I know how to derive it with the steady state assumptions but I do not know how to approach this question without steady state assumptions (when formation and dissociation of [ES] intermediate is no longer equal to 0 anymore). I can do simple derivations but I’m not too strong in math so if anyone could guide me through, I’d very much appreciate the intellectual help. Thanks. 
 

equation: 

 

[E] + (k-1)<—>(k1) [ES] —>(k2) [E] + [P]

There is an alternate derivation that assumes that E, S, and ES are in quasi-equilibrium.  It is actually an easier derivation than steady-state.

1 hour ago, BabcockHall said:

There is an alternate derivation that assumes that E, S, and ES are in quasi-equilibrium.  It is actually an easier derivation than steady-state.

No we have to integrate the equation d[ES]/dt = k1[E] - k-1[ES] - k2[ES] which would have been equal to zero if in steady state approx but in this case there’s no assumptions made whatsoever so the only way to do this is interagency this equation to maybe isolate ES. And I’m having trouble with that......

19 hours ago, Biochemstudneedshelp said:

now how to derive it with the steady state assumptions but I do not know how to approach this question without steady state assumptions (when formation and dissociation of [ES] intermediate is no longer equal to 0 anymore). I can do simple derivations but I’m not too strong in math so if anyone could guide me through, I’d very much appreciate the intellectual help.

 

15 hours ago, Biochemstudneedshelp said:

No we have to integrate the equation d[ES]/dt = k1[E]

You have missed a bit, an

The full equation is

[math]\frac{{d\left[ {ES} \right]}}{{dt}} = {k_1}\left[ E \right]\left[ S \right] - {k_{ - 1}}\left[ {ES} \right] - {k_2}\left[ {ES} \right][/math]

Now in the case of excess substrate compared to enzyme

>> [E]

means that must also be much greater than [ES]

So try setting [ES] to zero in the above equation.

 

I got the problem solved, its based upon rate laws.

Try expressing [ES] in terms of [E] and then proceeding with the differential equation.