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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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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\(\text{0.342 M}\)
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 \(0.342\ \text{M}\).
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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.
\(\text{Find}\ T\ \text{when}\ \Delta G = 0: \)
\(\Delta H-T \Delta S =0\)
\(T= \dfrac{\Delta H}{\Delta S}= \dfrac{-196.0}{0.125}=1568.0\ \text{K} \)
→ As \(T\) decreases, \(\Delta G\) decreases.
→ Reaction is spontaneous when \(\Delta G < 0\).
→ The reaction is spontaneous at temperatures below 1568.0 K.
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.
\(\text{Find}\ T\ \text{when}\ \Delta G = 0: \)
\(\Delta H-T \Delta S\)\(=0\)
→ 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.
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) --- 6 WORK AREA LINES (style=lined) ---
where \(\Delta H = -198.2\ \text{kJ}\) and \(\Delta S = -188\ \text{J K}^{-1} \) (3 marks)
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Prove \(\dfrac{1+\cot \theta}{1+\tan \theta}=\cot \theta\). (3 marks) --- 8 WORK AREA LINES (style=lined) ---
\(\text{LHS}\) | \(=\dfrac{1+\dfrac{\cos \theta}{\sin \theta}}{1+\dfrac{\sin \theta}{\cos \theta}}\) | |
\(=\dfrac{\dfrac{\sin \theta+\cos \theta}{\sin \theta}}{\dfrac{\cos \theta+\sin \theta}{\cos \theta}}\) | ||
\(=\dfrac{\sin \theta+\cos \theta}{\sin \theta} \times \dfrac{\cos \theta}{\cos \theta+\sin \theta}\) | ||
\(=\dfrac{\cos \theta}{\sin \theta}\) | ||
\(=\cot \theta\) | ||
\(\ =\text{ RHS}\) |
\(\text{LHS}\) | \(=\dfrac{1+\dfrac{\cos \theta}{\sin \theta}}{1+\dfrac{\sin \theta}{\cos \theta}}\) | |
\(=\dfrac{\dfrac{\sin \theta+\cos \theta}{\sin \theta}}{\dfrac{\cos \theta+\sin \theta}{\cos \theta}}\) | ||
\(=\dfrac{\sin \theta+\cos \theta}{\sin \theta} \times \dfrac{\cos \theta}{\cos \theta+\sin \theta}\) | ||
\(=\dfrac{\cos \theta}{\sin \theta}\) | ||
\(=\cot \theta\) | ||
\(\ =\text{ RHS}\) |
The displacement \(x\) metres from the origin at time, \(t\) seconds, of a particle travelling in a straight line is given by \(x=t^3-9 t^2+9 t, \quad t \geqslant 0\) --- 5 WORK AREA LINES (style=lined) --- --- 0 WORK AREA LINES (style=blank) --- --- 6 WORK AREA LINES (style=lined) ---
