## on equilibrium dyalisis - please help a desperate man...

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MrMistery
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### on equilibrium dyalisis - please help a desperate man...

Ok, so here it is. I have a problem on equilibrium dyalisis, from the theoretical part of IBO 2005. I have tried and tried to figure it out, but with no luck. I know what the dissociation constant is, but can't figure it out how the hell to answer these questions. Can someone please help? I don't want the answers(although they would be nice, since i haven't been able to find them on the internet), it would mean the world to me if someone could explain how exactly to solve it. Thanks in advance, here is the problem.

Question 67-69. Equilibrium dialysis is a method often used to determine dissociation constant KD for a ligand-binding protein. In this method, a protein at a know concentration is put into several dialysis tubes and each dialysis tube containing the protein is dialyzed against solutions containing the ligand at various ligand concentrations. Because the protein can not move cross the dialysis tube membrane while the ligand can, the ligand is "trapped" by the protein inside the dialysis tube and it creates a higher concentration of the ligand in dialysis tube than that outside the dialysis tube. The dissociation constant of the ligand can thus be determined

according to the following formula:
Kd=[M][L]/ [ML]

Where [M] is the concentration of free protein (no bound ligand) in dialysis tube, [L] is the concentration of the ligand and [ML] is the concentration of the protein with bound ligand. Therefore, KD is the ligand concentration when [M] equals [ML]. [MT] = [M] + [ML].
Where [MT] is the total concentration of the protein
The table below shows the measurement result of a calcium-binding protein. The protein has a molecular mass of 20 kDa and the concentration of the protein in equilibrium dialysis is 1 mg.ml-1.

[Ca] in dialysis solution (μM) [Ca] in dialysis tube (μM) [M]/[MT]
20 30
50 68
100 129
200 237
400 442
600 647
1000 1050
1500 1548
2000 2049

From my calculations, the molar concentration of the protein is 50 μM. Which can't possibly be right, unless the protein has about 400 binding sites for calcium.

67. How many calcium molecules does one protein molecule bind? (1 point)
A. 1
B. 2
C. 3
D. 4
E. It can not be determined
68. What is the KD of the protein? (3 point)

A. 30 μM
B. 78 μM
C. 95 μM
D. 104 μM
E. 200 μM

No idea how to do this, i think you need to blot the [Mt]/[M] and deduce from there. but since i haven't been able to come up with any results on the first questions, i am also at a standstil.
Again, thank you to anyone who can help me
"As a biologist, I firmly believe that when you're dead, you're dead. Except for what you live behind in history. That's the only afterlife" - J. Craig Venter

blcr11
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Posts: 672
Joined: Fri Mar 30, 2007 4:23 am
This looks like a Scatchard problem to me. For equilibrium dialysis, the difference between the outer and inner concentrations of Ca represents the bound concentration and the outer concentration is the free Ca concentration. Since the concentration difference “saturates” at 50 micromolar, which just so happens to be the same as the total protein concentration, I would say that the protein only binds 1 Ca per molecule. You have two (at least) ways to determine the K. You can either plot Ca(b)/Ca(f) as a function of Ca(b), or Ca(b)/(Ca(f)M(t)) as a function of Ca(b)/M(t). Either way, the slope is -1/K. The plot will flatten out as the bound concentration approaches 50 micromolar, so you can only use the data between 20 and roughly 400 micromolar free Ca. I estimated the slope using the point-slope formula and got something slightly less than 100, so my guess would be 95 for K.

MrMistery
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Joined: Thu Mar 03, 2005 10:18 pm
Location: Romania(small and unimportant country)
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I think i finally got it. Thank you very much...
"As a biologist, I firmly believe that when you're dead, you're dead. Except for what you live behind in history. That's the only afterlife" - J. Craig Venter

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