Medical University of Gdańsk Entrance Exam Practice Paper with Answers (2027)

Answer: B

Secretory proteins are made on ribosomes bound to the rough endoplasmic reticulum, enter its lumen as they are made, and travel on to the Golgi apparatus for packaging and exocytosis. The smooth ER has no ribosomes: it makes lipids and steroids and detoxifies drugs. Lysosomes digest material with hydrolytic enzymes, and peroxisomes break down fatty acids and hydrogen peroxide; neither makes proteins.

Answer: C

Every cell makes proteins, so every cell has ribosomes: 70S in prokaryotes, 80S in the cytoplasm of eukaryotes. Prokaryotes have no membrane-bound organelles: their DNA lies in a nucleoid without a nuclear envelope, and they have neither mitochondria (their respiration happens on the plasma membrane) nor endoplasmic reticulum.

Answer: A

In anaphase the cohesin holding sister chromatids together is cleaved and the spindle pulls each chromatid, now a chromosome, to an opposite pole. In metaphase the chromosomes line up at the equator; in prophase they condense and the spindle forms; in telophase nuclear envelopes re-form around the two sets that have already arrived at the poles.

Answer: D

Meiosis I separates homologous chromosomes, not sister chromatids, so each daughter cell receives one chromosome of each pair, 23 in all, and each of these still consists of two chromatids: 46 chromatids. Only meiosis II separates the sister chromatids, giving cells with 23 chromosomes of one chromatid each (option B describes those). Forty-six chromosomes with 92 chromatids is the parent cell after S phase, before division.

Answer: A

Each parental strand serves as a template for a new complementary strand, so every daughter double helix keeps one old strand and gains one new one, as Meselson and Stahl showed with heavy nitrogen in 1958. The second option is the conservative model and the third the dispersive model, both ruled out by that experiment; DNA is never broken down and rebuilt to be copied.

Answer: B

Codon and anticodon pair antiparallel and complementary: 5′-A U G-3′ pairs with 3′-U A C-5′, which written 5′ to 3′ is CAU. 5′-UAC-3′ is the right bases in the wrong direction, the commonest slip. TAC contains thymine, which occurs in DNA, not in RNA. GUA does not pair with AUG in either direction.

Answer: C

The genes assort independently (Mendel's second law), so the probabilities multiply: round is 3/4 (RR or Rr) and yellow is 3/4, and round and yellow together is 3/4 × 3/4 = 9/16, the familiar 9 : 3 : 3 : 1 ratio. 3/16 is each of the two mixed classes (round green, wrinkled yellow), 1/16 is wrinkled green, and 3/4 is the fraction for one trait alone.

Answer: D

The father's genotype is XʰY and the mother's XᴴXᴴ. Every daughter receives the father's only X, which carries the allele, and a normal X from her mother: all daughters are carriers (XᴴXʰ). Every son receives the father's Y and a normal X from the mother: no son is affected, and a son cannot be a carrier of an X-linked allele he does not have. Affected sons would need a carrier or affected mother.

Answer: B

The zygote first divides by cleavage (rapid mitoses without growth) into a ball of cells that hollows into the blastula. Gastrulation then moves cells inwards to form the three germ layers, and only then does neurulation fold the ectoderm into the neural tube. Every other option puts the blastula, the product of cleavage, before cleavage, or gastrulation before the blastula it reorganises.

Answer: C

The ectoderm forms the nervous system (through the neural tube made at neurulation) and the epidermis of the skin. The gut lining comes from the endoderm; the kidneys, skeletal muscle, bone, blood and heart come from the mesoderm.

Answer: A

Gas exchange needs the shortest possible diffusion path, which a single layer of flat cells gives. Stratified squamous epithelium protects surfaces that are rubbed, such as the skin and the oesophagus; simple columnar lines the stomach and intestine, where it secretes and absorbs; transitional epithelium lines the bladder and stretches as it fills.

Answer: B

Connective tissues are cells set in an extracellular matrix; in blood the cells (erythrocytes, leucocytes, platelets) are suspended in a liquid matrix, the plasma, so blood is a fluid connective tissue, like lymph; bone and cartilage are connective tissues with solid matrices. Epithelium consists of cells packed together with almost no matrix, and muscle and nervous tissue are defined by contraction and by electrical signalling.

