Download Result- Reach Your Potential

Module: 9.2 Maintaining a balance—Practice Exam Questions
Question and mark allocation
Dot point(s)
addressed
Verb(s)
addressed
Q1
A student investigated the effect of temperature on enzyme action. He
noted that when the temperature of an enzyme/substrate mixture was
5ºC the rate of reaction was low. The activity increased as the
temperature increased until a maximum activity was reached at 37ºC. In
a similar experiment, the student heated a substrate to 70ºC, then added
the enzyme. The student recorded the activity as the mixture cooled.
The rate of reaction was zero at 70ºC. When the student cooled the
mixture to 35ºC, there was still no reaction. Explain the observations
made by the student and account for the enzyme activity at different
temperatures. (6 marks)
9.2.1.1 9.2.1.10
Explain
H14.1 a, b, c
Account for
Higher-level answer
Higher-level
answer
explanation
The student approached 37ºC from two Strengths:
different directions to show the nature Student has
of enzymes. Enzymes do not work
accounted for the
very well at low temperatures as there action of enzymes
is little kinetic energy in the reaction
at high and low
mixture. Enzymes and substrates need temperatures. The
energy for collisions to occur and the
answer
reaction to take place. As the
differentiates
temperature increases, the movement
between the
of particles increases and more
situation at cold
collisions between the substrate
temperatures and
molecules and enzyme molecules
that at high
occur, thus the activity increases. At
temperatures,
35ºC, maximum activity occurs. This
explaining why
was shown in the student’s
there is a
observations. When temperatures are
difference.
very hot, the enzymes denature. This
means that the structural shape of the
protein is destroyed and the substrate
molecule and enzyme molecule won’t
‘fit’ together, so the enzyme can’t
perform its role. As the student
decreased the temperature, he found
that the reaction didn’t occur when the
temperature fell to 35ºC. This was
because the enzymes were denatured
at 70ºC. Once this happens, they will
no longer work. So, enzymes can work
if the temperature is cold and they are
heated to 35ºC, however, if the
enzymes are heated to 70ºC, they are
destroyed, never to work again.
Copyright © Pearson Australia 2010
(a division of Pearson Australia Group Pty Ltd)
Lower-level
answer
Enzymes don’t
react at cold
temperatures
because they are
not at body
temperature 37ºC,
which is the
optimum for
enzyme action.
When enzymes get
too hot, they don’t
react either
because they are
not at body
temperature which
is the best
temperature for
enzyme action.
When it is too cold
or too hot, the
enzymes are
destroyed.
Lower-level
answer
explanation
Strengths:
Student recognises
that body
temperature is 37ºC
and shows an
understanding of
optimum reaction
rate. Student also
recognises that hot
temperatures will
retard enzyme
action.
Areas for
improvement:
Student has not
explained the
difference between
hot and cold
temperatures and
enzyme action.
Student has not
shown
understanding of
the different nature
of enzymes in cold
and hot
temperatures, and in
fact, is incorrect
saying that enzymes
are destroyed at
cold temperatures.
Student has not
linked observations
to explanations
satisfactorily.
Heinemann Biology Third Edition HSC
ISBN 978-1-4425-2819-2
Page 1 of 13
Question and mark allocation
Dot point(s)
addressed
Verb(s)
addressed
Q2
During your Biology course, you planned and performed a first-hand
investigation to test the effect of change in pH on the activity of a
named enzyme.
a Outline your procedure. (3 marks)
b Describe your results. (2 marks)
c Explain your results. (2 marks)
9.2.1.1 9.2.1.2
Outline
9.2.1.10
Describe
H11.2 a, b
Explain
Higher-level answer
Higher-level
answer
explanation
We investigated the effect of pH on
Strengths:
the action of the enzyme catalase
a • Student has
(from potatoes) on the substrate
clearly stated
hydrogen peroxide (H2O2).
the enzyme
a Solutions of varying pH (1, 3, 5, 7,
and substrate
9 and 11) were prepared by our lab
tested in the
assistant. To 5 mL of each
investigation.
solution, 5 mL H2O2 was added. A
Procedure is
small piece of potato was placed in
logical and
the test tube. A drop of detergent
coherent and
was added to each test tube.
written in the
Independent variable: pH of the
past tense.
substrate solution
• Independent,
Dependent variable: reaction rate
dependent and
measured by the height of bubbles
controlled
in the test tube.
variables are
Controlled variables:
outlined
Size of test tube, cores of potatoes
correctly. The
(3 cm long), volume of substrate (5
controls are
mL), volume of pH buffer (5 mL)
explained.
solution, amount of detergent
• Reliability
added, temperature of test tubes.
was gained
Control test tubes:
through class
Six test tubes containing one of
repeats of the
each of the pH solutions + H2O2
experiment.
