Title: InkBiology / Enzymes and metabolic control
On every note of this deck, unless a note says otherwise:
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Front: SPOT THE ERROR
Say what is wrong with this claim:
Enzymes speed up reactions by giving them energy, a push to get them going.
Back: An enzyme supplies no energy: it sits in solution at the same temperature as everything around it. It offers an alternative route with a lower activation energy, so a far larger fraction of molecules already have enough thermal energy to react.

Front: DEFINE
Define: Catalyst.
Back: A substance that increases the rate of a reaction by lowering its activation energy, and which is not used up in the process.

Front: DEFINE
Define: Limiting factor.
Back: The factor that, of all those affecting a rate, is the one restricting it at that moment.

Front: DEFINE
Define: Allosteric site.
Back: A binding site on an enzyme away from the active site, at which a regulatory molecule binds.

Front: DEFINE
Define: Enzyme.
Back: A globular protein that acts as a biological catalyst, with an active site complementary to its substrate.

Front: SPOT THE ERROR
Say what is wrong with this claim:
A denatured enzyme is one that has been switched off, so cooling it down should switch it back on.
Back: Cooling reverses slowing, not denaturation. Below the optimum the molecules simply collide less often. Denaturation is the tertiary structure coming apart as hydrogen and ionic bonds break, and a cooled protein does not refold reliably into a working active site.

Front: CHECK YOURSELF
A company has two candidate molecules that both slow the same enzyme in cultured cells. When the substrate concentration in the culture is raised tenfold, molecule A stops having any measurable effect while molecule B works exactly as well as before. Identify each type of inhibition, and advise the company on which molecule is the better starting point for a drug that must work in a tissue where substrate concentration rises and falls sharply after meals.
Back: Molecule A is a competitive inhibitor. Its effect disappears at high substrate concentration because it binds the active site and is displaced once substrate molecules greatly outnumber it, so Vmax is reached in spite of the inhibitor being present.
Molecule B is a non-competitive inhibitor. It binds away from the active site and alters the enzyme's tertiary structure, so the site is no longer complementary to the substrate. Substrate cannot compete for a place it does not occupy, and the maximum rate is lower however much is available.
For the tissue described, B is the better starting point: a drug based on A would lose its effect exactly when substrate concentration peaked after a meal, so the dose would in effect vary with what the patient had eaten. The advice needs a caveat, though. Non-competitive inhibition cannot be reversed by raising the substrate concentration, so an overdose is harder to counteract, and a molecule that binds one allosteric site may bind others. Check B's specificity across related enzymes, and confirm its binding is reversible, before committing.

Front: DEFINE
Define: Optimum.
Back: The value of a variable at which an enzyme's rate is highest.

Front: DEFINE
Define: Active site.
Back: The region of an enzyme, formed by its tertiary structure, to which the substrate binds.

Front: DEFINE
Define: Non-competitive inhibitor.
Back: A molecule that binds to a site other than the active site, altering the enzyme's tertiary structure so that the active site is no longer complementary to the substrate.

Front: DEFINE
Define: Activation energy.
Back: The minimum energy that colliding particles must have for a reaction to take place.

Front: CHECK YOURSELF
Two students investigate how temperature affects amylase. One times how long a starch solution takes to stop giving a blue-black colour with iodine and calculates 1/time. The other measures the mass of reducing sugar formed after ten minutes. They disagree about whether 45 °C or 35 °C is better. Explain how the same enzyme can give them opposite answers, and say which method you would trust.
Back: The two methods are measuring different things, and the difference matters most at exactly the temperature they disagree about.
Timing to an early end point is close to an initial-rate measurement. It samples the first part of the run, before much enzyme has denatured, so at 45 °C it records the genuinely high rate that the extra kinetic energy delivers. That method will report 45 °C as the better temperature.
Measuring product after ten minutes gives a mean rate over the whole period. At 45 °C the enzyme denatures steadily during those ten minutes, so the reaction that began quickly slows and may stop altogether, while at 35 °C the enzyme survives and keeps working. The mean rate over ten minutes can therefore be higher at 35 °C even though the initial rate is higher at 45 °C. Both students have measured correctly; neither has measured the same quantity.
Which to trust depends on what is being asked. For how temperature affects the enzyme's activity, the initial-rate method is the fair comparison. For which temperature to run a ten-minute process at, the second student's answer is the useful one. The full response is that stability over time is a second variable, and that a proper investigation would take a progress curve at each temperature.

Front: CHECK YOURSELF
An industrial process converts substrate S into product P using an enzyme. At 25 °C the reaction reaches equilibrium with 90% P and 10% S. A colleague suggests doubling the amount of enzyme in order to push the yield above 90%. Evaluate that suggestion.
Back: It will not work, and the reason is worth stating precisely: an enzyme changes the rate at which equilibrium is reached, not where equilibrium lies.
The 90:10 split is fixed by the energy difference between S and P. An enzyme lowers the activation energy of the route between them, and it lowers it by the same amount in both directions, so the forward and reverse reactions are both speeded up in the same proportion and the balance point does not move. Doubling the enzyme will get the mixture to 90% P sooner, which may well be worth doing, but it will stop there.
To change the yield you have to change something the equilibrium responds to: remove P as it forms, raise the concentration of S, or alter the temperature, remembering that a temperature high enough to help the equilibrium may be high enough to denature the enzyme. More enzyme is a rate decision; yield is a different question.

Front: DEFINE
Define: End-product inhibition.
Back: The inhibition of an enzyme early in a metabolic pathway by the final product of that pathway, so that the rate of the pathway matches demand.

Front: DEFINE
Define: Specificity.
Back: The property of an enzyme that it catalyses only one reaction, or one class of reaction, because only a substrate complementary to its active site can bind.

Front: DEFINE
Define: Substrate.
Back: The molecule on which an enzyme acts.

Front: DEFINE
Define: Competitive inhibitor.
Back: A molecule with a shape similar to the substrate that binds to the active site, preventing the substrate from binding.

Front: DEFINE
Define: Enzyme-substrate complex.
Back: The temporary structure formed when a substrate binds to the active site.

Front: DEFINE
Define: Initial rate.
Back: The rate of reaction at time zero, found from the gradient of the tangent to a progress curve at the origin.

Front: SPOT THE ERROR
Say what is wrong with this claim:
An inhibitor must be a poison, and a cell would never make one on purpose.
Back: Inhibition is how metabolism is regulated, not only how it is poisoned. Cells inhibit their own enzymes deliberately and reversibly to stop making what they already have in surplus, and ATP is doing most of that inhibiting right now.