Vmax, Km and Enzyme Inhibition

AS · 17 min

When substrate concentration is increased, the rate of an enzyme-catalysed reaction rises towards a maximum. Two numbers taken from that curve describe an enzyme precisely: its maximum rate, $V_{\max}$ , and the Michaelis–Menten constant, $K_{\text{m}}$ , which measures how strongly the enzyme attracts its substrate. Inhibitors change these numbers in characteristic ways, and that is how biochemists tell a competitive inhibitor from a non-competitive one. Expect to read $V_{\max}$ and $K_{\text{m}}$ off graphs, compare enzymes using $K_{\text{m}}$ , and explain inhibition curves in Paper 2.

The maximum rate, Vmax

Recall the shape of a graph of initial rate against substrate concentration, with a fixed concentration of enzyme (see Factors affecting enzyme activity).

y = 10x/(2 + x) (0, 10) -- (20, 10) (0, 5) -- (2, 5) (2, 5) -- (2, 0)

(Horizontal axis: substrate concentration / mmol dm⁻³; vertical axis: initial rate of reaction / arbitrary units. The curve rises steeply, bends, and approaches the horizontal line at a rate of 10, which is $V_{\max}$ . The construction lines read across from half of $V_{\max}$ (a rate of 5) to the curve and down to the $x$ -axis, giving $K_{\text{m}} = 2\ \text{mmol dm}^{-3}$ .)

As substrate concentration increases, more and more active sites are occupied at any one time. Eventually, every active site is occupied as soon as it releases its products: the enzyme is saturated with substrate.

Definition

$V_{\max}$ is the maximum (theoretical) rate of an enzyme-catalysed reaction, reached when all of the enzyme molecules are involved in enzyme–substrate complexes (all the active sites are occupied), so that adding more substrate cannot increase the rate.

Three features of $V_{\max}$ are worth noting:

$V_{\max}$ depends on the concentration of enzyme. Double the enzyme concentration, and you double $V_{\max}$ , because there are twice as many active sites.
$V_{\max}$ is approached but, in practice, never quite reached: the curve gets closer and closer to it. It is estimated from the plateau of the graph.
$V_{\max}$ is a rate, so it has rate units (for example $\text{cm}^3\ \text{s}^{-1}$ , $\text{mmol dm}^{-3}\ \text{min}^{-1}$ , or arbitrary units).

The Michaelis–Menten constant, Km

Different enzymes need very different substrate concentrations before they work fast. Some reach high rates even at very low substrate concentrations; others need a lot of substrate. The Michaelis–Menten constant puts a number on this.

Definition

The Michaelis–Menten constant, $K_{\text{m}}$ , is the substrate concentration at which an enzyme works at half its maximum rate ( $\tfrac{1}{2}V_{\max}$ ). It is used as a measure of the affinity of an enzyme for its substrate.

Key result

Low $K_{\text{m}}$ = high affinity: the enzyme reaches half its maximum rate at a low substrate concentration. Substrate binds readily to the active site and ES complexes form easily even when substrate is scarce.
High $K_{\text{m}}$ = low affinity: a high substrate concentration is needed to reach half the maximum rate.
$K_{\text{m}}$ has units of concentration (e.g. $\text{mmol dm}^{-3}$ or $\text{mol dm}^{-3}$ ).
$K_{\text{m}}$ is a property of the enzyme and its substrate. It does not depend on the enzyme concentration (doubling the enzyme doubles $V_{\max}$ , so half of the new $V_{\max}$ is reached at the same substrate concentration).

Why use $K_{\text{m}}$ rather than $V_{\max}$ to compare enzymes? $V_{\max}$ depends on how much enzyme is present, which varies from experiment to experiment. $K_{\text{m}}$ does not, so it allows a fair comparison of different enzymes (or of the same enzyme with different substrates).

