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Catalonia PAU Biology study materials
Catalonia PAU Biology study resources

Catalonia PAU Biology practice questions

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Question 1Biological inquiry and interpretation of evidence

A researcher predicts that tomato plants given a nitrogen-enriched fertilizer will produce more fruit than plants given a standard fertilizer. In this study, the amount of fruit produced is best described as which of the following?

  1. The hypothesis
  2. A controlled variable
  3. The dependent variable Correct answer.
  4. The independent variable

Explanation

The dependent variable is the measured outcome that responds to the manipulation, which here is fruit yield. The fertilizer type is the factor the researcher deliberately changes, so it is the independent variable rather than the measured result.

Question 2Biological inquiry and interpretation of evidence

A student measures the mass of the same sample five times and obtains nearly identical values, but the balance was never zeroed and reads 2.0 g too high. How should these measurements be described?

  1. Both accurate and precise
  2. Precise but not accurate Correct answer.
  3. Accurate but not precise
  4. Neither accurate nor precise

Explanation

Precision refers to how closely repeated measurements agree with one another, while accuracy refers to how close they are to the true value. Because the readings cluster tightly but are systematically offset by the unzeroed balance, they are precise but inaccurate.

Question 3Biological inquiry and interpretation of evidence

Which of the following best distinguishes a scientific hypothesis from a scientific theory?

  1. A hypothesis is a testable proposed explanation, while a theory is a well-supported explanatory framework built from extensive evidence Correct answer.
  2. A hypothesis is always correct once tested, while a theory remains an untested guess
  3. A hypothesis is a broad framework, while a theory is a single testable prediction
  4. A hypothesis applies only to biology, while a theory applies only to physics

Explanation

In science, a hypothesis is a testable proposed explanation for a specific observation, whereas a theory is a broad, well-substantiated explanation supported by many lines of evidence. The common usage of theory as a mere guess is incorrect in scientific contexts.

Question 4Biological inquiry and interpretation of evidence

A scientist repeats an experiment several times and obtains consistent results. Which conclusion is most justified?

  1. The findings are reliable and reproducible under the tested conditions Correct answer.
  2. The results now apply to all organisms everywhere
  3. The hypothesis has been proven true beyond any doubt
  4. The experiment has no remaining sources of error

Explanation

Reproducible results indicate reliability, meaning the procedure yields consistent outcomes under the tested conditions. Science does not establish absolute proof, and replication does not eliminate error or guarantee universal applicability.

Question 5Biological inquiry and interpretation of evidence

A graph plots plant height on the y-axis against days of growth on the x-axis. Which statement correctly identifies the placement of these variables?

  1. Days of growth is the independent variable on the x-axis, and height is the dependent variable on the y-axis Correct answer.
  2. Both variables are dependent and should share the x-axis
  3. Days of growth is the dependent variable because it is measured over time
  4. Height is the independent variable on the y-axis, and days of growth is the dependent variable on the x-axis

Explanation

By convention, the independent variable is placed on the x-axis and the dependent variable on the y-axis. Height responds to elapsed time, so time is independent and height is dependent.

Question 6Biological inquiry and interpretation of evidence

A student notices that mold grows on bread left in a warm, humid place but not on bread stored in a dry, cool place. Which of the following is the most appropriate next step in the scientific process?

  1. Formulate a testable hypothesis and design a controlled experiment to test it Correct answer.
  2. Conclude immediately that warmth and moisture cause mold growth
  3. Publish the observation as a confirmed scientific law
  4. Discard the observation because it was not made in a laboratory

Explanation

An observation generates questions that should be framed as a testable hypothesis and examined through a controlled experiment. Reaching a firm causal conclusion from a single uncontrolled observation is premature.

Question 7Biochemistry

Which property of water is most directly responsible for its ability to moderate temperature in living organisms and large bodies of water?

  1. Its low specific heat
  2. Its inability to form hydrogen bonds
  3. Its nonpolar covalent bonds
  4. Its high specific heat Correct answer.

Explanation

Water has a high specific heat because hydrogen bonds must absorb or release substantial energy before water molecules change temperature. This allows water and water-rich organisms to resist large temperature swings, helping maintain homeostasis.

Question 8Cell structure, membranes and division

Which structure is found in a typical plant cell but NOT in a typical animal cell?

  1. Mitochondrion
  2. Nucleus
  3. Ribosome
  4. Cell wall made of cellulose Correct answer.

Explanation

Plant cells have a rigid cell wall composed primarily of cellulose that surrounds the plasma membrane, providing structural support. Animal cells lack a cell wall, relying on the cytoskeleton and extracellular matrix instead.

