AP Biology
Hücreden evrime: konu anlatımı, biyoistatistik ve hata teşhisli sorular.
Kritik Biyoloji Eşitlikleri (Tıkla ve Kopyala)
p: Dominant alel frekansı, q: Resesif alel frekansı, 2pq: Heterozigot birey oranı. Şart: Mutasyon, göç, doğal seçilim olmamalı; rastgele eşleşme.
o: Gözlenen değer (Observed), e: Beklenen Mendel teorik değeri (Expected). Kritik tablo değeriyle karşılaştırılarak genlerin bağlı olup olmadığı anlaşılır.
Su daima yüksek su potansiyelinden (daha az negatif) düşük su potansiyeline (daha negatif) doğru akar.
İki deney grubunun ±2 SEM hata çubukları üst üste biniyorsa (overlap), aradaki fark istatistiksel olarak anlamlı değildir!
AP Biyoloji Ünite Dağılımı
AP Biology Theoretical Foundations & Molecular Systems
Biochemical mechanisms, cellular signal transduction cascades, gene regulation circuits, and evolutionary biostatistics required for College Board 5/5 mastery and European biomedical standards:
Enzyme Catalysis & Induced-Fit Model: Enzymes lower activation energy ($E_a$) without altering free energy ($\Delta G$).
• Competitive Inhibition: Inhibitor binds reversibly to the active site. Can be overcome by increasing substrate concentration ($[S]$); $V_{max}$ remains unchanged, but apparent $K_m$ increases.
• Noncompetitive (Allosteric) Inhibition: Inhibitor binds to an allosteric site, altering the enzyme's 3D conformational shape. Cannot be overcome by adding substrate; $V_{max}$ decreases, while $K_m$ remains unchanged.
The 3-Step Signal Transduction Model:
1. Reception: A signaling ligand (hydrophilic e.g. peptide hormone, or hydrophobic e.g. steroid) binds specifically to a cell-surface receptor (GPCR, Receptor Tyrosine Kinase) or intracellular receptor.
2. Transduction: Multi-step phosphorylation cascade involving protein kinases (which add phosphate groups) and protein phosphatases (which remove them). Second messengers (such as cyclic AMP (cAMP) and $\text{Ca}^{2+}$) amplify the signal exponentially.
3. Response: Target cellular activity is modulated, such as activating gene transcription in the nucleus or activating metabolic enzymes in the cytoplasm.
The Operon Model in Bacteria: A cluster of functional genes transcribed under a single promoter and regulated by an operator:
• Lac Operon (Inducible): Typically OFF. In the absence of lactose, the active repressor protein binds to the operator, blocking RNA polymerase. When lactose is present, allolactose acts as an inducer, binding to the repressor and inactivating it.
• Trp Operon (Repressible): Typically ON. When tryptophan levels are abundant, tryptophan acts as a corepressor, binding to the inactive repressor to activate it, which then blocks transcription.
Hardy-Weinberg Equilibrium: Allele frequencies ($p + q = 1$) and genotype frequencies ($p^2 + 2pq + q^2 = 1$) remain constant over generations under 5 strict conditions:
1. Extremely large population (no genetic drift)
2. No gene flow (no immigration/emigration)
3. No net mutations
4. Random mating
5. No natural selection
• Null Hypothesis ($H_0$): There is no statistically significant difference between observed and expected frequencies; any observed difference is purely due to chance.
• If $\chi^2_{calc} > \chi^2_{crit}$ at $p = 0.05$, reject $H_0$ (the difference is statistically significant). If $\chi^2_{calc} \le \chi^2_{crit}$, fail to reject $H_0$.
Unit 1: Chemistry of Life
Polarity of Water: Water ($\text{H}_2\text{O}$) is an asymmetric polar molecule. Oxygen is significantly more electronegative than hydrogen ($\Delta EN \approx 1.24$), creating permanent partial negative charges ($\delta^-$) on oxygen and partial positive charges ($\delta^+$) on hydrogens.
\(\text{Hydrogen Bond: Electrostatic attraction between } \delta^- \text{ O of one molecule and } \delta^+ \text{ H of a neighboring molecule.}\)
1. Cohesion & Adhesion (Capillary Action): Cohesion (intermolecular H-bonding between water molecules) and Adhesion (H-bonding between water and polar xylem cell walls) generate transpirational pull, allowing water to ascend trees against gravity without breaking the continuous water column.
2. High Specific Heat Capacity & Heat of Vaporization: Breaking extensive hydrogen-bond networks requires substantial thermal energy ($4.184 \text{ J/g}^\circ\text{C}$). This buffers aquatic and cellular environments against lethal temperature fluctuations and enables evaporative cooling (sweating/transpiration).
3. Density Anomaly (Ice Floats): Below $4^\circ\text{C}$, hydrogen bonds stabilize into a rigid crystalline lattice, spacing water molecules further apart than in liquid state ($\rho_{\text{ice}} \approx 0.92\text{ g/cm}^3 < \rho_{\text{liquid}} \approx 1.00\text{ g/cm}^3$). Floating surface ice insulates underlying aquatic ecosystems in winter.
4. Universal Aqueous Solvent: Water forms hydration shells around polar molecules and dissolved ions (ion-dipole attractions), mediating metabolic biochemical reactions.
The Elemental Backbone of Biomolecules (CHNOPS):
• Carbon ($C$): Tetravalent (forms 4 stable covalent bonds); builds linear, branched, and ring-shaped skeletons.
• Nitrogen ($N$): Found mainly in proteins (amino groups $-\text{NH}_2$) and nucleic acids (nitrogenous bases).
• Phosphorus ($P$): Found in nucleic acids (sugar-phosphate backbone) and phospholipids (cell membranes); absent in standard proteins.
• Sulfur ($S$): Found exclusively in proteins (cysteine/methionine residues forming covalent disulfide bridges); absent in nucleic acids.
• Dehydration Synthesis: Monomers are covalently linked into polymers by removing an $-\text{OH}$ from one monomer and an $-\text{H}$ from another, releasing one molecule of $\text{H}_2\text{O}$ per bond formed. To polymerize $N$ monomers into a linear polymer, exactly $(N - 1)$ molecules of $\text{H}_2\text{O}$ are produced!
• Hydrolysis: Covalent bonds connecting monomers within a polymer are cleaved by the enzymatic addition of $\text{H}_2\text{O}$ ($-\text{OH}$ to one monomer, $-\text{H}$ to the other). Digestion of a polymer of $N$ units requires $(N - 1)$ molecules of $\text{H}_2\text{O}$.
Carbohydrates: Monosaccharide monomers ($\text{CH}_2\text{O}$ ratio) linked by glycosidic linkages.
• Energy Storage Polymers: Starch (plants) and Glycogen (animals) feature $\alpha$-glucose monomers forming helical chains easily cleaved by digestive enzymes.
• Structural Polymers: Cellulose (plant cell walls) features $\beta$-glucose monomers linked by $\beta\text{-1,4-glycosidic bonds}$. Straight, unbranched chains form dense hydrogen-bonded microfibrils that most animals cannot hydrolyze due to lack of cellulase.
• Saturated vs. Unsaturated Fatty Acids: Saturated fatty acids have no double bonds, maximum hydrogens, straight hydrocarbon chains that pack tightly (solid at room temperature; animal fats). Unsaturated fatty acids possess one or more cis-double bonds creating rigid kinks that prevent tight packing (liquid at room temperature; plant oils).
• Phospholipids (Amphipathic Molecules): A glycerol backbone attached to two hydrophobic nonpolar fatty acid tails and one hydrophilic polar phosphate head. In aqueous environments, they spontaneously assemble into a bilayer, with hydrophilic heads exposed to water and hydrophobic tails shielded in the interior.
Proteins are polypeptides composed of 20 standard amino acid monomers, each featuring a central alpha-carbon ($\text{C}_\alpha$), an amino group ($-\text{NH}_2$), a carboxyl group ($-\text{COOH}$), a hydrogen atom, and a variable chemical side chain ($R\text{-group}$).
1. Primary ($1^\circ$) Structure: Unique linear sequence of amino acids joined by covalent peptide bonds formed via dehydration synthesis between carboxyl and amino groups. Directionality: $\text{N-terminus}$ (amino) to $\text{C-terminus}$ (carboxyl).
2. Secondary ($2^\circ$) Structure: Regular local coils ($\alpha$-helices) and folds ($\beta$-pleated sheets) stabilized strictly by hydrogen bonds between the polypeptide backbone atoms ($\text{C=O}$ and $\text{N-H}$ groups), completely independent of $R$-groups!
3. Tertiary ($3^\circ$) Structure: Complete 3D globular or fibrous conformation determined by interactions between $R$-groups:
• Hydrophobic interactions (nonpolar $R$-groups cluster in the interior away from water).
• Hydrogen bonds (between polar uncharged $R$-groups).
• Ionic bonds / salt bridges (between positively and negatively charged $R$-groups).
• Covalent disulfide bridges ($-\text{S}-\text{S}-$) between adjacent cysteine residues.
4. Quaternary ($4^\circ$) Structure: Overall protein structure resulting from the aggregation of two or more distinct polypeptide subunits (e.g., Hemoglobin composed of $2\alpha$ and $2\beta$ subunits).
Nucleotide Monomers: Composed of three components: (1) a 5-carbon pentose sugar (Deoxyribose in DNA with $2'\text{-H}$, Ribose in RNA with $2'\text{-OH}$), (2) an inorganic phosphate group attached to the $5'$ carbon, and (3) a nitrogenous base attached to the $1'$ carbon.
DNA vs. RNA Architecture:
• DNA: Double-stranded antiparallel double helix ($5' \to 3'$ strand runs opposite to $3' \to 5'$ strand). Purines (Adenine, Guanine; double-ring) pair with Pyrimidines (Thymine, Cytosine; single-ring).
• Base-Pairing Thermodynamics: $\text{A}=\text{T}$ pairs share 2 hydrogen bonds; $\text{G}\equiv\text{C}$ pairs share 3 hydrogen bonds. Consequently, DNA sequences with higher $\text{G-C}$ percentage possess greater thermal stability and require a higher melting temperature ($T_m$) to denature!
Unit 1 Chemistry of Life: İnteraktif Soru İstasyonu
Özgün AP tarzı sorular (geçmiş sınav sorusu değil). Şık seç; anında çözüm ve hata türü gelir.
Unit 2: Cell Structure and Function
Subcellular Compartments: Non-membrane-bound ribosomes (composed of catalytic rRNA and ribosomal proteins) perform mRNA translation. Free ribosomes in the cytosol synthesize intracellular enzymes; bound ribosomes on the Rough Endoplasmic Reticulum (RER) synthesize integral membrane proteins, secretory proteins, and lysosomal hydrolases.