(and no enzyme).
• Safety
Repetition ensured reliability as the
considerations
different groups in the class
are justified.
performed exactly the same
b • Results are
experiment and results from 6
presented in a
groups were averaged.
table.
Safety considerations: care was
• A statement
taken using the acid and base
after the table
solutions and the H2O2. Aprons and
summarises
goggles were worn to protect
the results.
clothing and eyes. Care was also
c Student relates
taken using the potato corers to
results to an
make sure we didn’t cut our
explanation of
fingers.
the behaviour
of enzymes.
There is a
Copyright © Pearson Australia 2010
(a division of Pearson Australia Group Pty Ltd)
Lower-level
answer
Lower-level
answer
explanation
a We got 3
a
different test
Strengths:
tubes and put
Student outlines a
acid in one,
valid experiment.
water in one
There is an attempt
and base in the to control the
third. We then variables with detail
added a small
about volume of
(pea sized)
H2O2 and size of
piece of liver
piece liver.
to each test
Areas for
tube. Then we improvement:
added 5 mL
• Student needs to
H2O2 to each
name the enzyme,
test tube. We
describe the
measured the
independent,
reaction and
dependent and
ranked each
controlled variables
test tube on
and control. There
reaction rate.
is no mention of
b The highest
what the student is
reaction rate
actually measuring
was in the one to see the rate of
with water.
reaction.
The one in acid • Needs to consider
and the one in
safety and
base didn’t
repetition. A greater
react as much. range of pH values
c The optimum
would also be
pH for the
better, not just 3.
enzyme we
b
investigated
Strengths:
was 7. The
Student has shown
enzyme didn’t that the reaction
work as well in rate is greatest
acid or base.
in water.
Enzymes have Areas for
their own
improvement:
optimum range There is no
where they
measurement of
work the best
reaction rate, just a
Heinemann Biology Third Edition HSC
ISBN 978-1-4425-2819-2
Page 2 of 13
Question and mark allocation
Dot point(s)
addressed
Verb(s)
addressed
Higher-level answer
b
Figure 1 effect of pH on action of catalase
The results show that the maximum
rate of reaction occurred when the pH
of the substrate solution is 7. The
reaction rate is low at pHs furthest
from 7, in the acidic and basic regions
of the pH scale (1 and 11), so we
could conclude from our experiment
that the optimum pH for the enzyme
catalase is 7.
c pH is the acidity of a solution.
Enzymes work best within small pH
ranges. In acidic or basic regions
outside their optimum, the enzymes
are denatured. The protein structural
shape is destroyed, so the enzyme’s
active sites are not functional. This
was reflected in our experiment. The
optimum pH for catalase is 7. Either
side of this pH, the reaction rate (as
measured by the height of bubbles)
decreased.
Copyright © Pearson Australia 2010
(a division of Pearson Australia Group Pty Ltd)
Higher-level
answer
explanation
conclusion
which states
the optimum
pH range for
the named
enzyme.
Lower-level
answer
Lower-level
answer
explanation
and when they general qualitative
are out of those statement. More
conditions the specific results
enzymes don’t should be presented
work.
in a table.
c
Strengths:
Student recognises
that the optimum
pH value was 7.
Areas for
improvement:
Simply saying
enzymes don’t
work is not specific
enough. Student
needs to say that
enzymes are
denatured outside
of their optimum
range and therefore
the protein loses its
shape and cannot
function.
Heinemann Biology Third Edition HSC
ISBN 978-1-4425-2819-2
Page 3 of 13
Question and mark allocation
Dot point(s)
addressed
Verb(s)
addressed
Higher-level answer
Q3
9.2.1.1 9.2.1.10
Explain
9.2.1.4
Evaluate
9.2.1.5
Compare
We tested the effect of liver (a source
of the enzyme catalase) on the
substrate hydrogen peroxide (H2O2).
