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Aussie Maths & Science Teachers: Save your time with SmarterEd

BIOLOGY, M2 2015 HSC 27

  1. Outline TWO differences between whole blood and plasma.  (2 marks)

The steps below show the preparation and use of blood products in the treatment of Ebola Virus Disease. This disease is characterised by significant blood loss.

  1. Explain why this protocol produces an effective treatment for Ebola Virus Disease.  (3 marks)

Show Answers Only

a.    Whole blood and plasma differences:

→ Whole blood contains cells including WBC’s, RBC’s and platelets while plasma does not.

→ Plasma is a straw coloured liquid and whole blood is a thick red liquid.
 

b.   Answers could include:

→ The plasma will contain antibodies for this disease because it has been taken from someone who has survived the disease. This will help to immobilise the virus in recipients blood stream.

→ Screening blood prevents the spread of Ebola and other blood-borne diseases from donor to recipient.

→ The plasma is separated from the whole blood meaning no blood type match is needed as there are no cells in the plasma but it still contains the beneficial antibodies.

Show Worked Solution

a.    Whole blood and plasma differences:

→ Whole blood contains cells including WBC’s, RBC’s and platelets while plasma does not.

→ Plasma is a straw coloured liquid and whole blood is a thick red liquid.

♦ Mean mark (a) 49%.

b.   Answers could include:

→ The plasma will contain antibodies for this disease because it has been taken from someone who has survived the disease. This will help to immobilise the virus in recipients blood stream.

→ Screening blood prevents the spread of Ebola and other blood-borne diseases from donor to recipient.

→ The plasma is separated from the whole blood meaning no blood type match is needed as there are no cells in the plasma but it still contains the beneficial antibodies.

BIOLOGY, M2 2016 HSC 31

As altitude increases, the partial pressure of oxygen in air decreases. 

Species A and B are closely related endotherms that live in different habitats in Asia. The minimum required for 100% blood oxygen saturation differs in these species because of differences in their haemoglobin structure. Data related to these two species are shown below.

Explain how the differences in these species could have arisen, using the Darwin/Wallace theory of evolution and your understanding of the adaptive advantage of haemoglobin.  (8 marks)

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→ Haemoglobin is a protein that provides a mechanism for transport of oxygen around the body. As it is a protein, it’s structure is dependant on the individual’s genotype.

→ Species A and Species B are able to reach 100% saturation at differing partial pressures of oxygen, meaning they have different DNA which codes for different haemoglobin structures.

→ Species A and B are likely to have diverged from a common ancestor because of differing environmental pressures resulting in two different species. Within the ancestral population there was variation which resulted from random mutations. One mutation would have resulted in haemoglobin that is able to reach 100% saturation at a lower partial pressure of oxygen.

→ When members of the ancestral species moved to a higher altitude the ability of their haemoglobin to reach saturation at a lower gave them a survival advantage. These individuals were then more likely to reproduce and pass on their favourable genes.

→ For individuals living at lower altitudes, there is no survival advantage to being able to reach 100% saturation at lower which means this trait was not selected for.

→ Over time, due to the isolation at a higher altitude a new species evolved.

Show Worked Solution

→ Haemoglobin is a protein that provides a mechanism for transport of oxygen around the body. As it is a protein, it’s structure is dependant on the individual’s genotype.

→ Species A and Species B are able to reach 100% saturation at differing partial pressures of oxygen, meaning they have different DNA which codes for different haemoglobin structures.

→ Species A and B are likely to have diverged from a common ancestor because of differing environmental pressures resulting in two different species. Within the ancestral population there was variation which resulted from random mutations. One mutation would have resulted in haemoglobin that is able to reach 100% saturation at a lower partial pressure of oxygen.

→ When members of the ancestral species moved to a higher altitude the ability of their haemoglobin to reach saturation at a lower gave them a survival advantage. These individuals were then more likely to reproduce and pass on their favourable genes.

→ For individuals living at lower altitudes, there is no survival advantage to being able to reach 100% saturation at lower which means this trait was not selected for.

→ Over time, due to the isolation at a higher altitude a new species evolved.

♦♦ Mean mark 41%.

BIOLOGY, M2 2017 HSC 27

  1. Draw labelled diagrams to distinguish between transverse sections of a xylem vessel and a phloem vessel.  (2 marks)

  2. Describe the process that transports sugars through a plant.  (3 marks)

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a.   
       

b.   → Sugars produced in the leaves are transported into the phloem by active transport.

→ Water from the xylem then follows into the phloem by osmosis, which increases the pressure.

→ This creates a pressure gradient in the phloem allowing the sugar to move up or down the plant.

→ At the ‘sink’, sugars are then removed via active transport to required locations.

Show Worked Solution
a.   
       
♦ Mean mark (a) 47%.

b.   Transport of sugars through a plant

→ Sugars produced in the leaves are transported into the phloem by active transport.

→ Water from the xylem then follows by osmosis, which increases the pressure in the phloem.

→ This creates a pressure gradient in the phloem allowing the sugar to move up or down the plant.

→ At the sink, sugars are removed actively.

♦ Mean mark (b) 42%.

BIOLOGY, M1 2014 HSC 28

Rennin is an enzyme found in the stomach of young mammals. Rennin curdles the milk drunk by the mammal and allows the milk solids to stay longer in the stomach to be further digested.

Students conducted an investigation into rennin activity. They bubbled different volumes of carbon dioxide gas into milk samples. Each sample was 50mL and was kept at a constant temperature. The students then added rennin to each milk sample and recorded the time taken for the milk to curdle.
 

  1. Account for the students' calculated average time for 300 bubbles of .  (2 marks)

  2. Explain the results of this experiment.  (4 marks)

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a.    Calculating the average:

→ The outlier (311 seconds) was removed from the data set, then the remaining data points at 300 bubbles of were averaged.

→ This was necessary as it is significantly different to other values for time taken to curdle at that volume and allowed the calculated average to fall within the predetermined trend.
 

b.    → As bubbles of increase, time to curdle decreases.

→ This scenario is explicitly seen between 100-200 bubbles, where the curdling time reduces by 20 seconds at each interval.

→ This means increased concentration of also increases activity of the enzyme.

→ As increases acidity of a solution, these results show us that rennin activity increases at reduced pH levels.

→ The slower increase of enzyme activity after 250 bubbles of is due to the enzyme being close to its optimum pH where the enzyme activity graph flattens off at the peak of the curve.

Show Worked Solution

a.    Calculating the average:

→ The outlier (311 seconds) was removed from the data set, then the remaining data points at 300 bubbles of were averaged.

→ This was necessary as it is significantly different to other values for time taken to curdle at that volume and allowed the calculated average to fall within the predetermined trend.

♦ Mean mark (a) 46%.

b.    → As bubbles of increase, time to curdle decreases.

→ This scenario is explicitly seen between 100-200 bubbles, where the curdling time reduces by 20 seconds at each interval.

→ This means increased concentration of also increases activity of the enzyme.

→ As increases acidity of a solution, these results show us that rennin activity increases at reduced pH levels.

→ The slower increase of enzyme activity after 250 bubbles of is due to the enzyme being close to its optimum pH where the enzyme activity graph flattens off at the peak of the curve.

♦♦♦ Mean mark (b) 35%.

BIOLOGY, M1 2013 HSC 25

The graph below shows the results obtained from testing the activity of a bacterial enzyme.
 

  1. Name ONE variable, other than temperature, that would have been controlled in the experiment.  (1 mark)

  2. For what temperature range does the enzyme display the maximum rate of change in activity?   (1 mark)

  3. Account for the activity of the enzyme at the parts of the graph labelled A, B, C and D.  (4 marks)

  4. Based on the information in the graph, suggest the type of environment in which these bacteria might survive. (1 mark)

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a.    Answers could include:

→ pH

→ Enzyme concentration

→ Substrate concentration 

b.    89 – 93°C (D).

c.    Enzyme activity at each point:

A: A constant gradient indicates a constant increasing rate of enzyme activity as temperature increases.

B: Graph level at a maximum (10 μmol/s). Enzyme-substrate complex operating at maximum even though temperature is increasing.

C: Graph decreasing indicating a reduction in enzyme efficiency due to increasing temperature denaturing enzyme.

D: Graph rapidly goes to 0 indicating enzyme activity stops because the enzyme is denatured.

d.    Hotter environments, such as desserts or hot, geothermal springs.

Show Worked Solution

a.    Answers could include:

→ pH

→ Enzyme concentration

→ Substrate concentration 

b.    89 – 93°C (D).

c.    Enzyme activity at each point:

A: A constant gradient indicates a constant increasing rate of enzyme activity as temperature increases.

B: Graph level at a maximum (10 μmol/s). Enzyme-substrate complex operating at maximum even though temperature is increasing.

C: Graph decreasing indicating a reduction in enzyme efficiency due to increasing temperature denaturing enzyme.

D: Graph rapidly goes to 0 indicating enzyme activity stops because the enzyme is denatured.

♦♦♦ Mean mark (b) 22%.

d.    Hotter environments, such as desserts or hot, geothermal springs.

BIOLOGY, M1 2015 HSC 22

Explain why insects excrete uric acid as their principal nitrogenous waste.  (2 marks)

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→ Insects need to conserve water within their bodies by reducing water loss.

→ This is done through their creation of uric acid which low toxicity and therefore it can be excreted in a highly concentrated form, reducing insects need for large amounts of water to excrete their nitrogenous wastes.

Show Worked Solution

→ Insects need to conserve water within their bodies by reducing water loss.

→ This is done through their creation of uric acid which low toxicity and therefore it can be excreted in a highly concentrated form, reducing insects need for large amounts of water to excrete their nitrogenous wastes.

♦ Mean mark 48%.

BIOLOGY, M1 2016 HSC 23

  1. Explain ONE reason why the concentration of water in cells should be maintained within a narrow range for optimal cell function.  (2 marks)

  2. A person has consumed large amounts of water. Complete the table to show the effect on each of the variables listed.  (3 marks)

 

Show Answers Only

a.    Correct answers could include:

→ Concentration of water in cells is maintained to regulate concentrations of solutes in cells.

→  Concentration of water in a narrow range provides appropriate substrate concentrations for metabolic function.

b.

Show Worked Solution

a.    Correct answers could include:

→ Concentration of water in cells is maintained to regulate concentrations of solutes in cells.

→  Concentration of water in a narrow range provides appropriate substrate concentrations for metabolic function.

♦ Mean mark 43%.

b.

BIOLOGY, M7 SM-Bank 27

Influenza is an infectious respiratory disease. In humans, it can be caused by the influenza A or influenza B viruses.

