Charles University Third Faculty (LF3) Entrance Exam Sample Test 1, with Answers

Answer: D

Oxidative phosphorylation uses the NADH and FADH2 made in the earlier stages to drive ATP synthase and yields roughly 26 to 28 of the 30 to 32 ATP made per glucose. Glycolysis gives a net 2 ATP and the citric acid cycle 2 (as ATP or GTP) per glucose. The link reaction makes no ATP directly; its value is the NADH it passes on.

Answer: B

Bacteria make their proteins on 70S ribosomes, smaller than the 80S ribosomes in the cytoplasm of eukaryotic cells. As prokaryotes they have no nuclear envelope (their DNA lies in a nucleoid) and no membrane-bound organelles such as mitochondria or endoplasmic reticulum.

Answer: C

Reverse transcriptase, carried in the virus particle, is an RNA-dependent DNA polymerase: it builds DNA on an RNA template, which is what makes HIV a retrovirus. RNA polymerase II works the other way, making RNA from DNA, and DNA polymerase I is a bacterial enzyme that copies DNA from DNA. DNA ligase only joins DNA fragments.

Answer: A

A 0.3% solution is hypotonic to the cell contents, so water moves into the cells by osmosis; red blood cells have no cell wall, so they swell and can burst (haemolysis). Shrinking (crenation) happens in a hypertonic solution, and only a solution of about 0.9% NaCl would be isotonic.

Answer: A

The pump moves 3 sodium ions out and 2 potassium ions in per ATP, both against their concentration gradients. This keeps Na+ high outside and K+ high inside the cell, and because more positive charge leaves than enters, the pump also makes a small direct contribution to the negative resting potential. The other options reverse the numbers or the directions.

Answer: B

In double-stranded DNA adenine pairs with thymine and guanine with cytosine, so A = T = 18% and together they make 36%. The remaining 64% is shared equally between G and C, giving 32% guanine. 36% is A + T, and 64% is G + C together.

Answer: A

The mRNA is complementary and antiparallel to the template strand, with uracil in place of thymine: T pairs with A, A with U, C with G and G with C, so 3'-TAC GGA-5' gives 5'-AUG CCU-3'. 5'-ATG CCT-3' is the right order of bases but contains thymine, so it is the coding DNA strand, not RNA. The other two copy the template's bases instead of pairing with them.

Answer: C

In anaphase the cohesion holding the sister chromatids together is cleaved and the spindle pulls each chromatid, now a chromosome, to an opposite pole. In metaphase the chromosomes are still paired chromatids lined up at the equator, and in telophase the separated sets are already at the poles as nuclear envelopes re-form.

Answer: A

Both meiotic divisions in oogenesis are unequal: almost all the cytoplasm goes to one cell, and the other products are small polar bodies that degenerate. One primary oocyte therefore gives one ovum. Four is the number of functional sperm cells from one primary spermatocyte, which divides equally.

Answer: C

The AaBb parent makes four gamete types, AB, Ab, aB and ab, in equal proportions, and the aabb parent makes only ab. The offspring therefore come in four equal classes, and aabb is one of them: 1/2 (aa) x 1/2 (bb) = 1/4. 1/16 is the proportion of aabb from a cross of two AaBb parents.

Answer: D

In transduction a phage accidentally packages a piece of its host's DNA and injects it into the next bacterium it infects. Conjugation is transfer through direct contact via a pilus, and transformation is the uptake of free DNA from the surroundings. Transcription is making RNA from DNA and moves no DNA between cells.

Answer: B

The mother is XHXh and the father XHY. The child is a boy with probability 1/2, and a boy inherits his single X from his mother, which carries the h allele with probability 1/2, so the probability is 1/2 x 1/2 = 1/4. 1/2 is the tempting answer, but it is the probability for a child already known to be a son, not for the next child of either sex.

Answer: B

A trait that appears in a child of two unaffected parents must be recessive, with both parents carriers. It cannot be X-linked recessive, because an affected daughter would need an affected allele from her father, and a father with that allele on his only X would himself be affected. A dominant trait, autosomal or X-linked, would show in at least one parent.