Answer: D

Ca²⁺ binds to troponin, which changes shape and pulls tropomyosin away from the myosin-binding sites on actin; myosin heads can then form cross-bridges and the filaments slide. Ca²⁺ does not bind myosin or actin directly in skeletal muscle, and tropomyosin is the strand that blocks the binding sites until troponin moves it.

Answer: C

The epidermis, a stratified squamous epithelium, is avascular: its cells are nourished by diffusion from capillaries in the dermis beneath, which is why a shallow scratch does not bleed. The papillary layer is the uppermost part of the dermis and carries the capillary loops that do this, and the hypodermis below holds larger vessels and fat.

Answer: A

At threshold, voltage-gated Na⁺ channels open and Na⁺ rushes in down its concentration and electrical gradient, driving the membrane from about −70 mV towards +30 mV. K⁺ flows out, not in, during repolarisation, through voltage-gated K⁺ channels. The Na⁺/K⁺-ATPase pumps Na⁺ out of the cell and K⁺ in; it maintains the gradients but does not produce the spike.

Answer: B

In myopia the eyeball is too long or the lens system too strong, so light from distant objects is focused in front of the retina. A diverging (concave) lens spreads the rays a little before they enter the eye, moving the focus back onto the retina. An image behind the retina is hypermetropia (long-sightedness), corrected with a converging lens.

Answer: D

Insulin, from the β cells of the pancreatic islets, increases glucose uptake into muscle and fat cells and promotes glycogen synthesis in the liver, lowering blood glucose. Glucagon (from the α cells), adrenaline and cortisol all raise blood glucose, by glycogen breakdown and gluconeogenesis.

Answer: D

In the hypothalamus–pituitary–thyroid axis, TRH from the hypothalamus stimulates TSH release from the anterior pituitary, and TSH stimulates the thyroid. High thyroxine inhibits both the hypothalamus and the pituitary, so less TRH and TSH are secreted and thyroxine output falls back towards normal. Increasing TRH or TSH would be positive feedback, pushing the level higher still.

Answer: B

The pulmonary veins return oxygenated blood from the lungs to the left atrium. They are veins because they carry blood towards the heart, whatever its oxygen content. The pulmonary artery carries deoxygenated blood from the right ventricle to the lungs, the superior vena cava returns deoxygenated blood from the upper body to the right atrium, and the aorta carries oxygenated blood away from the heart to the body.

Answer: A

A person of group A has A antigens on their red cells and anti-B antibodies in their plasma (they cannot have anti-A, which would attack their own cells). Group AB red cells carry B antigens, which their anti-B antibodies would agglutinate, so AB blood is not safe for them; as an Rh-negative patient they should also receive Rh-negative blood.

Answer: C

Antibodies are made by plasma cells, which are B lymphocytes that have been activated by their antigen (usually with help from helper T cells) and differentiated into antibody factories; this is humoral immunity. Cytotoxic T cells kill infected cells directly (cellular immunity), neutrophils are phagocytes of innate immunity, and erythrocytes carry oxygen.

Answer: D

The child's own immune system responds to the toxoid by making antibodies and memory cells, so the immunity is active, and it was induced deliberately by a vaccine, so it is artificial. Passive immunity means receiving ready-made antibodies: naturally across the placenta or in breast milk, or artificially as an injection of antitoxin or immunoglobulin, which protects at once but leaves no memory. Natural active immunity follows recovery from the infection itself.

Answer: B

About 70% of CO₂ is carried as hydrogencarbonate: in the red cells carbonic anhydrase converts CO₂ and water to carbonic acid, which dissociates to H⁺ and HCO₃⁻, and the HCO₃⁻ moves out into the plasma in exchange for Cl⁻ (the chloride shift). Roughly 20–25% is bound to haemoglobin as carbaminohaemoglobin and only about 7–10% is dissolved. Carbonic acid is short-lived and is not carried on albumin.

Answer: A

At rest, inspiration is active: the diaphragm contracts and flattens and the external intercostals lift the ribs, the thoracic volume increases, the pressure inside the lungs falls below atmospheric (Boyle's law) and air flows in. Relaxation of the diaphragm is quiet expiration. Air always flows from higher to lower pressure, so it cannot enter when lung pressure is above atmospheric, and the internal intercostals pull the ribs down in forced expiration.

