Campus Bio-Medico Admission Test Practice Paper 1, with Answers

Answer: C

Each water molecule can form hydrogen bonds with its neighbours, and energy must be spent breaking these weak bonds before the molecules can move faster or separate at the surface. The covalent O–H bonds are much stronger, but they hold each molecule together rather than linking molecules to one another, and they are not broken when water is heated or stretched.

Answer: A

A competitive inhibitor resembles the substrate and competes with it for the active site, so a large excess of substrate wins most of the binding events and the enzyme approaches its normal maximum rate. Binding to a separate allosteric site and lowering the maximum rate whatever the substrate concentration describe a non-competitive inhibitor instead.

Answer: E

All cells make proteins, so all have ribosomes: prokaryotes have smaller 70S ribosomes and eukaryotes 80S ribosomes in their cytoplasm. Prokaryotes have no membrane-bound organelles, so they lack a nuclear envelope, mitochondria, endoplasmic reticulum and Golgi apparatus; their DNA lies in a nucleoid region, not a nucleus.

Answer: B

The 0.3% solution is hypotonic to the cells, so water moves by osmosis from the solution, where it is more concentrated, into the cells; with no cell wall to resist, they swell and can burst (haemolysis). Shrinking would happen in a hypertonic solution, and aquaporins are channels for water, not for salt.

Answer: D

A secreted protein is built by ribosomes on the rough endoplasmic reticulum, carried in transport vesicles to the Golgi apparatus to be modified and sorted, and then packed into secretory vesicles that release it by exocytosis. The nucleus holds the gene and makes the mRNA, but amino acids are joined in the cytoplasm, so the labelled amino acid never passes through it; the Golgi comes before, not after, the secretory vesicles.

Answer: D

The DNA doubles to 16 pg in S phase; meiosis I separates homologous chromosomes, giving cells with 6 chromosomes of two chromatids each and 8 pg, and meiosis II separates the chromatids, giving gametes with 6 chromosomes and 4 pg. Six chromosomes with 8 pg describes a cell at the end of meiosis I, and 2 pg comes from halving the doubled amount three times instead of twice.

Answer: B

Recombination frequency grows with the distance between genes (1% = 1 map unit), so the two genes furthest apart, A and B at 12 units, are at the ends and C lies between them: 7 + 5 = 12. Genes on different chromosomes assort independently and recombine at 50%, so 12% shows that A and B are linked, not that they are separate.

Answer: A

At the end of the chain in the inner mitochondrial membrane, electrons are passed to oxygen, which combines with H⁺ ions to form water. NAD⁺ accepts electrons earlier, in glycolysis and the Krebs cycle, and hands them to the start of the chain as NADH, so it is a carrier rather than the final acceptor.

Answer: E

The O₂ given off comes from the splitting of water in the light-dependent reactions, so it carries no oxygen from CO₂. The CO₂ is fixed in the Calvin cycle, and its oxygen atoms end up in the sugar and in the water formed by the overall reaction 6CO₂ + 12H₂O → C₆H₁₂O₆ + 6O₂ + 6H₂O; the label would appear in the O₂ only if the water had been labelled instead.

Answer: C

The cross Pp × pp gives Pp and pp offspring in equal numbers, so half are purple and half white; a homozygous PP plant would give only purple offspring, which is how the test cross tells the two apart. The 3 : 1 ratio is what two heterozygotes give when crossed with each other.

Answer: A

In plants, meiosis takes place in the diploid sporophyte and produces haploid spores, which grow by mitosis into the haploid gametophyte; that gametophyte, already haploid, makes its gametes by mitosis. Meiosis cannot halve a cell that is already haploid, and making gametes directly by meiosis is the animal pattern, not the plant one.

Answer: E

The genes assort independently, so the probabilities multiply: Aa × aa gives the dominant phenotype with probability 1/2, and Bb × Bb and Cc × Cc each give it with probability 3/4, so 1/2 × 3/4 × 3/4 = 9/32. The figure 27/64 would apply only if both parents were heterozygous at all three loci, which ignores that the second parent is aa.

