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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.