Grade 12 Biology (SBI4U) Practice Quiz, With Worked Answers

The 28 questions at a glance

The table shows what each strand asks of a student, in plain terms, which questions practise it, and the curriculum expectations those questions touch.

Fourteen questions are multiple choice, and the answer explains why each wrong option fails. The other fourteen are short-answer or calculation questions, and the answer shows the reasoning or the formula, the numbers and the units.

Is SBI4U still the Grade 12 biology course in 2026-27?

Yes. The ministry's course page, updated in August 2026, still lists SBI4U as Biology, Grade 12, University, with the prerequisite Grade 11 Biology, University. The ministry says the 2008 science curriculum for Grade 10 and the 2008 science curriculum for Grades 11 and 12 both remain in effect, and every secondary science course from Grade 10 to Grade 12 is still based on those two documents. Grade 9 science is the one course that moved to a newer curriculum. We found no announcement of a new Grade 12 biology course as of 26 October 2026.

SBI4U is described in The Ontario Curriculum, Grades 11 and 12: Science, 2008 (Revised). The ministry's own description says students "will study theory and conduct investigations in the areas of biochemistry, metabolic processes, molecular genetics, homeostasis, and population dynamics". Those are the course's five content strands. A sixth strand, scientific investigation skills and career exploration, runs through all five; this quiz does not test it.

SBI3U is the only prerequisite for SBI4U. There is no Grade 12 college biology course in Ontario, and SBI3C, the Grade 11 college course, does not lead to SBI4U. Genie's SBI4U explainer already covers what the course is, who needs it and how it differs from SBI3U, so this page sticks to practice.

How the SBI4U exam counts from 2026-27

The rules changed this school year. The ministry's Growing Success page, updated on 14 October 2026, says: "Beginning in the 2026-2027 school year, most secondary courses will require written exams as a percentage of the final course mark."

From the 2026-27 school year, a Grade 11 or 12 student's final mark is 65% classroom work, 25% final evaluation and 10% attendance and participation.

For SBI4U, the 25% final evaluation is a written exam. The ministry's Grade 11 and 12 table puts English, mathematics and science on the same row, with a written exam and no culminating task. The same page exempts Grade 11 and 12 workplace preparation (E) and open (O) courses from mandatory written exams, so the rule covers SBI4U, a university preparation course, but not every Grade 12 science course. Written final exams from Grades 9 to 12 take place during the designated exam period.

This replaces the older split. For 2026-27, the ministry changed the Grade 11 and 12 split to 65, 25 and 10, with a 25% written exam for Grade 11 and 12 science courses other than E and O courses. The earlier rule, set in Growing Success (2010), was 70% coursework and a 30% final evaluation. Guides written before this year may still show 70 and 30, so for this school year use the 65, 25 and 10 above.

The ministry policy lists no provincial exam for SBI4U. The one exception it makes to the school exam period is the EQAO Grade 9 mathematics assessment. The ministry page does not say who writes the paper for each course, and no ministry document we read names an SBI4U exam format, mark total or length, so ask the teacher what the exam will look like. Online and remote courses offered through school boards write the exam in person at the student's home school, or another location if required, unless the board can provide secure online proctoring.

Our page on Ontario's new exam and attendance rules covers the rest of the change, and what changed in Ontario schools in September 2026 puts it next to the other changes this year.

Conventions used in this quiz

Biology questions need a few fixed values, and textbooks do not all print them the same way. This quiz uses the values below and gives them again inside each question that needs them, so no answer depends on a school's own handout.

Codons used in this quiz. The table follows the standard genetic code. mRNA is written 5' to 3', and U replaces T.

Neuron values. A resting neuron sits at about -70 mV, and an action potential starts at a threshold of about -55 mV. OpenStax notes that the resting value varies by neuron type and species.

Population equations. Exponential growth: dN/dt = rN. Logistic growth: dN/dt = rmax × N × (K - N) / K, where K is the carrying capacity.

Trophic level transfer efficiency. TLTE = (production at the present trophic level ÷ production at the previous trophic level) × 100.

How to use this quiz

Do all 28 on paper before reading a single answer. Cover the answer under each question with a sheet of paper and slide it down only once the question is done. Keep the conventions section open, and skip the timer on the first pass. A basic calculator helps for Questions 4, 24 to 26 and 28.

Twenty-eight questions cannot predict an exam mark. What they can show is which strand is short, and which idea inside it. For the written questions, answer in full sentences and name the structure, molecule or ion each time, because a vague answer hides where the reasoning broke.

