Microbiology lab (BIO 3126)

Teaching Staff Lab coordinator: John Basso Lecturer: Benoît Pagé Email: jbasso@uottawa.ca Office: Bioscience 102 Tel.: 613-562-5800 Ext. 6358 Lecturer: Benoît Pagé Email: bpage051@uOttawa.ca

2 bonus points for 100% on 4/8 quizzes Course Evaluation Quiz 2 bonus points for 100% on 4/8 quizzes Assignments 20% Midterm exam 30% Final practical exam 10% Final theoretical exam 40%

Working in the microbiology lab

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

Before starting –At the end Disinfect you work area Helps prevent contamination of cultures with microorganisms from the environment

Before leaving the lab Wash your hands before leaving the lab Helps prevent contamination of the environment

Working with solutions

Definitions Solution Mixture of 2 or more substances in a single phase Solutions include 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

Concentrations Concentration = Quantity of solute Quantity of solution (Not solvent) Four ways to express concentrations: Molar concentration (Molarity) Percentages Mass per volume Ratios

Molarity No 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

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

Percentages (Cont’d) %v/v %m/v % m/m 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%

Mass per volume A mass amount per a given 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 100 ml? 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 What is the total number of parts?

Reducing a Concentration A fraction Dilutions Reducing a Concentration A fraction

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

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)

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; 10 parts).

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”.

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 of dye + 8 parts of 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 Two parts of blood are diluted with five parts of saline What is the dilution? 10 ml of saline are added to 0.05 L of water

Problem: Multiple ingredients One part of saline and three parts of sugar are added to 6 parts of water What are the dilutions? How would you prepare 15mL of this solution?

Serial Dilutions Dilutions made from dilutions Dilutions are multiplicative Ex. A1: A2: A3: The final dilution of the series = (A1 X A2 X A3) = 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. multiplied by the dilution factor = initial conc.

Determining the required fraction (the dilution) What I have What I want Determine the reduction factor (The dilution factor) = 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 Prepare 25mL of a 2mM solution from a 0.1M stock What is the dilution factor required? What is the dilution required? What volumes of solvent and solute are required?

Problem What volume of a 10M solution of HCl should you add to 18mL of water to obtain a 1M solution? What is the dilution required? What is the volume of one part? What volumes of solvent and solute are required?

Tonicity and Osmolarity Terms used to describe the relationship between the relative concentrations of solute particles on both sides of a semi-permeable membrane and the movement of water Tonicity only takes into consideration the concentration of impermeable solutes particles Osmolarity takes into consideration the total concentration of all solute particles Permeable et non permeables

Tonicity and Osmolarity

Tonicity Impermeable solute Isotonic solution Hypotonic solution Hypertonic Solution Impermeable solute

Tonicity Impermeable solute Permeable solute Isotonic solution Hypotonic solution Isotonic solution Hypertonic Solution

Osmolarity Impermeable solute Isosmotic solution Hyperosmotic solution Hyposmotic solution Isosmotic solution Hyperosmotic solution Impermeable solute

Osmolarity Impermeable solute Permeable solute Hyperosmotic solution Hyposmotic solution Hyperosmotic solution Isosmotic solution Impermeable solute Permeable solute

Tonicity and Osmolarity Determine the concentration of solute particles Expressed as the number of osmoles (Osm) or osmolarity (Osm/L: OsM) Ex. 1 molar (1M) NaCl = 1 mole of NaCl per liter of solution 1 molecule of NaCl = 2 particles (1 Na + 1 Cl) Therefore 1M NaCl = (1 moles Na + 1 moles Cl)/L Which is equal to 2 OsM