I have:

$$\begin{array}{c|c} \dfrac{v }{ [S]}/\pu{s^{-1}} & v \cdot 10^2/\pu{mol dm^-3 s^-1} \\ \hline 0.257\ & 5.15\\ 0.895 & 4.48\\ 2.00\ & 3.35\\ 3.59\ & 1.8\\ 4.82\ & 0.48\\ \hline \end{array}$$

And I want to calculate $K_M$ graphically.

I know that I should use the equation:


and that I should plot $1/v$ against $1/[S]$, and then the slope will equal $K_M$/$v_\mathrm{max}$.

I began with converting the data given from $v/[S]$ to $1/[S]$ by dividing each data point with $v$. This gave:

$$\begin{array}{c|c} \dfrac{1}{[S]}/\pu{dm^3 mol^{-1}} & v \cdot 10^2/\pu{mol dm^-3 s^-1} \\ \hline 4.99*10^{-4}\ & 5.15\\ 0.0019 & 4.48\\ 0.0059\ & 3.35\\ 0.0199\ & 1.8\\ 0.1004\ & 0.48\\ \hline \end{array}$$

Plotting $1/v$ against $1/[S]$ gives the graph:

Plot of 1/v vs 1/[S]

But when I want to calculate $K_M$ through: $$\text{slope} \times V_\mathrm{max} = \text{slope} \times \frac{1}{\text{intercept}}$$ I don't get the answer which is in our answer key. Unfortunately, only the answer for $K_M$ is given, and not how to get the answer. The answer is $K_M = \pu{0.0102 mol dm-3}$. Could someone explain where I am going wrong?

  • 1
    $\begingroup$ Your calculations are wrong on $\frac{1}{\ce{[S]}}$. $\endgroup$ May 26 at 17:32
  • 2
    $\begingroup$ You somehow reversed the order of the data points, too. That's why there is a hyperbole and not a straight line. Because of the format of the data, it is easier to do an Eadie Hofstee fit instead of the Lineweaver-Burk you attempted. $\endgroup$ May 26 at 18:37
  • $\begingroup$ When I am editing your question, I realized you have plot $\frac{1}{[S]}$ vs $v$ instead of $\frac{1}{[S]} $ vs $\frac{1}{v} $. That's why you got a hyperbole instead of straight-line as Karsten Theis suggested. $\endgroup$ May 26 at 20:43
  • $\begingroup$ @MathewMahindaratne I think the values plotted represent 1/v (off by a factor 100), but are in the wrong order, i.e. the first substrate concentration is plotted against the last rate etc. $\endgroup$ May 27 at 1:03

What you have given is: $$\begin{array}{c|c} \hline \dfrac{v}{[S]}, \ \pu{s^{-1}} & v, \ \pu{mol dm^-3 s^-1} \\ \hline 0.257 & 5.15 \times 10^{-2}\\ 0.895 & 4.48 \times 10^{-2}\\ 2.00 & 3.35 \times 10^{-2}\\ 3.59 & 1.8 \times 10^{-2}\\ 4.82 & 0.48 \times 10^{-2}\\ \hline \end{array}$$

Michaelis-Menten equation for enzyme kinetics is:

$$v = \frac{V_\mathrm{max} \cdot [S]}{K_M + [S]} \tag1$$

When rearrange this equation (reciprocal) to get Lineweaver-Burk relationship:

$$\frac{1}{v} = \frac{1}{V_\mathrm{max}} + \frac{K_M}{V_\mathrm{max}[S]} \tag2$$

When multiply the equation $(2)$ by $v \times V_\mathrm{max}$, you get:

$$V_\mathrm{max} = v + \frac{vK_M}{[S]} \tag3$$

When rearrange the equation $(3)$, you get $y = mx + c$ type equation (straight-line equation):

$$v = - K_M \cdot \frac{v}{[S]} + V_\mathrm{max} \tag4$$

The data you have is $\frac{v}{[S]}$ and $v$, which fix the relationship given in the equation $(4)$. Thus, if you can plot them against each other ($\frac{v}{[S]}$ versus $v$) you'd get a straight line with negative lope. The numerical value of the slope is $K_M$ so you can calculate K$_M$ graphically. Also, the $y$-intercept of the graph is equal to $V_\mathrm{max}$:

Plot of v/[S] vs v

From the equation of the graph: $\text{|The slope|} = 0.0102$ and $y\text{-Intercept} = 0.054$

Thus, $K_M = \pu{0.0102 mol L-1}$ and $V_\mathrm{max} = \pu{0.054 mol L-1 s-1}$

Once I found OP's mistake of plotting, I was curious to see how this set of data would show the Lineweaver-Burk relationship. Thus, I make the table for $\frac{1}{[S]}$ and $\frac{1}{v}$:

$$\begin{array}{c|c} \hline \dfrac{1}{[S]}, \ \pu{L mol-1} & \dfrac{1}{v}, \ \pu{L s mol-1} \\ \hline 4.99 & 19.42 \times 10^{-2}\\ 19.98 & 22.32 \times 10^{-2}\\ 59.70 & 29.85 \times 10^{-2}\\ 199.44 & 55.56 \times 10^{-2}\\ 1004.17 & 208.33 \times 10^{-2}\\ \hline \end{array}$$

As predicted by Lineweaver-Burk, the plot of $\frac{1}{[S]}$ versus $\frac{1}{v}$ is a straight-line with a positive slope:

Plot of 1/[S] vs 1/v

From the equation $(2)$, which is the equation of the graph: $\text{The slope} = 0.1891 = \frac{K_M}{V_\mathrm{max}}$ and $y\text{-Intercept} = 18.379 = \frac{1}{V_\mathrm{max}}$

Thus, $V_\mathrm{max} = \frac{1}{18.379} = \pu{0.0544 mol L-1 s-1}$ and $K_M = 0.1891 \times V_\mathrm{max} = 0.1891 \times 0.0544 = \pu{0.0103 mol L-1}$

Both graph give the same values for $K_M$ and $V_\mathrm{max}$.

  • 1
    $\begingroup$ Very nice answer indeed: went the extra mile! Glad I was the first upvoter! It should get the green checkmark ASAP! $\endgroup$
    – Ed V
    May 27 at 0:05
  • $\begingroup$ Thank you very much! Truly appreciate it! $\endgroup$
    – confused
    May 27 at 10:03

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