Masaryk University Entrance Exam Sample Test with Answers (2027)

Answer: A

The atomic number gives the protons: 26. Neutrons = mass number minus atomic number = 56 − 26 = 30. A 3+ charge means three electrons have been lost, so 26 − 3 = 23 electrons. 29 electrons would be a 3− anion, 23 protons would change the element, and 56 is the mass number (protons plus neutrons), not the neutron count.

Answer: C

Fe is [Ar] 3d⁶ 4s². A transition metal loses its 4s electrons first, then 3d, so Fe³⁺ is [Ar] 3d⁵. By Hund's rule the five d electrons occupy the five d orbitals singly, giving 5 unpaired electrons. 4 is the answer for the Fe atom or Fe²⁺ (3d⁶: one pair and four single electrons); removing the three electrons from 3d alone would give 3d³4s², which is not how the ion forms.

Answer: A

The highest shell is n = 3 (period 3), the last electrons enter a p sublevel (p-block), and the outer shell holds 3s² 3p⁴ = 6 valence electrons, which places it in group 16: this is sulfur, a nonmetal. It has six valence electrons, not four (the 3p⁴ alone is only part of the valence shell), and it tends to gain two electrons to form S²⁻, not lose two.

Answer: B

M(CO₂) = 12 + 2 × 16 = 44 g/mol, so n = 8.8 ÷ 44 = 0.20 mol and V = 0.20 × 22.4 = 4.48 L. 2.24 L corresponds to 0.10 mol, 8.96 L to 0.40 mol, and 44.8 L to 2 mol (it comes from multiplying instead of dividing by the molar mass somewhere along the way).

Answer: E

M(CaCO₃) = 40 + 12 + 48 = 100 g/mol, so n = 25 ÷ 100 = 0.25 mol. The equation is 1 : 1, so 0.25 mol CaO forms, and M(CaO) = 56 g/mol: 0.25 × 56 = 14 g. No option gives 14 g. 11 g is the mass of CO₂ given off (0.25 × 44), 25 g ignores the loss of CO₂, and 7.0 g and 28 g come from halving or doubling the amount.

Answer: D

n(Al) = 5.4 ÷ 27 = 0.20 mol. To react completely it would need 0.20 × 3 = 0.60 mol HCl, but only 0.40 mol is present, so HCl is the limiting reactant. 6 mol HCl gives 3 mol H₂, so 0.40 mol HCl gives 0.20 mol H₂: 0.20 × 22.4 = 4.48 L. 6.72 L (0.30 mol H₂) is what you get by assuming all the aluminium reacts; 13.4 L takes the 0.60 mol of HCl that the aluminium would need as if it were the amount of H₂.

Answer: C

Each C=O bond in CO₂ is polar, but the molecule is linear, so the two bond dipoles point in opposite directions and cancel: the molecule has no overall dipole. H₂O (bent) and NH₃ (trigonal pyramidal) have bond dipoles that do not cancel, and HCl has a single polar bond, so all three are polar molecules.

Answer: E

In NH₃ the nitrogen has three bonding pairs and one lone pair: four electron domains, so sp³, and with one position taken by the lone pair the atoms form a trigonal pyramid. CH₄ is sp³ but tetrahedral (no lone pair); H₂O is sp³ but bent (two lone pairs); BF₃ is sp² and trigonal planar; CO₂ is sp and linear.

Answer: A

Every bond contains exactly one σ bond; a double bond adds one π bond and a triple bond adds two. The bonds are three C–H, one C=C, one C–C and one C≡N: six bonds, so 6 σ. π bonds: 1 (in C=C) + 2 (in C≡N) = 3. Counting only the two π bonds of C≡N gives 2 π; forgetting one C–H gives 5 σ; and 3 σ and 6 π simply reverses the two counts.

Answer: B

Mass fraction = mass of solute ÷ mass of solution = 20 ÷ (20 + 180) = 20 ÷ 200 = 0.10, or 10 %. 11.1 % divides by the mass of the water alone (20 ÷ 180), which is the common mistake; 20 % treats the solute mass as a percentage directly.

Answer: D

M(NaOH) = 23 + 16 + 1 = 40 g/mol, so n = 4.0 ÷ 40 = 0.10 mol. c = n ÷ V = 0.10 ÷ 0.250 = 0.40 mol/L. 0.10 mol/L forgets that the volume is a quarter of a litre; 1.6 and 0.016 mol/L come from dividing the mass in grams by the volume instead of the amount in moles.

Answer: E

Al₂(SO₄)₃ gives 3 SO₄²⁻ per formula unit: 0.100 L × 0.20 mol/L = 0.020 mol salt, so 0.060 mol sulfate. Na₂SO₄ gives one: 0.020 mol sulfate. Total 0.080 mol in 0.200 L, so [SO₄²⁻] = 0.40 mol/L, which no option gives. 0.80 mol/L adds the two sulfate concentrations (0.60 + 0.20) without allowing for the dilution on mixing; 0.60 mol/L is the sulfate concentration of the aluminium sulfate solution alone, before mixing.

Answer: C

A conjugate base is what is left when an acid has given away one proton (H⁺): H₂PO₄⁻ → H⁺ + HPO₄²⁻. H₃PO₄ is its conjugate acid (one proton more), PO₄³⁻ has lost two protons, and H₃O⁺ is the conjugate acid of water.

Answer: D

Sodium acetate is the salt of a strong base and a weak acid. The acetate ion is a weak base and is hydrolysed: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻, so the solution is alkaline. NH₄Cl (strong acid, weak base) gives an acidic solution because NH₄⁺ donates protons; KNO₃ and NaCl are salts of a strong acid and a strong base, so their solutions are essentially neutral.

