Enzymes — lock-and-key vs induced fit; Michaelis-Menten kinetics: Exercises

Level 1 — Recognition

L1·Q1

Which model of enzyme action describes a rigid active site that fits only one exactly-shaped substrate, "like a key in a lock"?

Recall Solution — L1·Q1

The lock-and-key model (Emil Fischer, 1894). "Rigid pocket, one exact fit" is its signature. Contrast with induced fit, where the site is flexible and moulds around the substrate.

L1·Q2

An enzyme lowers the reaction's activation energy $E_{a}$ . Does it also shift the position of the equilibrium toward products?

Recall Solution — L1·Q2

No. An enzyme is a catalyst: it lowers $E_{a}$ (speeds up both forward and reverse directions equally), so you reach equilibrium faster but land at the same equilibrium position. See Chemical Kinetics for why a catalyst changes rate, not thermodynamics.

L1·Q3

On a graph of $v_{0}$ (y-axis) against $[S]$ (x-axis), what special shape does Michaelis–Menten predict, and what value does the curve level off toward?

Recall Solution — L1·Q3

A rectangular hyperbola: steep rise at low $[S]$ , then flattening. It levels off (saturates) toward $V_{ma x}$ , because there is a fixed amount of enzyme — once all enzyme molecules are occupied, more substrate cannot speed things up. See the shape in the figure below.

Figure — Enzymes — lock-and-key vs induced fit; Michaelis-Menten kinetics

Level 2 — Application

L2·Q1

An enzyme has $V_{ma x} = 20 μ M/s$ and $K_{m} = 4 mM$ . Find $v_{0}$ at $[S] = 4 mM$ .

Recall Solution — L2·Q1

Forecast first: $[S] = K_{m}$ , so by definition we should get exactly half of $V_{ma x}$ . $v_{0} = \frac{V _{ma x} [ S ]}{K _{m} + [ S ]} = \frac{20 \times 4}{4 + 4} = \frac{80}{8} = 10 μ M/s .$ Indeed $10 = \frac{1}{2} (20)$ . ✔ Forecast verified.

L2·Q2

Same enzyme ( $V_{ma x} = 20 μ M/s$ , $K_{m} = 4 mM$ ). Find $v_{0}$ at $[S] = 12 mM$ .

Recall Solution — L2·Q2

Forecast: $[S] = 3 K_{m}$ , well above $K_{m}$ , so we expect to be a fair way up toward $V_{ma x}$ but not there yet. $v_{0} = \frac{20 \times 12}{4 + 12} = \frac{240}{16} = 15 μ M/s .$ That is $75%$ of $V_{ma x}$ — consistent with the rule $\frac{v _{0}}{V _{ma x}} = \frac{[ S ]}{K _{m} + [ S ]} = \frac{3 K _{m}}{4 K _{m}} = \frac{3}{4}$ .

L2·Q3

Same enzyme. What substrate concentration $[S]$ makes $v_{0} = 16 μ M/s$ ?

Recall Solution — L2·Q3

Why cross-multiply? To free $[S]$ from inside the fraction. $16 = \frac{20 [ S ]}{4 + [ S ]} \Rightarrow 16 (4 + [S]) = 20 [S] \Rightarrow 64 + 16 [S] = 20 [S] .$ $64 = 4 [S] \Rightarrow [S] = 16 mM .$ Note $16 mM = 4 K_{m}$ gives $80%$ of $V_{ma x}$ — saturation is slow, the hallmark of the hyperbola.

Level 3 — Analysis

L3·Q1

Show algebraically that reaching $90%$ of $V_{ma x}$ requires $[S] = 9 K_{m}$ . Then state in words why "the last bit of speed is the most expensive."

Recall Solution — L3·Q1

Set $v_{0} = 0.9 V_{ma x}$ : $0.9 V_{ma x} = \frac{V _{ma x} [ S ]}{K _{m} + [ S ]} \Rightarrow 0.9 (K_{m} + [S]) = [S] .$ $0.9 K_{m} + 0.9 [S] = [S] \Rightarrow 0.9 K_{m} = 0.1 [S] \Rightarrow [S] = 9 K_{m} .$ In words: to go from $50%$ ( $[S] = K_{m}$ ) to $90%$ you must raise $[S]$ nine-fold. The curve is a hyperbola, so equal gains in rate demand ever-larger jumps in substrate — the top speed is asymptotic and the "last bit" costs enormous amounts of substrate.

L3·Q2

Two enzymes act on the same substrate. Enzyme A has $K_{m} = 0.2 mM$ ; Enzyme B has $K_{m} = 8 mM$ . Both have the same $V_{ma x}$ . At a low substrate level $[S] = 0.2 mM$ , which enzyme is faster, and by what factor?

Recall Solution — L3·Q2

Why $K_{m}$ decides here: at low $[S]$ , the reaction is roughly first order, $v_{0} \approx \frac{V _{ma x}}{K _{m}} [S]$ , so smaller $K_{m}$ → faster.