Question 9Genetics, inheritance and molecular information

Meiosis differs from mitosis in that meiosis produces

  1. Two genetically identical diploid cells
  2. One diploid cell with double the chromosomes
  3. Two haploid cells identical to the parent
  4. Four genetically varied haploid cells Correct answer.

Explanation

Meiosis consists of two divisions that produce four haploid gametes, each with half the chromosome number of the parent. Crossing over and independent assortment generate genetic variation, distinguishing meiosis from the identical-diploid products of mitosis.

Question 10Metabolism

An enzyme lowers the activation energy of a reaction. Which statement best describes how this affects the reaction?

  1. It increases the reaction rate without being consumed in the reaction Correct answer.
  2. It changes the overall free-energy difference between reactants and products
  3. It supplies the energy needed to make an endergonic reaction exergonic
  4. It permanently binds to the product to stop the reaction

Explanation

Enzymes are biological catalysts that lower activation energy, speeding up reactions without being consumed. They do not change the net free-energy difference (delta G) between reactants and products; they only make the reaction reach equilibrium faster.

Question 11Microbiology Immunology

Gram-positive bacteria differ structurally from Gram-negative bacteria in their cell wall composition. Which characteristic feature distinguishes Gram-positive bacterial cell walls upon Gram staining?

  1. Thick multi-layered peptidoglycan meshwork retaining crystal violet-iodine complex (purple) Correct answer.
  2. Thin single layer of peptidoglycan enclosed by an outer membrane containing lipopolysaccharide (LPS)
  3. Absence of peptidoglycan, composed entirely of chitin
  4. Lipid envelope derived from host plasma membrane

Explanation

Gram-positive bacteria possess a thick, multi-layered peptidoglycan (murein) cell wall rich in teichoic and lipoteichoic acids. This dense structure dehydrates during alcohol decolorization, trapping the crystal violet-iodine complex and staining purple. Gram-negative cell walls have thin peptidoglycan and an outer LPS membrane (staining pink with safranin).

Question 12Biotechnology

The polymerase chain reaction, or PCR, is mainly used to do what?

  1. Amplify a selected DNA sequence Correct answer.
  2. Destroy all proteins in a sample
  3. Make plant roots bend downward
  4. Measure animal breathing rate directly

Explanation

PCR uses primers, DNA polymerase, nucleotides, and temperature cycling to make many copies of a target DNA region. It is widely used in testing and research.

Question 13Biochemistry

Which physicochemical property of water allows humans and other mammals to release excess body heat effectively during sweating?

  1. High heat of vaporization resulting from extensive intermolecular hydrogen bonding. Correct answer.
  2. Low boiling point caused by nonpolar covalent bonds within the water molecule.
  3. High density in the solid phase compared to the liquid phase.
  4. Low surface tension facilitating rapid hydrophobic evaporation.

Explanation

Water has a high heat of vaporization (approximately 540 cal/g at 100 °C) because a significant amount of thermal energy is required to disrupt the extensive network of intermolecular hydrogen bonds before liquid water transitions into vapor. When sweat evaporates from the skin surface, it absorbs heat from the body, providing a highly efficient thermoregulatory mechanism.

Question 14Biochemistry

What happens to human red blood cells (erythrocytes) when placed in a hypertonic sodium chloride solution?

  1. Water moves out of the cells by osmosis, causing them to shrink and undergo crenation. Correct answer.
  2. Water enters the cells by osmosis until they swell and burst in a process called hemolysis.
  3. Sodium ions rapidly diffuse into the cells by simple diffusion, causing cell swelling.
  4. The cells maintain a constant volume because the plasma membrane is impermeable to water.

Explanation

In a hypertonic solution, the solute concentration outside the red blood cell is higher than inside the cytoplasm. Water leaves the cell via osmosis down its water potential gradient, reducing cellular volume and causing the plasma membrane to wrinkle, a process known as crenation.

Question 15Biochemistry

Which structural feature distinguishes D-glucose from D-fructose?

  1. D-glucose is an aldohexose with a carbonyl group at C1, whereas D-fructose is a ketohexose with a carbonyl group at C2. Correct answer.
  2. D-glucose is a pentose sugar found in RNA, whereas D-fructose is a hexose sugar found in plant cell walls.
  3. D-glucose contains beta-1,4-glycosidic linkages, whereas D-fructose contains alpha-1,4-glycosidic linkages.
  4. D-glucose is a non-reducing sugar, whereas D-fructose cannot undergo ring closure in aqueous solution.

Explanation

D-glucose and D-fructose are structural isomers with the molecular formula C6H12O6. D-glucose is an aldohexose featuring an aldehyde group at carbon 1 (C1), while D-fructose is a ketohexose featuring a ketone functional group at carbon 2 (C2).