Answer: C

Pepsin is secreted by the chief cells of the gastric glands as inactive pepsinogen, activated by the stomach's hydrochloric acid, and works best at about pH 2. Trypsin also digests proteins, but in the small intestine, from the pancreas. Amylase digests starch, and lipase digests fats.

Answer: A

Bile is produced by the liver and stored and concentrated in the gall bladder, which releases it into the duodenum. Its bile salts emulsify fats into small droplets, giving pancreatic lipase a much larger surface to work on, but bile itself contains no digestive enzymes, so it neither hydrolyses triglycerides nor digests proteins.

Answer: A

Glucose is reabsorbed in the proximal convoluted tubule by Na⁺-coupled cotransporters (SGLT), and in health all of it is taken back, so none appears in the urine. Glucose appears in the urine only when the blood level is so high that these carriers are saturated, as in uncontrolled diabetes. The loop of Henle concentrates the medulla, and the distal tubule and collecting duct adjust salt and water.

Answer: B

Osmoreceptors in the hypothalamus detect the rise in osmolality, and the posterior pituitary releases ADH. ADH inserts aquaporin water channels into the cells of the collecting ducts, so more water is reabsorbed into the concentrated medulla and a small volume of concentrated urine is produced, conserving body water. The first option describes the absence of ADH (as in diabetes insipidus), and ADH acts on water permeability, not on Na⁺ secretion or filtration rate.

Answer: D

Malaria is caused by Plasmodium species, single-celled eukaryotic parasites (protozoa), passed on in the saliva of an infected female Anopheles mosquito when she bites. They multiply in the liver and then in red blood cells, whose bursting causes the cycles of fever. Aedes mosquitoes transmit viruses such as dengue and yellow fever, and fleas transmit the plague bacterium.

Answer: C

Antibiotics act on structures found in bacteria but not in human cells: penicillin blocks the synthesis of the peptidoglycan cell wall, and others target 70S ribosomes or bacterial enzymes. A virus has no cell wall, no ribosomes and no metabolism of its own; it replicates using the host cell's machinery, so antibiotics have nothing to act on. Plasmids are found in bacteria, antibiotics reach the lungs well enough to treat bacterial pneumonia, and influenza is caused by a virus.

Answer: B

The mass number (37) is the number of protons plus neutrons, and the atomic number (17) is the number of protons, so the neutrons are 37 − 17 = 20. 17 is the number of protons (and electrons in the neutral atom), 37 is the mass number itself, and 54 adds the two numbers instead of subtracting them.

Answer: A

The mole (mol) is one of the seven SI base units, the unit of amount of substance. The gram is a unit of mass (the SI base unit of mass is the kilogram), the litre is a unit of volume derived from the cubic metre, and the dalton (unified atomic mass unit) is a unit of mass used for atoms and molecules, not of amount.

Answer: C

Let x be the fraction of ³⁵Cl: 35x + 37(1 − x) = 35.5, so 37 − 2x = 35.5 and x = 0.75, that is 75%. A 50:50 mixture would average 36, and 25% ³⁵Cl would give 36.5; the average of 35.5 lies closer to 35, so the lighter isotope must be the more common one.

Answer: A

A neutral iron atom has 26 electrons, equal to its 26 protons; the 3+ charge means three electrons have been lost, leaving 23. 26 is the electron count of the neutral atom (and the number of protons), 29 adds the three instead of subtracting them, and 30 is the number of neutrons (56 − 26).

Answer: D

The larger the difference in electronegativity, the more ionic the bond: Na–Cl differs by 3.0 − 0.9 = 2.1, against 1.4 for O–H, 1.0 for C–O and 0.4 for C–H. A difference above about 1.7 is usually taken as mainly ionic, so only Na–Cl qualifies; the others are polar (O–H, C–O) or nearly non-polar (C–H) covalent bonds.

Answer: B

The electronegativity difference is 3.5 − 2.1 = 1.4: large enough for the shared electrons to sit closer to oxygen, giving a polar covalent bond (δ− on O, δ+ on H), but below the roughly 1.7 at which a bond is usually called ionic. A non-polar covalent bond needs a difference close to zero, and metallic bonding occurs between metal atoms, not between two non-metals.

Answer: D

Atomic radius increases down a group, because each element down the group has its outer electrons in a new, more distant shell. Rubidium (period 5) is the lowest of the four in Group 1, so its atoms are the largest; lithium (period 2) has the smallest.

