bio315114 - assessment report 2015 · 2015 assessment report page 3 of 23 - age – young children...

2015 Assessment Report of 23

2015 ASSESSMENT REPORT

OFFICE OF TASMANIAN

ASSESSMENT, STANDARDS

& CERTIFICATION

Biology Course Code: BIO315114 Part 1 was well done and the median was 26. Candidates did well in Part 2 and the median was 23. Part 3 proved to be a more difficult for some candidates and median was 20. Part 4 was also well done and median was 23. Part 5 also saw a range of answers and 19 was the median. 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 candidates’ 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. Section 1-Criterion 2 Question 1 a) To determine the optimal pH for this enzyme (1).

OR To determine the effects of different pH levels on enzyme activity (1).

b) There is no control (½). A control should give a baseline comparison (½) by not including the independent variable (pH) (½) and it is not possible to have a treatment with no pH (½). In this case, a test tube without the enzyme (at each pH level) could be used for comparison (to see if pH alone affects the result) (½). OR Not directly (½) but there are ‘implied controls’ (½). There is a range of pH values that can all act as comparisons (to establish the optimum pH) (1) and there are duplicate tubes at each pH that can be used as a reference for each other (½).

c) Any 2 reasonable responses including: -‐ The time that each of the test tubes is at the pH/temperature needs to be consistent; -‐ A control that does not include the enzyme solution could be included at each pH level; -‐ The experiment could be conducted with each of the eight groups of candidates testing each

pH value, rather than just test the enzyme action at one pH level; -‐ Have more tests (increased sample size; more replicates) at each pH; -‐ Increase the range of pH levels being tested.

Comments

The majority of candidates (about 90%) were able to get full marks (1 mark) for part (a) and about 65% were able to get full marks (2 marks) for part (c) with the most common suggestions to improve the


validity of the data being ‘increase the sample size’ and ‘record the appearance of the suspension at regular time intervals’.

b) This was more discriminating and very few candidates were able to obtain full marks. Most candidates were able to obtain part marks because they could identify the independent variable (pH) and clearly understood the difference between the control and controlling extraneous variables (eg. temperature plus enzyme & starch concentrations). However, a surprising number of answers indicated a very poor understanding of pH (e.g. stating a pH of 5 is neutral; recommending a control at pH 0; or ‘adding no pH’). In the given scenario, very few candidates were able to correctly identify either what the control should be or both aspects of the implied controls.

c) The most common error was the suggestion that the experiment should be repeated. No mark could be given for this as repetition only confirms that the initial results were ‘reliable’ and does not increase validity. Improving the validity of an investigation means improving the experimental design as ‘validity’ is the degree to which the procedure adequately tests the hypothesis.

Question 2 a) i)The average food energy intake decreases(½) as the local mean temperature increases.(½)

ii) Clear description of IV and clear description of DV (1½) cause and effect (½).

Higher average food energy is required in climates of lower mean temperature to provide energy for cells to produce enough heat to maintain homeostasis.

OR Increased exposure of people to cold conditions increases appetites.

1 ½ marks for the following answers as the cause and effect is not mentioned-

The colder the climate, the higher the energy intake. OR People from warmer climates will have a lower energy intake.

b) It could be concluded that the colder the climate, the higher the energy intake (½)

In terms of validity, this conclusion is not certain (½); the sample size was quite small with only 20 individuals in each region being tested; more data would need to be collected and analysed to give more valid results. OR People use more energy from food to maintain body temperature in colder conditions. Validity questionable due to lack of control; many other variables not controlled e.g. amount of food, body size, health, weight etc. of individuals, how long, too many unknowns. OR Food energy is used to produce body heat to maintain average body temperature of 370 C. Yes valid; a decrease in trend is shown. OR No; sample size is only 20 and human genetics and lifestyle differences require a larger sample size for statistical validity.

c) Two factors need to be considered from the following: - Individual variance of subjects being tested, such as their initial weight/height, - reduces variance - Amount of exercise undertaken – activity levels could affect intake of food. - Whether the individuals spent much time outside in the weather conditions, or in artificial climates

(such as air conditioning) – similar hours would provide some consistency - The sample size would need to be increased considerably - to account for individual variation - Seasonal variation and more climate locations could be chosen – to increase reliability of the

conclusion regarding link between temperature and energy intake - There may be an ethical concern with restricting people to a limited menu - Health and disease – unhealthy people may eat more / less depending on their condition - Lifestyle – smoking / drugs / alcohol consumption could impact on energy intake - Gender – males generally eat more than females


- Age – young children consume less than young adults - Occupation – people with physically active jobs have a higher energy intake - Time – over how long the study was conducted. Would need to be a long period to make the

investigation valid - Physiological effects – people may eat less than normal because they were part of an investigation.

1.5 marks were assigned for each reason. Comments a) (i) Some candidates just described the relationship as linear, with no other explanation.

(ii) Most candidates did not describe the cause and effect for the hypothesis and consequently were not able to gain full marks

b) Many candidates were confused between validity and accuracy, thinking they are the same. Several candidates did not answer the question, to say what the conclusion is, and lost an easy 1 mark. 2 marks were assigned for reasoning why it was valid or not.

c) In many cases only one answer was provided. Question 3 a) A placebo acts as a control (½) so that a comparison can be made with the results of the treatment

group (those receiving the drug AZT) (½) and those who think they are being given a treatment (the placebo group)(½). The purpose of a placebo in this investigation is that it validates the results by removing bias and subjectivity (1).

b) It is unethical (½) to withhold a possible treatment from a fatal disease (½) and to give no treatment at all is socially unacceptable (½). In this case (where the clinical trial showed a reduction in HIV infection), many pregnant women were deliberately denied a possible treatment (½) and the transmission of HIV to many newborn babies could have been prevented (½).

Comments a) The majority of candidates were able to get part marks (1 mark) for making very general statements about

the nature of a placebo and the necessity of controlling extraneous variables in an investigation. However, only about 50% addressed the question and discussed the purpose of using a placebo in these investigations (i.e. to determine the effectiveness of the AZT treatments). Very few candidates used the words ‘bias’ or ‘subjectivity’ (or words to that effect).

b) This was answered very well with most candidates obtaining between 1 & 2 marks. Most candidates indicated that they fully understood this question and the majority of answers stated that the clinical trials had been ‘unethical’. However, a number of candidates suggested that the drug was a ‘cure’ for the mother with HIV and did not realise (or state) that the actual benefit of the treatment was the prevention of transmission of HIV to the newborn.

Question 4 a) There were a variety of responses for this question, the most commonly accepted were:

-‐ Fixed variables of food, temperature, humidity and volume of solution were controlled (1) -‐ Using different concentrations of Aloe Vera (IV) (1) -‐ A very small sample size as only 5 heads and tails per treatment (40 specimens in total) (1) -‐ The use of qualitative measurements - no growth, partial growth or full growth – is not very

accurate. (1) -‐ The scientist wanted to test for healing properties on burns, not on cuts as were given to the

Planaria (1)

b) There were two main ways candidates responded, either were equally valid:

It is unethical to subject humans to unproven/untested treatments (1) as they may have harmful/unintended consequences such as long term ill effects (1).

