GATE BT

Reaction and enzyme kinetics

Rate laws, Michaelis-Menten and inhibition

Reaction kinetics expresses the rate of a conversion as a function of concentration and temperature. Enzyme kinetics applies this to enzyme-catalysed reactions through the Michaelis–Menten equation and its inhibited forms. The standard calculations are rate constants, half-lives and batch times from integrated rate laws; activation energies from the Arrhenius equation; $K_{m}$ and $V_{ma x}$ from linear plots; apparent parameters and rates with an inhibitor present; and identifying the inhibition type from how a plot changes.

9 previous-year questions · 9 solved · 5 practice questions

Rates, rate laws and reaction order

Consider a reaction $a A + b B \to pP$ in a well-mixed, constant-volume batch reactor. The rate of disappearance of A is $- r_{A} = - \frac{d C _{A}}{d t}$ (positive when A is consumed), and the rates of the different species are tied together by stoichiometry.

A rate law writes $- r_{A}$ as a function of concentrations at a fixed temperature, usually in power-law form. The exponents are the orders with respect to each species, and their sum is the overall order $n$ . Orders are found by experiment. Only for an elementary (single-step) reaction must they equal the stoichiometric coefficients; for a non-elementary reaction they need not, and they can be fractional or zero.

RELATION BETWEEN SPECIES RATES

\frac{- r _{A}}{a} = \frac{- r _{B}}{b} = \frac{r _{P}}{p}

For $a A + b B \to pP$ , all rates on the same volume basis. A rate constant belongs to the species it is written for, so check whether a given $k$ refers to $- r_{A}$ or to the overall reaction rate.

POWER-LAW RATE LAW

- r_{A} = - \frac{d C _{A}}{d t} = k C_{A}^{α} C_{B}^{β}, n = α + β

Constant-volume batch. $k$ does not depend on concentration; it varies with temperature. Its units are $(mol/L)^{1 - n} time^{- 1}$ .

FRACTIONAL CONVERSION

X_{A} = \frac{C _{A 0} - C _{A}}{C _{A 0}}, C_{A} = C_{A 0} (1 - X_{A})

Constant volume. $C_{A 0}$ is the initial concentration of reactant A.

EXAM TIP

The units of $k$ reveal the order: mol L $^{- 1}$ s $^{- 1}$ means zero order, s $^{- 1}$ means first order, and L mol $^{- 1}$ s $^{- 1}$ means second order (any time unit in place of s). When a question gives $k$ with units, read them before choosing an integrated form.

Integrated rate laws and half-lives

Integrating the rate law for a constant-volume batch reactor gives concentration as a function of time. Each order has its own straight-line plot and its own dependence of the half-life $t_{1/2}$ (the time for $C_{A}$ to fall to $C_{A 0} /2$ ) on the starting concentration. The results below are for a single reactant with $- r_{A} = k C_{A}^{n}$ , where $k$ is defined for the disappearance of A, unless stated otherwise. All logarithms are natural logarithms.

ZERO ORDER

C_{A} = C_{A 0} - k t (t \leq \frac{C _{A 0}}{k}), t_{1/2} = \frac{C _{A 0}}{2 k}

The rate does not depend on concentration. Valid only while $C_{A} > 0$ : A is used up completely at $t = C_{A 0} / k$ and stays at zero after that.

FIRST ORDER

ln \frac{C _{A 0}}{C _{A}} = k t, t_{1/2} = \frac{ln 2}{k}

Equivalently $C_{A} = C_{A 0} e^{- k t}$ , or $- ln (1 - X_{A}) = k t$ in terms of conversion. The half-life does not depend on $C_{A 0}$ , and A never quite reaches zero.

SECOND ORDER (SINGLE REACTANT, OR EQUAL FEEDS)

\frac{1}{C _{A}} - \frac{1}{C _{A 0}} = k t, t_{1/2} = \frac{1}{k C _{A 0}}

Applies when $- r_{A} = k C_{A}^{2}$ , for example $2 A \to P$ with $k$ defined on A, or $A + B \to P$ with $- r_{A} = k C_{A} C_{B}$ and $C_{A 0} = C_{B 0}$ (so $C_{B} = C_{A}$ throughout). The half-life is inversely proportional to $C_{A 0}$ .

SECOND ORDER, A + B → P WITH UNEQUAL FEEDS

ln \frac{C _{B} C _{A 0}}{C _{B 0} C _{A}} = (C_{B 0} - C_{A 0}) k t

Applies when $- r_{A} = k C_{A} C_{B}$ with 1:1 stoichiometry, so $C_{B} = C_{B 0} - C_{A 0} + C_{A}$ . Requires $C_{B 0} \neq = C_{A 0}$ .

NTH ORDER, INTEGRATED FORM

C_{A}^{1 - n} - C_{A 0}^{1 - n} = (n - 1) k t

Any $n \neq = 1$ . Gives the zero-order result for $n = 0$ and the second-order result for $n = 2$ . For $n < 1$ it holds only until A is used up, at $t = \frac{C _{A 0}^{1 - n}}{( 1 - n ) k}$ .

NTH ORDER, HALF-LIFE

t_{1/2} = \frac{2 ^{n - 1} - 1}{( n - 1 ) k C _{A 0}^{n - 1}} \Rightarrow t_{1/2} \propto C_{A 0}^{1 - n}

Any $n \neq = 1$ . The proportionality is the basis of the half-life method for finding the order.

