Band 4 – Page 24

CHEMISTRY, M8 EQ-Bank 24

A common antacid tablet contains aluminium hydroxide to neutralise stomach acid. In order for the antacid to be effective, each 500 mg tablet must contain a minimum of 200 mg of aluminium hydroxide.

Two antacid tablets were crushed and reacted with 70 mL of 0.60 mol L¯ ¹ hydrochloric acid. After the antacid had reacted with the acid, the remaining hydrochloric acid was titrated against 0.60 mol L¯ ¹ sodium hydroxide. The average volume of sodium hydroxide used was 35 mL.

Calculate the amount of aluminium hydroxide present in each tablet and justify whether the tablets will be effective as an antacid. (4 marks)

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$\text{Tablets will be effective (See Worked Solutions)}$

Show Worked Solution

$\ce{3HCl(aq) + Al(OH)3(aq) \rightarrow AlCl3(s) + 3H2O(l)} $

$\ce{n(HCl^-) = c \times V = 0.6 \times 0.070 = 0.042 mol} $

$\ce{n(Na(OH)) = c \times V = 0.6 \times 0.035 = 0.021 mol} $

$\ce{Moles reacted with antacid = 0.042 – 0.021 = 0.021 mol} $

$\ce{Mole ratio = 3:1}$

\[\ce{Moles of Al(OH)3 in 2 tablets = \frac{0.021}{3} = 0.007 mol}\]

$\ce{Formula weight of Al(OH)3 = 27 + 3 \times (1 + 16) = 78 g}$

$\ce{m(Al(OH)3) in 2 tablets = 0.007 \times 78 = 0.546 g}$

\[\ce{m(Al(OH)3) in 1 tablet = \frac{0.546}{2} = 273 g}\]

$\text{Since 273 > 200, tablets will be effective.}$

CHEMISTRY, M8 EQ-Bank 28

Limestone $\ce{(CaCO_3)}$ contributes to the hardness of water by releasing $\ce{Ca^2^+}$ ions. The following chemical equation represents this reaction.

$\ce{CaCO3($s$) + H_2O($l$) + CO_2($g$) \rightleftharpoons Ca^2^+($aq$) + 2HCO3^-($aq$)}$ $(\Delta H<0)$

It has been suggested that heating water reduces its hardness.

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

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→ 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 $\ce{Ca^{2+}}$ in heated and non-heated samples of water and any difference in the relative concentrations of $\ce{Ca^{2+}}$ can be used to verify the suggestion.

→ It should be noted that a reduced concentration of $\ce{Ca^{2+}}$ indicates that the water hardness is reduced.

Methodology of testing

→ Prepare standard solutions with known concentrations of $\ce{Ca^{2+}}$ 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 $\ce{Ca^{2+}}$ 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 $\ce{Na+}$ and $\ce{Ca^{2+}}$).

→ 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 $\ce{Ca^{2+}}$ in the heated water samples are consistently lower than the concentrations of $\ce{Ca^{2+}}$ 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 should be noted that a reduced concentration of $\ce{Ca^{2+}}$ indicates that the water hardness is reduced.

Methodology of testing

→ Water used in the experiment should be de-ionised (normal drinking water has an abundance of $\ce{Na+}$ and $\ce{Ca^{2+}}$).

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

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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$\ce{N2(g) + 3H2(g) \rightleftharpoons 2NH3(g) \ \ \ \ ΔH = -92 kJ mol}^{-1} $

From $ t_0$ to $t_1: $ the system is in equilibrium.

At $t_1:$

→ 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 $t_2:$

→ 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 $t_3:$

→ 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

$\ce{N2(g) + 3H2(g) \rightleftharpoons 2NH3(g) \ \ \ \ ΔH = -92 kJ mol}^{-1} $

From $ t_0$ to $t_1: $ the system is in equilibrium.

At $t_1:$

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

At $t_2:$

→ 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.

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

At $t_3:$

→ 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.

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 (`U`) of the bullet has also decreased to zero as `U=-(GMm)/(r)`.

→ 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, `v_(esc)=sqrt((2GM)/(r))` will be different.

→ The initial values of kinetic and potential energy, `U=-(GMm)/(r)` 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.

→ The potential energy (`U`) of the bullet has also decreased to zero as `U=-(GMm)/(r)`.

→ The initial values of kinetic and potential energy, `U=-(GMm)/(r)` will also be different.

PHYSICS, M5 EQ-Bank 23

A rocket carrying a satellite is launched from Earth. Once the rocket engine is switched off the satellite continues in an elliptical orbit.

Explain the satellite's changes in energy during this journey. (3 marks)

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→ During the rocket’s launch, its fuel is used up and its chemical potential energy decreases as its potential and kinetic energy both increase. In this stage the total mechanical energy of the rocket increases.

→ Once the satellite is in orbit, its total mechanical energy is constant.

→ As the satellite moves along its orbit away from Earth and gains altitude, its kinetic energy is converted into potential energy.

→ Once the satellite reaches its maximum distance from Earth within its elliptical orbit, its altitude will then begin to decrease. At this stage, its potential energy will be converted into kinetic energy with total energy remaining constant.

Show Worked Solution

→ Once the satellite is in orbit, its total mechanical energy is constant.

→ As the satellite moves along its orbit away from Earth and gains altitude, its kinetic energy is converted into potential energy.

PHYSICS, M5 EQ-Bank 21

Comet Hale-Bopp is a long period comet that is believed to come from the Oort cloud which lies far beyond the outermost planets. In our solar system, Hale-Bopp travels in an elliptical orbit around the Sun and spends most of its time beyond the outermost planets.

Using a diagram, explain how the motion of Comet Hale-Bopp in its orbit supports Kepler's second law. (3 marks)

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→ Kepler’s Second Law states that a line between the comet and the sun will sweep an equal area in equal time.

→ When the comet is closer to the sun, it must travel with a greater velocity in its orbit compared to when it is far away to sweep the same area in the same time (i.e. such that `A_1=A_2`.