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) --- 6 WORK AREA LINES (style=lined) --- \(\text {Vertical asymptote at } \ x=-1 \ \ \Rightarrow p=1\) \(\text {Horizontal asymptote at } \ y=\dfrac{3}{2} \ \ \Rightarrow q=\dfrac{3}{2}\) \(y=\dfrac{1}{x+1}+\dfrac{3}{2}\) \(x\text{-intercept when} \ \ y=0:\) \(\text {Vertical asymptote at } \ x=-1 \ \ \Rightarrow p=1\) \(\text {Horizontal asymptote at } \ y=\dfrac{3}{2} \ \ \Rightarrow q=\dfrac{3}{2}\) \(y=\dfrac{1}{x+1}+\dfrac{3}{2}\) \(x\text{-intercept when} \ \ y=0:\)
\(\dfrac{1}{x+1}+\dfrac{3}{2}\)
\(=0\)
\(\dfrac{1}{x+1}\)
\(=-\dfrac{3}{2}\)
\(3 x+3\)
\(=-2\)
\(3x\)
\(=-5\)
\(x\)
\(=-\dfrac{5}{3}\)
\(\dfrac{1}{x+1}+\dfrac{3}{2}\)
\(=0\)
\(\dfrac{1}{x+1}\)
\(=-\dfrac{3}{2}\)
\(3 x+3\)
\(=-2\)
\(3x\)
\(=-5\)
\(x\)
\(=-\dfrac{5}{3}\)
Prove that the line between \((1,-1)\) and \((4,-3)\) is perpendicular to the line \(3x-2y-4=0\) (2 marks) --- 6 WORK AREA LINES (style=lined) ---
\(\text {Perpendicular lines}\ \ \Rightarrow\ m_1 \times m_2 = -1\)
\(\text {Line 1 gradient:}\)
\(P_1 (1,-1), P_2(4,-3) \)
\(m_1=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{-3+1}{4-1}=-\dfrac{2}{3}\)
\(\text {Line 2 gradient:}\)
\(3x-2y-4=0\ \ \Rightarrow \ y= \dfrac{3}{2}x-2\ \ \Rightarrow m_2=\dfrac{3}{2}\)
\(m_1 \times m_2 = -\dfrac{2}{3} \times \dfrac{3}{2} = -1\)
\(\therefore\ \text{Lines are perpendicular.}\)
\(\text {Perpendicular lines}\ \ \Rightarrow\ m_1 \times m_2 = -1\)
\(\text {Line 1 gradient:}\)
\(P_1 (1,-1), P_2(4,-3) \)
\(m_1=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{-3+1}{4-1}=-\dfrac{2}{3}\)
\(\text {Line 2 gradient:}\)
\(3x-2y-4=0\ \ \Rightarrow \ y= \dfrac{3}{2}x-2\ \ \Rightarrow m_2=\dfrac{3}{2}\)
\(m_1 \times m_2 = -\dfrac{2}{3} \times \dfrac{3}{2} = -1\)
\(\therefore\ \text{Lines are perpendicular.}\)
The triangle \(A B C\) has an exact area of \((12-\sqrt{3})\) cm\(^{2}\).
The exact value of \(x\) is
\(B\)
\(12-\sqrt{3}\) | \(=\dfrac{1}{2} \cdot 6 \cdot x \cdot \sin 120^{\circ}\) | |
\(12-\sqrt{3}\) | \(=3x \cdot \sin 120^{\circ}\) | |
\(12-\sqrt{3}\) | \(=\dfrac{3\sqrt{3}x}{2}\) | |
\(x\) | \(=\dfrac{2(12-\sqrt{3})}{3\sqrt{3}} \times \dfrac {\sqrt{3}}{\sqrt{3}}\) | |
\(=\dfrac{24\sqrt{3}-6}{9} \) | ||
\(=\dfrac{8\sqrt{3}-2}{3} \) |
\(\Rightarrow B\)
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 \(x=2\) ?