Question 16Biochemistry

Which chemical reaction forms a triglyceride (triacylglycerol) from glycerol and fatty acid molecules?

  1. Esterification of three fatty acid carboxyl groups with three glycerol hydroxyl groups, releasing three water molecules. Correct answer.
  2. Hydrolysis of peptide bonds between amino acid residues in the presence of acidic catalysts.
  3. Condensation of three phosphate groups with the nitrogenous base adenosine to form high-energy bonds.
  4. Hydrogenation of unsaturated carbon-carbon double bonds to produce branched sphingolipids.

Explanation

Triglycerides are saponifiable lipids synthesized when the three hydroxyl (-OH) groups of a glycerol molecule undergo esterification with the carboxyl (-COOH) groups of three fatty acids. Each ester bond formation releases one molecule of water, yielding three water molecules in total.

Question 17Biochemistry

Which functional groups react during the formation of a peptide bond between two amino acids?

  1. The alpha-carboxyl group of one amino acid and the alpha-amino group of another amino acid. Correct answer.
  2. The variable R-group side chain of a hydrophobic amino acid and a phosphate group.
  3. The hydroxyl group of a serine residue and the methyl group of an alanine residue.
  4. The aldehyde group of a reduced sugar and the imidazole ring of a histidine residue.

Explanation

A peptide bond is a covalent amide linkage formed through a dehydration condensation reaction between the alpha-carboxyl group (-COOH) of one amino acid and the alpha-amino group (-NH2) of an adjacent amino acid, eliminating a molecule of water.

Question 18Biochemistry

According to Chargaff's rules and the Watson-Crick DNA model, how do nitrogenous bases pair across antiparallel strands?

  1. Adenine pairs with thymine via two hydrogen bonds; guanine pairs with cytosine via three hydrogen bonds. Correct answer.
  2. Adenine pairs with uracil via three hydrogen bonds; guanine pairs with thymine via two hydrogen bonds.
  3. Cytosine pairs with thymine via covalent phosphodiester linkages between sugar rings.
  4. Guanine pairs with adenine via ionic bonds between purine ring nitrogen atoms.

Explanation

In double-stranded DNA, complementary base pairing occurs between purines and pyrimidines: Adenine (A) forms two hydrogen bonds with Thymine (T), while Guanine (G) forms three hydrogen bonds with Cytosine (C). This complementary geometry stabilizes the uniform diameter of the double helix.

Question 19Biochemistry

How does the carbonic acid-bicarbonate buffer system maintain human arterial blood plasma pH near 7.40 when lactic acid enters the bloodstream?

  1. Bicarbonate ions (HCO3-) accept excess protons (H+) to form carbonic acid (H2CO3), which dissociates into water and carbon dioxide exhaled by the lungs. Correct answer.
  2. Carbonic acid (H2CO3) donates hydroxide ions (OH-) directly to neutralize lactic acid, raising plasma osmolarity.
  3. Lactic acid reacts with dissolved nitrogen gas to form urea, preventing changes in hydrogen ion concentration.
  4. Hemoglobin precipitates out of red blood cells to absorb excess H+ ions permanently in the spleen.

Explanation

When lactic acid releases H+ into plasma, bicarbonate ions (HCO3-) combine with excess H+ to generate carbonic acid (H2CO3). Carbonic anhydrase catalyzes the reversible breakdown of H2CO3 into H2O and CO2, and the CO2 is eliminated by pulmonary respiration, maintaining physiological blood pH within the tight 7.35–7.45 range.

Question 20Biochemistry

Enzymatic hydrolysis of one molecule of sucrose yields which combination of monosaccharide products?

  1. One molecule of D-glucose and one molecule of D-fructose. Correct answer.
  2. Two molecules of D-glucose.
  3. One molecule of D-galactose and one molecule of D-glucose.
  4. Two molecules of D-galactose.

Explanation

Sucrose is a non-reducing disaccharide composed of an alpha-D-glucopyranose unit and a beta-D-fructofuranose unit linked by an alpha-1,beta-2-glycosidic bond. Hydrolysis of sucrose by the enzyme sucrase (invertase) breaks this bond, producing one molecule of D-glucose and one molecule of D-fructose.

Question 21Biochemistry

Why is plant cellulose indigestible to human digestive enzymes, whereas plant starch (amylose) is readily digested?

  1. Cellulose consists of glucose units joined by beta-1,4-glycosidic bonds forming straight chains, whereas human alpha-amylase only hydrolyzes alpha-1,4-glycosidic bonds. Correct answer.
  2. Cellulose is composed exclusively of D-galactose monomers, whereas starch is composed of D-fructose monomers.
  3. Cellulose contains high amounts of sulfur cross-links that covalently block enzymatic binding sites.
  4. Starch is a lipid-soluble polymer, whereas cellulose is an insolubilized lipid-carbohydrate complex.

