Band 4 – Page 10

CHEMISTRY, M2 EQ-Bank 2

A 500 mL solution contains 10.0 g of \(\ce{NaCl}\). Calculate the molarity of the \(\ce{NaCl}\) solution. Show all relevant calculations. (3 marks)

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\(\text{0.342 M}\)

Show Worked Solution

The molar mass of \(\ce{NaCl}\) is calculated as follows:

\( \ce{M(NaCl) = 22.99 + 35.45 = 58.44\ \text{g/mol}} \)

Calculate the moles of \(\ce{NaCl}\):

\( \ce{n(NaCl) =\dfrac{\text{m}}{\text{M}}= \dfrac{10.0}{58.44} = 0.171\ \text{mol}} \)

Finally, calculate the molarity of the solution using the volume in litres:

\(c = \dfrac{\text{n}}{\text{V}} = \dfrac{0.171}{0.500} = 0.342\ \text{M}\)

Therefore, the molarity of the \(\ce{NaCl}\) solution is \(0.342\ \text{M}\).

Trigonometry, 2ADV T3 EQ-Bank 6

Sketch the function \(y=\cos \left(\dfrac{x}{2}\right)\) from \(0 \leqslant x \leqslant 2 \pi\) (1 mark)
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Find the value of \(x\) when \(y=0.5\), for \(0 \leqslant x \leqslant 2 \pi\) (2 marks)
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b. \(x=\dfrac{2\pi}{3}\)

Show Worked Solution

b.	\(\quad y\)	\(=\cos \Big(\dfrac{x}{2}\Big)\)
	\(0.5\)	\(=\cos \Big(\dfrac{x}{2}\Big) \)
	\(\dfrac{x}{2}\)	\(=\cos^{-1} (0.5) \)
	\(\dfrac{x}{2}\)	\(=\dfrac{\pi}{3}, \ \dfrac{5\pi}{3}\)
	\(x\)	\(=\dfrac{2\pi}{3}\ \ (0 \leqslant x \leqslant 2 \pi) \)

CHEMISTRY, M4 EQ-Bank 10

Calculate the temperature range where the following reaction will occur spontaneously.

\(\ce{2H2O2(l) -> 2H2O(l) + O2(g)}\)

where \(\Delta H = -196.0\ \text{kJ}\) and \(\Delta S = 125.0\ \text{J K}^{-1} \) (3 marks)

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→ The reaction is spontaneous at temperatures below 1568.0 K.

Show Worked Solution

\(\text{Find}\ T\ \text{when}\ \Delta G = 0: \)

\(\Delta H-T \Delta S =0\)

\(T= \dfrac{\Delta H}{\Delta S}= \dfrac{-196.0}{0.125}=1568.0\ \text{K} \)

→ As \(T\) decreases, \(\Delta G\) decreases.

→ Reaction is spontaneous when \(\Delta G < 0\).

→ The reaction is spontaneous at temperatures below 1568.0 K.

CHEMISTRY, M4 EQ-Bank 9

At what temperature range would the following reaction occur spontaneously?

\(\ce{C(s) + O2(g) -> CO2(g)}\)

where \(\Delta H = -393.5\ \text{kJ}\) and \(\Delta S = 213.6\ \text{J K}^{-1} \) (3 marks)

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→ The reaction is spontaneous at all temperatures.

Show Worked Solution

\(\text{Find}\ T\ \text{when}\ \Delta G = 0: \)

\(\Delta H-T \Delta S\)\(=0\)

\(T= \dfrac{\Delta H}{\Delta S} = \dfrac{-393.5}{0.2136}= -1842.52\ \text{K}\)

→ Since negative temperature is not physically meaningful in this context, the reaction is spontaneous at all temperatures.

→ As \(T\) increases, \(\Delta G\) becomes more negative.

→ Reaction is spontaneous when \(\Delta G < 0\).

→ The reaction is spontaneous at all temperatures.

CHEMISTRY, M4 EQ-Bank 8

Determine the temperature range at which the following reaction will occur spontaneously.

\(\ce{N2(g) + 3H2(g) -> 2NH3(g)}\)

where \(\Delta H = -92.4\ \text{kJ}\) and \(\Delta S = -198.3\ \text{J K}^{-1} \) (3 marks)

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→ The reaction is spontaneous at temperatures below 465.94 K.

Show Worked Solution

\(\text{Find}\ T\ \text{when}\ \Delta G = 0: \)

\(\Delta H-T \Delta S\)\(=0\)

\(T= \dfrac{\Delta H}{\Delta S} = \dfrac{-92.4}{-0.1983} =465.94\ \text{K} \)

→ As \(T\) decreases, \(\Delta G\) decreases.

→ Reaction is spontaneous when \(\Delta G < 0\).

→ Therefore the reaction is spontaneous at temperatures below 465.94 K.

CHEMISTRY, M4 EQ-Bank 7

At what temperature range would the following reaction occur spontaneously?

\(\ce{2SO2(g) + O2(g) -> 2SO3(g)}\)

where \(\Delta H = -198.2\ \text{kJ}\) and \(\Delta S = -188\ \text{J K}^{-1} \) (3 marks)

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→ The reaction is spontaneous at temperatures below 1053.19 K.

Show Worked Solution

\(\text{Find}\ T\ \text{when}\ \Delta G = 0: \)

\(\Delta H-T \Delta S\)\(=0\)

\(T= \dfrac{\Delta H}{\Delta S}= \dfrac{-198.2}{-0.188} =1053.19\ \text{K} \)

→ As \(T\) decreases, \(\Delta G\) decreases.

→ Reaction is spontaneous when \(\Delta G < 0\).

→ The reaction is spontaneous at temperatures below 1053.19 K.

Trigonometry, 2ADV T2 EQ-Bank 7

Prove \(\dfrac{1+\cot \theta}{1+\tan \theta}=\cot \theta\). (3 marks)

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	\(\text{LHS}\)	\(=\dfrac{1+\dfrac{\cos \theta}{\sin \theta}}{1+\dfrac{\sin \theta}{\cos \theta}}\)
		\(=\dfrac{\dfrac{\sin \theta+\cos \theta}{\sin \theta}}{\dfrac{\cos \theta+\sin \theta}{\cos \theta}}\)
		\(=\dfrac{\sin \theta+\cos \theta}{\sin \theta} \times \dfrac{\cos \theta}{\cos \theta+\sin \theta}\)
		\(=\dfrac{\cos \theta}{\sin \theta}\)
		\(=\cot \theta\)
		\(\ =\text{ RHS}\)

Show Worked Solution

	\(\text{LHS}\)	\(=\dfrac{1+\dfrac{\cos \theta}{\sin \theta}}{1+\dfrac{\sin \theta}{\cos \theta}}\)
		\(=\dfrac{\dfrac{\sin \theta+\cos \theta}{\sin \theta}}{\dfrac{\cos \theta+\sin \theta}{\cos \theta}}\)
		\(=\dfrac{\sin \theta+\cos \theta}{\sin \theta} \times \dfrac{\cos \theta}{\cos \theta+\sin \theta}\)
		\(=\dfrac{\cos \theta}{\sin \theta}\)
		\(=\cot \theta\)
		\(\ =\text{ RHS}\)

