Aim for FiveBiology

Unit 3 · Topic 3.1

3.1 Enzymes

Enzymes are biological catalysts: they make reactions in cells go faster by reducing the energy needed to get a reaction started (the activation energy), and they aren't used up. Each enzyme works only on substrates whose shape and charge fit its active site. Enzymes are how cells control which reactions happen, and how fast.

Key terms

  • enzyme
  • catalyst
  • activation energy
  • active site
  • substrate
  • enzyme-substrate complex

What enzymes do

A catalyst is something that speeds up a chemical reaction without being changed or used up by it. Enzymes are catalysts made by cells, and they're proteins. (A few RNA molecules can also act as catalysts, which matters for the origin of life in 7.12.)

Many reactions a cell needs would happen far too slowly on their own. Enzymes make them fast enough to keep the cell alive. Because each enzyme speeds up only one reaction or a few closely related ones, the cell can control its chemistry by deciding which enzymes to make and when to switch them on or off. That's how enzymes help regulate biological processes.

Activation energy

Even reactions that release energy need a small push to get started. That starting energy is the activation energy: the energy needed to strain and break the bonds in the reactants so new bonds can form.

Enzymes work by lowering the activation energy. Picture an energy diagram with 'progress of reaction' on the x-axis and 'energy' on the y-axis. The curve starts at the reactants' energy level, rises over a hill and drops down to the products' energy level. The height of the hill above the reactants is the activation energy. With an enzyme, the hill is lower, but the starting and ending levels are exactly the same.

That last point is important. An enzyme doesn't change how much energy a reaction releases or absorbs, and it can't make an energy-requiring reaction happen on its own. It only makes the reaction go faster.

Active site and substrate

The molecule an enzyme acts on is its substrate. The substrate binds to a specific region of the enzyme called the active site, a pocket or groove formed by the enzyme's folded shape (1.7).

For a reaction to happen, the substrate's shape and charge have to match up with the active site. The R groups lining the active site have specific shapes and charges: for example, a positively charged R group in the active site may attract a negatively charged part of the substrate. That's why enzymes are specific.

When the substrate binds, the two form an enzyme-substrate complex. The active site often shifts slightly to grip the substrate more snugly, an idea called induced fit. While bound, the enzyme helps the reaction along, for example by holding two substrates close together in the right orientation or by straining a bond. Then the products are released, and the unchanged enzyme can bind another substrate. One enzyme molecule can catalyze the same reaction thousands or even millions of times.

Measuring reaction rate

Enzyme experiments usually measure how fast product appears or substrate disappears. The rate formula on the formula sheet is rate = dY/dt: the change in the amount (dY) divided by the change in time (dt). On a graph of product (y-axis) against time (x-axis), the rate is the slope.

The steepest part of the curve is usually at the start, when substrate is most plentiful. As substrate is used up, the curve flattens. That's why experiments compare initial rates.

Worked examples

Try each one yourself first, then open the solution.

  1. Example 1Calculator allowed

    Calculating rate from data

    The enzyme catalase breaks down hydrogen peroxide (H₂O₂) into water and O₂. A student measures the total O₂ produced: 0 mL at 0 s, 3.0 mL at 30 s, 5.6 mL at 60 s, 7.4 mL at 90 s and 8.4 mL at 120 s. Calculate the rate for the first 30 s and for the last 30 s, and explain the difference.

    Show the solution
    1. Step 1: Use rate = dY/dt, the change in O₂ divided by the change in time.
    2. Step 2: First 30 s: (3.0 mL − 0 mL) ÷ (30 s − 0 s) = 0.10 mL/s.
    3. Step 3: Last 30 s: (8.4 mL − 7.4 mL) ÷ (120 s − 90 s) = 1.0 mL ÷ 30 s ≈ 0.033 mL/s.
    4. Step 4: Explain: the enzyme is still present and unchanged, but the substrate (H₂O₂) is being used up. With fewer substrate molecules, they collide with active sites less often, so the rate drops.

    Answer: 0.10 mL O₂/s for 0–30 s and about 0.033 mL O₂/s for 90–120 s. The rate falls because the substrate is being used up.

  2. Example 2

    What an enzyme can't change (classic trap)

    A student claims that adding an enzyme to a reaction makes the reaction release more energy, because the reaction 'goes better.' Use an energy diagram to evaluate the claim.

    Show the solution
    1. Step 1: Sketch it in words: the reactants start at one energy level and the products end at a lower level. The difference between them is the energy released.
    2. Step 2: The enzyme lowers only the hill between them, the activation energy.
    3. Step 3: The reactant and product energy levels don't move, so the energy released stays the same.
    4. Step 4: What changes is the speed: with a lower hill, many more molecules can get over it each second.

    Answer: The claim is wrong. An enzyme lowers the activation energy so the reaction goes faster, but the energy released (the difference between reactants and products) is unchanged.

Common mistakes

  • Saying enzymes provide energy or 'add activation energy.' They lower the activation energy.
  • Saying enzymes are used up or changed by the reaction. They're released unchanged and reused.
  • Describing specificity only by shape. Charge matters too: the substrate's charges must be compatible with those in the active site.
  • Calculating the overall average rate when the question asks for the initial rate. Use the first time interval.

On the exam

  • Expect energy diagrams where you identify the activation energy and the curve for the catalyzed reaction (the lower hill, same start and end).
  • Rate calculations from tables or graphs are common. Show the formula, the values with units, and say whether the rate is increasing or decreasing and why.

Connected topics

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Check yourself

4 questions on 3.1 Enzymes. Pick an answer to see if you got it, and why.

Question 1 of 4

A reaction-energy diagram is described as follows. For the uncatalyzed reaction, the energy rises 80 kJ/mol above the reactants before falling to products that are 30 kJ/mol below the reactants. With an enzyme present, the energy rises only 40 kJ/mol above the reactants before falling to the same products. Which of the following correctly describes the enzyme's effect?

Question 2 of 4

An enzyme breaks down maltose but has no effect on sucrose, even though both are disaccharides with the same chemical formula. Which of the following best explains this?

Question 3 of 4

A small amount of an enzyme is added to a large amount of substrate. After the reaction is complete, nearly all of the substrate has been converted to product, and the enzyme can be recovered unchanged. Which property of enzymes does this demonstrate?

Question 4 of 4

The part of a substrate that binds an enzyme's active site carries a negative charge. A mutation replaces an amino acid in the active site that has a positively charged R group with one that has a nonpolar R group. Which of the following is the most likely effect?

0 of 4 answered