\(C\)
\(\text {If continuous at}\ x=2:\)
\(3-(x-2)^2\) | \(=mx+5\) | |
\(3-(2-2)^2\) | \(=2m+5\) | |
\(2m\) | \(=-2\) | |
\(m\) | \(=-1\) |
\(\Rightarrow C\)
Solve the equation \(8^{n+3}=\dfrac{1}{2}\) (2 marks) --- 5 WORK AREA LINES (style=lined) --- \(n=-\dfrac{10}{3} \)
\(8^{n+3}\)
\(=\dfrac{1}{2}\)
\(2^{3(n+3)}\)
\(=2^{-1}\)
\(3n+9\)
\(=-1\)
\(3n\)
\(=-10\)
\(n\)
\(=-\dfrac{10}{3} \)
Differentiate `x(1+3x)^5` with respect to `x`. (2 marks) --- 5 WORK AREA LINES (style=lined) --- `y^{′}=(1+3x)^4(18 x+1)` `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)`
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) --- 11 WORK AREA LINES (style=lined) --- `f(x)=5x^2-2x` `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`
`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`
Simplify \(\sin \, \theta \, \cos \theta \,\operatorname{cosec}^2 \, \theta\). (2 marks) --- 4 WORK AREA LINES (style=lined) ---
\(\cot \theta\)
\(\sin \theta \cos \theta \times \dfrac{1}{\sin ^2 \theta}\)
\(=\dfrac{\cos \, \theta}{\sin \, \theta}\)
\(=\cot \, \theta\)
Find the value of \(k\) if \(4kx^2-(3-4k) x+k=0\) has one root. (2 marks) --- 7 WORK AREA LINES (style=lined) --- \(k=\dfrac{3}{8}\) \(4kx^2-(3-4k) x+k=0\) \(\text{1 root}\ \Rightarrow \Delta=0\)
\(\Delta\)
\(=b^2-4 a c\)
\(0\)
\(=\left[ -\left( 3-4k \right)\right]^2-4\times 4k \times k\)
\(0\)
\(=9-24k + 16k^2-16k^2\)
\(24k\)
\(=9\)
\(k\)
\(=\dfrac{9}{24}=\dfrac{3}{8}\)
\(R\left(r, r^2\right), S\left(s, s^2\right)\) 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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\(R\left(r, r^2\right), S\left(s, s^2\right), T\left(t, t^2\right)\)
\(m_{S O}=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{s^2-0}{s-0}=s\)
\(m_{R T}=\dfrac{t^2-r^2}{t-r}=\dfrac{(t-r)(t+r)}{(t-r)}=t+r\)
\(\text{Given}\ R T \ \| \ SO \ \Rightarrow \ m_{SO}=m_{R T}\)
\(\therefore s=r+t\ \ …\ \text{as required} \)
\(R\left(r, r^2\right), S\left(s, s^2\right), T\left(t, t^2\right)\)
\(m_{S O}=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{s^2-0}{s-0}=s\)
\(m_{R T}=\dfrac{t^2-r^2}{t-r}=\dfrac{(t-r)(t+r)}{(t-r)}=t+r\)
\(\text{Given}\ R T \ \| \ SO \ \Rightarrow \ m_{SO}=m_{R T}\)
\(\therefore s=r+t\ \ …\ \text{as required} \)
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 \(M=7+20t-3t^2\)
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a. \(\text{Population}\ =35 \times 100=3500\)
b. \(M\ \text{is increasing at 800 per month}\)
c. \(\text{Population starts to decrease in May.}\)
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. \(\text{Find}\ t\ \text{when}\ \dfrac{dM}{dt}=0: \)
