Transport across boundaries
Transport
Across
Boundaries
10.
2
FT Diffusion is the spreading of the particles of a gas, or of any substance in
and solution, resulting in a net movement from a region where they are at a
HT higher concentration to a region where they are at a lower concentration.
The greater the difference in concentration, the faster the rate of diffusion.
Oxygen required for respiration passes through cell membranes and
through gas exchange surfaces, such as alveoli in the lungs, by diffusion.
Carbon dioxide enters leaves and leaf cells by diffusion.
Osmosis is the diffusion of water from a dilute to a more concentrated
solution through a partially permeable membrane that allows the passage of
water molecules but not solute molecules.
Other substances such as sugar and ions can also pass through cell
membranes.
Many organ systems are specialised for exchanging materials:
• in humans the surface area of the lungs is increased by the alveoli, and
that of the small intestine by villi;
• In plants the surface area of the roots is increased by root hairs and the
surface area of leaves by the flattened shape and internal air spaces.
HT Substances are sometimes absorbed against a concentration gradient. This
requires the use of energy from respiration. The process is called active
transport. It enables plants to absorb ions from very dilute solutions.
Similarly sugar may be absorbed from low concentrations in the intestine
and from low concentrations in the kidney tubules.
Candidates should be able, when provided with appropriate information, to
explain how other gas and solute exchange surfaces in humans and other
organisms are adapted to maximise effectiveness.
MOLECULES IN MOTION
Living things need molecules of different sizes, e.g. oxygen, water, sugars, amino-acids, etc. These
molecules need to move, or be moved, so that they can enter or leave cells, or be distributed around the
body. Similarly, cell waste materials (CO2, urea, etc.) have to be moved from the tissues which produce
them to the organs which remove them. Molecules move, or are moved, in several ways:
A
Diffusion: this is the random movement of particles (atoms, molecules, ions) in a gas or liquid,
from regions of high concentration to regions of low concentration. It results in the particles being
evenly spread in the space they occupy.
B
Osmosis: this is the diffusion of water molecules in a solution of low concentration to a solution
of high concentration, through a selectively permeable (semi-permeable) membrane. It is by
osmosis that cells gain or loose water.
C
Active uptake: This is the rapid uptake of particles against a diffusion gradient, with the
expenditure of energy (from respiration), e.g. uptake of ions from the soil by root hair cells.
A: DIFFUSION
TO DEMONSTRATE DIFFUSION IN A GAS
cotton wool
soaked in
ammonia
moist red litmus paper
glass tube
bung
1. Put strips of moist red litmus paper at 2 cm intervals along the length of a glass tube,
2. Seal one end with a rubber bung, and seal the other end with a rubber bung, which has a pad of
cotton wool soaked in ammonia solution pinned to it.
3. Time low long it takes for the red litmus paper to turn blue (ammonia turns red litmus blue).
4. By timing how long it takes for the red litmus to turn blue, we can work out the speed of diffusion
of ammonia molecules.
TO DEMONSTRATE DIFFUSION IN A LIQUID
Fill a beaker with water, then drop in a crystal of potassium permanganate.
The purple colour spreads through the water until the water is the same shade of purple throughout.
The potassium permanganate particles move from an area where they are in high concentration (the
crystal) to an area where they are in low concentration (the water), until the particles are evenly
distributed throughout.
SPEED OF DIFFUSION
Diffusion is rather slow (as demonstrated by the experiments). It can be speeded up in several ways:
1. Moving the molecules: e.g. by stirring a liquid or a gentle breeze in air.
2. Heating: particles move faster when warmed.
3. Increasing Diffusion Gradient: i.e. by increasing the concentration of the diffusing molecules.The
greater the difference in concentration, the faster the speed of diffusion.
4. Active Transport: energy is used moving particles across cell membranes. This only occurs in
living things.
EXAMPLES OF DIFFUSION IN BIOLOGY
1.
Gas exchange: oxygen, required for respiration, from air or water diffuses inwards through the gas
exchange surface and CO2 diffuses out.
Gas exchange happens:
(i) through alveoli in the lungs of Humans and other animals
(ii) through fish gills, frog skin, etc.
(iii)through leaf stomata for photosynthesis, when CO2 diffuses inwards and oxygen out.
2.
The placenta: oxygen, salts, soluble foods, etc, diffuse from the mother’s blood into the baby’s
blood across the placenta surface, while the baby is growing in the uterus. Carbon dioxide, urea,
etc diffuse in the opposite direction.
3.
Root hairs: Dissolved salts can diffuse into the root hair cells from the soil, although active
transport is more important.