Calculus, 2ADV C1 EQ-Bank 8

The displacement \(x\) metres from the origin at time, \(t\) seconds, of a particle travelling in a straight line is given by

\(x=t^3-9 t^2+9 t, \quad t \geqslant 0\)

Find the time(s) when the particle is at the origin. (2 marks)
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On the graph below, sketch the displacement, \(x\) metres, with respect to time \(t\). (2 marks)

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Find the acceleration of the particle when \(t=2\). (2 marks)
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a. \(\text{Particle at origin when}\ \ t=0, t=3.\)

c. \(a=-6\ \operatorname{m/s^{2}}\)

Show Worked Solution

a.	\(x\)	\(=t^3-9 t^2+9 t\)
		\(=t\left(t^2-9 t+9\right)\)
		\(=t(t-3)^2\)

\(\text{Particle at origin when}\ \ t=0, t=3.\)

c. \(x=t^3-9 t^2+9 t\)

\(v= \dfrac{dx}{dt} = 3 t^2-18 t+9\)

\(a= \dfrac{dv}{dt}=6 t-18\)

\(\text {When } t=2 \text { : }\)

\(a=6 \times 2-18=-6\ \operatorname{m/s^{2}}\)

Functions, 2ADV F2 EQ-Bank 2

A graph of the hyperbola \(y=\dfrac{1}{x+p}+q\) is shown, where \(p\) and \(q\) are constants.

Find the values of \(p\) and \(q\) and hence the graphs \(x\)-axis intercept. (2 marks)

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\(\text {Vertical asymptote at } \ x=-1 \ \ \Rightarrow p=1\)

\(\text {Horizontal asymptote at } \ y=\dfrac{3}{2} \ \ \Rightarrow q=\dfrac{3}{2}\)

\(y=\dfrac{1}{x+1}+\dfrac{3}{2}\)

\(x\text{-intercept when} \ \ y=0:\)

	\(\dfrac{1}{x+1}+\dfrac{3}{2}\)	\(=0\)
	\(\dfrac{1}{x+1}\)	\(=-\dfrac{3}{2}\)
	\(3 x+3\)	\(=-2\)
	\(3x\)	\(=-5\)
	\(x\)	\(=-\dfrac{5}{3}\)

Show Worked Solution

\(\text {Vertical asymptote at } \ x=-1 \ \ \Rightarrow p=1\)

\(\text {Horizontal asymptote at } \ y=\dfrac{3}{2} \ \ \Rightarrow q=\dfrac{3}{2}\)

\(y=\dfrac{1}{x+1}+\dfrac{3}{2}\)

\(x\text{-intercept when} \ \ y=0:\)

	\(\dfrac{1}{x+1}+\dfrac{3}{2}\)	\(=0\)
	\(\dfrac{1}{x+1}\)	\(=-\dfrac{3}{2}\)
	\(3 x+3\)	\(=-2\)
	\(3x\)	\(=-5\)
	\(x\)	\(=-\dfrac{5}{3}\)

Functions, 2ADV F1 EQ-Bank 25

Prove that the line between \((1,-1)\) and \((4,-3)\) is perpendicular to the line \(3x-2y-4=0\) (2 marks)

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\(\text {Perpendicular lines}\ \ \Rightarrow\ m_1 \times m_2 = -1\)

\(\text {Line 1 gradient:}\)

\(P_1 (1,-1), P_2(4,-3) \)

\(m_1=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{-3+1}{4-1}=-\dfrac{2}{3}\)

\(\text {Line 2 gradient:}\)

\(3x-2y-4=0\ \ \Rightarrow \ y= \dfrac{3}{2}x-2\ \ \Rightarrow m_2=\dfrac{3}{2}\)

\(m_1 \times m_2 = -\dfrac{2}{3} \times \dfrac{3}{2} = -1\)

\(\therefore\ \text{Lines are perpendicular.}\)

Show Worked Solution

\(\text {Perpendicular lines}\ \ \Rightarrow\ m_1 \times m_2 = -1\)

\(\text {Line 1 gradient:}\)

\(P_1 (1,-1), P_2(4,-3) \)

\(m_1=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{-3+1}{4-1}=-\dfrac{2}{3}\)

\(\text {Line 2 gradient:}\)

\(3x-2y-4=0\ \ \Rightarrow \ y= \dfrac{3}{2}x-2\ \ \Rightarrow m_2=\dfrac{3}{2}\)

\(m_1 \times m_2 = -\dfrac{2}{3} \times \dfrac{3}{2} = -1\)

\(\therefore\ \text{Lines are perpendicular.}\)

Trigonometry, 2ADV T1 EQ-Bank 6 MC

The triangle \(A B C\) has an exact area of \((12-\sqrt{3})\) cm\(^{2}\).

The exact value of \(x\) is

\(\dfrac{24-2\sqrt{3}}{3}\)
\(\dfrac{8\sqrt{3}-2}{3} \)
\(8\sqrt{3}-1\)
\(12\sqrt{3}+1\)

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\(B\)

Show Worked Solution

	\(12-\sqrt{3}\)	\(=\dfrac{1}{2} \cdot 6 \cdot x \cdot \sin 120^{\circ}\)
	\(12-\sqrt{3}\)	\(=3x \cdot \sin 120^{\circ}\)
	\(12-\sqrt{3}\)	\(=\dfrac{3\sqrt{3}x}{2}\)
	\(x\)	\(=\dfrac{2(12-\sqrt{3})}{3\sqrt{3}} \times \dfrac {\sqrt{3}}{\sqrt{3}}\)
		\(=\dfrac{24\sqrt{3}-6}{9} \)
		\(=\dfrac{8\sqrt{3}-2}{3} \)

\(\Rightarrow B\)

Functions, 2ADV F1 EQ-Bank 24 MC

Given that \(f(x)=\left\{\begin{array}{ll}3-(x-2)^2, & \text { for } x \leqslant 2 \\ m x+5, & \text { for } x>2\end{array}\right.\)

What is the value of \(m\) for which \(f(x)\) is continuous at \(x=2\) ?

\(1\)
\(2\)
\(-1\)
\(-2\)

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\(C\)

Show Worked Solution

\(\text {If continuous at}\ x=2:\)

	\(3-(x-2)^2\)	\(=mx+5\)
	\(3-(2-2)^2\)	\(=2m+5\)
	\(2m\)	\(=-2\)
	\(m\)	\(=-1\)

\(\Rightarrow C\)

L&E, 2ADV E1 EQ-Bank 5

Solve the equation \(8^{n+3}=\dfrac{1}{2}\) (2 marks)

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\(n=-\dfrac{10}{3} \)

Show Worked Solution

\(8^{n+3}\)	\(=\dfrac{1}{2}\)
\(2^{3(n+3)}\)	\(=2^{-1}\)
\(3n+9\)	\(=-1\)
\(3n\)	\(=-10\)
\(n\)	\(=-\dfrac{10}{3} \)

Calculus, 2ADV C1 EQ-Bank 7

Differentiate `x(1+3x)^5` with respect to `x`. (2 marks)

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`y^{′}=(1+3x)^4(18 x+1)`

Show Worked Solution

`y=x(1+3x)^5`

`text{Using the product and chain rules:}`

`y^{′}`	`=1 xx (1+3x)^5+15x(1+3x)^4`
	`=(1+3x)^4(1+3x+15 x)`
	`=(1+3x)^4(18 x+1)`