\(\dfrac{dM}{dt}=20-6t = 0\ \ \Rightarrow \ t= 3\ \dfrac{1}{3}\)
\(\dfrac{dM}{dt}<0\ \ \text{when}\ \ t>3\ \dfrac{1}{3} \)
\(\therefore\ \text{Population starts to decrease in May.}\)
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 \(C\). --- 3 WORK AREA LINES (style=lined) --- --- 3 WORK AREA LINES (style=lined) --- --- 5 WORK AREA LINES (style=lined) --- a. \(\text{See Worked Solutions}\) b. \(\text{See Worked Solutions}\) c. \(\text{225 metres}\) a. \(\text{In}\ \Delta TBC\ \Rightarrow \angle CTB=90-24=66^{\circ}\) b. \(\text{In}\ \Delta TAC\ \Rightarrow \angle CTA=90-13=77^{\circ}\) \(\text{By Pythagoras:}\)
\(\tan 66^{\circ}\)
\(=\dfrac{BC}{h} \)
\(BC\)
\(=h \times \tan 66^{\circ}\)
\(\tan 77^{\circ}\)
\(=\dfrac{AC}{h} \)
\(AC\)
\(=h \times \tan 77^{\circ}\)
c. \(\Delta ACB\ \text{is right-angled.}\)
\(AC^{2}+BC^{2}\)
\(=1100^{2}\)
\(h^2 \times \tan^{2} 77^{\circ} + h^2 \times \tan^{2}66^{\circ}\)
\(=1100^2\)
\(h^2(\tan^{2} 77^{\circ}+\tan^{2} 66^{\circ})\)
\(=1100^2\)
\(h^2\)
\(=\dfrac{1100^2}{(\tan^{2} 77^{\circ}+\tan^{2} 66^{\circ})}\)
\(h\)
\(=225.44…\)
\(=225\ \text{m (nearest m)}\)
Solve \(\sin x-\cos x=0 \quad-\pi \leqslant x \leqslant \pi\) (2 marks)
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\(x=\dfrac{\pi}{4}, -\dfrac{3\pi}{4}\)
\(\sin x-\cos x\) | \(=0\) | |
\(\dfrac{\sin x}{\cos x}-\dfrac{\cos x}{\cos x}\) | \(=0\) | |
\(\tan x-1\) | \(=0\) | |
\(\tan x\) | \(=1\) | |
\(x\) | \(=\tan ^{-1}(1)\) |
\(\therefore x=\dfrac{\pi}{4}, -\dfrac{3\pi}{4}\)
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) --- 5 WORK AREA LINES (style=lined) ---
\(a=16, b=9, c=24\)
\(\text {Amplitude}\ =\dfrac{40-8}{2}=16\ \ \Rightarrow a=16\)
\(\text {Centre of motion}\ =24\ \ \Rightarrow c=24\)
\(\text {Period }=\dfrac{360}{n}=40\ \ \Rightarrow n=b=9\)
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) --- 8 WORK AREA LINES (style=lined) --- \(\dfrac{-3+5 \sin ^2 x}{\sin x}\) \(3 \operatorname{cosec}(180+x)+5 \cos (90-x)\) \(=\dfrac{3}{\sin \left(180^{\circ}+x\right)}+5 \sin x\) \(=\dfrac{3}{-\sin x}+5 \sin x\) \(=\dfrac{-3}{\sin x}+\dfrac{5 \sin ^2 x}{\sin x}\) \(=\dfrac{-3+5 \sin ^2 x}{\sin x}\)
If \(f(x)=x^2-3\) and \(g(x)=\sqrt{x-2}\), --- 3 WORK AREA LINES (style=lined) --- --- 2 WORK AREA LINES (style=lined) --- a. \(f(g(x))=x-5\) b. \(\text{Restriction for}\ \ g(x)\ \Rightarrow \ x-2\geqslant 0 \ \Rightarrow \ x \geqslant 2\) \(\text{Domain}\ \ f(g(x)):\ x \geqslant 2\) b. \(\text{Restriction for}\ \ g(x)\ \Rightarrow \ x-2\geqslant 0 \ \Rightarrow \ x \geqslant 2\) \(\text{Domain}\ \ f(g(x)):\ x \geqslant 2\)
a.
