Free practice paper · 2027 format
Both tests in the format of the entry tests of the Medical University of Varna: Biology and Chemistry, 90 minutes each, in all five sections, from 35 multiple-choice questions to definitions in your own words. MU-Varna's sample papers come with a filled-in answer sheet but no explanations; this paper explains every answer. Written by IvyTest; it is not an official MU-Varna paper.
Before you start
| Section | Questions | Time | Marks (Entry Tests in Biology and Chemistry) |
|---|---|---|---|
| Biology | 35 | 90 min | +1 correct, 0 wrong, 0 blank |
| Chemistry | 35 | 90 min | +1 correct, 0 wrong, 0 blank |
Sit the two tests as on the exam day, 90 minutes each, without a calculator or periodic table (MU-Varna does not say whether either is allowed, so every value you need is given). Answer Section A here; write Sections B to E on paper and mark them against the model answers.
Section 1 of 2
35 questions · 90 minutes
Which nitrogenous base occurs in RNA but not in DNA?
Answer: C
RNA uses uracil where DNA uses thymine, so uracil pairs with adenine in RNA. Thymine is the base found only in DNA, while adenine and cytosine (like guanine) occur in both nucleic acids.
Which statement about enzymes is NOT correct?
Answer: B
Enzymes are biological catalysts: they come out of the reaction unchanged and are used again and again, which is why small amounts are enough. The other three statements are true: they speed reactions by lowering the activation energy, they bind the substrate at a complementary active site, and their activity depends on temperature and pH, which can denature them.
A triplet on the template strand of DNA reads 3'-CGT-5'. Which anticodon of tRNA pairs with the mRNA codon transcribed from it?
Answer: D
Transcription gives the complementary, antiparallel codon 5'-GCA-3'; the anticodon pairs with it antiparallel, 3'-CGU-5', which written 5'→3' is 5'-UGC-3' (the template's own letters with U for T). 5'-GCA-3' is the codon itself, 5'-ACG-3' puts the letters in the wrong direction (G would face G), and 5'-CGT-3' contains thymine, which RNA does not.
What does every virus particle contain?
Answer: A
A virion is a nucleic acid, DNA or RNA but never both, packed in a protein capsid, sometimes with a lipid envelope outside. Viruses have no ribosomes, which is why they can multiply only inside a host cell; a peptidoglycan wall belongs to bacteria.
Which of these structures is found in both prokaryotic and eukaryotic cells?
Answer: C
Every cell is bounded by a plasma membrane, and every cell also has cytoplasm, ribosomes and DNA. Prokaryotes have no membrane-bound organelles: no nuclear envelope (their DNA lies in a nucleoid), no mitochondria and no endoplasmic reticulum.
Insulin is made in the beta cells of the pancreas and secreted into the blood. Which route does it take through the cell?
Answer: D
Proteins for export are made by ribosomes on the rough ER and enter its lumen, then travel in vesicles to the Golgi apparatus, which modifies and packages them into secretory vesicles that fuse with the plasma membrane (exocytosis). The smooth ER makes lipids, mitochondria make ATP, and lysosomes hold digestive enzymes, not secretions.
Bacteria whose DNA contains only heavy nitrogen (¹⁵N) are moved to a medium containing only light nitrogen (¹⁴N) and replicate their DNA once. What does each daughter DNA molecule contain?
Answer: B
Replication is semi-conservative: the two parental strands separate and each serves as the template for one new strand, so every daughter molecule is one old ¹⁵N strand paired with one new ¹⁴N strand. One all-heavy and one all-light molecule would be conservative replication, and patchwork strands would be dispersive; neither is what happens.
In which phase of mitosis do the sister chromatids separate and move to opposite poles of the cell?
Answer: A
In anaphase the centromeres split and the spindle fibres pull the sister chromatids, now separate chromosomes, to opposite poles. In prophase the chromosomes condense, in metaphase they line up at the equator, and in telophase new nuclear envelopes form around the two sets.
Which event takes place in meiosis but not in mitosis?
Answer: B
Only in prophase I of meiosis do homologous chromosomes pair (forming bivalents) and exchange segments by crossing over at chiasmata. DNA is replicated in interphase before both divisions, sister chromatids separate in mitotic anaphase and in anaphase II, and both divisions use a spindle.
The body cells of an animal species have 2n = 24 chromosomes. How many chromosomes and chromatids are there in one cell at metaphase II of meiosis?
Answer: D
Meiosis I separates the homologous chromosomes, so each cell entering meiosis II is haploid, with 12 chromosomes; their sister chromatids do not part until anaphase II, so there are still 24 chromatids. 24 chromosomes with 48 chromatids is metaphase of mitosis or metaphase I, and 12 single-chromatid chromosomes is the state after meiosis II is complete.
Which stage of aerobic respiration produces most of the ATP obtained from one molecule of glucose?
Answer: C
Oxidative phosphorylation, driven by the electron transport chain on the inner mitochondrial membrane and ATP synthase, makes the large majority of the roughly 30 ATP per glucose. Glycolysis gives a net 2 ATP and the Krebs cycle 2 more directly; the link reaction makes none, though it, like the Krebs cycle, supplies NADH to the chain.
During intense exercise, muscle cells short of oxygen respire anaerobically. What is pyruvate converted to?
Answer: A
In human muscle, pyruvate is reduced to lactate, which regenerates the NAD⁺ that glycolysis needs to keep making ATP. Ethanol and carbon dioxide are the products in yeast and plants, and acetyl-CoA is formed in the mitochondria only when oxygen is available; glucose is remade from lactate later, in the liver.
How many molecules of carbon dioxide are released in total when one molecule of glucose is completely oxidised in aerobic respiration?
