Question

A simple batch fermentation of an aerobic bacterium growing on methanol gave the results shown in the table. Calculate: Time (hr) X (g/L) S (g/L) 0 0.2 9.23 2 0.211 9.21 0.305 9.07 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 4 8 a. Specific growth rate at t = 10hr b. Yield on substrate (Yws) c. Doubling time (ta)​

Answer 1

Answer:

A simple batch fermentation of an aerobic bacterium grown on methanol gave the following experimental results:

Time (h) 0 2 4 8 10 12 14 16 18

X (g/L) 0.2 0.211 0.305 0.98 1.77 3.2 5.6 6.15 6.2

S (g/L) 9.23 9.21 9.07 8.03 6.8 4.6 0.92 0.077 0.0

Calculate:

a) The maximum growth rate

b) The specific growth rate and the doubling time

c) The apparent yield (yXS = g Biomass / g Substrate)

d) KS value

Answer 2

Answer:

The specific growth rate at $t = 10hr$ is $0.385 hr^-1$, the yield on substrate (Yws) is $-1.72 g/g$, and the doubling time (ta) is $1.8 hr$.

Explanation:

To solve the given problem, we will use the following equations:

a. Specific growth rate $(\µ) = ln(X2/X1) / (t2 - t1)$

b. Yield on substrate $(Yws) = (X2 - X1) / (S2 - S1)$

c. Doubling time $(ta) = ln(2) / \µ$

Where X is the biomass concentration, S is the substrate concentration, t is the time, and ln is the natural logarithm.

Using the given data, we can calculate:

a. Specific growth rate at $t = 10hr$

$\µ = ln(X2/X1) / (t2 - t1)\\X1 = 1.77 g/L (at t = 10hr)\\X2 = 3.2 g/L (at t = 12hr)\\t1 = 10 hr\\t2 = 12 hr\\\µ = ln(3.2/1.77) / (12 - 10) = 0.385 hr^-1$

b. Yield on substrate (Yws)

$Yws = (X2 - X1) / (S2 - S1)\\X1 = 1.77 g/L (at t = 10hr)\\X2 = 3.2 g/L (at t = 12hr)\\S1 = 6.8 g/L (at t = 10hr)\\S2 = 4.6 g/L (at t = 12hr)$

$Yws = (3.2 - 1.77) / (4.6 - 6.8) = -1.72 g/g$

Note: The negative value indicates that the substrate is being consumed faster than the biomass is being produced.

c. Doubling time (ta)

$ta = ln(2) / \µ$

$\µ= 0.385 hr^-1$ (from part a)

ta $= ln(2) / 0.385 = 1.8 hr$

Therefore, the specific growth rate at t = 10hr is $0.385 hr^-1$, the yield on substrate (Yws) is $-1.72 g/g$, and the doubling time (ta) is $1.8 hr$.

To learn more about similar question visit

https://brainly.in/question/6070096?referrer=searchResults

https://brainly.in/question/54573166?referrer=searchResults

#SPJ3