Factors Affecting Enzyme Activity

AS · 19 min

The rate of an enzyme-catalysed reaction depends on how often substrate molecules meet active sites and on whether those active sites keep their shape. Five factors control this, and the syllabus requires you to investigate and explain each one: temperature, pH, enzyme concentration, substrate concentration and inhibitor concentration. Graphs of rate against each factor appear on every series of papers, together with practical questions on how to control the other variables. Every explanation in this note comes back to two ideas: the frequency of successful collisions forming enzyme–substrate complexes, and the shape of the active site.

The two ideas behind every explanation

Before looking at each factor, fix the two things that can change the rate.

  1. How many enzyme–substrate complexes form per unit time. This depends on how often substrate molecules collide with active sites in the correct orientation. More molecules, faster-moving molecules, or more active sites all mean more collisions.
  2. Whether the active site is still complementary to the substrate. Anything that changes the tertiary structure of the enzyme changes the active site. If the substrate no longer fits, no ES complexes form, however many collisions there are.

When you explain a graph, say which of these is changing in each region of the graph.

Temperature

y = 2(2^(x/10))/(1 + exp(0.35(x - 50)))

(Horizontal axis: temperature / °C; vertical axis: rate of reaction. The rate rises gradually as temperature increases, roughly doubling every 10 °C, reaches a maximum at the optimum temperature (here about 46 °C), then falls steeply to zero by about 65 °C. The curve is not symmetrical: the fall is much steeper than the rise.)

Below the optimum

As temperature increases, enzyme and substrate molecules gain kinetic energy and move faster. They collide more often, and more collisions have enough energy to react, so more enzyme–substrate complexes form per unit time and the rate increases. Over this region the rate roughly doubles for every 10 °C rise.

At low temperatures (for example 00–5 ∘C5\ ^\circ\text{C}) the enzyme is inactive but not denatured: molecules move slowly and few ES complexes form. If the temperature is raised again, activity returns. This is why food is kept in a refrigerator and why enzymes are stored cold.

At the optimum

The optimum temperature is the temperature at which the rate is greatest. It is a balance between the increasing kinetic energy (more collisions) and the start of loss of shape of the enzyme. For most human enzymes the optimum is about 3737–40 ∘C40\ ^\circ\text{C}; for enzymes from bacteria living in hot springs it can be 70 ∘C70\ ^\circ\text{C} or more.

Above the optimum

The enzyme molecules vibrate more and more strongly. The vibrations break hydrogen bonds and ionic bonds (and disrupt hydrophobic interactions) that hold the tertiary structure in place. The tertiary structure changes, so the active site changes shape and is no longer complementary to the substrate. Fewer and fewer ES complexes can form, so the rate falls rapidly. The enzyme is denatured. Denaturation is permanent: cooling the enzyme does not restore its activity.

Key result

Temperature: the marking points

  • Rising part: more kinetic energy; more frequent collisions; more ES complexes formed per unit time.
  • Optimum: maximum rate.
  • Falling part: increased vibration breaks hydrogen and ionic bonds; tertiary structure and active site change shape; substrate no longer complementary / cannot bind; fewer ES complexes; enzyme denatured.
  • Low temperature: enzyme inactive (low kinetic energy), not denatured.
Extension: the temperature coefficient

The temperature coefficient, Q10Q_{10}, compares the rate at one temperature with the rate 10 ∘C10\ ^\circ\text{C} lower: Q10=rate at (T+10) ∘Crate at T ∘CQ_{10} = \dfrac{\text{rate at } (T + 10)\ ^\circ\text{C}}{\text{rate at } T\ ^\circ\text{C}}. For enzyme-catalysed reactions below the optimum, Q10Q_{10} is usually about 2. You will not be asked to recall this, but data questions sometimes give it to you to use.

pH

y = 10exp(-(x - 2)^2/1.2) y = 10exp(-(x - 8)^2/1.2)

(Horizontal axis: pH; vertical axis: rate of reaction. The left curve is pepsin, a protease from the stomach, with an optimum pH of about 2. The right curve is trypsin, a protease from the small intestine, with an optimum pH of about 8. Each enzyme works over a narrow range of pH on either side of its optimum.)

pH is a measure of the concentration of hydrogen ions, H+\text{H}^+. Changing the pH changes the number of H+\text{H}^+ ions available to bind to charged groups.

