The Lineweaver-Burk linearisation

Concept

The Lineweaver-Burk linearisation

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Statement

For a single-substrate enzyme obeying the Michaelis–Menten rate law, plotting the reciprocal of the initial rate against the reciprocal of the substrate concentration gives an exactly straight line whose gradient and intercepts fix both kinetic parameters — at the cost of a severely distorted error structure. Formally: let \(v_0 = V_{max}[S]/(K_M+[S])\) with \(V_{max}\gt 0\), \(K_M\gt 0\) and \([S]\gt 0\); then the double-reciprocal map \((x,y)=(1/[S],\,1/v_0)\) carries the rate curve exactly onto the straight line \(y = (K_M/V_{max})\,x + 1/V_{max}\), whose slope is \(K_M/V_{max}\), whose ordinate intercept is \(1/V_{max}\) and whose abscissa intercept is \(-1/K_M\); the map is an algebraic identity, not an approximation, but it is not variance-preserving — if \(v_0\) carries additive measurement error of constant variance \(\sigma^2\) then \(\operatorname{Var}(y)=\sigma^2 y^4\) to first order, so ordinary unweighted least squares fitted to the transformed points is a misspecified estimator and the correct weights are \(w_i \propto v_i^{4}\).

Why it matters

The Michaelis–Menten rate law is a rectangular hyperbola, and a hyperbola is an awkward object to read by eye: its two parameters live in the curvature and in an asymptote that the data never reach. Every practical question asked of an enzyme — what is \(K_M\), what is \(V_{max}\), has an inhibitor changed one, the other, or both — is a question about those two numbers, and before nonlinear curve fitting was a keystroke away the only reliable way to extract them was to find a change of variable that turned the hyperbola into something a ruler could measure. The double-reciprocal plot introduced by Lineweaver and Burk in 1934 is that change of variable, and it is still the display in which enzyme inhibition is taught, published and diagnosed, because the three classical inhibition mechanisms produce three visually unmistakable families of straight lines.

It also matters as a cautionary example, and this is the half of the topic that is usually taught badly. The transformation is exact on the mathematics and destructive on the statistics: it takes the least precise measurements in the experiment — the slow rates at low substrate, where the signal is smallest — and throws them to the far right of the plot with enormously inflated leverage and enormously inflated error, while compressing the precise, high-substrate measurements into a crowd near the intercept where they can say almost nothing. Understanding exactly how a nonlinear change of variable redistributes error is a transferable skill: the same argument governs Scatchard plots in ligand binding, semi-log plots in pharmacokinetics, and every other linearisation a biologist is tempted to reach for.

Hypotheses

The underlying rate law really is the Michaelis–Menten hyperbola: one substrate, one intermediate, steady state, initial velocity.The linearity of the transformed plot is a theorem about that specific rate law and nothing else. A cooperative enzyme following a Hill law with \(n=2\) gives \(1/v_0 = (1/V_{max})\left(1 + (K_{0.5}/[S])^{2}\right)\), which is a parabola in \(x=1/[S]\), not a line; fitting a ruler to it returns parameters that depend entirely on which stretch of the plot was fitted, and can return a negative \(V_{max}\) (see Problem 5).

Every measured \([S]\) and every measured \(v_0\) is strictly positive.The map \(t \mapsto 1/t\) is undefined at \(t=0\) and discontinuous across it. The point \([S]=0\), which anchors the hyperbola at the origin and is the one point of the rate curve known exactly without any measurement, is sent to infinity and is lost. Worse, a noisy rate that happens to come out zero or slightly negative at very low substrate cannot be plotted at all, and the usual response — deleting it — selectively removes downward fluctuations and biases the fitted line.

The substrate concentrations are known accurately, so that all the error is in the ordinate.Least squares in either the transformed or the untransformed variable assumes the abscissa is exact. This is a mild assumption when the experimenter pipettes a dilution series, but it fails in two distinct ways with opposite signs. If \([S]\) is itself inferred and so carries random error, the result is the classical errors-in-variables attenuation of the gradient, biasing \(K_M\) downwards. If instead appreciable substrate is consumed during the assay window, the error is systematic — the nominal \([S]\) overstates the true mean — and it inflates the gradient instead, biasing \(K_M\) upwards (Fails without).

The measurement error in \(v_0\) is additive with constant variance \(\sigma^2\), and \(\sigma \ll v_0\) over the range fitted.Both clauses are needed for the variance statement in the Result. Constancy is what makes the transformation harmful, because with constant absolute variance the untransformed hyperbola needs no weights at all, so the entire weighting problem is manufactured by the change of variable. If instead the error were purely proportional, with constant coefficient of variation, then \(1/v_0\) would inherit exactly the same coefficient of variation and the transformation would neither create nor remove the need to weight — the untransformed fit would then need \(w_i\propto v_i^{-2}\) and the transformed fit \(w_i\propto v_i^{2}\), two spellings of the same correction, and unweighted regression would be wrong in either variable. The second clause is needed because the delta method is a first-order Taylor expansion; if \(v_0\) has non-negligible probability of lying near zero, \(1/v_0\) has no finite mean or variance at all and the leading-order formula is describing a quantity that does not exist.

