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Spectrophotometer

A project log for Medical tricorder

Using artificial intelligence to identify a disease by its symptoms

M. BindhammerM. Bindhammer 05/21/2015 at 12:182 Comments

I should have thought about the Beer-Lambert Law earlier!

\color{White} \large A_{\lambda}=\lg \frac{I_{0}}{I_{1}}=  \varepsilon_{\lambda} \times c \times lwhere

Spectrophotometers are commonly used to measure the concentration of a solution from its light absorbing properties. Some other examples of how they are used include:

For RGB spectrophotometry we can use a RGB LED and the TCS3200 color sensor, which output is a square wave (50% duty cycle) with frequency directly proportional to light intensity.

Proposed design, directly mountable on a PCB (complete copper fill without traces on both layers of the board in the area of the cuvette bottom to avoid exposure to light through the PCB):

As initially mentioned one use of the spectrophotometer is to determinate microbial population growth. How we do that? Let's start with a exponential bacteria growth model. Let donate the increase in cells numbers ΔN per time interval Δt, then this ratio is proportional to the actual number of cells N. If for example a population of 10000 cells produces 1000 new cells per hour, a 3 times bigger population of the same microorganism will produce 3000 new cells per hour.

\color{White} \large (1)~ \frac{\Delta N}{\Delta t}    \sim NWritten as a differential:

\color{White} \large (2)~\frac{dN}{dt} \sim NUsing a proportionality factor μ < 0, which is called specific growth rate, yields to the first-order ordinary differential equation:

\color{White} \large (3)~\frac{dN}{dt} = \mu \times N Separating the variables

\color{White} \large (4)~\frac{dN}{N} = \mu \times dtIntegrating both sides

\color{White} \large (5)~  \int  \frac{1}{N}~dN = \int \mu~dt\color{White} \large (6)~\ln(N) = \mu \times t + CDetermining the integration constant C using the initial condition t = 0:

\color{White} \large (7)~ \ln(N_{0}) = \mu \times 0 + C = CSubstituting C in equation (6):

\color{White} \large (8)~ \ln(N) =  \mu  \times t +  \ln(N_{0})Solving for N:

\color{White} \large (9)~N = N_{0} \times e^{\mu  \times t}Let donate the generation time tg, where exactly

\color{White} \large N = 2 \times N_{0}

the equation (8) yields

\color{White} \large (10)~ t_{g} = \frac{\ln(2)}{\mu}

Now, the optical density or in this case more correctly the turbidity is directly proportional to the cell density. However, the proportionality between the optical density OD and the cell density exists only for OD ≤ 0.4.

The cell density [cells/mL or cells/L] is given by

\color{White} \large (11)~cell~density =  \frac{N_{sample}}{V_{sample}} =  \frac{N}{V}where N_sample is a proportion of the total cell number N and V_sample a proportion of the total culture volume V.

Hence, equation (9) can be written as

\color{White} \large (12)~\frac{N}{V} =  \frac{N_{0}}{V} \times e^{\mu  \times t} =  \frac{N_{0_{sample}}}{V_{sample}} \times e^{\mu  \times t}Using a proportionality factor a equation (11) yields

\color{White} \large (13)~ \frac{N_{sample}}{V_{sample}} = a \times ODOnce the proportionality factor a is determined, we can calculate the cell density from any measured OD.

If we just want to calculate the specific growth rate μ of the bacterium, we don't need a proportionality factor at all. Let donate OD_0 as the optical density at t = 0, then

\color{White} \large (14)~OD=OD_{0}  \times e^{\mu \times t}Solving for μ:

\color{White} \large (15)~  \mu= \frac{\ln \big(\frac{OD}{OD_{0}}\big)}{t}

Discussions

M. Bindhammer wrote 08/14/2015 at 19:29 point

Very well! Thanks 

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Radu Motisan wrote 08/14/2015 at 16:52 point

This is very interesting , looking forward to first practical tests. How is the project progressing?

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