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100+ Free Catalonia PAU Biology Practice Questions
Catalonia PAU Biology Examination — CIC / Generalitat de Catalunya (Biologia 2n Batxillerat 2026) practice questions are available now; exam metadata is being verified.
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2026 Statistics
Key Facts: Catalonia PAU Biology Exam
90 Minutes
Official examination duration
CIC Catalonia PAU Guidelines
Min 4.0
Minimum Access Phase score required
Generalitat de Catalunya PAU Rules
5 Blocks
Biochemistry, Cell Biology, Metabolism, Genetics, Microbiology & Immunology
2n Batxillerat Biology Syllabus
EUR 110.00+ access phase (EUR 41.30 exam right + EUR 68.70 access phase; plus EUR 13.80 per admission exercise, Generalitat de Catalunya 2026)
Base registration fee set by Generalitat de Catalunya
Generalitat de Catalunya Fees
100
Practice questions in this adaptation bank
OpenExamPrep
Catalonia PAU Biology is a 90-minute entrance exam for 2n Batxillerat students, adapted here into 100 English MCQ practice questions across 5 syllabus blocks.
Sample Catalonia PAU Biology Practice Questions
Try these sample questions to test your Catalonia PAU Biology exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.
1Which physicochemical property of water allows humans and other mammals to release excess body heat effectively during sweating?
A.High heat of vaporization resulting from extensive intermolecular hydrogen bonding.
B.Low boiling point caused by nonpolar covalent bonds within the water molecule.
C.High density in the solid phase compared to the liquid phase.
D.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.
2What happens to human red blood cells (erythrocytes) when placed in a hypertonic sodium chloride solution?
A.Water moves out of the cells by osmosis, causing them to shrink and undergo crenation.
B.Water enters the cells by osmosis until they swell and burst in a process called hemolysis.
C.Sodium ions rapidly diffuse into the cells by simple diffusion, causing cell swelling.
D.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.
3Which structural feature distinguishes D-glucose from D-fructose?
A.D-glucose is an aldohexose with a carbonyl group at C1, whereas D-fructose is a ketohexose with a carbonyl group at C2.
B.D-glucose is a pentose sugar found in RNA, whereas D-fructose is a hexose sugar found in plant cell walls.
C.D-glucose contains beta-1,4-glycosidic linkages, whereas D-fructose contains alpha-1,4-glycosidic linkages.
D.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).
4Which chemical reaction forms a triglyceride (triacylglycerol) from glycerol and fatty acid molecules?
A.Esterification of three fatty acid carboxyl groups with three glycerol hydroxyl groups, releasing three water molecules.
B.Hydrolysis of peptide bonds between amino acid residues in the presence of acidic catalysts.
C.Condensation of three phosphate groups with the nitrogenous base adenosine to form high-energy bonds.
D.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.
5Which functional groups react during the formation of a peptide bond between two amino acids?
A.The alpha-carboxyl group of one amino acid and the alpha-amino group of another amino acid.
B.The variable R-group side chain of a hydrophobic amino acid and a phosphate group.
C.The hydroxyl group of a serine residue and the methyl group of an alanine residue.
D.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.
6According to Chargaff's rules and the Watson-Crick DNA model, how do nitrogenous bases pair across antiparallel strands?
A.Adenine pairs with thymine via two hydrogen bonds; guanine pairs with cytosine via three hydrogen bonds.
B.Adenine pairs with uracil via three hydrogen bonds; guanine pairs with thymine via two hydrogen bonds.
C.Cytosine pairs with thymine via covalent phosphodiester linkages between sugar rings.
D.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.
7How does the carbonic acid-bicarbonate buffer system maintain human arterial blood plasma pH near 7.40 when lactic acid enters the bloodstream?
A.Bicarbonate ions (HCO3-) accept excess protons (H+) to form carbonic acid (H2CO3), which dissociates into water and carbon dioxide exhaled by the lungs.
B.Carbonic acid (H2CO3) donates hydroxide ions (OH-) directly to neutralize lactic acid, raising plasma osmolarity.
C.Lactic acid reacts with dissolved nitrogen gas to form urea, preventing changes in hydrogen ion concentration.
D.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.
8Enzymatic hydrolysis of one molecule of sucrose yields which combination of monosaccharide products?
A.One molecule of D-glucose and one molecule of D-fructose.
B.Two molecules of D-glucose.
C.One molecule of D-galactose and one molecule of D-glucose.
D.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.
9Why is plant cellulose indigestible to human digestive enzymes, whereas plant starch (amylose) is readily digested?
A.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.
B.Cellulose is composed exclusively of D-galactose monomers, whereas starch is composed of D-fructose monomers.
C.Cellulose contains high amounts of sulfur cross-links that covalently block enzymatic binding sites.
D.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.
10What structural property allows glycerophospholipids to spontaneously arrange into lipid bilayers in aqueous cellular environments?
A.Their amphipathic nature, featuring a hydrophilic polar head group and two hydrophobic fatty acid tails.
B.Their completely nonpolar, hydrophobic structure that repels water molecules from both surfaces.
C.The presence of covalent disulfide bridges linking adjacent phosphate groups across the membrane.
D.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.
About the Catalonia PAU Biology Practice Questions
Verified exam format metadata for Catalonia PAU Biology Examination — CIC / Generalitat de Catalunya (Biologia 2n Batxillerat 2026) is pending. The practice questions above remain available while official exam length, timing, passing score, fee, and administrator details are reviewed.