Ligand Binding Flashcards
Whag is Ka and Kd
Ka = kon/off = RL/R•L
Kd=Koff/Kon=R•L/RL
What is the equation at equilibrium and why
The forward rate equals the reverse rate
Kon[R][L]=koff[RL]
Explain how the variables when taking about Kd vs fractional saturation are changing
Intially 4 variables, (Kd,R,L,RL from equation)
When doing fractional saturation you now have three variables in each case and lose one (fractional saturation= RL/Rtotal = L/kd+L)
Then when setting both forms of the equation equal to each other you go back to four variables
So by talking about fractional saturation we lose some information (variables since dropping by 1)
Whag is the fractional saturation equation
What does the modified RL complex equation turn into
[L]/Kd+L
RL= Rtotal [L]/ Kd+L
Whag usually is kept constant in the ligand binding experiments
Whag is the RL equation similar to
Usually RTotal is constant (same as Ptotal)
And varying L concentration
Similar to the michealis menten equation
How do you find the fraction of free receptor, how is it measures
Whag is special about it
Since fractional saturation is is fraction of receptor bound in a complex, 1-alpha is the fraction of free receptor
Hold R total constant and vary L and measure R
Also Turns the fractional saturation equation which had 3 variables back to 4 variables
The fraction of free receptor and the receptor ligand complex equation all turn is back to four variables
Whag are binding isotherm and why is it called that
How do we actually find Kd
What are the assumptions we make and what lets us make that assumption
This is a way to solve Kd under equilibrium and at constant temperature this is why it’s called a binding isotherm (constant temp)
The Kd is the concentration of ligand at half saturation (similar to how Km is found)
We assume the RL «< L so that L is equal to L total (since Ltotal = RL + L)
We can say insignificant amount of L is in the complex due to these things:
Rtotal is «< L, this means that barely any receptor and lots of unbound L (substrate) in the solution
We also say that Ka is very small (Kd very big) meaning barely any L in the complex due
Whag is special when we are solving for Kd under quallibrium condition
We don’t have to have the receptor fully saturated with ligand to make the curve and find Kd
In finding Kd, if the concentration of ligand and receptor is similar what happens
So Whag are the three situation we have
The assumption we made before don’t hold and we have to use a different equation to fit the data, since now considering the amount of ligand that went into the complex
Rtotal «Kd (so regular fractional saturation fits to the data)
Rtotal ~ Kd (need diff equation to fit to the data)
Rtotal»_space; Kd (need to fit two straight lines instead of a curve, the protien (receptor) is already saturated with the ligand , can only get the stoich and not the Kd)
How does the k and fractional saturation equation change when there are two equivalent binding sites on the receptor
Whag do we need to do to this equation
The fractional saturation becomes the number of ligands bound divided by the number of binding sites
So fractional saturation equation changes (on sheet)
Don’t want to keep K1 and K2 in the equation so we add in the K as the intrinsic dissociation constant
Whag is K (intrinsic dissociation constant)
Explain the difference between form kt can take
The constant at a given site
K= 2K1 because there is two ways for the ligand to bind intially to the receptor (protien) and one way to detach from RL
2K = K2 because now there is only one way for L to bind to RL an now two ways to detach from RL2
How does the fractional saturation equation change when there are a n number of eqauivalent sites
Aplha = n[L]/[L] + K
K is the intrinsic dissociation constant
And n is the number of binding sites
Whag are the different plots you can get with n number of equivalent sites
Direct
Double reciprocal (hughklotz)
Scatchard
Explain equilibrium dialysis in ligand binding
There’s two species across a membrane (in on side there is a macromolecule and a micro molecule and in the other just micromolecule)
The macromolecule stays on one side and the micro molecules can diffuse to reach equilibrium for the same chemical potential
But the macromolecules cannot move
If the small molcule binds to the protien, the total concentration of the small molcule will be higher on the side with the macromolecule
This is because it is free in solution and also bound to the macromolecule and its concentration be become higher to match the concentration in the outside
Can label the small molcule to find the where it’s binding (higher radioactivity on the side bound)