Answer: E

n(H⁺) = 0.050 × 0.10 = 0.0050 mol; n(OH⁻) = 0.050 × 0.060 = 0.0030 mol. After neutralisation 0.0020 mol H⁺ remains in 0.100 L, so [H⁺] = 0.020 mol/L and pH = −log 0.020 = 2 − log 2 = 1.70, which no option gives. 1.00 is the pH of the acid before mixing, 1.30 is the pH of the acid diluted to 100 mL with no base at all (0.050 mol/L), and 2.70 forgets the dilution to 100 mL ([H⁺] taken as 0.0020 mol/L).

Answer: B

The oxidation numbers sum to zero: 2(+1) + 2x + 7(−2) = 0, so 2x = 12 and x = +6. +12 is the total for both chromium atoms, not for one; +3 is chromium's oxidation number in Cr³⁺, the product when dichromate acts as an oxidising agent; +7 is manganese's in permanganate.

Answer: B

Iron goes from +3 to +2: each Fe³⁺ gains one electron and is reduced, so it is the oxidising agent. Iodine goes from −1 (in I⁻) to 0 (in I₂): iodide loses electrons, is oxidised and is the reducing agent, so the change is −1 to 0, not 0 to −1. Potassium stays +1 throughout and is a spectator ion.

Answer: C

Mn goes from +7 to +2 (gains 5 electrons) while each Fe²⁺ loses one, so 1 mol MnO₄⁻ oxidises 5 mol Fe²⁺. n(MnO₄⁻) = 0.010 ÷ 5 = 0.0020 mol, and V = 0.0020 ÷ 0.020 = 0.10 L = 100 mL. 500 mL assumes a 1 : 1 ratio; 2.5 L multiplies by 5 instead of dividing; 50 mL assumes each permanganate takes ten electrons instead of five.

Answer: E

Four moles of gas become two, so raising the pressure favours the side with fewer gas molecules: ammonia. The forward reaction is exothermic, so raising the temperature shifts the equilibrium back towards N₂ and H₂. A catalyst speeds up both directions equally and only helps equilibrium be reached sooner. Increasing the volume lowers the pressure and favours the side with more gas molecules, and removing a reactant shifts the equilibrium to the left.

Answer: A

By Hess's law, take the first equation as written and reverse the second (CO₂ → CO + ½ O₂, ΔH = +283 kJ/mol). Adding them gives C + ½ O₂ → CO with ΔH = −394 + 283 = −111 kJ/mol. −677 kJ/mol adds the two values without reversing the second equation, and +111 kJ/mol has the sign the wrong way round: forming CO from carbon is exothermic.

Answer: D

Forming 1.6 mol HI uses 0.8 mol H₂ and 0.8 mol I₂, leaving 0.2 mol of each. In 1.0 L: Kc = [HI]² ÷ ([H₂][I₂]) = 1.6² ÷ (0.2 × 0.2) = 2.56 ÷ 0.04 = 64. 40 forgets to square [HI] (1.6 ÷ 0.04); 8.0 divides by only one reactant concentration; 16 squares 1.6 ÷ 0.4, adding the two reactant concentrations (0.2 + 0.2) instead of multiplying them.

Answer: C

Aluminium oxide dissolves in acids (Al₂O₃ + 6 HCl → 2 AlCl₃ + 3 H₂O) and in alkalis, where it forms the tetrahydroxidoaluminate ion [Al(OH)₄]⁻: it is amphoteric. Na₂O and CaO are basic oxides that react only with acids, and SO₃ is an acid-forming oxide that reacts only with bases.

Answer: B

Copper is below hydrogen in the reactivity series, so it cannot displace H₂ from acids; nitric acid dissolves it because the nitrate ion is a strong oxidising agent. Dilute nitric acid is reduced to nitrogen monoxide: 3 Cu + 8 HNO₃ → 3 Cu(NO₃)₂ + 2 NO + 4 H₂O. NO₂ is the main product with concentrated nitric acid, and oxygen is not released.

Answer: E

Calcium hydride is an ionic hydride containing the H⁻ ion, which reacts with water as a strong base: CaH₂ + 2 H₂O → Ca(OH)₂ + 2 H₂, and the calcium hydroxide makes the solution alkaline. The other four are covalent hydrides and release no hydrogen: CH₄ does not react with water, H₂S and HCl dissolve to give acidic solutions, and NH₃ gives an alkaline solution but no hydrogen gas.

Answer: A

Both have the same unbranched four-carbon chain and the same functional group (a C=C double bond); only the position of the double bond differs, which is position isomerism, a kind of structural isomerism. Chain isomers differ in the branching of the carbon skeleton (as butane and 2-methylpropane do), functional group isomers contain different functional groups, and neither molecule has a chiral carbon, so optical isomerism does not apply.

Answer: D

In butan-2-ol, CH₃–CH(OH)–CH₂–CH₃, carbon 2 carries four different groups (H, OH, CH₃ and C₂H₅), so it is chiral and the alcohol exists as two enantiomers. In propan-2-ol, CH₃–CH(OH)–CH₃, carbon 2 carries two identical methyl groups; in 2-methylpropan-2-ol it carries three; and no carbon in ethanol has four different substituents.

Answer: E

The longest chain is not the five carbons written in a row: running through the ethyl group gives six, CH₃–CH(CH₃)–CH₂–CH(CH₃)–CH₂–CH₃, a hexane with two methyl branches. Numbering from the end that gives the lowest locants puts them on carbons 2 and 4, so the name is 2,4-dimethylhexane, which no option gives. Any pentane name has missed the longest chain (an ethyl group on carbon 2 of a chain is always a sign of this), and 3,5-dimethylhexane numbers the chain from the wrong end.

Answer: B

The bromine molecule adds across the C=C double bond: CH₂=CH₂ + Br₂ → CH₂Br–CH₂Br. Two molecules combine into one with nothing lost, which is addition. Substitution would replace a hydrogen atom (as in alkanes with halogens in light), elimination would form a double bond, and condensation would release a small molecule such as water.