Proof

The first half (Steps 1–6) is pure algebra and is exact. The second half (Steps 7–10) is the statistical content: what the change of variable does to the error, and what has to be done about it.

\[ v_0([S]) = \frac{V_{max}\,[S]}{K_M+[S]},\qquad [S]\in(0,\infty),\ V_{max}\gt 0,\ K_M\gt 0 \]

The hypothesis, stated with its domain. \(K_M\) and \(V_{max}\) are the unknowns; \([S]\) is set by the experimenter and \(v_0\) is measured. A

\[ [S]\gt 0 \ \Longrightarrow\ 0 \lt v_0 \lt V_{max} \]

Both numerator and denominator are strictly positive, and \([S]/(K_M+[S])\lt 1\) because \(K_M\gt 0\). Strict positivity is what licenses the next step: \(t\mapsto 1/t\) is a smooth, strictly decreasing bijection of \((0,\infty)\) onto itself, so taking reciprocals loses no information and introduces no spurious solutions. B

\[ \frac{1}{v_0} = \frac{K_M+[S]}{V_{max}\,[S]} \]

Reciprocate both sides of Step 1, legal by Step 2. A

\[ \frac{K_M+[S]}{V_{max}[S]} = \frac{K_M}{V_{max}[S]} + \frac{[S]}{V_{max}[S]} = \frac{K_M}{V_{max}}\cdot\frac{1}{[S]} + \frac{1}{V_{max}} \]

Split the single fraction over its two-term numerator and cancel \([S]\) in the second piece. This is the whole trick: the sum in the denominator of the hyperbola becomes a sum of two terms, one proportional to \(1/[S]\) and one constant. B

\[ y = m\,x + c,\qquad x \equiv \frac{1}{[S]},\quad y \equiv \frac{1}{v_0},\quad m = \frac{K_M}{V_{max}},\quad c = \frac{1}{V_{max}} \]

Naming the transformed coordinates turns Step 4 into an affine relation. Note the direction of the parameter map: \((K_M,V_{max})\mapsto(m,c)\) sends \((0,\infty)^2\) onto \((0,\infty)^2\) and is inverted by \(V_{max}=1/c\), \(K_M=m/c\), so the two descriptions carry exactly the same information — no parameter is lost and none is created. A

\[ y(0)=\frac{1}{V_{max}},\qquad y(x)=0 \iff x = -\frac{c}{m} = -\frac{1}{K_M},\qquad \text{but } x\gt 0 \text{ on the data} \]

Both intercepts are outside the image of the physical domain: \([S]\in(0,\infty)\) maps to \(x\in(0,\infty)\) and \(y\in(1/V_{max},\infty)\), an open ray that touches neither axis. Every parameter read off a double-reciprocal plot is therefore an extrapolation beyond the data, and the abscissa intercept sits on the side of the axis where no experiment can ever be done. That is a statement about leverage, and Step 8 makes it quantitative. C

\[ y = g(v_0)=\frac{1}{v_0},\qquad g'(v_0) = -\frac{1}{v_0^{2}},\qquad \operatorname{Var}(y) \approx \left(g'(v_0)\right)^{2}\operatorname{Var}(v_0) = \frac{\sigma^{2}}{v_0^{4}} \]

The delta method: expand \(g\) to first order about the true rate and take the variance. Valid when \(\sigma \ll v_0\) (Hypotheses, t3). B

\[ \operatorname{Var}(y) \approx \sigma^{2}y^{4} = \frac{\sigma^{2}}{V_{max}^{4}}\left(1+K_M x\right)^{4} \]

Substituting \(y=1/v_0\) into Step 7 gives the first form; substituting \(v_0 = V_{max}/(1+K_M x)\), which is Step 5 read backwards, gives the second. The variance of a transformed point therefore grows as the fourth power of its own ordinate, i.e. as \(x^4\) far from the origin. The points at the right-hand end of the plot — the low-substrate points, already the noisiest measurements in the experiment — are simultaneously the least reliable and the highest-leverage, which is the precise sense in which the transformation is statistically hostile. C

\[ w_i = \frac{1}{\operatorname{Var}(y_i)} \propto v_i^{4}, \qquad \hat m = \frac{S_w S_{wxy}-S_{wx}S_{wy}}{S_w S_{wxx}-S_{wx}^{2}},\qquad \hat c = \frac{S_{wxx}S_{wy}-S_{wx}S_{wxy}}{S_w S_{wxx}-S_{wx}^{2}} \]

Gauss–Markov: among linear unbiased estimators the minimum-variance one weights each observation by the reciprocal of its variance, and with Gaussian errors this is also the maximum-likelihood fit. The sums are \(S_w=\sum w_i\), \(S_{wx}=\sum w_i x_i\), \(S_{wy}=\sum w_i y_i\), \(S_{wxx}=\sum w_i x_i^{2}\), \(S_{wxy}=\sum w_i x_i y_i\). Only ratios of weights matter, so \(w_i=(v_i/v_{\max,\text{obs}})^4\) is as good as \(w_i=v_i^4\) and keeps the arithmetic on a human scale. Setting every \(w_i=1\) — the default in every spreadsheet — is the misspecification. B

\[ y_i-\hat y_i \approx -\frac{v_i-\hat v_i}{v_i^{2}} \ \Longrightarrow\ \sum_i w_i\left(y_i-\hat y_i\right)^{2} \approx \sum_i \frac{v_i^{4}\left(v_i-\hat v_i\right)^{2}}{v_i^{4}} = \sum_i \left(v_i-\hat v_i\right)^{2} \]