Antigenic drift can result in small changes to the structure of the antigens on the surface of the influenza virus, as shown in the diagram below.
 

  1. What change would have occurred within the virus to bring about the change in the structure of the antigen?  (1 mark)

  2. Vaccines against influenza are available and it is recommended that people are vaccinated each year.
  3. Explain, in terms of antigenic drift, why vaccinations are recommended yearly for influenza rather than once every few years.  (2 marks)

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a.   Mutation.

b.   Correct answers could include:

→ Vaccines contain new antigens resulting in the production of new antibodies OR memory cells specific to the antigen.

→ Memory cells allow for a faster OR larger immune response.

→ Any existing memory cells specific to the previous antigens would no longer be effective.

Show Worked Solution

a.   Mutation.

b.   Correct answers could include:

→ Vaccines contain new antigens resulting in the production of new antibodies OR memory cells specific to the antigen.

→ Memory cells allow for a faster OR larger immune response.

→ Any existing memory cells specific to the previous antigens would no longer be effective.

BIOLOGY, M6 SM-Bank 25

Over time, the South African cheetah population has suffered drastic reduction due to periodic droughts, disease and hunting. Currently, only small, isolated populations of cheetahs exist in the wild.

Explain, in terms of genetic diversity, why cheetah populations are now on the verge of extinction.  (3 marks)

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→ The cheetah population has suffered a severe bottleneck. This means that (by chance) certain alleles may be over- or under-represented in the surviving population, and some (potentially ‘fitter’) alleles may have been lost altogether.

→ This reduces genetic variability and means that the small surviving population’s gene pool is not representative of the original population.

→ If the environment suddenly changes, the selection pressures on the cheetahs will also change. Having a limited gene pool may mean the population cannot adapt to the changing conditions and are therefore at increased risk of extinction.

Show Worked Solution

→ The cheetah population has suffered a severe bottleneck. This means that (by chance) certain alleles may be over- or under-represented in the surviving population, and some (potentially ‘fitter’) alleles may have been lost altogether.

→ This reduces genetic variability and means that the small surviving population’s gene pool is not representative of the original population.

→ If the environment suddenly changes, the selection pressures on the cheetahs will also change. Having a limited gene pool may mean the population cannot adapt to the changing conditions and are therefore at increased risk of extinction.

CHEMISTRY, M5 EQ-Bank 29

The information in the table shows how the solubility of lead chloride is affected by temperature.  
 

Using a graph, calculate the solubility product of the dissolution of lead chloride at 50°C. Include a fully labelled graph and a relevant chemical equation in your answer   (6 marks)
 

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Show Worked Solution

 


 


 

CHEMISTRY, M5 EQ-Bank 28

A 100 mL saturated solution of calcium hydroxide at 25°C contains 0.173 g of calcium hydroxide.

  1. Calculate the solubility product of this salt at 25°C.  (3 marks)

  2. Explain why the undissolved solid is not included in the expression for the solubility product constant.  (1 marks)

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a.    

b.    → is constant throughout the reaction.

→ Since it does not change, it is not included in the equilibrium expression.

Show Worked Solution

a.   


 

 


 

 

b.    → is constant throughout the reaction.

→ Since it does not change, it is not included in the equilibrium expression.

CHEMISTRY, M6 EQ-Bank 14 MC

Equal volumes of four different acids are titrated with the same base at 25°.

Information about these acids is given in the table.
 

Which acid requires the greatest volume of base for complete reaction?

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Show Worked Solution

→ The volume of base required for complete reaction is independent of the strength of the acid as all neutralisation reactions go to completion.

→ Phosphoric acid () has three acidic hydrogens whereas the other given bases only have one. So, phosphoric acid requires three times the amount of base as any other acid given.

CHEMISTRY, M6 EQ-Bank 28

The flowchart shown outlines the sequence of steps used to determine the concentration of an unknown hydrochloric acid solution.
 

Describe steps A, B and C including correct techniques, equipment and appropriate calculations. Determine the concentration of the hydrochloric acid.  (8 marks)

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Step A

→ Prepare by drying. Protect solid from moisture in the air by storing in a desiccator.

→ Calculate mass of dried required and weigh accurately.

→ Clean and rinse a 500 mL volumetric flask with distilled water.

→ Add 5.30 grams of to the volumetric flask using a funnel and wash funnel using distilled water. Add distilled water to the flask to the bottom of the meniscus.

Step B

→ Clean and rinse a 50 mL burette. Fill burette with the unknown acid and place on a retort stand.

→ Clean and rinse a 250 mL conical flask with distilled water.

→ Clean a 25 mL pipette and rinse with 0.1 M solution. Fill pipette with solution to bottom of meniscus.

→ Transfer all pipette solution into conical flask and add an appropriate indicator. A white background (tile) should be placed under the flask to highlight any colour changes in the solution.

→ Slowly add acid solution from the burette into the conical flask and record the volume used when the indicator changes colour.

Step C

→ The initial titration represents a test run to establish an indicative volume. Three subsequent titrations should be performed with the average titration forming the basis of concentration calculations.

→ Calculate the concentration of

Show Worked Solution

Step A

→ Prepare by drying. Protect solid from moisture in the air by storing in a desiccator.

→ Calculate mass of dried required and weigh accurately.

→ Clean and rinse a 500 mL volumetric flask with distilled water.

→ Add 5.30 grams of to the volumetric flask using a funnel and wash funnel using distilled water. Add distilled water to the flask to the bottom of the meniscus.

Step B

→ Clean and rinse a 50 mL burette. Fill burette with the unknown acid and place on a retort stand.

→ Clean and rinse a 250 mL conical flask with distilled water.

→ Clean a 25 mL pipette and rinse with 0.1 M solution. Fill pipette with solution to bottom of meniscus.

→ Transfer all pipette solution into conical flask and add an appropriate indicator. A white background (tile) should be placed under the flask to highlight any colour changes in the solution.

→ Slowly add acid solution from the burette into the conical flask and record the volume used when the indicator changes colour.

Step C

→ The initial titration represents a test run to establish an indicative volume. Three subsequent titrations should be performed with the average titration forming the basis of concentration calculations.

→ Calculate the concentration of

CHEMISTRY, M6 EQ-Bank 29

Explain why a mixture of acetic acid (1 M) and sodium acetate (1 M) can act as a buffer while a mixture of hydrochloric acid (1 M) and sodium chloride (1 M) cannot.   (3 marks)

Show Answers Only

→ A buffer is a mixture of a weak acid and its conjugate base which is able to counteract changes in pH when small amounts of acid or base is added.

→ In this case acetic acid and sodium acetate can act as a buffer. The following equilibrium is established:

→ Any addition of acid to the mixture increases the and shifts the above equilibrium to the left, removing ions and minimising a change in pH. 

→ Likewise, any addition of base causes ions to react with in solution. This shifts the above equilibrium to the right, generating more and maintaining a relatively constant pH.

→ Hydrochloric acid is a strong acid, so its conjugate base, chloride has negligible basic activity.

→ The dissociation of hydrochloric acid goes to completion, so addition of base simply lowers the pH of the solution as there is no aqueous hydrochloric acid able to generate . Addition of acid will raise the pH as chloride ions are unable to react with ions.

Show Worked Solution

→ A buffer is a mixture of a weak acid and its conjugate base which is able to counteract changes in pH when small amounts of acid or base is added.

→ In this case acetic acid and sodium acetate can act as a buffer. The following equilibrium is established:

→ Any addition of acid to the mixture increases the and shifts the above equilibrium to the left, removing ions and minimising a change in pH. 

→ Likewise, any addition of base causes ions to react with in solution. This shifts the above equilibrium to the right, generating more and maintaining a relatively constant pH.

→ Hydrochloric acid is a strong acid, so its conjugate base, chloride has negligible basic activity.

→ The dissociation of hydrochloric acid goes to completion, so addition of base simply lowers the pH of the solution as there is no aqueous hydrochloric acid able to generate . Addition of acid will raise the pH as chloride ions are unable to react with ions.

CHEMISTRY, M7 EQ-Bank 27

Contrast ONE addition polymer and ONE condensation polymer in terms of their structures, properties and uses. Include structural formulae in your answers.   (7 marks)

Show Answers Only

Addition polymer – polyethylene. Condensation polymer – nylon.

Structures

→ Polyethylene is made by the addition of ethylene monomers with the following structural formula:

→ Nylon is a condensation polymer made from 1,6-diaminohexane and adipic acid, producing a by-product of water molecules. It has the following structural formula.

Properties

→ Polyethylene is inexpensive, weatherproof and relatively resistant to chemicals.

→ There are two main types of polyethylene which have different properties. Low-density polyethylene (LDPE) is semi-rigid while high density polyethylene (HDPE) is fluid.

→ Nylon is strong and relatively resistant to moisture absorptivity. It is longer lasting than polyethylene, resistant to chemicals and is used to make nylon fibre.

Applications

→ LDPE is used to produce products such as plastic soft drink bottles, flexible water pipes and cling wrap. HDPE’s fluidity make it an appropriate material for producing shopping bags, plastic crates and drums for storage.

→ Nylon is very versatile in its manufacturing uses. It is a common input for clothing, can be used for injection-moulded parts for vehicles and also as reinforcement for rubber tyres. 

→ Nylon is also used as the main material input for nylon thread which has a myriad of uses, including stitching for clothes and the production of rope.

Show Worked Solution

Addition polymer – polyethylene. Condensation polymer – nylon.

Structures

→ Polyethylene is made by the addition of ethylene monomers with the following structural formula:

→ Nylon is a condensation polymer made from 1,6-diaminohexane and adipic acid, producing a by-product of water molecules. It has the following structural formula.

Properties

→ Polyethylene is inexpensive, weatherproof and relatively resistant to chemicals.

→ There are two main types of polyethylene which have different properties. Low-density polyethylene (LDPE) is semi-rigid while high density polyethylene (HDPE) is fluid.

→ Nylon is strong and relatively resistant to moisture absorptivity. It is longer lasting than polyethylene, resistant to chemicals and is used to make nylon fibre.

Applications

→ LDPE is used to produce products such as plastic soft drink bottles, flexible water pipes and cling wrap. HDPE’s fluidity make it an appropriate material for producing shopping bags, plastic crates and drums for storage.

→ Nylon is very versatile in its manufacturing uses. It is a common input for clothing, can be used for injection-moulded parts for vehicles and also as reinforcement for rubber tyres. 

→ Nylon is also used as the main material input for nylon thread which has a myriad of uses, including stitching for clothes and the production of rope.

