor binds only the enzyme-substrate complex.

Question 10 · Enzyme kinetics and inhibition

s⁻¹

Show solution

kcat = Vmax / [E]total = (2.0 µM/s) / (0.010 µM) = 200 s⁻¹. Each enzyme molecule turns over 200 substrate molecules per second at saturation.

Question 11 · Carbohydrates, lipids and membranesWhich of these sugars is nonreducing?

Show solution

A sugar is reducing if it has a free anomeric carbon that can open to an aldehyde. Sucrose’s glycosidic bond uses the anomeric carbon of both glucose (C1) and fructose (C2), so it is nonreducing.

Question 12 · Carbohydrates, lipids and membranesWhich change makes a membrane more fluid?

Show solution

Fluidity goes up when tails pack poorly: more cis double bonds, shorter tails, and higher temperature. Cholesterol dampens changes in both directions.

Question 13 · Free energy and ATP

kJ/mol

Show solution

ΔG°′ values of coupled reactions add: +12.0 + (−30.5) = −18.5 kJ/mol.

Question 14 · Free energy and ATPWhat does an enzyme change about the reaction it catalyzes?

Show solution

An enzyme lowers the activation energy, so equilibrium is reached faster. ΔG, ΔG°′ and Keq stay the same.

Question 15 · Glycolysis and the citric acid cycleWhich glycolysis enzymes catalyze essentially irreversible steps in the cell? Select all that apply.

Show solution

The three irreversible steps are hexokinase (step 1), PFK-1 (step 3) and pyruvate kinase (step 10). They have large negative ΔG in the cell, they are the control points, and gluconeogenesis has to bypass each one.

Question 16 · Glycolysis and the citric acid cycleWhy do muscle cells convert pyruvate to lactate during intense exercise?

Show solution

During intense exercise, glycolysis can produce NADH faster than mitochondria reoxidize it. Lactate dehydrogenase uses NADH to reduce pyruvate to lactate, giving back NAD⁺ so glycolysis keeps making its 2 ATP per glucose. Lactate can form even when oxygen is available.

Question 17 · Electron transport and ATP yield

ATP per glucose

Show solution

Per glucose: 10 NADH (2 glycolysis, 2 pyruvate dehydrogenase, 6 cycle) × 2.5 = 25; 2 FADH₂ × 1.5 = 3; plus 2 ATP from glycolysis and 2 GTP from the cycle = 4. Total 32.

Question 18 · Electron transport and ATP yield2,4-Dinitrophenol (DNP) makes the inner mitochondrial membrane leaky to protons. What happens in the mitochondria?

Show solution

An uncoupler separates electron flow from ATP synthesis. The proton gradient collapses, so the chain runs faster (more O₂ used) while ATP synthase makes less ATP. The lost energy comes out as heat.

Your score by unit

Answer as many questions as you can, then grade the whole test. Unanswered questions count as missed.

Frequently asked questions

What is on a biochemistry final exam?

Most first-semester biochemistry finals cover water and buffers, amino acids, protein structure, enzyme kinetics and inhibition, carbohydrates and lipids, free energy, and the start of metabolism: glycolysis, the citric acid cycle and oxidative phosphorylation. Second-semester courses go deeper into metabolism and its regulation. Your syllabus is the final word on what is in scope.

How should I study for a biochemistry final?

Start with a practice test to find your weak units, then study those first. Work problems instead of rereading: pI, Henderson-Hasselbalch, Michaelis-Menten and ATP counting are skills you build by doing. Save memorizing pathway steps for short daily review sessions rather than one long cram.

Do I need to memorize every glycolysis and citric acid cycle step?

It depends on your instructor, but most finals expect the order of intermediates, the enzymes at the irreversible and regulated steps, and where NADH, FADH₂, ATP and CO₂ are made. Knowing why the control points are regulated usually earns more points than memorizing structures.

How many ATP does one glucose make?

About 30 to 32 ATP using the current values of 2.5 per NADH and 1.5 per FADH₂. The answer is 32 with the malate-aspartate shuttle and 30 with the glycerol-3-phosphate shuttle. Older textbooks give 36 to 38 because they used 3 and 2.

Is biochemistry harder than organic chemistry?

Many students find biochemistry less about drawing mechanisms and more about connecting ideas and remembering pathways. The math is light: logs, ratios and the Michaelis-Menten equation. If you can explain why each step happens, the memorizing gets much easier.

Your final is coming either way. Get feedback on every pathway step and every kinetics problem before it counts.

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