Explanation

Starch (amylose) consists of D-glucose monomers connected by alpha-1,4-glycosidic bonds, which are easily cleaved by human digestive enzymes like alpha-amylase. In contrast, cellulose consists of glucose monomers joined by beta-1,4-glycosidic bonds. Humans lack the enzyme cellulase needed to hydrolyze beta-1,4 linkages, making cellulose insoluble dietary fiber.

Question 22Biochemistry

What structural property allows glycerophospholipids to spontaneously arrange into lipid bilayers in aqueous cellular environments?

  1. Their amphipathic nature, featuring a hydrophilic polar head group and two hydrophobic fatty acid tails. Correct answer.
  2. Their completely nonpolar, hydrophobic structure that repels water molecules from both surfaces.
  3. The presence of covalent disulfide bridges linking adjacent phosphate groups across the membrane.
  4. Their high solubility in water driven by ionic bonds between hydrocarbon chains.

Explanation

Glycerophospholipids are amphipathic molecules containing a hydrophilic polar head group (phosphate plus an alcohol like choline) and two hydrophobic nonpolar fatty acyl tails. In water, they spontaneously assemble into bilayers with their polar heads facing the aqueous intracellular and extracellular fluids and their hydrophobic tails sequestered inside.

Question 23Biochemistry

What dual role does cholesterol play in regulating the fluidity of eukaryotic plasma membranes?

  1. It restricts phospholipid fatty acid movement at high temperatures and prevents tight packing of hydrocarbon chains at low temperatures. Correct answer.
  2. It dissolves membrane proteins at high temperatures and increases lipid saturation at low temperatures.
  3. It synthesizes ATP within the bilayer at low temperatures and hydrolyzes lipids at high temperatures.
  4. It forms covalent peptide bonds between phospholipids to solidify the membrane regardless of temperature.

Explanation

Cholesterol acts as a membrane fluidity buffer. At warm temperatures (e.g., 37 °C), its rigid steroid ring structure restrains the movement of phospholipid fatty acid tails, reducing excessive membrane fluidity. At cool temperatures, it interferes with the dense packing of hydrocarbon chains, preventing membrane gelation/solidification.

Question 24Biochemistry

Which chemical bonds stabilize the secondary structure motifs (alpha-helices and beta-pleated sheets) of proteins?

  1. Hydrogen bonds formed between polypeptide backbone carbonyl oxygens and amide hydrogens. Correct answer.
  2. Covalent disulfide bonds formed between cysteine side-chain thiol groups.
  3. Ionic bonds formed between basic lysine and acidic aspartate side chains.
  4. Phosphodiester bonds formed between adjacent ribose sugar rings.

Explanation

Secondary structure elements like alpha-helices and beta-sheets are stabilized by repetitive hydrogen bonding along the peptide backbone. Specifically, hydrogen bonds form between the carbonyl oxygen (C=O) of one amino acid residue and the amide hydrogen (N-H) of another residue.

Question 25Biochemistry

What structural change occurs during thermal or pH-induced denaturation of a functional globular protein?

  1. Disruption of non-covalent interactions (hydrogen bonds, ionic bonds, hydrophobic interactions) leading to loss of tertiary structure while leaving primary structure intact. Correct answer.
  2. Cleavage of covalent peptide bonds along the backbone, breaking the protein into individual free amino acids.
  3. Conversion of all amino acid L-stereoisomers into D-stereoisomers without altering overall shape.
  4. Condensation of adjacent polypeptide chains into double-stranded nucleic acid helices.

Explanation

Protein denaturation involves the unfolding of the native three-dimensional conformation due to the disruption of weak non-covalent bonds (hydrogen bonds, ionic interactions, hydrophobic interactions) and disulfide bridges. Because covalent peptide bonds remain unbroken, the primary amino acid sequence remains intact.

Question 26Biochemistry

How do biological enzymes increase the rate of chemical reactions in metabolic pathways?

  1. By lowering the activation energy barrier required to reach the transition state without changing overall reaction Gibbs free energy (delta G). Correct answer.
  2. By increasing the overall free energy change (delta G) to make endergonic reactions spontaneously exergonic.
  3. By raising the temperature of the microenvironment inside the active site to 100 °C.
  4. By consuming substrate molecules permanently to shift the chemical equilibrium constant (Keq) toward reactants.

Explanation

Enzymes function as biological catalysts by binding substrates at their active site and stabilizing the high-energy transition state, thereby lowering the activation energy (Ea). Enzymes accelerate reaction rates without altering the initial or final energy states (delta G) or changing the equilibrium position (Keq).