REVERSIBLE FIRST ORDER, A ⇌ B

ln \frac{C _{A 0} - C _{A e}}{C _{A} - C _{A e}} = (k_{1} + k_{- 1}) t, C_{A e} = \frac{k _{- 1} ( C _{A 0} + C _{B 0} )}{k _{1} + k _{- 1}}

For $A ⇌ B$ with $- r_{A} = k_{1} C_{A} - k_{- 1} C_{B}$ , constant volume. $C_{A e}$ is the equilibrium concentration of A, and $K_{e q} = k_{1} / k_{- 1} = C_{B e} / C_{A e}$ . The approach to equilibrium is first order with rate constant $k_{1} + k_{- 1}$ , not $k_{1}$ .

Order	Rate law	Integrated form	Straight-line plot	Half-life	Units of k
0	$- r_{A} = k$	$C_{A} = C_{A 0} - k t$	$C_{A}$ vs $t$ : slope $- k$	$C_{A 0} / (2 k)$ , proportional to $C_{A 0}$	mol L $^{- 1}$ s $^{- 1}$
1	$- r_{A} = k C_{A}$	$ln (C_{A 0} / C_{A}) = k t$	$ln C_{A}$ vs $t$ : slope $- k$	$ln 2/ k$ , independent of $C_{A 0}$	s $^{- 1}$
2	$- r_{A} = k C_{A}^{2}$	$1/ C_{A} - 1/ C_{A 0} = k t$	$1/ C_{A}$ vs $t$ : slope $+ k$	$1/ (k C_{A 0})$ , proportional to $1/ C_{A 0}$	L mol $^{- 1}$ s $^{- 1}$

Integral method: assume an order and plot the matching linearised integrated form (table above). The order that gives a straight line is correct, and $k$ comes from the slope.
Differential method: estimate $- d C_{A} / d t$ from the slope of the $C_{A}$ – $t$ curve at several points, then plot $ln (- r_{A})$ against $ln C_{A}$ . The slope is $n$ and the intercept is $ln k$ .
Initial-rate method: measure the initial rate at two starting concentrations of A, with all other concentrations held fixed; the order in A is $n = \frac{ln ( r _{2} / r _{1} )}{ln ( C _{A 0, 2} / C _{A 0, 1} )}$ .
Half-life method (single reactant, $- r_{A} = k C_{A}^{n}$ ): measure $t_{1/2}$ at two starting concentrations. Since $t_{1/2} \propto C_{A 0}^{1 - n}$ , $n = 1 - \frac{ln ( t _{1/2, 2} / t _{1/2, 1} )}{ln ( C _{A 0, 2} / C _{A 0, 1} )}$ .

EXAM TIP

Half-life test: if doubling $C_{A 0}$ doubles $t_{1/2}$ , the reaction is zero order. If $t_{1/2}$ is unchanged, it is first order. If $t_{1/2}$ halves, it is second order.

EXAM TIP

For first order, the time to reach any fixed conversion, $t = - ln (1 - X_{A}) / k$ , does not depend on $C_{A 0}$ . Useful multiples: $t_{90%} = ln 10/ k \approx 3.32 t_{1/2}$ , and $t_{99%} = 2 t_{90%}$ .

Temperature dependence: the Arrhenius equation

For a positive activation energy the rate constant rises with absolute temperature. Over a moderate range this follows the Arrhenius equation, where $A$ (the pre-exponential or frequency factor) has the same units as $k$ and $E_{a}$ is the activation energy.

The observed rate of an enzyme-catalysed reaction follows this rise only up to an optimum temperature. The catalytic constant $k_{c a t}$ of the active enzyme keeps rising with temperature, but thermal denaturation removes active enzyme, and above the optimum the loss outweighs the gain, so the observed rate (and $V_{ma x} = k_{c a t} [E]_{a c t i v e}$ ) falls.

ARRHENIUS EQUATION

k = A e^{- E_{a} / (R T)}, ln k = ln A - \frac{E _{a}}{R} \cdot \frac{1}{T}

$T$ in kelvin, $R = 8.314 J mol^{- 1} K^{- 1}$ , $E_{a}$ in J/mol. A plot of $ln k$ against $1/ T$ is a straight line of slope $- E_{a} / R$ .

ARRHENIUS, TWO TEMPERATURES

ln \frac{k _{2}}{k _{1}} = \frac{E _{a}}{R} (\frac{1}{T _{1}} - \frac{1}{T _{2}})

Eliminates $A$ . When $T_{2} > T_{1}$ and $E_{a} > 0$ , $k_{2} > k_{1}$ .

TEMPERATURE COEFFICIENT Q10

Q_{10} = \frac{k ( T + 10 )}{k ( T )} = exp [\frac{10 E _{a}}{R T ( T + 10 )}]

$T$ in K. The ratio of rate constants for a 10 K (equivalently 10 °C) rise. It depends on $T$ , so a quoted $Q_{10}$ holds only near its stated temperature.

FIRST-ORDER ENZYME DEACTIVATION

[E] = [E]_{0} e^{- k_{d} t}, t_{1/2, d} = \frac{ln 2}{k _{d}}

Loss of active enzyme at constant temperature with deactivation constant $k_{d}$ (e.g. in h $^{- 1}$ ), which itself follows the Arrhenius equation. Because $V_{ma x} = k_{c a t} [E]$ , $V_{ma x}$ decays by the same exponential.

EXAM TIP

Convert °C to K before using the Arrhenius equation, and keep $E_{a}$ and $R$ in the same energy unit (J with $R = 8.314$ , or kJ with $R = 8.314 \times 1 0^{- 3}$ ). To avoid losing precision on a calculator when the temperatures are close, compute $\frac{1}{T _{1}} - \frac{1}{T _{2}}$ as $\frac{T _{2} - T _{1}}{T _{1} T _{2}}$ .

Michaelis–Menten kinetics

The simplest enzyme mechanism has one substrate forming one enzyme–substrate complex, which breaks down irreversibly to product: $E + S k_{- 1} ⇌ k_{1} E S k_{2} E + P$ .