Answer: C

Human genes are interrupted by introns, which eukaryotic cells splice out of the pre-mRNA; bacteria have no splicing machinery, so they would translate the introns too. Mature mRNA has already been spliced, so its cDNA copy is an uninterrupted coding sequence. The genetic code is nearly universal, so bacterial ribosomes read human codons perfectly well.

Answer: B

Reading in triplets from the first AUG gives AUG–GCU–UAC–GAA and then UGA, a stop codon, so the polypeptide has 4 amino acids; the stop codon itself codes for none. Counting 6 means reading past UGA to the final UAA, and the second AUG in the sequence lies in a different reading frame, so it is never read as a codon.

Answer: D

The woman received her colour-blind father's X, so she is a carrier (XᴺXⁿ); the man is XⁿY. A daughter always receives her father's Xⁿ and has a 1/2 chance of receiving her mother's Xⁿ, so the probability is 1/2. The figure 1/4 is the chance that a child of unknown sex is a colour-blind daughter, not the chance for a daughter.

Answer: E

UAA is a stop codon, so translation ends at this point and a shortened, usually non-functional, protein is made: a nonsense mutation. A missense mutation swaps one amino acid for another, and a frameshift needs an insertion or deletion of bases that is not a multiple of three, not a substitution.

Answer: C

The recessive phenotype frequency is q² = 0.16, so q = 0.4 and p = 0.6, and the heterozygotes are 2pq = 2 × 0.6 × 0.4 = 0.48, or 48%. The figure 36% is p², the homozygous dominant individuals, and 84% is everyone with the dominant phenotype, heterozygotes and homozygotes together.

Answer: D

The pulmonary veins return oxygen-rich blood from the lungs to the left atrium. The pulmonary arteries carry blood the other way, from the right ventricle to the lungs, and it is deoxygenated: arteries and veins are named by the direction of flow relative to the heart, not by their oxygen content.

Answer: A

Osmoreceptors in the hypothalamus detect the rise, and the posterior pituitary releases ADH, which inserts aquaporins into the collecting ducts so more water returns to the blood: a negative feedback loop. ADH acts on reabsorption in the collecting ducts, not on filtration at the glomerulus, and its effect is less, more concentrated urine.

Answer: B

Simple squamous epithelium is a single layer of very flat cells, giving the shortest possible diffusion path for gases and solutes. Ciliated columnar epithelium lines the larger airways, where it moves mucus, and its tall cells would slow diffusion; stratified squamous epithelium has many layers and protects surfaces such as the skin.

Answer: C

Dissolved sodium chloride is spread evenly through the water as ions, so the solution has the same composition everywhere and shows a single phase: a homogeneous mixture. Distilled water is also uniform, but it is a pure compound, not a mixture; granite, sand in water and oil with vinegar all have visibly separate phases.

Answer: A

From the combined gas law, V₂ = V₁ × (T₂/T₁) × (p₁/p₂) = 6.0 × (450/300) × (100/150) = 6.0 × 1.5 × 2/3 = 6.0 L: the expansion caused by heating is exactly cancelled by the compression. Applying only the temperature change gives 9.0 L, and multiplying by both ratios instead of dividing by the pressure ratio gives 13.5 L.

Answer: E

The atomic number fixes 17 protons; the mass number 37 is protons plus neutrons, so there are 37 − 17 = 20 neutrons; and the single negative charge means one electron more than the protons, 18. Keeping 17 electrons ignores the charge, which changes only the number of electrons, never the number of protons or neutrons.

Answer: B

Across a period the nuclear charge rises while the electrons are added to the same shell, so they are pulled closer (smaller radius) and are harder to remove (higher ionisation energy, with small dips such as at aluminium and sulfur). Thinking that more electrons make a bigger atom gives the opposite trend; a new shell is started only at the beginning of the next period, and metallic character decreases from sodium to chlorine.

Answer: D

Potassium, an alkali metal of very low electronegativity, transfers its outer electron to fluorine, the most electronegative element, forming K⁺ and F⁻ ions held in a lattice. Hydrogen chloride is tempting because it ionises in water, but the pure substance consists of molecules with a polar covalent H–Cl bond; CH₄, Cl₂ and NH₃ are covalent molecules too.