Answer: D

Each PCR cycle heats the mixture to about 95 °C to separate the DNA strands, which would denature most polymerases. Taq polymerase comes from a bacterium living in hot springs and survives these cycles, so it need not be added again each time. Like every DNA polymerase it still needs primers, and it copies DNA, not RNA; copying RNA into DNA is the job of reverse transcriptase.

Answer: C

Affected people are aa, so q² = 1/400 and q = 1/20 = 0.05; then p = 1 - 0.05 = 0.95. Carriers are 2pq = 2 x 0.95 x 0.05 = 0.095, or 9.5%. 0.25% is q² (the affected), 5% is q (the allele frequency), and 19% doubles the answer by mistake.

Answer: A

The human arm and the bat wing are built from the same set of bones (humerus, radius, ulna, carpals, metacarpals, phalanges) inherited from a common ancestor, even though they do different jobs. The other pairs are analogous: they serve a similar function but evolved independently from different structures.

Answer: C

The neural tube, which becomes the brain and spinal cord, forms from a fold of the ectoderm, and the neural crest cells that form much of the peripheral nervous system come from it too. The endoderm lines the gut and its glands and the mesoderm gives muscle, bone and blood. The trophoblast is not a germ layer but the outer layer of the blastocyst that helps form the placenta.

Answer: D

Calcium binds to troponin, which changes shape and pulls tropomyosin away from the myosin-binding sites on actin, so cross-bridges can form. Tropomyosin is the protein that is moved, not the one that binds calcium, and actin and myosin are the filaments whose interaction calcium switches on.

Answer: A

The pulmonary veins return oxygenated blood from the lungs to the left atrium. The pulmonary trunk 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 coronary sinus drains deoxygenated blood from the heart wall.

Answer: C

More carbon dioxide lowers the pH and also binds haemoglobin directly, reducing its affinity for oxygen (the Bohr effect), which is how actively respiring tissues receive more oxygen. A fall in temperature, a rise in pH and a fall in 2,3-BPG all do the opposite: they raise the affinity and shift the curve to the left.

Answer: C

Bile salts break large fat droplets into small ones, so pancreatic lipase has far more surface to work on. Bile is made by the liver and only stored and concentrated in the gallbladder, it is released into the duodenum, not the stomach, and it contains no digestive enzymes of its own.

Answer: C

Glucose is filtered freely and then taken back by sodium-glucose cotransporters in the proximal tubule; it appears in urine only when blood glucose is high enough to exceed their capacity, as in uncontrolled diabetes. Urea is only partly reabsorbed and creatinine hardly at all, so both are normal parts of urine. Albumin is too large to be freely filtered in the first place.

Answer: B

Rising oestrogen from the maturing follicle triggers a surge of LH from the anterior pituitary, and ovulation follows about a day later. FSH stimulates follicle growth earlier in the cycle, progesterone is made mainly by the corpus luteum after ovulation, and oxytocin acts in labour and milk ejection.

Answer: A

Insulin is made by the beta cells of the pancreatic islets and lowers blood glucose. The alpha cells of the same islets secrete glucagon, which raises it. Acinar cells secrete digestive enzymes into the pancreatic duct, and the adrenal cortex secretes steroid hormones such as cortisol.

Answer: B

The failing gland makes too little T4. Thyroid hormones normally hold back TSH release by negative feedback, so with little T4 the healthy pituitary secretes more TSH. Low T4 with low TSH would point to a pituitary or hypothalamic cause, and high T4 with low TSH is the pattern of an overactive thyroid.

Answer: D

When the membrane reaches threshold, voltage-gated sodium channels open and Na+ rushes in down its electrochemical gradient, driving the inside of the cell positive. Potassium leaving the cell is what repolarises the membrane afterwards, and chloride entering would make the inside more negative, not less.