The same first-order expansion as Step 7, applied to the residual rather than the error. The \(v_i^4\) introduced by the weights cancels the \(v_i^{-4}\) introduced by the transformation exactly, so the weighted least-squares problem on the double-reciprocal plot has, to leading order, the same objective function as ordinary least squares on the untransformed hyperbola. This is the theorem that reconciles the two practices: correctly weighted Lineweaver–Burk and direct nonlinear fitting are the same estimator to first order, and it is only the missing weights that make the plot notorious. C

Result

\[ \frac{1}{v_0} = \frac{K_M}{V_{max}}\cdot\frac{1}{[S]} + \frac{1}{V_{max}}, \qquad \operatorname{Var}\!\left(\frac{1}{v_0}\right) \approx \frac{\sigma^{2}}{v_0^{4}} \]

Reading. Plot \(1/v_0\) against \(1/[S]\) and the Michaelis–Menten hyperbola becomes a straight line: the intercept on the ordinate is \(1/V_{max}\), the gradient is \(K_M/V_{max}\), so \(V_{max}=1/c\) and \(K_M=m/c\), and the line cuts the abscissa at \(-1/K_M\). The second formula is the price: reciprocation multiplies the variance of a rate by the fourth power of the reciprocal of that rate, so points must be weighted by \(v_i^4\) before a line is fitted.

Units check. With \([S]\) in \(\mu\mathrm{M}\) and \(v_0\) in \(\mu\mathrm{M\,s^{-1}}\), \(x\) is in \(\mu\mathrm{M^{-1}}\) and \(y\) in \(\mathrm{s}\,\mu\mathrm{M^{-1}}\); the gradient \(m=K_M/V_{max}\) is then in seconds and the intercept \(c=1/V_{max}\) in \(\mathrm{s}\,\mu\mathrm{M^{-1}}\), so \(m/c\) returns a concentration and \(1/c\) a concentration per unit time, as required.

Scope. Exact for any single-substrate enzyme obeying the hyperbolic rate law with strictly positive \([S]\) and \(v_0\); the variance statement is first-order and assumes additive, constant-variance, small error in \(v_0\) alone.

Corollaries & converses

Parameter recovery. \(V_{max}=1/c\) and \(K_M=m/c\); equivalently \(K_M\) is minus the reciprocal of the abscissa intercept. Because both formulas divide by \(c\), the fractional error in \(V_{max}\) and in \(K_M\) is at least the fractional error in the intercept, which is the most heavily extrapolated quantity on the plot.
Competitive inhibition pivots the line about its ordinate intercept. A competitive inhibitor at concentration \([I]\) replaces \(K_M\) by \(\alpha K_M\) with \(\alpha = 1+[I]/K_i\), giving \(y = (\alpha K_M/V_{max})x + 1/V_{max}\). The gradient is multiplied by \(\alpha\) and the intercept is untouched, so a family of lines at increasing \([I]\) fans out from a common point on the ordinate axis, and \(K_i=[I]/(\alpha-1)\).
Uncompetitive inhibition shifts the line without turning it. An inhibitor binding only the \(ES\) complex divides both \(K_M\) and \(V_{max}\) by \(\alpha' = 1+[I]/K_i'\), giving \(y = (K_M/V_{max})x + \alpha'/V_{max}\). The gradient is untouched and the intercept is multiplied by \(\alpha'\): the lines are parallel.
Pure non-competitive inhibition pivots the line about its abscissa intercept. When \(\alpha=\alpha'\) the line is \(y=(\alpha/V_{max})(K_M x + 1)\), which vanishes at \(x=-1/K_M\) for every \([I]\), so the whole family crosses at one point on the abscissa. Mixed inhibition, with \(\alpha\ne\alpha'\), crosses above or below the abscissa according to whether \(\alpha\gt\alpha'\) or \(\alpha\lt\alpha'\).
Sister linearisations. The same algebra rearranged differently gives Hanes–Woolf, \([S]/v_0 = (1/V_{max})[S] + K_M/V_{max}\), and Eadie–Hofstee, \(v_0 = V_{max} - K_M\,(v_0/[S])\). Under constant-variance error the Hanes–Woolf ordinate has variance \([S]^2\sigma^2/v_0^4\), which diverges only as \([S]^{-2}\) as \([S]\to 0\) rather than the \([S]^{-4}\) of the double-reciprocal plot, so it is the better-conditioned of the two; Eadie–Hofstee is worse behaved in a different way, since \(v_0\) appears on both axes and the errors in abscissa and ordinate are perfectly correlated.
Converse (with a caveat). If the transformed points are collinear within experimental error across a wide range of \(1/[S]\), the underlying rate law is hyperbolic, since the map is a bijection and only the hyperbola has a straight image. The caveat is that the test is weak where it matters: the high-substrate points, which are the precise ones, are crushed together near the intercept and would look collinear for almost any rate law.

