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Balearic Islands PAU Biology Examination — UIB (Biologia 2º Bachillerato UIB 2026) practice questions are available now; exam metadata is being verified.
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2026 Statistics
Key Facts: Balearic PAU Biology Exam
90 Minutes
Exam time duration
UIB PAU Commission
EUR 69.21
Ordinary registration fee for Access Phase
UIB 2026 Fees
0–10 Scale
Grading scale for Bachillerato and PAU exams
Spanish Ministry of Education
5 Core Blocks
Biochemistry, Cell Biology, Metabolism, Genetics, Microbiology & Immunology
UIB Syllabus
100 Questions
Practice bank size in OpenExamPrep
OpenExamPrep
Balearic Islands PAU Biology (UIB 2026) is a 90-minute university entrance exam assessing 2º Bachillerato Biology across 5 core subject blocks.
Sample Balearic PAU Biology Practice Questions
Try these sample questions to test your Balearic 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 accounts for its high specific heat capacity and high heat of vaporization?
A.Extensive hydrogen bonding between polar water molecules
B.The presence of covalent hydrogen-carbon bonds throughout the liquid matrix
C.The low dielectric constant preventing ionic dissociation
D.Van der Waals dispersion forces between nonpolar oxygen atoms
Explanation: Water molecules form a dense network of hydrogen bonds due to the strong dipole moment resulting from the electronegativity difference between oxygen and hydrogen. Breaking these bonds requires substantial thermal energy, giving water a high specific heat capacity and high heat of vaporization.
2Which group of chemical elements constitutes over 96% of total living organism mass (primary bioelements)?
A.Iron, Iodine, Zinc, and Copper
B.Carbon, Hydrogen, Oxygen, Nitrogen, Phosphorus, and Sulfur
C.Sodium, Potassium, Calcium, Magnesium, and Chlorine
D.Silicon, Aluminum, Titanium, and Boron
Explanation: Primary bioelements (C, H, O, N, P, S) form covalent bonds easily and make up more than 96% of the dry weight of living organisms, providing the backbone for all major biological macromolecules.
3D-Glucose and D-Galactose differ in configuration only at carbon-4 (C-4). What type of isomers are they?
A.Enantiomers
B.Anomers
C.Epimers
D.Structural chain isomers
Explanation: Epimers are stereoisomers that differ in absolute configuration at only one specific chiral center. Since D-glucose and D-galactose differ solely at C-4, they are C-4 epimers.
4Maltose is a disaccharide produced during starch digestion. Which glycosidic linkage connects its two D-glucose units?
A.beta(1->6) glycosidic bond
B.beta(1->4) glycosidic bond
C.alpha(1->2) glycosidic bond
D.alpha(1->4) glycosidic bond
Explanation: Maltose consists of two D-glucopyranose units joined by an alpha(1->4) glycosidic bond formed between the anomeric C-1 hydroxyl group in the alpha position of the first glucose and the C-4 hydroxyl group of the second glucose.
5Which polysaccharide serves as the principal energy storage molecule in animal liver and muscle tissue?
A.Glycogen
B.Amylose
C.Cellulose
D.Chitin
Explanation: Glycogen is a highly branched homopolysaccharide of glucose linked by alpha(1->4) chains with alpha(1->6) branches every 8-12 glucose residues, serving as the main glucose storage reserve in animal liver and skeletal muscle.
6Why do unsaturated fatty acids with cis double bonds have lower melting points than saturated fatty acids of equivalent chain length?
A.Unsaturated fatty acids form stronger covalent cross-links that destabilize membrane lipids
B.Cis double bonds create a rigid bend in the hydrocarbon tail, preventing tight crystalline packing
C.Saturated fatty acids contain polar carboxyl groups that lower their melting temperature
D.Cis double bonds increase overall molecular mass, reducing intermolecular kinetic energy
Explanation: The cis double bond introduces a 30-degree kink in the hydrocarbon chain. This structural bend disrupts linear hydrophobic alignment and prevents close packing into a solid crystal lattice, resulting in lower melting points and liquid state at room temperature.
7What products are generated when a triacylglycerol undergoes alkaline hydrolysis (saponification) with sodium hydroxide (NaOH)?
A.Sphingosine, phosphate, and choline
B.Three free fatty acids and ethanol
C.Glycerol and three sodium salts of fatty acids (soaps)
D.Glycerol-3-phosphate and three fatty alcohols
Explanation: Saponification is the base-catalyzed ester hydrolysis of triacylglycerols. Treatment with NaOH cleaves the three ester bonds, yielding one molecule of glycerol (propan-1,2,3-triol) and three sodium fatty acid carboxylate salts (soaps).
8What physical drive causes glycerophospholipids to spontaneously assemble into lipid bilayers in an aqueous environment?
A.Ionic repulsions between nonpolar hydrocarbon tails and water dipoles
B.Formation of covalent disulfide linkages between adjacent fatty acyl chains
C.Active ATP-dependent transport by membrane-bound translocases
D.Hydrophobic effect driven by the increase in entropy of surrounding water molecules
Explanation: The hydrophobic effect drives lipid bilayer self-assembly. Nonpolar fatty acyl tails force surrounding water molecules into ordered clathrate structures. Aggregation of nonpolar tails sequesters them from water, releasing caged water molecules into bulk solution and increasing overall thermodynamic entropy (Delta S > 0).
9How does cholesterol modulate cell membrane fluidity across varying temperature regimes?
A.It acts as a fluidity buffer: restricting phospholipid movement at high temperatures and preventing tail packing at low temperatures
B.It covalently cross-links phospholipids to permanently freeze the membrane in a gel phase
C.It enzymatically desaturates fatty acid tails when temperatures drop below freezing
D.It dissolves integral membrane proteins to increase lateral diffusion rates indiscriminately
Explanation: Cholesterol acts as a membrane fluidity buffer. At warm temperatures, its rigid steroid ring system interferes with phospholipid acyl chain movement, reducing fluidity. At cold temperatures, it inserts between acyl tails, preventing close packing and crystallization.
10An amino acid with a non-ionizable side chain (like alanine) reaches its isoelectric point (pI). What is its net electrical charge at this pH?
A.+2
B.0 (Zwitterion)
C.+1
D.-1
Explanation: At the isoelectric point (pI), the amino acid exists predominantly as a dipolar zwitterion with a protonated positively charged amino group (-NH3+) and a deprotonated negatively charged carboxyl group (-COO-), yielding a net electric charge of zero.
About the Balearic PAU Biology Practice Questions
Verified exam format metadata for Balearic Islands PAU Biology Examination — UIB (Biologia 2º Bachillerato UIB 2026) is pending. The practice questions above remain available while official exam length, timing, passing score, fee, and administrator details are reviewed.