For a spherical cell of radius $r$: $$\text{SA} = 4\pi r^2, \quad V = \frac{4}{3}\pi r^3 \implies \frac{\text{SA}}{V} = \frac{3}{r}$$ For a cuboidal cell of side length $s$: $$\text{SA} = 6s^2, \quad V = s^3 \implies \frac{\text{SA}}{V} = \frac{6}{s}$$ • Dimensional Scaling Paradox: As a cell grows linearly ($r \to 2r$), volume increases with the cube ($2^3 = 8\times$), whereas surface area increases only with the square ($2^2 = 4\times$). Consequently, the ratio $\text{SA}/V$ decreases inversely with radius ($\frac{1}{2}\times$).
• Diffusive Limits of Metabolism: Einstein's diffusion equation states that diffusion time scales with the square of distance: $t \approx \frac{x^2}{2D}$. When cell diameter exceeds $\sim 50-100\,\mu\text{m}$, the rate of passive nutrient influx ($\text{O}_2$, glucose) and metabolic waste removal ($\text{CO}_2$, urea) across the plasma membrane cannot satisfy the volume's metabolic rate.
• Morphological Adaptations: Cells overcome geometric limits through surface convolutions: microvilli in intestinal enterocytes, root hairs in vascular plants, thin flattened biconcave erythrocytes, and extensively folded inner organellar membranes (cristae, thylakoids).
The Endomembrane Transport Cascade:
Rough ER (translation & N-glycosylation) → Transport Vesicle → Golgi Cis Face → Medial Cisternae (carbohydrate modification) → Trans Golgi Network → Secretory Vesicle → Exocytosis
• Smooth ER (SER): Synthesizes lipids, phospholipids, and steroid hormones; metabolizes carbohydrates; detoxifies drugs and xenobiotics via Cytochrome P450; sequesters $\text{Ca}^{2+}$ in skeletal muscle (sarcoplasmic reticulum).
• Lysosomes & Autophagy: Contain acid hydrolases operating optimally at $\text{pH} \approx 4.5-5.0$, maintained by active vacuolar-type $\text{H}^+$-ATPase pumps. Dual protection: cytoplasmic neutral pH ($\sim 7.2$) rapidly inactivates leaked enzymes, preventing accidental cell autolysis.
• Phospholipid Bilayer: Amphipathic matrix held together strictly by noncovalent hydrophobic exclusion forces. Lipids undergo rapid lateral diffusion ($10^7\text{ times/s}$), while transverse "flip-flop" is energetically unfavorable ($< 1\text{ time/month}$) without ATP-dependent flippase/floppase enzymes.
• Fatty Acid Saturation & Membrane Fluidity: Saturated fatty acids have straight, fully hydrogenated chains that maximize Van der Waals interactions, producing dense packing and higher transition temperature ($T_m$). Unsaturated fatty acids contain rigid cis-double bonds that induce permanent $\sim 30^\circ$ kinks, sterically preventing tight packing and maintaining fluidity at low temperatures.
• Cholesterol as a Bidirectional Thermal Buffer:
1. At high physiological temperatures ($37^\circ\text{C}$): bulky, planar steroid rings hinder phospholipid hydrocarbon mobility, reducing excessive membrane fluidity.
2. At cold temperatures: cholesterol intercalates between fatty acid tails, disrupting regular crystalline lattice formation and preventing membrane freezing/solidification.
• Integral vs. Peripheral Proteins: Integral transmembrane proteins possess amphipathic structures: one or more hydrophobic $\alpha$-helices (rich in Leu, Ile, Val, Phe) embedded within the hydrophobic core, flanked by charged/polar residues facing aqueous cytosolic and extracellular environments.
The Hierarchy of Membrane Permeability:
1. Small Nonpolar Hydrophobic Molecules ($\text{O}_2, \text{CO}_2, \text{N}_2$, Steroid Hormones): Cross freely and rapidly via simple passive diffusion down concentration gradients ($\Delta G < 0$).
2. Small Uncharged Polar Molecules ($\text{H}_2\text{O}$, Glycerol, Urea, Ethanol): Cross slowly; water diffusion is drastically accelerated ($1000\times$) by specialized channel proteins called aquaporins.
3. Large Uncharged Polar Molecules (Glucose, Sucrose): Extremely low passive permeability; require carrier proteins (e.g., GLUT1-4).
4. Charged Ions ($\text{Na}^+, \text{K}^+, \text{Ca}^{2+}, \text{Cl}^-, \text{H}^+$): Completely impermeable to the lipid core. Stripping the surrounding hydration shell requires an enormous thermodynamic penalty ($\Delta G_{\text{desolvation}} \gg 0$), requiring gated ion channels or active pumps.
• Primary Active Transport ($\text{Na}^+/\text{K}^+$ ATPase): Direct ATP hydrolysis drives the conformational cycle exporting $3\,\text{Na}^+$ ions and importing $2\,\text{K}^+$ ions against their steep electrochemical gradients per ATP molecule consumed. Because it exports more positive charges than it imports, it is electrogenic, generating negative intracellular resting potential ($-70\,\text{mV}$).
• Secondary Active Transport (Cotransport / Symport & Antiport): Does not hydrolyze ATP directly; instead, it harnesses the potential energy stored in an ion gradient established by primary active transport:
• $\text{Na}^+/\text{Glucose}$ Symporter (SGLT1): Uses the downhill influx of $\text{Na}^+$ to pull glucose uphill into intestinal epithelial cells.
• $\text{H}^+/\text{Sucrose}$ Cotransporter: In plant phloem loading, a plasma membrane $\text{H}^+$-ATPase pumps protons out, creating a proton-motive force ($\Delta \text{pH} + \Delta \Psi$) that drives sucrose entry against its gradient.
• Bulk Transport: Endocytosis (phagocytosis, pinocytosis, clathrin-mediated receptor endocytosis) and Exocytosis (SNARE-mediated vesicular fusion).
The Master Water Potential Equation: Water potential ($\Psi$, measured in bars or $\text{MPa}$) represents the chemical potential of water per unit volume relative to pure water at standard temperature and pressure: $$\Psi = \Psi_s + \Psi_p$$ where $\Psi_s$ is the Solute (Osmotic) Potential and $\Psi_p$ is the Physical Pressure Potential.
$$\Psi_s = -i \cdot C \cdot R \cdot T$$ • $i$ = Ionization Constant: Number of ions the solute dissociates into in water (e.g., $i = 1.0$ for non-ionizing sucrose/glucose; $i = 2.0$ for $\text{NaCl} \to \text{Na}^+ + \text{Cl}^-$; $i = 3.0$ for $\text{CaCl}_2 \to \text{Ca}^{2+} + 2\text{Cl}^-$).
• $C$ = Molar Concentration ($\text{M} = \text{mol/L}$).
• $R$ = Pressure Constant ($0.0831 \frac{\text{L}\cdot\text{bar}}{\text{mol}\cdot\text{K}}$ or $0.00831 \frac{\text{L}\cdot\text{MPa}}{\text{mol}\cdot\text{K}}$).
• $T$ = Absolute Temperature in Kelvin ($T_K = ^\circ\text{C} + 273$).
• Thermodynamic Law of Osmosis: Pure water in an open container has $\Psi = 0\text{ bar}$. Solute addition always lowers water potential ($\Psi_s \le 0$). Water ALWAYS moves spontaneously from an area of HIGHER (less negative) water potential to an area of LOWER (more negative) water potential until dynamic equilibrium ($\Psi_{\text{in}} = \Psi_{\text{out}}$) is achieved.
• Hypotonic Environment ($\Psi_{\text{ext}} > \Psi_{\text{cell}}$): Animal cells swell and undergo lysis (bursting). Plant cells take in water, developing positive turgor pressure ($\Psi_p > 0$) pushing against the rigid cellulose wall until $\Psi_{\text{cell}} = \Psi_{\text{ext}}$ (Turgid State: normal, healthy plant support).
• Isotonic Environment ($\Psi_{\text{ext}} = \Psi_{\text{cell}}$): Animal cells maintain normal morphology. Plant cells lose positive wall pressure ($\Psi_p = 0$; Flaccid State).
• Hypertonic Environment ($\Psi_{\text{ext}} < \Psi_{\text{cell}}$): Animal cells lose water and shrivel (Crenation). Plant cells lose water, collapsing the protoplast away from the cell wall ($\Psi_p = 0$; Plasmolysis, lethal).
Evolutionary Advantages of Membrane Compartmentalization:
1. Separation of Competing Metabolic Reactions: Prevents futile cycles (e.g., simultaneous fatty acid synthesis in cytosol and fatty acid $\beta$-oxidation inside peroxisomes/mitochondria).
2. Localized Enzyme & Substrate Concentration: Substrate channeling increases reaction velocity ($V$) without requiring high whole-cell solute concentrations.
3. Localized Microenvironments: Maintenance of high electrochemical proton gradients across cristae/thylakoid membranes for ATP synthesis, and acidic pH ($\approx 4.8$) in lysosomes.
Mitochondria originated from an ancestral aerobic alpha-proteobacterium, and chloroplasts from an ancestral photosynthetic cyanobacterium, engulfed by a primitive nucleated archaeal/eukaryotic ancestor.
1. Double Membrane Envelope: The inner membrane retains prokaryotic lipid composition (e.g., cardiolipin) and prokaryotic electron transport complexes; the double-membrane envelope as a whole is consistent with engulfment of a bacterium by the host (the outer membrane still carries bacterial-type β-barrel porins).
2. Circular Non-Histone DNA: Mitochondria and chloroplasts contain autonomous closed circular, double-stranded DNA genomes (mtDNA and cpDNA) lacking eukaryotic histone proteins, homologous to bacterial chromosomes.
3. Bacterial $70\text{S}$ Ribosomes: Possess distinct $70\text{S}$ ribosomes (inhibited by bacterial translation inhibitors like chloramphenicol, erythromycin, and tetracycline, but unaffected by cycloheximide, which specifically inhibits eukaryotic $80\text{S}$ cytoplasmic ribosomes!).
4. Autonomous Binary Fission: Divide independently of nuclear mitosis through a contractile division ring utilizing bacterial FtsZ-homologous cytoskeletal machinery.
Unit 2 Cell Structure and Function: İnteraktif Soru İstasyonu
Özgün AP tarzı sorular (geçmiş sınav sorusu değil). Şık seç; anında çözüm ve hata türü gelir.