We changed the concentration of
substrate solution and added the same
amount of liver to each of 8 test tubes
containing the different concentrations
of H2O2. We recorded the height of
bubbles after two minutes for each test
tube then graphed the results.
(Concentration of substrate on the xaxis and height of bubbles (amount of
reaction) on the y-axis. We generated a
graph similar to that in figure 2. As
concentration of substrate increases, so
too does reaction rate. This is because
increased concentration means more
collisions between molecules of
substrate and enzyme, thus more
reaction and formation of substrateenzyme complexes. When the
concentration reaches a certain
amount, however, the reaction rate
increases no further. This is because
there are not enough active sites
available for any more substrate
molecules. The enzyme’s active sites
have become saturated. Increasing
substrate concentration any more will
not increase the rate of reaction.
Negative feedback occurs when the
output of a system acts to oppose
changes to the input of the system.
Figure 3 shows a good example of a
feedback mechanism. It is a helpful
model in gaining understanding of
how feedback mechanisms work. Like
in biological feedback mechanisms,
there is a detection phase (via the
sensor), a control phase (the
Figure 2 Effect of substrate concentration on enzyme action
Figure 2 shows the effect of substrate concentration on enzyme action.
Explain the shape of the graph making specific reference to a named
enzyme and substrate you investigated. (4 marks)
Q4
9.2.1.6
9.2.1.11
Figure 3 Negative feedback model
A student developed a model of a negative feedback mechanism. She
Copyright © Pearson Australia 2010
(a division of Pearson Australia Group Pty Ltd)
Higher-level
answer
explanation
Strengths:
• Student outlines
their first-hand
investigation
results.
• Student states
trend of graph and
comments that it
was the same for
their
investigation.
• Student explains
the trends in the
graph; i.e. the
increase and the
plateau of the
graph.
Lower-level
answer
Strengths:
• Student has
defined negative
feedback.
• Student has
identified the
various features of
a control system
(detection, control
and response).
When humans eat
too much sugar,
the sugar is stored
for later. A
message goes to
the brain to say
that levels in the
blood are high.
When the body
needs sugar, the
As substrate
concentration
increases, the rate
of reaction
increases, and then
levels off. We
studied rennin and
milk and had
different
concentrations of
milk. We found
that when the
concentration of
the milk was
greater, the rennin
took less time to
react and go
lumpy.
Lower-level
answer
explanation
Strengths:
Student has made
reference to their
first-hand
investigation and
made an
observation stating
that increased
concentration
results in increased
rate of reaction.
Areas for
improvement:
• Student has only
stated the trend of
the graph. There is
no attempt to
explain it.
• There is no
explanation of why
the graph plateaus
as concentration
continues to
increase.
• Student should
provide a more
detailed account of
their first-hand
investigation.
Strengths:
• Student has a
basic understanding
of feedback,
recognising the
detecting and
responding phases,
(although not
stating as much).
• An attempt has
Heinemann Biology Third Edition HSC
ISBN 978-1-4425-2819-2
Page 4 of 13
Question and mark allocation
Dot point(s)
addressed
Verb(s)
addressed
created the flowchart in figure 3 to show her understanding of the
thermostat in a heating system. Evaluate the use of this model in
understanding feedback systems in humans. Use an example of a
negative feedback mechanism you have studied in the human body and
compare it with the model in figure 3. (6 marks)
Q5
Discuss the adaptations of ONE named Australian endotherm and ONE
named Australian ectotherm to hot temperatures. (4 marks)
Higher-level answer
thermostat) and a response phase (the
heater being switched on). As the
temperature in the room increases
above the ‘set point’, information is
fed back to the sensor, which detects
an increase in temperature. That
increase causes the heater to stop.
Similarly, the temperature goes down,
is detected and the heater is switched
back on. In humans, temperature is
regulated in a similar fashion.
Temperature is detected by
thermoreceptors in the skin (not
thermometers). They send information
to the control centre–hypothalamus (as
opposed to the thermostat) and the
hypothalamus sends a message to the
effector–piloerector muscle (not an
electric switch) and a response is
initiated. The skin heats up and
information is sent back to the
hypothalamus via an electrochemical
signal to stop the response. The
messages in the thermostat are sent via
electricity in copper wires. In the body
messages are sent along
interconnecting nerve cells.