Answer: A

Iron(III) is Fe³⁺ and sulfate is SO₄²⁻; balancing the charges needs two Fe³⁺ (+6) for three SO₄²⁻ (−6), giving Fe₂(SO₄)₃. FeSO₄ is iron(II) sulfate, and Fe₂(SO₃)₃ has the right ratio but contains sulfite, SO₃²⁻, not sulfate.

Answer: C

The molar mass of CO₂ is 12 + 2 × 16 = 44 g/mol, so 11 g is 0.25 mol of molecules; each molecule has two oxygen atoms, giving 0.50 mol of O, that is 0.50 × 6.0 × 10²³ = 3.0 × 10²³ atoms. Stopping at the number of molecules gives 1.5 × 10²³, and counting every atom, carbon included, gives 4.5 × 10²³.

Answer: E

The amount needed is concentration × volume = 0.40 mol/L × 0.250 L = 0.10 mol, and with a molar mass of 23 + 16 + 1 = 40 g/mol that is 0.10 × 40 = 4.0 g. Multiplying the concentration by the molar mass alone gives 16 g, which would be right only for a full litre of solution.

Answer: B

CN⁻ is the conjugate base of a weak acid, so it undergoes hydrolysis, CN⁻ + H₂O ⇌ HCN + OH⁻, and the extra OH⁻ makes the solution basic; K⁺, from the strong base KOH, does not react with water. Assuming every salt is neutral holds only for salts of a strong acid with a strong base, such as NaCl, and KCN contains no hydroxide of its own.

Answer: D

The –CO– in the middle of the chain is a carbonyl group with a carbon atom on each side, which is what defines a ketone, and the -one ending of the name says so. An aldehyde also has a carbonyl group, but at the end of the chain, bonded to a hydrogen atom (–CHO).

Answer: D

Ksp = [Ba²⁺][SO₄²⁻]; the sulfate already present (0.010 mol/L) is far more than the little BaSO₄ adds, so s = Ksp / 0.010 = 1.0 × 10⁻¹⁰ / 1.0 × 10⁻² = 1.0 × 10⁻⁸ mol/L. The value 1.0 × 10⁻⁵ mol/L is the solubility in pure water: the common ion shifts the equilibrium towards the solid and lowers it a thousandfold.

Answer: A

Sulfur goes from −2 in H₂S to 0 (oxidised, so H₂S is the reducing agent) and from +4 in SO₂ to 0 (reduced, so SO₂ is the oxidising agent), while hydrogen stays at +1 throughout. Calling SO₂ the reducing agent confuses the species that is reduced with the one that does the reducing: the oxidising agent is the one that is itself reduced.

Answer: B

Metal oxides such as BaO react with water to give hydroxides (Ba(OH)₂) and are basic, while non-metal oxides such as N₂O₅ give oxyacids (HNO₃) and are acidic. CO₂ with CaO and Cl₂O₇ with Na₂O do contain one of each, but in the reverse order; SO₂ and P₄O₁₀ are both acidic, Li₂O and MgO both basic.

Answer: E

Magnesium and oxygen combine into one product, and in doing so magnesium goes from 0 to +2 and oxygen from 0 to −2, so electrons are transferred. The reaction of CaO with water is also a combination, but every element keeps its oxidation number (Ca +2, O −2, H +1), so it is not a redox reaction; the others are a precipitation, a decomposition and a neutralisation.

Answer: C

Bond polarity grows with the difference in electronegativity: 1.9 for H–F, against 1.4 for O–H, 0.9 for N–H, 0.5 for C–Cl and 0.4 for C–H. O–H is tempting because water is the familiar example of a polar molecule, but fluorine pulls the shared pair even more strongly than oxygen.

Answer: C

A catalyst opens a different reaction pathway whose energy barrier is lower, so a larger fraction of collisions has enough energy to react; it is recovered unchanged at the end. Raising the particles' kinetic energy is what heating does, and since a catalyst speeds up the forward and reverse reactions equally, it changes neither the equilibrium constant nor the enthalpy change.

