I always come across this question: since we measure the product concentration ([P]) at different times (t) and substrate concentrations ([S]), when plotting Lineweaver-Burk, what 1/v values are we plotting? I mean: if I measure [P] at five different times (t1-t5) starting from four different [S], I obtain a rate (v) for each t at different [S]? which one should be plotted?
Lineweaver Burk is independent of time as it takes the initial velocity for each substrate concentration. While it is true that you may need to take multiple readings at each individual substrate concentration in order to be able to calculate the gradient of the curve tangent, it is only the one velocity value per substrate concentration that is subsequently used in plotting the Lineweaver Burk plot.
Use the Michaelis-Menton Equation to calculate the missing values of [S] given below if Vmax = 5mmol/min. Plot V versus [S]. draw a line parallel to the x-axis at V max and extend your plotted line to show its approach to Vmax
A question: I have two purified enzymes with unknown concentrations. If I determine the Km and Vmax as you did in this video, are these values comparable to each other? Or I really must use equal amount of enzymes? I know Km is independent of enzyme amount but what if it comes to comparison? Thanks in advance!
An interesting question, the Vmax will depend on the concentration of active enzyme present so it is not really appropriate to directly compare, Some people try to get around this by expressing as per ug of protein, but then this is not strictly Vmax.
Sometimes, I get LB plot equation as y=ax-b. Not y=ax+b . Why is it negative? If i include more reaction (with differently substrate amount), I am getting so differently values. Why is it changing so easly?
@@zaysiz the short answer is that you cannot have a negative intercept if the enzyme is indeed michealian. Are you absolutely sure about your substrate concentrations? If you’re overestimating, especially at low concentrations this could shift the curve and give the result you describe. The LB plot has a weakness in that the distribution of points is uneven and there is a greater emphasis on lower concentrations, which are inherently more difficult to determine accurately. You could try Eadie or Hanes plots to check your answers.
@@NigelFrancis Thank you for your answer. MM graph looks normal. If I neglect some reactions with low substrate concentration, become LB plot equation to y=ax+b. Should I neglect them for y=ax+b equation? I learned on the internet how to calculate km vmax by Eadie and tried Eadie Hofstee plot. Equation looks normal but I am not sure of the accuracy. I wonder before everything, Do you say that I can not calculate Km and Vmax by y=ax-b equation. I should change substrate concentrations and have to fix the equation for y=ax+b, haven't I?
@@zaysiz You cannot calculate Vmax from a negative y-intercept as the intercept point is 1/Vmax. I suspect that there is either an error in the experimental procedure or that the concentration of the substrate is incorrect and is causing errors that are magnified at the lower concentrations due to the bias in the LB plot.
@@NigelFrancis Of course the advantage of your method is that one has the satisfaction of seeing a traditional L-B plot emerge. Thank you for a superb click-by-click demo of the procedure.
Vmax and Km have different units, Vmax is normally expressed as a unit of concentration/a unit of time (e.g.micromoles/second). Km on the other hand is expressed as a unit of concentration (e.g. micromoles). The Michaelis-Menten plot allows you to approximate Km from the 1/2 Vmax value e.g. if Vmax was 100 then the concentration that would be equivalent to Km would be that where Vmax is equal to 50. It does not mean that Km is exactly equal to the value of 1/2 Vmax. Hope that helps?
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I always come across this question: since we measure the product concentration ([P]) at different times (t) and substrate concentrations ([S]), when plotting Lineweaver-Burk, what 1/v values are we plotting? I mean: if I measure [P] at five different times (t1-t5) starting from four different [S], I obtain a rate (v) for each t at different [S]? which one should be plotted?
Lineweaver Burk is independent of time as it takes the initial velocity for each substrate concentration. While it is true that you may need to take multiple readings at each individual substrate concentration in order to be able to calculate the gradient of the curve tangent, it is only the one velocity value per substrate concentration that is subsequently used in plotting the Lineweaver Burk plot.
@@NigelFrancis Oh thank you! Now I understand it much better. Then, how is the initial velocity for each concentration calculated?
@@ramiroillanes8354 That's the gradient of the tangent at t=0 for each [S]
you made it so simple! Thank you!
Glad you found it useful
Thank you for this!
I'm glad it helped you
My prof gave us this homework and the units in the homework were mole for km, and mol s -1 for Vmax should I convert the units ?
