Question 1
(a)(i)-(iv): Subtopic: B15.4 Sexual reproduction in humans
(b)(i)-(iv): Subtopic: B15.4 Sexual reproduction in humans
(a) Fig. 1.1 is a photomicrograph of sperm surrounding an egg cell.
(i) Describe two visible differences between the sperm cells and the egg cell shown in Fig. 1.1.
▶️Answer/Explanation
1. Sperm are much smaller in size compared to the egg cell.
2. Sperm have an elongated shape with a tail (flagellum), while the egg cell is spherical.
Explanation: The egg cell is the largest human cell (about 0.1mm visible to naked eye), while sperm are among the smallest (about 0.05mm long). The sperm’s tail enables motility.
(ii) State one adaptive feature of sperm that is not visible in Fig. 1.1.
▶️Answer/Explanation
Presence of enzymes (e.g., acrosome contains digestive enzymes to penetrate egg)
Alternative: Mitochondria in mid-piece to provide energy for movement
(b) The fusion of the nuclei from a sperm and an egg is called fertilisation.
(i) State where fertilisation occurs in the female reproductive system.
▶️Answer/Explanation
Oviduct (Fallopian tube)
Note: Specifically occurs in the ampulla region of the oviduct
(ii) State two parts of the female reproductive system the sperm must pass through before fertilisation.
▶️Answer/Explanation
1. Vagina
2. Cervix
Alternative: Uterus (womb) is also acceptable
Explanation: Sperm travel through vagina → cervix → uterus → oviduct in that order
(iii) After fertilisation, the egg forms a layer that prevents more sperm entering the egg.
State the name of this adaptive feature.
▶️Answer/Explanation
Jelly coating (cortical reaction)
Explanation: This forms the fertilization membrane through cortical granules releasing chemicals that harden the zona pellucida
(iv) Complete the sentence to describe the nucleus in a gamete.
Gametes contain a …… nucleus containing a single set of …… chromosomes.
▶️Answer/Explanation
Gametes contain a haploid nucleus containing a single set of
unpaired chromosomes.
Explanation: Haploid (n) means half the normal chromosome number (23 in humans vs 46 in body cells)
Question 2
(a)-(b)(i)(ii): Subtopic: C2.2 Atomic structure and the Periodic Table
(c): Subtopic: C2.5 Simple molecules and covalent bonds
(d): Subtopic: C10.2 Air quality and climate
(e): Subtopic: C6.2 Rate of reaction
Atoms contain protons, neutrons and electrons.
(a) Complete Table 2.1 about protons, neutrons and electrons.
| relative charge | relative mass | location in an atom | |
|---|---|---|---|
| protons | …… | …… | in nucleus |
| neutrons | …… | 1 | …… |
| electrons | -1 | …… | …… |
▶️Answer/Explanation
| relative charge | relative mass | location in an atom | |
|---|---|---|---|
| protons | +1 | 1 | in nucleus |
| neutrons | 0 | 1 | in nucleus |
| electrons | -1 | 1/1840 (or ≈0) | in shells/orbits around nucleus |
Explanation: – Protons are positively charged with same mass as neutrons – Neutrons are neutral (no charge) – Electrons are negatively charged with negligible mass (about 1/1840th of a proton)
(b) Fig. 2.1 shows the structure of an atom of nitrogen.
(i) Write the electronic structure for a nitrogen atom.
▶️Answer/Explanation
2,5
Explanation: Nitrogen (atomic number 7) has electron arrangement: – First shell: 2 electrons – Second shell: 5 electrons Total: 2 + 5 = 7 electrons
(ii) Nitrogen is in Group V of the Periodic Table.
State how Fig. 2.1 shows that nitrogen is in Group V.
▶️Answer/Explanation
(Nitrogen has) five electrons in the outer shell
Explanation: Group number corresponds to the number of valence electrons. The diagram would show 5 electrons in the outermost shell.
(c) Nitrogen atoms bond together to form nitrogen molecules, N2. Draw a dot-and-cross diagram to show the bonding in a nitrogen molecule.