Fails without

Hyperbolic kinetics dropped (cooperativity): for a Hill law with \(K_{0.5}=500\ \mu\mathrm{M}\), \(V_{max}=60\ \mu\mathrm{M\,s^{-1}}\) and \(n=2\), the transformed chord gradients between successive points at \([S]=2000,1000,500,200,100\ \mu\mathrm{M}\) are \(6.25,\,12.5,\,29.2,\,62.5\ \mathrm{s}\) — a tenfold increase, i.e. gross upward curvature. Fitting a line to the three left-hand points returns \(V_{max}=87\ \mu\mathrm{M\,s^{-1}}\) and \(K_M=935\ \mu\mathrm{M}\); fitting the three right-hand points returns an intercept of \(-0.0947\ \mathrm{s}\,\mu\mathrm{M^{-1}}\), i.e. \(V_{max}=-10.6\ \mu\mathrm{M\,s^{-1}}\), a negative maximum rate. The plot does not warn you; it simply returns numbers.
Constant-variance weighting dropped (unweighted least squares retained): in Worked example 1 the same five points give \(V_{max}=64.9\ \mu\mathrm{M\,s^{-1}}\), \(K_M=579\ \mu\mathrm{M}\) unweighted but \(V_{max}=59.9\ \mu\mathrm{M\,s^{-1}}\), \(K_M=496\ \mu\mathrm{M}\) weighted by \(v_i^4\), against true values of \(60\) and \(500\). The unweighted fit is biased by \(+8\%\) in \(V_{max}\) and \(+16\%\) in \(K_M\) with no increase in the scatter to signal that anything is wrong.
Strict positivity of \(v_0\) dropped: at \([S]=0.1K_M\) the true rate is under a tenth of \(V_{max}\), so with realistic noise a replicate can read zero or slightly negative. That point has no image under \(y=1/v_0\); discarding it removes only downward fluctuations from the noisiest, highest-leverage end of the plot, which systematically lowers the fitted gradient and biases \(K_M\) downwards. The untransformed hyperbola has no such problem — a negative measured rate is simply a residual.
Exact abscissa dropped (substrate consumed during the assay): if \(20\%\) of the substrate is turned over during the measurement window, the concentration falls from \([S]\) to \(0.8[S]\) and its mean over the window is about \(0.9[S]\), so the true abscissa is about \(1/(0.9[S]) = 1.11\,x\) while the plotted one is \(x\) and the ordinate is unaffected. The fitted gradient is inflated by that factor and the intercept is untouched, so \(V_{max}\) survives but \(K_M\) is overestimated by roughly \(11\%\), and worse at low \([S]\), where a given absolute depletion is a larger fraction. Note the sign: this systematic error pushes \(K_M\) up, whereas random error in a merely inferred \([S]\) attenuates the gradient and pushes it down (Hypotheses). The remedy for both is the initial-rate hypothesis of the underlying Michaelis–Menten treatment: a window short enough that depletion is negligible.

Common errors

“The ordinate intercept gives \(K_M\).” It gives \(1/V_{max}\). \(K_M\) is the gradient divided by the intercept, or minus the reciprocal of the abscissa intercept; the intercept alone contains no information about \(K_M\) whatsoever.
“The \(x\)-intercept is a measurement of \(-1/K_M\).” It is an extrapolation to a negative substrate concentration, on the far side of an axis no data point can approach (Step 6). Its uncertainty is dominated by the ratio of two fitted quantities and is far larger than the visual tidiness of the plot suggests.
“\(r^2 = 0.998\), so the enzyme is Michaelis–Menten.” A high coefficient of determination on a double-reciprocal plot is nearly guaranteed: one or two low-substrate points sit far out along the abscissa and dominate the total sum of squares, so \(r^2\) is measuring the leverage of those points, not the adequacy of the model. Curvature must be judged from the residuals, or better, in the untransformed plot.
“The gradient changed, so the inhibitor is competitive.” The gradient also changes under mixed and pure non-competitive inhibition. The diagnosis requires both coefficients: gradient changed with intercept fixed is competitive, intercept changed with gradient fixed is uncompetitive, both changed is mixed (Corollaries).
“Weighting is a refinement for careful workers.” Unweighted regression on this plot is not a slightly worse estimator; it is a different estimator of different quantities, minimising \(\sum(v_i-\hat v_i)^2/v_i^4\) instead of \(\sum(v_i-\hat v_i)^2\) (Step 10). It is biased even when the model is exactly right and the data are plentiful.
“Just fit the line through the origin-most points.” Choosing a sub-range to fit is what makes a curved plot look straight; the two windows in Fails without give \(V_{max}=87\) and \(V_{max}=-10.6\ \mu\mathrm{M\,s^{-1}}\) from the same data set.

Discussion

Hans Lineweaver and Dean Burk published the double-reciprocal plot in 1934, and it was not the first such rearrangement: Barnet Woolf’s three linear forms were published by Haldane and Stern in 1932, and Charles Hanes used the \([S]/v_0\) form in the same period, with the \(v_0\) versus \(v_0/[S]\) plot later associated with Eadie and with Hofstee. That four different linearisations of a two-parameter curve should all have been thought worth naming tells you how badly a pre-computational biochemistry needed straight lines. Of the four, the one that won the textbooks is, on purely statistical grounds, the worst.

