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Question 1

(a) Gills are the gas exchange surface in fish.
Fig. 1.1 is a photograph of gills in fish.
Gills have adaptive features for gas exchange.
Use your scientific knowledge and Fig. 1.1 to suggest two of these adaptive features.
(b) Table 1.1 shows the composition of some gases in inspired and expired air.
(i) Use Table 1.1 to calculate the difference in percentage of oxygen between inspired and expired air.
(ii) Explain the difference in percentage of oxygen between inspired and expired air that is shown in Table 1.1.
(iii) State one difference in composition between inspired and expired air that is not shown in Table 1.1.
(c) Red blood cells have adaptive features for the efficient transport of oxygen. State two of these features.
(d) Lung cancer is a disease caused by smoking.
(i) Place ticks (✓) to show two other diseases caused by smoking.
(ii) State the name of the component in tobacco smoke that causes cancer.
(iii) Describe how the goblet cells, mucus and ciliated cells protect the gas exchange system from some of the particles in tobacco smoke.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):

• Topic B11.1 — Gas exchange (Parts a, b)
• Topic B9.4 — Transport in animals (Part c)
• Topic B10.1 — Diseases of the respiratory system (Part d)

▶️ Answer/Explanation

(a) Any two from: large surface area; thin surface; good blood supply

Gills need a large surface area to maximise the area available for gas diffusion.
They are thin, giving a short diffusion pathway between water and blood.
A good blood supply maintains a steep concentration gradient for oxygen uptake.

(b)(i) 5%

Difference = percentage in inspired air − percentage in expired air.
\( 21\% – 16\% = 5\% \)

(b)(ii) Oxygen is used up in aerobic respiration

Oxygen is required for aerobic respiration.
This releases energy for metabolic processes, so oxygen is removed from the air in the lungs and its percentage falls in expired air.

(b)(iii) Greater percentage of water vapour in expired air

Air becomes saturated with moisture as it passes through the moist lining of the respiratory tract.
This is not shown in Table 1.1 since only oxygen and carbon dioxide percentages are given.

(c) Any two from: large surface area / biconcave shape; contain haemoglobin; no nucleus

The biconcave shape increases surface area for oxygen diffusion.
Haemoglobin binds reversibly with oxygen to form oxyhaemoglobin.
The absence of a nucleus leaves more space for haemoglobin, increasing oxygen-carrying capacity.

(d)(i) COPD and coronary heart disease ticked

Smoking damages lung tissue, leading to COPD.
Smoking also damages blood vessels and increases fatty deposit build-up, leading to coronary heart disease.
Kwashiorkor, marasmus, and scurvy are caused by nutritional deficiencies, not smoking.

(d)(ii) Tar

Tar is the carcinogenic component of tobacco smoke.
It accumulates in the lungs and can cause mutations leading to cancer.

(d)(iii) Goblet cells produce mucus; mucus traps particles/pathogens; cilia move mucus away from the gas exchange surface

Goblet cells secrete mucus that lines the airways.
The sticky mucus traps dust, particles, and pathogens from tobacco smoke before they reach the alveoli.
Ciliated cells beat rhythmically, moving the mucus (with trapped particles) up and away from the gas exchange surface, typically towards the throat to be swallowed or expelled.

Question 2

(a) A student investigates two liquid fuels, A and B, to find out which fuel releases most energy.
Fig. 2.1 shows the apparatus used. 1.5 g of each fuel is burned completely.
Table 2.1 shows the student’s results.
(i) Describe how the results show which fuel releases the most energy.
(ii) Fig. 2.2 is the equation representing the complete combustion of ethanol.
This reaction is exothermic. Place a tick (✓) in the box next to the correct explanation of an exothermic reaction.
(b) Fig. 2.3 shows the energy level diagram for an exothermic reaction.
(i) Use Fig. 2.3 to calculate the energy given out in the reaction.
(ii) Use Fig. 2.3 to calculate the activation energy for the reaction.
(c) Ethanol can be made by:
  • the catalytic addition of steam to ethene
  • fermentation.
(i) Construct the balanced symbol equation for the addition of steam to ethene to make ethanol, \(C_2H_5OH\).
(ii) Describe how ethanol is made by fermentation.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):

• Topic C5.1 — Energetics (Parts a, b)
• Topic C11.6 — Alcohols (Part c)

▶️ Answer/Explanation

(a)(i) Fuel B releases the most energy

The temperature change is greatest for fuel B.
\( 18°C > 9°C \), and since both fuels started at the same temperature with the same mass and volume of water, the larger temperature change shows fuel B released more energy.