OR


It is unethical to conduct this type of research on humans, as we would not be able to cut/burn them. (½)-AND THEN A RANGE OF RESPONSES TO GIVE THE REST OF THE MARKS- So if using this model, would be unable to apply to humans (½); the cost of testing on humans is prohibitive (½); a large sample size is needed for humans due to large variations in genetics/lifestyle (½); approval would need to be obtained from an ethics committee (½); people wouldn’t volunteer to be burned (½); can’t test on humans before a clinical trial (½); humans have to give consent (½).

c) The results would show that more growth (full or partial growth) occurred to Planaria in higher concentrations of Aloe Vera when compared to the control (1). This would show a positive correlation between Aloe Vera and its effects on growth warranting further investigation (1)

d) There were many responses that were accepted, and in most cases were a combination of: -‐ That higher concentrations of Aloe Vera should be trialled (1) This would help find an optimal

level to use for the repairing of tissue (1) -‐ Different types of application could be tested (gels, solutions, creams) to see if one might be

better in healing tissue than another (1) -‐ Different types of tissue damage should be trialled – animals with skin closer to humans/ burns/

abrasions/sunburn (1) -‐ The quality of the tissue that is regrown should be tested to see if it is flexible/same as before.

(1) -‐ Different temperatures of treatment need to be investigated to see their effects on tissue

growth (1) Comments As a whole, this question was very well attempted, and very well answered (with a median of 7.5 out of a possible 10 marks). Candidates showed they had a good understanding of experimental design and could tease these elements out well in the parts of the question. a) Most candidates scored very well in this section gaining 3 or 4 marks generally – but common

misconceptions included: -‐ There was no control -‐ There were no replicas -‐ That it was a huge problem that Planaria could regenerate on its own -‐ That heads and tails would grow differently and this would make the experiment invalid -‐ That Planaria is a plant and without a head the roots won’t grow -‐ That it is important to control light intensity.

c) This question usually scored 1 out of 2, candidates were able to understand that it was not ethical, but not really understand why.

d) This was the most poorly attempted part of the question, as many candidates failed to understand the significance of “what results would support..” and just went on to list many improvements that could be made in a very generic way, thus getting 0 out of 2. The candidates who did relate to the results could say there should be an increase in growth, but did not elaborate in terms of compared to the control, or in which strengths of solution and so on.

Mostly well done, but the candidates who did poorly, did not relate to “....effects on tissue growth”, and again just rewrote the original experiment, or suggested improvements to the original experiment in such a way that did not show understanding of the question.


Question 5 1mark Identify independent

variable Addition of “Triple 7 Bio Concentrate”/concentration of “Triple 7 Bio-Concentrate”/concentration of oil (fixed volume of Triple 7/type of oil (fixed volume of Triple 7)

1 mark Identify the dependent variable

Clarity of water (removal of oil)/efficiency of clean-up (time taken to break down oil or similar)

1 mark Large sample size/appropriate replication

Several, 100, containers with 50L of water of varying compositions (fresh/salt water) with 5 varying mixtures of water – should be set up in duplicate to mimic environmental conditions of waterways. OR 100 containers to test varying concentrations of Concentrate / varying amounts in 5 sets, (0, + 4 treatments or some range) OR Large (>10000L) tanks with varying concentrations of Triple7 or oil and replicated in time (3+ replicates) OR Other sensible sample regime (taking into account cost and feasibility) – ½ mark taken off for unrealistic conditions (1000+ tanks etc.)

2 marks Describe treatment of groups

1L of crude oil / or differing amounts should be put into each container. With the duplicates of water containers, half will receive 50mL of “Triple 7 Bio Concentrate” / or differing amounts and one set will not. OR 100 containers of seawater divided into 5 groups – one set no treatment added / others with varying amounts of treatment. OR Field experiment (following actual oil spill) where affected area is divided into different zones (e.g. physical barrier such as booms) and treatments randomly allocated. OR Other sensible designs e.g. different types of oil tested with same amount of Triple 7 or different concentrations of oil tested with same amount of Triple 7.

1 mark Identify control The containers without the “Triple 7 Bio Concentrate”. Need this one for full marks ½ mark for testing against previous best treatment option

1 mark Identify controlled variables

Heat, light, humidity (volume of oil/volume of Triple 7 depending on design) will be kept the same and amount of sample water

1mark Identify dependent variable and how it will be measured

Testing the clarity of the water. The more oil that is removed the more effective the “Triple 7 Bio Concentrate”. OR Skimming the water and weighing the oil collected OR Other sensible method of measuring oil breakdown or dispersal

1 mark Shortcoming of the experiment

The impact of the variables like changing climate and other variable that cannot be controlled in a field experiment, is not clear in this lab experiment, effectiveness of Triple 7 against different types or oil or effectiveness at different temperatures (i.e. oil spills in different locations)

1 mark Hypothesis The addition of “Triple 7 Bio Concentrate” is effective in removal of oils in case of oil spills in marine environments.

½-1 mark

Ethical considerations Unethical to purposefully create an oil spill in a nature environment (½), consideration of impact of Triple 7 on marine organisms (½) Full marks for this question not given to unethical designs e.g. deliberately creating oil spill

More marks than the 8 are suggested and any 8 well explained will do but the Control need to be included.


Comments This question was generally well-answered with modal scores of 6-8. There was quite a wide scope in the question to design an experiment which would be able to get marks in one or more of the categories provided in the model answers above. Most candidates who scored well organised their answers in a similar manner started with hypothesis, IV, DV, experimental set-up, results & analysis and potential problems with the design. A significant number (>25) did not attempt this question at all. A common mistake was to have the independent variable and dependent variable the wrong way round. Similarly, some candidates often didn’t mention the hypothesis being tested or identify the variables. Some candidates did not score full marks due to lack of detail in the dot points about the experimental design (“repeat a few times” or similar) and lack of detail about how to measure the dependent variable (“measure the effectiveness of the clean-up”). Most candidates chose to use a laboratory design recognising the greater challenges of conducting a field experiment. However, some candidates managed to score well using a field experiment provided they explained in detail how to avoid the issues of environmental variation. Section 2-Criterion 5 Question 6

Organic Molecule Likely location in the cell ATP Mitochondrion Lipid Cell membrane mRNA Nucleus Polypetide Rough endoplasmic reticulum Monosaccharide Chloroplast 1 mark for each correct answer

Comments Candidates did very well on this question with the vast majority scoring 5. The most common mistake made was to think monosaccharide were found in cell membranes rather than the chloroplast. Question 7 a) As substrate concentration increases, the enzyme action also increases (as seen by the number of

bubbles per second). (1) Credit was also given for the following answer: Enzyme rate of reaction is limited by substrate concentration (1/2)

b) D. (1) c) The enzyme catalyses the substrate through the induced fit theory (1/2) The active site attracts the

substrate (1/2) and is (chemically) changed to accommodate the shape of the substrate (1/2). The bonds of the substrate are weakened and products A & B are formed (½) No marks for stating that it is the lock and the key theory as it is stated in the stem.