AP Biology — Unit 3: Cellular Energetics
Every living cell is an open thermodynamic system that must capture, convert and spend energy. Unit 3 follows that energy from the shape of an enzyme's active site to the proton gradients that drive ATP synthase in chloroplasts and mitochondria. The topics below follow the College Board Course and Exam Description (CED Topics 3.1–3.7):
Enzyme Structure: Most enzymes are globular proteins (some catalytic RNAs, called ribozymes, also exist). The active site is a small pocket formed by specific R-groups brought together by the tertiary (and sometimes quaternary) structure. Substrate specificity depends on complementarity of both shape and chemical properties (charge, polarity, hydrophobicity) between the active site and the substrate.
• Induced fit: Substrate binding causes a small conformational change that tightens the fit and strains substrate bonds, stabilizing the transition state.
• Cofactors and coenzymes: Inorganic ions (e.g., $\text{Mg}^{2+}$, $\text{Zn}^{2+}$) or organic molecules (e.g., NAD⁺, FAD, many derived from vitamins) are required by some enzymes for activity.
• Enzymes lower the activation energy $E_a = G_{\text{transition state}} - G_{\text{reactants}}$, so more molecules reach the transition state per unit time and the reaction rate increases.
• Enzymes do not change $\Delta G = G_{\text{products}} - G_{\text{reactants}}$, do not change the equilibrium position, and are not consumed — the same enzyme molecule catalyzes the reaction many times.
• Enzymes can lower $E_a$ by orienting substrates, straining bonds, providing a favorable microenvironment (e.g., an acidic R-group), or briefly forming covalent bonds with the substrate.
Temperature: Below the optimum, raising temperature increases molecular motion and the frequency of enzyme–substrate collisions, so the rate rises. Above the optimum, thermal motion disrupts hydrogen bonds, ionic bonds and hydrophobic interactions; the enzyme denatures, the active site changes shape and activity falls sharply. Low temperature slows the reaction but usually does not denature the enzyme (activity returns on warming).
pH: A change in $[\text{H}^+]$ changes the ionization of acidic and basic R-groups, disrupting ionic and hydrogen bonds in the active site and throughout the tertiary structure. Each enzyme has an optimum pH matching its environment (pepsin ≈ 2, trypsin ≈ 8).
Substrate and enzyme concentration: Rate rises with $[S]$ until all active sites are occupied (saturation, $V_{max}$). With unlimited substrate, rate is proportional to enzyme concentration.
• Competitive inhibitor: resembles the substrate and binds the active site (usually reversibly). Its effect is overcome by raising $[S]$: $V_{max}$ unchanged, apparent $K_m$ increased.
• Noncompetitive (allosteric) inhibitor: binds a different (allosteric) site and changes the enzyme's shape. Adding substrate does not overcome it: $V_{max}$ decreased.
• Irreversible inhibitors (e.g., some nerve agents and penicillin) bind covalently and permanently inactivate the enzyme.
• Feedback inhibition: the end product of a pathway allosterically inhibits the first committed enzyme (e.g., isoleucine inhibits threonine deaminase), preventing wasteful overproduction.
• ✗ "Enzymes make reactions more exergonic." ✓ They lower $E_a$ only; $\Delta G$ is unchanged.
• ✗ "Enzymes are used up in the reaction." ✓ They are regenerated unchanged; rates fall because substrate is depleted.
• ✗ "Denaturation breaks peptide bonds." ✓ Primary structure stays intact; only weak interactions are disrupted.
• ✗ "Cold denatures enzymes like heat does." ✓ Cold mainly slows molecular motion; it is usually reversible.
First law: Energy is neither created nor destroyed; cells only transform it (light → chemical; chemical → mechanical, electrical, heat).
Second law: Every energy transformation increases the entropy of the universe; some energy is always lost as heat. Organisms maintain internal order only because they are open systems that take in energy and release heat and simpler molecules ($\text{CO}_2$, $\text{H}_2\text{O}$) to the surroundings. If energy input stops, order is lost and the organism dies.
• Metabolic pathways are sequences of enzyme-catalyzed steps; the product of one reaction is the substrate of the next. Loss of a single enzyme (e.g., by mutation) can block an entire pathway.
$$6\,\text{CO}_2 + 6\,\text{H}_2\text{O} \xrightarrow{\text{light}} \text{C}_6\text{H}_{12}\text{O}_6 + 6\,\text{O}_2$$
Photosynthesis first evolved in prokaryotes (cyanobacteria); the chloroplasts of eukaryotes descend from them.
Light-dependent reactions (thylakoid membranes):
1. Pigments (chlorophyll a, chlorophyll b, carotenoids) absorb mainly blue-violet and red light and pass energy to reaction-center chlorophylls in photosystem II (P680) and photosystem I (P700).
2. Excited electrons leave PSII and are replaced by electrons from the splitting of water (photolysis): $2\text{H}_2\text{O} \to 4\text{H}^+ + 4e^- + \text{O}_2$. All released $\text{O}_2$ comes from water.
3. Electrons pass along an electron transport chain (plastoquinone → cytochrome complex → plastocyanin) to PSI; the energy released pumps $\text{H}^+$ from the stroma into the thylakoid lumen.
4. PSI re-energizes the electrons, which reduce $\text{NADP}^+$ to NADPH on the stroma side.
Calvin Cycle (stroma; light-independent, but needs ATP + NADPH from the light reactions):
• Carbon fixation: rubisco attaches $\text{CO}_2$ to RuBP (5C) → unstable 6C → 2 × 3-phosphoglycerate (3-PGA).
• Reduction: 3-PGA is phosphorylated by ATP and reduced by NADPH to G3P.
• Regeneration: 5 G3P are rearranged, using ATP, into 3 RuBP.
• Stoichiometry: 3 $\text{CO}_2$ → 1 net G3P costs 9 ATP + 6 NADPH (18 ATP + 12 NADPH per glucose equivalent).
• ✗ "Plants do photosynthesis instead of respiration." ✓ Plant cells have mitochondria and respire continuously, day and night.
• ✗ "The O₂ released comes from CO₂." ✓ It comes from water split at PSII.
• ✗ "The Calvin cycle is the 'dark reaction' and runs at night." ✓ It depends on ATP/NADPH from the light reactions and several of its enzymes are light-activated; it runs mainly in the light.
• ✗ "Plants absorb green light." ✓ Green is mostly reflected/transmitted, which is why leaves look green.
$$\text{C}_6\text{H}_{12}\text{O}_6 + 6\,\text{O}_2 \to 6\,\text{CO}_2 + 6\,\text{H}_2\text{O} + \text{energy (ATP + heat)}$$
1. Glycolysis (cytosol, no $\text{O}_2$ required): glucose (6C) → 2 pyruvate (3C); net 2 ATP (substrate-level phosphorylation) and 2 NADH.
2. Pyruvate oxidation (mitochondrial matrix): each pyruvate → acetyl-CoA + $\text{CO}_2$ + NADH (per glucose: 2 $\text{CO}_2$, 2 NADH).
3. Krebs (citric acid) cycle (matrix): per glucose (2 turns): 4 $\text{CO}_2$, 6 NADH, 2 $\text{FADH}_2$, 2 ATP (or GTP). All six glucose carbons have now left as $\text{CO}_2$.
4. Electron transport chain (inner mitochondrial membrane / cristae): NADH and $\text{FADH}_2$ donate electrons to protein complexes; exergonic electron flow pumps $\text{H}^+$ from the matrix into the intermembrane space. $\text{O}_2$ is the final electron acceptor, forming $\text{H}_2\text{O}$.
5. Oxidative phosphorylation (chemiosmosis): $\text{H}^+$ flows back into the matrix through ATP synthase, producing most of the cell's ATP.
Inhibitors vs. Uncouplers: Cyanide or carbon monoxide blocks complex IV → electron flow, $\text{O}_2$ consumption and ATP synthesis all stop. An uncoupler (DNP; natural UCP1 in brown fat) lets $\text{H}^+$ leak across the inner membrane → the gradient collapses, ATP synthesis falls, but $\text{O}_2$ consumption rises and energy is released as heat (thermogenesis).
Fermentation (no $\text{O}_2$): pyruvate (lactic acid fermentation) or acetaldehyde (alcohol fermentation, releasing $\text{CO}_2$) is reduced by NADH, regenerating NAD⁺ so glycolysis can continue. Yield: net 2 ATP per glucose.
• ✗ "The ETC makes ATP directly." ✓ The ETC builds the $\text{H}^+$ gradient; ATP synthase makes the ATP.
• ✗ "Fermentation produces extra ATP from pyruvate." ✓ Its only job is to regenerate NAD⁺; all ATP comes from glycolysis.
• ✗ "Glucose carbons end up in water." ✓ Carbon leaves as $\text{CO}_2$; the oxygen in metabolic water comes from $\text{O}_2$.
• ✗ "Respiration only happens in animals." ✓ Plants, fungi, protists and most prokaryotes respire too (prokaryotes use their plasma membrane for the ETC).
Variation at the molecular level gives organisms a wider range of functions and affects their fitness in a given environment. Examples: different photosynthetic pigments (chlorophyll a, chlorophyll b, carotenoids, phycobilins) let organisms capture a broader range of wavelengths; enzyme variants (alleles) with different temperature optima suit populations to cold or warm habitats (e.g., LDH-B alleles in killifish); fetal vs. adult hemoglobin differ in $\text{O}_2$ affinity; organisms that can switch between aerobic respiration and fermentation survive in fluctuating $\text{O}_2$ conditions. Natural selection acts on this existing variation; it does not create the mutations "on demand."
• Independent variable: the factor the investigator changes (pH, temperature, light intensity, inhibitor concentration).
• Dependent variable: what is measured (rate of O₂ production, ET₅₀ of floating disks, absorbance, mL O₂ consumed per minute).
• Controlled variables: kept constant (enzyme and substrate concentration, volume, temperature, time).
• Negative control: a group in which no effect is expected (boiled enzyme, chloroplasts kept in the dark, glass beads in a respirometer). Positive control: a group with a known effect.
• Hypothesis / null hypothesis: state a testable, directional prediction with a biological reason (e.g., "If pH moves away from 7, catalase activity will decrease because R-group ionization changes the active site").
• Replication: use multiple trials and report means with error bars.
• $SE_{\bar{x}} = \dfrac{s}{\sqrt{n}}$; error bars of $\bar{x} \pm 2\,SE$ represent roughly a 95% confidence interval for the mean.
• Bars that do not overlap → the difference is likely statistically significant. Bars that overlap substantially → the difference may not be significant (do not claim the treatment had an effect).
• Rate = slope = $\dfrac{\Delta y}{\Delta x}$ with units (e.g., mL O₂/min). The initial rate uses the first, linear part of the curve. Always subtract the control (e.g., glass-bead respirometer) before calculating.
• Use cautious language: "The data support / are consistent with…", never "prove."
Unit 3 Cellular Energetics: İnteraktif Soru İstasyonu
Özgün AP tarzı sorular (geçmiş sınav sorusu değil). Şık seç; anında çözüm ve hata türü gelir.
