2010 biology -...
TRANSCRIPT
FOR OFFICEUSE ONLY
SUPERVISORCHECK
RE-MARKED
ATTACH SACE REGISTRATION NUMBER LABELTO THIS BOX
External Examination 2010
2010 BIOLOGY
Monday 8 November: 9 a.m.
Time: 3 hours
Section A and Part 1 of Section B
Examination material: Question Booklet 1 (20 pages)Question Booklet 2 (8 pages)Question Booklet 3 (8 pages)one multiple-choice answer sheetone SACE registration number label
Approved dictionaries and calculators may be used.
Instructions to Students
1. You will have 10 minutes to read the paper. You must not write in your question booklets, or on your multiple-choice answer sheet, or use a calculator during this reading time but you may make notes on the scribbling paper provided.
2. This paper is in three sections: Section A and Part 1 of Section B are in Question Booklet 1; Part 2 of Section B is in Question Booklet 2; and Section C is in Question Booklet 3.
Section A: Multiple-choice Questions (Questions 1 to 25)Answer Section A on the separate multiple-choice answer sheet, using black or blue pen.Answer all questions in Section A.Section B: Short-answer Questions (Questions 26 to 33)Answer Part 1 of Section B (Questions 26 to 29) in the spaces provided in Question Booklet 1.Write on page 20 of Question Booklet 1 if you need more space.Answer Part 2 of Section B (Questions 30 to 33) in the spaces provided in Question Booklet 2.Write on page 8 of Question Booklet 2 if you need more space.Section C: Extended-response Questions (Questions 34 and 35)Answer both questions in Section C in Question Booklet 3.Write on page 8 of Question Booklet 3 if you need more space.
3. In Section B there is no need to fill all the space provided; clear, well-expressed answers are required. If you delete part or all of an answer you should clearly indicate your final answer and label it with the appropriate question number.
4. The allocation of marks and suggested allotment of time are as follows:Section A 50 marks 40 minutesSection B 120 marks 110 minutesSection C 30 marks 30 minutesTotal 200 marks 180 minutes
5. Attach your SACE registration number label to the box at the top of this page. Copy the information from your SACE registration number label into the boxes on your multiple-choice answer sheet and on the front covers of Question Booklets 2 and 3.
6. At the end of the examination, place Question Booklets 2 and 3, and your multiple-choice answer sheet, inside the back cover of Question Booklet 1.
QUESTIONBOOKLET
120 pages, 29 questions
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STUDENT’S DECLARATION ON THE USE OF CALCULATORS
By signing the examination attendance roll I declare that:my calculators have been cleared of all memory;• no external storage media are in use on these calculators.•
I understand that if I do not comply with the above conditions for the use of calculators I will:
be in breach of the rules;• have my marks for the examination cancelled or amended;• be liable to such further penalty, whether by exclusion from • future examinations or otherwise, as the SACE Board of South Australia determines.
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SECTION A: MULTIPLE-CHOICE QUESTIONS (Questions 1 to 25)
(50 marks)
Answer all questions in this section.
Each of the twenty-five multiple-choice questions in Section A involves choosing from four alternative answers. Read each question carefully. Then indicate the one alternative that you consider best answers the question by shading the bubble by the appropriate letter alongside the question number on the multiple-choice answer sheet. Use black or blue pen. It is in your interest to give an answer to every question in this section of the paper, as no marks are deducted for incorrect answers. Each question is worth 2 marks. You should spend about 40 minutes on this section.
1. According to the induced-fit model for enzyme action, an enzyme changes shape
J. when it binds to its substrate.K. and therefore it can work with many different substrates.L. to be the same as the shape of its substrate.M. to fit closely to the active site on the substrate.
2. A student used the following phrases in an assignment on DNA:
‘fragments of DNA are copied’ (1)
‘fragments of DNA are translated’ (2)
‘fragments of DNA are heated’ (3)
‘fragments of DNA are cut’ (4).
Which one of the following combinations of phrases identifies steps in the polymerase chain reaction (PCR)?