CHEMISTRY, M6 EQ-Bank 25

The graph shows changes in pH for the titrations of equal volumes of solutions of two monoprotic acids, Acid 1 and Acid 2.
 

Explain the differences between Acid 1 and Acid 2 in terms of their relative strengths and concentrations.   (3 marks)

Show Answers Only

→ Acid 1 is a strong acid. Its initial pH = 1 and its equivalence point is at pH = 7.

→ Acid 2 is a weaker acid. Its initial pH ~ 2 and its equivalence point is >7.

→ Acid 2 has a higher concentration than Acid 1 as it doesn’t take that much more to neutralise it.

Show Worked Solution

→ Acid 1 is a strong acid. Its initial pH = 1 and its equivalence point is at pH = 7.

→ Acid 2 is a weaker acid. Its initial pH ~ 2 and its equivalence point is >7.

→ Acid 2 has a higher concentration than Acid 1 as it doesn’t take that much more to neutralise it.

CHEMISTRY, M5 EQ-Bank 15 MC

What will happen when sulfuric acid is added to a saturated solution of sparingly soluble calcium sulfate?

  1. The concentration of calcium and sulfate ions will increase over time due to the presence of ions.
  2. The concentration of calcium and sulfate ions will decrease over time due to the presence of  ions.
  3. The concentration of calcium and sulfate ions will increase over time due to the presence of ions.
  4. The concentration of calcium and sulfate ions will decrease over time due to the presence of ions.
Show Answers Only

Show Worked Solution

→ The saturated solution of calcium sulfate is originally at equilibrium

→ The addition of sulfuric acid increases the concentration of (sulfate) ions in solution.

→ By Le Chatelier’s principle, the above equilibrium will shift left to counteract this, decreasing the concentration of calcium and sulfate ions over time.

CHEMISTRY, M5 EQ-Bank 13 MC

0.20 moles of phosphorus pentachloride were heated to 200°C in a 2 L container in the presence of a vanadium catalyst according to the following reaction.

At equilibrium, the mixture was found to contain 0.16 moles of chlorine.

Which of the following is the equilibrium constant for this reaction at this temperature?

  1. 0.32
  2. 0.64
  3. 1.56
  4. 3.13
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Show Worked Solution

 

Algebra, STD1 A3 2019 HSC 9 MC

The container shown is initially full of water.
 

Water leaks out of the bottom of the container at a constant rate.

Which graph best shows the depth of water in the container as time varies?
 

A. B.
C. D.
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Show Worked Solution

Financial Maths, STD1 F3 2021 HSC 30

Blake opens a new credit card account on 1 May. He uses it, for the first time, on 4 May to buy concert tickets for $850.

He makes no further purchases or repayments during the month of May.

A statement for the credit card is issued on the last day of each month.

The statement for May shows that interest is charged at 19.75% per annum, compounding daily, from 20 May (included) until 31 May (included).

  1. What is the compound interest shown on the statement issued on 31 May?  (3 marks)

  2. The minimum payment is calculated as 3% of the closing balance on 31 May. Calculate the minimum payment.  (1 mark)

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Show Worked Solution

a.   

 
 

 

 

 

b.   
   

CHEMISTRY, M5 EQ-Bank 21

Three gases and were mixed in a closed container and allowed to reach equilibrium. A change was imposed at time and the equilibrium was re-established. The concentration of each gas is plotted against time.
 

  1. What is a possible change that was imposed at time ?  (1 mark)

  2. Write a chemical reaction that is represented by the concentration graph above.  (2 marks)

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  1. An increase in the volume of the container.
Show Worked Solution

a.   At time :

→ The concentration of all species decreases by an amount proportional to their initial concentration.

→ This could be due to an increase in the volume of the container.

 

b.   Chemical reaction:

→ After the system change, equilibrium re-establishes by favouring the conversion of into and .

→ As change in concentration of all species is equal, they react in a 1:1:1 molar ratio.

→ Therefore the chemical reaction is: 

CHEMISTRY, M5 EQ-Bank 12 MC

Nitrogen dioxide (a brown gas) and dinitrogen tetroxide (a colourless gas) are both forms of oxides of nitrogen. They are in equilibrium according to the equation

.

An equilibrium mixture of the two gases at room temperature is light brown but at higher temperatures the colour becomes a much deeper brown.

What conclusion can be drawn from this observation?

  1. The reverse reaction in the equation is endothermic.
  2. The forward reaction in the equation is endothermic.
  3. The brown colour is due to the strong nitrogen–oxygen bonds in .
  4. The equilibrium concentration of is not dependent on temperature.
Show Answers Only

Show Worked Solution

→ Le Chatelier’s principle states that when an equilibrium system is subject to a change in conditions, it will shift such as to partially counteract the imposed change.

→ When the temperature of the system is increased, it will shift to favour the endothermic reaction, counteracting the increase in temperature.

→ In this case, the deeper brown colour shows the conversion to nitrogen dioxide is favoured following an increase in temperature. The reverse reaction is therefore endothermic.

CHEMISTRY, M7 EQ-Bank 26

This flow chart shows reactions involving six different organic compounds (A to F).
 

 

Draw the structures of compounds A to F, justifying your diagrams with reference to the information provided.   (7 marks)

Show Answers Only

A:  
       

B:  
       

C:  
           

D:  
       

E:  
       

F:  
       

Show Worked Solution

→ A combines with dilute    to produce two alcohols.

→ C doesn’t react with strong oxidant (i.e. it is a tertiary alcohol).

C:  
           

→ B must therefore be 2-methylbutan-1-ol or 3-methylbutan-2-ol.

→ Since B gives two products when heated with concentrated , it must be 3-methylbutan-2-ol, as 2-methylbutan-1-ol will only produce one product.

B:  
       

→ D is a ketone, produced by B reacting with a strong oxidant.

D:  
       

→ A can be dehydrated using concentrated    on either B or C.

A:  
       

→ B dehydrates to A or E. Given A‘s structure above,

E:  
       

→ C dehydrates to A or F. Again, given A’s structure above,

F:  
       

CHEMISTRY, M8 EQ-Bank 23

and react to form an ionic compound according to the general equation

where and are numbers representing the ratio in which and combine.

Spectrophotometry was used to determine the stoichiometric ratio between and . To do this, eight 10 mL samples were prepared by reacting solutions of with solutions of in varying ratios. All and solutions had the same concentration. The absorbance of the samples is tabulated below.
 

  1. On the grid, construct a graph of absorbance against volume of solution from 0.00 mL to 6.00 mL, and draw TWO lines of best fit.  (3 marks)
     

     

  1. The reaction proceeds according to the general equation
  2.       .
  3. Find the values of and . Justify your answer with reference to the data given and the graph in part (a).   (3 marks)

Show Answers Only

a.  
           

 

b.   Find values of and :

→ The graph shows that as the volume of compound decreases between 10mL and 7.5mL, there is an increase in ions. This indicates an excess of ions limiting the products.

→ The absorbance reaches a maximum when the 2.5 mL of compound is added to 7.5 mL of compound . Since the concentrations of the initial solutions are equal (given), equal volumes contain equal moles.

→ The ratio of volumes at peak absorbance = 2.5 : 7.5 = 1 : 3 (i.e. the correct stoichiometric ratio).

→ Hence and .

→ The curve then turns down sharply, indicating a reduction in .

Show Worked Solution

a.  
           

 

b.   Find values of and :

→ The graph shows that as the volume of compound decreases between 10mL and 7.5mL, there is an increase in ions. This indicates an excess of ions limiting the products.

→ The absorbance reaches a maximum when the 2.5 mL of compound is added to 7.5 mL of compound . Since the concentrations of the initial solutions are equal (given), equal volumes contain equal moles.

→ The ratio of volumes at peak absorbance = 2.5 : 7.5 = 1 : 3 (i.e. the correct stoichiometric ratio).

→ Hence and .

→ The curve then turns down sharply, indicating a reduction in .

CHEMISTRY, M8 EQ-Bank 28

Limestone contributes to the hardness of water by releasing ions. The following chemical equation represents this reaction.

     

It has been suggested that heating water reduces its hardness.

Explain how this suggestion can be tested accurately, validly and reliably.   (9 marks)

Show Answers Only

→ Atomic absorption spectroscopy (AAS) can be used to test if heating reduces water hardness.

→ It does this by calculating the concentrations of metal ions in solutions. AAS can calculate the concentration of in heated and non-heated samples of water and any difference in the relative concentrations of can be used to verify the suggestion.

→ It should be noted that a reduced concentration of indicates that the water hardness is reduced.

Methodology of testing

→ Prepare standard solutions with known concentrations of and measure their absorbance. Plot the concentrations against the absorbance of the standard solutions and draw a calibration curve (i.e. a line of best fit).

→ Measure the absorbance of a water sample before heating and another after heating. Using the absorbance and the calibration curve, calculate the concentration of in each sample and compare the concentrations between the heated and unheated samples.

→ The AAS should be calibrated, at which point the concentration of calcium ions can be calculated to an accuracy in the parts per million (ppm). To ensure accurate calibration of the AAS, the standard solutions need to be prepared precisely which will involve the accurate weighing of solids and the use of a pipette or a similar instrument to measure solution volumes.

→ Water used in the experiment should be de-ionised (normal drinking water has an abundance of and ).

→ The margin of experimental error decreases when sufficient calibration samples are used and the measurement of absorbance of these samples is repeated and averaged. 

→ The reliability of results increases when many samples of heated and non-heated water are used to confirm that the concentrations of in the heated water samples are consistently lower than the concentrations of in the unheated water samples.

→ AAS can also be used to test the validity of the results. A hollow cathode lamp for calcium can direct light through the solution. This light has a specific wavelength that will only be absorbed by calcium ions. In this way, accurate measurements are made which can then be compared against the results and provide evidence of the validity of the original suggestion.

Show Worked Solution

→ Atomic absorption spectroscopy (AAS) can be used to test if heating reduces water hardness.

→ It does this by calculating the concentrations of metal ions in solutions. AAS can calculate the concentration of in heated and non-heated samples of water and any difference in the relative concentrations of can be used to verify the suggestion.

→ It should be noted that a reduced concentration of indicates that the water hardness is reduced.

Methodology of testing

→ Prepare standard solutions with known concentrations of and measure their absorbance. Plot the concentrations against the absorbance of the standard solutions and draw a calibration curve (i.e. a line of best fit).

→ Measure the absorbance of a water sample before heating and another after heating. Using the absorbance and the calibration curve, calculate the concentration of in each sample and compare the concentrations between the heated and unheated samples.