AP Biology — Unit 4: Cell Communication and Cell Cycle
Cells constantly send, receive and interpret signals, and they divide only when internal and external conditions are right. Unit 4 connects these two ideas: a growth factor binding a receptor at the surface can switch on the cyclin–CDK machinery that drives a cell through its cycle, and a single mutation in that signaling pathway can produce cancer. The topics below follow the College Board Course and Exam Description (CED Topics 4.1–4.7):
Cells communicate by generating, transmitting, receiving and responding to chemical signals (ligands). The type of signaling is defined by the distance the signal travels:
• Direct contact: signals pass through cell junctions that connect the cytoplasm of adjacent cells (gap junctions in animals, plasmodesmata in plants), or membrane-bound molecules on one cell bind receptors on another (cell–cell recognition, e.g., an antigen-presenting cell displaying an MHC–antigen complex to a helper T cell).
• Local signaling (paracrine and synaptic): a secreted ligand acts on nearby target cells. Paracrine examples include growth factors; in synaptic signaling a neuron releases neurotransmitters across a narrow synaptic cleft.
• Long-distance (endocrine) signaling: hormones are released into the circulatory system (or plant vascular tissue) and act on distant target cells that carry the matching receptor (e.g., insulin, epinephrine, testosterone).
Every signal transduction pathway has three stages:
1. Reception: the ligand binds a specific receptor protein, changing the receptor's shape. Binding is specific (shape and chemical complementarity), like enzyme–substrate binding.
2. Transduction: the conformational change starts a sequence of molecular interactions (a signaling cascade) that relays and usually amplifies the signal inside the cell.
3. Response: the cascade changes cell behavior — activation of an enzyme, opening of a channel, rearrangement of the cytoskeleton, or (very often) a change in gene expression, cell growth, secretion or apoptosis.
• Hydrophilic, large or charged ligands (peptide hormones such as insulin, epinephrine, neurotransmitters) cannot cross the bilayer and bind plasma membrane receptors: G protein–coupled receptors (GPCRs), receptor tyrosine kinases (RTKs) and ligand-gated ion channels.
• Small nonpolar ligands (steroid hormones such as testosterone, estrogen, aldosterone; thyroid hormone; NO) diffuse through the membrane and bind intracellular receptors; the hormone–receptor complex usually acts as a transcription factor.
• The same ligand can cause different responses in different cells because the cells have different receptors or different transduction and response proteins (e.g., acetylcholine contracts skeletal muscle but slows heart rate).
G protein–coupled receptor (epinephrine → glycogen breakdown in liver):
1. Epinephrine binds the GPCR → the receptor activates a G protein, which exchanges GDP for GTP.
2. The active G protein activates adenylyl cyclase, which converts ATP into the second messenger cyclic AMP (cAMP).
3. cAMP activates protein kinase A (PKA), which starts a phosphorylation cascade (PKA → phosphorylase kinase → glycogen phosphorylase).
4. Response: glycogen is broken down to glucose-1-phosphate; glucose is released into the blood.
Switching off: the G protein hydrolyzes its own GTP to GDP (intrinsic GTPase activity); phosphodiesterase converts cAMP to AMP; protein phosphatases remove phosphates from proteins. Without these, the response would continue after the signal is gone.
• Receptor tyrosine kinases (e.g., insulin, growth factors): ligand binding → two receptor monomers dimerize → each phosphorylates tyrosines on the other (cross-phosphorylation) → relay proteins such as Ras bind and start a MAP kinase cascade that changes gene expression and cell division.
• Ca²⁺ and IP₃: phospholipase C cleaves PIP₂ into IP₃ and DAG; IP₃ opens ligand-gated Ca²⁺ channels in the ER, and the rise in cytosolic Ca²⁺ activates proteins such as calmodulin.
• Ligand-gated ion channels: neurotransmitter binding opens the channel, ions (Na⁺, Ca²⁺) flow, and membrane potential changes within milliseconds.
• Amplification: each step activates many molecules of the next. If 1 receptor activates 10 G proteins and each adenylyl cyclase makes 100 cAMP, one hormone molecule yields about 1000 cAMP — a tiny hormone concentration produces a large response.
• ✗ "The hormone enters the cell and does the job itself." ✓ Most hydrophilic ligands never enter; the receptor and second messengers relay the information.
• ✗ "Kinases remove phosphates." ✓ Kinases add phosphate groups (from ATP); phosphatases remove them.
• ✗ "cAMP is a hormone." ✓ cAMP is an intracellular second messenger, made inside the target cell.
A change in any component — receptor, relay protein, second messenger or response protein — changes the response. Typical AP reasoning: identify the blocked or overactive step, then predict every step downstream of it.
• Receptor mutations can prevent ligand binding (no response) or make the receptor active without ligand (constant response).
• Ras that cannot hydrolyze GTP stays "on": the MAP kinase cascade signals for division even without growth factor (an oncogene).
• Cholera toxin modifies the G protein so it cannot hydrolyze GTP → adenylyl cyclase stays active → cAMP stays high → intestinal Cl⁻ channels (CFTR) stay open → Cl⁻ and water are lost (severe diarrhea).
• Drugs act on specific steps: β-blockers block epinephrine receptors in the heart; some cancer drugs inhibit receptor tyrosine kinases; caffeine inhibits phosphodiesterase, prolonging the cAMP signal.
Negative feedback returns a variable to its set point and maintains homeostasis: the response opposes the stimulus. Examples: blood glucose regulation (high glucose → insulin → glucose uptake and glycogen synthesis → glucose falls; low glucose → glucagon → glycogen breakdown), body temperature (sweating, shivering), and hormone axes such as hypothalamus → TSH → thyroxine, where thyroxine inhibits further TSH release.
Positive feedback amplifies a response and moves the variable further from its starting point, until an outside event ends the loop. Examples: oxytocin and uterine contractions during childbirth, blood clotting (activated platelets recruit more platelets), fruit ripening (ethylene stimulates more ethylene production), lactation.
• Type 1 diabetes: autoimmune destruction of pancreatic β cells → no insulin → the negative feedback loop is broken; blood glucose stays high.
• Type 2 diabetes: target cells respond poorly to insulin (insulin resistance); insulin may be normal or high, but glucose is not lowered effectively.
• Removing a gland that produces an inhibitory hormone (e.g., thyroidectomy) removes the negative feedback, so the upstream stimulating hormone (TSH) rises.
Interphase (about 90% of the cycle):
• G₁: cell growth, protein synthesis, organelle duplication; DNA content = 2c.
• S: DNA replication; each chromosome now consists of two identical sister chromatids joined at the centromere (DNA content 2c → 4c, chromosome number unchanged).
• G₂: further growth and preparation for division (4c).
• G₀: a non-dividing state; many differentiated cells (neurons, most muscle cells) remain in G₀ indefinitely.
Mitosis (M phase): prophase (chromosomes condense, spindle forms) → prometaphase (nuclear envelope breaks down, kinetochores attach to spindle microtubules) → metaphase (chromosomes align at the metaphase plate) → anaphase (sister chromatids separate and move to opposite poles) → telophase (nuclear envelopes re-form). Cytokinesis divides the cytoplasm: a cleavage furrow in animal cells, a cell plate in plant cells.
Mitotic index: (cells in mitosis ÷ total cells counted). If the total cycle length is known, the time spent in a phase ≈ (fraction of cells in that phase) × (cycle length).
• ✗ "Chromosome number doubles in S phase." ✓ DNA content doubles; chromosome number stays the same until anaphase.
• ✗ "Interphase is a resting phase." ✓ It is the most metabolically active part of the cycle.
• ✗ "Homologous chromosomes pair and separate in mitosis." ✓ That happens in meiosis I (Unit 5); mitosis separates sister chromatids.
Checkpoints are internal control points that halt the cycle until conditions are met:
• G₁ checkpoint ("restriction point"): Is the cell large enough? Are nutrients and growth factors present? Is DNA undamaged? If not, the cell may enter G₀.
• G₂ checkpoint: Has DNA replication been completed correctly? Is DNA damaged?
• M (spindle assembly) checkpoint: Are all kinetochores attached to spindle microtubules? Anaphase begins only when they are.
Cyclins and CDKs: cyclin-dependent kinases (CDKs) are present at constant levels but are active only when bound to a cyclin. Cyclin concentrations rise and fall during the cycle (synthesis, then targeted degradation). Example: MPF (maturation/M-phase promoting factor, a cyclin–CDK complex) peaks at the G₂→M transition and triggers mitosis; cyclin degradation in anaphase switches MPF off. External signals (growth factors such as PDGF) and density-dependent inhibition also regulate division.
• p53 ("guardian of the genome") is a tumor-suppressor transcription factor activated by DNA damage. It induces p21, a CDK inhibitor, which halts the cycle at G₁ for repair; if damage is severe, p53 triggers apoptosis. Loss of p53 lets damaged cells keep dividing.
• Cancer results from accumulated mutations: gain-of-function mutations convert proto-oncogenes (e.g., ras, growth-factor receptors) into oncogenes; loss-of-function mutations inactivate tumor suppressors (e.g., p53, Rb). Cancer cells lose density-dependent inhibition and anchorage dependence, and often divide indefinitely.
• Apoptosis is programmed, regulated cell death: caspases (proteases) dismantle the cell from within and the fragments are engulfed by phagocytes without inflammation. It shapes development (removal of tissue between fingers; 131 cells in C. elegans), removes infected or damaged cells, and contrasts with necrosis (uncontrolled cell bursting).
• Describe = name the components in order: ligand → receptor → relay proteins/second messenger → response. Name the response explicitly (e.g., "glycogen phosphorylase is activated and glycogen is hydrolyzed").
• Predict the effect of a mutation or drug: state which step is blocked or stuck "on," then state the consequence downstream (and note that upstream steps are unaffected).
• Justify a prediction: link cause and effect with a mechanism ("because the G protein cannot hydrolyze GTP, adenylyl cyclase remains active and cAMP stays high").
• Feedback: identify the stimulus, the sensor/control center, the effector and whether the response opposes (negative) or reinforces (positive) the change.
• Flow cytometry / DNA content histograms: 2c peak = G₁ (or G₀), between 2c and 4c = S, 4c peak = G₂ or M. A drug that makes the 4c peak grow arrests cells in G₂/M; one that enlarges the 2c peak arrests in G₁.
• Mitotic index and phase duration: show the calculation with units; the longest phase contains the most cells.
• Cyclin/MPF graphs: describe the trend ("cyclin accumulates during interphase and falls sharply at anaphase") and connect it to CDK activation.
• Use "the data support…" language; propose controls (e.g., untreated cells, cells treated with the drug solvent only).
