• Quick retrieval from earlier learning (How enzymes work: active sites, temperature and pH). Students answer on mini whiteboards.
• Show answers together and note who needs support in the main task.

1. Why does a reaction speed up as temperature rises towards the optimum? (The enzyme gets bigger / The active site disappears / Molecules move faster, so enzyme and substrate collide more often / The substrate becomes an enzyme) → Molecules move faster, so enzyme and substrate collide more oftenPrompt: Think about kinetic energy.
2. Enzymes: which statement is NOT true? (Enzymes are not used up in the reactions they speed up. / Cold temperatures denature enzymes permanently. / Enzymes are proteins. / Most human enzymes work best at about 37 °C.) → Cold temperatures denature enzymes permanently.Prompt: Is the effect of cold permanent?
3. A human enzyme has an optimum temperature of about 37 °C. What is the enzyme most likely to be doing at 15 °C? (working slowly, but not denatured / working at or near its fastest rate / denatured and no longer working) → working slowly, but not denaturedPrompt: Cold slows molecules down; it does not change the active site's shape.

• Show a microscope image of cells and ask what students think they are looking at.
• Share the objective: Work out enzyme reaction rates from experimental data. Read the success criteria together.

• Work through the example out loud, saying why you do each step, not just what you do.
• Example: Catalase in liver broke down hydrogen peroxide and released 10 cm³ of oxygen in 20 s. What was the mean rate of reaction, in cm³/s?
• Step 1: 10 ÷ 20 = 0.5
• Step 2: So the answer is: 0.5 cm³/s
• Answer: 0.5 cm³/s
• Pause after each step and ask a student to say what comes next.

• Do these together. Students try each step on whiteboards before you reveal it.
• Use the hints as prompts rather than giving the answer.

1. Catalase in liver broke down hydrogen peroxide and released 5 cm³ of oxygen in 25 s. What was the mean rate of reaction, in cm³/s? → 0.2 cm³/sPrompt: rate = amount of product ÷ time
2. Amylase took 125 s to digest all the starch at pH 7. Calculate the rate using rate = 1000 ÷ time, in s⁻¹. → 8 s⁻¹Prompt: Substitute the time into the formula.

• Students work on the student sheet. Circulate and check the first two answers of every student early.
• Students who finish: write a question of their own for a partner, with the answer.

The 8 questions are on the student sheet below, with answers in the key.

• Two questions on their own. Sort answers into got it / nearly / not yet to plan the next lesson.

1. Amylase took 250 s to digest all the starch at pH 7. Calculate the rate using rate = 1000 ÷ time, in s⁻¹. → 4 s⁻¹Prompt: Substitute the time into the formula.
2. An enzyme was timed digesting the same amount of substrate at different pH values: pH 3: 121 s pH 5: 85 s pH 7: 36 s pH 9: 87 s pH 11: 145 s What is its optimum pH? → 7Prompt: The optimum gives the fastest reaction.

Free online practice with hints and printable worksheets for this skill: eurekami.com/science/grade-9.