Enzyme is conserved, so $[E]_{0} = [E] + [E S]$ , and the rate is $v = k_{2} [E S]$ . The unknown $[E S]$ is removed by one of two assumptions, compared in the table below. Both give the same hyperbolic rate equation; only the meaning of $K_{m}$ differs. Both also take the free substrate concentration as equal to the total, which needs $[E]_{0} ≪ [S]_{0}$ . Here $v$ is the initial rate, measured while product is negligible (no reverse reaction, no product inhibition) and $[S] \approx [S]_{0}$ .

	Quasi-steady state	Rapid equilibrium
Assumption	$d [E S] / d t \approx 0$ after a brief initial transient	The binding step stays at equilibrium; product formation is too slow to disturb it
Condition	$[E]_{0} ≪ [S]_{0}$ (more precisely $[E]_{0} ≪ [S]_{0} + K_{m}$ )	$k_{2} ≪ k_{- 1}$
$K_{m}$	$(k_{- 1} + k_{2}) / k_{1}$	$k_{- 1} / k_{1} = K_{S}$
Meaning of $K_{m}$	Substrate concentration at half of $V_{ma x}$ ; not a pure measure of affinity	Dissociation constant of ES; a true measure of binding (smaller $K_{S}$ means tighter binding)
Rate equation	$v = V_{ma x} [S] / (K_{m} + [S])$	Same

MICHAELIS–MENTEN EQUATION

v = - \frac{d [ S ]}{d t} = \frac{d [ P ]}{d t} = \frac{V _{ma x} [ S ]}{K _{m} + [ S ]}

$K_{m}$ has concentration units, the same as [S] (mM, mol/L or g/L). $v$ and $V_{ma x}$ are concentration per time (e.g. µmol L $^{- 1}$ min $^{- 1}$ ).

MICHAELIS CONSTANT (QUASI-STEADY STATE)

K_{m} = \frac{k _{- 1} + k _{2}}{k _{1}}

From $d [E S] / d t \approx 0$ . Because $k_{2} > 0$ , $K_{m}$ is always larger than the dissociation constant $k_{- 1} / k_{1}$ .

MICHAELIS CONSTANT (RAPID EQUILIBRIUM)

K_{m} \approx K_{S} = \frac{k _{- 1}}{k _{1}} = \frac{[ E ] [ S ]}{[ E S ]}

Valid when $k_{2} ≪ k_{- 1}$ . Only in this case is $K_{m}$ a true dissociation constant, so that a smaller $K_{m}$ means tighter binding.

MAXIMUM RATE AND TURNOVER NUMBER

V_{ma x} = k_{2} [E]_{0} = k_{c a t} [E]_{0}

$k_{c a t}$ (s $^{- 1}$ ) is the turnover number: substrate molecules converted per active site per unit time when the enzyme is saturated. $[E]_{0}$ is the molar concentration of active sites.

CATALYTIC EFFICIENCY

efficiency = \frac{k _{c a t}}{K _{m}}, v \approx \frac{k _{c a t}}{K _{m}} [E]_{0} [S] ([S] ≪ K_{m})

Units L mol $^{- 1}$ s $^{- 1}$ , a second-order rate constant. It compares substrates for the same enzyme. For the one-intermediate mechanism, $k_{c a t} / K_{m} = k_{1} k_{2} / (k_{- 1} + k_{2})$ , which is less than $k_{1}$ , so it cannot exceed the rate constant for enzyme and substrate meeting by diffusion, of the order of $1 0^{8}$ to $1 0^{9}$ L mol $^{- 1}$ s $^{- 1}$ .

LIMITING FORMS

v \approx \frac{V _{ma x}}{K _{m}} [S] ([S] ≪ K_{m}), v \approx V_{ma x} ([S] ≫ K_{m})

At low substrate the reaction is first order in S with rate constant $V_{ma x} / K_{m}$ . At high substrate it is zero order.

SUBSTRATE NEEDED FOR A GIVEN FRACTION OF VMAX

[S] = K_{m} \frac{f}{1 - f}, f = \frac{v}{V _{ma x}}

At $[S] = K_{m}$ , $v = V_{ma x} /2$ . This is the working definition of $K_{m}$ under either assumption.

Enzyme amounts are often given as activity. One unit (U) converts 1 µmol of substrate per minute under stated conditions, and 1 katal = 1 mol s $^{- 1}$ , so 1 U = 16.67 nkat. Specific activity is units per mg of protein.

EXAM TIP

$v / V_{ma x}$ is 0.5, 0.75, 0.8, 0.9 and 0.99 at $[S] = K_{m}$ , $3 K_{m}$ , $4 K_{m}$ , $9 K_{m}$ and $99 K_{m}$ . Raising the rate from 50% to 90% of $V_{ma x}$ takes nine times as much substrate.

EXAM TIP

$K_{m}$ does not depend on enzyme concentration, but $V_{ma x}$ is proportional to $[E]_{0}$ . Doubling the enzyme doubles $V_{ma x}$ and every rate, and leaves $K_{m}$ unchanged.

Linearised plots and their biases

$K_{m}$ and $V_{ma x}$ can be estimated by rearranging the Michaelis–Menten equation into a straight line. Each rearrangement weights experimental error differently. Fitting $v$ against [S] by nonlinear least squares is the statistically preferred method. The linear plots are still useful for reading parameters off a graph and for diagnosing the type of inhibition.

LINEWEAVER–BURK (DOUBLE RECIPROCAL)

\frac{1}{v} = \frac{K _{m}}{V _{ma x}} \cdot \frac{1}{[ S ]} + \frac{1}{V _{ma x}}

Plot $1/ v$ against $1/ [S]$ : slope $K_{m} / V_{ma x}$ , $y$ -intercept $1/ V_{ma x}$ , $x$ -intercept $- 1/ K_{m}$ .

