UniCamillus Admission Test Practice Paper 1, with Answers

Answer: B

If Marta were a nurse on Ward B she would have completed the course; she has not, so she cannot be one. The premises say nothing about where Marta works or whether she is a nurse at all, so 'Marta is a nurse on another ward' and 'Marta works on Ward B, but not as a nurse' are both possible but neither is certain.

Answer: D

The equivalent form is the contrapositive: if the test is not ordered, then the condition 'fever and rash' is false, and the negation of 'fever and rash' is 'no fever or no rash'. Saying the patient has neither is too strong: a patient with a fever but no rash may well get no test without breaking the rule. The first option is the converse, which does not follow.

Answer: C

With Q on Wednesday, P and R must take Monday–Tuesday or Thursday–Friday. If they took Monday and Tuesday, S and T would be left with Thursday and Friday, and S before T would put T on Friday, which is not allowed. So P is on Thursday and R on Friday, with S on Monday and T on Tuesday.

Answer: E

Bea is always in neurology: if Amir is in dermatology the fourth rule rules out paediatrics, and if Amir is in paediatrics that clinic is taken. With Carlo not in cardiology, either Carlo is in paediatrics (so Dina, who cannot be there, takes cardiology) or Carlo is in dermatology (which the last rule allows only if Dina is in cardiology). Both arrangements are possible, so Amir's and Carlo's clinics are not fixed, but Dina's is.

Answer: A

A mark of 26 is below 27 but at least 25, and €18,000 is below €20,000, so the second condition is met and there is no earlier grant. The student averaging 28 meets the mark condition but has already had a grant; the student averaging 25 fails only because an income of exactly €20,000 is not below €20,000.

Answer: E

75 is odd, so it becomes 76 ÷ 2 = 38; 38 is even and becomes 19; 19 is odd and becomes 20 ÷ 2 = 10; 10 is not less than 10, so the step is applied once more to give 5. Stopping at 10 forgets that the procedure only ends below 10.

Answer: A

The passage says that needless use exposes bacteria to the drugs, and that exposure kills the susceptible cells and leaves the resistant ones to multiply. The idea that antibiotics make bacteria mutate is exactly what the first sentence denies: the resistant cells were already there by chance, and the drug only selects them.

Answer: B

The passage's closing sentences say the study shows only that walking and less heart disease go together, not that walking causes the difference, and the councillor's plan assumes it does. The figures were adjusted for age and smoking, as the passage states, so the criticism is not that smoking was ignored; the study's size is never questioned.

Answer: D

75% of 24 beds is 18 occupied beds. After 4 discharges and 7 admissions there are 18 − 4 + 7 = 21 patients, and 21 ÷ 24 = 7/8 = 87.5%. Work in beds, not percentages: each bed is 1/24 of the ward, so the net gain of 3 patients adds 12.5 percentage points to the original 75%, not 3 or 5.

Answer: C

Out of every 100 adults, 40 are vaccinated and 4 of them catch flu; 60 are unvaccinated and 15 of them catch flu. So 19 adults catch flu, of whom 4 were vaccinated: 4/19. The tempting 4/15 compares the vaccinated cases with the unvaccinated cases instead of with all the cases.

Answer: B

Cellulose is a long, unbranched chain of glucose units joined by β-1,4-glycosidic bonds, and human amylases attack only the α bonds of starch and glycogen, so it reaches the colon intact as dietary fibre. Chitin is also a structural polysaccharide that resists digestion, but it builds the walls of fungi and the exoskeletons of arthropods, not plant cell walls. Amylose and amylopectin are the two components of starch, which plants store and we digest, and glycogen is the animal storage form.

Answer: D

The fatty-acid tails are non-polar and cannot form hydrogen bonds with water, so the arrangement that disturbs water least buries them inside the bilayer while the polar heads face the water on both sides: a weak, non-covalent effect, which is also why membranes are fluid. Ionic bonds between the heads are the tempting answer, but the negatively charged phosphate groups interact with water and ions around them rather than locking neighbouring heads together. No covalent or disulfide bonds join the molecules, and hydrocarbon tails have no groups able to form hydrogen bonds.

