Enzyme kinetics by numbers

What the equation says

A simple enzyme turns a substrate S into a product P. The substrate first binds the active site to make an enzyme-substrate complex, and the complex then releases the product. For many enzymes the rate follows the Michaelis-Menten equation.

• $v$ is the rate, in micromoles of product per minute (µmol/min).
• $[S]$ is the substrate concentration, in millimoles per litre (mmol/L).
• $V_{\max}$ is the greatest rate, reached when every enzyme molecule is busy.
• $K_m$, the Michaelis constant, is the substrate concentration at which $v = V_{\max}/2$. It is a concentration, with units of mmol/L.

Three regions of the curve

Look at the denominator, $K_m + [S]$.

• $[S]$ much smaller than $K_m$. The denominator is close to $K_m$, so $v$ is nearly proportional to $[S]$. Doubling the substrate doubles the rate.
• $[S] = K_m$. The rate is exactly half of $V_{\max}$.
• $[S]$ much larger than $K_m$. The denominator is close to $[S]$, so $v$ is close to $V_{\max}$. Extra substrate does almost nothing, because every active site is already busy.

Here are the rates for an enzyme with $V_{\max} = 120$ µmol/min and $K_m = 2.0$ mmol/L, at six substrate concentrations in mmol/L.

Even at 8 times $K_m$ the rate is still 11 % below $V_{\max}$. To reach 99 % of $V_{\max}$ you need $[S] = 99\,K_m$. This is why $V_{\max}$ is hard to read straight from a curve.

What Km and Vmax tell you

• $V_{\max}$ depends on the amount of enzyme. Doubling the enzyme doubles $V_{\max}$. In symbols, $V_{\max} = k_{cat}[E]_{\text{total}}$.
• $K_m$ does not depend on the amount of enzyme. It is a property of the enzyme and its substrate.
• A small $K_m$ usually means a tight grip. The enzyme reaches half speed at a low substrate concentration.
• $k_{cat}$, the turnover number, is the number of substrate molecules that one enzyme molecule converts each second when it is saturated. Carbonic anhydrase turns over about a million a second. Catalase manages tens of millions.
• $k_{cat}/K_m$ is the catalytic efficiency. It tells how well an enzyme works at low substrate. The best enzymes reach $10^{8}$ to $10^{9}$ L/(mol s), where only the slow meeting of enzyme and substrate in water holds them back.

Reading Km and Vmax from a table

A curve is hard to read. A straight line is easy. Take the reciprocal of both sides of the equation.

Plot $1/v$ against $1/[S]$. This is the Lineweaver-Burk or double reciprocal plot. It is a straight line of the form $y = mx + c$.

• The y-intercept is $1/V_{\max}$.
• The slope is $K_m/V_{\max}$.
• The x-intercept is $-1/K_m$.

For the table above, the points lie on a line with a slope of 0.0167 and a y-intercept of 0.0083. So $V_{\max} = 1/0.0083 = 120$ µmol/min. The x-intercept is $-0.5$, so $K_m = 2.0$ mmol/L.

![Two graphs of an enzyme with Vmax 120 micromoles per minute and Km 2.0 millimoles per litre. On the left, rate against substrate concentration is a curve that rises steeply and flattens towards 120. A red dot marks half of Vmax, 60, at a substrate concentration equal to Km, 2.0. On the right, 1/v against 1/[S] is a straight line sloping upward through six black data points. It meets the vertical axis at 1/Vmax, 0.0083, marked in red, and meets the horizontal axis at minus 1/Km, minus 0.5, marked in green.](https://tuygolavdykfydmasiqc.supabase.co/storage/v1/object/public/discussion-images/57b80fd5-0d05-40b0-bb7c-fdadd190c00d/figures/ch06-biomolecules/bm-m4-kinetics.png)

The plot has a weakness. The points at low substrate give large reciprocals, and a small error there bends the line. Today a computer fits the curve directly. The double reciprocal plot is still the best way to see how an inhibitor acts, as the next lesson shows.

Set $[S]$ equal to $K_m$ and read the share of $V_{\max}$. Then raise $[S]$ to ten times $K_m$.

Try it in Python

The program takes the table, takes reciprocals, fits the straight line, and reads $V_{\max}$ and $K_m$ from it. It then checks the fit against the measured rates.

Practise with new numbers

Common mistakes

• Reading $V_{\max}$ from the highest rate measured. The curve is still rising at 16 mmol/L, where the rate is 106.7 and $V_{\max}$ is 120.
• Saying that $K_m$ is the concentration that gives $V_{\max}$. It gives half of $V_{\max}$.
• Thinking that $K_m$ rises when you add more enzyme. More enzyme raises $V_{\max}$ only.
• Reading $K_m$ from the x-intercept without the sign and the reciprocal. The x-intercept is $-1/K_m$.
• Mixing the units. $V_{\max}$ is a rate. $K_m$ is a concentration.

Remember these

• $v = V_{\max}[S]/(K_m + [S])$. At $[S] = K_m$ the rate is $V_{\max}/2$.
• $V_{\max} = k_{cat}[E]_{\text{total}}$. $K_m$ does not depend on the amount of enzyme.
• On the double reciprocal plot: y-intercept $1/V_{\max}$, x-intercept $-1/K_m$, slope $K_m/V_{\max}$.
• At very high $[S]$ the enzyme is saturated, and the rate is close to $V_{\max}$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. An enzyme works at nine times its $K_m$. What share of $V_{\max}$ is its rate?

2. A student doubles the amount of enzyme at a substrate level far above $K_m$. What happens to the rate, and why?

3. A double reciprocal line has a y-intercept of 0.010 and an x-intercept of $-0.25$. Find $V_{\max}$ and $K_m$.

4. Why is it hard to find $V_{\max}$ from the curve alone?

5. Enzyme P has $K_m = 0.2$ mmol/L and enzyme Q has $K_m = 5$ mmol/L for the same substrate. Which usually grips the substrate more tightly?