→ Comet Hale-Bopp supports this as it spends most of its time beyond the outermost planets. It therefore travels at a slower speed compared to when it is close to the sun. This reduced velocity allows it to sweep the same area compared to when it is closer to the sun.

Show Worked Solution

→ Kepler’s Second Law states that a line between the comet and the sun will sweep an equal area in equal time.

→ When the comet is closer to the sun, it must travel with a greater velocity in its orbit compared to when it is far away to sweep the same area in the same time (i.e. such that `A_1=A_2`.

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:

`t`	`=(s)/(v)`
	`=(2xx8000)/(3.00 xx10^(8))`
	`=5.33 xx10^(-5) text{s}`

→ 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.

`omega`	`=(Delta theta)/(t)`
	`=((2pi)/(100))/(5.33 xx10^(-5))`
	`=1180 text{rad s}^(-1)`

→ 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

→ Calculating the time taken:

`t`	`=(s)/(v)`
	`=(2xx8000)/(3.00 xx10^(8))`
	`=5.33 xx10^(-5) text{s}`

`omega`	`=(Delta theta)/(t)`
	`=((2pi)/(100))/(5.33 xx10^(-5))`
	`=1180 text{rad s}^(-1)`

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.

Explain how any one of the dark bands forms on the screen. (3 marks)

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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.
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 `(lambda)/(2)` 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 `(lambda)/(2)` further than light from the other slit.

b. `0.0059 text{m}`

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.

→ For example, the dark bands either side of the central maximum form where light from one slit travels `(lambda)/(2)` further than light from the other slit.

b. `d sin theta=m lambda`

	`0.15 xx10^(-3) sin theta`	`=2xx590 xx10^(-9)`
	`sin theta`	`=0.007867`

→ Using trigonometry, `tan theta=(d)/(0.75)`, where `theta` is the angular separation of the second order bright band from the central maximum and `d` is the distance between the centre of the central maximum and the centre of the second order bright band.

→ As the angle, `theta` is small, the approximation `sin theta=tan theta` is valid:

`tan theta`	`=(d)/(0.75)`
`d`	`=0.007867 xx0.75`
	`=0.0059\ text{m}`

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 `K_(max)=hf-Phi`.

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.

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.

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.

→ 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.

Technology and Special Relativity

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

→ 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 `I_0` pass through two polarising filters `P_1` and `P_2` to a detector. The filters are initially aligned so that they produce the maximum amount of light, then filter `P_2` is slowly rotated through 180° as shown.

On the axes provided sketch a graph showing how the intensity of light at the detector, `I`, changes as `P_2` rotates from zero to 180°.

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`P_2` is now rotated to a position such that no light reaches the detector. Without moving `P_1` or `P_2`, a third polarising filter is inserted between `P_1` and `P_2` and rotated at an angle of 30° from `P_1`.
Explain, with the aid of calculations, why the light intensity at the detector is no longer zero.

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b. → `P_(2)` must be rotated 90° from `P_(1)` (no light reaches detector)

→ Calculate the light passing through the middle polariser:

`I`	`=I_(max)cos^2theta`
`I_(m)`	`=I_(1)cos^(2)30°`
	`=0.75 xxI_(1)`

→ `P_(2)` is a 60° rotation from the middle polariser:

`I`	`=I_(max)cos^2theta`
`I_(2)`	`=0.75xxI_(1)cos^(2)60°`
	`=0.1875xxI_(1)`

→ Light intensity at detector is therefore no longer zero.

Show Worked Solution

b. → `P_(2)` must be rotated 90° from `P_(1)` (no light reaches detector)

→ Calculate the light passing through the middle polariser:

`I`	`=I_(max)cos^2theta`
`I_(m)`	`=I_(1)cos^(2)30°`
	`=0.75 xxI_(1)`

→ `P_(2)` is a 60° rotation from the middle polariser:

`I`	`=I_(max)cos^2theta`
`I_(2)`	`=0.75xxI_(1)cos^(2)60°`
	`=0.1875xxI_(1)`

→ Light intensity at detector is therefore no longer zero.

PHYSICS, M6 EQ-Bank 25

The diagram shows a DC motor with a constant current flowing to the rotor.

Sketch graphs to compare the behaviour of the force `F` on wire `AB` and the torque `tau` on the rotor, as functions of time `t`. (4 marks)

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Show Worked Solution

PHYSICS, M7 EQ-Bank 32

Applying the law of conservation of energy, explain why `K_max = h f-phi`. (3 marks)

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→ The law of conservation of energy states that energy cannot be created or destroyed, only transferred or transformed into different forms.

→ The initial energy of a photon is equal to `hf`.

→ If this photon strikes a metal surface, a photoelectron may be released. Some energy is required to remove the electron from the metal surface which is equal to the work function (`phi`) of the metal.

→ The electron will also possess kinetic energy, `K_(max)`.

→ Applying the law of conservation of energy, the energy before equals the energy after, or `hf=K_(max)+phi`.

→ Rearranging this gives `K_(max)=hf-phi`.

Show Worked Solution

→ The law of conservation of energy states that energy cannot be created or destroyed, only transferred or transformed into different forms.

→ The initial energy of a photon is equal to `hf`.

→ The electron will also possess kinetic energy, `K_(max)`.

→ Applying the law of conservation of energy, the energy before equals the energy after, or `hf=K_(max)+phi`.

→ Rearranging this gives `K_(max)=hf-phi`.

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 `1.6 xx10^(-19) text{C}`. 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.

→ Millikan found that the value of the charge of an oil drop was always an integer multiple of `1.6 xx10^(-19) text{C}`. This was then said to be the fundamental charge on an electron.

PHYSICS, M7 EQ-Bank 21

A student was given a smartphone with a light sensor and an angle sensor, and a computer screen which emitted polarised light. A polariser was fixed over the top of the light sensor in the smartphone.

The student wants to use this equipment to investigate Malus's Law of polarised light. Describe a procedure that is suitable for carrying out this investigation. (3 marks)

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An experiment was conducted to demonstrate Malus's Law for plane polarisation of light. The results are shown in the graph.