Question 27Biochemistry

In Michaelis-Menten enzyme kinetics, how does a competitive inhibitor affect the apparent Km and Vmax values of an enzymatic reaction?

  1. It increases the apparent Km (decreases affinity) while leaving Vmax unchanged at high substrate concentrations. Correct answer.
  2. It decreases Vmax while leaving the apparent Km value completely unchanged.
  3. It decreases both Km and Vmax proportionally by binding exclusively to the enzyme-substrate complex.
  4. It increases Vmax and decreases Km by irreversibly modifying the active site.

Explanation

A competitive inhibitor structurally resembles the substrate and competes directly for binding to the free enzyme's active site. This increases the apparent Michaelis constant (Km), indicating reduced binding affinity. However, because increasing substrate concentration can outcompete the inhibitor, the maximal velocity (Vmax) remains unchanged.

Question 28Biochemistry

Which specific region of a transfer RNA (tRNA) molecule attaches to its corresponding specific amino acid?

  1. The 3' hydroxyl terminal CCA sequence (acceptor stem). Correct answer.
  2. The 5' phosphate cap containing 7-methylguanosine.
  3. The central anticodon loop containing methylated uracil residues.
  4. The T-pseudouridine-C loop that binds directly to DNA promoters.

Explanation

Transfer RNA (tRNA) features a characteristic cloverleaf secondary structure folding into an L-shaped tertiary structure. The 3' terminal acceptor stem ends in a single-stranded 5'-CCA-3' sequence, where aminoacyl-tRNA synthetase attaches the appropriate amino acid via an ester bond to the 3'-OH group.

Question 29Biochemistry

Phosphofructokinase-1 (PFK-1) exhibits sigmoidal substrate saturation kinetics. What metabolic signals allosterically inhibit and activate PFK-1 in human cells?

  1. Inhibited by high ATP and citrate levels; activated by high AMP and fructose-2,6-bisphosphate levels. Correct answer.
  2. Inhibited by high AMP and ADP; activated by high ATP and glucose-6-phosphate.
  3. Inhibited by fructose-1,6-bisphosphate; activated by high oxygen partial pressures.
  4. Inhibited by insulin; activated by high intracellular lactate concentrations.

Explanation

PFK-1 is the key rate-limiting enzyme of glycolysis. High concentrations of ATP and citrate indicate abundant cellular energy and biosynthetic precursors, binding to allosteric sites on PFK-1 to inhibit its activity. Conversely, high AMP (indicating energy depletion) and fructose-2,6-bisphosphate allosterically activate PFK-1 to stimulate glycolytic flux.

Question 30Biochemistry

How does oxygen binding to one subunit of the tetrameric hemoglobin molecule affect the remaining unoxygenated subunits?

  1. It triggers a conformational shift from the low-affinity T (tense) state to the high-affinity R (relaxed) state, increasing oxygen affinity in remaining subunits (cooperativity). Correct answer.
  2. It induces irreversible dissociation of the tetramer into four isolated monomeric myoglobin subunits.
  3. It locks all remaining subunits into the T state, preventing further oxygen binding.
  4. It degrades the heme iron prosthetic group from Fe2+ into Fe3+, blocking further oxidation.

Explanation

Hemoglobin displays positive cooperativity in oxygen binding. Binding of O2 to a heme iron in one subunit shifts the quaternary conformation from the constrained T (tense) state to the relaxed R (relaxed) state. This conformational change increases the oxygen binding affinity of the remaining unoccupied subunits, producing a sigmoidal oxygen dissociation curve.

Question 31Biochemistry

Which structural component forms the amino alcohol backbone of sphingomyelin, a key lipid in neuronal myelin sheaths?

  1. Sphingosine (an 18-carbon unsaturated amino alcohol). Correct answer.
  2. Glycerol-3-phosphate attached to three saturated fatty acids.
  3. Cholesterol fused with a branched polypeptide chain.
  4. Isoprene units joined head-to-tail to form a cyclic terpene.

Explanation

Sphingomyelin is a membrane sphingolipid containing sphingosine, a long-chain 18-carbon unsaturated amino alcohol. A fatty acid is linked via an amide bond to the amino group of sphingosine (forming a ceramide), and a phosphocholine head group is attached to the terminal hydroxyl group.

Question 32Biochemistry

Why does double-stranded DNA with a high guanine-cytosine (G+C) content require a higher melting temperature (Tm) to denature into single strands than DNA with high adenine-thymine (A+T) content?

  1. G-C pairs contain three hydrogen bonds and stronger base-stacking interactions, requiring more thermal energy to disrupt than two-hydrogen-bonded A-T pairs. Correct answer.
  2. G-C pairs form covalent peptide bonds across complementary strands, whereas A-T pairs interact only through weak ionic forces.
  3. High G+C content causes DNA to adopt a triple-stranded alpha-helix structure resistant to thermal heat.
  4. A-T pairs contain extra phosphodiester bonds that spontaneously hydrolyze at room temperature.

