Question 1
State two other main components of blood.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic B2.1 — Cell structure (Parts (a)(i)–(ii))
• Topic B9.4 — Blood (Parts (a)(iii), (b))
• Topic B9.3 — Blood vessels (Part (c))
▶️ Answer/Explanation
(a)(i) B – root hair cell; E – sperm cell
Cell B has a long thin projection that increases surface area for water and mineral ion uptake.
Cell E has a tail (flagellum) for movement and a streamlined head, adapting it to swim towards and fertilise an egg.
(a)(ii) Cilia allow removal of mucus
Cell A has cilia on its surface.
These beat in a coordinated way to sweep mucus (and trapped particles) away from the trachea/bronchi/gas exchange system.
(a)(iii) Biconcave shape; contains haemoglobin
Cell D (red blood cell) has a biconcave shape, increasing surface area for oxygen diffusion.
It contains haemoglobin, which binds reversibly to oxygen.
It also has no nucleus, allowing more space for haemoglobin.
(b) White blood cells; platelets; plasma
Blood is made up of red blood cells, white blood cells, platelets and plasma.
Any two of white blood cells, platelets or plasma are accepted.
(c)(i) To withstand high pressure
Arteries carry blood at high pressure away from the heart.
The thick elastic wall stretches and recoils to withstand this pressure without bursting.
(c)(ii) To prevent backflow of blood
Blood in veins is at low pressure, so valves stop blood flowing backwards, especially against gravity.
(c)(iii) Short diffusion distance
Capillary walls are one cell thick.
This gives a short diffusion distance for oxygen, glucose, and waste products between blood and tissues.
Question 2
Poly(ethene) is a polymer.
Describe what is meant by a saturated hydrocarbon.
Fig. 2.2 shows the equation for the reaction.
Complete the equation in Fig. 2.2 by drawing the structure of the compound formed.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic C11.5 — Alkenes (Parts (a), (b), (d), (f))
• Topic C11.7 — Synthetic polymers (Part (c))
• Topic C11.4 — Alkanes (Part (e))
▶️ Answer/Explanation
(a) alkenes
Ethene, \( C_2H_4 \), has a carbon-carbon double bond.
This double bond is the defining feature of the alkene family.
(b) cracking
Cracking breaks down large hydrocarbon molecules from petroleum into smaller, more useful ones.
It uses heat and often a catalyst.
(c)(i) A long chain molecule made of many repeating small units
A polymer is a long-chain molecule.
It is built up from many small repeating units called monomers, joined by covalent bonds.
(c)(ii) Repeat units completed
Poly(chloroethene) repeat unit: \( -CHCl-CH_2- \).
Poly(tetrafluoroethene) repeat unit: \( -CF_2-CF_2- \).
Each repeat unit reflects the atoms attached to the carbon backbone in the original monomer.
(d) \( H_2 \)
Ethene reacts with hydrogen gas in a hydrogenation (addition) reaction.
This converts the C=C double bond into a single bond, forming ethane.
(e) Contains only single bonds between carbon atoms
A saturated hydrocarbon has only single C–C bonds.
It therefore contains the maximum possible number of hydrogen atoms.
(f) 1,2-dibromoethane, \( BrCH_2CH_2Br \)
The C=C double bond breaks, and one bromine atom adds to each carbon.
This addition reaction is used as a chemical test for unsaturation.
Question 3
The gravitational field strength, g, is 10 N/kg.
State the unit for your answer.
Fig. 3.2 shows the circuit that includes the electric motor.
Calculate the resistance of the motor.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic P1.3 — Mass and weight (Part (a)(i))
• Topic P1.6.2 — Work (Part (a)(ii))
• Topic P4.2.3 — Potential difference (Part (b)(i))
• Topic P4.2.4 — Resistance (Part (b)(ii))
▶️ Answer/Explanation
(a)(i) 1400 N
Weight is calculated using \( W = m \times g \).
\( W = 140 \times 10 = 1400 \, \text{N} \).