EADIE–HOFSTEE

v = V_{ma x} - K_{m} \frac{v}{[ S ]}

Plot $v$ against $v / [S]$ : slope $- K_{m}$ , $y$ -intercept $V_{ma x}$ , $x$ -intercept $V_{ma x} / K_{m}$ .

HANES–WOOLF

\frac{[ S ]}{v} = \frac{1}{V _{ma x}} [S] + \frac{K _{m}}{V _{ma x}}

Plot $[S] / v$ against [S]: slope $1/ V_{ma x}$ , $y$ -intercept $K_{m} / V_{ma x}$ , $x$ -intercept $- K_{m}$ .

Plot	y vs x	Slope	y-intercept	x-intercept	Error behaviour
Lineweaver–Burk	$1/ v$ vs $1/ [S]$	$K_{m} / V_{ma x}$	$1/ V_{ma x}$	$- 1/ K_{m}$	Distorts error the most. Taking reciprocals magnifies errors in the small rates at low [S], so those points dominate the fit, while high-[S] points bunch near the $y$ -axis
Eadie–Hofstee	$v$ vs $v / [S]$	$- K_{m}$	$V_{ma x}$	$V_{ma x} / K_{m}$	$v$ appears on both axes, so an error in $v$ moves a point in both directions; less distortion than Lineweaver–Burk
Hanes–Woolf	$[S] / v$ vs $[S]$	$1/ V_{ma x}$	$K_{m} / V_{ma x}$	$- K_{m}$	When the error is mainly in $v$ , usually the most even weighting of the three; [S] appears on both axes but is usually known accurately

EXAM TIP

Watch the signs: the Lineweaver–Burk $x$ -intercept is $- 1/ K_{m}$ and the Hanes–Woolf $x$ -intercept is $- K_{m}$ , both on the negative axis. If a question gives such an intercept as a positive number, it means the magnitude. On a Lineweaver–Burk plot, $K_{m} = slope / intercept$ and $V_{ma x} = 1/ intercept$ .

Inhibition: competitive, uncompetitive, non-competitive and substrate inhibition

A reversible inhibitor I binds non-covalently and is described by a dissociation constant: $K_{i} = [E] [I] / [E I]$ for binding to free enzyme, and $K_{i}^{'} = [E S] [I] / [E S I]$ for binding to the enzyme–substrate complex. Define $α = 1 + [I] / K_{i}$ and $α^{'} = 1 + [I] / K_{i}^{'}$ . Inhibitor binding is taken to be at equilibrium, [I] is the free inhibitor concentration (equal to the total when $[I] ≫ [E]_{0}$ ), and EI and ESI are catalytically inactive. The general (mixed) rate law covers all three named cases.

MIXED INHIBITION, GENERAL RATE LAW

v = \frac{V _{ma x} [ S ]}{α K _{m} + α ^{'} [ S ]}

Competitive: $α^{'} = 1$ . Uncompetitive: $α = 1$ . Pure non-competitive: $K_{i} = K_{i}^{'}$ , so $α = α^{'}$ .

MIXED INHIBITION, APPARENT PARAMETERS

V_{ma x, a pp} = \frac{V _{ma x}}{α ^{'}}, K_{m, a pp} = \frac{α}{α ^{'}} K_{m}

The inhibited reaction still follows the Michaelis–Menten form, with these apparent constants at a fixed [I].

COMPETITIVE INHIBITION

v = \frac{V _{ma x} [ S ]}{K _{m} ( 1 + \frac{[ I ]}{K _{i}} ) + [ S ]}

I binds only free E, competing with S for the active site. $K_{m, a pp} = K_{m} (1 + [I] / K_{i})$ increases and $V_{ma x}$ is unchanged, so enough substrate overcomes the inhibition.

UNCOMPETITIVE INHIBITION

v = \frac{V _{ma x} [ S ]}{K _{m} + [ S ] ( 1 + \frac{[ I ]}{K _{i}^{'}} )}

I binds only ES. $V_{ma x, a pp}$ and $K_{m, a pp}$ both fall by the factor $1 + [I] / K_{i}^{'}$ , so their ratio $V_{ma x} / K_{m}$ is unchanged. The inhibition grows as [S] rises.

PURE NON-COMPETITIVE INHIBITION

v = \frac{V _{ma x} [ S ]}{( K _{m} + [ S ]) ( 1 + \frac{[ I ]}{K _{i}} )}

I binds E and ES with equal affinity, at a site other than the active site. $V_{ma x, a pp} = V_{ma x} / (1 + [I] / K_{i})$ and $K_{m}$ is unchanged. The fractional inhibition is the same at every [S].

Type	I binds	Apparent Km	Apparent Vmax	Lineweaver–Burk lines (with vs without I)	Overcome by high [S]?
Competitive	E only	$α K_{m}$ (increases)	Unchanged	Meet on the $1/ v$ axis (same intercept); inhibited line steeper	Yes
Uncompetitive	ES only	$K_{m} / α^{'}$ (decreases)	$V_{ma x} / α^{'}$ (decreases)	Parallel (same slope); inhibited line has a higher intercept	No; inhibition is largest at high [S]
Pure non-competitive	E and ES, $K_{i} = K_{i}^{'}$	Unchanged	$V_{ma x} / α$ (decreases)	Meet on the $1/ [S]$ axis at $- 1/ K_{m}$	No
Mixed	E and ES, $K_{i} \neq = K_{i}^{'}$	$α K_{m} / α^{'}$ (up if $K_{i} < K_{i}^{'}$ , down if $K_{i} > K_{i}^{'}$ )	$V_{ma x} / α^{'}$ (decreases)	Meet left of the $1/ v$ axis, at $1/ [S] = - K_{i} / (K_{i}^{'} K_{m})$ : above the $x$ -axis if $K_{i} < K_{i}^{'}$ , below it if $K_{i} > K_{i}^{'}$	No

FRACTIONAL INHIBITION (COMPETITIVE)

i = 1 - \frac{v _{i}}{v _{0}} = \frac{[ I ] / K _{i}}{1 + [ I ] / K _{i} + [ S ] / K _{m}}

$v_{i}$ and $v_{0}$ are the rates with and without inhibitor at the same [S]. Setting $i = 0.5$ gives $IC_{50} = K_{i} (1 + [S] / K_{m})$ . For pure non-competitive inhibition, $i = \frac{[ I ] / K _{i}}{1 + [ I ] / K _{i}}$ at every [S], so $IC_{50} = K_{i}$ . For uncompetitive inhibition, $IC_{50} = K_{i}^{'} (1 + K_{m} / [S])$ .

