Km is the amount of substrate needed to get the enzyme to half-maximum velocity. So if the brain enzyme needs 10 times less substrate to reach half-maximum velocity, you would see the apparent Km go down from 25, not up.
Any undergrad biochemistry text will have all this stuff in it. Or am I doing your homework assignments for you? On Nov 18, 2009, at 10:54 AM, james09 pruza wrote: > Dear Sir, > I need one more help. > If there are two isozymes at two different compartment, say one in brain and > one in serum. If the serum one has Km of 25 and that of brain has Km of 2.5. > If by chance, the brain one comes to the serum, What should be the Km of > the serum sample of the person --- Will it be close to 2.5 or will it be > 27.5 ? > Please help. > Thanks. > James > > On Wed, Nov 18, 2009 at 8:38 PM, William G. Scott < > wgsc...@chemistry.ucsc.edu> wrote: > >> Dear James: >> >> The problem is that the equation is for an equilibrium constant, and Km is >> not one. Kd is one. If you can measure Kd, what you describe I think is >> fine. If you are measuring Km, you have to find a way of justifying the Km >> ~ Kd approximation, which usually is only safe for enzymes that turn over >> very slowly. >> >> I've never done these kinds of experiments, so someone else needs to answer >> this. Sorry. I've always been content to push the deltaG button in COOT. >> >> All the best, >> >> Bill >> >> >> On Nov 18, 2009, at 10:18 AM, james09 pruza wrote: >> >>> Dear Sir, >>> >>> At equilibrium, how can one determine delH and delS, if temp. and km are >>> known. >>> >>> I have km value and temperatures. I have to calculate delH and delS in >>> equilibrium. >>> Should I plot *lnKm vs 1/T* to have a slope of -delH/R and intercept of >>> delS/R. Multyplying this value with R, will give the value of delH and >> delS >>> , respectively. >>> If delG= delH - TdelS >>> delG= -RTlnK >>> >>> So, -RTlnK = delH -TdelS >>> >>> lnK = -delH/RT + delS/R >>> >>> Please suggest If I am wrong. I have planned to do ITC with the mutants >> as >>> well to get the perfect parameters. >>> Thanks for help. >>> >>> On Wed, Nov 18, 2009 at 7:44 PM, William G. Scott < >>> wgsc...@chemistry.ucsc.edu> wrote: >>> >>>> The K is an equilibrium constant. Km is not an equilibrium constant. >> It >>>> is a collection of rate constants. Kd is an equilibrium constant, but >> you >>>> have to be careful to make sure that the enthalpy and entropy terms >>>> correspond directly to this as well. The equation you quote implies >> standard >>>> state. >>>> >>>> Km is not the reciprocal of Kd. It is not an equilibrium constant. >>>> >>>> >>>> On Wed, Nov 18, 2009 at 8:24 AM, james09 pruza <james09x...@gmail.com >>> wrote: >>>> >>>>> Dear All, >>>>> Sorry for the non-ccp4 query. >>>>> I have solved a crystal structure of an enzyme and woring on its >>>>> biochemical aspect. We have a mutant of this enzyme and we are >> comparing >>>>> some thermodynamic parameters of this enzyme with mutant( lke delH and >> delS, >>>>> delG). we have done the expt at different temp. and know the km value >> at >>>>> these temp. Now the question is how to get the value of delH and delS. >>>>> The relation of these parameters is:- >>>>> -RTlnK = delH - TdelS >>>>> >>>>> Is the K in this relation is km or kd. If it can be treated as kd, so >> it >>>>> should be 1/km. >>>>> >>>>> All suggestions are welcome. >>>>> Thanks >>>>> James >>>>> >>>> >>>> >>>> >>>> >> >>