Unit 4 Cell Communication and Cell Cycle: İnteraktif Soru İstasyonu
Özgün AP tarzı sorular (geçmiş sınav sorusu değil). Şık seç; anında çözüm ve hata türü gelir.
AP Biology — Unit 5: Heredity
Heredity explains how genetic information is packaged into gametes, recombined at fertilization and expressed as phenotype. Unit 5 starts with the chromosome mechanics of meiosis, builds probability rules from Mendel's laws, extends them to inheritance patterns that do not follow simple dominance, and ends with the chromosomal basis of genetic disorders and pedigree analysis. The topics below follow the College Board Course and Exam Description (CED Topics 5.1–5.6):
Meiosis is a pair of divisions (meiosis I and meiosis II) that follows a single round of DNA replication and produces four haploid (n) cells from one diploid (2n) cell. Fertilization of two haploid gametes restores the diploid number, so chromosome number stays constant from generation to generation.
• Meiosis I (reductional division): homologous chromosomes pair up (synapsis) in prophase I and form tetrads (bivalents); crossing over occurs; homologous pairs line up at the metaphase plate in metaphase I; homologs separate in anaphase I. Each daughter cell is haploid (n) but each chromosome still has two sister chromatids.
• Meiosis II (equational division): resembles mitosis; sister chromatids separate in anaphase II, giving four haploid cells with unreplicated chromosomes.
• G₁ of the parent cell: 8 chromosomes, 8 chromatids (DNA = 2c).
• After S phase / metaphase I: 8 chromosomes, 16 chromatids (4c), arranged as 4 tetrads.
• Each cell after meiosis I: 4 chromosomes, 8 chromatids (2c) — the chromosome number has already been halved.
• Each cell after meiosis II: 4 chromosomes, 4 chromatids (1c).
• Mitosis: one division, 2 genetically identical diploid cells, no synapsis, used for growth, repair and asexual reproduction.
• Meiosis: two divisions, 4 genetically different haploid cells, synapsis and crossing over in prophase I, used for sexual reproduction.
• ✗ "Chromosome number is halved in meiosis II." ✓ It is halved in meiosis I, when homologs separate.
Three processes make every gamete — and therefore every offspring — genetically unique:
1. Crossing over (prophase I): non-sister chromatids of homologous chromosomes exchange segments at chiasmata, producing recombinant chromatids that carry new combinations of alleles on the same chromosome.
2. Independent assortment (metaphase I): each homologous pair orients randomly at the metaphase plate, independently of the other pairs. The number of possible chromosome combinations in gametes is $2^n$, where $n$ is the haploid number. For humans ($n = 23$): $2^{23} \approx 8.4$ million combinations, before crossing over is even considered.
3. Random fertilization: any sperm can fuse with any egg, so one human couple can produce $2^{23} \times 2^{23} \approx 7 \times 10^{13}$ zygote combinations.
Law of segregation: the two alleles of a gene separate during gamete formation (anaphase I), so each gamete carries one allele. Law of independent assortment: alleles of genes on different chromosomes (or far apart on the same chromosome) segregate independently of each other.
• Monohybrid cross $Aa \times Aa$: genotypic ratio 1 $AA$ : 2 $Aa$ : 1 $aa$; phenotypic ratio 3 dominant : 1 recessive. Among offspring with the dominant phenotype, $\frac{2}{3}$ are heterozygous.
• Dihybrid cross $AaBb \times AaBb$: phenotypic ratio 9 : 3 : 3 : 1 ($A\_B\_$ : $A\_bb$ : $aaB\_$ : $aabb$).
• Test cross: an individual with the dominant phenotype is crossed with a homozygous recessive. All dominant offspring ⟹ the parent was homozygous ($AA$); a 1 : 1 ratio ⟹ heterozygous ($Aa$). For $AaBb \times aabb$ with unlinked genes, expect 1 : 1 : 1 : 1.
• Product rule ("and"): multiply probabilities of independent events. For $AaBbCc \times AaBbCc$, $P(AaBbcc) = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{4} = \frac{1}{16}$.
• Sum rule ("or"): add probabilities of mutually exclusive events. $P(\text{exactly one dominant trait})$ in a dihybrid cross $= \frac{3}{16} + \frac{3}{16} = \frac{3}{8}$.
• Several children: if order does not matter, multiply by the number of arrangements. $P(\text{exactly 2 of 3 children affected})$ with $p = \frac{1}{4}$: $3 \times \left(\frac{1}{4}\right)^2 \times \frac{3}{4} = \frac{9}{64}$.
• ✗ "Dominant alleles are more common in a population." ✓ Dominance describes the phenotype of the heterozygote, not allele frequency.
• ✗ "If the first three children are affected, the fourth is less likely to be affected." ✓ Each fertilization is an independent event; the probability is the same every time.
• ✗ "A_ means homozygous dominant." ✓ $A\_$ means $AA$ or $Aa$ (dominant phenotype).
Linked genes and recombination frequency: genes close together on the same chromosome tend to be inherited together, so a dihybrid test cross gives an excess of parental phenotypes and few recombinant phenotypes instead of 1 : 1 : 1 : 1.
• Recombination frequency $= \frac{\text{number of recombinant offspring}}{\text{total offspring}} \times 100\%$; 1% recombination = 1 map unit (centimorgan, cM).
• Example: test cross offspring $AB$ 420, $ab$ 400, $Ab$ 90, $aB$ 90 ⟹ RF $= \frac{180}{1000} = 18\%$ ⟹ 18 map units.
• Gene order: the two genes with the largest distance are at the ends (A–B 12, B–C 7, A–C 19 ⟹ order A–B–C). Maximum observable RF is 50% (equal to independent assortment).
Sex-linked (X-linked) inheritance: males ($XY$) are hemizygous for X-linked genes, so one recessive allele (e.g., hemophilia, red–green color blindness) produces the phenotype. Sons receive their X from their mother; fathers pass their X to every daughter and never to sons.
• $X^HX^h \times X^HY$: $\frac{1}{2}$ of sons affected, no daughters affected, $\frac{1}{2}$ of daughters carriers; $P(\text{child is an affected son}) = \frac{1}{4}$.
• Incomplete dominance: heterozygote has an intermediate phenotype (red × white snapdragon ⟹ pink; pink × pink ⟹ 1 red : 2 pink : 1 white — the phenotypic ratio equals the genotypic ratio).
• Codominance: both alleles are fully expressed in the heterozygote (blood type AB: $I^AI^B$ produces both A and B antigens).
• Multiple alleles: more than two alleles exist in the population ($I^A$, $I^B$, $i$); $I^A$ and $I^B$ are codominant, both dominant to $i$.
• Epistasis: one gene masks the expression of another. Labrador coat: $B$ (black) / $b$ (chocolate) pigment, but $ee$ blocks pigment deposition ⟹ yellow. $BbEe \times BbEe$ ⟹ 9 black : 3 chocolate : 4 yellow.
• Polygenic inheritance: many genes with additive effects produce continuous variation (height, skin color). For $AaBbCc \times AaBbCc$, the number of dominant alleles follows the binomial pattern 1 : 6 : 15 : 20 : 15 : 6 : 1 (out of 64).
• Mitochondrial (non-nuclear) inheritance: mitochondria and chloroplasts carry their own DNA and are inherited through the egg cytoplasm. An affected mother passes the trait to all her children; an affected father passes it to none.
• ✗ "Recombination frequency = recombinants ÷ parentals." ✓ Always divide by the total number of offspring.
• ✗ "Only one recombinant class counts." ✓ Add both recombinant classes ($Ab$ and $aB$).
• ✗ "Daughters can never show an X-linked recessive trait." ✓ A daughter is affected if her father is affected and her mother is at least a carrier.
The phenotype is the product of the genotype and the environment. The same genotype can produce different phenotypes in different environments (phenotypic plasticity):
• Hydrangea flower color depends on soil pH (acidic soil, more available aluminum ⟹ blue; alkaline soil ⟹ pink).
• Himalayan rabbits and Siamese cats: a temperature-sensitive enzyme for pigment synthesis works only in cooler body regions, so ears, paws and nose are dark.
• Sex determination by temperature in many reptiles (turtle egg incubation temperature).
• Phenylketonuria (PKU): individuals homozygous for the mutant allele develop normally on a low-phenylalanine diet — an environmental intervention changes the phenotype of a genetic disorder.
Nondisjunction is the failure of chromosomes to separate correctly:
• In meiosis I (homologs fail to separate): all four gametes are abnormal — two are $n+1$ and two are $n-1$.
• In meiosis II (sister chromatids fail to separate in one cell): two gametes are normal ($n$), one is $n+1$ and one is $n-1$.
• Fertilization of an abnormal gamete produces aneuploidy: trisomy 21 (Down syndrome), XXY (Klinefelter syndrome), XO (Turner syndrome). A karyotype detects these changes.
Pedigree analysis — reading the inheritance pattern:
• Autosomal recessive: can skip generations; two unaffected parents can have an affected child (both are carriers); males and females affected about equally (e.g., cystic fibrosis, sickle-cell disease).
• Autosomal dominant: every affected individual has at least one affected parent; does not skip generations (e.g., Huntington's disease).
• X-linked recessive: mostly males affected; affected males usually have unaffected carrier mothers; an affected daughter requires an affected father.
• Mitochondrial: passed only from mothers to all their children.
The chi-square test asks whether the difference between observed (O) and expected (E) offspring counts is small enough to be explained by random chance.
• Null hypothesis ($H_0$): there is no statistically significant difference between observed and expected counts; the cross follows the predicted ratio (e.g., 9 : 3 : 3 : 1 for unlinked genes).
• Formula: $\chi^2 = \sum \frac{(O - E)^2}{E}$. Expected counts are calculated from the total offspring and the predicted ratio.
• Degrees of freedom: $df = (\text{number of phenotypic classes}) - 1$.
• Compare with the critical value at $p = 0.05$: df = 1 → 3.84; df = 2 → 5.99; df = 3 → 7.81.
• If $\chi^2 \le$ critical value: fail to reject $H_0$ — the deviation can be explained by chance. If $\chi^2 >$ critical value: reject $H_0$ — the data do not fit the predicted ratio (for example, the genes may be linked).
Observed 98 : 28 : 26 : 8; expected (9 : 3 : 3 : 1) 90 : 30 : 30 : 10.
$\chi^2 = \frac{8^2}{90} + \frac{(-2)^2}{30} + \frac{(-4)^2}{30} + \frac{(-2)^2}{10} = 0.711 + 0.133 + 0.533 + 0.400 \approx 1.78$.
$df = 4 - 1 = 3$; critical value 7.81. Because $1.78 < 7.81$, we fail to reject $H_0$: the data are consistent with independent assortment.
