Free Interactive Activity
Enzyme Activity Lab
Catalase in liver cells breaks hydrogen peroxide into water and oxygen. Pick one factor to investigate, write a hypothesis, then run trials and compare the oxygen rate. SBI4U, Metabolic Processes.
Bench
Substrate only binds if it hits the active site (the notch). Bound substrate is converted and released as O₂, and new substrate is added so the concentration stays constant.
Independent: temperature. Dependent: O₂ produced per minute. Controlled: pH 7, substrate concentration 50 mM, enzyme concentration 5 units, inhibitor concentration 0 mM.
Controls
Hypothesis and data
| Trial | Temperature (°C) | O₂ (mL/min) |
|---|---|---|
| Run a trial to record data. | ||
Analyse your results
- At which temperature is the rate highest? Explain why it drops so sharply above that point.
- Compare 0 °C with 80 °C. Both give a low rate. Would the enzyme recover if returned to 37 °C in each case?
Check the science
Warming increases collisions and kinetic energy, so the rate climbs until the optimum. Beyond it, hydrogen and ionic bonds holding the tertiary structure break, the active site loses its shape and the enzyme denatures, which is not reversible. Cold only slows molecules; the enzyme keeps its shape and recovers.
Learning objectives
- Design a fair test: change one factor, keep the rest controlled.
- Predict and explain how temperature, pH, substrate, enzyme, and inhibitors affect rate.
- Read a rate-vs-factor curve and identify the optimum, Vmax, and Km.
- Distinguish competitive from non-competitive inhibition using data.
How the model works
- Catalase: 2 H₂O₂ → 2 H₂O + O₂. Rate is modelled with Q₁₀, a pH bell curve, Michaelis–Menten saturation, and denaturation.
- Denaturation: above ~44 °C the fold breaks (red enzymes), and it is not reversible.
- Inhibitors: competitive blocks the active site (outcompeted by substrate); non-competitive binds elsewhere and lowers Vmax.
Try this
- Temperature: sweep 0 → 80 °C. Find the optimum and watch enzymes turn red above it.
- Substrate: notice the curve level off at saturation — that plateau is Vmax.
- Inhibitor: run competitive vs non-competitive at 10 and 90 mM substrate and compare Vmax.
This is a teaching model of enzyme kinetics, not a full molecular-dynamics simulation. Want AI-built lessons, quizzes, and chatbots around it?