Sofia University Medicine Entrance Exam Practice Paper with Answers (2027)

Answer: B

Cartilage is avascular: its chondrocytes sit in lacunae and receive nutrients by diffusion through the matrix from the surrounding perichondrium or synovial fluid, which is why damaged cartilage repairs so slowly. Compact bone is supplied by vessels running in its central (Haversian) canals, skeletal muscle has a dense capillary network, and loose connective tissue carries the vessels that supply the epithelia above it.

Answer: D

About a third of the bone matrix is organic, mainly collagen, which resists bending and stretching; the rest is mineral, mostly calcium phosphate as hydroxyapatite, which resists compression. The first option swaps the two cell types: osteoblasts build bone and osteoclasts resorb it. A long bone grows in length at the epiphyseal (growth) plate between the epiphysis and the diaphysis, not at the articular cartilage.

Answer: A

Calcium binding to troponin moves tropomyosin off the binding sites on actin, and the filaments then slide past each other without changing length, so the I bands and H zone narrow while the A band keeps its width. Statement 2 is the usual misreading of the sliding filament theory. Statement 4 is wrong because the binding of a new ATP molecule is what releases the myosin head from actin, which is why muscles stay locked in rigor mortis once ATP runs out.

Answer: C

Group A plasma contains anti-B antibodies, so red cells carrying the B antigen (groups B and AB) would be agglutinated. Group O red cells carry neither the A nor the B antigen, and Rh-negative cells lack the D antigen, so they meet both conditions. Group A, Rh-positive cells match for ABO but would make an Rh-negative patient produce anti-D antibodies, endangering later transfusions and pregnancies.

Answer: C

The sinoatrial node is the heart's pacemaker: its cells depolarise spontaneously faster than any other cardiac tissue, so they set the rate. The atrioventricular node delays the impulse so that the atria finish contracting before the ventricles do, and the bundle of His and the Purkinje fibres carry it through the ventricles. These can fire on their own too, but more slowly, so normally they follow the sinoatrial node.

Answer: B

This is the Bohr effect: carbon dioxide lowers the pH, and together with the higher temperature this lowers haemoglobin's affinity for oxygen, shifting the dissociation curve to the right so that more oxygen is unloaded where it is needed. A lower affinity means a shift to the right, not to the left, so the third option contradicts itself. Carbon dioxide does not compete for the iron of haem; it binds to amino groups of the globin chains or travels mostly as hydrogencarbonate ions.

Answer: A

Bile contains no digestive enzymes: bile salts act like a detergent, breaking large fat globules into small droplets so that pancreatic lipase has more surface to work on. Hydrolysing triglycerides is lipase's own job, and pepsinogen is activated by hydrochloric acid in the stomach. Bile is made by the liver and only stored and concentrated in the gallbladder.

Answer: D

Statements 1, 3 and 4 describe the normal control of reproduction: LH stimulates the Leydig cells in men, its mid-cycle surge causes ovulation in women, and the corpus luteum's progesterone thickens and maintains the endometrium. Statement 2 is wrong because FSH, like LH, is made by the anterior pituitary gland; the ovaries respond to it but do not produce it. Every option containing 2 is therefore wrong.

Answer: B

The ectoderm gives rise to the epidermis and its derivatives and, through the neural tube, to the brain and spinal cord. Skeletal muscles, the heart and the blood vessels come from the mesoderm, so the first and last options are wrong. The epithelial lining of the gut and the respiratory tract comes from the endoderm, not the mesoderm.

Answer: A

Motor neuron cell bodies lie in the anterior horns of the grey matter, and their axons leave through the ventral roots. The dorsal root ganglia hold the cell bodies of sensory neurons, which is why they are tempting. The white matter consists of myelinated axons in ascending and descending tracts, and the posterior horns receive sensory fibres and contain interneurons.

Answer: A

The adrenal medulla is a modified sympathetic ganglion, so nerve impulses release its adrenaline, and aldosterone acts on the distal tubule and collecting duct to retain sodium (and with it water). Statement 3 is wrong because cortisol raises blood glucose, mainly by stimulating gluconeogenesis in the liver. Statement 4 is wrong because cortisol secretion is driven by adrenocorticotropic hormone (ACTH) from the anterior pituitary; TSH acts on the thyroid.

Answer: C

Rods are very sensitive to light but contain one pigment only, so they give vision in dim light without colour; the ciliary muscle adjusts the tension on the suspensory ligaments and so the curvature of the lens (accommodation). The organ of Corti in the cochlea turns sound vibrations into nerve impulses, while the fluid in the semicircular canals detects rotation of the head. The second option swaps the two parts of the inner ear, a common confusion because both lie in the same bony labyrinth.