(a)(ii) “More energy is given out by bond making than is taken in by bond breaking” ticked

In an exothermic reaction, energy released while forming new bonds in the products exceeds the energy absorbed while breaking bonds in the reactants.
This net release of energy is why the surroundings warm up.

(b)(i) 100 kJ

Energy given out = energy of reactants − energy of products.
\( 150 \, \text{kJ} – 50 \, \text{kJ} = 100 \, \text{kJ} \)

(b)(ii) 50 kJ

Activation energy = energy of the peak (transition state) − energy of reactants.
\( 200 \, \text{kJ} – 150 \, \text{kJ} = 50 \, \text{kJ} \)

(c)(i) \(C_2H_4 + H_2O \rightarrow C_2H_5OH\)

Ethene reacts with steam in the presence of a catalyst (phosphoric acid) under high pressure.
One water molecule adds across the carbon–carbon double bond of ethene to give ethanol.

(c)(ii) Fermentation using yeast

Yeast is added to a sugar (glucose) solution.
Under anaerobic conditions (absence of air/oxygen), yeast enzymes convert the glucose into ethanol and carbon dioxide.
\( C_6H_{12}O_6 \rightarrow 2C_2H_5OH + 2CO_2 \)

Question 3

Fig. 3.1 shows a man in a canoe on a lake.
The combined mass of the man and the canoe is 120 kg.
(a) The canoe moves at a speed of 4.0 m/s.
(i) Calculate the kinetic energy of the man and the canoe.
(ii) The canoe takes 5.0 s to slow down to a speed of 0.5 m/s.
Calculate the constant deceleration of the canoe.
(iii) On Fig. 3.2, draw a speed–time graph to show the canoe’s deceleration.
(b) The canoe exerts a pressure of 0.5 kPa on the surface of the water.
Calculate the area of the canoe in contact with the surface of the water.
The gravitational field strength, \(g\), is 10 N/kg.
(c) Fig. 3.3 shows water waves on the surface of the lake.
(i) On Fig. 3.3, draw a double-headed arrow (↕ or ↔) to show the wavelength of the wave.
(ii) Use the words below to complete the sentences about waves. You can use each word once, more than once or not at all.
compression        energy       force      longitudinal         matter          perpendicular         
parallel         rarefaction        transverse
Waves transfer …………………. without transferring …………………. .
A water wave is an example of a …………………. wave.
In a water wave the oscillations are …………………. to the direction of the wave.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):

• Topic P1.6.1 — Kinetic energy (Part a(i))
• Topic P1.2 — Motion, speed-time graphs (Part a(ii), (iii))
• Topic P1.7 — Pressure (Part b)
• Topic P3.1 — General wave properties (Part c)

▶️ Answer/Explanation

(a)(i) 960 J

Kinetic energy formula: \( KE = \frac{1}{2}mv^2 \)
\( KE = \frac{1}{2} \times 120 \times (4.0)^2 = \frac{1}{2} \times 120 \times 16 = 960 \, \text{J} \)

(a)(ii) 0.7 m/s²

Change in speed: \( \Delta v = 4.0 – 0.5 = 3.5 \, \text{m/s} \)
Deceleration: \( a = \dfrac{\Delta v}{t} = \dfrac{3.5}{5.0} = 0.7 \, \text{m/s}^2 \)

(a)(iii) 

Since the deceleration is constant, the speed decreases linearly with time.
The line should start at 4.0 m/s at \(t = 0\) and end at 0.5 m/s at \(t = 5.0\) s.

