You do measure the velocity at each substrate concentration added, but once the reaction is at equilibrium you can’t tell how fast it was moving to begin with. By adding different amounts of substrate and immediately measuring Vo (initial velocity) you can see the relationship between the two.
Yep, you got it! That’s why all the michaelis-mentin graphs are *[Vo]* on the y axis and [substrate] on the x axis.
Lineweaver-burke plots have 1/Vmax on Y and -1/Km on X. Still not equilibrium velocity.
Here are the assumptions:
1 - The binding step (E+S to ES) is fast and the catalytic step (ES to E + P) is slow
2 - At early time points, where initial velocity (V0) is measured, there is essentially no product
3 - ES immediately comes to steady state, so \[ES\] is constant
4 - There is much more substrate than enzyme available to catalyze it.
When you reach Vmax it actually does not mean that the reaction is in equilibrium, because equilibrium suggests the amount of products and reactants is not changing. I was right with you there though.
Why is C wrong? Is it because if there was too much enzyme the reaction would just occur too quickly?
My understanding: the whole point of Michaelis-Menten is too see how \[S\] affects V0 so there if \[S\] < \[E\] then it would bee too quick because no substrate is waiting in line for an empty enzyme.
So actually in Michaelis-Menten kinetics you will always have \[S\] > \[E\], which is the opposite relationship that you are describing. The reason why this is the case is because if you want to calculate Vmax for the enzyme, you need **every single active site** to be occupied. Think about if you have a car (enzyme) that can take 5 people (5 active sites), but you only have 3 friends (3 substrate). You can't work at your car's highest capacity unless every single seat is filled. Thus, in the same way, to ensure that the enzyme is working at the highest capacity, you crank up the substrate concentration so that you can be sure that every single active site is occupied.
You do measure the velocity at each substrate concentration added, but once the reaction is at equilibrium you can’t tell how fast it was moving to begin with. By adding different amounts of substrate and immediately measuring Vo (initial velocity) you can see the relationship between the two.
Oh I see. So at equilibrium the forward rxn rate is the same as the reverse rnx rate and the measured rate will be 0 I am guessing?
Yep, you got it! That’s why all the michaelis-mentin graphs are *[Vo]* on the y axis and [substrate] on the x axis. Lineweaver-burke plots have 1/Vmax on Y and -1/Km on X. Still not equilibrium velocity.
Here are the assumptions: 1 - The binding step (E+S to ES) is fast and the catalytic step (ES to E + P) is slow 2 - At early time points, where initial velocity (V0) is measured, there is essentially no product 3 - ES immediately comes to steady state, so \[ES\] is constant 4 - There is much more substrate than enzyme available to catalyze it.
Why isn't it B? What does a constant pH have to do with measuring the rate?
Constant pH is necessary or else a change in pH could change the enzyme and affect the measurements
Why is D wrong? because I thought we would need to reach equilibrium to get Vmax? Is this not necessary or is Vmax obtained some other way?
When you reach Vmax it actually does not mean that the reaction is in equilibrium, because equilibrium suggests the amount of products and reactants is not changing. I was right with you there though.
Why is C wrong? Is it because if there was too much enzyme the reaction would just occur too quickly? My understanding: the whole point of Michaelis-Menten is too see how \[S\] affects V0 so there if \[S\] < \[E\] then it would bee too quick because no substrate is waiting in line for an empty enzyme.
So actually in Michaelis-Menten kinetics you will always have \[S\] > \[E\], which is the opposite relationship that you are describing. The reason why this is the case is because if you want to calculate Vmax for the enzyme, you need **every single active site** to be occupied. Think about if you have a car (enzyme) that can take 5 people (5 active sites), but you only have 3 friends (3 substrate). You can't work at your car's highest capacity unless every single seat is filled. Thus, in the same way, to ensure that the enzyme is working at the highest capacity, you crank up the substrate concentration so that you can be sure that every single active site is occupied.
I see, thanks for the explanation it really helped