J. (1) and (3).K. (1) and (2) .L. (2), (3), and (4).M. (1), (2), and (3).
3. Erythromycin is an antibiotic that interferes with the process of protein synthesis in bacterial cells.
It is possible that erythromycin would affect the
J. transcription of mRNA in the nucleus of the bacterial cell.K. attachment of amino acids to mRNA.L. movement of ribosomes along mRNA.M. attachment of ribosomes to the endoplasmic reticulum of the bacterial cell.
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4. The human body’s immune system can normally distinguish the body’s own cells from other cells because of protein markers, called the major histocompatibility complex (MHC), on the surface of the cells. Each person’s cells have a unique MHC. In some circumstances the immune system recognises a person’s own cells as being foreign, and attacks them.
The attack on a person’s own cells could be due to
J. a change in the shape of the MHC gene(s).K. an increase in the number of MHC protein markers on the surface of the cells.L. a change in the nucleotide sequence of the MHC gene(s).M. an increase in temperature that causes the MHC protein markers to denature.
5. Refer to the following diagram, which shows six DNA fingerprints: the DNA of a man, a woman, and the woman’s four children.
Child 3 Child 4Man Child 2Woman Child 1DNAfragment
A
B
C
D
E
Which child is not the biological offspring of the man?
J. Child 1.K. Child 2.L. Child 3.M. Child 4.
6. Small pieces of RNA, called RNA interference (RNAi), can prevent the expression of specific genes. To do this they must first enter a cell.
RNA molecules are negatively charged and therefore cannot readily dissolve in lipids. RNAi molecules can be modified to be smaller and to have a reduced negative charge, thus making them lipid-soluble.
Modified RNAi, but not other RNA
J. could enter a cell through the membrane.K. could attach to ribosomes.L. would be produced by transcription.M. may contain thymine.
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7. Refer to the cell structure labelled A in the electron micrograph below:
A
Source: www.botany.unimelb.edu.au
Which one of the following combinations of name and cell type is associated with the cell structure labelled A?
Name Cell type
J. endoplasmic reticulum prokaryotic
K. Golgi body eukaryotic
L. endoplasmic reticulum eukaryotic
M. Golgi body prokaryotic
8. Hydrogen cyanide (HCN) is a highly toxic substance that can kill human beings within 60 seconds. HCN inhibits enzymes that are involved in the production of ATP in the presence of oxygen.
Which one of the following cell structures would be most affected by HCN?
J. Cell membrane.K. Nucleus.L. Chloroplast.M. Mitochondrion.
9. When undertaking an experiment it is important to
J. change the dependent variable while holding all other factors constant.K. hold factors other than the independent variable constant.L. eliminate random errors by using a large sample size.M. minimise systematic errors by using measuring instruments with a high resolution.
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10. Cultures of embryonic stem cells have been grown in laboratories so that their potential use in repairing damaged spinal tissue can be investigated.
For these cultures to be grown successfully, it is important that they contain
J. spinal cells at an optimal temperature. K. stem cells at an optimal pH.L. spinal cells in a sterile nutrient medium.M. stem cells in sterile anaerobic conditions.
11. Which one of the following combinations matches the concentration of glucose inside a cell, the concentration of glucose outside the cell, and the method by which glucose enters the cell under these conditions?
Concentration of glucose inside a cell (mmol L–1)
Concentration of glucose outside the cell (mmol L–1)
Method by which glucose enters the cell
J. 50 500 exocytosis
K. 60 700 osmosis
L. 400 60 diffusion
M. 350 70 active transport
12. Refer to the following diagram, which shows events in the cell cycle:
cell cycle checkpoint
G1 growth phase 1
S synthesis phase
G2 growth phase 2
M mitosis
C cytokinesis
Key
G1
G2
S
M C
In human beings, one of the known functions of a protein called p53 is to monitor the DNA and regulate cell division. If the DNA is damaged, p53 will prevent DNA replication and will allow the cell cycle to continue only after the DNA has been successfully repaired.
According to the information above, in a cell with damaged DNA the p53 protein is likely to
J. prevent progression of the cell from the G2 phase to the M phase.K. prevent progression of the cell from the C phase to the G1 phase.L. allow the cell to divide if it has reached the M phase.M. allow the cell to divide if it has reached the G1 phase.