Explanation

Guanine-Cytosine (G-C) base pairs are connected by three interstrand hydrogen bonds, whereas Adenine-Thymine (A-T) pairs are connected by only two. Additionally, G-C pairs engage in stronger base-stacking interactions with adjacent nucleotides in the helix, requiring higher thermal energy (higher Tm) to separate the strands.

Question 33Cell structure, membranes and division

Which type of ribosomes is present in prokaryotic cytoplasm and within the matrix of eukaryotic mitochondria?

  1. 70S ribosomes (composed of 50S and 30S subunits). Correct answer.
  2. 80S ribosomes (composed of 60S and 40S subunits).
  3. 100S ribosomes (composed of dual 50S catalytic subunits).
  4. 50S ribosomes (composed solely of small 16S rRNA fragments).

Explanation

Prokaryotes (bacteria and archaea) contain 70S ribosomes composed of a 50S large subunit and a 30S small subunit. In alignment with the endosymbiotic theory, mitochondria and chloroplasts also contain 70S ribosomes within their internal compartments.

Question 34Cell structure, membranes and division

Which of the following substances crosses the plasma membrane via simple diffusion down its concentration gradient without transport proteins or energy input?

  1. Oxygen gas (O2). Correct answer.
  2. Glucose (C6H12O6).
  3. Sodium ions (Na+).
  4. Amino acids like lysine.

Explanation

Small, nonpolar, uncharged molecules like oxygen (O2) and carbon dioxide (CO2) rapidly dissolve in the hydrophobic core of the phospholipid bilayer and cross the cell membrane by simple passive diffusion down their concentration gradient.

Question 35Cell structure, membranes and division

What primary function is associated with the rough endoplasmic reticulum (RER) in eukaryotic cells?

  1. Synthesis and preliminary N-linked glycosylation of membrane-bound and secreted proteins. Correct answer.
  2. Synthesis of steroid hormones and detoxification of lipid-soluble drugs.
  3. Direct conversion of light energy into chemical energy stored in glucose.
  4. Degradation of aged organelles by acidic hydrolytic enzymes.

Explanation

The rough endoplasmic reticulum (RER) is studded with ribosomes on its cytosolic face. It synthesizes proteins destined for insertion into cellular membranes, packaging into lysosomes, or extracellular secretion, initiating initial protein folding and N-linked glycosylation.

Question 36Cell structure, membranes and division

Which structures are contained within the mitochondrial matrix?

  1. Circular double-stranded mitochondrial DNA (mtDNA), 70S ribosomes, and soluble Krebs cycle enzymes. Correct answer.
  2. Linear chromosomes wrapped around histone octamers and 80S cytosolic ribosomes.
  3. Thylakoid membranes studded with Photosystem I and Photosystem II complexes.
  4. Hydrolytic enzymes operating at pH 4.5 for intracellular digestion.

Explanation

The mitochondrial matrix is the fluid-filled innermost compartment bounded by the inner mitochondrial membrane. It contains circular double-stranded mtDNA, 70S ribosomes, tRNA, metabolic enzymes for pyruvate oxidation and the Krebs cycle, and enzymes for mitochondrial gene expression.

Question 37Cell structure, membranes and division

What hallmark event characterizes metaphase during eukaryotic mitotic division?

  1. Chromosomes reach maximum condensation and align along the equatorial metaphase plate attached to spindle microtubules. Correct answer.
  2. Sister chromatids separate and move toward opposite spindle poles.
  3. The nuclear envelope reforms around decondensing chromatin at opposite poles.
  4. Homologous chromosomes undergo synapsis and genetic crossing over.

Explanation

During metaphase, mitotic spindle fibers attached to the kinetochores of sister chromatids maneuver chromosomes until all centromeres align along the cell's equatorial plane, known as the metaphase plate.

Question 38Cell structure, membranes and division

During which specific sub-phase of meiotic prophase I does genetic crossing over (recombination) take place?

  1. Pachytene. Correct answer.
  2. Leptotene.
  3. Diakinesis.
  4. Telophase II.

Explanation

Prophase I of meiosis is divided into five stages: leptotene, zygotene, pachytene, diplotene, and diakinesis. Genetic crossing over-the reciprocal exchange of non-sister chromatid segments between homologous chromosomes-occurs during pachytene after full synapsis is established.

Question 39Cell structure, membranes and division

How does the Na+/K+ ATPase pump maintain resting membrane potentials in animal cells?