Finding Vmax and Km from a graph

Draw (or look at) the graph of initial rate (y-axis) against substrate concentration (x-axis).
Estimate $V_{\max}$ : the value the curve levels off towards. Draw a horizontal line at this value.
Halve it: $\tfrac{1}{2}V_{\max}$ .
Read across from $\tfrac{1}{2}V_{\max}$ on the y-axis to the curve.
Read down from that point to the x-axis. The substrate concentration there is $K_{\text{m}}$ .
Give $K_{\text{m}}$ with concentration units and $V_{\max}$ with rate units.

Reading Vmax and Km (routine)

The initial rate of an enzyme-catalysed reaction was measured at different substrate concentrations, with a fixed enzyme concentration.

Substrate concentration / mmol dm⁻³	0.5	1.0	2.0	4.0	8.0	16.0	32.0
Initial rate / µmol min⁻¹	10.0	16.7	25.0	33.3	40.0	44.4	47.1

(a) Estimate $V_{\max}$ . (b) Use your answer to find $K_{\text{m}}$ . (c) Explain why increasing the substrate concentration from 16.0 to 32.0 mmol dm⁻³ increases the rate by only 2.7 µmol min⁻¹.

Solution

(a) The rate is levelling off and approaching about $50\ \mu\text{mol min}^{-1}$ . (Any value a little above the highest reading is acceptable, since $V_{\max}$ is never actually reached.)

(b) $\tfrac{1}{2}V_{\max} = 25\ \mu\text{mol min}^{-1}$ . This rate occurs at a substrate concentration of $2.0\ \text{mmol dm}^{-3}$ , so $K_{\text{m}} = 2.0\ \text{mmol dm}^{-3}$ .

(c) At 16 mmol dm⁻³ almost all the active sites are already occupied at any time; increasing substrate cannot greatly increase the number of enzyme–substrate complexes formed per unit time. The rate is close to $V_{\max}$ and the enzyme concentration is the limiting factor.

Comparing affinities with Km (moderate)

In human cells, glucose is phosphorylated by two different enzymes. Hexokinase, found in most tissues, has a $K_{\text{m}}$ for glucose of about $0.1\ \text{mmol dm}^{-3}$ . Glucokinase, found in liver cells, has a $K_{\text{m}}$ of about $8\ \text{mmol dm}^{-3}$ . Blood glucose concentration is normally about $5\ \text{mmol dm}^{-3}$ .

(a) State which enzyme has the higher affinity for glucose. Explain your answer. (b) Using $\text{rate} = \dfrac{V_{\max} \times [\text{S}]}{K_{\text{m}} + [\text{S}]}$ , calculate the rate of each enzyme at a glucose concentration of $5\ \text{mmol dm}^{-3}$ as a percentage of its $V_{\max}$ . (c) Suggest why it is an advantage that the liver enzyme has a high $K_{\text{m}}$ .

Solution

(a) Hexokinase: it has the lower $K_{\text{m}}$ , so it reaches half its maximum rate at a much lower glucose concentration; glucose binds to its active site more readily / ES complexes form at low substrate concentrations.

(b) Hexokinase: $\dfrac{5}{0.1 + 5} = 0.98$ , so 98% of $V_{\max}$ . Glucokinase: $\dfrac{5}{8 + 5} = 0.38$ , so 38% of $V_{\max}$ .

(c) Hexokinase is almost saturated at normal blood glucose, so its rate hardly changes when blood glucose rises; cells in most tissues take in glucose at a steady rate even when glucose is scarce. Glucokinase is far from saturated, so its rate increases as blood glucose rises (after a meal), so the liver takes up and stores more glucose (as glycogen) only when there is excess.

(The formula is given here to let you calculate. You do not need to learn it for this syllabus: $K_{\text{m}}$ questions are normally answered from a graph.)

Enzyme inhibitors

Definition

An inhibitor is a substance that reduces the rate of an enzyme-catalysed reaction by binding to the enzyme. A reversible inhibitor binds by weak, non-covalent interactions (such as hydrogen bonds and ionic bonds), so it can detach again; its effect depends on its concentration and is removed when the inhibitor is removed.

