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When the $\mathrm{pH}$ was $5.0$, the temperature was $\pu{25 °C}$,the $K_\mathrm{D}$ was $\pu{5 μM}$, and [L] was equal to $K_D$, the protein(s) were half bound.

If $K_\mathrm{D} = \pu{20 μM}$ and $\mathrm{pH} = 6.5$, what is the impact on binding affinity?

I understand that $K_\mathrm{D}$ is the inverse of $K_\mathrm{a}$ and that $K_\mathrm{D}$ describes the degree of dissociation at equilibrium for the reaction, thus a low $K_\mathrm{D}$ corresponds to a high binding affinity complex.

I don't understand how to account for $\mathrm{pH}$ change though.

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    $\begingroup$ It would be nice to define what RL and $K_\mathrm{D}$ are. $\endgroup$ – andselisk Apr 30 at 12:48
  • $\begingroup$ Whoops. Question considers half bound condition where [L]= $K_D$ $\endgroup$ – ThermoRestart Apr 30 at 12:52
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    $\begingroup$ Since you are not given (or you don't provide) more information about how pH affects the protein, other than Kd, I would assume it's not really important. The point would be that the affinity goes down at the higher pH. $\endgroup$ – Buck Thorn Apr 30 at 15:55
  • $\begingroup$ @NightWriter - Yes, that was all the information I was given. But don't some proteins do better @ higher pH than lower pH? $\endgroup$ – ThermoRestart Apr 30 at 15:57
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    $\begingroup$ Sure, but this is presumably just an exercise for you to reveal that you understand that lower Kd means higher affiinity and vice-versa... $\endgroup$ – Buck Thorn Apr 30 at 16:35

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