2023 Cambridge IGCSE Sciences - Co-ordinated (Double) (0654) Practice Paper with Answers

a) A dominant allele is an allele that is expressed in the phenotype if it is present in either the heterozygous or homozygous state. b) i) The white-flowered parent must be homozygous recessive, so its genotype is rr. ii) The parental gametes are R and r from the heterozygous parent, and r (or r and r) from the white-flowered parent. The offspring genotypes produced are Rr and rr. The corresponding offspring phenotypes are red-flowered (for Rr) and white-flowered (for rr). The resulting phenotypic ratio is 1:1 (or 50% red and 50% white). c) Homozygous is the term used to describe an organism with two identical alleles for a particular gene. d) A breeder uses a test cross to determine whether an individual showing a dominant phenotype is homozygous dominant or heterozygous. This is done by crossing the individual with a homozygous recessive individual. If any offspring show the recessive phenotype, the parent must be heterozygous.

a) [1] for stating it is an allele that is expressed/always visible; [1] for stating it is expressed if present / even with only one copy. b) i) [1] for rr. ii) [1] for correct parental gametes (R, r and r); [1] for correct offspring genotypes (Rr and rr); [1] for correct offspring phenotypes (red and white); [1] for correct phenotypic ratio (1:1 or 50% red, 50% white). c) [1] for homozygous. d) [1] for crossing with a homozygous recessive individual; [1] for explaining how offspring phenotypes reveal the parent's genotype (e.g., if recessive offspring appear, parent is heterozygous).

a) i) Isotopes of the same element have the same number of protons (similarity) but a different number of neutrons (difference). ii) Carbon-13 has atomic number 6 and mass number 13. Thus, it contains 6 protons, 13 - 6 = 7 neutrons, and, being a neutral atom, 6 electrons. b) i) In carbon dioxide, the central carbon atom shares a total of four electrons (two pairs) with each of the two oxygen atoms, forming two double covalent bonds. This allows all three atoms to achieve a stable outer shell of eight electrons. ii) Carbon dioxide consists of simple molecules. While the covalent bonds holding the atoms together inside the molecules are very strong, the forces of attraction between different molecules (intermolecular forces) are very weak and require very little thermal energy to overcome, leading to a low boiling point.

a) i) [1] for same number of protons (or same atomic number / number of electrons); [1] for different number of neutrons (or different nucleon/mass number). ii) [1] for 6 protons; [1] for 7 neutrons; [1] for 6 electrons. b) i) [1] for carbon sharing four electrons (or forming double bonds); [1] for two pairs of shared electrons in each C=O bond; [1] for all atoms achieving a full outer shell (or 8 outer electrons). ii) [1] for identifying weak intermolecular forces / weak forces between molecules; [1] for stating that little energy is needed to overcome these weak forces (accept: covalent bonds are not broken during boiling).

a) Electric current is defined as the rate of flow of electric charge. b) i) Using Ohm's law, V = I * R = 1.5 A * 6.0 ohms = 9.0 V. ii) In a series circuit, the total voltage of the power supply is shared between the components: V_total = V_resistor + V_lamp. Therefore, V_lamp = 12 V - 9.0 V = 3.0 V. iii) Using Ohm's law, R_lamp = V_lamp / I = 3.0 V / 1.5 A = 2.0 ohms. c) As the temperature of the filament increases, the positive metal ions in the lattice vibrate with greater amplitude. This increases the frequency of collisions between the free conduction electrons and the vibrating metal ions, which resists the flow of charge and causes the electrical resistance to increase.

a) [1] for 'rate of flow of charge' (or 'charge per unit time'). b) i) [1] for formula V = I * R or correct substitution (1.5 * 6.0); [1] for 9.0 V (allow 9 V). ii) [1] for subtracting resistor voltage from total EMF (12 - 9); [1] for 3.0 V (allow ECF from b i). iii) [1] for formula R = V / I or correct substitution (3.0 / 1.5); [1] for 2.0 ohms (allow ECF from b ii). c) [1] for stating that resistance increases; [1] for metal ions vibrating more (at higher temperatures); [1] for increased frequency of collisions between electrons and metal ions.