Calculus, 2ADV C1 EQ-Bank 6

Use the definition of the derivative, `f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}` to find `f^{\prime}(x)` if `f(x)=5x^2-2x`. (2 marks)

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`f(x)=5x^2-2x`

`f^{′}(x)`	`= \lim_{h->0} \frac{5(x+h)^2-2(x+h)-(5x^2-2x)}{h}`
	`= \lim_{h->0} \frac{5x^2+10xh+h^2-2x-2h-5x^2+2x}{h}`
	`= \lim_{h->0} \frac{10xh+h^2-2h}{h}`
	`= \lim_{h->0} \frac{h(10x+h-2)}{h}`
	`= \lim_{h->0} 10x+h-2`
	`=10x-2`

Show Worked Solution

`f(x)=5x^2-2x`

`f^{′}(x)`	`= \lim_{h->0} \frac{5(x+h)^2-2(x+h)-(5x^2-2x)}{h}`
	`= \lim_{h->0} \frac{5x^2+10xh+h^2-2x-2h-5x^2+2x}{h}`
	`= \lim_{h->0} \frac{10xh+h^2-2h}{h}`
	`= \lim_{h->0} \frac{h(10x+h-2)}{h}`
	`= \lim_{h->0} 10x+h-2`
	`=10x-2`

Trigonometry, 2ADV T2 EQ-Bank 6

Simplify \(\sin \, \theta \, \cos \theta \,\operatorname{cosec}^2 \, \theta\). (2 marks)

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\(\cot \theta\)

Show Worked Solution

\(\sin \theta \cos \theta \times \dfrac{1}{\sin ^2 \theta}\)

\(=\dfrac{\cos \, \theta}{\sin \, \theta}\)

\(=\cot \, \theta\)

Functions, 2ADV F1 EQ-Bank 20

Find the value of \(k\) if \(4kx^2-(3-4k) x+k=0\) has one root. (2 marks)

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\(k=\dfrac{3}{8}\)

Show Worked Solution

\(4kx^2-(3-4k) x+k=0\)

\(\text{1 root}\ \Rightarrow \Delta=0\)

	\(\Delta\)	\(=b^2-4 a c\)
	\(0\)	\(=\left[ -\left( 3-4k \right)\right]^2-4\times 4k \times k\)
	\(0\)	\(=9-24k + 16k^2-16k^2\)
	\(24k\)	\(=9\)
	\(k\)	\(=\dfrac{9}{24}=\dfrac{3}{8}\)

Functions, 2ADV F1 EQ-Bank 19

\(R\left(r, r^2\right), S\left(s, s^2\right)\) and \(T\left(t, t^2\right)\) are points on the parabola \(y=x^2\).

Given \(RT\) is parallel to \(SO\), show \(r+t=s\) (2 marks)

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\(R\left(r, r^2\right), S\left(s, s^2\right), T\left(t, t^2\right)\)

\(m_{S O}=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{s^2-0}{s-0}=s\)

\(m_{R T}=\dfrac{t^2-r^2}{t-r}=\dfrac{(t-r)(t+r)}{(t-r)}=t+r\)

\(\text{Given}\ R T \ \| \ SO \ \Rightarrow \ m_{SO}=m_{R T}\)

\(\therefore s=r+t\ \ …\ \text{as required} \)

Show Worked Solution

\(R\left(r, r^2\right), S\left(s, s^2\right), T\left(t, t^2\right)\)

\(m_{S O}=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{s^2-0}{s-0}=s\)

\(m_{R T}=\dfrac{t^2-r^2}{t-r}=\dfrac{(t-r)(t+r)}{(t-r)}=t+r\)

\(\text{Given}\ R T \ \| \ SO \ \Rightarrow \ m_{SO}=m_{R T}\)

\(\therefore s=r+t\ \ …\ \text{as required} \)

A magpie plague hit Raymond Terrace this year but was eventually brought under control. A bird researcher estimated that the magpie population \(M\), in hundreds, \(t\) months after 1st January, was given by \(M=7+20t-3t^2\)

Find the magpie population on 1st March. (1 mark)
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At what rate was the population changing at this time? (1 mark)
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In what month does the magpie population start to decrease? (2 marks)
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a. \(\text{Population}\ =35 \times 100=3500\)

b. \(M\ \text{is increasing at 800 per month}\)

c. \(\text{Population starts to decrease in May.}\)

Show Worked Solution

a. \(\text{Find}\ M\ \text{when}\ \ t=2:\)

\(M=7+20 \times 2-3 \times 2^2 = 35\)

\(\therefore \text{Population}\ =35 \times 100=3500\)

b. \(M=7+20t-3t^2\)

\(\dfrac{dM}{dt}=20-6t\)

\(\text{Find}\ \dfrac{dM}{dt}\ \text{when}\ \ t=2: \)

\(\dfrac{dM}{dt}=20-6 \times 2 = 8\)

\(\therefore M\ \text{is increasing at 800 per month}\)

c. \(\text{Find}\ t\ \text{when}\ \dfrac{dM}{dt}=0: \)

\(\dfrac{dM}{dt}=20-6t = 0\ \ \Rightarrow \ t= 3\ \dfrac{1}{3}\)

\(\dfrac{dM}{dt}<0\ \ \text{when}\ \ t>3\ \dfrac{1}{3} \)

\(\therefore\ \text{Population starts to decrease in May.}\)

Trigonometry, 2ADV T1 EQ-Bank 3

A tower \(T C\) is \(h\) metres high.

At point \(A\), due south of the tower, the angle of elevation to the top of the tower, point \(T\), is 13°.

Point \(B\) is due east of the tower, with an angle of elevation to the top of 24°, as shown in the diagram.

Point \(A\) is 1.1 kilometres from Point \(B\) and both points are on the same ground level as the base of the tower, point \(C\).

Show \(B C=h \times \tan 66^{\circ}\). (1 mark)

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Find a similar expression for \(A C\). (1 mark)

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Hence, or otherwise, determine the height, \(h\), of the tower. Give your answer correct to the nearest metre. (3 marks)

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a. \(\text{See Worked Solutions}\)

b. \(\text{See Worked Solutions}\)

c. \(\text{225 metres}\)

Show Worked Solution

a. \(\text{In}\ \Delta TBC\ \Rightarrow \angle CTB=90-24=66^{\circ}\)

\(\tan 66^{\circ}\)	\(=\dfrac{BC}{h} \)
\(BC\)	\(=h \times \tan 66^{\circ}\)

b. \(\text{In}\ \Delta TAC\ \Rightarrow \angle CTA=90-13=77^{\circ}\)

\(\tan 77^{\circ}\)	\(=\dfrac{AC}{h} \)
\(AC\)	\(=h \times \tan 77^{\circ}\)

c. \(\Delta ACB\ \text{is right-angled.}\)

\(\text{By Pythagoras:}\)

\(AC^{2}+BC^{2}\)	\(=1100^{2}\)
\(h^2 \times \tan^{2} 77^{\circ} + h^2 \times \tan^{2}66^{\circ}\)	\(=1100^2\)
\(h^2(\tan^{2} 77^{\circ}+\tan^{2} 66^{\circ})\)	\(=1100^2\)