Answer: D

The epidermis is a stratified squamous epithelium with no blood vessels; melanocytes sit in its basal layer and pass the melanin they make to the surrounding keratinocytes. Sebaceous glands, blood capillaries and the coiled secretory part of the sweat glands all lie in the dermis. Sweat reaches the surface through a duct that crosses the epidermis, which is what makes the third option tempting.

Answer: B

In each PCR cycle the DNA is denatured by heat, two short primers anneal on either side of the target sequence, and DNA polymerase extends them, so the number of copies roughly doubles and even a few viral genomes become detectable. The method uses a heat-stable polymerase (such as Taq polymerase from a hot-spring bacterium) precisely so that it survives the 95 °C steps. Separating fragments by size in an electric field is gel electrophoresis, a different technique often used afterwards, and PCR makes no proteins.

Answer: C

RNA polymerase reads the template 3'→5' and builds a complementary, antiparallel mRNA 5'→3' with uracil in place of thymine, giving 5'-AUG CCU AAG-3'. The anticodon pairs with the codon 5'-CCU-3', so it is 3'-GGA-5'; repeating the codon itself, as in the first option, is the commonest slip. The second option copies the template instead of pairing with it, and the fourth contains thymine, which RNA does not.

Answer: A

This is competitive inhibition: substrate and inhibitor compete for the same active site, so with much more substrate the enzyme is occupied by substrate most of the time and the rate approaches its maximum again. The second option describes non-competitive inhibition, whose effect extra substrate cannot overcome, and the fourth describes irreversible inhibition, which contradicts the reversible binding in the stem. The same principle explains why ethanol is given in methanol poisoning: it competes with methanol for alcohol dehydrogenase.

Answer: D

Penicillin works by blocking the building of the peptidoglycan cell wall of bacteria. Viruses are not cells but particles of nucleic acid in a protein coat, sometimes with an envelope, and they multiply only by using the host cell's machinery, so there is nothing for the antibiotic to act on. They have no cell wall at all and are much smaller than bacteria, and influenza is caused by a virus, not a bacterium.

Answer: B

Moving ions against their concentration gradients needs energy, and the pump, a carrier protein, takes it from the hydrolysis of ATP, moving three sodium ions out and two potassium ions in on each cycle. Simple and facilitated diffusion are passive and only move substances down a gradient, even when a protein channel is involved. Osmosis is the passive movement of water, not of ions.

Answer: C

The nucleolus is a dense region of the nucleus, with no membrane of its own, where rRNA genes are transcribed and the rRNA is combined with proteins into ribosomal subunits that leave through the nuclear pores. Mitochondrial DNA stays inside the mitochondria, and chromatin is DNA wound around histone proteins, not phospholipids. Proteins are made on ribosomes in the cytoplasm, free or on the rough endoplasmic reticulum, after the mRNA has left the nucleus.

Answer: D

The DNA was replicated in the S phase of interphase, so each of the 46 chromosomes consists of two sister chromatids joined at the centromere until anaphase. 23 chromosomes with 46 chromatids describes a cell in metaphase II of meiosis, and 92 chromosomes are counted only from anaphase until the cell divides, once the sister chromatids have separated. Karyotypes are prepared from cells stopped in metaphase because the chromosomes are then most condensed and easiest to count.

Answer: A

This is an analytical (test) cross: the green, wrinkled parent (yyrr) gives only yr gametes, so the offspring show directly which gametes the other parent made. All the offspring are yellow, so that parent gives only Y and is YY; round and wrinkled appear about 1 : 1, so it is Rr. YyRr would give four classes in a 1 : 1 : 1 : 1 ratio, YyRR would give yellow and green but no wrinkled seeds, and YYRR would give only yellow, round seeds.

Answer: D

Mutations are classed as gene (point) mutations, which change the base sequence of one gene, chromosomal mutations, which change the structure of chromosomes, and genomic mutations, which change their number. Statements 1, 2 and 3 fit this: trisomy 21 is a change in number, sickle-cell anaemia comes from one base substitution, and a single deletion causes a frameshift. A translocation changes the structure of chromosomes, so it is a chromosomal mutation, not a gene mutation, and every option containing 4 is wrong.

Answer: B

The cross Aa × Aa gives AA, Aa and aa in a 1 : 2 : 1 ratio, so each child has a 1/4 chance of being affected. A child known to be unaffected is AA or Aa, and two of those three equally likely outcomes are carriers, so the answer to b) is 2/3; 1/2 is the tempting figure for a child chosen before anyone knows whether it is affected. Each birth is independent, so the chance that two children are both affected is 1/4 × 1/4 = 1/16.

