# Calculate Michaelis-Menten constant of enzyme catalyzed reaction

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:

$$\frac{1}{v}=\frac{1}{v_\mathrm{max}}+\frac{K_M}{v_\mathrm{max}[S]},$$

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:

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?

• Your calculations are wrong on $\frac{1}{\ce{[S]}}$. May 26, 2021 at 17:32
• 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.
– Karsten
May 26, 2021 at 18:37
• 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. May 26, 2021 at 20:43
• @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.
– Karsten
May 27, 2021 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 $$V_\mathrm{max}$$:

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:

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 $$V_\mathrm{max}$$.

• Very nice answer indeed: went the extra mile! Glad I was the first upvoter! It should get the green checkmark ASAP!
– Ed V
May 27, 2021 at 0:05
• Thank you very much! Truly appreciate it! May 27, 2021 at 10:03