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Thread Subject:
Bacterial specific growth rate calculation.

Subject: Bacterial specific growth rate calculation.

From: smith

Date: 14 Mar, 2012 07:10:14

Message: 1 of 4

I need to calculate the bacterial specific growth rate for a batch process i.e I want to calculate the maximum growth rate at exponential growing condition. I have to calculate for a large set of varied cases.
In the plot of log of biomass versus time,I have been taking the mean of slopes at each point (x2-x1)/(t2-t1) from time t=0 to the end of exponential growth phase. This method is not accurate,gives a reduced growth rate. For each case one has to take the mean of the slopes those are at higher range and exclude the lower values of slope.
Is there any other way to get this?

Subject: Bacterial specific growth rate calculation.

From: Roger Stafford

Date: 14 Mar, 2012 08:11:12

Message: 2 of 4

"smith" wrote in message <jjpg8m$gej$1@newscl01ah.mathworks.com>...
> I need to calculate the bacterial specific growth rate for a batch process i.e I want to calculate the maximum growth rate at exponential growing condition. I have to calculate for a large set of varied cases.
> In the plot of log of biomass versus time,I have been taking the mean of slopes at each point (x2-x1)/(t2-t1) from time t=0 to the end of exponential growth phase. This method is not accurate,gives a reduced growth rate. For each case one has to take the mean of the slopes those are at higher range and exclude the lower values of slope.
> Is there any other way to get this?
- - - - - - - - - -
  If your time intervals are all equal you could get a second order approximation to the derivative using matlab's 'gradient' function.

  With unequal intervals you can use the formula

 (x2-x1)/(t2-t1)*(t3-t2)/(t3-t1) + (x3-x2)/(t3-t2)*(t2-t1)/(t3-t1)

as a second order approximation to the derivative at (t2,x2) with (t1,x1) and (t3,x3) as points on either side. Second order approximation here means the slope of a parabola at (t2,x2) which runs through all three points.

  Both methods assume that your plotted data is relatively noise-free.

Roger Stafford

Subject: Bacterial specific growth rate calculation.

From: Torsten

Date: 14 Mar, 2012 08:10:41

Message: 3 of 4

Am Mittwoch, 14. März 2012 08:10:14 UTC+1 schrieb smith :
> I need to calculate the bacterial specific growth rate for a batch process i.e I want to calculate the maximum growth rate at exponential growing condition. I have to calculate for a large set of varied cases.
> In the plot of log of biomass versus time,I have been taking the mean of slopes at each point (x2-x1)/(t2-t1) from time t=0 to the end of exponential growth phase. This method is not accurate,gives a reduced growth rate. For each case one has to take the mean of the slopes those are at higher range and exclude the lower values of slope.
> Is there any other way to get this?

If (t1,y1),...,(tn,yn) are your measurements of biomass versus time,
build an (nx2)-matrix A with rows [1 ti] (1<=i<=n) and an (nx1)-vector
b = [log(y1);log(y2);...;log(yn)].
If you now let
sol=A\b,
sol(2) gives you the fitted specific growth rate of the underlying
process.

Best wishes
Torsten.

Subject: Bacterial specific growth rate calculation.

From: Adam

Date: 5 Mar, 2013 13:04:08

Message: 4 of 4

It's my understanding that you need to take the slope of a log base 2 semi log plot.




"Torsten" wrote in message <19588632.928.1331712642039.JavaMail.geo-discussion-forums@vbyj18>...
> Am Mittwoch, 14. März 2012 08:10:14 UTC+1 schrieb smith :
> > I need to calculate the bacterial specific growth rate for a batch process i.e I want to calculate the maximum growth rate at exponential growing condition. I have to calculate for a large set of varied cases.
> > In the plot of log of biomass versus time,I have been taking the mean of slopes at each point (x2-x1)/(t2-t1) from time t=0 to the end of exponential growth phase. This method is not accurate,gives a reduced growth rate. For each case one has to take the mean of the slopes those are at higher range and exclude the lower values of slope.
> > Is there any other way to get this?
>
> If (t1,y1),...,(tn,yn) are your measurements of biomass versus time,
> build an (nx2)-matrix A with rows [1 ti] (1<=i<=n) and an (nx1)-vector
> b = [log(y1);log(y2);...;log(yn)].
> If you now let
> sol=A\b,
> sol(2) gives you the fitted specific growth rate of the underlying
> process.
>
> Best wishes
> Torsten.

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