→ The AAS should be calibrated, at which point the concentration of calcium ions can be calculated to an accuracy in the parts per million (ppm). To ensure accurate calibration of the AAS, the standard solutions need to be prepared precisely which will involve the accurate weighing of solids and the use of a pipette or a similar instrument to measure solution volumes.

→ Water used in the experiment should be de-ionised (normal drinking water has an abundance of and ).

→ The margin of experimental error decreases when sufficient calibration samples are used and the measurement of absorbance of these samples is repeated and averaged. 

→ The reliability of results increases when many samples of heated and non-heated water are used to confirm that the concentrations of in the heated water samples are consistently lower than the concentrations of in the unheated water samples.

→ AAS can also be used to test the validity of the results. A hollow cathode lamp for calcium can direct light through the solution. This light has a specific wavelength that will only be absorbed by calcium ions. In this way, accurate measurements are made which can then be compared against the results and provide evidence of the validity of the original suggestion.

CHEMISTRY, M5 EQ-Bank 12

An industrial plant makes ammonia from nitrogen gas and hydrogen gas. The reaction is exothermic.  

The graph shows the adjustments made to increase the yield of ammonia.

Account for the changes in conditions that have shaped the graph during the time the system was observed. Include a relevant chemical equation in your answer.   (5 marks)

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From to the system is in equilibrium.

At

→ Nitrogen is introduced to the system and its concentration increases sharply.

→ Le Chatelier’s principle states that when a system in equilibrium is disturbed, the equilibrium will shift in the direction that minimises the change. In this case, the equilibrium will shift to the right to use up more nitrogen. A greater yield of ammonia will result until equilibrium is re-established.

At

→ The concentration of both reactants and products increases. This effect could be caused by a decrease in volume of the reaction vessel which will result in an increase in pressure on the system.

→ The above equation shows that 4 moles of gas (on the left-hand side) react to form 2 moles of gas (on the right-hand side). Le Chatelier’s principle dictates that this increase in pressure will cause the system to again shift right, to the side with fewer moles of gas, to counteract the change.

→ This right shift will further increase the yield of ammonia until equilibrium is re-established.

At

→ There is a change to the system that shifts the reaction back to the left. The gradual change in concentrations indicate that this could be due to a change in temperature.

→ Since this reaction is exothermic, the reverse reaction (left shift) absorbs heat. An increase in temperature would cause this shift, lowering the yield of ammonia until equilibrium is again restored.

Show Worked Solution

From to the system is in equilibrium.

At

→ Nitrogen is introduced to the system and its concentration increases sharply.

→ Le Chatelier’s principle states that when a system in equilibrium is disturbed, the equilibrium will shift in the direction that minimises the change. In this case, the equilibrium will shift to the right to use up more nitrogen. A greater yield of ammonia will result until equilibrium is re-established.

At

→ The concentration of both reactants and products increases. This effect could be caused by a decrease in volume of the reaction vessel which will result in an increase in pressure on the system.

→ The above equation shows that 4 moles of gas (on the left-hand side) react to form 2 moles of gas (on the right-hand side). Le Chatelier’s principle dictates that this increase in pressure will cause the system to again shift right, to the side with fewer moles of gas, to counteract the change.

→ This right shift will further increase the yield of ammonia until equilibrium is re-established.

At

→ There is a change to the system that shifts the reaction back to the left. The gradual change in concentrations indicate that this could be due to a change in temperature.

→ Since this reaction is exothermic, the reverse reaction (left shift) absorbs heat. An increase in temperature would cause this shift, lowering the yield of ammonia until equilibrium is again restored.

PHYSICS, M5 EQ-Bank 29

A student used the following scale diagram to investigate orbital properties. The diagram shows a planet and two of its moons,   and . The distances between each of the moons and the planet are to scale while the sizes of the objects are not.
 

Complete the table to compare the orbital properties of Moon and Moon . Show relevant calculations in the space below the table.  (4 marks)

Show Answers Only

 
Comparing radii:


 

As moons and are both orbiting the same body, the ratio will be the same for both.

Comparing orbital periods:

 
 
 

 
Comparing orbital velocities.

 
Using the calculations for radius and period:

 
   
   
   
 

Show Worked Solution

 
Comparing radii:


 

As moons and are both orbiting the same body, the ratio will be the same for both.

Comparing orbital periods:

 
 
 

 
Comparing orbital velocities.

 
Using the calculations for radius and period:

 
   
   
   
 

PHYSICS, M5 EQ-Bank 28

A bullet is fired vertically from the surface of Mars, at the escape velocity of Mars. Another bullet is fired vertically from the surface of Earth, at the escape velocity of Earth.

Neglecting air resistance, compare the energy transformations of the two bullets.  (5 marks)

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→ The total energy of the bullet fired from the surface of Mars will remain constant. Its kinetic energy will transform into potential energy as it gains altitude. 

→ As the bullet fired from Mars reaches an ‘infinite’ distance away, it escapes Mars’ gravitational attraction. Here, the kinetic energy of the bullet is zero as it has expended all of its initial kinetic energy.

→ The potential energy () of the bullet has also decreased to zero as  .

→ A similar process occurs for the bullet fired from the surface of Earth at its escape velocity. As the mass and radius of Earth are different to that of Mars, the actual escape velocity, will be different.

→ The initial values of kinetic and potential energy, will also be different.

Show Worked Solution

→ The total energy of the bullet fired from the surface of Mars will remain constant. Its kinetic energy will transform into potential energy as it gains altitude. 

→ As the bullet fired from Mars reaches an ‘infinite’ distance away, it escapes Mars’ gravitational attraction. Here, the kinetic energy of the bullet is zero as it has expended all of its initial kinetic energy.

→ The potential energy () of the bullet has also decreased to zero as  .

→ A similar process occurs for the bullet fired from the surface of Earth at its escape velocity. As the mass and radius of Earth are different to that of Mars, the actual escape velocity, will be different.

→ The initial values of kinetic and potential energy, will also be different.

PHYSICS, M5 EQ-Bank 25

In the 1840s, French physicist, Hippolyte Fizeau performed an experiment to measure the speed of light. He shone an intense light source at a mirror 8 km away and broke up the light beam with a rotating cogwheel. He adjusted the speed of rotation of the wheel until the reflected light beam could no longer be seen returning through the gaps in the cogwheel.

The diagram shows a similar experiment. The cogwheel has 50 teeth and 50 gaps of the same width.
 

Explain why specific speeds of rotation of the cogwheel will completely block the returning light. Support your answer with calculations.  (5 marks)

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→ Light travelling through a gap in the cogwheel will be completely blocked by a tooth when it returns if a tooth moves exactly the width of a gap in the time it takes for light to travel to the mirror and back.

→ Calculating the time taken:

 
   
   

 
→ To completely block the light, the tooth will have moved into the path of a gap in this time. As there are 50 teeth and 50 gaps, the wheel will have undergone one hundredth of a rotation in this time.

 
   
   

 
→ Additionally, spinning the cogwheel at 3, 5, 7 or any odd multiple of this speed will also completely block returning light. This is as light will be blocked by subsequent teeth, i.e, if the cogwheel is spun at 3 times this rate, the cogwheel will have undergone three hundredths of a rotation in the time it takes for light to return and this returning light will be completely blocked by the second tooth.

Show Worked Solution

→ Light travelling through a gap in the cogwheel will be completely blocked by a tooth when it returns if a tooth moves exactly the width of a gap in the time it takes for light to travel to the mirror and back.

→ Calculating the time taken:

 
   
   

 
→ To completely block the light, the tooth will have moved into the path of a gap in this time. As there are 50 teeth and 50 gaps, the wheel will have undergone one hundredth of a rotation in this time.

 
   
   

 
→ Additionally, spinning the cogwheel at 3, 5, 7 or any odd multiple of this speed will also completely block returning light. This is as light will be blocked by subsequent teeth, i.e, if the cogwheel is spun at 3 times this rate, the cogwheel will have undergone three hundredths of a rotation in the time it takes for light to return and this returning light will be completely blocked by the second tooth.

PHYSICS, M7 EQ-Bank 26

The diagram shows a light source, slits and a translucent screen arranged for an experiment on light. Light and dark bands form on the screen. The light has a wavelength of 590 nm. The diagram is not to scale.
 

  1. Explain how any one of the dark bands forms on the screen.  (3 marks)
  1. The distance between the centres of the double slit is 0.15 mm, and the distance between the double slit and the screen is 0.75 m.
  2. Calculate the distance on the screen from the centre of the central maximum to the centre of a second-order bright band.  (3 marks)
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a.    → When light arrives at the double slit, it undergoes diffraction.

→ The two slits act as sources of secondary wavefronts and light travels from both slits to the screen.

→ When light from one slit travels any odd multiple of more than light from the other slit to the screen, it is out of phase and destructive interference occurs. Here waves superimpose and cancel each other out, creating dark bands.

→ For example, the dark bands either side of the central maximum form where light from one slit travels further than light from the other slit.
 

b.   

Show Worked Solution

a.    → When light arrives at the double slit, it undergoes diffraction.

→ The two slits act as sources of secondary wavefronts and light travels from both slits to the screen.

→ When light from one slit travels any odd multiple of more than light from the other slit to the screen, it is out of phase and destructive interference occurs. Here waves superimpose and cancel each other out, creating dark bands.

→ For example, the dark bands either side of the central maximum form where light from one slit travels further than light from the other slit.
 

b.   

 
 

 

  
→ Using trigonometry, , where is the angular separation of the second order bright band from the central maximum and is the distance between the centre of the central maximum and the centre of the second order bright band.

→ As the angle, is small, the approximation is valid:

 
 
   

PHYSICS, M7 EQ-Bank 28

In an experiment to investigate the photoelectric effect, a group of students used a piece of equipment containing a metal cathode inside a glass tube. The students were able to accurately measure both the current produced and the maximum energy of electrons in response to light hitting the cathode.

Explain how the choice of independent variable would give rise to different results. Sketch graphs to illustrate your answer.  (7 marks)

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Variables: the frequency of incident light (independent), and the maximum kinetic energy of ejected electrons (dependent).

→ Students would observe that below a certain frequency, no photoelectrons would be ejected. Photons with frequency less than the threshold frequency do not have enough energy to eject an electron.

→ Above this frequency, the students would observe that as the frequency increases, the kinetic energy of ejected electrons would increase linearly. This is because a specific amount of the photon’s energy is required to eject an electron, and any photon energy remaining is transferred to the electrons as kinetic energy, consistent with  .
 


 

Variables: intensity of incident light (independent) and the resultant photocurrent (dependent).

→ The frequency of light would be controlled and would be above the threshold frequency.

→ They would observe as the intensity of light increases the current produced would increase linearly.