Answer: B
Glycolysis releases no CO₂. Each of the two pyruvates loses one CO₂ in the link reaction (2), and each of the two acetyl groups releases two in the Krebs cycle (4), giving 6, as in C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O. Counting only the link reaction or only the Krebs cycle gives the tempting 2 or 4.
In peas, yellow seeds (Y) are dominant over green seeds (y). Two plants heterozygous for seed colour (Yy) are crossed. What proportion of the offspring have green seeds?
Answer: C
Yy × Yy gives 1 YY : 2 Yy : 1 yy, and only yy shows the recessive green colour, so 1/4. 3/4 is the proportion with yellow seeds, and 1/2 is the proportion that are heterozygous, not the proportion that are green.
In guinea pigs, black coat (B) is dominant over white (b). A black guinea pig crossed with a white one has 6 black and 5 white young. What is the genotype of the black parent?
Answer: A
This is a test cross. Each white young is bb and so received a b allele from both parents, which means the black parent carries b and must be Bb; Bb × bb gives the 1:1 ratio seen. A BB parent would have given only black young, and bb would itself be white, so the genotype can be determined.
Two individuals heterozygous for two unlinked genes (AaBb × AaBb) are crossed; each gene shows complete dominance. What fraction of the offspring show the dominant phenotype for the first gene and the recessive phenotype for the second?
Answer: D
The genes assort independently, so the probabilities multiply: dominant for A (A_) is 3/4 and recessive for B (bb) is 1/4, giving 3/16. This is one of the two 3/16 classes of the 9:3:3:1 ratio; 9/16 shows both dominant traits and 1/16 is aabb.
A woman who carries haemophilia A (XᴴXʰ) and a man with normal blood clotting (XᴴY) have a son. What is the probability that this son has haemophilia?
Answer: B
A son receives the Y chromosome from his father and one of his mother's two X chromosomes, so he has an even chance of receiving Xʰ and being affected. 1/4 is the chance that any child of this couple, before its sex is known, is an affected son; the father's Xᴴ goes only to daughters, so it cannot protect a son.
Which of these is a genomic mutation, a change in the number of chromosomes?
Answer: C
Genomic mutations change the number of chromosomes: whole sets (polyploidy, such as triploidy, 3n) or single chromosomes (aneuploidy). A deletion or an inversion alters the structure of one chromosome, so they are chromosomal mutations, and a base substitution changes one gene, so it is a gene mutation.
Genes A and B lie on the same chromosome, 10 map units (centimorgans) apart. A dihybrid with A and B on one chromosome and a and b on its homologue (AB/ab) is test-crossed with ab/ab. What percentage of the offspring have the genotype Ab/ab?
Answer: A
A distance of 10 map units means 10% of gametes are recombinant, and the recombinants come in two equal classes, Ab and aB, so 5% each. 10% is both recombinant classes together, 45% is each parental class (AB or ab), and 25% would hold only if the genes assorted independently.
Which of these human diseases is inherited as an autosomal recessive trait?
Answer: D
Phenylketonuria is caused by two recessive alleles of an autosomal gene for the enzyme phenylalanine hydroxylase, so both parents of an affected child are usually healthy carriers. Haemophilia A is X-linked recessive, Huntington's disease is autosomal dominant, and Down syndrome is a chromosomal disorder (trisomy 21), not a single-gene trait.
Where does fertilisation normally take place in the human female?
Answer: B
The secondary oocyte released at ovulation is fertilised in the uterine tube, usually in its widest part, the ampulla. The early embryo then takes several days to reach the uterus, where it implants; the ovary releases the oocyte, and the cervix is only a passage for sperm.
How many mature egg cells (ova) can develop from one primary oocyte in human oogenesis?
Answer: A
Both meiotic divisions of oogenesis divide the cytoplasm unequally: meiosis I gives a secondary oocyte and a small first polar body, and meiosis II, completed only after fertilisation, gives the ovum and a second polar body. The polar bodies degenerate, so one primary oocyte yields one ovum; four functional cells is the result of spermatogenesis.
Which cells secrete antibodies into the blood?
Answer: C
When a B lymphocyte meets its antigen and is activated, it multiplies and differentiates into plasma cells, which secrete antibodies (humoral immunity), and memory cells. Cytotoxic T cells kill infected cells directly, and macrophages and neutrophils are phagocytes; none of them makes antibodies.
Which of these bones is NOT part of the upper limb?
Answer: D
The fibula is the thin lateral bone of the leg, beside the tibia. The humerus is the bone of the arm, and the radius and ulna are the two bones of the forearm.
When a sarcomere of skeletal muscle shortens during contraction, which of these keeps the same length?
Answer: A
By the sliding filament theory, the thin actin filaments slide over the thick myosin filaments without either changing length. The A band is the length of the myosin filaments, so it stays the same, while the I band and the H zone narrow as the filaments overlap more and the Z lines move closer together.
Which valve prevents blood from flowing back from the left ventricle into the left atrium?
Answer: B
The left atrioventricular valve has two cusps and is called the bicuspid or mitral valve; it closes when the left ventricle contracts. The tricuspid valve does the same job on the right side, and the aortic and pulmonary semilunar valves stop blood returning from the arteries into the ventricles.
What happens during quiet inhalation?
Answer: C
In quiet inhalation the diaphragm contracts and flattens and the external intercostal muscles lift the ribs, so the thorax enlarges, the pressure in the lungs falls below atmospheric pressure, and air flows in. A relaxing, doming diaphragm and the internal intercostals belong to exhalation, and air always moves from higher to lower pressure, so a rise in lung pressure pushes air out.