Show only the outer shell electrons.
▶️Answer/Explanation
N≡N (with 3 pairs of shared electrons)
Detailed diagram explanation: – Each N atom has 5 outer electrons – They share 3 electron pairs (triple bond) – Each N gets a full outer shell of 8 electrons (octet) – Represented as :N:::N: or with dots/crosses for each atom’s electrons
(d) Nitrogen is one of the gases found in clean air.
Complete Table 2.2 about the gases in clean air.
| gas | percentage (%) in clean air |
|---|---|
| carbon dioxide | 0.041 |
| oxygen | …… |
| nitrogen | …… |
| …… | varies |
▶️Answer/Explanation
| gas | percentage (%) in clean air |
|---|---|
| carbon dioxide | 0.041 |
| oxygen | 21 |
| nitrogen | 78 |
| noble gases (or water vapor) | varies |
Explanation: Dry clean air composition is approximately: – Nitrogen: 78% – Oxygen: 21% – CO2: 0.04% – Noble gases (Argon, Neon etc.): 0.96% – Water vapor varies with humidity
(e) Nitrogen monoxide gas, NO, is an air pollutant. A catalytic converter removes nitrogen monoxide from car exhaust gases.
Write a balanced symbol equation for this reaction.
▶️Answer/Explanation
2CO + 2NO → N2 + 2CO2
Alternative: 2NO → N2 + O2
Explanation: Catalytic converters use redox reactions: – CO is oxidized to CO2 – NO is reduced to N2 – Requires platinum/palladium/rhodium catalyst
Question 3
(a)(i)(ii): Subtopic: P1.6.1 Energy
(b)(i)-(iii): Subtopic: P3.4 Sound
(c): Subtopic: P2.1.2 Particle model
(a) A golfer swings her golf club to hit a stationary golf ball of mass 0.05 kg.
(i) State the kinetic energy of the golf ball before it is hit.
▶️Answer/Explanation
0 (J)
Explanation: The ball is stationary (speed = 0 m/s) before being hit, and kinetic energy = ½mv². Since velocity (v) is zero, kinetic energy is zero.
(ii) The speed of the golf ball immediately after it has been hit is 35 m/s. Calculate the kinetic energy of the golf ball when it is moving at 35 m/s.
▶️Answer/Explanation
31 (J)
Calculation:
KE = ½mv²
= ½ × 0.05 kg × (35 m/s)²
= 0.5 × 0.05 × 1225
= 30.625 J → 31 J (to 2 significant figures)
Note: The mark scheme accepts 31 J, though precise calculation gives 30.6 J
(b) When the golfer hits the ball, she hears a sound. Sound waves are longitudinal waves and pass through the air as a series of compressions and rarefactions.
(i) State what is meant by a longitudinal wave.
▶️Answer/Explanation
A wave where particles oscillate parallel to the direction of wave energy transfer
Alternative: A wave where the vibration of particles is in the same direction as the wave travels
Key difference: Unlike transverse waves (e.g., light) where particles move perpendicular to wave direction
(ii) Describe one difference between a compression and a rarefaction.
▶️Answer/Explanation
Compression is a region of high pressure where particles are closer together, while rarefaction is a region of low pressure where particles are farther apart
Visualization: In sound waves, compressions are the “peaks” and rarefactions are the “troughs” of the wave
(iii) Fig. 3.1 shows a sound wave travelling through the air.
On Fig. 3.1, label a compression with the letter C and a rarefaction with the letter R.
▶️Answer/Explanation
Labels should be placed on:
– C: Where particle spacing is minimum (densely packed region)
– R: Where particle spacing is maximum (sparse region)
Diagram note: In a longitudinal wave diagram, compressions appear as darker bands where lines are close together, rarefactions as lighter bands where lines are farther apart
(c) Part of a golf club is made of solid metal. Explain why solids have a fixed shape. Use ideas about the forces between atoms in your answer.