\(f(g(x))\)
\(=(\sqrt{x-2})^2-3\)
\(=x-2-3\)
\(=x-5\)
Fully simplify the expression \(\dfrac{4}{x^2-9}-\dfrac{2x+1}{x+3}\) (3 marks)
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\(\dfrac{-(2x-7)(x+1)}{(x^2-9)}\)
\(\dfrac{4}{x^2-9}-\dfrac{2x+1}{x+3}\) | \(=\dfrac{4-(2x+1)(x-3)}{(x+3)(x-3)}\) | |
\(=\dfrac{4-(2x^2-5x-3)}{(x+3)(x-3)}\) | ||
\(=\dfrac{-(2x^2-5x-7)}{(x+3)(x-3)}\) | ||
\(=\dfrac{-(2x-7)(x+1)}{(x^2-9)}\) |
The displacement of a particle is given by \(x=3t^{3}-6t^{2}-15\) . The acceleration is zero at:
\(A\)
\(x=3t^{3}-6t^{2}-15\)
\(v=9t^{2}-12t\)
\(a=18t-12\)
\(\text{Find}\ t\ \text{when}\ \ a=0:\)
\(18t-12=0\ \ \Rightarrow\ \ t=\dfrac{2}{3} \)
\(\Rightarrow A\)
Determine whether the function \(f(x)=2x^3-5x\) is even, odd or neither. Show all working. (2 marks)
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\(f(x)=2x^{3}-5x\)
\(\text{Function is odd if:}\ \ f(-x)=-f(x) \)
\(f(-x)\) | \(=2(-x)^{3}-5(-x) \) | |
\(=-2x^{3}+5x \) | ||
\(=-(2x^{3}-5x)\) | ||
\(=-f(x)\) |
\(\therefore f(x)\ \text{is odd.}\)
\(f(x)=2x^{3}-5x\)
\(\text{Function is odd if:}\ \ f(-x)=-f(x) \)
\(f(-x)\) | \(=2(-x)^{3}-5(-x) \) | |
\(=-2x^{3}+5x \) | ||
\(=-(2x^{3}-5x)\) | ||
\(=-f(x)\) |
\(\therefore f(x)\ \text{is odd.}\)
The derivative of \(n x^{2n+1}\) can be expressed as
\(C\)
\(y\) | \(=n x^{2n+1}\) | |
\(y^{′}\) | \(=(2 n+1) n x^{2n+1-1}\) | |
\(=(2 n+1) n x^{2 n}\) |
\(\Rightarrow C\)
"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
→ 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
Which of the following best distinguishes a unicellular organism from a colonial organism?
\(D\)
→ 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.
\(\Rightarrow D\)
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.
→ 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.
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.
→ 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.
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.
→ 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.
Which of the following best distinguishes cellular waste from cellular secretions?
\(B\)
→ 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.
\(\Rightarrow B\)
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).
→ 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).
Which process is primarily responsible for removing waste materials from a cell?
\(C\)
→ 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.
\(\Rightarrow C\)
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.
→ 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.
"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.
→ 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.
Which of the following best describes movement of molecules in passive transport with respect to concentration gradients?
\(B\)
→ 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.
\(\Rightarrow B\)
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).
→ 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).
Exocytosis is critical for which of the following cellular functions?
\(B\)
→ 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.
\(\Rightarrow B\)
Which of the following best describes the process of endocytosis?
\(B\)\
→ 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.
\(\Rightarrow B\)
Explain how the cell arrangement supports the role and specific functions of the following cellular structures
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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.
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.
Which of the following correctly compares the structure and function of the mitochondria and the chloroplast?
\(C\)
→ 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.
\(\Rightarrow C\)
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.
→ 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.
According to the fluid mosaic model of the cell membrane
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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.
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.
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.
→ 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.
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.
→ 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.
Which of the following correctly explains why a scanning electron microscope is advantageous in studying living cells?
\(A\)
→ 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.
\(\Rightarrow A\)
Which of the following technologies was pivotal in discovering the internal structure of organelles such as mitochondria and the endoplasmic reticulum?
\(B\)
→ 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.
\(\Rightarrow B\)
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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.
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.
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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.
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.
Which of the following accurately explains why prokaryotic cells are generally smaller than eukaryotic cells?
\(B\)
→ 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.
\(\Rightarrow B\)
The diagram below shows the process of phagocytosis. This process is vital for immunity against extracellular infections.
What is happening at position 3 ?
\(A\)
→ 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.
\(\Rightarrow A\)
Monoclonal antibodies attaching to antigens on a cancer cell are shown in the diagram below.
Monoclonal antibodies
\(B\)
→ 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.
\(\Rightarrow B\)
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
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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.
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.
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.
→ 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.
The primary structure of a protein is important because it
\(C\)
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.
\(\Rightarrow C\)
Neutrophils
\(B\)
→ 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.
\(\Rightarrow B\)
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.
→ 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.