9.2.1.8 9.2.1.12
Discuss
The bilby (Macrotis lagotis) is an
endotherm meaning it maintains its
own body temperature. It has very
large ears. These ears are very thin and
highly vascularised and allow the bilby
to dump heat over a large surface area.
The bilby is nocturnal, which is a
behavioural adaptation to avoiding the
heat of the day. It feeds at night when
it is cooler.
The diamond python (Morelia spilota)
is an ectotherm meaning its
Copyright © Pearson Australia 2010
(a division of Pearson Australia Group Pty Ltd)
Higher-level
answer
explanation
• Student has
shown
understanding of
the above features
in both the
thermostat model
and temperature
control in the
human body.
• Student has
compared both
systems (i.e.
outlined
similarities and
differences).
• Student has
included a value
statement to show
whether the model
is valuable or not.
Strengths:
Student has given
two adaptations
for each animal
for two marks
each. Use of
scientific
language is
evident with the
scientific names
of the organisms
and terms such as
Lower-level
answer
Lower-level
answer
explanation
reserves are
been made to
broken down into
compare the
sugar again. The
thermostat model
model is like this
with the blood
because our brain
sugar mechanism.
is like the
Areas for
thermostat. The
improvement:
sugar levels are
• There is a lack of
like the
coherency and logic
temperature
in the first
(always going up
explanation of
and down).
blood sugar. The
student should work
on a more logical
explanation and
endeavour to use
more scientific
language.
• The comparison
of the two systems
shows the
similarities but not
the differences
between the
systems.
• No attempt has
been made to
include a value
statement.
The red kangaroo
Strengths:
licks its paws to
The student has
cool down during
correctly identified
the heat of the day. two adaptations of
The spinifex
named Australian
hopping mouse has animals, the red
a long Loop of
kangaroo and the
Henle which helps brown snake.
it to not lose too
Areas for
much water in the improvement:
desert where it is
• Whilst the student
hot. The brown
has given true
Heinemann Biology Third Edition HSC
ISBN 978-1-4425-2819-2
Page 5 of 13
Question and mark allocation
Dot point(s)
addressed
Verb(s)
addressed
Higher-level answer
temperature varies with the ambient
temperature. It also retreats from the
direct sun and seeks shade under rocks
if it is becoming too hot. It will also
hunt at night to avoid the heat of the
day. (4 marks)
Q6
You performed a first-hand investigation to demonstrate the effect of
carbon dioxide on the pH of water. Briefly outline your method and
discuss your results. (5 marks)
9.2.2.7
Outline
Discuss
A conical flask and delivery tube were
set up. 2 g calcium carbonate was
added to 20mL 1 M HCl in the conical
flask. The delivery tube was placed in
a solution of lime water. The
limewater went cloudy, showing that
carbon dioxide was present. Next, we
set up the same apparatus but replaced
the limewater with a solution of
universal indicator and water. This
time, the UI went orange then pink.
Copyright © Pearson Australia 2010
(a division of Pearson Australia Group Pty Ltd)
Higher-level
answer
explanation
vascularised and
nocturnal and
surface area.
Strengths:
• Student has
given a brief
outline of method,
including a
control. The other
very strong part to
the answer is a
series of chemical
equations which
shows the
Lower-level
answer
Lower-level
answer
explanation
snake hides under information about
rocks so that it is
the spinifex
out of the sun.
hopping mouse, the
(2 marks)
information is
irrelevant. The
adaptation is for
water balance NOT
temperature
regulation. This is a
common mistake
students make. Be
sure to answer the
question.
• Also, the question
is worth 4 marks,
which suggests
TWO pieces of
information are
required for each
named animal. The
student has only
given one for each.
The student should
also aim to use
more scientific
language. The
examples used
should be identified
as endotherm or
ectotherm.