▶️Answer/Explanation
Strong forces of attraction between atoms hold them in fixed positions
Extended explanation:
– In solids, particles are arranged in a regular lattice structure – Strong intermolecular forces prevent particles from moving freely – Particles can only vibrate about fixed positions – This gives solids definite shape and volume
Question 4
(a)(i)(ii): Subtopic: B19.1 Habitat destruction
(b): Subtopic: B19.1 Habitat destruction
(c): Subtopic: B11 Gas exchange in humans
(d): Subtopic: B11 Gas exchange in humans
(e): Subtopic: B9.2 Heart
(a) Researchers estimated the percentage of people that smoke tobacco in a country.
Fig. 4.1 shows the results.
(i) Describe two general trends shown by the data in Fig. 4.1.
▶️Answer/Explanation
1. Between 15-29 years, the percentage of smokers increases with age
2. Between 45-70+ years, the percentage of smokers decreases with age
Additional trends:
– Males consistently show higher smoking rates than females across all age groups
– Peak smoking prevalence occurs in the 35-44 age group
– Sharpest decline occurs after age 55
(ii) Calculate the percentage of females that do not smoke tobacco in the 45-49-year-old age group.
▶️Answer/Explanation
68%
Calculation:
If 32% of females smoke (from graph), then non-smokers = 100% – 32% = 68%
Graph reading skills: Requires accurate interpretation of the female line at 45-49 age group
(b) There is a smaller percentage of the population that smoke tobacco compared to 50 years ago. Suggest a reason for this.
▶️Answer/Explanation
Increased public education about health risks / Stronger legislation / Proven link with lung cancer / Smoking bans in public places
Historical context:
– Major anti-smoking campaigns began in 1960s after surgeon general’s report
– Taxation policies made cigarettes more expensive
– Social stigma increased over time
(c) Table 4.1 shows some of the major components of tobacco smoke and their effects on the body.
Complete Table 4.1.
| component of tobacco smoke | effect on the body |
|---|---|
| carbon monoxide | …… |
| …… | is an addictive substance |
| …… | causes cancer |
▶️Answer/Explanation
| component of tobacco smoke | effect on the body |
|---|---|
| carbon monoxide | binds to hemoglobin reducing oxygen transport |
| nicotine | is an addictive substance |
| tar | causes cancer |
Physiological effects:
– CO: Forms carboxyhemoglobin (reduces O2 by ~15% in smokers)
– Nicotine: Stimulates dopamine release creating addiction
– Tar: Contains 60+ carcinogens including benzopyrene
(d) One of the effects of smoking tobacco is that ciliated cells lining the airways begin to die. Explain the effects of this on the gas exchange system.
▶️Answer/Explanation
1. Mucus cannot be effectively removed due to loss of cilia movement
2. Leads to mucus accumulation in airways (“smoker’s cough”)
3. Reduced surface area for gas exchange as alveoli become damaged
4. Increased risk of infection as pathogens remain trapped
5. Chronic inflammation leads to conditions like bronchitis
Pathology: This process leads to COPD (Chronic Obstructive Pulmonary Disease) over time
(e) Smoking is one of the risk factors for coronary heart disease. State two other risk factors.
▶️Answer/Explanation
1. High blood pressure
2. High cholesterol diet
Other accepted answers:
– Obesity
– Diabetes
– Family history
– Physical inactivity
– Stress
Mechanism: These factors contribute to atherosclerosis (plaque buildup in arteries)
Question 5
(a): Subtopic: C8.4 Transition elements
(b)(i): Subtopic: C12.5 Identification of ions and gases
(b)(ii): Subtopic: C3.3 The mole and the Avogadro constant
(b)(iii): Subtopic: C3.3 The mole and the Avogadro constant
(c): Subtopic: C2.7 Metallic bonding
Copper is a transition metal. Transition metals form coloured compounds.
(a) Write down two other properties of transition metals that are not properties of all metals.
▶️Answer/Explanation
1. High density
2. Act as catalysts
Additional properties:
– Variable oxidation states
– Form complex ions
– High melting points (except mercury)
Example: Copper has density 8.96 g/cm³ (vs 2.7 for aluminum) and copper(II) sulfate is blue
(b) Copper carbonate, CuCO3, reacts with dilute hydrochloric acid, HCl.