Based on the graph shown, how effective was the experiment in meeting its aim? (3 marks)

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a. Procedure steps:

→ Secure the smartphone so it can rotate at a measured and fixed distance from the computer screen.

→ Set the computer to emit a constant intensity of light.

→ Find the position of the smartphone at which light of maximum intensity is obtained. Set this position to be 0° and record the intensity.

→ Rotate the smartphone through 360° and record the intensity at regular intervals of 10°.

→ Plot obtained results on a graph.

b. → Malus’ Law gives the equation `I=I_(max)cos^2theta`

→ The graph shape that is produced by the experimental data is consistent with the equation.

→ The experiment uses a sufficient range of data to show a relationship and uses a sufficient number of data points.

→ However, the equipment used must have been incorrectly calibrated, or the measurements were not taken correctly. Measurements appear to be offset by 30° as the maximum intensity of light should occur at 0° rather than 30°.

Show Worked Solution

a. Procedure steps:

→ Secure the smartphone so it can rotate at a measured and fixed distance from the computer screen.

→ Set the computer to emit a constant intensity of light.

→ Find the position of the smartphone at which light of maximum intensity is obtained. Set this position to be 0° and record the intensity.

→ Rotate the smartphone through 360° and record the intensity at regular intervals of 10°.

→ Plot obtained results on a graph.

b. → Malus’ Law gives the equation `I=I_(max)cos^2theta`

→ The graph shape that is produced by the experimental data is consistent with the equation.

→ The experiment uses a sufficient range of data to show a relationship and uses a sufficient number of data points.

PHYSICS, M7 EQ-Bank 23

What is the magnitude of the momentum `(\text{in kg m s}^(-1))` of an electron travelling at 0.75`c`? (2 marks)

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`3.10 xx10^(-22) text{kg m s}^(-1)`

Show Worked Solution

`p_(v)`	`=(m_(0)v)/(sqrt((1-(v^(2))/(c^(2)))))`
	`=((9.109 xx10^(-31))(0.75 xx3 xx10^8))/(sqrt((1-((0.75c)^(2))/(c^(2)))))`
	`=((9.109 xx10^(-31))(0.75 xx3 xx10^8))/(sqrt((1-0.75^(2)))`
	`= 3.10 xx10^(-22) text{kg m s}^(-1)`

COMMENT: Note the simplification of the denominator.

PHYSICS, M7 EQ-Bank 14 MC

The graph shows the electromagnetic radiation emitted from a black body cavity.

What is the best estimate of the temperature of this black body?

`5.9 × 10^(3) \ text{K}`
`7.2 × 10^(3) \ text{K}`
`1.7 × 10^(5) \ text{K}`
`5.9 × 10^(6) \ text{K}`

Show Answers Only

`A`

Show Worked Solution

→ The peak wavelength of this black body is approximately 490 nm.

`lambda_(max)`	`=(b)/(T)`
`T`	`=(2.898 xx10^(-3))/(490 xx10^(-9))`
	`=5.91 xx10^3 text{K}`

`=>A`

PHYSICS, M7 EQ-Bank 27

Monochromatic light of wavelength `lambda` strikes a double slit and produces bright and dark fringes on a screen. Light from slit `S_1` travels along path `P_1` and light from slit `S_2` travels along `P_2` to produce the dark fringe shown.

Calculate the difference in length between `P_1` and `P_2` ? (2 marks)

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`(3lambda)/2`

Show Worked Solution

→ Dark fringes occur when the path difference is an odd multiple of `(lambda)/2`. i.e at `(lambda)/2`, `(3lambda)/2`, `(5lambda)/2` etc.

→ This is the second dark fringe from the central maximum.

→ So, the difference in length between `P_1` and `P_2` is `(3lambda)/2`

PHYSICS, M7 EQ-Bank 5 MC

Anna and Bo carried out independent experiments to investigate Malus's Law. They graphed the results of their experiments. The graphs are shown below.

Based on the two graphs, which of the following is correct?

Anna has taken more measurements but Bo has used a better data range.
Bo's graph is more precise as the angles in Anna's graph are too small.
Anna's graph is more valid as Bo's graph shows a straight line relationship.
Anna's measurements are more reliable than Bo's as a line of best fit cannot be drawn for Bo's graph.

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`A`

Show Worked Solution

By Elimination:

→ Anna’s graph uses too narrow an angle range whereas Bo’s graph does not include enough measurements to show a valid relationship (possibly A).

→ It is not possible to determine precision or validity without knowing the equipment used or experimental design of both experiments (eliminate B and C).

→ Reliability refers to the consistency of results and has no impact on the ability to draw a line of best fit (eliminate D).

`=>A`

PHYSICS, M7 EQ-Bank 6 MC

Which statement describes how an electromagnetic wave is propagated?

An oscillating electric field causes a constant magnetic field parallel to the electric field.
An oscillating magnetic field causes an oscillating electric field parallel to the magnetic field.
An oscillating electric field causes an oscillating magnetic field perpendicular to the electric field.
An oscillating magnetic field causes a constant electric field perpendicular to the magnetic field.

Show Answers Only

`C`

Show Worked Solution

→ As per Maxwell’s predictions, an oscillating electric field induces an oscillating magnetic field perpendicular to the electric field.

→ This oscillating magnetic field induces an oscillating electric field perpendicular to it. This process continues and an electromagnetic wave is propagated.

`=>C`

PHYSICS, M7 EQ-Bank 22

Diagram 1 shows the absorption spectrum of light produced by an incandescent filament, after it has been shone through a quantity of hydrogen gas. Diagrams 2 and 3 show the spectra obtained from two stars: Croesus and Dromus. The dark lines in the diagrams are absorption bands.

Explain what the spectrum of each star, Croesus and Dromus, tells us about the motion of that star. (3 marks)

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→ The spectra from Croesus is red shifted which indicates it is moving away from Earth.

→ The thin spectra lines from Croesus also indicate that it is not rotating.