Answer: C

Propan-2-ol is a secondary alcohol, and oxidation of a secondary alcohol gives a ketone: CH₃–CH(OH)–CH₃ → CH₃–CO–CH₃, propanone (acetone). The ketone resists further oxidation. Propanal and then propanoic acid would come from oxidising the primary alcohol propan-1-ol, and propene is the product of dehydration, not oxidation.

Answer: A

Dehydration removes water to give propene, CH₂=CH–CH₃ (X). HBr adds by Markovnikov's rule: hydrogen goes to the CH₂ carbon that already carries more hydrogens and bromine to the middle carbon, giving 2-bromopropane (Y). Aqueous NaOH substitutes OH for Br, giving propan-2-ol (Z). Propan-1-ol would require the anti-Markovnikov product 1-bromopropane; propanone would need an oxidation step, which never occurs; propene is X.

Answer: D

RNA contains adenine, guanine, cytosine and uracil; DNA contains the same bases except that thymine (5-methyluracil) replaces uracil. Adenine and cytosine occur in both nucleic acids, and thymine is the one found in DNA but not in RNA.

Answer: B

Purine is a fused bicyclic heterocycle (a pyrimidine ring joined to an imidazole ring); adenine and guanine are its derivatives. Cytosine, thymine and uracil are pyrimidines, built on a single six-membered ring, so every other pair contains at least one pyrimidine.

Answer: C

The fragment has 2,000 bases. A = T = 30 % each, so A + T = 60 % = 1,200 bases = 600 A–T pairs; the remaining 40 % = 800 bases = 400 G–C pairs. A–T pairs are held by 2 hydrogen bonds and G–C pairs by 3: 600 × 2 + 400 × 3 = 1,200 + 1,200 = 2,400. 2,600 swaps the pair counts (400 A–T and 600 G–C), and 2,000 and 3,000 assume every pair has two or three bonds.

Answer: D

In sucrose the glycosidic bond joins the anomeric carbon of glucose (C1) to the anomeric carbon of fructose (C2), so neither unit has a free hemiacetal group that can open to an aldehyde or ketone form: it does not reduce Fehling's or Tollens' reagent. Maltose, lactose and cellobiose are all linked from C1 of one unit to C4 of the other, leaving a free anomeric carbon, so they are reducing sugars.

Answer: E

Lactose is β-D-galactopyranosyl-(1→4)-D-glucose: its hydrolysis (by lactase in the gut) gives galactose and glucose, a pair no option lists. Glucose and glucose come from maltose (or cellobiose), glucose and fructose from sucrose, and no common disaccharide contains galactose with fructose.

Answer: A

n(fat) = 8.07 ÷ 807 = 0.010 mol. A triacylglycerol has three ester bonds, and each one consumes one NaOH, giving glycerol and three sodium palmitate (soap) molecules. n(NaOH) = 3 × 0.010 = 0.030 mol, and m = 0.030 × 40 = 1.2 g. 0.40 g assumes one NaOH per fat molecule and 0.80 g two; 3.6 g multiplies by three twice.

Answer: C

Cysteine has a thiol side chain (–CH₂–SH). Two cysteine thiols can be oxidised to a disulfide bond (–S–S–), which cross-links parts of the same chain or different chains. Methionine also contains sulfur, but as a thioether (–CH₂–CH₂–S–CH₃) with no SH group, so it cannot form disulfides. Serine and tyrosine carry OH groups, which form hydrogen bonds, not covalent cross-links.

Answer: A

The α-helix is a secondary structure held by hydrogen bonds within the backbone of one chain: the C=O of residue n bonds to the N–H of residue n + 4, and the side chains point outwards. Disulfide and ionic bonds between side chains belong mainly to tertiary (and quaternary) structure, and hydrogen bonding between strands lying side by side describes the β-pleated sheet, not the helix.

Answer: B

All four digestive enzymes split their substrates (peptide bonds, ester bonds of fats, glycosidic bonds of starch) by adding water, which makes them hydrolases. Oxidoreductases catalyse redox reactions, transferases move a group from one molecule to another, and ligases join two molecules using the energy of ATP.

Answer: D

β-Oxidation runs in the mitochondrial matrix on the acyl-CoA. Each round includes two dehydrogenations (one reducing FAD to FADH₂, one reducing NAD⁺ to NADH) and ends by cleaving off acetyl-CoA, so the chain is two carbons shorter. The acetyl-CoA enters the citric acid cycle; it cannot be turned into glucose in humans, and pyruvate is the end product of glycolysis, not of β-oxidation. It is an oxidation and needs FAD and NAD⁺.

Answer: B

Hair and milk-producing mammary glands are present in all mammals, including the egg-laying monotremes, and in no other vertebrates. Monotremes (platypus, echidnas) have no true placenta and lay eggs, so neither the placenta nor live birth is found in every mammal; live birth also occurs in some sharks, fish and reptiles. A four-chambered heart is shared with birds and crocodilians.

Answer: E

All four named organisms are eukaryotes with a nucleus bounded by an envelope: Candida albicans is a yeast and Aspergillus fumigatus a mould (both fungi), while Plasmodium falciparum and Toxoplasma gondii are protozoa. A prokaryote would be a bacterium or archaeon, for example Escherichia coli or Staphylococcus aureus, and none is offered, so E is correct.

Answer: C

Viruses are acellular obligate intracellular parasites: they have no metabolism or ribosomes of their own and replicate only by using the machinery of a living host cell. A virus carries either DNA or RNA as its genome, never both. Only some viruses (for example influenza and HIV) have a lipid envelope taken from a host membrane; many, such as adenoviruses, are naked.

Answer: A

Malaria is caused by protozoa of the genus Plasmodium, transmitted by Anopheles mosquitoes. Tuberculosis is caused by a bacterium (Mycobacterium tuberculosis), influenza by an RNA virus, and candidiasis by the yeast Candida albicans, which is a fungus, not a protozoan.

