AP Biology · Unit 3 · Topic 3.2
Temperature, pH and concentration each have a range an enzyme tolerates and a point past which it does not. Push each one and watch the rate.
Learning objective 3.2.A Explain how environmental factors — temperature, pH and concentration — affect the structure and function of an enzyme.
Pick an enzyme, then move it away from the conditions it likes. Some of what you do here can be undone. One thing cannot — and finding out which is the whole point of the panel.
Try this: set the pH to 7 and step through all three enzymes. Then set it to 2 and step through them again. The human enzyme and pepsin swap places completely. There is no pH that is good for enzymes — there is only a pH that suits a particular enzyme.
On the way to the optimum, raising the temperature speeds the reaction up for an ordinary reason: molecules move faster, so enzyme and substrate collide more often and with more energy.
Past the optimum the rate falls for a different reason. Heat starts to disturb the interactions that hold the folded protein in shape. At first this is a slight, partial loosening — the site is a slightly poorer fit, so fewer collisions produce a reaction. Push further and the structure gives way entirely.
Denature an enzyme in the panel above, then press Put it back at its optimum. The conditions return; the rate does not. To get a working rate again you have to fetch enzyme that was never denatured — a different molecule, not the same one recovered.
Concentration changes nothing about the enzyme's structure — every change you make here is undoable. What it changes is how often an active site has something to work on.
Every one of these steps adds the same amount of substrate. The rate they buy gets smaller each time, because a larger and larger share of the active sites is already occupied when the substrate arrives.
Switch the enzyme amount above from 1× to 2× with the substrate slider left where it is. The rate does not just creep up toward the same ceiling — the ceiling moves, because you have added active sites rather than work for the existing ones.
This is the distinction an exam question is testing when it says “the rate levelled off” and asks what would raise it further. More substrate will not; more enzyme will.
Two inhibitors, both slowing the same enzyme. Read the rate at one substrate concentration and you cannot tell them apart. Run the right experiment and the difference is obvious.
A flask contains an unknown inhibitor. You are given the rate at an ordinary substrate concentration and the rate when the enzyme is flooded with substrate. Decide which kind it is.
Write an answer to each in your own words before you open the answers. The last two are the ones that carry the marks.
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Every claim on this page is a shape on a graph. These are the four shapes, and what each one commits you to saying.
| What the graph shows | What it tells you |
|---|---|
| A peak with a fall on both sides | An optimum. Below it the rate is limited by how often and how energetically molecules collide; above it by the active site beginning to lose its shape. |
| The rate reaches zero and does not come back | Denaturation. The active site has lost the shape that made it specific, and restoring the condition does not restore the shape. |
| A rise that flattens into a plateau | Saturation. Every active site is occupied; the enzyme, not the substrate, is now the limit. |
| A curve that needs more substrate but reaches the same plateau | Competitive inhibition. The maximum is unchanged; only the substrate needed to reach it has risen. |
| A curve that plateaus lower, at any substrate concentration | Non-competitive inhibition. The maximum itself is lowered, and substrate cannot recover it. |