→ The centre of spectral lines from Dromus are blue shifted which indicates it is moving towards Earth.

→ Dromus’ spectral lines are wide and blurry (simultaneously red and blue shifting), which indicates that it is rotating.

Show Worked Solution

→ The spectra from Croesus is red shifted which indicates it is moving away from Earth.

→ The thin spectra lines from Croesus also indicate that it is not rotating.

→ The centre of spectral lines from Dromus are blue shifted which indicates it is moving towards Earth.

→ Dromus’ spectral lines are wide and blurry (simultaneously red and blue shifting), which indicates that it is rotating.

PHYSICS, M7 EQ-Bank 4 MC

Betelgeuse is a red giant star in our galaxy. The following are facts about this star:

Fact 1: Its distance from us is 640 light years.

Fact 2: It has a surface temperature of 3500 K.

Fact 3: Its atmosphere contains titanium dioxide.

Fact 4: It is moving away from us at a speed of `21.9 \ text{km s}^(-1)`.

Which of the given facts about Betelgeuse CANNOT be determined from its spectrum?

Fact 1
Fact 2
Fact 3
Fact 4

Show Answers Only

`A`

Show Worked Solution

By elimination:

→ The star’s surface temperature can be found by determining the peak wavelength from its spectrum then using Wein’s Law. (Eliminate B)

→ The atmospheric composition of the star can be found by matching its spectra to the known spectra of gases. (Eliminate C)

→ Comparing the spectral lines of the star with those of corresponding compounds on Earth, the degree of red shift can be calculated. Applying the doppler effect, the recessional velocity of the star can be determined. (Eliminate D)

`=>A`

PHYSICS, M6 EQ-Bank 27

The diagram shows two rings `A` and `B`, connected to a balancing arm which swings freely on a pivot. Ring `A` 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 `B`, 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 `B`, a like pole is produced which causes the magnet and the ring to repel. When the magnet is pulled out of ring `B`, 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 `A`, 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 `B`, 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).

→ 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 `A`, 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 22

An 'electron gun' like that used by JJ Thomson is shown.

Electrons leave the cathode and are accelerated towards the anode.

Show that the acceleration of the electrons as they just leave the cathode is `4 × 10^(16) \ text{m s}^(-2)`. (2 marks)

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Calculate the velocity of an electron as it reaches the anode. (2 marks)

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`4 xx10^(16) text{m s}^(-2)`
`4 xx10^7 text{m s}^(-1)`

Show Worked Solution

a. The force on the electrons is given by `F=qE`.

The electric field strength is given by `E=(V)/(d)`:

`F`	`=(Vq)/(d)`
`a`	`=(F)/(m)`
	`=(Vq)/(dm)`

Calculate the acceleration:

`a`	`=(Vq)/(dm)`
	`=(5000 xx1.602 xx10^(-19))/(0.02 xx9.109 xx10^(-31))`
	`=4 xx10^(16) text{m s}^(-2)`

b. Using kinematic equations:

`v^(2)`	`=u^(2)+2as`
`v`	`=sqrt(2as),\ \ \ (u=0)`
	`=sqrt(2xx4xx10^(16)xx0.02)`
	`=4 xx10^7 text{m s}^(-1)`

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 `5.0 × 10^(-3) \ text {m}` 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)

Show Answers Only

→ 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:

`E=(V)/(d)=(100)/(0.005)=20\ 000\ text{V m}^(-1)`

`F=Eq=20\ 000 xx1.602 xx10^(-19)=3.2 xx10^(-15) text{N}`

`a=(F)/(m)=(3.2 xx10^(-15))/(9.109 xx10^(-31))=3.5 xx10^(15) text{m s}^(-2) text{down page.}`

→ 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:

`E=(V)/(d)=(100)/(0.005)=20\ 000\ text{V m}^(-1)`

`F=Eq=20\ 000 xx1.602 xx10^(-19)=3.2 xx10^(-15) text{N}`

`a=(F)/(m)=(3.2 xx10^(-15))/(9.109 xx10^(-31))=3.5 xx10^(15) text{m s}^(-2) text{down page.}`

→ 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 23

An electric motor is connected to a power supply of constant voltage. The motor runs at different speeds by adjusting a brake.

On the graph below, show the relationship between the current through the motor and its speed? (2 marks)

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Show Worked Solution

Points to note on graph shape:

→ As the speed of the motor increases, the rate of change of magnetic flux through its coil increases.

→ This increases the induced emf in the motor (Faraday’s Law).

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

→ The back emf opposes the current supplied to the motor and decreases the net current through the motor.

PHYSICS, M6 EQ-Bank 3 MC

A rectangular loop of wire passes between two magnets as shown and is free to rotate about `X Y`. The loop has a current flowing through it.

Without changing the current, which of the following would result in the greatest increase in torque?

Increase the thickness of the wire in the loop.
Decrease the thickness of the wire in the loop.
Extend the length of the loop in the `X Y` direction.
Extend the width of the loop towards the magnets.

Show Answers Only

`D`

Show Worked Solution

→ Consider `tau=rFsin theta`.

→ Extending the width of the loop towards the magnets increases the distance between the point of application of force to the coil and the axis of rotation (i.e. `r`).

→ By increasing `r`, the torque will increase.

`=>D`

PHYSICS, M6 EQ-Bank 6 MC

The diagram shows an ideal transformer.

When the switch is closed, the pointer on the galvanometer deflects.

How could the size of this deflection be increased?

Decrease the number of primary coils.
Decrease the number of secondary coils.
Replace the iron core with a copper core.
Place a resistor in series with the galvanometer.

Show Answers Only

`B`

Show Worked Solution

→ In order to increase the deflection of the galvanometer, the current through the secondary coil must increase.

→ The input power (`V_(p)I_(p)`) is fixed.

→ Since `V_(p)I_(p)=V_(s)I_(s)`, the voltage of the secondary coil (`V_(s)`) must decrease if `I_(s)` increases.

