2 years ago

Plz help ASAP it’s dude soon

Chemistry

1 answer:

Alex_Xolod [135]2 years ago

6 0

Answer:

Explanation:

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The $K_{m}$ of a substrate will be "10 μM".

Explanation:

The given values are:

$E_{t} = 20 \ nM$

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As we know,

⇒ $Vo=\frac{K_{cat}[E_{t}][S]}{K_{m}+[S]}$

On putting the estimated values, we get

⇒ $9.6=\frac{600\times 20\times 10^{-3}\times 40}{K_{m}+40}$

⇒ $K_{m}+40=\frac{600\times 20\times 10^{-3}\times 40}{9.6}$

⇒ $K_{m}+40=50$

On subtracting "40" from both sides, we get

⇒ $K_{m}+40-40=50-40$

⇒ $K_{m}=10 \ \mu M$

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