→ This is because an increasing intensity of light increases the number of photons. This increases the rate at which photons strike the metal surface which increases the rate of photoelectron emission which in turn increases the photocurrent.
 

Show Worked Solution

Variables: the frequency of incident light (independent), and the maximum kinetic energy of ejected electrons (dependent).

→ Students would observe that below a certain frequency, no photoelectrons would be ejected. Photons with frequency less than the threshold frequency do not have enough energy to eject an electron.

→ Above this frequency, the students would observe that as the frequency increases, the kinetic energy of ejected electrons would increase linearly. This is because a specific amount of the photon’s energy is required to eject an electron, and any photon energy remaining is transferred to the electrons as kinetic energy, consistent with  .
 


 

Variables: intensity of incident light (independent) and the resultant photocurrent (dependent).

→ The frequency of light would be controlled and would be above the threshold frequency.

→ They would observe as the intensity of light increases the current produced would increase linearly.

→ This is because an increasing intensity of light increases the number of photons. This increases the rate at which photons strike the metal surface which increases the rate of photoelectron emission which in turn increases the photocurrent.
 

PHYSICS, M7 EQ-Bank 29

A proton travels along a particle accelerator at 10 m s ¯1 less than the speed of light.

Compare its speed and momentum with a proton travelling at 99% the speed of light. Support your answer with calculations.  (4 marks)

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→ First proton speed = , second proton speed =

→ Comparing the two: .

→ The first proton is travelling 1% faster than the second proton.

→ Calculate the relativistic momentum of the first proton ():

 
   
   

 
→ Calculate the relativistic momentum of the second proton ():

 
   

 
→ Compare the two: .

→ The momentum of the first proton is 551 times greater than the momentum of the second proton, while only going 1% faster.

Show Worked Solution

→ First proton speed = , second proton speed =

→ Comparing the two: .

→ The first proton is travelling 1% faster than the second proton.

→ Calculate the relativistic momentum of the first proton ():

 
   
   

 
→ Calculate the relativistic momentum of the second proton ():

 
   

 
→ Compare the two: .

→ The momentum of the first proton is 551 times greater than the momentum of the second proton, while only going 1% faster.

PHYSICS, M8 EQ-Bank 28

Our understanding of matter is still incomplete and the Standard Model of matter is still being validated and tested. Technology plays a substantial role in this.

Explain the role of technology in developing both the Standard Model of matter and our understanding in ONE other area of physics.  (9 marks)

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Technology and the development of the Standard Model

→ Technology has played a significant role in developing the standard model of matter.

→ Scientists have used the technology of linear accelerators to accelerate a beam of electrons at stationary protons. Technology was then used to analyse the scattering patterns of the electrons which was inconsistent with protons being fundamental particles.

→ It was determined that protons were comprised of both positive and negative internal charges. This led to the discovery of quarks.

→ Further, the Large Hadron Collider (LHC) is technology which accelerates protons to speeds extremely close to the speed of light, and collides them with each other.

→ When these protons collide, their dilated kinetic energy is converted to mass in the form of new particles such as the Higgs Boson. This significantly develops our understanding of the standard model of matter.
 

Technology and Special Relativity

→ Another area of physics in which technology has played a vital role is special relativity.

→ Einstein’s prediction of time dilation has been validated by the Hafele-Keating experiment. Technology such as atomic clocks and high speed aeroplanes were used to demonstrate time differences recorded when atomic clocks were flown around the world.

→ In this instance, technology made it possible to validate Einstein’s predictions, improving our understanding of special relativity.

Show Worked Solution

Technology and the development of the Standard Model

→ Technology has played a significant role in developing the standard model of matter.

→ Scientists have used the technology of linear accelerators to accelerate a beam of electrons at stationary protons. Technology was then used to analyse the scattering patterns of the electrons which was inconsistent with protons being fundamental particles.

→ It was determined that protons were comprised of both positive and negative internal charges. This led to the discovery of quarks.

→ Further, the Large Hadron Collider (LHC) is technology which accelerates protons to speeds extremely close to the speed of light, and collides them with each other.

→ When these protons collide, their dilated kinetic energy is converted to mass in the form of new particles such as the Higgs Boson. This significantly develops our understanding of the standard model of matter.
 

Technology and Special Relativity

→ Another area of physics in which technology has played a vital role is special relativity.

→ Einstein’s prediction of time dilation has been validated by the Hafele-Keating experiment. Technology such as atomic clocks and high speed aeroplanes were used to demonstrate time differences recorded when atomic clocks were flown around the world.

→ In this instance, technology made it possible to validate Einstein’s predictions, improving our understanding of special relativity.

PHYSICS, M7 EQ-Bank 24

Parallel light rays of intensity pass through two polarising filters and to a detector. The filters are initially aligned so that they produce the maximum amount of light, then filter is slowly rotated through 180° as shown.
 

  1. On the axes provided sketch a graph showing how the intensity of light at the detector, , changes as rotates from zero to 180°.
     

  1. is now rotated to a position such that no light reaches the detector. Without moving or , a third polarising filter is inserted between and and rotated at an angle of 30° from .
  2. Explain, with the aid of calculations, why the light intensity at the detector is no longer zero.
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a.   

b.    → must be rotated 90° from (no light reaches detector)

→ Calculate the light passing through the middle polariser:

 
 
   

 
→ is a 60° rotation from the middle polariser:

 
 
   

  
→ Light intensity at detector is therefore no longer zero.

Show Worked Solution

a.   

b.    → must be rotated 90° from (no light reaches detector)

→ Calculate the light passing through the middle polariser:

 
 
   

 
→ is a 60° rotation from the middle polariser:

 
 
   

  
→ Light intensity at detector is therefore no longer zero.

PHYSICS, M8 EQ-Bank 25

An experiment was conducted to model Millikan's oil drop experiment. In the experiment, different numbers of dominoes were placed inside seven identical boxes. The boxes were then sealed and weighed. The table shows the mass of each sealed box and some preliminary analysis.
 

Analyse this experiment to assess its effectiveness in modelling Millikan's oil drop experiment.  (6 marks)

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→ Analysing the results shows that the smallest difference in masses between two boxes is 4.3 g, and all other differences are multiples of 4.3. 

→ If we assume that this difference is due to one domino, the mass of a single domino is 4.3 g. While this can’t be known for certain, repeated tests would improve the probability that this is true.

→ This experiment and analysis is similar to that done in Millikan’s oil drop experiment, where he balanced the electric and gravitational forces on charged oil drops. From this, Millikan was able to calculate the charge on a single oil drop.

→ Millikan found that the value of the charge of an oil drop was always an integer multiple of . This was then said to be the fundamental charge on an electron.

→ The domino experiment is effective in demonstrating Millikan’s analysis of his results to find a fundamental quantity. It also demonstrates assumptions and problems Millikan may have had, such as whether just one electron was being measured.

Show Worked Solution

→ Analysing the results shows that the smallest difference in masses between two boxes is 4.3 g, and all other differences are multiples of 4.3. 

→ If we assume that this difference is due to one domino, the mass of a single domino is 4.3 g. While this can’t be known for certain, repeated tests would improve the probability that this is true.

→ This experiment and analysis is similar to that done in Millikan’s oil drop experiment, where he balanced the electric and gravitational forces on charged oil drops. From this, Millikan was able to calculate the charge on a single oil drop.

→ Millikan found that the value of the charge of an oil drop was always an integer multiple of . This was then said to be the fundamental charge on an electron.

→ The domino experiment is effective in demonstrating Millikan’s analysis of his results to find a fundamental quantity. It also demonstrates assumptions and problems Millikan may have had, such as whether just one electron was being measured.

PHYSICS, M7 EQ-Bank 16 MC

In 1972 , four caesium clocks were flown twice around the world on commercial jet flights, once eastward and once westward. The travelling clocks were compared with reference clocks at the US Naval Observatory and the results were compared with predictions from Einstein's theory of special relativity.
 

Which of the following is correct about the observed results in relation to Einstein's theory?

  1. Both of the results are inconclusive.
  2. Both of the results support the theory.
  3. One of the results supports the theory and the other is inconclusive.
  4. One of the results supports the theory and the other rejects the theory.
Show Answers Only

Show Worked Solution

→ The tolerance range for the eastward journey is –63 to –17 nanoseconds. 

→ The tolerance range for the westward journey is 254 to 296 nanoseconds. 

→ Therefore, both of the results fall within the range predicted by Einstein’s theory of special relativity.

PHYSICS, M7 EQ-Bank 17 MC

The graph shows the maximum kinetic energy with which photoelectrons are emitted as a function of frequency for two different metals and .
 

The metals are illuminated with light of wavelength 450 nm.

What would be the effect of doubling the intensity of this light without changing the wavelength?

  1. For metal , the number of photoelectrons emitted would not change but the maximum kinetic energy would increase.
  2. For metal , the number of photoelectrons emitted would increase but the maximum kinetic energy remains unchanged.
  3. For metals and , the number of photoelectrons emitted would not change but the maximum kinetic energy would increase.
  4. For metals and , the number of photoelectrons emitted would increase but the maximum kinetic energy remains unchanged.
Show Answers Only

Show Worked Solution

→ Calculating the frequency of incident light:

 
 
   
   

 
→ This is greater than the work function of metal , but less than the work function of metal . So, when the light is incident upon both metals, photoelectron emission will occur for but not .

→ Increasing the intensity of the light increases the rate at which photons strike the surface of metal . This increases the rate of photoelectron emission.

→ The only way to increase the maximum kinetic energy is to increase the frequency of incident light. So, the maximum kinetic energy of photoelectrons remains unchanged. 

PHYSICS, M6 EQ-Bank 27

The diagram shows two rings and , connected to a balancing arm which swings freely on a pivot. Ring has a split in it as shown.
 

When a bar magnet is pushed into one of the rings, the whole balancing arm begins to rotate on the pivot. When the magnet is pulled out, the balancing arm begins to rotate in the opposite direction. When the magnet is pushed in and out of the other ring, the apparatus does not move at all.

Account for these observations using Lenz's Law and conservation of energy.  (5 marks)

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→ When the magnet is pushed into ring , a current is induced in the ring (Faraday’s Law). 

→ The induced current is in the direction such that the magnetic field it produces opposes the original change caused by the moving magnet (Lenz’s Law).

→ When the magnet is pushed into ring , a like pole is produced which causes the magnet and the ring to repel. When the magnet is pulled out of ring , an opposite pole is produced which causes the magnet and the ring to attract.