There are two kinds of reversible inhibitor on the syllabus.

Competitive inhibitors

A competitive inhibitor has a similar shape to the substrate. It is complementary to the active site, so it can bind to the active site and occupy it. While it is bound, the substrate cannot enter, so no ES complex can form at that active site. The inhibitor is not converted to products (or only very slowly), and eventually leaves.

Substrate and inhibitor are competing for the same active sites. Whether a given active site is occupied by substrate or by inhibitor depends on the relative concentrations of the two:

If substrate concentration is increased, a substrate molecule is more likely than an inhibitor molecule to collide with any free active site. The inhibition becomes less and less important.
At very high substrate concentrations, the inhibitor's effect is almost entirely overcome. The same $V_{\max}$ is eventually reached.
But a higher substrate concentration is needed to reach $\tfrac{1}{2}V_{\max}$ , so the apparent $K_{\text{m}}$ increases (the apparent affinity for substrate falls).

Example: malonate (malonic acid) has a similar shape to succinate, the substrate of the respiratory enzyme succinate dehydrogenase, and is a competitive inhibitor of that enzyme. A medical example: methanol poisoning is treated with ethanol, which competes with methanol for the active site of alcohol dehydrogenase, so less methanol is converted to toxic methanal.

Non-competitive inhibitors

A non-competitive inhibitor does not resemble the substrate and does not bind to the active site. It binds to a different site on the enzyme (sometimes called an allosteric site). This changes the tertiary structure of the enzyme, which changes the shape of the active site, so the substrate can no longer bind (or the reaction can no longer be catalysed even if it does).

The inhibitor and substrate are not competing for the same site, so increasing the substrate concentration does not overcome the inhibition.
The inhibited enzyme molecules are effectively removed from action: it is as if there were less enzyme. So $V_{\max}$ is reduced.
The remaining, uninhibited enzyme molecules work normally, so (for a simple non-competitive inhibitor) $K_{\text{m}}$ is unchanged.

Example: cyanide is a non-competitive inhibitor of cytochrome c oxidase, an enzyme of the electron transport chain in mitochondria; it binds away from the active site, which is why cyanide is so toxic. Many heavy metal ions, such as $\text{Hg}^{2+}$ and $\text{Ag}^+$ , inhibit enzymes non-competitively by binding to R groups (such as the $-\text{SH}$ of cysteine) away from the active site, although their binding is often irreversible.

Comparing the graphs

y = 10x/(2 + x) y = 10x/(6 + x) y = 5x/(2 + x) (0, 10) -- (40, 10)

(Horizontal axis: substrate concentration / mmol dm⁻³; vertical axis: initial rate. The top curve, levelling towards the line at 10, is the enzyme with no inhibitor ( $V_{\max} = 10$ , $K_{\text{m}} = 2$ ). The middle curve is with a competitive inhibitor: it rises more slowly but continues to approach the same $V_{\max}$ of 10 at high substrate concentrations ( $K_{\text{m}}$ is now 6). The lowest curve is with a non-competitive inhibitor: it levels off at a lower maximum of 5, and reaches half of that at the same substrate concentration, 2, as the uninhibited enzyme.)

Key result

Feature	Competitive inhibitor	Non-competitive inhibitor
Shape	similar to substrate	not similar to substrate
Where it binds	active site	a site other than the active site (allosteric site)
Effect on active site	blocks it; substrate cannot enter	changes its shape; substrate no longer complementary / cannot bind
Effect of increasing substrate concentration	inhibition reduced; can be overcome	inhibition not reduced
$V_{\max}$	unchanged (eventually reached at high [S])	reduced
$K_{\text{m}}$	increased	unchanged
Example	malonate on succinate dehydrogenase	cyanide on cytochrome c oxidase

Not examined: end-product inhibition and irreversible inhibitors

In many metabolic pathways, the final product acts as a non-competitive inhibitor of an enzyme early in the pathway. When product builds up, the pathway slows; when product is used up, the inhibitor detaches and the pathway speeds up again. This is end-product inhibition, a form of negative feedback. Irreversible inhibitors bind permanently, usually by covalent bonds; many nerve gases and some insecticides inhibit acetylcholinesterase in this way. Neither idea is a named learning outcome at AS, but both are useful context.