→ `(V_(p))/(V_(s))`	`=(N_(p))/(N_(s))`
`(V_(s))/(V_(p))`	`=(N_(s))/(N_(p))`
`V_(s)`	`=(V_(p)N_(s))/(N_(p))`

→ Reducing the number of secondary coils (`N_(s)`) will therefore decrease the secondary voltage (`V_(s)`) and consequently increase the current through the secondary coil.

`=>B`

PHYSICS, M6 EQ-Bank 5 MC

The total flux in the core of an electrical machine is 30 mWb and its flux density is 0.6 T.

What is the cross-sectional area of the core?

`0.018 \ text{m}^(2)`
`0.05 \ text{m}^(2)`
`20 \ text{m}^(2)`
`50 \ text{m}^(2)`

Show Answers Only

`B`

Show Worked Solution

`Phi`	`=BA`
`A`	`=(Phi)/(B)`
	`=(30 xx10^(-3))/(0.6)`
	`=50 xx10^(-3)\ text{m}^(2)`
	`=0.05\ text{m}^(2)`

`=>B`

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.

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

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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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Show Worked Solution

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)

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→ 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 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

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

→ 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.

→ 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

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

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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.

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, M8 EQ-Bank 12

Polio is a disease that can both kill people and leave others scarred and disabled for life.

The poster below is part of an educational program that targets polio.

What are the benefits of such campaigns to the broader society. (3 marks)

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→ Educational programs can take the form of school lessons, television advertisements or posters like the one above. They aim to reduce incidence of a disease by targeting and promoting the uptake of polio vaccines.

→ The poster highlights vulnerable babies, encouraging their parents to vaccinate them against the disease. It emphasises the zero cost involved to remove hurdles for positive action.

→ The benefit of the poster is that it is cheap to produce and easy to target a large group of people if placed in high foot-traffic areas.

→ Reducing polio incidence will benefit families of potential sufferers that do not become ill by allowing them to contribute to the economy and not require ongoing medical assistance.

→ Reducing polio incidence will also benefit the government through reduced health costs.

Show Worked Solution

→ The poster highlights vulnerable babies, encouraging their parents to vaccinate them against the disease. It emphasises the zero cost involved to remove hurdles for positive action.

→ The benefit of the poster is that it is cheap to produce and easy to target a large group of people if placed in high foot-traffic areas.

→ Reducing polio incidence will benefit families of potential sufferers that do not become ill by allowing them to contribute to the economy and not require ongoing medical assistance.

→ Reducing polio incidence will also benefit the government through reduced health costs.

BIOLOGY, M7 EQ-Bank 23

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

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On the diagram, continue the graph to show the immune response upon secondary exposure to the same pathogen. (2 marks)
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)

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→ 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 7 MC

A patient has experienced an injury that affected the retina of her eye. She asked her doctor if LASIK eye surgery could help restore her sight. The doctor said that LASIK eye surgery would have no impact on her condition.

Why did the doctor give this advice?

LASIK eye surgery is a replacement of the lens in the eye and cannot repair a retina
LASIK eye surgery will only temporarily repair her retina and her vision problems would return
LASIK eye surgery is only used for people who have a damaged cornea and cannot repair a retina
LASIK eye surgery is only used to change the shape of the cornea to enhance vision and cannot repair a retina

Show Answers Only

`D`

Show Worked Solution

`=>D`

BIOLOGY, M8 EQ-Bank 6 MC

A patient has been diagnosed with severe hearing loss in the inner ear.

Which type of hearing technology can be used to help restore the patient's hearing?

Hearing aid
Cochlear implant
Artificial ear drum
Bone conduction implant

Show Answers Only

`B`

Show Worked Solution

→ The cochlear implant can bypass the cochlear and send electric impulses directly to the auditory nerve.

`=>B`

BIOLOGY, M8 EQ-Bank 7 MC

For many years, some cigarette companies have denied that there were increased risks of lung cancer as a result of cigarette smoking.

How can an epidemiological study into lung cancer be useful in this situation?

It can show that the chemicals in cigarette smoke cause cancer.
It can demonstrate that second-hand smoke has no impact on lung cancer rates.
It can provide evidence that people who smoke are more likely to develop lung cancer.
It can show that the cigarette companies have been lying about the cause of lung cancer.

Show Answers Only

`C`

Show Worked Solution

→ An epidemiological study will gather statistics surrounding smoking and lung cancer.

→ The data will then be analysed and conclusions drawn to show that a link between them exists.

`=>C`

BIOLOGY, M8 EQ-Bank 1 MC

The diagram shows a homeostatic mechanism in a mammal.

What does X represent in the diagram?

The brain
The heart
A thermoreceptor in the skin
A pressure receptor in a blood vessel

Show Answers Only

`A`

Show Worked Solution

→ The brain (more specifically the hypothalamus) is the organ responsible for the detection of a change in body temperature.

`=>A`

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.

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

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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.

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

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.

→ 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.

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

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Consider another plant with a different response to a specific pathogen.
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 11

Complete the following table to show the distinguishing characteristic of each pathogen and a disease caused by each. (3 marks)

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BIOLOGY, M7 EQ-Bank 7 MC

How do vaccinations prevent disease?

They increase the inflammation process.
They enable the infected cells to seal off the pathogen.
They increase the number of antibodies against the pathogen.
They decrease the number of antigens that trigger the immune response.

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`C`

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→ Vaccines stimulate the immune response, encouraging the division of B cells and therefore producing more antibodies.

`=>C`

BIOLOGY, M7 EQ-Bank 9 MC

The map shown was drawn by Dr John Snow during the 1854 London cholera epidemic.

The dots indicate people who died from cholera and the crosses indicate the location of water pumps.

Which of the following is the most likely hypothesis for which Dr John Snow was gathering evidence?

That the outbreak of cholera was caused by people living near each other
That the people who died from cholera drank water from the Broad Street pump
That cholera was caused by an infectious agent that can be found and transmitted in water
That the cause of a disease can be determined by mapping the location of infected patients

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`C`

Show Worked Solution

→ The outbreak was mainly centralised around the Broad Street water pump.

