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Microbiology lab (BIO 3126) 1 My Coordinates • • • • Instructor : Benoît Pagé Email : [email protected] Office : Bioscience 102 Web page: http://mysite.science.uottawa.ca/jbasso/microlab/home.htm 2 My Availability • By e-mail: – All week including weekends • Office hours: – Monday to Tuesday : 10h00am – 12h00pm – Thursday to Friday : 2h00pm – 4h30pm – Also available by appointment 3 Course Evaluation Quiz 2 bonus points for 100% on 4/8 quizzes Pre and post labs 5% Assignments 20% Midterm Exam 30% Final Exam* 45% *The final exam will consist of a practical and theoretical component Overview of web page http://mysite.science.uottawa.ca/jbasso/microlab/home.htm 5 Microbiology Working in a microbiology lab At the beginning of the lab • As soon as you enter the lab wash your hands – Helps avoid the contamination of cultures with microorganisms from your natural flora Before starting and at the end of the lab • Disinfect you work area – Helps prevent the contamination of cultures with microorganisms from the environment Before leaving the lab • Wash your hands before leaving the lab – Helps prevent the contamination of the environment Working in a Microbiology Lab Sterile Technique The Material • The material used for the growth and handling of microorganisms must be sterile and remain sterile – – – – – Growth media Tubes Petri dishes Inoculation loop Etc… Maintaining Sterility • Use sterile technique for all transfers of microorganisms – Prevents the contamination of your cultures – Prevents the contamination of the environment – Prevents self contamination • All bacteria are opportunistic Transfers Using Sterile Technique Test tube to test tube • Sterilize the inoculation loop with the Bunsen burner – The entire length of the wire must become Red/Orange • Do not deposit the loop on the table! • Allow it to cool down Boucle d’ensemencement Transfers Using Sterile Technique Test tube to test tube • Remove the cap with the small finger from the hand holding the inoculation loop – Do not put the cap on the table! Transfers Using Sterile Technique Test tube to test tube • Heat the mouth of the test tube with the Bunsen burner – Keep the test tube as close to horizontal as possible – Keep the opening of the cap downward Flame mouth of tube Transfers Using Sterile Technique Test tube to test tube • Use the sterile loop to remove inoculum – Liquid from broths – Solid from plates – Solid from slants Transfers Using Sterile Technique Test tube to test tube • Heat the mouth of the tube once again – Keep the test tube as close to horizontal as possible Flame mouth of tube Transfers Using Sterile Technique Test tube to test tube • Put the cap back on the pure culture test tube (test tube containing the inoculum) • Return the test tube to the rack Transfers Using Sterile Technique Test tube to test tube • Repeat the same steps to inoculate a new tube – – – – – Remove cap Flame mouth of tube Inoculate Flame mouth of tube Close tube Inoculation Transfers Using Sterile Technique • All transfers should be done using sterile technique – – – – Test tube to plate Plate to test tube Plate to plate Etc… • Under certain circumstances such as transfers done from plates (or to plates), the sterile technique should be slightly modified Working in a Microbiology Lab Working with solutions 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 (volume or mass) • Solvent (OR Diluent) – Part of solution in which solute is dissolved – Usually greater volume Concentrations • Concentration = Quantity of solute Quantity of solution (Not solvent) • Four ways to express concentrations: – – – – Molar concentration (Molarity) Percentages Mass per volume Ratios Molarity • # of Moles of solute/Liter of solution – Mass of solute: given in grams (g) – Molecular weight (MW): give in grams per mole (g/mole) 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 represented as a fraction of 100 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! 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? 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 are there in this solution? Dilutions Reducing a Concentration Dilutions • Dilution = making less concentrated solutions from more concentrated ones • Example: Making orange juice from frozen concentrate. You mix one can of frozen orange juice with three (3) cans of water. 