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Madrid PAU Biology Examination — UCM & Public Universities (Biología 2º Bachillerato 2026) practice questions are available now; exam metadata is being verified.
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2026 Statistics
Key Facts: Madrid PAU Biology Exam
90 Minutes
Exam duration
Madrid PAU Commission / UCM
EUR 93.02
Base registration fee for Access Phase
Comunidad de Madrid
0–10 Scale
Grading scale for Bachillerato and PAU exams
Spanish Ministry of Education
5 Core Blocks
Biochemistry (25%), Cell Bio (25%), Metabolism (20%), Genetics (15%), Microbiology & Immunology (15%)
LOMLOE Biology Syllabus
100 Questions
Practice bank size in OpenExamPrep
OpenExamPrep
Madrid PAU Biology (UCM 2026) is a 90-minute university entrance exam assessing 2º Bachillerato Biology across 5 core subject blocks.
Sample Madrid PAU Biology Practice Questions
Try these sample questions to test your Madrid 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 aquatic organisms to survive under ice-covered lakes during cold winter seasons?
A.Maximum density of water occurs in the liquid state at 4 °C, causing ice to float and insulate the underlying water
B.High specific heat capacity causes water to freeze from the bottom of the lake upward
C.Low surface tension prevents ice crystals from forming on the surface of the lake
D.High latent heat of vaporization forces liquid water to contract upon freezing
Explanation: Water reaches its maximum density at 4 °C in the liquid state. As water freezes at 0 °C, hydrogen bonds form a open crystalline lattice that makes ice less dense than liquid water, allowing ice to float on the surface and act as a thermal insulator for aquatic life below.
2When red blood cells (erythrocytes) are placed in a hypertonic NaCl solution, what cellular phenomenon occurs?
A.Water leaves the cells by osmosis, causing them to shrink and undergo crenation
B.Water enters the cells by osmosis, causing them to swell and undergo osmotic lysis (hemolysis)
C.Solutes actively pump out of the cells until turgor pressure equals osmotic pressure
D.The cell wall exerts wall pressure to prevent any net movement of water
Explanation: In a hypertonic solution (higher solute concentration outside than inside the cytoplasm), water moves out of the cell along its concentration gradient by osmosis. Lacking a rigid cell wall, animal cells shrink and contract, a process termed crenation.
3Calculate the osmotic pressure (π) at 27 °C (300 K) for a 0.2 M aqueous solution of sodium chloride (NaCl), assuming complete dissociation (van 't Hoff factor i = 2) and R = 0.0821 L·atm/(mol·K).
A.9.85 atm
B.4.93 atm
C.2.46 atm
D.19.70 atm
Explanation: Using the van 't Hoff equation π = i · M · R · T: π = 2 × 0.2 mol/L × 0.0821 L·atm/(mol·K) × 300 K = 9.852 atm. Rounding to two decimal places gives 9.85 atm.
4D-Glucose and D-Galactose are structural isomers that differ in configuration specifically around carbon-4. What term precisely describes their structural relationship?
A.C-4 Epimers
B.Enantiomers
C.Anomers
D.Ketose-aldose tautomers
Explanation: Epimers are stereoisomers that differ in absolute configuration at only one specific chiral center. Because D-glucose and D-galactose differ only at C-4, they are C-4 epimers.
5Which of the following disaccharides is a non-reducing sugar because both of its anomeric carbons are involved in the glycosidic bond?
A.Sucrose (α-D-glucopyranosyl-(1→2)-β-D-fructofuranoside)
B.Maltose (α-D-glucopyranosyl-(1→4)-D-glucopyranose)
C.Lactose (β-D-galactopyranosyl-(1→4)-D-glucopyranose)
D.Cellobiose (β-D-glucopyranosyl-(1→4)-D-glucopyranose)
Explanation: Sucrose is formed by a glycosidic bond between C-1 of α-D-glucose and C-2 of β-D-fructose. Since both anomeric carbon atoms are tied up in the linkage, no free hemiacetal or hemiketal group remains to reduce Benedict's or Fehling's reagent.
6What structural feature of cellulose makes it resistant to human digestive enzymatic hydrolysis while starch is readily digested?
A.Cellulose consists of unbranched chains of glucose monomers linked by β-(1→4) glycosidic bonds
B.Cellulose contains highly branched chains linked by α-(1→6) glycosidic bonds
C.Cellulose is composed of galactose and fructose disaccharides linked by peptide bonds
D.Cellulose monomer residues are joined by ester bonds that require lipase breakdown
Explanation: Cellulose consists of linear unbranched polymer chains of D-glucose linked by β-(1→4) glycosidic bonds. Human digestive enzymes such as α-amylase can hydrolyze α-(1→4) bonds in starch but lack the specificity to cleavage β-(1→4) linkages.
7Which of the following lipids belongs to the class of non-saponifiable lipids because it lacks esterified fatty acid chains?
A.Cholesterol (a steroid derivative)
B.Triacylglycerol (triglyceride)
C.Phosphatidylcholine (lecithin)
D.Sphingomyelin
Explanation: Non-saponifiable lipids do not contain fatty acid residues joined by ester or amide bonds and therefore cannot undergo alkaline hydrolysis (saponification) to yield soap salts. Cholesterol is a steroid composed of a tetracyclic cyclopentanoperhydrophenanthrene core.
8Why do unsaturated fatty acids with cis double bonds have lower melting points than saturated fatty acids of equivalent chain length?
A.Cis double bonds introduce rigid kinks in the hydrocarbon tail that inhibit tight molecular packing and weaken van der Waals interactions
B.Cis double bonds increase hydrogen bonding capacity with surrounding water molecules
C.Saturated fatty acids contain polar head groups that increase electrostatic repulsion
D.Unsaturated fatty acids form covalent cross-links that destabilize solid crystal formation
Explanation: The cis double bond creates a fixed ~30° bend (kink) in the hydrocarbon chain. This geometric disruption prevents molecules from packing tightly into a crystalline lattice, reducing intermolecular van der Waals dispersion forces and lowering the melting point.
9What physical property causes phospholipids to spontaneously self-assemble into lipid bilayers when introduced into aqueous solutions?
A.Amphipathic nature with hydrophilic polar head groups facing water and hydrophobic fatty acyl tails sequestered inside
B.High solubility of fatty acid chains in polar solvents like water
C.Formation of covalent disulfide bridges between adjacent phosphate groups
D.Ionic repulsion between nonpolar hydrocarbon tails and water dipoles
Explanation: Phospholipids are amphipathic: they possess a polar, hydrophilic phosphate head group and two nonpolar, hydrophobic fatty acid tails. Thermodynamically, water molecules force the hydrophobic tails to cluster together inside to minimize entropy loss (hydrophobic effect) while heads interact with water.
10At its isoelectric point (pI), an amino acid in aqueous solution exhibits which predominant electrical state?
A.A dipolar zwitterion with a net electrical charge of zero
B.A positively charged cation that migrates toward the cathode
C.A negatively charged anion that migrates toward the anode
D.A fully un-ionized neutral molecule lacking any charged groups
Explanation: The isoelectric point (pI) is the pH at which an amino acid carries equal numbers of positive and negative charges (protonated -NH3+ and deprotonated -COO-), existing as a zwitterion with a net charge of zero.
About the Madrid PAU Biology Practice Questions
Verified exam format metadata for Madrid PAU Biology Examination — UCM & Public Universities (Biología 2º Bachillerato 2026) is pending. The practice questions above remain available while official exam length, timing, passing score, fee, and administrator details are reviewed.