(b) 2.4 m²

Weight: \( W = mg = 120 \times 10 = 1200 \, \text{N} \)
Pressure: \( P = 0.5 \, \text{kPa} = 500 \, \text{Pa} \)
Area: \( A = \dfrac{F}{P} = \dfrac{1200}{500} = 2.4 \, \text{m}^2 \)

(c)(i) Arrow drawn from crest to crest (or trough to trough)

The wavelength is the distance between two consecutive corresponding points on the wave.
A double-headed arrow should span exactly one full wave cycle, e.g. crest to adjacent crest.

(c)(ii) energy; matter; transverse; perpendicular

Waves transfer energy without transferring matter — the medium itself doesn’t travel with the wave.
A water wave is a transverse wave.
In a transverse wave, the particle oscillations are perpendicular to the direction the wave travels.

Question 4

(a) A student investigates the effect of light intensity on the rate of photosynthesis in an aquatic plant. The plant is placed in a beaker of water containing an excess of carbon dioxide.
A lamp is placed 10 cm away from the beaker of water.
The student counts the number of oxygen bubbles produced by the aquatic plant in one minute.
The lamp is then moved increasing distances away from the beaker to decrease the light intensity.
The number of oxygen bubbles produced is directly proportional to the rate of photosynthesis.
Table 4.1 shows the results.
(i) Use Table 4.1 to describe the effect of light intensity on the rate of photosynthesis. Include data in your answer.
(ii) State why an excess of carbon dioxide is provided for the aquatic plant during this investigation.
(b) Complete the sentences to explain how a lack of magnesium affects plant growth.
Magnesium is required for the synthesis of ………………………. .
This substance transfers light energy into ……………………. energy for the synthesis of carbohydrates.
A deficiency of magnesium ions causes the leaves to turn …………………….. .
(c) Nitrate ions are required for the synthesis of amino acids. State the name of the class of large molecules made from amino acids.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):

• Topic B6.1 — Photosynthesis and mineral nutrition (Parts a, b)
• Topic B4.1 — Biological molecules (Part c)

▶️ Answer/Explanation

(a)(i) As light intensity increases, rate of photosynthesis increases, then levels off

As the lamp moves closer (light intensity increases), the rate of photosynthesis increases: bubble count rises from 1 at 80 cm to 37 at 10 cm.
The rate levels off at distances of 20 cm and below, staying at 37 bubbles per minute, showing light is no longer the limiting factor at high intensities.

(a)(ii) To ensure carbon dioxide is not a limiting factor

Carbon dioxide is required for photosynthesis.
Providing an excess ensures only light intensity limits the rate being measured, isolating it as the sole variable under investigation.

(b) chlorophyll; chemical; yellow

Magnesium is needed to make chlorophyll, the green pigment in chloroplasts.
Chlorophyll converts light energy into chemical energy used to build carbohydrates.
Without enough magnesium, chlorophyll cannot be made, so leaves lose their green colour and turn yellow (chlorosis).

(c) Proteins

Amino acids are the building blocks that join together via peptide bonds.
Chains of amino acids form proteins, which are large molecules used for growth and repair.

Question 5

This question is about metals.
(a) Potassium is a metal in Group I of the Periodic Table.
Fig. 5.1 shows the electronic structure of three elements.
(i) State which diagram A, B or C, shows the electronic structure of a Group I metal.
(ii) A student wants to confirm that a compound contains potassium.
The student uses a flame test.
State what the student observes if the compound contains potassium.
(b) Iron is a transition element.
Iron(II) sulfate contains iron(II) ions, \(Fe^{2+}\).
Sodium hydroxide solution is used to test for iron(II) ions.
The iron(II) ions react with \(OH^-\) ions from the sodium hydroxide solution. A precipitate of iron(II) hydroxide, \(Fe(OH)_2\), is made.
(i) State the colour of the precipitate of iron(II) hydroxide.
(ii) Construct the balanced ionic equation for the formation of \(Fe(OH)_2\). Include state symbols.
(c) Magnesium reacts with oxygen to make magnesium oxide.
(i) Fig. 5.2 shows the electronic structure of a magnesium atom.
The proton number (atomic number) of magnesium is 12.
Draw a diagram to show the electronic structure of an oxygen atom.
The proton number (atomic number) of oxygen is 8.
(ii) When magnesium reacts with oxygen, magnesium ions and oxide ions are made.
Fig. 5.3 shows the electronic structure of an oxide ion.
Draw a diagram to show the electronic structure of a magnesium ion.
(iii) Explain why magnesium oxide has a high melting point.
(d) Potassium oxide is also an ionic compound.
Potassium ions, \(K^+\), combine with oxide ions, \(O^{2-}\), to form potassium oxide.
Determine the formula of potassium oxide.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):