  • Many R groups in an enzyme are ionisable: carboxyl groups (−COO−-\text{COO}^- / −COOH-\text{COOH}) and amine groups (−NH3+-\text{NH}_3^+ / −NH2-\text{NH}_2). At low pH, the excess H+\text{H}^+ ions bind to −COO−-\text{COO}^- groups, making them uncharged; at high pH, H+\text{H}^+ ions are removed from −NH3+-\text{NH}_3^+ groups, making them uncharged.
  • Ionic bonds and hydrogen bonds between R groups are broken or altered, so the tertiary structure and the shape of the active site change.
  • Even a small change in pH can alter the charges on R groups in the active site (and on the substrate), so the substrate is attracted to or held by the active site less well, and fewer ES complexes form.
  • At extreme pH the change in shape is large and the enzyme is denatured. Small changes either side of the optimum are often reversible.
Definition

The optimum pH is the pH at which an enzyme catalyses a reaction at the maximum rate. It reflects the environment in which the enzyme normally works: pepsin about pH 2 (stomach), salivary amylase about pH 7 (mouth), trypsin about pH 8 (small intestine).

In investigations, pH is controlled using buffer solutions. A buffer resists changes in pH when small amounts of acid or alkali are added (or produced by the reaction), so the pH stays at the chosen value throughout.

Enzyme concentration

y = 2x y = 8tanh(x/4)

(Horizontal axis: enzyme concentration; vertical axis: initial rate of reaction. The straight line through the origin shows the result when substrate is in excess: the rate is directly proportional to enzyme concentration. The curve that levels off at a rate of 8 shows what happens if the amount of substrate is limited: at high enzyme concentrations, substrate becomes the limiting factor.)

  • When substrate is in excess, increasing the enzyme concentration increases the number of active sites available. More ES complexes form per unit time, so the initial rate increases in direct proportion to enzyme concentration.
  • If substrate is limited, at high enzyme concentrations there are more active sites than substrate molecules can fill at any moment. Adding more enzyme then has no effect: substrate concentration has become the limiting factor, and the graph levels off.

Substrate concentration

y = 10x/(2 + x) (0, 10) -- (20, 10)

(Horizontal axis: substrate concentration; vertical axis: initial rate of reaction. The rate rises steeply at first, then more slowly, and approaches a maximum (the horizontal line at a rate of 10), called Vmax⁡V_{\max}. The enzyme concentration is the same for every point.)

  • At low substrate concentrations, many active sites are empty at any moment. Increasing substrate concentration increases the frequency of collisions with active sites, so more ES complexes form per unit time and the rate rises almost in proportion. Substrate concentration is the limiting factor.
  • As substrate concentration increases further, more and more active sites are occupied at any one time, so each extra substrate molecule makes less difference. The curve bends.
  • At high substrate concentrations, all the active sites are occupied all the time: the enzyme molecules are working as fast as they can. The rate reaches its maximum, Vmax⁡V_{\max}, and increasing substrate concentration has no further effect. Enzyme concentration is now the limiting factor (or some other factor, such as temperature).

This curve is the basis of Vmax⁡V_{\max} and the Michaelis–Menten constant, KmK_{\text{m}}, covered in Vmax, Km and enzyme inhibition.

Key result

Limiting factors

The factor that is limiting the rate is the one that, if increased, would increase the rate. On a rising part of a graph, the variable on the xx-axis is limiting. On a plateau, something else is limiting.

