Unit 1: Biochemistry — Unit Test

Assessment OF Learning · Strand B
Graded — Counts Toward 70% Term Mark
Duration: 75 minutes  |  Total: /60 marks  |  Show all work. Open answer keys after attempting each question.
K/U
/15
Thinking
/15
Comm.
/15
Applic.
/15
Part A: Knowledge & Understanding [15 marks]
1
[3]
Match each macromolecule with its monomer and one major function: carbohydrate, lipid, protein, nucleic acid.
Answer Key

Carbohydrate: monosaccharide (e.g., glucose), energy & structural (cellulose); Lipid: glycerol + fatty acids (no true monomer), energy storage / membranes / hormones; Protein: amino acid, structure / catalysis / transport; Nucleic acid: nucleotide (sugar + phosphate + base), information storage (DNA) / expression (RNA).

2
[2]
Which is true of phospholipids?
Answer Key

B. Phosphate-containing head is polar/charged (hydrophilic); two fatty-acid tails are non-polar (hydrophobic). This amphipathic property drives bilayer formation.

3
[2]
An enzyme catalyzes a reaction by:
Answer Key

B. Enzymes provide an alternate pathway with lower E_a so more substrate molecules have enough energy to react at body temperature. They do not change ΔG or K_eq.

4
[3]
Label the following on a Michaelis-Menten v vs [S] curve: V_max, K_m, the saturation region. Explain what each tells you about the enzyme.
Answer Key

V_max = horizontal asymptote; rate when all sites occupied. K_m = [S] at v = ½V_max; lower K_m means higher affinity. Saturation region = plateau where adding [S] no longer increases rate (rate-limited by enzyme amount).

5
[2]
A protein loses function above 60 °C because:
Answer Key

B. Heat disrupts H-bonds and hydrophobic interactions that hold tertiary structure; the active site collapses (denaturation). Primary sequence is intact (peptide bonds are covalent and stable).

6
[3]
List the components of the cell membrane in the fluid mosaic model and state one role for each.
Answer Key

Phospholipid bilayer (semipermeable barrier); cholesterol (fluidity buffer); integral/transmembrane proteins (transport, receptors); peripheral proteins (signaling, structure); glycoproteins/glycolipids (cell recognition, immunity).

Part B: Thinking & Investigation [15 marks]
7
[5]
A scientist measures initial reaction velocity at 5 substrate concentrations and finds V_max = 100 µmol/min, K_m = 2 mM. Predict the rate at [S] = 2 mM, [S] = 4 mM, and [S] = 20 mM. Show calculations using v = V_max[S]/(K_m + [S]).
Answer Key

v = 100·[S]/(2+[S]). At [S]=2: v = 100·2/4 = 50 µmol/min. At [S]=4: v = 100·4/6 ≈ 66.7 µmol/min. At [S]=20: v = 100·20/22 ≈ 90.9 µmol/min (approaching V_max).

8
[5]
Design a controlled experiment to determine whether temperature affects amylase activity. Identify variables, controls, and how you would measure rate.
Answer Key

IV: temperature (e.g., 0, 20, 37, 60, 80 °C). DV: rate of starch breakdown (time for iodine test to stop turning blue-black, or absorbance over time). Controlled: amylase concentration, starch concentration, pH (buffer), volume, mixing. Hypothesis: rate increases up to ~37 °C optimum then declines as enzyme denatures. Replicate ≥3 trials/condition; plot rate vs T.

9
[5]
A non-competitive inhibitor I is added to an enzyme. On a single graph, sketch v vs [S] curves for: (a) no inhibitor, (b) with inhibitor. Label V_max and K_m for each. Explain the shapes.
Answer Key

Without I: hyperbolic curve to V_max with K_m = some value. With non-competitive I: lower V_max (because some enzyme molecules are inactive regardless of [S]); K_m unchanged (the enzymes that remain active have unchanged affinity). The curve plateaus at lower height. (Contrast with competitive: same V_max, higher apparent K_m.)

Part C: Communication [15 marks]
10
[5]
Explain the induced-fit model of enzyme action and contrast it with the lock-and-key model. Use proper terminology: active site, substrate, conformational change, transition state.
Answer Key

Lock-and-key (Fischer, 1894): rigid active site fits one substrate exactly. Induced-fit (Koshland, 1958): active site is flexible — substrate binding induces a conformational change that brings catalytic residues into optimal alignment, stabilizes the transition state, and lowers E_a. Induced-fit better explains promiscuous enzymes and allosteric regulation.

11
[5]
Draw and label a labeled diagram of a phospholipid and a labeled cross-section of a fluid-mosaic membrane (description acceptable). Show: phosphate head, fatty acid tails, cholesterol, integral and peripheral proteins, glycolipid, glycoprotein.
Answer Key

Phospholipid: glycerol backbone, phosphate head (polar), 2 fatty acid tails (non-polar). Bilayer with heads outward, tails inward. Cholesterol nestled among tails. Integral (transmembrane) proteins span the bilayer; peripheral proteins on one face only. Glycolipids and glycoproteins on extracellular side carry oligosaccharide chains for cell recognition.

12
[5]
Compare and contrast simple diffusion, facilitated diffusion, and active transport in a 3-row table. Include: ATP requirement, gradient direction, example.
Answer Key

Simple diffusion: no ATP, with gradient, example O₂/CO₂. Facilitated diffusion: no ATP, with gradient, requires channel/carrier protein, example glucose via GLUT, water via aquaporin. Active transport: ATP required, against gradient, example Na+/K+ ATPase pumping 3 Na+ out / 2 K+ in.

Part D: Application [15 marks]
13
[5]
Case Study — Lactose Intolerance: A patient cannot digest dairy. Explain the molecular mechanism of the deficiency, what happens in the gut when undigested lactose reaches the colon, and how an enzyme supplement (lactase) restores function.
Answer Key

Lactase, a brush-border enzyme on small intestine villi, hydrolyzes lactose → glucose + galactose for absorption. Deficient lactase means lactose persists, drawing water osmotically (diarrhea) and is fermented by colon bacteria producing H₂, CO₂, methane (gas, cramps). Oral lactase supplement provides exogenous enzyme that hydrolyzes lactose before bacteria can ferment it.

14
[5]
Case Study — IV Saline: A dehydrated patient receives 0.9% NaCl IV (isotonic). Explain why this concentration is chosen rather than pure water or 5% NaCl, using osmosis and tonicity.
Answer Key

0.9% NaCl matches blood plasma osmolarity (~300 mOsm), so RBCs experience no net water movement and remain normal. Pure water (hypotonic) → water rushes into RBCs → swell & hemolyse. 5% NaCl (hypertonic) → water leaves RBCs → crenate (shrivel), losing function. Isotonic prevents both.

15
[5]
STSE — Trans-Fats: Health Canada limits trans-fat content. Explain (i) why trans-fats are produced industrially, (ii) why their structure makes them solid at room temperature, and (iii) the cardiovascular health concern.
Answer Key

(i) Partial hydrogenation of vegetable oils converts cis-double bonds (kinked) to trans (straight), increasing shelf life and giving spreadable texture. (ii) Trans tails pack like saturated fats, raising melting point. (iii) Trans-fats raise LDL ("bad") cholesterol and lower HDL ("good"), accelerating atherosclerosis and heart-disease risk; hence regulatory limits.