Every question in this quiz is original and was written for this page. None is copied from a textbook, a board exam, a teacher's review package or the ministry's curriculum.

Search for an SBI4U practice test and the first page, checked on 26 October 2026, is mostly uploaded documents on note-sharing sites, a paid exam listing, a quiz app and a parked domain. The dated documents on that page are from 2020 and the 2021-22 school year, before the 2026-27 exam rules. None of the pages we could read puts a worked answer in the page text under each question. Old exams are still useful practice once you have finished this set.

Biochemistry: questions 1 to 6

Question 1. Two glucose molecules join to form the disaccharide maltose. Which statement describes the reaction?

• (A) It is hydrolysis, and one water molecule is used up.
• (B) It is a condensation (dehydration synthesis) reaction, and one water molecule is released.
• (C) It is a condensation reaction, and one water molecule is used up.
• (D) It is a neutralisation reaction, and a salt and water form.

Answer: (B). When two monomers join, an -OH from one and an -H from the other leave together as water, and the remaining pieces bond. The formulas show it: 2 C₆H₁₂O₆ → C₁₂H₂₂O₁₁ + H₂O. Two glucose molecules hold 24 hydrogens and 12 oxygens, maltose holds 22 and 11, and the missing pair of hydrogens and one oxygen is the water. (A) describes the reverse reaction: hydrolysis splits maltose back into two glucose molecules and uses a water molecule to do it. (C) has the right reaction type and the water on the wrong side. (D) needs an acid and a base, and neither sugar is one.

Question 2. A short peptide is made of 4 amino acids joined in a single chain. How many peptide bonds does it contain, and how many water molecules were released in making it? How many water molecules would it take to break it back into free amino acids?

Answer. Three peptide bonds, three water molecules released, and three water molecules to break it down.

• Each peptide bond forms between the carboxyl group of one amino acid and the amino group of the next, and each one releases a water molecule.
• A chain of n amino acids has n - 1 links. With 4 amino acids, that is 3 bonds and 3 water molecules.
• Hydrolysis reverses each bond and adds one water molecule per bond, so it takes 3.

The slip to avoid is counting one bond per amino acid. Draw four circles in a row and count the links between them: there are three.

Question 3. A student measures the rate of an enzyme reaction (in µmol/min) at four substrate concentrations, with no inhibitor and with each of two inhibitors, A and B. Each inhibitor is used at the same fixed concentration across its whole row.

Which conclusion fits the data?

• (A) A is noncompetitive and B is competitive.
• (B) Both inhibitors are competitive.
• (C) Both inhibitors are noncompetitive.
• (D) A is competitive, and B is noncompetitive and binds at an allosteric site.

Answer: (D). Look at the highest substrate concentration, where the enzyme is closest to its maximum rate. With inhibitor A the rate climbs to 38, close to the 39 with no inhibitor. Extra substrate outcompetes A for the active site, which is how a competitive inhibitor behaves: it affects the rate at low substrate and does not change the maximal rate. With inhibitor B the rate is half the uninhibited value at every concentration, to the nearest whole number, and tops out at 20. Adding substrate cannot overcome B, so B lowers the maximal rate, which is what a noncompetitive inhibitor does. A noncompetitive inhibitor binds somewhere other than the active site, at an allosteric site. (A) swaps the two. (B) cannot explain why B's rate stays at half however much substrate is added. (C) cannot explain why A's rate almost catches up. Both inhibitors give 10 at 2 mM, which is why the low-substrate column alone cannot tell them apart.

Question 4. Cores cut from a potato each have a mass of 5.00 g. One core sits in a salt solution for 2 hours and then has a mass of 4.35 g. Assume the whole change in mass is water. Calculate the percent change in mass, say which way water moved, and say whether the solution was hypotonic or hypertonic to the potato cells.

Answer. -13.0%. Water left the cells, so the solution was hypertonic to them.

• Percent change = (final mass - initial mass) ÷ initial mass × 100.
• (4.35 g - 5.00 g) ÷ 5.00 g × 100 = -0.65 ÷ 5.00 × 100 = -13.0%.

The negative sign means the core lost mass. Water moves by osmosis toward the side with the higher solute concentration. The cells lost water, so the solution outside had the higher solute concentration: it was hypertonic to the cells. A red blood cell in a hypertonic solution shrivels for the same reason. In a hypotonic solution, the core would have gained mass.