→ This is consistent with the law of conservation of energy as follows:

    • If the current were in the opposite direction then the field produced would cause a movement that increases the change in flux through the ring.
    • This would thereby produce an even greater induced current which in turn would accelerate the ring even more, in turn leading to an even greater change in flux through the ring.
    • This cycle would continue producing more kinetic and heat energy in the ring than was initially in the system thereby violating the law of conservation of energy. 

→ When the magnet is pushed into ring , no movement of the ring is observed because the gap in the ring prevents a current from being induced.

→ In this instance, no magnetic field is created which means there is no attractive or repulsive force between the ring and the magnet.

Show Worked Solution

→ When the magnet is pushed into ring , a current is induced in the ring (Faraday’s Law). 

→ The induced current is in the direction such that the magnetic field it produces opposes the original change caused by the moving magnet (Lenz’s Law).

→ When the magnet is pushed into ring , a like pole is produced which causes the magnet and the ring to repel. When the magnet is pulled out of ring , an opposite pole is produced which causes the magnet and the ring to attract.

→ This is consistent with the law of conservation of energy as follows:

    • If the current were in the opposite direction then the field produced would cause a movement that increases the change in flux through the ring.
    • This would thereby produce an even greater induced current which in turn would accelerate the ring even more, in turn leading to an even greater change in flux through the ring.
    • This cycle would continue producing more kinetic and heat energy in the ring than was initially in the system thereby violating the law of conservation of energy. 

→ When the magnet is pushed into ring , no movement of the ring is observed because the gap in the ring prevents a current from being induced.

→ In this instance, no magnetic field is created which means there is no attractive or repulsive force between the ring and the magnet.

PHYSICS, M6 EQ-Bank 26

A solenoid was connected to a data logger to measure voltage. A magnet was dropped through the solenoid from above as shown.
 

On the axes provided, sketch a graph showing the change in voltage as the magnet falls completely through the solenoid.  (3 marks)
 

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Points to note on graph:

→ First peak negative, longer wavelength and shorter amplitude.

→ Second peak positive, shorter wavelength and higher amplitude.

Show Worked Solution

Points to note on graph:

→ First peak negative, longer wavelength and shorter amplitude.

→ Second peak positive, shorter wavelength and higher amplitude.

PHYSICS, M6 EQ-Bank 24

Negatively charged particles were accelerated from rest between a pair of parallel metal plates. The potential difference between the plates was varied, and the final velocity of the particles was measured for each variation.
 

The data in the table show the potential difference between the plates and the square of the corresponding final velocity of the particles.

  1. Plot the data on the grid provided and draw a line of best fit.
     

   

  1. A student hypothesised that the charged particles are electrons. Justify whether the student's hypothesis is correct or not. Support your answer using the data provided and relevant calculations.
Show Answers Only

a.   
         

     

b.    → The gradient of the line =

→ Calculate the gradient using two points on the line of best fit (choose values close to limits):

 
   

 
→ The change in kinetic energy is equal to the work done by the electric field on the charged particles:

 
 
 

 
→ Calculate the charge to mass ratio of the particles:

 
   
   

 
 → Calculate the charge to mass ratio of an electron:

 
   

  
→ Therefore, the charged particles are not electrons.

Show Worked Solution

a.   
         

     

b.    → The gradient of the line =

→ Calculate the gradient using two points on the line of best fit (choose values close to limits):

 
   

 
→ The change in kinetic energy is equal to the work done by the electric field on the charged particles:

 
 
 

 
→ Calculate the charge to mass ratio of the particles:

 
   
   

 
 → Calculate the charge to mass ratio of an electron:

 
   

  
→ Therefore, the charged particles are not electrons.

PHYSICS, M6 EQ-Bank 26

The diagram shows a stationary electron in a magnetic field. The magnetic field is surrounded by two parallel plates separated by a distance of and connected to a power supply and a switch.
 

The switch is initially open. At a later time the switch is closed.

Analyse the effects of the magnetic and electric fields on the acceleration of the electron both before and immediately after the switch is closed. In your answer, include calculation of the acceleration of the electron immediately after the switch is closed.   (5 marks)

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→ Before the switch is closed, there is no electric field and so no electric force causing the electron to accelerate. As the electron is stationary, the magnetic field has no effect on it.

→ Once the switch is closed, the electric field causes the electron to accelerate down the page (towards the positive plate).

→ Calculating the acceleration of the electron immediately after the switch is closed:

→ Now that the electron is moving, the magnetic field exerts a force on it towards the right (using the right hand palm rule).

→ The force and acceleration on the electron will increase due to its increasing velocity.

Show Worked Solution

→ Before the switch is closed, there is no electric field and so no electric force causing the electron to accelerate. As the electron is stationary, the magnetic field has no effect on it.

→ Once the switch is closed, the electric field causes the electron to accelerate down the page (towards the positive plate).

→ Calculating the acceleration of the electron immediately after the switch is closed:

→ Now that the electron is moving, the magnetic field exerts a force on it towards the right (using the right hand palm rule).

→ The force and acceleration on the electron will increase due to its increasing velocity.

PHYSICS, M6 EQ-Bank 16 MC

An experiment was carried out to investigate the change in torque for a DC motor with a radial magnetic field. The data from start up to operating speed were graphed.

Which graph is most likely to represent this set of data?
 

Show Answers Only

Show Worked Solution

→ The DC motor has a radial magnetic field, so the coil is always perpendicular to the magnetic field. This means the torque for the motor does not depend on the angle of the coil.

→  The torque for this DC motor is therefore given by  .

→ As the speed of the motor increases from start to operating speed, the rate of change of magnetic flux through its coil increases, increasing the induced emf in the motor (Faraday’s Law).

→ This induced back emf acts to oppose the rotation of the motor (Lenz’s Law). 

→ The back emf opposes the current supplied to the motor, decreasing the net current through the motor.

→ Therefore, the torque produced by the motor decreases as motor speed increases.

PHYSICS, M6 EQ-Bank 23

The diagram shows a circuit containing two ideal transformers connected with an ammeter. The current through the load is 5.0 A.
 

What is the reading on the ammeter?   (3 marks)

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Show Worked Solution

→ The total power () of the circuit is constant. This can be calculated from the data given regarding the load.

 
   
   

 
→ Calculating the voltage in the secondary coil of transformer 1:

 
 
   
   

 
→ Calculating the reading on the ammeter:

 
 
 

PHYSICS, M6 EQ-Bank 8 MC

A positively-charged ion travelling at 250 ms ¯1 is fired between two parallel charged plates, and . There is also a magnetic field present in the region between the two plates. The direction of the magnetic field is into the page as shown. The ion is travelling perpendicular to both the electric and the magnetic fields.
 

The electric field between the plates has a magnitude of 200 V m ¯1. The magnetic field is adjusted so that the ion passes through undeflected.

What is the magnitude of the adjusted magnetic field, and the polarity of the terminal relative to the terminal?

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Show Worked Solution

→ The ion passes through undeflected, so the magnitude of the force it experiences due to the magnetic field is equal to the magnitude of the force it experiences due to the electric field. 

→ As the ion travels perpendicular to the magnetic field:

 
 
 
   
   
   

 
→ Using the right hand palm rule, the magnetic field exerts a force up the page on the positive ion. The electric field must therefore exert a force down the page on the ion.

→ must be positive relative to .

BIOLOGY, M8 EQ-Bank 14

Glucose is a chemical that must be maintained at concentrations between 70 to 130 mg/dL in the blood in order for the body to function normally.

  1. Draw a diagram that illustrates how the body maintains blood glucose within this range.  (3 marks)

  2. Sketch a graph on the axes provided showing the expected blood glucose levels of both a healthy person and a diabetic person after consuming a fruit juice. On the same graph, show what would happen when the diabetic person injects themself with insulin 20 minutes after consuming the fruit juice.  (3 marks)
     
     

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a.  

 
b.

   

Show Worked Solution

a.  

 
b.

   

BIOLOGY, M8 EQ-Bank 15

The diagram shows a rural coastal area and the towns, rivers and associated industry for each of the townships.
 

An epidemic of a disease has broken out in Nanavale. The symptoms are stomach ache, vomiting and tiredness. Many families in Nanavale have only one member with the disease, therefore it appears to be non-infectious. The symptoms are worse in infants than in adults.

Isolated cases of this disease have occurred in the nearby towns of Dairyville and Beefville. No cases have been reported on Gull Island.

Design an epidemiological study to investigate the origin of the disease. Refer to features of validity and reliability in your answer.   (7 marks)

Show Answers Only

→ When planning an epidemiological study it is important to first analyse all the initial evidence to better construct an effective study.

→ The disease is most likely infectious as an outbreak that affects many people is very unlikely to be due to a previously masked non-infectious disease.

→ The fact that the disease is not present on gull island also supports this and may indicate that the disease is not waterborne and may spread through physical touch, close proximity, food or radioactive toxic elements.

→ The disease also affects children more severely. This fact must be addressed in the study and measures taken to protect and monitor this vulnerable group.

→ The study should survey affected families to try to pinpoint the transmission of the disease.

→ Appropriate fact finding questions should include

    • Where have you travelled to?
    • What have you eaten/drunk and where did you get it from?
    • Who else have you been in contact with and where are they from?

→ These results should be analysed for common factors and then compared to results from the same set of questions asked of unaffected families, thus increasing the study’s validity.

→ The more people that can be reached and questioned, the more accurate the findings of the study will be.

→ Geiger readings, screening and soil extraction may pinpoint whether the disease is caused by a carcinogen on the coastal area.

→ Common factors found in the study may also reveal an antidote or treatment for affected individuals.

Show Worked Solution

→ When planning an epidemiological study it is important to first analyse all the initial evidence to better construct an effective study.

→ The disease is most likely infectious as an outbreak that affects many people is very unlikely to be due to a previously masked non-infectious disease.

→ The fact that the disease is not present on gull island also supports this and may indicate that the disease is not waterborne and may spread through physical touch, close proximity, food or radioactive toxic elements.

→ The disease also affects children more severely. This fact must be addressed in the study and measures taken to protect and monitor this vulnerable group.

→ The study should survey affected families to try to pinpoint the transmission of the disease.

→ Appropriate fact finding questions should include

    • Where have you travelled to?
    • What have you eaten/drunk and where did you get it from?
    • Who else have you been in contact with and where are they from?

→ These results should be analysed for common factors and then compared to results from the same set of questions asked of unaffected families, thus increasing the study’s validity.

→ The more people that can be reached and questioned, the more accurate the findings of the study will be.

→ Geiger readings, screening and soil extraction may pinpoint whether the disease is caused by a carcinogen on the coastal area.

→ Common factors found in the study may also reveal an antidote or treatment for affected individuals.