The reason it won anyway is diagnostic, not estimative. The three inhibition mechanisms are distinguished by which coefficient of the line moves, and coefficient-moving is something the eye is extremely good at judging: a fan of lines through a common point on the ordinate axis, a ladder of parallel lines, a fan through a common point on the abscissa. No comparable visual signature exists on the untransformed hyperbola, where three sets of slightly different saturation curves look much the same. This is why the plot survives in the inhibition literature long after it was abandoned for measuring \(K_M\), and why the honest modern practice is to obtain the parameters by nonlinear least squares on the hyperbola and then display them as a double-reciprocal plot with the fitted lines overlaid.

Step 10 explains why that division of labour is not a contradiction. Weighted least squares on the transformed data and ordinary least squares on the untransformed data agree to first order, because the Jacobian factor \(v^{-2}\) that the transformation introduces into the residual is exactly undone by the \(v^{4}\) in the weights. The agreement is only to first order: the two estimators differ at second order in \(\sigma/v\), and they differ more than that in their tails, since \(1/v_0\) has heavy tails whenever \(v_0\) can approach zero — formally, if \(v_0\) is modelled as Gaussian, \(1/v_0\) has no finite moments at all, and the delta-method variance in Step 7 is an asymptotic device rather than a property of the actual distribution. The practical reading is that the transformation is safe wherever the relative error in the rate is small and dangerous exactly where it is not, which on this plot is the region carrying most of the leverage.

Common misconceptions. That the plot is an approximation to the hyperbola — it is an exact identity, and every criticism of it is statistical rather than algebraic. That a Lineweaver–Burk plot can detect cooperativity reliably — it curves, but the curvature lives at low substrate where the error bars are largest, so a Hill plot or a direct fit is a far more sensitive test. And that the “\(-1/K_M\)” label on the abscissa means \(K_M\) has been measured; it has been extrapolated, from data that all lie in the opposite quadrant.

Worked examples

Example 1. A purified hydrolase is assayed at five substrate concentrations under initial-rate conditions. The data below were generated from \(K_M=500\ \mu\mathrm{M}\) and \(V_{max}=60\ \mu\mathrm{M\,s^{-1}}\) with a few per cent of added noise, so that the two fits can be scored against known truth. Fit the double-reciprocal plot (a) unweighted and (b) weighted by \(v_i^4\), and compare.

\[ \begin{array}{c|ccccc} [S]/\mu\mathrm{M} & 100 & 200 & 500 & 1000 & 2000 \\ v_0/\mu\mathrm{M\,s^{-1}} & 9.4 & 17.8 & 29.2 & 40.9 & 47.6 \end{array} \]

The raw assay. Note that the rate at the lowest concentration is under a fifth of the rate at the highest, so at fixed absolute noise its relative error is five times worse. A

\[ \begin{array}{c|ccccc} x=1/[S] \ /\mu\mathrm{M^{-1}} & 0.01000 & 0.00500 & 0.00200 & 0.00100 & 0.00050 \\ y=1/v_0 \ /\mathrm{s}\,\mu\mathrm{M^{-1}} & 0.10638 & 0.05618 & 0.03425 & 0.02445 & 0.02101 \end{array} \]

Apply the transformation of Step 5 to each pair. The abscissa now spans a factor of twenty, and the single point at \(x=0.01\) sits alone at one end of that span. A

\[ n=5,\quad S_x = 0.01850,\quad S_y = 0.24227,\quad S_{xx}=1.3025\times10^{-4},\quad S_{xy}=1.44818\times10^{-3} \]

The five sums needed for an unweighted straight-line fit. A

\[ \Delta = nS_{xx}-S_x^{2} = 6.5125\times10^{-4}-3.4225\times10^{-4}=3.0900\times10^{-4} \]

The common denominator of the unweighted normal equations, a scalar that must be non-zero — it vanishes only if every \(x_i\) is equal, i.e. if the experiment was done at a single substrate concentration. A

\[ m = \frac{nS_{xy}-S_xS_y}{\Delta} = \frac{2.75890\times10^{-3}}{3.0900\times10^{-4}} = 8.93\ \mathrm{s},\qquad c = \frac{S_{xx}S_y-S_xS_{xy}}{\Delta} = \frac{4.7643\times10^{-6}}{3.0900\times10^{-4}} = 0.015418\ \mathrm{s}\,\mu\mathrm{M^{-1}} \]

The unweighted least-squares line. Units confirm the Result: a gradient in seconds and an intercept in \(\mathrm{s}\,\mu\mathrm{M^{-1}}\). A

\[ V_{max}=\frac{1}{c}=64.9\ \mu\mathrm{M\,s^{-1}},\qquad K_M=\frac{m}{c}=\frac{8.93}{0.015418}=579\ \mu\mathrm{M} \]

Inverting the parameter map of Step 5. Both are high: \(+8\%\) on \(V_{max}\) and \(+16\%\) on \(K_M\) against the generating values. A

\[ w_i=\left(\frac{v_i}{47.6}\right)^{4}:\qquad 0.00152,\ \ 0.01955,\ \ 0.14161,\ \ 0.54509,\ \ 1.00000 \]

The Step 9 weights, normalised to the largest rate so the sums stay near unity; only ratios matter. The lowest-substrate point is now worth about \(0.15\%\) of the highest — which is what “its variance is \(660\) times larger” means in practice. B

\[ S_w=1.70777,\quad S_{wx}=1.44126\times10^{-3},\quad S_{wy}=4.04476\times10^{-2},\quad S_{wxx}=2.00250\times10^{-6},\quad S_{wxy}=4.06412\times10^{-5} \]