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13. White blood cells are derived from stem cells in the bone marrow of healthy human beings. It has been estimated that the mass of DNA in a newly formed white blood cell is close to 7 micrograms (μg).
Which one of the following graphs most closely represents the mass of DNA in a stem cell as it grows and divides to form a white blood cell?
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m
time
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7
m
time
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7
m
time
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m
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14. The nervous system and the hormonal system work together in the human body.
The table below compares aspects of the nervous system and the hormonal system.
Which one of the following comparisons is correct?
Nervous system Hormonal system
J. Mode of transmission chemical electrochemical
K. Specificity usually specific always specific
L. Speed of transmission faster slower
M. Duration of response longer shorter
15. Carbohydrates, but not lipids
J. consist of amino acids.K. are a component of DNA.L. contribute to the energy reserves in cells.M. contribute to the structural components of organisms.
J. L.
K. M.
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16. In a human being, cells in
J. different organs have different functions and different sets of genes.K. a tissue have similar functions but different sets of genes.L. different organs have identical functions and different sets of genes.M. a tissue have similar functions and the same set of genes.
17. Refer to the following diagram, which shows structures in the small intestine:
A
B
Source: Adapted from B. LeCornu & T. Diercks, Biology: Levels of Life, 2nd edn (rev.), Briton Books, Adelaide, 2005, p. 100
Which one of the following statements is correct?
J. Structure B returns nutrients to the small intestine.K. Structure A is called a villus.L. Structure A carries out filtration and reabsorption.M. Structure B transports lipids.
18. Facultative bacteria do not need oxygen in order to grow. Obligate bacteria cannot grow in the presence of oxygen.
From this information it can be concluded that
J. facultative bacteria undergo fermentation.K. facultative bacteria undergo only aerobic respiration.L. obligate bacteria undergo both aerobic respiration and fermentation.M. both facultative and obligate bacteria undergo only anaerobic respiration.
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19. Refer to the following diagram, which shows the formation of two planarians from one. Planarians are non-parasitic flatworms that live in many environments, including freshwater rivers, saltwater, and soil.
planarian breaksinto two parts
each part regeneratesinto a new individual
(process X)
Which one of the following combinations identifies the type of cell division involved in process X and a source of genetic variation in the offspring?
Type of cell division in process X
Source of genetic variation in the offspring
J. meiotic independent assortment
K. mitotic mutation
L. meiotic fertilisation
M. mitotic crossing over
20. Refer to the following diagram, which shows a pair of homologous chromosomes before and during a crossing-over event:
before crossing over during crossing over
Which one of the following diagrams represents a structure that could not be formed after a single crossing-over event between the pair of chromosomes shown in the diagram above?
J. K. L. M.
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21. An ecologist describes an area of bushland as a ‘river red gum community’.
The term ‘river red gum community’ refers to all the
J. organisms living on a single river red gum tree.K. river red gum trees in the area of bushland.L. organisms in the area of bushland.M. living and non-living components in the area of bushland.
22. A student observed the following changes on the surface of a slice of bread that was exposed to the air.
The bread became hard. The first fungi to appear were fuzzy green spots. Stringy mould then developed and grew over the fuzzy green spots, covering the whole surface of the bread. Larger fungi then began to replace the stringy mould. After a few weeks a variety of brown and yellow fungi dominated the surface of the bread.
These changes illustrate
J. succession.K. speciation.L. decomposition.M. natural selection.
23. Refer to the following continuum, which represents extremes of r-selected species and K-selected species, and the position of species A on the continuum:
r-selected species K-selected species
species A
Which one of the following combinations best matches the reproductive effort, the number of offspring, and the type of community of species A?
Reproductive effort Number of offspring Type of community
J. high many disturbed
K. low many disturbed
L. high few stable
M. low few stable
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24. The Galapagos Rift, which receives no light, is a deep-sea boundary between oceanic plates. Sulfate in the sea water is converted into hydrogen sulfide at high temperatures. Chemosynthetic bacteria obtain energy from the hydrogen sulfide; they use this energy to convert carbon dioxide dissolved in the water into organic molecules. In the same area clams that feed on the chemosynthetic bacteria are an energy source for crabs and octopods. Chemoheterotrophic bacteria return resources to the community.
