Difficulty: Intermediate | Prerequisites: Enzyme Structure and Function study notes (Doc 1), comfort with slopes and simple graphs.
Once you know what enzymes are and how they work, the next question is: how fast do they work, and what controls that speed? This topic covers how to measure enzyme activity (reaction rate), how enzyme and substrate concentrations affect that rate, and how inhibitors slow enzymes down. You will also learn the critical distinction between competitive and noncompetitive inhibition, which is one of the most commonly tested concepts in introductory biology. This material builds directly on the enzyme-substrate complex model from the fundamentals.
Enzyme activity is measured as reaction rate (change in substrate or product concentration over time). Increasing enzyme concentration increases rate proportionally (when substrate is abundant). Increasing substrate concentration increases rate up to a plateau (enzyme saturation). Inhibitors reduce enzyme activity: competitive inhibitors bind the active site and can be outcompeted by more substrate; noncompetitive inhibitors bind elsewhere and cannot.
Reaction rate (enzyme activity)
The change in concentration of substrate or product per unit time. In the lab, this is measured as the change in absorbance per minute (AU/min). Think of it as how fast the enzyme is working.
Initial reaction rate (initial rate)
The reaction rate measured at the very beginning of the reaction, before substrate depletion or product accumulation begins to slow things down. This is the slope of the absorbance-vs-time graph during its linear phase.
Absorbance (AU)
A spectrophotometer reading that is proportional to the concentration of a coloured molecule in solution. More coloured product means higher absorbance.
Spectrophotometry
A technique that measures the amount of light absorbed by a solution at a specific wavelength. Used here to track the formation of coloured products over time.
Serial dilution
A stepwise dilution technique in which each step reduces concentration by a fixed factor (commonly 1/2). Starting from stock, successive two-fold dilutions give concentrations of 1/2, 1/4, 1/8, 1/16, and so on.
Inhibitor
A chemical that decreases the rate of an enzymatic reaction. Inhibitors are categorised by how they interact with the enzyme.
Competitive inhibitor
An inhibitor that resembles the substrate and binds to the enzyme's active site, blocking substrate access. Its effect can be overcome by increasing substrate concentration. Think of it as a molecule that pretends to be the substrate.
Noncompetitive inhibitor
An inhibitor that binds to a site other than the active site, reducing the enzyme's ability to process substrate without affecting substrate binding. Its effect cannot be overcome by adding more substrate.
Percent inhibition
The percentage reduction in enzyme activity caused by an inhibitor, calculated from the rates with and without the inhibitor present.
Enzyme saturation
The state in which every enzyme molecule is occupied by substrate. At saturation, adding more substrate does not increase the reaction rate because there are no free enzyme molecules available.
Enzyme activity = reaction rate = change in concentration of substrate or product over time.
In this lab, we measure the formation of coloured products using a spectrophotometer set to 525 nm.
Absorbance increases as more coloured product forms. The rate of absorbance increase over time tells us the reaction rate.
Initial rate is measured from the linear (straight-line) portion of the absorbance-vs-time graph, typically the first 2 to 3 minutes.
The slope of this linear region = initial rate, expressed in AU/min.
As the reaction proceeds, substrate is consumed and product accumulates. Both of these slow the reaction.
Initial rate captures the enzyme's activity before these complicating factors set in.
Comparing initial rates across conditions (different concentrations, with or without inhibitor) gives a clean measure of each variable's effect.
When substrate is in excess, increasing enzyme concentration increases the initial rate roughly proportionally. Double the enzyme, roughly double the rate.
This is because more enzyme molecules means more active sites available to bind substrate simultaneously.
If the rate levels off at high enzyme concentration, it would suggest substrate has become limiting: all available substrate is already being processed.
Increasing substrate concentration increases the initial rate, but the relationship is not linear at higher concentrations.
At low substrate concentrations, many enzyme molecules are unoccupied, so adding more substrate gives each enzyme molecule something to do. The rate rises steeply.
At high substrate concentrations, the enzyme becomes saturated: every active site is occupied. Adding more substrate cannot speed things up because there are no free enzyme molecules.
The resulting graph (rate vs. substrate concentration) is a curve that rises steeply at first, then levels off to a plateau. This plateau is called Vmax (maximum velocity).
A competitive inhibitor has a molecular structure similar to the substrate.
It binds to the active site, blocking the substrate from entering.
The binding is reversible: the inhibitor does not permanently attach.
The key experimental signature: percent inhibition decreases when substrate concentration increases, because more substrate molecules outcompete the inhibitor for the active site.
At very high substrate concentrations, a competitive inhibitor can be entirely overcome.
Example from the lab: para-hydroxybenzoic acid (PHBA) has a structure similar to catechol and is predicted to be a competitive inhibitor of polyphenoloxidase.
A noncompetitive inhibitor does not bind to the active site. It binds to a different location on the enzyme.
It reduces the enzyme's catalytic ability without preventing substrate binding.
Mechanisms vary: some remove a cofactor, some change the enzyme's shape, some reduce its flexibility.
The key experimental signature: percent inhibition remains constant regardless of substrate concentration. Adding more substrate does not help because the problem is not competition for the active site.
