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

Before anything, one reminder of the only equation we need. From Chemical Kinetics we track how fast product appears — the rate $v_{0}$ — as we change how much substrate $[S]$ we pour in:

Definition What

$k_{2}$ and $[E]_{T}$ mean The formula $V_{ma x} = k_{2} [E]_{T}$ hides two ideas that deserve their own plain words:

$[E]_{T}$ = total enzyme concentration — all the enzyme molecules you put in the tube, both the free ones and the ones currently gripping a substrate. It is a fixed amount you choose when you set up the experiment (units: concentration, e.g. $μ M$ ).
$k_{2}$ = the turnover number — the rate constant of the final step $E S \to E + P$ . In plain words, it is how many substrate molecules one single enzyme can convert to product per second when it is working flat out (units: per second, $s^{- 1}$ ). So $V_{ma x} = k_{2} [E]_{T}$ literally reads: (work rate of one enzyme) × (how many enzymes there are) = fastest possible combined output. More enzymes, or a faster-turning enzyme, raises the ceiling.

Look at the curve of this equation before we compute anything — every example below is just one dot somewhere on this shape.

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

What to look for in figure s01: the magenta curve is the MM equation. Trace it left-to-right: it leaves the origin as a steep straight line (the first-order zone, Example 2), bends through the orange dotted cross-hairs at $[S] = K_{m}$ where it reaches exactly $V_{ma x} /2$ (Example 1), then flattens toward the violet dashed ceiling $V_{ma x}$ that it never quite touches (Examples 3 and 10). Every worked answer below is one point on this single curve.

The scenario matrix

Every MM problem you will ever see falls into one of these case classes. Each row is one "cell"; the last column names the example that lands on it.

#	Case class	What is unusual about the input	Covered by
A	$[S] = K_{m}$ (the exact-half point)	ratio is exactly 1	Example 1
B	$[S] ≪ K_{m}$ (first-order regime)	substrate tiny	Example 2
C	$[S] ≫ K_{m}$ (zero-order / saturated)	substrate huge	Example 3
D	Degenerate input $[S] = 0$	nothing to react	Example 4
E	Inverse problem: find $[S]$ for a target $v_{0}$	solve backwards	Example 5
F	Compare two enzymes (affinity)	two $K_{m}$ values	Example 6
G	Change enzyme amount $[E]_{T}$	$V_{ma x}$ shifts, $K_{m}$ fixed	Example 7
H	Real-world word problem (units!)	translate words → numbers	Example 8
I	Exam twist: read $K_{m}, V_{ma x}$ off a Lineweaver–Burk line	reciprocal graph	Example 9
J	Limiting behaviour as $[S] \to \infty$	can you ever reach $V_{ma x}$ ?	Example 10

Ten examples, ten cells. Let's go.

Example 1 — Cell A: the exact-half point

Example 2 — Cell B: $[S] ≪ K_{m}$ (first order)

Worked example Statement

Same enzyme ( $V_{ma x} = 10 μ M / s$ , $K_{m} = 2 m M$ ). Find $v_{0}$ at $[S] = 0.02 m M$ (one-hundredth of $K_{m}$ ).

Forecast: With almost no substrate, doubling $[S]$ should roughly double the rate — rate behaves like a straight line through the origin. Guess: much less than half of $V_{ma x}$ .

Step 1 — Compare $[S]$ to $K_{m}$ . $0.02 ≪ 2$ , so the $[S]$ in the denominator is negligible next to $K_{m}$ . Why this step? When one number is tiny beside another, dropping it barely changes the answer but hugely simplifies it. Step 2 — Approximate the denominator as $K_{m} + [S] \approx K_{m}$ : $v_{0} \approx \frac{V _{ma x}}{K _{m}} [S] = \frac{10}{2} \times 0.02 = 0.1 μ M / s .$ Why this step? This is the first-order form $v_{0} \propto [S]$ — the rate is proportional to substrate, exactly the low- $[S]$ regime of the curve. Step 3 — Check against the exact formula: $v_{0} = \frac{10 \times 0.02}{2 + 0.02} = \frac{0.2}{2.02} = 0.0990 μ M / s .$ Why this step? To prove the approximation is trustworthy — it differs from $0.1$ by only about $1%$ .

Verify: $0.0990 \approx 0.1$ ; both far below $V_{ma x} /2 = 5$ . On figure s01 this is the steep straight part near the origin (bottom-left). ✔

Example 3 — Cell C: $[S] ≫ K_{m}$ (saturated, zero order)

Worked example Statement

Same enzyme. Find $v_{0}$ at $[S] = 200 m M$ (one hundred times $K_{m}$ ).

Forecast: So much substrate that every enzyme is always busy. Guess: essentially $V_{ma x}$ , and doubling $[S]$ again changes almost nothing.