\(h^2\)	\(=\dfrac{1100^2}{(\tan^{2} 77^{\circ}+\tan^{2} 66^{\circ})}\)
\(h\)	\(=225.44…\)
	\(=225\ \text{m (nearest m)}\)

Trigonometry, 2ADV T2 EQ-Bank 5

Solve \(\sin x-\cos x=0 \quad-\pi \leqslant x \leqslant \pi\) (2 marks)

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\(x=\dfrac{\pi}{4}, -\dfrac{3\pi}{4}\)

Show Worked Solution

	\(\sin x-\cos x\)	\(=0\)
	\(\dfrac{\sin x}{\cos x}-\dfrac{\cos x}{\cos x}\)	\(=0\)
	\(\tan x-1\)	\(=0\)
	\(\tan x\)	\(=1\)
	\(x\)	\(=\tan ^{-1}(1)\)

\(\therefore x=\dfrac{\pi}{4}, -\dfrac{3\pi}{4}\)

Trigonometry, 2ADV T3 EQ-Bank 4

The graph below has the equation \(y=a \sin (b x)+c\) for \(0 \leqslant x \leqslant 50\).

Determine the values of \(a, b\) and \(c\). (3 marks)

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\(a=16, b=9, c=24\)

Show Worked Solution

\(\text {Amplitude}\ =\dfrac{40-8}{2}=16\ \ \Rightarrow a=16\)

\(\text {Centre of motion}\ =24\ \ \Rightarrow c=24\)

\(\text {Period }=\dfrac{360}{n}=40\ \ \Rightarrow n=b=9\)

Trigonometry, 2ADV T2 EQ-Bank 2

Express \(3 \operatorname{cosec}(180+x)+5 \cos (90-x)\) as a single fraction in terms of \(\sin x\), given all angles are measured in degrees. (3 marks)

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\(\dfrac{-3+5 \sin ^2 x}{\sin x}\)

Show Worked Solution

\(3 \operatorname{cosec}(180+x)+5 \cos (90-x)\)

\(=\dfrac{3}{\sin \left(180^{\circ}+x\right)}+5 \sin x\)

\(=\dfrac{3}{-\sin x}+5 \sin x\)

\(=\dfrac{-3}{\sin x}+\dfrac{5 \sin ^2 x}{\sin x}\)

\(=\dfrac{-3+5 \sin ^2 x}{\sin x}\)

Functions, 2ADV F1 EQ-Bank 18

If \(f(x)=x^2-3\) and \(g(x)=\sqrt{x-2}\),

Find \(f(g(x))\). (1 mark)
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What is the domain of \(f(g(x))\) ? (1 mark)
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a. \(f(g(x))=x-5\)

b. \(\text{Restriction for}\ \ g(x)\ \Rightarrow \ x-2\geqslant 0 \ \Rightarrow \ x \geqslant 2\)

\(\text{Domain}\ \ f(g(x)):\ x \geqslant 2\)

Show Worked Solution

a.	\(f(g(x))\)	\(=(\sqrt{x-2})^2-3\)
		\(=x-2-3\)
		\(=x-5\)

b. \(\text{Restriction for}\ \ g(x)\ \Rightarrow \ x-2\geqslant 0 \ \Rightarrow \ x \geqslant 2\)

\(\text{Domain}\ \ f(g(x)):\ x \geqslant 2\)

Functions, 2ADV F1 EQ-Bank 10

Fully simplify the expression \(\dfrac{4}{x^2-9}-\dfrac{2x+1}{x+3}\) (3 marks)

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\(\dfrac{-(2x-7)(x+1)}{(x^2-9)}\)

Show Worked Solution

\(\dfrac{4}{x^2-9}-\dfrac{2x+1}{x+3}\)	\(=\dfrac{4-(2x+1)(x-3)}{(x+3)(x-3)}\)
	\(=\dfrac{4-(2x^2-5x-3)}{(x+3)(x-3)}\)
	\(=\dfrac{-(2x^2-5x-7)}{(x+3)(x-3)}\)
	\(=\dfrac{-(2x-7)(x+1)}{(x^2-9)}\)

Calculus, 2ADV C1 EQ-Bank 2 MC

The displacement of a particle is given by \(x=3t^{3}-6t^{2}-15\) . The acceleration is zero at:

\(t=\dfrac{2}{3}\)
\(t=\dfrac{4}{3}\)
\(t=\dfrac{5}{2}\)
\(\text{never}\)

Show Answers Only

\(A\)

Show Worked Solution

\(x=3t^{3}-6t^{2}-15\)

\(v=9t^{2}-12t\)

\(a=18t-12\)

\(\text{Find}\ t\ \text{when}\ \ a=0:\)

\(18t-12=0\ \ \Rightarrow\ \ t=\dfrac{2}{3} \)

\(\Rightarrow A\)

Functions, 2ADV F1 EQ-Bank 45

Determine whether the function \(f(x)=2x^3-5x\) is even, odd or neither. Show all working. (2 marks)

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\(f(x)=2x^{3}-5x\)

\(\text{Function is odd if:}\ \ f(-x)=-f(x) \)

\(f(-x)\)	\(=2(-x)^{3}-5(-x) \)
	\(=-2x^{3}+5x \)
	\(=-(2x^{3}-5x)\)
	\(=-f(x)\)

\(\therefore f(x)\ \text{is odd.}\)

Show Worked Solution

\(f(x)=2x^{3}-5x\)

\(\text{Function is odd if:}\ \ f(-x)=-f(x) \)

\(f(-x)\)	\(=2(-x)^{3}-5(-x) \)
	\(=-2x^{3}+5x \)
	\(=-(2x^{3}-5x)\)
	\(=-f(x)\)

\(\therefore f(x)\ \text{is odd.}\)

Calculus, 2ADV C1 EQ-Bank 1 MC

The derivative of \(n x^{2n+1}\) can be expressed as

\(2 n^2 x^{2 n+1}\)
\(2 n^2 x^{2 n}\)
\((2 n+1) n x^{2 n}\)
\((2 n+1) n x^{2 n+1}\)

Show Answers Only

\(C\)

Show Worked Solution

\(y\)	\(=n x^{2n+1}\)
\(y^{′}\)	\(=(2 n+1) n x^{2n+1-1}\)
	\(=(2 n+1) n x^{2 n}\)

\(\Rightarrow C\)

BIOLOGY, M2 EQ-Bank 1

"Multicellular organisms exhibit different levels of cell complexity, from simple cells to highly specialised ones."

Justify this statement, providing examples to support your answer. (4 marks)

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→ Multicellular organisms show different levels of cell complexity, starting with simple cells that perform basic functions, like skin cells, to highly specialised cells, such as neurons, which have unique structures for transmitting signals.

→ Cell groups within multicellular organisms cannot survive without other cell types performing their role.

→ A hierarchy arises, organising cells into tissues, organs and other systems, each with distinct functions that work together to support life.

→ For example, muscle cells form muscle tissue, which contracts to allow movement, while red blood cells transport oxygen through the circulatory system but cannot reproduce.

→ The specialisation of cells allows multicellular organisms to perform complex tasks more efficiently, contributing to the organism’s overall survival and adaptability.