Answer: C

In oogenesis both meiotic divisions split the cytoplasm unequally, so one large egg cell keeps most of the cytoplasm and nutrients while the small polar bodies degenerate. In spermatogenesis the divisions are equal and give four spermatids, each of which matures into a spermatozoon. Answers giving four or two eggs overlook the unequal division, and one spermatozoon overlooks that both daughter cells of each division survive.

Answer: A

Statements 1 and 2 describe the two arms of specific immunity: antibodies from plasma cells (humoral) and cytotoxic T cells that kill infected cells (cell-mediated). A vaccine contains antigens, not antibodies, and makes the body produce its own antibodies and memory cells, so it gives artificial active immunity. Maternal antibodies are ready-made, so they give natural passive immunity, which fades within months because the baby makes no memory cells from them.

Answer: A

The atomic number gives the protons: 17. The mass number counts protons plus neutrons, so the neutrons are 37 − 17 = 20. A neutral atom has as many electrons as protons, 17; 18 electrons would be the chloride ion, Cl⁻.

Answer: C

X has 16 electrons, so it is sulfur: three shells put it in period 3, and six outer electrons (3s² 3p⁴) in group 16 (VIA), not 14. Its highest oxidation state is +6, so its highest oxide is XO₃ (SO₃); SO₂ exists but is not the highest. Gaining two electrons gives it −2 in H₂X (H₂S), so a correct answer is listed.

Answer: B

In N₂ both atoms are the same element, so they have equal electronegativity and share the bonding electrons equally. The H-Br and N-H bonds join atoms of different electronegativity and are polar covalent. KCl is ionic: potassium gives its electron to chlorine.

Answer: D

The compound is neutral: 3 × (+1) for potassium + x + 6 × (−1) for the cyanide ligands = 0, so x = +3. Six CN⁻ ligands are bonded to the iron, so its coordination number is 6. The name is potassium hexacyanidoferrate(III); +2 would be K₄[Fe(CN)₆].

Answer: C

Manganese goes from +7 in KMnO₄ to +2 in MnCl₂, gaining 5 electrons: it is reduced, so KMnO₄ is the oxidising agent. Ten Cl⁻ are oxidised from −1 to 0 (5Cl₂), losing 10 electrons to match the 2 × 5 gained; going from −1 to 0 is oxidation, not reduction. The other 6 chlorine atoms stay as Cl⁻ in 2KCl and 2MnCl₂, so only 10 of the 16 HCl act as the reducing agent.

Answer: B

At the cathode Cu²⁺ + 2e⁻ → Cu: 6.4 / 64 = 0.10 mol Cu takes 0.20 mol of electrons. At the anode water is oxidised rather than sulfate: 2H₂O → O₂ + 4H⁺ + 4e⁻, so 0.20 / 4 = 0.050 mol O₂ forms. Hydrogen would form at the cathode, not the anode, and sulfate is not oxidised to SO₂.

Answer: D

SO₃ is an acidic oxide: SO₃ + H₂O → H₂SO₄. Na₂O and CaO are basic oxides and give the hydroxides NaOH and Ca(OH)₂. CO is a neutral (non-salt-forming) oxide and gives no acid with water.

Answer: A

HClO₄ is perchloric acid, the oxoacid of chlorine in its +7 state. H₂SO₃ is sulfurous acid (sulfuric acid is H₂SO₄); NaNO₂ is sodium nitrite (nitrate is NO₃⁻); PO₄³⁻ is phosphate, so Ca₃(PO₄)₂ is calcium phosphate.

Answer: C

Na₂CO₃ is the salt of a strong base and a weak acid, and its anion is hydrolysed: CO₃²⁻ + H₂O ⇌ HCO₃⁻ + OH⁻, so the solution is alkaline. NH₄Cl, from a weak base and a strong acid, gives an acidic solution (NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺). NaCl and KNO₃ come from a strong acid and a strong base and give neutral solutions.

Answer: B

The rate depends on the square of [NO]: v′ = k(2[NO])²[O₂] = 4k[NO]²[O₂] = 4v. The rate constant k does not change because the temperature is constant. An eightfold increase would need [O₂] doubled as well.

Answer: D

The forward reaction is exothermic, so raising the temperature shifts the equilibrium to the left (Le Chatelier). Four moles of gas become two, so lowering the pressure favours the side with more moles, again the left. A catalyst speeds up both directions equally: equilibrium is reached sooner but its position does not change, so none of the three changes helps.

Answer: A

n(NaOH) = 0.40 / 40 = 0.010 mol in 1.0 L, so c = 0.010 mol/L. NaOH is a strong electrolyte and dissociates completely, so [OH⁻] = 1 × 10⁻² mol/L and pOH = 2. pH = 14 − 2 = 12; 2 is the pOH, not the pH.