SUBSTRATE INHIBITION, RATE LAW

v = \frac{V _{ma x} [ S ]}{K _{m} + [ S ] + [ S ] ^{2} / K _{S I}}

A second substrate molecule binds ES to form an inactive $E S_{2}$ , with $K_{S I} = [E S] [S] / [E S_{2}]$ . At low [S] this reduces to Michaelis–Menten. At very high [S], $v \approx V_{ma x} K_{S I} / [S]$ , so the rate falls. On a Lineweaver–Burk plot the points curve upward near the $1/ v$ axis.

SUBSTRATE INHIBITION, OPTIMUM

[S]_{o pt} = K_{m} K_{S I}, v_{p e ak} = \frac{V _{ma x}}{1 + 2 K _{m} / K _{S I}}

From $d v / d [S] = 0$ . $v_{p e ak}$ is the highest rate the enzyme can reach and is always below $V_{ma x}$ .

WORKED EXAMPLE · WORKED EXAMPLE: ONE INHIBITOR, THREE MECHANISMS

An enzyme has $K_{m} = 4$ mM and $V_{ma x} = 120$ µmol L $^{- 1}$ min $^{- 1}$ . An inhibitor at $[I] = 1$ mM has a dissociation constant of 0.5 mM for each complex it forms. At $[S] = 6$ mM, find the initial rate if the inhibitor is (a) competitive, (b) uncompetitive, (c) pure non-competitive. (d) For the competitive case, find the substrate concentration that restores the uninhibited rate at 6 mM.

Solution

Uninhibited: $v_{0} = \frac{120 \times 6}{4 + 6} = 72$ µmol L $^{- 1}$ min $^{- 1}$ .
$[I] / K_{i} = 1/0.5 = 2$ , so $α = α^{'} = 3$ wherever they apply.
(a) Competitive: $K_{m, a pp} = 4 \times 3 = 12$ mM, so $v = \frac{120 \times 6}{12 + 6} = 40$ µmol L $^{- 1}$ min $^{- 1}$ (44.4% inhibition).
(b) Uncompetitive: $v = \frac{120 \times 6}{4 + 3 \times 6} = \frac{720}{22} = 32.7$ µmol L $^{- 1}$ min $^{- 1}$ (54.5% inhibition).
(c) Pure non-competitive: $v = v_{0} /3 = 72/3 = 24$ µmol L $^{- 1}$ min $^{- 1}$ (66.7% inhibition).
(d) Competitive, find [S] for $v = 72$ : $\frac{120 [ S ]}{12 + [ S ]} = 72 \Rightarrow 120 [S] = 864 + 72 [S] \Rightarrow [S] = 18$ mM. Without inhibitor the rate of 72 occurs at $[S] = 1.5 K_{m}$ ; with it, at $1.5 K_{m, a pp}$ , which is three times as much substrate.

Answer: (a) 40, (b) 32.7, (c) 24 µmol L $^{- 1}$ min $^{- 1}$ ; (d) 18 mM.

EXAM TIP

The quantity that stays fixed tells you the type. Competitive keeps $V_{ma x}$ (same $1/ v$ intercept). Uncompetitive keeps $V_{ma x} / K_{m}$ (same Lineweaver–Burk slope). Pure non-competitive keeps $K_{m}$ (same $x$ -intercept).

EXAM TIP

Names vary: some problems say 'non-competitive' for the mixed case with $K_{i} \neq = K_{i}^{'}$ . If two different inhibition constants are given, use the general mixed form. If one constant is given for binding to both E and ES, the inhibition is pure non-competitive.

EXAM TIP

An irreversible inhibitor binds covalently and permanently removes active enzyme. The enzyme that remains has its normal $K_{m}$ , so the result is a lower $V_{ma x}$ with $K_{m}$ unchanged, the same pattern as pure non-competitive inhibition. It is told apart by not being reversed on dilution or dialysis, and often by inhibition that grows with the time enzyme and inhibitor are incubated together.

Integrated Michaelis–Menten equation for batch conversion

In a batch enzyme reactor the substrate falls over time, so the Michaelis–Menten rate changes continuously. Integrating $- d [S] / d t = V_{ma x} [S] / (K_{m} + [S])$ from $[S]_{0}$ at $t = 0$ gives the batch time directly. Assumptions: constant volume, constant enzyme activity (no deactivation), irreversible reaction and no product inhibition.

INTEGRATED MICHAELIS–MENTEN EQUATION

V_{ma x} t = ([S]_{0} - [S]) + K_{m} ln \frac{[ S ] _{0}}{[ S ]}

Natural log. $V_{ma x}$ in concentration per time; $K_{m}$ and [S] in the same concentration unit.

INTEGRATED FORM IN TERMS OF CONVERSION

V_{ma x} t = [S]_{0} X + K_{m} ln \frac{1}{1 - X}, X = \frac{[ S ] _{0} - [ S ]}{[ S ] _{0}}

Same assumptions. Gives the batch time for a target conversion $X$ .