d) This is an example of non-competitive inhibition (1) it appears that product B (from the breakdown of the initial substrate), binds to a site on the enzyme other than the active site to bring about a conformational change to the active site (1). This would stop the substrate from binding to the enzyme, thus preventing the reaction (1). OR Product B acts as a non-competitive inhibitor (1) it appears that product B (from the breakdown of the initial substrate), binds to a site on the enzyme other than the active site (1) to bring about a conformational change to the active site (½) This would stop the substrate from binding to the enzyme, thus preventing the reaction(½) OR Product B acts as a non-competitive inhibitor binding to the enzyme at another place(not the active site)(1) and in doing so causes the shape of the active site to change (1)so that the substrate no longer fits in enough to allow the reaction to occur(1)


Comments Overall this question was a well answered question with the majority of candidates gaining more than 4 out of 7 marks. Every single student answered at least 1 part of the question and very few gained 0 out of 7. a) A common error in part a was candidates misinterpreting the graph and simply stating that enzymes speed

up reactions which was not relevant to the question. b) Very well done with over 90% of candidates selecting graph D. Typically, candidates that got this incorrect

selected graph B. c) A common theme was that candidates did not mention that the enzyme needs to split the substrate up to

form the two products. Most candidates indicated it used the induced fit model. A common error was to refer to the lowering of activation energy instead of the induced fit model.

d) This was very well answered with most candidates gaining at least 2 out of 3. Number of candidates simply forgot to mention that once the active site has changed, the substrate can no longer bond to it.

Question 8 a) Cellular Respiration: ATP (adenosine triphosphate) (½). Photosynthesis: Glucose (C6H12O6) (½) b) Plants are autotrophs, and as they make their own food by converting the sun’s light energy into

chemical energy glucose in photosynthesis (1); This glucose is then broken down in plant cells via respiration to make ATP, as ATP is the type of energy needed for metabolic processes (1). Answers should show a clear understanding of the concept of plants as autotrophs and producing glucose as their source of energy and the link with respiration.

c) The high energy bond between phosphates

- It is able to be recycled /rechargable easily (ADPàATPàADP) - Carbohydrate rich diets and breakdown via mitochondria make it readily available - Is relatively small - Allows controlled many staged release of energy from glucose so not all

released at once. - ATP can be relatively easily stored and transported unlike many larger molecules.

Two well explained ideas. Comments This question proved to be challenging for candidates and showed a distinct lack of understanding in regards to photosynthesis and respiration. a) Typically answered well with the only common mistake being that candidates wrote out all the products

of respiration and photosynthesis instead of just the energy rich products. b) Very poorly answered; in fact majority of candidates gained less than 1 mark out of 2. This issue was

common in which many candidates believed that photosynthesis happens during the day to make energy and then when there is no light the plant starts doing respiration to make energy. Appears to be a very common misconception

c) A common mistake was candidates writing about how respiration produces ATP and the difference between anaerobic and aerobic respiration.

Question 9 a) i) Area (a):

Glucose produced during photosynthesis (1). The carbohydrate balance increases with maximum light (1) the rate of CO2 used (½)

ii) Area (b): Glucose being used in cellular respiration (1). CO2 produced (½)

b) Yes - Photosynthesis produces glucose and oxygen (½), while cellular respiration uses glucose and oxygen (½) As shown by the graph, during daylight hours (½), the rate of photosynthesis is higher (½) than the rate of respiration (½)therefore the amount of oxygen produced (½) is greater than the amount of oxygen used/ consumed (½)


Comments Overall if candidates understood the graph they did well with this question. Candidates needed to differentiate the activity during the day or when the sun is out and night time as during the night the oxygen production is less than oxygen consumed. Without this maximum marks were 1½. Candidates also needed to mention a link between carbohydrate balance and oxygen. Some candidates didn’t relate the graph to glucose production in photosynthesis or glucose used in respiration and then just stated one or more of the labels on the graph. Question 10 a) tRNA / (transfer RNA) (1) Ribosome (1) anticodon (½) b) Translation (1) Protein synthesis (½) c) To synthesize a polypeptide, a gene in the DNA (located in the nucleus) is unzipped;

Free nucleotides move in to create a complimentary strand of mRNA (to the original/coding DNA strand).This is the process of transcription(1).The mRNA leaves the nucleus through a pore in the nuclear membrane, and ‘attaches’ to a ribosome (in rough ER, located outside nucleus).At the ribosome, the mRNA is decoded (every 3 base pairs forms a codon)(1) Each codon attracts a tRNA molecule with the anticodon and the corresponding amino acid attached This is the process of translation(1). The amino acids are left behind, forming peptide bonds between them, creating a polypeptide chain, that has been specifically coded for from the original DNA strand (1).

OR A section of the DNA is unzipped (½) and a complimentary strand of mRNA is formed (½)by copying the section of DNA by matching the nucleotide bases. mRNA has uracil instead of thymine (½). This is the process of transcription. The mRNA leaves the nucleus (½) and travels to ribosomes in the cytoplasm (½) either free or on Endoplasmic Reticulum. At the ribosome the mRNA codons (three nucleotide bases) (½) are read by the ribosome and the tRNA with the complementary anticodon (½) brings the specific amino acid (½)that has been coded for by the mRNA. As each tRNA moves into the ribosome the amino acids are joined to make a polypeptide (½) which is specific to the origional code on the DNA (½)

Comments A number of candidates mentioned extra details like the DNA is unzipped by the enzyme helicase or The initial copying of the DNA included coding and non-coding sections. The non-coding sections are spiced out to make mRNA. Almost all candidates attempted this question. Candidates wrote fluently on the topic of protein synthesis showing that they understood both transcription and translation and the specificity of the process. Candidates were awarded ½ marks for each point that built their explanation to a maximum of 8 half marks. Question 11 a) Glu – Gly – Phe

b) Although two of the bases have changed GAA – GGG – TTT the first codon still codes for the same amino acid Glu, while the second codon changes the amino acid from Arg to Gly (1). A change in the amino acid sequence will mean a different polypeptide/protein is formed, (1) which could make the resulting protein/enzyme non-functional. (1)

c) If the TSD mutation has occurred and Hex A has been affected then it will not be able to break down the glycolipid into compound Q.(1) This means that the glycolipid would then accumulate in a Tay Sachs patient, (1) so compounds Q, S, T are not produced. (1) OR Glycolipids cannot be converted to compound Q therefore glycolipids will accumulate. (1) Without Q, compound S cannot be formed and nor can T. (1) The pathway stops at glycolipid therefore compound Q is not produced.(1) The plasma membrane may therefore be structurally compromised.(1) OR


