RESULTS tor Exércisé 14: Measurement of Bacterial Cell Number Name VIABLE COUNTf

ID: 257795 • Letter: R

Question

RESULTS tor Exércisé 14: Measurement of Bacterial Cell Number Name VIABLE COUNTfill tonectus PLATE DILUTION NUMBER OF COLONY FORMING UNITS Plate 1 Plate 2 Average 10-4 103 104 TMTC THTC 12. 1 ? (o 5 at Co 107 li5 10* * TNTC ( TFTC (too few to count)-fewer than 30 colonies Number of CFU/ml of the undiluted original culture (This is calculated by multiplying the number of colonies times the dilution factor. The dilution factor is the inverse of the plate dilution. For example, the 10 plate dilution is the dilution factor 10) REALTIONSHIP OF ABSORBANCE TO VIABLE COUNT too numerous to count- more than 300 colonies A dett Tube Dilution Absorbance CFU/ml 1:32 0.02 1:16 0103 1:8 ?.IHI 0121 1:4 1:2 Undiluted 1o calculate the CFU/ml for each sample, first write the number of CFU/ml of the undiluted original culture colculated above in the undiluted box. Divide this number by 2 and record the new number in the 1:2 box. Divide the 1:2 CFU/mi by 2 and record the new number in the 1:4 box. Continue in the same manner until you have colculated all the dilutions Exercise 14 Measurement of Bacterial Cell Number 95

Explanation / Answer

Average CFU for 10-7 dilution are 96

CFU/mL = cfu/ml = (no. of colonies x dilution factor) / volume of culture plate

Dilution factor is inverse of the plate dilution.

Here for 10-7 plate dilution, dilution factor is 107

CFU/mL = =96 x 107/0.1

CFU/mL =9.6 x 1010

Number of CFU/ml of the undiluted original culture = 9.6 x 1010

Relationship of absorbance to viable count

Tube dilution

Absorbance

CFU/mL

1:32

Divide 1:8 CFU/mL/2 =0.6 x1010/2 = 0.3 x1010

1:16

Divide 1:8 CFU/mL/2 = 1.2 x1010/2 = 0.6 x1010

1:8

Divide 1:4 CFU/mL/2 = 2.4 x1010/2 = 1.2 x1010

1:4

Divide 1:2 CFU/mL/2 = 4.8 x1010/2 = 2.4 x1010

1:2

Undiluted /2 = 9.6x1010 /2 = 4.8 x1010

Undiluted

9.6 x 1010