Copper chloride, CuCl2, is made.
CuCO3 + 2HCl → CuCl2 + H2O + CO2
(i) Copper chloride contains copper ions, Cu2+, and chloride ions, Cl–. Describe the test and its positive result for chloride ions.
▶️Answer/Explanation
Test: Add nitric acid followed by silver nitrate solution
Result: White precipitate forms
Chemical equation:
Ag+(aq) + Cl–(aq) → AgCl(s)↓
Note: Nitric acid is added first to remove carbonate impurities that would also give precipitate
(ii) In an experiment, 4.0 g of copper carbonate reacts with excess dilute hydrochloric acid.
Calculate the maximum mass of copper chloride that can be made. [Ar; C, 12; Cl, 35.5; Cu, 64; O, 16]
▶️Answer/Explanation
4.4 g
Calculation:
1. Calculate molar masses:
– CuCO3 = 64 + 12 + (3×16) = 124 g/mol
– CuCl2 = 64 + (2×35.5) = 135 g/mol
2. Moles of CuCO3 = 4.0 g ÷ 124 g/mol = 0.03226 mol
3. 1:1 mole ratio → 0.03226 mol CuCl2 produced
4. Mass = 0.03226 mol × 135 g/mol = 4.354 g → 4.4 g (2 sig figs)
(iii) In another experiment, 8.8 g of carbon dioxide gas is made. Calculate the volume of carbon dioxide gas in cm3 at 25°C.
The molar gas volume at 25°C is 24 dm3.
[Ar; C, 12; O, 16]
▶️Answer/Explanation
4800 cm3
Calculation steps:
1. Molar mass CO2 = 12 + (2×16) = 44 g/mol
2. Moles of CO2 = 8.8 g ÷ 44 g/mol = 0.2 mol
3. Volume at 25°C = 0.2 mol × 24 dm3/mol = 4.8 dm3
4. Convert to cm3: 4.8 × 1000 = 4800 cm3
Key points:
– 1 dm3 = 1000 cm3
– Temperature must be in Kelvin for exact calculations (25°C = 298K)
(c) Explain why copper is a conductor of electricity. Use ideas about metallic bonding.
▶️Answer/Explanation
1. Copper has delocalized electrons that can move freely through the metal lattice
2. These mobile electrons carry charge when potential difference is applied
Metallic bonding details:
– Cu atoms form a “sea of electrons”
– Outer shell electrons (4s1 in copper) become delocalized
– Applied voltage causes net electron flow
Question 6
(a)(i)(ii): Subtopic: P1.5.1 Effects of forces
(b): Subtopic: P1.7 Pressure
(c): Subtopic: P4.1 Simple phenomena of magnetism
(d): Subtopic: P3.4 Sound
(e): Subtopic: P5.2.3 Radioactive decay
(a) A horse of mass 450 kg accelerates constantly from rest and reaches a maximum speed of 9 m/s after 3 seconds. In this time, the horse has travelled 13.5 m.
(i) Show that the force that causes the acceleration of the horse is 1350 N.
▶️Answer/Explanation
Calculation:
1. Calculate acceleration (a):
a = Δv/Δt = (9 m/s – 0)/3 s = 3 m/s²
2. Apply Newton’s Second Law:
F = ma = 450 kg × 3 m/s² = 1350 N
Key concepts:
– Acceleration is rate of change of velocity
– Force is directly proportional to acceleration (F=ma)
– Units: 1 N = 1 kg·m/s²
(ii) Calculate the work done by the horse in travelling 13.5 m.
▶️Answer/Explanation
18,225 J (or 18,200 J to 3 sig figs)
Calculation:
Work done = Force × Distance
= 1350 N × 13.5 m
= 18,225 J
Alternative method:
Using kinetic energy KE=½mv²=½×450×9²=18,225 J
(valid since all work converts to kinetic energy)
(b) The horse stands with all four hooves in contact with the ground. The horse exerts a force of 4500 N on the ground. Each hoof of the horse has an area of 90 cm2.