Answer: B

Nitrogen has four electron pairs around it: three bonding pairs and one lone pair. They point to the corners of a tetrahedron, but the shape is named from the atoms only, so the three hydrogens and the nitrogen form a trigonal pyramid (bond angle about 107°). Trigonal planar would need three pairs and no lone pair (as in BF₃); tetrahedral describes the electron pairs, or a molecule like CH₄ with four bonded atoms.

Answer: B

n = m/M = 9.0 g ÷ 18 g mol⁻¹ = 0.50 mol. 2.0 mol inverts the division (18 ÷ 9.0), 162 multiplies mass by molar mass, and 0.25 mol halves the right answer again.

Answer: D

V = n × Vm = 0.50 mol × 22.4 dm³ mol⁻¹ = 11.2 dm³. 22.4 dm³ is the volume of one mole, 44.8 dm³ that of two, and 5.6 dm³ that of a quarter of a mole.

Answer: C

22 g of propane is 22 ÷ 44 = 0.50 mol. The equation gives 3 mol of CO₂ per mole of propane, so 1.5 mol of CO₂ forms, and 1.5 mol × 44 g mol⁻¹ = 66 g. 22 g ignores the 1:3 ratio (as if one CO₂ formed per propane), 44 g is the mass of one mole, and 132 g is the CO₂ from a whole mole of propane.

Answer: D

At constant pressure and amount of gas, volume is proportional to absolute temperature (Charles's law, from pV = nRT): V₂ = V₁ × T₂/T₁ = 2.0 L × 450/300 = 3.0 L. 1.3 L uses the ratio upside down, 2.0 L assumes no change, and 4.5 L has no basis in the data.

Answer: B

n = 5.85 ÷ 58.5 = 0.100 mol, and c = n/V = 0.100 mol ÷ 0.500 L = 0.20 mol L⁻¹. 0.10 mol L⁻¹ forgets to divide by the volume in litres (it is the amount in moles), 0.50 mol L⁻¹ uses the volume figure, and 11.7 divides grams by litres, a mass concentration in g L⁻¹, not a molar one.

Answer: C

Normality is the number of equivalents per litre; in an acid–base reaction one equivalent of an acid supplies one mole of H⁺. Each mole of H₂SO₄ gives two moles of H⁺, so a 0.10 mol L⁻¹ solution is 2 × 0.10 = 0.20 N (0.20 eq L⁻¹). 0.10 N would be right for a monoprotic acid such as HCl, 0.05 N divides instead of multiplying, and 0.40 N counts the protons twice.

Answer: B

Mixing silver nitrate with sodium chloride brings Ag⁺ and Cl⁻ together, and silver chloride is insoluble: Ag⁺ + Cl⁻ → AgCl↓, a white precipitate. In the other three mixtures every possible combination of ions (NaNO₃, KCl, KNO₃, NaCl, K₂SO₄, Na₂SO₄) is soluble, so the ions simply stay in solution.

Answer: A

Hydrochloric acid is a strong acid and dissociates completely, so [H⁺] = 1.0 × 10⁻³ mol L⁻¹ and pH = −log[H⁺] = 3. A pH of 11 is that of a 0.0010 mol L⁻¹ strong base such as NaOH (pOH 3), 1 would need [H⁺] = 0.10 mol L⁻¹, and 13 is a strongly basic value.

Answer: D

Non-metal oxides are generally acidic: sulfur trioxide reacts with water to give sulfuric acid, SO₃ + H₂O → H₂SO₄. Na₂O, CaO and MgO are metal oxides and basic: Na₂O and CaO react with water to form the hydroxides NaOH and Ca(OH)₂, and MgO is only slightly soluble but neutralises acids.

Answer: C

The acid contains 0.020 L × 0.10 mol L⁻¹ = 2.0 × 10⁻³ mol of H₂SO₄. Each mole needs 2 mol of NaOH, so 4.0 × 10⁻³ mol of NaOH is needed, which is 4.0 × 10⁻³ mol ÷ 0.10 mol L⁻¹ = 0.040 L = 40 mL. 20 mL treats the acid as monoprotic, 10 mL halves instead of doubling, and 80 mL doubles twice.

