Nov 6, 2019
23
2
Status
  1. Pre-Medical
I had a few questions about the mathematics behind the Lineweaver-Burk Plot for enzyme kinetics. To get the best answer, I wrote out all derivations of the different parts of the Plot, so that people can see my logic. So to start off, Lineweaver is the inverse of the Michaelis-Menten Equation, where this -

1. v = VmaxS/(Km + S)

Becomes this...

2. 1/Vo = (Km + S)/(VmaxS) ->
3. 1/Vo = Km/VmaxS + S/(VmaxS) ->
4. 1/Vo = Km/Vmax * 1/S + 1/Vmax

To get the same format as the linear equation y = mx + b with y = 1/Vo, m = Km/Vmax (slope), x = 1/S, and b = 1/Vmax (y-intercept, where x = 0), so that the Michaelis-Menten Equation can be plotted on a graph.

An image of the graph can be seen here: Lineweaver–Burk plot - Wikipedia

This formula can be plotted as a straight line with 1/S on the x-axis, meaning (since we are using the substrate concentration as the denominator) as negative S gets large in value (less negative), x becomes less negative when it is on the left half of the y-axis and approaches where the x-axis equals zero (goes in the right direction from x = negative infinity).

My first question is, since I'd assume there can never be a negative amount of substrate, S, would all these numbers be hypothetical (not real)? If so, then this would be my second question. Km is defined as substrate concentration (I guess in moles?) at 1/2 the reaction max rate, or Vmax. Therefore, using our derived equation, we would input 1/2Vmax for Vo (since that would be the rate at that y-point), seen bellow

4. 1/Vo = Km/Vmax * 1/s + 1/Vmax
5. 1/(Vmax (1/2)) -1/Vmax = Km/Vmax * 1/s + 1/Vmax -1/Vmax
6. 2/Vmax - 1/Vmax = Km/Vmax * 1/s
7. 1/Vmax = Km/Vmax * 1/s
8. S = Km

This would tell us that where ever on the graph 1/2 Vmax is located, that's where you can plot Km on the x-axis, since this is the only place on the graph one could input Km for S.

I get how 1/Vmax is the x-intercept, since if you have abundant enough of substrate to reach max velocity, you will have S going to infinity and with the plot being 1/S, then x = 0. (one can see S going to infinity as it approaches 0 from the right side of the axis, since S will get smaller going in the positive direction to produce a larger x (x = 1/S) value, with 1/0 = undefined, and S will get larger going in the positive direction to produce a smaller x (= 1/S) value, until S reaches infinity to reach x =0).

However, you would need to make the same mathematical logic for y-intercept to be -1/Km, which would mean that Vo would have to go to infinity, since the y-axis is 1/Vo. This would give you the following derivation.

4. 1/Vo = Km/Vmax * 1/s + 1/Vmax
5. 1/∞ = Km/Vmax * 1/s + 1/Vmax
6. 0 - 1/Vmax = -1/Vmax = Km/Vmax * 1/s
7. -1 = Km/S or S = - Km

Therefore, if y = 0 (Vo = infinity), then S = - Km. So, based on the equation, it makes sense, but that would be saying that the velocity of the reaction at Km is infinity, which would be saying that Vmax would be double infinity (in moles/sec?) (since Vmax is double the rate of Km). This would be impossible and does not follow the idea that it should be 1/2 rate of Vmax (for example, if the max rate is 4, then rate at Km would 2, or 1/2 for this plot, not the y-intercept). Could someone explain this aspect of the plot, in terms of enzyme kinetics?
 
Last edited:
Nov 6, 2019
23
2
Status
  1. Pre-Medical
I guess what I was asking was in terms of physiology, I get why one would want the enzyme to be saturated with substrate (S = infinity) to achieve Vmax. Why would one want Vo to be infinity, when Km is half the rate of Vmax (since 1/vo = 0, when Vo = infinity)?
 
D

deleted936470

Km is proportionate to 1/V0, so when V0 = ∞, then Km is at its theoretical minimum for the scope enzyme kinetics. An enzyme with a low Km has a high affinity for its substrate and so to achieve Vmax this is why you want V0 = ∞, i.e. it binds its substrates at the fastest rate.
 

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