Step 1 — Compare. $200 ≫ 2$ , so now $K_{m}$ is negligible in the denominator. Why this step? Same "drop the tiny number" move — but this time it is $K_{m}$ that is small. Step 2 — Approximate $K_{m} + [S] \approx [S]$ : $v_{0} \approx \frac{V _{ma x} [ S ]}{[ S ]} = V_{ma x} = 10 μ M / s .$ Why this step? The $[S]$ 's cancel — the rate no longer depends on $[S]$ at all. This is zero order: the flat plateau of the curve. Step 3 — Exact check: $v_{0} = \frac{10 \times 200}{2 + 200} = \frac{2000}{202} = 9.901 μ M / s .$ Why this step? Confirms we are within $1%$ of the ceiling.

Verify: $9.90 \approx 10 = V_{ma x}$ ; never exceeds it. On figure s01 this is the flat right-hand part just under the violet dashed ceiling. ✔

Example 4 — Cell D: degenerate input $[S] = 0$

Example 5 — Cell E: inverse problem (find $[S]$ )

Worked example Statement

Same enzyme ( $V_{ma x} = 10 μ M / s$ , $K_{m} = 2 m M$ ). What $[S]$ makes $v_{0} = 8 μ M / s$ ?

Forecast: $8$ is $80%$ of $V_{ma x}$ . Because the hyperbola saturates slowly, reaching $80%$ should demand $[S]$ much bigger than $K_{m}$ — guess a multiple of $K_{m}$ .

Step 1 — Write the equation with the target rate: $8 = \frac{10 [ S ]}{2 + [ S ]} .$ Why this step? Now $[S]$ is the unknown, so we set $v_{0}$ to the value we want and solve for $[S]$ . Step 2 — Cross-multiply to clear the fraction: $8 (2 + [S]) = 10 [S] \Rightarrow 16 + 8 [S] = 10 [S] .$ Why this step? Fractions are hard to isolate variables in; multiplying out gives a plain linear equation. Step 3 — Collect and solve: $16 = 2 [S] \Rightarrow [S] = 8 m M .$ Why this step? Gather the $[S]$ terms on one side; divide.

Verify: Put $[S] = 8$ back: $v_{0} = \frac{10 \times 8}{2 + 8} = \frac{80}{10} = 8 μ M / s$ . ✔ And $8 m M = 4 K_{m}$ — confirming that $80%$ speed needs four times $K_{m}$ . Saturation really is slow. On figure s01 this point sits high on the shoulder, well right of the cross-hairs.

Example 6 — Cell F: comparing affinity (two enzymes)

Worked example Statement

Enzyme A: $K_{m} = 0.1 m M$ . Enzyme B: $K_{m} = 5 m M$ . Same substrate, same $V_{ma x} = 10 μ M / s$ . (i) Which binds tighter? (ii) At $[S] = 0.1 m M$ , find $v_{0}$ for each.

Forecast: Smaller $K_{m}$ means half-speed is reached at lower $[S]$ — so it grabs substrate more eagerly. Guess A binds tighter, and A should be much faster at low $[S]$ .

Step 1 — Compare $K_{m}$ values. A has $K_{m} = 0.1$ , B has $K_{m} = 5$ . Why this step? $K_{m}$ is the affinity yardstick — low $K_{m}$ = high affinity (linked to Equilibrium: a smaller $K_{m}$ means the ES complex forms readily). So A binds tighter. Step 2 — Compute $v_{0}$ for A at $[S] = 0.1 m M$ (this is $[S] = K_{m, A}$ , the half point): $v_{0, A} = \frac{10 \times 0.1}{0.1 + 0.1} = \frac{1}{0.2} = 5 μ M / s .$ Why this step? At $[S] = K_{m, A}$ we are at A's exact half-saturation point, so the rate must land at $V_{ma x} /2 = 5$ . Step 3 — Compute $v_{0}$ for B at the same $[S] = 0.1 m M$ : $v_{0, B} = \frac{10 \times 0.1}{5 + 0.1} = \frac{1}{5.1} = 0.196 μ M / s .$ Why this step? Same substrate crowding, but B's big $K_{m}$ swamps the small $[S]$ in the denominator, so B is barely moving — this is the contrast we wanted to expose.

Verify: $v_{0, A} = 5$ (half speed) vs $v_{0, B} = 0.196$ — A is about $25 \times$ faster here. Lower $K_{m}$ ⇒ faster at low $[S]$ . ✔ Figure s02 shows the two curves.