Identifying the type of inhibitor from data (moderate)

The initial rate of an enzyme reaction was measured with no inhibitor, with inhibitor X and with inhibitor Y.

Substrate concentration / mmol dm⁻³	1	2	4	8	16	32	64
No inhibitor	16.7	25.0	33.3	40.0	44.4	47.1	48.5
Inhibitor X	7.1	12.5	20.0	28.6	36.4	42.1	45.7
Inhibitor Y	8.3	12.5	16.7	20.0	22.2	23.5	24.2

(a) Calculate the percentage reduction in rate caused by each inhibitor at $2\ \text{mmol dm}^{-3}$ and at $64\ \text{mmol dm}^{-3}$ substrate. (b) Identify the type of inhibitor X and Y. Explain your answers using the data.

Solution

(a) At $2\ \text{mmol dm}^{-3}$ : X: $\dfrac{25.0 - 12.5}{25.0} \times 100 = 50\%$ ; Y: $\dfrac{25.0 - 12.5}{25.0} \times 100 = 50\%$ .

At $64\ \text{mmol dm}^{-3}$ : X: $\dfrac{48.5 - 45.7}{48.5} \times 100 = 5.8\%$ ; Y: $\dfrac{48.5 - 24.2}{48.5} \times 100 = 50\%$ .

(b) X is competitive. Its inhibition falls from 50% to under 6% as substrate concentration increases; at high substrate concentration the rate approaches the same maximum (about 50) as without inhibitor. Substrate molecules outcompete the inhibitor for the active sites.

Y is non-competitive. Its inhibition stays at 50% at all substrate concentrations; the rate levels off at about half the uninhibited value (about 25), so $V_{\max}$ is reduced. Increasing substrate cannot overcome it because Y does not bind to the active site.

(A further check: without inhibitor and with Y, half of the respective $V_{\max}$ (25 and 12.5) is reached at the same substrate concentration, $2\ \text{mmol dm}^{-3}$ , so $K_{\text{m}}$ is unchanged by Y. With X, half of $V_{\max}$ (25) is reached at about $6\ \text{mmol dm}^{-3}$ , so the apparent $K_{\text{m}}$ is increased.)

Explaining competitive inhibition (moderate, 4 marks)

Malonate is a competitive inhibitor of succinate dehydrogenase. Explain why adding a large amount of succinate reduces the inhibition caused by malonate.

Solution

Malonate has a similar shape to succinate / is complementary to the active site;
it binds to the active site so succinate cannot bind / prevents ES complex formation;
malonate and succinate compete for the active site; binding is reversible so malonate leaves again;
with more succinate, succinate molecules are more likely to collide with / occupy free active sites than malonate, so more ES complexes form and the rate approaches $V_{\max}$ .

Exam-hard: sketching and explaining inhibition (6 marks)

(a) On one set of axes, sketch curves of initial rate against substrate concentration for an enzyme with no inhibitor, with a fixed concentration of a competitive inhibitor and with a fixed concentration of a non-competitive inhibitor. Label $V_{\max}$ for each. (b) Explain the differences between the three curves.

Solution

(a) All three curves start at the origin. The uninhibited curve rises steeply and levels off at $V_{\max}$ . The competitive curve rises less steeply but, at high substrate concentrations, approaches the same $V_{\max}$ . The non-competitive curve levels off at a lower $V_{\max}$ . (Points for: correct axes labels; competitive curve below uninhibited but converging; non-competitive plateau clearly lower.)

