SYLLABUS REQUIREMENTS:

3.2 Cell structure and function

3.2.3 Movement of molecules across membranes.

Diffusion, osmosis, facilitated diffusion, primary and secondary active transport, Endocytosis including receptor-mediated endocytosis and exocytosis. (The use of the equation = s + p is required; description of hypertonic, isotonic and hypotonic solutions on the effect of cells is expected).

TOPIC OUTLINE

A) ACTIVE AND PASSIVE PROCESSES OF MEMBRANE TRANSPORT

B) SIMPLE DIFFUSION C) FACILITATED DIFFUSION D) OSMOSIS E) ACTIVE TRANSPORT F) BULK TRANSPORT – ENDOCYTOSIS &

EXOCYTOSIS G) WATER RELATIONS H) WORKING OUT DILUTIONS

Which term is better to describe a plasma membrane?

semi-permeable

selectively/differentially permeable

polar

head

nonpolar

tails

hydrophobic

molecules hydrophilic

molecules cytosol

Biological membranes allow some substances but not others, to pass through them

This characteristic of membranes is called selective permeability

The plasma membrane is differentially permeable

not simply semi-permeable since:

1. substances e.g. amino acids, glycerol, glucose and ions can diffuse slowly through

2. it can control actively what substances enter

Substances cross biological membranes by TWO fundamentally different processes:

1. PASSIVE TRANSPORT

ATP is not needed

simple diffusion, facilitated diffusion & osmosis

high

low

Weeee!!!

Substances cross biological membranes by TWO fundamentally different processes:

2. ACTIVE TRANSPORT

ATP is needed

endocytosis & exocytosis

high

low

This is going to be hard

work!!

TOPIC OUTLINE

A) ACTIVE AND PASSIVE PROCESSES OF MEMBRANE TRANSPORT

B) SIMPLE DIFFUSION C) FACILITATED DIFFUSION D) OSMOSIS E) ACTIVE TRANSPORT F) BULK TRANSPORT – ENDOCYTOSIS &

EXOCYTOSIS G) WATER RELATIONS H) WORKING OUT DILUTIONS

DIFFUSION is the: movement of molecules or ions from a region of

high concentration to region of their low concentration down a diffusion gradient

Net movement driven by diffusion will continue until the concentration is the same in all regions

The major barrier to crossing a biological membrane is the hydrophobic interior that:

repels polar molecules

nonpolar molecules are not repelled

If a concentration difference exists for a nonpolar molecule

it will move across the membrane until the concentration is equal on both sides

At this point, movement in

both directions still occurs, but there is no net

change in either

direction

Relative permeability of cell membranes

Low High permeability

Ions e.g. Na+

Hydrophobic molecules

e.g. O2, N2 Small uncharged

polar molecules e.g. e.g. H2O, CO2, urea,

glycerol, amino acids

Large uncharged polar molecules e.g.

e.g. glucose, sucrose

THREE factors determine how fast a substance diffuses:

1. The diameter of the molecule or ion: smaller molecules diffuse faster

2. The temperature of the solution: higher temperatures lead to faster diffusion

as molecules or ions have more energy

3. The concentration gradient in the system: the greater the concentration gradient,

the more rapidly the substance diffuses

TOPIC OUTLINE

A) ACTIVE AND PASSIVE PROCESSES OF MEMBRANE TRANSPORT

B) SIMPLE DIFFUSION C) FACILITATED DIFFUSION D) OSMOSIS E) ACTIVE TRANSPORT F) BULK TRANSPORT – ENDOCYTOSIS &

EXOCYTOSIS G) WATER RELATIONS H) WORKING OUT DILUTIONS

Substances which enter by facilitated diffusion:

those that cannot diffuse through the phospholipid bilayer such as:

1. Large particles: glucose amino acids proteins

2. Some ions: Na+ ions Cl- ions

Some ions and polar molecules:

can diffuse through special transport proteins called:

channel proteins Carrier proteins

Extracellular space

Intracellular space

and

Channel proteins have a hydrophilic interior

that provides an aqueous channel through which polar molecules can pass when the channel is open

aqueous channel

selective filter

Potassium channel

Carrier proteins bind specifically:

to the molecule they assist, much like an enzyme binds to its substrate

Difference between channel & carrier proteins:

Channel proteins

fixed shape

Carrier proteins

undergo rapid changes in shape

Intracellular space

Extracellular space

Facilitated diffusion of ions through channels

transport proteins which allow the passage of ions

possess a hydrated interior that spans the

membrane

ions can diffuse through the channel in either direction, depending on their relative concentration across the membrane

ion channels:

Ion channels may be closed or open

Gated channels

can be opened or closed in response to a stimulus

the stimulus can be:

the stimulus causes a change in the 3D shape of the channel

electrical

chemical (ligand)

Two types of gated channels, depending on the stimulus:

1. Ligand-gated channels: chemical stimulus

2. Voltage-gated channels: electrical stimulus

important in the conduction of nerve impulses

Two types of diffusion compared

Simple diffusion: Facilitated diffusion:

small molecules cross the lipid bilayer

substances cross the bilayer aided by:

channel or

carrier proteins

Facilitated

diffusion

Simple

diffusion

Both types occur down a concentration gradient, BUT rate of uptake varies

Why has a maximum been reached for facilitated diffusion?

Facilitated

diffusion

Simple

diffusion

As carriers become saturated.

Facilitated diffusion has three essential characteristics:

it is:

1. Specific

2. Passive

3. Saturates

Facilitated diffusion in red blood cells:

glucose enters red blood cells via glucose transporters

movement of:

chloride

hydrogen carbonate ions

TOPIC OUTLINE

A) ACTIVE AND PASSIVE PROCESSES OF MEMBRANE TRANSPORT

B) SIMPLE DIFFUSION C) FACILITATED DIFFUSION D) OSMOSIS E) ACTIVE TRANSPORT F) BULK TRANSPORT – ENDOCYTOSIS &

EXOCYTOSIS G) WATER RELATIONS H) WORKING OUT DILUTIONS

Osmosis

is the passage of water molecules from a region of their high concentration to a region of their low concentration through a partially permeable membrane

Water continues to flow until:

equilibrium is reached

Three terms are used to compare the solute concentrations of two

solutions separated by a membrane:

Isotonic

Hypotonic

Hypertonic

Isotonic solutions

[same concentration]

Hypotonic solution

Hypertonic solution

[lower conc. of solute]

[higher conc. of solute]

Aquaporins: water channels

the transport of water across membranes is complex

studies on artificial membranes show that water:

despite its polarity, can cross the membrane - but this flow is limited

Water molecules cross plasma membranes:

Diffusion across the lipid bilayer to some extent: as water is a

small molecule.

The main routes of water diffusion, are protein-lined pores:

Aquaporins are selective water-channels in cell membranes.

1

2

TOPIC OUTLINE

A) ACTIVE AND PASSIVE PROCESSES OF MEMBRANE TRANSPORT

B) SIMPLE DIFFUSION C) FACILITATED DIFFUSION D) OSMOSIS E) ACTIVE TRANSPORT F) BULK TRANSPORT – ENDOCYTOSIS &

EXOCYTOSIS G) WATER RELATIONS H) WORKING OUT DILUTIONS

Summary: Four basic mechanisms for transport across cell membranes:

1. Diffusion

2. Osmosis

3. Active transport

4. Bulk transport (endocytosis or exocytosis)

PASSIVE

ACTIVE

Active Transport is the

energy-consuming transport of molecules or ions across a membrane against a concentration gradient

achieved by carrier proteins present in the cell surface membrane

Movement of substances by active transport and diffusion

In diffusion:

Direction is reversible in diffusion.

In active transport: Movement is usually in

one direction.

Unlike in facilitated diffusion, the carrier proteins need energy to keep changing shape

Uniport transporters

Three types of proteins are involved in active transport:

Symport

transporters

Antiport transporters

Coupled transport

1. Uniport transporters [uniporters]:

move a single solute (e.g. Ca 2+) in one direction

move two solutes in the same direction

2. Symport transporters [symporters]:

A protein transporting:

Na+ ions & sugar molecules (coupled transport)

amino acids & Na+ : bind to the same carrier protein in intestine to

take up amino acids

move two solutes in opposite directions:

one into the cell

the other out of the cell

Na+-K+ pump

3. Antiport transporters [antiporters]:

Symporters and antiporters

are also known as coupled transporters because they move two substances at once

The use of energy from ATP in active transport may be:

Direct Indirect called primary active transport

called secondary active transport

only cations are transported e.g. sodium-potassium pump

aid in the uptake of amino acids and glucose

Two basic types of active transport processes:

Primary active transport

Secondary active transport

The sodium-potassium pump

most animal cells have a:

Inside cell: Low conc. of Na+;

High conc. of K+

These concentration differences are maintained by actively pumping:

Na+ out of the cell

K+ into the cell

The Na+/K+ pump:

uses the energy stored in ATP to move Na+ and K+

the energy is used to:

change the conformation of the carrier protein, which changes its affinity for either Na+ or K+ ions

Na+-K+ pump

The Na+-K+ pump:

seems to exist in all animal cells

moves: 3 Na+ out of the cell

2 K+ into the cell

The Na+-K+ pump is important to:

control the osmotic balance of animal cells

maintain electrical activity in: nerve & muscle cells [interior of nerve cell:

-70 mV]

The Na+-K+ pump is important to:

drive active transport of substances like:

sugars & amino acids

Coupled transport uses ATP indirectly

some molecules are moved against their concentration gradient by using the energy stored in a gradient of a different molecule

in this process, called coupled transport: the energy released as one molecule moves down its

concentration gradient is captured and used to move a different molecule against its gradient

as explained already, the energy stored in ATP molecules can be used to create a gradient of Na+ and K+ across the membrane

these gradients can then be used to power the transport of other

molecules across the membrane

Plant cells have a proton pump instead of a Na+-K+ pump

Secondary active transport :

does not use ATP directly

H+ Proton pump Symport carrier

Sucrose

Sucrose

TOPIC OUTLINE

A) ACTIVE AND PASSIVE PROCESSES OF MEMBRANE TRANSPORT

B) SIMPLE DIFFUSION C) FACILITATED DIFFUSION D) OSMOSIS E) ACTIVE TRANSPORT F) BULK TRANSPORT – ENDOCYTOSIS &

EXOCYTOSIS G) WATER RELATIONS H) WORKING OUT DILUTIONS

Endocytosis & Exocytosis

are active processes involving the bulk transport of materials through membranes,

Exocytosis

Transport out of cells

Endocytosis

Transport into cells

vesicle (small vacuole)

or

vacuole (fluid-filled membrane-bound sac)

Endocytosis occurs by:

an infolding or extension of the cell

surface membrane to form a:

Two types of endocytosis:

Phagocytosis

‘cell eating’:

A solid is taken up

Pinocytosis ‘cell drinking’

A liquid is taken up

Cells specialising in phagocytosis are called:

Phagosome

[Phagosome: a membrane-bound vesicle in a phagocyte containing the phagocytized material]

Phagocytosis in Amoeba

Pinocytosis: vesicles formed are very small

Pinocytosis is very common in both:

plant

animal cells

Pinocytosis is used by the human egg cell to take up nutrients from the surrounding follicle cells.

Exocytosis occurs when: 1. waste materials are removed from the cells

(e.g. solid, undigested remains from phagocytic vacuoles)

2. useful chemicals are secreted (e.g. enzymes)

3. plants export substances needed to form the cell walls

Receptor-mediated endocytosis

A coated pit in the membrane contains a protein receptor.

When triggered by a specific molecule, the pit closes over &

traps the molecule within a vesicle.

The receptors then return to the cell surface.

1 2

3

Cholesterol is taken up by Receptor-mediated endocytosis

Phospholipid

outer layer

Protein

Cholesterol

Plasma

membrane CYTOPLASM

Receptor protein

Vesicle

TOPIC OUTLINE

A) ACTIVE AND PASSIVE PROCESSES OF MEMBRANE TRANSPORT

B) SIMPLE DIFFUSION C) FACILITATED DIFFUSION D) OSMOSIS E) ACTIVE TRANSPORT F) BULK TRANSPORT – ENDOCYTOSIS &

EXOCYTOSIS G) WATER RELATIONS H) WORKING OUT DILUTIONS

Terms used in plant water relations

the term water potential is used to describe water movement through a membrane in a plant cell

water potential of plant cells is affected by:

solute concentration

[expressed as solute potential]

the pressure generated when water enters and inflates plant cells

[expressed as pressure potential]

1 2

Water potential (symbol , Greek letter psi)

water molecules: possess kinetic energy move very rapidly in

random directions when in liquid or gaseous form

the greater the concentration of water molecules in a system:

the greater the total kinetic energy of water molecules in that system

and the higher its so-called water potential

Water potential ( - psi)

is a measure of the free kinetic energy of the water molecules in the solution

is usually expressed in pressure units, kPa

PURE WATER:

has the maximum water potential ( = 0)

The water potential of a solution is the tendency of a solution to take up

water from pure water

the concentration of water molecules is reduced

and therefore the water potential is lowered.

What happens to the water potential when a solute is dissolved

in pure water?