LINEARISED PROGRESS CURVE

\frac{1}{t} ln \frac{[ S ] _{0}}{[ S ]} = \frac{V _{ma x}}{K _{m}} - \frac{1}{K _{m}} \cdot \frac{[ S ] _{0} - [ S ]}{t}

Fits $K_{m}$ and $V_{ma x}$ from a single batch run, under the same assumptions. Plot $\frac{1}{t} ln \frac{[ S ] _{0}}{[ S ]}$ against $\frac{[ S ] _{0} - [ S ]}{t}$ : slope $- 1/ K_{m}$ , $y$ -intercept $V_{ma x} / K_{m}$ .

Two limits check the result. If $[S] ≫ K_{m}$ throughout, the log term is negligible and $[S] = [S]_{0} - V_{ma x} t$ (zero order). If $[S]_{0} ≪ K_{m}$ , the linear term is negligible and $ln ([S]_{0} / [S]) = (V_{ma x} / K_{m}) t$ (first order with $k = V_{ma x} / K_{m}$ ). A competitive inhibitor held at constant [I] is handled by replacing $K_{m}$ with $K_{m} (1 + [I] / K_{i})$ .

STEADY CONTINUOUS ENZYME REACTORS

PFR: V_{ma x} τ = ([S]_{0} - [S]) + K_{m} ln \frac{[ S ] _{0}}{[ S ]}, CSTR: τ = \frac{([ S ] _{0} - [ S ]) ( K _{m} + [ S ])}{V _{ma x} [ S ]}

$τ = V_{R} / F$ is the space time (reactor volume over volumetric flow rate), with constant density and the same enzyme activity everywhere. $[S]_{0}$ is the feed and [S] the outlet concentration. A plug-flow reactor behaves like the batch reactor with $t$ replaced by $τ$ . A CSTR runs entirely at the outlet concentration, so its rate is evaluated at [S] and it needs a longer $τ$ for the same conversion.

WORKED EXAMPLE · WORKED EXAMPLE: PARAMETERS FROM DATA, THEN BATCH TIME

An enzyme at $[E]_{0} = 0.01$ µmol/L follows Michaelis–Menten kinetics. Its initial rates are $v = 20$ µmol L $^{- 1}$ min $^{- 1}$ at $[S] = 1$ mM and $v = 40$ µmol L $^{- 1}$ min $^{- 1}$ at $[S] = 4$ mM. Find (a) $K_{m}$ and $V_{ma x}$ from a Lineweaver–Burk plot, (b) $k_{c a t}$ and $k_{c a t} / K_{m}$ , and (c) the time for 90% conversion when the same enzyme concentration acts on $[S]_{0} = 10$ mM in a batch reactor.

Solution

Reciprocals: $(1/ [S], 1/ v) = (1, 0.05)$ and $(0.25, 0.025)$ , with $1/ [S]$ in mM $^{- 1}$ and $1/ v$ in L min µmol $^{- 1}$ .
Slope $= \frac{0.05 - 0.025}{1 - 0.25} = \frac{1}{30}$ . Intercept $= 0.05 - \frac{1}{30} = \frac{1}{60} = \frac{1}{V _{ma x}}$ , so $V_{ma x} = 60$ µmol L $^{- 1}$ min $^{- 1}$ .
$K_{m} = slope \times V_{ma x} = \frac{60}{30} = 2$ mM. Check: $x$ -intercept $= - \frac{1/60}{1/30} = - 0.5$ mM $^{- 1} = - 1/ K_{m}$ .
(b) $V_{ma x} = 60$ µmol L $^{- 1}$ min $^{- 1}$ $= 1 \times 1 0^{- 6}$ mol L $^{- 1}$ s $^{- 1}$ and $[E]_{0} = 1 \times 1 0^{- 8}$ mol/L, so $k_{c a t} = V_{ma x} / [E]_{0} = 100 s^{- 1}$ .
$k_{c a t} / K_{m} = \frac{100}{2 \times 1 0 ^{- 3}} = 5 \times 1 0^{4}$ L mol $^{- 1}$ s $^{- 1}$ .
(c) $V_{ma x} t = [S]_{0} X + K_{m} ln \frac{1}{1 - X} = 10 (0.9) + 2 ln 10 = 9 + 4.605 = 13.605$ mM.
$V_{ma x} = 0.060$ mM min $^{- 1}$ , so $t = 13.605/0.060 = 226.8$ min, about 3.78 h.
For comparison, the zero-order estimate $[S]_{0} X / V_{ma x} = 9/0.060 = 150$ min is about one-third too short.

Answer: (a) $K_{m} = 2$ mM, $V_{ma x} = 60$ µmol L $^{- 1}$ min $^{- 1}$ ; (b) $k_{c a t} = 100$ s $^{- 1}$ , $k_{c a t} / K_{m} = 5 \times 1 0^{4}$ L mol $^{- 1}$ s $^{- 1}$ ; (c) about 227 min (3.78 h).

EXAM TIP

Don't calculate batch time as $[S]_{0} X / V_{ma x}$ (or as initial rate × time) unless [S] stays well above $K_{m}$ until the end. That shortcut ignores the slowdown as substrate runs out and always underestimates the time. Use the integrated equation.

PREVIOUS-YEAR QUESTIONS

9 questions from the BT papers

Try each question first, then open the answer. We placed each question under a topic ourselves; the exam paper does not label topics.