Answer: D

In myopia the eye focuses light too strongly, usually because the eyeball is too long, so distant objects are focused in front of the retina. A diverging lens spreads the rays slightly before they enter the eye and moves the focus back onto the retina. Focusing behind the retina is long sight (hypermetropia), corrected with a converging lens.

Answer: D

A person with group AB carries both A and B antigens on their red blood cells, so their plasma has neither antibody, as either would attack their own cells. Group A plasma has anti-B, group B plasma has anti-A, and only group O plasma has both.

Answer: D

The protection comes from antibodies made by another body and given by injection, so it is passive (the patient's own immune system makes nothing) and artificial (given medically). It acts at once but fades within weeks. Vaccination with tetanus toxoid would be artificial active immunity, and antibodies passing from mother to baby across the placenta are natural passive.

Answer: B

Energy passes from producers to primary consumers (2,000 kJ), then to secondary consumers (200 kJ), then to tertiary consumers (20 kJ): three transfers, each keeping 10%. 2,000 kJ and 200 kJ stop one or two levels too early, and 2 kJ takes one transfer too many.

Answer: B

The atomic number gives 26 protons, and the neutrons are the mass number minus the atomic number, 56 − 26 = 30. A 2+ charge means two electrons have been lost, so 26 − 2 = 24 electrons. Twenty-eight electrons would be a 2− ion, and 56 is the total of protons and neutrons, not the neutrons alone.

Answer: C

Ionisation energy generally rises across period 2 as the nuclear charge increases, so lithium is lowest. Oxygen breaks the trend: its fourth 2p electron is paired in an orbital that already holds one, and the repulsion makes it easier to remove than an electron from nitrogen's half-filled 2p subshell. Nitrogen is therefore highest of the four.

Answer: B

The anion H₂PO₄⁻ is phosphate (PO₄³⁻) carrying two hydrogens, so it is dihydrogen phosphate, and two of them balance one Ca²⁺. Calcium hydrogen phosphate is CaHPO₄ and calcium phosphate is Ca₃(PO₄)₂. A phosphite contains phosphorus in the +3 state, whereas here phosphorus is +5.

Answer: D

Ammonia is trigonal pyramidal because of the lone pair on nitrogen, so its three N–H bond dipoles do not cancel and the molecule has a net dipole. CO₂ is linear, BF₃ is trigonal planar and CCl₄ is tetrahedral: each has polar bonds, but they are arranged symmetrically and their dipoles cancel.

Answer: A

ΔH = ΣΔHf(products) − ΣΔHf(reactants) = [−394 + 2 × (−286)] − (−75) = −966 + 75 = −891 kJ/mol; O₂ is an element in its standard state, so its value is zero. −605 kJ/mol comes from forgetting the 2 in front of water, and −1041 kJ/mol from adding the methane term instead of subtracting it. Combustion releases heat, so the sign must be negative.

Answer: C

In an exothermic reaction the products have less enthalpy than the reactants, so ΔH is negative and the difference is released as heat to the surroundings. A positive ΔH and surroundings that cool describe an endothermic reaction. If the products had more enthalpy, the reaction would have absorbed energy, not released it.

Answer: D

0.9% (w/v) is 0.9 g per 100 mL, or 9 g/L, and 9 ÷ 58.5 ≈ 0.154 mol/L. Each NaCl gives two particles, Na⁺ and Cl⁻, so the osmolarity is about 2 × 154 = 308 mOsm/L, close to that of plasma. 154 mOsm/L is the molarity with dissociation ignored, and 462 mOsm/L would be for a salt giving three ions, such as CaCl₂.

Answer: A

Distilled water is hypotonic to the cytoplasm, so water moves into the cell down its water potential gradient, and with no cell wall the red cell swells and can burst. Shrinking happens in a hypertonic solution such as concentrated salt. Osmosis is the movement of water, not of solutes, and the membrane is permeable to water, partly through aquaporins.