LOWER water potentials than pure water and

so have negative values of

ALL solutions have:

Pure water = 0 kPa

Dilute solution = -200 kPa

Concentrated solution = -500 kPa

Osmosis can be defined as:

the movement of water from a higher water potential to a lower potential through a

partially permeable membrane.

HIGHER LOWER

Water molecule

In which direction does water move? Pure water = 0 kPa

Dilute solution = -200 kPa

Concentrated solution = -500 kPa

From a less negative (higher) to a more negative (lower)

Water potential is applicable to two systems in contact

water potential can be regarded as the tendency of water molecules to move from one place to another

the two systems do not need to be separated by a membrane

thus the term may be used to measure the tendency for water to move from:

the soil to the roots

the leaves to the air

water can be said to move down a gradient of water potential from soil to air

Solute potential (s) of a solution is a measure of the change in

water potential of a system due to the presence of solute molecules

The effect of dissolving solute molecules in pure water is to:

s = -100 kPa

s =

-400 kPa

reduce the concentration of water molecules

hence lower the water potential

is always negative

the more solute molecules present:

the lower (more negative) is s

Solute potential (s)

Pure water s = 0 kPa

Dilute solution s = -200 kPa

Concentrated solution s = -500 kPa

the greater the tendency of water to flow into it from another solution of lower solute

concentration

The greater the solute concentration of a solution:

Pure water s = 0 kPa

Dilute solution s = -200 kPa

Concentrated solution s = -500 kPa

Solute potential (s)

was previously referred to as:

‘osmotic pressure’

or

‘osmotic potential’

Explain why there is no net

water movement.

Solute potentials are equal.

s = -200 kPa

Pressure potential (p) is the pressure acting outwards from a cell

opposing the inward pressure of the cell wall

p is usually positive in most cells e.g. turgid cells

Wat

er

po

ten

tial

So

lute

po

ten

tial

Pre

ssure

po

ten

tial

p is usually positive e.g. turgid cells

BUT p may be negative in certain circumstances

As in xylem when water is under tension (negative pressure)

Mechanical pressure in cell water relations was previously known as:

‘turgor pressure’

p

Equation

the water potential is the sum of the solute potential and the pressure potential:

= s + p

water

potential solute

potential pressure potential

Question The diagram shows two neighbouring plant cells that are in direct contact. The water potential (Ψ), pressure potential (Ψp) and solute potential (Ψs) are given.

a) Which cell has the higher water potential? (1) Cell B

b) In which direction will water move by

osmosis? (1) Cell B to A

c) What will be the water potential of the cells at equilibrium? (1)

The average of the two i.e. -1000 kPa.

d) Assuming that the solute potential does not change significantly, what would be Ψp at equilibrium in cell A and cell B? (1)

Cell A at equilibrium:

= s + p

-1000 = -2000 + p

-1000 + 2000 = p

1000 kPa = p

Cell B at equilibrium:

= s + p

-1000 = -1400 + p

-1000 + 1400 = p

400 kPa = p

Movement of water between solutions by osmosis

the partially permeable membranes of importance in the water relations of plant cells are the:

The cell wall is: usually freely permeable to

substances in solution so is not important in

osmosis

Effect of different solutions on animal and plant cells.

isotonic

solution

hypertonic

solution

hypotonic

solution

equal movement of water into and out of cells

net water movement out of cells

net water movement into cells

Crenated cells: have spikes

Why is it that plant cells do

not burst?

Presence of a cell wall.

Plant cells in a hypotonic solution:

Water enters the cell and fills the vacuole.

The plasma membrane pushes against the cell

wall making the cell very inflexible.

Cells in this state are TURGID.

Plant cells in a hypertonic solution:

The cell loses water and goes FLACCID.

This state is called PLASMOLYSIS.

A cell at this stage is said to be plasmolysed.

REASON: the vacuole becomes flaccid the cytoplasm stops pushing against the cell

wall

Shrinkage of the protoplasts where they just begin to pull away from the cell walls:

cell is flaccid as Ψp = 0

Incipient plasmolysis is the point when plasmolysis is just about to happen

At incipient plasmolysis p = 0

i.e. no pressure is exerted by protoplast against the cell wall

and so:

of the cell = s of the cell = s + p

The process of plasmolysis is usually reversible without permanent damage

to the cell

(2 3) Recovered. Water is once more available. The leaf cells take up water by osmosis and become turgid. The leaves are now firm and are held horizontally. Woody plants such as trees and shrubs do not collapse when they wilt but their leaves become limp.

