Environmental Impacts on Enzyme Function: Study Guide (Biology 3.2)

Shape is everything

In Topic 3.1 you saw that an enzyme catalyses a reaction only when the substrate fits its active site in shape and charge. That shape is not fixed by strong covalent bonds alone. The tertiary structure of a protein is held by many weaker interactions between R groups: hydrogen bonds, ionic interactions between charged groups, and hydrophobic interactions (Topic 1.7).

Weak interactions are easy to disturb. When the conditions around an enzyme change, some of these interactions break or form in new places. The protein changes shape, and the active site may no longer fit the substrate. This topic is about how the environment of a cell (its temperature, pH, chemicals, concentrations of substrate and product, and inhibitors) changes enzyme activity.

Denaturation: losing the working shape

Denaturation is a change in a protein’s three-dimensional structure that removes its ability to function. For an enzyme, denaturation means the active site is distorted, so the substrate can no longer bind properly and the reaction is no longer catalysed.

Key points:

• Denaturation breaks the weak interactions (especially hydrogen bonds) that hold the secondary and tertiary structure. It does not break the peptide bonds, so the amino acid sequence stays the same.
• It can be caused by temperature, by pH and by the chemical environment, for example high salt concentrations, some metal ions or organic solvents.
• Conditions only a little outside the best range change the shape slightly and make the enzyme less efficient. More extreme conditions unfold it so much that it stops working.

Temperature: two effects pulling in opposite directions

Temperature affects enzyme activity in two ways at once.

1. More collisions. At higher temperatures, molecules in solution move faster on average. Substrate molecules meet active sites more often, and more of them arrive with enough energy to react. So, from low temperatures up to the optimum, the rate rises.
2. Loss of shape. Heat also makes the atoms in the enzyme vibrate more strongly. Above the optimum, hydrogen bonds and other weak interactions in the enzyme start to break. The active site changes shape, fewer enzyme–substrate complexes form, and the rate falls steeply. With enough heating, the enzyme is fully denatured.

The optimum temperature is where these two effects balance and the rate is highest. Many enzymes in the human body work best close to body temperature (about 37 °C). Enzymes from organisms that live in hot springs keep their shape at much higher temperatures, because their structures contain more stabilising interactions. Each enzyme has its own optimum.

Notice the shape of Figure 1: the rise is gradual, but the fall above the optimum is steep. Once heat starts to unfold the protein, activity is lost quickly.

pH: changing the charges

pH measures the concentration of hydrogen ions (H⁺). Many R groups can gain or lose H⁺. For example, a carboxyl group (–COO⁻) can pick up H⁺ in acid and lose its negative charge; an amine group (–NH₃⁺) can lose H⁺ in alkaline conditions and lose its positive charge.

When the charges on R groups change:

• ionic interactions and hydrogen bonds that hold the fold are disrupted, so the shape of the enzyme, including its active site, changes;
• the charge pattern of the active site changes, so a substrate that used to be attracted may now be ignored or repelled.

Each enzyme has an optimum pH, which usually matches the place where it works.

For example, pepsin, a protease made in the stomach, works best in strongly acidic conditions (about pH 1.5 to 2.5), matching stomach acid. Enzymes that work in the small intestine work best in slightly alkaline conditions. Moving either enzyme to the other’s environment changes the charges on its R groups and lowers its activity.

Can denaturation be reversed?

Not always, but sometimes.

• Small changes in pH or the removal of a disrupting chemical can let some enzymes refold into their working shape, because the amino acid sequence (which decides the fold) is unchanged. Activity returns.
• Strong heating usually causes permanent damage: the unfolded chains tangle and stick to each other, so they cannot find their correct shape again.
• Cold is different again. Low temperatures slow the reaction because molecules collide less often, but they do not usually denature the enzyme. Warm it back up and the rate recovers.

The only way to know which case applies is to test: expose the enzyme to the condition, return it to its optimum, and measure activity again. Worked example 1 shows how.

Substrate and product concentrations

The cellular environment also includes the molecules the enzyme acts on and produces.

Substrate concentration. When there is little substrate, many active sites are empty, so adding substrate increases the rate. As substrate concentration rises, more active sites are occupied at any moment. Eventually almost every active site is busy all the time: the enzyme is saturated, and adding more substrate barely raises the rate. The curve levels off (look at the solid line in Figure 4). At this point only more enzyme (more active sites) would raise the rate further.

Product concentration. Many enzyme reactions can run in either direction. The relative concentrations of substrate and product decide which way the overall (net) reaction goes and how quickly:

• high substrate and low product: the net reaction runs forward efficiently;
• as product builds up, the reverse reaction becomes more frequent, so the net forward rate falls;
• if product is removed (for example, used straight away by the next enzyme in a pathway), the forward reaction keeps going.

A clear example is carbonic anhydrase in red blood cells. It catalyses CO₂ + H₂O ⇌ H₂CO₃. In body tissues, where CO₂ concentration is high, the reaction runs forward and CO₂ is taken into the blood. In the lungs, where CO₂ concentration is low, the same enzyme drives the reverse reaction and CO₂ is released. The enzyme is the same; the concentrations decide the direction.

Inhibitors

An inhibitor is a molecule that binds to an enzyme and reduces its activity. The inhibitors in this topic bind reversibly: they attach and detach.

Competitive inhibitors

• have a shape (and charge) similar to the substrate;
• bind to the active site, so substrate and inhibitor compete for the same place;
• while an inhibitor occupies the active site, the substrate cannot bind;
• their effect can be overcome by adding more substrate: with far more substrate than inhibitor, the substrate wins most of the competition.

Noncompetitive inhibitors

• bind to a different place on the enzyme, an allosteric site;
• change the shape of the enzyme, so the active site works less well or not at all;
• their effect is not overcome by more substrate, because substrate and inhibitor are not competing for the same site.