That depends on the format in which the answer is required.
thank you so much my man
You're welcome
Use the Michaelis-Menton Equation to calculate the missing values of [S] given
below if Vmax = 5mmol/min. Plot V versus [S]. draw a line parallel to the x-axis at
V max and extend your plotted line to show its approach to Vmax
Except this is only an approximation as MM plots never truly plateau
A question: I have two purified enzymes with unknown concentrations. If I determine the Km and Vmax as you did in this video, are these values comparable to each other? Or I really must use equal amount of enzymes? I know Km is independent of enzyme amount but what if it comes to comparison? Thanks in advance!
An interesting question, the Vmax will depend on the concentration of active enzyme present so it is not really appropriate to directly compare, Some people try to get around this by expressing as per ug of protein, but then this is not strictly Vmax.
@@NigelFrancis Thank you!
Sometimes, I get LB plot equation as y=ax-b. Not y=ax+b . Why is it negative? If i include more reaction (with differently substrate amount), I am getting so differently values. Why is it changing so easly?
For example;
When I use;
My substrate column; 1, 2, 3, 4, 5, 6 mM.
Velocity column; 0.25, 0.55, 0.75, 0.92, 1.05, 1.138, 1,14
Vmax: 5,042 Km: 18,59
When I use;
My substrate column; 0.1, 0.3, 0.5, 1, 2, 3, 4, 5, 6 mM.
Velocity column; 0.06, 0.09, 0.11, 0.25, 0.55, 0.75, 0.92, 1.05, 1.138, 1,14
Vmax : 0,567 Km: 0,954
So differently, Why?
@@zaysiz the short answer is that you cannot have a negative intercept if the enzyme is indeed michealian. Are you absolutely sure about your substrate concentrations? If you’re overestimating, especially at low concentrations this could shift the curve and give the result you describe.
The LB plot has a weakness in that the distribution of points is uneven and there is a greater emphasis on lower concentrations, which are inherently more difficult to determine accurately. You could try Eadie or Hanes plots to check your answers.
@@NigelFrancis Thank you for your answer.
MM graph looks normal. If I neglect some reactions with low substrate concentration, become LB plot equation to y=ax+b. Should I neglect them for y=ax+b equation?
I learned on the internet how to calculate km vmax by Eadie and tried Eadie Hofstee plot. Equation looks normal but I am not sure of the accuracy.
I wonder before everything, Do you say that I can not calculate Km and Vmax by y=ax-b equation. I should change substrate concentrations and have to fix the equation for y=ax+b, haven't I?
@@zaysiz You cannot calculate Vmax from a negative y-intercept as the intercept point is 1/Vmax. I suspect that there is either an error in the experimental procedure or that the concentration of the substrate is incorrect and is causing errors that are magnified at the lower concentrations due to the bias in the LB plot.
@@NigelFrancis Thank you so much for your interest.
What are the units your using for velocity? mines are umol/min*gcells because my enzyme is whole-cell
Off top of my head is either nanomoles/min or micromoles/min. This example data was generated using purified enzyme.
@@NigelFrancis my enzymes came in wet cells extract thats why
Thanks Sir. This really help me
Glad it helped you
Very nice!!!!
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How did you find the gradient on the equation without using excel please help
The gradient ‘m’ can be calculated using the equation (y2 - y1)/(x2 - x1). I hope that answers your question?
@@NigelFrancis thank you so much
can you please show me how (Km=(-1)/Km) will be like?
Hi there, do you mean an example of how to perform the calculation or how the relationship itself is derived?
Bravo my friend
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Thank you so much :)
You’re welcome, hope you found it useful
You can also get the exact values of the intercepts by setting x equal to zero, and then setting y equal to zero, in the trendline equation.
Yes, you can and this is how I calculated the values for the forecasts.
@@NigelFrancis Of course the advantage of your method is that one has the satisfaction of seeing a traditional L-B plot emerge. Thank you for a superb click-by-click demo of the procedure.
thank you very much
You’re very welcome! I’m glad you found the video useful
nice very nice
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why is the Km higher than Vmax, I thought km was 1/2 max?
Vmax and Km have different units, Vmax is normally expressed as a unit of concentration/a unit of time (e.g.micromoles/second). Km on the other hand is expressed as a unit of concentration (e.g. micromoles).
The Michaelis-Menten plot allows you to approximate Km from the 1/2 Vmax value e.g. if Vmax was 100 then the concentration that would be equivalent to Km would be that where Vmax is equal to 50. It does not mean that Km is exactly equal to the value of 1/2 Vmax.
Hope that helps?
Thank you
You’re welcome, hope you found the video helpful