Question 5. Name the type of biochemical reaction in each case:

• (a) NAD⁺ accepts electrons and becomes NADH.
• (b) A protein is broken into amino acids by adding water.
• (c) An acid reacts with a base to give a salt and water.
• (d) Two amino acids join and release a water molecule.

Answer. (a) oxidation-reduction (redox); (b) hydrolysis; (c) neutralisation; (d) condensation.

These are the four main types of biochemical reaction the curriculum names. In (a), NAD⁺ gains electrons, so it is reduced, and whatever gave up those electrons was oxidised at the same time; the two always happen together, which is why the type is called oxidation-reduction. (b) and (d) are the same bond run in opposite directions: condensation releases water as a bond forms, and hydrolysis uses water to break it. (c) is the one defined by its products: an acid and a base react to give a salt and water.

Question 6. A cell in the pancreas makes a digestive enzyme that it releases outside the cell. Which sequence shows the enzyme's route from where it is made to where it leaves?

• (A) ribosome on the rough endoplasmic reticulum → transport vesicle → Golgi apparatus → secretory vesicle → plasma membrane
• (B) ribosome on the rough endoplasmic reticulum → Golgi apparatus → smooth endoplasmic reticulum → lysosome → plasma membrane
• (C) free ribosome in the cytoplasm → mitochondrion → Golgi apparatus → plasma membrane
• (D) Golgi apparatus → rough endoplasmic reticulum → transport vesicle → nucleus → plasma membrane

Answer: (A). Proteins that leave the cell are made on ribosomes attached to the rough endoplasmic reticulum. Vesicles bud from the ER and carry the protein to the Golgi apparatus, which sorts, tags and packages it. A vesicle from the Golgi then fuses with the plasma membrane and releases the enzyme outside the cell, which is exocytosis. (B) sends the protein backward from the Golgi to the smooth ER and into a lysosome; lysosomes hold enzymes for digestion inside the cell. (C) sends it through a mitochondrion, which is not on the route. (D) starts at the Golgi, before the protein has been made, and passes it through the nucleus.

Metabolic processes: questions 7 to 12

Question 7. For one molecule of glucose going through glycolysis, which statement is correct?

• (A) 2 ATP are used and 2 are made, for a net gain of zero, in the mitochondrial matrix.
• (B) 4 ATP are made and none are used, for a net gain of 4 ATP, in the cytoplasm.
• (C) 2 ATP are used and 4 are made, for a net gain of 2 ATP and 2 NADH, in the cytoplasm.
• (D) 2 ATP are used and 4 are made, for a net gain of 2 ATP and 2 NADH, in the mitochondrial matrix.

Answer: (C). The first half of glycolysis spends 2 ATP to prepare the glucose. The second half makes 4 ATP and 2 NADH as the two three-carbon pieces become two pyruvate molecules. Net: 4 - 2 = 2 ATP, plus 2 NADH. Glycolysis happens in the cytoplasm, in prokaryotic and eukaryotic cells alike, which is why it runs even without mitochondria. (A) has the second half making only 2 ATP when it makes 4, and puts glycolysis in the wrong place. (B) counts the 4 made and forgets the 2 spent. (D) has the accounting right and the location wrong.

Question 8. Give the location of each stage in a eukaryotic cell: (a) glycolysis; (b) pyruvate oxidation and the citric acid (Krebs) cycle; (c) the electron transport chain in cellular respiration; (d) the Calvin cycle.

Answer.

• (a) Glycolysis: the cytoplasm.
• (b) Pyruvate oxidation and the citric acid cycle: the mitochondrial matrix.
• (c) The electron transport chain: the inner mitochondrial membrane.
• (d) The Calvin cycle: the stroma of the chloroplast.

A useful check is to ask what each stage needs. The electron transport chain pumps protons across a membrane, so it must sit in one; the inner membrane is where the protons build up on one side. The Calvin cycle is a set of enzyme reactions in solution, so it runs in the fluid stroma, around the thylakoids where the light-dependent reactions happen.

Question 9. In aerobic cellular respiration, what is the final electron acceptor at the end of the electron transport chain, and what does it become?

• (A) Oxygen, which is reduced to water.
• (B) NAD⁺, which becomes NADH.
• (C) Carbon dioxide, which is released from the cell.
• (D) Pyruvate, which becomes lactate.

