Inhibition and the control of pathways

Two kinds of inhibitor

An inhibitor slows an enzyme. Many drugs and poisons are inhibitors. The two kinds that you meet first differ in where they bind.

A competitive inhibitor looks like the substrate. It binds the active site, so substrate and inhibitor compete for the same place. A high substrate level can win the competition.

• $V_{\max}$ stays the same.
• The apparent $K_m$ rises: $K_{m,\text{app}} = K_m\left(1 + \dfrac{[I]}{K_i}\right)$.

A non-competitive inhibitor binds somewhere else, on the enzyme or on the enzyme-substrate complex. The substrate still binds, but the complex makes no product. More substrate cannot help.

• The apparent $V_{\max}$ falls: $V_{\max,\text{app}} = \dfrac{V_{\max}}{1 + [I]/K_i}$.
• $K_m$ stays the same.

Here $K_i$ is the inhibition constant. An inhibitor at $[I] = K_i$ doubles the apparent $K_m$ in the competitive case, and halves $V_{\max}$ in the non-competitive case. Write $i = [I]/K_i$.

Numbers for one enzyme

Take the enzyme of the last lesson, with $V_{\max} = 120$ µmol/min and $K_m = 2.0$ mmol/L, and an inhibitor at $i = 3$.

• Competitive: $V_{\max}$ stays 120. The apparent $K_m$ is $2.0 \times 4 = 8.0$ mmol/L.
• Non-competitive: $K_m$ stays 2.0. The apparent $V_{\max}$ is $120/4 = 30$ µmol/min.

At $[S] = 8$ mmol/L the enzyme alone gives 96 µmol/min. With the competitive inhibitor it gives 60. With the non-competitive inhibitor it gives only 24. Now raise $[S]$ to 64 mmol/L. The competitive case climbs to 106.7, but the non-competitive case reaches only 29.1.

Telling them apart from a graph

On a plot of rate against substrate, a competitive inhibitor gives a curve that rises more slowly but reaches the same top. A non-competitive inhibitor gives a curve with a lower top.

The double reciprocal plot is even clearer.

• Competitive: the lines cross on the y-axis. They share $1/V_{\max}$, and the inhibited line is steeper.
• Non-competitive: the lines cross on the x-axis. They share $-1/K_m$, and the inhibited line is steeper and higher.

Some inhibitors change both $K_m$ and $V_{\max}$. They are called mixed inhibitors, and their lines cross to the left of the y-axis.

Examples that you should know

• Malonate resembles succinate and competes with it for succinate dehydrogenase. Statins compete with HMG-CoA for HMG-CoA reductase, an enzyme of cholesterol synthesis. Methotrexate competes with the substrate of dihydrofolate reductase. All are competitive.
• Mercury and lead ions bind –SH groups on enzymes, far from the active site. They act as non-competitive inhibitors.
• Aspirin adds an acetyl group to the enzyme cyclooxygenase, for good. This is an irreversible inhibitor, and it cannot be washed out.

Allosteric enzymes

Some enzymes have a second site, away from the active site. A small molecule that binds there, an effector, switches the enzyme between a more active and a less active shape. This is allosteric regulation.

Allosteric enzymes usually have several subunits that help each other, which is called cooperativity. Their plot of $v$ against $[S]$ is S-shaped, or sigmoid. It does not follow the Michaelis-Menten equation, and it gives no straight double reciprocal line. An activator moves the curve to the left. An inhibitor moves it to the right.

Feedback and the rate-limiting step

A pathway is a chain of enzymes, each one making the substrate for the next. The rate-limiting step is the slowest one, and it sets the speed of the whole chain. Making the other enzymes faster changes nothing.

Cells control a pathway at its first committed step, which is usually slow and allosteric. The end product acts as the inhibitor. This is feedback, or end-product inhibition. When the product is plentiful it switches off the supply line, and no raw material is wasted.

• In bacteria, isoleucine inhibits threonine deaminase, the first enzyme of its own pathway.
• In the making of pyrimidines, CTP inhibits aspartate transcarbamoylase, the first enzyme, and ATP activates it.

In a branched pathway each end product inhibits the first enzyme of its own branch, so one product does not starve the other.

Glycolysis and ATP

Glycolysis turns glucose into pyruvate and makes ATP. Its main control point is phosphofructokinase-1, the enzyme of its first committed step.

• ATP and citrate inhibit it. Plenty of ATP, or plenty of fuel from other sources, means that the cell needs less glycolysis.
• AMP, ADP and fructose 2,6-bisphosphate activate it. In a working muscle ATP is used up and AMP rises, so glycolysis speeds up.

Look at this one enzyme and you see everything in the lesson: an allosteric site, a sigmoid curve, an effector, and feedback from the product that the cell is short of.

Try it in Python

The first program tabulates the rate for the enzyme alone, with a competitive inhibitor and with a non-competitive inhibitor.

The second program is a pathway that makes a product P. The cell uses P at a steady rate. With feedback, P slows the first enzyme.

Practise with new numbers

Common mistakes

• Saying that a competitive inhibitor lowers $V_{\max}$. At a high enough substrate level the enzyme still reaches $V_{\max}$.
• Saying that a non-competitive inhibitor raises $K_m$. $K_m$ does not change. Only $V_{\max}$ falls.
• Mixing up where the lines cross. Competitive lines cross on the y-axis. Non-competitive lines cross on the x-axis.
• Expecting an allosteric enzyme to follow Michaelis-Menten. Its curve is S-shaped.
• Putting the feedback at the last step. A cell saves the most raw material by switching off the first committed step.

Remember these

• Competitive: $K_{m,\text{app}} = K_m(1 + i)$, $V_{\max}$ unchanged, lines cross on the y-axis, more substrate overcomes it.
• Non-competitive: $V_{\max,\text{app}} = V_{\max}/(1 + i)$, $K_m$ unchanged, lines cross on the x-axis, more substrate does not help.
• Allosteric enzymes have a sigmoid curve. Effectors bind away from the active site.
• Feedback from the end product usually acts on the first committed step, which is rate-limiting.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A drug raises the apparent $K_m$ of an enzyme and leaves $V_{\max}$ alone. What kind of inhibitor is it, and how could you reduce its effect?

2. The double reciprocal lines with and without an inhibitor cross on the x-axis. What kind of inhibitor is it?

3. Why does an allosteric enzyme not give a straight Lineweaver-Burk line?

4. In a pathway of four steps that makes an amino acid, where does the amino acid usually act as an inhibitor, and why?

5. Why does a high level of ATP slow phosphofructokinase-1?