Download Maintaining a Balance - The Bored of Studies Community

Maintaining a Balance
1. Most organisms are active in a limited temperature range
Identify the role of enzymes in metabolism, describe their chemical composition and use a simple model to
describe their specificity on substrates
o
Role of enzymes
Enzymes are biological catalysts

The manufacture of enzymes are controlled by the nucleus

Metabolism- chemical and physical processes by which energy and matter available for the organism’s use.

To regulate the chemical activity in the cells of the organisms

They lower the energy required to start a chemical reaction within a cell without getting used up.
o
 Allow reactions to occur at lower temperature
Chemical composition-
o
 Proteins called Enzymes
Simple model
Enzymes are globular proteins that have specialised shapes so that other chemicals (substrates) can form
a temporary bond with them.

The lock and key model-
Identify the pH as a way of describing the acidity of a substance

pH is a the amount of hydrogen ions measured in a substance (sol n)

pH of 7 is neutral e.g. water, 0-6 is acidic .e.g. lemon juice, 8-14 is basic .e.g. sodium bicarbonate
Explain why the maintenance of a constant internal environment is important for optimal metabolic
efficiency

Cells work best under their optimum conditions- right pH, temperature and the best concentration.

Enzymes control all metabolic processes in the body

Enzymes work in an environment where they’re optimum temperatures and pH conditions are met. At
temperatures and pH values other than the optimum, the enzymes fail to work as efficiently as they should
or not at all.
Describe homeostasis as the process by which organisms maintain a relatively stable internal environment

In order to maintain maximum efficiency of an enzyme, the body must maintain a stable/ constant internal
environment. This is known as homeostasis.

Homeostasis is the process by which the internal environment is kept within normal limits regardless, of the
external environmental conditions. This includes temperature, pH, gas levels, water and salt concentrations.
This allows the enzyme’s optimal conditions to be met and the body to work efficiently and kept stable
Explain that homeostasis consists of two stages:
o
Detecting changes
o
Detecting changes from the stable state
Counteracting changes from the stable state
Receptors detect changes to the normal internal environment. e.g. sensory neurons in the skin pick up a
decrease or increase in temperature of air surrounding the body.
Counteracting changes
After the receptors have detected these changes, action is taken to counteract these so that homeostasis is
maintained. These are done by effectors. e.g. shivering to generate heat in muscles.
Outline the role of the nervous system in detecting and responding to environmental changes

The co-ordinating system in humans is the nervous system.

The nervous system consists of the central nervous system (CNS) and the peripheral nervous system
(PNS). The CNS consists of the brain and spinal chord and the PNS consists of the sensory nerves and the
effector nerves. When the environmental temperature begins to exceed a comfortable level for the body.
The stimulus is detected and the sensory neurons send an impulse to the brain where the information is
interpreted and a response is initiated. This response is then sent to the effectors.

Example- Temperature sensors in the skin detect the temperature change (the stimuli) and a sensory
neuron conducts a nervous impulse to the hypothalamus found in the brain. Nerve impulses pass this
response from the receptors to effector neurons then onto effectors, such as blood vessels, sweat glands,
endocrine glands and muscles. This is when we shiver to try and generate heat.(counteract the stimuli)
Identify the broad range of temperatures over which life is found compared with the narrow limits for
individual species

Life forms can found in the temperature range of –40ºC to 120ºC

Majority of life forms are in the –2ºC to 40ºC temperature range and is narrower for each individual species.

Below 0ºC cell risk ice crystals forming and above 45ºC causes them to denature.
Compare responses of named Australian ectothermic and endothermic organisms to changes in the
ambient temperature and explain how these responses assist temperature regulation
o
Endotherms

Regulate body temperature using metabolism homeostasis. (birds and mammals)

Insulation- control of blood flow (capillaries dilate to keep you cool), evaporation (humans sweat to keep
cool), counter-current exchange (blood vessels placed together and chilled bloody returning from veins
gets heated up) and metabolic activity (during hotter weather metabolism slows down {by product of
this HEAT}) are ways that endotherms control their body temperature.

In hot conditions~ Example (1) Red kangaroo licks the inside of its paws, where skin is thinner, and
blood supply is closer to the surface, so that heat can be easily dumped to the outside. Evaporation
from saliva promotes the loss of heat from the blood. Example (2) The large ears of the rabbit-earedbandicoot provide a large surface area to pass excess heat when it is burrowing during the heat of day
and when it is active at dusk.
o
Ectotherms-

Body temperature fluctuates, according to ambient temperatures. (reptiles)

Hibernation- nocturnal activity, controlling exposure and migration are ways that ectotherms control
their body temperature.