→ This organisation enhances both the functionality and efficiency of biological processes

Show Worked Solution

→ Cell groups within multicellular organisms cannot survive without other cell types performing their role.

→ A hierarchy arises, organising cells into tissues, organs and other systems, each with distinct functions that work together to support life.

→ For example, muscle cells form muscle tissue, which contracts to allow movement, while red blood cells transport oxygen through the circulatory system but cannot reproduce.

→ The specialisation of cells allows multicellular organisms to perform complex tasks more efficiently, contributing to the organism’s overall survival and adaptability.

→ This organisation enhances both the functionality and efficiency of biological processes

BIOLOGY, M2 EQ-Bank 1 MC

Which of the following best distinguishes a unicellular organism from a colonial organism?

A unicellular organism is made up of multiple cells, while a colonial organism is a single cell.
A unicellular organism can survive independently, while a colonial organism involves many specialised cells that must live as a group to survive.
Unicellular organisms rely on other cells to perform essential functions, while colonial organisms do not.
Unicellular organisms consist of one cell that performs all life functions, while colonial organisms are groups of identical cells that function together but can survive independently.

Show Answers Only

\(D\)

Show Worked Solution

→ In unicellular organisms, a single cell carries out all necessary life processes.

→ Colonial organisms consist of individual cells that are physically connected or grouped, but each cell can perform essential functions independently, allowing survival even if separated from the colony.

\(\Rightarrow D\)

BIOLOGY, M1 EQ-Bank 10

Explain why lysosomes store acidic enzymes, including one way in which this storage process benefits the cell. (2 marks)

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→ Lysosomes store acidic enzymes because these type of enzymes are most effective in helping the organelle perform its major role of breaking down waste and cellular debris.

→ This storage process plays a crucial role in protecting the cell by keeping these enzymes contained.

→ If the acidic enzymes were released into the cytoplasm, they would lower the pH in the cytosol, adversely affecting other enzymes and potentially damaging other cellular components.

Show Worked Solution

→ Lysosomes store acidic enzymes because these type of enzymes are most effective in helping the organelle perform its major role of breaking down waste and cellular debris.

→ This storage process plays a crucial role in protecting the cell by keeping these enzymes contained.

→ If the acidic enzymes were released into the cytoplasm, they would lower the pH in the cytosol, adversely affecting other enzymes and potentially damaging other cellular components.

BIOLOGY, M1 EQ-Bank 8

Discuss how the structure of the phospholipid bilayer contributes to the selective permeability of a cell membrane. (3 marks)

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→ The phospholipid bilayer’s structure, with hydrophilic heads facing outward and hydrophobic tails facing inward, forms a barrier that allows only small, nonpolar molecules to pass through easily, while blocking larger or charged molecules.

→ This selective permeability is further controlled by embedded proteins, which act as channels or carriers to facilitate the movement of ions and larger molecules that cannot cross the bilayer on their own.

→ This structure helps the cell regulate the entry and exit of specific substances, maintaining internal balance.

Show Worked Solution

→ This structure helps the cell regulate the entry and exit of specific substances, maintaining internal balance.

BIOLOGY, M1 EQ-Bank 7

How do temperature and pH affect enzyme activity? In your answer, briefly explain how extreme conditions of each factor influence enzyme function. (4 marks)

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→ Temperature and pH significantly impact enzyme activity.

→ At high temperatures, enzymes can denature, losing their shape and functionality. At extreme heat, the active site can be destroyed or made inaccessible.

→ At low temperatures, enzyme activity slows down due to reduced molecular movement.

→ Certain enzymes have evolved to be most effective in specific pH environments (such as the acidity of the stomach) and are much less effective in more alkaline settings.

→ Extreme pH levels can alter the enzyme’s structure, affecting the active site’s ability to bind with substrates.

→ Optimal enzyme activity occurs within specific temperature and pH ranges for each enzyme.

Show Worked Solution

→ Temperature and pH significantly impact enzyme activity.

→ At high temperatures, enzymes can denature, losing their shape and functionality. At extreme heat, the active site can be destroyed or made inaccessible.

→ At low temperatures, enzyme activity slows down due to reduced molecular movement.

→ Certain enzymes have evolved to be most effective in specific pH environments (such as the acidity of the stomach) and are much less effective in more alkaline settings.

→ Extreme pH levels can alter the enzyme’s structure, affecting the active site’s ability to bind with substrates.

→ Optimal enzyme activity occurs within specific temperature and pH ranges for each enzyme.

BIOLOGY, M1 EQ-Bank 6 MC

Which of the following best distinguishes cellular waste from cellular secretions?

Both cellular waste and secretions are expelled by the cell for communication with other cells.
Cellular waste is produced as a byproduct of metabolic reactions, while secretions are released by cells to perform specific functions.
Cellular waste is stored in vesicles, while secretions are immediately broken down within the cell.
Secretions are only produced by prokaryotic cells, while waste is produced by all cell types.

Show Answers Only

\(B\)

Show Worked Solution

→ Cellular waste includes substances like carbon dioxide or urea, which result from metabolic processes and must be removed.

→ Secretions, on the other hand, are purposeful substances like hormones or enzymes that are actively produced and released by cells for specific roles in the body.

\(\Rightarrow B\)

BIOLOGY, M1 EQ-Bank 5

Explain how cells remove waste products to maintain proper function and provide an example of a specific waste product that cells eliminate. (3 marks)

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→ Cells remove waste products through exocytosis, where waste is trapped within vesicles inside the cell.

→ These vesicles then fuse with the cell membrane, releasing the waste to the outside environment.

→ Examples of a waste product: could include carbon dioxide, lactic acid or urea (and other nitrogen based waste compounds).

Show Worked Solution

→ Cells remove waste products through exocytosis, where waste is trapped within vesicles inside the cell.

→ These vesicles then fuse with the cell membrane, releasing the waste to the outside environment.

→ Examples of a waste product: could include carbon dioxide, lactic acid or urea (and other nitrogen based waste compounds).

BIOLOGY, M1 EQ-Bank 5 MC

Which process is primarily responsible for removing waste materials from a cell?

Phagocytosis
Endocytosis
Exocytosis
Osmosis

Show Answers Only

\(C\)

Show Worked Solution

→ Exocytosis is the process by which cells expel waste materials and other substances by transporting vesicles to the cell membrane, where they fuse and release their contents outside the cell.

\(\Rightarrow C\)

BIOLOGY, M1 EQ-Bank 3

Discuss how concentration gradients affect the transport of molecules across the cell membrane? (3 marks)

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→ Concentration gradients determine the direction of molecule movement across the cell membrane.

→ Molecules naturally move from areas of high concentration to low concentration in processes like diffusion and osmosis (passive transport). The higher the gradient, the faster the movement.

→ In contrast, active transport requires energy to move substances against the gradient, from low to high concentration. The higher the gradient, the more energy that is required.

→ These gradients are crucial for maintaining cellular balance and function.

Show Worked Solution

→ Concentration gradients determine the direction of molecule movement across the cell membrane.

→ Molecules naturally move from areas of high concentration to low concentration in processes like diffusion and osmosis (passive transport). The higher the gradient, the faster the movement.

