Free AP Biology Cellular Energetics

AP Biology · Cellular Energetics · free lesson

Enzymes

The idea, then worked examples solved step by step, then practice questions whose numbers change every time you try them.

The idea

Enzymes: protein catalysts that speed up the cell's reactions

A cell carries out thousands of chemical reactions, and many of them would be far too slow at body temperature if left alone. Sugar in a bowl can sit for years in air even though burning it releases energy. Cells get around the problem with enzymes. Enzymes are proteins that act as biological catalysts: they make a chemical reaction in a cell go faster by lowering its activation energy. Because a cell can change how much of an enzyme it has, or how active the enzyme is, enzymes are also points of control: the structure and function of enzymes contribute to the regulation of biological processes.

Activation energy: the hill a reaction must climb

Activation energy is the input of energy a reaction needs to reach the transition state before it can proceed, even when the reaction releases energy overall. Its size varies, and for many reactions it is large enough that they are far too slow without help. Think of a cart at the top of a slope with a small bump just in front of it: the cart will roll down, but first it must be nudged over the bump. The reactants must reach a strained, in-between arrangement (the transition state) before they can turn into products, and getting there costs energy.

One way to speed a reaction is to heat it, but heating a cell would damage it. An enzyme does the job differently. It holds the reactants in a position that makes the strained arrangement easier to reach, so the bump gets lower. More of the colliding molecules have enough energy to get over a lower bump, and the reaction runs faster at the cell's own temperature.

Read the picture from left to right. Both curves start at the same height (the reactants) and end at the same lower height (the products). The difference is the height of the bump in between. The distance from the reactants up to the top of the bump is the activation energy, and the enzyme's curve has a much shorter climb.

Warning.

Three misconceptions to drop here. An enzyme does not supply energy to the reaction; it lowers the energy barrier. An enzyme does not change the free-energy change of a reaction, which is the difference between the products' height and the reactants' height, and so it cannot make an energy-requiring (endergonic) reaction proceed on its own. It speeds up a reaction that is already favorable. And an enzyme is not used up: it is left unchanged by the reaction it catalyzes, so one enzyme molecule can work on substrate molecule after substrate molecule.

The active site: where the substrate fits

An enzyme is a long chain of amino acids folded into a particular three-dimensional shape. On its surface sits a small pocket or groove called the active site. The reactant that an enzyme acts on is its substrate, and the substrate binds at the active site. For the reaction to take place, the shape and the charge of the substrate must be compatible with the active site: the right shape to fit, and charges and polar regions that attract rather than repel. A cell holds many different molecules, and an enzyme usually works on one substrate or a few closely related ones, because other molecules do not fit its pocket or do not attract it.

When the substrate is bound, the pair is called the enzyme-substrate complex. This is the model the course uses to explain how an enzyme works, and it has three steps: the substrate enters the active site, the enzyme holds it in a position that favors the reaction while bonds are broken or formed, and then the products leave the active site and the enzyme is free to take another substrate.

The picture shows the three steps with the same enzyme at the start and the end: the enzyme comes out the same as it went in. Real active sites are not rigid. In the induced-fit picture, the enzyme and substrate both change shape slightly as they bind, so that the fit becomes best for the reaction. A substrate whose shape or charge does not suit the pocket does not bind well, and that is why an enzyme that breaks down one molecule leaves a differently shaped one alone.

Rate: what an enzyme changes, and what it does not

The rate of a reaction is how much product forms (or how much substrate is used) in a given time: the change in amount divided by the change in time. On a graph of product against time, the steeper the line, the faster the reaction. An enzyme raises the rate. It does not change which way the reaction goes, and when the supply of substrate is fixed, the same total amount of product forms with or without the enzyme; with the enzyme it simply gets there sooner.

Both curves climb toward the same final amount, because the same amount of substrate was supplied. The curve with the enzyme is much steeper at the start, so its rate is higher. Reading the steepness, not the height at one moment, is how you compare rates.

Adapted from Biology by OpenStax (Rice University), licensed under CC BY 4.0. Access for free at https://openstax.org/details/books/biology

Worked examples

Worked examples

Each example is the kind of question the AP exam asks on this topic, worked step by step.

Example.