Which one of the following combinations identifies the producer, the first-order consumer, the second-order consumer, and the decomposer for the community described above?
Producer First-order consumer
Second-order consumer Decomposer
J. photosyntheticbacteria clams octopods chemosynthetic
bacteria
K. chemosyntheticbacteria clams crabs chemoheterotrophic
bacteria
L. photosyntheticbacteria octopods crabs chemosynthetic
bacteria
M. chemosyntheticbacteria crabs octopods chemoheterotrophic
bacteria
25. Refer to the following diagram, which shows four populations of frogs (A, B, C, and D). These populations are represented by circles. The circles that overlap indicate interbreeding populations. The circles that do not overlap indicate populations that cannot interbreed to produce fertile offspring.
A B C D
It would be reasonable to conclude that:
J. population D will evolve into a new species.K. populations A and C have the same predators.L. three different species would remain if population C died out.M. two different species are represented by populations A, B, C, and D.
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SECTION B: SHORT-ANSWER QUESTIONS (Questions 26 to 33)
(120 marks)
You should spend about 110 minutes on this section. Answers may be in note form. The allocation of marks is shown in brackets at the end of each part of each question. Answer all questions in the spaces provided.
Part 1 (Questions 26 to 29)
(60 marks)
26. Atrial natriuretic peptide (ANP) is a protein released naturally by human heart cells. It reduces blood pressure by dilating blood vessels and lowering blood volume.
Scientists have introduced the gene for the production of ANP into cultured human skin cells.
People with high blood pressure can be treated with injections of ANP. Alternatively, ANP could be administered to patients by means of grafts of genetically modified skin cells.
(a) Name the molecular tools that would be used to locate and remove the gene for the production of ANP from human DNA.
Locate: ______________________________________________________________________________________________
Remove: __________________________________________________________________________________ (4 marks)
(b) Describe one method by which the gene for the production of ANP could be introduced into cultured human skin cells.
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(c) State one benefit of using grafts of genetically modified skin cells rather than injections to administer medication.
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(d) State one benefit to society of having an effective treatment for people with high blood pressure.
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27. An experiment was conducted to determine the effect of light intensity on the rate of photosynthesis in a sample of the aquatic plant Elodea. The apparatus that was used is shown in the following diagram. This apparatus was placed in a range of light intensities at a constant temperature of 25ºC. The rate of photosynthesis was measured by the volume of oxygen, in millilitres (mL), collected in 10 minutes. The results are shown in the table below.
oxygen collected
test tube
funnel
beaker
Elodea plant
clamp stand
hydrogencarbonate solution
Relative light intensity (arbitrary units)
Volume of oxygen collected in 10 minutes (mL × 10–3)
50 3
100 7
200 17
250 20
(a) (i) Write a balanced chemical equation for photosynthesis.
(2 marks)
(ii) State the site of photosynthesis in a plant cell.
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(b) On the grid below, graph the data in the table opposite. Use the values 0 to 250 arbitrary units for relative light intensity.
(4 marks)
(c) The ‘compensation point’ is defined as the light intensity that produces a rate of photosynthesis equal to the rate of respiration in a plant sample. At this point the volume of oxygen collected is zero.
Using your graph or the data in the table, estimate the value of the compensation point in this experiment.
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(d) State the dependent variable in this experiment.
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(e) The experiment was conducted at 25ºC.
State two factors other than temperature that would need to be kept constant in this experiment.
Factor 1: ____________________________________________________________________________________________
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Factor 2: ____________________________________________________________________________________________
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(f ) The experiment was then conducted at 15ºC instead of at 25ºC.
Explain the most likely effect that this change in temperature would have on the rate of photosynthesis.
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(g) It is thought that the organelle responsible for photosynthesis evolved from a free-living organism as a result of an endosymbiotic event.
Describe two pieces of evidence that support this hypothesis.
Evidence 1: _________________________________________________________________________________________
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Evidence 2: _________________________________________________________________________________________
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28. A genetic code, based on reading the mRNA sequence four bases at a time, has recently been developed.
The four bases are called a ‘quadruplet codon’. Each quadruplet codon codes for an ‘artificial’ amino acid.