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) A Second 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). A third 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”. 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 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. Problem 1. Two parts of blood are diluted with five parts of saline – What is the dilution? 2/(2+5) = 2/7 =1/3.5 2. 10 ml of saline are added to 0.05 L of water – What is the dilution? 10/(10+50) = 10/60=1/6 Problem : More than one ingredient 1. 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 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 Note: Change pipettes between each dilution to avoid carryover 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. * the dilution factor = initial conc. Note: The denominator is the dilution factor only when the numerator is 1. Determining the Required Fraction (The Dilution) Determine the reduction factor (The dilution factor) = What I have What I want Ex. You have a solution at 25 mg/ml and want to obtain a solution at 5mg/ml Therefore the reduction factor is: 25mg/ml 5mg/ml = 5 (Dilution factor) The fraction is equal to 1/the dilution factor = 1/5 (the dilution) 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 Problem #1 • Prepare 25mL of a 2mM solution from a stock of 0.1M – What is the dilution factor required? 50 – What is the dilution required? 1/50 – What volumes of solvent and solute are required? Solute 0.5ml Solvent 24.5ml Solution #1 • Fractions : 1) 2mM = 0.002M (what I want) Stock = 0.1M (what I have) Dilution factor = (what I want)/(what I have) Dilution factor = 0.1/0.002 = 50x 2) Required dilution = 1/Dilution factor = 1/50 3) Volume of a part = (Final volume)/(# of parts) Volume of a part = 25mL/50 parts = 0.5mL/part Volume of solute = 1 part * 0.5mL/part = 0.5mL Volume of solvent = (50 – 1) parts * 0.5mL/part Volume of solvent = 24.5mL Solution #1 (cont’d) • C1V1 = C2V2 1) See previous slide 2) See previous slide 3) C1 = 0.1M; C2 = 0.002M; V1 = ?; V2 = 25mL; C1V1=C2V2 V1 = C2V2/C1 = 0.002M * 25mL / 0.1M = 0.5mL Volume of solute = V1 = 0.5mL Volume of solvent = V2–V1=25mL–0.5mL=24.5mL Problem #2 • 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? 1/10 – What volumes of solvent and solute are required? Solvent (water) = 18mL and solute (HCl) = 2mL Solution #2 • Fractions: 1) What I want = 1M What I have = 10M Dilution factor = (what I have) / (what I want) Dilution factor = 10/1 = 10x Required dilution = 1/10 2) Volume solvent = 18mL Dilution = 1/10 = 1/ (9 parts solvent + 1 part solute) Volume 1 part = Volume Solvent / # of parts solvent Volume 1 part = 18mL / 9 parts = 2mL/part Volume of solute = # of parts solute * (volume/part) Volume of solute = 1 part * 2mL/part = 2mL Solution #2 (cont’d) • C1V1 = C2V2 1) See previous slide 2) C1 = 10M; C2 = 1M; V1 = ?; V2 = 18mL + V1; C1V1=C2V2 10M * V1 = 1M * (18mL + V1) 10V1 = 18 + V1 10V1 – V1 = 18 9V1 = 18 V1 = 18/9 = 2mL Osmolarity • Number of osmoles (Osm, solute particles) per litre of solution (Osm/L = OsM) – Ex. 1 molar (1M) NaCl = 1 mole of solute molecules (NaCl) per liter of solution – 1 osmolar (OsM) NaCl = 1 mole of solute particles Na + Cl) per liter of solution • 1 molecule NaCl = 2 particles (1 Na + 1 Cl) • Therefore 1 OsM NaCl = (0.5 moles Na + 0.5 moles Cl)/L • 1 Molar NaCl is equal to what osmolarity? Microbiology The Study of Microorganisms Definition of a Microorganism • Derived from the Greek: Mikros, «small» and Organismos, “organism” – Microscopic organism which is single celled (unicellular) or a mass of identical (undifferentiated) cells – Includes bacteria, fungi, algae, viruses, and protozoans 50 Microorganisms in the Lab Growth Media Goals • • • • Growth under controlled conditions Maintenance Isolation of pure cultures Metabolic testing Types • Liquid (Broths) – Allows growth in suspension – Uniform distribution of nutrients, environmental parameters and others – Allows growth of large volumes • Solid media – Same as liquid media + solidification agent • Agar: Polysaccharide derived from an algae Growth in Broths Non inoculated clear Turbid + sediment Turbid Clear + sediment Growth on Agar • Growth on solid surface • Isolated growth • Allows isolation of single colonies • Allows isolation of pure cultures Single colony Solid Media (Cont’d) • Slants – Growth on surface and in depth – Different availabilities of oxygen – Long term storage • Stab – Semi-solid medium – Long term storage – Low availability of oxygen