Answer: D

In cysticercosis the human takes the place of the pig as intermediate host: swallowed eggs hatch in the gut, and the larvae spread through the blood and encyst in muscle, eye and brain. The eggs come from the faeces of a human carrier, through contaminated food, water or hands. Eating undercooked pork with cysticerci is the tempting answer, but it gives taeniasis, an adult tapeworm in the intestine, not larval cysts. Skin penetration is the route of hookworms, Strongyloides and schistosomes, and tsetse flies transmit trypanosomes (sleeping sickness).

Answer: C

The carboxyl group of one amino acid and the amino group of the next react in a condensation, giving a peptide (amide) bond, and the sequence of amino acids so joined is the primary structure. Glycosidic bonds join monosaccharides, ester bonds join fatty acids to glycerol in lipids, and phosphodiester bonds join the nucleotides of DNA and RNA.

Answer: B

In double-stranded DNA adenine pairs with thymine and guanine with cytosine, so A = T and G = C (Chargaff's rules). A + T = 30% + 30% = 60%, which leaves 40% for G + C, shared equally: cytosine is 20%. 30% is the thymine figure, 40% is G and C together, and 70% wrongly subtracts adenine alone from 100%.

Answer: E

A triacylglycerol is glycerol esterified with three fatty acids and contains only carbon, hydrogen and oxygen. ATP contains the nitrogenous base adenine. Phosphatidylcholine is a phospholipid whose head group, choline, carries a quaternary ammonium nitrogen, which is easy to overlook. Chlorophyll has a porphyrin ring of four nitrogen-containing pyrrole rings around magnesium, and haemoglobin is a protein (every amino acid has an amino group) with nitrogen-containing haem groups.

Answer: D

Every cell needs ribosomes to make proteins: prokaryotes have smaller 70S ribosomes, eukaryotes 80S ribosomes in the cytoplasm. The nuclear envelope, mitochondria and endoplasmic reticulum are membrane-bounded structures found only in eukaryotic cells; a prokaryote's DNA lies free in the cytoplasm as a nucleoid.

Answer: A

Proteins for secretion are made by ribosomes bound to the rough ER and pass into its lumen, travel in transport vesicles to the Golgi apparatus for modification and sorting, and leave in secretory vesicles that fuse with the plasma membrane (exocytosis). The Golgi comes after the ER, not before it, and lysosomes receive digestive enzymes rather than exporting proteins. The smooth ER makes lipids, and proteins do not pass through the nucleus or mitochondria on their way out; free ribosomes make proteins that stay in the cytosol.

Answer: B

The axoneme of a eukaryotic cilium or flagellum has nine outer microtubule doublets around a central pair (9 + 2), and dynein arms bend it. Microtubules are made of tubulin, while actin forms microfilaments. In muscle it is actin microfilaments, not intermediate filaments, that slide relative to myosin; intermediate filaments (keratins, lamins) give mechanical strength. The mitotic spindle is made of microtubules.

Answer: C

The sodium-potassium pump is primary active transport: one ATP drives 3 Na+ out and 2 K+ in, both against their concentration gradients. This keeps Na+ high outside and K+ high inside, and because more positive charge leaves than enters it contributes slightly to the negative resting potential. The other options reverse the direction or the numbers, or describe passive diffusion, which needs no ATP.

Answer: E

0.3% NaCl is hypotonic to the cell contents, so water enters the cells by osmosis (largely through aquaporins, passively); the cells swell and many burst (haemolysis). No option says this. Shrinking happens in a hypertonic solution, and 0.9% NaCl, not 0.3%, is isotonic. Option C names swelling but gives a false cause: the pump moves Na+ out of the cell, and the swelling is due to water entering by osmosis. The membrane is freely permeable to water, so D is false too.

Answer: B

DNA is replicated during the S (synthesis) phase of interphase, after which every chromosome consists of two sister chromatids. G1 is growth before replication, G2 is preparation for division after it, and in anaphase the already replicated sister chromatids are separated.

Answer: D

Homologous chromosomes pair (synapsis) in prophase I of meiosis, and non-sister chromatids exchange segments at chiasmata, recombining alleles. In mitosis homologues do not pair, so there is no regular crossing over. By anaphase I homologues are already separating, and in meiosis II there are no homologous pairs left in the cell.

Answer: A

DNA is replicated in S phase before meiosis begins, so a cell in prophase I holds twice the G1 amount, 12 pg (4c). Meiosis I halves it to 6 pg per cell (2c, haploid but with two chromatids per chromosome), and meiosis II halves it again to 3 pg (1c) in each gamete, half the amount of the diploid G1 cell. Answers with 6 pg at prophase I forget replication; 6 pg for the gamete stops after meiosis I.

Answer: B

Transcription is carried out by RNA polymerase, which reads the template strand of DNA and builds a complementary RNA strand. DNA polymerase makes DNA during replication, reverse transcriptase makes DNA from an RNA template (as in retroviruses), and DNA ligase joins DNA fragments with phosphodiester bonds.

Answer: A

The mRNA is complementary and antiparallel to the template, with uracil in place of thymine: T pairs with A, A with U, C with G and G with C, so 3'-TAC GGA TTC-5' gives 5'-AUG CCU AAG-3'. Option B copies the template instead of pairing with it, option C contains thymine and so is DNA, and option D has the right bases written with the wrong polarity.

Answer: C

Replication is semiconservative, and on the lagging strand the Okazaki fragments are sealed by DNA ligase once their primers have been replaced. DNA polymerase can add nucleotides only to the free 3'-OH end, so strands grow 5' to 3'. That is why the leading strand is made continuously and the lagging strand, not the leading one, discontinuously. Primers are short RNA sequences made by primase; helicase unwinds the double helix.

Answer: A

This is a test cross. The AaBb parent makes four kinds of gamete in equal proportions (AB, Ab, aB, ab), and the aabb parent only ab, so the offspring are AaBb, Aabb, aaBb and aabb in a 1:1:1:1 ratio. Equivalently, P(aa) × P(bb) = 1/2 × 1/2 = 1/4. 1/16 and 3/16 are fractions from a dihybrid cross of two AaBb parents, not a test cross.