Answer: (A). Electrons pass along the chain to its end, where they reduce oxygen, and together with protons that makes water. Without oxygen to take the electrons, the chain backs up and stops, oxidative phosphorylation stops with it, and the cell is left with the 2 ATP per glucose that glycolysis makes, kept running by fermentation. (B) is backwards: NADH brings electrons to the start of the chain, and NAD⁺ is what is left after it hands them over. (C) names a waste product: carbon dioxide is released by pyruvate oxidation and the citric acid cycle, and it accepts no electrons. (D) is lactic acid fermentation, which happens when there is no oxygen and the electron transport chain is not running.

Question 10. Yeast cells ferment 10 mmol of glucose with no oxygen present. Assume all of it goes through alcohol fermentation. How much ethanol, carbon dioxide and net ATP are made, and why must the cells ferment at all?

Answer. 20 mmol of ethanol, 20 mmol of carbon dioxide and 20 mmol of ATP (net).

• Glycolysis splits each glucose into 2 pyruvate, with a net gain of 2 ATP.
• In alcohol fermentation, each pyruvate loses one CO₂ and the remaining two-carbon piece is reduced to ethanol. So each glucose gives 2 ethanol and 2 CO₂: C₆H₁₂O₆ → 2 C₂H₅OH + 2 CO₂.
• For 10 mmol of glucose: 2 × 10 = 20 mmol of ethanol, 20 mmol of CO₂, and 2 × 10 = 20 mmol of ATP.

Fermentation makes no extra ATP. Its job is to turn NADH back into NAD⁺. Glycolysis needs NAD⁺ to keep going, and without oxygen the electron transport chain cannot recycle it. A student who adds ATP from the fermentation step has the right products and the wrong energy count.

Question 11. A leaf in the light releases 6 molecules of O₂ from photosynthesis. How many water molecules were split to release them, and where does the oxygen come from?

• (A) 6 water molecules; the oxygen comes from water.
• (B) 12 water molecules; the oxygen comes from carbon dioxide.
• (C) 3 water molecules; the oxygen comes from water.
• (D) 12 water molecules; the oxygen comes from water.

Answer: (D). In the light-dependent reactions, splitting one water molecule releases two electrons, two hydrogen ions (protons) and one oxygen atom. An O₂ molecule has two oxygen atoms, so it takes two split water molecules to make one O₂. For 6 O₂: 6 × 2 = 12 water molecules. The splitting of water by light energy is called photolysis. (A) counts one water molecule per O₂ and forgets that each O₂ holds two oxygen atoms. Do not take the count from the summary equation for photosynthesis: it gives the net water used overall, not the water split in the light-dependent reactions. (B) has the right count and the wrong source: the O₂ released comes from the water split in the light-dependent reactions, while carbon dioxide is taken up by the Calvin cycle. (C) divides by 2 where it should multiply.

Question 12. (a) Explain why the Calvin cycle stops soon after a plant is moved into the dark. (b) Use the second law of thermodynamics to explain why a cell needs a constant supply of energy.

Answer.

(a) The Calvin cycle runs on the ATP and NADPH that the light-dependent reactions make. In the dark, the light-dependent reactions stop, so the supply of ATP and NADPH runs out, and the Calvin cycle stops soon after. The name "dark reactions", which some notes still use, means only that the Calvin cycle does not use light directly. It still depends on the light reactions.

(b) The second law of thermodynamics says every energy transfer or transformation increases the entropy of the universe. Some usable energy is lost, mostly as heat, each time energy changes form, so no conversion in a cell is 100% efficient. A cell is a highly ordered system that keeps losing usable energy as it works, so it needs a steady input: light for a plant, food for an animal.

Molecular genetics: questions 13 to 18

Question 13. A sample of double-stranded DNA is 28% adenine. What are the percentages of the other three bases?

• (A) thymine 28%, guanine 28%, cytosine 16%
• (B) thymine 28%, guanine 22%, cytosine 22%
• (C) thymine 22%, guanine 28%, cytosine 22%
• (D) thymine 28%, guanine 44%, cytosine 44%

Answer: (B). In double-stranded DNA, adenine pairs with thymine and guanine pairs with cytosine, so A = T and G = C. Thymine is 28%. Together, A and T make 56%, which leaves 100% - 56% = 44% for G and C, split evenly: 22% each. (A) pairs adenine with guanine. (C) pairs adenine with guanine and thymine with cytosine. (D) forgets to split the 44% between two bases, and its total comes to 144%.

