Question

7. (10 points) E. Coli bacteria were grown in a 20 L batch cell culture reactor, with the following data collected: Inoculation Time ---> Time 11:27 am 11:43 am 12:16 pm 1:19 pm 1:27 pm 1:42 pm 1:54 pm 2:05 pm 2:17 pm 2:41 pm Cell Count (per mL) 10,277,000 10,313,000 29,750,000 224,750,000 229,216,000 229,115,000 227,260,000 204,935,000 183,745,000 146,103,000 A. Draw a log10 X vs. time graph and label the phases for batch cell growth.

Answer 1

Incubation time	Time (min)	Cell count /ml (X)	log10X
11:27 am	0	10,277,000	7.011866
11:43 am	16	10,313,000	7.013385
12:16 pm	49	29,750,000	7.473487
1:19 pm	112	224,750,000	8.3517
1:27 pm	120	229,216,000	8.360245
1:42 pm	135	229,115,000	8.360054
1:54 pm	147	227,260,000	8.356523
2:05 pm	158	204,935,000	8.311616
2:17 pm	167	183,745,000	8.264216
2:41 pm	194	146,103,000	8.164659

The graph is as followed:

do is do it (xol boy) ynopna 2 vann gott 2.4 gilar Decline death phase Exponentials log phase stationary 7.4 phase 7 & phase L 6.80 150 100 150 Time (min)

2. The equation is C_f = C_i * e^µ(t_f – ^t_i⁾

• (C_f = final cell count of the exponential phase, C_i= initial cell count of the exponential phase, µ = growth rate constant, t_f = final time point, t_i = initial time point)
• C_f = C_i * e^µ(t_f – ^t_i⁾
• or C_f / C_i = e^µ(t_f – ^t_i⁾
• ln (C_f / C_i) = µ(t_f – t_i)

$\rightarrow$The exponential phase data look like this:

Incubation time	Time (min)	Cell count /ml (X)	log10X
11:43 am	16	10,313,000	7.013385
12:16 pm	49	29,750,000	7.473487
1:19 pm	112	224,750,000	8.3517

• Plotting the data in the equation we get,
• ln(224,750,000 / 10,313,000) = µ(112 – 16) min
• or 3.08 = 96 µ min
• or µ = 19.18/96 min
• or growth constant µ = 0.0321 min  
• Doubling time t_1/2 = ln2 / µ = 0.693/ 0.0321 min Or Doubling time t_1/2 = 21.59 min

C):

• The equation is C_f = C_i * e^k_d^(t_f – ^t_i⁾

(C_f = final cell count of the exponential phase, C_i= initial cell count of the exponential phase, k_d= the specific rate of cell mass loss, t_f = final time point, t_i = initial time point)

C_f = C_i * e^k_d^(t_f – ^t_i⁾

or C_f / C_i = e^k_d^(t_f – ^t_i⁾

ln (C_f / C_i) = k_d (t_f – t_i)

The exponential phase data look like this:

Incubation time	Time (min)	Cell count /ml (X)	log10X
1:54 pm	147	227,260,000	8.356523
2:05 pm	158	204,935,000	8.311616
2:17 pm	167	183,745,000	8.264216
2:41 pm	194	146,103,000	8.164659

• Plotting the data in the equation we get,
• ln (146,103,000 / 227,260,000) = k_d (194 – 147) min
• Or - 0.44 = 47 k_d min
• Or k_d = - 0.44/47 min
• Or cell mass loss rate k_d = - 0094 min

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