Answer: C

O₂ is a small, non-polar molecule, so it dissolves in the hydrophobic core of the bilayer and diffuses straight through down its concentration gradient. Glucose is the tempting answer because it is small and enters cells passively, but it is polar and needs carrier proteins (facilitated diffusion). Na⁺ is charged and moves only through channels or pumps, and ATP and glycine are charged or polar as well.

Answer: A

The pump is primary active transport: each cycle uses one ATP to expel 3 Na⁺ and bring in 2 K⁺, against both gradients, keeping Na⁺ high outside the cell and K⁺ high inside. Because three positive charges leave for every two that enter, the pump is electrogenic and adds a little to the negative resting potential. The reverse ratio, 2 out and 3 in, is the usual slip; moving Na⁺ in and K⁺ out would be going down their gradients, which needs no ATP.

Answer: E

Both homologues go to the same pole in meiosis I, so one secondary spermatocyte has 24 chromosomes and the other 22; meiosis II then separates sister chromatids normally, so each gives two sperm with its own number: two with 24 and two with 22, and no normal gamete. One 24, one 22 and two 23 is what happens when the error is in meiosis II instead, where only one of the two secondary cells is affected. 47 and 45 are the counts of the zygotes such sperm would produce, not of the sperm themselves.

Answer: B

On the thylakoid membranes light energy makes ATP and reduces NADP⁺ to NADPH; in the stroma the Calvin cycle spends that ATP and uses NADPH as the reducing agent that turns fixed CO₂ into sugar. ADP and NADP⁺ go the other way: they are what the Calvin cycle returns to the thylakoids to be recharged. O₂ is a by-product released from the splitting of water, not something the cycle uses, and NADH and FADH₂ are the carriers of cellular respiration.

Answer: C

Complex IV is where electrons are handed to O₂; blocked, the whole chain backs up, its carriers stay reduced, NADH can no longer be reoxidised and oxygen is not used, so proton pumping and oxidative phosphorylation stop. Glycolysis stopping is the tempting answer, but glycolysis takes place in the cytosol and uses no oxygen: it carries on, regenerating NAD⁺ by making lactate, which is why cyanide poisoning causes lactic acidosis. With no pumping the proton gradient runs down rather than steepening, and there is no route from NADH to O₂ that bypasses the chain.

Answer: A

Males are hemizygous for X-linked genes (XY): there is no second X to carry a normal allele, so a single recessive allele is expressed, while a female needs two copies, one from each parent. A Y-linked gene would affect only males and pass from father to every son, which is not the pattern here. A son receives his X from his mother, not his father, and X inactivation in females is random, cell by cell.

Answer: D

The two template strands are antiparallel, and DNA polymerase can only extend a 3′ end; on one template it follows the fork continuously (leading strand), but on the other it has to work away from the fork, so it makes a short piece each time a new stretch is exposed, and DNA ligase later seals the pieces. Primers on only one template is the tempting answer, but the leading strand needs a primer too: the lagging strand simply needs a new one for every fragment. Both templates are DNA, and the helicase unwinds the double helix continuously.

Answer: E

An affected child shows that both parents are carriers (Aa × Aa), whose children are AA, Aa and aa in the ratio 1 : 2 : 1. The second child is known to be healthy, which rules out aa and leaves AA and Aa in the ratio 1 : 2, so the probability of being a carrier is 2/3. 1/2 is the trap: it is the chance that a child of these parents is a carrier before we know whether the child is healthy.

Answer: B

Without lactose the repressor sits on the operator and blocks RNA polymerase; allolactose acts as an inducer, binding the repressor and changing its shape so that it falls off the operator, and with glucose absent the cAMP–CAP complex further boosts transcription. Assuming the repressor is no longer made is the common slip: its gene is expressed all the time, and it is the protein's activity that changes. A repressor, by definition, does not activate transcription, and no enzyme removes genes from the chromosome.

