Download BIO315109 Assessment Report

TASMANIAN QUALIFICATIONS AUTHORITY
T A S M A N I A N
Biology
C E R T I F I C A T E
Subject Code: BIO315109
O F
2011 Assessment Report
E D U C A T I O N
The feedback from the final examination paper was positive and seen as one that had a good
range of questions and enough longer questions to give the better candidates a chance to
show their abilities. Candidates completed Part A very easily with the 70% of candidates
scoring 17.5 and above. In Part B, a much larger % of candidates found this section one of
the more challenging and 17% of candidates scored less than 10. Candidates did well in Part
C with 70% scoring 15.5 and above. Part D was also a good discriminating section whilst
challenging for some candidates, 16% of candidates scored 11 and below. Part E also saw a
good spread. Candidates should read questions carefully and make sure their answers relate
to the question being asked. Many “A” standard answers require detailed explanations to
show understanding of the concepts and linking of ideas that are being asked in the question
rather than a general overview.
Suggested Marking Scheme and Comments
Suggested answers with mark allocations for each question are given in the following section
along with comments on candidate’s performance in the exam. Marking examiners have
provided specific comments on aspects such as how the question was assessed, where
candidates gained and lost marks and where candidates misinterpreted questions. Comments
on the open-ended questions may necessarily be limited to general comments rather than
specific details.
The suggested answers are by no means prescriptive and a number of them go into a greater
detail than would be required to gain full marks. Candidates providing different but valid
answers were given credit for any points that addressed the criterion and relevant to the
question.
Question 1
(a)
i)Which surface was covered with petroleum jelly (1) OR
Number of exposed stomata on upper compared to lower surface (1)
Part credit (½) was given for: petroleum jelly; amount/application of petroleum jelly;
where petroleum jelly was placed.
ii) Rate of water loss (1) OR transpiration (1) OR distance moved by bubble (1).
Part credit (½) was given for photosynthesis rate.
(b)
The upper and lower leaf surfaces both covered with petroleum jelly (1). Water loss
through either/or both uncovered leaf surfaces can be compared with a minimal loss set
up (1).
OR neither upper or lower surfaces covered (1). Water loss through neither/or both leaf
covered surfaces can be compared with normal leaf (1).
Part credit (½) was given for: so changes can be observed; for comparisons with other
results.
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(c)
The allocation of marks was: ½ for identifying the sources of error and ½ for
descriptions of the effect each would have.
Examples of complete answers are:
- Variation in air currents around shoot that would affect the rate of transpiration (1)
- Thickness of petroleum jelly over leaf surfaces affects the amount of water loss (1)
- Handling of leaves may alter their temperature and increase the rate of diffusion (1)
- Changing air humidity that would affect the rate of transpiration (1)
Examples of possible sources of error (½ mark each):
- Health/ age/ size of the leaves
- Variation in light exposure/ type of petroleum jelly/ time allowed/ water level in the
reservoir/time before readings were taken/ amount of stem in the water
- Potometer not level
- Petroleum jelly not covering the whole leaf surface
(d)
While using one plant would reduce a number of experimental variables (½), the nature
of the treatment is not really reversible so you can only apply it to one or the other and
then both (1) as petroleum jelly would probably enter some of the stomata (½) and it
would not be possible to remove it completely (½) so removing it from one surface and
covering the other would not give a reliable result (½). (Up to 2 marks)
Part credit (½) was given for: to avoid confusion when no recording is made/ zero
written down.
Examiners comments
It was apparent that many were not familiar with the purpose of a photometer but were still
able to give suitable responses concerning the design of the experiment. Only about 3% of
candidates stated that transpiration rate is the dependent variable but most were able to
suggest that temperature, light intensity, humidity, leaf size etc would influence the
movement of the air bubble.
(a) This was done well by most candidates with a relative few getting them around the wrong
way.
(b) This was also answered very well. Nearly all candidates correctly identified the control.
Lack of clear understanding of the purpose of the control was the main reason for
candidates not scoring full marks on this part.
(c) This was not answered well because, although the question asked candidates to discuss
three sources of error, most candidates simply stated three sources of error and thus only
scored 1½ marks.
Half a mark was given for an adequate description of how each error would affect the
experiment. It was given for ‘it affects the transpiration rate/ diffusion rate/ how far the
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bubble moves’ but not for ‘it affects the results’ (which had been written by many
candidates).
(d)
This was confusing for the majority of candidates. They did not see that once
petroleum jelly had been applied to one leaf surface it would be difficult to remove in
order to get a reliable reading for water loss from that surface. Many candidates did not
score any marks for this part as marks were not given for suggestions about eliminating
out-lying results and results which did not support the hypothesis.
This question sorted out candidates into those who could think through the experiment
and those who could not. Using n/a confused many candidates and there were many
suggestions about ‘it is better to write n/a than to make up results’.
Question 2
(a)
Any six of the following:
•
Amount of sunlight (1/2)
•
Average daily temperature
(1/2)
•
Genetic differences (1/2)
•
Local rainfall (1/2)
•
Soil moisture (1/2)
•
Soil depth (1/2)
•
Soil type
(rocky/sandy/loamy) (1/2)
•
Soil nutrient levels (1/2)
•
Wind exposure (1/2)
•
Wind speed (1/2)
•
Wind strength (1/2)
•
•
Intensity of light (1/2)
Age of the plants (1/2)
•
•
Amount of frost/snow (1/2)
The number of browsing animals (1/2)
•
•
•
•
•
Types of browsing animals (1/2)
Competition with other plants (1/2)
Human impacts (1/2)
Proximity to walking tracks (1/2)
The previous addition of fertilisers (1/2)
•
•
•
•
Toxins present in the soil (1/2)
Salinity of the soil (1/2)
Exposure to salt spray (1/2)
CO2/O2 concentrations of the surrounding air
(1/2)
Aspect of the site (1/2)
The presence of beneficial soil
microorganisms (1/2)
•
•
OR any other factors that potentially could influence the average height of Achillea
lanulosa plants across the mountain range, making sure that that each suggested factor was
significantly different (eg. soil nutrient levels really encompasses soil nitrogen levels and so
if both were mentioned the candidate only received half a mark. Similarly – the amount of
sunlight is very similar to the amount of shade).
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(b)
•
•
•
•
Any of the following:
Increases in average daily temperature cause the average height of Achillea lanulosa
plants to increase. (3)
Increases in soil depth cause the average height of Achillea lanulosa plants to increase.
(3)
Increased exposure of Achillea lanulosa plants to strong winds decreases their average
height. (3)
The average height of Achillea lanulosa plants is reduced as soil salinity increases. (3)
Alternatively, any plausible hypothesis that accounts for the change in the dependent
variable.
Mark allocation:
-­‐
1 mark for identifying a measurable and plausible independent variable.
-­‐
1 mark for identifying the dependent variable and a clear, testable cause/effect
relationship between the independent and dependent variables.
-­‐
1 mark for the hypothesis being in the proper form of a short direct statement including
the specific plant in question (Achillea lanulosa).
(c)
•
Any of the following:
No, variation could be due to a number of factors as there could be many uncontrolled
variables (1/2). Each variable would need to be controlled or investigated (1/2) and then
the experiment would need to be repeated several times (1/2) to validate the results (1/2).