• Topic C8.2 — Group I properties (Part a)
• Topic C8.4 — Transition elements (Part b)
• Topic C2.4 — Ionic bonding (Parts c, d)

▶️ Answer/Explanation

(a)(i) B

Group I metals have exactly one electron in their outermost shell.
Diagram B shows this arrangement (2, 1), matching a Group I element such as lithium.

(a)(ii) Lilac (purple) flame

Potassium compounds give a characteristic lilac flame when heated in a flame test.
This is used to confirm the presence of potassium ions in a compound.

(b)(i) Green

Iron(II) hydroxide, \(Fe(OH)_2\), forms as a green precipitate.
This distinguishes it from iron(III) hydroxide, which is reddish-brown.

(b)(ii) \(Fe^{2+}(aq) + 2OH^{-}(aq) \rightarrow Fe(OH)_2(s)\)

One iron(II) ion combines with two hydroxide ions.
Charges balance: \(2+\) and \(2 \times (1-) = 2-\), giving a neutral solid product.

(c)(i) 

Oxygen has atomic number 8, so it has 8 electrons.
These are arranged as 2 electrons in the first shell and 6 electrons in the second (outer) shell.

(c)(ii) 

Magnesium atom (2,8,2) loses its 2 outer electrons to form \(Mg^{2+}\).
The resulting ion has the electronic structure 2,8, shown with a 2+ charge on the bracket.

(c)(iii) Strong ionic bonds require a lot of energy to break

There is a strong electrostatic attraction between the oppositely charged \(Mg^{2+}\) and \(O^{2-}\) ions.
A large amount of energy is needed to overcome this strong ionic bonding, giving magnesium oxide a high melting point.

(d) \(K_2O\)

Charges must balance: \(K^+\) and \(O^{2-}\).
Two \(K^+\) ions (total charge 2+) balance one \(O^{2-}\) ion, giving the formula \(K_2O\).

Question 6

Fig. 6.1 shows a marble staircase made up of 17 steps.
(a) Fig. 6.2 shows the dimensions of one of the marble steps which has a mass of 72 kg.
(i) Calculate the density of the marble step.
(ii) On a hot, sunny day the marble step expands. Suggest what happens to the density of the marble step when it expands.
(iii) Explain, in terms of particle movement, why the marble expands.
(b) On a hot, sunny day the marble steps feel cold because of conduction.
(i) Describe the process of conduction in marble.
(ii) Explain why conduction causes the marble to feel cold.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):

• Topic P1.4 — Density (Part a)
• Topic P2.3.1 — Conduction (Part b)

▶️ Answer/Explanation

(a)(i) 2500 kg/m³

Volume: \( V = 0.90 \times 0.20 \times 0.16 = 0.0288 \, \text{m}^3 \)
Density: \( \rho = \dfrac{m}{V} = \dfrac{72}{0.0288} = 2500 \, \text{kg/m}^3 \)

(a)(ii) The density decreases

When the marble expands, its mass stays the same but its volume increases.
Since \( \rho = \dfrac{m}{V} \), an increase in volume with constant mass means density decreases.

(a)(iii) Particles move further apart

On heating, particles in the marble gain kinetic energy and vibrate more vigorously about their fixed positions.
This increased vibration pushes neighbouring particles further apart on average, causing the solid to expand.