What to look for in figure s02: two curves share the same violet-dashed $V_{ma x}$ ceiling but rise at very different speeds. The magenta curve (Enzyme A, tiny $K_{m}$ ) shoots up almost immediately — its filled dot at $[S] = 0.1$ already sits at $v_{0} = 5$ (half speed). The violet curve (Enzyme B, big $K_{m}$ ) crawls along the bottom — its dot at the same $[S] = 0.1$ is barely off the axis at $v_{0} = 0.196$ . The horizontal gap between the two curves is the affinity difference.

Example 7 — Cell G: change enzyme amount

Worked example Statement

An enzyme runs with $V_{ma x} = 10 μ M / s$ , $K_{m} = 2 m M$ . You triple the amount of enzyme $[E]_{T}$ . Find the new $V_{ma x}$ , the new $K_{m}$ , and the new $v_{0}$ at $[S] = 2 m M$ .

Forecast: Recall $V_{ma x} = k_{2} [E]_{T}$ — the turnover number times the total enzyme concentration — but the parent note's definition of $K_{m}$ has no $[E]_{T}$ in it. Guess which one changes.

Step 1 — New $V_{ma x}$ . Since $V_{ma x} = k_{2} [E]_{T}$ and $[E]_{T}$ (total enzyme) triples while $k_{2}$ (turnover number, a fixed per-enzyme speed) stays the same, the top speed triples: $V_{ma x}^{n e w} = 3 \times 10 = 30 μ M / s .$ Why this step? More machines, each still turning over at the same $k_{2}$ ⇒ higher combined flat-out speed. Step 2 — New $K_{m}$ . From the parent note, $K_{m}$ is built only from the reaction's rate constants (the speeds of the binding and conversion steps) and contains no $[E]_{T}$ , so it is unchanged: $K_{m} = 2 m M$ . Why this step? $K_{m}$ is an intrinsic property of the enzyme–substrate pair (from Catalysis), not of how much enzyme you use. Step 3 — New $v_{0}$ at $[S] = 2 m M$ (still the half point): $v_{0} = \frac{30 \times 2}{2 + 2} = \frac{60}{4} = 15 μ M / s .$ Why this step? At $[S] = K_{m}$ the rate is half of the new $V_{ma x}$ , so $15$ — exactly triple the old $5$ .

Verify: $15 = 30/2 = V_{ma x}^{n e w} /2$ , and $15 = 3 \times 5$ (old answer). $K_{m}$ fixed, $V_{ma x}$ scaled. ✔

Example 8 — Cell H: real-world word problem

Worked example Statement

Lactase in your gut breaks down lactose (the substrate). In a lab tube it has $V_{ma x} = 6 μ m o l / (L \cdot min)$ and $K_{m} = 1.5 mm o l / L$ . A patient's meal gives a gut lactose concentration of $[S] = 4.5 mm o l / L$ . What is the initial breakdown rate?

Forecast: $[S] = 4.5$ is $3 \times K_{m}$ — well above half-saturation but not enormous. Guess: comfortably above $V_{ma x} /2 = 3$ but still below the $6$ ceiling.

Step 1 — Match the words to symbols. "top speed" $= V_{ma x} = 6$ ; "half-saturation" $= K_{m} = 1.5$ ; "gut lactose" $= [S] = 4.5$ . Why this step? Word problems fail on translation, not algebra — pin every phrase to a symbol and check units line up ( $mm o l / L$ throughout). Step 2 — Substitute: $v_{0} = \frac{6 \times 4.5}{1.5 + 4.5} = \frac{27}{6} = 4.5 μ m o l / (L \cdot min) .$ Why this step? Same one tool; numbers already share consistent units so no conversion needed.

Verify: $4.5$ lies between $V_{ma x} /2 = 3$ and $V_{ma x} = 6$ — matches the forecast. Units: amount per volume per time, a rate. ✔ (Aside: this is why lactose-intolerant people, with low $[E]_{T}$ hence low $V_{ma x}$ , digest lactose too slowly — see Enzyme Inhibition for the related "blocked enzyme" case.)

Example 9 — Cell I: exam twist — read a Lineweaver–Burk line

Worked example Statement

A double-reciprocal (Lineweaver–Burk) plot of $\frac{1}{v _{0}}$ against $\frac{1}{[ S ]}$ gives a straight line with y-intercept $= 0.1 (s / μ M)$ and slope $= 0.2 (s / μ M) \cdot m M$ . Find $V_{ma x}$ and $K_{m}$ .

Forecast: The intercept encodes $V_{ma x}$ and the slope-to-intercept ratio encodes $K_{m}$ . Guess $V_{ma x}$ is the reciprocal of a small intercept — so a large number.