• Dividing by $O$ instead of $E$, or forgetting to divide at all.
• Using the number of classes instead of (classes − 1) as df.
• Saying "we accept / prove $H_0$." ✓ The correct wording is "fail to reject $H_0$."
• A large $\chi^2$ does not tell you why the data deviate; you must propose a biological explanation (linkage, lethal genotype, epistasis).
Unit 5 Heredity: İnteraktif Soru İstasyonu
Özgün AP tarzı sorular (geçmiş sınav sorusu değil). Şık seç; anında çözüm ve hata türü gelir.
AP Biology — Unit 6: Gene Expression and Regulation
Unit 6 follows genetic information from its chemical structure to its use. DNA is copied faithfully before every division, transcribed into RNA, processed, and translated into polypeptides; cells then decide which genes to express, when, and how strongly. Mutations change the information, and biotechnology lets us read, copy and move it. The topics below follow the College Board Course and Exam Description (CED Topics 6.1–6.8):
DNA is a double helix of two antiparallel polynucleotide strands: one runs 5′→3′, the other 3′→5′. Each nucleotide contains a deoxyribose sugar, a phosphate group and a nitrogenous base. The sugar–phosphate backbone is held together by covalent phosphodiester bonds; the two strands are held together by hydrogen bonds between complementary bases.
• Purines (two rings): adenine (A) and guanine (G). Pyrimidines (one ring): cytosine (C), thymine (T) in DNA and uracil (U) in RNA.
• A pairs with T (2 hydrogen bonds); G pairs with C (3 hydrogen bonds). Because a purine always pairs with a pyrimidine, the helix has a uniform diameter.
• Chargaff's rules (double-stranded DNA): %A = %T and %G = %C. If A = 18%, then T = 18% and G = C = (100 − 36) / 2 = 32%.
• RNA is usually single-stranded, contains ribose (with a 2′-OH group) and uses uracil instead of thymine. Major types: mRNA (carries the code), tRNA (carries amino acids), rRNA (structural and catalytic core of the ribosome).
• Prokaryotes: usually one circular chromosome in the nucleoid, plus small circular plasmids that often carry antibiotic-resistance genes.
• Eukaryotes: multiple linear chromosomes wrapped around histone proteins (chromatin) inside the nucleus; mitochondria and chloroplasts keep their own circular DNA.
• ✗ "Chargaff's rules apply to every nucleic acid." ✓ They apply only to double-stranded molecules; a single-stranded RNA or DNA virus can have A ≠ U (or T) and G ≠ C.
• ✗ "A–T pairs are held by three hydrogen bonds." ✓ A–T has two, G–C has three, so GC-rich DNA needs more heat to separate.
DNA replication is semiconservative: each daughter molecule contains one original (parental) strand and one newly synthesized strand. This was shown by the Meselson–Stahl experiment (¹⁵N → ¹⁴N density gradients).
The enzymes, in order of action:
• Helicase unwinds the double helix at the replication fork by breaking hydrogen bonds between base pairs.
• Topoisomerase relieves the supercoiling (twisting strain) ahead of the fork by cutting and rejoining the backbone.
• Single-strand binding proteins keep the separated strands apart.
• Primase builds a short RNA primer, because DNA polymerase cannot start a strand from scratch.
• DNA polymerase adds nucleotides only to a free 3′-OH end, so new DNA always grows 5′→3′. It also proofreads.
• DNA ligase seals the gaps (forms the final phosphodiester bonds) between DNA fragments.
• The leading strand is synthesized continuously toward the fork from a single primer.
• The lagging strand is synthesized discontinuously, away from the fork, as short Okazaki fragments, each needing its own primer. The RNA primers are later replaced with DNA and the fragments are joined by ligase.
• Meselson–Stahl prediction for semiconservative replication: after 1 generation in ¹⁴N, 100% hybrid (¹⁵N/¹⁴N) DNA; after 2 generations, 50% hybrid and 50% light; after n generations, the hybrid fraction is 2 / 2ⁿ.
• ✗ "The lagging strand is synthesized 3′→5′." ✓ Both strands are synthesized 5′→3′; the lagging strand is discontinuous because it must be made away from the fork.
• ✗ "Ligase joins the two parental strands." ✓ Ligase joins Okazaki fragments (new DNA to new DNA) on the same strand.
Transcription copies the information of a gene into RNA. RNA polymerase binds the promoter (in eukaryotes, with the help of transcription factors, often at a TATA box), unwinds the DNA and reads the template (antisense) strand 3′→5′, building RNA 5′→3′. No primer is needed.
• The RNA sequence is complementary to the template strand and identical to the coding (sense, non-template) strand, with U in place of T.
• Example: template 3′-TACGGATTC-5′ ⟹ mRNA 5′-AUGCCUAAG-3′; coding strand 5′-ATGCCTAAG-3′.
Eukaryotic RNA processing (in the nucleus, before export):
• A modified guanine 5′ cap is added: it protects the mRNA from degradation and helps the ribosome attach.
• A poly-A tail (50–250 adenines) is added to the 3′ end: it increases stability and helps export.
• Splicing: the spliceosome removes introns and joins exons. A gene with exons of 120, 85 and 200 nucleotides and introns of 300 and 150 nucleotides gives a pre-mRNA of 855 nt but a mature coding region of only 405 nt.
• ✗ "mRNA has the same sequence as the template strand." ✓ It matches the coding strand (T → U).
• ✗ "Introns are removed in the cytoplasm." ✓ Splicing occurs in the nucleus; only processed mRNA is exported.
Translation converts the mRNA nucleotide sequence into an amino acid sequence at the ribosome. The mRNA is read 5′→3′ in non-overlapping triplets called codons.
• Initiation: the small subunit binds the mRNA and the initiator tRNA (Met) pairs with the start codon AUG; the large subunit joins.
• Elongation: charged tRNAs enter the A site; their anticodon pairs antiparallel with the codon. The rRNA of the large subunit catalyzes the peptide bond (the growing chain moves to the new tRNA in the A site); the ribosome translocates so tRNAs move A → P → E.
• Termination: a stop codon (UAA, UAG, UGA) enters the A site; a release factor binds and the polypeptide is released. Stop codons do not code for an amino acid.
• 64 codons: 61 specify amino acids, 3 are stops. The code is redundant (most amino acids have several codons, often differing at the third base) but not ambiguous (each codon has one meaning), and it is nearly universal — which is why a human gene can be expressed in bacteria.
• Example: 5′-AUG UUC GGA CAU UGG UAG-3′ ⟹ Met-Phe-Gly-His-Trp (UAG = stop).
• Anticodon to codon: anticodon 5′-GCA-3′ pairs with codon 5′-UGC-3′ (complement and reverse) ⟹ this tRNA carries Cys.
• Polypeptide length: (first base of stop codon − first base of AUG) ÷ 3. AUG at nucleotide 51 and stop starting at 951 ⟹ 900 ÷ 3 = 300 amino acids.
• ✗ "Reading the anticodon in the codon table gives the amino acid." ✓ The codon table is written for mRNA codons (5′→3′).
• ✗ "The stop codon adds a final amino acid." ✓ No tRNA matches a stop codon; a release factor ends translation.
Prokaryotic operons: an operon is a cluster of genes under one promoter and one operator (the repressor binding site), transcribed as a single mRNA.
• lac operon (inducible): the repressor (product of lacI) is active by itself and binds the operator, blocking RNA polymerase. When lactose is present, allolactose binds the repressor and changes its shape, so it leaves the operator and the genes are transcribed. Positive control: when glucose is low, cAMP rises, cAMP–CAP binds near the promoter and strongly increases transcription. Maximum expression = lactose present and glucose absent.
• trp operon (repressible): the repressor is inactive by itself. When tryptophan is abundant, tryptophan acts as a corepressor: it binds the repressor, activates it, and the repressor blocks the operator — the cell stops making an amino acid it already has.
Eukaryotic regulation acts at many levels:
• Chromatin / epigenetics: histone acetylation loosens chromatin and increases transcription; DNA methylation (often at promoters) generally silences genes. These marks can be inherited by daughter cells without changing the DNA sequence.
• Transcription factors bind regulatory sequences (enhancers, proximal control elements) and recruit or block RNA polymerase.
• Post-transcriptional: alternative splicing, mRNA stability, and small regulatory RNAs (miRNA, siRNA) that bind mRNA and block translation or trigger degradation.
• Repressor that cannot bind the operator (or an operator that cannot be bound) ⟹ constitutive expression (always on).
• lac repressor that cannot bind allolactose ⟹ always off ("super-repressor").
• trp repressor that cannot bind tryptophan ⟹ always on, even when tryptophan is high.
• Data clue: if mRNA levels stay the same but protein levels fall, regulation is after transcription (e.g., a miRNA).
• ✗ "Lactose activates the lac repressor." ✓ Allolactose inactivates it; tryptophan activates the trp repressor.
• ✗ "Correlation proves causation." ✓ High promoter methylation in a non-expressing tissue is correlation; a demethylating-drug experiment tests causation.
Almost every somatic cell of a multicellular organism carries the same genome. Cells become different (liver cell, neuron, muscle cell) because they express different subsets of genes — differential gene expression.
• Each gene has a particular combination of control elements. A cell type expresses a gene only when it contains the right combination of specific transcription factors (combinatorial control). A small number of transcription factors can therefore produce many unique expression patterns.
• Example: Gene 1 needs TF1 + TF3, Gene 2 needs TF2 + TF4, Gene 3 needs TF1 + TF2. A cell containing TF1, TF2 and TF3 expresses Genes 1 and 3 only.
• During development, signals and cytoplasmic determinants switch on master regulatory genes (e.g., MyoD for muscle, Hox genes for body-plan identity) that turn on whole sets of downstream genes.
A mutation is a change in the DNA sequence. Mutations arise from replication errors, radiation (UV, X-rays) and chemical mutagens. Their effect on the phenotype depends on how they change the protein:
• Silent: the codon changes but codes for the same amino acid (GAG → GAA, both Glu).
• Missense: one amino acid is replaced (GAG → GUG, Glu → Val in sickle-cell hemoglobin).
• Nonsense: a premature stop codon (UGG → UGA) gives a truncated, usually nonfunctional protein.
• Frameshift: an insertion or deletion of a number of bases that is not a multiple of 3 shifts the reading frame; every codon after the change is read differently. An in-frame deletion of 3 bases removes only one amino acid.
Mutations are the original source of new alleles and therefore of genetic variation; whether a mutation is harmful, neutral or beneficial depends on the environment.
• Transformation: uptake of naked DNA from the environment.
• Transduction: DNA moved from one bacterium to another by a bacteriophage.
• Conjugation: direct transfer of DNA (often a plasmid) through a pilus.
