Factors affecting enzyme activityAQA A-Level Biology: Revision notes
Section 1
Temperature
As temperature rises from low values, the rate of reaction increases because enzyme and substrate molecules have more kinetic energy, so there are more frequent collisions and more enzyme-substrate complexes form each second. At the optimum temperature the rate is greatest.
Above the optimum, the hydrogen and ionic bonds that hold the tertiary structure break. The active site changes shape and the substrate is no longer complementary: the enzyme is denatured, and the rate falls sharply. Denaturation is permanent.
Enzymes are not 'killed' by heat, because they are not alive. Say they are denatured.
Section 2
pH and the pH calculation
Each enzyme has an optimum pH. Changes in hydrogen ion concentration disrupt the ionic and hydrogen bonds in the tertiary structure, changing the active site, so the enzyme is denatured away from its optimum.
pH = -log₁₀[H⁺], where [H⁺] is in mol dm⁻³. A low pH means a high hydrogen ion concentration.
Worked example: [H⁺] = 6.3 × 10⁻³ mol dm⁻³, so pH = -log₁₀(6.3 × 10⁻³) = 2.2. Reversed: a pH of 8.4 means [H⁺] = 4.0 × 10⁻⁹ mol dm⁻³.
Pepsin (stomach, pH about 2) and trypsin (small intestine, pH about 8) show that optimum pH matches where the enzyme works.
Section 3
Enzyme and substrate concentration
Substrate concentration: at low concentration the rate increases in proportion to it, because substrate is the limiting factor and more substrate means more collisions with active sites. At high concentration all active sites are occupied (saturated) and the rate levels off, because enzyme concentration is now limiting.
Enzyme concentration: if substrate is in excess, the rate increases in proportion to enzyme concentration, as there are more active sites. If substrate is limiting, increasing enzyme concentration makes no difference.
Section 4
Competitive inhibitors
A competitive inhibitor has a shape similar to the substrate. It is complementary to the active site, binds there and forms an enzyme-inhibitor complex, so the substrate cannot bind. Fewer enzyme-substrate complexes form and the rate falls.
Increasing substrate concentration reduces the effect, because more substrate molecules compete successfully for the active sites, so at very high concentration the rate nears the uninhibited rate. Example: malonate inhibits succinate dehydrogenase.
Section 5
Non-competitive inhibitors
A non-competitive inhibitor binds to the enzyme at a site other than the active site. This changes the tertiary structure, so the shape of the active site changes and the substrate is no longer complementary.
Because the inhibitor and substrate do not compete for the same site, increasing substrate concentration does not overcome the inhibition: the rate stays below the uninhibited rate, as a proportion of the enzyme molecules cannot work.
Do not say a competitive inhibitor 'blocks' the enzyme everywhere. It binds specifically at the active site.
Section 6
Required practical 1
Investigate the effect of one named variable (temperature, pH, enzyme or substrate concentration) on the rate of an enzyme-controlled reaction, for example catalase and hydrogen peroxide, or amylase and starch.
- Independent variable: the one you change. Dependent variable: the rate, e.g. volume of oxygen per unit time.
- Control the other variables: temperature (water bath), pH (buffer), enzyme and substrate concentration and volumes.
- Take repeats and calculate a mean; find rate as volume ÷ time (cm³ s⁻¹), ideally the initial rate.
- Safety: hydrogen peroxide is an irritant, so wear eye protection.
That's the notes covered.
Carry on to the next subtopic.
Exam questions on Factors affecting enzyme activity
- A student investigates the effect of temperature on the rate of the reaction catalysed by catalase in potato tissue. She adds 5 g of potato tissue to 20 cm³ of hydrogen peroxide solution at each temperature and collects the oxygen produced for 60 seconds. The volumes of oxygen collected at 20 °C, 30 °C, 40 °C, 50 °C and 60 °C were 4.0 cm³, 9.0 cm³, 15.0 cm³, 3.5 cm³ and 0 cm³ respectively.Explain why less oxygen was collected at 50 °C than at 40 °C.2 marks
- Pepsin is a protease secreted into the stomach, where the hydrogen ion concentration of the gastric juice is 6.3 × 10⁻³ mol dm⁻³. Its optimum pH is about 2. Trypsin is a protease secreted into the small intestine, where the pH is about 8, and its optimum pH is about 8.Explain why pepsin is inactive when it passes into the small intestine.2 marks
- Succinate dehydrogenase catalyses the oxidation of succinate in respiration. Malonate has a molecular shape very similar to succinate and reduces the rate of this reaction. A student also tests inhibitor Q, which binds to a site on the enzyme away from the active site. With malonate present, increasing the succinate concentration to a very high value raises the rate almost to the rate with no inhibitor. With inhibitor Q present, the rate stays well below the rate with no inhibitor, even at very high succinate concentration.Explain how malonate reduces the rate of the reaction catalysed by succinate dehydrogenase.3 marks
Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).