See these papers in full: GATE BT 2026 · GATE BT 2025 · GATE BT 2024 · GATE BT 2023 · GATE BT 2022

GATE 2026 BT Q.21

GATE 2026 · BTQ.21 · 1 mark · MCQ

The catalytic efficiency of an enzyme following Michaelis-Menten kinetics is defined by

(A)
$k_{Cat}$
(B)
$V_{m a x} / k_{Cat}$
(C)
$k_{Cat} / K_{M}$
(D)
$k_{Cat} / V_{m a x}$

Answer and solution

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GATE 2026 BT Q.32

GATE 2026 · BTQ.32 · 1 mark · NAT

You are characterizing a new enzyme isolated and purified in the laboratory. If the maximum velocity of the enzyme is 1800 $μ moles L^{- 1} min^{- 1}$ and the total concentration of the enzyme in the reaction mixture is 1.5 $μ M$ , then the turnover number of the enzyme is ______ $s^{- 1}$ . (answer in integer)

Answer and solution

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GATE 2026 BT Q.60

GATE 2026 · BTQ.60 · 2 marks · NAT

A substrate is consumed in a zero order reaction such that its concentration falls from 42 g L $^{- 1}$ to 14 g L $^{- 1}$ in 4 hours. The total time taken for complete utilization of substrate will be ________ hours. (answer in integer)

Answer and solution

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GATE 2025 BT Q.32

GATE 2025 · BTQ.32 · 1 mark · NAT

Within the Michaelis-Menten framework, the ratio of $v_{0} / V_{m a x}$ when $[S] = 20 \times K_{m}$ is ______. (Round off to two decimal places)

Answer and solution

GATE 2025 BT Q.32: answer and solution

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GATE 2024 BT Q.44

GATE 2024 · BTQ.44 · 2 marks · MSQ

Which of the following statements is/are correct about an uncompetitive inhibitor of an enzyme?

(A)
It binds to the substrate binding site of the enzyme only
(B)
It binds to the enzyme-substrate complex only
(C)
It reduces the V $_{ma x}$ of the enzyme
(D)
It binds to both free enzyme and enzyme-substrate complex

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GATE 2023 BT Q.47

GATE 2023 · BTQ.47 · 2 marks · NAT

Temperature of a reaction with an activation energy value of $15 kcal mol^{- 1}$ is increased from $300 K$ to $310 K$ . If the value of the ideal gas constant ( $R$ ) is $1.9872 cal mol^{- 1} K^{- 1}$ , the ratio of the reaction rate constants $(k_{310} / k_{300})$ (rounded off to two decimal places) is ________.

Answer and solution

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GATE 2023 BT Q.54

GATE 2023 · BTQ.54 · 2 marks · NAT

An enzyme (E) catalyzes the biochemical reaction $A \to B$ with $k_{c a t}$ equal to $500 s^{- 1}$ . If the initial reaction velocity ( $V_{0}$ ) is $10 μ M.s^{- 1}$ at the total enzyme concentration $[E_{t}]$ of $30 nM$ and substrate concentration [A] of $40 μ M$ , the value of $K_{m}$ (in $μ M$ ) is ________.

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GATE 2022 BT Q.19

GATE 2022 · BTQ.19 · 1 mark · MCQ

Which of the following statements about reversible enzyme inhibitors are CORRECT?

P. Uncompetitive inhibitors bind only to the enzyme-substrate complex

Q. Non-competitive inhibitors bind only at a different site from the substrate

R. Competitive inhibitors bind to the same site as the substrate

(A)
P and Q only
(B)
P and R only
(C)
Q and R only
(D)
P, Q and R

Answer and solution

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GATE 2022 BT Q.50

GATE 2022 · BTQ.50 · 2 marks · MSQ

Which of the following plot(s) is(are) CORRECT for an enzyme that obeys Michaelis-Menten kinetics, assuming $[S] ≪ K_{m}$ ?

$[S]$ is the concentration of the substrate, $K_{m}$ is the Michaelis constant, and $v_{0}$ is the initial reaction velocity.

GATE 2022 BT question 50: Each option is a sketch graph with no numbers on the axes. — Figure from the GATE 2022 BT paper. It is part of the question.

(A)
Figure — $v_{0}$ (vertical) against $[S]$ (horizontal): a straight line through the origin with a constant positive slope.
(B)
Figure — $v_{0}$ (vertical) against $[S]$ (horizontal): a sigmoidal curve, flat at first, rising steeply, then levelling off to a plateau.
(C)
Figure — $v_{0}$ (vertical) against $[S]$ (horizontal): a concave-up curve from near the origin that gets steeper as $[S]$ increases.
(D)
Figure — $[S]$ (vertical) against time (horizontal): a curve starting at a nonzero $[S]$ at time zero and decaying, steeply at first and then more slowly, toward the time axis.

Answer and solution

GATE 2022 BT Q.50: answer and solution

PRACTICE

5 questions to test yourself

PRACTICE 11 mark · NAT

An enzyme is inhibited by its own substrate and follows $v = \frac{V _{ma x} [ S ]}{K _{m} + [ S ] + [ S ] ^{2} / K _{S I}}$ with $V_{ma x} = 10$ µmol L $^{- 1}$ min $^{- 1}$ , $K_{m} = 1$ mM and $K_{S I} = 16$ mM. The highest initial rate that can be obtained by choosing [S] is ______ µmol L $^{- 1}$ min $^{- 1}$ (rounded to two decimal places).

Answer and solution

Answer: $6.67$ (accepted range $\pm 0.02$ )

Divide numerator and denominator by [S]: $v = \frac{V _{ma x}}{K _{m} / [ S ] + 1 + [ S ] / K _{S I}}$ . The rate is highest where the denominator is smallest, i.e. where $- K_{m} / [S]^{2} + 1/ K_{S I} = 0$ , giving $[S]_{o pt} = K_{m} K_{S I} = 1 \times 16 = 4$ mM.
$v = \frac{10 \times 4}{1 + 4 + 16/16} = \frac{40}{6} = 6.67$ µmol L $^{- 1}$ min $^{- 1}$ .
Check: $v_{p e ak} = V_{ma x} / (1 + 2 K_{m} / K_{S I}) = 10/ (1 + 0.5) = 6.67$ .