OR
•
No, data obtained from a single experiment will help support a hypothesis (1/2), but
despite efforts to control all variables, there may be unidentified confounding variables
(1/2) and so a large sample size (1/2) and repetition of the experiment is needed to
minimise their influence (1/2).
OR
•
No, even if all variables were controlled (1/2) and the experiment was conducted with a
significantly large sample size (1/2) and repeated many times with fresh biological
material (1/2), it is still difficult to say that the change in height was definitely caused by
the change in the independent variable (1/2).
Examiners comments
Most candidates attempted all parts of this question and did reasonably well.
(a)
Lists of factors that needed to be taken into account when investigating the variation in
height of Achillea lanulosa plants were usually complete. A few candidates struggled to
identify all six factors and a number of candidates stated variables such as ‘the altitude
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of the plants’ and ‘the distance the plants are from the ocean’ which weren’t accepted
as these factors were detailed in the graphs. A number of candidates were very brief
with their answers and stated factors such as – wind, light, temperature etc., without
providing extra details such as whether or not it was wind exposure, wind direction,
wind strength/speed etc. These candidates were not penalised, as each factor was only
worth half a mark. No marks were awarded for general terms such as ‘weather’,
‘climate’ or ‘the environment’. Candidates need to be careful to answer the question
asked as some candidates went to great lengths to explain every factor in detail that was
identified and its potential impact on the growth of Achillea lanulosa plants. This was
unnecessary and would have consumed a large amount of time.
(b)
Very few candidates failed to write anything for this question. To receive full marks,
candidates needed the hypothesis to be in the appropriate format and include an
independent and dependent variable; a clear testable relationship between the two; be
specific (i.e. the Achillea lanulosa plants) and concise (i.e. no unnecessary extra
information). The hypothesis should not be in the form of a question, should show a
clear ‘direction’ for the DV (eg. ‘increased’, ‘ decreased’, not simply affected) and
should not involve more than one sentence of writing. Candidates scored less for each
breach of convention.
A significant number of candidates had difficulty identifying a plausible cause for the
change in the average height of Achillea lanulosa plants and merely restated the initial
observation ‘average height decreased with increasing altitude’ or they were very
general with their independent variable and made largely untestable statements such as
‘the climate influences plant growth’ or ‘the environment influences the height of
Achillea lanulosa plants’. Candidates were scored less for such general and untestable
statements.
(c)
Many candidates stated that if the average height of the plants did vary as the
independent variable was manipulated that they would be confident that the
independent variable was the cause of the change in height. No marks were awarded for
such a statement. Partial marks were awarded if they added that all variables would
need to be controlled, or a large sample size and repetition would be required.
Question 3
(a)
Candidates were required to mention that the controlled environment of the laboratory
is significantly different from the variable environment within the field AND identify
the need for field testing and/or some of the problems that field tests help to identify.
For example:
‘The conditions in a laboratory are highly controlled and quite different from the
conditions in the field (1). Field tests are needed in addition to laboratory tests as they
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more accurately reflect the variety of conditions to which these plants are exposed (1).
The interaction of uncontrolled variables in the field might result in detrimental effects
on soil microorganisms or ecosystems (1).’
OR
‘The highly uncontrolled nature of the field compared to the controlled environment of
the laboratory (1) allows for unforseen impacts on the natural environment to be
observed (1) such as toxicity to native wildlife or the build-up of residues within the
soil (1).’
•
•
•
•
•
•
•
•
Other accepted points:
Uncontrolled variables such as rain may cause fertiliser X to be ineffective (1).
Runoff may have a detrimental effect on stream organisms (1).
Interactions with the different variables in the soil may make the fertiliser more or less
effective (1).
It allows the long-term effects of the fertiliser on the environment and its organisms to
be studied (1).
Parts of the fertiliser molecule may accumulate in tissues and be biomagnified through
the natural food chains that occur in the field (1).
The fertiliser may be toxic to animals, plants and humans (1).
Field tests allow much larger sample sizes to be used as they are not confined by the
structure of the laboratory (1)
To make sure that the fertiliser works in real world conditions that are largely
uncontrolled (1).
(b)
•
Any four of the following variables:
Aspect (1/2)
•
•
Exposure to frost (1/2)
•
•
•
•
•
•
Prevailing winds (1/2)
Drainage (1/2)
Proximity to the stream (1/2)
Proximity to the road (1/2)
Runoff into the stream (1/2)
•
•
•
•
•
Leaf litter from the neighbouring
bushland (1/2)
Cattle adding nutrients to the field
through their waste (1/2)
Slope of the land (1/2)
Exposure to the sun (1/2)
Excess nitrates from the pea field (1/2)
Impact of native animals (1/2)
Contact with other organisms (1/2)
OR any other significant variable that was uncontrolled and could potentially influence
the growth of cabbages.
(c)
Any of the following answers received full marks:
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A large number of replicates of each fertiliser treatment would be required. Each group
of treatments would need to be placed in the field so that they are exposed to the
different variables (1). For example, each strength of fertiliser should be should be used
on cabbages that are near the road, near the hilltop, near the cattle and near the stream
etc. (1). If all fertiliser treatments are exposed to all variables equally, treatment groups
can be averaged and compared (1).
OR
Divide the paddock into a large number of square sections. Randomly allocate each of
the three strengths of fertiliser to all of the square sections (1). By having a large
sample size and the treatments randomly allocated, no one treatment should be exposed
to more or less of a particular variable than any other treatment (1). All treatment
groups will include cabbages exposed to conditions near the stream, near cattle faecal
runoff, the road etc, allowing a comparison of the effectiveness of the different fertiliser
strengths to be made (1).
OR
To minimise the influence of the road, stream, cattle, bush, peas, broccoli- reduce the
effective size of the paddock by not planting cabbages too near these variables, so that
their influence is minimised (1). Plant the cabbages in an area on top of the hill so that
they receive similar drainage, wind and sun exposure (1). Divide the area into three
sections in the north/south direction and apply each different strength fertiliser to a
different section (1).
OR
Using an area of the paddock away from the road, stream, bush, cattle and other cropsconstruct a greenhouse over the cabbages (1). Divide the cabbages into three even plots
and add a different strength of fertiliser to each plot (1). The greenhouse will help
control the influence of leaf litter from the bush adding nutrients to the soil, the
exposure to wind and grazing by native animals (1).
OR
Divide the paddock into four 12.5m strips running north/south. Add a different strength
of fertiliser to the cabbages in each strip (1). Each group of cabbages will experience
similar exposure to most variables (1), i.e. each treatment group will contain cabbages
that have been grown close to the road and stream and exposed to similar light and
drainage conditions (1). However, exposure to the wind, bush, cattle and other crops
will still be variable (1).
Examiners comments
(a)
A large number of candidates misread or misinterpreted the question and stated that
field trials are necessary as they act as a control – either for the laboratory trials or for
the trials conducted without fertiliser. No marks were given for this statement alone.
A number of candidates were able to identify how field trials allow detrimental effects
on the environment and other organisms to be investigated, but failed to indicate why
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these would not be picked up in laboratory trials. If crops are toxic to animals, that
could be picked up in the laboratory. It is the interactions of uncontrolled variables and
the fertiliser in the field which often leads to the unpredictable outcomes for organisms
and the environment, and it is these uncontrolled interactions that are difficult to
replicate in the laboratory.