Cells use both kinds of binding to regulate their pathways. For example, the final product of a pathway can bind to an allosteric site on the first enzyme and slow the whole pathway when enough product has been made.

Worked example 1: temperature, optimum and reversibility

Question. A student measures the initial rate of an enzyme extracted from a mammal at six temperatures (fictional data):

Temperature / °C	10	20	30	40	50	60
Rate / µmol min⁻¹	2.1	4.0	7.6	11.8	6.0	0.4

She then does two extra tests. Sample P is held at 60 °C for 10 minutes, then cooled to 30 °C and tested: rate 0.5 µmol min⁻¹. Sample Q is held at 5 °C for 10 minutes, then warmed to 30 °C and tested: rate 7.5 µmol min⁻¹.

(a) Calculate the percentage change in rate from 20 to 30 °C and from 40 to 50 °C. (b) Estimate the optimum and say how the estimate could be improved. (c) Explain the change from 20 to 30 °C and from 40 to 50 °C. (d) Interpret the results for samples P and Q.

(a) Percentage changes.

1. 20 → 30 °C: (7.6 − 4.0) ÷ 4.0 × 100 = 3.6 ÷ 4.0 × 100 = +90%.
2. 40 → 50 °C: (6.0 − 11.8) ÷ 11.8 × 100 = −5.8 ÷ 11.8 × 100 = −49%.

(b) Optimum. The highest measured rate is at 40 °C, so the optimum is near 40 °C, somewhere between 30 and 50 °C. With 10 °C steps the true peak could be a few degrees either side. Testing at 2 °C intervals between 34 and 46 °C would give a better estimate.

(c) Explanation.

1. 20 → 30 °C: molecules move faster, so enzyme and substrate collide more often and more collisions lead to reaction. The rate nearly doubles.
2. 40 → 50 °C: hydrogen bonds and other weak interactions in the enzyme break. The active site changes shape, fewer enzyme–substrate complexes form, and the rate falls by about half.

(d) Reversibility.

1. Sample P: 0.5 ÷ 7.6 × 100 = 6.6% of the normal rate at 30 °C. Returning to 30 °C did not restore activity, so the enzyme was permanently denatured at 60 °C.
2. Sample Q: 7.5 ÷ 7.6 × 100 = 98.7% of the normal rate. Cold slowed the enzyme only while it was cold; its shape was kept, and activity returned on warming.

Check. Both percentage changes use the starting value as the denominator. The conclusion about P relies on one sample; repeats would make it more secure.

Worked example 2: identifying the type of inhibitor

Question. An enzyme’s initial rate is measured at six substrate concentrations, with no inhibitor, with inhibitor X and with inhibitor Y. Both inhibitors are used at the same concentration. The fictional results are in the table and in Figure 4.

Substrate / mmol L⁻¹	0.5	1	2	5	10	20
No inhibitor / µmol min⁻¹	12.5	20.0	28.6	38.5	43.5	46.5
Inhibitor X / µmol min⁻¹	5.0	9.1	15.4	26.3	34.5	40.8
Inhibitor Y / µmol min⁻¹	6.3	10.0	14.3	19.2	21.7	23.3

(a) For each inhibitor, calculate the rate as a percentage of the uninhibited rate at 0.5 and at 20 mmol L⁻¹. (b) Identify each inhibitor as competitive or noncompetitive, and justify your answer.

(a) Percentages.

1. X at 0.5 mmol L⁻¹: 5.0 ÷ 12.5 × 100 = 40%. X at 20 mmol L⁻¹: 40.8 ÷ 46.5 × 100 = 88%.
2. Y at 0.5 mmol L⁻¹: 6.3 ÷ 12.5 × 100 = 50%. Y at 20 mmol L⁻¹: 23.3 ÷ 46.5 × 100 = 50%.

(b) Identification.

1. X is competitive. Its effect shrinks as substrate increases (from 40% to 88% of normal). With more substrate, substrate molecules win the active site more often. The X line approaches the uninhibited line.
2. Y is noncompetitive. It halves the rate at every substrate concentration, and the curve levels off at a lower value. Adding substrate does not help, because Y binds an allosteric site and changes the active site of the enzymes it binds, whatever the substrate concentration.

Interpretation. To reduce the effect of a competitive inhibitor in a cell, raise the substrate concentration. That strategy fails for a noncompetitive inhibitor; there, only removing the inhibitor or making more enzyme helps.

Common misconceptions

• “High temperature kills the enzyme.” Enzymes are molecules, not living things. They are denatured, not killed.
• “Denaturation breaks the enzyme into amino acids.” Peptide bonds stay intact. Hydrogen bonds and other weak interactions break, so the fold is lost.
• “Cold denatures enzymes.” Cold usually just slows collisions; activity returns on warming.
• “Denaturation is always permanent.” Sometimes an enzyme can refold and regain activity when normal conditions return.
• “Every enzyme has an optimum of 37 °C and pH 7.” Optima vary with where the enzyme works, such as pH 2 for a stomach protease.
• “The rate levels off at high substrate because the enzyme is used up.” It levels off because the active sites are all occupied (saturated).
• “Competitive inhibitors permanently destroy the active site.” They bind reversibly; enough substrate can outcompete them.
• “A noncompetitive inhibitor competes with the substrate.” It binds an allosteric site, not the active site, so it does not compete.

Where this leads

Topic 3.3 looks at how cells capture and use energy through linked enzyme pathways, where everything on this page (temperature, pH, concentrations and allosteric control) decides how fast energy flows: see Topic 3.3, Cellular Energy. Test yourself with the practice questions, then use the revision notes and the checklist to consolidate.