DIFFUSION AND SURFACE AREA
More molecules can diffuse through larger surface areas.
Small organisms have a large surface area compared with their volume, so they can obtain oxygen by
diffusion through surface area alone.
Large organisms have a small surface area : volume ratio, so they need specialised surfaces to
increase area, while occupying the same volume.
Living things increase surface area for molecules to diffuse by:
1.
Root Hairs: Roots have many root hair cells growing out from the root surface, so greatly
increasing surface area for absorbing water and salts (which are ions in solution).
2.
3.
Leaves:
4.
Lung Alveoli:
5.
Villi:
Flattened shape and internal air spaces makes large surface areas for gas exchange.
The numerous air-sacs (alveoli) of the lungs in Humans increase surface area for
gas exchange.
These projections from the lining of the small intestine in Humans make the surface
area larger for absorbing the soluble products of digestion. There are similar
projections in the placenta, where dissolved substances pass from the Mother’s blood
to the baby’s blood while the baby is growing in the uterus.
Root hairs
Lung alveoli
Villi of small intestine
Villi in placenta
B. OSMOSIS
Osmosis is the process by which water molecules move from a region of high concentration of
water molecules to one of a lower concentration of water molecules, through a selectively
(differentially/partially) permeable membrane in order to make both the concentrations equal,
i.e. Osmosis is really a specialised form of diffusion.
EXAMPLE – THE OSMOMETER
Set up the experiment as shown in the diagram. What will happen and why?
EXPLANATION
The visking tubing is a selectively permeable membrane, which contains small pores. These are small
enough to prevent sugar molecules from passing through, but large enough to allow water molecules
through. Water molecules pass through in both directions, but more move into the bag than move out,
i.e. there is a net movement of water into the bag, Therefore the liquid in the capillary tubing will rise.
IMPORTANCE OF OSMOSIS
1.
Water control in all animals (Osmoregulation)
If we sweat a lot, e.g. on a hot day, the solutions in our body will become more concentrated. If we
drink a lot of fluid on a cold day, our body solutions will become much weaker. Both of these situations
will stop our body cells from working properly. So, we use our kidneys to excrete more, or less, water
to keep the concentration of our body solutions constant.
2. Water control in animals which live in water
Animals living in fresh water tend to take in more water than they need, whilst animals living in sea
water tend to lose more than they need. These creatures have evolved special ways to either gain or lose
water in order to live.
Example
a. Animal cell in pure water
Osmosis occurs. Water diffuses into the cell through the selectively permeable membrane. The cell
swells and may burst.
b. Animal cell in a concentrated solution
Osmosis occurs. Water diffuses out of the cell through the selectively permeable membrane. The cell
shrinks.
3.Water and plant cells
a. Plant cell in pure water
Osmosis occurs. Water diffuses into the cytoplasm and vacuole through the selectively permeable
membrane. The cell swells and becomes turgid.
b. Plant cell in a concentrated solution
Osmosis occurs. Water diffuses out of the cytoplasm and vacuole through their selectively permeable
membranes. First, the cell shrinks slightly and becomes flaccid. Then the cell membrane pulls away
from the cell wall and the cell is said to be plasmolysed.
IMPORTANCE OF WATER
1. 70-95% of living organisms is water.
2. It dissolves things, e.g. for transport of food, minerals and oxygen.
3. All chemical reactions take place in solution.
Osmosis is important to the plant for several reasons:1. To get water into the plant, replacing that lost during transpiration.
2. Turgor pressure built up during osmosis helps to support the plant.
3. Osmosis plays a part in opening and closing of stomata (see notes on transpiration).
C: ACTIVE TRANSPORT
Active transport is an important process:
a) In Plants: Because substances such as ionic salts are present in the soil in very dilute solutions.
A higher concentration of these salts is present inside the cells.
Plants must absorb these salts, so they are absorbed from an area of low concentration
(the soil) to an area of higher concentration (inside the cells).
b)In Humans: The blood sugar concentration must be kept constant to allow brain cells
to function correctly.
Sugar may be present in the small intestine in only very low concentration, so the sugar
molecules move from an area of low concentration (the intestine) to an area of higher
concentration (the blood).
Similarly, sugar which is filtered out of the blood by the kidney, is reabsorbed back into
the blood, against a concentration gradient.
Active transport differs from diffusion in the following ways:
1. It is rapid.
2. It is selective, i.e. only what is required is actively absorbed.
3. Substances can be absorbed against a concentration gradient.
4. Living surfaces or membranes are essential.
5. Energy from respiration is needed for active uptake.
The diagram below contrasts diffusion with active uptake.
