Question of Kd

[Anders Fugelli]

In article <48qtcj$2jb at sun0.urz.uni-heidelberg.de>, Thomas W Bluem
<tbluem at ix.urz.uni-heidelberg.de> wrote:

> Help ... Help ... Help ... Help ...
> 
> What does this mean:
> 
>   K  = 1 mM  ???
>    D
> 
> What is K   ?
>          D
> 
> Thanks in advance.
> 
> 
> T. W. Bluem, Dept. of Anatomy and Cellbiology, University of Heidelberg, 
> Germany

The definition of this, can be found in most textbooks of pharmacology
(under Hill equation),
but in brief it relates to the dissociation constant of an Agent, A to its
receptor.
It is often refered to as the [A]50, which simply relates to the
concentration at which there
is a 50% effect.  This latter is most appropriatly used in indirect
measurements, whereas Kd is
measured in more direct ways, such as by radioligand binding experiments.
For a quick reference, see; Jenkinson, D.H. et al, Pharmacological Reviews
47: 255-266

In your case, Kd=1 mM, means that at a concentration of 1 mM, half your
receptors will be occupied 
(A fairly high Kd value in terms of affinity)

Anders
-- 
Anders Fugelli
anders.fugelli at basalmed.uio.no
Institute of Neurophysiology
University of Oslo
Norway