For full marks, candidates needed to discuss how field trials differ to laboratory trials in
terms of the controlled variables.
A number of candidates stated that field trials were necessary to make sure that the
fertiliser works. They received half a mark for this and received the other half mark for
stating to see if it worked in the uncontrolled natural environment in which it will
ultimately be used.
(b)
(c)
Candidates received a maximum of 1 mark for merely stating that the fertiliser might
have detrimental effects on the ecosystem, native animals or local waterways. They
needed to elaborate on these points to obtain 2 marks.
Candidates answered this question well. Each significant variable was worth only 1/2 a
mark, and in most instances no real explanation was required. However, the variable
needed to be significant. Many candidates merely stated ‘cattle’ as a significant
variable. The cattle were fenced from the paddock and so their influence was not
obvious unless further explanation was provided - eg. ‘cattle provide faeces that may
work its way into the field influencing nutrient levels’ or ‘cattle may get through the
fence and decimate crops’. General terms such as weather, climate and the
environment were not accepted.
This was a difficult question for a number of candidates. So many variables needed to
be controlled and there were very few solutions that dealt with all variables. The detail
required when answering this question should have earned the candidates more than
three marks. Many candidates drew pictures of how they would split the field and
allocate fertiliser strengths, but failed to link it to how that helped control the variables
that they mentioned in part b. Or they incorrectly stated that it controlled variables that
it didn’t. For full marks, candidates really had to address at least a couple of the
variables that they had mentioned in part b; however, if candidates discussed a couple
of variables that they hadn’t previously mentioned, they weren’t penalised.
Conversely, many candidates talked at length about controlling the variables but failed
to mention how the three different strengths of fertiliser were incorporated into their
design. A maximum of 2 marks were allocated for these responses.
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Question 4
a)
As the drug can cause death (harm ½) it would have been unethical (½) to use it on
humans.
Mice are mammals (½) so they would respond in a similar (½) way to humans.
A lot is known about the biology/neurological systems of mice (½)
It is easy to breed large numbers of mice with comparable genetic backgrounds and
minimise related experimental variables (1) or Mice are easy to keep in the laboratory
and the variables to which they are exposed can be controlled effectively (1)
Variables can be tightly controlled using mice subjects (1)
Other mouse studies can be used for comparison (1)
b)
The temperature spiked immediately after administration of the MDMA indicating the
link between MDMA and core body temperature. (½)
The temperature in the rectum was consistently higher in the MDMA group than the
control group, (1) which could be due to increased heat generation or failure to lose
heat through the skin as the result of vasodilation or sweating. (1)
As no clear pattern of reduced temperature of the skin on the tail of the MDMA group
is shown, (1) loss of heat appears not to be altered. (1)
Thus their conclusion is highly likely.
The respiration rate in mice is high therefore mice can generate heat at a greater rate.
(1)
Statistical validity is low. (1)
Some candidates questioned the likelihood that a 1oC temperature rise would cause
death given that human subjects are regularly exposed to core temperatures in excess of
2 degrees above normal (1)
The small size of the mice meant a large SA:vol. and therefore they would lose heat
rapidly so there size may mask the heat gain potential in the human body (1)
c)
The sample size is too small only 10 mice per group were used. (1)
Only 1 dose (1experimental group) of MDMA was given. (1)
Dosage may be different for the mouse because of its higher metabolism. (1)
The MDMA was injected in the mice whereas humans took the drug as a tablet and the
digestive system may break the drug down. (1)
Mice may react differently to MDMA. (1)
Other factors may be causing the fatalities in humans (½), for example, cutting
(diluting) materials such as amphetamines and caffeine or the consumption of alcohol.
(½)
The length of time prior to injecting mice should be monitored to examine normal
thermoregulation. (1)
Ambient temperature and humidity are not mentioned and may influence heat loss. (1)
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Examiners comments
Candidates generally performed well in part a) and c). Candidates performed poorly in part
b).
Candidates failed to see that part b) was asking them to evaluate the conclusion which meant
that candidates needed to assess the strengths and weakness of the claim made by the
researchers that; MDMA was responsible for hyperthermia induced death.
No candidate appeared to recognise that the control mouse would have been given an
injection of saline as part of the experiment. The small change in temperature and the
thermoregulation evident from graph 1 was in response to this insult. Candidates need to be
aware of the heat loss mechanisms available to mammals. A coherent argument needs to be
developed in answering these type of questions using all the information given in the
question including graphs.
Many candidates reported on the use of wild mice because they were abundant and people
didn’t much care for mice as they were a pest (clearly scientists would use genetically similar
mice produced under highly regulated and expensive conditions).
Variables one would expect to be regulated; and could reasonably assume to be regulated
under normal experimental conditions, were not given much credit in part c) e.g. species of
mouse, weight of mouse.
Part 2
Question 5
a)
Amino acid (1) – part marks also given for: peptide (1/2) and for the abbreviation a.a’s
(1/2)
b)
i) The stem of the name “prot” indicates that they both act on protein molecules (1) OR
because they both act on protein molecules (1) OR they both catalyse the reaction that
hydrolyses peptide bonds of proteins (1) – part marks were given for: because they
both break peptide bonds (1/2) OR because they have the same function (1/2).
ii) The graph illustrates that the activity of the enzymes is affected by pH (1/2) The
optimum pH of the two enzymes is very different (pepsin around pH 2 and trypsin
around pH 8) (1). The further the pH is away from the optimal values, the less active
each is, and neither is active at the optimal pH of the other (1). The stomach is acidic,
so pepsin is active there while the small intestine is alkali, so trypsin is active there (1).
ALSO The overlap in enzyme activity at pH 5 in minimal, with each enzyme being
basically ineffective in this range (1), as this is the only pH that they share with neither
being very effective they therefore cannot be active in the same area of the gut (1); each
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enzyme is denatured at the others optimum so they cannot be effective in the same area
(1/2)
c)
As the temperature rose in the 10˚ – 30˚C range, the kinetic energy of the molecules
increased (1) resulting in more collisions between molecules which resulted in an
increased reaction rate (1). ALSO: a certain amount of heat energy is required (1/2) to
provide the activation energy so the reaction can proceed (1); enzymes require an
optimal temperature to react (1/2); enzymes have inflexible bonds at low temperatures
making them unable to react (1/2).
ii) Above a temperature of about 40˚C human enzymes rapidly denature (1/2). Bonds
which hold the molecule in shape break, changing the active site rendering the
molecule ineffective (1). The higher the temperature, the less effective the enzyme
molecule (1/2); as the temperature increases, more enzyme is denatured and the
reaction rate decreases (1/2).
Examiners comments
a)
This question was well answered.
b)
i) This question was very well done, most candidates made the link between the name
and the function. The most common error was candidates stating that they were the
same because they ended in ‘psin’.
c)
ii) There were a lot of misconceptions and errors in the responses to this question. A
lot of candidates made no reference to the graph; many stated that the enzymes were
acidic/alkali ; many did not detail why the enzymes would not be able to function in the
same part of the gut; many candidates just explained a digestive process with the two
enzymes, but did not address the question; some candidates stated if they mixed the
two enzymes together they would cancel each other out. Many candidates interpreted
the crossover point on the graph as pH 4.5 and not 5 (and the other points similarly).
d)
i) This part of the question was the most poorly attempted by candidates. Most only
identified that enzymes required an optimal temperature, but were unable to explain
why the reaction rate changes.
ii) Many candidates only responded that the enzyme was denatured without offering
explanation as to how this would affect reaction rate.