Calculate the pressure, in N/m2, exerted by the horse on the ground.
▶️Answer/Explanation
125,000 N/m² (or 1.25×105 Pa)
Calculation steps:
1. Convert cm² to m²: 90 cm² = 90 × (0.01)² = 0.009 m² per hoof
2. Total area = 4 hooves × 0.009 m² = 0.036 m²
3. Pressure = Force/Area = 4500 N / 0.036 m² = 125,000 Pa
Real-world comparison:
– Human foot pressure ~80,000 Pa when standing
– Car tire pressure ~200,000 Pa
(c) Horseshoes are usually made from either iron or steel.
Describe one difference between the magnetic properties of iron and steel.
▶️Answer/Explanation
Iron magnetizes and demagnetizes more easily than steel
Alternative answers:
– Iron is a soft magnetic material (loses magnetism quickly)
– Steel is a hard magnetic material (retains magnetism)
– Steel has higher coercivity (resists demagnetization)
Application: Iron used for temporary magnets, steel for permanent magnets
(d) The audible frequency range for horses is from 14 Hz to 25,000 Hz. Compare this range to that of a human.
▶️Answer/Explanation
Horses have a wider audible range, especially for higher frequencies (humans: 20 Hz – 20,000 Hz)
Biological significance:
– Horses can hear higher-pitched sounds important for communication
– Human hearing typically declines above 15,000 Hz with age
– Both species are most sensitive in 1,000-5,000 Hz range
(e) A horse is treated for cancer using the isotope iridium-192. The iridium-192 is injected into the cancer. Iridium-192 decays by β-emission to produce an isotope of platinum.
Use nuclide notation to complete the symbol equation for the β-decay process.
▶️Answer/Explanation
19277Ir → 19278Pt + 0-1β
Nuclear rules:
1. Mass number (A) stays same (192)
2. Atomic number (Z) increases by 1 (77→78)
3. β-particle is electron with A=0, Z=-1
Medical use: Iridium-192 is used in brachytherapy (γ-rays kill cancer cells)
Question 7
(a)(i)(ii): Subtopic: B6.1 Photosynthesis
(b): Subtopic: B16.3 Monohybrid inheritance
(c)(i)(ii): Subtopic: B6.1 Photosynthesis
(a) Table 7.1 shows the effects of using fertilisers containing nitrate ions on the yield of pea plants. The yield is the mass of peas produced per square metre.
| application of fertiliser | yield/g per m2 |
|---|---|
| fertiliser containing nitrate ions used | 340 |
| no fertiliser used | 120 |
(i) Calculate the percentage increase in yield when using fertilisers containing nitrate ions. Give your answer to the nearest whole number.
▶️Answer/Explanation
183%
Calculation:
1. Difference = 340 – 120 = 220 g/m²
2. Percentage increase = (220/120) × 100 = 183.33% → 183%
Agricultural context:
Nitrate fertilisers typically increase crop yields by 50-200% depending on soil conditions
(ii) Explain why adding nitrate to pea plants increases the yield of peas.
▶️Answer/Explanation
1. Nitrates are needed to make amino acids and proteins
2. Proteins are essential for plant growth and development
3. More growth leads to higher pea production (yield)
Biological process:
– Nitrogen cycle: NO3– → NH4+ → amino acids
– Peas are legumes but still benefit from added nitrates
– Proteins used for enzymes and cellular structures
(b) Peas can be wrinkled or round. Fig. 7.1 is a photograph of a wrinkled pea and a round pea.
Peas inherit the wrinkled or round feature from their parent plants.
- The dominant allele for round peas is R.
- The recessive allele for wrinkled peas is r.