(a) Sixty-four different codons are possible in the triplet (three-base) codon system.
How many different codons are possible in the quadruplet codon system?
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(b) Explain why the production of mRNA would be the same if either a triplet codon system or a quadruplet codon system were used.
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(c) How would tRNA molecules need to be modified in order to read a quadruplet codon?
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(d) The properties of proteins formed from quadruplet codons are different from the properties of proteins formed from triplet codons. The bonds formed between artificial amino acids are more stable than those formed between natural amino acids. As a result, proteins that are produced using artificial amino acids maintain their shape in a wider-than-normal range of environmental conditions.
Explain why some people may oppose the widespread use of proteins formed from quadruplet codons.
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29. Refer to the following diagram, which shows a cell that is undergoing division:
R
b B
r
R
r
b B
(a) State the diploid number of this cell.
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(b) Using the information in the diagram above, draw all possible combinations of chromosomes in the daughter cells. Assume that independent assortment did occur and that crossing over did not occur.
Chromosome combination 1 Chromosome combination 2
Chromosome combination 3 Chromosome combination 4
(4 marks)
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(c) State the effect that crossing over would have on the daughter cells produced by this division.
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(d) Explain the importance for a species of crossing over and independent assortment.
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You may write on this page if you need more space to finish your answers to any questions in Part 1 of Section B. Make sure to label each answer carefully (e.g. 26(b) continued).
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© SACE Board of South Australia 2010
FOR OFFICEUSE ONLY
SUPERVISORCHECK
RE-MARKED
SACE REGISTRATION NUMBER
FIGURES
BIOLOGY
CHECKLETTER
SEQ BIN
External Examination 2010
2010 BIOLOGY
Monday 8 November: 9 a.m.
Part 2 of Section B
Write your answers to Part 2 of Section B in this question booklet.
QUESTIONBOOKLET
28 pages, 4 questions
2
SECTION B: SHORT-ANSWER QUESTIONS
Part 2 (Questions 30 to 33)
(60 marks)
Answer all questions in the spaces provided.
30. Refer to the following flow chart, which shows one mechanism for the control of levels of copper in human cells:
activatesprotein B onsome cells in thesmall intestine
normal levelsof copper
low levelsof copper hormone Y
activatesprotein A onsome cells in thesmall intestine
high levelsof copper hormone X
normal levelsof copper
prevents absorption
increases absorption
(a) Identify the stimulus in this flow chart that results in the activation of protein B.
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(b) Name one effector in this flow chart.
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(c) Protein A is known to be a receptor molecule in the control of levels of copper. It is located in the cell membrane of some cells in the lining of the small intestine, but is not found on other cells in the small intestine.
Explain why not all cells in the small intestine would be affected by hormone X.
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(d) A number of conditions, including heart disease, anaemia, and high blood cholesterol, have been linked to copper deficiencies in the human diet.
Discuss, using an example other than copper, how choices about nutrition may affect the well-being of an individual.
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(e) Other mechanisms in human beings control body temperature. A decrease in body temperature below normal triggers a number of responses that restore it to normal (37ºC).
State two responses that would restore the body temperature of a human being to normal if it fell below 37ºC, and explain how each one works.
Response 1: _________________________________________________________________________________________
Explanation: _________________________________________________________________________________________
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Response 2: _________________________________________________________________________________________
Explanation: _________________________________________________________________________________________
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31. Vinblastine is a drug that is used to treat some types of cancer. This drug inhibits the formation of microtubules in the cytoskeleton of eukaryotic cells and prevents cell division.
(a) Explain how vinblastine prevents cell division.
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(b) Explain how vinblastine could cause harmful side effects in a patient.
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32. Populations of a species of fish were found living in several small shallow lakes up to 50 kilometres apart in an area well known for its hot springs. An initial investigation showed that the temperature range of each of the lakes was between 30ºC and 35ºC. A small number of this species of fish from each of the lakes were tested in the laboratory to find the temperatures they could tolerate. None of those tested survived at temperatures below 27ºC.
Over several years it was observed that the temperature of one of the lakes (Lake A) was falling. A study made 50 years after the initial investigation showed that some members of this species of fish were living in Lake A even though its temperature was now between 21ºC and 26ºC.