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Question 6
(a)
i
ii
(b)
3 marks were awarded for any combination of answers below that gave a complete
overview of how amino acid sequence can be used to make a gene.
A gene is a length of DNA which codes for a protein molecule (1).
The DNA code is a sequence of bases (½), with three bases forming a codon and
each codon coding for a specific amino acid (½).
Amino acids are coded for by codons on mRNA (½).mRNA coded for by DNA
base sequence. (½)
mRNA is formed by transcription (½) where DNA strands unwind and complimentary
bases attach to form a single mRNA strand (½).
-The mRNA codes for the complimentary anticodon on tRNA (½).
Translation results in amino acids being joined together according to relevant
codons/anticodons (½).A codon is a three base sequence on mRNA (½).
The link DNA code → codons → amino acid sequence → protein can be read
backwards to determine the gene (1).
-
(c)
-
Ribosome (1)
amino acid (1)
3 marks were awarded for any combination of answers below that addressed how
different genes can code for the same protein.
-No (½) or not necessarily (½).
-There are several combinations of the four bases in DNA (A, T, G, C) that can code
for most amino acids (1). Bases can be combined to form 64 different codons (½) so
there are about three times the number of codons compared to amino acids found in
proteins(½) .
Different scientists may choose different codons for a particular amino acid, resulting
in different versions of the required gene (1).
Although the above answer is more likely to be true, credit also had to be given for
answers saying that the genes would be identical as in theory this is possible. Answers
in the affirmative attracted up to one and a half (1½) marks. To achieve this answers
needed to indicate an understanding of how specific proteins are made up of set amino
acid sequences and then relate this back to genetic structure.
Examiners Comments
The bulk of candidates achieved marks in the three and a half (3½) to five and a half (5½)
mark range. Generally a lack of detail in explanations or simple misuse of biological
terminology prevented many from achieving higher marks than this. Some common errors
are set out below.
(a)
A number of candidates incorrectly interpreted the diagram and so identified the
organelle X and/or substance Y as a nucleus or cytoplasm or enzyme or polypeptide or
protein. Although related, no credit was given for peptide (double unit of amino acids)
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or protein (multiple units of amino acids) as the diagram clearly shows a single entity
carried on tRNA.
(b)
Many candidates struggled with structuring their answers, getting cause and effect the
wrong way around. Where this was the case, candidates were stating that amino acids
code for DNA, rather than the other way around.
Many candidate answers lacked detail, just stating that DNA codes for mRNA which
codes for tRNA. Candidates needed to explain how nucleotide bases are transcribed
and translated even though they did not need to use these terms for full marks.
Misusing ‘codon’ with regard to tRNA instead of ‘anticodon’ was a common problem.
Also stating that mRNA codons are replicas of DNA codons rather than complimentary
strands was another problem that occurred
Candidates may have discussed mRNA and tRNA coding relationships but failed to
discuss the coding relationship between mRNA and DNA or make the final link back to
how DNA and amino acids are related.
(c)
A significant number of candidates stated that the genes would be identical, although
statistically this is unlikely. If candidates gave sound reasoning behind this assertion
they were credited with up to one and a half marks for this question.
Many stated that the same codon can code for different amino acids, whereas
candidates should have stated that different codons can code for the same amino acid.
By limiting discussion to how the protein would be identical as long as it contained the
same amino acid sequence meant that many candidates missed the focus of the question
asked, which was about whether the genes were identical.
Question 7
(a)
As rate of exercise increases, O2 consumption increases (½) because the rate of aerobic
respiration increases (½) to provide the increased energy required (½). However, at
high rates of exercise, O2 consumption levels off (½) as O2 can’t be delivered to
muscles any faster (½). Anaerobic respiration (½) provides the additional energy
required (½), hence the concentration of lactic acid increases (½), as lactic acid is a
product of anaerobic respiration.
(b)
Graph 1 shows that at high rates of exercise, anaerobic respiration contributes
significantly to supplying energy (1). The lactic acid that is produced as a result of
anaerobic respiration (1) causes fatigue and muscle cramping as it accumulates in the
body (1). Also, anaerobic respiration yields only 2 ATP molecules for each glucose
molecule (compared with 38 for aerobic respiration) (1). This means that glucose
stores will be depleted rapidly during anaerobic respiration (1). Graph 2 shows that as
the length of a race increases the average speed drops off rapidly; only 100m and 200m
have very high average speeds; above this, as race length increases the average speed
drops off steeply (1). The more sustained the activity (the longer the race) the more an
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athlete must rely on aerobic respiration for his/her energy need, so the rate of exercise
(speed) must be lower (1). In short, high speed races, sprinters usually don’t breathe
during the race to replenish their O2 levels (1).
Examiners Comments
(a)
This was generally well done, with a pleasing number of candidates scoring full marks.
Up to 1 mark was awarded for answers which simply described the data shown in the
graph. Many candidates reversed the cause and effect relationship between O2
consumption and lactic acid, suggesting that O2 consumption must increase to
“overcome” the increasing concentration of lactic acid (i.e. more O2 consumption due
to the need to use O2 to convert lactic acid to glucose). This is not the case, and the
graph shows O2 consumption is increasing even when lactic acid levels are low. Only
very strong candidates realised that aerobic respiration continued at high rates of
exercise, and that anaerobic respiration was required to provide the additional energy
required. A large number of candidates suggested that, at high rates of exercise, the
concentration of lactic acid exceeded the rate of O2 consumption. However, as no scale
is given on the graph, this is not necessarily the case (and even if it was, it is not
significant/relevant).
(b)
The majority of candidates were able to gain 2-3 marks by recognising that anaerobic
respiration is occurring in short (100-200 m) races, and then providing a reason why
such high speeds can’t be sustained. A large number of candidates thought that Graph
2 was showing the average speed of a single runner, as they ran 45 000 m, starting off
at a fast pace and then decreasing their speed. They were not penalised for this, and
generally their answers contained many correct statements that were awarded marks.
Even candidates who suggested that sprints are conducted aerobically, and that it is
during longer races that anaerobic respiration “kicks in”, were generally able to gain 2
or 3 marks for correct statements about lactic acid build up and the relative energy
outputs derived from anaerobic and aerobic respiration.
Overall, candidates performed quite well on this question. However, incorrect
statements like “Respiration requires energy from glucose” and “Respiration produces
O2 for working muscles” are a concern. It was felt that in many cases such statements
are a result of a poor use of language, with some candidates believing terms like
“produce” and “provide” have the same meaning.
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Question 8
In “normal” light conditions, photosynthesis occurs at a faster rate than respiration (1).
Therefore, more glucose is produced by photosynthesis that is used in respiration (1). This
extra glucose is:
•
converted to tissue – growth, repair (1)
•
converted to molecules required by cells – e.g. enzymes, cellulose (1)
•
stored for future use – e.g. as starch (1)
Up to 4 marks for discussing the above.