Use your knowledge and this information to complete Table 7.2.
| genotype for wrinkled peas | …… |
|---|---|
| phenotype of a pea with a heterozygous genotype | …… |
| the type of breeding if two wrinkled pea plants were crossed | …… |
▶️Answer/Explanation
| genotype for wrinkled peas | rr |
|---|---|
| phenotype of a pea with a heterozygous genotype | round |
| the type of breeding if two wrinkled pea plants were crossed | pure breeding |
Genetic principles:
– Recessive traits only show in homozygous (rr) state
– Heterozygous (Rr) shows dominant phenotype
– Pure breeding = homozygous parents → identical offspring
(c) Peas contain a store of carbohydrates made during photosynthesis.
(i) Describe two other uses of the carbohydrates made during photosynthesis.
▶️Answer/Explanation
1. Energy source: Used in respiration to produce ATP for cellular processes
2. Structural material: Converted to cellulose for cell walls
Additional uses:
– Form starch for energy storage
– Make sucrose for transport in phloem
– Produce nectar to attract pollinators
(ii) List the three chemical elements present in carbohydrates.
▶️Answer/Explanation
Carbon, Hydrogen, Oxygen
Ratio: Typically 1:2:1 (e.g., glucose C6H12O6)
Note: All carbohydrates contain these three elements, though some derivatives may contain others (e.g., chitin has nitrogen)
Question 8
(a)-(c): Subtopic: C6.2 Rate of reaction
A student investigates the reaction between sodium thiosulfate solution and dilute hydrochloric acid.
Fig. 8.1 shows the apparatus the student uses.
The student looks down at the cross drawn on the paper. A solid is made during the reaction and the mixture in the flask becomes cloudy. At the moment she adds the dilute acid, the student starts a stop-watch.
She measures the time it takes until she can no longer see the cross.
The student does four experiments with different concentrations (A, B, C, D) of sodium thiosulfate.
Table 8.1 shows her results.
| concentration | time taken for cross to disappear/s |
|---|---|
| A | 39 |
| B | 78 |
| C | 127 |
| D | 61 |
(a) Look at the student’s results. State which is the most concentrated solution of sodium thiosulfate (A, B, C or D). Explain your answer.
▶️Answer/Explanation
Answer: A
Explanation: The cross disappears fastest (39s) indicating highest reaction rate, which occurs with highest concentration
Chemical principle: Na2S2O3(aq) + 2HCl(aq) → 2NaCl(aq) + S(s)↓ + SO2(g) + H2O(l)
The sulfur precipitate causes cloudiness. Higher concentration → more collisions → faster reaction
(b) The rate of the reaction can be increased by increasing the temperature of the reaction mixture to 45 °C. Explain why. Use ideas about collisions between particles.
▶️Answer/Explanation
1. Particles gain kinetic energy and move faster
2. More frequent successful collisions per second
3. More particles have energy ≥ activation energy
Quantitative effect: Typically reaction rate doubles every 10°C rise (Q10 = 2)
Practical note: Above 60°C, the reaction becomes too fast to measure accurately
(c) The reaction between sodium thiosulfate solution and dilute hydrochloric acid is exothermic. Explain why. Use ideas about bond forming and bond breaking.
▶️Answer/Explanation
1. Bond breaking in reactants absorbs energy (endothermic)
2. Bond making in products releases energy (exothermic)
3. More energy is released than absorbed (net energy release)
Energy values:
– Breaking S-S and S-O bonds: +ve ΔH
– Forming NaCl and H2O: -ve ΔH
– Overall ΔH ≈ -50 kJ/mol (exothermic)
Question 9
(a): Subtopic: P1.3 Mass and weight
(b)(i)(ii): Subtopic: P4.2.5 Electrical energy and electrical power
(c): Subtopic: P2.3.3 Radiation
(d): Subtopic: P2.2.2 Melting, boiling and evaporation
(a) The information booklet about an oven states that the weight of the oven is 45 kg. Explain why this statement is incorrect.
▶️Answer/Explanation
Weight should be measured in newtons (N), while mass is measured in kilograms (kg)
Physics principle: Weight = mass × gravitational acceleration (W=mg)
On Earth, 45 kg mass would weigh ≈ 450 N (using g=9.81 m/s²)
(b) (i) Fig. 9.1 shows information on a label attached to the electric oven.
Calculate the maximum working current of the oven.