Question 14. At a replication fork, name the enzyme that does each job: (a) unwinds the double helix; (b) makes the short RNA primer; (c) adds DNA nucleotides to the new strand; (d) seals the gaps between Okazaki fragments. Then explain why the lagging strand is made in fragments.

Answer.

• (a) Helicase.
• (b) Primase.
• (c) DNA polymerase (DNA polymerase III in bacteria).
• (d) DNA ligase, which forms the phosphodiester bond that closes each gap.

DNA polymerase can add nucleotides only in the 5' to 3' direction, and the two template strands run in opposite directions. On one template, the new strand can grow continuously toward the fork; that is the leading strand. On the other, the new strand has to grow away from the fork, so it is made in short pieces, each started from a new primer as the fork opens. Those pieces are the Okazaki fragments, and ligase joins them once the RNA primers have been replaced with DNA.

Question 15. Bacteria are grown for many generations with heavy nitrogen (¹⁵N) until all their DNA is heavy. They are then moved to light nitrogen (¹⁴N) and divide three times. Assume every DNA molecule is copied once per division. After the third division there are 8 DNA molecules for each original one. If replication is semi-conservative, what do those 8 molecules look like when spun in a density gradient?

• (A) 4 hybrid (half heavy, half light) and 4 light
• (B) all 8 at a single density between heavy and light
• (C) 2 hybrid and 6 light, with no fully heavy molecules
• (D) 1 heavy and 7 light

Answer: (C). In semi-conservative replication, each new molecule keeps one old strand and builds one new strand. Follow the two original heavy strands. After the first division, both molecules are hybrid: one heavy strand, one light strand. After that, the two heavy strands are never lost and never joined to each other again, so there are always exactly 2 hybrid molecules, and every other molecule is light. After 3 divisions: 2 hybrid and 8 - 2 = 6 light, so 25% hybrid and 75% light. (A) has 4 hybrid molecules, but the two original heavy strands can sit in at most 2 molecules, so there are never more than 2 hybrid. (B) is what the dispersive model predicts, with old and new DNA mixed through every molecule. (D) is the conservative model, where the original double helix stays intact.

Question 16. A DNA template strand reads 3'-TAC GGA CTT ATC-5'. Write the mRNA made from it and the coding (non-template) strand of the DNA, then use the codon table to give the amino acid sequence.

Answer.

• mRNA: 5'-AUG CCU GAA UAG-3'.
• Coding strand: 5'-ATG CCT GAA TAG-3'.
• Amino acids: methionine, proline, glutamic acid, then the stop codon UAG. The result is a chain of three amino acids: Met-Pro-Glu.

The mRNA is complementary to the template and runs the opposite way, so read the template 3' to 5' and pair each base: T with A, A with U, C with G and G with C. The coding strand has the same sequence as the mRNA with T in place of U, which is a quick check on the work. AUG is both the start codon and the codon for methionine. UAG codes for no amino acid, so translation stops there.

Question 17. The mRNA 5'-AUG GCA GAA ACU UGG-3' codes for Met-Ala-Glu-Thr-Trp. Classify each change to the third codon, GAA, and give the new amino acid sequence:

• (a) GAA becomes GAG.
• (b) GAA becomes UAA.
• (c) GAA becomes GUA.
• (d) The G of GAA is deleted.

Answer.

• (a) Silent mutation. GAG also codes for glutamic acid, so the protein is unchanged: Met-Ala-Glu-Thr-Trp.
• (b) Nonsense mutation. UAA is a stop codon, so the chain ends early: Met-Ala.
• (c) Missense mutation. GUA codes for valine, so one amino acid changes: Met-Ala-Val-Thr-Trp.
• (d) Frameshift mutation. Removing one base shifts every codon after it. The mRNA now reads AUG GCA AAA CUU GG, which gives Met-Ala-Lys-Leu, with two leftover bases that do not make a full codon.

A substitution changes one codon at most. A deletion or insertion of one base changes the reading frame, so every codon from that point on is read differently, which is why a frameshift usually leaves the protein nonfunctional. The silent change in (a) is in the third base of the codon, where many substitutions leave the amino acid the same.

Question 18. In E. coli, the lac operon carries genes for using lactose. Assume no glucose is available, so only the lac repressor decides whether the genes are transcribed. Say whether the genes are transcribed in each case, and why:

• (i) lactose absent, normal cell;
• (ii) lactose present, normal cell;
• (iii) lactose absent, in a mutant cell whose repressor cannot bind the operator;
• (iv) lactose present, in the same mutant cell.

