here is an easy one for you - Biology Forum

last updated by

oppox 18 years, 1 month ago

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- October 19, 2006 at 10:07 am #6047
  oppox
  Participant
  Im trying to fully understand this. I got a table with different enzymes and inhibitors.
  ……..x…….. y………..z.
  1…..0,67…..42……..0,17
  2…..0,53…..13……..0,29
  where 1 and 2 are different inhibitors and x, y, z are different enzymes. the figures are Ki (microM). Its been some time ago I read the course covering this so im alittle lost.
  Does these result tell me that enzyme x and z are inhibited fairly good whilst
  y is not. Is there anything more I can say about these result?
  thanks in advance
- October 19, 2006 at 8:57 pm #56993
  sdekivit
  Participant
  Ki tells you something about the binding affinity for the inhibitor to the enzyme, accoriding to the following equilibrium:
  I + E <–> EI with Ki = [E][I]/[EI]
  in other words: the lower the Ki, the higher the affinity of the inihibitor to the enzyme.
  So indeed: enzyme x and z are potent target of the inhibitors 1 and 2.
- October 19, 2006 at 11:19 pm #57001
  oppox
  Participant
  Then I was thinking in the right direction, thank you for making it more clear.
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