The five weighted sums of Step 9, each term multiplied by its \(w_i\) before summing. A

\[ D = S_wS_{wxx}-S_{wx}^{2} = 3.41981\times10^{-6}-2.07723\times10^{-6}=1.34258\times10^{-6} \]

The weighted analogue of \(\Delta\). A

\[ \hat m = \frac{1.11104\times10^{-5}}{1.34258\times10^{-6}} = 8.28\ \mathrm{s},\qquad \hat c = \frac{2.24215\times10^{-8}}{1.34258\times10^{-6}} = 0.016700\ \mathrm{s}\,\mu\mathrm{M^{-1}} \]

Substituting the Step 8 sums into the two formulas of Step 9. B

\[ V_{max}=\frac{1}{0.016700}=59.9\ \mu\mathrm{M\,s^{-1}},\qquad K_M=\frac{8.28}{0.016700}=496\ \mu\mathrm{M} \]

Inverting the parameter map again. Both are now within \(1\%\) of the generating values, from exactly the same five measurements. A

\[ \text{unweighted: } V_{max}=64.9,\ K_M=579 \qquad\text{weighted: } V_{max}=59.9,\ K_M=496 \qquad(\text{true } 60,\ 500) \]

Reading. The unweighted fit is wrong by \(+8\%\) and \(+16\%\); the weighted fit is right to better than \(1\%\). Nothing about the appearance of the plot distinguishes them — both lines pass convincingly through the five points — so the error is invisible to inspection and is removed only by putting the \(v_i^4\) weights in.

Scope. Rates in \(\mu\mathrm{M\,s^{-1}}\), concentrations in \(\mu\mathrm{M}\); the weighting is the one appropriate to additive, constant-variance error in \(v_0\) (Hypotheses, t3). Under constant relative error the weights would instead be \(w_i \propto v_i^{2}\).

Example 2. The same enzyme is re-assayed at the same five substrate concentrations in the presence of \([I]=200\ \mu\mathrm{M}\) of a reversible inhibitor. Identify the mechanism and obtain \(K_i\), using the weighted control line \(m=8.28\ \mathrm{s}\), \(c=0.01670\ \mathrm{s}\,\mu\mathrm{M^{-1}}\) from Example 1.

\[ \begin{array}{c|ccccc} [S]/\mu\mathrm{M} & 100 & 200 & 500 & 1000 & 2000 \\ v_0/\mu\mathrm{M\,s^{-1}} & 3.75 & 7.06 & 15.00 & 24.00 & 34.29 \end{array} \]

The inhibited assay. Every rate is depressed, but by a factor that shrinks as \([S]\) rises — from \(2.5\) at \(100\ \mu\mathrm{M}\) to \(1.4\) at \(2000\ \mu\mathrm{M}\) — which already hints that the inhibitor is being outcompeted by substrate. A

\[ \begin{array}{c|ccccc} x \ /\mu\mathrm{M^{-1}} & 0.01000 & 0.00500 & 0.00200 & 0.00100 & 0.00050 \\ y \ /\mathrm{s}\,\mu\mathrm{M^{-1}} & 0.26667 & 0.14164 & 0.06667 & 0.04167 & 0.02916 \end{array} \]

Transform as before. These five points are collinear to within the rounding of the table, so weighted and unweighted fits coincide here and the arithmetic can be done unweighted without prejudice. A

\[ S_x=0.01850,\quad S_y=0.54581,\quad S_{xx}=1.3025\times10^{-4},\quad S_{xy}=3.56449\times10^{-3},\quad \Delta = 3.0900\times10^{-4} \]

The same five sums as Example 1, Step 3, for the inhibited data. A

\[ m_I = 25.00\ \mathrm{s},\qquad c_I = 0.016662\ \mathrm{s}\,\mu\mathrm{M^{-1}} \]

The fitted line, from the two formulas of Example 1, Step 5. A

\[ \frac{c_I}{c} = \frac{0.016662}{0.016700}=0.998 \approx 1, \qquad \frac{m_I}{m} = \frac{25.00}{8.28}=3.02 \gg 1 \]

The diagnostic comparison. The ordinate intercept is unchanged within \(0.2\%\) while the gradient has trebled: the two lines cross on the ordinate axis. By the Corollaries this is competitive inhibition — \(V_{max}\) untouched, apparent \(K_M\) raised. B

\[ V_{max}^{app}=\frac{1}{c_I}=60.0\ \mu\mathrm{M\,s^{-1}},\qquad K_M^{app}=\frac{m_I}{c_I}=\frac{25.00}{0.016662}=1500\ \mu\mathrm{M} \]

Reading the inhibited line as if it were an uninhibited enzyme confirms the diagnosis numerically: \(V_{max}\) is recovered at its control value, while \(K_M\) has gone from \(496\) to \(1500\ \mu\mathrm{M}\), a factor of \(3.02\). A

\[ \alpha = 1+\frac{[I]}{K_i} = 3.02 \ \Longrightarrow\ K_i = \frac{[I]}{\alpha-1} = \frac{200\ \mu\mathrm{M}}{2.02} = 99\ \mu\mathrm{M} \]

Solving the competitive relation of the Corollaries for \(K_i\). Symbols first, numbers last: \(K_i\) is the inhibitor concentration that would double the apparent \(K_M\). B

\[ \text{competitive inhibition},\qquad \alpha = 3.02,\qquad K_i = 99\ \mu\mathrm{M} \]

Reading. The inhibitor competes with substrate for the free enzyme: it raises the substrate concentration needed for half-maximal rate threefold but leaves the saturated rate alone, so enough substrate always overcomes it. \(K_i\approx 99\ \mu\mathrm{M}\) is about a fifth of \(K_M\), so the inhibitor binds the free enzyme about five times more tightly than the substrate does.

