SBI4U Final Examination — Cumulative

All five strands · Ontario Curriculum 2008 (Revised)
Counts toward 30% Final Evaluation
Duration: 3 hours  |  Total: /100 marks  |  Coverage: Strands B–F. Show all work. Worked answer keys appear in collapsible sections after each question.
K/U
/25
Thinking
/25
Comm.
/25
Applic.
/25
Part A — Knowledge & Understanding [25 marks]
1
[3]B · Biochem
Match each macromolecule to its monomer and one major function: carbohydrate, lipid, protein, nucleic acid.
Answer Key

Carbohydrate: monosaccharide; energy and structure (cellulose, glycogen). Lipid: glycerol + fatty acids; energy storage, membranes, hormones. Protein: amino acid; enzymes, structure, transport. Nucleic acid: nucleotide; information storage and expression.

2
[3]C · Metabolism
State the location and net products of: glycolysis, Krebs cycle, electron transport chain.
Answer Key

Glycolysis: cytoplasm; net 2 ATP, 2 NADH, 2 pyruvate. Krebs (×2 turns/glucose): mitochondrial matrix; 2 ATP, 6 NADH, 2 FADH₂, 4 CO₂. ETC + chemiosmosis: inner mitochondrial membrane; ~26–28 ATP from oxphos, H₂O as final product.

3
[3]D · Genetics
List the four DNA bases, the pairing rules, and the number of H-bonds in each pair.
Answer Key

A, T, G, C. A=T (2 H-bonds); G≡C (3 H-bonds). Antiparallel strands; sugar = deoxyribose.

4
[3]E · Homeostasis
Identify the four phases of an action potential and the dominant ion movement in each.
Answer Key

Resting (~−70 mV; Na/K pump). Depolarization (Na+ in via voltage-gated Na+ channels). Repolarization (K+ out via voltage-gated K+ channels; Na+ channels inactivate). Hyperpolarization & refractory period (overshoot of K+ efflux; pump restores).

5
[3]F · Pop Dynamics
Define carrying capacity, intrinsic growth rate, and the difference between density-dependent and density-independent factors.
Answer Key

K = max sustainable population given resources. r = per-capita growth rate when N << K. Density-dependent (disease, predation, competition) intensify with crowding; density-independent (weather, fire) act regardless of density.

6
[2]B
A non-competitive inhibitor of an enzyme:
Answer Key

B. Binds an allosteric site, inactivating some enzyme molecules → V_max ↓; remaining active enzymes still bind substrate the same way → K_m unchanged.

7
[2]C
The final electron acceptor in aerobic respiration is:
Answer Key

B. O₂ accepts electrons at Complex IV → forms H₂O.

8
[3]D
Define and give an effect on protein for: silent, missense, nonsense mutation.
Answer Key

Silent: same amino acid (degeneracy of code) — no protein change. Missense: different amino acid (e.g., Glu→Val in sickle-cell). Nonsense: introduces stop codon → truncated protein.

9
[3]E
Match each hormone with its action: insulin, glucagon, ADH, aldosterone.
Answer Key

Insulin (β-cells): lowers blood glucose by promoting uptake/storage. Glucagon (α-cells): raises blood glucose via glycogenolysis. ADH (post. pituitary): increases water reabsorption in collecting duct. Aldosterone (adrenal cortex): increases Na+ (and water) reabsorption in distal tubule, raising BP.

Part B — Thinking & Investigation [25 marks]
10
[6]B · Macromolecule Analysis
A nutritionist has an unknown sample. Tests show: positive Biuret (purple), negative Benedict's, negative Sudan III, positive ninhydrin. (a) Identify the macromolecule class. (b) Predict whether it is a polymer or a small molecule. (c) Design one additional test that would confirm.
Answer Key

(a) Protein: Biuret detects peptide bonds (purple); ninhydrin detects free amine groups. (b) Polymer (Biuret detects peptide bonds, plural). (c) Confirmation: gel electrophoresis (SDS-PAGE) to estimate molecular weight, or denaturation test (heat, acid → coagulation), or specific enzymatic digest (trypsin) followed by mass-spec.

11
[6]C · Metabolic Pathway Integration
Calculate total ATP per glucose using P/O ratios NADH = 2.5, FADH₂ = 1.5. Assume cytosolic NADH uses the glycerol-3-phosphate shuttle (effectively becomes FADH₂ → 1.5 ATP). Show the contribution of each stage.
Answer Key

Glycolysis: 2 ATP + 2 NADH (cytosolic) × 1.5 = 2 + 3 = 5. Pyruvate oxidation: 2 NADH × 2.5 = 5. Krebs: 2 ATP + 6 NADH × 2.5 + 2 FADH₂ × 1.5 = 2 + 15 + 3 = 20. Total = 30 ATP per glucose. (32 if malate-aspartate shuttle is used: cytosolic NADH yields 2.5 each.)

