Enzyme Kinetics and Substrate Inhibition: 20-Bio-A3 Biomechanics, December 2018, Q2

Question 2: Enzyme Kinetics and Substrate Inhibition (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Given data
Quantity	Value
Maximum reaction rate	$V_{max} = 85\ \mu\text{mol/s}$
Michaelis constant	$K_M = 0.8\ \text{mM}$
Substrate concentration (a)	$[S] = 0.008\ \text{mM}$
Substrate concentration (b)	$[S] = 0.8\ \text{mM}$
Substrate concentration (c)	$[S] = 80\ \text{mM}$

Find. The reaction rate $v$ at each of the three substrate concentrations (a)–(c); a labelled $v$ vs $[S]$ curve showing $V_{max}$, $K_M$ and the three computed points (d); and a general rate law for the case where a second substrate molecule can bind the ES complex to form a catalytically dead-end ESS species (e).

Approach. Parts (a)–(c) substitute the given $[S]$ values directly into the Michaelis–Menten equation; part (d) plots the full curve; part (e) re-derives the rate law from a steady-state mass balance on total enzyme, exactly as in the Michaelis–Menten derivation, but now with THREE enzyme species (E, ES, ESS) instead of two.

Michaelis–Menten rate law. For a simple one-substrate enzyme reaction, $$ v = \frac{V_{max}[S]}{K_M + [S]} $$ with $V_{max}=85\ \mu\text{mol/s}$ and $K_M=0.8\ \text{mM}$.
(a) Rate at $[S]=0.008$ mM. $$ v(0.008) = \frac{(85)(0.008)}{0.8 + 0.008} = \frac{0.68}{0.808} = \boxed{0.84\ \mu\text{mol/s}} $$ $[S]\ll K_M$ here (100× below $K_M$), so the enzyme is operating in the linear, first-order regime where $v\approx (V_{max}/K_M)[S]$ — the rate is essentially proportional to substrate concentration and far below saturation.
(b) Rate at $[S]=0.8$ mM. $$ v(0.8) = \frac{(85)(0.8)}{0.8+0.8} = \frac{68}{1.6} = \boxed{42.5\ \mu\text{mol/s}} $$ Because $[S]=K_M$ exactly, this point is by definition the half-saturation point, $v=V_{max}/2=42.5\ \mu\text{mol/s}$ — this is precisely how $K_M$ is defined and read off a Michaelis–Menten plot.
(c) Rate at $[S]=80$ mM. $$ v(80) = \frac{(85)(80)}{0.8+80} = \frac{6800}{80.8} = \boxed{84.2\ \mu\text{mol/s}} $$ $[S]\gg K_M$ here (100× above $K_M$), so the enzyme is nearly saturated and $v$ approaches (but never reaches) $V_{max}=85\ \mu\text{mol/s}$; the reaction rate is now essentially independent of further increases in $[S]$ (zero-order in substrate).
(d) Graph and key features. Because the three required substrate concentrations span four orders of magnitude (0.008 to 80 mM), the curve below uses a logarithmic $[S]$ axis so all three points are visible and clearly separated; on this axis the classic Michaelis–Menten hyperbola appears as a sigmoid-looking rise from the linear low-$[S]$ regime, through the $K_M$ half-saturation point, to the $V_{max}$ plateau at high $[S]$. All three computed points sit exactly on the curve as an internal consistency check.

Figure 2.1 — Michaelis–Menten plot of $v$ vs $[S]$ (log-scale $x$-axis) for $V_{max}=85\ \mu\text{mol/s}$, $K_M=0.8\ \text{mM}$, with the three computed points (a)–(c) marked.

Question 2(a)–(c) — final results
Quantity	Value
$v$ at $[S]=0.008$ mM	0.84 µmol/s
$v$ at $[S]=0.8$ mM	42.5 µmol/s
$v$ at $[S]=80$ mM	84.2 µmol/s
$V_{max}$ (asymptote)	85 µmol/s
$K_M$ (half-saturation $[S]$)	0.8 mM (at $v=42.5\ \mu\text{mol/s}$)

(e) Substrate-inhibition rate law. The reaction scheme adds a second substrate-binding equilibrium on top of the standard Michaelis–Menten scheme: the productive complex ES can bind a SECOND substrate molecule to form ESS, a catalytically dead-end complex (it does not proceed to product; only ES does, via the same rate-determining step as ordinary Michaelis–Menten). At high $[S]$ this side reaction increasingly diverts enzyme into the unproductive ESS pool, which is why substrate inhibition characteristically causes $v$ to rise, peak, and then FALL again at very high $[S]$ — unlike ordinary Michaelis–Menten kinetics, which simply saturates.