We got a straw and Strengths:
blew into a beaker Student has
of water and
outlined a valid
universal indicator. experiment and has
The water went
briefly outlined
yellow after a
results.
minute. Yellow is Areas for
a pH of 6.
improvement:
(3 marks)
• Student needs to
include a control (a
beaker of water and
Heinemann Biology Third Edition HSC
ISBN 978-1-4425-2819-2
Page 6 of 13
Question and mark allocation
Dot point(s)
addressed
Verb(s)
addressed
Higher-level answer
This showed that carbon dioxide
makes water become acidic with a low
pH of 3–4.
Our control was a test tube of distilled
water with UI. This remained green
(neutral pH 7).
Reaction in flask: CaCO3 +2HCl →
CaCl2 + H2O + CO2
Reaction in limewater: CO2 + Ca(OH)2
→ CaCO3 + H2O
Reaction of carbon dioxide with water
to form carbonic acid:
CO2+ H2O → H2CO3
(5 marks)
Q7
9.2.2.8
Estimate
Show
working
Count of the number of red blood cells
across the labelled line of Field of
View for the image shown = 15.
Field of view = 120 µm
120/15 = 8
Therefore the RBC is 8 μm wide.
The WBC is wider than the RBC so I
estimate it as 11–12 μm, using the
scale below the diagram.
(3 marks)
120 µm
Figure 4 Human blood
Copyright © Pearson Australia 2010
(a division of Pearson Australia Group Pty Ltd)
Higher-level
answer
explanation
production of
carbon dioxide
and the
production of
acid. (Chemical
equations are not
necessary in
Biology to get full
marks if enough
specific detail is
given.)
• The student has
discussed the
results showing
clear
understanding of
colour change of
UI, pH and acidity
of the resulting
solution.
Areas for
improvement:
Student could
include a diagram
of apparatus—
conical flask and
delivery tube.
Strength:
Student has
shown how she
calculated the size
of each cell. She
has calculated the
sizes correctly
using the scale
provided and
shows
understanding of
field of view.
Areas for
improvement:
Lower-level
answer
Lower-level
answer
explanation
UI). Student also
needs to state that
exhaled air contains
carbon dioxide,
therefore the
method is valid.
• Student states that
the pH is 6 but does
not mention that
this is acidic, which
is he whole point of
the investigation. It
is very important to
explain cause and
effect.
The red blood cells
are 8 microns and
the white blood
cells are 15
microns. (1 mark)
Strength:
Student knows the
size of red and
white blood cells.
Areas for
improvement:
Student has not
answered the
question showing
working. There is
no indication that
the student knows
how to calculate
field of view, count
Heinemann Biology Third Edition HSC
ISBN 978-1-4425-2819-2
Page 7 of 13
Question and mark allocation
Dot point(s)
addressed
Verb(s)
addressed
Higher-level answer
9.2.2.4
Outline
a Lungs
blood into lungs – (low in oxygen,
high in carbon dioxide)
blood out of lungs – (high in O2, low
in CO2)
b Kidneys
Blood into – (high in urea)
Blood out of – (low in urea)
c Digestive tract
Blood into tract – (low in digestive
products: glucose, amino acids)
Blood out of tract – (high in glucose,
amino acids)
d Muscles
Blood into muscles – (high in glucose
and O2)
Respiration occurs in muscle cells
(C6H12O6 + 6O2 → 6H2O + 6 CO2 )
Blood out of muscles – (low in
glucose, high in CO2)
(8 marks)
Using figure 4 above, estimate the size of the red and white blood cells.
Show all working. (3 marks)
Q8
Outline the main changes in the chemical composition of the
blood as it moves through
a the lungs
b the kidneys
c the digestive system
d the muscles
(8 marks)
Copyright © Pearson Australia 2010
(a division of Pearson Australia Group Pty Ltd)
Higher-level
answer
explanation
Student should do
a few estimates to
get a more reliable
result. For
example
trial 1: 120/16
trial 2: 120/13
trial 3: 120/15
average = 8.2 µm.
It is important to
take a calculator
to the exam for
questions like this.
Strengths:
Student has
outlined in an
excellent
summary, the
changes to the
blood in and out
of each organ.
The use of the
respiration
equation is very
good to show why
there is the change
in the muscle
cells.
Areas for
improvement:
Perhaps answer
could be better
set-out in a table
or a fully labelled
diagram.
Lower-level
answer
Lower-level
answer
explanation
cells and estimate
sizes of cells under
a microscope.