• Viruses with RNA genomes mutate quickly because RNA replicases lack proofreading. Errors in meiosis (nondisjunction, Topic 5.6) and polyploidy also change genetic information.
• ✗ "A frameshift changes only one amino acid." ✓ It changes the whole reading frame downstream and often introduces an early stop codon.
• ✗ "Organisms mutate because they need to adapt." ✓ Mutations are random; selection acts on them afterward.
Polymerase chain reaction (PCR) amplifies a target DNA region. Each cycle has three steps: denaturation (~95 °C, strands separate), annealing (~55–65 °C, primers bind the flanking sequences) and extension (~72 °C, heat-stable Taq polymerase extends the primers). Each cycle doubles the number of target copies: copies $= N_0 \times 2^n$. Five starting molecules after 20 cycles give $5 \times 2^{20} = 5\,242\,880$ copies.
Gel electrophoresis separates DNA fragments by size. DNA is negatively charged (phosphate backbone), so it moves toward the positive electrode (anode); smaller fragments travel farther through the agarose matrix. Fragment size is estimated from a DNA ladder; distance traveled is roughly linear with the logarithm of size.
Restriction enzymes cut DNA at specific (often palindromic) recognition sequences, producing sticky ends. A circular plasmid cut at k sites gives k fragments; a linear molecule cut at k sites gives k + 1 fragments.
Bacterial transformation introduces a recombinant plasmid into bacteria. Only transformed cells grow on a plate with the antibiotic for which the plasmid carries a resistance gene. DNA sequencing determines the exact nucleotide order, used for diagnosis, phylogeny and forensics.
• ✗ "Large fragments move faster because they carry more charge." ✓ Charge per length is constant; the gel slows large fragments more.
• ✗ "PCR copies the entire genome." ✓ Only the region between the two primers is amplified exponentially.
Unit 6 free-response questions usually give a short experiment and ask you to identify variables, predict results and justify them with a molecular mechanism.
• Operon experiments: state the condition (lactose / glucose / tryptophan), the state of the repressor (bound or not bound to the operator), and the outcome for transcription. One complete chain earns the point: "Allolactose binds the repressor ⟹ repressor leaves the operator ⟹ RNA polymerase transcribes lacZ."
• Gene-expression data: separate mRNA levels (transcription) from protein levels (translation / protein stability). Unchanged mRNA with lower protein points to post-transcriptional control.
• Mutation questions: write the normal and mutant codons, translate both with the codon table, then name the mutation type and predict the effect on protein structure and function.
• Gel questions: state that DNA moves toward the positive electrode, that smaller fragments move farther, and compare unknown bands with the ladder or with reference samples (paternity, forensics).
• Independent variable: the factor you change (e.g., presence of a demethylating drug).
• Dependent variable: what you measure (mRNA by RT-PCR, protein by Western blot, colony color).
• Negative control: e.g., bacteria given no plasmid, which should not grow on ampicillin. Positive control: e.g., bacteria plated without ampicillin, which show that the cells were alive.
• Constants: same strain, temperature, incubation time and DNA amount; use replicates.
• "The gene is turned on" earns no point without the mechanism (which molecule binds where).
• Predicting a result without justification earns at most half the credit in "predict and justify" prompts.
Unit 6 Gene Expression and Regulation: İnteraktif Soru İstasyonu
Özgün AP tarzı sorular (geçmiş sınav sorusu değil). Şık seç; anında çözüm ve hata türü gelir.
AP Biology — Unit 7: Natural Selection
Unit 7 is the largest unit on the AP Biology exam (13–20%). It explains how populations change over time: natural selection acts on heritable variation, random processes such as genetic drift shift allele frequencies, and the Hardy–Weinberg model gives the null hypothesis against which real populations are tested. You will also read evidence for evolution, build and interpret phylogenetic trees, explain how new species arise, and describe the scientific evidence for the origin of life. The topics below follow the College Board Course and Exam Description (CED effective Fall 2025, Topics 7.1–7.12). Throughout, remember the CED's warning: natural selection acts on individuals, but it is populations that evolve, and every claim must avoid Lamarckian language ("the giraffe stretched its neck, so its offspring had long necks").
Natural selection is a major mechanism of evolution. Darwin's argument rests on four observations that can be checked in any population:
• Overproduction: organisms produce more offspring than the environment can support.
• Competition: resources (food, mates, nesting sites) are limited, so individuals compete.
• Heritable variation: individuals differ in traits, and many of these differences are passed to offspring.
• Differential survival and reproduction: individuals with more favorable phenotypes are more likely to survive and leave more offspring, so the favorable alleles become more common in the next generation.
• Fitness is measured by the number of viable, fertile offspring an individual contributes to the next generation, not by strength, size or lifespan.
• Example: a male bird that lives 9 years and fathers 4 surviving chicks has lower fitness than a male that lives 3 years and fathers 11 surviving chicks.
• Biotic and abiotic environments fluctuate (drought years, new predators, disease), so the direction and rate of selection can change from one generation to the next. A phenotype favored this year may be neutral or harmful next year.
• ✗ "Survival of the fittest means the strongest survive." ✓ Fitness is reproductive success in a particular environment.
• ✗ "Individuals evolve during their lifetime." ✓ Individuals are selected; the allele frequencies of the population change across generations.
Natural selection acts on phenotypic variation. A changing environment applies selective pressures (a new predator, a pesticide, a shift in flowering season under climate change). Some phenotypic variations raise fitness in a particular environment, others lower it.
• Molecular variation matters too: variation in the number and types of molecules within cells (for example, several forms of an enzyme, or different pigments) gives a population a greater ability to survive and reproduce in different environments.
• Sickle-cell allele (HbS): homozygotes (HbSHbS) suffer sickle-cell disease, but heterozygotes (HbAHbS) are more resistant to malaria. Where malaria is common, heterozygotes have the highest fitness (heterozygote advantage), so the allele stays at a relatively high frequency.
• DDT resistance in insects: resistant individuals already existed before spraying; the pesticide removed the susceptible ones, so the resistant allele increased in frequency.
• Directional: one extreme is favored; the mean shifts (e.g., darker moths on soot-covered trees).
• Stabilizing: intermediate phenotypes are favored; the mean stays, the variance narrows (e.g., human birth weight).
• Disruptive: both extremes are favored over the intermediate; the distribution becomes bimodal (e.g., seed-cracking finches with small or large beaks when only small and large seeds are available).
• Sexual selection favors traits that increase mating success (bright plumage, large antlers), even when those traits reduce survival.
• ✗ "The pesticide caused the mutations for resistance." ✓ The variation existed first; the pesticide is the selective agent.
• ✗ "Natural selection produces perfect organisms." ✓ It can only act on existing variation and is limited by trade-offs and by changing environments.
In artificial selection, humans choose which individuals reproduce, so humans (not the environment) decide which phenotypes have high fitness. Through artificial selection, humans change the variation in other species.
• Wild mustard (Brassica oleracea) was selected into cabbage (terminal bud), broccoli (flower clusters), kale (leaves) and kohlrabi (stem).
• All dog breeds descend from domesticated wolves; their great diversity in size and shape arose in a few thousand years of breeding.
• Artificial selection can reduce genetic diversity: modern crop varieties are often genetically uniform, which makes them vulnerable to a single pathogen (Topic 7.11).
Generation 1 has a mean of 3.0 leaf-margin hairs per plant. Students cross only the hairiest 10% of plants. Generation 2 has a mean of 5.1 hairs.
• Percent change $= \dfrac{5.1 - 3.0}{3.0} \times 100 = 70\%$ (always divide by the original value).
• The shift shows that hair number is at least partly heritable; if it were purely environmental, the offspring mean would not change.
• Chi-square analysis can test whether observed counts differ from those expected under a null hypothesis (see the FRQ notes below).
• ✗ "Artificial selection creates new genes on demand." ✓ Breeders select among variants already present (or arising by random mutation).
• ✗ Percent change $= (\text{new} - \text{old}) / \text{new}$. ✓ Divide by the old (initial) value.
Evolution is also driven by random occurrences. A population evolves whenever its allele frequencies change from one generation to the next; such changes are the evidence that evolution is occurring.
• Mutation is a random process that adds new genetic variation (new alleles), providing the phenotypes on which natural selection acts.
• Genetic drift is a change in allele frequencies caused by chance (a nonselective process). Its effect is strongest in small populations, where an allele can be lost or fixed simply by luck. Drift can make a small population diverge from other populations of the same species.
• Bottleneck effect: the population size is reduced to a few individuals for at least one generation (natural disaster, overhunting). The survivors' alleles may not represent the original population, and diversity is lost — northern elephant seals were reduced to about 20 individuals in the 1890s and today show very low genetic variation.
• Founder effect: a few individuals separate from the main population and start a new one; allele frequencies in the new population reflect only the founders (e.g., a high frequency of a rare recessive condition in an isolated human community).
• Gene flow (migration) adds or removes alleles. Gene flow between two populations makes them more similar and prevents them from diverging into separate species.
• Selection: the change is consistent across replicate populations in the same environment and is linked to a phenotype that affects survival or reproduction.
• Drift: replicate small populations change in different, unpredictable directions; large populations change little.
• ✗ "Genetic drift always removes harmful alleles." ✓ Drift is random; it can fix a harmful allele and lose a beneficial one.
• ✗ "Gene flow causes speciation." ✓ Gene flow homogenizes populations; reproductive isolation (stopping gene flow) is required for speciation.
The Hardy–Weinberg equilibrium is a model for describing and predicting allele frequencies in a non-evolving population. For one gene with two alleles, let $p$ = frequency of the dominant allele and $q$ = frequency of the recessive allele.
• Allele frequencies: $p + q = 1$.
• Genotype frequencies at equilibrium: $p^2$ (homozygous dominant) $+\; 2pq$ (heterozygous) $+\; q^2$ (homozygous recessive) $= 1$.
Conditions (all five must hold for allele frequencies to stay constant): (1) a large population size, (2) no migration, (3) no new mutations, (4) random mating, (5) no natural selection. These conditions are never fully met, but they provide a valuable null hypothesis: if observed genotype frequencies differ significantly from $p^2 : 2pq : q^2$, at least one condition is being violated and the population may be evolving.
In a large, randomly mating population, 16% of individuals show the recessive phenotype.
1. $q^2 = 0.16 \implies q = \sqrt{0.16} = 0.40$.
2. $p = 1 - 0.40 = 0.60$.
3. Heterozygotes: $2pq = 2(0.60)(0.40) = 0.48$ (48%); homozygous dominant: $p^2 = 0.36$ (36%).
4. Check: $0.36 + 0.48 + 0.16 = 1.00$. The dominant phenotype is $0.36 + 0.48 = 84\%$.
A sample of 1000 individuals contains 360 AA, 240 Aa and 400 aa.