Answer.

• (i) Not transcribed. The repressor is bound to the operator and blocks RNA polymerase.
• (ii) Transcribed. Lactose is converted to allolactose, which binds the repressor and changes its shape, so the repressor lets go of the operator.
• (iii) Transcribed. Nothing holds the operator, so the genes are read even with no lactose present.
• (iv) Transcribed, for the same reason as (iii).

The normal cell makes the lactose enzymes only when lactose is there to use, which saves energy. The mutant in (iii) and (iv) has lost that control and makes the enzymes all the time. Glucose is left out on purpose: when glucose is present, a second control lowers transcription, and that is a separate question.

Homeostasis: questions 19 to 23

Question 19. Blood glucose rises after a meal. Identify the stimulus, the hormone released, the cells that release it and the response. Then explain why this is negative feedback.

Answer.

• Stimulus: blood glucose rising above its normal level.
• Hormone: insulin.
• Released by: the beta cells of the pancreas.
• Response: body cells take up more glucose, and some of it is stored as glycogen, so blood glucose falls back toward normal.

This is negative feedback because the response works against the original change: glucose went up, and the response brings it down. As glucose falls, insulin release slows down. When glucose drops too low, between meals for example, the alpha cells of the pancreas release glucagon, which raises it. The two hormones push in opposite directions around the same set point.

Question 20. A neuron rests at -70 mV, and its threshold is -55 mV. One stimulus takes the membrane to -62 mV. A second, separate stimulus takes it to -50 mV. Which statement is correct?

• (A) Only the second stimulus triggers an action potential. Na⁺ enters the cell during depolarisation, and K⁺ leaves during repolarisation.
• (B) Both stimuli trigger action potentials, and the first one is smaller.
• (C) Only the second stimulus triggers an action potential. K⁺ enters the cell during depolarisation, and Na⁺ leaves during repolarisation.
• (D) Neither stimulus triggers an action potential, because the membrane stays below 0 mV both times.

Answer: (A). An action potential starts only when the membrane reaches threshold. -62 mV is below the -55 mV threshold, so the first stimulus does not trigger one. -50 mV is past it, so the second does. Once threshold is reached, voltage-gated sodium channels open, Na⁺ flows in and the inside becomes positive, to about +40 mV. Then K⁺ flows out and the membrane returns toward its negative resting value. (B) ignores the threshold. An action potential is all-or-nothing: a stimulus below threshold gives no action potential at all, and one that reaches threshold gives a full one. (C) swaps the two ions. (D) misreads the rule: the trigger is reaching -55 mV, and the membrane never has to cross 0 mV for that.

Question 21. Name the three steps of urine formation in the nephron, in order, and say what each step does to the fluid in the tubule (the filtrate).

Answer.

• Glomerular filtration: blood pressure pushes water and small solutes out of the blood in the glomerulus and into the nephron. Blood cells and large proteins stay in the blood.
• Tubular reabsorption: most of the water and useful solutes, such as glucose and many ions, move from the filtrate back into the blood.
• Tubular secretion: extra wastes and ions move from the blood into the filtrate.

What is left at the end is urine. Reabsorption and secretion run in opposite directions, so it helps to name the direction each time: reabsorption returns things to the blood, and secretion adds things to the filtrate. Adjusting how much water is reabsorbed is how the kidney helps keep the body's water balance.

Question 22. Testosterone levels in the blood rise and act on the hypothalamus and the anterior pituitary. What happens next?

• (A) Release of GnRH, FSH and LH increases. This is positive feedback.
• (B) Release of GnRH, FSH and LH is inhibited. This is negative feedback.
• (C) GnRH release rises while FSH and LH fall. This is negative feedback.
• (D) Nothing changes, because testosterone acts only on the testes.

Answer: (B). GnRH from the hypothalamus tells the anterior pituitary to release FSH and LH, and LH tells the testes to make testosterone. When testosterone rises, it acts back on the hypothalamus and anterior pituitary to inhibit the release of GnRH, FSH and LH. Less LH means less testosterone, so the level is held near its set point. That is negative feedback. (A) runs the loop the wrong way: if testosterone raised its own trigger hormones, the level would keep climbing. (C) splits the hormones, but rising testosterone inhibits all three. (D) ignores the effect on the hypothalamus and the anterior pituitary, which is the whole feedback loop.

Question 23. Which statement about feedback in homeostasis is correct?