(a)(ii) 2100 J
Work done = force × distance moved in direction of force.
\( \text{Work} = 1400 \times 1.5 = 2100 \, \text{J} \).
The unit is the joule (J).
(b)(i) 11.5 V
Total supply voltage is 12 V, and the voltmeter reads 0.50 V across the variable resistor.
p.d. across motor \( = 12 – 0.50 = 11.5 \, \text{V} \).
(b)(ii) 1.25 Ω
Resistance is found using \( R = \dfrac{V}{I} \).
\( R = \dfrac{11.5}{9.20} = 1.25 \, \Omega \).
Question 4
Person E and person F are ……………………. by Tay-Sachs disease.
Person E and person F both have the genotype ……….. .
Person G has Tay-Sachs disease. They have the genotype ……….. .
Person G will have inherited one ……………………. allele from each parent.
The box on the left contains the term mitosis.
The boxes on the right contain some sentence endings.
Draw three lines from the word mitosis to the boxes on the right to make three correct sentences about mitosis.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic B16.3 — Genetic diagrams/pedigrees (Part (a))
• Topic B16.2 — Mitosis (Part (b))
• Topic B16.1 — Chromosomes, genes, mutation (Parts (c), (d))
▶️ Answer/Explanation
(a)(i) 4
Counting the unshaded (unaffected) male squares in the pedigree gives 4 unaffected males.
(a)(ii) unaffected; Tt; tt; recessive
E and F show no symptoms, so they are unaffected by Tay-Sachs disease.
Since their child G is affected (tt), E and F must each carry one recessive allele, so are Tt.
G has Tay-Sachs disease, so must be homozygous recessive, tt.
G inherits one recessive (t) allele from each parent.
(a)(iii) 0 (0%)
Both parents are TT, so they can only pass on the dominant T allele.
No child can inherit two t alleles, so the probability of Tay-Sachs disease is 0.
(b) Mitosis joins to: occurs after exact duplication of chromosomes; produces cells with diploid nuclei; produces nuclei with paired chromosomes
Mitosis follows DNA replication, so chromosomes are exactly duplicated beforehand.
It produces two genetically identical diploid daughter cells.
These cells contain the full set of paired (homologous) chromosomes.
(c) 46 (23 pairs)
Human body (somatic) cells are diploid, containing 23 pairs of chromosomes, giving 46 in total.
(d) mutation
A mutation is any change in the base sequence of a gene, or in the structure/number of chromosomes.
Question 5
Calculate the relative formula mass, \( M_r \), of ammonium sulfate.
[\( A_r \): H, 1; N, 14; O, 16; S, 32]
Ammonia is made by the Haber process.
Nitrogen, \( N_2 \), and hydrogen, \( H_2 \), are the starting materials.
Look at the equation for the reaction:
\[ N_2 + 3H_2 \rightleftharpoons 2NH_3 \]
You should include:
- the sources of nitrogen and hydrogen gas
- the conditions used.
Ammonium nitrate, \( NH_4NO_3 \), is made.
Look at the equation:
Show your working.
[\( A_r \): H, 1; N, 14; O, 16]
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic C7.1 — Acids and bases (Part (a))
• Topic C3.2 — Relative formula mass and moles (Parts (b), (c)(ii))
• Topic C5.1 — Haber process (Part (c)(i))
▶️ Answer/Explanation
(a)(i) pH increases
As alkali is added, \( H^+ \) ions in the acid are neutralised by \( OH^- \) ions.
The solution becomes less acidic, so the pH increases.
(a)(ii) salt
Neutralisation of an acid by an alkali always produces a salt and water.
(a)(iii) \( H_2SO_4 + 2KOH \rightarrow K_2SO_4 + 2H_2O \)
Sulfuric acid is diprotic, so it needs two moles of KOH for complete neutralisation.
The products are potassium sulfate and water, and the equation is balanced with 2 water molecules.
(a)(iv) \( H^+ \)
All acids release hydrogen ions, \( H^+ \), when dissolved in water.