Photosynthesis converts light energy into chemical energy (glucose) that cells can use/access
(1). This energy is released by respiration (1) and stored in ATP molecules (½). This energy
is used for processes such as active transport (1) and protein synthesis (1) [up to 2 marks for
valid examples of processes requiring energy].
Up to 4 marks for discussing the above.
Examiners Comments
Candidates found this question challenging, and many did not get beyond restating the
information given in the stem. The most common approach used was to compare the
equations for photosynthesis and respiration and note that respiration “produced energy” and
this represented a gain for the plant. Often some discussion of ATP/ADP followed, but
unfortunately uses made of the energy were often not mentioned. Candidates who recognised
that plants gain glucose through the imbalance between respiration and photosynthesis were
generally able to state at least one use for that glucose.
An alarming number of candidates believe that plants photosynthesise during the day and
respire at night (not true – respiration occurs all the time!).
Part 3
Question 9
a)
A
golgi body or golgi apparatus
(1)
B
rough endoplasmic reticulum (1) endoplasmic reticulum with ribosomes (1)
endoplasmic reticulum (½)?
C
nucleolus(1) nucleus (½)
(b)
Any 2 of the following:
Cell membrane involved in exocytosis (1)
Nucleus involved in directing which enzymes are produced and where they go. (1)
Mitochondria supply energy for exocytosis (1)
Rough endoplasmic reticulum transports proteins/enzymes towards the Golgi bodies (1)
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Golgi body packages enzymes into secretory vesicles or produces secretory vesicles
which contain enzymes (1)
Secretory vesicles transports enzymes to the cell membrane for secretion (1)
1/2 a mark only if they explained how enzymes are made but did not explain how it is
secreted.
(c)
Any 1 of the following:
It lacks a cell wall, it lacks a large vacuole, it lacks chloroplasts. (1)
½ mark for non angular/rounded shape
(d)
Aerobic respiration, which is more efficient than anaerobic respiration, is carried out in
the mitochondria (1/2) it produces energy in the form of ATP which is needed for use
in exercise (1/2) . People with mitochondrial disease would produce less ATP (½) &
would thus only be able to exercise for a short time. (1/2)
1 mark for describing what mitochondrion does (aerobic respiration) and 1 mark for the
describing why the disease affects energy production.
Examiners comments
Overall the question was done well by most candidates with the average mark being between
5 ½ and 7 out of 8. A good C standard question where most candidates could answer it and
get strong marks for it.
Common mistakes:
b)
c)
d)
Several candidates lost ½ marks for only saying endoplasmic reticulum instead of
rough endoplasmic reticulum or nucleus instead of nucleolus.
Several people answered how the enzymes are formed and coded for and only received
½ instead of focusing on how they are secreted after manufacture.
Most candidates did well but lost a ½ mark for not mentioning that the ribosome is the
site of aerobic respiration and hence for efficient producers of ATP than anaerobic
respiration.
Question 10
(a)
Cell A would have been placed in a hypertonic solution (1/2) , a solution of lower
osmotic concentration than that of the cell (1/2). As a result of osmosis (1) there would
have been a net movement (1/2) of water out of the cell (1/2) across a selectively
permeable cell membrane (½) to even out the concentrations (½). Turgidity decreases
& the protoplast shrinks. (1/2) The cell membrane has pulled away from the cell wall.
(1/2)
(b)
Plasmolysed (1) or flaccid (1/2).
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(c)
No (1/2) , there would have been a net diffusion of water into the cell & it would
become turgid (1/2) . The cell wall (1/2), which is thick, rigid & composed of cellulose
(1/2), would prevent the cell from bursting (½).
Examiners Comments
(a)
This question was quite well done, although some candidates misinterpreted the
diagrams, and gave their answer assuming that cell A was plasmolysed, and cell B was
turgid. These received a maximum mark of 3.
Many other candidates confused the terms hypotonic and hypertonic. They were
penalised half a mark for incorrect use.
(b)
Many candidates described the cell as flaccid, which is a term better used to describe a
whole plant. They received half a mark. Relatively few gave the correct answer.
(c)
Those who had misinterpreted part (a) were given the full mark for the answer turgid.
This question was well done, with many candidates scoring one and a half to two
marks.
Question 11
(a)
contractile vacuole
(1)
(b)
The concentration gradient between the amoeba & its external environment is greater
(1) & so the rate of water moving into the amoeba has increased (1) due to osmosis (1).
The contractile vacuole needs to work harder & contract faster to remove this extra
water. (1)
Examiners comments
Overall the candidates did well in this question.
(a)
Most identified the organelle as the contractile vacuole. Candidates were awarded ½
mark if they answered just vacuole for part (a).
(b)
Candidates lost a mark for not mentioning the term osmosis and the direction of
movement of the water molecules. The concept of concentration gradient caused some
confusion.
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Question 12
(a)
16 (½). Male drones are produced from a haploid egg (½) that is not fertilised (1).
Therefore it contains half the number of chromosomes of the female worker (½), who is
produced by the fertilisation of both a haploid egg and sperm (1).
(b)
Drone is more closely related (½), as it gets all its genes/chromosomes from the queen
(1). Whereas the female worker gets only half its chromosomes from the queen (½) and
half from an unrelated male (½).
(c)
Mitosis (1), whereby the haploid male produces haploid sperm (1). It cannot be
meiosis, as this is a reduction division (½) and would result in half the number of
chromosomes needed, i.e. 8 (½) and this would mean that at fertilisation the zygote
would have less than the diploid number (24) (½).
(d)
No, Drones born from the same queen won’t be identical (½). They are formed from
eggs/gametes produced by meiosis that introduces genetic variation (1) through
processes such as random assortment (½), crossing over (½) or mutation (½). The
males only get half of the queen’s 32 chromosomes, so there will be a mix of
chromosomes from each of the pairs (1).
Examiners Comments
Virtually all the candidates attempted the question; however, many struggled to apply the
basic principles of cell division, gamete production and fertilisation to this unusual situation.
As a result, many candidates struggled to get more than half marks. On the whole the first
two parts were better done than the last two, but there was a lot of inconsistency between
parts. There were still a lot of candidates (at least 10%) who scored 6.5 – 7 marks and
showed an excellent understanding of the concepts. On the whole this question was a good
discriminator, especially of the A candidates.
(a)
While there was a wide range of possible chromosome numbers, varying from 13 – 64,
this part was generally well done. Some candidates were distracted over role the sex
chromosomes and how they would affect the chromosome numbers.
(b)
A healthy percentage got full marks for this section and included a contrast between the
male drones who got all their chromosomes from the queen with the worker females
who got half theirs from an unrelated male. Some credit (up to 1 mark) was given
where candidates argued that they had an equal inheritance of the queen’s
chromosomes.
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(c)
Candidates who generally did well on this part were focussed on the chromosome
numbers produced in terms of what was needed for future fertilisation. However, many
were still relating their answers to the queen’s production of drones. The greatest
problems arose when the candidates tried to apply human gamete production based on
meiosis to this situation or got it in their minds that this was simply asexual
reproduction and then answered the last two parts of the questions according to their
general understanding of this concept. Simple information dumps on gamete production
by meiosis was given no marks, but ones showing understanding of the need to restore
the diploid number of chromosomes during fertilisation got ½ a mark.