Answer: C

mRNA is read in consecutive triplets from a fixed start, so losing one nucleotide shifts the reading frame and every codon after it is read differently (a frameshift), usually until a premature stop codon appears. Deleting or inserting three nucleotides is the tempting answer, but it removes or adds a single codon's worth and leaves the frame intact, so the rest of the protein is unchanged. A substitution affects one codon at most, and at the third position it is often silent because the code is degenerate.

Answer: D

A few founders carry only a chance sample of the mainland's alleles, and in a small population allele frequencies keep changing at random from one generation to the next: this is genetic drift, here starting from the founder effect. Natural selection is the tempting answer, but it needs the allele to affect survival or reproduction, and this one is neutral. Gene flow is exchange with other populations, which isolation prevents and which would make the island more like the mainland, not less.

Answer: E

Stretching of the cervix triggers oxytocin release from the posterior pituitary, oxytocin strengthens the contractions, and stronger contractions stretch the cervix further: the response amplifies the stimulus until birth ends the loop. The other four are negative feedback, in which the response opposes the change. Thyroxine reducing TSH may look different because it acts on the hormone's own control, but it too damps the original rise.

Answer: A

The plateau of the cardiac action potential, kept up by Ca²⁺ entering through slow channels, makes the refractory period last almost until the muscle has relaxed, so a new impulse cannot arrive in time to add to a contraction already under way, and the ventricles always relax and can refill. Intercalated discs are real and let the cells contract together as a unit, but that explains coordination, not the absence of tetanus. Cardiac muscle is striated with sarcomeres, uses Ca²⁺ to trigger contraction and is innervated by both sympathetic and parasympathetic fibres.

Answer: C

Diamond is the element carbon, and carbon dioxide and glucose are compounds of fixed composition, so all three are pure substances. Every other list hides at least one mixture: air and steel, brass (an alloy of copper and zinc), sea water and milk all have a variable composition. Brass is the most tempting, because it looks like a single metal, but it is a homogeneous mixture.

Answer: D

At the same volume and temperature, pV = nRT makes pressure proportional to the amount of gas. 8.0 g of O₂ (M = 32 g/mol) is 0.25 mol, while 8.0 g of CH₄ (M = 16 g/mol) is 0.50 mol, so the methane exerts twice the pressure. Answering 1 : 1 assumes equal masses mean equal pressures, and 1 : 2 inverts the ratio of the molar masses.

Answer: B

Atomic radius grows down a group, as a new electron shell is added, and shrinks across a period, as the nuclear charge rises while electrons enter the same shell. Potassium and calcium are both in period 4, one shell more than sodium, magnesium and chlorine, and potassium comes first in that period. Calcium is the tempting choice, but its extra proton pulls the same shells in more tightly, so it is smaller than potassium.

Answer: E

With similar molecular masses the London (dispersion) forces are comparable, so the deciding factor is the strongest intermolecular force present. Only ethanol has a hydrogen atom bonded to oxygen, so only its molecules form hydrogen bonds with one another, and it boils at about 78 °C. Methoxymethane and ethanal contain oxygen and are polar, but neither has an O–H bond, so they have only dipole–dipole and dispersion forces and boil far lower.

Answer: A

The hydrogen carbonate ion is HCO₃⁻, so two of them balance one Ca²⁺: Ca(HCO₃)₂. Hypochlorite is ClO⁻ (NaClO₂ is sodium chlorite), nitrate is NO₃⁻ (NH₄NO₂ is ammonium nitrite), and permanganate is MnO₄⁻, giving KMnO₄ (K₂MnO₄ is potassium manganate). FePO₄ pairs one Fe with one PO₄³⁻, so the iron is +3: it is iron(III) phosphate, and iron(II) phosphate would be Fe₃(PO₄)₂.