(b)

Competitive inhibitor has a similar shape to the substrate and binds to the active site, so fewer active sites are available for substrate and fewer ES complexes form at low substrate concentration.
At high substrate concentration, substrate outcompetes the inhibitor (more likely to occupy the active site), so the same $V_{\max}$ is reached; $K_{\text{m}}$ increases.
Non-competitive inhibitor binds to a site other than the active site;
this changes the tertiary structure / shape of the active site, so substrate cannot bind;
increasing substrate concentration does not affect this, so a proportion of enzyme molecules is always out of action;
$V_{\max}$ is lower; $K_{\text{m}}$ is unchanged because the unaffected enzyme molecules bind substrate normally.

Watch out

$K_{\text{m}}$ is the substrate concentration, not a rate. Students often read the $y$ -value ( $\tfrac{1}{2}V_{\max}$ ) and call it $K_{\text{m}}$ . Always read down to the $x$ -axis.
Low $K_{\text{m}}$ means high affinity. The reverse is a very common slip.
Do not say a non-competitive inhibitor "binds to the active site and changes its shape". It binds elsewhere and the active site changes shape as a result.
Do not say a competitive inhibitor "changes the shape of the active site". It simply occupies it.
"Adding more substrate overcomes non-competitive inhibition" is wrong. Only competitive inhibition can be overcome by more substrate.
$V_{\max}$ depends on enzyme concentration; $K_{\text{m}}$ does not.

Exam tip

Learn the syllabus statement of $V_{\max}$ : reached when all of the enzyme molecules are involved in ES complex formation / all active sites occupied.
For $K_{\text{m}}$ , two marks are usually available: "substrate concentration at which rate is half $V_{\max}$ " and "low $K_{\text{m}}$ = high affinity".
When comparing inhibitors from a graph, refer to the data: "with inhibitor A, the rate at high substrate concentration reaches the same maximum of 48 units as without inhibitor".
When asked "suggest the type of inhibitor", look at the high substrate end of the graph: same maximum = competitive; lower maximum = non-competitive.
A common question gives an inhibitor's structure and the substrate's structure: if they look similar, it is competitive.

Summary

$V_{\max}$ : the maximum rate, reached when all enzyme molecules are in ES complexes (all active sites occupied); it depends on enzyme concentration.
$K_{\text{m}}$ : substrate concentration at which the rate is $\tfrac{1}{2}V_{\max}$ ; a measure of affinity; low $K_{\text{m}}$ = high affinity; independent of enzyme concentration.
Find $K_{\text{m}}$ by halving $V_{\max}$ , reading across to the curve, then down to the substrate axis.
Reversible inhibitors bind by weak bonds and can detach.
Competitive: similar shape to substrate; binds active site; overcome by high substrate; $V_{\max}$ unchanged, $K_{\text{m}}$ increased.
Non-competitive: binds elsewhere, changing the active site shape; not overcome by substrate; $V_{\max}$ reduced, $K_{\text{m}}$ unchanged.

Practice questions

Question

Define $V_{\max}$ .
Define the Michaelis–Menten constant and state what it indicates about an enzyme.
Enzyme A has a $K_{\text{m}}$ of $0.5\ \text{mmol dm}^{-3}$ and enzyme B has a $K_{\text{m}}$ of $12\ \text{mmol dm}^{-3}$ for the same substrate. Which has the greater affinity for the substrate? Explain.
A student doubled the concentration of an enzyme. State and explain the effect on (a) $V_{\max}$ and (b) $K_{\text{m}}$ .
An enzyme has a $V_{\max}$ of $80\ \text{mmol dm}^{-3}\ \text{s}^{-1}$ . A graph shows a rate of $40\ \text{mmol dm}^{-3}\ \text{s}^{-1}$ at a substrate concentration of $3.5\ \text{mmol dm}^{-3}$ . State $K_{\text{m}}$ .
Explain why a competitive inhibitor does not change $V_{\max}$ . (3 marks)
Explain how a non-competitive inhibitor reduces the rate of an enzyme-catalysed reaction. (3 marks)
Using the equation in the hexokinase example, calculate the rate of an enzyme with $V_{\max} = 60$ arbitrary units and $K_{\text{m}} = 4.0\ \text{mmol dm}^{-3}$ at a substrate concentration of $12\ \text{mmol dm}^{-3}$ . Then explain what the answer shows about the occupancy of active sites.
A drug is designed to inhibit an enzyme in a parasite. In tests, the inhibition was 70% at low substrate concentration and 68% at high substrate concentration. Identify the type of inhibition and explain why this might make a more effective drug than a competitive inhibitor when the substrate accumulates in the parasite. (4 marks)
Sulfanilamide, an antibacterial drug, has a structure similar to para-aminobenzoic acid (PABA), the substrate of a bacterial enzyme that makes folic acid. Humans obtain folic acid from their diet and do not have this enzyme. (a) State the type of inhibition. (b) Predict the effect of sulfanilamide on $V_{\max}$ and $K_{\text{m}}$ of the bacterial enzyme. (c) Explain why high concentrations of PABA in the bacterium would reduce the effectiveness of the drug, and why the drug is relatively harmless to the patient. (6 marks)