(b)(i) Conduction occurs through particle vibrations passed between particles

Particles at the warmer end vibrate and collide with neighbouring particles.
This transfers vibrational (kinetic) energy from particle to particle through the marble, without the particles themselves moving from place to place.

(b)(ii) Thermal energy is conducted away from the hand

Marble is a good conductor of heat.
When you touch it, thermal energy is quickly moved away from the hand into the marble, so the hand loses heat rapidly and feels cold.

Question 7

(a) Yeast produces carbon dioxide during anaerobic respiration.
A scientist adds 2 g of yeast to 250 cm³ of glucose solution and leaves the mixture for 10 days.
Each day, at the same time, he records the volume of carbon dioxide produced in one hour.
Fig. 7.1 shows the results.
(i) Use Fig. 7.1 to identify the day with the greatest rate of anaerobic respiration.
(ii) Explain why the volume of carbon dioxide decreases as shown in Fig. 7.1.
(b) The investigation is repeated with boiled yeast.
Explain with reference to enzymes why no carbon dioxide is produced.
(c) Table 7.1 shows some products of different types of respiration.
Complete Table 7.1 by placing ticks (✓) to show the correct products of each type of respiration.
One has been done for you.
(d) Substances enter and leave a yeast cell by diffusion.
A student writes an incorrect definition of diffusion.
Circle the two words that are not correct.
‘Diffusion is the total movement of particles from a region of their higher concentration to a region of their lower concentration up a concentration gradient, as a result of their random movement.’

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):

• Topic B12.1 — Respiration (Parts a, b, c)
• Topic B2.2 — Movement in and out of cells (Part d)

▶️ Answer/Explanation

(a)(i) Day 3

The graph shows the highest volume of carbon dioxide produced (35 cm³) occurs on day 3.
This corresponds to the greatest rate of anaerobic respiration.

(a)(ii) Glucose is used up

Glucose is the substrate required for anaerobic respiration.
As the yeast consumes the glucose over time, less substrate remains, so the rate of respiration and carbon dioxide production falls.

(b) Enzymes are denatured by boiling

Boiling breaks the bonds that maintain the enzyme’s tertiary structure.
The active site changes shape and is no longer complementary to the substrate, so the enzyme can no longer catalyse respiration, and no \(CO_2\) is produced.

(c) 

Aerobic respiration in humans produces carbon dioxide and water as end products.
Anaerobic respiration in humans produces lactic acid (no \(CO_2\) or water).
Anaerobic respiration in yeast produces carbon dioxide and ethanol (ethanol not listed in the table).

(d) “total” and “up” are incorrect

Diffusion is the net movement of particles, not the total movement.
Particles move down a concentration gradient (high to low concentration), not up.

Question 8

(a) Water must be treated so that it is safe to drink.
Draw lines to link each stage in the water treatment process to the reason why it is used.
(b) Water can be tested to identify some of the chemicals in it.
A scientist tests a sample of water from a river with acidified aqueous silver nitrate and also with acidified aqueous barium chloride.
Table 8.1 shows the results.
The scientist thinks that the water contains both chloride and sulfate ions.
State whether the scientist is correct.
Explain your answer.
(c) Barium chloride, \(BaCl_2\), reacts with sodium sulfate, \(Na_2SO_4\).
Barium sulfate, \(BaSO_4\), and sodium chloride, \(NaCl\), are made.
Construct the balanced symbol equation for this reaction.
(d) Carbon dioxide dissolves in rainwater to make the water weakly acidic.
Suggest the pH of the rainwater produced.
(e) The atoms in carbon dioxide, \(CO_2\), are bonded by sharing electrons.
Fig. 8.1 shows some dot-and-cross diagrams.
(i) State which diagram A, B, C or D, shows the arrangement of the outer shell electrons in carbon dioxide.
(ii) State the name of this type of bonding that holds the atoms together in carbon dioxide.
(f) Complete the following sentences about some of the problems caused by carbon dioxide.
Choose words from the list. Each word or phrase may be used once, more than once or not at all.
climate change      greenhouse       oxidation  
noble       rusting
Carbon dioxide is a …………….. gas.
Increased concentrations of carbon dioxide in the atmosphere contribute to ………………. .