Answer: D

One C₂H₆ gives 2CO₂ and 3H₂O, which need 4 + 3 = 7 oxygen atoms, that is 3.5 O₂. Doubling everything to clear the fraction gives 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O, with 14 oxygen atoms on each side. The 14 counts oxygen atoms rather than O₂ molecules.

Answer: E

Each formula unit gives two Al³⁺ and three SO₄²⁻, five ions in all: Al₂(SO₄)₃ → 2Al³⁺ + 3SO₄²⁻. So 0.10 mol releases 0.20 mol of Al³⁺ plus 0.30 mol of SO₄²⁻, 0.50 mol of ions. The 1.7 mol figure counts the 17 atoms in the formula, but each sulfate ion stays together as one ion.

Answer: C

The 8.8 g lost is CO₂: 8.8 ÷ 44 = 0.20 mol. The equation is 1 : 1, so 0.20 mol of CaCO₃ decomposed, which is 0.20 × 100 = 20 g, and 20 ÷ 25 × 100 = 80%. Dividing the 8.8 g loss by 25 g (35.2%) takes the carbon dioxide's mass for the carbonate's, and 20% is the carbonate's mass in grams mistaken for a percentage.

Answer: A

Doubling [A] multiplies the rate by 2² = 4, because the reaction is second order in A, and halving [B] multiplies it by ½, because it is first order in B. Overall 4 × ½ = 2, so the rate doubles; k does not change, as the temperature is constant. Answering that it stays the same treats both orders as 1, ignoring the square on [A].

Answer: B

500 mL is five times 100 mL, so the bag holds 5 × 0.90 g = 4.5 g of NaCl. The 9.0 g figure would be the amount in a full litre, and 0.45 g divides by 1000 mL instead of 100 mL.

Answer: E

pOH = −log(1.0 × 10⁻³) = 3, and at 25 °C pH + pOH = 14, so pH = 14 − 3 = 11, a basic solution. Equivalently, [H₃O⁺] = Kw ÷ [OH⁻] = 1.0 × 10⁻¹¹ mol/L. Answering 3 gives the pOH, not the pH.

Answer: C

The ethanoate ion is the buffer's base: it takes up the added H₃O⁺ and turns it into the weak acid CH₃COOH, which barely ionises, so [H₃O⁺] stays almost unchanged. The reaction of CH₃COOH with OH⁻ is real, but it is how the buffer takes up an added base, not an added acid. Na⁺ and Cl⁻ are spectator ions that stay dissolved, and HCl is a strong acid that does not re-form in water.

Answer: A

Nitrogen is +5 in HNO₃ and +2 in NO, so only the nitrogen that ends up in the 2 NO molecules is reduced: two HNO₃ act as the oxidant, gaining 2 × 3 = 6 electrons, exactly the 3 × 2 = 6 electrons lost as three Cu go from 0 to +2. The other six HNO₃ supply the nitrate ions of Cu(NO₃)₂, where nitrogen is still +5, so they act only as an acid. Answering 8 assumes every nitric acid molecule is reduced.

Answer: B

Both are C₃H₆O, but propanal has its C=O at the end of the chain as an aldehyde group (–CHO), while propanone has it on the middle carbon as a ketone: structural isomers with different functional groups. Members of a homologous series differ by CH₂ in formula, which these do not, and cis–trans isomerism needs a C=C double bond with two different groups on each carbon. Neither compound has the –COOH group of a carboxylic acid.

Answer: D

C₄H₈ has one double bond or one ring. The open-chain alkenes are but-1-ene, cis-but-2-ene, trans-but-2-ene and 2-methylpropene (four), and the alicyclic ones are cyclobutane and methylcyclopropane (two), six in all. Neither but-1-ene nor 2-methylpropene has cis and trans forms, because one carbon of their C=C carries two identical hydrogen atoms. Five forgets to split but-2-ene into its cis and trans forms, and four leaves out the rings.

