Questions on factors affecting enzyme rate: temperature, pH and concentration

Question 1

Explain why the rate of an enzyme-catalysed reaction climbs gradually as the temperature is raised towards the optimum, and then falls steeply above it.

Accepted Answer

B1 raising the temperature gives the molecules more kinetic energy, so substrate and active site collide more often and with more energy B1 more of those collisions are successful, so more enzyme-substrate complexes form and the rate roughly doubles for each 10 °C rise B1 above the optimum the increased vibration breaks the hydrogen bonds and ionic bonds holding the tertiary structure, so the enzyme denatures B1 the active site is then no longer complementary to the substrate, and denaturation proceeds far faster than a doubling per 10 °C, which is why the fall is steeper than the climb

Question 2

A student collects the oxygen released by catalase every thirty seconds for ten minutes and finds that the rate falls steadily throughout the run. Suggest three reasons for the fall, and suggest what she should measure instead if she wants to compare different temperatures fairly.

Accepted Answer

B1 substrate is being used up, so collisions between hydrogen peroxide and active sites become less frequent B1 hydrogen peroxide is a reactive oxidant and inactivates some of the catalase as the run goes on; product inhibition does not apply, since the products are water and oxygen B1 a fraction of the enzyme denatures over a long warm run B1 she should compare initial rates, taken from the gradient of the tangent at time zero or from 1/time to a fixed early end point

Question 3

Compare a graph of initial rate against substrate concentration with a graph of initial rate against enzyme concentration when substrate is in excess, referring to the overall shape and to the limiting factor in each case.

Accepted Answer

B1 the substrate-concentration graph rises and then plateaus, whereas the enzyme-concentration graph, with substrate in excess, is a straight line through the origin with no plateau B1 the substrate-concentration graph plateaus because every active site becomes occupied as soon as it is vacated, so the number of active sites becomes limiting B1 on the enzyme-concentration graph, with substrate in excess, substrate concentration never becomes limiting, so doubling the enzyme keeps doubling the rate B1 both graphs pass through the origin, since no enzyme or no substrate gives no reaction

Question 4

A student adds pepsin to a test tube of egg albumen at pH 8 and observes almost no digestion. Suggest why, and suggest what she should do differently to demonstrate pepsin's activity.

Accepted Answer

B1 pepsin's optimum pH is around 2, the acidity of the stomach B1 at pH 8 the hydrogen ion concentration is far from this optimum, so the ionic and hydrogen bonds holding pepsin's tertiary structure fail to hold it in shape, and the active site is not complementary to albumen B1 few or no enzyme-substrate complexes form, so almost no digestion occurs B1 she should repeat the investigation with the reaction mixture adjusted to a low pH, around pH 2, using dilute hydrochloric acid, to bring the enzyme close to its optimum

Question 5

Explain why cooling an enzyme from 37 °C to 4 °C and then rewarming it to 37 °C restores its original rate of reaction, whereas heating the same enzyme to 70 °C and then cooling it back to 37 °C usually does not.

Accepted Answer

B1 cooling to 4 °C only reduces the kinetic energy of the enzyme and substrate molecules, so collisions between them become less frequent and less energetic, and the rate falls B1 the hydrogen and ionic bonds holding the enzyme's tertiary structure are not broken by cooling, so the active site keeps its shape and rewarming restores the original rate B1 heating to 70 °C breaks those hydrogen and ionic bonds, so the tertiary structure unfolds and the active site is no longer complementary to the substrate: the enzyme is denatured B1 cooling a denatured enzyme does not reform the original bonds in most cases, so the tertiary structure, and with it activity, is usually not recovered

Question 6

An enzyme has an initial rate of 3.0 arbitrary units at 30 °C, and its Q10 over the range below its optimum is 2.0. Calculate the rate you would predict at 20 °C, and calculate the rate you would predict at 10 °C.

Accepted Answer

M1 going down ten degrees means dividing by Q10 rather than multiplying: rate at 20 °C = 3.0 ÷ 2.0 A1 1.5 arbitrary units M1 rate at 10 °C = 1.5 ÷ 2.0 A1 0.75 arbitrary units

Question 7

A biologist wants to identify a bacterial enzyme's optimum pH by adding a fixed volume of enzyme extract to substrate solutions buffered at pH 4, 6, 8 and 10, and comparing the volume of product formed after twenty minutes at each pH. Evaluate this experimental design as a way of finding the true optimum pH.

Accepted Answer

B1 twenty minutes is not an initial rate, so a lower reading at an extreme pH could reflect a slower start, ongoing denaturation over the twenty minutes, or both, and the design cannot separate these B1 near the optimum the substrate may be used up well before twenty minutes, so several pH values give the same final volume and the readings cannot show which is fastest B1 testing only four pH values, two units apart, could miss an optimum that falls between them, for example at pH 7 if only 6 and 8 are tested B1 judgement: the design is a reasonable first survey to locate roughly where the optimum lies, but should be refined with initial-rate measurements and pH values spaced more closely once that rough region is found

Question 8

A student investigated the effect of temperature on the initial rate at which trypsin clears a milk suspension. Explain why she kept the trypsin and the milk in separate tubes in the water bath for ten minutes before mixing them, why both were made up in a buffer, why she set up a tube of milk with boiled trypsin, and why she set up a tube of milk with buffer in place of trypsin.

