BIOL 2301 Lecture Notes - Lecture 14: Petri Dish, Antimicrobial Resistance, Bacterial Growth

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skybear452

9 Mar 2018

School

Northeastern University

Department

Biology

Course

BIOL 2301

Professor

Eric J.Stewart

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Document Summary

Techniques: growth media contains amino acids and sugars that support bacterial growth. Plates also contain agar, a solid, jello-like support substance. Growing them in a liquid flask to get a large number of bacteria. There is a multiplication rule for dilution. (calculating cell concentration by sell dilution) Tenth of a ml in the petri dish. Multiply by 10 get answer in 1 ml. To determine how many cells were in the original culture, grow 0. 1 ml of the diluted culture, then multiply the # of colonies by the reciprocal of the plating dilution factor. Count between 30 to 300 colonies on a plate. We count the colonies which are nearest to. These are called met-. minus because it is absent. Alternatively, picking up dna from the outside environment. Tolerant to antibiotics but not resistant is different. Tolerant means they can"t grow but they"re not dying. They are in a dormant, metabolically inactive state.

Related Questions

Viable cell counts are often performed on exponentially growing cells in order to relate the absorbance reading (AU) with actual cell number. For this purpose, a sample of the culture is removed during the exponential phase after the absorbance reading is recorded. Spread plates are prepared in order to perform a viable cell count. The goal is to have a spread plate containing between 30 and 300 cells (colonies) when plating a volume of 0.1 ml (100 Î¼l). To determine the correct dilution, you will need to estimate the number of cells/ml present from the AU (absorbance units) value of your sample. For this calculation, we will assume that 1 AU is equivalent to 10⁷ cells /ml. So:

AU of sample x [(10⁷ cells/ml)/1 AU] x 0.1 ml plated = total # of cells expected

QUESTION 1:

You have removed a volume of a culture of E. coli during exponential growth in order to measure the viable cell count. If the absorbance reading was 0.3, estimate the maximum dilution factor you would need in order to obtain between 30 and 300 colonies on a plate spread with 0.1 ml of the dilution. Assume that there are 10⁷ cells/ml in one absorbance unit.

QUESTION 2:

You have removed a volume of a culture of E. coli during exponential growth in order to measure the viable cell count. If the absorbance reading was 0.3, estimate the minimum dilution factor you would need in order to obtain between 30 and 300 colonies on a plate spread with 0.1 ml of the dilution. Assume that there are 10⁷ cells/ml in one absorbance unit. â

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Determining Dilution Scheme

Viable cell counts are often performed on exponentially growing cells in order to relate the absorbance reading (AU) with actual cell number. For this purpose, a sample of the culture is removed during exponential phase after the absorbance reading is recorded. Spread plates are prepared in order to perform a viable cell count. The goal is to have a spread plate containing between 30 and 300 cells (colonies) when plating a volume of 0.1 ml (100 ?l). To determine the correct dilution, you will need to estimate the number of cells/ml present from the AU (absorbance units) value of your sample. For this calculation, we will assume that 1 AU is equivalent to 10⁷ cells /ml. So:

AU of sample x [(10⁷ cells/ml)/1 AU] x 0.1 ml plated = total # of cells expected

You want to calculate two different dilutions that will give you between 30 and 300 colonies (cells). In order to determine the maximum dilution factor, divide the total # of expected cells by 30. In order to determine the minimum dilution factor, divide the total # of expected cells by 300. For example if the AU was 0.3, the total number of expected cells would be 3.0 x 10⁵. In this case the maximum dilution factor would be 10⁴ (3.0 x 10⁵/30) or a dilution of 10^-4 and should produce approximately 30 colonies. The minimum dilution factor would be 10³ (3.0 x 10⁵/300) or a dilution of 10^-3, which should produce ~300 colonies. Since performing a single dilution of 10^-4 or 10^-3 would be highly inaccurate, serial dilutions are performed. Typically, the spread plates are prepared in duplicate for greater accuracy.

(NOTE: Do not put units in your answers, if you do the computer will grade the answer as wrong and you will lose the points. Answers should be written in numerical form or if you want to write as scientific notation you can write as 1.22E8 or 1.22E-8. The system will only take up to 15 significant figures.) (Another NOTE: Your answer must be within 5% of the correct value to get credit for this problem.)

You have removed a volume of a culture of E. coli during exponential growth in order to measure the viable cell count. If the absorbance reading was 0.8, estimate the maximum dilution factor you would need in order to obtain between 30 and 300 colonies on a plate spread with 0.1 ml of the dilution. Assume that there are 10⁷ cells/ml in one absorbance unit. Answer:

You have removed a volume of a culture of E. coli during exponential growth in order to measure the viable cell count. If the absorbance reading was 0.8, estimate the minimum dilution factor you would need in order to obtain between 30 and 300 colonies on a plate spread with 0.1 ml of the dilution. Assume that there are 10⁷ cells/ml in one absorbance unit. Answer:

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A student grows a culture of E. coli in 50 ml of liquid medium contained in a 250 ml Erlenmeyer flask and decides to pour plate samples of the culture onto LB agar plates. (LB stands for Luria Broth. Itâs simply a typeof broth on which E. coli grow well.) Before plating, she sets up several dilutions of the culture (indicated in the table) and plates 0.2 ml of each by pour plating. She does this in duplicate for each dilution. She incubates all of her LB agar plates in a 37 ÂºC incubator and counts the colonies on each plate later. Her results are as follows:

Total Dilution Factor: 10^6 PlateA: 756 PlateB: 349

TDF: 10^7 PlateA: 73 PlateB: 76

TDF: 10^8 PlateA: 7 PlateB: 8

Determine the average density in cfu/ml of the culture based on the relevant results, also significantly fewer colonies grew on Plate B at the 10^ dilution as compared toPlate A at the same dilution. Which plate (Plate A or Plate B) is more likely to be a true indication of the number cells that were present in the culture at the 10^6 dilution?

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BIOL 2301 Lecture Notes - Lecture 14: Petri Dish, Antimicrobial Resistance, Bacterial Growth

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