BIOLOGY, M8 EQ-Bank 16

How effective is renal dialysis in compensating for the loss of kidney function?   (7 marks)

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Normal kidney function

→ The kidneys are the main components of the mammalian urinary system. They are organs which filter blood and maintain water, pH, ion and salt concentration in the body through varying concentrations of each in excreted urine dependent on the body’s needs.

→ Each kidney contains 1 million nephrons, the main unit responsible for filtration.

→ Each one also contains a Bowman’s capsule, proximal and distal tubules as well as the Loop of Henle which acts as a site for selective re-absorption of certain components of the blood. This is controlled by both passive diffusion of unwanted substances through a concentration gradient (e.g. urea) or by hormonal control.

→ Aldosterone and ADH are hormones secreted by the hypothalamus which increase the permeability of the distal convoluted tubule to salt and water respectively.
 

Kidney disfunction and dialysis

→ Kidney function can however be impaired by diseases or disorders (such as polycystic kidney disease), many of which can kill affected individuals in a number of months if left untreated.

→ When kidney function drops below 80%, haemodialysis is an effective treatment to replace kidney function.

→ Haemodialysis involves the removal of blood from the body into the dialysis machine, which will clean the blood before returning it to the body. This is achieved by running a fluid known as dialysate, countercurrent to the blood.

→ The dialysate contains a similar composition to the blood with low urea and toxins to allow the passive diffusion of the substances via the concentration gradient into the dialysate. This is then removed and constantly replenished during a session. The countercurrent direction also improves effectiveness of this process. The dialysate can also be altered to have varying amounts of salt and ions depending on the concentration in the patients body.

→ Haemodialysis can provide an effective treatment for individuals until death or an effective transplant can be found, however the process often requires 3-4 sessions per week each of which is 4 hours long.

→ Without haemodialysis loss of kidney function is often fatal, but this life-saving technology is extremely effective in preventing many deaths despite its inconvenience.

Show Worked Solution

Normal kidney function

→ The kidneys are the main components of the mammalian urinary system. They are organs which filter blood and maintain water, pH, ion and salt concentration in the body through varying concentrations of each in excreted urine dependent on the body’s needs.

→ Each kidney contains 1 million nephrons, the main unit responsible for filtration.

→ Each one also contains a Bowman’s capsule, proximal and distal tubules as well as the Loop of Henle which acts as a site for selective re-absorption of certain components of the blood. This is controlled by both passive diffusion of unwanted substances through a concentration gradient (e.g. urea) or by hormonal control.

→ Aldosterone and ADH are hormones secreted by the hypothalamus which increase the permeability of the distal convoluted tubule to salt and water respectively.
 

Kidney disfunction and dialysis

→ Kidney function can however be impaired by diseases or disorders (such as polycystic kidney disease), many of which can kill affected individuals in a number of months if left untreated.

→ When kidney function drops below 80%, haemodialysis is an effective treatment to replace kidney function.

→ Haemodialysis involves the removal of blood from the body into the dialysis machine, which will clean the blood before returning it to the body. This is achieved by running a fluid known as dialysate, countercurrent to the blood.

→ The dialysate contains a similar composition to the blood with low urea and toxins to allow the passive diffusion of the substances via the concentration gradient into the dialysate. This is then removed and constantly replenished during a session. The countercurrent direction also improves effectiveness of this process. The dialysate can also be altered to have varying amounts of salt and ions depending on the concentration in the patients body.

→ Haemodialysis can provide an effective treatment for individuals until death or an effective transplant can be found, however the process often requires 3-4 sessions per week each of which is 4 hours long.

→ Without haemodialysis loss of kidney function is often fatal, but this life-saving technology is extremely effective in preventing many deaths despite its inconvenience.

BIOLOGY, M8 EQ-Bank 14

  1. Identify a disorder or disease, and describe how it affects the normal function of an organ.   (2 marks)

  2. Evaluate the effectiveness of a technology in managing the disorder or disease described in part (a).   (4 marks)

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a.   Polycystic kidney disease

→ A disease in which cysts grow in the kidney.

→ This interferes with filtration and overtime, adversely affects the overall function of the kidney.
 

Other disorders/disease can include

→ Astigmatism, myopia or hyperopia.

→ Single sided deafness, hearing loss.
  

b.   Use of technology in treating polycystic kidney disease

→ Polycystic kidney disease will often progress in patients until kidney failure has occurred.

→ When 80% of kidney function is lost and a transplant cannot be found, dialysis is often the only viable option.

→ There are two types of dialysis, both of which replace the function of the kidney and clean the blood.

    • Haemodialysis is where blood is removed from the body and passed through a dialysis machine. The main component of the machine runs a special fluid known as dialysate countercurrent to the blood next to a permeable membrane. This allows the blood to effectively remove the toxins into the dialyse by diffusion. While this is extremely effective, it involves multiple visits per week and is very time consuming for the patients.
       
    • Peritoneal dialysis involves the flushing of dialysate directly into the abdominal cavity through a catheter. The peritoneal membrane then acts as the filter and blood is cleaned via diffusion. The dialysate containing the waste products then exits the body into another section of the catheter. In contrast to haemodialysis, this method requires a more involved initial surgery but provides the benefit of patients being able to do this by themselves at night.

→ Both forms of dialysis are extremely effective in replacing kidney function. Despite being time consuming and uncomfortable, they allow patients suffering from polycystic kidney disease to survive for extended periods, during which a transplant may become available.
 

Other technologies can include

→ Glasses.

→ Any form of hearing aid dependent on the ear disease/disorder, including standard hearing aids, bone conduction implants or cochlear implant.

Show Worked Solution

a.   Polycystic kidney disease

→ A disease in which cysts grow in the kidney.

→ This interferes with filtration and overtime, adversely affects the overall function of the kidney.
 

Other disorders/disease can include

→ Astigmatism, myopia or hyperopia.

→ Single sided deafness, hearing loss.
  

b.   Use of technology in treating polycystic kidney disease

→ Polycystic kidney disease will often progress in patients until kidney failure has occurred.

→ When 80% of kidney function is lost and a transplant cannot be found, dialysis is often the only viable option.

→ There are two types of dialysis, both of which replace the function of the kidney and clean the blood.

    • Haemodialysis is where blood is removed from the body and passed through a dialysis machine. The main component of the machine runs a special fluid known as dialysate countercurrent to the blood next to a permeable membrane. This allows the blood to effectively remove the toxins into the dialyse by diffusion. While this is extremely effective, it involves multiple visits per week and is very time consuming for the patients.
       
    • Peritoneal dialysis involves the flushing of dialysate directly into the abdominal cavity through a catheter. The peritoneal membrane then acts as the filter and blood is cleaned via diffusion. The dialysate containing the waste products then exits the body into another section of the catheter. In contrast to haemodialysis, this method requires a more involved initial surgery but provides the benefit of patients being able to do this by themselves at night.

→ Both forms of dialysis are extremely effective in replacing kidney function. Despite being time consuming and uncomfortable, they allow patients suffering from polycystic kidney disease to survive for extended periods, during which a transplant may become available.
 

Other technologies can include

→ Glasses.

→ Any form of hearing aid dependent on the ear disease/disorder, including standard hearing aids, bone conduction implants or cochlear implant.

BIOLOGY, M7 EQ-Bank 23

The diagram shows the immune response after primary exposure to a pathogen.
 

  1. On the diagram, continue the graph to show the immune response upon secondary exposure to the same pathogen.   (2 marks)
  2. Using annotations on the diagram, explain the shape of the entire graph.   (4 marks)

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a. & b.

Show Worked Solution

a. & b.

BIOLOGY, M8 EQ-Bank 13

A scientist performed an epidemiological study to investigate the cause and effect relationship of smoking and lung cancer as follows:

    1. Handed out a scientifically valid questionnaire to all employees at her daughter's high school (n=96)
    2. Checked that there were an equal number of male and female respondents
    3. Discovered that there were more non-smoking respondents than smoking respondents. Removed some of the non-smokers until both groups had equal numbers
    4. Required all respondents to have a general check up from their doctor before participating 
    5. Analysed data, wrote the paper and published it in a scientific blog

From the information provided, assess the suitability of the methodology for this investigation.   (5 marks)

Show Answers Only

→ This study cannot be referred to as a suitable comparison between smoking and lung cancer due to several flaws in its methodology.

→ The sample size of the study is too small and confined to one workplace.

→ The participants should represent a number of equal categories such as age and ethnicity, not just equal representation of males and females.

→ The fact that participants were removed in order to create equal smoking/non-smoking categories reduces the validity of the findings.

→ The medical check ups should not be general and instead be in relation to the study such as cancer and lung checks.

→ The study will also benefit from peer review.

Show Worked Solution

→ This study cannot be referred to as a suitable comparison between smoking and lung cancer due to several flaws in its methodology.

→ The sample size of the study is too small and confined to one workplace.

→ The participants should represent a number of equal categories such as age and ethnicity, not just equal representation of males and females.

→ The fact that participants were removed in order to create equal smoking/non-smoking categories reduces the validity of the findings.

→ The medical check ups should not be general and instead be in relation to the study such as cancer and lung checks.

→ The study will also benefit from peer review.

BIOLOGY, M8 EQ-Bank 3 MC

A student wanted to test the claim, made by an agricultural seed company, that their variety of wheat was more salt tolerant than other wheat varieties on the market. The company explained that their variety can better maintain water balance by increasing the organic salt concentration in the roots of the plants which increases osmotic pressure.

Which row of the table shows the independent and dependent variables that the student should use to test the company's claim?
 

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Show Worked Solution

By Elimination

→ When testing this aim, the variety of wheat plant should vary and the salt concentration given to each must remain the same. (Eliminate A and B).

→ Each variety’s turgor will give an accurate representation of ability to maintain water balance, the initial claim by the company as to its wheat’s benefit. (Eliminate C).

BIOLOGY, M7 EQ-Bank 26

The immune system's primary role is to defend against pathogens. For this to be effective the immune system must be able to recognise cells that belong to the body and cells that do not.

  1. Describe the mechanism that the immune system uses to distinguish between body cells and potential pathogens. Support your answer with an example.   (3 marks)

  2. Explain why this mechanism means that patients who receive an organ donation require immune suppression drugs.   (3 marks)

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a.    Immune system blood type mechanisms

→ Every cell has proteins on its surface which the body can use to distinguish them from self and non-self.

→ The body’s own cells produce a specific composition of surface proteins referred to as the major histocompatibility complex (MHC), as opposed to other proteins which are not recognised by the immune system and are called antigens.

→ This provides the basis of blood types, where A and B are specific surface proteins while O is blood with no surface proteins.