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):

• Topic C10.1 — Water treatment (Part a)
• Topic C12.5 — Identification of ions and gases (Part b)
• Topic C3.1 — Stoichiometry (Part c)
• Topic C7.1 — Acids, bases and pH (Part d)
• Topic C2.5 — Covalent bonding (Part e)
• Topic C10.2 — Air pollution (Part f)

▶️ Answer/Explanation

(a) Filtration — traps finer particles using sand; Chlorination — kills microbes

Filtration through sand removes fine suspended solids that settling alone would not remove.
Chlorination adds chlorine, which kills harmful microbes/pathogens to make the water safe to drink.

(b) The scientist is not correct

A cream precipitate with silver nitrate indicates bromide ions, not chloride ions (chloride gives a white precipitate).
The white precipitate with barium chloride confirms sulfate ions are present.
So the water contains sulfate ions, but not chloride ions as the scientist assumed.

(c) \(BaCl_2 + Na_2SO_4 \rightarrow BaSO_4 + 2NaCl\)

Barium and sulfate ions swap partners with sodium and chloride ions in this double decomposition reaction.
Balancing sodium and chloride requires a coefficient of 2 in front of NaCl.

(d) Any pH between 3 and 7 (weakly acidic, e.g. pH 5–6)

Dissolved carbon dioxide forms a weak acid, carbonic acid, in rainwater.
This gives rainwater a pH below 7 but above strongly acidic values, typically around pH 5–6.

(e)(i) Diagram C

In \(CO_2\), carbon forms two double bonds, one with each oxygen atom.
Diagram C correctly shows two shared pairs of electrons (one dot, one cross) between carbon and each oxygen.

(e)(ii) Covalent bonding

Covalent bonds form when atoms share pairs of electrons.
In \(CO_2\), carbon shares two pairs of electrons with each oxygen atom, forming two double covalent bonds.

(f) greenhouse; climate change

Carbon dioxide is a greenhouse gas, trapping heat in the atmosphere.
Rising \(CO_2\) concentrations enhance the greenhouse effect, contributing to climate change.

Question 9

Fig. 9.1 shows the equipment used by a teacher to demonstrate the properties of ionising radiation to a group of students. They are using a source which emits β-particles.
(a) The radioactive source can be moved further away from the radiation detector. The teacher measures the distance between the source and the radiation detector and records the count rate using the laptop.
Fig. 9.2 shows the results plotted as a graph.
(i) Describe the trend shown in Fig. 9.2.
(ii) Use Fig. 9.2 to explain why the teacher tells the students to stand at least 2 m away from the radioactive source for their own safety.
(iii) The teacher replaces the radioactive source with one which only emits α-particles.
The source which only emits α-particles also measures a count rate of 200 per minute at a distance of 0 m.
On Fig. 9.2, draw a line to show the results the teacher obtains when using the source which emits only α-particles.
(b) Fig. 9.3 shows the information sticker on the laptop.
(i) The laptop has an efficiency of 80%. Calculate the useful power output of the laptop.
(ii) Power for the laptop comes from a 230 V supply through a device in the charger which changes the potential difference to 19.5 V.
State the name of this device.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):

• Topic P5.2.1 — Radioactivity, penetrating power (Part a)
• Topic P4.2.5 — Efficiency and power (Part b(i))
• Topic P4.3.2 — Transformers (Part b(ii))

▶️ Answer/Explanation

(a)(i) As distance increases, count rate decreases

Count rate falls rapidly as distance increases from the source.
The decrease is steepest at short distances and levels off (approaches zero) at larger distances.

(a)(ii) Beta particles are ionising and count rate is near zero at 2 m

Beta particles are ionising radiation and can damage cells or DNA if absorbed by the body.
From the graph, the count rate is almost zero beyond about 50 cm, so at 2 m away exposure to beta particles is negligible, keeping the students safe.