→ In contrast, active transport requires energy to move substances against the gradient, from low to high concentration. The higher the gradient, the more energy that is required.

→ These gradients are crucial for maintaining cellular balance and function.

BIOLOGY, M1 EQ-Bank 2

"The surface area to volume ratio is crucial for efficient cellular transport."

Evaluate the above statement, providing an example of how cells adapt their shape to maintain an effective ratio. (3 marks)

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→ The surface area to volume ratio is vital for efficient cellular transport because a higher ratio allows for quicker exchange of materials, such as nutrients and waste, across the cell membrane.

→ As cells grow larger, their volume increases faster than their surface area, reducing efficiency.

→ To compensate, some cells increase surface area without significantly adding volume.

→ An example is red blood cells that have adapted to be biconcave in shape. This maximises surface area relative to volume to enhance oxygen exchange.

Show Worked Solution

→ The surface area to volume ratio is vital for efficient cellular transport because a higher ratio allows for quicker exchange of materials, such as nutrients and waste, across the cell membrane.

→ As cells grow larger, their volume increases faster than their surface area, reducing efficiency.

→ To compensate, some cells increase surface area without significantly adding volume.

→ An example is red blood cells that have adapted to be biconcave in shape. This maximises surface area relative to volume to enhance oxygen exchange.

BIOLOGY, M1 EQ-Bank 3 MC

Which of the following best describes movement of molecules in passive transport with respect to concentration gradients?

Molecules move against the concentration gradient, from low to high concentration.
Molecules move down the concentration gradient, from high to low concentration.
Molecules remain stationary when the concentration gradient is equal.
Molecules move randomly, regardless of the concentration gradient.

Show Answers Only

\(B\)

Show Worked Solution

→ Passive transport relies on molecules moving naturally from an area of high concentration to an area of low concentration without the need for energy.

→ This process includes diffusion and osmosis, allowing molecules to equalise concentration across membranes.

\(\Rightarrow B\)

BIOLOGY, M1 EQ-Bank 1

Describe how active transport and passive transport differ in their mechanisms for moving substances into and out of cells, and give an example of each. (2 marks)

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→ Active transport requires energy (usually in the form of ATP) to move or create the passage for substances into or out of cells.

→ Endocytosis and exocytosis are examples of active transport (only one example required).

→ In contrast, passive transport does not require energy to move substances into or out of cells.

→ Diffusion and osmosis are examples of passive transport (only one example required).

Show Worked Solution

→ Active transport requires energy (usually in the form of ATP) to move or create the passage for substances into or out of cells.

→ Endocytosis and exocytosis are examples of active transport (only one example required).

→ In contrast, passive transport does not require energy to move substances into or out of cells.

→ Diffusion and osmosis are examples of passive transport (only one example required).

BIOLOGY, M1 EQ-Bank 2 MC

Exocytosis is critical for which of the following cellular functions?

Importing large food particles into the cell.
Exporting proteins and other materials out of the cell.
Transporting ions directly through the phospholipid bilayer.
Facilitating the movement of oxygen into the cell.

Show Answers Only

\(B\)

Show Worked Solution

→ Exocytosis is a process where vesicles fuse with the plasma membrane to expel materials, such as proteins or waste, from the cell.

→ This process is crucial for cellular secretion and maintaining cell health.

\(\Rightarrow B\)

BIOLOGY, M1 EQ-Bank 1 MC

Which of the following best describes the process of endocytosis?

The movement of water molecules through a membrane by osmosis.
The transport of large molecules into the cell by engulfing them in a vesicle.
The passive diffusion of small molecules across the cell membrane.
The removal of cellular waste through the membrane via a vesicle.

Show Answers Only

\(B\)\

Show Worked Solution

→ Endocytosis is a process by which the cell membrane folds inward to form a vesicle, bringing large molecules like nutrients or particles into the cell.

\(\Rightarrow B\)

BIOLOGY, M1 EQ-Bank 7

Explain how the cell arrangement supports the role and specific functions of the following cellular structures

Cytoplasm (2 marks)
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Ribosomes (2 marks)
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a. Cytoplasm:

→ The cytoplasm is a gel-like substance that fills the cell, providing a medium in which organelles are suspended.

→ It facilitates the movement of materials within the cell and allows biochemical reactions to occur efficiently.

b. Ribosomes:

→ Ribosomes can be found either floating freely in the cytoplasm or attached to the rough endoplasmic reticulum.

→ Their arrangement allows them to efficiently translate mRNA into proteins, with free ribosomes synthesising proteins for use within the cell and membrane-bound ribosomes manufacturing proteins for export or use in the cell membrane.

Show Worked Solution

a. Cytoplasm:

→ The cytoplasm is a gel-like substance that fills the cell, providing a medium in which organelles are suspended.

→ It facilitates the movement of materials within the cell and allows biochemical reactions to occur efficiently.

b. Ribosomes:

→ Ribosomes can be found either floating freely in the cytoplasm or attached to the rough endoplasmic reticulum.

BIOLOGY, M1 EQ-Bank 8 MC

Which of the following correctly compares the structure and function of the mitochondria and the chloroplast?

Both mitochondria and chloroplasts produce energy through photosynthesis.
Mitochondria have a single membrane, while chloroplasts have a double membrane.
Mitochondria generate ATP through cellular respiration, while chloroplasts convert light energy into chemical energy via photosynthesis.
Both mitochondria and chloroplasts are responsible for protein synthesis.

Show Answers Only

\(C\)

Show Worked Solution

→ Mitochondria are responsible for producing ATP through the process of cellular respiration.

→ Chloroplasts, found in plant cells, convert light energy into glucose via photosynthesis.

→ Both organelles have distinct functions related to energy conversion in cells.

\(\Rightarrow C\)

BIOLOGY, M1 EQ-Bank 6

Describe the structure and role of phospholipids in the cell membrane based on the fluid mosaic model. (3 marks)

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→ In the fluid mosaic model, phospholipids form a bilayer that acts as the fundamental structure of the cell membrane.

→ Each phospholipid has a hydrophilic head that faces outward and a hydrophobic tail that faces inward, creating a barrier that separates the cell’s internal and external environments.

→ This bilayer is semi-permeable, allowing selective substances to pass through while blocking others.

Show Worked Solution

→ In the fluid mosaic model, phospholipids form a bilayer that acts as the fundamental structure of the cell membrane.

→ Each phospholipid has a hydrophilic head that faces outward and a hydrophobic tail that faces inward, creating a barrier that separates the cell’s internal and external environments.

→ This bilayer is semi-permeable, allowing selective substances to pass through while blocking others.

BIOLOGY, M1 EQ-Bank 5

According to the fluid mosaic model of the cell membrane

describe the structure of the phospholipid bilayer. (2 marks)

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explain the role of proteins. (1 mark)

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a. The phospholipid bilayer:

→ Consists of two layers of phospholipids.

→ One layer involves hydrophilic (water-attracting) heads facing outward toward the water-based environment while the opposite layer has hydrophobic (water-repelling) tails facing inward, away from water.

b. Proteins:

→ are embedded in the double phospholid layer and play a role in the cell structure and communication.

Show Worked Solution

a. The phospholipid bilayer:

→ Consists of two layers of phospholipids.

b. Proteins:

→ are embedded in the double phospholid layer and play a role in the cell structure and communication.