Answers

The maximum rate of an enzyme-catalysed reaction, reached when all the enzyme molecules are involved in enzyme–substrate complexes (all active sites are occupied), so that increasing substrate concentration cannot increase the rate further.
The substrate concentration at which an enzyme catalyses the reaction at half its maximum rate ( $\tfrac{1}{2}V_{\max}$ ); it indicates the affinity of the enzyme for its substrate (low $K_{\text{m}}$ = high affinity).
Enzyme A: lower $K_{\text{m}}$ ; reaches half its maximum rate at a much lower substrate concentration, so the substrate binds to its active site / forms ES complexes more readily.
(a) $V_{\max}$ doubles: twice as many active sites, so twice as many ES complexes can form per unit time when all are occupied. (b) $K_{\text{m}}$ is unchanged: it is a property of the enzyme's affinity for its substrate; half of the new $V_{\max}$ is reached at the same substrate concentration.
$40 = \tfrac{1}{2} \times 80$ , so $K_{\text{m}} = 3.5\ \text{mmol dm}^{-3}$ .
The competitive inhibitor binds reversibly to the active site and competes with substrate; at very high substrate concentration, substrate molecules are far more likely to occupy any free active site than the inhibitor; so (virtually) all active sites are occupied by substrate and the same maximum rate is reached.
It binds to the enzyme at a site other than the active site (allosteric site); this changes the tertiary structure of the enzyme and so the shape of the active site; substrate is no longer complementary/cannot bind, so fewer ES complexes form (the affected enzyme molecules are out of action).
Rate $= \dfrac{60 \times 12}{4.0 + 12} = \dfrac{720}{16} = 45$ arbitrary units, which is 75% of $V_{\max}$ . It shows that at three times the $K_{\text{m}}$ , about three-quarters of the active sites are occupied at any moment; a further increase in substrate gives smaller and smaller increases in rate.
Non-competitive inhibition: the percentage inhibition is (almost) the same at low and high substrate concentrations, so it is not overcome by substrate. If substrate accumulates (because the enzyme is inhibited), a competitive inhibitor would be outcompeted and its effect would decline; a non-competitive inhibitor binds elsewhere and keeps the active site in a changed shape whatever the substrate concentration, so inhibition is maintained.
(a) Competitive inhibition. (b) $V_{\max}$ unchanged; $K_{\text{m}}$ increased. (c) Sulfanilamide has a similar shape to PABA and competes with it for the active site; a high concentration of PABA makes it more likely that PABA, not the drug, occupies the active sites, so more ES complexes form and folic acid is still made. Human cells do not have this enzyme (they take in folic acid from food), so there is no human enzyme with a complementary active site for the drug to inhibit in this pathway.

The maximum rate, Vmax

The Michaelis–Menten constant, Km

Enzyme inhibitors

Competitive inhibitors

Non-competitive inhibitors

Comparing the graphs

Practice questions

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