Question

You have been given 6 new strains of Bacillus sp. recently isolated from a mine site. In order to gain some understanding about these strains you set up growth experiments to determine their growth over 24 hours. A spread plate viable cell count technique was utilised. The bacteria were grown at 37^oC in a nutrient broth medium and samples were taken after 24 hrs. Each sample was serially diluted from 10^-1 to 10^-4 and 100 µl aliquots were plated onto nutrient agar plates to be counted. The plates were incubated overnight and the colony counts recorded in the table below.

Strain number	Dilution
	10-1	10-2	10-3	10-4
B1	TMTC	TMTC	956	210
B2	540	72	5	0
B3	TMTC	480	59	4
B4	TMTC	TMTC	670	37
B5	156	22	4	0
B6	985	99	18	0

Note: TMTC = Too Many To Count

Complete the table below by calculating the number of CFU/ml for each of strains listed above (hint: use the appropriate dilution to calculate CFU/ml)

Strain number	CFU/ml
B1
B2
B3
B4
B5
B6

Answer 1

For CFU calculation generally plates having colony number in between 30 to 300 are chosen. Because more than 300colonies in plate will generate a difficulty in count and less than 30 colonies is too small sample size to represent the original sample size accurately.

So, CFU/ml = Number of colonies/ml * dilution factor

For CFU/ml calculation of B1, B2, B3, B4,B5 and B6, dilution of 10-4, 10-2, 10-3, 10-4, 10-1 and 10-2 are chosen respectively.

CFU/ml of B1 = (210*1000/100) * 10^4 = 2.1*10^6

CFU/ml of B2 = (72*1000/100) * 10^2 = 7.2*10^4

CFU/ml of B3 = (59*1000/100) * 10^3 = 5.9*10^5

CFU/ml of B4 = (37*1000/100)*10^4 = 3.7*10^6

CFU/ml of B5 = (156*1000/100)*10^1 = 1.56*10^4

CFU/ml of B6 = (99*1000/100)*10^2 = 9.9*10^4