Define the two rapid-equilibrium constants exactly as given. $$ K_M = \frac{[E][S]}{[ES]} \qquad K_{IS} = \frac{[ES][S]}{[ESS]} $$ Rearranged for the two complexes in terms of free enzyme $[E]$ (or of $[ES]$, chosen below because it is the productive species): $$ [E] = \frac{K_M[ES]}{[S]} \qquad [ESS] = \frac{[ES][S]}{K_{IS}} $$
Total-enzyme mass balance. The total enzyme concentration $E_t$ is partitioned among all three species: $$ E_t = [E] + [ES] + [ESS] $$ Substituting the two relations from Step 1 (everything expressed in terms of $[ES]$): $$ E_t = \frac{K_M[ES]}{[S]} + [ES] + \frac{[ES][S]}{K_{IS}} = [ES]\left(\frac{K_M}{[S]} + 1 + \frac{[S]}{K_{IS}}\right) $$
Rate of product formation. Only the ES complex proceeds to product, at the same rate-determining step as ordinary Michaelis–Menten kinetics, so $v = k_{cat}[ES]$, and by definition $V_{max}=k_{cat}E_t$ (the rate if every enzyme molecule were in the productive ES form). Solving Step 2 for $[ES]$ and substituting: $$ [ES] = \frac{E_t}{\dfrac{K_M}{[S]} + 1 + \dfrac{[S]}{K_{IS}}} \quad\Rightarrow\quad v = k_{cat}[ES] = \frac{k_{cat}E_t}{\dfrac{K_M}{[S]} + 1 + \dfrac{[S]}{K_{IS}}} $$ Multiplying numerator and denominator by $[S]$ to clear the fraction inside the denominator gives the final closed form: $$ v = \boxed{\dfrac{V_{max}[S]}{K_M + [S] + \dfrac{[S]^2}{K_{IS}}}} $$
Consistency check. As $K_{IS}\to\infty$ (the second substrate-binding step essentially never happens), the $[S]^2/K_{IS}$ term vanishes and the equation collapses exactly back to the ordinary Michaelis–Menten rate law from Step 1 of parts (a)–(c), as it must. At low-to-moderate $[S]$ the extra $[S]^2/K_{IS}$ term in the denominator is negligible and $v$ behaves like ordinary Michaelis–Menten; at high $[S]$ that term dominates the denominator, so $v\approx V_{max}K_{IS}/[S]$ — the rate now DECREASES as $[S]$ increases further, producing the characteristic rate peak (at $[S]=\sqrt{K_MK_{IS}}$, found by setting $dv/d[S]=0$) that distinguishes substrate inhibition from simple saturation.

Topic: Michaelis–Menten enzyme kinetics and its extension to substrate inhibition.

Key relations: $v=V_{max}[S]/(K_M+[S])$; substrate inhibition $v=V_{max}[S]/(K_M+[S]+[S]^2/K_{IS})$; $K_M=[S]$ at $v=V_{max}/2$; rate peak at $[S]=\sqrt{K_MK_{IS}}$.

Why this works: Both derivations use the identical two-step logic — write the rapid-equilibrium expressions for every enzyme-containing species in terms of one reference species (here $[ES]$), sum them into a total-enzyme mass balance, then substitute $v=k_{cat}\times(\text{productive species})$ — substrate inhibition is not a new kinetic model but the ordinary Michaelis–Menten derivation with one additional (unproductive) equilibrium species added to the mass balance.

Common pitfall: forgetting that ESS does NOT contribute to $v$ (only ES does) — a common error is including a $k_{cat}[ESS]$ term as if ESS also produced product; also, mixing up which equilibrium constant goes with which complex ($K_M$ for E+S⇌ES, $K_{IS}$ for ES+S⇌ESS) will flip the algebra and produce a rate law that increases without bound at high $[S]$ instead of correctly turning over and falling.

Source: Nelson & Cox, Lehninger Principles of Biochemistry, ch. 6 (Enzymes); Alberts et al., Molecular Biology of the Cell, ch. 3 (Proteins, catalysis).

Question 2 of 6: Enzyme Kinetics and Substrate Inhibition

Question 2: Enzyme Kinetics and Substrate Inhibition (20 marks)