(1) wilting (2) recovering (3) recovered

(1) Wilting. Pant is short of water. The leaf cells are no longer turgid and the leaves curl up and droop.

Onion epidermal cells were placed in a 1M sucrose solution.

What fills the space between the cell wall and the protoplast?

Sucrose solution

Potato strips were first placed in either a hypotonic or a hypertonic solution, than in water. Explain why potato

strip P floated while Q sank.

Was previously placed in a hypotonic solution. TURGID

Was previously placed in a hypertonic solution. PLASMOLYSED

The water potential of a solution is the tendency of a solution to take up water from pure water

The solute potential of a solution is a measure of the change in water potential of a system due to the presence of solute molecules

Recall:

The pressure potential is the pressure acting outwards from a cell opposing the inward pressure of the cell wall

EXPERIMENT Aim: To find the water potential of potato cells

For a solution at atmospheric pressure = s

The principle underlying this experiment is to discover a solution, of known water potential, in which the tissue being examined neither gains nor loses water.

Samples of the tissue are allowed to come to equilibrium in a range of sucrose solutions of different concentrations and the solution that induces neither an increase nor a decrease in mass or volume of the tissue would have the same water potential as the tissue.

Procedure 0 M

0.25 M

0.10 M

0.50 M

1.00 M

0.75 M Precaution: Cover with lid.

40 cm3

Six Petri dishes are prepared as shown.

12 potato strips are cut:

70mm x 4mm x 4mm

[important to be long & thin]

Using potato cylinders cut by a cork borer is better

than strips. Why?

Uniform thickness.

Precaution:

Work quickly while measuring to avoid water loss.

Potato strips are blotted dry.

Length of each strip is measured.

Two strips are weighed at a time.

To remove water results from tissue damage (through cutting).

Two potato strips are placed in each petri dish.

Strips are completely immersed. Why?

In an evaluation of the experiment, the importance of a “standardised drying technique” was emphasised. Explain what is meant by a “standardised drying technique” and its importance to the accuracy of the result. [2] A standard/the same pressure is applied to the filter paper when drying prior to both initial and final weighings; if over-dried/under-dried prior to final weighing then weight loss/gain will be overestimated/underestimated;

After exactly 1 hour each strip is removed, blotted dry, and again measured.

Molarity of solution (M)

0.00

0.10

0.25

0.50

0.75

1.00

Water potential (kPa)

0

-260

-680

-1450

-2370

-3510

Original length of strip 1 (mm)

Final length of strip 1 (mm)

Change in length* of strip 1 (mm)

Percentage change in length** of strip 1 (%)

Original length of strip 2 (mm)

Final length of strip 2 (mm)

Change in length of strip 2 (mm)

Percentage change in length * of strip 2 (%)

Mean percentage change in length (%)

* Change in length = Final – Original length ** Percentage change in length = Change in length x 100 Original length

QUESTION:

Explain why it is important to calculate change in mass as a percentage in this experiment. [1]

As the initial mass of the discs is different.

The average percentage change in length & weight are calculated: graphs are plotted:

1. % change in length vs water potential

2. % change in mass vs water potential

Water potential (kPa)

Use graph to read the water potential

of the sucrose solution which causes

no change in length/mass

990 kPa

Give a reason for this statement:

Value of the water potential of the sucrose solution which causes no change in mass is

more accurate than in length.

When weighing: all the potato

cells are considered.

When taking length: only one

dimension is considered.

List two precautions

1. All 12 strips used were taken from the same potato to reduce probability of cells having a different .

2. Strips may not have been evenly thick, affecting the amount of water that could be exchanged.

3. Strips were prevented from sticking to each other – this reduces surface area for exchange.

4. The same pressure was applied during blotting to avoid over-drying.

List two sources of error 1. Loss of water from potato strips while length and

weight readings.

2. Strips may not have been evenly thick, affecting the amount of water that could be exchanged.

3. Time potato strips were left in the sugar solutions may not be enough for equilibrium to be reached between the tissues and the external solution.

4. Water potential differences may be present between the cells in the centre of the tuber and the cells in the periphery of the tuber.

EXPERIMENT Aim: To find the solute potential of a plant tissue.

Background:

In this experiment samples of the tissue being investigated will be allowed to come to equilibrium in a range of solutions of different concentrations (water potentials).

The aim is to find out which solution causes incipient plasmolysis, that is, shrinkage of the protoplasts to the point where they just begin to pull away from the cell walls. At this point, turgor pressure is zero since no pressure is exerted by the protoplasts against the cell walls and it can be said that:

of cell = s of cell = of sol. = s of sol.

