Assignment.
Bacterial Sample Hour 1 Hour 2 Hour 3 Hour 4 Hour 5 Hour 6
Hour 7
1
16
64
256 1024 4096
16,384
65,536
2
97
291
873 2619 7857
23,571
70,713
3
112
784 5488 38,416 268,912 1,882,384 13,176,688
4
7
63
567 5103 45,927 413,343 3,720,087
5
143
286
572 1144 2288
4576
9152
Directions: Assuming that the growth pattern for each bacterial sample follows a geometric sequence,
determine the following:
1. Determine the rate at which the culture grows in a hour. This rate will be the factor r by which the
number of bacterial cultures has increased since the last recorded observation.
2. Write a formula that represents the growth of this bacteria based upon your observations. Your
formula will be based upon the basic format for a geometric sequence:
3. Using the formula you’ve developed, determine the number of cultures you would expect to see in
the Petri dish on the 8th, 10th, and 12th hour of your observations.
4. Compute the total number of bacterial cultures observed after 24 hours of growth assuming that the
growth follows a geometric series.
5. Repeat steps 1–4 for all five bacterial samples.
Solutions:
1. Determine the rate at which the culture grows in an hour. This rate will be the factor r by which the
number of bacterial cultures has increased since the last recorded observation.
a=number of bacterial at the beginning of time interval
b=number of bacterial at the end of time interval
Bn=a*rn-1
Time (n) =1 hour
Rate of growth per hour(r ) =..?
Sample(n)
1
2
3
4
a(beginning)
16
97
112
7
b(end)
64
291
784
63
b/a
64/16
291/97
784/112
63/7
r
4
3
7
9
5
143
286
286/143
2
2. Write a formula that represents the growth of this bacteria based upon your observations. Your formula
will be based upon the basic format for a geometric sequence:
Bacteria sample 1:
Bn=a*rn-1
a=16
r=4=22
Bn =16(4n-1)
Bn =24*(22(n-1))
Bn =24+2n-2)
Bn =22n+2
Bacteria sample 2:
Bn=a*rn-1
a =97
r=3
Bn =97(3 n-1)
Bacteria sample 3:
Bn=a*rn-1
a1 =112
r=7
Bn =112(7n-1)
Bacteria sample 4:
Bn=a*rn-1
a1 =7
r=9
Bn =7(9n-1)
Bacteria sample 5:
Bn=a*rn-1
a1 =143
r=2
Bn =143(2n-1)
2. Using the formula you’ve developed, determine the number of cultures you would expect to see in
the Petri dish on the 8th, 10th, and 12th hour of your observations.
Sample
1st
term
Ratio
nth term
8th
n=8
10th
n=10
12th
n=12
1
16
4
Bn =22n+2
2
97
3
Bn =97(3 n-1)
3
112
7
Bn =112(7n-1)
4
7
9
Bn =7(9n-1)
5
143
2
Bn =143(2n-1)
Bn=22(8)+2
Bn=216+2
Bn=218
=262,144
Bn=22(10)+2
Bn=220+2
Bn=222
=4,194,304
97(38-1)
=97(37)
=212,139
112(78-1)
=112(77)
=92,236,816
7(98-1)
=7(97)
=33,480,783
143(28-1)
=143(27)
=18,304
97(310-1)
=97(39)
=1,909,251
112(710-1)
7(910-1)
143(210-1)
9
9
=112(7 )
=7(9 )
=143(29)
=4,519,603,984 =2,711,943,423 =73,216
Bn=22(12)+2
97(312-1)
112(712-1)
Bn=224+2
=97(311)
=112(711)
26
Bn=2
=17,183,259 =2.215*1011
=67,108,864
7(912-1)
=7(911)
=2.197*1011
143(212-1)
=143(211)
=292,864
4. Compute the total number of bacterial cultures observed after 24 hours of growth assuming that the
growth follows a geometric series.
Sample
1st
term
Ratio
nth term
n=24
24th
1
16
4
Bn =22n+2
2
97
3
Bn =97(3 n-1)
Bn=22(24)+2
97(324-1)
Bn=248+2
=97(323)
Bn=250
=9.132*1012
15
=1.126*10
3
112
7
Bn =112(7n-1)
4
7
9
Bn =7(9n-1)
5
143
2
Bn =143(2n-1)
112(724-1)
=112(723)
=3.065*1021
7(924-1)
=7(923)
=6.204*1022
143(224-1)
=143(223)
=1.2*109
Report the results of the calculations you performed above.
i.
Which strain of E. coli exhibited the highest growth rate?
Sample 4 with a growth rate of 9 bacterial cultures per hour.
ii.
Which strain of E. coli exhibited the lowest growth rate?
Sample 5 with a growth rate of 2 bacterial cultures per hour.
iii.
Assuming that all five of the E. coli strains present a high toxicity danger to humans,
which do you suppose would be the most manageable based upon growth? Why?
Since the common ratios are all positive, there will be a tendency of exponential growth towards infinity.
However, sample 1 is the most manageable strains since it has a low growth rate of 4 and has a small
manageable size of 16.
iv.
Consider how you’ve modeled the growth of the E. coli strains using the concept of
geometric sequence. Is this a realistic approach to modeling bacterial growth?
Normally, under perfect temperature conditions, bacterial growth doubles at regular intervals. The
exponential growth by geometric progressions is part of the bacterial life cycle. However the modeled
growth rate of E. coli strains is not representative of the normal pattern of growth of bacteria in nature
and hence is not a realistic approach in modeling the bacterial growth.
v.
What other factors do you think should be considered when modeling the growth of
bacteria such as E. coli?
The Environmental conditions conducive for growth of E.coli include:
i)
Temperatures.
Increase in temperatures decreases the time taken to exterminate the microorganisms. Also, increase in
temperatures leads to an increase in the toxicity levels of the E.coli bacteria. Increase in temperatures
decreases the time taken to exterminate the microorganisms. The vice versa is true.
NB: applying high temperatures over a short period of time is considered fit rather than applying low
levels of temperatures for a considerable long period.
ii)
pH levels
Increase in acidity levels decreases the time needed to kill the microorganisms. There are basically 3
levels of pH which are:



