Question

6.1. A simple, batch fermentation of an aerobic bacterium growing on methanol gave the results shown in the table. Calculate: a. Maximum growth rate (max) b. Yield on substrate (Yx/s) c. Mass doubling time (ta) Time (h) x (g/l) S (g/l) 9.23 9.21 9.07 0.2 0.211 0.305 0.98 1.77 3.2 5.6 6.15 8.03 6.8 4.6 0.92 0.077 6.2

Answer 1

Answer-

According to the given question-

Time	X (g/ l)	S (g / l)
0	0.2	9.23
2	0.211	9.21
4	0.305	9.07
8	0.98	8.03
10	1.77	6.8
12	3.2	4.6
14	5.6	0.92
16	6.15	0.077
18	6.2	0

we have to calculate

(a) Maximum growth rate (µ_max ), (b) Yield on substrate( _{Y x/s} ), (c) Mass doubling time (t_d)

first

(a) Maximum growth rate (µ_max )-

here from the table we can easily observed that the Maximum growth rate is between t = 10 h and t = 12 h . and we know that

X₂ = X₁ e^µt

taking natural log on both the side= ln X₂ =ln (X₁ e^µt)

ln X₂ = (ln X₁)  * (ln e^µt)

ln (X₂ / X₁ )  = µ∆t

putting the value in the above equation we get,

ln (3.2 / 1.77) = $\mu$ * (12 - 10)

0.592= $\mu$ * 2

$\mu$ = 0.592 / 2 = 0.296

so the Maximum growth rate (µ_max )= 0.296

(b) Yield on substrate( _{Y x/s} )-

from the table we can observed the value

Initial biomass concentration X₀ = 0.2 h^-1

Initial substrate concentration S₀ = 9.23 h^-1

Final biomass concentration Xf= 6.2 h^-1

final substrate concentration Sf= 0 h^-1

we know that the equation for yield on substrate

Y x/s = Xf - X₀  / S₀ - S_f

putting the value in above equation we get-

Y x/s = 6.2 - 0.2 / 9.23 - 0= 6 / 9.23 = 0.65

so the Yield on substrate( _{Y x/s} ) = 0.65

(c) Mass doubling time (t_d)-

we know that the mass is double only during the exponential phase of growth

so considering above information we can calculate the doubling time by using following formula-

t_d = ln 2 / µ_max

so putting the valye in above equation we get-

t_d = ln 2 / 0.296= 0.6931 / 0.296= 2.34 h

so the Mass doubling time (t_d) = 2.34 h