Question

In: Biology

1. a.If a bioreactor starts with 10,000 cells in a culture that has a generation time...

1.

a.If a bioreactor starts with 10,000 cells in a culture that has a generation time of 2 h, how many cells will be in the culture after 4, 24, and 48 h? (30 points)

b.How long would it take for an initial population of 106 cells/ml to reach a size of 108 cells/ml if the generation time is 30 min? (20 points)

c. In a picnic, the dosa was contaminated with 46 cells of Propionibacterium acnes. If P. acnes has a generation time of 1.5 h and a lag phase of 1.5 h, how many cells of this bacterium will be present in the paella after 10 h? (20 points)

d. There are two special cases where the Monod equation can be simplified. Describe those cases and the simplify the Monod equation. (20 points)

e.Write down the equation to determine the growth rate at exponential growth in a batch culture system? (5 points)

f.Write the Monod equation and define each of the components. (5 points)

please give answers to all because these are the one question dear experts

Solutions

Expert Solution

1. Using he equation X = 2nX0 where

x = initial number of cells

n = number of generations

X​​​​​​0​​​​​ = number of cells after n generations

After 4h, n = 4h/2h per generation = 2 generations

X= 22(104) = 4.0 * 104 cells

After 24h, n = 24h/2h per generation = 12 generations

X = 212(104) = 4.1 * 107 cells

After 48h, n = 48h/2h per generation = 24 generations

X = 224(104) = 1.7 * 1011 cells

B. The formula is G = t/3.3logb/B

Where, G is the generation time

t is the time required

logb is the final population

LogB is the initial population

30= t/3.3log 108 /106

30 = t /3.3*2

30 = t/6.6

T= 198minutes i.e 3.3h

C. Initial number of cells N0 are 46, generation time is 1.5h, lag phase is 1.5h

t = 10h - 1.5h = 8.5h

Log N - log N​​​​​​​​​​​​0 = /2.303.(t-t​0)

=0.693/g= 0.462h-1

log N = [/2.303.(t-t​​​​0​​​t-t0)]+logN​​​0 = [0.462/2 303 (8.5)]+ 1.663 = 3.368

2.33*103 cells

D. The first special case is when the substrate concentration is very much less than K​​​​​s. In this case, monod equation is simplified to

=mS/K​​​​​​s

The second case is when the substrate concentration S Is much greater than K​​​​​​s . In this case the monod equation is simplified to =m

E. Expotential phase is the log phase of bacterial growth. The cells divide at constant rate. The rate of expotential growth can be expressed as generation time or doubling time i.e. G. Generation time is defined as the time per generation. Hence, G = t/n, where n = number of generations, t= time

F. Monod equation =mS/K​​​​​s+S

Where, m= maximum growth rate achieved by microbe at a defined temperature and for a particular substrate.

K​​​​​​S= half saturation constant defined as substrate concentration at which growth occurs at ½m

S= substrate concentration

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