1. Reading an energy profile. An energy profile for one reaction gives these free-energy values, in kilojoules per mole: the reactants at 50, the top of the bump without an enzyme at 90, the top of the bump with the enzyme at 62, and the products at 20. How much does the enzyme lower the activation energy, and what does it do to the free-energy change?

  1. The activation energy is the climb from the reactants to the top of the bump. Without the enzyme it is 9090 - 5050 = 4040 kilojoules per mole.
  2. With the enzyme it is 6262 - 5050 = 1212 kilojoules per mole.
  3. The enzyme lowers the activation energy by 4040 - 1212 = 2828 kilojoules per mole.
  4. The free-energy change is the products minus the reactants, 2020 - 5050 = −30-30 kilojoules per mole. The products and the reactants sit at the same heights with or without the enzyme, so the value is the same either way.
Result.

The enzyme lowers the activation energy by 2828 kilojoules per mole, from 4040 to 1212, and leaves the free-energy change at −30-30 kilojoules per mole.

Example.

2. A rate from a table. A student measures the product formed in two tubes holding the same amount of substrate. The tube with the enzyme: 0, 15, 27, 33 and 35 micromoles at 0, 1, 2, 3 and 4 minutes. The tube without the enzyme: 0, 3, 6, 9 and 12 micromoles at the same times. Compare the reaction rates over the first minute, and say why the rate in the enzyme tube changes.

  1. A rate is the change in product divided by the change in time. For the enzyme tube in the first minute, (1515 - 00) / (11 - 00) = 1515 micromoles per minute.
  2. For the tube without the enzyme, (33 - 00) / (11 - 00) = 33 micromoles per minute.
  3. The enzyme tube reacts 1515 / 33 = 55 times as fast over the first minute.
  4. The enzyme tube's rate then falls: 1515 in the first minute, 1212 in the second and 22 in the fourth. The substrate is being used up, so there are fewer substrate molecules left to meet the enzyme. The tube without the enzyme holds a constant rate of 33 over the four minutes because it has not yet used up much of its substrate.
Result.

1515 micromoles per minute with the enzyme and 33 without it, so the enzyme makes the reaction 55 times as fast at the start.

Example.

3. Why an enzyme is specific. The enzyme lactase speeds up the breakdown of lactose but has no effect on sucrose, a sugar of similar size. Explain this with the enzyme-substrate complex model.

  1. For an enzyme-catalyzed reaction to occur, the shape and the charge of the substrate must be compatible with the active site of the enzyme.
  2. Lactose has a shape and a pattern of charges that suit the active site of lactase, so it binds there and an enzyme-substrate complex forms. Bonds in the lactose are then broken while it is held in the active site, and the products leave.
  3. Sucrose has a different shape and a different arrangement of charged and polar groups, so it does not fit the active site of lactase and does not bind well. No complex forms, so the reaction is not sped up.
Result.

The active site of lactase suits the shape and charge of lactose, not of sucrose, so the enzyme speeds up one reaction and not the other.

Example.

4. Designing the test. A student wants to find out whether an extract of liver speeds up the breakdown of hydrogen peroxide into water and oxygen. She puts hydrogen peroxide solution in several tubes, adds 1, 2, 3 or 4 milliliters of extract to them, and measures the oxygen gas released in the first minute. Name the independent and dependent variables, and describe and justify a control.

  1. The independent variable is the one the student changes on purpose: the volume of extract added. The dependent variable is the one she measures in response: the volume of oxygen released in the first minute.
  2. A control is a tube that is treated the same way except for the factor being tested. Here it is a tube with the same volume of hydrogen peroxide at the same temperature, with water added in place of the extract.
  3. The control is justified because hydrogen peroxide breaks down slowly even without an enzyme. If the tubes with extract release much more oxygen than the water tube, the extra oxygen can be credited to something in the extract and not to the tube, the temperature or the time.
Result.

Independent variable: volume of extract. Dependent variable: oxygen released in the first minute. Control: peroxide with water in place of extract, kept at the same temperature.

Practice

Answer AP-style questions on enzymes. Each answer is checked, with the full solution, and every try brings new numbers. Practice needs a free account: it keeps your progress.

Practise enzymes with new numbers