Glycolipids cannot be converted to compound Q therefore glycolipids will accumulate in a Tay Sachs patient. (1) As compound Q is not formed then compound R does not combine with it to produce S and then T.(1) Both compound R and glycolipid may build up due to mutation of Hex A.(1)

Comments a) Most candidates attempted this question and majority answered it correctly allowing for an easy mark. b) Many candidates simply copied the information from part (a) by writing out the amino acid sequences

again. This did not gain any marks. Only half a mark was given for stating the enzyme is faulty (stated in the stem of the question). The key word here was effect of the mutation on Hex A i.e. the question does not ask ‘what is’ the mutation (i.e. substitution/point). Many candidates simply stated that the mutation will change the shape of the enzyme without any reference to the change in sequence of amino acids which affects the polypeptide chain which could therefore produce a different shaped protein/enzyme. Those candidates who mentioned that the plasma membrane may be structurally compromised gained marks over candidates who simply copied information from the diagram without stating any cause and effect relationship. Majority of candidates did not recognise that GAG and GAA code for the same amino acid and therefore could not gain the full three marks.

c) Most candidates understood there was a pathway that was being affected by the lack of Hex A and so gained some marks by stating that compound Q would not be produced. To gain full marks the candidates had to state that the glycolipid would accumulate due to the mutation of Hex A. Majority of candidates found it difficult to relate to compound R and many thought compound T would still be produced as compound R was being converted to S. In fact compound R would usually combine with compound Q to create S and then T but as Q would be absent (in a Tay Sachs patient) then compound S would not be produced.

Section 3-Criterion 6 Question 12 a)

Diagram Name

A Animal Cell/ Eukaryote Cell

B Prion

C Prokaryote Cell

D Virus

E. Eukaryote Cell

F. Plant cell/ Eukaryote Cell


NOTE: ½ a mark per correct response b) Diagram C – the Prokaryote cell; Bacteria are single celled prokaryotes, and contain circles of DNA which

are plasmids. c) Any cell with 1 mark for a structural feature/s and 1 mark for a detailed explanation of how this/these

relate/s to its function. Examples include:

- Prokaryote Cell (bacterium): has pili and flagellum; as is a single cell and needs to be able to move freely /has a large SA: V ratio; making exchange of materials with the environment more efficient.

- Animal cell (sperm): has a tail; so that this gamete can find its corresponding gamete for fertilisation; OR has a mitochondrion in close proximity to the tail; to provide the energy for movement; OR has an arrow shaped head for streamlining; OR has an acrosome to help break through ovum membrane.

- Virus: tail fibre and base plate; so that the viral particle can attach to cells plus small size. -‐ Prion: misfolded protein binds to other proteins of the same type and induce them to change their

conformation as well, producing new infectious material. -‐ Plant cell: Chloroplast for photosynthesis, large central vacuole for turgor and to push chloroplasts

to outer edges and cell wall for support. -‐ Eukaryote cell; small for better SA : V ratio, flagellum for movement, eyespot to detect light,

chloroplasts for photosynthesising

d) 1. Prion 2. Virus 3. Bacterium 4. Paramecium

1/2 mark if only one part is correct.1 mark for all correct.

Comments a) Most student got 3/3 for this section. A few confused the sperm cell and paramecium. Note that it is

possible to choose eukaryote cell for plant or animal, but then neither of these are appropriate for E. Only use each name once.

b) Many candidates lost marks on this question even if they knew the answer as they were not able to explain it clearly. They needed to recognise that plasmids are found in bacteria ad many prokaryotes are bacteria. A lot of candidates used the information sheet to write about plasmids but didn’t connect it to diagram C.

c) Most candidates did well on this section and gained 2/2. Most found at least 2 structural features and linked them to the appropriate function. The most common choices were the sperm cell or the plant. With the plant the function of strength for the cell and plant should have had the cell wall and the large vacuole together providing the function.

d) Many candidates got this correct. The most common error was to make the bacteria smallest.

Question 13 a) (i) A = hydrophyllic (water loving) head/glycerol/phosphate (½) (ii) B = fatty acid/hydrophobic (water hating) tail/lipid (½) (iii) C = bilayer/membrane/phospholipid membrane (layer) (½) (iv) D = protein/pore/channel/cholesterol/ion pump (½)

A = head with B = tail gained ½ mark b) As a channel/pore for fac i l i tated d i f fus ion (1) OR for Act ive Transport (1) OR for the movement

of large particles (1) OR to provide structural support for the cell (1) General answers such as “for the movement of particles in/out of the cell” gained no marks; whilst answers that mentioned selective movement of particles gained ½ mark.


c) Select ive transport of molecules OR to regulate movement of materials in/out of the cell (1) – lipid bilayer is a barrier to the movement of non-lipid soluble molecules/water/charged particles, whilst protein channels allow movement of large/charged particles (1). OR Lipid bilayer is very flexible (1) which enables endocytosis (invagination of the membrane to encapsulate large solids/liquid droplets) and exocytosis (secretion of large substances when vesicles fuse with the cell membrane) to occur (1).

Comments a) A = head with B = tail gained ½ mark

This part was straightforward and done well by most candidates. A few candidates had not read the stem and thought that the diagram represented a cross-section through a leaf.

b) To gain full marks, candidates need to specify how the function of pore/channel membrane differed from that of the phospholipid bilayer. Identifying D as an antigen or receptor and describing their function also gained ½ to 1 marks, depending on the clarity of the response.

c) This part was the most challenging, with many candidates only providing general statements about cell membrane being “responsible for the transport of materials”, without referring to the fluid mosaic model and the differing roles of the various components of membranes.

Question 14 There was an error on the graph for this question. The “after” for the vinegar treatment on the “Change in circumference of the egg” histogram should have been 18 cm. As a result of this error, it was decided that complete explanation of the corn syrup and water treatments was all that would be required for full marks. In all cases osmosis (½) – the movement of water in/out of the egg has occurred (½). This must be mentioned somewhere for full marks. Corn Syrup: The egg is in a hypertonic/more concentrated solution (1), so there is net/overall movement of water (½) into the corn syrup (½), which decreases the mass and circumference (½) of the egg, from 100g to 40g and from 18cm to 16 cm (½) respectively. Use of data/numbers from the graph is required for full marks. Water: The egg is in a hypotonic/less concentrated solution (1), so there is net/overall movement of water (½) into the cell (½), which increases the mass and circumference (½) of the egg, from 40g to 80g and from 16cm to 19cm (½) respectively. Use of data/numbers from the graph is required for full marks. Where candidates failed to gain full marks, their answer to the effect of vinegar on the egg was considered and ½ mark given for either recognition that the graph was incorrect OR any sensible explanation for the fact that the circumference of the egg was unchanged (e.g. the vinegar dissolved the egg shell/the egg swelled, filling up air space inside the shell). Comments This question was very well done, with a large number of candidates gaining 5 or more marks. The most common reason for loss of marks was the incorrect application of the terms “hypotonic” and “hypertonic”. Where candidates used these terms incorrectly, but also correctly referred to the solutions as having a “strong/high concentration” or “weak/low concentration”, they were not penalised. However, incorrect use of “hypotonic” to refer to a highly concentrated (in terms of solute) or “strong” solution meant a loss of 1 mark (and another 1 mark was lost if the equivalent error was made with the term “hypertonic”). Whilst the use of correct biological terminology is encouraged, no marks are generally lost for using simple, everyday terms to describe situations! There was also some confusion regarding the term “concentration” with a number of candidates using it to refer the relative amounts of solvent/water, rather than solute/dissolved stuff. Other things that should be noted:

-‐ Changes in volume/mass are a lways due to the movement of water , not ions/particles/molecules/solute

-‐ Solutions consist of a solute dissolved in a solvent (usually water). Make sure you specifically state what is moving – water, ions, and solute – rather than saying “it/stuff/the solution/liquid etc. moved into/out of the cell”.