Students must show
working. It is easy
to rote learn sizes of
blood cells but this
question is not a
recall question but a
skills question
testing microscope
skills.
In the lungs, the
blood gets oxygen.
The kidneys filter
the blood and the
kidney gets rid of
the urea. In the
digestive system,
that is where all of
the food gets
processed. In the
muscles, glucose
and oxygen
combine to
produce energy
and carbon
dioxide. So the
blood going to the
muscles is high in
oxygen and
glucose and the
blood going back
to the heart from
the muscles is high
in carbon dioxide.
(4 marks)
Strengths:
Student shows
limited
understanding of
changes to the
blood across the
lungs and kidney so
receives only 1 of a
possible 2 marks for
a and b. Student
shows thorough
understanding of
blood changes
occurring in muscle
tissue and receives
2 marks for this
section of the
answer.
Areas for
improvement:
• Student needs to
set out answer more
coherently with a, b
answers delineated.
• Student needs to
outline what
happens to the
content of the blood
Heinemann Biology Third Edition HSC
ISBN 978-1-4425-2819-2
Page 8 of 13
Question and mark allocation
Dot point(s)
addressed
Verb(s)
addressed
Higher-level answer
Higher-level
answer
explanation
Q9
Explain why the removal of metabolic wastes is essential for continued
metabolic activity and outline how metabolic wastes are discharged
from the body. (6 marks)
9.2.2.5
Explain
9.2.3.2
Outline
Two main waste products that affect
metabolism are carbon dioxide and
ammonia.
Carbon dioxide makes the blood acidic
by producing hydrogen ions according
to the equations:
CO2 + H2O → H2CO3 ↔H+ + HCO3–
Nitrogenous waste – ammonia makes
the blood basic by producing
hydroxide ions according to the
equation:
NH3 + H2O → NH4+ + OH–
If the blood becomes too acidic or
basic, enzymes may become denatured
and cease to work, thus affecting
metabolic reactions and decreasing
metabolic activity.
Carbon dioxide is removed from the
blood in the lungs when carbon
dioxide diffuses across the alveoli.
Ammonia is quickly converted to the
less toxic compound urea in the liver,
filtered by the kidneys, stored in the
bladder as urine and periodically
Strengths:
Student has, using
relevant
equations, shown
why it is essential
to remove two
important
metabolic waste
products from the
blood by linking
production of acid
or base to the
function of
enzymes
responsible for
cellular reactions.
Copyright © Pearson Australia 2010
(a division of Pearson Australia Group Pty Ltd)
Lower-level
answer
Lower-level
answer
explanation
entering and
leaving each organ,
rather than
summarising the
function of the
organ.
• Student must be
more specific for
the digestive
function than ‘that’s
where the food is
processed’ Student
must discuss
concentration of
food products such
as amino acids and
glucose.
If metabolic
Strengths:
wastes stay in the
This student knows
blood, the activity a lot about carbon
of cells can
dioxide transport
decrease. Carbon
through the blood.
dioxide causes the She recognises that
blood to become
too much carbon
acidic, which can
dioxide causes an
change the activity increase in acidity
of enzymes in
and relates this to
cellular reactions.
enzyme function
Carbon dioxide is
and metabolic
removed from the activity.
blood at the lungs. Areas for
The carbon
improvement:
dioxide travels
The discussion
through the blood
about transport of
as the hydrogen
carbon dioxide in
carbonate ion and
the blood, although
some is dissolved
correct, is
in the blood
irrelevant. The
plasma. It travels
student has only
in capillaries to the outlined how
lungs where it is
carbon dioxide
Heinemann Biology Third Edition HSC
ISBN 978-1-4425-2819-2
Page 9 of 13
Question and mark allocation
Dot point(s)
addressed
Verb(s)
addressed
Higher-level answer
Higher-level
answer
explanation
discharged through the urethra.
(6 marks)
Q10
Figure 5 Glasswort (Sarcocornia quinqueflora), a succulent
9.2.3.8
Define
9.2.3.15
Describe
Enantiostasis is the maintenance of
metabolic and physiological functions
in response to variations in the
environment.
The glasswort (Sarcocornia
quinqueflora) accumulates salt in the
swollen leaf bases which are
progressively sloughed off, thus
removing excess salt.