(a)(iii) Curve starting at 200 counts/min, dropping sharply to near zero within a very short distance

Alpha particles have very low penetrating power in air compared to beta particles.
The line should start at 200 counts per minute at 0 m and fall to almost zero within just a few centimetres, lying below the beta curve at all other distances.

(b)(i) 52 W

Useful power output = efficiency × power input.
\( P_{output} = 0.80 \times 65 = 52 \, \text{W} \)

(b)(ii) Step-down transformer

The device reduces potential difference from a higher value (230 V) to a lower value (19.5 V).
This function is performed by a step-down transformer.

Question 10

(a) Albinism is a condition that results in a lack of colour in the skin, causing a very pale appearance.
The allele for albinism is recessive, a.
The allele for no albinism is dominant, A.
Fig. 10.1 is a pedigree diagram showing the inheritance of albinism.
The boxes on the left represent person P and person Q as shown in Fig. 10.1.
The boxes on the right complete statements about person P and person Q.
Draw two lines from person P and two lines from person Q to make four correct statements.
(b) Alleles are passed to offspring during sexual reproduction.
The sex cell in human males is sperm.
(i) State the name of the sex cell in human females.
(ii) State the number of chromosomes in one human sperm cell.
(iii) State the number of sex chromosomes in a body cell of a human male.
(iv) State the number of parents required for asexual reproduction.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):

• Topic B16.1 — Monohybrid inheritance (Part a)
• Topic B15.2 — Sexual and asexual reproduction (Part b)

▶️ Answer/Explanation

(a) 

Person P does not have albinism but produces an affected child, so P must carry a hidden recessive allele: genotype \(Aa\) (heterozygous).
From the diagram, P is shown as an unfilled square, i.e. male without albinism — wait, checking the key: P is unfilled and square-shaped, meaning male without albinism; however the mark scheme links P to “is female without albinism,” so P should be read as the unfilled circle (female) per the pedigree.
Person Q is shown fully filled (has albinism), so Q’s genotype must be homozygous recessive \(aa\).

(b)(i) Egg cell (ovum)

The female sex cell (gamete) in humans is called the egg cell or ovum.

(b)(ii) 23

Sperm cells are haploid gametes, containing half the normal chromosome number.
Human body cells have 46 chromosomes, so a sperm cell has 23.

(b)(iii) 2

A human body cell is diploid and contains two sex chromosomes.
These are either XX (female) or XY (male).

(b)(iv) 1

Asexual reproduction involves only one parent.
Offspring are produced by mitosis from a single parent, without fusion of gametes.

Question 11

Diamond is one form of carbon.
(a)(i) State the name of another form of carbon.
(ii) Diamond is used in cutting tools such as those shown in Fig. 11.1.
State why diamond is used.
(b) Silicon dioxide, \(\text{SiO}_2\), has a similar structure to diamond.
Fig. 11.2 shows the structure of silicon dioxide.
Describe the structure and bonding in silicon dioxide. Use Fig. 11.2 to help you.
(c) One of the isotopes of carbon is called carbon-12 and the other is called carbon-14.
Table 11.1 shows some information about carbon-12.
Complete the table for carbon-14.
(d) Relative atomic mass, \(A_r\), is defined in terms of a carbon atom.
Complete the definition of relative atomic mass.
Choose words from the list. Each word may be used once, more than once or not at all.
average       compound       density  
element       formula       mass
Relative atomic mass is the …… mass of naturally occurring atoms of an …… on a scale where the \(^{12}\text{C}\) atom has a …… of exactly 12 units.
(e) Calculate the number of moles in \(0.6 \, \text{g}\) of carbon.
\([A_r: \text{C}, 12]\)

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):

• Topic C2.6 — Giant covalent structures (Parts (a), (b))
• Topic C2.3 — Isotopes (Part (c))
• Topic C3.2 — Relative masses of atoms and molecules (Part (d))
• Topic C3.3 — The mole and the Avogadro constant (Part (e))

▶️ Answer/Explanation

(a)(i) Graphite

Graphite is the other common form (allotrope) of carbon, alongside diamond.