-‐ If data is provided in a graph or table, make sure you refer to it/use it in your answer. -‐ Some candidates used the words “endocytosis” and exocytosis” to describe the d irect ion of

movement of particles. However, these terms only relate to the movement of very large particles (like proteins) that are floating”/suspended in water rather than being dissolved in it, and are encapsulated in a membrane vesicle.

Question 15 a) The cell must use energy to actively transport these ions. (1 mark) b) Active transport requires energy (ATP)

A respiratory inhibitor would interfere with cellular respiration; thus there would be less ATP produced, and less energy available for active transport; So no active transport of Na+ out of cell and K+ into the cell So Na+/K+ ion levels would only be maintained by diffusion, Na+ will enter cell along concentration gradient K+ will leave cell along concentration gradient thus achieving an equilibrium (not maintaining a difference). (½ each point)

Comments a) A very high percentage of candidates got 1/1 for this section. But some candidates looked at the table

incorrectly and read down instead of across to determine the direction of the concentration gradient. This meant that they answered “passive diffusion”.

b) Very few candidates gained full marks for this section as they wrote extensively about cellular respiration with and without oxygen and there resultant ATP molecules. It was necessary to point out that the respiratory inhibitor would reduce the amount of ATP but the main point of the answer needed to be its effect on the cell and the actual measureable outcome (i.e. the concentration of Na+/K+ inside and outside the cell after the inhibitor).

Question 16 a)

*Crossing over must occur between AA and aa chromatids to swap alleles, resulting in chromatids with Aa and Aa (½) *Independent Assortment (lining up across equator) should be shown to indicate resulting cells

½

½

*One of these should be a lower case ‘a’ allele


b) Up to one mark for each clear description of a mechanism causing genetic variation during meiosis and

fertilisation. -‐ During gamete formation in meiosis crossing over occurs between the arms of homologous

chromosomes leading to mixing of genes in chromatids (1) -‐ random assortment occurs in the second stage of meiosis, so there is a chance mixing of chromatid

combinations in the gamete (1) -‐ random pairing of gametes during fertilisation (1)

Credit also given to answers describing how non-segregation can lead to variation. The word ‘mutation’ on its own did not attract credit. Comments a) Generally well answered. Most candidates were able to draw paired homologs with the correct alleles side

by side. ½ mark was lost if crossing over was not shown in the first diagram. The second division would normally show the chromatid carrying the ‘A’ allele ending up in the left gamete, but as the corresponding diagram showed the ‘A’ chromatid in the right gamete, either option was accepted. Correct use of upper and lowercase labelling was required.

b) Generally answered poorly which resulted in a significant number of candidates achieving total marks in

the 1 ½ – 2 ½ range. Understanding of crossing over and random assortment was limited. Many mentioned the terms but gave no explanatory comment that expanded on the terms. Candidates commonly referred to swapping of chromosomes instead of swapping of alleles during crossing over. It was clear that understanding of the terms homologous and chromatid was limited.

Question 17 Candidates needed to address both points to attain full marks.

- What makes cells the ideal basic unit of life? - How do viruses and prions “fit” into this theory?

Responses regarding ce l ls that attracted a mark each included: -‐ Cells are the smallest unit that can sustain themselves in their environment -‐ Cells can undertake mitosis to be able to replace cells, and meiosis in specialised regions to form

gametes. Therefore, they are capable of reproducing themselves -‐ Cells have a very high SA:V ratio making exchange of materials with the environment an efficient

process (decreased diffusion distance) -‐ Cells can specialise by having different organelles, shapes etc. to suit particular functions -‐ Cells can form complex arrangements (cells à tissues à organs à organ systems) that exist to

give rise to multicellular organisms. -‐ Cells can carry out metabolic processes such as respiration that sustain life.

Responses regarding v iruses that attracted a mark each included:

-‐ Outside a cell, viruses can crystallize, behave like inert chemicals -‐ Viruses do not show growth or carry out metabolic processes. -‐ Viruses do not have a need for nutrition or are unable to reproduce. -‐ Viruses are only capable of replication/reproduction when inside a host cell. -‐ Viruses are considered non-living. -‐ They contain DNA or RNA and have a protein coat. -‐ They have different ‘species’ and can exhibit mutations (which is why some consider them to be

living) -‐ Are not cellular in structure (½)

Responses regarding pr ions that attracted a mark each included:


-‐ Prions are proteins that have a fault (abnormally folded) -‐ They act as a template for other prions to be formed upon contact with other proteins -‐ When they collide with other proteins they cause them to change their shape to prion form. -‐ Prions cannot reproduce independently. -‐ They lack cellular structure and cannot metabolise. -‐ Do not require nutrients or produce wastes -‐ Do not have genetic material and cannot be classified as living

Comments This question was not answered well by the majority of candidates. Many spent time re-writing the stem of the question and provided sweeping statements about how all life is based on cells and how life cannot exist without cells and how all cells come from pre-existing cells. Unless new information about cell characteristics was given, no marks were able to be awarded. A number of candidates gained some credit by dissecting information about prions and viruses from the Examination Information Sheet, but this typically only resulted in mark totals in the 2-3½ range. Approximately 40 candidates did not attempt the question. Section 4-Criterion 7 Question 18 a) i) Autosomal recessive 1 (½ autosomal, ½ recessive);

Reason for being autosomal (1) (or not X linked) include one of: -‐ each parent must have one recessive allele to pass to the offspring, but are not affected

themselves as they have a dominant allele to mask its expression; -‐ could not be X linked as I-2 would have to have the trait to pass it on to II-1.

Reason for being recessive include: This is due to both II-1 and III-5 inheriting the disorder, while neither of the parents has the disorder or skips a generation (1).

ii) For sickle cell disease – DD = normal (homozygous dominant); Dd = Normal/Carrier (heterozygous); dd = affected (homozygous recessive).