→ Snow therefore hypothesised that cholera may be caused by a pathogen in the water, hence mainly affecting people that used and drank the water from the infected pump.

`=>C`

BIOLOGY, M7 EQ-Bank 4 MC

A student was vaccinated for rubella when they were 13 . Three years later, they were exposed to the active rubella virus.

Which graph best represents the student's production of antibodies over time?

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`D`

Show Worked Solution

→ When exposed to a pathogen for a second time, memory B cells can produce more antibodies in a shorter amount of time, providing the basis of immunity.

→ The initial exposure represents the vaccine, and second represents the exposure 3 years later.

`=>D`

BIOLOGY, M7 EQ-Bank 5 MC

When a foreign body breaches the first line of defence the mast cells produce histamines in response.

What is the role of histamines in the defence of the body?

To attack the invading pathogen
To activate B and T lymphocytes of the specific immune response
To activate the inflammation response and increase blood flow to the affected area
To retain information on a pathogen's antigen so that the immune system can respond quickly to any subsequent infection

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`=>C`

Show Worked Solution

→ Histamines are the chemicals released by a mast cell in response to tissue damage.

→ They increase blood flow by dilating blood vessels and increasing permeability of capillaries.

`=>C`

BIOLOGY, M7 EQ-Bank 3 MC

The diagram shows a model of disease transmission.

A pathogen was identified as being unadapted to dry conditions and as having the gastrointestinal tract as the 'route of entry' and the 'route of exit'.

Using this information, what is the most likely mode of transmission?

Skin to skin contact
Coughing or sneezing
Contaminated water supplies
Transmission of infected blood products

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`C`

Show Worked Solution

→ The pathogen is most likely to have adapted to be transmitted through conditions which are moist and can be more effective at entering through the gastrointestinal tract.

`=>C`

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)

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→ 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.

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)

BIOLOGY, M6 EQ-Bank 25

A woman recently conceived a baby guaranteed to be free from hereditary breast cancer. Doctors screened for an embryo that was free from a gene that can cause breast cancer.

The screening was performed due to the long history of this form of cancer in the family and the fact that any daughter born with the gene would have a 50-80% chance of developing breast cancer.

Explain the possible impact of this reproductive technology on the genetic composition of the population. (2 marks)

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Discuss the use of this genetic technology in the treatment of medical conditions. (3 marks)

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a. Impact on genetic composition of population

→ This use of this technology will reduce the frequency of the gene that is likely to produce breast cancer in the population.

→ In the longer term, proceeding generations of offspring will also not have the gene, resulting in fewer people suffering from breast cancer.

b. Genetic technology treatments

→ Genetic technologies are a great tool against preventing life-threatening genetic diseases, as well as being used as a treatment for other serious conditions such as cancer.

→ These genetic technologies can treat and potentially cure many genetic conditions, improving life expectancy, quality of life and reducing incidence.

→ Although many of the treatments are expensive upfront, they can reduce future ongoing medical costs.

→ Genetic technologies are relatively new and experimental on humans. This poses a risk of unknown side-effects later on in the patient’s life.

→ Scepticism surrounding many genetic technologies has brought up several ethical considerations including the unnatural attributes of the procedures and concerns of personal liberty and choice.

Show Worked Solution

a. Impact on genetic composition of population

→ This use of this technology will reduce the frequency of the gene that is likely to produce breast cancer in the population.

→ In the longer term, proceeding generations of offspring will also not have the gene, resulting in fewer people suffering from breast cancer.

b. Genetic technology treatments

→ Genetic technologies are a great tool against preventing life-threatening genetic diseases, as well as being used as a treatment for other serious conditions such as cancer.

→ These genetic technologies can treat and potentially cure many genetic conditions, improving life expectancy, quality of life and reducing incidence.

→ Although many of the treatments are expensive upfront, they can reduce future ongoing medical costs.

→ Genetic technologies are relatively new and experimental on humans. This poses a risk of unknown side-effects later on in the patient’s life.

→ Scepticism surrounding many genetic technologies has brought up several ethical considerations including the unnatural attributes of the procedures and concerns of personal liberty and choice.

BIOLOGY, M6 EQ-Bank 24

The yeast Saccharomyces cerevisiae cannot naturally ferment the sugar xylose. Low value biomass, such as straw and wood fibres, contains up to 20% xylose. S. cerevisiae was modified to enable it to produce ethanol from xylose. Information on the two species involved in making the modified S. cerevisiae is shown in the table.

Explain why biotechnology was needed to modify S. cerevisiae. (2 marks)

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Two strains of genetically modified S.cerevisiae were produced. The two strains were compared under the same conditions. The results are shown.
Justify which of these two strains would be better to use to produce commercial quantities of ethanol using low value biomass. In your answer, refer to information from the graph. (3 marks)

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“

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a. → The table shows that the organisms are from different genera.

→ Genetic material generally can’t be transferred between organisms of different genera.

→ Genetic technology was therefore needed to cut out and insert the required genes from B. cenocepacia to create the genetically modified S. cerevisiae as this process could not occur naturally.

b. Best strain for commercial production

→ Strain B would be the more effective strain to use as it consistently produces double the ethanol of Strain A.

→ This can be seen even at the plateau of both strains, where at 30 hrs Strain B produced 20g/L of ethanol whilst Strain A only produced 10g/L.

→ These data in the graph indicates that Strain B will be more efficient at producing commercial ethanol in any given time-frame.

BIOLOGY, M6 EQ-Bank 23

'The application of reproductive technologies in plant and animal breeding limits genetic diversity.'

To what extent is this statement correct? (6 marks)

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→ Reproductive technologies intervene in natural reproduction and allow offspring with a higher chance of favourable characteristics.

→ Examples includes artificial insemination in animals, artificial pollination in plants, as well as IVF and selective breeding.

→ In all the above processes, male gametes with favourable traits are isolated and delivered into the reproductive system of selected females using technology.