(a) State the name given to all the genetic information within the population of this species of fish in Lake A.
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(b) Describe the process that resulted in some members of this species of fish in Lake A having the ability to survive at lower temperatures (21ºC to 26ºC) 50 years after the initial investigation.
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(c) Recent human activities are threatening the survival of this species of fish and may lead to its extinction.
Explain why the best way to preserve this species of fish is to preserve its habitat.
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33. An experiment was conducted to determine the suitability of three species of plant for growth in arid regions: species X (Atriplex vesicaria), species Y (Danthonia caespitosa), and species Z (Atriplex nummularia). Dry conditions were simulated by adding the chemical polyethylene glycol (PEG) to a nutrient solution in which the plants were grown. PEG slows the uptake of water by the roots of the plants, which are submerged in the nutrient solution.
Ten plants of each of the three species (X, Y, and Z) were grown in the presence of PEG and ten plants of each species were grown in the absence of PEG. All plants were kept in a nutrient solution and continuously illuminated. The mass of the plants from each of the six groups was measured at weekly intervals and was plotted as a percentage of the initial mass.
The results of the experiment are presented in the graphs below:
160150140130120110100
900 1 2 3 4 5
perc
enta
ge o
f ini
tial m
ass
time (weeks)
Absence of PEG
X Y Z160150140130120110100
900 1 2 3 4 5
perc
enta
ge o
f ini
tial m
ass
Presence of PEG
Z
XY
time (weeks)
(a) The results of this experiment for species X in the presence of PEG were originally recorded as shown below:
Using the raw data above, construct a table that could be used to plot the graph for species X in the presence of PEG. Show the week, the mass in grams, and the percentage of the initial mass.
(4 marks)
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(b) For species Y, compare the pattern of results in the presence of PEG with the pattern of results in the absence of PEG.
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(c) With reference to the data in the graphs, explain which one of the three species of plant is most suitable for growth in arid regions.
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(d) State one hypothesis about the effect of PEG on the growth of species Z that is not supported by the results of this experiment.
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(e) State one possible source of systematic error in this experiment, and explain how it could be identified.
Source: ______________________________________________________________________________________________
Explanation: _________________________________________________________________________________________
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(f ) State why ten plants were used in each of the six groups.
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You may write on this page if you need more space to finish your answers to any questions in Part 2 of Section B. Make sure to label each answer carefully (e.g. 30(d) continued).
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© SACE Board of South Australia 2010
FOR OFFICEUSE ONLY
SUPERVISORCHECK
RE-MARKED
SACE REGISTRATION NUMBER
FIGURES
BIOLOGY
CHECKLETTER
SEQ BIN
External Examination 2010
2010 BIOLOGY
Monday 8 November: 9 a.m.
Section C
Write your answers to Section C in this question booklet.
QUESTIONBOOKLET
38 pages, 2 questions
2
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SECTION C: EXTENDED-RESPONSE QUESTIONS (Questions 34 and 35)
(30 marks)
Answer both questions in this section.
Write your answers in this question booklet:Question 34, on pages 4 and 5, is worth 15 marks.• Question 35, on pages 6 and 7, is worth 15 marks.•
You should spend about 30 minutes on this section, 5 to 10 minutes planning and 20 to 25 minutes writing.
Credit will be given for clear, well-expressed answers that are well organised and relevant to the questions.
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34. The maintenance of the internal composition of a eukaryotic cell is critical to its survival.
Describe:the structure of the cell membrane• the role of the cell membrane in the regulation of the internal composition of a cell• how a large molecule, such as a protein, could leave a cell.•
(15 marks)
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6
35. There are many types of community. Energy enters most of them in the form of sunlight. Some of the organisms in a community are able to convert light energy into chemical energy, which is needed so that cellular processes can maintain life.
Describe how:environmental factors may determine the type of the community• chemical energy is made available for cellular processes.•
(15 marks)
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7 PLEASE TURN OVER
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8
You may write on this page if you need more space to fi nish your answers to Questions 34 and 35. Make sure to label each answer carefully (e.g. 35 continued).
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© SACE Board of South Australia 2010