Therefore: dd x Dd (1)

d d D Dd Dd d dd dd

This cross shows that there is: 50% chance of Dd – being heterozygous for the trait. This would be a normal child, but a carrier for the trait. 50% chance of dd – being homozygous recessive and being affected by the trait. Therefore, the chance of having an affected child is 50% (1).

b) As alleles IA and IB are codominant to allele iO ; This means that both allele A and allele B are expressed, while allele O is recessive and thus only expressed when there are two copies (iOiO).Hence, there can be four blood groups given by 6 possible genotypes: (1)

Group O – iOiO Group A – IAIA or IAiO

Group B – IBIB or IBiO

Group AB - IAIB c) Children must have type O, A, B and AB. As they inherit one allele for blood group from each parent, the

OO group child must have one O from each parent, the AB group child must also have an A from one


parent and a B from the other parent .So, one parent must be heterozygous for Group A (IAiO), the other heterozygous for Group B (IBiO). (1)

Possible offsrping for this combination of parents are (1)

IA iO IB IAIB IBiO iO IAiO iOiO

Therefore, possible offspring from these parent:(1)

25% chance IAIB or group AB 25% chance IBiO or group B 25% chance IAiO or group A 25% chance iOiO or group O

There is an equal chance of each blood type in the resulting offspring. Therefore, one parent’s phenotype is heterozygous for Group A and the other is heterozygous for Group B. Parents genotype and explanation for offspring (2 marks), Working (1) Comments Overall candidates made good attempts at this question. There seems to be some confusion in pedigrees around X linked terminology. The impression candidates seem to have is that some inheritance is just male or female based forgetting that X linked inheritance can at times affect both male and female. a) i) Most candidates were able to recognise and explain that it was recessively inherited disease although

they thought the difference between recessive and dominant is its frequency (i.e. it is recessive because less people expressed it). There were logical presentations of the reasoning and also quite obscure but accurate reasoning. If the reasoning was correct but not necessarily sequenced correctly then they were given the marks. ii) This was answered satisfactorily, the main error was trying to use X and Y in the punnet square after just having said in the previous question that it was autosomal inheritance.

b) Many candidates failed to give all possible genotypes and lost part marks for this. A lot did not understand codominance or remember the word or also forgot to mention that O was recessive as part of their reasoning.

c) This was an easy question if they understood inheritance. Many candidates could do the punnet square correctly and identify the parents alleles, but had a little more trouble explaining their reasoning even as simply as saying something like’ parents had to be heterozygous for all alleles to be present’. Or not indicating parent phenotypes just showing genotype.

Question 19 a) A number of reasonable responses were accepted, including:

-‐ Villi/Microvilli (½) increase the surface area (1) for absorption of digestive products -‐ Muscle layers in the digestive tract (½) help to move food along by peristaltic contractions (1) -‐ Cells contain many mitochondria (½) for the production of ATP (energy) to facilitate absorption (1) -‐ Blood vessels in close contact with digestive surface (½) allow for the products of digestion to be

transported. Single layer of cells of epithelium (½)-small diffusion distance. (½) OR Maintenance of concentration gradient by removal of absorbed nutrients (1)

½ mark for structure 1 mark for explaining how structure relates to function. Stating a structure without clearly indicating its relationship with the function yielded no marks. b) The increase in water in the waste product (i.e. diarrhoea) indicates that water is not being absorbed

into the bloodstream (1); Water reabsorption mainly occurs in the large intestine (1), so this must be


the area most affected. (intestine, small intestine were also accepted however whilst this is true the majority of water is absorbed into the body via the colon)

Comments Most candidates scored full or near full marks for this question. Candidates lost marks for repeating what they had said about villi and increased SA by then writing about microvilli and increased SA instead of mentioning something else. Question 20 a) They work together to maintain a high concentration gradient between body cells and tissue fluids; so that

carbon dioxide and oxygen can move passively to provide O2 and remove CO2 from cells. The above details are required for full marks. OR The two systems work together, the respiratory system obtains oxygen from the atmosphere into the body, and expels carbon dioxide from the body to the atmosphere (½), while the circulatory system transports the oxygen to the cells and the carbon dioxide away from the cells (1).

b) i) That smoking increases heart rate and number of red blood cells (1) ii) Smoke exposure decreases the lungs ability to exchange gases(1)so this needs to be compensated by increasing capacity of blood system to carry oxygen to cells (1) carbon monoxide reduces carrying capacity, higher red blood cell count high viscosity and blood pressure (1). The above details are required for full marks. OR If carbon monoxide is taken up by red blood cells instead of oxygen in smokers (1) then the body would make more red blood cells to try and get more oxygen (1). OR More red blood cells in same volume of plasma makes blood thicker (1) the heart has to work harder to push the thicker blood around, leading to increased heart rate (1).

Comments a) Most candidates lost a mark for not mentioning the high concentration gradient that makes exchange of

gases possible. b) i) Most candidates understood the information in the table and scored full marks.

ii) This part was more discriminating and most candidates lost marks for not stating the decrease in ability of lungs to exchange gases as a cause of changes in the heart rate and blood cell count. Most candidates showed good understanding of the general information about the negative impact of smoking but the details required to obtain full marks were missing.

Question 21 a) Plants have two systems for the transportation of substances - using two different types of transport tissue

– xylem and phloem. Together they are known as vascular bundle


Transpiration Translocation Xylem vessels are involved in the movement of water (and dissolved minerals) through a plant - from its roots to its leaves via the stem. • Transport in the xylem is by pass ive

processes. • Water is absorbed from the soil through

root hair cells. • Water moves by osmosis from root cell to

root cell until it reaches the xylem. • It is transported through the xylem vessels

up the stem to the leaves. • The processes of capillary action and

transpirational pull are involved in moving water passively through the xylem

• Cohesive forces between water molecules assist the movement of water through the xylem.

• Water evaporates from the leaves (transpirat ion) through the stomata.

Phloem vessels are involved in moving d isso lved sugars , produced during photosynthesis, and other soluble food molecules are moved from the leaves to growing tissues (e.g. the tips of the roots and shoots) and storage tissues (e.g. in the roots). • Transport in the phloem is act ive . • Substances can move against a

concentration gradient. • Substances can be moved both up and

down the plant. • The companion cell provides the ATP

necessary for active transport.

½ mark for each re levant piece of information. The focus of the question is on the mechanisms of transport rather than the structure of the tissues involved. b) ½ mark for naming a feature, ½ mark for explaining. Many possible answers including:

-‐ Thick cuticle layer (½): provides a waterproof layer to help minimise water being evaporated from the leaf surface (½).

-‐ Sunken stomata (½): by having stomata below the surface, no direct sunlight is acting on the stomata (½); creates a small pocket of humidity to equalise water gradient thus reducing water lost by evaporation (½)

-‐ Reduced number of stomata (½): less opportunity for water to be lost to atmosphere (½) -‐ Thick layer of parenchyma (½) provides insulation in both extremes of temperature (½) -‐ Thick layer of parenchyma (½) allows for the storage of water (½) -‐ Slightly curled leaves (½) trap moist air and provide a microclimate that slows the transpiration

rate and thus the rate of water loss (½) -‐ The angle and shape of the leaves (½) directs water down towards the roots of the plant (½) -‐ Spikes on the leaves (½) deter predators (½).

