SME
Biochemistry · Problem type
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.”
Chart
v(60) = 2.5
Half bound at [S] = K = 60: v(K) = 2.5, half of 5
This is an example. Type your own values to start a new problem.
type
Inhibitor type
example
Vₘₐₓ
Maximum rate
example · μM/s
Kₘ
Michaelis constant
example · μM
[I]
Inhibitor concentration
example · μM
Kᵢ
Inhibition constant
example · μM
α
Inhibition factor
calculated
Kₘ′
Apparent Michaelis constant
calculated · μM
Vₘₐₓ′
Apparent maximum rate
calculated · μM/s
[S]
Substrate concentration
example · μM
v
Rate with the inhibitor
calculated · μM/s
Formulas
α
=
1
+
[I]
÷
Kᵢ
3
=
1
+
10
5
Kₘ′
=
α
×
Kₘ
60 = 3 × 20
Vₘₐₓ′ = Vₘₐₓ
5 = 5
v
=
Vₘₐₓ′
×
[S]
÷
(Kₘ′
+
[S])
2.5
=
5
×
60
60 + 60
Assumptions
•
The inhibitor binds and leaves quickly (reversible), and α = 1 + [I] ÷ Kᵢ says how much it shifts the enzyme.
•
Competitive raises Kₘ to αKₘ; uncompetitive divides both Kₘ and Vₘₐₓ by α; noncompetitive divides only Vₘₐₓ by α.
•
Noncompetitive here is the pure kind: the inhibitor binds the enzyme and the complex equally (one Kᵢ).
•
Then v = Vₘₐₓ′[S] ÷ (Kₘ′ + [S]), the Michaelis–Menten rule with the apparent constants.
Step-by-step
WE KNOW
type: competitive, Vₘₐₓ = 5 μM/s, Kₘ = 20 μM, [I] = 10 μM, Kᵢ = 5 μM, [S] = 60 μM
FIND
inhibition factor (α), apparent Michaelis constant (Kₘ′), apparent maximum rate (Vₘₐₓ′), rate with the inhibitor (v)
Step 1 · Find inhibition factor (α)
Use α = 1 + [I] ÷ Kᵢ
The inhibitor level measured in Kᵢ, plus 1 for the enzyme it leaves free.
α
=
1
+
10
5
α
=
1
+
2
α
=
3
Step 2 · Find apparent Michaelis constant (Kₘ′)
Use Kₘ′ = α × Kₘ
A
competitive
inhibitor
takes
the
active
site,
so
it
takes
α
times
the
substrate
to
fill
half
the
enzyme.
competitive chosen: competitive
Kₘ′ = 3 × 20
Kₘ′ = 60 μM
Step 3 · Find apparent maximum rate (Vₘₐₓ′)
Use Vₘₐₓ′ = Vₘₐₓ
Enough substrate outcompetes the inhibitor, so the top rate is unchanged.
Vₘₐₓ′ = 5 μM/s
Step 4 · Find rate with the inhibitor (v)
Use v = Vₘₐₓ′ × [S] ÷ (Kₘ′ + [S])
Michaelis–Menten with the apparent constants in place of the true ones.
v
=
5
×
60
60 + 60
v
=
5
×
60
120
v
=
300
120
v
=
2.5
μM/s
→
Without
the
inhibitor
(the
dashed
curve):
v₀
=
3.75
μM/s.
Check
Put the numbers back in. Both sides should match.
3
=
1
+
10
5
✓
60 = 3 × 20 ✓
5 = 5 ✓
2.5
=
5
×
60
60 + 60
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
Related lessons
Enzyme kinetics (Michaelis–Menten)
Main lesson
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.”
Which kind of inhibitor?
Use this for “With the inhibitor, Kₘ doubled but Vₘₐₓ did not change. What kind of inhibitor is it?”