Scope. A single inhibitor concentration is the minimum for this calculation; a real determination repeats the assay at three or four values of \([I]\) and checks that the lines share one intercept before quoting \(K_i\) from the gradient of \(m_I\) against \([I]\).

Problems

A double-reciprocal fit of an assay with \([S]\) in \(\mu\mathrm{M}\) and \(v_0\) in \(\mu\mathrm{M\,s^{-1}}\) gives gradient \(12.5\ \mathrm{s}\) and ordinate intercept \(0.0250\ \mathrm{s}\,\mu\mathrm{M^{-1}}\). Find \(V_{max}\), \(K_M\), the abscissa intercept, and the rate expected at \([S]=200\ \mu\mathrm{M}\).
Solution
From the Result, \(V_{max}=1/c = 1/0.0250 = 40.0\ \mu\mathrm{M\,s^{-1}}\) and \(K_M=m/c = 12.5/0.0250 = 500\ \mu\mathrm{M}\). The abscissa intercept is \(-1/K_M = -2.00\times10^{-3}\ \mu\mathrm{M^{-1}}\). Substituting into the hyperbola, \(v_0 = (40.0)(200)/(500+200) = 8000/700 = 11.4\ \mu\mathrm{M\,s^{-1}}\). As a check, \(1/v_0\) should equal \(m x + c = 12.5(0.00500)+0.0250 = 0.0875\ \mathrm{s}\,\mu\mathrm{M^{-1}}\), and indeed \(1/11.43 = 0.0875\).
Three assay points are obtained: \(([S],v_0) = (250,\,30.8),\ (500,\,44.4),\ (2000,\,66.7)\) in \(\mu\mathrm{M}\) and \(\mu\mathrm{M\,s^{-1}}\). Transform them, fit an unweighted line, and report \(K_M\) and \(V_{max}\).
Solution
Transforming: \(x = 4.00\times10^{-3},\ 2.00\times10^{-3},\ 5.00\times10^{-4}\ \mu\mathrm{M^{-1}}\) and \(y = 0.032468,\ 0.022523,\ 0.014993\ \mathrm{s}\,\mu\mathrm{M^{-1}}\). Sums: \(n=3\), \(S_x = 6.50\times10^{-3}\), \(S_y = 0.069984\), \(S_{xx}=2.025\times10^{-5}\), \(S_{xy}=1.82411\times10^{-4}\). Then \(\Delta = 3(2.025\times10^{-5})-(6.50\times10^{-3})^2 = 6.075\times10^{-5}-4.225\times10^{-5} = 1.850\times10^{-5}\). Gradient \(m = [3(1.82411\times10^{-4}) - (6.50\times10^{-3})(0.069984)]/\Delta = (5.47233\times10^{-4}-4.54896\times10^{-4})/1.850\times10^{-5} = 4.99\ \mathrm{s}\). Intercept \(c = [(2.025\times10^{-5})(0.069984)-(6.50\times10^{-3})(1.82411\times10^{-4})]/\Delta = (1.41717\times10^{-6}-1.18567\times10^{-6})/1.850\times10^{-5} = 0.012512\ \mathrm{s}\,\mu\mathrm{M^{-1}}\). Hence \(V_{max}=1/c = 79.9\ \mu\mathrm{M\,s^{-1}}\) and \(K_M = m/c = 4.99/0.012512 = 399\ \mu\mathrm{M}\). (The data were generated from \(80\) and \(400\); with only three points spanning a modest range the unweighted bias is small.)
Rates are measured with a constant standard deviation \(\sigma = 1.5\ \mu\mathrm{M\,s^{-1}}\), independent of the rate. Using the data of Worked example 1, find the standard deviation of \(y=1/v_0\) at \(v_0 = 9.4\) and at \(v_0=47.6\ \mu\mathrm{M\,s^{-1}}\), express each as a fraction of \(y\) itself, and state the ratio of the two variances.
Solution
By Step 7, \(\sigma_y = \sigma/v_0^{2}\). At \(v_0=9.4\): \(\sigma_y = 1.5/9.4^{2} = 1.5/88.36 = 1.70\times10^{-2}\ \mathrm{s}\,\mu\mathrm{M^{-1}}\), against \(y = 0.10638\), a relative error of \(16.0\%\). At \(v_0=47.6\): \(\sigma_y = 1.5/2265.76 = 6.62\times10^{-4}\ \mathrm{s}\,\mu\mathrm{M^{-1}}\), against \(y=0.02101\), a relative error of \(3.15\%\). The ratio of variances is \((47.6/9.4)^{4} = 5.064^{4} = 658\). Note the relative errors stand in the ratio \(47.6/9.4 = 5.06\), exactly the ratio of the rates, as they must: reciprocation preserves relative error to first order but the constant-\(\sigma\) assumption means the relative error in \(v_0\) itself was already five times worse at the low point.