Accepted Answer

B1 separate tubes bring enzyme and substrate to the test temperature before the reaction starts, so the initial rate is measured at that temperature and not during warming or cooling B1 hydrolysis can change the pH, and pH affects the enzyme's tertiary structure and rate, so the buffer keeps pH constant as a control variable B1 boiled trypsin is denatured, so no clearing in that tube shows that clearing in the test tubes is caused by active enzyme B1 the tube with buffer instead of trypsin shows whether the milk clears or settles by itself, so any change in absorbance without enzyme can be allowed for

Question 9

Explain how a change in pH away from the optimum reduces the rate of an enzyme-catalysed reaction.

Accepted Answer

B1 a change in hydrogen ion concentration protonates or deprotonates the side chains, altering the charge they carry B1 ionic bonds between those side chains fail and hydrogen bonds are disrupted, so the tertiary structure shifts B1 the active site is a product of that structure, so its shape changes and the substrate is no longer complementary to it

Question 10

An enzyme has an initial rate of 0.8 µmol per minute at 10 °C and 1.8 µmol per minute at 20 °C. Calculate the Q10 for this enzyme, and calculate the rate you would predict for it at 30 °C.

Accepted Answer

M1 Q10 is the rate at one temperature divided by the rate ten degrees lower, so 1.8 divided by 0.8 A1 a Q10 of 2.25, or 2.3 to two significant figures A1 a predicted rate of 1.8 × 2.25 = 4.05, so about 4.1 µmol per minute at 30 °C

Question 11

Describe the shape of a graph of initial rate against substrate concentration, and describe what is limiting the rate in each part of that graph.

Accepted Answer

B1 the rate climbs steeply from the origin and the climb then eases, before the curve levels off at a plateau B1 on the rising part the limiting factor is substrate concentration, because active sites are sitting empty and each extra substrate molecule finds one B1 at the plateau every active site is occupied, so the limiting factor is the number of active sites, set by the enzyme concentration

Question 12

A disc of potato was placed in hydrogen peroxide solution and the oxygen released was collected. Calculate the mean rate of oxygen production over the first 20 s of the run, giving the unit.

Accepted Answer

M1 reads 24 cm³ from the curve at 20 s M1 mean rate = volume collected ÷ time taken = 24 ÷ 20 A1 1.2 cm³ s⁻¹, with the unit

Question 13

The rate of oxygen production over the whole 90 s of the run is lower than the rate over the first 20 s. Explain why, and explain why only an initial rate should be used to compare two enzymes.

Accepted Answer

B1 the gradient falls throughout the run, so the rate at any later moment is lower than the rate at the start B1 hydrogen peroxide is being used up, so fewer substrate molecules collide with active sites in a given time and the rate falls B1 the initial rate is the only one measured with the substrate concentration still at its starting value, so it is the only rate that compares the enzymes rather than how much substrate happened to be left

Question 14

Describe the shape of a graph of initial rate against enzyme concentration when substrate is present in excess, and describe what would happen to that graph if the substrate concentration became limiting instead.

Accepted Answer

B1 the graph is a straight line through the origin: rate is directly proportional to enzyme concentration B1 this is because each extra enzyme molecule adds more active sites, and with excess substrate every site can find a substrate molecule B1 if substrate became limiting, the graph would level off at higher enzyme concentrations instead of climbing indefinitely, because there would not be enough substrate for the added active sites to bind

Question 15

A student mixed trypsin with a milk suspension and recorded the absorbance as the suspension cleared. A tangent drawn to the absorbance-time curve at time zero passes through an absorbance of 0.90 at 0 s and 0.30 at 120 s. After 300 s the absorbance was 0.40. The data are idealised for this question. Calculate the initial rate, and calculate the mean rate over the first 300 s.

Accepted Answer

M1 initial rate is the gradient of the tangent: (0.90 − 0.30) ÷ 120 A1 0.0050 absorbance units per second, or s⁻¹; accept 0.30 per minute B1 mean rate (0.90 − 0.40) ÷ 300 = 0.0017 absorbanceunits per second, or s⁻¹; accept 0.00167

Question 16

A cross behind a tube of milk suspension and trypsin became visible after 150 s at 30 °C and after 70 s at 40 °C. The data are idealised for this question. Taking rate as 1/time, calculate the rate at each temperature and the Q10 between 30 and 40 °C.

Accepted Answer

M1 rates 1 ÷ 150 = 0.0067 s⁻¹ at 30 °C and 1 ÷ 70 = 0.014 s⁻¹ at 40 °C M1 Q10 = rate at 40 °C ÷ rate at 30 °C, or 150 ÷ 70 A1 Q10 = 2.1; accept 2.14

Question 17

Give the optimum pH of pepsin and the optimum pH of trypsin, naming where in the gut each of these enzymes works.

Accepted Answer

B1 pepsin has an optimum of about pH 2 and works in the stomach B1 trypsin has an optimum of about pH 8 and works in the small intestine

Question 18

State the approximate optimum temperature of most human enzymes, and state what is meant by an enzyme's optimum temperature.

Accepted Answer

B1 most human enzymes have an optimum near 37 to 40 °C B1 the optimum temperature is the temperature at which the enzyme's rate of reaction is highest