Answer: E

NaOH converts 0.10 mol of HA into A⁻, leaving 0.10 mol HA and 0.10 mol A⁻: a buffer in which [H⁺] = Ka × [HA]/[A⁻] = 1.0 × 10⁻⁴ × 1 = 1.0 × 10⁻⁴ mol/L, so pH = 4.0. Expecting pH 7.0 assumes the base has neutralised all the acid, but only half of it has reacted; treating the remaining HA as a strong acid gives 1.0.

Answer: A

Isomers share a molecular formula but differ in structure: cyclohexane (a ring) and hex-1-ene (a chain with one C=C) are both C₆H₁₂. Cyclohexane and benzene both have six-carbon rings, but benzene is C₆H₆; the other pairs differ in hydrogen or oxygen atoms (C₂H₆O and C₂H₄O, C₄H₁₀ and C₄H₈, C₃H₈O and C₃H₆O₂).

Answer: D

4.0 g of H₂ is 2.0 mol and 16 g of O₂ is 0.50 mol; 2.0 mol of H₂ would need 1.0 mol of O₂, so oxygen is the limiting reagent, and 0.50 mol O₂ gives 1.0 mol H₂O, that is 18 g. Basing the answer on the hydrogen gives 36 g, and adding the two masses gives 20 g, which ignores the 2.0 g of hydrogen left unreacted.

Answer: B

The atom has 26 electrons (iron) and its last electrons enter the 3d subshell, which makes it a d-block transition element; its outermost shell is n = 4, so it sits in period 4. The 4s² is tempting, but it is filled before the 3d in every period-4 transition element; the configuration gives the number of protons, not of neutrons, and the outermost shell (n = 4) holds only two electrons.

Answer: C

The seven SI base quantities are length, mass, time, electric current, thermodynamic temperature, amount of substance (unit: the mole) and luminous intensity. Electric charge is tempting, but it is derived: the coulomb is one ampere flowing for one second, and it is the current that is the base quantity.

Answer: A

1 J = 1 kg·m²/s², and 1 kg = 10³ g while 1 m² = (10² cm)² = 10⁴ cm², so 1 J = 10³ × 10⁴ g·cm²/s² = 10⁷ erg. The answer 10⁵ comes from converting the metre only once instead of squaring it (it is the number of dynes in a newton), and 10⁻⁷ is the conversion turned upside down.

Answer: D

Pressure × volume has units (N/m²) × m³ = N·m = J, so its dimensions are those of energy, [M L² T⁻²]. Force × velocity is tempting because it contains a force, but a force times a distance per unit time is energy per unit time, which is power; mass × velocity is momentum and force ÷ area is pressure.

Answer: B

With constant deceleration the average velocity is (12 + 0)/2 = 6.0 m/s, so the distance is 6.0 m/s × 4.0 s = 24 m. The answer 48 m treats the initial 12 m/s as if it lasted the whole 4.0 s, ignoring that the speed falls steadily to zero.

Answer: E

The angular velocity is ω = 2π × 2.0 = 4π rad/s, so a = ω²r = 16π² × 0.25 = 4π² m/s² (about 39 m/s²). Equivalently v = ωr = π m/s and a = v²/r = π²/0.25 = 4π² m/s²; stopping at v² gives π² m/s², and forgetting to multiply ω² by the radius gives 16π² m/s².

Answer: B

In simple harmonic motion v = ω√(A² − x²) = 10 × √(0.010 − 0.0025) = 10 × √0.0075 = 10 × 0.05√3 = (√3/2) m/s. The answer 0.50 m/s assumes the speed falls in proportion to the distance from the centre, but the maximum speed Aω = 1.0 m/s falls off with the square root, so halfway out the particle still has most of its speed.

Answer: D

Surface gravity is g = GM/R², so doubling M doubles g but doubling R divides it by four: g on the planet is 10 × 2/4 = 5.0 m/s², and the weight is 80 × 5.0 = 400 N. Mass is the amount of matter and stays 80 kg everywhere; 800 N comes from assuming the doubled mass and doubled radius cancel, forgetting that the radius is squared.