  1. It hydrolyzes one ATP molecule to pump 3 Na+ ions out of the cell and 2 K+ ions into the cell against their electrochemical gradients. Correct answer.
  2. It uses proton gradients to pump 2 Na+ ions into the cell and 3 K+ ions out of the cell passively.
  3. It transports 3 Na+ ions and 3 K+ ions into the cell simultaneously by facilitated diffusion.
  4. It allows free movement of Na+ and K+ until intracellular and extracellular concentrations become equal.

Explanation

The sodium-potassium pump (Na+/K+ ATPase) is an electrogenic primary active transporter. For every ATP molecule hydrolyzed, it pumps 3 Na+ ions out of the cell and 2 K+ ions into the cell, both against their concentration gradients, maintaining negative resting membrane potentials.

Question 40Cell structure, membranes and division

Which cytosolic coat protein coat drives invagination of membrane pits during receptor-mediated endocytosis of LDL particles?

  1. Clathrin. Correct answer.
  2. COPI.
  3. COPII.
  4. Kinesin.

Explanation

Receptor-mediated endocytosis of low-density lipoprotein (LDL) particles relies on clathrin-coated pits. Clathrin triskelions polymerize on the inner cytosolic face of the plasma membrane, driving membrane curvature and vesicle budding.

Question 41Cell structure, membranes and division

What path do newly synthesized secretory proteins follow through the Golgi apparatus?

  1. They enter at the cis-Golgi network (receiving face) from RER vesicles, progress through medial cisternae, and exit at the trans-Golgi network (shipping face). Correct answer.
  2. They enter at the trans-Golgi network from the plasma membrane and travel backwards to exit at the cis-Golgi into mitochondria.
  3. They bypass Golgi cisternae entirely, passing through nuclear pores directly into secretory vesicles.
  4. They are degraded in the cis-Golgi into free fatty acids for lipid bilayer storage.

Explanation

Proteins synthesized in the RER enter the Golgi apparatus at the cis face via transport vesicles. As they pass through medial cisternae to the trans face, they undergo post-translational modifications (such as oligosaccharide processing) before being sorted into secretory or lysosomal vesicles at the trans-Golgi network.

Question 42Cell structure, membranes and division

How do lysosomes maintain an internal acidic pH (~4.5 to 5.0) required for optimal acid hydrolase activity?

  1. By active pumping of protons (H+) into the lysosomal lumen by membrane-bound V-type H+-ATPases. Correct answer.
  2. By passive diffusion of bicarbonate ions into the lumen through open sodium channels.
  3. By consuming oxygen during aerobic respiration inside the lysosomal matrix.
  4. By exporting hydrogen ions continuously into the cytosol via sodium-potassium exchangers.

Explanation

Lysosomes maintain their acidic lumen (pH ~4.5–5.0) through vacuolar-type H+-ATPases (V-ATPases) embedded in their membrane. These proton pumps actively transport H+ from the cytosol into the lysosome using ATP, creating an optimal environment for acid hydrolases.

Question 43Cell structure, membranes and division

Which enzyme in peroxisomes protects cells by decomposing toxic hydrogen peroxide (H2O2) into water and oxygen?

  1. Catalase. Correct answer.
  2. Ribulose-1,5-bisphosphate carboxylase/oxygenase (RuBisCO).
  3. DNA polymerase III.
  4. Cytochrome c oxidase.

Explanation

Peroxisomes contain oxidative enzymes that produce hydrogen peroxide (H2O2) during fatty acid oxidation and amino acid catabolism. The enzyme catalase neutralizes this toxic byproduct by catalyzing the reaction 2 H2O2 -> 2 H2O + O2.

Question 44Cell structure, membranes and division

What internal structural arrangement of microtubules characterizes eukaryotic cilia and flagella (axoneme)?

  1. A 9+2 arrangement consisting of 9 peripheral doublets and 2 central single microtubules. Correct answer.
  2. A 9+0 arrangement consisting of 9 peripheral triplets and no central microtubules.
  3. A 7+2 arrangement of actin filaments wrapped around intermediate filaments.
  4. A random network of microfilaments cross-linked by myosin heads.

Explanation

The axoneme of motile eukaryotic cilia and flagella features a classic 9+2 pattern: nine outer doublets of alpha/beta-tubulin microtubules surrounding two central single microtubules, connected by radial spokes and dynein arms.

Question 45Cell structure, membranes and division

How does the retinoblastoma protein (Rb) regulate entry into S phase at the cell cycle G1 restriction point?

  1. Unphosphorylated Rb binds and inhibits E2F transcription factors; phosphorylation of Rb by Cyclin D-CDK4/6 releases E2F to transcribe S-phase genes. Correct answer.
  2. Rb directly synthesizes RNA primers to initiate DNA replication.
  3. Phosphorylated Rb degrades spindle fibers to halt mitosis during telophase.
  4. Rb binds to p53 to stimulate immediate apoptosis in healthy G1 cells.