a) Root hair cells have a long hair-like projection that greatly increases their surface area for water absorption. They also have thin cell walls, which reduces the distance over which water must diffuse/osmose into the cell. b) Water evaporates from the mesophyll cells into the air spaces and diffuses out of the stomata (transpiration). This evaporation creates a tension or suction force called transpiration pull, drawing water up the xylem. Water molecules are polar and stick to one another via hydrogen bonds (cohesion) and also stick to the cellulose walls of the xylem vessels (adhesion). This allows a continuous, unbroken column of water to be pulled up the stem. c) i) An increase in wind speed increases the rate of transpiration. This is because the moving air sweeps away the humid air and accumulated water vapor from the surface of the leaf, maintaining a steep concentration gradient of water vapor between the inside of the leaf and the external air, which speeds up diffusion. ii) Other factors include: an increase in temperature, a decrease in humidity (dry air), or an increase in light intensity (which opens stomata).

a) [1] for 'large surface area' (due to projection/extension); [1] for 'thin cell wall' (or short diffusion distance). b) [1] for transpiration pull / tension created by evaporation from leaves; [1] for cohesion defined as water molecules sticking together; [1] for adhesion defined as water molecules sticking to xylem walls; [1] for maintaining a continuous, unbroken column of water. c) i) [1] for stating transpiration rate increases; [1] for wind moving water vapor away from the leaf surface/stomata; [1] for maintaining a steep concentration gradient (of water vapor). ii) [1] for high temperature / high light intensity / low humidity (dry conditions).

a) i) Reactivity increases as you go down Group 1, because the outer shell electron is further from the nucleus and more shielded, making it easier to lose. ii) Solid sodium reacts with liquid water to produce aqueous sodium hydroxide and hydrogen gas: 2Na(s) + 2H2O(l) -> 2NaOH(aq) + H2(g). b) i) Chlorine is a pale green (or yellow-green) gas. Bromine is a red-brown (or brown) liquid. ii) Chlorine is more reactive than bromine, so it displaces bromide ions. The observation is that the colorless solution of potassium bromide turns yellow, orange, or brown due to the formation of aqueous bromine. The ionic equation is: Cl2(aq) + 2Br-(aq) -> 2Cl-(aq) + Br2(aq).

a) i) [1] for 'reactivity increases (down the group)'. ii) [1] for correct formulas of all reactants and products (Na, H2O, NaOH, H2); [1] for correct balancing (2, 2, 2, 1); [1] for correct state symbols: Na(s), H2O(l), NaOH(aq), H2(g). b) i) [1] for chlorine is a green/pale-green gas; [1] for bromine is a red-brown liquid (accept brown liquid). ii) [1] for observation: colorless solution turns yellow/orange/brown; [1] for stating bromine is formed / displaced; [1] for correct formulas in ionic equation (Cl2 + 2Br- -> 2Cl- + Br2); [1] for correct balancing of ionic equation.

a) i) The speed of all electromagnetic waves in a vacuum is 3.0 * 10^8 m/s (or 3 * 10^8 m/s). ii) Components of the EM spectrum with higher frequencies than visible light are ultraviolet (UV), X-rays, and gamma rays. b) i) The refractive index formula is n = sin(i) / sin(r). ii) Substituting the values: 1.5 = sin(35) / sin(r). Rearranging: sin(r) = sin(35) / 1.5 = 0.5736 / 1.5 = 0.3824. Therefore, r = sin^-1(0.3824) = 22.5 degrees (accept 22 to 23 degrees depending on rounding). iii) When light enters a more optically dense medium like glass, its speed decreases and its wavelength decreases, while its frequency remains constant. c) The critical angle is the angle of incidence in a denser medium that results in an angle of refraction of 90 degrees in the less dense medium.

a) i) [1] for 3.0 * 10^8 m/s (accept 3 * 10^8 m/s, must include correct units). ii) [1] each for any two of: ultraviolet, X-rays, gamma rays. b) i) [1] for n = sin(i) / sin(r). ii) [1] for correct substitution (1.5 = sin(35) / sin(r)); [1] for sin(r) = 0.38; [1] for final answer 22.5 degrees (accept range 22 to 23 degrees). iii) [1] for wavelength decreases; [1] for speed decreases. c) [1] for angle of incidence in denser medium where angle of refraction is 90 degrees (or beyond which total internal reflection occurs).

a) The word equation is: calcium carbonate + hydrochloric acid -> calcium chloride + water + carbon dioxide. b) i) To ensure a fair test, the mass of calcium carbonate, the surface area / particle size of calcium carbonate, and the volume (and initial concentration) of hydrochloric acid must be kept constant. ii) Increasing the concentration of hydrochloric acid means there are more acid particles (H+ ions) per unit volume. This increases the frequency of collisions between the reactant particles, which leads to a greater number of successful collisions per unit time, thereby increasing the rate of reaction. c) i) A catalyst is a substance that increases the rate of a chemical reaction, but remains chemically unchanged and is not used up at the end of the reaction. ii) A catalyst works by providing an alternative reaction pathway that has a lower activation energy. As a result, a larger proportion of colliding particles have energy equal to or greater than this lower activation energy, increasing the rate of successful collisions.