Answer: B

Colloidal particles are large enough to scatter light, so a beam shows up from the side: the Tyndall effect, which a true solution does not show. They are too small to settle out quickly, unlike the particles of a suspension. A dialysis membrane holds colloidal particles back while letting small ions and molecules through, which is how dialysis separates them.

Answer: C

Three carbon skeletons are possible: pentane, CH₃CH₂CH₂CH₂CH₃; 2-methylbutane, CH₃CH(CH₃)CH₂CH₃; and 2,2-dimethylpropane, C(CH₃)₄. They share one molecular formula but differ in the order the atoms are joined, which is what structural theory means by isomers. Any other drawing is one of these three turned or bent.

Answer: D

Light splits Cl₂ into chlorine radicals, which take a hydrogen atom from methane; the methyl radical then takes a chlorine atom from Cl₂, and the chain continues. One hydrogen atom is replaced by one chlorine atom, so it is a substitution, by a radical mechanism. Alkanes have no double bond to add to.

Answer: A

By Markovnikov's rule the hydrogen adds to the carbon of the double bond that already has more hydrogens (C1), and the bromine to C2, through the more stable secondary carbocation. 1,2-Dibromopropane would come from Br₂ and propane from H₂. 1-Bromopropane is the main product only when peroxides are present.

Answer: B

Sulfuric acid generates the nitronium ion, NO₂⁺, an electrophile that replaces one hydrogen of the ring: C₆H₆ + HNO₃ → C₆H₅NO₂ + H₂O. Substitution keeps the stable aromatic ring intact, which addition would destroy. The mechanism is ionic, not radical.

Answer: D

In water, OH⁻ acts as a nucleophile and replaces the bromine: CH₃CH₂Br + NaOH → CH₃CH₂OH + NaBr. In ethanol, KOH acts mainly as a base and removes HBr: CH₃CH₂Br + KOH → CH₂=CH₂ + KBr + H₂O, an elimination. Ethane would need a reduction, which neither reagent carries out.

Answer: C

The phenoxide ion's charge is spread over the benzene ring, so phenol is the stronger acid and is neutralised by NaOH: C₆H₅OH + NaOH → C₆H₅ONa + H₂O. Ethanol is too weak an acid to react with NaOH. Both react with sodium, releasing hydrogen: 2C₂H₅OH + 2Na → 2C₂H₅ONa + H₂.

Answer: A

Amines are classed by how many carbon groups are bonded to the nitrogen. (CH₃)₂NH, N-methylmethanamine, has two, so it is secondary; CH₃NH₂ (methanamine) is primary and (CH₃)₃N tertiary. CH₃CONH₂ is ethanamide, an amide, not an amine.

Answer: B

Adding one H₂O across the triple bond gives the enol CH₂=CH–OH, which is unstable and rearranges at once to the aldehyde: HC≡CH + H₂O → CH₃CHO (the Kucherov reaction). The formulae balance: C₂H₂ + H₂O = C₂H₄O. Ethanol would need two more hydrogen atoms, and ethanoic acid an oxidation.

Answer: D

M = 7.4 / 0.10 = 74 g/mol. For CₙH₂ₙ₊₁COOH, M = 14n + 1 + 45 = 14n + 46, so 14n = 28 and n = 2: C₂H₅COOH, propanoic acid (C₃H₆O₂ = 36 + 6 + 32 = 74). Methanoic acid is 46, ethanoic 60 and butanoic 88 g/mol.

Answer: C

CH₃COOH + CH₃OH ⇌ CH₃COOCH₃ + H₂O: the acyl part comes from the acid (ethanoate) and the alkyl part from the alcohol (methyl). Ethyl methanoate has the same formula, C₃H₆O₂, but comes from methanoic acid and ethanol. Methoxymethane is an ether, not an ester.

Answer: A

At low pH there is a large excess of H⁺, so both groups are protonated: the amino group as –NH₃⁺ and the carboxyl group as –COOH, giving a cation. Near its isoelectric point (about pH 6) glycine is the zwitterion ⁺H₃N–CH₂–COO⁻, and in strongly alkaline solution the anion H₂N–CH₂–COO⁻. The uncharged form is negligible in water at any pH.

Answer: C

In sucrose the glycosidic bond joins C1 of glucose to C2 of fructose, using both anomeric carbons, so neither ring can open to a free carbonyl group. Maltose (α-1,4) and lactose (β-1,4) keep a free anomeric carbon on one unit and are reducing. Sucrose is listed, so there is a correct answer.