In cold conditions~ Example (1) magnetic termites (Amitermes meridionalis) pack the walls of their
mounds with insulating wood pulp and align their mounds north-south to maximise exposure to the sun
in the mornings and the evenings when the air is cooler and to minimise exposure during heat of day.
Example (2) Bogong moths are able to avoid their bodies freezing by supercooling their tissues. This
process involves reducing the temperature of the body fluids below their usual point of freezing and as
a result, ice crystals do not form and destroy the cells.

Characteristics of ectotherms and endotherms:
Behavioural adaptations:
-
Insulation
-
Migration
Hibernation and aestivation
Physiological adaptations:
- Metabolic activity
-
Sheltering
Nocturnal activity
-
Control of blood flow
Counter-current exchange
-
Controlling exposure
-
Evaporation
Structural adaptations:
Identify some responses of plants to temperature change

Plants can be damaged at temperature extremes when enzyme structures are altered or
membranes change their proteins. As many enzymes are involved in photosynthesis and
respiration temperature extremes can be a major problem.

In cold conditions, extracellular ice formation causes dehydration. Some plants ca
tolerate temperatures as low as
-50ºC by altering the solute concentrations and through lack of ice-nucleating sites in
cells to prevent intracellular freezing.

In hot desert conditions, plants develop a compromise between access to gases for
photosynthesis and access to gases for respiration by keeping their stomates open and
cooling by evaporation. This risks dehydration.

In hot or cold plants may die but leave dormant seeds so to keep the flow going.

They may die above the soil but leave their roots embedded in the ground so that they
can keep living when conditions are good.

Too high temperatures during flower formation may cause poor crop

Some seeds may only germinate after a fire.

Example (1)- eucalyptus leaves grow vertically to reduce exposure to the sun. Example
(2)- porcupine grass-sand in central Australia have curled leaves that opens up after rainexposing the stomata. In dry conditions the leaf curls up, burying the stomata and
reducing transpiration ... conserving water.
Identify data sources, plan, choose equipment or resources and perform a first-hand
investigation to test the effect of:
- Increased temperature
-
Change in pH
-
Change in substrate concentrations on the activity of named enzyme(s)
FACTOR
EFFECT ON ENZYME ACTIVITY
Increasing temperature
Increases activity of the enzyme until it
denatures.
They
have
an
optimum
temperature.
Change in pH
Depends on the enzyme. Each enzyme has
they’re own optimum pH and it may denature.
Change in concentration of the substrate
The higher the concentration the faster the
reaction rate. The rate of reaction is limited by
the amount on enzyme present. (still works
after saturation point)
Lipase investigation
1. Effect on temperature
Aim: to determine the relationship the relationship between the effect of temperature and the
activity of the enzyme, lipase
Hypothesis: the enzyme lipase works best at a temperature of 35-40ºC.
Apparatus:

Full fat milk

8 test tubes

Test tube rack

5% lipase solution

0.05M sodium bicarbonate solution

Phenolphthalein

Thermometer

Large beaker with water

Hot plate

Measuring cylinders

Stopwatch

Pipettes
Method:
1. Place 50g of ice into a large beaker and test tubes 1 and 2 which contain 1ml of milk in
each test tube. Record the temperature.
2. Then add 11mls of sodium bicarbonate solution and 5 drops of phenolphthalein to each
test tube.
3. Add 1ml of lipase solution to test tube 1 and start timing. Note the colour changes of the
solutions at I minute intervals for 5 minutes from the time of addition of Lipase. Note: test
tubes 2,4,6 and 8 are controls (.i.e. they don’t have the enzyme added to them.)
4. Repeat steps 1-3 for test tubes 3 and 4 but this time use water at room temperature
approx. 20ºC
5. Repeat steps 1-3 for test tubes 5 and 6 but this time use a water bath set at 35ºC
6. Repeat steps 1-3 for test tubes 7 and 8 but this time use a water bath set at 50ºC
7. Repeat steps 1-6 three times and average your results.
8. Collate data in a table and graph format.
N.B- for the experiment to be fair the following things are to be kept constant:

Lipase %: a higher lipase concentration will obviously speed up the time it takes to break
down the fat.