Answer: B

Mass = amount × molar mass = 0.25 mol × 180 g/mol = 45 g. 720 g comes from dividing the molar mass by the amount instead of multiplying. 18 g and 72 g correspond to 0.1 mol and 0.4 mol.

Answer: C

The amount of HCl does not change on dilution: 0.020 L × 0.50 mol/L = 0.010 mol. Spread over 0.250 L, that is 0.010 ÷ 0.250 = 0.040 mol/L, which also follows from c₁V₁ = c₂V₂. 0.40 mol/L is a slip of a factor of ten, and 0.10 mol/L would be the result of diluting to 100 mL rather than 250 mL.

Answer: D

M(C₃H₈) = 3 × 12 + 8 × 1 = 44 g/mol, so 11 g is 0.25 mol. Each mole of propane needs 5 mol O₂, so 0.25 × 5 = 1.25 mol, and 1.25 × 22.4 = 28 L. 5.6 L ignores the 1 : 5 ratio, and 112 L is the oxygen for a whole mole of propane.

Answer: C

There are 4 mol of gas on the left and 2 on the right, so raising the pressure shifts the equilibrium towards ammonia, the side with fewer gas molecules. The forward reaction is exothermic, so raising the temperature shifts it back towards the reactants. A catalyst only speeds up reaching equilibrium without moving it, and removing nitrogen shifts the equilibrium to the left.

Answer: D

The reaction is second order in A, so doubling [A] multiplies the rate by 2² = 4, and first order in B, so halving [B] multiplies it by 0.5. Overall 4 × 0.5 = 2, so the rate doubles. Quadrupling ignores the change in B, and 'unchanged' treats both changes as first order.

Answer: A

HCl is a strong acid and dissociates completely, so [H₃O⁺] = 0.01 = 10⁻² mol/L and pH = −log 10⁻² = 2. A pH of 12 would be a 0.01 mol/L strong base, and 0.01 is the concentration itself, not its negative logarithm.

Answer: B

By the Henderson–Hasselbalch equation, pH = pKa + log([A⁻]/[HA]) = 4.76 + log 10 = 4.76 + 1 = 5.76. 4.76 is the pH only when acid and salt are equal, and 3.76 is what you get with the ratio inverted, as if the acid were ten times more concentrated.

Answer: D

The acid supplies 0.050 L × 0.20 = 10 mmol H⁺ and the base 0.050 L × 0.10 = 5 mmol OH⁻, so 5 mmol H⁺ remain in 100 mL: [H⁺] = 0.05 mol/L. pH = −log(5 × 10⁻²) = 2 − log 5 = 2 − 0.70 = 1.3, using log 5 = 1 − log 2. 1.0 forgets that the volume doubled on mixing, 0.7 ignores the NaOH, and 2.0 would need [H⁺] of only 0.01 mol/L.

Answer: C

In K₂Cr₂O₇, 2(+1) + 2x + 7(−2) = 0 gives 2x = 12, so chromium is +6. Manganese in KMnO₄ is +7 (1 + x − 8 = 0), sulfur in H₂SO₃ is +4 (2 + x − 6 = 0) and chlorine in HClO₄ is +7 (1 + x − 8 = 0).

Answer: D

Manganese goes from +7 to +2, gaining 5 electrons, while each Fe²⁺ loses only 1, so the balanced equation is MnO₄⁻ + 5 Fe²⁺ + 8 H⁺ → Mn²⁺ + 5 Fe³⁺ + 4 H₂O. 0.020 mol MnO₄⁻ therefore oxidises 5 × 0.020 = 0.10 mol Fe²⁺. 0.004 mol uses the ratio upside down, and 0.16 mol is the hydrogen ion consumed, not the iron.

Answer: C

The alcohols are butan-1-ol, butan-2-ol, 2-methylpropan-1-ol and 2-methylpropan-2-ol: four. Seven is the total for C₄H₁₀O once the three ethers (ethoxyethane, 1-methoxypropane and 2-methoxypropane) are included. Butan-2-ol also exists as two enantiomers, but those are stereoisomers, not structural isomers.

