Lesson 3 of 4 / Enzymes and investigations
Measure a rate, explain a plateau
What becomes limiting when more substrate no longer increases rate?
In this lesson: Explain concentration effects and calculate rates using products or substrate loss.
About 7 min
The key ideaAt high substrate concentration, occupied active sites limit rate; measure initial slopes or a justified reciprocal-time proxy.
Explore the idea
Find the limiting factor
At low substrate, additional substrate increases productive encounters. Near the plateau, available active sites and cycle speed limit rate.
Original qualitative model, not a fitted experimental curve. More enzyme changes rate; equal starting substrate can still give the same final product amount.
Try a product-rate or endpoint calculation
Explore the idea
Make the rate measure explicit
From 0 to 40 s: (8 - 0)/(40 - 0) = 0.20 cm3 s-1. From 40 to 60 s the mean rate is lower: (10 - 8)/20 = 0.10 cm3 s-1.
Original data for calculation practice. The catalase values are 0, 4, 8 and 10 cm3 at 0, 20, 40 and 60 s. Real investigations need calibrated measurements, controlled conditions and repeats; this does not certify practical skill.
Explanation
At fixed enzyme concentration, more substrate initially increases enzyme-substrate complex formation. At sufficiently high substrate concentration, most active sites are occupied much of the time. The enzyme population's catalytic turnover limits the maximum rate, producing a plateau.
Increasing enzyme concentration raises the number of active sites and can raise rate if enough substrate is available. The relationship need not remain proportional if substrate becomes limiting. Specify what is held constant rather than saying concentration always increases rate.
For catalase, record oxygen volume over time and estimate the initial slope in cm3 per minute. A gas syringe gives volume more directly than counting bubbles of variable size. Later slopes may decrease as substrate is depleted or products accumulate, even without enzyme denaturation.
For amylase, sample the reaction into iodine at fixed intervals until starch is no longer detected. If the initial starch amount and endpoint are standardised, 1/time is a relative-rate proxy. It is not automatically an absolute mass-per-time rate, and the endpoint interval limits timing precision.
Step by step
- 1
Define the rate measure
State product volume per time or a relative reciprocal-time measure.
- 2
Use comparable intervals
Prefer the initial linear region for product formation.
- 3
Identify the limiting factor
Explain the plateau using occupied active sites.
Worked example
Work through the evidence
Catalase produces 12 cm3 oxygen in the initial 30 s. Another assay gives 18 cm3 in 30 s. Calculate and compare initial average rates.
One way to explain it
The rates are 24 and 36 cm3 min-1. The second is 1.5 times the first, a 50% increase, over the stated initial interval.
Why this answer works
- 30 s is 0.5 min.
- The percentage increase uses the first rate as the denominator.
Is this true? "A plateau means the substrate molecules have stopped moving."
Substrates still move; enzyme availability and turnover limit further increases in rate.