The grey mangrove (Avicennia
marina) in estuarine areas uses
exclusion and secretion of salt. Roots
exclude most salt from seawater.
Leaves remove excess salt with special
salt secreting glands. Salt crystals
build up on leaves and are washed
away with the rain. (4 marks)
Strengths:
Student has
correctly defined
enantiostasis.
Student has
described in detail
the adaptations to
salt regulation of
two plants in
saline
environments.
Lower-level
answer
Lower-level
answer
explanation
breathed out.
affects metabolism.
(3 marks)
The question asks
for wastes (plural)
so ammonia, or
some other waste
product must be
discussed in order
to gain full marks.
Enantiostasis is the Strengths:
maintenance of
Student has
metabolic and
correctly defined
physiological
enantiostasis and
functions in
given one
response to
adaptation to life in
variations in the
a saline
environment.
environment.
Mangroves are
Areas for
adapted to living
improvement:
in saline
• Student should
environments.
name the species of
They excrete salt
mangrove, as
across the surface
different species
of their leaves to
have different
get rid of salt. (2
adaptations to
marks)
coping with salt.
• Student needs to
outline adaptations
of another plant, not
just mangroves, as
the question says
plants.
Figure 6 The grey mangrove (Avicennia marina)
Figures 5 and 6 show plants found on the coastline of NSW. Define
enantiostasis and for some of the different plants you studied while
investigating enantiostasis, describe the processes used for salt
regulation in saline environments. (4 marks)
Copyright © Pearson Australia 2010
(a division of Pearson Australia Group Pty Ltd)
Heinemann Biology Third Edition HSC
ISBN 978-1-4425-2819-2
Page 10 of 13
Question and mark allocation
Dot point(s)
addressed
Verb(s)
addressed
Q11
Explain how the processes of filtration and reabsorption in the
mammalian nephron regulate body fluid composition. (4 marks)
9.2.3.6
Explain
Q12
During your HSC studies in Biology, you have processed information
about the use of various technologies in medicine. Using two examples
you have studied, explain how technologies are used to monitor and/or
treat patients in hospitals and evaluate their use. (7 marks)
9.2.2.9
Explain
9.2.2.10
Evaluate
9.2.2.11
9.2.3.11
9.2.3.12
Higher-level answer
Higher-level
answer
explanation
Materials are filtered from the blood in Strengths:
the glomerulus (where the blood is
• Student has
under pressure). Everything in the
explained how the
blood except for large proteins and
nephron works in
blood cells is forced under pressure
filtering the blood.
into the Bowman’s capsule.
He has outlined
As the fluid moves along the nephron
where filtration
proximal tubule, glucose, amino acids, and reabsorption
sodium, chloride and phosphate ions
occur and has
are actively reabsorbed into the blood. given details of
Water follows passively by diffusion.
the chemicals
Water is absorbed passively in the
reabsorbed and
Loop of Henle. In the distal tubule,
where along the
potassium and hydrogen ions and
nephron the
drugs like penicillin are actively
absorption takes
secreted into the collecting tubule. The place.
fluid leaving the kidney, via the
• Student has
collecting duct varies from person to
given a good
person in composition. For example,
example of how
somebody with a diet high in protein
urine composition
will have more urea in their urine; a
can vary between
well hydrated person’s urine will be
individuals.
very dilute. (4 marks)
Areas for
improvement:
Use a labelled
diagram to give
the same
information.
Two current technologies that are in
Strengths:
use in NSW hospitals are pulse
Student has
oximeters (used to monitor blood
chosen two
oxygen levels) and renal dialysis
examples of
machines (used to treat kidney failure). technology used
Pulse oximeters are ‘peg-like’ devices in medicine.
that are placed on the fingers, ear lobes Student has
or toes of patients to measure heart
explained how
rate and oxygen concentration in the
and where each
blood. A light on one side of the ‘clip’ technology works.
shines through the finger. The other
Student concludes
side of the device reads absorption of
with a good value
the light through the skin and feeds
statement,
Copyright © Pearson Australia 2010
(a division of Pearson Australia Group Pty Ltd)
Lower-level
answer
Lower-level
answer
explanation
Blood entering the Strengths:
kidney is squeezed Student shows a
out into the
limited
Bowman’s
understanding of
capsule. It travels
the processes
along the
occurring in the
collecting tubules
mammalian
and along the way, nephron. There is
substances that the use of scientific
body needs are
terms such as
reabsorbed into the collecting tubules
blood. Urea and
and Bowman’s
water flows
capsule.
through the
Areas for
collecting tubes
improvement:
and is known as
Student needs to
urine. Some water explain filtration
is reabsorbed into
and reabsorption.
the blood.