[explained in next two slides]

The & s for a solution are equal

This is an open container, so the ψp = 0

This makes the ψ = ψs

= s + p

water

potential

solute

potential

pressure

potential

At incipient plasmolysis Ψp = 0: osmosis stops

CELL

of cell = s of cell of sol. = s of sol. =

= s + p

= s

= s

SOLUTION

= = = s

Method

Lid

Strip of leaf epidermis [1cm x 1cm]:

immersed in 2 cm3 of distilled water

left for 30 minutes

Watch glass

Repeated with:

0.1M; 0.3M; 0.35M; 0.4M; 0.45M; 1M

Epidermis is placed on a slide in a drop of the same solution it was in.

Percentage of plasmolysed cells for each solution is worked out: a graph is plotted:

The number of plasmolysed cells present in the first 50 cells observed is noted.

After 30 minutes, leaf epidermis is observed under microscope.

Results:

Molarity of solution (M)

0.00

0.10

0.30

0.35

0.40

0.45

1.00

Solute potential (kPa)

0

-260

-820

-970

-1120

-1280

-3510

Total number of cells counted

Number of plasmolysed cells

Percentage plasmolysis (%)

Incipient plasmolysis is said to occur when 50% of the cells are plasmolysed

List TWO precautions taken during the course of this experiment to ensure

accuracy. Give a reason for each precaution.

1. The onion epidermis was not handled with bare hands for too long. To avoid undue evaporation from the tissue samples.

2. The epidermal tissue was submerged completely in the solution. To ensure that all cells were subject to the same water potential of the test solution.

Give TWO possible sources of error.

1. Low resolution of light microscopes does not allow a thorough examination of incipient plasmolysis. Some cells might not have been counted.

2. Artefacts (such as air bubbles introduced during mounting) may have interfered with the counting of the cells.

Question: MAY, 2007 [Paper 3]

A number of strips of onion epidermis, each with a thickness of one layer of cells, have been peeled from an onion bulb. Briefly describe how one would determine the solute potential of the epidermal cells. [6]

Samples of epidermal tissue are placed for 30 minutes in a range of sucrose solutions of different concentrations. The onion epidermis is viewed under medium power of the light microscope and the number of plasmolysed cells of the first 50 viewed is recorded. The percentage of plasmolysed cells for each solution is determined.

A graph of % plasmolysed cells against solute potential for each sucrose solution is plotted. The solution causing 50% plasmolysis is read from the graph. This solution causes incipient plasmolysis, that is shrinkage of the protoplasts to the point where they just begin to pull away from the cell walls. At this point the pressure potential is zero since no pressure is exerted by the protoplasts against the cell walls. The solution causing incipient plasmolysis has the same solute potential as the cell sap.

TOPIC OUTLINE

A) ACTIVE AND PASSIVE PROCESSES OF MEMBRANE TRANSPORT

B) SIMPLE DIFFUSION C) FACILITATED DIFFUSION D) OSMOSIS E) ACTIVE TRANSPORT F) BULK TRANSPORT – ENDOCYTOSIS &

EXOCYTOSIS G) WATER RELATIONS H) WORKING OUT DILUTIONS

LEARN how to work out dilutions starting from a:

1. given % stock solution

2. 1M sucrose solution.

How to work out % concentration starting from 5% detergent

e.g. if you want 20 cm3 of a 0.3% detergent solution:

0.3 x 20

5

Required %

Original % X volume needed

= 1.2 cm3 of 5% detergent

Add 18.8 cm3 of distilled water

Table of dilutions

Final

Concentration /

%

Volume of

detergent /

cm3

Volume of

distilled water /

cm3

1 20 0

0.8 16 4

0.6 12 8

0.4 8 12

0 0 20

Or more simply

Final

Concentration /

%

Volume of

detergent /

cm3

Volume of

distilled water /

cm3

1 20 0

0.5 10 10

0.25 5 15

0.125 2.5 17.5

0 0 20

2. How much volume of a 1M sucrose solution and distilled water need to be mixed to

produce 20 cm3 of 0.6M sucrose solution?

1M sucrose solution ? volume

Distilled water

? volume

20 cm3 of 0.6M sucrose solution

Work out CONC x VOLUME to find volume of 1M sucrose solution needed.

0.6 x 20 = 12 cm3

20 – 12 = 8 cm3

1M sucrose solution: 12 cm3

Distilled water 8 cm3

20 cm3 of 0.6M sucrose solution

THE END