IB Biology: Enzymes and Kinetics — Tutor Lesson Plan (activation energy, Michaelis-Menten, Q10)

2026-11-05

This chapter is not tested on definitions but on why the curve flattens. A student who cannot explain saturation cannot answer most of the questions.

Objectives — the student can explain catalysis through activation energy, describe saturation and affinity with Vmax and Km, and use Q10 to quantify the effect of a ten degree rise.

01 Prerequisites (3-minute check)

If any of these fails, fix it before the lesson starts.

Skill	Quick check	Pass mark
Percentage	A rate rising from 12 to 24 is an increase of what percentage?	100 percent
Protein structure	Which bonds hold the three dimensional shape?	Hydrogen bonds, ionic bonds, hydrophobic interactions
Graph reading	What does a flattening curve mean?	One factor has become limiting

02 Lesson flow (60 minutes)

Time	Stage	Tutor does	Student does
0-8 min	What an enzyme is	Three features of a protein catalyst	State that it is not used up
8-20 min	Activation energy	Draw both energy profiles	Show lower Ea, unchanged delta G
20-34 min	Active site and fit	Lock and key against induced fit	Explain specificity
34-46 min	Michaelis-Menten	v = Vmax S over (Km plus S)	Get 75 and 66.7
46-54 min	Temperature and pH	Q10 and denaturation	Get Q10 = 2.0
54-60 min	Exit check	Three questions marked	Finish alone

03 Teaching script (read it out)

Scene 1: Enzymes lower activation energy, not delta G

Draw two profiles: a high peak without the enzyme and a low peak with it, both starting and ending at the same energy. So the enzyme changes how fast equilibrium is reached, not where it lies, and delta G is untouched. Students often write that the enzyme supplies energy, which loses the mark.

Student explains what the enzyme changes and what it does not.

Scene 2: Michaelis-Menten: one equation for saturation

v = Vmax times substrate concentration divided by Km plus that concentration. With Vmax 100 and Km 5: at 5 the rate is 50, at 15 the rate is 1500 over 20 = 75, and at 10 the rate is about 66.7.
Doubling the substrate from 5 to 10 only moves the rate from 50 to 66.7, far less than double. That is the arithmetic behind the flattening curve.
Most common misconception: that adding substrate always speeds the reaction up.

Student explains whether a smaller Km means tighter or looser binding.

Scene 3: Temperature: Q10 and denaturation are two stories

Q10 is the rate after a ten degree rise divided by the rate before it, so 12 to 24 gives 24 over 12 = 2.0.
Above the optimum, hydrogen bonds break, the active site distorts and the rate falls to zero irreversibly. Cold only slows molecular motion, so warming restores activity. A fridge does not kill an enzyme, boiling water does.

Student explains why cold and heat act differently.

04 Question types and worked steps

Type	Example	Steps	Answer
Rate calculation	Vmax 100, Km 5, substrate 15, find v	Substitute into Vmax S over (Km plus S)	75 micromol per minute
Q10 calculation	Rate 12 at 18 degrees, 24 at 28 degrees	Later rate divided by earlier rate	2.0
Curve explanation	Why the rate curve flattens	Active sites saturated, Vmax is the ceiling	Every enzyme occupied
Inhibitor comparison	Effect of a competitive inhibitor on Km and Vmax	Km rises, Vmax unchanged	Overcome by more substrate

Sanity check (teach this): At any substrate concentration the rate stays below Vmax, and 75 is below 100. Km is the concentration giving half of Vmax, so Km 5 must give a rate of 50, which it does. Q10 normally sits between 2 and 3, so 2.0 is plausible and a value of 10 would signal a unit error.

05 Independent practice (3 questions)

Question	Answer	What to watch
Vmax 80, Km 4, substrate 12, find v	60 micromol per minute	Formula written before substitution
A rate rises from 15 to 33 when temperature goes from 18 to 28, find Q10	2.2	Same units used for both rates
Explain why the rate falls to almost zero at 70 degrees	Heat breaks hydrogen bonds, the active site distorts and denaturation is irreversible	The words denaturation and irreversible both appear

06 Common mistakes and what to say

Mistake	Student writes	Say this
Believing the enzyme is used up	The enzyme runs out, that is why the rate falls	Say: an enzyme is a catalyst and is reused, the rate falls because substrate runs low or the enzyme denatures.
Calling cold denaturation	Putting it in the fridge kills the enzyme	Say: cold only lowers kinetic energy, warming restores it, denaturation means the shape is broken for good.
Reading Km the wrong way round	A large Km means tight binding	Say: Km is the substrate needed for half speed, so a small Km means tight binding.
Assuming substrate always helps	More substrate always means a faster rate	Say: there are only so many active sites, once saturated extra substrate changes nothing.

07 Homework (paper, section and question numbers)

IB biology chapter index — Use the three practice sets linked there for homework checking

08 Exit check (last 5 minutes)

Question	Answer	Decision
Vmax 60, Km 6, substrate 6, find v	30 micromol per minute	One wrong means reteach the equation
Why does heat destroy an enzyme permanently?	Denaturation, the three dimensional active site is destroyed	One wrong means reteach temperature
What happens to Km and Vmax under competitive inhibition?	Km rises, Vmax unchanged	Two wrong means return to act two

09 Tutor FAQ

Must the equation be memorised?

Yes, and it must be used to interpret data, not merely to substitute numbers.

Are Km and Vmax always found experimentally?

Most IB questions supply values or a graph and ask you to read Km at half saturation and Vmax at the plateau.

How does this link to metabolism?

Pathway rate is set by the activity of a key enzyme, so temperature, pH and inhibitors act through that point.

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A TutorZone lesson-prep guide. Written by the TutorZone team; questions and lesson flow are original and are not taken from any school or publisher material.