English Deutsch Français Italiano Español Português 繁體中文 Bahasa Indonesia Tiếng Việt ภาษาไทย
All categories

can anyone explain how to use the log function to model bacteria growth. Plz explain how & why u would use it and it's relevance to exponential functions

2007-06-02 11:05:37 · 1 answers · asked by QK 1 in Science & Mathematics Biology

1 answers

An exponential growth function is characterized by a growth rate proportional to the current size; mathematically,
dy/dt = ky, where k is the "rate constant" (reciprocal of time constant).
The exponential growth equation expresses growth over a given interval t:
Y = Y0 * e^(kt), or as a function of the time constant TC,
Y = Y0 * e^(t/TC)
(A very similar equation, with only the sign of the exponent changed, describes the inverse of exponential growth, exponential decay.)
Thus you can use logarithms to find the rate or time constant given the observed initial Y0, Y and t:
1/k = TC = t / log(y/Y0).
The rate and time constants above are parameters for growth by a factor of e. Other growth factors are also easily calculated. For bacterial growth a common assumption is that every cell divides at a constant interval TC; this is a population doubling rate. Assuming a population small enough not to be affected by realities such as nutrient supply limitations, the growth will be exponential,
P = P0 * 2^(t/TC). And here you would find the doubling time constant as
TC = t / log-base-2(P/P0).
[Note that log-base-x(y) = log(y)/log(x).]
Eventually "logistical" factors prevent unlimited exponential growth. Suppose we have a bacterial layer of constant thickness (population proportional to area), and nutrients are now only available at the perimeter. Then the growth rate is proportional to the perimeter, not the population or area. In other words, the logistic problem prevents population doubling forever. You have to use judgment and measurement data to find when the exponential rule no longer applies.

2007-06-02 13:39:24 · answer #1 · answered by kirchwey 7 · 0 0

fedest.com, questions and answers