Inhibitor concentration

An inhibitor is a substance that reduces the rate of an enzyme-catalysed reaction. As the concentration of inhibitor increases, more enzyme molecules are affected, so fewer active sites are available to form ES complexes and the rate falls.

y = 10/(1 + 0.6x)

(Horizontal axis: inhibitor concentration; vertical axis: rate of reaction, with enzyme and substrate concentrations kept constant. The rate falls steeply at first, then more gradually.)

The two kinds of reversible inhibitor, competitive and non-competitive, behave differently when the substrate concentration is changed. That is explained in Vmax, Km and enzyme inhibition.

Describing and explaining a temperature curve (routine)

The table shows the initial rate of an enzyme-catalysed reaction at different temperatures.

Temperature / °C102030405060
Initial rate / arbitrary units4.08.016.031.132.03.8

(a) Calculate the percentage decrease in rate between 50 ∘C50\ ^\circ\text{C} and 60 ∘C60\ ^\circ\text{C}. (b) Explain the change in rate between 50 ∘C50\ ^\circ\text{C} and 60 ∘C60\ ^\circ\text{C}. (3 marks) (c) Suggest how the investigation could be modified to find the optimum temperature more precisely.

Solution

(a) Percentage decrease =32.0−3.832.0×100=88%= \dfrac{32.0 - 3.8}{32.0} \times 100 = 88\%.

(b)

  1. Above the optimum, the increased kinetic energy makes the enzyme vibrate more;
  2. hydrogen bonds and ionic bonds holding the tertiary structure break, so the active site changes shape (the enzyme is denatured);
  3. the substrate is no longer complementary / cannot bind, so fewer enzyme–substrate complexes form.

(c) The optimum lies between 40 ∘C40\ ^\circ\text{C} and 50 ∘C50\ ^\circ\text{C} (or 4040–55 ∘C55\ ^\circ\text{C}), so repeat with smaller intervals in that range, e.g. every 2 ∘C2\ ^\circ\text{C} from 4040 to 50 ∘C50\ ^\circ\text{C}, with replicates at each temperature.

Rates from times at different pH values (moderate)

Amylase was added to starch solution in buffers of different pH, and the time to the achromic point was recorded.

pH456789
Time / s21095524061130

(a) Calculate the rate at each pH as 1000/t1000/t, giving units. (b) State the optimum pH as precisely as the data allow. (c) Explain why the rate at pH 4 is lower than at pH 7. (3 marks)

Solution

(a) Rate =1000/t= 1000/t in 10−3 s−110^{-3}\ \text{s}^{-1} (i.e. s−1×1000\text{s}^{-1} \times 1000):

pH456789
Rate / 10−3 s−110^{-3}\ \text{s}^{-1}4.810.519.225.016.47.7

(b) The fastest rate in the data is at pH 7, so the optimum is between pH 6 and pH 8; the data cannot show it more precisely than "about pH 7". (Because the rate at pH 6 is higher than at pH 8, the true optimum is probably slightly below 7.)

(c)

  1. At pH 4 there is a higher concentration of H+\text{H}^+ ions;
  2. these alter the charges on R groups (e.g. −COO−-\text{COO}^- becomes −COOH-\text{COOH}), breaking ionic and hydrogen bonds;
  3. the shape of the active site changes / charges in the active site change, so starch binds less well and fewer ES complexes form.
Enzyme concentration and limiting factors (moderate)

A student measured the time taken for trypsin to clear a cloudy suspension of milk protein (casein) at different trypsin concentrations, with the casein in excess.

Trypsin concentration / %0.20.40.60.81.0
Time to clear / s250125846250

(a) Calculate the rate (1/t1/t) at each concentration and describe the relationship. (b) Explain the relationship. (c) Predict and explain the result if the experiment were repeated with 2.0% trypsin but with only one-tenth of the casein concentration.

Solution

(a) Rates in 10−3 s−110^{-3}\ \text{s}^{-1}: 4.0, 8.0, 11.9, 16.1, 20.0. The rate is directly proportional to trypsin concentration: doubling the concentration doubles the rate (a straight line through the origin).

(b) Casein is in excess, so every extra enzyme molecule adds active sites that can be filled. More active sites mean more enzyme–substrate complexes formed per unit time, so the rate increases in proportion.