Answer: E

Each partner's parents must both be carriers (Aa × Aa), since they had an affected son. A healthy child of such parents is AA or Aa in the ratio 1:2, so each partner is a carrier with probability 2/3. The child is affected only if both are carriers and the child receives two recessive alleles: 2/3 × 2/3 × 1/4 = 1/9. That value is not offered, so E is correct. 1/4 assumes both parents are certainly carriers, 4/9 is the probability that both are carriers, and 2/3 is one partner's carrier risk.

Answer: D

An affected father passes his only X chromosome, carrying the allele, to every daughter, so the woman is certainly a carrier (XHXh). With an unaffected husband (XHY), half the children are boys, and half of the boys receive her Xh: 1/2 × 1/2 = 1/4. 1/2 is the risk for a son already known to be a boy, not for any child. Daughters receive their father's XH and are not affected.

Answer: C

Free trisomy 21 gives 47 chromosomes with an extra chromosome 21, and a boy has XY: 47,XY,+21. 45,X is Turner syndrome, 47,XXY is Klinefelter syndrome, and 47,XX,+18 is a girl with Edwards syndrome.

Answer: A

UAA is a stop codon, so translation ends at this point and the protein is truncated: a nonsense mutation. A silent mutation would give a different codon for the same amino acid, and a missense mutation a codon for a different amino acid. A frameshift is caused by inserting or deleting a number of bases that is not a multiple of three; a substitution leaves the reading frame intact.

Answer: E

Each cycle (denaturation, primer annealing, extension by a heat-stable DNA polymerase) doubles the number of molecules, so after n cycles one molecule becomes 2^n: 2^5 = 32. 10 assumes two copies are added per cycle rather than doubling, 16 is the number after 4 cycles, 64 after 6 cycles, and 25 is 5 squared rather than 2 to the power 5.

Answer: D

Skeletal muscle consists of long multinucleate striated fibres controlled voluntarily through motor neurones. Cardiac muscle is also striated but contracts involuntarily, set by the heart's own pacemaker, and smooth muscle is neither striated nor voluntary.

Answer: C

In the sliding filament model, myosin heads pull the thin actin filaments towards the centre of the sarcomere. The filaments keep their length; they overlap more. The Z lines therefore move closer together, and the I band (actin only) and H zone (myosin only) shorten, while the A band, which is the length of the myosin filaments, stays the same.

Answer: C

Antibodies (immunoglobulins) are secreted by plasma cells, which develop from B lymphocytes activated by their antigen. Helper T cells help to activate B cells with cytokines but make no antibodies, neutrophils are phagocytes, and erythrocytes carry oxygen.

Answer: D

A group O person has anti-A and anti-B antibodies in plasma, so red cells must carry neither A nor B antigen, which leaves group O. An Rh-negative patient should not receive RhD-positive cells, which could sensitise them (or, if already sensitised, cause a haemolytic reaction). Only O, Rh-negative cells fit both conditions. A and AB cells would be agglutinated by the patient's antibodies.

Answer: E

The disease arises when an Rh-negative mother, sensitised by fetal Rh-positive red cells (usually at an earlier delivery), carries an Rh-positive fetus: her IgG anti-D antibodies cross the placenta and destroy the fetal red cells. No option states this, so E is correct. A reverses the Rh types, and a first pregnancy is usually spared because sensitisation has not yet happened. IgM is too large to cross the placenta. The father must be Rh-positive for the fetus to be Rh-positive. Anti-D immunoglobulin given to the mother removes fetal Rh-positive cells from her own circulation before she makes antibodies; it does not act in the newborn.

Answer: B

Salivary amylase begins the hydrolysis of starch in the mouth. Pepsin works in the acid of the stomach (around pH 2) and is inactivated at the alkaline pH of the small intestine. Fat-digesting lipase comes mainly from the pancreas; the gall bladder stores bile, which emulsifies fat but contains no enzymes. Trypsin is secreted by the pancreas as inactive trypsinogen, whereas gastric chief cells secrete pepsinogen.

Answer: A

ADH inserts aquaporins into the membranes of collecting duct cells, so more water is reabsorbed into the hypertonic medulla and the urine becomes smaller in volume and more concentrated. It is made in the hypothalamus and released from the posterior pituitary, not the adrenal cortex. Its secretion rises when blood osmolarity rises (dehydration) and falls when it falls. Stimulating Na+ reabsorption, in the distal tubule and collecting duct, is mainly the work of aldosterone; the proximal tubule reabsorbs most Na+ without hormonal control of this kind.

Answer: B

When the membrane reaches threshold, voltage-gated Na+ channels open and Na+ rushes in down its electrochemical gradient, driving the inside towards positive values. Repolarisation follows as these channels inactivate and voltage-gated K+ channels let K+ flow out, not in. Chloride movement does not produce the action potential, and the Na+/K+-ATPase restores ion gradients slowly; it does not generate the spike.

Answer: E

Oxytocin (like ADH) is made by neurosecretory cells of the hypothalamus, carried down their axons and released from the posterior pituitary (neurohypophysis). Growth hormone, TSH and prolactin are made by the anterior pituitary (adenohypophysis) itself, under control of hypothalamic releasing hormones, and cortisol is made by the adrenal cortex.

Answer: D

Fertilisation normally happens in the ampulla of the uterine tube, shortly after ovulation. The early embryo then travels down the tube and implants in the wall of the uterus about a week later as a blastocyst. The ovary releases the secondary oocyte, and the cervix is only a passage for sperm.

Answer: A

In spermatogenesis both meiotic divisions are equal, so one primary spermatocyte yields four haploid spermatids that mature into spermatozoa. In oogenesis the cytoplasm is divided unequally: one primary oocyte gives a single large egg cell, and the small polar bodies degenerate. The other options either give the oocyte four gametes or the spermatocyte fewer than four.