Example from the lab: potassium arsenite disrupts disulfide bonds in the protein's tertiary structure. This is a structural effect, not an active-site competition, so it is predicted to be noncompetitive.
Run the reaction at several different substrate concentrations, with and without the inhibitor.
Calculate percent inhibition at each substrate concentration.
If percent inhibition varies with substrate concentration (lower at high substrate), the inhibitor is competitive.
If percent inhibition stays the same across all substrate concentrations, the inhibitor is noncompetitive.
Percent inhibition:
\text{percent inhibition} = \frac{(\text{rate without inhibitor}) - (\text{rate with inhibitor})}{\text{rate without inhibitor}} \times 100Initial reaction rate:
The slope of the absorbance-vs-time graph during its linear phase.
\text{initial rate (AU/min)} = \frac{\Delta \text{Absorbance}}{\Delta \text{Time (min)}}Serial dilution (two-fold):
Each step: 1 ml of previous solution + 1 ml diluent = 1/2 concentration. After n steps, the concentration is (1/2)^n of the stock.
Drug design relies heavily on enzyme inhibition. Many pharmaceuticals work as competitive or noncompetitive inhibitors of specific enzymes in disease pathways. For instance, statins (cholesterol-lowering drugs) are competitive inhibitors of HMG-CoA reductase. Understanding how inhibitor type determines whether the effect can be overcome by substrate concentration is central to pharmacology and dosage design.
Students often think that competitive inhibitors permanently block the active site. They do not. The binding is reversible, and the inhibitor can be displaced by increasing substrate concentration.
A frequent mistake is assuming that noncompetitive inhibitors bind to the active site from a different angle. They bind to a completely different region of the enzyme, away from the active site.
Students sometimes confuse "reaction rate" with "total product formed." Rate is the speed at a given moment (the slope), not the cumulative amount.
Some students believe that doubling substrate always doubles the rate. This is only true at low substrate concentrations. At high concentrations, the enzyme is saturated and the rate plateaus.
⚠️ The percent inhibition formula is a near certainty on exams. Know it cold, and be able to apply it to a set of rate data.
⚠️ You must be able to state the hallmark difference between competitive and noncompetitive inhibition: competitive inhibition varies with substrate concentration; noncompetitive does not.
⚠️ Know why we measure initial rates rather than rates later in the reaction. This comes up in both exam questions and lab report requirements.
⚠️ Be prepared to interpret an absorbance-vs-time graph: identify the linear phase, calculate its slope, and explain why the curve levels off.
True or False: The initial reaction rate is measured after the reaction has been running for 10 minutes.
Fill in the blank: A competitive inhibitor binds to the enzyme's ______.
True or False: Noncompetitive inhibition can be overcome by adding more substrate.
Fill in the blank: In a two-fold serial dilution, after three steps the concentration is ______ of the original.
True or False: If the rate-vs-substrate graph levels off at high substrate concentrations, the enzyme has become saturated.
Answers: 1. False (measured during the initial linear phase). 2. Active site. 3. False. 4. 1/8. 5. True.
Q: An enzyme reaction is run with and without an inhibitor at three different substrate concentrations. The percent inhibition is 60%, 45%, and 30% at low, medium, and high substrate concentrations respectively. What type of inhibitor is this? Explain.
A: This is a competitive inhibitor. The percent inhibition decreases as substrate concentration increases, which means the substrate is outcompeting the inhibitor for the active site. This is the hallmark of competitive inhibition.
Q: Why does the absorbance-vs-time curve for an enzyme reaction eventually level off, even when enzyme is still present?
A: The reaction slows because substrate is being consumed. As substrate concentration falls, fewer enzyme-substrate complexes form per unit time, so the rate decreases. Product accumulation may also contribute. Eventually, when substrate is depleted (or equilibrium is reached), the reaction effectively stops.
Q: You double the enzyme concentration while keeping substrate concentration constant. Under what conditions would you expect the initial rate to roughly double?
A: When substrate is present in excess relative to enzyme. Under these conditions, adding more enzyme means more active sites are available, and there is enough substrate to keep them all occupied. If substrate were limiting, doubling enzyme would produce less than a doubling of rate.
Q: Calculate the percent inhibition if the rate without inhibitor is 0.12 AU/min and the rate with inhibitor is 0.04 AU/min.
A: Percent inhibition = (0.12 - 0.04) / 0.12 x 100 = 66.7%.
Q: What would a levelling off of the initial-rate-vs-substrate-concentration graph signify?
A: Enzyme saturation. All enzyme molecules are occupied by substrate, so adding more substrate cannot increase the rate. The rate at this plateau is Vmax.
This material connects to the polyphenoloxidase lab experiments (Doc 3), where you apply all of these concepts hands-on: measuring initial rates with a spectrophotometer, running serial dilutions, and testing specific inhibitors. The competitive-vs-noncompetitive framework also appears in pharmacology, metabolic regulation, and advanced biochemistry courses. Understanding enzyme saturation here lays the groundwork for Michaelis-Menten kinetics if you continue in biochemistry.
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