→ AB blood type individuals are therefore universal recipients, as their immune system has recognised both A and B proteins as self. This also explains why type O individuals are universal donors as their blood has no surface proteins, hence type O, A, B and AB individuals will not recognise it as foreign.
 

b.   → Immune systems detect new donated organs as foreign bodies.

→ When foreign cells/material are detected by the body, an immune response will attempt to kill or remove it from the body.

→ The same scenario occurs during an organ transplant, as the donor’s MHC complex is different to that of the recipient.

→ To counteract this, specialists run genetic testing on donors as well as the recipient to find people with a similar chromosome 6 genetic sequence. MHC is coded by genes found on this chromosome and compatible organ donors are often family members.

→ As a precaution, specialists administer immune suppression drugs which will slow the patient’s immune response, reducing the chance of an attack on the donated organ.

Show Worked Solution

a.    Immune system blood type mechanisms

→ Every cell has proteins on its surface which the body can use to distinguish them from self and non-self.

→ The body’s own cells produce a specific composition of surface proteins referred to as the major histocompatibility complex (MHC), as opposed to other proteins which are not recognised by the immune system and are called antigens.

→ This provides the basis of blood types, where A and B are specific surface proteins while O is blood with no surface proteins.

→ AB blood type individuals are therefore universal recipients, as their immune system has recognised both A and B proteins as self. This also explains why type O individuals are universal donors as their blood has no surface proteins, hence type O, A, B and AB individuals will not recognise it as foreign.
 

b.   → Immune systems detect new donated organs as foreign bodies.

→ When foreign cells/material are detected by the body, an immune response will attempt to kill or remove it from the body.

→ The same scenario occurs during an organ transplant, as the donor’s MHC complex is different to that of the recipient.

→ To counteract this, specialists run genetic testing on donors as well as the recipient to find people with a similar chromosome 6 genetic sequence. MHC is coded by genes found on this chromosome and compatible organ donors are often family members.

→ As a precaution, specialists administer immune suppression drugs which will slow the patient’s immune response, reducing the chance of an attack on the donated organ.

BIOLOGY, M7 EQ-Bank 25

A silver birch tree, Butula pendula, has been attacked by a fungal stem canker. Fungal cankers are opportunistic plant pathogens that gain access to the inner layers of the stem as a result of damage to the protective outer layer of bark. The inner layers provide suitable conditions for the fungus to grow. Once established it destroys the bark cells that protect the tree. The tree responds to the presence of the canker under its bark by producing an excess of resinous sap at the wound site.
 

  1. Using the information provided, suggest a hypothesis to explain how the tree is responding to the presence of the canker.   (2 marks)

  2. Consider another plant with a different response to a specific pathogen.
  3. Compare the necessity and limitations of this plant's response with the response of the birch tree described above.   (4 marks)

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a.    Hypothesis of birch tree response:

→ Excess sap production alters the growth conditions of the canker.

→ The sap will change the chemical environment around the wound site and inflame the site to reduce the spread of the pathogen, much similar to inflammation in humans.
 

b.   Red river gum comparison

→ The river red gum responds to halo leaf spot cankers by dropping its leaves.

→ While this means the tree will have to expend energy and resources regrowing leaves, it simultaneously counteracts the canker as it blocks the ability of the leaves to photosynthesise.

→ By dropping the affected leaves, the tree can grow them quicker than if the leaves were left to die.

→ The birch tree will counteract a stem canker by producing excess sap. While this may not completely get rid of any pathogens, the tree cannot simply remove its own stems like the river red gum.

→ This method isolates the infected area, similar to inflammation, retarding its growth and potentially killing the fungus.

→ The isolation reduces the canker’s ability to adversely effect the birch tree’s supply of water and nutrients via the xylem and phloem of the tree.

→ This process occurs with the canker still attached to the tree, making the fungus relatively more difficult to kill.

Show Worked Solution

a.    Hypothesis of birch tree response:

→ Excess sap production alters the growth conditions of the canker.

→ The sap will change the chemical environment around the wound site and inflame the site to reduce the spread of the pathogen, much similar to inflammation in humans.
 

b.   Red river gum comparison

→ The river red gum responds to halo leaf spot cankers by dropping its leaves.

→ While this means the tree will have to expend energy and resources regrowing leaves, it simultaneously counteracts the canker as it blocks the ability of the leaves to photosynthesise.

→ By dropping the affected leaves, the tree can grow them quicker than if the leaves were left to die.

→ The birch tree will counteract a stem canker by producing excess sap. While this may not completely get rid of any pathogens, the tree cannot simply remove its own stems like the river red gum.

→ This method isolates the infected area, similar to inflammation, retarding its growth and potentially killing the fungus.

→ The isolation reduces the canker’s ability to adversely effect the birch tree’s supply of water and nutrients via the xylem and phloem of the tree.

→ This process occurs with the canker still attached to the tree, making the fungus relatively more difficult to kill.

BIOLOGY, M7 EQ-Bank 8 MC

Eight sick animals had the same symptoms. Blood tests showed that they were infected with the same type of bacterium.

Which of the following would be the best course of action to determine if this particular type of bacterium is the cause of the symptoms?

  1. Treat all eight animals with the antibiotic known to kill this type of bacterium. Check if they recover.
  2. Find other animals with the same symptoms. Attempt to isolate the same type of bacterium from their blood.
  3. Inject blood from animals with the symptoms into suitable host individuals. Check if they develop the same symptoms.
  4. Use bacteria from the blood of affected animals to inoculate healthy animals. If these healthy animals develop the symptoms, attempt to isolate the same bacterium from their blood.
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Show Worked Solution

→ The blood from an affected animal will contain bacteria that is presumed to be responsible for the disease.

→ If it is injected into healthy organisms that subsequently develop the same symptoms, and their blood is isolated and observed to have the same bacteria, then it is certain that the bacteria is responsible for the disease.

→ This follows Koch’s postulates.

BIOLOGY, M7 EQ-Bank 9 MC

Melanomas are characterised by uncontrolled cell division caused by mutations that continue to occur once the tumour has developed. Scientists have discovered that vaccines produced using antigens extracted from the patient's own melanoma cells can be useful in treating melanoma. When injected, the vaccines stimulate an immune response.

The effect of the melanoma vaccine is to stimulate

  1. T cells which produce antibodies.
  2. cytotoxic T cells which activate B cells.
  3. cell division to produce more lymphocytes.
  4. production of B cells which destroy melanoma cells.
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Show Worked Solution

→ The vaccine will encourage the division of lymphocytes (B and T cells) as a response to melanoma antigens.

BIOLOGY, M7 EQ-Bank 7 MC

Melanomas are characterised by uncontrolled cell division caused by mutations that continue to occur once the tumour has developed. Scientists have discovered that vaccines produced using antigens extracted from the patient's own melanoma cells can be useful in treating melanoma. When injected, the vaccines stimulate an immune response.

What can be inferred from the scientists' discovery?

  1. Cancer cells carry unique antigens.
  2. Self-antigens are not present on cancer cells.
  3. The melanoma patient has a dysfunctional immune system.
  4. The body cannot mount an immune response against cancer cells.
Show Answers Only

Show Worked Solution

→ To provide a vaccine, there must be antigens present on cancer cells in order to trigger an immune response.

BIOLOGY, M7 EQ-Bank 7 MC

The diagram shows a model of disease transmission.
 

An epidemiologist suspected that bats were acting as a reservoir for an infectious disease in humans.

Which condition would need to be met to confirm the epidemiologist's suspicion?

  1. The infectious agent would need to have a mode of entry into humans.
  2. The infectious agent would need a mode of transmission from bats to humans.
  3. The bats would have to be able to transmit the infectious agent between each other.
  4. The susceptible human host must be able to transmit the infectious agent to the reservoir of bats.
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Show Worked Solution

→ For the assumption to be true, the pathogen would need to have developed an adaptation allowing it to transfer from bats to humans.

BIOLOGY, M6 EQ-Bank 28

Describe how technological developments led to the advancement of our knowledge and understanding of inheritance. Support your answer with examples.   (7 marks)

Show Answers Only

→ Artificial pollination is the process by which pollen with known characteristics is manually applied to another plant with known genotypes anther.

→ This is used in agriculture to preserve favourable characteristics in plants. However, it can also be used to study inheritance patterns and genotype/phenotype relationships.

→ Mendel used this process to investigate ‘pure-breeding’ individuals/plants (ones which always produce the same offspring) to develop the idea of recessive and dominant alleles and the offspring they produced depending on the genotype. This laid the basis for the concept of ‘Mendelian ratios’.

→ Mendel used artificial pollination on purple and white pea plants which were known pure-breeding plants and found that all the offspring (F1 Generation) were purple, determining that the purple allele was dominant.

→ When he again used artificial pollination between F1 Generation plants, he found that it always produced 3/4 of purple plants and 1/4 white. This showed that F1 Generation purple plants were not pure-breed and instead ‘heterozygous’ and contained a masked allele for the white colour. This 3:1 ratio is what became known as a Mendelian ratio.
 

Other suggested examples include:

→ Gene editing technology CRISPR

→ Microscopy and staining – behaviour of chromosomes (Sutton and Boveri)

→ X‐ray crystallography and DNA structure (Rosalind Franklin)

→ Radiation – one gene, one polypeptide (Beadle and Tatum)

Show Worked Solution

→ Artificial pollination is the process by which pollen with known characteristics is manually applied to another plant with known genotypes anther.

→ This is used in agriculture to preserve favourable characteristics in plants. However, it can also be used to study inheritance patterns and genotype/phenotype relationships.

→ Mendel used this process to investigate ‘pure-breeding’ individuals/plants (ones which always produce the same offspring) to develop the idea of recessive and dominant alleles and the offspring they produced depending on the genotype. This laid the basis for the concept of ‘Mendelian ratios’.

→ Mendel used artificial pollination on purple and white pea plants which were known pure-breeding plants and found that all the offspring (F1 Generation) were purple, determining that the purple allele was dominant.

→ When he again used artificial pollination between F1 Generation plants, he found that it always produced 3/4 of purple plants and 1/4 white. This showed that F1 Generation purple plants were not pure-breed and instead ‘heterozygous’ and contained a masked allele for the white colour. This 3:1 ratio is what became known as a Mendelian ratio.
 

Other suggested examples include:

→ Gene editing technology CRISPR

→ Microscopy and staining – behaviour of chromosomes (Sutton and Boveri)

→ X‐ray crystallography and DNA structure (Rosalind Franklin)

→ Radiation – one gene, one polypeptide (Beadle and Tatum)