(c) The rate would not be doubled compared with 1.0% trypsin (it would level off). With so little casein, there are more active sites than substrate molecules to fill them; substrate concentration becomes the limiting factor, so adding more enzyme does not increase the rate. (The suspension also clears quickly simply because there is less casein to digest, so a time-to-clear method would need to be interpreted with care.)

Exam-hard: explaining a substrate concentration curve (5 marks)

Describe and explain the effect of increasing substrate concentration on the rate of an enzyme-catalysed reaction, when the enzyme concentration is kept constant.

Solution

Description:

  1. At low substrate concentrations the rate increases (almost) in proportion to substrate concentration;
  2. the rate then increases more slowly and levels off at a maximum (Vmax⁡V_{\max}).

Explanation:

  1. At low concentrations, there are free active sites; more substrate means more frequent collisions and more enzyme–substrate complexes formed per unit time; substrate concentration is the limiting factor.
  2. At high concentrations all active sites are occupied (saturated) at any one time; extra substrate cannot bind until an active site is free.
  3. Enzyme concentration (number of active sites) is now the limiting factor.

A good answer uses both "collisions/ES complexes" for the rising part and "all active sites occupied" for the plateau. Saying only "the enzyme is saturated" without explaining what that means usually earns one mark, not two.

Exam-hard: planning a pH investigation (6 marks)

Outline a method to investigate the effect of pH on the activity of catalase in a yeast suspension. Include how you would control variables and how you would obtain reliable results.

Solution
  1. Independent variable: pH, using buffer solutions at five or more values, e.g. pH 4, 5, 6, 7, 8, 9.
  2. Mix a fixed volume (e.g. 5 cm35\ \text{cm}^3) of buffer with a fixed volume of hydrogen peroxide of fixed concentration (e.g. 10 cm310\ \text{cm}^3 of 3%) in a conical flask.
  3. Place the flask and a separate tube of yeast suspension in a thermostatically controlled water bath at e.g. 30 ∘C30\ ^\circ\text{C} for 5 minutes to equilibrate.
  4. Add a fixed volume of yeast suspension (e.g. 2 cm32\ \text{cm}^3, stirred before sampling so the concentration is the same), seal the flask and connect to a gas syringe.
  5. Dependent variable: volume of oxygen collected; record every 10 s for 2 minutes, plot volume against time and find the initial rate from the gradient of a tangent at t=0t = 0 (or record the volume after a fixed time such as 30 s).
  6. Control variables: temperature, volume and concentration of H2O2\text{H}_2\text{O}_2, volume and concentration of yeast suspension, total volume, size of flask.
  7. Control experiment: boiled yeast suspension at each pH, to show that oxygen production is due to catalase.
  8. Reliability: carry out at least three replicates at each pH, calculate a mean, identify and repeat anomalous results; then repeat with smaller intervals around the apparent optimum.
Practical skills

Investigating the effect of temperature on enzyme activity

  • Temperature control: use a thermostatically controlled water bath (or a beaker of water heated with a Bunsen and monitored with a thermometer, adding hot or cold water as needed). Use at least five temperatures across a wide range, e.g. 1010, 2020, 3030, 4040, 5050, 60 ∘C60\ ^\circ\text{C}; ice water for low values.
  • Equilibration: put the enzyme and substrate solutions in separate tubes in the water bath for about 5 minutes before mixing, so the reaction starts at the chosen temperature. Mixing first would start the reaction at room temperature.
  • pH control: add a buffer at the optimum pH (or a fixed pH) to every tube.
  • Measurements: initial rate from a progress curve (tangent at t=0t = 0), or 1/t1/t from an end point (achromic point with amylase; disc rising time with catalase; time for milk suspension to clear with trypsin).
  • Reliability: three or more replicates per temperature, calculate means; identify anomalous results and repeat them.
  • Common limitations: temperature fluctuates in a hand-maintained water bath; temperature of the mixture changes once removed from the bath; subjective end points; enzyme starts to denature during the equilibration time at high temperatures.
  • Graph: rate (y-axis) against temperature (x-axis); plot points with sharp crosses and join with a smooth curve or ruled straight lines point to point; do not extrapolate beyond the data.
Practical skills

