Enzymes and enzyme kinetics
แนะนำใหม่| English | ความหมาย |
|---|---|
| activation energy/ˌæktɪˈveɪʃn ˈenədʒi/ | The minimum energy that reacting particles need for a successful reaction. |
| Michaelis–Menten kinetics | Michaelis–Menten kinetics |
| Michaelis constant | Michaelis constant |
| saturated/ˈsætʃəreɪtɪd/ | Describing a solution unable to dissolve more solute under stated conditions, or an organic molecule with no carbon-carbon double or triple bonds. |
| Lineweaver–Burk plot | Lineweaver–Burk plot |
| competitive inhibition | competitive inhibition |
The lab report is enzymes
- Many results on a blood panel are enzyme activities, and most drugs are enzyme inhibitors.
- Kinetics is how biochemistry becomes something you can measure, and how you will read the lab.
Lowering the hill
- An enzyme lowers the activation energy 活化能 of a reaction, so it runs millions of times faster. It does not change where the reaction ends up.
- The enzyme comes out unchanged, so one molecule can turn over thousands of substrate molecules.
An enzyme makes a reaction go further towards products at equilibrium.
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An enzyme speeds both directions equally; it reaches the same equilibrium sooner but does not move it.
The Michaelis–Menten curve
- The rate follows Michaelis–Menten kinetics 米氏动力学:
- $V_{max}$ is the top speed. The Michaelis constant 米氏常数 $K_m$ is the substrate concentration at half of $V_{max}$.
What is $K_m$?
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Put $[S] = K_m$ into the equation: $v = V_{max}K_m / 2K_m = V_{max}/2$.
Why the curve flattens
- E + S ⇌ ES → E + P. At low [S] most enzyme is free, so $v \approx (V_{max}/K_m)[S]$: double the substrate, double the rate.
- At high [S] every enzyme is busy, so $v$ approaches $V_{max}$ and more substrate barely helps. The enzyme is saturated 饱和.
An enzyme has $V_{max}$ = 100 µmol per min and $K_m$ = 2 mM. What is v at [S] = 6 mM, in µmol per min?
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v = 100 × 6 / (2 + 6) = 75 µmol per min. Three times $K_m$ gives three quarters of $V_{max}$: saturation is slow.
The straight-line trick
- Take reciprocals of both sides: $\dfrac{1}{v} = \dfrac{K_m}{V_{max}}\cdot\dfrac{1}{[S]} + \dfrac{1}{V_{max}}$. The curve becomes a straight line.
- On this Lineweaver–Burk plot 双倒数作图, the y-intercept is $1/V_{max}$ and the x-intercept is $-1/K_m$.
A Lineweaver–Burk line crosses the x-axis at −0.25 per mM. What is $K_m$, in mM?
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The x-intercept is −1/$K_m$, so $K_m$ = 1/0.25 = 4 mM.
Worked example: reading two points
- $v = 33$ µmol per min at 1 mM and $v = 60$ at 4 mM. Then $V_{max} = 33(K_m + 1) = 15(K_m + 4)$, so $K_m = 1.5$ mM and $V_{max} = 82.5$ µmol per min.
- Check a third point: at $[S] = 1.5$ mM, $v = 82.5 \times 1.5/3 = 41.25$, exactly half of $V_{max}$.
An enzyme gives v = 40 µmol per min at 2 mM and v = 60 at 6 mM. What is $K_m$?
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$V_{max}$ = 40(K + 2)/2 = 20(K + 2) and $V_{max}$ = 60(K + 6)/6 = 10(K + 6). So 20K + 40 = 10K + 60, $K_m$ = 2 mM and $V_{max}$ = 80. Check: at 2 mM, v = 80 × 2/4 = 40.
Worked example: two kinds of blocker
- (Fictional patient on a statin-like drug) The drug looks like the substrate and sits in the active site: competitive inhibition 竞争性抑制. Apparent $K_m$ rises, $V_{max}$ stays, and enough substrate overcomes it.
- A non-competitive inhibitor binds elsewhere and lowers $V_{max}$, leaving $K_m$ alone. On the Lineweaver–Burk plot the competitive lines meet on the y-axis, the non-competitive ones on the x-axis.
A competitive inhibitor changes the kinetics how?
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It competes for the active site, so more substrate is needed for half speed, but enough substrate still reaches $V_{max}$.
Match each change to what it does to the kinetics.
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Reading which constant moved tells you the mechanism. More enzyme raises $V_{max}$ because there are more active sites.
Three common slips
- "$K_m$ is an affinity." It is a concentration; a small $K_m$ only suggests high apparent affinity.
- Trusting two points. Two points always fit a curve; check a third.
- "Inhibitors remove enzyme." They change activity; changing the amount of enzyme is induction or repression.
A drug that lowers an enzyme's $V_{max}$ must have destroyed some of the enzyme.
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A non-competitive inhibitor lowers $V_{max}$ by disabling bound enzyme without destroying it; fewer enzyme molecules would also lower it.
Recap
- $v = V_{max}[S]/(K_m + [S])$; $K_m$ is the [S] at half $V_{max}$.
- Competitive: $K_m$ up, $V_{max}$ the same. Non-competitive: $V_{max}$ down, $K_m$ the same.
- The Lineweaver–Burk intercepts show which one moved.