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Check the residual plot after fitting — a random scatter confirms MM is appropriate; a systematic S-shaped residual means the wrong model was used.",{"type":25,"tag":34,"props":866,"children":867},{},[868,873,875,880,882,886],{"type":25,"tag":40,"props":869,"children":870},{},[871],{"type":31,"value":872},"My Km and Vmax have very wide confidence intervals — what is wrong?",{"type":31,"value":874},"\nWide confidence intervals almost always mean the ",{"type":25,"tag":40,"props":876,"children":877},{},[878],{"type":31,"value":879},"substrate concentration range is too narrow",{"type":31,"value":881},". If you never have ",{"type":25,"tag":53,"props":883,"children":884},{},[885],{"type":31,"value":57},{"type":31,"value":887}," >> Km (data doesn't plateau), Vmax cannot be estimated from the data — it requires extrapolation. Add data points at higher substrate concentrations. If that is not possible, report Vmax and Km as rough estimates and note the extrapolation uncertainty.",{"type":25,"tag":34,"props":889,"children":890},{},[891,896,900,902,907],{"type":25,"tag":40,"props":892,"children":893},{},[894],{"type":31,"value":895},"What is the difference between Km and Kd?",{"type":25,"tag":40,"props":897,"children":898},{},[899],{"type":31,"value":95},{"type":31,"value":901}," is a kinetic constant: the substrate concentration at V = Vmax/2. It reflects both the binding equilibrium and the catalytic step. ",{"type":25,"tag":40,"props":903,"children":904},{},[905],{"type":31,"value":906},"Kd",{"type":31,"value":908}," is a true thermodynamic dissociation constant: the equilibrium concentration at which half the enzyme is bound. For slow enzymes (kcat \u003C\u003C k₋₁), Km ≈ Kd. For fast enzymes, Km > Kd. They are not interchangeable.",{"type":25,"tag":34,"props":910,"children":911},{},[912,917,928,929,933,935,939,941,946],{"type":25,"tag":40,"props":913,"children":914},{},[915],{"type":31,"value":916},"How do I compute kcat from Vmax?",{"type":25,"tag":40,"props":918,"children":919},{},[920,922,926],{"type":31,"value":921},"kcat = Vmax / ",{"type":25,"tag":53,"props":923,"children":924},{},[925],{"type":31,"value":482},{"type":31,"value":927},"total",{"type":31,"value":67},{"type":25,"tag":53,"props":930,"children":931},{},[932],{"type":31,"value":482},{"type":31,"value":934},"total is the total enzyme concentration in the same units as Vmax/time. For example, if Vmax = 50 µmol/min and ",{"type":25,"tag":53,"props":936,"children":937},{},[938],{"type":31,"value":482},{"type":31,"value":940},"total = 5 nM, then kcat = 50 µmol/min ÷ 5 nM = 10⁷ min⁻¹ (after unit conversion). Ask the AI to ",{"type":25,"tag":177,"props":942,"children":943},{},[944],{"type":31,"value":945},"\"compute kcat given enzyme concentration of 5 nM\"",{"type":31,"value":947},".",{"type":25,"tag":34,"props":949,"children":950},{},[951,956,958,962],{"type":25,"tag":40,"props":952,"children":953},{},[954],{"type":31,"value":955},"Can I fit inhibited data with the Michaelis-Menten tool?",{"type":31,"value":957},"\nYes — run separate MM fits for each inhibitor concentration. The pattern tells you the inhibition type: if Km increases but Vmax stays constant → competitive inhibition. If Vmax decreases but Km stays constant → noncompetitive/uncompetitive inhibition. Use the ",{"type":25,"tag":199,"props":959,"children":960},{"href":776},[961],{"type":31,"value":779},{"type":31,"value":963}," to visualize all curves together, which makes the inhibition pattern immediately visible.",{"title":7,"searchDepth":965,"depth":965,"links":966},2,[967,968,969,970,971,972,973,974],{"id":28,"depth":965,"text":32},{"id":133,"depth":965,"text":136},{"id":229,"depth":965,"text":232},{"id":406,"depth":965,"text":409},{"id":551,"depth":965,"text":554},{"id":681,"depth":965,"text":684},{"id":765,"depth":965,"text":768},{"id":800,"depth":965,"text":803},"markdown","content:tools:036.michaelis-menten-fit.md","content","tools/036.michaelis-menten-fit.md","tools/036.michaelis-menten-fit","md",{"loc":4},1775502468196]