Amount of milk/ lipase/ sodium bicarbonate/ water/ ice

Use the same stopwatch for each test
 Repeat the experiment 3 times
Results:
Effect of temperature on the enzyme, lipase
Time (mins):
Temperature
Test tubes:
(ºC)
1-
0
2- control
34-control
56- control
7-
0
20
20
35
35
50
0
1
2
3
4
5
Deep
Deep
Deep
Deep
Deep
Deep
milky
pink
milky
pink
milky
pink
milky
pink
milky pink
milky pink
Deep
Deep
Deep
Deep
Deep
Deep
milky
pink
milky
pink
milky
pink
milky
pink
milky pink
milky pink
Deep
Deep
Pale
Lighter
Lighter
Lighter
pink
pink
pink
pink
pink
pink
Deep
Deep
Deep
Deep
Deep
Deep
milky
milky
milky
milky
milky pink
milky pink
pink
pink
pink
pink
Deep
Pale
Lighter
Lighter
Lighter
White
pink
pink
pink
pink
pink
Deep
milky
Deep
milky
Deep
milky
Deep
milky
Deep
milky pink
Deep
milky pink
pink
pink
pink
pink
Deep
Deep
Deep
Deep
Deep
Deep
8- control
50
milky
milky
milky
milky
milky pink
milky pink
pink
pink
pink
pink
Deep
Deep
Deep
milky
milky
milky
Deep
Deep
Deep
milky
milky pink
milky pink
pink
pink
pink
pink
Conclusion: the milk solutions turn white when the lipase and milk come in contact to produce
fatty acids. Sodium bicarbonate makes the solution alkaline (pink) to start with but as more acids
are produced the pH drops and the solution turns white. Lipase works best at a temperature of
35ºC but beyond this denature, and becomes inactive. Active sites of enzymes have a particular
shape and because of this only one substrate molecule will fit into it. When an enzyme has lost its
shape because of heat, the shape of the active site changes so a substrate molecule will no
longer fit.
2. Substrate Concentration
Aim: to determine the relationship the relationship between substrate concentration and the effect
of the enzyme, lipase
Hypothesis: higher concentration milk will break down more quickly into fatty acids by the action
of the enzyme lipase.
Apparatus:

Range of 3 different milk concentrations

Large beaker

6 test tubes

Hot plate

Test tube rack

Measuring cylinders

5% lipase solution

Stopwatch

0.05M sodium bicarbonate solution

Pipettes

Phenolphthalein
The different milks are:
Type of milk:
Test Tube 1 and 2
1ml full fat milk
Test Tube 3 and 4
0.5ml of full fat milk and 0.5ml water
Test Tube 5 and 6
0.25ml of full fat milk and 0.75ml water
Method:
1. Set up a water bath at 35ºC with test tubes 1 and 2
2. Then add 11mls of sodium bicarbonate solution and 5 drops of phenolphthalein to each
test tube.
3. Add 1ml of lipase solution to test tube 1 and start timing. Note the colour changes of the
solutions at I minute intervals for 5 minutes from the time of addition of Lipase. Note: test
tubes 2,4 and 6 are controls (.i.e. they don’t have the enzyme added to them.)
4. Repeat steps 1-3 for test tubes 3 and 4.
5. Repeat steps 1-3 for test tubes 5 and 6.
6.
Repeat steps 1-5 three times and average your results.
7. Collate data in a table and graph format.
N.B- for the experiment to be fair the following things are to be kept constant:

Water bath at 37ºC

Lipase %: a higher lipase concentration will obviously speed up the time it takes to break
down the fat.

Amount of milk/ lipase/ sodium bicarbonate

Use the same stopwatch for each test
 Repeat the experiment 3 times
Results:
Effect of temperature on the enzyme, lipase
Time (mins):
0
1
2
3
4
5
Test tubes:
1- high conc.
of milk
Deep
pink
Pale
pink
Lighter
pink
Lighter
pink
Lighter
pink
White
2- control
Deep
Deep
Deep
Deep
Deep
Deep
milky
pink
milky
pink
milky
pink
milky
pink
milky pink
milky pink
3- med. Con.
Deep
Deep
Pale
Lighter
Lighter
Lighter
of milk
pink
pink
pink
pink
pink
pink
4-control
Deep
Deep
Deep
Deep
Deep
Deep
milky
milky
milky
milky
milky pink
milky pink
pink
pink
pink
pink
5- low conc. of
Deep
Deep
Pale
Pale
Pale pink
Pale pink
milk
pink
pink
pink
pink
6- control
Deep
milky
Deep
milky
Deep
milky
Deep
milky
Deep
milky pink
Deep
milky pink
pink
pink
pink
pink
Conclusion: the milk solutions turn white when the lipase and milk come in contact to produce
fatty acids. Sodium bicarbonate makes the solution alkaline (pink) to start with but more acids
are produced the pH drops and the solution turns white. The results indicate that the higher
the concentration of the substrate milk, the quicker the reaction of the enzyme, lipase.
Gather, process and analyse information from secondary sources and use available
evidence to develop a model of a feedback mechanism

In a feedback system, the response alters the stimulus. Feedback can be negative (when
the effect of the stimulus is reduced) or positive (when the effect of the stimulus is
increased). E.g. the control of hormone levels in the body, in which an increase in the
level of the hormone in the blood decreases the output by the gland.

