Microbiology lab (BIO 3126) 1 Teaching Staff Lab coordinator: John Basso Email: jbasso@uottawa.ca Office: Bioscience 102 Tel.: 613-562-5800 Ext. 6358 Lecturer: Benoît Pagé Email: bpage051@uottawa.ca Office: Bioscience 102 Tel.: 613-562-5800 Ext. 6358 2 My Availability By e-mail: Any day or the week or weekend Office hours: Monday to Tuesday : 10h00am 12h00pm Thursday : 2h00pm 4h30pm Friday : 10h00am 12h00pm Also available by appointment 3 1
Quiz Course Evaluation 2 bonus points for 100% on 4/8 quizzes Assignments 20% Midterm exam 30% Final practical exam 10% Final theoretical exam 40% 4 Overview of web page http://mysite.science.uottawa.ca/jbasso/microlab/home.htm 5 Microbiology Working in a microbiology lab 2
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 your 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 3
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 4
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 5
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 6
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 7
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 1) Moles of solute = 2) Molarity = Mass of solute MWof solute Moles of solute volume in L of solution Mass of solute: given in grams (g) Molecular weight (MW): give in grams per mole (g/mole) 8
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? 100g 100% 9
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? 50 Dilutions Reducing a Concentration (A Fraction) 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. 10
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. 11
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 12
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. 13
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 14
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) C 1 V 1 = C 2 V 2 1) See previous slide 2) See previous slide 3) C 1 = 0.1M; C 2 = 0.002M; V 1 =?; V 2 = 25mL; C 1 V 1 =C 2 V 2 V 1 = C 2 V 2 /C 1 = 0.002M * 25mL / 0.1M = 0.5mL Volume of solute = V 1 = 0.5mL Volume of solvent = V 2 V 1 =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 15
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) C 1 V 1 = C 2 V 2 1) See previous slide 2) C 1 = 10M; C 2 = 1M; V 1 =?; V 2 = 18mL + V 1 ; C 1 V 1 =C 2 V 2 10M * V 1 = 1M * (18mL + V 1 ) 10V 1 = 18 + V 1 10V 1 V 1 = 18 9V 1 = 18 V 1 = 18/9 = 2mL Tonicity and Osmolarity 16
Tonicity and Osmolarity Terms used to describe the relationship between the relative concentrations of solute particles on both sides of a semi-permeable membrane (it also helps describe the movement of water between the membrane) Tonicity only takes into consideration the concentration of impermeable solutes particles Osmolarity takes into consideration the total concentration of all solute particles Permeable and impermeables 49 Tonicity and Osmolarity 50 Tonicity Isotonic solution Hypotonic solution Hypertonic Solution Impermeable solute 51 17
Tonicity Isotonic solution Hypotonic solution Hypertonic Solution Impermeable solute Permeable solute 52 Osmolarity Isosmotic solution Hyposmotic solution Hyperosmotic solution Impermeable solute 53 Osmolarity Hyperosmotic solution Hyposmotic solution Isosmotic solution Impermeable solute Permeable solute 54 18
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? 19