How do enzymes speed up reactions, and what controls the rate of an enzyme-catalyzed reaction?
Topic 3.2 Enzyme Catalysis: explain how enzymes lower activation energy and how substrate concentration, enzyme concentration and inhibitors affect the rate of an enzyme-catalyzed reaction.
A focused answer to AP Biology Topic 3.2, covering activation energy, the transition state, saturation, the effect of substrate and enzyme concentration, and competitive versus noncompetitive inhibition, with a worked rate calculation.
Reviewed by: AI editorial process; not yet individually human-reviewed
Have a quick question? Jump to the Q&A page
Jump to a section
What this topic is asking
The College Board (Topic 3.2) wants you to explain how enzymes speed up reactions by lowering activation energy, and to explain how substrate concentration, enzyme concentration, and inhibitors change the rate of an enzyme-catalyzed reaction. Many exam questions give you rate data and ask you to describe, explain or predict from a graph.
Lowering activation energy
A useful image is an energy hill: the products may sit lower than the reactants (energy released), but the molecules must first climb over a hill (activation energy) to get there. An enzyme does not move the start or end points; it lowers the hill.
What controls the rate
Reaction rate depends on how often substrate molecules occupy active sites and are converted to product.
- Substrate concentration. At low concentration, rate increases with substrate because more active sites are occupied. Eventually every active site is busy: the enzyme is saturated, the rate reaches a maximum, and adding more substrate has no further effect.
- Enzyme concentration. Adding enzyme provides more active sites, so the maximum rate is higher (assuming plenty of substrate).
- Temperature and pH. These change the enzyme's shape and therefore its activity; that is the focus of Topic 3.3.
Inhibitors
The key distinction the exam tests is whether adding substrate reverses the inhibition:
- Competitive: more substrate outcompetes the inhibitor for the active site, so a high substrate concentration can restore the maximum rate.
- Noncompetitive: the inhibitor does not compete for the active site, so adding substrate does not help; the maximum rate stays lowered.
Try this
Q1. Explain why the rate of an enzyme-catalyzed reaction plateaus at high substrate concentration. [2 points]
- Cue. The enzyme is saturated, meaning all active sites are occupied, so adding more substrate cannot increase the rate; enzyme availability is now limiting.
Q2. Identify whether the effect of a competitive inhibitor can be reduced by adding more substrate, and justify. [2 points]
- Cue. Yes; substrate and inhibitor compete for the same active site, so more substrate outcompetes the inhibitor and restores rate.
Exam-style practice questions
Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.
AP 2018 (style)4 marksSection II (long FRQ excerpt, data). In an experiment the rate of an enzyme reaction is measured at increasing substrate concentrations. The rate increases steeply at first, then levels off at a maximum. (a) Describe the relationship shown. (b) Explain why the rate levels off. (c) Predict the effect on the curve of adding more enzyme.Show worked answer →
A 4-point describe-explain-predict FRQ on reaction rate.
(a) Describe (1 point): as substrate concentration increases, rate increases and then plateaus at a maximum rate.
(b) Explain (2 points): (1 point) at low substrate concentration, rate rises because more active sites are occupied as substrate increases; (1 point) at high concentration the enzyme is saturated, meaning all active sites are occupied, so adding more substrate cannot increase rate (enzyme availability is now limiting).
(c) Predict (1 point): adding more enzyme provides more active sites, so the plateau (maximum rate) is reached at a higher rate.
Markers reward linking the plateau to saturation of active sites and predicting a higher maximum rate when enzyme is added.
AP 2022 (style)3 marksSection II (short FRQ). A drug binds to the active site of an enzyme and competes with the normal substrate. (a) Identify the type of inhibition. (b) Predict the effect of greatly increasing substrate concentration and justify your prediction.Show worked answer →
A 3-point identify-predict-justify FRQ on inhibition.
(a) Identify (1 point): competitive inhibition.
(b) Predict (1 point): increasing substrate concentration reduces the inhibitor's effect, so the reaction rate can rise back toward its maximum. Justify (1 point): substrate and inhibitor compete for the same active site, so a high concentration of substrate outcompetes the inhibitor and occupies more active sites.
Markers reward identifying competitive inhibition and explaining the prediction in terms of competition for the same active site.
Related dot points
- Topic 3.1 Enzyme Structure: describe the structure of enzymes, the role of the active site, and how the structure of an enzyme determines its specificity for a substrate.
A focused answer to AP Biology Topic 3.1, covering enzymes as protein catalysts, the active site, the induced-fit model, enzyme-substrate specificity, and how three-dimensional shape determines which reaction is catalyzed.
- Topic 3.3 Environmental Impacts on Enzyme Function: explain how changes in temperature and pH affect enzyme structure and the rate of an enzyme-catalyzed reaction, including denaturation and optimum conditions.
A focused answer to AP Biology Topic 3.3, covering the optimum temperature and pH of enzymes, why activity rises then falls with temperature, denaturation, and how to read enzyme-rate graphs.
- Topic 3.4 Cellular Energy: explain how cells use free energy, ATP and coupled reactions to drive endergonic processes, and how energy flows into and out of biological systems.
A focused answer to AP Biology Topic 3.4, covering free energy, exergonic and endergonic reactions, ATP as the energy currency, energy coupling, and why living systems require a constant input of free energy.
- Topic 3.6 Cellular Respiration: explain how glycolysis, the Krebs cycle and oxidative phosphorylation release energy from glucose to make ATP, and how fermentation allows ATP production without oxygen.
A focused answer to AP Biology Topic 3.6, covering glycolysis, the Krebs cycle, oxidative phosphorylation, chemiosmosis, the role of oxygen, and fermentation, with the link back to photosynthesis.
- Topic 1.4 Properties of Biological Macromolecules: describe the properties of carbohydrates, lipids and proteins, including the directionality of their structures and how their subunits and bonding give rise to their functions.
A focused answer to AP Biology Topic 1.4, covering carbohydrates, lipids and proteins, the four levels of protein structure, saturated versus unsaturated fats, and how subunits and bonding determine properties and function.
Sources & how we know this
- AP Biology Course and Exam Description — College Board (2020)