Brendan recently planted a 10 ft x 10 ft garden and wants to know more about the

Question

Brendan recently planted a 10 ft x 10 ft garden and wants to know more about the microbes in the soil near his plants. He bought 2000 lbs of sterile soil which covered the top three inches of his garden. At the end of the growing season, Brendan collected 5 grams of soil from roots of different plants and suspended his samples in 100 ml of rich growth media and prepared serial dilutions from his solution. Next, he plated 100 jul of each dilution and determined the number of colony forming units of bacteria on agar plates. His data is shown below. 10-6 0 6 Soil source 102 103 10-4 105 control soil (no roots) | TMTC | 5255 bean roots corn roots tomato root TMTC TMTC TMTC TMTC TMTC TMTC 563 231 329 57 24 a. For each soil sample, determine the number of bacteria per gram of soil. (2 points) b. Brendan also incubated the 100 ml solutions containing 5 grams of soil and media for 24 hours and repeated his experiment by plating 100 ul of each dilution. His data is below. Soil source 102 103104 10-5 10-6 control soil (no roots) TMTC TMTC 105 101 TMTC TMTC TMTC TMTO tomato root TMTC TMTC383 TMTC TMTC bean roots corn roots 129 63 39 14 4 Assume that the bacteria are growing exponentially and determine the average growth rate of each bacterial population. (4 points) Please write your final answers here and show your work on the next page.

Explanation / Answer

Answer a-

To calculate concentration of microbes in the solution we use following formula.

= Numbers of colonies/ (Volume x Dilution factor)

In control soils- For this we consider 10-5 dilution because in 10-6 the colonies are zero.

100 ul = 0.1ml

Therefore, number of colonies in the original sample = 5/(0.1 ml x 10-5)

= 5 x106 colonies per ml

In 100 ml solution, the colonies are= 5 x106 x 100

In 100 ml solution, the colonies are= 5 x108

We know that sample was made by mixing 5 grams of soil in 100 ml water. Therefore, the concentration of colonies in 100 ml represents concentration in 5grams.

Now, colonies concentration in 1 gram= 5 x108/5

Colonies concentration in 1 gram= 1x108

Therefore, in bean roots

Colonies concentration = Numbers of colonies/ (Volume x Dilution factor)

Colonies concentration =6/(0.1 ml x 10-6)

Colonies concentration =6 x 107

In 100 ml solution, the colonies are= 6 x 109

Colonies concentration in 1 gram= 1.2 x109

For, Corn roots sample

Colonies concentration =2 x 107

In 100 ml solution, the colonies are= 2 x 109

Colonies concentration in 1 gram= 4 x108

For tomato roots,

Colonies concentration =3 x 107

In 100 ml solution, the colonies are= 3 x 109

Colonies concentration in 1 gram= 6 x108

Answer b-

For control sample

Colonies concentration =1 x 107

In 100 ml solution, the colonies are= 1 x 109

Colonies concentration in 1 gram= 2 x108

For bean root sample

Colonies concentration =14 x 107

In 100 ml solution, the colonies are= 14 x 109

Colonies concentration in 1 gram= 28 x108

For corn root sample

Colonies concentration =5 x 107

In 100 ml solution, the colonies are= 5 x 109

Colonies concentration in 1 gram= 1 x109

For tomato root sample,

Colonies concentration =4 x 107

In 100 ml solution, the colonies are= 4 x 109

Colonies concentration in 1 gram= 8 x108

Now calculate generation number by using formula Nt =N0 X 2n

Here Nt is the colonies number after time t and N0 is the initial colonies numbers. Here n is the numbers of generations.

For control

Nt =N0 X 2n

2 x108 =1x108 X 2n

2n = 21

Therefore, n =1

Here generation is one.

Now growth rate k =n/t

Here k is the growth rate.

k= 1/24 hrs-1

k= 0.04 colonies hrs-1

For bean roots

Nt =N0 X 2n

28 x108 =1.2x109 X 2n

2n = 28/12

2n = 7/3

Taking log in both sides

nlog2 = log7- log 3

n x 0.3= 0.37

n= 1.23

Therefore, growth rate k =n/t

k =1.23/24 hrs-1

k= 0.051 colonies hrs-1

For corn roots

Nt =N0 X 2n

1 x109=4 x108X 2n

2n = 10/4

2n = 5/2

Taking log in both sides

nlog2 = log5- log2

n x 0.3= 0.4

n= 1.3

Therefore, growth rate k =n/t

k =1.3/24 hrs-1

k= 0.054 colonies hrs-1

For tomato roots

Nt =N0 X 2n

8 x108=6 x108X 2n

2n = 8/6

2n = 4/3

Taking log in both sides

nlog2 = log4- log3

n x 0.3= 0.12

n= 0.4

Therefore, growth rate k =n/t

k =0.4/24 hrs-1

k= 0.016 colonies hrs-1

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