A model of a feedback mechanism-
-
Analyse information from secondary sources to describe adaptations and responses that
have occurred in Australian organisms to assist temperature regulation
Sweat
Increase in
temperature
Decrease in
temperature
Hypothalamus
(the brain)
Decrease in
temperature
Increase in
temperature
Shivering
Australian organism
Endotherm
Adaptation or response to temperature regulation
or ectotherm
Red kangaroo
Endotherm
Licks the inside of its paws, where skin is thinner, and blood
supply is closer to the surface, so that heat can be easily
dumped to the outside. Evaporation from saliva promotes
the loss of heat from the blood.
The rabbit-earedbandicoot
Endotherm
Magnetic termites
(Amitermes
Ectotherm
The large ears of the rabbit-eared-bandicoot provide a
large surface area to pass excess heat when it is burrowing
during the heat of day and when it is active at dusk.
meridionalis)
Pack the walls of their mounds with insulating wood pulp
and align their mounds north-south to maximise exposure
to the sun in the mornings and the evenings when the air is
cooler and to minimise exposure during heat of day.
Bogong moth
Ectotherm
Able to avoid their bodies freezing by supercooling their
tissues. This process involves reducing the temperature of
the body fluids below their usual point of freezing and as a
result, ice crystals do not form and destroy the cells.
2. Plants and animals transport dissolved nutrients and gases in a fluid medium
Identify the form(s) in which each of the following is carried in mammalian blood:
-
Carbon dioxide
Oxygen
-
Water
Salts
-
Lipids
-
Nitrogenous waste
Other products of digestion
Substance
From
To
Form
Carried by
Oxygen
Lungs
Body cells
Oxyhaemoglobin
RBC’s
Carbon Dioxide
Body cells
Lungs
Hydrogen carbonate
ions, bicarbonate ions
RBC’s and
plasma
Water
Digestive
Body cells
Water molecules
Plasma
Body cells
As ions in the plasma
Plasma
Kidneys
Mostly as urea,
Plasma
system and
body cells
Salts
Digestive
system and
body cells
Lipids
Nitrogenous
Liver and body
waste
cells
Other products of
Digestive
digestion
system and
liver
sometimes ammonia or
uric acid
Body cells
As separate molecules,
Plasma
.e.g. glucose, amino
acids
Explain the adaptive advantage of haemoglobin
o
Haemoglobin
Large protein molecules found in RBC’s

Oxygen isn’t very soluble in water, which is why it’s carried in the haemoglobin

It increases the oxygen carrying capacity of RBC’s by about four times.

Mammals require a constant and large supply of oxygen to produce enough heat to
maintain homeostasis.

Supply can be adjusted to suit altitude

Can bind to oxygen loosely-therefore release it quickly

And advantage to be carried in and RBC. If just dissolved in plasma it would upset the
osmotic balance of the blood.

The development of RBC’s without a nucleus leaves more room for haemoglobin.
Compare the structures of arteries capillaries and veins in relation to their function
Artery
Vein
Thick, elastic,
Bigger
Capillary
Sketch
Description
muscular walls
Function
Reason
in
diameter
One
cell
thick
...
than arteries, but their
diameter
muscular wall is much
thinner.
larger surface area to
volume ratio.
Carry oxygen rich
blood from lungs to
Carry
blood
Exchange materials
between blood and
the entire body
heart.
body cells.
Blood
is
under
pressure and needs to
They are not under as
much pressure and
Large SA:V allows for
the easy exchange of
be pumped around to
only travel one way.
nutrients (needed by
the
body.
muscular
They are pushed up
through valves.
cells
and
products)
The
walls
deoxygenated
back to the
has
in
a
waste
expand and contract
to push the blood
through.
Describe the main changes in the chemical composition of the blood as it moves around
the body and identify tissues in which these changes occur

The blood circulates through two systems in the body: the pulmonary system and the
systemic system.
o
The pulmonary system-

Blood flows from the heart to the lungs and then back to the heart. Blood travels in the
pulmonary artery from the right ventricle to the lungs where carbon dioxide is released
into the alveoli of the lungs. This is then ultimately released out of the body. Oxygen is
picked up from the alveoli and diffused into the red blood cells to then be taken back to
the heart. So via the pulmonary system, carbon dioxide is decreased and oxygen levels
o

increased.
The systemic systemBlood flows from the heart to the rest of the body, except the lungs, and then returns. The
left ventricle pumps oxygenated blood to the rest of the body, and as this blood circulates
in capillaries, oxygen is delivered to the cells and carbon dioxide is picked up. Other
waste products, such as urea, are also picked up from the liver and transported in the
blood to the kidneys. Blood flowing to the small intestines collects the products of
digestion and transports them to the liver. Glucose is circulated in the blood stream to all
cells in the body for respiration. Deoxygenated blood returns to the heart via the inferior
and superior vena cava.
Outline the need for oxygen in living cells and explain why removal of carbon dioxide from
cells is essential
o
Oxygen
o
Needed for aerobic respiration to release energy
 A constant supply of oxygen is needed for the cells, otherwise they’ll die
Carbon dioxide

A bi-product of respiration is carbon dioxide

Increased carbon dioxide in blood stimulates the breathing centre in the brain, which is
why we pant after exercise. We take in more oxygen than we give out.

