1. Microbiology lab (BIO 3126)

2. My Coordinates Instructor : Benoît Pagé :
Office : Bioscience 102
Web page:

3. My Availability By e-mail: Office hours: All week including weekends
Monday to Tuesday : 10h00am – 12h00pm
Thursday to Friday : 2h00pm – 4h30pm
Also available by appointment

4. 2 bonus points for 100% on 4/8 quizzes
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

5. Overview of web page

6. Working in a microbiology lab

7. 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

8. Before starting and at the end of the lab
Disinfect you work area
Helps prevent the contamination of cultures with microorganisms from the environment

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

10. Working in a Microbiology Lab
Sterile Technique

11. 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…

12. 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

13. 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

14. 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!

15. 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

16. 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

17. 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

18. 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

19. 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
Close tube
Inoculation

20. 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

21. Working in a Microbiology Lab
Working with solutions

22. 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

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

24. 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)

25. 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

26. 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%

27. 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?

28. 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?

29. Reducing a Concentration
Dilutions
Reducing a Concentration

30. 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.

31. 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)

32. 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).

33. 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 parts water). The dilution is written as 1/101 or said “one in one hundred and one”.

34. 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

35. 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.

36. 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
2/(2+5) = 2/7 =1/3.5
10/(10+50) = 10/60=1/6

37. Problem : More than one ingredient
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
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

38. 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

39. 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.

40. 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: mg/ml
5mg/ml =
5 (Dilution factor)
The fraction is equal to 1/the dilution factor = 1/5 (the dilution)

41. 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

42. Problem #1 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? 50
1/50
Solute 0.5ml
Solvent 24.5ml

43. 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

44. Solution #1 (cont’d) C1V1 = C2V2 1) 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

45. 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?
What volumes of solvent and solute are required?
1/10
Solvent (water) = 18mL and solute (HCl) = 2mL

46. 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

47. 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

48. 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 moles Cl)/L
1 Molar NaCl is equal to what osmolarity?

49. The Study of Microorganisms
Microbiology
The Study of Microorganisms

50. 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

51. Microorganisms in the Lab
Growth Media

52. Goals Growth under controlled conditions Maintenance
Isolation of pure cultures
Metabolic testing

53. Types Liquid (Broths) Solid media 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

54. Growth in Broths Non inoculated clear Turbid + sediment Turbid
Clear + sediment

55. Growth on Agar Growth on solid surface Isolated growth
Allows isolation of single colonies
Allows isolation of pure cultures
Easier for counting colonies and observing morphology of colonies.
Plates are used to:
grow and culture bacteria, fungi, animal tissues, or plant tissues
obtain separated pure cultures of bacteria (plate streaking)
count colonies from serial dilutions
test for growth and reactions on certain materials (such as manitol-salt, or blood agar)
test for bacteria viruses (bacteriophages)
test for resistance to materials (such as antibiotics) or nutritional needs (this is also used to select for bacteria with certain properties or favor the growth of one type/strain over others)
Single colony

56. Solid Media (Cont’d) Slants Stab Growth on surface and in depth
Different availabilities of oxygen
Long term storage
Stab
Semi-solid medium
Low availability of oxygen