(b) 132
\( M_r = (2 \times 14) + (8 \times 1) + 32 + (4 \times 16) \).
\( M_r = 28 + 8 + 32 + 64 = 132 \).
(c)(i) Nitrogen from air; hydrogen from natural gas; 200 atm, 450 °C, iron catalyst
Nitrogen gas is obtained from the air.
Hydrogen gas is obtained from natural gas (methane).
The reaction uses a pressure of about 200 atmospheres, a temperature of about 450 °C, and an iron catalyst.
(c)(ii) 240 kg
\( M_r(NH_3) = 17 \), \( M_r(NH_4NO_3) = 80 \).
The mole ratio of \( NH_3 : NH_4NO_3 \) is 1:1.
Mass \( = \dfrac{80}{17} \times 51 = 240 \, \text{kg} \).
Question 6
Calculate the speed of the water flow.
The lamp emits light with a wavelength of \( 4.0 \times 10^{-7} \, \text{m} \).
Calculate the frequency of the light.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic P1.6.3 — Sources of energy (Part (a)(i))
• Topic P1.6.1 — Kinetic energy (Part (a)(ii))
• Topic P4.5.2 — a.c. generator (Part (b))
• Topic P3.3 — Electromagnetic spectrum (Part (c))
▶️ Answer/Explanation
(a)(i) The Moon
Tides are caused mainly by the gravitational pull of the Moon on the Earth’s oceans.
(a)(ii) 1.8 m/s
Using \( KE = \tfrac{1}{2}mv^2 \), so \( v = \sqrt{\dfrac{2KE}{m}} \).
\( v = \sqrt{2 \times 1.62 / 1} = \sqrt{3.24} = 1.8 \, \text{m/s} \).
(b)(i) Arrow from N to S
Magnetic field lines run from the North pole to the South pole outside the magnet.
(b)(ii) Slip rings; maintain electrical contact while allowing rotation
X are the slip rings.
They rotate with the coil and stay in contact with fixed brushes, allowing current to flow to the external circuit without the wires twisting.
(b)(iii) Sinusoidal wave, constant amplitude and period
The output is a smooth sine wave because the induced e.m.f. depends on the rate of change of magnetic flux as the coil rotates at constant speed.
(c) \( 7.5 \times 10^{14} \, \text{Hz} \)
Using \( c = f\lambda \), so \( f = \dfrac{c}{\lambda} \).
\( f = \dfrac{3 \times 10^8}{4.0 \times 10^{-7}} = 7.5 \times 10^{14} \, \text{Hz} \).
Question 7
Describe the disadvantages to a plant in the wild of reproducing asexually.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic B15.3 — Pollination, fertilisation, germination (Parts (a), (b), (c), (e))
• Topic B15.1 — Sexual and asexual reproduction (Part (d))
▶️ Answer/Explanation
(a) Stigma
In wind-pollinated flowers, the stigma is feathery and hangs outside the flower to catch airborne pollen.
(b) Insect pollen larger/sticky, petals large/coloured/scented
Insect-pollinated pollen is larger and stickier, so it attaches to visiting insects.
Insect-pollinated petals are large, brightly coloured and often scented to attract insects, unlike the small, dull petals of wind-pollinated flowers.
(c) Ovule / ovary
The pollen tube grows down the style, delivering the male nucleus to the ovule in the ovary, where fertilisation occurs.
(d) No genetic variation; population vulnerable to disease/environmental change
Asexual reproduction produces genetically identical offspring (clones).
This means there is no variation, so the whole population could be wiped out by the same disease or environmental change.
(e) Water; oxygen; suitable temperature (any two)
Water rehydrates the seed and activates enzymes.
Oxygen is needed for aerobic respiration to release energy.
A suitable temperature is needed for enzymes to work effectively.
Question 8
Each of the atoms is an isotope of chlorine.
Explain why argon is very unreactive.
Use ideas about electronic structure.
Describe the bonding in a metal.
You may draw a diagram to help your answer.