Comments Most candidates attempted this question and most gained some marks in doing so. There was some confusion about the information that was required in part (a). This was caused by the mention of the heart in the stem of the question which led candidates to assume that they needed to make a comparison between animal and plant transport systems. No marks were given for information on animal transport. Candidates who mentioned animals still managed to give enough information on plants to get some marks. Some candidates assumed that the word ‘fluid’ in the question referred just to water and thus only gave detail on transport in the xylem. There were also a number of candidates who thought that the mechanism of movement of water in the xylem was simple diffusion. Candidates should be careful in using the information sheet and ensure they are answering the question asked. The question asks specifically about the mechanisms of transport and so chunks of information about the structure of the tissues involved attracted no marks.


a) Asks for a comparison of systems of plant transport. However, although many candidates presented quite detailed information about the different tissues, few actively compared them.

b) Generally well answered with many candidates gaining the full 2 marks. Marks were lost for in this part for features that were described that had no relevance to survival in an extreme environment. Quite a number of candidates claimed that the leaves of the Agave plant were reduced – narrow and thin – which they clearly are not. Their thinking in saying this was that a reduced leaf would have a small surface area to volume ratio and thus would be less prone to dehydration in times of water stress.

Question 22 a)

One right ½ mark, 2 right 1 mark, 3 right 1 ½ marks, 5 right 2 marks. b) Alcohol is A (½) Alcohol causes a decrease in the amount of ADH released. A decrease in ADH will lead to less reabsorption of water, therefore more urine being produced (1). Therefore patient A, with urinary volumes of between 2.8 and 3.5L per day is more likely to be under the effects of alcohol (½).

Ecstasy is C (½) Ecstasy causes an increase in the amount of ADH released. An increase in ADH will lead to more absorption of water and less urine being produced ( 1 ). Therefore, patient C, with urinary volumes of between 0.2 and 0.4L per is likely to be under the effects of Ecstasy ( ½ ). Comments a) This question was well answered. b) A number of candidates did not make the correct connection between ADH, water absorption and urine

amount. It was disappointing that many candidates referred to patient B as their answer when it was stated that Patient B had normal results. A common mistake was the lack of reference to the data in the graph. Marks were also allocated to candidates who further explained the process of absorption in the kidney. Many candidates did well in this part of the question.

Blood concentration returns to normal

Large volumes of dilute urine produced

Renal tubule reabsorbs less water

Detected by receptors in the hypothalamus

Pituitary releases less ADH


Section 5 - Criterion 8 Question 23 a) *½ mark for each correct response

(i) Detritivore: Arthropods (ii) Primary producer: Higher plants OR algae (iii) Is both a tertiary and quaternary consumer: Top carnivore algae à mollusc à bird à mammal à top carnivore higher plants à birds à mammals à top carnivore (iv) Herbivore: annelid OR mollusc OR arthropod OR birds OR mammals

b) As they are heterotrophs (rely on a food source), they would most likely be decomposers; Decomposers (½) (have a significant role in the recycling of matter (1), breaking down organic material into its inorganic components (1).

c) There are a variety of ways this could be explained, marks could be awarded for responses including:

The organisms in the table belong to a food chain within the web: Algae à mollusc à fish à bird (½)

- Cadmium levels appear within normal levels at the lowest trophic levels at 2.9µg in the algae, however levels can then be seen to increase at each consecutive trophic level (½)

- This must be due to bioaccumulation(½) of the cadmium in the fatty tissues of each organism; Cd is not biodegradable(½)

- As a mollusc eats many algae, then a fish eats many molluscs (and so on), the cadmium concentrations are increased at each step of the food chain, (½)which is the process of biomagnification (½)/ body conc. levels can increase by around 10x each trophic level as each consumer eats many times its own weight in food; (½), 10% of the energy gained from the previous trophic level is passed on to the next level, (½) the total biomass of organisms occupying each trophic level decrease but Cd content will pass on 100% (½).

- Higher order consumers (such as top carnivores and humans), would be exposed to many times the recommended level of cadmium if eating organisms in this from this food web (½).

Comments a) This question was well attempted and most candidates gained some marks. Lost marks were typically due

to not identifying the detritivore as the organisms that lives on detritus, the most common answer was detritus. Identifying the 3th and 4th order consumer, less than half candidates choose the correct answer.

b) Most attempted, candidates who could relate the function with that of decomposers did well. Some copied the definition from the information sheet.

c) Most candidates answered the question and stated bioaccumulation as the account for the result. There were some very good and well detailed responses as well as some very general responses with no real explanation as for why it was happening.

Question 24 a) Due to energy losses, only 10% of the biomass of one trophic level is transferred from one level to the next

(1), so each trophic level must consume up to 10 times their own biomass to sustain life (1). Thus after 4 or 5 trophic levels there would not be sufficient biomass to support a higher level (1).

b) Organisms need energy for cellular processes such as respiration (½) so some energy is changed from chemical to heat energy (½) and lost from the system. Energy is also used in tissues that are not consumed and passed on (such as bones and fur), and so is not passed on any further in the food chain(1).


c) i) Energy in an ecosystem comes from the sun (1). Producers trap this energy and along with CO2 from the carbon cycle and convert it to chemical energy in glucose (1). The process of photosynthesis combines energy flow from the sun, with the cycling of matter (carbon) from the atmosphere (1).

ii) Carbon is an element which cycles between the environment and the food web. It changes form (½) but is never lost from the system (½). Energy enters the living system from the sun (½) during photosynthesis and flows in one direction through the trophic levels (½). It is gradually lost as heat and cannot be recycled through living things (½). (Any 2 of these answers for a total of 1 mark for energy flow).

Comments a) This question was answered well, with most candidates having a good understanding of the loss of energy

between trophic levels in a food chain. b) Most candidates identified heat loss due to metabolic processes and scored 1 mark for this question. Very

few mentioned loss of energy due to indigestible body parts. c) i) Candidates had difficulty explaining the link between the carbon cycle and energy movement. Many

candidates gained ½ or 1 mark for mentioning photosynthesis and/or glucose without going on to explain the link between carbon and energy. Some discussed respiration/decomposition, with release of energy and carbon dioxide, and were given credit for that. ii) Most candidates understood that carbon is cycled, whereas energy is a one way flow through ecosystems. This basic answer gained 1 mark, with more detail required for a higher marks.

Question 25 a) The graph shows a predator-prey relationship (1); the rabbit has the high numbers of the prey organism, its

numbers peak (possibly due to food abundance/good seasonal conditions), then decline rapidly (1). The fox has the lower numbers of the predator organism, its population peak follows the rabbits, after rabbit numbers increase (½), foxes have more to eat so their numbers increase (1). When they increase by too many, rabbit numbers decline. Once there is less to eat, fox numbers decline (1). This pattern is repeated (½).

b) Structural (1) - Ability to borrow provides protection for young. Ability to see at night increases hunting efficiency. Red colour offers camouflage in the Australian desert. Digestive system that enabled them to digest both plant and animal material.