Which enzyme in saliva begins the digestion of starch in the mouth?
Answer: D
Salivary amylase (ptyalin) breaks starch down into maltose and shorter chains while food is chewed. Pepsin digests proteins in the stomach, trypsin digests proteins in the small intestine (from the pancreas), and lipase digests fats.
Which statement about the small intestine is NOT correct?
Answer: A
Hydrochloric acid is made by the parietal cells of the stomach, where it activates pepsin; the contents of the small intestine are made neutral to alkaline by bicarbonate from the pancreas. The other statements are true: villi and microvilli give an enormous absorbing surface, most absorption happens there, and fats leave the cells as chylomicrons into the lacteals.
Which hormone is secreted by the alpha cells of the islets of Langerhans and raises the concentration of glucose in the blood?
Answer: B
Glucagon, from the alpha cells, makes the liver break glycogen down and release glucose. Insulin comes from the beta cells and lowers blood glucose; adrenaline also raises it but is made by the adrenal medulla, and thyroxine, from the thyroid, raises the metabolic rate.
Which effect is produced by the parasympathetic division of the autonomic nervous system?
Answer: C
The parasympathetic division (largely through the vagus nerve) prepares the body for rest and digestion: it slows the heart, constricts the pupils and stimulates digestion. Dilated pupils, dilated bronchi and release of glucose prepare for action, so they are effects of the sympathetic division.
The arm of a human, the flipper of a whale and the wing of a bat do different jobs but are built on the same pattern of bones. What are they called, and what do they show?
Answer: A
Organs with the same basic structure and embryonic origin but different functions are homologous; they are comparative anatomical evidence that the species inherited the pattern from a common ancestor and then diverged. Analogous organs, such as the wings of a bird and an insect, share a function but not an origin, and these limbs are fully functional, not vestigial.
Which sequence lists these hominins in the order in which they appeared, from the earliest to the latest?
Answer: D
The australopithecines walked upright with a small brain from about 4 million years ago; Homo habilis, with a larger brain and the first stone tools, appeared about 2.4 million years ago; Homo erectus, which used fire and left Africa, followed about 1.9 million years ago; and Homo sapiens appeared about 300,000 years ago. Putting Homo erectus before Homo habilis is the commonest slip.
Which of these is an abiotic factor of the environment?
Answer: B
Abiotic factors are the non-living physical and chemical conditions, such as temperature, light, water and soil pH. Predation, competition and parasitism are interactions between living organisms, so they are biotic factors.
Arctic foxes have shorter ears and a more compact body than foxes of hot deserts. What is the main advantage of this in the cold?
Answer: C
Heat is lost through the body surface, so a compact body with small extremities, having a low surface area to volume ratio, conserves heat in the cold. Large ears do the opposite, which is why desert foxes have them; they help to lose heat, and the Arctic fox's short ears are not an adaptation for hearing.
Answer on paper, then mark your work against the model answer.
Fill in the four missing words or terms (1 point each).
A molecule of ATP is built from the base adenine, the five-carbon sugar (1) ______ and three phosphate groups. When its last phosphate group is removed by hydrolysis, energy is released for the cell's work and ATP becomes (2) ______. Glycolysis, the first stage of respiration, takes place in the (3) ______ of the cell; it splits one molecule of glucose into two molecules of pyruvate, with a net gain of (4) ______ molecules of ATP.
(1) ribose. (2) ADP (adenosine diphosphate; 'ADP and a phosphate' also earns the point). (3) cytoplasm (also accepted: cytosol). (4) two (2).
Why: ATP is a ribonucleotide, so its sugar is ribose, not deoxyribose. Removing the terminal phosphate leaves adenosine diphosphate, which respiration turns back into ATP. Glycolysis needs no organelle and runs in the cytosol; it uses 2 ATP and makes 4, a net gain of 2. 'Four' is the gross yield and does not earn the point.
Fill in the four missing words or terms (1 point each).
The skin has three layers. The outer layer, the (1) ______, is a stratified squamous epithelium whose surface cells are dead and filled with keratin. Beneath it lies the (2) ______, a layer of connective tissue that holds blood vessels, nerve endings, sweat and sebaceous glands and hair follicles. The deepest layer, the (3) ______, stores fat and insulates the body. Cells called melanocytes make the pigment (4) ______, which protects the deeper tissues from ultraviolet radiation.
(1) epidermis. (2) dermis. (3) hypodermis (also accepted: subcutaneous tissue, subcutis). (4) melanin.
Why: the epidermis is the keratinised epithelium with no blood vessels of its own; the dermis is the connective-tissue layer that carries the vessels, glands and follicles; the hypodermis beneath it is mostly fatty tissue. Melanin is made by melanocytes in the basal layer of the epidermis and passed to the surrounding cells, where it absorbs ultraviolet light; 'keratin' is the protein of the surface cells, not the pigment.
Match each structure of the eye or ear in Column 1 with its function in Column 2 (1 point per correct pair). Use each letter once.
Column 1: 1. cornea 2. lens 3. retina 4. cochlea 5. semicircular canals
Column 2: a. detect rotation of the head, for balance b. contains the rods and cones, which turn light into nerve impulses c. transparent front of the eye, which does most of the bending of light d. changes shape to focus on near or distant objects e. contains the organ of Corti, whose hair cells turn sound vibrations into nerve impulses
1 – c. The cornea is the clear front part of the outer coat of the eye, and its curved surface refracts light more than any other part.
2 – d. The lens is made rounder or flatter by the ciliary muscle (accommodation), fine-tuning the focus for near or distant objects.
3 – b. The retina is the inner layer that holds the photoreceptors: rods for dim light and cones for colour and detail.