Answer: B

Miller and Urey passed electric discharges through a mixture modelling a reducing early atmosphere and found amino acids and other organic molecules forming, support for chemical evolution. They produced no cells or living things. The experiment says nothing about which molecule carried heredity first (the RNA world hypothesis favours RNA), and their mixture deliberately contained no free oxygen, which would have destroyed the products.

Answer: E

Homologous structures are inherited from a common ancestor and share a basic plan even when their uses differ, for example a human arm, a bat's wing and a whale's flipper, all built on the pentadactyl limb. Every pair offered is instead analogous: similar in function or form because of convergent evolution, not shared ancestry. Bat and butterfly wings, octopus and human eyes, shark fins and dolphin flippers, and the body shapes of tuna and dolphins evolved independently. No option is homologous, so E is correct.

Answer: C

Producers are autotrophs that make organic matter from inorganic substances, here green plants using light energy in photosynthesis. Herbivores are primary consumers, carnivores secondary or higher consumers, and decomposers break down dead matter; all are heterotrophs.

Answer: D

Secondary consumers are two transfers above the producers: 100,000 kJ to primary consumers gives about 10,000 kJ, and to secondary consumers about 1,000 kJ. 10,000 kJ is what reaches the primary consumers, 100 kJ is the tertiary consumers' share, and 50,000 kJ assumes half is passed on. The rest is lost mainly as heat in respiration, and in material that is not eaten or digested.

Answer: C

The seven SI base units are the metre, kilogram, second, ampere, kelvin, mole and candela, so the kelvin is a base unit. The newton (kg m s⁻²), the joule (kg m² s⁻²) and the volt (kg m² s⁻³ A⁻¹) are all derived units, built from base units.

Answer: A

1 μm = 10⁻⁶ m and 1 nm = 10⁻⁹ m, so 1 μm = 10³ nm and 7.5 μm = 7.5 × 10³ nm (7500 nm). 7.5 × 10⁻³ nm divides by 1000 instead of multiplying. 7.5 × 10⁶ nm treats the micrometre as a millimetre. 75 nm uses a factor of 10 instead of 1000.

Answer: D

The forces are perpendicular, so the resultant is √(5.0² + 12²) = √169 = 13 N. By Newton's second law a = F/m = 13/2.0 = 6.5 m/s². 3.5 m/s² subtracts the forces (7 N / 2) and 8.5 m/s² adds them as scalars (17 N / 2); neither is correct for perpendicular vectors. 13 m/s² is the resultant force in newtons, not divided by the mass.

Answer: E

Every set hides one scalar. Pressure is a scalar (it acts equally in all directions at a point; the force it produces has a direction, the pressure does not). Electric potential is a scalar (energy per unit charge). Work is a scalar (the scalar product of force and displacement). Magnetic flux is a scalar (the product of flux density and an area, Φ = BS cos α), even though magnetic flux density B is a vector. Force, impulse, momentum, torque, displacement, weight, velocity and magnetic flux density are all vectors, but no set consists only of them, so none of the answers is correct.

Answer: B

In free fall from rest v = gt = 10 × 3.0 = 30 m/s. 15 m/s is the average speed over the fall (half the final speed). 45 m/s is the number for the distance fallen, s = ½gt² = 45 m, not a speed. 90 m/s multiplies g by t².

Answer: D

Take moments about the elbow. The load's weight is mg = 2.0 × 10 = 20 N, acting 0.32 m from the elbow, so its torque is 20 × 0.32 = 6.4 N m. The biceps acts 0.040 m from the elbow, so F × 0.040 = 6.4 and F = 160 N. 20 N is the load's weight, ignoring the lever arms. 16 N uses the mass (2.0) instead of the weight. 2.5 N inverts the ratio of the arms (20 × 4/32).

Answer: A

The useful work is the gain in potential energy, mgh = 50 × 10 × 12 = 6000 J, done in 20 s, so the useful output power is 6000/20 = 300 W. Efficiency = output/input, so the input is 300/0.60 = 500 W. 300 W is the output, not the input. 180 W multiplies by the efficiency instead of dividing. 10 kW is the input energy, 6000/0.60 = 10 000 J, mistaken for a power.

Answer: C

Area: 50 mm² = 50 × 10⁻⁶ m² = 5.0 × 10⁻⁵ m². Normal stress σ = F/S = 500 / 5.0 × 10⁻⁵ = 1.0 × 10⁷ Pa. Strain ε = σ/E = 1.0 × 10⁷ / 1.0 × 10⁹ = 0.010. Extension Δl = ε l = 0.010 × 0.20 m = 2.0 × 10⁻³ m = 2.0 mm. 0.20 mm and 20 mm come from converting mm² to m² with the wrong power of ten (10⁻⁵ and 10⁻⁷ per mm²). 0.010 mm is the strain itself, which has no unit, read as a length.

Answer: E

In a longitudinal wave the particles of the medium oscillate along the direction the wave travels, in alternating compressions and rarefactions. Sound and ultrasound in fluids and soft tissue are longitudinal. Light, X-rays and radio waves are electromagnetic waves, which are transverse, and the wave along a string is transverse too: the string moves at right angles to the direction the wave travels.

Answer: B

λ = v/f = 1540 / (5.0 × 10⁶) = 3.08 × 10⁻⁴ m ≈ 0.31 mm. 3.1 mm and 3.1 μm are power-of-ten slips with the megahertz (taking 5.0 × 10⁵ Hz or 5.0 × 10⁸ Hz). 7.7 × 10⁹ m multiplies speed by frequency instead of dividing.

Answer: C

For a mass on a spring T = 2π√(m/k) = 2π√(0.40/40) = 2π√0.010 = 2π × 0.10 = 0.63 s. 0.10 s leaves out the 2π. 63 s inverts the ratio inside the root, 2π√(k/m) = 2π × 10 ≈ 63. 1.6 s is the frequency, f = 1/T = 1.6 Hz, given in seconds.

