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Microbiology lab (BIO 3126)
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My Coordinates
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Instructor : Benoît Pagé
Email : [email protected]
Office : Bioscience 102
Web page:
http://mysite.science.uottawa.ca/jbasso/microlab/home.htm
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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
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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
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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
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–
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–
–
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
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Microorganisms in the Lab
Growth Media
Goals
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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
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