01/08/2018. Microbiology Lab. General Information. My Web Page 3 BIO3126

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1 Microbiology Lab BIO General Information 2 Instructor: John Basso jbasso@uottawa.ca Tel Poste 6358 Office: BSC102 My web page: Page web du cours: My Web Page 3 1

2 Course Evaluation 4 Quiz 2 bonus points for 100% on 4/8 quiz Pre-labs 5% Assignments 20% Midterm exam 25% Practical final exam 10% Theoretical final exam 40% Solutions 5 Definitions 6 Solution Mixture of 2 or more ingredients in a single phase Solutions are composed of two constituents Solute (ingredient) A solid that is being dissolved Or a stock solution that is being diluted Solvent (OR Diluent) Part of solution in which solute is dissolved or stock solution is being diluted 2

3 Preparing Solutions 7 2 ways to create a solution By dissolving a solid By diluting a more concentrated solution Do your calculations Final volume required Mass of solid (solutes) Volume of stock solutions Volume of solvent Add solvent first (usually water) Add other ingredients to solvent Preparing Solutions 8 Working with concentrations Dilutions Amounts Concentrations 9 Concentration = Quantity of solute Quantity of solution Ways to express concentrations: Molar concentration (Molarity) Percentages Mass per volume Ratios 3

4 Molarity 10 N o of Moles of solute/liter of solution Moles of solute = Molarity = Mass of solute MM of solute Moles of solute Volume in L of solution Mass of solute : given in grams (g) Molecular mass (MM) : given as grams per mole (g/mole) Percentages 11 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 fractions of 100 Percentages (Cont d) 12 %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! 4

5 Mass per Volume 13 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 14 A way to express the relationship between different constituents Expressed according to the number of parts of each component Ex. 24 ml of chloroform + 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 15 Reducing a concentration A fraction 5

6 Dilutions 16 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) 17 Dilutions are expressed as the volume of the solution being diluted per the total final volume of the dilution In the orange juice example, the dilution would be expressed as 1/4, for one can of O.J. to a TOTAL of four cans of diluted O.J. When saying the dilution, you would say, in the O.J. example: one in four. Dilutions (Cont d) 18 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). 6

7 Dilutions (Cont d) 19 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 parts water). The dilution is written as 1/101 or said one in one hundred and one. Dilutions (Cont d) 20 Dilutions do NOT have units (cans, ml, or parts) but are expressed as the number parts to the total number of parts Dilutions are always expressed as a fraction of one 1 part / total number of parts Example: 1/10 or one in ten OR: 1/(1+9) OR 1 part solute/1 part solute + 9 parts solvent Dilutions (Cont d) 21 Dilutions are always expressed with the original substance being diluted as one (1). If more than one part of original substance is initially used, it is necessary to convert the original substance part to one (1) when the dilution is expressed. 7

8 Dilutions (Cont d) 22 Example: Two (2) parts of dye are diluted with eight (8) parts of solvent. Total volume of solution = 2 parts dye + 8 parts diluent = 10 parts The dilution is initially represented as 2/10 Convert so that the numerator is one (1) Therefore 1/5 The Dilution Factor 23 Represents the inverse of the dilution Expresed as the denominator followed by X EX. A 1/10 dilution represents a dilution factor of 10X The dilution factor allows one to determine the original concentration Final conc. X Dilution factor = Original conc. Problem 24 Two parts of blood are diluted with five parts of saline What is the dilution? 2/(2+5) = 2/7 =1/ ml of saline are added to 0.05 L of water What is the dilution? 10/(10+50) = 10/60=1/6 8

9 Working with Parts 25 Preparation of 110 ml solution representing a 1/10 dilution 1/10 th of final volume must be stock solution 1/10 th of 110 ml = 11 ml = 1 parts 9/10 th of final volume must be solvent 9/10 th of 110 ml = 99 ml = 9 parts Working with Parts 26 A solution is prepared by adding 15 ml of a stock solution to 75 ml of solvent. What is the dilution and the volume of 1 part? Fraction: 15mL stock/(15ml stock + 75mL solvent) = 15/90 = dilution of 1/6 Volume of one part Total volume X the dilution = 90 ml X 1/6 = 15 ml Problem : More than one ingredient 27 Want to prepare 15 ml of a solution containing two ingredients (solutes) Need the following dilutions Solute a : 1/10 Solute b : 1/3 9