→ The processes greatly increase the number of offspring generated by one parent, thus decreasing the genetic variability amongst them.

→ The repetitive use of the same male gametes across generations will eventually reduce the genetic diversity amongst the entire species, as the favoured characteristics will increase and less favourable alleles will be lost.

→ It should be noted that these technologies can also increase genetic diversity of a species through the collection and transportation of male gametes. This process helps genes to overcome geographical and time barriers, allowing for genotypes and hybrids which would otherwise not be possible by natural processes.

Show Worked Solution

→ Reproductive technologies intervene in natural reproduction and allow offspring with a higher chance of favourable characteristics.

→ Examples includes artificial insemination in animals, artificial pollination in plants, as well as IVF and selective breeding.

→ In all the above processes, male gametes with favourable traits are isolated and delivered into the reproductive system of selected females using technology.

→ The processes greatly increase the number of offspring generated by one parent, thus decreasing the genetic variability amongst them.

BIOLOGY, M6 EQ-Bank 27

Compare the processes and effects of point mutations and chromosomal mutations. Include examples in your answer. (8 marks)

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→ Mutations are the change in an individuals genome. Their effects are dependent on

- the location of the mutation. Mutations in coding DNA can affect proteins and have adverse physiological affects while mutations in non-coding DNA often have no affect, however may slightly impact cell activity.
- the type of cell it occurs in. Somatic cell mutations only affect the individual, while germ-line mutations occur in germ-line cells which go on to produce gametes, thus potentially affecting offspring.

→ Point mutations are mutations which change only a single base in a DNA sequence. They can be a single base substitution or the deletion or insertion of an extra base. Deletion or insertions are referred to as ‘frameshift’ mutations as they impact all codons proceeding it.

→ Sometimes substitution mutations are neutral and occur in non-coding DNA, code for an amino acid of very similar chemical composition or even code for the same amino acid. When a mutation does code for a dissimilar amino acid or a stop codon causing premature termination, this interferes with the shape of the polypeptide chain and the protein it becomes part of, rendering it faulty or completely unusable (e.g. sickle cell anaemia, NRAS mutation). The nature of frameshift mutations affecting every codon proceeding it are often more severe, as there is more potential to cause a faulty polypeptide chain (e.g. cystic fibrosis).

→ Chromosomal mutations can refer to the removal, insertion, translocation or inversion of large sections of a chromosome. These often affect genes by either splitting them up or relocating them to a new chromosome.

→ Chromosomal mutations can also refer to the relocation of entire chromosomes during meiosis of germ-line cells where they fail to separate. This is called non-disjunction and results in aneuploidy of two gametes, one which will result in having an extra chromosome (trisomy) and one will lack a chromosome (monosomy). Conditions such as Down’s syndrome (trisomy-21) or Turner syndrome (XO) are aneuploidy conditions but most often cause a miscarriage or unsuccessful zygote development.

→ Polyploidy is the complete development of an entirely new set of chromosomes. This is known to occur in strawberries which have 8 sets of chromosomes but cannot occur in humans.

→ While both can cause radical health defects, it is important to note that these mutations are the basis of evolution. Beneficial mutations are more likely to be carried through a species through generations according to natural selection.

Show Worked Solution

→ Mutations are the change in an individuals genome. Their effects are dependent on

- the location of the mutation. Mutations in coding DNA can affect proteins and have adverse physiological affects while mutations in non-coding DNA often have no affect, however may slightly impact cell activity.
- the type of cell it occurs in. Somatic cell mutations only affect the individual, while germ-line mutations occur in germ-line cells which go on to produce gametes, thus potentially affecting offspring.

→ Polyploidy is the complete development of an entirely new set of chromosomes. This is known to occur in strawberries which have 8 sets of chromosomes but cannot occur in humans.

BIOLOGY, M6 EQ-Bank 22

The flow chart illustrates the effect of a point mutation on an organism.

Outline the series of events from stages 2 to 4 that resulted in the faulty protein. (3 marks)

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Describe how a type of mutagen may have caused the changes observed in stage 2. (2 marks)

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Given the information in the chart shown, describe the effect caused by the mutation in stage 4 and the effect this would have on the organism. (3 marks)

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a. → Stage 2: A mutagen causes a point mutation in the DNA.

→ Stage 3: The mutated DNA transcription into mRNA.

→ Stage 4: The translation of the mRNA into a polypeptide chain which is not functional due to the mutation.

b. Possible causes of Stage 2 observed changes:

→ High frequency photons including gamma and X-rays are regarded as ionising radiation due to their extremely high energy.

→ They can interfere with DNA undergoing replication causing misalignment and a possible substation change.

→ High energy gamma radiation may also cause mutation when directly striking a nitrogenous base.

Answers could also include

→ Chemical mutagens (alkylating, delaminating and structurally similar chemicals).

→ Carcinogenic metals/elements (radon, cadmium)

c. Effects of Stage 4 mutation

→ When considering the above sequence as 2 codons, the chart shows us that this mutation results in an arginine amino becoming a stop codon.

→ This mutation will hence cause a premature termination of the polypeptide and result in a non-functional polypeptide/protein it forms a part of.

→ Due to the severity of this type of mutation it usually results in extensive health complications for affected individuals.

Show Worked Solution

a. → Stage 2: A mutagen causes a point mutation in the DNA.

→ Stage 3: The mutated DNA transcription into mRNA.

→ Stage 4: The translation of the mRNA into a polypeptide chain which is not functional due to the mutation.

b. Possible causes of Stage 2 observed changes:

→ High frequency photons including gamma and X-rays are regarded as ionising radiation due to their extremely high energy.

→ They can interfere with DNA undergoing replication causing misalignment and a possible substation change.

→ High energy gamma radiation may also cause mutation when directly striking a nitrogenous base.

Answers could also include

→ Chemical mutagens (alkylating, delaminating and structurally similar chemicals).

→ Carcinogenic metals/elements (radon, cadmium)

c. Effects of Stage 4 mutation

→ When considering the above sequence as 2 codons, the chart shows us that this mutation results in an arginine amino becoming a stop codon.