Using buffers in pH investigations

  • Buffer solutions are provided at set pH values (e.g. pH 3, 5, 7, 9, 11). Add the same volume of buffer to every tube.
  • Check the pH with a pH meter or narrow-range indicator paper.
  • A buffer is needed because the reaction itself may produce acids (e.g. fatty acids from lipase) that would otherwise change the pH during the experiment.
  • Do not change the temperature in a pH investigation; carry it out at a constant temperature in a water bath, below the optimum so the enzyme does not denature during the experiment.
Watch out
  • "Enzymes are killed at high temperature" loses the mark. Use denatured: the active site changes shape so the substrate is no longer complementary.
  • Low temperature does not denature enzymes; it makes them inactive. Activity returns on warming.
  • Do not say "heat breaks peptide bonds". Denaturation breaks hydrogen bonds, ionic bonds and hydrophobic interactions, not the peptide bonds of the primary structure.
  • On the substrate concentration graph, the plateau is not because "the substrate has run out". There is plenty of substrate: it is the active sites that are all occupied.
  • "pH affects the active site" is too vague. Say how: H+\text{H}^+ ions alter the charges on R groups, breaking ionic and hydrogen bonds, changing the shape of the active site.
Exam tip
  • Command words matter. Describe means say what the graph shows, using numbers from the graph (e.g. "the rate increases from 4 to 31 arbitrary units between 10 and 40 °C"). Explain means give the biological reason. Many questions ask for both; give both, in that order.
  • When you quote data, include units and paired values ("at 40 °C the rate is 31.1").
  • For the temperature curve, examiners want kinetic energy, collisions, ES complexes, then bonds broken, active site shape, denatured. Learn this sequence as a chain.
  • For "suggest how the investigation could be improved", typical creditworthy answers are: smaller intervals around the optimum; more replicates; thermostatic water bath; buffer; colorimeter instead of judging colour by eye; measure initial rate.
  • If you are asked to identify the limiting factor at a point on a graph, look at whether the graph is still rising: rising means the xx-axis variable is limiting.
Summary
  • Temperature: rate increases with kinetic energy (more collisions, more ES complexes) up to the optimum; above it, bonds break, active site changes shape, enzyme denatured; low temperatures make enzymes inactive, not denatured.
  • pH: H+\text{H}^+ ions alter charges on R groups, breaking ionic and hydrogen bonds, changing the active site; each enzyme has an optimum pH; buffers control pH in experiments.
  • Enzyme concentration: with substrate in excess, rate is proportional to enzyme concentration (more active sites).
  • Substrate concentration: rate rises then levels off at Vmax⁡V_{\max} when all active sites are occupied; enzyme concentration then limits.
  • Inhibitor concentration: higher concentration, lower rate.
  • A limiting factor is the one that increases the rate if increased.
  • In practicals: equilibrate solutions separately, use water baths and buffers, measure initial rates, replicate and use means.