An enzyme with control parameters \(m = 8.33\ \mathrm{s}\), \(c = 0.01667\ \mathrm{s}\,\mu\mathrm{M^{-1}}\) is assayed with \([I]=300\ \mu\mathrm{M}\) of a second inhibitor, giving \(m_I = 8.33\ \mathrm{s}\), \(c_I = 0.02500\ \mathrm{s}\,\mu\mathrm{M^{-1}}\). Classify the inhibition, obtain the apparent parameters and the inhibition constant, and say why this mechanism cannot be overcome by adding substrate.
Solution
The gradient is unchanged and the intercept has risen: the two lines are parallel, which by the Corollaries is uncompetitive inhibition (the inhibitor binds only the \(ES\) complex). Control values: \(V_{max}=1/0.01667 = 60.0\ \mu\mathrm{M\,s^{-1}}\), \(K_M = 8.33/0.01667 = 500\ \mu\mathrm{M}\). Inhibited: \(V_{max}^{app} = 1/0.02500 = 40.0\ \mu\mathrm{M\,s^{-1}}\), \(K_M^{app} = 8.33/0.02500 = 333\ \mu\mathrm{M}\). Both fall by the same factor \(\alpha' = c_I/c = 0.02500/0.01667 = 1.500\), as the uncompetitive form requires. Hence \(1+[I]/K_i' = 1.500\) and \(K_i' = 300/0.500 = 600\ \mu\mathrm{M}\). Adding substrate does not help: the inhibitor binds \(ES\), so raising \([S]\) raises the concentration of the very species the inhibitor targets. Both \(V_{max}\) and \(K_M\) fall together, their ratio \(k_{cat}/K_M\) is unchanged, and no substrate concentration restores the control rate.
An enzyme in fact obeys a Hill law \(v_0 = V_{max}[S]^{2}/(K_{0.5}^{2}+[S]^{2})\) with \(V_{max}=60\ \mu\mathrm{M\,s^{-1}}\) and \(K_{0.5}=500\ \mu\mathrm{M}\). Compute the double-reciprocal points at \([S]=2000,1000,500,200,100\ \mu\mathrm{M}\), demonstrate the curvature from the chord gradients, and fit straight lines to the three left-hand and the three right-hand points. Comment.
Solution
Rates: at \(2000\), \(v_0 = 60(4\times10^{6})/(2.5\times10^{5}+4\times10^{6}) = 56.471\); at \(1000\), \(48.000\); at \(500\), \(30.000\); at \(200\), \(8.276\); at \(100\), \(2.308\ \mu\mathrm{M\,s^{-1}}\). Transformed, \((x,y)\) in \(\mu\mathrm{M^{-1}}\) and \(\mathrm{s}\,\mu\mathrm{M^{-1}}\): \((5.0\times10^{-4},\,0.01771)\), \((1.0\times10^{-3},\,0.02083)\), \((2.0\times10^{-3},\,0.03333)\), \((5.0\times10^{-3},\,0.12083)\), \((1.0\times10^{-2},\,0.43333)\).
Chord gradients between successive points, left to right: \((0.02083-0.01771)/(5.0\times10^{-4}) = 6.25\ \mathrm{s}\); then \(12.5\), \(29.2\) and \(62.5\ \mathrm{s}\). A straight line has one gradient; these increase tenfold, so the plot is strongly concave upwards.
Fitting the three left-hand (high-substrate) points \(x = 5\times10^{-4},10^{-3},2\times10^{-3}\): \(\Delta = 3(5.25\times10^{-6})-(3.5\times10^{-3})^{2} = 1.575\times10^{-5}-1.225\times10^{-5}=3.50\times10^{-6}\), giving \(m = 10.71\ \mathrm{s}\) and \(c = 0.011458\ \mathrm{s}\,\mu\mathrm{M^{-1}}\), i.e. \(V_{max}=87.3\ \mu\mathrm{M\,s^{-1}}\) and \(K_M = 935\ \mu\mathrm{M}\). Fitting the three right-hand (low-substrate) points \(x = 2\times10^{-3},5\times10^{-3},10^{-2}\): \(\Delta = 3(129\times10^{-6})-(1.7\times10^{-2})^{2} = 3.87\times10^{-4}-2.89\times10^{-4}=9.80\times10^{-5}\), giving \(m = 51.28\ \mathrm{s}\) and \(c = -0.09473\ \mathrm{s}\,\mu\mathrm{M^{-1}}\), i.e. \(V_{max} = -10.6\ \mu\mathrm{M\,s^{-1}}\) and \(K_M = -541\ \mu\mathrm{M}\).
Comment: neither pair of numbers means anything, and they are not even close to one another, yet each individual fit would look acceptable plotted on its own. A negative \(V_{max}\) is the plot's only honest signal that the model is wrong, and it appears only in the window a careful experimenter is most likely to discard as noisy. The correct diagnosis is made before any linearisation: the untransformed rate curve is sigmoid, not hyperbolic, and \(n\) should be estimated directly.