Enzyme A: $v_{0}^{A} = \frac{V _{ma x} ( 0.2 )}{0.2 + 0.2} = \frac{V _{ma x}}{2} = 0.5 V_{ma x} .$
Enzyme B: $v_{0}^{B} = \frac{V _{ma x} ( 0.2 )}{8 + 0.2} = \frac{0.2}{8.2} V_{ma x} \approx 0.02439 V_{ma x} .$

Ratio $\frac{v _{0}^{A}}{v _{0}^{B}} = \frac{0.5}{0.02439} = 20.5$ . Enzyme A is about $20.5 \times$ faster at this low $[S]$ , because its low $K_{m}$ means high affinity — it grabs substrate even when scarce.

L3·Q3

Starting from the Michaelis–Menten equation, take reciprocals to derive the Lineweaver–Burk straight line, and read off its slope, y-intercept and x-intercept.

Recall Solution — L3·Q3

Why reciprocals? A hyperbola hides $V_{ma x}$ (an asymptote you can't pin down) — a straight line makes both constants readable. $\frac{1}{v _{0}} = \frac{K _{m} + [ S ]}{V _{ma x} [ S ]} = \frac{K _{m}}{V _{ma x}} \cdot \frac{1}{[ S ]} + \frac{1}{V _{ma x}} .$ Compare with $y = m x + c$ where $y = \frac{1}{v _{0}}$ , $x = \frac{1}{[ S ]}$ :

slope $m = \frac{K _{m}}{V _{ma x}}$
y-intercept $c = \frac{1}{V _{ma x}}$ (set $x = 0$ )
x-intercept: set $y = 0$ → $0 = \frac{K _{m}}{V _{ma x}} x + \frac{1}{V _{ma x}}$ → $x = - \frac{1}{K _{m}}$ .

Level 4 — Synthesis

L4·Q1

For an enzyme, $k_{1} = 1 0^{7} M^{- 1} s^{- 1}$ , $k_{- 1} = 800 s^{- 1}$ , $k_{2} = 200 s^{- 1}$ , and total enzyme $[E]_{T} = 5 nM$ . Compute (a) $K_{m}$ and (b) $V_{ma x}$ .

Recall Solution — L4·Q1

(a) $K_{m} = \frac{k _{- 1} + k _{2}}{k _{1}} = \frac{800 + 200}{1 0 ^{7}} = \frac{1000}{1 0 ^{7}} = 1 0^{- 4} M = 0.1 mM .$ Units check: $\frac{s ^{- 1}}{M ^{- 1} s ^{- 1}} = M$ ✔ — $K_{m}$ is a concentration, as it must be.

(b) $V_{ma x} = k_{2} [E]_{T} = 200 s^{- 1} \times 5 \times 1 0^{- 9} M = 1.0 \times 1 0^{- 6} M/s = 1 μ M/s .$

These are the two lab-measurable constants extracted purely from the underlying rate constants.

L4·Q2

A classmate claims: "If I add twice as much enzyme, $K_{m}$ doubles." Is this true? Justify using the definitions, and state what actually doubles.

Recall Solution — L4·Q2

False. $K_{m} = \frac{k _{- 1} + k _{2}}{k _{1}}$ contains only rate constants — properties of the enzyme–substrate pair, not of how much enzyme you added. So $K_{m}$ is unchanged. What does scale with enzyme amount is $V_{ma x} = k_{2} [E]_{T}$ : doubling $[E]_{T}$ doubles $V_{ma x}$ . This is the parent note's key warning — $V_{ma x}$ is not an intrinsic enzyme property, $K_{m}$ is.

L4·Q3

Using the $V_{ma x}$ and $K_{m}$ from L4·Q1 ( $V_{ma x} = 1 μ M/s$ , $K_{m} = 0.1 mM$ ), find $v_{0}$ at $[S] = 0.3 mM$ , and state the reaction order regime.

Recall Solution — L4·Q3

$[S] = 0.3 mM = 3 K_{m}$ , so this is the crossover region (neither $≪$ nor $≫ K_{m}$ ). $v_{0} = \frac{V _{ma x} [ S ]}{K _{m} + [ S ]} = \frac{1 \times 0.3}{0.1 + 0.3} = \frac{0.3}{0.4} = 0.75 μ M/s .$ That is $75%$ of $V_{ma x}$ . Since $[S]$ is comparable to $K_{m}$ , the rate depends on $[S]$ but is bending toward saturation — mixed order, transitioning from first order (below $K_{m}$ ) toward zero order (well above $K_{m}$ ).

Level 5 — Mastery

L5·Q1

A competitive inhibitor is added. It raises the apparent $K_{m}$ but leaves $V_{ma x}$ unchanged. Explain physically why each of these happens, and describe what the Lineweaver–Burk lines (with vs without inhibitor) look like.

