Question

An enzyme is discovered that catalyzes the chemical reaction SAD ↔ HAPPY A team of motivated researchers sets out to study the enzyme, which they call happyase. They find that the kcat for happyase is 600 s-1. They carry out several experiments and found that, when [Et] = 20 nM and [SAD] = 40 μM, the reaction velocity, V0, is 9.6 μM s-1. The Km for the substrate SAD is __________ μM .

Answer 1

Answer:

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

Explanation:

The given values are:

$E_{t} = 20 \ nM$

$[Substract] = 40 \ \mu M$

$K_{cat}=600 \ s^{-1}$

Reaction velocity, $Vo=9.6 \ \mu M s^{-1}$

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

⇒  $K_{m}+40-40=50-40$

⇒  $K_{m}=10 \ \mu M$