4 – e. The cochlea in the inner ear holds the organ of Corti, where vibrations of the fluid bend hair cells and start impulses in the auditory nerve.
5 – a. The three semicircular canals, at right angles to each other, sense rotation of the head through movement of their fluid; they serve balance, not hearing.
Column 1 lists five stages of spermatogenesis out of order. Match each with its place in the sequence in Column 2, from first to fifth (1 point per correct pair).
Column 1: 1. spermatid 2. primary spermatocyte 3. spermatozoon 4. spermatogonium 5. secondary spermatocyte
Column 2: a. first b. second c. third d. fourth e. fifth
4 – a (first). The spermatogonium is the diploid stem cell in the wall of the seminiferous tubule, which multiplies by mitosis.
2 – b (second). A spermatogonium grows into a primary spermatocyte, still diploid, which enters meiosis I.
5 – c (third). Meiosis I gives two secondary spermatocytes, haploid, each chromosome still made of two chromatids.
1 – d (fourth). Meiosis II gives four haploid spermatids from each primary spermatocyte.
3 – e (fifth). Each spermatid matures, without dividing, into a spermatozoon with a head, midpiece and tail (spermiogenesis).
Write True or False for each statement (1 point each).
1. In an ecosystem, energy flows in one direction along the food chains, whereas chemical elements such as carbon and nitrogen are recycled.
2. Mutualism is a relationship between two species in which one species benefits and the other is harmed.
1. True. Energy enters as light captured by producers and is lost as heat at every trophic level, so it cannot be reused, while decomposers return carbon, nitrogen and other elements to the soil, water and air, where producers take them up again.
2. False. In mutualism both species benefit, as in the bacteria of the human large intestine, which are fed by their host and make vitamin K for it; a relationship in which one benefits and the other is harmed is parasitism.
Define active and passive immunity. Explain how each can be acquired naturally and artificially, giving an example of each, and say which kind lasts longer and why.
Active immunity is produced by a person's own immune system after it meets an antigen: lymphocytes are activated, plasma cells make antibodies and memory cells form. Passive immunity is protection by ready-made antibodies from another organism, without stimulating the person's own immune system.
Active immunity is acquired naturally by recovering from an infection, such as measles, and artificially by vaccination with a harmless form of the antigen, as in the tetanus vaccine. Passive immunity is acquired naturally when antibodies cross the placenta or pass to the baby in breast milk, and artificially by injecting antibodies, such as tetanus immunoglobulin after a dirty wound.
Active immunity takes days to develop but lasts years, because memory cells remain and respond quickly to a second exposure. Passive immunity acts at once but fades within weeks, because the borrowed antibodies are broken down and no memory cells are made.
Marking: 1 point for a correct definition of active immunity (own immune system, response to an antigen); 1 point for a correct definition of passive immunity (antibodies made elsewhere); 1 point for natural active and artificial active with an example of each; 1 point for natural passive and artificial passive with an example of each; 1 point for saying that active immunity lasts longer because of memory cells.
What is a gene (point) mutation? Name its main types, explain how they change the protein the gene codes for, and give an example of a human inherited disease caused by one.
A gene (point) mutation is a change in the nucleotide sequence of a single gene, affecting one or a few base pairs. Its main types are substitution (one base replaced by another), insertion (bases added) and deletion (bases lost).
A substitution changes one codon: it may be silent (same amino acid, because the code is degenerate), missense (a different amino acid) or nonsense (a stop codon, giving a shortened protein). Inserting or deleting a number of bases not divisible by three shifts the reading frame, so every later codon is misread (a frameshift) and the protein is usually useless.
Example: in sickle-cell anaemia a substitution in the β-globin gene changes the mRNA codon GAG to GUG, so valine replaces glutamic acid at position 6. The haemoglobin (HbS) clumps at low oxygen and red cells become sickle-shaped. It is autosomal recessive.
Marking: 1 point for the definition (a change in the nucleotide sequence of one gene); 1 point for naming substitution, insertion and deletion; 1 point for the effect of a substitution (silent, missense or nonsense, at least two named); 1 point for explaining the frameshift caused by an insertion or deletion; 1 point for a correct example, such as sickle-cell anaemia or cystic fibrosis (most often a deletion of three bases that removes one phenylalanine from the CFTR protein), with the change it causes.
Section 2 of 2
35 questions · 90 minutes
Which pair of atoms are isotopes of the same element?
Answer: A
Isotopes have the same atomic number (the same number of protons) but different numbers of neutrons: carbon-12 has 6 protons and 6 neutrons, carbon-14 has 6 protons and 8 neutrons. Carbon-14 and nitrogen-14, like argon-40 and calcium-40, share a mass number but are different elements. O₂ and O₃ are allotropes, two molecules of the same element, not atoms with different numbers of neutrons.
How many unpaired electrons does an oxygen atom (Z = 8) have in its ground state?
Answer: C
The configuration is 1s² 2s² 2p⁴. By Hund's rule the 2p electrons first occupy the three 2p orbitals singly, and only the fourth pairs up (↑↓ ↑ ↑), so two electrons are unpaired. Four unpaired electrons would need four 2p orbitals, but a p sublevel has only three.
Which particle contains a donor-acceptor (coordinate) bond?
Answer: B
The ammonium ion forms when nitrogen's lone pair is donated into the empty orbital of a proton: NH₃ + H⁺ → NH₄⁺. Once formed, the four N–H bonds are identical. In NH₃, CH₄ and H₂O each bond is an ordinary covalent bond, to which each atom gives one electron.
Which of these substances is made of ions held together by ionic bonds?