Answer: A

Only the component of the force perpendicular to the handle turns the bolt: M = F·r·sin θ = 40 × 0.50 × 0.50 = 10 N·m. Using cos 30° (about 17 N·m) takes the component along the handle, which pulls straight through the bolt and has no turning effect; 20 N·m ignores the angle altogether.

Answer: C

Momentum is a vector, so the velocity changes from +15 m/s to −10 m/s, a change of 25 m/s: the impulse is 0.20 × 25 = 5.0 kg·m/s, and F = Δp/Δt = 5.0/0.010 = 500 N. The answer 100 N subtracts the speeds (15 − 10 = 5 m/s) as if the ball had kept moving the same way.

Answer: E

Energy is conserved: ½mv² = mgh, so the mass cancels and h = v²/2g = 100/20 = 5.0 m. Forgetting the ½ in the kinetic energy gives 10 m, and the mass of 0.50 kg is irrelevant to the height reached.

Answer: A

By Pascal's law the pressure is the same on both pistons, so F₁/A₁ = F₂/A₂ and F₁ = 15 000 × 10/500 = 300 N. Multiplying by the area ratio instead of dividing gives 750 000 N, and 15 000 N assumes the press passes on the force rather than the pressure.

Answer: C

By Stevin's law p = p₀ + ρgh = 1.0 × 10⁵ + 1000 × 10 × 20 = 3.0 × 10⁵ Pa, about 3 atm: every 10 m of water adds about 1 atm. The answer 2 atm is the pressure due to the water alone (the gauge pressure), leaving out the atmosphere pressing on the lake's surface.

Answer: E

The upthrust is the loss of apparent weight: 2.0 N in water and 1.6 N in liquid X. The stone displaces the same volume in both, and by Archimedes' principle the upthrust is proportional to the liquid's density, so ρ = 1000 × 1.6/2.0 = 800 kg/m³. Taking the ratio the other way up gives 1250 kg/m³, a denser liquid, which cannot be right because liquid X buoys the stone less than water does.

Answer: D

Convection transfers energy by the bulk movement of a fluid, driven by the lower density of the warmer part. Conduction also happens in the water, but it passes energy from particle to particle without the material itself flowing, which is not what the rising and sinking currents describe.

Answer: B

Melting takes mL = 0.20 × 330 = 66 kJ with no change in temperature, and warming the water takes mcΔT = 0.20 × 4.2 × 20 = 16.8 kJ, a total of 82.8 kJ. Stopping at 66 kJ forgets the warming, and 16.8 kJ forgets that melting needs latent heat even though the temperature stays at 0 °C.

Answer: B

From pV/T = constant, with temperatures in kelvin (300 K and 600 K), p₂ = 1.0 × 10⁵ × (6.0/2.0) × (600/300) = 6.0 × 10⁵ Pa. Using the Celsius temperatures instead gives a factor of 327/27 ≈ 12 in place of 2, and so 3.6 × 10⁶ Pa; 3.0 × 10⁵ Pa allows for the compression but not the heating.

Answer: E

R = ρL/A, and the cross-sectional area grows with the square of the diameter, so doubling the diameter makes the area four times larger: R = 8.0 × 2/4 = 4.0 Ω. The answer 8.0 Ω assumes doubling the diameter only doubles the area, so that it cancels the doubled length.

Answer: A

In series 1/C = 1/3.0 + 1/6.0 = 1/2.0, so the combination is 2.0 µF and Q = CV = 2.0 × 12 = 24 µC; capacitors in series all carry the same charge. The answer 108 µC adds the capacitances (9.0 µF), which is the rule for capacitors in parallel, not in series.

Answer: D

The two fields point in opposite directions between like charges, and they cancel where k(4.0)/x² = k(1.0)/(3.0 − x)², that is where x/(3.0 − x) = 2, so x = 2.0 m from the larger charge (1.0 m from the smaller). The answer 2.4 m comes from making the field fall with the distance rather than its square, and the point cannot be nearer the larger charge, whose field is stronger there.

Answer: C

By Faraday's law of electromagnetic induction the induced emf equals the rate of change of magnetic flux through the coil; with the magnet at rest the flux is constant, so there is no emf and no current. Being in a magnetic field is not enough: a current is induced only while the magnet, or the coil, moves or the field changes.