(d)
This was the most poorly done part. The key was to consider that the drones came from
an unfertilised egg produced by meiosis and then consider the variations related to
meiosis. Many candidates who had convinced themselves that this was a classic case of
asexual reproduction then went on to say that this would mean no genetic variation,
which got no marks unless they included mutations as an exception (½ mark). A
disappointing number wrongly were convinced that “meiosis cannot occur without
fertilisation” saying that the eggs must have been produced by mitosis and asexual
reproduction, forgetting that the eggs could also fertilised to become worker females as
well as being haploid from the diploid queen and so a reduction division, i.e. meiosis, is
required.
Question 13
Metabolism is the sum of all the chemical reactions occurring in a cell. (1) The speed of the
reactions is largely determined by the ability of the raw materials e.g. oxygen (1/2) to enter
(1/2) & wastes e.g. carbon dioxide (½) to leave the cell (1/2). This in turn depends on having
a high surface area to volume ratio to allow for rapid exchange with the cell’s surrounds (1)
as this increases the rates of processes like diffusion and osmosis as well as active transport
(1).
Cells lining the proximal convoluted tubule in the kidney (1/2) are small and flat with
numerous microvilli (1/2) & have a very high surface area to volume ratio (1/2). They can
exchange materials with their environment faster (1/2) & can achieve a higher metabolic rate
(1/2) necessary to reabsorb needed materials faster than could larger more spherical cells.
(1/2)
Examples can be any cell that has a high metabolic rate. Other suggestions include:
The muscle cells help in movement and therefore require a high metabolic rate to perform
their function(1). The shape of muscle cells tends to be long and thin,(1) which provides a
favourable surface area-to-volume ratio for rapid exchange of gases required for a high
metabolic rate.
OR
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Most of the digested food molecules are absorbed in the small intestine. Microvilli border in
the cells lining the small intestine increases the surface area for absorption of digested
foods(1).Some of the food molecules require active absorption and a large surface area will
enable the cells to increase the metabolic rate required for active transport .
OR
The root hair cells absorb water and mineral ions from the soil .Some these ions are absorbed
by diffusion while some require active absorption. The root hair cells have long 'finger-like'
projections with very thin wall, which gives a large surface area for diffusion(1/2) and a
favourable surface area to volume ratio for active transport.(1)
Examiners comments
Candidates struggled to attain above 3 marks in this question. About 10% of the candidates
did not attempt the question. To obtain full marks the answers had to explain how the size
and shape of the cell(s) used as e.g.(s) relates to greater surface area to volume ratio and
hence more efficient exchange of nutrients.
A lot of candidates used a dictionary definition to explain the meaning of metabolism and lost
marks if they were unable to relate it to the size and shape of the cell. Some candidates did
not read the question properly and ended up explaining the surface area and volume ratio in a
small and large animal (mouse and an elephant) and its effect on metabolism. Candidates in
general were not clear about the rate of metabolism and how it is related to the exchange of
material between the cell and its surroundings.
Note: Red blood cells would be a poor example, as they do not have a high metabolic rate,
although they do show how increased surface area can be used to increase the rate of
exchange.
Part 4
Question 14
(a)
Recessive, (1).
Individual 4 (or 6) (1/2) has it & neither of their parents have it. (1/2)
It has skipped a generation (1/2)
If it was dominant then some of individual 1 &2’s children would not have it.(1)
If it was dominant then girls of individual 4 and 6 would have it but they do not (1)
Candidates must refer to pedigree in some way to get full marks and not just recite note
on sex- linked recessive genes.
(b)
(i)
(ii)
(c)
Ind. 3
XBXb
XBXB or XBXb
(1)
(1)
Ind. 4
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XBXb
x
XbY
Working (1mark total- including genotypes can be part of table only)
XB
Xb
Xb
XB Xb
Xb Xb
Y
XB Y
Xb Y
genotype
1
/4 XBXb +
1
/4XbXb
phenotype
1
/4 non colour
blind female
or
(50%, ½) normal, (50%, ½ ) colour blind
probability of colour blind child = 50%
+
1
/4 XB Y + 1/4 Xb Y
1
+
/4 colour
blind female
1
+
/4 non colour
blind male
1
+
/4 colour
blind male (max 1)
or 1:1 ratio
(1)
(1)
If candidates did not add to X or Y chromosome and just wrote Bb x bb punnet max ½
If candidates did not use a Y or added chromosome to Y (e.g. Xb Yb) max ½ for other part
correct.
If a candidate wrote only 50% max ½ mark as no working and relating to colour blind.
If candidate only works out that individual 3 is heterozygous XBXb, ½ mark given.
Examiners Comments
Question was very well answered on the whole. A good ‘C’ type standard for the start of the
booklet where many candidates were able to get 6 or 7 out of 7.
Common errors were:
Many candidates were messy with their Y’s and it was hard to tell if they were meant to be
Y’s or they in fact were incorrectly X’s for male genotypes.
Many had males being XX as well as females
1:2 does not mean a 1 in 2 chance. It should have been 1:1
Candidates did not read the question and use XB and Xb and made up their own letters even
though it was given in the question. Candidates were not penalised if they had the correct
patterns though.
Question 15
(a)
Glucose is a small molecule that can fit through the through the glomerulus wall(1/2) is
used in the body(1/2) for aerobic respiration(1/2) therefore is not excreted into the
bladder but is reabsorbed(1/2) back into the blood vessels surrounding the nephron(1/2)
using active transport(1/2) This happens at the proximal convoluted tubule (1/2) thus
no glucose is present in the urine(1/2).
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(b)
Protein is not usually present in the urine as it is too large (1/2) to pass through the
walls of the glomerulus membrane. (1/2) The glomerulus membrane must be damaged
so that protein passes through into the filtrate in Bowman’s Capsule. (1)
Or (1) There is no mechanism to reabsorb protein once it has passed into the kidney
nephron.
(c)
Lower levels of ADH in the blood cause the walls of the kidney tubule (distal
convoluted tubule & collecting duct) to be less permeable to water. (1/2) Thus less
water is reabsorbed (1/2) back into the blood (1/2) & more is excreted (1/2) thus a
larger volume of urine is produced. (1)
Examiners comments
(a)
Generally answered well. Most candidates who scored poorly failed to include enough
detail on why the glucose was not present. Many failed to mention that the process of
reabsorption was an active one. Poor scorers often had no idea about bio molecules
saying glucose was a protein, or even, protein polysaccharide and amino acid
monosaccharide! ). A significant number of candidates said that glucose was a big
molecule and therefore too big to go through the glomerulus and would stay in the
blood and not enter the nephron at all!
(b)
Main problem with this question was failure to realise that protein is too big to cross
into the nephron therefore there must be a problem with the glomerulus membrane.
(c)
Many candidates did not get full marks as they failed to mention that ADH increases
water absorption and just mentioned absorption generally.