Minimum pH levels. At this level, E. coli strains cannot thrive.
Maximum pH levels: At this level, E.coli strains cannot survive.
Optimum pH levels: In this level of pH, E.coli strains would survive since they thrive best under at
optimal levels of pH.
However, between the range of minimum and optimum ranges of pH levels there will be an increase in
growth rate of the E.coli strains. Also, between optimum and maximum pH levels, a decrease in growth
rate will be evident. The diagram below illustrates the pH scale against various pH levels.
12
10
pH scale
8
alkaliphiles
6
acidophiles
4
neutrophiles
2
0
basic
neutral
acidic
pH levels
Inferences:
At acidic levels i.e. (pH below 7.0/ pH<7), acidophiles thrive best at high.
At neutral levels i.e. (pH is equal to 7.0), neutrophiles thrive best at neutral
At basic levels i.e. (pH above 7.0/ pH>7.0), alkaliphiles thrive best and are seen to have the highest
Oxygen levels
This condition has two types; that is ,


Obligate organisms. This type of E.coli requires oxygen to thrive.
Aerophobes- This type of bacteria do not require oxygen as a nutrient to thrive.
Controlling growth of E.coli bacteria.
To control growth of E.coli bacteria, several methods are used. The best way of curbing the growth of
microorganisms is by:



Complete termination of microorganisms.
Sterilization-it refers to doing away with organisms inside or on the material being sterilized. It often
utilizes the use of thermal heat, radiation or chemicals.
Preventing the growth of microorganisms.
In controlling their growth, both chemical and physical agents have to be used.
However, for complete termination, 2 methods are used:


Cidal agents. These agents completely kill the microorganisms.
Static agents. Static agents on the other hand inhibit growth without killing the microorganisms.
In Sterilization method, it will involve the use of:
Heat.
To ensure destruction of the microorganisms, the right amount of heat ought to be used. A good example
of microorganisms to curb using this method includes the endospores.
Incineration.
This is another option which entails destroying the organisms by burning them.
Boiling.
To completely terminate the E.coli especially the endospores, it requires you to boil the solution for period
longer than 6 hours in order to sterilize the solution completely.