• (A) Positive feedback is the main way the body keeps a variable near its set point.
• (B) In negative feedback, the response pushes the variable further in the direction of the original change.
• (C) In mammals, homeostasis is controlled mainly by the digestive and excretory systems.
• (D) Negative feedback reverses a change and moves the variable back toward its set point, while positive feedback keeps the change going in the same direction.

Answer: (D). Homeostasis is maintained by negative feedback loops: a change is detected, and the response opposes it. Positive feedback works the other way. It keeps the change going, and can speed it up, until some end point is reached. OpenStax names blood clotting as an example, where each activated clotting factor activates more. OpenStax notes that few such loops exist in animal bodies, and that they push the organism further out of homeostasis. (A) gives positive feedback the job that negative feedback does. (B) describes positive feedback and calls it negative. (C) names the wrong systems: in mammals, homeostasis is controlled by the nervous and endocrine systems.

Population dynamics: questions 24 to 28

Question 24. A population grows exponentially with r = 0.20 per day, using dN/dt = rN. What is its growth rate when N = 500 and when N = 2000?

• (A) 100 per day at both sizes
• (B) 0.20 per day at both sizes
• (C) 100 per day at N = 500, and 400 per day at N = 2000
• (D) 2500 per day at N = 500, and 10 000 per day at N = 2000

Answer: (C). Substitute into dN/dt = rN. At N = 500: 0.20 × 500 = 100 individuals per day. At N = 2000: 0.20 × 2000 = 400 individuals per day. The per capita rate, r, stays the same, but the number of new individuals each day grows with the population, which is why an exponential curve gets steeper and steeper. (A) calculates the first rate and assumes it stays fixed. (B) gives r itself, the per capita rate, rather than the population's growth rate. (D) divides N by r instead of multiplying.

Question 25. A population follows logistic growth with rmax = 0.30 per day and a carrying capacity K = 1000. Use dN/dt = rmax × N × (K - N) / K to find the growth rate at N = 100, 500, 900 and 1000. Describe the shape of the growth curve.

Answer. 27, 75, 27 and 0 individuals per day.

• N = 100: 0.30 × 100 × (1000 - 100) / 1000 = 0.30 × 100 × 0.9 = 27 per day.
• N = 500: 0.30 × 500 × (1000 - 500) / 1000 = 0.30 × 500 × 0.5 = 75 per day.
• N = 900: 0.30 × 900 × (1000 - 900) / 1000 = 0.30 × 900 × 0.1 = 27 per day.
• N = 1000: 0.30 × 1000 × 0 / 1000 = 0 per day.

The growth rate is highest at half the carrying capacity, K/2 = 500, and falls to zero when the population reaches K. Plotted over time, the population makes an S-shaped (sigmoid) curve: slow at first, fastest in the middle, then levelling off at the carrying capacity as resources become scarce. Compare N = 100 with exponential growth at r = 0.30: 0.30 × 100 = 30 per day. Even at a tenth of K, the logistic model is already slightly slower.

Question 26. In a lake, 60 fish are caught, tagged and released. A week later, a second sample of 45 fish includes 9 tagged fish. Assume the tagged fish mixed evenly through the population, no tags were lost, and no fish were born, died, arrived or left between the two samples. Estimate the number of fish in the lake.

• (A) 300
• (B) 2700
• (C) 105
• (D) 12

Answer: (A). If the tagged fish mixed evenly, the share of tagged fish in the second sample matches their share in the whole lake. In the sample, 9 of 45 are tagged, which is 1 in 5. So the 60 tagged fish are about 1 in 5 of the population: N = 60 × 45 ÷ 9 = 300 fish. (B) multiplies 60 by 45 and never divides by the 9 recaptured. (C) adds the two samples together. (D) turns the proportion upside down: 60 × 9 ÷ 45 = 12, which is fewer fish than the 60 already tagged, so it cannot be right. The assumptions in the question matter. If tagged fish were easier to catch the second time, there would be more of them in the sample, and the estimate would come out too low.

Question 27. Classify each factor as density-dependent or density-independent, and give the reason: (a) a flu outbreak spreading through a crowded herd; (b) a late spring frost; (c) competition among birds for nest sites; (d) a wildfire.

Answer.

• (a) Density-dependent. Disease spreads more easily when animals are crowded together, so its effect grows with density.
• (b) Density-independent. A frost kills about the same share of a population whether it is sparse or crowded.
• (c) Density-dependent. The more birds there are, the more of them compete for the same number of nest sites.
• (d) Density-independent. A fire kills animals in its path regardless of how many live there.