It shows a cell, which has a semi-permeable membrane, and dots which represent particles of dissolved
substances.
At A: The arrow shows movement of particles from an area of high concentration to an area of lower
concentration.
This is diffusion of particles into the cell.
At B: The arrow shows movement of particles from an area of high concentration to an area of lower
concentration.
This is diffusion of particles out of the cell.
At C: The arrow shows movement of particles from an area of low concentration to an area of higher
concentration.
This is active transport of particles into the cell.
At D: The arrow shows movement of particles from an area of low concentration to an area of higher
concentration.
This is active transport of particles out of the cell.
EVIDENCE FOR ACTIVE TRANSPORT
Active transport only occurs through a living surface, happens against a concentration gradient, and
needs energy (from respiration).
Aerobic respiration needs oxygen to produce energy.
Diffusion happens without the need for energy, but can only happen by movement of particles from an
area of high concentration to an area of lower concentration.
Energy is needed for active transport, so when oxygen is available, active transport can happen.
If there is no oxygen, there is no energy produced, so there is no active transport.
This is shown in the graph below, which shows the transport of sulphate ions by root hair cells, in the
presence of oxygen or the absence of oxygen.
We see from the graph that in the absence of oxygen at first uptake occurs by diffusion. Then, when the
concentration of sulphate ions inside the root hairs and outside the root hairs is the same, no further
uptake occurs. But in the presence of oxygen, because there is respiratory energy, not only is uptake
quicker, but more ions are also absorbed.
MOLECULES IN MOTION – COMPREHENSION
1. Name the type of molecular movement where particles may enter cells without using energy from
respiration.
2. a) Name three types of substance which can enter cells by the process you have mentioned above.
b) Name two types of substance which leaves cells by the process mentioned above.
3. By what process does water enter or leave cells?
4. a) Name one other way substances can be moved, and
b) Give an example of each.
5. Two plants of the same species, size and age, growing in identical conditions of temperature and
CO2 availability, were put in solutions which were identical in salt composition and concentration,
apart from the concentration of phosphate ions. Solution A had phosphate ions at the concentration
of 100 parts per million and solution B and phosphate ions at 5000 parts per million.
a) Into which plant, A or B, did the phosphate ions enter the quickest?
b) How could the process of entry be slowed down?
6. Give three examples where diffusion is important in living things.
7. Why is surface area so important to a living thing which relies on diffusion for uptake of atoms,
molecules or ions?
8. How is the surface area of the small intestine increased for the absorption of sugars, etc?
9. A cube has dimensions of 2 cm x 2 cm x 2 cm
a) What is its surface area : volume ratio?
b) If its dimensions doubled, what would be its surface area : volume ratio be?
10. The table below shows the rate at which two simple sugars are absorbed through the lining of the
small intestine of a
mammal. The rate is shown for both living intestine and intestine which has been poisoned with a
substance which stops respiration in the cells lining the intestine.
SIMPLE SUGAR
Arabinose
Glucose
a)
b)
c)
d)
e)
RATE OF ABSORPTION (arbitrary units)
LIVING INTESTINE
POISONED INTESTINE
29
100
29
32
If respiration has been stopped, what is not available for the plant?
By what process do you think that arabinose passes through the intestine?
Explain your answer.
What does the data suggest about the absorption of glucose in the living intestine?
Explain your answer.
INVESTIGATING THE EFFECT OF OSMOSIS ON PLANT CELLS
MATERIALS AND APPARATUS
Rhubarb stem (or onion scales)
1M sucrose solution
Distilled water
Microscope
Filter paper
Microscope slides and coverslips
Forceps
Scissors
Scalpel
Mounted needle
BACKGROUND INFORMATION
Cytoplasm contains an aqueous solution. Cell membranes are partially permeable. Thus, if cells are put
in water or a strong solution osmosis should occur, i.e. water should enter cells if placed in water, or
leave the cells if placed in a strong solution. In which case the cytoplasm should expand and the cells
become turgid, or shrink and the cells become plasmolysed. If the cells contain coloured cytoplasm,
plasmolysis is easy to see. Rhubarb and some other plant materials have epidermal cells with coloured
cytoplasm, therefore they are useful to use when studying osmosis in plant cells.
METHOD
1. With forceps, strip off a piece of the red epidermis of the rhubarb stem. Make sure that you have
only epidermis. Any green tissue will be additional to the epidermis and will obscure epidermal
cells when you come to look at them under the microscope.