Answer: A

The level is L = 10 log(I/I₀) dB. An increase of 10 dB means log(I₂/I₁) = 1, so the intensity is 10 times greater. Since the intensities of independent sources add, 10 ventilators are needed. 2 ventilators double the intensity and raise the level by only 10 log 2 ≈ 3 dB, to 63 dB. 100 ventilators raise it by 20 dB, to 80 dB. 20 ventilators give 60 + 10 log 20 ≈ 73 dB.

Answer: D

p = ρgh = 1000 × 10 × 2.0 = 20 000 Pa = 20 kPa. The other values are power-of-ten slips: 2.0 kPa and 200 Pa are too small, 200 kPa too large. (The total pressure at that depth would also include atmospheric pressure, about 100 kPa, but the question asks only for the pressure due to the water.)

Answer: A

By the equation of continuity S₁v₁ = S₂v₂, and the area of a circular cross-section is πr². Halving the radius divides the area by 4, so the speed is multiplied by 4: v₂ = 0.20 × (4.0/2.0)² = 0.80 m/s. 0.40 m/s scales by the ratio of the radii instead of the areas. 0.10 m/s and 0.050 m/s have the blood slowing down in the narrowing, the reverse of continuity.

Answer: E

A floating body is in equilibrium, so its weight equals the buoyant force: ρ_block V g = ρ_water V_sub g, giving V_sub/V = 600/1000 = 0.60. So 60% is submerged and 40% of the volume is above the surface. No option gives 40%. 60% is the submerged fraction, not the fraction above. 67% comes from 1000/600 − 1. 20% and 30% have no physical basis here.

Answer: B

For a sphere with one liquid surface Laplace's formula gives Δp = 2σ/r = 2 × 0.050 / (0.10 × 10⁻³) = 0.10 / 1.0 × 10⁻⁴ = 1000 Pa. 2000 Pa uses 4σ/r, the formula for a soap bubble, which has two surfaces. 500 Pa uses σ/r. 0.50 Pa uses σ/r with the radius left in millimetres.

Answer: C

T = t + 273.15 = 37 + 273.15 ≈ 310 K. 236 K subtracts 37 from 273 instead of adding. 37 K takes the Celsius number unchanged. 99 K is the Fahrenheit value, 98.6 °F, wrongly given in kelvin.

Answer: B

In an isothermal process pV is constant (Boyle's law): p₂ = p₁V₁/V₂ = 100 × 6.0 / 2.0 = 300 kPa. 33 kPa inverts the ratio of volumes. 600 kPa multiplies by the initial volume only. 100 kPa leaves the pressure unchanged, which would be true of an isobaric process, not an isothermal compression.

Answer: D

Q = mcΔT = 0.50 × 4200 × (80 − 20) = 0.50 × 4200 × 60 = 126 000 J = 126 kJ. 168 kJ uses the final temperature, 80 °C, as ΔT. 42 kJ uses the initial temperature, 20 °C. 252 kJ uses a mass of 1.0 kg.

Answer: A

Carnot efficiency uses thermodynamic temperatures: T₁ = 527 + 273 = 800 K, T₂ = 127 + 273 = 400 K, so η = 1 − T₂/T₁ = 1 − 400/800 = 0.50. Work W = ηQ₁ = 0.50 × 2000 = 1000 J (the other 1000 J goes to the cold reservoir). 1520 J uses the Celsius temperatures, η = 1 − 127/527 ≈ 0.76. 500 J uses η = 0.25. 2000 J would be 100% efficiency, which the second law forbids.

Answer: A

By Ohm's law I = U/R = 12/4.0 = 3.0 A. 48 A multiplies voltage by resistance. 0.33 A divides resistance by voltage. 16 A adds the two numbers.

Answer: E

The parallel pair: 1/R = 1/6.0 + 1/3.0 = 3/6, so R = 2.0 Ω. In series with 4.0 Ω the total is 2.0 + 4.0 = 6.0 Ω, which no option gives. 13 Ω puts all three in series. 2.0 Ω is the parallel pair alone. 1.3 Ω puts all three in parallel. 2.8 Ω puts the 6.0 Ω and 3.0 Ω in series and that pair in parallel with the 4.0 Ω (9 × 4/13).

Answer: C

For capacitors in series 1/C = 1/C₁ + 1/C₂ = 1/2.0 + 1/3.0 = 5/6 μF⁻¹, so C = 6/5 = 1.2 μF (equivalently C₁C₂/(C₁ + C₂) = 6/5). 5.0 μF is the parallel combination. 6.0 μF is the product alone. 0.83 μF is 1/C, not inverted back.

Answer: D

Charge Q = It = 2.0 × 965 = 1930 C. Moles of electrons = Q/F = 1930/96 500 = 0.020 mol. Each Cu²⁺ needs 2 electrons, so 0.010 mol of copper is deposited: m = 0.010 × 63.5 = 0.635 g ≈ 0.64 g (Faraday's law, m = MQ/(zF)). 1.27 g forgets the charge number z = 2. 0.32 g divides by 4 instead of 2. 63.5 g is a whole mole of copper.

Answer: B

For an ideal transformer U₂/U₁ = N₂/N₁, so U₂ = 230 × 50/1000 = 11.5 V: a step-down transformer. 4600 V uses the inverted turns ratio. 230 V ignores the turns ratio. 50 V is the number of secondary turns, not a voltage.

Answer: C

F = BIl sin α = 0.40 × 5.0 × 0.20 × 0.50 = 0.20 N, where α is the angle between the wire and the field. 0.40 N leaves out sin α (true only for a wire at right angles to the field). 0.35 N uses cos 30° instead of sin 30°. 2.0 N is a power-of-ten slip.

Answer: E

The effective (rms) voltage is U_ef = U_max/√2 = 325/1.414 ≈ 230 V, and the mean power is P = U_ef²/R = 230²/100 ≈ 528 W (equivalently U_max²/(2R) = 105 625/200 ≈ 528 W). 1056 W uses the peak voltage as if it were the effective value. 2.3 W is the effective current (2.3 A) given as a power. 0 W confuses the mean voltage over a cycle, which is zero, with the mean power, which is not. 264 W divides by 2 twice.

