microbiologie - university of ottawapermeable membrane and the movement of water •tonicity only...

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Microbiology Lab

BIO3126

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General Information

• Instructor: John Basso

• Email: [email protected]

• Tel. 613-562-5800 Poste 6358

• Office: BSC102

• My web page:

http://mysite.science.uottawa.ca/jbasso/home.htm

• Page web du cours:

http://mysite.science.uottawa.ca/jbasso/microlab/home.htm

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My Web Page 3

mailto:[email protected]

http://mysite.science.uottawa.ca/jbasso/home.htm

http://mysite.science.uottawa.ca/jbasso/micro/accueil.htm

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Course Evaluation 4

Quiz2 bonus points for

100% on 4/8 quiz

Pre-labs 5%

Assignments 20%

Midterm exam 25%

Practical final exam 10%

Theoretical final exam 40%

Solutions5

Definitions

• 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

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Preparing Solutions

• 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

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Preparing Solutions

• Working with concentrations

• Dilutions

• Amounts

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Concentrations

• Ways to express concentrations:• Molar concentration (Molarity)

• Percentages

• Mass per volume

• Ratios

• Concentration = Quantity of solute

Quantity of solution

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Molarity

• No of Moles of solute/Liter of solution

• Mass of solute : given in grams (g)

• Molecular mass (MM) : given as grams per mole (g/mole)

Moles of solute = Mass of solute

MM of solute

Molarity = Moles of solute

Volume in L of solution

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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 fractions of 100

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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!

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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?

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Ratios

• A way to express the relationship betweendifferent 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?

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Dilutions

Reducing a concentration

A fraction

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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.

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Dilutions (Cont’d)

• 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”.

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• 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).

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• 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”.

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• 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

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• 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.

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• 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

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The Dilution Factor

• 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.

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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

• What is the dilution?

2/(2+5) = 2/7 =1/3.5

10/(10+50) = 10/60=1/6

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Working with Parts

• Preparation of 110 mL solution representing a 1/10 dilution

• 1/10th of final volume must be stock solution

• 9/10th of final volume must be solvent

1/10th of 110 mL = 11 mL = 1 parts

9/10th of 110 mL = 99 mL = 9 parts

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Working with Parts

• 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?

• Volume of one part

• Fraction: 15mL stock/(15mL stock + 75mL solvent)

• = 15/90 = dilution of 1/6

• Total volume X the dilution = 90 mL X 1/6 = 15 mL

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Problem : More than one ingredient

• Want to prepare 15 mL of a solution containing two ingredients (solutes)

• Need the following dilutions

• Solute “a”: 1/10

• Solute “b”: 1/3

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• Express each dilution over a commondenominator

• 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

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• 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 = 14.11 mL

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Determining the Required Fraction:(The dilution)

Ex. You have a solution at 25 mg/ml and you wantto obtain a solution at 5mg/ml

The fraction is equal to 1/dilution factor = 1/5 (the dilution)

What I have

What I want

Determine the reduction factor (Dilution factor) =

Reduction factor is: 25mg/ml

5mg/ml = 5 (Dilution factor)

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Determining the Required Volumes

• Ex. You want 55 mL of a solution whichrepresents 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

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Problem 1

• Prepare 25mL of a 2mM solution from a stock of 0.1M

• What is the dilution required?

• What volumes of solvent and stock solution are required?

2mM/100mM =1/50

Solute:1/50 X 25 mL = 0.5 mL = 1 partSolvent: 49 parts X 0.5 mL OR 25-0.5 mL = 24.5 mL

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Problem 2

• 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?

• Volumes of solute required?

2 mM/100mM = 1/50

1/50 dilution = 1 part stock/50 parts solution= 1 part stock/1 part stock + 49 parts solvent

Volume of solvent = 25 mL = 49 partsTherefore volume of one part = 25 mL/49 = 0.51mL0.51/(25 +0.51) = 0.51/25.51 = 1/50

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Serial Dilutions

• 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

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Tonicity & Osmolarity35

Tonicity & Osmolarity

• Terms used to describe the relationshipbetween the relative concentrations of solute particles on both sides of a semi-permeable membrane and the movementof 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

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Measures of Osmolarity & Tonicity

• 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

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Osmolarity vs Tonicity 38

Solute Tonicity Osmolarity

1M Sucrose 1 OsM 1 OsM

1M MgCl2 3 OsM 3 OsM

1M sucrose + 1M MgCl2 4 OsM 4 OsM

1M Urea 0 OsM 1 OsM

1M Urea + 1M MgCl2 3 OsM 4 OsM

Urea is a permeable solute and does not contribute to tonicity

Osmotic Relationship 39

Inside Outside

1M Sucrose1 OsM

0.5M Urea0.5 OsM

Cell is hyperosmoticWater will move in

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Inside Outside

0.1M Sucrose0.1 OsM

0.5M Urea0.5 OsM

Cell is hypoosmoticWater will move out


Inside Outside

0.5M Sucrose0.5 OsM

0.5M Urea0.5 OsM

Cell is isoosmoticNo water movement

Tonic Relationship 42

Inside Outside

1M Sucrose1 OsM

1M Urea0 OsM

Cell is hypertonicWater will move in

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Inside Outside

0.5M Sucrose + 0.5M Urea0.5 OsM

1M Sucrose1 OsM

Cell is HypotonicWater will move in


Inside Outside

0.5M Sucrose + 0.5M Urea0.5 OsM

0.25M NaCl + 0.5M Urea0.5 OsM

Cell is IsotonicNo water movement

microbiologie - university of ottawapermeable membrane and the movement of water •tonicity only...

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