Recall Solution — L5·Q1

A competitive inhibitor resembles the substrate and binds the same active site, so it competes with $S$ .

$K_{m}$ rises (apparent): because the inhibitor blocks some active sites, you now need more substrate to reach half top speed — the enzyme appears to bind substrate less well. Affinity looks lower, so apparent $K_{m}$ increases.
$V_{ma x}$ unchanged: if you flood the system with enough substrate, $S$ outcompetes the inhibitor and every enzyme still ends up busy — so the true ceiling $V_{ma x} = k_{2} [E]_{T}$ is still reachable.
Lineweaver–Burk: both lines share the same y-intercept ( $1/ V_{ma x}$ unchanged) but the inhibited line has a steeper slope ( $K_{m} / V_{ma x}$ larger) and an x-intercept closer to the origin (since $- 1/ K_{m}$ moves right as $K_{m}$ grows).

Deeper mechanism in Enzyme Inhibition.

L5·Q2

Derive, from the mechanism $E + S k_{- 1} ⇌ k_{1} E S \to k_{2} E + P$ , the full Michaelis–Menten equation, stating why the steady-state assumption is legitimate.

Recall Solution — L5·Q2

Step 1 — product rate. Product only leaves via $E S \to E + P$ , so $v_{0} = k_{2} [E S]$ .

Step 2 — steady state. $E S$ is a short-lived intermediate: it's built up almost instantly and then its concentration barely changes during the initial-rate window. So $\frac{d [ E S ]}{d t} \approx 0$ , i.e. formation = breakdown: $k_{1} [E] [S] = (k_{- 1} + k_{2}) [E S] .$ Why legitimate: over the measurement, enzyme is far scarcer than substrate, so $[E S]$ reaches a plateau fast and holds it — the approximation error is tiny.

Step 3 — solve for $[E S]$ and define $K_{m}$ . $[E S] = \frac{[ E ] [ S ]}{K _{m}}, K_{m} \equiv \frac{k _{- 1} + k _{2}}{k _{1}} .$

Step 4 — enzyme conservation. We can't measure free $[E]$ , but total is fixed: $[E]_{T} = [E] + [E S]$ , so $[E] = [E]_{T} - [E S]$ . Substitute: $[E S] = \frac{([ E ] _{T} - [ E S ]) [ S ]}{K _{m}} \Rightarrow [E S] K_{m} = [E]_{T} [S] - [E S] [S] .$ $[E S] (K_{m} + [S]) = [E]_{T} [S] \Rightarrow [E S] = \frac{[ E ] _{T} [ S ]}{K _{m} + [ S ]} .$

Step 5 — assemble. Put into $v_{0} = k_{2} [E S]$ and use $V_{ma x} = k_{2} [E]_{T}$ : $v_{0} = \frac{V _{ma x} [ S ]}{K _{m} + [ S ]} .$

L5·Q3

An enzyme obeys Michaelis–Menten. At $[S] = 1 mM$ , $v_{0} = 30 μ M/s$ ; at $[S] = 4 mM$ , $v_{0} = 60 μ M/s$ . Solve for $K_{m}$ and $V_{ma x}$ .

Recall Solution — L5·Q3

Why two equations? Two unknowns ( $K_{m}, V_{ma x}$ ) need two data points. $30 = \frac{V _{ma x} ( 1 )}{K _{m} + 1}, 60 = \frac{V _{ma x} ( 4 )}{K _{m} + 4} .$ From the first: $V_{ma x} = 30 (K_{m} + 1)$ . Substitute into the second: $60 = \frac{30 ( K _{m} + 1 ) \cdot 4}{K _{m} + 4} \Rightarrow 60 (K_{m} + 4) = 120 (K_{m} + 1) .$ $60 K_{m} + 240 = 120 K_{m} + 120 \Rightarrow 120 = 60 K_{m} \Rightarrow K_{m} = 2 mM .$ Then $V_{ma x} = 30 (2 + 1) = 90 μ M/s .$ Check: at $[S] = 4$ : $v_{0} = \frac{90 \times 4}{2 + 4} = \frac{360}{6} = 60$ ✔.

Recall One-line takeaways

Half-saturation happens at $[S] = K_{m}$ ::: $v_{0} = V_{ma x} /2$ there, by definition. To reach $90%$ of $V_{ma x}$ you need $[S] = ?$ ::: $9 K_{m}$ — saturation is slow. Which constant depends on enzyme amount? ::: $V_{ma x} = k_{2} [E]_{T}$ ; $K_{m}$ does not. Competitive inhibitor changes which constant? ::: Raises apparent $K_{m}$ , leaves $V_{ma x}$ unchanged. Lineweaver–Burk y-intercept and slope? ::: $1/ V_{ma x}$ and $K_{m} / V_{ma x}$ ; x-intercept $= - 1/ K_{m}$ .