1. Count alleles: total $= 2 \times 1000 = 2000$. Allele a $= 2(400) + 240 = 1040 \implies q = 1040/2000 = 0.52$; $p = 0.48$.
2. Expected aa under Hardy–Weinberg: $q^2 \times 1000 = 0.2704 \times 1000 \approx 270$. Observed: 400 — far more homozygotes than expected, as happens with non-random mating (e.g., inbreeding).
3. Here $\sqrt{0.400} = 0.63$ would be wrong: the square-root shortcut is valid only if the population is already at equilibrium.
• ✗ "$q$ is the frequency of the recessive phenotype." ✓ $q^2$ is; take the square root to get $q$.
• ✗ "Hardy–Weinberg means dominant alleles become more common." ✓ Dominance does not change allele frequency; only the five evolutionary forces do.
• ✗ "Allele frequencies in nature are static." ✓ The model tells you what would happen if nothing acted on the gene; it does not claim that this is the usual case.
Evolution is supported by evidence from many disciplines: geographical, geological, physical, biochemical and mathematical data. Evidence from both extant (living) and extinct organisms contributes.
• Fossils are dated by (1) the age of the rock layers in which they are found (deeper layers are generally older), (2) the decay rate of radioactive isotopes such as carbon-14, and (3) geographical data.
• Morphological homologies: structures with the same underlying anatomy but different functions (the forelimb bones of humans, whales, bats and cats) indicate common ancestry. Vestigial structures (the human appendix and tailbone, pelvic bones in whales) are reduced remnants of structures that were functional in ancestors.
• Molecular evidence: comparisons of DNA nucleotide sequences and protein amino acid sequences (e.g., cytochrome c) show that species with fewer differences shared a more recent common ancestor.
Carbon-14 has a half-life of about 5730 years. A bone retains 12.5% of the carbon-14 found in living tissue.
• $100\% \to 50\% \to 25\% \to 12.5\%$: three half-lives.
• Age $\approx 3 \times 5730 = 17\,190$ years. (In general, number of half-lives $n$ satisfies $(1/2)^n$ = fraction remaining.)
• Carbon-14 is useful only up to roughly 50 000 years; older rocks are dated with isotopes of much longer half-life.
• ✗ "Similar-looking structures always prove close relationship." ✓ Wings of birds and insects are analogous (convergent evolution, Topic 7.10), not homologous.
• ✗ "12.5% remaining means 8 half-lives." ✓ $(1/2)^3 = 1/8 = 12.5\%$, so it is 3 half-lives.
Structural and functional evidence at the cellular and molecular levels indicates that all eukaryotes share a common ancestor. The CED lists three features found in eukaryotes but not in prokaryotes:
• Membrane-bound organelles (nucleus, endoplasmic reticulum, Golgi apparatus, mitochondria).
• Linear chromosomes (prokaryotes usually have one circular chromosome).
• Genes that contain introns (removed by splicing, Topic 6.3).
At a deeper level, all living organisms — prokaryotes included — share DNA as genetic material, a nearly universal genetic code, ribosomes, and core metabolic pathways such as glycolysis. These shared features point to a single origin of life.
All species have evolved and continue to evolve. Evolution is observable today:
• Genomic changes over time can be measured directly by sequencing samples collected years apart.
• The fossil record shows continuous change in lineages.
• Resistance to antibiotics, pesticides, herbicides and chemotherapy drugs evolves quickly because these chemicals are strong selective agents and microbes, insects and cancer cells have short generation times and huge population sizes.
• Pathogens evolve and cause emerging diseases (new influenza strains, SARS-CoV-2 variants).
1. Within the bacterial population, a few cells already carry a resistance allele (from random mutation or horizontal gene transfer).
2. The antibiotic kills susceptible cells.
3. Resistant cells survive and reproduce, passing on the resistance allele.
4. The frequency of the resistance allele increases in the population. Stopping a course of antibiotics early leaves partially resistant survivors and speeds this process.
Phylogenetic trees and cladograms show hypothetical evolutionary relationships among lineages; they are hypotheses that are constantly revised as new evidence appears.
• A phylogenetic tree shows the amount of change over time, calibrated by fossils or a molecular clock; a cladogram shows only the branching order, with no time scale and no measure of evolutionary difference.
• Each node (branch point) represents the most recent common ancestor of the lineages that branch from it. Two taxa are more closely related if they share a more recent node — not because they sit next to each other on the page.
• Shared derived characters (traits gained or lost in a common ancestor and present in its descendants) are used to group lineages. Shared ancestral characters (present in all taxa) are not informative.
• The outgroup is the lineage least closely related to all the others; it lacks the derived characters and helps root the tree.
• Molecular data (DNA and protein sequences) usually provide more accurate and reliable evidence than morphological traits, which can be similar because of convergent evolution.
Characters: vertebral column (V), jaws (J), four limbs (L), amniotic egg (E), hair (H).
• Lancelet: none. Lamprey: V. Shark: V, J. Frog: V, J, L. Lizard: V, J, L, E. Mouse: V, J, L, E, H.
• Order of branching: lancelet (outgroup) → lamprey → shark → frog → lizard → mouse. Lizard and mouse share the most derived characters (V, J, L, E), so they share the most recent common ancestor of any pair here.
Molecular clock: if two lineages accumulate differences in a gene at a combined rate of 2 differences per million years and now differ at 24 sites, they diverged about $24 / 2 = 12$ million years ago.
• ✗ "The taxon at the right edge is the most advanced." ✓ All living tips have evolved for the same length of time.
• ✗ "Rotating branches around a node changes the relationships." ✓ Branches can rotate freely; only the nodes matter.
The biological species concept defines a species as a group of organisms capable of interbreeding and exchanging genetic information to produce viable, fertile offspring. Speciation occurs when two populations become reproductively isolated.
• Allopatric speciation: populations are geographically isolated (a river, mountain range or new island); drift, mutation and different selective pressures make them diverge.
• Sympatric speciation: populations share the same area but stop interbreeding — through polyploidy (common in plants), habitat or host shifts (apple maggot fly Rhagoletis on apples vs. hawthorns), or sexual selection.
Reproductive isolating mechanisms:
• Prezygotic (prevent fertilization): habitat, temporal (different breeding seasons or times of day), behavioral (different courtship songs), mechanical (incompatible reproductive structures) and gametic (sperm cannot fertilize the egg).
• Postzygotic (act after a hybrid zygote forms): reduced hybrid viability, reduced hybrid fertility (a mule is sterile) and hybrid breakdown (second-generation hybrids are weak or sterile).
• Punctuated equilibrium: long periods of stasis interrupted by rapid change. Gradualism: slow, steady change over hundreds of thousands to millions of years.
• Divergent evolution and adaptive radiation: one ancestral lineage diversifies into many species as new habitats become available (Hawaiian Drosophila, Caribbean Anolis lizards). Speciation rates can be especially rapid during adaptive radiation.
• Convergent evolution: similar selective pressures produce similar phenotypes in unrelated lineages (streamlined bodies of sharks and dolphins; the wings of birds and insects).
• ✗ "A mule proves horses and donkeys are one species." ✓ Mules are sterile — a postzygotic barrier — so horses and donkeys are separate species.
• ✗ Mixing up the categories: temporal, behavioral and mechanical isolation are all prezygotic; only barriers acting on the hybrid are postzygotic.
The level of variation in a population affects its dynamics. The ability of a population to respond to environmental change depends on its genetic diversity.
• Species and populations with little genetic diversity are at risk of decline or extinction (California condors, black-footed ferrets, prairie chickens).
• Genetically diverse populations are more resilient to environmental perturbation because they are more likely to contain individuals that can withstand the new pressure — not all individuals in a diverse population are susceptible to the same disease outbreak.
• Alleles that are adaptive in one environment may be deleterious in another (the sickle-cell allele with and without malaria).
The origin of life on Earth is supported by scientific evidence, and geological evidence reinforces the models.
• Earth formed about 4.6 billion years ago (bya). The environment was too hostile for life until about 3.9 bya, and the earliest fossil evidence of life dates to about 3.5 bya. Together these dates give a plausible window for the origin of life.
• Laboratory models (e.g., Miller–Urey type experiments) show that organic monomers such as amino acids can form from inorganic molecules under early-Earth conditions with an energy source.
RNA world hypothesis: RNA could have been the earliest genetic material. Its three assumptions are:
1. At some point, genetic continuity was assured by the replication of RNA.
2. Base pairing is necessary for replication.
3. Genetically encoded proteins were not involved as catalysts.
Supporting evidence: RNA can both store information and act as a catalyst (ribozymes), and the catalytic core of today's ribosome is rRNA.
Chi-square ($\chi^2$) goodness-of-fit test — compares observed counts ($o$) with counts expected ($e$) under a null hypothesis:
$\chi^2 = \sum \dfrac{(o - e)^2}{e}$, degrees of freedom (df) = number of categories − 1. Compare $\chi^2$ with the critical value at $p = 0.05$ from the AP formula sheet: df 1 = 3.84, df 2 = 5.99, df 3 = 7.81, df 4 = 9.49.
• If $\chi^2$ > critical value: reject the null hypothesis — the difference is unlikely to be due to chance alone.
• If $\chi^2$ < critical value: fail to reject the null hypothesis (never write "the null hypothesis is proven").
$\chi^2 = \dfrac{(68 - 50)^2}{50} + \dfrac{(32 - 50)^2}{50} = 6.48 + 6.48 = 12.96$; df $= 2 - 1 = 1$; critical value 3.84.
Because $12.96 > 3.84$, reject the null hypothesis: predation depends on moth color, which is a selective pressure favoring dark moths on dark bark.
• Name the variation that already exists, the selective pressure, the differential survival or reproduction and the change in allele frequency in the population.
• For Hardy–Weinberg questions, show $q^2 \to q \to p \to 2pq$ explicitly and state which condition is violated if observed and expected values differ.
• For cladograms, cite the shared derived character or the number of sequence differences that justifies each grouping.
• For speciation, name the specific isolating mechanism and say whether it is prezygotic or postzygotic.
• "The species adapted because it needed to" earns no credit (Lamarckian, need-based language).
• Using "fitness" without saying "more surviving offspring" is a buzzword without explanation.
• Writing "accept the null hypothesis" instead of "fail to reject".
Unit 7 Natural Selection: İnteraktif Soru İstasyonu
Özgün AP tarzı sorular (geçmiş sınav sorusu değil). Şık seç; anında çözüm ve hata türü gelir.
Unit 8 Ecology: İnteraktif Soru İstasyonu
Özgün AP tarzı sorular (geçmiş sınav sorusu değil). Şık seç; anında çözüm ve hata türü gelir.