BIOLOGY, M1 EQ-Bank 4

Describe the progression of microscopy technologies and discuss how these advancements have enhanced our knowledge of cell structure and function. (4 marks)

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→ The progression of microscopy technologies has revolutionised our understanding of cell structure and function.

→ Early light microscopes allowed scientists to observe basic cell components, such as the nucleus, but were limited in resolution.

→ The invention of the transmission electron microscope brought a major leap, enabling the detailed 2-D visualisation of organelles like mitochondria and the endoplasmic reticulum.

→ The development of the scanning electron microscope created high resolution 3-D images of cells. This allowed for the study of bacteria and cell-surface structures such as cilia.

→ More recently, confocal and fluorescence microscopy have allowed researchers to study living cells in real-time, providing dynamic insights into cellular processes.

→ These advancements have deepened our knowledge of cell biology, revealing intricate structures and functions previously unseen.

Show Worked Solution

→ The progression of microscopy technologies has revolutionised our understanding of cell structure and function.

→ Early light microscopes allowed scientists to observe basic cell components, such as the nucleus, but were limited in resolution.

→ The invention of the transmission electron microscope brought a major leap, enabling the detailed 2-D visualisation of organelles like mitochondria and the endoplasmic reticulum.

→ The development of the scanning electron microscope created high resolution 3-D images of cells. This allowed for the study of bacteria and cell-surface structures such as cilia.

→ More recently, confocal and fluorescence microscopy have allowed researchers to study living cells in real-time, providing dynamic insights into cellular processes.

→ These advancements have deepened our knowledge of cell biology, revealing intricate structures and functions previously unseen.

BIOLOGY, M1 EQ-Bank 3

Describe how the development of electron microscopy has improved our understanding of cell structure and function.

In your answer, compare the capabilities of transmission electron microscopes (TEM) and scanning electron microscopes (SEM), and give examples of how each is used in biological research. (4 marks)

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→ The development of electron microscopy has significantly enhanced our understanding of cell structure and function by providing much higher resolution images than light microscopes.

→ Transmission electron microscopes (TEM) allow scientists to view internal structures of cells in great detail by transmitting electrons through thin slices of specimens, revealing organelles like the mitochondria and endoplasmic reticulum.

→ In contrast, scanning electron microscopes (SEM) produce three-dimensional images by scanning the surface of a specimen with electrons, which is useful for studying cell surfaces and structures such as cilia or bacterial shapes.

→ Both TEM and SEM have been pivotal in biological research, allowing the discovery of cellular organelles and understanding of cell interactions at the microscopic level.

Show Worked Solution

→ The development of electron microscopy has significantly enhanced our understanding of cell structure and function by providing much higher resolution images than light microscopes.

→ Both TEM and SEM have been pivotal in biological research, allowing the discovery of cellular organelles and understanding of cell interactions at the microscopic level.

BIOLOGY, M1 EQ-Bank 7 MC

Which of the following correctly explains why a scanning electron microscope is advantageous in studying living cells?

It provides three-dimensional imaging of cells with high resolution.
It uses X-rays to visualise internal structures in detail.
It only detects non-living tissue with electron beams.
It relies on staining methods that permanently kill cells before imaging.

Show Answers Only

\(A\)

Show Worked Solution

→ Scanning electron microscopes capture detailed three-dimensional images of living cells, allowing researchers to observe cellular processes in real time with high resolution.

→ Unlike electron microscopes, this technology can be used to study live cells without extensive sample preparation that might kill the cells.

\(\Rightarrow A\)

BIOLOGY, M1 EQ-Bank 6 MC

Which of the following technologies was pivotal in discovering the internal structure of organelles such as mitochondria and the endoplasmic reticulum?

Light microscope
Transmission electron microscope
X-ray diffraction
Gel electrophoresis

Show Answers Only

\(B\)

Show Worked Solution

→ The transmission electron microscope (TEM) allows for the visualisation of internal cell structures at high resolution by passing electrons through thin sections of a sample.

→ This technology was crucial in discovering detailed organelle structures like mitochondria and the endoplasmic reticulum, which are not visible with light microscopes.

\(\Rightarrow B\)

BIOLOGY, M1 EQ-Bank 2

Describe the role of two cell organelles found in eukaryotic cells. (2 marks)

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How do specialised organelles contribute to the efficient functioning of eukaryotic cells, compared to prokaryotic cells? (2 marks)

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a. Answers could include any two of the following:

→ Nucleus – the nucleus contains the cell’s genetic material and regulates gene expression.

→ Mitochondria – generates ATP through cellular respiration, providing energy for the cell’s functions.

→ Endoplasmic reticulum (ER) – helps synthesise proteins (rough ER) and lipids (smooth ER), contributing to various cell processes.

b. → Specialised organelles allow eukaryotic cells to compartmentalise tasks.

→ This increases metabolic efficiency and enables complex processes to occur within the cell.

→ In contrast, prokaryotic cells, lacking organelles, must perform all functions in the cytoplasm, limiting their ability to handle as many simultaneous or specialised activities as eukaryotic cells.

Show Worked Solution

a. Answers could include any two of the following:

→ Nucleus – the nucleus contains the cell’s genetic material and regulates gene expression.

→ Mitochondria – generates ATP through cellular respiration, providing energy for the cell’s functions.

→ Endoplasmic reticulum (ER) – helps synthesise proteins (rough ER) and lipids (smooth ER), contributing to various cell processes.

b. → Specialised organelles allow eukaryotic cells to compartmentalise tasks.

→ This increases metabolic efficiency and enables complex processes to occur within the cell.

BIOLOGY, M1 EQ-Bank 1

Compare the structural and functional differences between prokaryotic and eukaryotic cells. (2 marks)

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Discuss how these differences influence the complexity and organisation of organisms that consist of these cell types. (2 marks)

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a. Structural and functional differences:

→ Prokaryotic cells lack membrane-bound organelles and a true nucleus, with their DNA free-floating in the cytoplasm.

→ Eukaryotic cells, on the other hand, have a membrane-bound nucleus and various organelles such as mitochondria, which allow compartmentalisation of functions.

→ This structural difference enables eukaryotes to perform more complex processes, while prokaryotes are limited to simpler metabolic activities.

b. → Eukarytic organisms can manage multiple, specialised functions simultaneously.

→ This supports greater complexity in multicellular organisms and leads to highly organised systems with differentiated tissues and organs.

→ Prokaryotic organisms remain unicellular or simple multicellular forms, with all functions occurring in the same space.

Show Worked Solution

a. Structural and functional differences:

→ Prokaryotic cells lack membrane-bound organelles and a true nucleus, with their DNA free-floating in the cytoplasm.

→ Eukaryotic cells, on the other hand, have a membrane-bound nucleus and various organelles such as mitochondria, which allow compartmentalisation of functions.

→ This structural difference enables eukaryotes to perform more complex processes, while prokaryotes are limited to simpler metabolic activities.

b. → Eukarytic organisms can manage multiple, specialised functions simultaneously.

→ This supports greater complexity in multicellular organisms and leads to highly organised systems with differentiated tissues and organs.

→ Prokaryotic organisms remain unicellular or simple multicellular forms, with all functions occurring in the same space.