Carbon dioxide reacts with plasma (mostly water) to form carbonic acid (which is how it’s
carried around the body). If this becomes too much, the carbonic acid upsets the pH
level-making it more acidic, poisonous. However, we have special buffer systems to stop
this.
Describe current theories about processes responsible for the movement of materials
through plants in xylem and phloem tissue
o
Xylem
The transpiration-cohesion-tension mechanism is currently the theory that accounts for
the ascent of xylem sap. This sap is mainly pulled by transpiration rather than pushed by
root pressure. Cohesion is the “sticking” together of water molecules so that they form a
continuous stream of molecules extending from the leaves down to the roots. Water
molecules also adhere to the cellulose molecules in the walls of the xylem. As water
molecules are removed by transpiration in the leaf, the next molecule moves upwards to
o
take its place, pulling the stream of molecules continuously along.
Phloem
The pressure-flow mechanism is a model for phloem transport now widely accepted.
The model has the following steps.
Step 1: Sugar is loaded into the phloem tube from the sugar source, e.g. the leaf (active
transport)
Step 2: Water enters by osmosis due to a high solute concentration in the phloem tube.
Water pressure is now raised at this end of the tube.
Step 3: At the sugar sink, where sugar is taken to be used or stored, it leaves the phloem
tube. Water follows the sugar, leaving by osmosis and thus the water pressure in the tube
drops.
The building up of pressure at the source end, and the reduction of pressure at the sink
end, causes water to flow from source to sink. As sugar is dissolved in the water, it flows
at the same rate as the water. Sieve tubes between phloem cells allow the movement of
the phloem sap to continue relatively unimpeded.
Perform a first-hand investigation to demonstrate the effect of dissolved carbon dioxide
on the pH of water
Changing pH
Aim: to demonstrate the effect of dissolved carbon dioxide on the pH of water.
Apparatus:

Hydrochloric acid (0.1M)

Stopper

Calcium carbonate- powered

Measuring cylinder

Water

Stop watch

Conical

Universal indicator

scales
flask
with
side
arm
connecting tubes
and
 Beaker
Method:
1. set up apparatus as shown
2. measure 30ml of hydrochloric acid and pour it into the conical flask
3. Measure 40ml of water and place it in the beaker, add a few drops of universal indicator
and record pH.
4. Weigh 10g of calcium carbonate. Slowly pour into flask and quickly cover with stopper.
Once the calcium carbonate is in contact with acid begin timing. Make sure the tube is
placed in water.
5. Record pH, measure at one minute intervals for five minutes
6. Repeats steps 1-5 three times and record total average of results
7. Repeat steps 1-5. Controlled reaction.
8. Repeat step 7 three times and record total average of results.
Results:
Experiment 1
Time
pH
Experiment 2
Experiment 3
1
2
3
4
5
1
2
3
4
5
1
2
3
4
5
6.5
6.5
6
5.5
5.5
6.5
6
5.5
5
5
6
6
5.5
6
5
Conclusion:
When carbon dioxide dissolves in water, it produced carbonic acid, which causes a decrease in
pH, from pH of 6.5 to 5.5 in distilled water.
Perform a first-hand investigation using the light microscope and prepared slides to
gather information to estimate the size of red and white blood cells and draw scaled
diagrams of each
Analyse information from secondary sources to identify current technologies that allow
measurement of oxygen saturation and carbon dioxide concentrations in blood and
describe and explain the conditions under which these technologies are used
o
Biosensers
Made analysing blood gases quicker and more accurate.

A biosensor is a device with a transducer and a bioreceptor, usually one that recognises
specific biochemical molecules. The bioreceptor reacts specifically with the substance to
be detected and the transducer, which may be electrochemical, optical or thermal,
converts the biochemical signal into an electrical signal
o
Sensors
A sensor is a device that translates a physical or chemical property into an electrical
signal that can be measured. The key component is the transducer or signal-converting
element that converts the poverty to be measured into a signal. Sensors usually use
either optical or electrochemical technologies. Optical fibres are now replacing electrical
wire for carrying signals in many sensors.
o
Pulse oximeter (in hospitals) - a peg is attached to the finger of the patient where a light is
transmitted through to the other side of the finger. A photo detector on the other side
measures how much light has been transmitted through. The amount is directly proportional
to the amount of oxygen in arterial blood.
o
Arterial Blood Gas Analysis
Blood gas analysis, also called arterial blood gas (ABG) analysis, is a test which
measures the amounts of oxygen and carbon dioxide in the blood, as well as the acidity
(pH) of the blood.

