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Offline Nescafe

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kobserved
« on: November 12, 2012, 12:11:33 PM »
Hi,

In certain binding experiments I see that they calculate kobs, then use it to calculate kon. What I find strange is that in this technique (SPR) they  show the part of the graph where they calculate the kobs from to be region that demonstrates binding (kon). Why get kobs then kon? why can't we just calculate kon directly?

Nescafe.

Offline Yggdrasil

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Re: kobserved
« Reply #1 on: November 12, 2012, 01:04:24 PM »
In SPR, the region of the response curve where you see an increase in signal represents both binding of your protein from solution as well as competition from dissociation of protein that is already bound.  Therefore, to get an accurate measure of k_on, you need more information such as the binding constant and the off rate, both of which can be measured directly.

Offline Nescafe

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Re: kobserved
« Reply #2 on: November 12, 2012, 01:09:22 PM »
In SPR, the region of the response curve where you see an increase in signal represents both binding of your protein from solution as well as competition from dissociation of protein that is already bound.  Therefore, to get an accurate measure of k_on, you need more information such as the binding constant and the off rate, both of which can be measured directly.

So that initial part of the curve represents both A+B -> AB as well as AB -> A + B and this is given by kobs which is not a true reflection of kon which is only A + B -> AB. Did I understand it correctly?

Nescafe.

Offline Nescafe

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Re: kobserved
« Reply #3 on: December 03, 2012, 12:54:56 PM »
In SPR, the region of the response curve where you see an increase in signal represents both binding of your protein from solution as well as competition from dissociation of protein that is already bound.  Therefore, to get an accurate measure of k_on, you need more information such as the binding constant and the off rate, both of which can be measured directly.

Can you elaborate a bit more Yggsdrill, I would really like to understand this a bit better. So the reason is the kobserved is because we are interested in obtaining kon which is the rate of association of A and B, whereas the curve represents not only A binding to B, but also competition of AB and B (if B is what is free in solution and A immobilized on to the surface). Am I understanding this correctly?

Thanks,

Nescafe.

Offline Yggdrasil

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Re: kobserved
« Reply #4 on: December 03, 2012, 11:56:26 PM »
During the association phase, the free ligand (L) is binding to the immobilized receptor (R) to form the ligand-receptor complex (RL), and the binding is governed by the constant k_on.  At the same time, ligand-receptor complexes are dissociating to free ligand and free receptor, governed by the constant k_off.  Binding continues until the rate of free ligand binding to receptors reaches an equilibrium with the rate of RL complex dissociation.  In the spr experiment, we measure time required for binding and dissociation to come to equilibrium, given by the rate constant k_obs.

Based on this info, we can write the following differential equation to represent the change in [RL] over time:
[tex]\frac{d[RL]}{dt} = k_{on}[R][L] - k_{off}[RL][/tex]
Since the total concentration of receptors is given by [R]_T = [R] + [RL], we can write:
[tex]\frac{d[RL]}{dt} = k_{on}[L]([R]_T-[RL]) - k_{off}[RL] = k_{on}[L][R]_T - (k_{on}[L]+k_{off})[RL][/tex]
where k_on, k_off, [R]_T and [L] are constants ([L] is constant because the ligand is continuously being replenished by the flow in the spr experiment).

Solving the differential equation yields [RL] as a fuction of time:
[tex][RL] = A(1-e^{-k_{obs}t})[/tex]
where A represents the fraction of receptors bound at equilibrium and k_obs = k_on [L] + k_off

Thus the observed rate constant during the association phase depends on both k_on and k_off.  Determining k_on requires additional experiments to determine k_off.

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