12
[7]D · Genetic Engineering
A biotech firm wants to produce human Factor IX (clotting factor) in E. coli for treating hemophilia B. Outline the workflow: (a) source of gene, (b) tools to insert into plasmid, (c) selection of transformed bacteria, (d) protein purification, (e) one safety/quality concern.
Answer Key

(a) Reverse-transcribe Factor IX mRNA from human liver cells using reverse transcriptase to produce cDNA (no introns — bacteria cannot splice). (b) Cut cDNA and plasmid with the same restriction enzyme (matching sticky ends); ligate using DNA ligase. (c) Transform competent E. coli (heat-shock or electroporation); plate on antibiotic agar — only those carrying the plasmid (with antibiotic-resistance marker) grow; screen for insert (blue/white α-complementation or PCR). (d) Induce expression (IPTG); lyse cells; purify by affinity chromatography (e.g., His-tag); refold and verify purity by SDS-PAGE. (e) Concern: endotoxin (LPS) contamination from bacterial membranes — must be rigorously removed before clinical use; also post-translational modifications differ from human cells (Factor IX needs γ-carboxylation), so production sometimes uses CHO mammalian cells instead.

13
[6]F · Population Calculation
A protected sea otter population grows logistically with K = 5000, r = 0.10/year, and starts at N₀ = 100. (a) Use dN/dt = rN(K−N)/K to compute initial dN/dt. (b) Compute dN/dt at N = 2500 (=K/2). (c) State which N gives the maximum dN/dt and explain.
Answer Key

(a) dN/dt = 0.10 × 100 × (5000−100)/5000 = 0.10 × 100 × 0.98 = 9.8 otters/year. (b) dN/dt = 0.10 × 2500 × (5000−2500)/5000 = 0.10 × 2500 × 0.5 = 125 otters/year. (c) Maximum dN/dt occurs at N = K/2 = 2500 — the inflection point of the logistic S-curve, where the product N(K−N) is maximized.

Part C — Communication [25 marks]
14
[6]B
Draw and describe a labeled diagram of a fluid-mosaic membrane. Identify: phospholipid, cholesterol, transmembrane (integral) protein, peripheral protein, glycoprotein, glycolipid. State the role of each.
Answer Key

Bilayer of phospholipids (polar heads outward, non-polar tails inward) — selective barrier. Cholesterol — fluidity buffer. Transmembrane proteins — channels/carriers/receptors. Peripheral proteins — anchored to one face; signaling, structure. Glycoproteins/glycolipids — sugar chains on extracellular surface; recognition, immunity (e.g., ABO blood types).

15
[6]C
Compare photosynthesis and aerobic cellular respiration in a 5-row table: location, inputs, outputs, energy conversion, and net direction.
Answer Key

Photosynthesis: chloroplast (thylakoid + stroma); inputs CO₂ + H₂O + light; outputs C₆H₁₂O₆ + O₂; light energy → chemical energy; anabolic. Respiration: cytoplasm + mitochondria; inputs C₆H₁₂O₆ + O₂; outputs CO₂ + H₂O + ATP; chemical energy of glucose → chemical energy of ATP (with heat); catabolic. The two are reciprocal — the basis of energy flow in ecosystems.

16
[7]D
Describe the central dogma in detail. For each arrow (DNA → RNA → protein), name the process, the key enzyme(s), and the location in a eukaryote.
Answer Key

DNA → RNA: transcription. RNA polymerase II reads template 3'→5', synthesizes pre-mRNA 5'→3'. Nucleus. Then 5' cap, poly-A, splicing → mature mRNA. mRNA → protein: translation. Ribosome (rRNA + proteins; large + small subunits) reads codons 5'→3'; tRNAs deliver amino acids matching anticodons; peptidyl transferase forms peptide bonds. Cytoplasm or rough ER. Termination at stop codons by release factors. (Reverse transcriptase exists in retroviruses: RNA → DNA.)

17
[6]E
Explain the negative-feedback loop that maintains body temperature. Include receptor, control center, and effectors for both hot and cold deviations.
Answer Key

Receptors: thermoreceptors in skin and hypothalamus. Control center: hypothalamus (set point ~37 °C). Cold deviation: vasoconstriction of skin arterioles (less heat loss); shivering (skeletal muscle contractions generate heat); piloerection; thyroid hormone release (longer term). Hot deviation: vasodilation; sweating (evaporative cooling); reduced muscle activity. All effects oppose the original change (negative feedback).

