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Michaelis Menten Function Lineweaver Burk Plot

Okay, so the kinetics of enzymatic velocity of catalysis with a substrate is usually measured with a hyperbolic model known as the Michaelis Menten model. That is, v = (VMax * [S]) / ([S] + KM), where "KM = one-half the substrate concentration at VMax, and VMax is the expected upper plateau of the hyperbola.

So I have a nonlinear regression model and I want the 95% confidence bands accounting for the standard errors. How do I get that?

S = c(2.5, 5.0, 10.0, 15.0, 20.0)
v = c(0.024, 0.036, 0.053, 0.060, 0.064)
mm <- data.frame(cbind(S,v))
#View(mm)
mm

S v
1 2.5 0.024
2 5.0 0.036
3 10.0 0.053
4 15.0 0.060
5 20.0 0.064

plot(v ~ S, data = mm, main = "Michealis Menten Nonlinear Regression Fitting", 
     xlab = "Substrate Concentration (mM)", ylab = "Rate (mM / sec)", 
     cex.lab = 1.4, cex.main = 1, pch = 16)

model <- nls(v ~ S*Vm/(S + K), data = mm, start = list(K = max(mm$v)/2, Vm = max(mm$v)))
summary(model)
#----
Formula: v ~ S * Vm/(S + K)

Parameters:
Estimate Std. Error t value Pr(>|t|) 
K 6.561892 0.478720 13.71 0.00084 ***
Vm 0.085857 0.002353 36.49 4.52e-05 ***
---
Signif. codes: 
0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.001035 on 3 degrees of freedom

Number of iterations to convergence: 6 
Achieved convergence tolerance: 2.298e-06

curve(x * 0.085857 /(x + 6.561892), col = "darkorchid3", add = TRUE) 

I've also double reciprocal LineWeaver Burk Model for calculating VMax & KM with linear regression. But in my opinion, the nonlinear regression model computes the parameters better.

mm$Reciprocal_S <- 1 / mm$S
mm$Reciprocal_v <- 1 / mm$v
mm
#---------
S v Reciprocal_S Reciprocal_v
1 2.5 0.024 0.40000000 41.66667
2 5.0 0.036 0.20000000 27.77778
3 10.0 0.053 0.10000000 18.86792
4 15.0 0.060 0.06666667 16.66667
5 20.0 0.064 0.05000000 15.62500
#-----
plot(Reciprocal_v ~ Reciprocal_S, data = mm, pch = 16, xlab = "(1 / [S])", ylab = "(1 / v)",
     cex.main = 2, cex.lab = 1.4, main = "Experimental Lineweaver Burk Plot")

linear_model <- lm(mm$Reciprocal_v ~ mm$Reciprocal_S)
summary(linear_model)
#--------
Call:
lm(formula = mm$Reciprocal_v ~ mm$Reciprocal_S)

Residuals:
1 2 3 4 5 
-0.30625 0.89115 -0.47556 -0.16243 0.05309

Coefficients:
Estimate Std. Error t value Pr(>|t|) 
(Intercept) 11.8003 0.4449 26.53 0.000118 ***
mm$Reciprocal_S 75.4314 2.1357 35.32 4.99e-05 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.6173 on 3 degrees of freedom
Multiple R-squared: 0.9976, Adjusted R-squared: 0.9968 
F-statistic: 1248 on 1 and 3 DF, p-value: 4.991e-05
#-----
 abline(linear_model) 

Now how do I get the 95% confidence bands for this meaningful biochemical model, the hyperbolic one?

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  • 2
    $\begingroup$ This is really a methods/theory question rather than a coding question. After you resolve your current lack of an algorithm, then it will become a fairly easy coding question. $\endgroup$ – DWin Feb 28 '17 at 18:15
  • $\begingroup$ I want R to make the model so I think it's a coding question. How do you think I made the hyperbola? With the nls R function. $\endgroup$ – xyz123 Feb 28 '17 at 18:16
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    $\begingroup$ I agree with @42-. It's unclear on what method you want to use to calculate confidence intervals. You need some modeling assumption. Just because you want to do this in R doesn't make it a coding question. You still at the part where you don't know what to do, not how to do it. There are no magic CI functions in stats that work for everything. $\endgroup$ – MrFlick Feb 28 '17 at 18:23
  • $\begingroup$ This question could be improved with an explanation of how standard error and confidence are typically calculated for the Michales Menten model. Experienced R programmers may not be familiar with this particular model, but could likely help you implement the algorithm you want to use if you describe it clearly. $\endgroup$ – Joe Feb 28 '17 at 18:28
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    $\begingroup$ First of all, whether Lineweaver-Burk or the non-linear fit is better depends on the uncertainty strukture. You should examine the residuals carefully. Anyway, I'd bootstrap the predictions. $\endgroup$ – Roland Feb 28 '17 at 18:30
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I think the propagate package can do what you are looking for.

require(propagate)

pred_model <- predictNLS(model, newdata=mm)
conf_model <- pred$summary
plot(v~S)

lines(conf_model$Prop.Mean.1 ~ S, lwd=2)
lines(conf_model$"Sim.2.5%" ~ S, lwd=1)
lines(conf_model$"Sim.97.5%" ~ S, lwd=1)

enter image description here

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  • $\begingroup$ Wow! Thank you very much. I like to see that the points are within the confidence intervals. I wonder why the main hyperbola has lost its curve, though. It seems to come to a few points. $\endgroup$ – xyz123 Mar 1 '17 at 7:41

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