Answer: C
Potassium is a group 1 metal of low electronegativity (0.8) and chlorine a very electronegative non-metal (3.0); the difference of 2.2 is large enough for potassium to transfer its electron, giving K⁺ and Cl⁻. HCl and CCl₄ have polar covalent bonds, and Cl₂ a non-polar covalent bond between identical atoms.
Which oxide reacts with water to form an acid?
Answer: D
Sulfur trioxide is the acidic oxide of a non-metal: SO₃ + H₂O → H₂SO₄. Na₂O and CaO are basic oxides that react with water to form the bases NaOH and Ca(OH)₂, and MgO is a basic oxide that reacts with acids.
An iron nail is placed in a solution of copper(II) sulfate. What happens?
Answer: A
Iron stands before copper in the activity series, so it displaces copper from its salt: Fe + CuSO₄ → FeSO₄ + Cu. Iron is oxidised to Fe²⁺ and Cu²⁺ is reduced to red-brown copper metal, while the blue colour of the solution fades. No hydrogen forms, because there is no acid for the iron to react with.
Copper reacts with dilute nitric acid: Cu + HNO₃ → Cu(NO₃)₂ + NO + H₂O. When the equation is balanced with the smallest whole-number coefficients, what is the coefficient of HNO₃?
Answer: C
Copper goes from 0 to +2 (loses 2e⁻) and nitrogen from +5 to +2 in NO (gains 3e⁻), so 6 electrons are exchanged by 3Cu and 2NO. The 3Cu(NO₃)₂ need 6 nitrate groups, so HNO₃ = 6 + 2 = 8, and its 8 H atoms give 4H₂O: 3Cu + 8HNO₃ → 3Cu(NO₃)₂ + 2NO + 4H₂O. Oxygen checks: 24 on the left, 18 + 2 + 4 = 24 on the right.
The rate of a reaction doubles for every 10 °C rise in temperature. By what factor does the rate increase when the temperature is raised from 20 °C to 50 °C?
Answer: D
The rise is 50 − 20 = 30 °C, which is three steps of 10 °C. The rate doubles at each step: 2 × 2 × 2 = 2³ = 8. Six comes from multiplying 2 by 3 instead of raising 2 to the power 3.
For the equilibrium N₂O₄(g) ⇌ 2NO₂(g), ΔH = +57 kJ, which pair of changes both shift the equilibrium towards NO₂?
Answer: B
The forward reaction is endothermic, so by Le Chatelier's principle raising the temperature favours it, as the system absorbs the added heat. The forward reaction also turns 1 mole of gas into 2, so lowering the pressure favours it, towards the side with more gas molecules. Only raising the temperature together with lowering the pressure moves the equilibrium to the right on both counts.
25 g of glucose is dissolved in 100 g of water. What is the mass fraction of glucose in the solution?
Answer: A
The mass of the solution is solute plus solvent: 25 + 100 = 125 g. The mass fraction is 25 / 125 × 100% = 20%. Dividing by the mass of water alone gives the wrong 25%, and 80% is the mass fraction of the water.
What is the pH of a 0.005 mol/L solution of sulfuric acid, H₂SO₄, at 25 °C, assuming both protons dissociate completely?
Answer: B
Each formula unit gives two hydrogen ions: H₂SO₄ → 2H⁺ + SO₄²⁻, so [H⁺] = 2 × 0.005 = 0.01 = 10⁻² mol/L. pH = −log[H⁺] = −log 10⁻² = 2. The value 2.3 counts only one proton per molecule (−log 0.005).
Which particle can act both as a Brønsted-Lowry acid and as a Brønsted-Lowry base?
Answer: A
The hydrogencarbonate ion can donate a proton (HCO₃⁻ + H₂O ⇌ CO₃²⁻ + H₃O⁺) or accept one (HCO₃⁻ + H₃O⁺ ⇌ H₂CO₃ + H₂O), so it is amphiprotic. CO₃²⁻ has no proton to give and can only accept one; NH₄⁺ can only donate one; Na⁺ does neither.
Neighbouring members of a homologous series differ in composition by:
Answer: D
Methanol CH₃OH, ethanol C₂H₅OH and propan-1-ol C₃H₇OH each differ from the next by CH₂ (14 g/mol). Members of a series keep the same functional group and general formula, so their chemical properties are similar while their physical properties change gradually.
Which compound is a functional isomer of ethanol, CH₃–CH₂–OH?
Answer: B
Ethanol and methoxymethane (dimethyl ether) are both C₂H₆O, but one is an alcohol and the other an ether, so they are functional isomers. Propan-1-ol (C₃H₈O) is the next homologue, ethanal is C₂H₄O and ethane-1,2-diol is C₂H₆O₂: none has ethanol's molecular formula.
What is the IUPAC name of CH₃–CH(CH₃)–CH₂–CH₂–OH?
Answer: A
The longest chain that carries the OH group has four carbons (butanol), numbered from the end nearer the OH, so the OH is on C1 and the methyl branch on C3: 3-methylbutan-1-ol. Numbering from the other end would give the OH the locant 4, which is not the lowest. 2-methylbutan-1-ol is a different isomer, CH₃–CH₂–CH(CH₃)–CH₂–OH, and pentan-1-ol is the unbranched one.
How many structural isomers with the formula C₄H₉OH are alcohols?
Answer: C
There are four: butan-1-ol CH₃–CH₂–CH₂–CH₂–OH, butan-2-ol CH₃–CH(OH)–CH₂–CH₃, 2-methylpropan-1-ol CH₃–CH(CH₃)–CH₂–OH and 2-methylpropan-2-ol (CH₃)₃C–OH. Ethers such as ethoxyethane also have the formula C₄H₁₀O but are not alcohols. The two mirror-image forms of butan-2-ol are stereoisomers, not structural isomers, so they count once.
