How to Estimate Km from Reaction Velocities Using Michaelis-Menten Kinetics

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April 14, 2025
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6. Two experiments were conducted with an enzyme following MichaelisMenten kinetics at substrate
concentrations of 0.5 g/l and 1 g/l. If the enzymatic reaction velocity increases approximately 2-fold at the
higher substrate concentration, the Km for the enzyme would be around:
1. 0.001 g/l
2. 0.01 g/l
3. 0.1 g/l
4. 1 g/

The Michaelis constant (Km) is a key parameter in enzyme kinetics that indicates the substrate concentration at which the reaction velocity is half of Vmax. By comparing reaction rates at two different substrate concentrations, we can estimate Km without needing full kinetic data.

Question:

Two experiments were conducted using an enzyme that follows Michaelis-Menten kinetics. The substrate concentrations used were 0.5 g/L and 1 g/L. The reaction velocity approximately doubled at the higher substrate concentration.

What is the approximate Km value for the enzyme?

Options:

0.001 g/L
0.01 g/L
0.1 g/L
1 g/L

Step-by-Step Explanation

The Michaelis-Menten equation is:

$\frac{V_{max} \cdot [S]}{K_m + [S]}$

Let’s denote:

$v_1$ at $g/L[S_1] = 0.5 \, \text{g/L}$
$v_2$ at $g/L[S_2] = 1.0 \, \text{g/L}$
It’s given that $v2≈2⋅v1v_2 \approx 2 \cdot v_1$

Let’s apply this in the ratio form:

$v2v1=Vmax⋅1Km+1÷Vmax⋅0.5Km+0.5=1(Km+0.5)0.5(Km+1)=2\frac{v_2}{v_1} = \frac{V_{max} \cdot 1}{K_m + 1} \div \frac{V_{max} \cdot 0.5}{K_m + 0.5} = \frac{1(K_m + 0.5)}{0.5(K_m + 1)} = 2$

Now solve this equation:

$Km+0.50.5(Km+1)=2⇒Km+0.50.5Km+0.5=2⇒(Km+0.5)=2(0.5Km+0.5)⇒Km+0.5=Km+1⇒0.5=1\frac{K_m + 0.5}{0.5(K_m + 1)} = 2 \Rightarrow \frac{K_m + 0.5}{0.5K_m + 0.5} = 2 \Rightarrow (K_m + 0.5) = 2(0.5K_m + 0.5) \Rightarrow K_m + 0.5 = K_m + 1 \Rightarrow 0.5 = 1$

Wait—that’s not consistent. Let’s double-check and go back to the math more carefully:

$Km+0.50.5Km+0.5=2⇒Km+0.5=2(0.5Km+0.5)⇒Km+0.5=Km+1⇒0.5=1\frac{K_m + 0.5}{0.5K_m + 0.5} = 2 \Rightarrow K_m + 0.5 = 2(0.5K_m + 0.5) \Rightarrow K_m + 0.5 = K_m + 1 \Rightarrow 0.5 = 1$

Still inconsistent. That tells us that the assumption of exactly double is only approximately true. So let’s estimate instead by choosing values of Km and seeing what gives a 2-fold change between 0.5 and 1 g/L.

Try Km = 1 g/L

$v1=Vmax⋅0.51+0.5=0.51.5=0.33Vmaxv_1 = \frac{V_{max} \cdot 0.5}{1 + 0.5} = \frac{0.5}{1.5} = 0.33 V_{max}$ $v2=Vmax⋅11+1=12=0.5Vmaxv_2 = \frac{V_{max} \cdot 1}{1 + 1} = \frac{1}{2} = 0.5 V_{max}$ $double)\frac{v_2}{v_1} = \frac{0.5}{0.33} \approx 1.5 \text{ (Not double)}$

Try Km = 0.1 g/L

$v1=0.50.1+0.5=0.50.6≈0.83Vmaxv_1 = \frac{0.5}{0.1 + 0.5} = \frac{0.5}{0.6} \approx 0.83 V_{max}$ $v2=10.1+1=11.1≈0.91Vmaxv_2 = \frac{1}{0.1 + 1} = \frac{1}{1.1} \approx 0.91 V_{max}$ $small)\frac{v_2}{v_1} \approx \frac{0.91}{0.83} \approx 1.1 \text{ (Too small)}$

Try Km = 0.5 g/L

$v1=0.50.5+0.5=0.5Vmaxv_1 = \frac{0.5}{0.5 + 0.5} = 0.5 V_{max}$ $v2=10.5+1=11.5≈0.67Vmaxv_2 = \frac{1}{0.5 + 1} = \frac{1}{1.5} \approx 0.67 V_{max}$ $2-fold)\frac{v_2}{v_1} = \frac{0.67}{0.5} = 1.34 \text{ (Still not 2-fold)}$

Try Km = 0.1 g/L

Earlier we saw that doesn’t give a 2-fold change either.

Try Km = 0.01 g/L

$v1=0.50.5+0.01=0.50.51≈0.98Vmaxv_1 = \frac{0.5}{0.5 + 0.01} = \frac{0.5}{0.51} \approx 0.98 V_{max}$ $v2=11+0.01=11.01≈0.99Vmaxv_2 = \frac{1}{1 + 0.01} = \frac{1}{1.01} \approx 0.99 V_{max}$ $small)\frac{v_2}{v_1} \approx \frac{0.99}{0.98} \approx 1.01 \text{ (Way too small)}$

Try Km = 0.1 g/L

It seems that Km = 0.1 g/L gives about:

$v1≈0.83v_1 \approx 0.83$
$v2≈0.91v_2 \approx 0.91$
$v2v1≈1.1\frac{v_2}{v_1} \approx 1.1$

Still too small.

Let’s go back to Km = 0.25 g/L as a test:

$v1=0.50.75≈0.67v_1 = \frac{0.5}{0.75} \approx 0.67$ $v2=11.25=0.8v_2 = \frac{1}{1.25} = 0.8$ $(Closer)\frac{v_2}{v_1} = \frac{0.8}{0.67} \approx 1.19 \text{ (Closer)}$

Finally, only Km = 1 g/L gives a near 2-fold increase:

$v1=0.33,v2=0.5⇒v2v1=1.5v_1 = 0.33, \quad v_2 = 0.5 \Rightarrow \frac{v_2}{v_1} = 1.5$

Now try Km = 2 g/L

$v1=0.52.5=0.2v_1 = \frac{0.5}{2.5} = 0.2$ $v2=13=0.33v_2 = \frac{1}{3} = 0.33$ $v2v1=0.33/0.2=1.65⇒toohigh\frac{v_2}{v_1} = 0.33 / 0.2 = 1.65 \Rightarrow too high$

Eventually, we find that Km ≈ 1 g/L is the best approximation to produce a 2-fold increase in rate when doubling the substrate from 0.5 g/L to 1 g/L.

✅ Correct Answer: (4) 1 g/L

Conclusion

If the velocity doubles when the substrate concentration doubles from 0.5 g/L to 1 g/L, it suggests that the Km is approximately equal to the lower concentration — around 1 g/L.

This reflects a situation where the enzyme is not yet saturated, and increasing substrate significantly increases reaction rate.