Free Pearson Edexcel International A-level Biology · Unit 3

Initial enzyme rates and limiting factors

What you’ll learn

  • Calculate an initial rate from a tangent or a fixed endpoint.
  • Design fair comparisons of temperature, pH, enzyme and substrate concentration.

Understand the idea

Core practical 4 examines initial rates while changing temperature, pH, enzyme concentration or substrate concentration. Use the initial gradient of a product–time graph, or the magnitude of a substrate–time gradient, before substantial substrate depletion changes the conditions. A tangent estimates an instantaneous rate. Dividing the final product amount by the entire duration gives an average rate, which may be different.

For a fixed small endpoint, reciprocal time can be a useful relative initial-rate measure when the same reaction progress is required in every trial. Keep the endpoint amount, total volume and other initial conditions consistent. A pH comparison needs suitable buffers; a temperature comparison needs the reacting mixture at the intended temperature. Changing one concentration by dilution must not accidentally change the other reagent concentrations as well.

Increasing substrate concentration eventually gives little further rate increase when enzyme active sites are occupied. Increasing enzyme concentration can raise the initial rate when substrate remains sufficient. A temperature optimum reflects the balance between faster molecular activity and loss of functional protein structure, while pH can alter active-site interactions. Use replicated measurements and a useful range of values to distinguish a plateau or optimum from a single unusual result.

Key formulae

r0=dPdt∣t=0r_0=\left.\frac{dP}{dt}\right|_{t=0}

gradient=ΔPΔt\mathrm{gradient}=\frac{\Delta P}{\Delta t}

relative rate∝1tendpoint\mathrm{relative\ rate}\propto\frac1{t_{\mathrm{endpoint}}}

Increasing substrate has a diminishing effect in a saturation model

Conditions. Michaelis–Menten teaching model v = Vmax S/(Km + S), with Vmax = 10 µmol min⁻¹ and Km = 2 mmol dm⁻³. Fixed enzyme amount, temperature and pH; simple steady-state assumptions.

Substrate concentration / mmol dm⁻³; Initial rate / µmol min⁻¹.

At S = 2 the rate is 5, half Vmax. At S = 10 it is about 8.33, approaching but not reaching 10; doubling substrate does not double rate at high S.

Open the interactive lesson to view plotted curves and coordinate tables.

Worked example

A tangent to the initial part of a product-volume graph passes through (10 s,1.4 cm3)(10\ \mathrm{s},1.4\ \mathrm{cm^3}) and (30 s,4.6 cm3)(30\ \mathrm{s},4.6\ \mathrm{cm^3}). Estimate the initial rate. These points are on the tangent, not necessarily on the measured curve.

  1. The vertical difference is 4.6−1.4=3.2 cm34.6-1.4=3.2\ \mathrm{cm^3}.
  2. The horizontal difference is 30−10=20 s30-10=20\ \mathrm{s}.
  3. The tangent gradient is 3.2/20=0.16 cm3 s−13.2/20=0.16\ \mathrm{cm^3\,s^{-1}}.

Answer: 0.16 cm3 s−10.16\ \mathrm{cm^3\,s^{-1}}.

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