Discussion of all aspects of biological molecules, biochemical processes and laboratory procedures in the field.
You are plotting the reciprocals of v and s, right? The data given here looks pretty linear when you plot 1/v on the y-axis vs 1/s on the x-axis. (Plotting the raw numbers of v vs s will give you a sigmoidal plot; not at all linear.) If you have a calculator that will do linear regression, enter your values of (1/s,1/v) as your (x,y) pairs. The slope of the regression will be Vm while the y-intercept is -1/Km. You can estimate them graphically if you prefer, but regression is a bit more accurate, plus you can estimate the errors in the parameters from the residual sums of squares - with a little math, anyway. Sometimes, the highest substrate concetrations (those at or near saturation levels) will give reciprocals that fall off the line. When that happens, you probably shouldn't use those data points to estimate the kinetic parameters. They won't be reliable.
Sorry, that should be X-intercept = -1/Km, not y-intercept. And the slope is Km/Vmax not Vmax. Memory didn't serve me very well this time and I was too lazy to derive it from the rate expression-though that's a useful excercise if you've never done it before.
This link has a small, but straightforward graph of the rate law and the double-reciprocal (Lineweaver-Burk) plots.
http://www.graphpad.com/help/Prism5/pri ... enzyme.htm
the ques is pretty simple in fact....
1/v=Km/(Vm[S]) + 1/Vm
so by plotting the 1/[V] by 1/[S] graph, we may find the intercept on y axis....which is 1/Vm in this case....and then do the reciprocal to find out Vmax.....and then put this value in the above equation.....at a particular [S] and v....and find the value of Km....
Who is online
Users browsing this forum: No registered users and 1 guest