Luria-delbruck experiment question

daniel6874 · Oct 31, 2011

My question is about the classic study cited below (*) and available as a free pdf download from

http://www.genetics.org/content/28/6/491.full.pdf+html

At page 495 the authors use the "average division time" of bacteria dt, divided by ln(2), as the time unit in the equations

(1) dN_t /dt = N_t , (2) N_t = N_o e^t.

Can anyone tell me why this division by ln(2) was done? N_o is the original number of bacteria present. N_t is the number at time t.

Obviously it has something to do with integration/differentiation, but I am missing the point. Thanks.

*Luria, S. E.; Delbrück, M. (1943). "Mutations of Bacteria from Virus Sensitivity to Virus Resistance". Genetics 28 (6): 491–511.

Ygggdrasil · Oct 31, 2011

The doubling time divided by ln(2) gives the appropriate time constant for an exponential growth. Writing the equation for exponential growth in terms of the doubling time (t_d) gives:

N(t) = N_o 2^(t/t_d)

Equivalently, this can be written as:

N(t) = N_o e^(t ln(2) / t_d) = N_o e^(t/t_c)

Where t_c = t_d/ln(2). This makes use of the fact that e^ln(2) = 2.

bobze · Oct 31, 2011

Bacteria double exponentially; 2 become 4, 4 become 8, etc. This describes Log in base 2 (2², 2³, etc).

The application then is used to describe things that grow or decay exponentially. A more detailed explanation can be found http://logbase2.blogspot.com/2007/12/log-base-2.html" [Broken]

daniel6874 · Oct 31, 2011

Ygggdrasil said: ↑

...
N(t) = N_o 2^(t/t_d)

Equivalently, this can be written as:

N(t) = N_o e^(t ln(2) / t_d) ...

This is crystal clear. I forgot he was starting with 2^t/t_d as a growth law. Thanks!

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Luria-delbruck experiment question

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