# Boltzmann Probability

1. Nov 1, 2008

### Jim Newt

1. The problem statement, all variables and given/known data

Simplistically, hemoglobin can exist in 5 states (0,1,2,3 or 4 bound O2). Name each state accordingly as UB (unbound) and B1, B2, B3, and B4 (for the four bound states). Using Boltzmann's Law, write down the probability that the molecule exists in a bound form.

2. Relevant equations

Here's Boltzmann's Law as I know it:

Pa/Pb = 1/z * exp[Ui/KbT]

where:

Pa: Probability A
Pb: Probability B
z: partition function
Ui: energy state
Kb: Boltzmann's constant
T: Temperature in K

3. The attempt at a solution

This problem is on a study guide for an exam on Monday morning. This is tough, because the instructor did not explain very well how to use Boltzmann's Law for this sort of problem. If someone could point me in the right direction, it would be a giant help.

2. Nov 1, 2008

### f95toli

The equation is actually an example of a probability distribution (the Boltzmann distribution). In this case you can think of it as the the probability that the system will be in a state with a specific energy.

First of all, remember that the probability that the system is in SOME state must be 1 (so what is Pb?) and the probability for each of the other states should be a number between 0 and 1.

What is the partition function in this case?
Also, to keep your answer as simple as possible: what is the probability that the molecule exists in the UB form?

3. Nov 1, 2008

### Jim Newt

Hi F95toli,

Thanks greatly for your suggestions. So do I treat the B1, B2,...etc. and UB as energy states? What do I do with the KbT?

So to start, I would take out the Pb and set Pa=1?

z=5?

I'm lost on this one...

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4. Nov 2, 2008

### f95toli

*Look at eq. 1 on this wiki page

http://en.wikipedia.org/wiki/Maxwell-Boltzmann_distribution

This is essentially the equation above. However, look at what is written instead of your "Pb"

also look at

http://en.wikipedia.org/wiki/Boltzmann_factor

which is perhaps easier to understand and relates to the point I made above about the total probability being 1.

And yes, the different states must have different energies; otherwise they would have the same probability.

kBT will be a parameter in the final expression since you don't know the temperature.

5. Nov 2, 2008

### Jim Newt

F95toli,

Thanks again for all your help. So based on all the great info that you've provided me, here's what I think the final solution would be:

4/5 = exp(-UB/KT) / [exp(-B1/KT) + exp(-B2/KT) + exp(-B3/KT) + exp(-B4/KT)]

Where UB= energy state of Unboud
B!, B2...etc = engergy states of Bound

Is this correct?

6. Nov 2, 2008

### f95toli

No, not quite. Although you are quite close.
First of all: there is no way to get a numerical answer from the information you have been provided with, so I don't know where you get the "4/5" from.

The right hand side is close. However, remember that you need to normalize with the sum of ALL Boltzmann factors (there is one factor missing).
Also, the question was what the probability is that the molecule exists in bound form.

7. Nov 2, 2008

### Jim Newt

P = exp(-B/KT) / [exp(-B1/KT) + exp(-B2/KT) + exp(-B3/KT) + exp(-B4/KT)+exp(-UB/KT)]

F95toli,

How does this look? Thanks,

Jim

8. Nov 2, 2008