A batch fermentation of a bacteria growing on sucrose gave the results shown in
ID: 182988 • Letter: A
Question
A batch fermentation of a bacteria growing on sucrose gave the results shown in the table below. Time (h) 0 Biomass, X(e/l) 8 10 12 14 16 18 0.5 1.0 2.1 4.8 7.7 9.6 10.4 10.7 8.9 Sucrose, S(e/l) 100 95 85 58 30 12 5 0 Answer the following questions based on the data collected above: a) When would the maximum growth rate (Hinex) occurs? Calculate the Hms of the bacteria culture. During exponential phase. Ln (4.8/0.5) umax(8) umax 0.282 hr-1 b) What is the definition of yield on substrate (Yus)? Determine the Ys The amount of biomass produced over the amount of substrate consumed. Y·(107-05)/(100. 2), 0.104 What is the doubling time (ta) of the culture? Td In2/0.282-2.45 hr c) d) If the culture can be considered to be substrate limited growth, what is the saturation constantExplanation / Answer
The maximum growth rate occurs at 16 hrs interval, when the biomass is 10.7g/l and the substrate concentration is 2g/l.
The maximum growth rate during exponential growth phase is calculated by the following formula
=In X2- In X1/t2-t1
µ=In 10.7- In 2.1/16-4
µ= 2.37-0.74/12
µ=0.141h-1
2. In single cell protein production, temerature optimization to maximize the yield coefficient (Yx/s) is critical. When the temperature increased above the optimum temerature, the maintenance requirement of cell increase.
Yx/s=10.7-0.5/100-2
Yx/s=10.2/98 =0.104g cells/g substrate
3. Doubling time of the culture is calculated by the following formula
Td =In2/ µ
Td =In2/ 0.282=0.69/0.282=2.45hr
4. If the culture can be considered to be substrate limited growth then the saturation constant (Ks) can be calculated by the following formula
µ=µmax x S/ Ks+S
where s= substrate
0.141=0.282x2/Ks+2
0.141= 0.564/Ks+2
Ks=(0.564/0.141) - 2
Ks=4-2=2
5. If t=15, then specific growth rate is calculated by the following formula
X=(X0) x (eµt)
10.7=(0.5)x (eµx15)
By using we can calculate this specific growth rate.
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