Question 16
(a)
P is an artery (1) as it carries blood away from the heart (1/2) towards the gills. (1/2)
(b)
Concentration similar (1/2) as at P it is yet to be oxygenated at gills (1/2) and at Q it has
just passed through 2 capillary beds (1/2) and has lost most oxygen by diffusion into
surrounding tissue (1/2)
P lower than Q (1/2) as it has travelled further through the circulatory system (1/2) and
may have lost more oxygen (1/2) by diffusion into cells(1/2) on the way. (Less likely
but was accepted as a reasonable response if explanation was correct)
The concentration of oxygen would somewhat higher at P compared to Q (1) as the
blood at P has passed though the gills and head. It could be argued that less O2 has been
used at P whilst more O2 has been used by the trunk and tail resulting in lower levels of
O2
Or
Or
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(c)
The blood pressure at P would be much higher (1/2) than at Q. Blood has just been
pumped out of the heart into P (1/2) so blood pressure is high. (1/2) At Q the blood has
passed through two capillary networks so blood pressure would be much lower. (1/2)
Examiners comments
(a)
Generally very well answered.
(b)
Many candidates failed to understand the diagram in this question. A surprisingly large
number of candidates thought that the heart supplies oxygen to the blood and this didn’t
help their score. Several candidates thought that Q would have higher oxygen than P as
the blood had passed through the gills (failed to recognise the removal of oxygen in
trunk and tail capillary beds). Many candidates failed to mention the reason for oxygen
loss.
Generally well answered although many candidates failed to get full marks as they
didn’t mention that the pressure is lower at Q because it has passed through 2 capillary
beds. Most just mentioned it is a vein and therefore the pressure is lower.
(c)
Question 17
(a) Any 3 of the following:
The alveoli in a bat’s lungs may be smaller in size (1/2), which would increase their
surface area to volume ratio (1/2) so oxygen is absorbed more rapidly.
The layers of cells that separate the lumen of each alveolus from the blood may be
thinner (1/2) thus the diffusion distance is less (1/2) so oxygen is absorbed faster.
Many more capillaries may surround each alveolus (1/2) (or faster blood flow) thus a
higher concentration gradient is maintained (1/2) so more oxygen is absorbed into the
blood.
The bats may have larger lungs/more alveoli (1/2) so surface area is larger (1/2) so
oxygen is absorbed more rapidly.
Blood in capillaries may have higher red blood cell count or have higher levels of
haemoglobin (1/2) so that oxygen is diffused faster as a result of the higher
concentration gradient. (1/2)
(b)
(i)
(ii)
Ventilation removes air with a low concentration of oxygen (1/2) and high
concentration CO2 (1/2) from the alveoli & replaces it with air with a high
concentration of oxygen (1/2) & this maintains a high concentration gradient between
the lungs and capillaries. (1/2)
Blood flowing (1/2) through capillaries that surround the alveoli carry the oxygenated
blood away. (1/2)
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Examiners comments
(a) Most candidates answered this question by stating a feature without a supporting
explanation and consequently received only half marks. Eg “thin membranes” or “moist
surfaces”. Many wrote that a bats gas exchange system needed a large surface area to
volume ratio but didn’t give a suitable structure, ie many more / smaller alveoli. Common
problems: Some candidates forgot (or didn’t know) that bats have lungs for breathing and
suggested bat wings that were thinner or had a large surface area would enable better
rates of gas exchange! Others wrote about “thin cell walls” in the lungs instead of using
the term membrane.
(b)
(i) A large number of candidates didn’t seem to know about “ventilation” and
consequently didn’t answer the question.
(ii) A number of candidates answered this question with reference to gas exchange at
muscle sites when it was actually about lung capillaries.
Question 18
Negative feedback is where the response (increase in glucagon or increase in insulin) alters
the stimulus (low or high plasma glucose concentration) (1/2) returning conditions (plasma
glucose concentration) back to normal thus maintaining homeostasis. (1/2)
Following the meal the graph shows plasma glucose concentration was high. (1/2)
Consequently the pancreas (1/2) released insulin (1/2) that stimulated the liver and muscle
cells (1/2) to take up glucose and convert it to glycogen (1/2) so that glucose levels fell (1/2) end of 1st section of graph. The lower glucose levels became a stimulus for the release of
glucagon (1/2) by the pancreas (1/2) that stimulated the liver and muscle cells (1/2) to convert
glycogen back to glucose (1/2) so that glucose levels were raised toward normal. (1/2)
Examiners Comments
The majority of candidates had a reasonable to excellent understanding of glucose balance
and the role of negative feedback to maintain it. They were able to refer to the graph and
relate high blood sugar levels and the pancreas’ response in releasing insulin to counter this
and likewise the role of glucagon in combatting low glucose levels. Common problems were
linked to incorrect spelling of glucose, glucagon or glycogen or the inaccurate use of these
terms. Many incorrectly suggested that insulin converted glucose to glycogen and glucagon
converted glycogen to glucose with some referring to insulin and glucagon as enzymes
instead of hormones.
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Part 5
Question 19
(a)
Phytoplankton (1)
(b)
Herbivorous zooplankton feed on the phytoplankton, which is an autotroph or plant
source of food.(1)
Carnivorous phytoplankton feed on the herbivorous zooplankton and krill, which are
heterotrophs or animal sources of food. (1)
(c)
Leopard seals are a secondary consumer when they eat krill, which is a primary
consumer (1) and they are tertiary consumers when they eat fish, other birds or
penguins, which are secondary consumers when they prey on krill. (1)
Or
•
•
•
(d)
In the food chain:
Phytoplankton → krill → leopard seal, the leopard seal is a secondary consumer, (1)
and in the food chain
AND
Phytoplankton → krill → fish → leopard seal, the leopard seal is a tertiary consumer.
(1)
Marks were given for:
Live off krill and fish etc, which have consumed krill. (1½)
Because they eat organisms from both the second and third trophic levels. (1½)
Because it eats on different levels (1)
Only about 10% of the energy available at a particular trophic level is available as food
for the level above it (1), due to energy being lost (½) via respiration and wastes, or
used by the organism. (½)
By feeding on krill which are primary consumers, the blue whale is a secondary
consumer (½) and is able to consume a significant amount of the original energy
captured by the phytoplankton(½), allowing it to make efficient use of food sources and
grow to a much larger size. (½)
Killer whales feed on higher order consumers such as penguins and leopard seals ( are
4th, 5th and 6th trophic level consumers) (½) and therefore have less energy available for
them to consume and grow, as it has passed through many trophic levels and much of it
has been lost. (½)
Examiners Comments
Candidates did well on this question, most receiving an average of 6½ to 7½ marks.
(b)
Candidates had to do more than just state information included on the food web, such
as “herbivorous phytoplankton eat phytoplankton so are herbivores” (Max 1)
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Many said that carnivores ate organisms and herbivores don’t, suggesting that
phytoplankton aren’t organisms. (0 marks)
Max of 1 if candidates stated “carnivores eat meat and herbivores eat plants”, as they
needed to relate this to the food web as requested.
(c)
The question was marked with a focus on recognising that animals may feed on several
different trophic levels. Many candidates (25%) mislabelled the consumers. Eg.
Labelled secondary consumers as tertiary consumers, however if they demonstrated the
idea of the different trophic levels, up to a max of 1½ was awarded. Candidates had to
include an example of what the leopard seal was feeding on to be classified at that
trophic level.