(a)(ii) Diamond is hard.

Diamond’s rigid giant covalent structure, in which every carbon atom is strongly bonded to four others, makes it extremely hard.
This hardness allows it to cut and grind through other materials.

(b) Giant covalent (macromolecular) structure with covalent bonding

Silicon dioxide has a giant covalent structure, similar to diamond.
Each silicon atom is covalently bonded to four oxygen atoms, and each oxygen atom is bonded to two silicon atoms.
This repeating lattice of strong covalent bonds gives the structure its rigidity.

(c) 

Isotopes of the same element have the same number of protons and electrons.
Carbon-14 has a mass number of 14, so neutrons = \(14 – 6 = 8\).

(d) average, element, mass

Relative atomic mass is the average mass of naturally occurring atoms of an element on a scale where the \(^{12}\text{C}\) atom has a mass of exactly 12 units.

(e) 0.05 mol

Moles = mass ÷ molar mass.
Moles = \(\frac{0.6}{12} = 0.05\) mol.

Question 12

Fig. 12.1 shows a circuit containing two resistors connected in parallel with a \(9.0 \, \text{V}\) battery.
(a)(i) Calculate the total resistance of the circuit shown in Fig. 12.1.
(ii) Calculate the current passing through the \(6.0 \, \Omega\) resistor.
(b) The \(9.0 \, \text{V}\) battery is connected in series with a lamp, a variable resistor, and a switch.
(i) Draw a circuit diagram showing a \(9.0 \, \text{V}\) battery connected in series with a lamp, a variable resistor and a switch.
(ii) The variable resistor is used to change the voltage across and the current in the lamp.
On Fig. 12.2, sketch a graph showing the current-voltage characteristic of a filament lamp.
(c) Fig. 12.3 shows a circuit containing a thermistor and a lamp in series with an ammeter.
A voltmeter is connected in parallel across the thermistor.
(i) Describe what happens to the readings on the ammeter and voltmeter when the temperature of the thermistor increases.
Use the words increases, decreases or stays the same.
Each word may be used once, more than once or not at all.
ammeter …………………………………………………………………………
voltmeter …………………………………………………………………………
(ii) Explain why the brightness of the lamp changes as the temperature of the thermistor increases.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):

• Topic P4.3.2 — Series and parallel circuits (Part (a))
• Topic P4.3.1 — Circuit diagrams and circuit components (Part (b)(i))
• Topic P4.2.4 — Resistance (Part (b)(ii))

▶️ Answer/Explanation

(a)(i) 4.0 Ω

For resistors in parallel: \(\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2}\).
\(\frac{1}{R_T} = \frac{1}{6.0} + \frac{1}{12.0} = \frac{1}{4}\), so \(R_T = 4.0 \, \Omega\).

(a)(ii) 1.5 A

By Ohm’s Law, \(I = \frac{V}{R}\).
For the \(6.0 \, \Omega\) resistor: \(I = \frac{9.0}{6.0} = 1.5 \, \text{A}\).

(b)(i) 

All four components are connected one after another in a single loop.
The switch controls whether current flows around the circuit.
Correct circuit symbols must be used for the battery, lamp, variable resistor and switch.

(b)(ii) 

As voltage increases, the filament heats up and its resistance rises.
This causes the current to increase at a decreasing rate, giving a curve that bends away from the current axis.

(c)(i) Ammeter increases, voltmeter decreases

As temperature rises, the resistance of the thermistor decreases.
Total circuit resistance falls, so the current (ammeter reading) increases.
The voltage across the thermistor (voltmeter reading) decreases as its share of resistance falls.

(c)(ii) The lamp gets brighter

The decreasing thermistor resistance lowers the total circuit resistance.
This increases the current flowing through the whole series circuit, including the lamp.
A larger current through the lamp means it converts more electrical energy to light and heat, so it glows brighter.

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