Explanation

In early G1 phase, hypophosphorylated Rb binds E2F transcription factors, inhibiting S-phase gene expression. In response to mitogenic signals, Cyclin D-CDK4/6 complexes hyperphosphorylate Rb, releasing active E2F to transcribe genes required for DNA replication.

Question 46Cell structure, membranes and division

Which enzyme cleaves cohesin protein complexes to trigger the transition from metaphase to anaphase during mitosis?

  1. Separase (activated after APC/C ubiquitinates and degrades securin). Correct answer.
  2. DNA ligase III.
  3. Topoisomerase II.
  4. Caspase-3.

Explanation

At the metaphase-to-anaphase transition, the Anaphase-Promoting Complex/Cyclosome (APC/C) targets securin for proteasomal degradation. Destruction of securin releases active separase, a protease that cleaves cohesin rings holding sister chromatids together, initiating anaphase.

Question 47Cell structure, membranes and division

How does plant cell cytokinesis differ fundamentally from animal cell cytokinesis?

  1. Plant cells construct a cell plate from Golgi-derived vesicles coalescing along a phragmoplast, whereas animal cells form a cleavage furrow via an actin-myosin contractile ring. Correct answer.
  2. Plant cells pinch inward from the plasma membrane using centrioles, whereas animal cells form a rigid cell plate.
  3. Plant cells undergo nuclear division without cytoplasmic division, forming coenocytic multinucleate cells permanently.
  4. Animal cells synthesize a temporary cellulose wall before undergoing cell division.

Explanation

Because plant cells are enclosed by a rigid cell wall, cytokinesis occurs from the inside out: Golgi-derived vesicles containing cell wall precursors accumulate along microtubules of the phragmoplast, fusing to form a cell plate. Animal cells lack cell walls and divide from the outside in via a contractile ring of actin and myosin filaments.

Question 48Cell structure, membranes and division

What key difference distinguishes anaphase I of meiosis from anaphase of mitosis?

  1. In anaphase I of meiosis, homologous chromosome pairs separate while sister chromatids remain attached; in mitotic anaphase, sister chromatids separate. Correct answer.
  2. In anaphase I of meiosis, individual sister chromatids separate completely to form haploid gametes directly.
  3. Mitotic anaphase reduces chromosome number from 2n to 1n, whereas meiotic anaphase I maintains diploidy.
  4. DNA replication occurs continuously during anaphase I of meiosis.

Explanation

In anaphase I of meiosis, spindle fibers pull intact homologous chromosome pairs (bivalents) to opposite poles, while sister chromatids remain held together by centromeric cohesin. In mitotic anaphase, centromeric cohesin is cleaved, allowing sister chromatids to separate.

Question 49Cell structure, membranes and division

Which set of empirical evidence strongly supports the endosymbiotic origin of mitochondria and chloroplasts from primitive prokaryotic ancestors?

  1. Presence of circular double-stranded DNA unassociated with histones, 70S ribosomes, binary fission division, and double membranes. Correct answer.
  2. Presence of linear histone-bound chromosomes, 80S ribosomes, and a single surrounding membrane layer.
  3. Sensitivity to human peptide hormones and capacity to survive independently outside host cells in sterile water indefinitely.
  4. Ability to synthesize eukaryotic mRNA with 5' caps and poly-A tails within the cytoplasm.

Explanation

The endosymbiotic theory (formulated by Lynn Margulis) is supported by multiple prokaryotic characteristics retained by mitochondria and chloroplasts: circular naked DNA, 70S ribosomes inhibited by bacterial antibiotics, reproduction via binary fission, and double membrane systems where the inner membrane reflects the original ancestral prokaryotic membrane.

Question 50Cell structure, membranes and division

How does the SGLT1 symporter transport glucose across the apical membrane of intestinal epithelial cells?

  1. It uses energy from sodium ions moving down their electrochemical gradient into the cell to cotransport glucose against its concentration gradient (secondary active transport). Correct answer.
  2. It directly hydrolyzes ATP to pump glucose against its gradient without involvement of sodium ions.
  3. It allows glucose to diffuse freely out of the cell through open ion channels down its concentration gradient.
  4. It engulfs extracellular fluid containing glucose by phagocytosis.

Explanation

SGLT1 is a secondary active symporter. It utilizes the favorable inward sodium electrochemical gradient (maintained by the basolateral Na+/K+ ATPase) to transport two Na+ ions into the cell down their gradient, driving the uphill transport of one glucose molecule into the cell against its concentration gradient.