BIOLOGY, M1 EQ-Bank 5 MC

Which of the following accurately explains why prokaryotic cells are generally smaller than eukaryotic cells?

Prokaryotic cells lack a cell membrane, limiting their size.
The absence of internal compartmentalisation in prokaryotic cells restricts their metabolic efficiency.
Prokaryotic cells do not perform cellular respiration, which reduces their need for space.
Eukaryotic cells require less genetic material, allowing them to be larger than prokaryotic cells.

Show Answers Only

\(B\)

Show Worked Solution

→ Prokaryotic cells lack membrane-bound organelles, which limits their ability to compartmentalize functions.

→ This results in lower metabolic efficiency, constraining their size compared to eukaryotic cells.

\(\Rightarrow B\)

BIOLOGY, M7 2019 VCE 20 MC

The diagram below shows the process of phagocytosis. This process is vital for immunity against extracellular infections.

What is happening at position 3 ?

Enzymes that break down the microorganism are released into the vesicle.
Antibodies are added to the vesicle to kill the microorganism.
The cell is sampling the vesicle for antigen presentation.
Intracellular microbes are attacking the microorganism.

Show Answers Only

\(A\)

Show Worked Solution

→ When the phagosome and lysosome fuse, they form a phagolysosome (vesicle), an acidic and oxidising environment that can effectively kill and degrade the ingested microorganism.

\(\Rightarrow A\)

BIOLOGY, M7 2020 VCE 35 MC

Monoclonal antibodies attaching to antigens on a cancer cell are shown in the diagram below.

Monoclonal antibodies

are used to suppress B cells acting on cancer cells.
make it easier for cells of the immune system to detect cancer cells.
can bind to dendritic cells to stimulate them to destroy cancer cells.
can attach to many structurally different proteins found on the surface of cancer cells.

Show Answers Only

\(B\)

Show Worked Solution

→ Monoclonal antibodies bind to proteins on the surface of cancer cells, marking them for detection and destruction by the immune system.

→ By making the cancer cells more visible to the immune system, the monoclonal antibodies facilitate the immune system’s ability to recognise and attack the cancer cells.

\(\Rightarrow B\)

BIOLOGY, M6 2020 VCE 27*

Tasmanian devils (Sarcophilus harrisii) were originally broadly distributed across Australia. When sea levels rose 12 000 years ago, an island, now referred to as Tasmania, was formed. The small number of Tasmanian devils on Tasmania was cut off from the Australian mainland populations. The population in Tasmania showed less genetic variation than the mainland populations. Mainland populations became extinct approximately 3000 years ago.

Over the last 20 years, the total Tasmanian devil population on Tasmania has halved. Many of the deaths have been the result of Tasmanian devil facial tumour disease (DFTD). Scientists have taken some Tasmanian devils that do not have DFTD to mainland Australia to set up a conservation program. The scientists have shown that greater genetic diversity among offspring in this program is observed when the Tasmanian devils are kept in isolated male-female pairs rather than in larger groups.

Giving reasons, describe if the conservation program for Tasmanian devils is an example of

allopatric speciation (1 mark)

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selective breeding (1 mark)

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natural selection (1 mark)

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i. Allopatric speciation – No

→ Involves the formation of new species due to geographic isolation whereas the program focuses on preserving the existing Tasmanian devil species.

ii. Selective breeding – Yes

→ Scientists are intentionally choosing which Tasmanian devils to breed together based on their genetic diversity.

iii. Natural selection – No

→ Natural selection is the differential survival and reproduction based on inherited traits.

→ The program involves intentional breeding decisions by scientists.

Show Worked Solution

i. Allopatric speciation – No

→ Involves the formation of new species due to geographic isolation whereas the program focuses on preserving the existing Tasmanian devil species.

ii. Selective breeding – Yes

→ Scientists are intentionally choosing which Tasmanian devils to breed together based on their genetic diversity.

iii. Natural selection – No

→ Natural selection is the differential survival and reproduction based on inherited traits.

→ The program involves intentional breeding decisions by scientists.

BIOLOGY, M7 2020 VCE 18*

Some human cells produce proteins called cytokines.

Describe the major function of cytokines. (2 marks)

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→ The major function of cytokines is to regulate the immune response.

→ Cytokines do this by signalling between cells, which helps coordinate the body’s defense against infections and inflammation.

Show Worked Solution

→ The major function of cytokines is to regulate the immune response.

→ Cytokines do this by signalling between cells, which helps coordinate the body’s defense against infections and inflammation.

BIOLOGY, M5 2020 VCE 8 MC

The primary structure of a protein is important because it

is the active, functional form of the protein.
has a very specific three-dimensional shape.
influences the way that the polypeptide folds.
directly controls the way proteins are transported into a cell.

Show Answers Only

\(C\)

Show Worked Solution

Consider each option:

Option A: Incorrect – The active form depends on the 3D structure, not just the primary sequence.

Option B: Incorrect – The primary structure itself is linear, not 3D.

Option C: Correct – The primary structure (linear sequence of amino acids) influences how the
polypeptide chain folds into its 3D conformation.

Option D: Incorrect – Protein transport is controlled by signal sequences, not directly by the
primary structure.

\(\Rightarrow C\)

BIOLOGY, M7 2021 VCE 15 MC

Neutrophils

secrete complement proteins
migrate to the site of infection and engulf invading pathogens.
increase the diameter of blood vessels to make them more permeable.
release histamine following the binding of the antigen to receptors on the neutrophil's surface.

Show Answers Only

\(B\)

Show Worked Solution

→ Neutrophils use phagocytosis to engulf and destroy invading pathogens, such as bacteria, fungi, and other microorganisms.

→ They recognise and bind to pathogens, then internalize and digest them using various antimicrobial mechanisms.

\(\Rightarrow B\)

BIOLOGY, M7 2021 VCE 4a

Varicella (chickenpox) is a highly contagious disease caused by the varicella zoster virus. A live, attenuated varicella zoster virus vaccine is recommended for children at age 18 months. In Australia, this vaccine is provided free of charge under the National Immunisation Program.

Once the varicella zoster virus vaccine is injected into the arm of a child, an immune response occurs.

Summarise the immune response that occurs within the child to result in long-term protection from chickenpox. (5 marks)

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→ The varicella zoster virus vaccine is taken up by cells in the body, such as macrophages.

→ These cells then display the viral antigens on their surface.

→ Antigen-presenting cells then migrate to the lymph nodes, where they interact with naive B cells.

→ The interaction with the antigen-presenting cells stimulates the naive B cells to divide and differentiate into plasma cells and memory B and T cells.

→ Plasma cells produce specific antibodies.

→ Some of the activated B cells and T cells become memory cells.

→ These memory cells provide long-term immunity against the virus.

Show Worked Solution

→ The varicella zoster virus vaccine is taken up by cells in the body, such as macrophages.

→ These cells then display the viral antigens on their surface.

→ Antigen-presenting cells then migrate to the lymph nodes, where they interact with naive B cells.

→ The interaction with the antigen-presenting cells stimulates the naive B cells to divide and differentiate into plasma cells and memory B and T cells.

→ Plasma cells produce specific antibodies.

→ Some of the activated B cells and T cells become memory cells.

→ These memory cells provide long-term immunity against the virus.

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