Student needs to be
(2 marks)
specific about
which substances
are filtered and
reabsorbed and
explain where along
the nephron, these
processes occur.
Blood products
such as plasma are
used to treat burns
victims.
Technology at the
blood bank has
meant that it is
now possible to
get a range of
products from one
donation,
including red
blood cells, plasma
Strengths:
Student has shown
two examples of
technologies used
to treat patients and
has given some
valid points to
outline how the
technologies are
used.
Areas for
improvement:
• Student needs to
Heinemann Biology Third Edition HSC
ISBN 978-1-4425-2819-2
Page 11 of 13
Question and mark allocation
Dot point(s)
addressed
Verb(s)
addressed
Higher-level answer
data into a computer. The blood
absorbs different amounts of light,
depending on the percentage of
haemoglobin saturated with oxygen.
This device is useful for monitoring
pulse and oxygen saturation levels in
patients during and after surgery. It is
also used in monitoring oxygen levels
in premature babies to ensure their
oxygen levels are safe.
Renal haemodialysis is required when
a patient’s kidneys have limited or no
function. The patient’s blood flows out
of the body and into a tube with a
semipermeable membrane, which
flows through a dialysing solution. As
the blood passes through the tube,
wastes pass from the blood into the
surrounding dialysing fluid. The
dialysing fluid is constantly removed
to ensure a large concentration
gradient, maximising flow of waste
products from the blood into the fluid.
Both pulse oximeters and dialysis
machines are very valuable for patients
in hospitals. Pulse oximeters are a
cheap, quick, painless and noninvasive way to monitor oxygen levels.
Dialysis machines are very expensive;
however, enable a person with kidney
failure to survive, when they would
otherwise die without this life support.
Dialysis takes 3–4 hours and must be
done three times a week. (7 marks)
Copyright © Pearson Australia 2010
(a division of Pearson Australia Group Pty Ltd)
Higher-level
answer
explanation
commenting on
the benefits and
costs of both
technologies for
society.
Lower-level
answer
and factor 8.
Florinef®
(fludrocortisone
acetate) is a drug
used to treat
sufferers of
Addison’s disease.
The
pharmaceutical
industry has
developed a
synthetic drug to
replace
aldosterone in
people who have
Addison’s disease.
(4 marks)
Lower-level
answer
explanation
be more specific
with examples.
Which technology
at the blood bank?
(Perhaps discuss the
fact that as blood is
collected it is
constantly moving
to prevent clotting;
a centrifuge is used
to separate the
components of the
blood; different
blood components
are stored at
different
temperatures; etc.)
Again, information
for technology
helping Addison’s
sufferers is too
brief. It is perhaps a
poor choice of
technology, or
needs to discuss
drugs on the whole,
perhaps discussing
others such as
Ventolin® inhalers
used in asthma
sufferers to expand
airways and
increase oxygen
delivery to the
lungs. (Here you
could discuss the
drug and the inhaler
as a piece of
technology.)
• The student has
not addressed the
Heinemann Biology Third Edition HSC
ISBN 978-1-4425-2819-2
Page 12 of 13
Question and mark allocation
Dot point(s)
addressed
Verb(s)
addressed
Higher-level answer
Copyright © Pearson Australia 2010
(a division of Pearson Australia Group Pty Ltd)
Higher-level
answer
explanation
Lower-level
answer
Lower-level
answer
explanation
“evaluate” part of
the question. There
needs to be a value
statement to
complete the
question and
maximise marks.
The blood bank
technologies help
many burns victims,
sufferers of
haemophilia, etc. by
allowing the
collection so that
safe transfusions of
many blood
products can help
save the lives of
crash victims,
surgery patients,
etc.
Heinemann Biology Third Edition HSC
ISBN 978-1-4425-2819-2
Page 13 of 13