Answer: C

A metal with a more negative electrode potential reduces the ions of a metal with a more positive one: zinc (−0.76 V) displaces copper from Cu²⁺ (+0.34 V), and the cell potential is +0.34 − (−0.76) = +1.10 V, positive and so spontaneous. The other three run the wrong way: in each, a metal higher in potential (copper, silver) is asked to reduce the ions of a more reactive metal (zinc, iron), which gives a negative cell potential.

Answer: D

Potassium is +1 and each oxygen −2, and the oxidation numbers in a neutral compound add up to zero: +1 + x + 4(−2) = 0, so x = +7. +2 is manganese in Mn²⁺ (the product when permanganate is reduced in acid), +4 in MnO₂ and +6 in the manganate ion MnO₄²⁻.

Answer: A

Manganese goes from +7 to +2, gaining 5 electrons; each Fe²⁺ loses one electron, so 5 Fe²⁺ are needed. The 4 oxygen atoms of MnO₄⁻ become 4 H₂O, which needs 8 H⁺: MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O. Charges check: −1 + 10 + 8 = +17 on the left and +2 + 15 = +17 on the right. '5 and 4' confuses the H⁺ with the water, and 3 or 1 Fe²⁺ would not balance the five electrons.

Answer: A

A triple bond removes four hydrogen atoms from the corresponding alkane, CₙH₂ₙ₊₂, giving CₙH₂ₙ₋₂ (ethyne, C₂H₂; propyne, C₃H₄). CₙH₂ₙ₊₂ is the alkanes, CₙH₂ₙ the alkenes (and cycloalkanes), and CₙHₙ fits ethyne and benzene but not the series as a whole.

Answer: A

The catalyst polarises Br₂ into an electrophile that attacks the electron-rich ring; a hydrogen is then replaced by bromine, giving bromobenzene and HBr. Benzene substitutes rather than adds, because substitution keeps the stable delocalised (aromatic) π system intact. Nucleophilic addition is typical of carbonyl groups, and an elimination removes atoms to form a multiple bond.

Answer: A

The C=C double bond of an alkene adds bromine quickly at room temperature, so the orange bromine water turns colourless; an alkane reacts with bromine only by slow radical substitution in UV light. Both hydrocarbons burn and both are insoluble in water, and limewater detects carbon dioxide, not a hydrocarbon.

Answer: C

Propan-2-ol is a secondary alcohol, and a secondary alcohol is oxidised to a ketone, here propanone (acetone); the orange dichromate turns green. Propanal and propanoic acid come from oxidising the primary alcohol propan-1-ol. Propene would come from dehydration (an elimination with concentrated acid), not from oxidation.

Answer: B

An ester is named alcohol part first, acid part second: methanol gives the methyl group and ethanoic acid the ethanoate, so the product is methyl ethanoate, CH₃COOCH₃, with water. Ethyl methanoate is the isomer made from ethanol and methanoic acid; ethyl ethanoate comes from ethanol and ethanoic acid, and methyl methanoate from methanol and methanoic acid.

Answer: B

Phenol is a weak acid, much stronger than ethanol, because the negative charge of the phenoxide ion is spread over the benzene ring; it reacts with sodium hydroxide to give sodium phenoxide and water. Ethanol is a weaker acid than water and is not deprotonated by aqueous NaOH (it forms an alkoxide only with sodium metal). So only phenol forms a salt with NaOH.

Answer: D

Lysine's side chain ends in an amino group, which is basic and is protonated to –NH₃⁺ at physiological pH, so lysine is a basic, positively charged amino acid. Aspartic acid's side-chain carboxyl group loses its proton and is negatively charged at pH 7; serine's –OH is polar but uncharged, and glycine's hydrogen side chain is non-polar.

Answer: A

A triglyceride is the triester of glycerol (propane-1,2,3-triol) with three fatty acids. Hydrolysis by NaOH breaks the three ester bonds, giving one molecule of glycerol and three carboxylate salts, which are soap. There is only one glycerol per triglyceride, glucose belongs to carbohydrates, and amino acids come from hydrolysing proteins.

Answer: C

Sucrose (table sugar) is a disaccharide of glucose and fructose joined by a glycosidic bond, so hydrolysis gives one of each. Two glucose units come from maltose, and glucose plus galactose from lactose; no common disaccharide gives fructose and galactose.