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

The
`K_(m)` of a substrate will be "10 μM".

Step-by-step 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=(K_(cat)[E_(t)][S])/(K_(m)+[S])`

On putting the estimated values, we get

⇒
`9.6=(600* 20* 10^(-3)* 40)/(K_(m)+40)`

⇒
`K_(m)+40=(600* 20* 10^(-3)* 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`