2. Trim the piece of epidermis with scissors until it is about 0.5 cm square.
3. Put the piece of epidermis in a drop of water on a slide and cover it with a cover-slip.
4. After a few minutes look at the preparation under low power of the microscope. Look at 20 cells
and describe their appearance. Then focus under high power and draw two or three cells.
5. With a teat pipette, place a drop of sucrose solution against one side of the cover-slip, then put a
piece of filter paper against the other side of the cover-slip and draw the sucrose solution across.
6. After 5 minutes look at the epidermal cells under the low power of the microscope. Examine 20
cells and describe their appearance compared with the preparation in water. Under high power,
draw two or three of the cells.
RESULTS
Present your results in the form of written descriptions of what you see under the microscope and
drawings of cells in each of the two liquids.
CONCLUSION
Describe what effect each solution has on the epidermal cells and explain your observations.
EVALUATION
Evaluate your investigation, mentioning any difficulties you encountered, how they were overcome,
and how accurate you think your results were. What safety procedures do you think are necessary in
this investigation?
OSMOSIS – COMPREHENSION EXERCISE
1. How does the term osmosis differ in meaning from the term diffusion?
2. Water, chloride ions, the sugar arabinose and potassium ions are all capable of passing through the
gut lining. State which of these materials pass through by diffusion, and which passes through by
osmosis.
3. a)
b)
c)
d)
What is meant by the term ‘partially permeable membrane’?
Give two examples of partially permeable membranes.
Give an example of a surface that is fully permeable.
Give an example, from a living organism, of an impermeable surface.
4.
a) What causes plasmolysis?
b) Why do animal cells not show plasmolysis?
c) Make a diagram of a plasmolysed cell.
5.
a)
b)
6.
Rhubarb epidermis from the same plant was cut into two 1cm square pieces, A and B. Epidermis
A was put in 0.5M sucrose solution, epidermis B was put in 0.05M sucrose solution. Which of the
two, A or B, would show the highest number of plasmolysed cells?
7.
List three factors that affect the rate of osmosis.
8.
Give two reasons why osmosis is important to plants.
9.
Suggest an explanation for each of the following statements:a) If certain kinds of lettuce get floppy, they can be made firm and crisp by putting them in cold
water for a while.
b) If you sprinkle sugar on a bowl of strawberries, the juice comes out of them.
Why do animals need to osmo-regulate?
Why is osmo-regulation less important to marine creatures?
10. A pupil in a school carried out the following experiment. With a cork-borer he cut two cylinders
of potato tissue from a fresh potato, and two cylinders from a boiled potato. He trimmed all four
cylinders to exactly 5 cm long. He then put a fresh potato cylinder and a boiled potato cylinder into
distilled water, and fresh and boiled cylinders into a strong sugar solution. After 4 hours he
measured the four cylinders. His results were as follows:-
Fresh Potato
Boiled Potato
LENGTH OF POTATO CYLINDER AFTER 4 HOURS
(cm)
IN SUGAR SOLUTION
IN DISTILLED WATER
4.2
5.8
5.0
Explain these results from your knowledge of osmosis.
5.0
11. A student wanted to find out the concentration of water inside carrot tissue (this is called the water
potential). She decided on a clever method. This involved cutting discs of carrot tissue, all of the
same diameter and thickness. She weighed the discs out into five 10 g batches, A, B, C, D and E.
She put batch A into distilled water, B into 0.05M sucrose solution, C into 0.1M sucrose solution,
D into 0.3Msucrose solution and E into 0.5M sucrose solution. She left them for 4 hours, quickly
blotted off surplus liquid and re-weighed them. Her results were as follows:BATCH
A
MASS AT START (g)
10
MASS AT END (g)
11.1
B
10
10.9
C
10
10.6
D
10
9.8
E
10
9.2
She then plotted these data on a graph.
She then used the graph to find out the concentration of water in the carrot tissue.
a)
b)
c)
Plot the above data on a graph.
Use the graph to find the concentration of water in the carrot tissue. This will be the point on
the graph where there is no mass change.
Explain carefully why the fact that there is no mass change tells you that this is the water
concentration inside the carrot tissue.
12. Three Visking tubing bags of the same size were prepared as follows:Bag A was filled with strong sucrose solution
Bag B was filled with starch ‘solution’
Bag C was filled with water.
All three bags were weighed.
Bags A and B were put in a beaker of water, bag C was put in a beaker of strong sucrose solution.
All three bags were left for 4 hours. After this time they were weighed again.
a) What change in mass (if any) would you expect from these three bags?
b) Give a scientific explanation to your answer above.
DIFFUSION DIAGRAM