microbiology lab (bio 3126) 1. my coordinates instructor : john basso email : [email protected]...
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Microbiology lab (BIO 3126)
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My coordinates
• Instructor : John Basso• Email : [email protected] • Office : Bioscience 102• Tel. : 613-562-5800 Ext. 6358• Web page:http://mysite.science.uottawa.ca/jbasso/home.htm
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My Availability
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•Email :•All week before 5
•Tel. :•Mon - Fri : 9-5
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Course Evaluation• Quiz
– 2 bonus points for 100% on 4/8 quizzes • Assignments 20% • 2 Reports 10% • Midterm Exam 25% • 3MT presentations 10% • Practical Exam
– In lab 5%– Job Interview 5%
• Final Exam 25%
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Overview of web page
• http://mysite.science.uottawa.ca/jbasso/home.htm
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Microbiology
Working in the microbiology lab
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At the beginning of the lab
• Wash your hands as soon as you enter the lab– Helps to avoid
contamination of the cultures with microorganisms from your natural flora
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Before starting –At the end
• Disinfect your work area– Helps prevent contamination of cultures with
microorganisms from the environment
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Before leaving the lab
• Wash your hands before leaving the lab– Helps prevent
contamination of the environment
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Working in Microbiology
Sterile Technique
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The Material
• The material used for the growth and handling of microorganisms must be sterile and maintained sterile– Growth media– Tubes– Petri dishes– Inoculation loop– Etc.
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• Use sterile technique for all transfers of microorganisms– Prevents contamination of your cultures– Prevents contamination of the environment– Prevents contamination of self
• All bacteria are opportunistic
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Transfers Using Sterile Technique
• Sterilize the inoculation loop with the Bunsen burner– The whole length of the wire
must become Red• Do not deposit on the table!• Allow to cool down
Boucle d’ensemencement
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Transfers Using Sterile Technique
• Remove cap with your small finger of the inoculation loop hand– Do not put the cap on the table!
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Transfers Using Sterile Technique
• Heat the mouth of the tube with the Bunsen burner– Keep the orientation of the tube
as close as possible to the horizontal
– Keep the opening of the cap downward
Flame mouth of tube
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Transfers Using Sterile Technique
• Use the sterile loop to remove inoculum– Liquid from broths– Solid from plates– Solid from slants
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Transfers Using Sterile Technique
• Heat once again the mouth of the tube!– Keep the orientation of the tube
as close as possible to the horizontal
Flame mouth of tube
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Transfers Using Sterile Technique
• Put the cap back on the tube of pure culture
• Return the tube to the rack
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Transfers Using Sterile Technique
• Repeat the same steps to inoculate a new tube– Remove cap– Flame mouth of tube– Inoculate– Flame mouth of tube– Close tube Inoculation
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Working with solutions
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Definitions
• Solution– Mixture of 2 or more substances in a single phase– Solutions are composed of two constituents
• Solute– Part that is being dissolved or diluted – Usually smaller
amount
• Solvent (OR Diluent)– Part of solution in which solute is dissolved – Usually greater
volume
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Concentrations
• Concentration = Quantity of solute Quantity of solution (Not solvent)
• Four ways to express concentrations:– Molar concentration (Molarity)– Percentages– Mass per volume– Ratios
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Molarity
• # of Moles of solute/Liter of solution
– Mass of solute/MW of solute = Moles of solute
– Moles of solute/vol. in L of solution = Molarity
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Percentages
• Percentage concentrations can be expressed as either:– V/V – volume of solute/100 ml of solution– m/m – mass of solute/100g of solution– m/V – Mass of solute/100ml of solution
• All represent fractions of 100
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Percentages (Cont’d)
• %V/V– Ex. 4.1L solute/55L solution =7.5%
• Must have same units top and bottom!
• %m/V– Ex. 16g solute/50mL solution =32%
• Must have units of same order of magnitude top and bottom!
• % m/m– Ex. 1.7g solute/35g solution =4.9%
• Must have same units top and bottom!
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Mass per volume
• A mass amount per a volume– Ex. 1kg/L– Know the difference between an amount and a
concentration!• In the above example 1 litre contains 1kg (an amount)
– What amount would be contained in 100ml? – What is the percentage of this solution?
