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D value – The time requires to inactivate 1 log scale (90%) of bacterial population at a particular temperature. Z value – The difference of temperature.

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Presentation on theme: "D value – The time requires to inactivate 1 log scale (90%) of bacterial population at a particular temperature. Z value – The difference of temperature."— Presentation transcript:

1 D value – The time requires to inactivate 1 log scale (90%) of bacterial population at a particular temperature. Z value – The difference of temperature between 1 log scale of D-values. D and Z values determination

2 Materials and Methods Bacteria: Escherichia coli Medium: TSA Dilution buffer: Butterfield’s phosphate buffer (BPB, pH 7.2) Water bath at 60, 70, 80°C Pipette Petri-dish incubator

3 Procedures 1.Incubate two flasks of E. coli culture. 2.Turn on water bathes before class begins. Make sure water bathes have enough water. 3.Each group has one tube containing phosphate buffer saline (PBS) into water baths. 4.Place the tube into the water bath. 5.Aliquot the E. coli culture into tubes. Each table has one tube. (teaching assistant) 6.Take the PBS tube from water bath. 7.Transfer 1 mL E. coli culture into a tube containing 9 mL PBS. Place the culture into one tube immediately. 8.Bacterial density in each tube should be at 10 8 CFU/mL.

4 7.Start timing. Gently shake the tubes during heating. 8.When heating time is achieved, place the tubes under running tap water. Gently rotating the tube during cooling for one minute. 9.Decimally serial dilution 10.For 0 dilution using pour plate :transfer one mL of dilutant into a petri-dish. Gently mixed agar and bacterial suspension. 11.For other dilutions using spread plate: transfer 0.1 mL onto the agar surface. 12.Spread the plates, DO NOT pack the plates vertically. 13.Place the plates into 37C incubator. Incubate at 37C for 24 hour. Count colonies.

5 grouptemperaturetimedilutions 160°C12 min0-2-3 260°C8 min -2-3-4 360°C6 min-2-3-4-5 460°C4 min-3-4-5-6 570°C6 min0-2 670°C6 min0-2 770°C4 min 0 -2-3 870°C2 min-2-3-4 975°C4 min0-2 1075°C2 min0-2 1175°C1 min0-2 值日組 (1) No heatingcontrol-6-7-8

6 1 mL Heating 0.1 mL 0, pour plate Duplicate plates/dilution Bacterial culture 9 mL -2-3-7-6-5-4 After cooling 0.1 mL

7 -2-6-5-4-3 0 0.1 mL Duplicate plates/dilution Bacterial culture Control -6-7 1 mL 9 mL -7 -8

8 Record colony number: 25-250 Calculate average number of the duplicate plates Average number x dilution = bacterial population – Example: 42 x 10 3 = 4.2 x 10 4 =4.62 log 10 CFU/mL Calculate bacterial population of control Example: 42 x 10 8 = 4.2 x 10 9 =9.62 log 10 CFU/mL – Therefore, bacterial population in the first dilution tube (the same dilution used for heating process) should be = 4.2 x 10 8 =8.62 log 10 CFU/mL – The control number is time 0 number (the number on Y axie)

9 Record bacterial population. Draw a figure. Use regression function. Y axle: bacterial population (log CFU/mL) X axle: heating time. Obtain an equation. Obtain D value.

10

11 y = -1.0155x + 8.3753 when y= 6, x=2.339 When y=5, x= 3.234 When y=4, x=4.308 D value = 0.98 minute Small D values mean more sensitive to heat Time to reduce 90% (log scale) population Different temperature different D values –Higher temperature, smaller D values 12D –Reduce 12 log scale

12 min Population (log CFU/mL) 60 ℃ 70 ℃ 75 ℃

13 Z values Combine data from all groups. Draw a figure. Use regression function. Y axle: D values(log scale) X axle: temperatuare Obtain an equation. Obtain Z value.

14 Z values ( ℃ or ℉ ) log D values vs. temperature The time between 1 log scale D values temperatures log D values Z value


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