▶️Answer/Explanation
25 A
Calculation:
P = IV → I = P/V
= 6000 W / 240 V
= 25 A
Safety note: This high current requires special wiring (typically 30A circuit)
(ii) The oven has its own fuse. Use your answer to (b)(i) to explain why a fuse rated at 13 A is not suitable for use in the oven circuit.
▶️Answer/Explanation
The working current (25A) exceeds the fuse rating (13A), so it would blow immediately
Electrical standards: UK plugs typically have 3A or 13A fuses – this oven requires hardwiring with higher-rated protection
(c) A thermocouple is used to measure the temperature inside the oven. Describe the structure of a thermocouple. You may draw a diagram if it helps your answer.
▶️Answer/Explanation
1. Two different metals joined together at a measuring junction
2. Connected to a voltmeter at the reference junctions
Common pairs: Copper-constantan (Type T), iron-constantan (Type J), chromel-alumel (Type K)
Working principle: Temperature difference creates voltage (Seebeck effect)
(d) Some water is heated in a dish in the oven. As the water is heated, some of the water evaporates. Eventually the water begins to boil. Describe two differences between evaporation and boiling.
▶️Answer/Explanation
1. Temperature: Evaporation occurs at any temperature, while boiling only at boiling point
2. Location: Evaporation only at surface, boiling occurs throughout liquid
Additional differences:
– Evaporation is slow, boiling is rapid
– Bubbles form only during boiling
– Evaporation causes cooling, boiling doesn’t
Question 10
(a)(i)(ii): Subtopic: B13.3 Homeostasis
(b): Subtopic: B13.3 Homeostasis
(c): Subtopic: B13.1 Coordination and response
(a) The temperature of a person’s skin is recorded in different environments.
Fig. 10.1 shows two body parts (A and B) where readings are taken.
Table 10.1 shows the results.
(i) Describe how the skin responds to cold temperatures in order to maintain a constant internal body temperature.
▶️Answer/Explanation
1. Vasoconstriction: Arterioles narrow to reduce blood flow to skin surface
2. Shivering: Muscle contractions generate heat
3. Hair erection: Piloerection traps insulating air layer
Physiological details:
– Can reduce skin blood flow to just 1% of cardiac output
– May increase metabolic heat production by 5x
– Controlled by hypothalamus through sympathetic nervous system
(ii) Suggest why the temperature range of the skin on part A is less than on part B.
▶️Answer/Explanation
1. Part A covers vital organs needing stable temperature for enzyme function
2. Part B is more peripheral (likely extremities) where temperature can vary more
Anatomical context:
– Part A is probably torso (core temperature 36.5-37.5°C)
– Part B is likely hands/feet (can safely drop to 15°C in cold)
– Core organs maintain ±0.5°C while skin can vary 20+°C
(b) Body temperature is controlled to keep it within set limits. Name the term used to describe this.
▶️Answer/Explanation
Homeostasis
Key features:
– Negative feedback loops
– Receptors → control center → effectors
– Maintains core temperature at ~37°C in humans
(c) Temperature control of the body shows that humans have sensitivity to their environment. Define sensitivity.
▶️Answer/Explanation
The ability to detect and respond to changes in the environment
Components:
1. Sensory receptors detect stimuli (temperature, light, etc.)
2. Effectors produce appropriate responses
Evolutionary advantage: Enables adaptation to changing conditions
Question 11
(a): Subtopic: C11.4 Alkanes
(b): Subtopic: C11.5 Alkenes
(c): Subtopic: C11.3 Fuels
(d): Subtopic: C11.7 Polymers
Propane and propene are both hydrocarbons. Fig. 11.1 shows the structure of a propene molecule.
(a) Draw the structure of a propane molecule.
▶️Answer/Explanation
H H H
| | |
H-C-C-C-H
| | |
H H H
Key features:
– 3 carbon single bonds (C-C)
– 8 hydrogen atoms total
– No double bonds (vs propene’s C=C)
– Molecular formula C3H8
(b) Aqueous bromine is used to tell the difference between propane and propene. Describe the colour change, if any, that is seen when aqueous bromine is added to propane and to propene.