Answer: A

1 L = 10⁻³ m³ and 1 h = 3600 s, so 3.6 L/h = 3.6 × 10⁻³ m³ ÷ 3600 s = 1.0 × 10⁻⁶ m³/s (that is, 1 cm³ every second). 1.0 × 10⁻³ m³/s comes from dividing 3.6 by 3600 but treating litres as cubic metres; 6.0 × 10⁻⁵ m³/s converts the hour into 60 instead of 3600 seconds.

Answer: C

Pressure is force per area, [M L T⁻²]/[L²] = [M L⁻¹ T⁻²], and energy per volume is [M L² T⁻²]/[L³] = [M L⁻¹ T⁻²], the same; indeed 1 Pa = 1 N/m² = 1 J/m³. Density and specific gravity are tempting because they share a number, but specific gravity is a ratio of two densities and is dimensionless, while density is [M L⁻³]. Impulse carries an extra time factor compared with force, and power is work per unit time.

Answer: E

With constant acceleration the average speed is the mean of the initial and final speeds, (10 + 30) ÷ 2 = 20 m/s, so t = 100 m ÷ 20 m/s = 5.0 s (check: a = (30² − 10²) ÷ (2 × 100) = 4.0 m/s², and 20 ÷ 4.0 = 5.0 s). 10 s divides the distance by the initial speed only, and 4.0 is the acceleration in m/s², not a time.

Answer: B

If the fall lasts t seconds, the distance covered in the last second is ½gt² − ½g(t − 1)² = gt − ½g = 10t − 5, so 10t − 5 = 35 and t = 4.0 s; the height is ½ × 10 × 4.0² = 80 m. 45 m is how far the stone has fallen after 3.0 s, before its last second, and 61 m wrongly treats 35 m/s as the final speed (35² ÷ 20).

Answer: D

A specific gravity of 0.80 means a density of 0.80 × 1000 kg/m³ = 0.80 kg/L, so the mass is 0.80 × 2.5 = 2.0 kg and the weight is 2.0 kg × 10 m/s² = 20 N. 2.0 kg is the mass, not the weight, and is not even in units of force; 25 N treats the oil as if it were water.

Answer: A

Taking moments about the right-hand end: the person is 3.0 m from it and the plank's weight acts at its midpoint, 2.0 m from it, so F × 4.0 = 600 × 3.0 + 200 × 2.0 = 2200 N·m and F = 550 N (the right support then gives 800 − 550 = 250 N). 250 N is the right-hand support's force, and 450 N leaves out the plank's own weight.

Answer: E

The total momentum stays zero, so the fragments have momenta of equal size p. Kinetic energy is p²/(2m), so with equal p it is inversely proportional to mass: the ratio is 6.0 : 2.0 = 3 : 1 (the lighter fragment moves three times as fast). 1 : 1 assumes kinetic energy is shared equally, which only momentum is; 9 : 1 squares the speed ratio but forgets the mass ratio.

Answer: C

Gravity supplies the centripetal force: GMm/r² = mv²/r, so v = √(GM/r) and the speed varies as 1/√r. Quadrupling the radius halves the speed, so the inner satellite is twice as fast: 2 : 1. 4 : 1 takes the speed as inversely proportional to r rather than to √r, and 1 : 2 assumes the higher orbit is the faster one.

Answer: B

By Stevin's law the pressure of a liquid column is p = ρgh, so h = p ÷ (ρg) = 1.0 × 10⁵ ÷ (1000 × 10) = 10 m. 0.76 m is the height of the mercury column (760 mmHg) that balances one atmosphere; water is about 13.6 times less dense, so its column is that much taller.

Answer: D

A floating body displaces its own weight, so the submerged fraction equals ρ(body)/ρ(liquid). In water 60% is submerged, so the wood's density is 0.60 × 1000 = 600 kg/m³; in the oil the submerged fraction is 600 ÷ 800 = 0.75, that is 75%. 48% multiplies 0.60 by 0.80 instead of dividing, which would wrongly make the block float higher in the less dense liquid; 25% is the part above the oil's surface.