a) [1] for correct reactants and products (calcium carbonate + hydrochloric acid -> calcium chloride + carbon dioxide + water). b) i) [1] each for any two of: mass of calcium carbonate, surface area/particle size of calcium carbonate, volume of acid, concentration of acid. ii) [1] for stating there are more particles per unit volume; [1] for increased frequency of collisions / more collisions per unit time; [1] for more successful/fruitful collisions per unit time. c) i) [1] for 'substance that increases rate of reaction'; [1] for 'remains chemically unchanged' / 'not used up'. ii) [1] for providing an alternative pathway with lower activation energy; [1] for 'more particles have energy greater than or equal to activation energy' / 'more successful collisions'.

a) i) Acceleration a = (v - u) / t = (4.0 m/s - 0 m/s) / 2.5 s = 1.6 m/s^2. ii) Resultant force F = m * a = 0.80 kg * 1.6 m/s^2 = 1.28 N (or 1.3 N). b) Kinetic Energy E_k = 0.5 * m * v^2 = 0.5 * 0.80 kg * (4.0 m/s)^2 = 0.4 * 16 = 6.4 J. c) Work done W = F * d = 1.2 N * 3.0 m = 3.6 J.

a) i) [1] for formula a = (v-u)/t or correct substitution (4.0 / 2.5); [1] for 1.6; [1] for unit m/s^2 (or m s^-2). ii) [1] for formula F = m * a or correct substitution (0.80 * 1.6); [1] for 1.28 N (accept 1.3 N). b) [1] for formula E_k = 0.5 * m * v^2 or correct substitution (0.5 * 0.80 * 4^2); [1] for 6.4 J (must include unit J / Joules). c) [1] for formula W = F * d; [1] for correct substitution (1.2 * 3.0); [1] for 3.6 J (must include unit J / Joules).

*(a)* A gene is defined as a length of DNA that codes for a specific protein.
*(b)(i)* The child has blue eyes, which is a recessive phenotype. Therefore, the child's genotype must be homozygous recessive, \(bb\). The child must inherit one \(b\) allele from each parent. Since both parents have brown eyes (the dominant phenotype), they must each also possess a dominant \(B\) allele. Thus, both parents are heterozygous, with the genotype \(Bb\).
*(b)(ii)* Parents: \(Bb \times Bb\). Gametes: \(B, b \times B, b\). Possible offspring genotypes: \(BB\) (brown eyes), \(Bb\) (brown eyes), \(Bb\) (brown eyes), and \(bb\) (blue eyes). The offspring phenotypes are 3 brown eyes to 1 blue eyes. The probability of having a child with brown eyes is \(3/4\) or \(0.75\) or \(75\\%\).
*(c)* A phenotype is the physical or observable features of an organism, determined by its genotype and interaction with the environment.

*(a)*
- length of DNA [1]
- that codes for a specific protein [1]

*(b)(i)*
- parents' genotype: both are \(Bb\) / heterozygous [1]
- child must be \(bb\) (to have blue eyes) / inherits one \(b\) allele from each parent [1]
- parents must have dominant \(B\) allele to have brown eyes [1]

*(b)(ii)*
- parental gametes identified correctly as \(B\) and \(b\) [1]
- correct offspring genotypes shown (\(BB\), \(Bb\), \(Bb\), \(bb\)) [1]
- probability stated as \(3/4\) / \(0.75\) / \(75\\%\) / 3 in 4 [1]

*(c)*
- observable / physical features [1]
- of an organism (resulting from genotype / interaction with environment) [1]

*(a)(i)* In the notation \({}_{16}^{32}\text{S}\), the atomic number (bottom) is 16, which represents the number of protons. For a neutral atom, the number of electrons equals the number of protons, which is 16. The mass number (top) is 32, so the number of neutrons is \(32 - 16 = 16\).
*(a)(ii)* With 16 electrons, the electronic configuration is 2 electrons in the first shell, 8 in the second, and 6 in the third shell: 2,8,6.
*(b)(i)* A covalent bond is formed when a pair of electrons is shared between two atoms. This results in an electrostatic attraction between the shared pair of electrons and the positively charged nuclei of the bonded atoms.
*(b)(ii)* In a water molecule (\(H_2O\)), oxygen shares one electron with each of the two hydrogen atoms, forming 2 covalent bonds (2 shared pairs of electrons). Oxygen starts with 6 outer-shell electrons; after sharing 2, it has 4 unshared outer-shell electrons remaining.
*(c)* Substance X has a high melting point and conducts electricity only when liquid (molten), which is characteristic of ionic bonding. Substance Y has a very low melting point and does not conduct electricity in either solid or liquid state, which is characteristic of simple covalent bonding.

