SME
Biochemistry
Enzyme kinetics (Michaelis–Menten)
Use this for “An enzyme at 0.010 μM has kcat = 500 s⁻¹ and Kₘ = 20 μM. Find Vₘₐₓ, the rate at 60 μM substrate, and kcat/Kₘ.”
Chart
Horizontal asymptote v = 5
v(60) = 3.75
Half bound at [S] = K = 20: v(K) = 2.5, half of 5
This is an example. Type your own values to start a new problem.
v (μM/s)
=
kcat (s⁻¹)
×
[E]ₜ (μM)
×
[S] (μM)
Kₘ (μM)
+
[S] (μM)
kcat × [E]ₜ is Vₘₐₓ, the level the curve approaches: kcat is each enzyme’s turnovers a second and [E]ₜ all the enzyme present.
Vₘₐₓ
Maximum rate
calculated · μM/s
kcat/Kₘ
Specificity constant
calculated · M⁻¹s⁻¹
Formulas
Vₘₐₓ = kcat × [E]ₜ
5 = 500 × 0.01
v
=
Vₘₐₓ
×
[S]
÷
(Kₘ
+
[S])
3.75
=
5
×
60
20 + 60
kcat/Kₘ
=
kcat
Kₘ
×
10
6
2.50 ×
10
7
=
500
20
×
10
6
Assumptions
•
Initial rates only: little substrate is used up and no product has built up.
•
Substrate far outnumbers enzyme ([S] ≫ [E]ₜ), and the enzyme–substrate complex stays at a steady level.
•
At [S] = Kₘ the rate is Vₘₐₓ ÷ 2; well above Kₘ it levels off at Vₘₐₓ.
•
kcat/Kₘ uses Kₘ in M (μM × 10⁻⁶), so it comes out in M⁻¹s⁻¹.
Step-by-step
WE KNOW
[E]ₜ = 0.01 μM, kcat = 500 s⁻¹, Kₘ = 20 μM, [S] = 60 μM
FIND
maximum rate (Vₘₐₓ), initial rate (v), specificity constant (kcat/Kₘ)
Step 1 · Find maximum rate (Vₘₐₓ)
Use Vₘₐₓ = kcat × [E]ₜ
Every enzyme molecule turns over kcat substrates a second when saturated.
Vₘₐₓ = 500 × 0.01
Vₘₐₓ = 5 μM/s
Step 2 · Find initial rate (v)
Use v = Vₘₐₓ × [S] ÷ (Kₘ + [S])
The share of enzyme holding substrate, [S] ÷ (Kₘ + [S]), times the top rate.
v
=
5
×
60
20 + 60
v
=
5
×
60
80
v
=
300
80
v
=
3.75
μM/s
→
The
enzyme
runs
at
75%
of
Vₘₐₓ;
at
[S]
=
Kₘ
it
would
run
at
50%.
Step 3 · Find specificity constant (kcat/Kₘ)
Use kcat/Kₘ = kcat ÷ Kₘ × 10⁶
Divide
kcat
by
Kₘ;
×
10
6
turns
per
μM
into
per
M.
kcat/Kₘ
=
500
20
×
10
6
kcat/Kₘ
=
2.50 ×
10
7
M
-1
s
-1
Check
Put the numbers back in. Both sides should match.
5 = 500 × 0.01 ✓
3.75
=
5
×
60
20 + 60
✓
2.50 ×
10
7
=
500
20
×
10
6
✓
← Previous lesson
Levels of protein structure
Next lesson →
Kₘ and Vₘₐₓ from two rates (Lineweaver–Burk)
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
More problem types
Kₘ and Vₘₐₓ from two rates (Lineweaver–Burk)
Use this for “An enzyme runs at 2.5 μM/s with 20 μM substrate and 4 μM/s with 80 μM. Find Kₘ and Vₘₐₓ.”
Rate with an inhibitor
Use this for “A competitive inhibitor with Kᵢ = 5 μM is added at 10 μM to an enzyme with Vₘₐₓ = 5 μM/s and Kₘ = 20 μM. Find the rate at 60 μM substrate.”
Which kind of inhibitor?
Use this for “With the inhibitor, Kₘ doubled but Vₘₐₓ did not change. What kind of inhibitor is it?”