Equilibrium Protein Binding (BIO)

# Equilibrium Protein Binding (BIO)

Here’s a problem to do some equilibrium calculations
on a biological system where in the table we’re given equilibrium constants for a couple
proteins at different temperatures, and these are binding to the same DNA site and we want
to compare these by calculating the delta H for the reaction of binding and the delta
S, and so, a couple things to point out, very large equilibrium constants and a very narrow
temperature range because of where we can operate with the proteins and not change them,
and we also notice that as we raise the temperature for protein Q equilibrium constant increases,
we raise the temperature for protein W the equilibrium constant decreases, so that means
that this reaction of protein Q at the DNA site is endothermic, and this one is exothermic.
So to calculate the heat of reaction we use the Van’t Hoff equation, and this equation
assumes that the heat of reaction is a constant, because there is a very narrow temperature
range this is actually a very good assumption, so in the form that we normally use it, the
integrated form, it looks like so, and this relates equilibrium constant K2 at temperature
T2, equilibrium constant K1, temperature T1, heat of reaction, the gas constant, so with
just two data point we could calculate delta H if we want to be more accurate and use the
three data points we could use the form that this equation was derived from and we say
it’s at standard conditions, gas constant, and remember this is the heat of reaction.
So this says if I make a plot of the log of the equilibrium constant versus one over temperature
then I expect to have a straight line. So what I’ve done is create a table here where
I repeated the temperature values and equilibrium constant and I just calculate the natural
log of these equilibrium constants and I calculate the inverse temperature and I multiplied it
by ten to the third, so this is a thousand divided by the temperature and that’s just
to make it easier to display, so keep in mind I got 3.57 by taking a thousand divided by
7 plus 273 to make this absolute temperature and that’s critical that this is in Kelvin.
So what I’ve done then is on a spreadsheet entered the values and have a plot of log
of K versus inverse temperature so I can get an equation from a least squares fit, and
this is a comma here, 3730 over absolute temperature plus 33.2, so this is a fit that of course
from our equation, of how log K changes with inverse temperature means that 3730 is equal
to delta H reaction, standard conditions, divided by the gas constant. and 33.2 is equal
to delta S standard conditions for the reaction divided by R, and this results if we look
at the equations delta G is delta H minus delta TS and so then naturally follows from
that to show that the intercept is delta S so let’s look at this, the log of the equilibrium
constant is related to minus delta G standard conditions over RT and delta G from our definition,
so delta H minus delta TS over RT if we’re evaluating this at a temperature, so we have
some equilibrium constant, you know at 280 Kelvin for example, then minus delta H over
R, one over T and plus, so T now is a given value, delta S over R, and this then is the
intercept, this was the slope, and so delta H then, minus sign here, so this ends up being
positive, positive sign here, so entropy change is positive, delta H is positive, as we expected
right, endothermic reaction, entropy change is positive even though we’re binding a protein
to a site, so other things must be happening, the shape of the protein is changing, water
molecules are released so that delta S is positive. So I’ve just made the substitution
for the values to calculate delta H reaction, 3730 times the gas constant, so 31 kilojoules
per mole, and again remember this is positive so endothermic.So again, even though we’re
making chemical bonds, we might expect that might be exothermic, we’re forming a bond
between these two molecules, other things are happening, overall reaction is endothermic.
Well no need to do the details of the calculation for the other protein, for protein W, right
this was for protein Q. I can just show the values which corresponds to 29.8 kilojoules
per mole. So for protein W, exothermic, and delta S from the intercept, 7.8 so that was
delta S over R so delta S is 64.8 joules per mole Kelvin. So one reaction is endothermic
the other is exothermic and notice that the delta S change, and I guess I forgot to calculate,
so let’s look at the delta S change. So the delta S change, this is for protein Q is much
larger than for protein W and that’s necessary in the sense that if we expect delta G, which
is a driving force, to be negative. So let’s do a quick comparison, so delta G at some
temperature, and lets just pick 300 Kelvin, delta G, T, delta S so if we do this for each,
and I’ll just do the calculations. So I’ve just calculated delta G so this is delta G
for protein Q, delta H minus T delta S at 300 Kelvin, so it’s negative even though it’s
endothermic because of the large entropy change delta G for protein W is also negative, this
is because it’s exothermic even though it has a smaller entropy change you’ll notice
that the delta Gs are really quite similar and negative for both of these, but for quite
different reasons, and so the changes with the binding, the entropy increases even though
we’re binding a protein to DNA because of things like changes in conformation of the
molecules and also the potential release of bound water that would cause these significant
entropy increases.