Behavioural (1) - Being nocturnal provides darkness for protection when hunting for food.

c) A number of possible responses. Examples include:

- They are in direct competition for resources (1) - The Dingo is a better competitor for resources than the fox (1) - They inhabit the same niche (½)

d) Any plausible response (1 mark per point). Examples include:

-‐ No, Tasmanian Devils are not native to that area (1) and thus may have the same impact as an invasive species (1). This would cause predation on and extinction of native species (1).

-‐ No, As yet it has not been demonstrated that Devils control fox numbers in Tasmania. Introducing another predator into the Victoria would increase competition and predation on native animals (1) and reduce those populations to extinction (1).

-‐ Yes, Devils could be used as a biological control (1). They are a native Australian animal (½) and as such it would be a better option (½) to have them outcompete the fox for resources (1).

-‐ Yes, The climate (abiotic factors) in Victoria are more like Tasmania (1) and the Devils are more likely to control fox numbers than the Dingoes (1). This would increase numbers of native fauna species (1).


Any rational explanation accepted. Comments This question was quite well done with most candidates scoring more than half marks. Most candidates knew about predator–prey cycles and they were marked on the degree to which they could explain this concept. Most candidates understood adaptations although they were less able to differentiate behavioural and structural adaptations. The relationships question was well understood but candidates should really have just said “Competition” to name the relationship. The last part was the most discriminating in that it tested the candidates’ ability to apply the relevant concepts. a) Some candidates pointed out the relative biomass difference because of the trophic levels and they were

rewarded for this. Many thought that the fox and the rabbit were only introduced recently and/or at the same time. Many used the concept of carrying capacity quite wrongly. The oscillations in the populations are around the carrying capacity. The carrying capacity is not at the peaks or troughs. E.g. “The foxes brought the rabbits back down to their carrying capacity.” Or “The carrying capacity is too large for the devils.” A few answers talked about environmental resistance and some used this term well but others used it in more formulaic manner indicating a degree of rote learning rather than application to the question being asked. Answers that spelt out the fact that foxes take longer to reproduce were rare but were rewarded.

b) Many did not differentiate between a behavioural and a structural adaptation. Some mixed up structure and function and there was some overlap between the two.

Most candidates used the examples of burrowing and being nocturnal but few talked about the structures that would enable them to do these things.

c) Only about half had that the fox and dingo were in competition.

Many didn’t name the relationship but could describe it adequately to gain their point. The majority of the rest had a predator prey relationship the dingo preying on the fox. Since there was nothing to discount this, they were given the mark. Often candidates mentioned niches but unless they supplied a bit more detail this only earned half marks. Parasitism was a common wrong answer.

d) The main criticism of answers to this question was that candidates did not make three points or did not make them fully. Candidates need to explain how their various points are linked together in a logical sequence and should be able to express themselves in more specific biological language rather than general, vague terms.

There were a lot of answers that mentioned the TD Facial Tumour Disease. This was irrelevant to this question and scored no marks. Although a few good answers did state that “The Devil population may become established again on the mainland and this would act as an insurance population.” This was rewarded.

Question 26 a) There were a number of possible answers for this part:

-‐ The dingo is more closely related to the Asian wolf than the domestic dog (1). -‐ The Asian Wolf and the dingo are the same species (1). -‐ The dingo couldn’t produce fertile offspring with the dog, but could produce fertile offspring

with the Asian wolf (1).


-‐ The dingoes are still undergoing speciation and are now more similar to the Asian wolf (1). -‐ The dingo can produce fertile offspring with the Asian wolf (1). -‐ The dingo is now thought to have originated in Asia (½) and is descended from the Asian wolf

(½). -‐ The dingo shares a recent common ancestor with the Asian wolf (½), but not the dog (½). -‐ The dingo and Asian wolf are both wild (not domesticated) (½) and therefore closely related

(½). b) Physical/physiological variation existed in the initial native dog population of Indonesia and New Guinea

caused by genetic variation (½). Genetic variation was caused by random mutation, crossing over, independent assortment, or sexual recombination (½) (only need to mention one cause of variation). Geographic isolation occurred between populations of native dogs (½) due to continental drift, sea levels rising, and distance due to migrating to Australia, or another physical barrier such as rivers or mountain ranges forming (½) (only need to mention one example of an isolating barrier). OR Behavioural isolation occurred between populations of original native dogs (½) due to different mating rituals/ development of nocturnal habits etc.(½). Isolation restricts gene flow between the populations and causes inbreeding within each population (½). The smaller isolated populations are more susceptible to genetic drift (½). Different selective pressures exist in each environment inhabited by the different populations (½). Different traits were selected for in each environment because those with the suitable trait (selective advantage) were more likely to survive in that environment, breed and pass on the trait (if it had a genetic basis) to future generations (½). The frequency of the allele/alleles that confer the selective advantage would increase in the population (½). OR Genetic differences would increase between the different native dog populations (½). After extended periods of time (a large number of generations) with continual selection in the one direction (½the native dog populations became so different that they could not interbreed and produce fertile offspring (½)suggesting that the population of dogs residing in Australia was now a new species – The Dingo (½)

Comments Most candidates attempted this question with some success. Candidates most frequently received between 3 and 31/2 marks for their responses. a) Generally well answered, but many responses received only partial marks due to a lack of clarity. Many

candidates stated the reason for the name change was ‘because they were more closely related’, but since the question discussed three types of ‘dogs’ it was unclear which of the dogs were the more closely related. Many candidates stated that the reason for the name change was due to the dingo being related to the Asian wolf. However, since all three ‘dogs’ share the same genus, this is insufficient as they are all related. A large number of candidates confused genus with family, genus with species or sub-species with species. However, if candidates had a correct answer but then confused the components of the binomial names, they were not penalised.

b) Generally well answered; however, a considerable number of candidates tended to reproduce material

from the information sheet without relating their answers to the context of the question. Candidates who did so did not receive full marks. Very little information on natural selection is included in the information sheet and, as a result, a description of natural selection was often missing from those responses based solely on the information sheet. In many instances candidates just listed the ways the ‘dogs’ could become isolated without mentioning the importance of isolation in preventing gene flow. A significant number of candidates stated that isolation would cause mutation or that mutation alone was enough to create a new species. A clear mechanism of how the frequency of genes in each population was changed needed to be included in each response.


Genetic drift was often used in the wrong context. Several candidates stated that natural selection would result in genetic drift, when genetic drift is independent of natural selection. Another common omission was not mentioning the extended periods of time needed for selection to bring about the gradual change in the genetic makeup of a population.

Most candidates could state the requirements for determining a separate species, i.e. the inability for the different populations to interbreed and produce fertile offspring. However, some candidates did not state whether the inability to interbreed made the populations the same species or a different species and so lost partial marks. Candidates received a maximum of 4 out of 5 marks for Part B if they did not include a description of how it could be confirmed that speciation had occurred.

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