The test is whether the factor's effect on the birth or death rate changes with the population's density. Most density-dependent factors are biological, such as predation, competition, waste build-up and disease. Most density-independent factors are physical or chemical, such as weather, natural disasters and pollution.

Question 28. In a grassland, producers make 20 000 kJ of energy per square metre per year, primary consumers make 2600 kJ/m²/year, and secondary consumers make 260 kJ/m²/year. What is the trophic level transfer efficiency from producers to primary consumers, and from primary to secondary consumers?

• (A) 13.0% and 1.3%
• (B) 87.0% and 90.0%
• (C) 10.0% at both steps
• (D) 13.0% and 10.0%

Answer: (D). TLTE = (production at the present trophic level ÷ production at the previous trophic level) × 100. From producers to primary consumers: 2600 ÷ 20 000 × 100 = 13.0%. From primary to secondary consumers: 260 ÷ 2600 × 100 = 10.0%. (A) compares the secondary consumers with the producers rather than with the level just below them. (B) gives the share of energy that does not pass to the next level, 100% - 13.0% and 100% - 10.0%. (C) assumes a 10% rule instead of using the numbers given. Because so little energy passes up each step, there is not enough left after a few levels to support another one. That low efficiency is usually the main factor limiting how long a food chain can be.

Answer key at a glance

Use this table to mark the quiz by strand. The skill column is what a wrong answer points to.

What to do with your score

Count the wrong answers by strand, not in total. Twenty-eight questions spread over five strands is five or six per strand, which is enough to show a pattern and too few to grade anyone.

A few wrong in each strand. The biology is probably fine. Go back to each miss and find the word or step that broke: an ion going the wrong way, a strand read in the wrong direction, a ratio turned upside down. Those are the cheapest marks to win back.

One strand clearly weaker. That is the useful result, and it is what the quiz is for. Take that strand back to the class notes and the textbook chapter, then redo its questions a day later without looking at the working.

Metabolic processes weak, especially questions 7 to 9. Learn where each stage happens and what it makes before learning the details inside it. Most respiration questions get easier once the map is fixed: cytoplasm, matrix, inner membrane.

Molecular genetics weak, especially questions 14, 16 and 17. Questions 14 and 16 depend on reading strands 5' to 3' or 3' to 5'. In Question 17, parts (a) to (c) ask you to classify a single-base substitution from the codon table, and only part (d) depends on keeping the reading frame. A reversed strand or a shifted frame gives a wrong answer with every later step done correctly.

Homeostasis weak, especially questions 19, 22 and 23. Draw each loop as a flow chart: stimulus, sensor, signal, response, and then the arrow back to the start. The arrow back is the feedback, and its direction decides whether it is negative or positive.

Nearly everything wrong. Stop doing practice tests and find out what is missing underneath. SBI3U is the prerequisite for SBI4U, so start with the Grade 11 notes. If the trouble is how to prepare rather than the biology, our guide on how to study for a test covers the method. Students taking chemistry too can use the same format in Genie's SCH4U practice quiz.

Where the course goes next

Ontario's diploma requires 2 credits in science, and no biology course is compulsory. Whether a student needs SBI4U depends on the university programs they are aiming for, so check each program's own admission page for its exact list. Applications to Ontario universities go through OUAC, and our OUAC application guide explains the process.

Our SBI4U explainer covers what programs ask for and how SBI4U compares with SCH4U and SPH4U, so this page does not repeat it.

When a teacher helps more than another practice test

A practice quiz shows which strand is short. It cannot show why, and the why decides what to do next.

A student who misses Question 20 might not know what threshold means. Or they might know it and swap sodium and potassium at the last line. A student who misses Question 16 might not understand transcription, or might understand it and read the template in the wrong direction. Those look the same on a mark sheet and need different teaching.

A teacher who reads the working can see which it was. Genie's online biology tutor page says a teacher can work through "lab write-ups, biological diagrams and Punnett squares, past unit tests, and exam review using the student's own course material". Genie works only with Canadian Certified Teachers, licensed by a provincial governing body such as the Ontario College of Teachers. Each teacher's profile lists the teaching certifications that teacher has on file, so families can check them before choosing.

A certified biology teacher can also chart which courses a student needs to reach a target program, then focus the teaching on the subject areas that get them there. Our page on what a provincially certified teacher is explains what that certification means.