What is the general formula of the open-chain alkenes with one C=C double bond?
Answer: B
Ethene C₂H₄ and propene C₃H₆ fit CₙH₂ₙ: the double bond leaves two fewer hydrogen atoms than in the alkane with the same number of carbons, CₙH₂ₙ₊₂. CₙH₂ₙ₋₆ is the general formula of benzene and its homologues.
What is the main product when hydrogen bromide adds to propene, CH₂=CH–CH₃?
Answer: B
By Markovnikov's rule the hydrogen adds to the carbon of the double bond that already holds more hydrogen atoms (the CH₂ end) and the bromine to the middle carbon: CH₂=CH–CH₃ + HBr → CH₃–CHBr–CH₃. 1,2-dibromopropane is what bromine, Br₂, gives, and propane what hydrogen gives.
Benzene reacts with bromine in the presence of iron(III) bromide. Which describes this reaction?
Answer: D
C₆H₆ + Br₂ → C₆H₅Br + HBr, with FeBr₃ as catalyst. The delocalised π system makes the benzene ring stable, so it reacts by substitution, keeping the ring, rather than by addition. One bromine molecule replaces one hydrogen atom, which leaves as HBr, not as H₂.
What volume of oxygen is needed to burn 2 dm³ of propane completely, with both gases measured at the same temperature and pressure?
Answer: C
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O. By Avogadro's law equal volumes of gases at the same temperature and pressure hold equal numbers of molecules, so the volume ratio is the mole ratio, 1 : 5. 2 dm³ × 5 = 10 dm³; 5 dm³ would burn only 1 dm³ of propane, and 6 dm³ is the volume of CO₂ formed.
How is the alcohol CH₃–CH(OH)–CH₃ classified?
Answer: A
The carbon carrying the OH group is bonded to two other carbon atoms, so propan-2-ol is a secondary alcohol. In a primary alcohol (propan-1-ol) that carbon is bonded to one carbon, in a tertiary to three. It has only one OH group, so it is not a diol.
Which compound reacts with aqueous sodium hydroxide to form a salt?
Answer: D
Phenol is a weak acid, stronger than water and the alcohols, because the negative charge of the phenoxide ion spreads into the benzene ring: C₆H₅OH + NaOH → C₆H₅ONa + H₂O. Alcohols are no stronger acids than water (methanol is about as weak, ethanol and propan-1-ol weaker): they react with sodium metal but not with sodium hydroxide solution.
Which compound dissolves freshly precipitated copper(II) hydroxide, giving a deep blue solution?
Answer: B
Polyhydric alcohols with OH groups on neighbouring carbon atoms form a soluble deep blue copper(II) complex, which is the test that tells glycerol from monohydric alcohols. Ethanol and propan-2-ol have only one OH group, and propanone has none.
Which alcohol is not oxidised by acidified potassium dichromate(VI) under ordinary conditions?
Answer: D
In 2-methylpropan-2-ol, (CH₃)₃C–OH, the carbon carrying the OH group has no hydrogen atom, so a tertiary alcohol cannot be oxidised to a carbonyl compound without breaking a C–C bond, and the dichromate stays orange. Butan-1-ol and 2-methylpropan-1-ol are primary (oxidised to aldehydes, then acids); butan-2-ol is secondary (oxidised to butanone).
Which compound gives a silver mirror when warmed with Tollens' reagent (ammoniacal silver nitrate)?
Answer: B
Aldehydes are easily oxidised, and they reduce Ag⁺ to metallic silver: CH₃CHO + 2[Ag(NH₃)₂]OH → CH₃COONH₄ + 2Ag + 3NH₃ + H₂O. Ketones such as propanone and butan-2-one, and alcohols such as ethanol, give no silver mirror.
Propanone, CH₃–CO–CH₃, is reduced with hydrogen over a nickel catalyst. What is the product?
Answer: C
CH₃–CO–CH₃ + H₂ → CH₃–CH(OH)–CH₃. Reducing a ketone gives a secondary alcohol with the OH on the carbon that carried the C=O, here C2. Propan-1-ol is what propanal, the aldehyde, gives on reduction.
What is the IUPAC name of CH₃–CH₂–CO–CH(CH₃)–CH₃?
Answer: A
The main chain has five carbons including the C=O carbon, so the compound is a pentanone, and the C=O is on C3 from either end. The tie is broken by the branch: numbering from the end nearer the methyl group puts it on C2, not C4, so the name is 2-methylpentan-3-one. 3-methylpentan-2-one and hexan-3-one are different isomers of C₆H₁₂O.
Ethanoic acid is heated with ethanol and a little concentrated sulfuric acid. What are the products?
Answer: A
This is esterification: CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O. The reaction is reversible, and the sulfuric acid acts as the catalyst. Ethyl methanoate would need methanoic acid, and ethoxyethane is the ether formed from ethanol alone.
What is the IUPAC name of CH₃–CH(CH₃)–COOH?
Answer: D
The carbon of the COOH group is always C1, and the longest chain containing it has three carbons, so the parent is propanoic acid with a methyl group on C2: 2-methylpropanoic acid. Butanoic acid, CH₃–CH₂–CH₂–COOH, is its unbranched isomer; 2-methylbutanoic acid has five carbons, and 3-methylpropanoic acid would simply be butanoic acid misnamed.
To which class does dimethylamine, (CH₃)₂NH, belong?