Answer: A

Carbon 2 of lactic acid carries four different groups, H, OH, CH₃ and COOH, so it is a stereocentre and the acid exists as D and L forms. In propan-2-ol the central carbon has two identical CH₃ groups, in glycine the α-carbon has two hydrogens, and in 2-methylpropan-2-ol the central carbon has three CH₃ groups.

Answer: C

Cyclohexene has a C=C double bond, which adds bromine across it by electrophilic addition, so the orange colour disappears. Cyclohexane and hexane are saturated and react with bromine only by radical substitution in UV light. Benzene's delocalised ring is too stable to add bromine, and it reacts only by substitution with a catalyst such as FeBr₃.

Answer: B

H⁺ adds first to the CH₂ end of the double bond, giving the more stable secondary carbocation on carbon 2, which then takes up Br⁻ (Markovnikov's rule). 1-bromopropane would need the less stable primary carbocation and is only a minor product. 1,2-dibromopropane is the product with Br₂, and 2-bromopropene would require a substitution that keeps the double bond.

Answer: A

Propan-2-ol is a secondary alcohol, and oxidising a secondary alcohol gives a ketone, here propanone, which resists further oxidation. Propanal and then propanoic acid come from the primary alcohol propan-1-ol. Propene forms by dehydration with a hot concentrated acid, which is not an oxidation.

Answer: B

Mild oxidation removes a hydrogen from each thiol and joins the two sulfur atoms: 2 R–SH → R–S–S–R + 2 H⁺ + 2 e⁻. These disulfide bridges stabilise the tertiary structure of proteins such as insulin and keratin. A sulfonic acid needs much stronger oxidation, a thioether has one sulfur between two carbons (as in methionine), and a thioester involves a carbonyl group.

Answer: B

An ester is named alkyl from the alcohol, then -oate from the acid, so methanol and propanoic acid give methyl propanoate, CH₃CH₂COOCH₃. Propyl methanoate would come from propan-1-ol and methanoic acid. Ethyl ethanoate is an isomer with the same formula but a different structure, and methoxypropane is an ether, not an ester.

Answer: A

Ethanol is barely acidic because its alkoxide ion has no resonance stabilisation, whereas the carboxylate ion spreads its charge over two oxygens. Electron-withdrawing chlorine atoms stabilise the carboxylate further, and three of them more than one, so the acids strengthen from ethanoic to chloroethanoic to trichloroethanoic acid. The other orders either place ethanol above a carboxylic acid or reverse the effect of chlorine.

Answer: C

The methyl group pushes electron density onto nitrogen, making its lone pair more available to accept a proton, so methylamine is a stronger base than ammonia. In phenylamine the lone pair is partly delocalised into the benzene ring, making it a weaker base than ammonia. In an amide the lone pair is delocalised onto the carbonyl oxygen, so ethanamide is practically neutral.

Answer: D

Adenine and guanine are purines, with a six-membered pyrimidine ring fused to a five-membered imidazole ring. Cytosine, thymine and uracil are pyrimidines, with a single six-membered ring containing two nitrogen atoms. Each of the other pairs contains at least one pyrimidine.

Answer: B

In sucrose the glycosidic bond joins carbon 1 of glucose to carbon 2 of fructose, the anomeric carbons of both units, so neither ring can open to give a free aldehyde or ketone group, and Fehling's or Benedict's test is negative. Maltose, lactose and cellobiose are all joined through carbon 1 of one unit and carbon 4 of the other, leaving one anomeric carbon free, so they are reducing sugars.

Answer: A

At pH 1, below both pKa values, both groups are protonated: the carboxyl group is COOH and the amino group is NH₃⁺, so glycine is a cation with a net charge of +1. The zwitterion H₃N⁺–CH₂–COO⁻ dominates between the two pKa values, around the isoelectric point near pH 6, and the anion H₂N–CH₂–COO⁻ above pH 9.6. The fully uncharged form is never the main species in water.