An ABG analysis evaluates how effectively the lungs are delivering oxygen to the blood
and how efficiently they are eliminating carbon dioxide from it. The test also indicates
how well the lungs and kidneys are interacting to maintain normal blood pH (acid-base
balance).

Done to assess respiratory disease and other conditions that may affect the lungs, and to
manage patients receiving oxygen therapy (respiratory therapy).

The acid-base component of the test provides information on kidney function.

Blood gas analysis is performed on blood from an artery. It measures the partial
pressures of oxygen and carbon dioxide in the blood, as well as oxygen content, oxygen
saturation, bicarbonate content, and blood pH.

Testing the partial pressure of oxygen is actually measuring how much oxygen the lungs
are delivering to the blood. Carbon dioxide is released into the blood as a by-product of
cell metabolism. The partial carbon dioxide pressure indicates how well the lungs are
eliminating this carbon dioxide.

A related value is the oxygen saturation, which compares the amount of oxygen actually
combined with hemoglobin to the total amount of oxygen that the hemoglobin is capable
of combining with.

Procedure-

Oxygen Concentration In Blood
 Sample of blood taken
 Diffuses through a gas permeable membrane

This produces an electrochemical reaction

Which produces a current
This current of proportional to oxygen concentration


Carbon Dioxide Concentration In Blood
 Sample of blood is taken


Diffused through a gas permeable membrane
This changes the pH level in the solution

The change in pH is proportional to carbon dioxide concentration.
Analyse information from secondary sources to identify the products extracted from
donated blood and discuss the uses of these products
o
Donated blood
Broken down into:
 RBC’s- used to carry oxygen. Given to people with anaemia whose people don’t

make enough RBC’s or people who’ve lost a lot of blood
WBC’s- used to combat infection. Given to people with cancer of the blood e.g.leukaemia. Used vary rarely-usually antibiotics are used

Platelets-used for blood clotting. Given to people with cancer of the blood because
they don’t make enough platelets

Plasma-also used for blood clotting. Used to treat people with haemophilia. Used to
adjust osmotic pressure of blood and to pull fluids out of tissues.

Immunoglobins-infection fighting parts in plasma. Used to treat people with difficulty
fighting infection.

Whole blood-only given when >20% of blood is lost
Analyse and present information from secondary sources to report on progress in the
production of artificial blood and use available evidence to propose reasons why such
research is needed
o
Artificial Blood
Need to be stored at room temp. and have a prolonged shelf life; 3-4 weeks

World wide shortage of donor blood due to HIV/AIDS, Mad Cow Disease, new
screening etc

Need a safe and effective way to get a new blood source (considering there’s no donor
blood anymore)

Used to:
 Increase plasma volume- artificial plasma expanders are used for severe burns- so
the blood can clot and heal


Carry oxygen and carbon dioxide- not other substitutes for any other nutrient yet
Types

Perflurochemicals (PFC’s)
 Can dissolve about 50 times more oxygen than blood plasma
 Cheap and free of biological materials- no risk of infection
 To work must combine with other materials to mix with the blood stream-usually
lipids.

Haemoglobin- based oxygen carriers (HBOCs)
 Oxygen bonds chemically but only dissolves in PFC’s
 Not contained in a membrane-don’t require blood matching

PROBLEMS
 Doesn’t stop haemoglobin from oxidising (doesn’t have the enzymes to stop it)- once
oxidised it can’t carry oxygen

Membrane protects the haemoglobin from degradation and toxic effects of
haemoglobin

Alters blood flow through smallest vessels
Only stay in circulation for 20-30 hours, instead of RBC’s-100days


If in an accident you’ll usually be given saline (sodium chloride- same concentration as
blood and other tissues- 0.9%) OR dextrose- 4% glucose and 0.18% saline solution.
Choose equipment or resources to perform a first-hand investigation to gather first-hand
data to draw transverse and longitudinal sections of phloem and xylem tissue

Xylem carries water and minerals upward from the root hairs (where the water comes
from). Due to capillarity and the transpiration stream. They are dead tissues and narrow.

Phloem carries minerals produced by photosynthesis; mainly sugar, up and down the
plant.

Symplastic Loading- materials travel in the cytoplasm from the mesophyll cells to the
sieve element (the phloem) through plasmodesmata. This means a lot of
plasmodesmata are required.