100g100%
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Ratios
• A way to express the relationship between different constituents
• Expressed according to the number of parts of each component– Ex. 24 ml of chloroforme + 25 ml of phenol + 1 ml
isoamyl alcohol• Therefore 24 parts + 25 parts + 1 part• Ratio: 24:25:1
How many parts total?
50
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Dilutions
Reducing a ConcentrationA Fraction
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Dilutions
• Dilution = making weaker solutions from stronger ones
• Example: Making orange juice from frozen concentrate. You mix one can of frozen orange juice with three (3) cans of water.
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Dilutions (cont’d)
• Dilutions are expressed as a fraction of the number of parts of solute over the total number of parts of the solution (parts of solute + parts of solvant)
• In the orange juice example, the dilution would be expressed as 1/4, for one can of O.J. ( 1 part) for a TOTAL of four parts of solution (1 part juice + 3 parts water)
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Another example:
• If you dilute 1 ml of serum with 9 ml of saline, the dilution would be written 1/10 or said “one in ten”, because you express the volume of the solution being diluted (1 ml of serum) per the TOTAL final volume of the dilution (10 ml total).
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Another example:
• One (1) part of concentrated acid is diluted with 100 parts of water. The total solution volume is 101 parts (1 part acid + 100 parts water). The dilution is written as 1/101 or said “one in one hundred and one”.
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Dilutions (cont’d)
• Dilutions are always a fraction expressing the relationship between ONE part of solute over a total number of parts of solution– Therefore the numerator of the fraction must be 1– If more than one part of solute is diluted you must
transform the fraction
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Example:
• Two (2) parts of dye are diluted with eight (8) parts of solvent– The total number of parts of the solution is 10
parts (2 parts dye + 8 parts solvent)– The dilution is initially expresses as 2/10– To transform the fraction in order to have a
numerator of one, use an equation of ratios
• The dilution is expressed as 1/5.
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Problem
1. Two parts of blood are diluted with five parts of saline– What is the dilution?
2. 10 ml of saline are added to 0.05 L of water– What is the dilution?
2/(2+5) = 2/7 =1/3.5
10/(10+50) = 10/60=1/6
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Problem : More than one ingredient1. One part of saline and three parts of sugar
are added to 6 parts of water– What are the dilutions?
Saline: 1/(1+3+6) = 1/10 Sugar: 3/(1+3+6) 3/10 = 1/3.3
2. How would you prepare 15mL of this solution? – Express each component being diluted over the
same common denominator!
Saline: 1/10 + Sugar 3/10 = 1.5/15 + 4.5/15
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Serial Dilutions
• Dilutions made from dilutions • Dilutions are multiplicative
– Ex.
– A1: 1/10– A2: 1/4– A3: 0.5/1.5 = 1/3– The final dilution of the series = (A1 X A2 X A3) = 1/120
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The Dilution Factor
• Represents the inverse of the dilution• Expressed as the denominator of the fraction
followed by “X”– EX. A dilution of 1/10 represents a dilution factor of 10X
• The dilution factor allows one to determine the original concentration– Final conc. X the dilution factor = initial conc.
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Determining the required fraction (the dilution)
Ex. You have a solution at 25 mg/ml and want to obtain a solution at 5mg/ml
The fraction is equal to 1/the dilution factor = 1/5 (the dilution)
What I haveWhat I want
Determine the reduction factor (The dilution factor) =
Therefore the reduction factor is: 25mg/ml5mg/ml = 5 (Dilution factor)
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Determining the amounts required
• Ex. You want 55 ml of a solution which represents a dilution of 1/5– Use a ratio equation:– 1/5 = x/55 = 11/55
• Therefore 11 ml of solute / (55 ml – 11 ml) of solvent
• = 11 ml of solute / 44 ml of solvent
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Problem
• Prepare 25mL of a 2mM solution from a stock of 0.1M– What is the dilution factor required?
– What is the dilution required?
– What volumes of solvent and solute are required?
501/50
Solute 0.5mlSolvent 24.5ml
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Problem
• How much of a 10M solution of HCl would you add to 18mL of water to obtain a 1M solution?– What is the dilution required?
– What volumes of solvent and solute are required?
1/10
1 part Solute / 10 parts of Solution1 part Solute / 1 part Solute = 9 parts SolventTherefore 9 parts solvent = 18mL or 1 part = 2mL