10 Problem : More than one ingredient 28 Express each dilution over a common denominator Solute «a» : 1/10 = 3/30 Solute «b» : 1/3 = 10/30 Therefore need 3 parts of «a» + 10 parts of «b» + 17 parts of solvent Total of 13 parts of solute/30 parts of solution Volume of one part 30 parts of solution = 25mL, therefore 1 part = 0.83 ml Problem : More than one ingredient 29 3 parts of solute a : 3 X 0.83 ml = 2.49 ml 10 parts of solute b : 10 X 0.83 ml = 8.3 ml 17 parts of solvent: 17 X 0.83 ml = ml Determining the Required Fraction: (The dilution) 30 Determine the reduction factor (Dilution factor) = What I have What I want Ex. You have a solution at 25 mg/ml and you want to obtain a solution at 5mg/ml Reduction factor is: 25mg/ml 5mg/ml = 5 (Dilution factor) The fraction is equal to 1/dilution factor = 1/5 (the dilution) 10

11 Determining the Required Volumes 31 Ex. You want 55 ml of a solution which represents a 1/5 dilution Use a ratio equation: 1/5 = x/55 = 11/55 Therefore 11 ml of stock/ (55 ml 11 ml) of solvent = 11 ml of stock/ 44 ml of solvent Problem 1 32 Prepare 25mL of a 2mM solution from a stock of 0.1M What is the dilution required? 2mM/100mM =1/50 What volumes of solvent and stock solution are required? Solute:1/50 X 25 ml = 0.5 ml = 1 part Solvent: 49 parts X 0.5 ml OR ml = 24.5 ml Problem 2 33 What volume of a 0.1M stock solution should be added to 25 ml of water to obtain a 2mM final concentration? What is the dilution required? 2 mm/100mm = 1/50 Volumes of solute required? 1/50 dilution = 1 part stock/50 parts solution = 1 part stock/1 part stock + 49 parts solvent Volume of solvent = 25 ml = 49 parts Therefore volume of one part = 25 ml/49 = 0.51mL 0.51/( ) = 0.51/25.51 = 1/50 11

12 Serial Dilutions 34 Dilutions made from dilutions The dilutions are multiplied Ex. A1: 1/10 A2: 1/4 A3: 0.5/1.5 = 1/3 Final dilution of the series = (A1 X A2 X A3) = 1/120 Tonicity & Osmolarity 35 Tonicity & Osmolarity 36 Terms used to describe the relationship between the relative concentrations of solute particles on both sides of a semipermeable 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 and impermeable 12

13 Measures of Osmolarity & Tonicity 37 Same units Number of osmoles (Osm) of particles of solute per liter of solution (Osm/L) Ex. 1 molar (1M) NaCl = 1 mole of NaCl per litre 1 mole of Na+ and 1 mole of Cl- Therefore 2 moles of ions per liter Equivalent to 2 Osm/litre = 2 OsM Osmolarity vs Tonicity 38 Solute Tonicity Osmolarity 1M Sucrose 1 OsM 1 OsM 1M MgCl 2 3 OsM 3 OsM 1M sucrose + 1M MgCl 2 4 OsM 4 OsM 1M Urea 0 OsM 1 OsM 1M Urea + 1M MgCl 2 3 OsM 4 OsM Urea is a permeable solute and does not contribute to tonicity Osmotic Relationship 39 Inside Outside 1M Sucrose 1 OsM 0.5M Urea 0.5 OsM Cell is hyperosmotic Water will move in 13

14 Osmotic Relationship 40 Inside Outside 0.1M Sucrose 0.1 OsM 0.5M Urea 0.5 OsM Cell is hypoosmotic Water will move out Osmotic Relationship 41 Inside Outside 0.5M Sucrose 0.5 OsM 0.5M Urea 0.5 OsM Cell is isoosmotic No water movement Tonic Relationship 42 Inside Outside 1M Sucrose 1 OsM 1M Urea 0 OsM Cell is hypertonic Water will move in 14

15 Tonic Relationship 43 Inside Outside 0.5M Sucrose + 0.5M Urea 0.5 OsM 1M Sucrose 1 OsM Cell is Hypotonic Water will move in Tonic Relationship 44 Inside Outside 0.5M Sucrose + 0.5M Urea 0.5 OsM 0.25M NaCl + 0.5M Urea 0.5 OsM Cell is Isotonic No water movement 15