Answer: B
Amines are classified by the number of carbon groups bonded to the nitrogen atom: two methyl groups make (CH₃)₂NH a secondary amine. Methylamine, CH₃NH₂, is primary and trimethylamine, (CH₃)₃N, tertiary. Amides contain a nitrogen atom bonded to a carbonyl group (–CO–NH₂ in the simplest), which this compound lacks.
Which of these is the weakest base in water?
Answer: C
A base accepts a proton with the lone pair on nitrogen. In phenylamine the lone pair is partly delocalised into the benzene ring, so it is less available and phenylamine is a weaker base than ammonia. Methyl groups push electron density towards nitrogen, which makes methylamine and dimethylamine stronger bases than ammonia.
Which group is the peptide bond that links two amino acids?
Answer: C
The carboxyl group of one amino acid condenses with the amino group of the next, losing water: H₂N–CH₂–COOH + H₂N–CH(CH₃)–COOH → H₂N–CH₂–CO–NH–CH(CH₃)–COOH + H₂O (glycylalanine). –CO–O– is an ester linkage, –O– an ether linkage and –NH₂ the free amino group.
Which of these α-amino acids has no chiral (asymmetric) carbon atom?
Answer: D
In glycine, H₂N–CH₂–COOH, the α-carbon carries two hydrogen atoms, so it does not hold four different groups and glycine is not optically active. In alanine (side chain CH₃), serine (CH₂OH) and cysteine (CH₂SH) the α-carbon carries H, NH₂, COOH and the side chain, four different groups.
Which statement about glucose and fructose is correct?
Answer: C
Both are hexoses with the formula C₆H₁₂O₆. In the open-chain form glucose has a CHO group on C1 (an aldohexose) and fructose a C=O group on C2 (a ketohexose), so they are isomers. They are monosaccharides; sucrose is the disaccharide made from one of each, and C₅H₁₀O₅ is a pentose such as ribose.
Which statement about starch and cellulose is correct?
Answer: A
Both are polysaccharides, (C₆H₁₀O₅)ₙ, and both give only glucose on complete hydrolysis. Starch is made of α-glucose units, joined α-1,4 (and α-1,6 at the branches of amylopectin); cellulose of β-glucose units joined β-1,4 into straight chains that hydrogen-bond into fibres. It is starch, not cellulose, that turns iodine solution blue-black.
Answer on paper, then mark your work against the model answer.
Fill in the four missing words or terms (1 point each).
A substance whose solution or melt conducts an electric current because it contains free ions is called an (1) ______. Sodium chloride and hydrochloric acid are almost completely broken into ions in water, whereas in a solution of ethanoic acid only a small share of the molecules ionise; that share, the number of dissociated molecules divided by the total number dissolved, is called the (2) ______ and is written α. A solution that holds as much solute as can dissolve at a given temperature, and stays in equilibrium with undissolved solid, is called a (3) ______ solution. When such a solution of copper(II) chloride is electrolysed, the positively charged Cu²⁺ ions move towards the cathode and are therefore called (4) ______.
(1) electrolyte. (2) degree of dissociation. (3) saturated. (4) cations.
Also accepted: (2) degree of ionisation, or degree of electrolytic dissociation. Nothing else for (1), (3) or (4): 'strong electrolyte' is wrong for (1), because the sentence defines every electrolyte, weak ones included; 'concentrated' is wrong for (3), since a concentrated solution need not be saturated; 'anions' for (4) is the wrong sign, as anions move to the anode.
Fill in the four missing words, terms or expressions (1 point each).
For the elementary (one-step) reaction A + 2B → C, the law of mass action gives the rate equation v = (1) ______. If the concentration of B is doubled while that of A is kept the same, the reaction becomes (2) ______ times faster. Colliding particles react only if they have at least a minimum energy, called the (3) ______ energy, which a catalyst lowers by opening a different reaction pathway. A catalyst that is in the same phase as the reactants, such as the H⁺ ions that speed up the hydrolysis of an ester in aqueous solution, is called a (4) ______ catalyst.
(1) k[A][B]², the rate constant times the concentration of A times the square of the concentration of B. (2) four. (3) activation. (4) homogeneous.
Also accepted: (1) k·c(A)·c²(B), or the same with any clear notation for concentration; the square on B is required, because two molecules of B take part in the single step. (2) 4: doubling [B] multiplies the rate by 2² = 4. (4) Not 'heterogeneous', which is a catalyst in a different phase, such as solid nickel in the hydrogenation of an alkene.
Match each substance in column 1 with the type of bonding it contains in column 2. Each letter is used once (1 point each).
Column 1 1. Copper, Cu 2. Ammonium chloride, NH₄Cl 3. Chlorine, Cl₂ 4. Sodium fluoride, NaF 5. Hydrogen chloride, HCl
Column 2 a. Ionic bond only b. Non-polar covalent bond c. Metallic bond d. Polar covalent bond e. Ionic bond together with polar covalent and donor-acceptor bonds
1 c; 2 e; 3 b; 4 a; 5 d.
Why: copper's atoms are held by a sea of delocalised electrons (metallic). NH₄Cl is ionic between NH₄⁺ and Cl⁻, and inside NH₄⁺ the N–H bonds are polar covalent, one of them formed by donor-acceptor from nitrogen's lone pair to H⁺. Cl₂ shares a pair equally between identical atoms (non-polar covalent). NaF joins the most electronegative element to a group 1 metal (electronegativity difference about 3.1), so it is ionic. In HCl chlorine pulls the shared pair towards itself (polar covalent).
Match each reaction in column 1 with its name or the observation it gives in column 2. Each letter is used once (1 point each).