▶️Answer/Explanation
Propane: Bromine water remains orange (no reaction)
Propene: Bromine water decolorizes (turns colorless)
Chemical reaction:
CH3CH=CH2 + Br2 → CH3CHBrCH2Br
Mechanism: Electrophilic addition across the C=C double bond
(c) Cracking is used to make propene. Describe cracking. Include what happens to the large hydrocarbon molecules during cracking, and the conditions needed for cracking.
▶️Answer/Explanation
1. Process: Breaking long hydrocarbon chains into shorter molecules
2. Products: Makes alkenes (like propene) + smaller alkanes
3. Conditions:
– High temperature (600-900°C)
– Catalyst (zeolite/alumina/silica)
– Optional: Steam (steam cracking)
Industrial context: Converts heavy fractions from crude oil into more valuable chemicals
(d) Fig. 11.2 shows the structure of a chloropropene molecule.
Chloropropene is used to make the polymer poly(chloropropene). Draw the structure of poly(chloropropene).
▶️Answer/Explanation
Cl
|
[-CH2-C=CH-CH2-]n
|
CH3
Polymerization: Addition polymerization of chloropropene monomers
Properties: Weather-resistant rubber used in cables and hoses
Question 12
(a)(i)(ii): Subtopic: P2.2.1 Thermal expansion of solids, liquids and gases
(b): Subtopic: P4.2.1 Electrical charge
(c): Subtopic: P3.2.2 Refraction of light
(d)(i)-(iii): Subtopic: P4.5.2 The a.c. generator
(a) Fig. 12.1 shows a truck crossing a bridge.
The bridge is designed with gaps in the road surface as shown in Fig. 12.2.
The temperature of the road surface increases on a hot day.
(i) Describe what happens to the gaps in the road surface when the temperature increases. Explain your answer.
▶️Answer/Explanation
1. Observation: Gaps become narrower as the road expands
2. Explanation: Thermal expansion causes metal/road materials to increase in length (ΔL = αLΔT)
Engineering details: – Typical steel expansion coefficient (α) = 12×10-6 °C-1 – A 50m bridge expands ~6cm when heated 100°C
(ii) Suggest what may happen to the bridge if there were no gaps in the road surface.
▶️Answer/Explanation
The bridge would buckle/warp/crack from thermal stress
Structural impact: – Compression forces can exceed 70 MPa – May cause joint failures or pavement blowouts
(b) Fig. 12.3 shows the fuel tank of the truck being filled with diesel fuel.
Explain why the diesel fuel becomes positively charged.
▶️Answer/Explanation
1. Mechanism: Friction between fuel and pipe transfers electrons
2. Result: Fuel loses electrons → becomes positively charged
Safety note: – Requires grounding to prevent sparks – Flow rates kept below 7 m/s to minimize charge
(c) The truck has a warning triangle to alert other drivers. Fig. 12.4 shows the warning triangle.
Many tiny prisms are contained in the warning triangle.
Fig. 12.5 shows one ray of light entering a prism.
The ray undergoes total internal reflection inside the prism.
Complete Fig. 12.5 to show the path of the ray of light through the prism and the ray of light leaving the prism.
▶️Answer/Explanation
1. First surface: Refraction entering prism
2. Second surface: Total internal reflection (critical angle ~42° for glass)
3. Third surface: Refraction exiting parallel to incident ray
Optical principle: – Retroreflection returns light to source – Works for wide range of incident angles
(d) The truck has a generator. Fig. 12.6 shows a simple generator producing an alternating voltage.
(i) On Fig. 12.6, label the coil C.
(ii) On Fig. 12.6, label the slip rings S.
(iii) Describe how turning the coil induces an alternating voltage.
▶️Answer/Explanation
(d)(i) coil correctly labelled ;
(d)(ii) slip rings correctly labelled ;
(d)(iii) magnetic field ;
rotating coil cuts magnetic field or flux / experiences a changing magnetic field ;
e.m.f. / current reverses every half turn ;