Approximately 25% of candidates incorrectly labelled the leopard seal as a secondary
consumer because it was eaten by a tertiary consumer (Orcas), suggesting that
organisms are labelled by what eats them as opposed to what they eat.
Many candidates wrongly suggested that a tertiary consumer was the highest order
consumer and included organisms on the 4th, 5th, and 6th trophic levels in this group.
(d)
Maximum of 3 marks given if the answer was fully explained.
Candidates that just focussed on the whale, what it ate and the fact that it had an
unlimited food supply which was easily assessed, then the Orca, what it ate, and that its
food supply was not as plentiful and energy was used to capture it, gained a maximum
of 2 marks.
Some candidates incorrectly focused on competition being the main cause of the size
difference.
Marks were not deducted for not obviously stating the principle involved as requested
in the question. Some candidates incorrectly answered from a focus of
biomagnification.
Question 20
(a)
i
ii
F (1)
B (1)
(b)
Denitrifying bacteria convert nitrates and other nitrogen compounds into nitrogen
which is released into the air (1).
Bacteria releasing nitrogen (½)
(c)
Peas are legumes which have nodules of nitrogen fixing bacteria on their roots (1).
Nitrogen fixed by a legume crop becomes available this increasing nitrogen levels in
the soil (1). If increase or decrease not specified (½)
Only peas are harvested from the pea crop, rather than wheat where most is harvested,
so peas would have more waste material to be put into the soil (1)
(d)
- there would be no equivalent to a crop harvest (½), so nitrogen wouldn't be removed
from the ecosystem in such large amounts (½).
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- no fertiliser would be added (½); plants (and animals) would have to rely on natural
recycling of nitrogen (½)
- Far less nitrogen being lost due to leaching (½). The soil would have a slow, steady
conversion of nitrogen into nitrates, not the sudden addition of large amounts added at a
time, much of which would be washed away (½) more animal activity in the native
vegetation, could lead to nitrogen input (½).
Examiners comments
(a)
(i) was often incorrect
(b)
Common answers focussed on bacteria using nitrogen for energy which was a direct
take from a dictionary. Many failed to make link to path D.
(c)
Few received full marks, with very good answers. Candidates commonly lost ½ as
didn’t state whether nitrogen would increase in the soil, therefore didn’t fully answer
the question. Many candidates focussed on the difference between the crops, and did
not mention the nitrifying bacteria present in peas.
(d)
Many ½’s given; candidates commonly listed factors, but gave no explanation of why
that was significant. Clear confusion in many cases about what ‘native vegetation’
meant in comparison to the crops.
Question 21
(a)
i) colder temperatures (1) or, water availability (1) Change of season (1)
ii) increased predation (1), or increased disease (1), or food shortages (1).
2011 Assessment Report
Biology
28
Subject Code: BIO315109
(b)
Graph
(c)
Population change depends on the relative birth, death, immigration and emigration
rates.(1). These depend on the corrective measures which are the density dependent
environmental resistance factors, availability of food/water, availability of nesting
sites/space, predation, parasitism, rate of spread of disease. (2.5) Homeostasis means
that a population maintains a size close to the optimum for its environment (1).
Negative feedback results in environmental responses to deviations from this optimum
which result in the change being reversed (1).
Under favourable conditions a population will increase. Corrective mechanism, such as
increased competition for resources (food, light, water, space, nesting sites), increases
in predator numbers, more rapid spread of disease and immigration will eventually
cause the population to peak, and then decrease. (1.5)
Falls in population below the optimum result in the opposite applying, and provide the
conditions which allow the population to increase. Over time any increase in population
is followed by a decrease, and a decrease is followed by an increase (1).
Examiners comments
(a)
i) and ii) Most candidates did well on these sections. The main errors were from mixing
the terms density dependent and independent.
(b)
The main error was not placing the carrying capacity line directly in the middle of the
fluctuating curve. Many candidates got 3/3 from these sections.
(c)
Most candidates didn’t link the idea that population depends on birth, death,
immigration and emigration rates. Most did determine that the corrective measures
were the density dependent environmental resistance factors, availability of food/water,
availability of nesting sites/space, predation, parasitism, rate of spread of disease. But
many only mentioned one or two of these. Most candidates made the connection
between a corrective measure and the reduction or increase in population. Some only
addressed the decrease caused by overcrowding. Many candidates didn’t specifically
address the negative feedback part of the question, even though they had implied it with
their descriptions of the processes that brought about the decrease or increase.
Question 22
All the tiger snakes belong to the one genus which indicates that they are closely related, and
have all evolved from a common ancestor relatively recently (1). Earlier in time, when sea
levels were lower, Tasmania and surrounding island were connected to the mainland.
Populations of the common ancestor would have spread to various parts of southern
Australia, including Tasmania (1). Various populations became reproductively isolated from
each other by rising sea levels and other geographical barriers (1). Because environmental
2011 Assessment Report
Biology
29
Subject Code: BIO315109
conditions, such as the type of food available, differed for the isolated populations natural
selection lead to differing characteristics being selected. (1). This would have led to
adaptations and caused changes in the gene allele frequency leading to survival of the fittest
(1) Over time gene flow would have been prevented between the groups and quite possibly
the groups would not have been able to interbreed resulting in 6 different groups.
Or
The differences between Notechis scutatus and the rest became sufficiently great that
interbreeding was not possible even when placed in a common environment, so this group is
considered to be a new species (1). The remaining subspecies have diverged from each other
to the extent that they can be distinguished from each other but are capable of interbreeding
(1).
2-3 marks were given for describing isolation changes
2- 3 marks were given for describing the process of natural selection
1-2 marks were given for describing the divergence of the species and the incapable of
breeding aspects.
Note for teachers: Recent studies suggest that despite geographic and morphological
differences there is a low level of genetic divergence in the genus and that all tiger snakes be
treated as belonging to a single species, Notechis scutatus.
Examiners comments
Too many candidates got caught up with describing the spread off the species across the
continent and missed explaining the process of natural selection and speciation. Very few
candidates understood the concept of sub species. On the other hand some candidates also
gave a perfect description of natural selection which they had learnt but then made no
reference to the question asked and no reference to the snakes either. Consequently their
answers were not given full marks. Candidates must refer to the context of the question in
their answer. It was interesting to note that some candidates who wrote well beyond the lines
tended to say very little and scored only 1-1 ½ marks were other who contained their answer
to the lines were able to score 4 and above marks.
2011 Assessment Report
TASMANIAN QUALIFICATIONS AUTHORITY
BIO3 15109 Biology
ASSES SMENT PANEL REPORT
Award Distribution
EA
This year
Last year
L3% Q2)
L4%
$00)
HA
CA
20% (L38)
41% (28s',)
18% (13 1)
40%
(28s)
SA
27% (L86)
70L
(20L)
7L7
28%
Last year (all
LL%
20%
39%
30%
Previous 5 years
L6%
20%
37%
28%
Previous 5 years
(all examined
LL%
L9%
40%
30%
examined
subjects)
subjects)
Student Distribution (SA or better)
Male
Female
Year 11
Year 12
This year
36% (2s4)
64% (447',)
Ls%
(t07)
8s% (s94)
Last year
36% (2s8)
64% (459',)
20%
(L4s)
80%
33%
67%
Previous 5 years
t4%
Total
(s72)
86%