*(a)(i)*
- protons = 16 [1]
- neutrons = 16 [1]
- electrons = 16 [1]

*(a)(ii)*
- 2,8,6 [1]

*(b)(i)*
- shared pair of electrons [1]
- electrostatic attraction between shared electrons and nuclei [1]

*(b)(ii)*
- two shared pairs / two single bonds [1]
- four unshared outer-shell electrons / two lone pairs on oxygen [1]

*(c)*
- Substance X: ionic (bonding) [1]
- Substance Y: covalent / simple molecular (bonding) [1]

*(a)(i)* In a longitudinal wave, the particles of the medium vibrate or oscillate back and forth about a fixed position. This movement is parallel to the direction in which the wave/energy is travelling.
*(a)(ii)* The range of audible frequencies for a healthy human ear is approximately 20 Hz to 20,000 Hz (or 20 kHz).
*(b)(i)* Using the wave equation: \(v = f \lambda\), we rearrange to find wavelength \(\lambda = v / f\). Substitute the given values: \(\lambda = 1500 / 440 = 3.41\text{ m}\) (or 3.4 m).
*(b)(ii)* The frequency of the wave is determined by the source and does not change when transitioning between media; therefore, frequency remains 440 Hz. Since the speed decreases from 1500 m/s to 330 m/s and frequency remains constant, the wavelength must decrease, which can be calculated as \(\lambda = v / f = 330 / 440 = 0.75\text{ m}\).

*(a)(i)*
- particles vibrate / oscillate [1]
- parallel to the direction of energy transfer / wave travel [1]

*(a)(ii)*
- lower limit: 20 Hz [1]
- upper limit: 20,000 Hz / 20 kHz [1]

*(b)(i)*
- formula: \(\lambda = v/f\) or \(v = f\lambda\) [1]
- calculation: \(1500 / 440 = 3.4\) or \(3.41\) [1]
- unit: m / metres [1]

*(b)(ii)*
- frequency does not change / remains 440 Hz [1]
- wavelength decreases [1]
- explanation: wavelength is proportional to speed / calculation showing \(\lambda = 0.75\text{ m}\) [1]

*(a)(i)* Acceleration is defined as the rate of change of velocity: \(a = (v - u) / t\). Since the car starts from rest, \(u = 0\text{ m/s}\), \(v = 15\text{ m/s}\), and \(t = 6.0\text{ s}\) or \(t = 6.0\text{ s}\). Therefore, \(a = (15 - 0) / 6.0 = 2.5\text{ m/s}^2\).
*(a)(ii)* Using Newton's second law: \(F = m a\). Substituting the mass \(m = 1200\text{ kg}\) and \(a = 2.5\text{ m/s}^2\), we get \(F = 1200 \times 2.5 = 3000\text{ N}\).
*(a)(iii)* Kinetic energy is given by \(KE = \frac{1}{2} m v^2\). Substitute the values: \(KE = 0.5 \times 1200 \times 15^2 = 0.5 \times 1200 \times 225 = 135,000\text{ J}\) or \(135\text{ kJ}\).
*(b)* When the car is travelling at a constant speed of 15 m/s, its acceleration is zero. Since \(F = ma\), if \(a = 0\), the resultant force \(F\) must also be \(0\text{ N}\). This means the forward driving force is exactly balanced by the resistive forces (friction and air resistance).

*(a)(i)*
- formula: \(a = \Delta v / t\) or \((v-u)/t\) [1]
- calculation: \(15 / 6.0 = 2.5\text{ m/s}^2\) [1]

*(a)(ii)*
- formula: \(F = ma\) [1]
- calculation: \(1200 \times 2.5 = 3000\) [1]
- unit: N / Newtons [1]

*(a)(iii)*
- formula: \(KE = \frac{1}{2}mv^2\) [1]
- calculation: \(0.5 \times 1200 \times 15^2 = 135,000\text{ J}\) (or \(135\text{ kJ}\)) [1]

*(b)*
- resultant force = 0 / zero [1]
- explanation: constant speed means acceleration is zero [1]
- reference to Newton's First Law / driving force equals resistive forces [1]