Vmax, Km and inhibitors
| English | Chinese | Pinyin |
|---|---|---|
| maximum rate | 最大速率 | zuì dà sù lǜ |
| Michaelis–Menten constant | 米氏常数 | mǐ shì cháng shù |
| affinity | 亲和力 | qīn hé lì |
| competitive | 竞争性 | jìng zhēng xìng |
| non-competitive | 非竞争性 | fēi jìng zhēng xìng |
| immobilised enzyme | 固定化酶 | gù dìng huà méi |
Vmax and Km
- When every active site is full, the reaction hits a top speed — the maximum rate 最大速率, $V_{max}$.
- The curve of rate against substrate concentration climbs and then flattens at $V_{max}$.
What does V_max represent?
V_max is the top rate; the curve flattens there because every active site is busy.
What Km tells you
- The Michaelis–Menten constant 米氏常数, $K_m$, is the substrate concentration that gives half of $V_{max}$.
- It measures the enzyme's affinity 亲和力 (pulling power) for its substrate:
- low $K_m$ → reaches half-speed at low substrate → high affinity.
- high $K_m$ → needs lots of substrate → low affinity.
- So $K_m$ lets you compare how strongly different enzymes hold their substrates.
Both reach the same $V_{max}$, but the low-$K_m$ enzyme gets to half-speed on far less substrate — it grips its substrate more tightly.
Vmax and Km
rate = Vmax·[S] / (Km + [S])
Drag the substrate. The rate approaches Vmax; the substrate that gives half of Vmax is Km — a measure of how tightly the enzyme binds.
An enzyme has V_max = 80 units. At a substrate concentration equal to its Km, what is the rate?
By definition Km is the substrate concentration that gives half of V_max, so the rate is 80 ÷ 2 = 40 units.
A LOW Km means the enzyme only reaches half its maximum rate at a HIGH substrate concentration.
The opposite: low Km means half-speed is reached at LOW substrate — a high affinity, gripping the substrate tightly.
Reversible inhibitors
- An inhibitor slows an enzyme; a reversible one can leave again. Two types:
| Type | Binds | Add more substrate? | $V_{max}$ | $K_m$ |
|---|---|---|---|---|
| competitive 竞争性 | the active site (similar shape to substrate) | out-competes it — effect drops | unchanged | rises |
| non-competitive 非竞争性 | a different site (changes the active site) | does not help | falls | unchanged |
Match each inhibitor effect to the right outcome.
A competitive inhibitor blocks the active site (raises Km, same Vmax); a non-competitive one binds elsewhere and lowers Vmax (Km unchanged).
Adding more substrate overcomes a competitive inhibitor but not a non-competitive one. Why?
A competitive inhibitor blocks the active site, so flooding with substrate wins the competition. A non-competitive inhibitor binds elsewhere, so substrate cannot displace it.
Immobilised enzymes 固定化酶
- An immobilised enzyme is fixed in place — e.g. trapped in alginate beads — while substrate flows past.
- A free enzyme works a little faster, but immobilising it brings big practical wins:
- it is not washed away, so it can be reused.
- the product is pure (not mixed with enzyme).
- it is more stable to changes in temperature and pH.
- the process can run continuously.
- Used industrially — e.g. lactase beads make lactose-free milk.
Select all the advantages of immobilising an enzyme (e.g. in alginate beads).
Immobilised enzymes are reusable, give a pure product and are more stable — but a free enzyme actually works a little faster.
An enzyme fixed in place (e.g. trapped in alginate beads) so it can be reused is described as ______.
Immobilising the enzyme lets it be recovered and reused, and keeps the product enzyme-free.
You've got it
- $V_{max}$ = top rate (all active sites full); $K_m$ = substrate concentration giving half $V_{max}$
- low $K_m$ = high affinity; high $K_m$ = low affinity
- competitive inhibitor: active site, raises $K_m$, $V_{max}$ unchanged (more substrate overcomes it)
- non-competitive inhibitor: other site, lowers $V_{max}$, $K_m$ unchanged
- immobilised enzymes: reusable, pure product, more stable, continuous