PRACTICE 21 mark · MCQ

A reaction A → P is zero order in A with rate constant $k = 0.5$ mol L $^{- 1}$ h $^{- 1}$ . It is run in a constant-volume batch reactor with $C_{A 0} = 2$ mol/L. Which statement is correct?

(A)
$t_{1/2} = 1.39$ h, and $C_{A}$ approaches zero only asymptotically
(B)
$t_{1/2} = 2$ h, and A is completely consumed at $t = 4$ h
(C)
$t_{1/2} = 1$ h, and A is completely consumed at $t = 2$ h
(D)
$t_{1/2} = 2$ h, and $C_{A}$ approaches zero only asymptotically

Answer and solution

Answer: (B)

Zero order: $t_{1/2} = C_{A 0} / (2 k) = 2/ (2 \times 0.5) = 2$ h.
$C_{A} = C_{A 0} - k t$ reaches zero at $t = C_{A 0} / k = 2/0.5 = 4$ h, so a zero-order reactant is used up in finite time. That rules out D.
Option A uses the first-order result $ln 2/ k$ , and option C uses the second-order result $1/ (k C_{A 0})$ .

PRACTICE 32 marks · NAT

A first-order reaction has rate constant $k = 0.02$ min $^{- 1}$ at 300 K and activation energy 50 kJ/mol. Assume the Arrhenius equation holds, with $R = 8.314$ J mol $^{- 1}$ K $^{- 1}$ . The half-life of the reaction at 320 K is ______ min (rounded to two decimal places).

Answer and solution

Answer: $9.9$ (accepted range $\pm 0.05$ )

$ln \frac{k _{2}}{k _{1}} = \frac{E _{a}}{R} \cdot \frac{T _{2} - T _{1}}{T _{1} T _{2}} = \frac{50000}{8.314} \times \frac{20}{300 \times 320} = 6013.95 \times 2.0833 \times 1 0^{- 4} = 1.2529$ .
$k_{2} = 0.02 e^{1.2529} = 0.02 \times 3.5005 = 0.07001$ min $^{- 1}$ .
$t_{1/2} = ln 2/ k_{2} = 0.6931/0.07001 = 9.90$ min.

PRACTICE 42 marks · MSQ

Without inhibitor, an enzyme follows Michaelis–Menten kinetics with parameters $K_{m}$ and $V_{ma x}$ . A reversible inhibitor is added at concentration $[I] = K_{i}$ , where $K_{i}$ is the dissociation constant of every enzyme–inhibitor complex the inhibitor forms (EI for a competitive inhibitor, ESI for an uncompetitive inhibitor, both EI and ESI for a pure non-competitive inhibitor). Which of the following statements is/are correct?

(A)
If the inhibitor is competitive, the apparent $K_{m}$ is $2 K_{m}$ and $V_{ma x}$ is unchanged.
(B)
If the inhibitor is uncompetitive, the apparent $K_{m}$ is $K_{m} /2$ and the apparent $V_{ma x}$ is $V_{ma x} /2$ .
(C)
If the inhibitor is pure non-competitive, the Lineweaver–Burk lines with and without inhibitor intersect on the $1/ v$ axis.
(D)
If the inhibitor is uncompetitive, the fractional reduction in rate is greater at $[S] ≪ K_{m}$ than at $[S] ≫ K_{m}$ .

Answer and solution

Answer: (A), (B)

$[I] / K_{i} = 1$ , so the factor $1 + [I] / K_{i} = 2$ .
A: competitive inhibition gives $K_{m, a pp} = 2 K_{m}$ with $V_{ma x}$ unchanged, so A is correct.
B: uncompetitive inhibition divides both $K_{m}$ and $V_{ma x}$ by 2, so B is correct.
C: pure non-competitive inhibition leaves $K_{m}$ unchanged and lowers $V_{ma x}$ . The lines share the $x$ -intercept $- 1/ K_{m}$ and meet on the $1/ [S]$ axis, not the $1/ v$ axis, so C is incorrect.
D: for uncompetitive inhibition, $v_{i} / v_{0} = (K_{m} + [S]) / (K_{m} + 2 [S])$ . This tends to 1 (no inhibition) for $[S] ≪ K_{m}$ and to 1/2 for $[S] ≫ K_{m}$ , so the inhibition is larger at high [S] and D is incorrect.

PRACTICE 52 marks · NAT

An enzyme obeying Michaelis–Menten kinetics, with $K_{m} = 5$ g/L and $V_{ma x} = 2$ g L $^{- 1}$ h $^{- 1}$ , converts a substrate in a constant-volume batch reactor. The enzyme does not deactivate, the reaction is irreversible and there is no product inhibition. Starting from $[S]_{0} = 20$ g/L, the time for the substrate concentration to fall to 2 g/L is ______ h (rounded to two decimal places).

Answer and solution

Answer: $14.76$ (accepted range $\pm 0.05$ )

Integrated Michaelis–Menten equation: $V_{ma x} t = ([S]_{0} - [S]) + K_{m} ln ([S]_{0} / [S])$ .
$2 t = (20 - 2) + 5 ln 10 = 18 + 11.513 = 29.513$ g/L.
$t = 29.513/2 = 14.76$ h.

OFFICIAL GATE 2027 SYLLABUS WORDING

Copied from the GATE 2027 syllabus; see all 2027 syllabi.

GATE BT

Bioreaction Engineering: Rate law, zero and first order kinetics; (More advanced kinetic models, including Michaelis-Menten kinetics for enzyme reactions. Inhibition kinetics: Competitive, non-competitive, and uncompetitive inhibition in enzyme catalysis)

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