Apoplastic Loading- nutrients travel through the cell walls until they get to the sieve
element. They then cross the cell membrane into the phloem. The sieve element
becomes loaded with sugars (nutrients) and dumps them into a sink cell. This keeps
pressure of the phloem constant.
3. Plants and animals regulate the concentration of gases, water and waste products of
metabolism in cells in interstitial fluid.
Explain why the concentration of water in cells should be maintained within a narrow
range for optimal function.
Water concentration in cells is critical for most living organisms. It must remain constant as slight
changes may lead to cell death. This is because:

Water is essential for life. Water is the solvent for all the metabolic reactions in living
cells. It takes part directly in many of them such as photosynthesis and is formed as a
product in many others including respiration.

Living cells function best in an isotonic environment (one in which the solute
concentration is the same both inside and outside the cell). They are very sensitive to
changes in solute concentration and as consequence may lose or take in large amounts
of water by osmosis.

Changes in concentration of water in the cell will affect the concentration of dissolve
substances, which in turn can affect the metabolic function.
Explain why the removal of wastes is essential for continued metabolic activity.
Wastes products are constantly being formed as a result of metabolic processes that occur in
cells. However, the removals of these wastes are essential because:
 Some of the wastes build up as toxins and could poison the cells. For example:
o
o

Carbon dioxide
Nitrogenous wastes
If allowed to accumulate in cells and tissues, these wastes could disrupt or slow down
metabolic reaction rates
Different animals excrete different wastes products
Aquatic animals, fish and invertebrates mostly excrete ammonia. Ammonia is toxic, but can be
released continuously and directly into the water and is quickly dispersed
Terrestrial animals excrete nitrogenous waste as either urea or uric acid. This is because
terrestrial animals need to conserve water by converting ammonia into less toxic forms and
excrete it periodically. Urea is soluble and is released in urine while uric acid is almost insoluble
and non-toxic.
Identify the role of the kidney in the excretory system of fish and mammals.
Kidney is an organ of filtration, reabsorption and secretion. The primary role of the kidneys is
osmoregulation, the regulation of the water and salt concentrations in the body.
The kidney forms urine by removing wastes such as nitrogenous wastes, salts, other unwanted
metabolic products and excess water, from the blood. The kidney maintains the balance of salts
and water in the body, and so has a vital role in homeostasis.
NB: The structural and functional unit of the kidney is the nephron. There are approximately 1
million nephrons in each human kidney. Each nephron is made up of glomerulus and tubules.
Explain why the processes of diffusion and osmosis are inadequate in removing dissolved
nitrogenous wastes in some organisms.
The processes of diffusion and osmosis are inadequate for removal of wastes because:


Diffusion is too slow and non-selective of solutes
Diffusion would mean all salts would be eliminated along with glucose and vitamins;
whereas the body needs to retain some salts and nutrients

Osmosis would mean that wastes would stay in the body and that water would leave
Distinguish between active and passive transport and relate these to processes occurring
in the mammalian kidney.
In the kidneys both forms of transport are used in regulating the body fluid composition.
Passive transport requires no energy. Passive transport occurs in filtration and in the osmosis of
water back into the blood.
Active transport requires energy from metabolism (ATP). Active transport occurs in the secretion
of substances into the nephron, the active transport of nutrients back into the blood, and the
selective reabsorption of salts required by the body. These processes require energy, as they
would have to go against the concentration gradient.
Explain how the processes of filtration and reabsorption in the mammalian nephron
regulate fluid composition.
Filtration and reabsorption occurs in the nephron.
Filtration involves the removal of substances from blood if they are small enough to be forced
through the glomerulus and into the Bowman’s capsule. The glomerulus acts like an ultra filter
and particles that are too large such as proteins cannot pass through the Bowman’s capsule.
Filtered blood then moves along the tubules. Useful substances such as water, glucose, amino
acids, vitamins, hormones and inorganic salts are reabsorbed through diffusion and osmosis.
Hence, the processes of filtration and reabsorption regulate body fluid composition as it perform
the complex balancing of retaining essential substances and removing toxic wastes from blood to
maintain homeostasis.
Outline the role of the hormones, aldosterone and ADH (anti-diuretic hormone) in the
regulation of water and salts levels in blood.
Adolsterone is a steroid hormone produced by the adrenal cortex of the kidney. Its role is to
maintain a balance of water and salts in the body. It stimulates the nephron to increase the
concentration of sodium ions leading to a decrease in reabsorption of potassium ions and more
water diffusing into blood at the nephron. This causes a rise in blood pressure and volume.
Anti-diuretic Hormone (ADH) is a hormone produced by the hypothalamus and stored in the
pituitary gland that stimulates the nephrons to absorb more water. This acts to decrease urine
volume.
Define enantiostasis as the maintenance of metabolic and physiological functions in
response to variations in the environment and discuss its importance to estuarine
organisms in maintaining appropriate salt concentrations.
Enantiostasis is the maintenance of metabolic and physiological functions in response to
variations in the environment.
To estuarine organisms the maintenance of salt concentrations is important. Many of the
organisms cannot control salt and water levels, instead they exhibit enantiostasis in order to
survive the daily change in salinity.
Ways in which some estuarine organisms function to overcome daily change in salinity:


Fast-swimming organisms can move away from area with high salinity
Molluscs can close their shells


Bottom dwellers burrow or dig deep into mud or sand
Halophytes tolerates changes in salinity by having a special mechanism to control their

level of slat
Saltbushes have special salt excretion glands on their leaves

Some mangroves excrete salt from special glands in their leaves
Describe adaptations of a range of terrestrial Australian plants that assist in minimizing
water loss
Some adaptations to limit water loss that Australian plants exhibit include:
 Hard or thick waxy cuticles on leaves, such as eucalypts

Hairy leaves, stems and even flowers to restrict air flow and evaporation, such as alpine

groundsel
Leaves that droop or roll to reduce the exposure of stomates, such as spinifex

A tough, woody structure that prevents plants wilting even when they lose water, such as
alpine groundsel


Small leaves, such as saltbushes
Leaves with a reduced number of stomates

Widely spreading or deep root systems to obtain more water, such as mulga
Perform a first-hand investigation of the structure of a mammalian kidney dissection, use
a model or visual resource and identify regions involved in the excretion of waste
products.
Gather, process and analyse information from secondary sources to compare the process
of renal dialysis with the function of the kidney.
Kidney function
Renal dialysis
A natural body process
An artificial process replacing damage kidney
Performed by two fist-sized organs
Performed by a large machine attached to a
variety of computer and other equipment
Removes wastes continuously
Performed repeatedly under hospital conditions
(two or three times each week, for several
hours each time)
Varies output automatically, depending on
concentrations of wastes in blood
Concentrations of substances in blood and
dialysis fluid are monitored by computers so
that most wastes are removed during treatment
Wastes may be removed by both diffusion and
active transport
Wastes removed by diffusion
Present information to outline the general use of hormone replacement therapy in people
who cannot secrete adolsterone.
Aldosterone is used to regulate water and salt reabsorption. When aldosterone cannot be
secreted, the excretory system will not be as efficient. The person cannot maintain homeostasis
and become severely dehydrated. The hormone replacement is taken on a regular basis, to
maintain balance of salts. The main artificial substitute for adolsterone is called fludrocortisone.
Appropriate hormone replacement therapy can enable patients to manage symptoms such as
fluid retention and high blood pressure and lead normal lives.
Analyse information from secondary source to compare and explain the differences in
urine concentration of terrestrial mammals, marine fish and freshwater fish.
Type of animal
Urine components and
concentration
Explanation
Terrestrial mammal
Concentrated urine, usually
Excess salts and other wastes are excreted
(e.g. bilby)
composed of urea, salts,
other wastes and water.
dissolved in water. Water needs to be
conserves, urine produced is concentrated.
Nitrogenous wastes present as urea - it is
less toxic than ammonia and can be present
in higher concentration
Freshwater fish
Large quantities of very
Freshwater fish absorb large volumes of
(e.g. native bass)
dilute urine, usually
composed of ammonia,
water through gills and mouth thus much
water must be excreted. Ammonia is suitable
small amounts of salts and
large amount of water
- sufficient water to dilute it. Salts - low
concentration in fresh water, therefore fish
take up salts from water as replacement.
Marine fish
(e.g. whiting)
Small quantities of
concentrated urine, usually
Marine fish constantly lose water - high salt
environment. Excrete little water in
composed of
trimethylamine oxide, other
concentrated urine containing high levels of
non-toxic trimethylamine oxide and salts.
wastes and small volumes
of water
Use available evidence to explain the relationship between the conservation of water and
the production and excretion of concentrated nitrogenous wastes in a range of Australian
insects and terrestrial mammals.
Type of nitrogenous wastes (uric acid or urea) and its high concentration enable these organisms
to reduce the amount of water they lose to remove wastes. This helps them to conserve water in
harsh and dry environments.
Process and analyse information from secondary sources and use available evidence to
discuss processes used by different plants for salt regulation in saline environments.
Mangrove:
 Shrubby tree that grows in estuaries

Its roots have a layer of cells that actively restrict the movement of salt into xylem vessels

Able to excrete salt through the underside of its leaves. Salt crystals accumulate on
leaves and so salt is lost when older leaves fall from plant
Saltbush:
 Tolerate salinity levels that kill most other plants

Excrete large amount of salt through their leaves
In general plants removed salt for regulation by:

Salt can be redirected towards drying leaves, so when drop off the plants, the salt is
removed

Salt excretion glands actively excrete salt by allowing it to crystallize and be blown or
washed away

Osmotic adjustment
Perform a first-hand investigation to gather information about structures in plants that
assist in the conservation of water.