Answer: D

The magnetic force provides the centripetal force: evB = mv²/r, so r = mv/(eB) = (9.1 × 10⁻³¹ × 2.0 × 10⁷) / (1.6 × 10⁻¹⁹ × 1.0 × 10⁻³) = 1.82 × 10⁻²³ / 1.6 × 10⁻²² ≈ 0.11 m = 11 cm. 1.1 mm and 1.1 m are power-of-ten slips. 8.8 m inverts the fraction, eB/(mv) ≈ 8.8 m⁻¹, which is not even a length.

Answer: D

Dioptric power is the reciprocal of the focal length in metres: φ = 1/f = 1/0.25 m = +4.0 D, positive for a converging lens. +0.25 D is the focal length in metres, not its reciprocal. −4.0 D has the sign of a diverging lens. +25 D uses the focal length in centimetres.

Answer: A

At the critical angle the refracted ray grazes the surface (90°), so by Snell's law sin α_c = n₂/n₁ = 1.00/1.50 = 0.667 and α_c ≈ 41.8° ≈ 42°. Rays meeting the core wall at more than 42° are totally reflected, which is how the fibre guides light. 48° is the complement, 90° − 42°. 34° and 60° do not satisfy sin α_c = 0.667 (sin 34° ≈ 0.56, sin 60° ≈ 0.87).

Answer: B

Lens equation: 1/a + 1/a′ = 1/f, so 1/a′ = 1/20 − 1/30 = (3 − 2)/60 = 1/60 and a′ = 60 cm, a real, inverted image behind the lens. Magnification |Z| = a′/a = 60/30 = 2.0. 12 cm comes from adding 1/20 and 1/30 instead of subtracting. 0.50 inverts the magnification ratio (a/a′).

Answer: E

The lens must take an object at a = 0.25 m and form a virtual image at the near point, on the same side, so a′ = −1.0 m. Then φ = 1/a + 1/a′ = 1/0.25 − 1/1.0 = 4.0 − 1.0 = +3.0 D, a converging lens. No option gives +3.0 D. +4.0 D ignores the patient's own near point. −3.0 D has the sign of a diverging lens, which corrects short sight. +1.0 D is the reciprocal of the near point alone. +5.0 D adds the two terms, treating the image as real.

Answer: A

One of the postulates of special relativity is that the speed of light in a vacuum is the same, c, for every inertial observer, whatever the motion of the source. 1.5c adds the speeds classically. 0.50c subtracts them. 1.25c has no basis; no observer measures light faster than c.

Answer: C

Time dilation: Δt = Δt₀/√(1 − v²/c²) = 2.0/√(1 − 0.36) = 2.0/0.80 = 2.5 μs. The moving particle's clock runs slow as seen from the laboratory, so the measured lifetime is longer than 2.0 μs. 1.6 μs multiplies by 0.80 instead of dividing (a contraction, not a dilation). 1.2 μs multiplies by 0.60. 3.3 μs divides by 0.60, using v/c in place of √(1 − v²/c²).

Answer: E

E = mc² = 1.0 × 10⁻³ kg × (3.0 × 10⁸)² = 1.0 × 10⁻³ × 9.0 × 10¹⁶ = 9.0 × 10¹³ J. 9.0 × 10¹⁶ J uses 1 kg. 3.0 × 10⁵ J uses mc, with c not squared. 9.0 × 10¹⁰ J takes 1 g as 10⁻⁶ kg. 4.5 × 10¹³ J uses ½mc², the classical kinetic energy formula, which is not the rest energy.

Answer: B

Relativistic addition of velocities: u = (u′ + v)/(1 + u′v/c²) = (0.60c + 0.60c)/(1 + 0.36) = 1.20c/1.36 ≈ 0.88c. 1.20c is the classical sum, faster than light, which is impossible. 0.60c ignores the ship's motion. 0.75c divides by 1.60, adding v/c rather than u′v/c² in the denominator.

Answer: D

Isotopes are nuclides of the same element: they have the same proton number Z (6 for carbon) and so, as neutral atoms, the same number of electrons, but different mass numbers A. Carbon-12 has 12 − 6 = 6 neutrons and carbon-14 has 14 − 6 = 8, so they differ in the number of neutrons. The other three options are the same for both.

Answer: E

24 h is 24/6.0 = 4 half-lives, so the activity falls by a factor of 2⁴ = 16: A = 800/16 = 50 MBq. No option gives 50 MBq. 400 MBq is after one half-life, 200 MBq after two, 100 MBq after three and 25 MBq after five.

Answer: C

The shortest wavelength comes from an electron giving all its kinetic energy eU to one photon: eU = hc/λ_min, so λ_min = hc/(eU) = (6.63 × 10⁻³⁴ × 3.0 × 10⁸)/(1.6 × 10⁻¹⁹ × 5.0 × 10⁴) = 1.99 × 10⁻²⁵ / 8.0 × 10⁻¹⁵ ≈ 2.5 × 10⁻¹¹ m = 25 pm. 25 nm uses 50 V instead of 50 kV. 2.5 nm and 0.25 pm are power-of-ten slips.

Answer: B

Photon energy E = hc/λ = (6.63 × 10⁻³⁴ × 3.0 × 10⁸)/(250 × 10⁻⁹) = 7.96 × 10⁻¹⁹ J = 7.96 × 10⁻¹⁹ / 1.6 × 10⁻¹⁹ ≈ 5.0 eV. By Einstein's photoelectric equation E_k,max = hf − W = 5.0 − 3.0 = 2.0 eV. 5.0 eV is the photon energy, forgetting the work function. 8.0 eV adds the work function instead of subtracting it. Electrons are emitted, because the photon energy (5.0 eV) exceeds the work function (3.0 eV).