Determine the formula of magnesium chloride.
Construct the balanced symbol equation for the reaction.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic C2.2 — Atomic structure (Parts (a), (c))
• Topic C2.3 — Isotopes (Part (b))
• Topic C2.7 — Metallic bonding (Part (d))
• Topic C2.4 — Ionic bonding (Part (e))
• Topic C8.3 — Reactivity of halogens (Part (f))
▶️ Answer/Explanation
(a) Argon proton number = 18; Magnesium electronic structure = 2.8.2
Argon’s proton (atomic) number is 18.
Magnesium has 12 electrons, arranged as 2 in the first shell, 8 in the second, and 2 in the outer shell.
(b)(i) Same proton number, different nucleon number
Isotopes are atoms of the same element with the same number of protons.
They have different numbers of neutrons, giving different nucleon numbers.
(b)(ii) Same electronic structure
Chemical properties depend on the number and arrangement of outer-shell electrons, which is identical for both isotopes.
(c) Full outer shell of electrons
Argon has a complete outer electron shell.
This makes it energetically stable, so it has no tendency to gain, lose or share electrons.
(d) Lattice of positive ions in a sea of delocalised electrons
Metal atoms lose their outer electrons to form a regular lattice of positive ions.
The delocalised electrons move freely throughout the structure.
Strong electrostatic attraction between the positive ions and the electron sea holds the metal together.
(e) \( MgCl_2 \)
The charges must balance: one \( Mg^{2+} \) ion needs two \( Cl^{-} \) ions.
(f) \( F_2 + 2NaCl \rightarrow Cl_2 + 2NaF \)
Fluorine is more reactive than chlorine, so it displaces chlorine from sodium chloride.
This forms chlorine gas and sodium fluoride.
Question 9
Fig. 9.3 shows a thin converging lens.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic P2.1.2 — Particle Model (Part (a))
• Topic P4.2.4 — Resistance (Part (b))
• Topic P3.2.2 — Lenses and refraction (Part (c))
▶️ Answer/Explanation
(a) Collisions with fast-moving, invisible air particles
Smoke particles are constantly hit by fast-moving air molecules.
These collisions are random and unequal, causing the smoke particles to change direction randomly.
(b) Resistance increases as p.d. increases, because the filament heats up
The graph curves towards the p.d. axis, showing a decreasing gradient.
This means resistance increases as p.d. increases.
As current increases, the filament heats up, and the increased vibration of ions impedes electron flow, raising resistance.
(c)(i) Ray diagram showing a real, inverted image
A ray parallel to the axis refracts through the principal focus.
A ray through the centre of the lens continues undeviated.
Where these rays cross (beyond the focal point) marks the real, inverted image.
(c)(ii) 1.63
Refractive index \( n = \dfrac{\sin i}{\sin r} \).
\( n = \dfrac{\sin 50°}{\sin 28°} \approx 1.63 \).
Question 10
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic B13.1 — Coordination and response (Parts (a), (b))
• Topic B13.3 — Homeostasis, blood glucose (Part (c))
• Topic B13.2 — Hormones (Part (d))
▶️ Answer/Explanation
(a) Coughing and sneezing ticked
Coughing and sneezing are automatic reflex actions.
Cycling, reading and talking are voluntary, requiring conscious control.
(b) Sensitivity
Sensitivity is the characteristic of living organisms that allows them to detect and respond to changes (stimuli) in their environment.
(c)(i) 60 minutes
The peak glucose concentration occurs at 20 minutes.
It returns to its starting value at around 80 minutes.
\( 80 – 20 = 60 \) minutes.
(c)(ii) Insulin release lowers blood glucose (negative feedback)
The rise in blood glucose is detected by the pancreas.
Insulin is released into the blood.
Insulin causes liver and muscle cells to convert glucose into glycogen for storage.
This causes blood glucose concentration to fall between 20–30 minutes.
(c)(iii) Negative feedback
The response corrects the deviation from the normal set point, bringing glucose concentration back down towards its starting level.