Practice questions

Question
  1. State what is meant by the optimum temperature of an enzyme.
  2. Explain why the rate of an enzyme-catalysed reaction increases between 10 ∘C10\ ^\circ\text{C} and 30 ∘C30\ ^\circ\text{C}. (3 marks)
  3. An enzyme kept at 4 ∘C4\ ^\circ\text{C} for a week still works at its normal rate when warmed to 37 ∘C37\ ^\circ\text{C}. An enzyme kept at 80 ∘C80\ ^\circ\text{C} for 10 minutes does not work when cooled to 37 ∘C37\ ^\circ\text{C}. Explain the difference.
  4. State the function of a buffer in an enzyme investigation.
  5. Suggest why pepsin, which works in the stomach, has an optimum pH of about 2, while trypsin has an optimum of about 8.
  6. On a graph of initial rate against substrate concentration, name the limiting factor (a) on the steep rising part and (b) on the plateau.
  7. In an investigation of enzyme concentration, the time for the reaction to reach an end point was 80 s with 1% enzyme. Assuming substrate is in excess, predict the time with 2% enzyme and calculate both rates.
  8. Explain why enzyme and substrate solutions should be placed in the water bath separately for several minutes before they are mixed. (2 marks)
  9. A student found these rates for an enzyme: 20 ∘C20\ ^\circ\text{C}, 12 units; 30 ∘C30\ ^\circ\text{C}, 23 units; 40 ∘C40\ ^\circ\text{C}, 41 units; 50 ∘C50\ ^\circ\text{C}, 18 units. (a) Calculate the percentage increase in rate between 20 and 30 ∘C30\ ^\circ\text{C}. (b) Explain why the optimum temperature cannot be stated as 40 ∘C40\ ^\circ\text{C} from these data.
  10. Hydrogen peroxide was added to catalase in excess at pH 7 and the volume of oxygen recorded. The experiment was repeated at pH 7 with three times the concentration of hydrogen peroxide. Sketch the two progress curves on the same axes and explain the differences in (a) initial gradient and (b) final volume. (5 marks)
Answers
  1. The temperature at which the enzyme catalyses the reaction at the maximum rate.
  2. Molecules gain kinetic energy and move faster; enzyme and substrate molecules collide more frequently (and with more energy); more enzyme–substrate complexes form per unit time.
  3. At 4 ∘C4\ ^\circ\text{C} the enzyme has little kinetic energy, so few collisions occur and it is inactive, but its tertiary structure and active site are unchanged; on warming, normal collisions resume. At 80 ∘C80\ ^\circ\text{C} the enzyme vibrates so much that hydrogen and ionic bonds break; the tertiary structure and active site change shape (denaturation); this is irreversible, so the substrate cannot bind even after cooling.
  4. To keep the pH constant (resist changes in pH), at a known value, throughout the reaction.
  5. Each enzyme's optimum matches the environment it works in: the stomach is acidic (hydrochloric acid secreted by the stomach lining) and the small intestine is slightly alkaline (pancreatic juice and bile contain hydrogencarbonate). At these pH values the R groups in each enzyme have charges that maintain the shape of the active site and allow the substrate to bind.
  6. (a) Substrate concentration. (b) Enzyme concentration (number of active sites), or another factor such as temperature.
  7. Doubling enzyme concentration doubles the rate, so the time halves to 40 s. Rates: 1/80=0.0125 s−11/80 = 0.0125\ \text{s}^{-1} and 1/40=0.025 s−11/40 = 0.025\ \text{s}^{-1}.
  8. So that both solutions reach the experimental temperature before the reaction starts; if they were mixed first, the reaction would begin at room temperature and the initial rate would not be the rate at the chosen temperature.
  9. (a) (23−12)/12×100=92%(23 - 12)/12 \times 100 = 92\% (91.7%). (b) Readings were taken only every 10 ∘C10\ ^\circ\text{C}. The highest measured rate is at 40 ∘C40\ ^\circ\text{C}, but the true peak could lie anywhere between 3030 and 50 ∘C50\ ^\circ\text{C} (for example at 43 ∘C43\ ^\circ\text{C}), between the readings. Smaller intervals (e.g. every 2 ∘C2\ ^\circ\text{C} from 30 to 50 ∘C50\ ^\circ\text{C}) with replicates are needed.
  10. Sketch: both curves start at the origin; the higher-concentration curve has a steeper initial gradient and levels off at three times the final volume. (a) The initial gradient is steeper because the higher substrate concentration gives more frequent collisions with active sites, so more ES complexes form per unit time (as long as the active sites are not already saturated; if they were, the initial rates would be the same). (b) The final volume is three times higher because there is three times as much substrate to be converted to oxygen; the final volume depends on the amount of substrate, not the enzyme. The curve takes longer to level off because more substrate must be used up.

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