→ This mutation will hence cause a premature termination of the polypeptide and result in a non-functional polypeptide/protein it forms a part of.

→ Due to the severity of this type of mutation it usually results in extensive health complications for affected individuals.

BIOLOGY, M6 EQ-Bank 6 MC

Glofish are a genetically-modified organism in which the gene that causes fluorescence in jellyfish has been inserted into a tropical fish species, typically Zebra fish. These fish are sold commercially for home aquariums. Some sectors of the community have said that humans do not have the right to make genetically-modified organisms for this purpose.

What is the main nature of their concern?

The limited application the Glofish have in society
The risks to the biodiversity of the Zebra fish species
The ethics of manipulating an organism's genes for commercial gain
That the Glofish may interbreed with other species causing serious mutations in the future

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`C`

Show Worked Solution

→ Ethical and moral dilemmas are always primary concerns when evaluating the use of transgenic organisms.

`=>C`

BIOLOGY, M6 EQ-Bank 7 MC

The following events occur after DNA is subjected to radiation. The events are listed in no specific order.

1: Mutation
2: Change in cell activity
3: Change in protein structure
4: Change in polypeptide sequence

What is the correct sequence of steps?

1, 2, 3, 4
4, 2, 1, 3
4, 3, 2, 1
1, 4, 3, 2

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`D`

Show Worked Solution

`=>D`

BIOLOGY, M6 EQ-Bank 12 MC

The bread mould, Neurospora crassa, normally produces its own amino acids from raw materials through a system of enzymes.

If a mutation occurred in gene B, the bread mould would still produce arginine if supplied with

citrulline.
ornithine.
enzyme C.
raw materials.

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`A`

Show Worked Solution

→ Enzyme B only aids the production of ornithine to citrulline.

→ If the mould was directly supplied with citrulline, this would bypass enzyme B and it would be able to still produce arginine.

`=>A`

BIOLOGY, M5 EQ-Bank 25

A student plans to investigate whether the development of insulin has affected the prevalence of Type 1 diabetes in the human population and subsequently influenced human evolution. She has access to data on Australians with diabetes extending back to 1973 .

Propose a suitable hypothesis for this investigation. (2 marks)

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Identify TWO variables that need to be controlled for this investigation and explain their importance. (4 marks)

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a. Two possible hypotheses include:

→ The development of insulin has led to a decrease in the prevalence of Type 1 diabetics.

→ The use of insulin by people who have Type 1 diabetes has increased the prevalence of Type 1 diabetes in the human population.

b. Variables to be controlled

→ The growth of the Australian population in general will statistically lead to a bias in more recent years, where higher a number of individuals with Type 1 diabetes is to be expected.

→ To control this variable, the investigation should be measured using a fixed rate such as ‘per 100 000’.

→ The time of diagnosis and length of time an individual has used insulin can impact the trend of the obtained data.

→ This data must be obtained and accounted for to be able to increase the validity of findings.

Show Worked Solution

a. Two possible hypotheses include:

→ The development of insulin has led to a decrease in the prevalence of Type 1 diabetics.

→ The use of insulin by people who have Type 1 diabetes has increased the prevalence of Type 1 diabetes in the human population.

b. Variables to be controlled and their importance

→ The growth of the Australian population in general will statistically lead to a bias in more recent years, where higher a number of individuals with Type 1 diabetes is to be expected.

→ To control this variable, the investigation should be measured using a fixed rate such as ‘per 100 000’.

→ The time of diagnosis and length of time an individual has used insulin can impact the trend of the obtained data.

→ This data must be obtained and accounted for to be able to increase the validity of findings.

BIOLOGY, M5 EQ-Bank 24

A non-infectious disease was observed in a mother and her four sons who live with her. She has no daughters. The father of these children does not have the disease and does not live with them. The woman's parents and her two sisters who live overseas do not have the disease.

A geneticist suspects that the disease is inherited.

Draw the family pedigree for this disease. (3 marks)
From the evidence, what indicates that the disease could be the result of a recessive allele and not be sex-linked? (2 marks)

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b. Evidence disease is recessive and not sex-linked:

→ Generation 1 parents do not have the disease but it is present in their daughter, meaning it must be recessive.

→ If the disease was sex-linked, the same daughter who has the disease must have a father with the disease, however this is not the case.

Show Worked Solution

b. Evidence disease is recessive and not sex-linked:

→ Generation 1 parents do not have the disease but it is present in their daughter, meaning it must be recessive.

→ If the disease was sex-linked, the same daughter who has the disease must have a father with the disease, however this is not the case.

BIOLOGY, M5 EQ-Bank 23

Students conducted preliminary experiments to analyse the DNA base composition of five different individuals.

The table shows the experimental data collected.

On the grid provided, plot the % Guanine against % Adenine of the individuals analysed and draw a suitable line of best fit. (3 marks)
Identify the relationship shown by the data. (1 mark)

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Explain the relationship shown by the data. (3 marks)

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b. Relationship: As the % of Adenine increases, the % of Guanine decreases.

c. Adenine/Guanine relationship

→ Because Adenine pairs with Thymine and Cytosine pairs with Guanine occurring to DNA base pairing rules, each should directly correlate to each other and account for the same % within the DNA composition of any species.

→ There are the only 4 components that make up all DNA composition. This also means that when comparing % of bases that do not pair, such as Adenine and Guanine, they must combine to form 50% of all DNA. This is evident when looking at individuals A and B, as well as C, D and E which are only off by 1-2%, most likely due to human error.

→ This always explains the relationship where an increase in Adenine (and Thymine) % will result in a decrease in Guanine (and Cytosine) % as the A and T pairs now occupy more space in the genome.

Show Worked Solution

b. Relationship: As the % of Adenine increases, the % of Guanine decreases.

c. Adenine/Guanine relationship

→ This always explains the relationship where an increase in Adenine (and Thymine) % will result in a decrease in Guanine (and Cytosine) % as the A and T pairs now occupy more space in the genome.