(d) Glucagon; adrenaline
Glucagon stimulates the conversion of glycogen back into glucose.
Adrenaline also raises blood glucose concentration as part of the “fight or flight” response.
Question 11
State which two experiments show this.
Increasing the concentration of the acid also affects the rate of reaction.
For each factor (temperature and concentration):
- describe how the rate of reaction changes
- explain why the rate of reaction changes, using ideas about particles.
Calculate the concentration of the dilute hydrochloric acid in g/dm³.
[\( A_r \): H, 1; Cl, 35.5]
Explain why. Use ideas about bond breaking and bond making.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic C6.2 — Rates of reaction (Parts (a), (b))
• Topic C3.2 — Moles and concentration (Part (c))
• Topic C5.1 — Exothermic/endothermic reactions (Part (d))
▶️ Answer/Explanation
(a) Experiments 2 and 3
Both use concentrated acid at 20 °C but different volumes (50 cm³ vs 100 cm³).
Since both give the same time (66 s), volume has no effect on rate.
(b) Temperature: rate increases; concentration: rate increases
As temperature increases, the rate of reaction increases (Experiment 4 vs 2: 32 s vs 66 s).
Particles have more kinetic energy, so collisions are more frequent and more energetic, exceeding activation energy more often.
As concentration increases, the rate of reaction increases (Experiment 2 vs 1: 66 s vs 131 s).
There are more particles per unit volume, so collisions between reacting particles occur more frequently.
(c) 3.65 g/dm³
Molar mass of HCl \( = 1 + 35.5 = 36.5 \, \text{g/mol} \).
Concentration \( = 0.1 \times 36.5 = 3.65 \, \text{g/dm}^3 \).
(d) Energy released in bond making exceeds energy absorbed in bond breaking
Breaking bonds in the reactants requires energy (endothermic).
Forming new bonds in the products releases energy (exothermic).
Since more energy is released on bond making than is absorbed on bond breaking, there is a net release of heat.
Question 12
The satellite orbits at a height of 2000 km above the surface of the Earth.
The satellite takes 125 minutes to complete one orbit.
The satellite travels at an average speed of 7.1 km/s.
This radiation includes α-particles, β-particles and γ-rays.
Use the correct nuclide notation to complete the decay equation for iodine-131.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic P1.5.1 — Effects of forces (Part (a)(i)–(ii))
• Topic P1.3 — Mass and weight (Part (a)(iii))
• Topic P1.2 — Motion (Part (b))
• Topic P5.2.3 — Radioactive decay (Part (c))
▶️ Answer/Explanation
(a)(i) \( 7.9 \times 10^6 \, \text{N} \)
Resultant force = thrust − weight.
\( 15.7 \times 10^6 – 7.8 \times 10^6 = 7.9 \times 10^6 \, \text{N} \) upwards.
(a)(ii) Accelerates upwards
Since thrust exceeds weight, there is a net upward force.
This causes the rocket to accelerate upwards, gaining speed.
(a)(iii) Gravitational field strength decreases with distance
Weight depends on gravitational field strength, which decreases as distance from Earth increases, so weight decreases.
(b) 6475 km
Time for one orbit \( = 125 \times 60 = 7500 \, \text{s} \).
Orbit circumference \( = 7.1 \times 7500 = 53\,250 \, \text{km} \).
Orbital radius \( = \dfrac{53\,250}{2\pi} \approx 8475 \, \text{km} \).
Earth’s radius \( = 8475 – 2000 = 6475 \, \text{km} \).
(c)(i) Alpha and beta particles
Alpha and beta particles are charged, so they experience a force in a magnetic field and are deflected.
Gamma rays are uncharged electromagnetic waves, so they are not deflected.
(c)(ii) \( ^{131}_{53}I \rightarrow \, ^{131}_{54}Xe + \, ^{0}_{-1}\beta \)
Beta decay converts a neutron into a proton, increasing the proton number by 1 while the nucleon number stays the same.