Column 1 1. CH₄ + Cl₂ → CH₃Cl + HCl (ultraviolet light) 2. CH₂=CH–CH₃ + H₂ → CH₃–CH₂–CH₃ (nickel catalyst, heat) 3. CH₂=CH₂ + Br₂ → CH₂Br–CH₂Br (bromine in an inert solvent, room temperature) 4. C₆H₆ + HNO₃ → C₆H₅NO₂ + H₂O (concentrated H₂SO₄, about 50 °C) 5. CH₂=CH₂ + H₂O → CH₃–CH₂–OH (H₃PO₄ catalyst, heat and pressure)
Column 2 a. Hydration b. Nitration, an electrophilic substitution on the benzene ring c. Substitution by a halogen in light (chlorination of an alkane) d. Addition of bromine: the orange-brown colour disappears, the test for a C=C double bond e. Hydrogenation
1 c; 2 e; 3 d; 4 b; 5 a.
Why: in 1 a chlorine atom replaces a hydrogen atom of the alkane, which needs light to start, and HCl is the by-product. In 2 hydrogen adds across the double bond, turning propene into propane. In 3 bromine adds across the double bond, so its colour disappears; an alkane would not decolourise bromine in the dark. In 4 a nitro group, –NO₂, replaces a hydrogen atom of the ring and the ring is kept. In 5 water adds across the double bond, which is how ethanol is made industrially.
Write T if the statement is true and F if it is false (1 point each).
1. Adding a catalyst to a reaction at equilibrium increases the value of its equilibrium constant. 2. For an exothermic reaction, raising the temperature decreases the value of the equilibrium constant.
1 F; 2 T.
Why: 1, a catalyst speeds up the forward and the reverse reaction equally, so equilibrium is reached sooner but its position and its constant are unchanged. 2, raising the temperature favours the endothermic direction, which for an exothermic reaction is the reverse one, so less product is present at the new equilibrium and the constant falls. Temperature is the only condition that changes the value of the equilibrium constant.
Define oxidation and reduction, in terms of electrons and of oxidation numbers. Explain what an oxidising agent and a reducing agent are, and illustrate your answer with one example and its equation.
Model answer: Oxidation is the loss of electrons by a particle, so its oxidation number increases; reduction is the gain of electrons, so its oxidation number decreases. The two always happen together, because electrons lost by one particle are gained by another, and the number lost equals the number gained. The oxidising agent is the substance that accepts the electrons and is itself reduced; the reducing agent gives the electrons and is itself oxidised. For example, when zinc is placed in copper(II) sulfate solution, Zn + CuSO₄ → ZnSO₄ + Cu. Zinc goes from 0 to +2 (Zn → Zn²⁺ + 2e⁻) and is oxidised, so zinc is the reducing agent; copper goes from +2 to 0 (Cu²⁺ + 2e⁻ → Cu) and is reduced, so Cu²⁺ is the oxidising agent. The sulfate ion takes no part and keeps its oxidation numbers.
Marking: 1 point for oxidation as loss of electrons with an increase in oxidation number; 1 for reduction as gain of electrons with a decrease; 1 for the oxidising and the reducing agent, each named by what it does and what happens to it; 1 for a correct, balanced redox equation; 1 for the oxidation numbers or half-equations in the example, with the oxidising and reducing agents identified. Any correct redox example earns the last two points (for instance 2Na + Cl₂ → 2NaCl).
Define an α-amino acid. Explain why amino acids are amphoteric, and give an example with equations for its reaction with an acid and with a base.
Model answer: An α-amino acid is a carboxylic acid that carries an amino group, –NH₂, on the α-carbon, the carbon next to the carboxyl group; its general formula is H₂N–CH(R)–COOH, where R is the side chain. Amino acids are amphoteric because each molecule has both an acidic group, –COOH, which can donate a proton, and a basic group, –NH₂, which can accept one. In the solid and in neutral solution the proton passes from one group to the other, giving the dipolar ion (zwitterion) H₃N⁺–CH(R)–COO⁻, which explains why amino acids are crystalline solids with high melting points that dissolve in water. For example, glycine reacts with hydrochloric acid at its amino group: H₂N–CH₂–COOH + HCl → [H₃N⁺–CH₂–COOH]Cl⁻; and with sodium hydroxide at its carboxyl group: H₂N–CH₂–COOH + NaOH → H₂N–CH₂–COONa + H₂O.
Marking: 1 point for the definition with the amino group on the α-carbon (a correct general formula earns it); 1 for naming both groups and the role of each in amphoterism; 1 for the zwitterion; 1 for a correct equation with an acid; 1 for a correct equation with a base. Any of the five syllabus amino acids (glycine, alanine, serine, phenylalanine, cysteine) may serve as the example.
About this paper
The Medical University of Varna publishes no past papers. Its Entry Tests page links free PDFs instead: one sample test in Biology and one in Chemistry, each with a filled-in answer sheet that works as the key, and the official topic list for each subject. Each sample has all five sections of the real paper, 35 multiple-choice questions in Section A and 8 more in Sections B to E. For more practice the university's bookshop sells A Handbook of Biology and Chemistry Test Items (fourth revised edition; eBook €10, print €18) and a Chemistry Textbook for the entrance tests (eBook €10). IvyTest's free practice paper on this page has both tests in all five sections, every answer explained.
This paper keeps the five sections of MU-Varna's tests and the shape of its 2018 sample papers: 35 four-option questions (Section A), two passages with four gaps each (B), two matching sets of five (C), two true-or-false statements (D) and two definitions with examples (E). MU-Varna publishes no points per section, so the points on Sections B to E are IvyTest's marking guide: one per gap, pair or statement, five per definition. Every question is new, and every one has its answer explained.
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