Part D — Application [25 marks]
18
[6]B/E · Homeostatic Disorder
Case Study — Cystic Fibrosis: A child has CF caused by ΔF508 mutation in the CFTR gene. Explain (a) the molecular defect, (b) effect on lungs and pancreas, (c) why a high-calorie diet with pancreatic-enzyme replacement is required.
Answer Key

(a) ΔF508 is a 3-nucleotide deletion removing phenylalanine at position 508. The CFTR chloride channel misfolds and is degraded in the ER, so the apical membrane has too few channels. (b) Lungs: thick mucus traps bacteria → recurrent infections (Pseudomonas), inflammation, bronchiectasis. Pancreas: pancreatic ducts blocked → digestive enzymes don't reach intestine → malabsorption of fats and proteins → failure to thrive. (c) Without enzymes, fats and proteins are poorly digested; calories lost. High-calorie, high-fat diet plus oral pancreatic-enzyme capsules (lipase, protease, amylase) at meals and fat-soluble vitamins (A, D, E, K) supplementation maintain growth and nutrition.

19
[6]D · Genetic Engineering Scenario
STSE — CRISPR for Sickle-Cell: In 2023, Casgevy (exa-cel) was approved as a CRISPR therapy for sickle-cell disease. (a) Explain the molecular target. (b) Outline the procedure. (c) Discuss one benefit and two ethical/access concerns.
Answer Key

(a) Casgevy edits the BCL11A enhancer in patient HSCs (hematopoietic stem cells). Disrupting BCL11A re-activates fetal hemoglobin (HbF), which doesn't sickle, compensating for defective adult HbS. (b) Patient HSCs collected by apheresis → ex vivo CRISPR-Cas9 editing → patient receives myeloablative chemotherapy → edited cells re-infused → engraft in marrow → produce HbF-rich red cells. (c) Benefit: potentially curative — reduces vaso-occlusive crises dramatically; one-time therapy. Concerns: cost (~$2 million USD per patient) — equity/access; long-term safety unknown (off-target effects, secondary malignancies); chemotherapy carries significant short-term risk and infertility issues; somatic-cell editing only — does not affect germline (this is generally accepted, in contrast to germline editing).

20
[7]E · Homeostatic Disorder
Case Study — Diabetic Ketoacidosis (DKA): A teen with previously undiagnosed Type 1 diabetes presents with: blood glucose 28 mM, blood pH 7.20, fruity breath, rapid deep breathing, dehydration. (a) Explain why glucose is so high. (b) Explain the source of acidosis and breath odor. (c) Explain Kussmaul breathing. (d) Outline emergency treatment.
Answer Key

(a) No insulin → cells cannot import glucose → hyperglycemia; liver also continues gluconeogenesis. Glucose exceeds renal threshold → glucosuria → osmotic diuresis → dehydration. (b) Cells switch to fatty-acid oxidation → liver produces ketone bodies (acetoacetate, β-hydroxybutyrate, acetone) → ketonemia → metabolic acidosis. Acetone exhaled gives the fruity breath. (c) Kussmaul respirations: deep, rapid breathing — respiratory compensation that blows off CO₂ to raise blood pH. (d) Treatment: IV isotonic saline (rehydrate), IV regular insulin infusion, careful K+ replacement (insulin shifts K+ into cells; total-body K+ may be depleted), monitor pH and electrolytes, treat precipitating cause (often infection), then transition to subcutaneous insulin and patient education.

21
[6]F · Population Dynamics Calculation
Case — Conservation: Vancouver Island marmot population N₀ = 30, captive breeding adds 8/year, mortality 4/year, no migration. (a) Calculate r. (b) Predict population in 5 years if exponential. (c) If habitat carrying capacity K = 200, predict whether logistic growth will reach K within 30 years (qualitative reasoning sufficient).
Answer Key

(a) r = (births − deaths)/N₀ = (8 − 4)/30 = 4/30 ≈ 0.133/year. (b) N(5) = 30 × e^(0.133 × 5) = 30 × e^0.67 ≈ 30 × 1.95 ≈ 58 marmots. (c) With K = 200 and r = 0.133, the logistic S-curve approaches but rarely "reaches" K asymptotically. Time to ~95% of K (N ≈ 190) ≈ ln(K/N₀ × 19) / r ≈ ln(126.7)/0.133 ≈ 4.84/0.133 ≈ 36 years. So within 30 years the population would approach but not quite reach K — perhaps ~150–180. Real conservation also faces stochastic events that may slow recovery further.