Answer: C

Heat lost by the hot water equals heat gained by the cold water, and the specific heat cancels: 200 × (80 − T) = 300 × (T − 20), so 16 000 + 6000 = 500T and T = 44 °C. 50 °C is the plain average, which would only be right for equal masses; 56 °C swaps the masses, giving the larger share to the hot water.

Answer: A

The work done by the gas at constant pressure is W = pΔV = 1.0 × 10⁵ Pa × 3.0 × 10⁻³ m³ = 300 J. By the first law, ΔU = Q − W = 800 − 300 = +500 J. 1100 J adds the work instead of subtracting it: the gas spends energy pushing back its surroundings; 800 J ignores the expansion work altogether.

Answer: B

The power is P = VI = 220 × 5.0 = 1100 W, and 10 minutes is 600 s, so E = Pt = 1100 × 600 = 660 000 J = 660 kJ. 11 kJ multiplies the power by 10 without converting minutes into seconds, and 66 kJ is the energy for one minute only.

Answer: E

The parallel pair is equivalent to (6.0 × 12) ÷ (6.0 + 12) = 4.0 Ω, so the total resistance is 8.0 Ω and the battery current is 24 ÷ 8.0 = 3.0 A. The pair then has 3.0 × 4.0 = 12 V across it, so the 6.0 Ω resistor carries 12 ÷ 6.0 = 2.0 A (and the 12 Ω resistor the other 1.0 A). 3.0 A is the total current, which divides between the branches; 4.0 A wrongly applies the whole 24 V to the 6.0 Ω resistor.

Answer: D

The charge Q = CV = 4.0 µF × 100 V = 400 µC is conserved and shares between the two capacitors, now in parallel (8.0 µF), so the common voltage is 50 V. The total energy is ½ × 8.0 × 10⁻⁶ × 50² = 1.0 × 10⁻² J = 10 mJ, half of the initial ½ × 4.0 × 10⁻⁶ × 100² = 20 mJ. 20 mJ assumes the energy is conserved, but charge is, not energy (the rest is dissipated in the wires as the charge flows); 5.0 mJ is what each capacitor holds, not the total.

Answer: C

Cholera is caused by the bacterium Vibrio cholerae, which is passed out in the faeces of infected people and spreads when sewage gets into drinking water or food (the faecal–oral route). That is why safe water and sanitation prevent it. Mosquito bites spread diseases such as malaria and dengue, not cholera.

Answer: A

Dunant witnessed the aftermath of the Battle of Solferino in 1859 and described it in 'A Memory of Solferino', which called for trained volunteer relief societies and an international agreement to protect the wounded; the ICRC and the first Geneva Convention followed. Balaclava, a Crimean War battle, is the tempting choice, but it was not in Italy, and Verdun came half a century after the ICRC already existed.

Answer: E

The Declaration of Alma-Ata (1978, in today's Kazakhstan) set out primary health care, close to where people live and work, as the foundation of health systems; the Astana Declaration of 2018 renewed that commitment. The Declaration of Helsinki is also about health, but it sets ethical principles for medical research on human subjects, not for organising health care.

Answer: B

In 2003 Urbani identified SARS, a new coronavirus disease, and his early warning helped WHO launch measures that contained the outbreak; he caught the infection while treating patients and died that year. MERS is also caused by a coronavirus, which makes it tempting, but it was first identified in 2012, and COVID-19 appeared only in 2019.

Answer: D

Camillo de Lellis founded the Order of the Ministers of the Sick (the Camillians), whose members wear a red cross on their habit, and set out a way of caring for hospital patients built on respect for their dignity; he is a patron saint of the sick, nurses and hospitals. The red cross makes the ICRC tempting, but that was founded by Henry Dunant in 1863, more than two centuries later.