Step 1 — Recall the linear form (from the parent note): $\frac{1}{v _{0}} = slope \frac{K _{m}}{V _{ma x}} \cdot \frac{1}{[ S ]} + y-intercept \frac{1}{V _{ma x}} .$ Why this step? Turning the hyperbola into $y = m x + c$ lets us read $V_{ma x}, K_{m}$ straight off a ruler-straight line. Step 2 — Get $V_{ma x}$ from the y-intercept: $\frac{1}{V _{ma x}} = 0.1 \Rightarrow V_{ma x} = \frac{1}{0.1} = 10 μ M / s .$ Why this step? Where the line hits the vertical axis, $\frac{1}{[ S ]} = 0$ , so only the constant term survives. Step 3 — Get $K_{m}$ from slope ÷ intercept: $\frac{slope}{intercept} = \frac{K _{m} / V _{ma x}}{1/ V _{ma x}} = K_{m} = \frac{0.2}{0.1} = 2 m M .$ Why this step? Dividing cancels $V_{ma x}$ and leaves $K_{m}$ alone.

Verify: These reproduce our running enzyme ( $V_{ma x} = 10, K_{m} = 2$ ). Cross-check the x-intercept: it should be $- 1/ K_{m} = - 0.5 m M^{- 1}$ . ✔ Figure s03 shows the line and its intercepts.

What to look for in figure s03: a single straight magenta line. Focus on the two dots where it crosses the axes. The violet dot on the vertical axis sits at height $0.1$ — that height is $1/ V_{ma x}$ , so $V_{ma x} = 10$ . The orange dot on the horizontal axis sits at $- 0.5$ — that is $- 1/ K_{m}$ , so $K_{m} = 2$ . The whole point of the reciprocal plot is that these two crossing points hand you $V_{ma x}$ and $K_{m}$ directly, no curve-fitting needed.

Example 10 — Cell J: limiting behaviour as $[S] \to \infty$

Worked example Statement

Can the rate ever equal $V_{ma x}$ ? Same enzyme. What does $v_{0}$ approach as $[S] \to \infty$ , and what is $v_{0}$ at $[S] = 1000 K_{m}$ ?

Forecast: The plateau in figure s01 looks like it touches $V_{ma x}$ . Guess whether it actually reaches it or only creeps up forever.

Step 1 — Take the limit. Divide top and bottom by $[S]$ : $v_{0} = \frac{V _{ma x}}{\frac{K _{m}}{[ S ]} + 1} \to [S \infty] \frac{V _{ma x}}{0 + 1} = V_{ma x} .$ Why this step? Writing $K_{m} / [S]$ shows the only $[S]$ -dependent piece shrinks to $0$ — the honest way to test a limit. Step 2 — But at any finite $[S]$ , $K_{m} / [S] > 0$ , so the denominator is $> 1$ and $v_{0} < V_{ma x}$ strictly. Why this step? A limit is approached, not attained — $V_{ma x}$ is a ceiling you never quite hit. Step 3 — Numerical taste at $[S] = 1000 K_{m} = 2000 m M$ : $v_{0} = \frac{10 \times 2000}{2 + 2000} = \frac{20000}{2002} = 9.990 μ M / s .$ Why this step? Even a thousandfold $K_{m}$ leaves a $0.1%$ gap — proving how slowly the last sliver of speed arrives.

Verify: $9.990 < 10$ always; limit is exactly $10$ . The curve is asymptotic to $V_{ma x}$ — on figure s01, the flat part hugs the violet dashed line but never merges with it. ✔

Recap of the whole matrix

Recall Cover the answer, name the regime

$[S] = K_{m}$ gives which rate? ::: Exactly $V_{ma x} /2$ (Example 1). $[S] ≪ K_{m}$ : order and shape? ::: First order, straight line, $v_{0} \approx (V_{ma x} / K_{m}) [S]$ (Example 2). $[S] ≫ K_{m}$ : order and shape? ::: Zero order, flat plateau, $v_{0} \approx V_{ma x}$ (Example 3). $[S] = 0$ gives? ::: $v_{0} = 0$ , curve through origin (Example 4). Lower $K_{m}$ means? ::: Higher affinity, faster at low $[S]$ (Example 6). Tripling $[E]_{T}$ changes what? ::: $V_{ma x}$ triples; $K_{m}$ unchanged (Example 7). What is $k_{2}$ ? ::: The turnover number — molecules of substrate one enzyme converts to product per second (units $s^{- 1}$ ). What is $[E]_{T}$ ? ::: The total enzyme concentration — all enzyme molecules, free plus bound. Lineweaver–Burk y-intercept and slope give? ::: $1/ V_{ma x}$ and $K_{m} / V_{ma x}$ (Example 9). Is $V_{ma x}$ ever reached at finite $[S]$ ? ::: No — only approached as $[S] \to \infty$ (Example 10).

Back to the parent topic · related: Chemical Kinetics, Catalysis, Enzyme Inhibition, Proteins, Vitamins and Coenzymes.