Question

For an E.coli strain, the specific growth rate changes as a function of dissolved oxygen according to the following equation:

Ko+DO

Where ?, ? _max, DO, and K_O refer to specific growth rate, maximum specific growth rate, dissolved oxygen, and cell-oxygen affinity constant, respectively. The values of ?_max and KO are 1.0 hr^-1 and 0.05 mg/l, respectively. Make a plot of ? as a function of DO with DO ranging from 0 to 8 mg/l. If the saturation solubility of air in water is 8 mg/l (note that saturation solubility corresponds to 100% set point), which percentage set point would you choose so that cells grow at a cellular growth rate that is 95% of its maximum value?

Answer 1

If different values of DO ranging from 0 to 8mg/L are taken at an interval of 1 each, i.e. 0, 1, 2, 3, 4 and so on, the calculation for the above given equation will be:

DO=0, u = 0

DO=1, u = 1*1/0.05+1 or 1/1.05 or 0.952 units.

In this way, the values of u can be calculated for values of DO according to following table:

u (mg/L)	DO
0	0
1	0.952
2	0.975
3	0.983
4	0.987
5	0.990
6	0.991
7	0.992
8	0.993

The graph between different values of u and DO can be obtained as below:

1.2 0.8 E 0.6 O 0.4 0.2 1 2 34 5 6 78 9 Value of μ

If the saturation solubility of air in water is 8mg/l we can choose the value of DO between 0.8 and 1 for which the cell growth will maximum. This is because the above graph shows that the growth rate already reaches a plateu maxima at DO values neary 1. Increase the DO value above it will not cause any further increment in growth rate of microbe. Practically, achieving growth rates of nearly 95% is considered good. According to the tabulated data above, it can be seen that nearly 95.2% of growth rate is achieved at DO value 1. So, this validates that DO values should be neary unity.