Biology · Introductory biology · Worked example

Find the substrate level for a target rate

An enzyme has $K_{m} = 4 mM$ and $V_{\max} = 20 µmol / \min$ . At what substrate concentration does it run at $16 µmol / \min$ , $80 %$ of $V_{\max}$ ? How much substrate does $90 %$ take?

Write the target as a fraction

Set the rate to $16 µmol / \min$ in $v = 20 S / (4 + S)$ , writing $S$ for $[S]$ in $mM$ . Dividing both sides by $20$ turns the target into a fraction of full speed, $0.8$ .

0.8 = \frac{S}{4 + S}

Clear the fraction

Multiply both sides by $4 + S$ , which is positive for any real concentration, so nothing is lost.

0.8 (4 + S) = S

Expand

Multiply out the left side.

3.2 + 0.8 S = S

Collect the $S$ terms

Subtract $0.8 S$ from both sides, so $S$ appears on one side only.

3.2 = 0.2 S

Divide

Divide both sides by $0.2$ .

S = 16

Check, and compare with $90 %$

At $16 mM$ the rate is $20 \times 16 / (4 + 16)$ $= 16 µmol / \min$ , as required. The same steps for $90 %$ give $S = 0.9 \times 4 / (1 - 0.9)$ $= 36 mM$ . Going from $80 %$ to $90 %$ of $V_{\max}$ takes more than twice the substrate.

\begin{array}{l} \frac{20 (16)}{4 + 16} = 16 \\ \frac{0.9 (4)}{1 - 0.9} = 36 \end{array}

Result

$16 mM$ of substrate, four times $K_{m}$ , gives $80 %$ of $V_{\max}$ . Reaching $90 %$ takes $36 mM$ , nine times $K_{m}$ .

Your turn

1. One like the example

An enzyme has $K_{m} = 2.5 mM$ . At what substrate concentration does it run at $60 %$ of $V_{\max}$ ?

Show the answer and explanation

$3.75 mM$ .

$0.6 = S / (2.5 + S)$ gives $0.6 (2.5 + S) = S$ , so $1.5 + 0.6 S = S$ , then $1.5 = 0.4 S$ and $S = 3.75$ . Above half speed, so above $K_{m}$ , as it should be.

\begin{array}{l} 0.6 = \frac{S}{2.5 + S} \\ 0.6 (2.5 + S) = S \\ 1.5 + 0.6 S = S \\ 1.5 = 0.4 S \\ S = 3.75 \end{array}

2. A variation

An enzyme with $V_{\max} = 50 µmol / \min$ runs at $30 µmol / \min$ when the substrate is $6 mM$ . Find $K_{m}$ , writing $K$ for it.

Show the answer and explanation

$K_{m} = 4 mM$ .

$30 = 50 \times 6 / (K + 6)$ . Multiply by $K + 6$ to get $30 (K + 6) = 300$ , so $30 K + 180 = 300$ , $30 K = 120$ and $K = 4$ . The rate is $60 %$ of $V_{\max}$ at $6 mM$ , above half, so $K_{m}$ must be below $6 mM$ .

\begin{array}{l} 30 = \frac{50 (6)}{K + 6} \\ 30 (K + 6) = 300 \\ 30 K + 180 = 300 \\ 30 K = 120 \\ K = 4 \end{array}

3. Somewhere new

Show that the substrate concentration giving a fraction $f$ of $V_{\max}$ is $S = f K / (1 - f)$ , where $K$ is $K_{m}$ . Then find how many times $K_{m}$ it takes to reach $99 %$ of $V_{\max}$ .

Show the answer and explanation

$S = f K / (1 - f)$ , and $99 %$ takes $99$ times $K_{m}$ .

From $f = S / (K + S)$ , multiply out to $f K + f S = S$ , collect to $f K = S (1 - f)$ , and divide by $1 - f$ , which needs $f \neq 1$ . At $f = 0.99$ , $S = 0.99 K / 0.01 = 99 K$ . As $f$ nears $1$ , $S$ grows without limit, which is why $V_{\max}$ is never reached.

\begin{array}{l} f = \frac{S}{K + S} \\ f (K + S) = S \\ f K + f S = S \\ f K = S - f S \\ f K = S (1 - f) \\ S = \frac{f K}{1 - f}, f \neq 1 \\ \frac{0.99}{1 - 0.99} = 99 \end{array}

Keep exploring

Enzyme kinetics opens in Biology with rates that follow $V_{\max} = 20 µmol / \min$ and $K_{m} = 4 mM$ exactly. The fitted curve passes $16 µmol / \min$ at $16 mM$ .

Return to the concept →

Sources and scope

Authored study material. Tool results depend on the stated inputs and model assumptions.

After reading

Study it, then try it

Practice the key terms as flashcards, then open the example in its tool, change a value and keep a useful result on your board.

Study the 8 key terms Open the enzyme kinetics tool Solve for S in Math Open worked example on a board Biomolecules reference

Your existing work stays on this device. Examples open as editable copies.

Write the target as a fraction

Clear the fraction

Expand

Collect the SS terms

Divide

Check, and compare with 90⁢%90%

Result

Your turn

1. One like the example

3.75 mM3.75 mM.

2. A variation

Km=4 mMKm​=4 mM.

3. Somewhere new

S=fK⁢/⁢(1−f)S=fK/(1−f), and 99⁢%99% takes 9999 times KmKm​.

Keep exploring

Study it, then try it

Collect the $S$ terms

Check, and compare with $90 %$

$3.75 mM$ .

$K_{m} = 4 mM$ .

$S = f K / (1 - f)$ , and $99 %$ takes $99$ times $K_{m}$ .