Final Study Guide Bonus PDF

1 Final Study Guide * 150 points, 114 ques ons, 2 hour me limit star ng at the beginning of class *Exams are taken in class on laptops. Bring your charger. *See the announcements on D2L for...

1 Final Study Guide * 150 points, 114 ques ons, 2 hour me limit star ng at the beginning of class *Exams are taken in class on laptops. Bring your charger. *See the announcements on D2L for the speci c me and date of your nal. It may occur outside of regular class hours, as decided by the Lone Star College’s Final Exam Schedule. *This is 5 point bonus assignment. If it is completed thoroughly and submi ed by 11:59 pm on the due date you will receive a 5-point bonus. Late work is not accepted. All the informa on in it can be found in the textbook, the presenta ons (the “eye”), your notes, and prac ce problems from the Peardeck presenta ons. *There is no par al credit for this bonus, so complete it all. The points are not awarded based on correct answers, but comple on. It will not be given back to you with correc ons. *DO NOT Google the answers. Google will provide answers, but they will not be the speci c answers you need for this test. Do NOT use AI. *You are not only asked to memorize informa on but to understand rela onships and apply what you know to novel situa ons. For example, if you know how a process works, you may be asked to apply it to a di erent situa on than the one we have used as an example in class. *The items below represent possible exam ques ons, but do not indicate the ques on format. For example, the informa on could appear in a mul ple choice, a ll-in or matching. *There are no essays on this exam. *Tests are set to auto-submit, meaning they will close when me is up, even if you have not nished. Be prepared to move consistently through the ques ons so you do not run out of me. *Need help? Click on the Need Help? Folder under Content in D2L for informa on about free tutoring and how to study. Sign up to meet with for extra help at SignUpGenius.com. *THERE ARE NO EXAM MAKEUPS. Chapter 1 – Evolu on and the Founda ons of Biology Chapter 2 – The Chemical Context of Life Chapter 3 – Carbon and the Molecular Diversity of Life Chapter 4 – A Tour of the Cell Chapter 17 – Viruses Chapter 5 – Transport and Signaling fi ti ti ti ff ti ti ti ti ti ti ti ti ti fi ti ti ti ti ti ti fi fi ti tt ti ti ti ti ti fi ti  2 Chapter 9, 16.3 - Cell Cycle, Mitosis, and Cancer Chapter 6 - Energy and ATP Chapter 7 - Cellular Respira on and Fermenta on Chapter 8 – Photosynthesis Chapter 13 DNA Replica on Chapter 14 Transcrip on and Transla on Chapter 15 Gene Regula on Chapter 10 Meiosis DNA Technology Virtual Labs – CRISPR, DNA Fingerprin ng, PCR, Gel Electrophoresis Chapter 11 Mendel and the Gene Chapter 12 Chromosomal Basis of Inheritance Chapter 19 Natural Selec on and Evolu on Chapter 1 – Evolu on and the Founda ons of Biology 1. Dis nguish a hypothesis from a theory. Hypothesis: A testable predic on, To be tested by experiments. Theory: A well-substan ated explana on, To explain phenomena based on evidence. 2. De ne controlled experiment. An experiment where only one variable is changed at a time to test its effect, while all other variables are kept constant. 3. Determine the dependent and independent variable of a hypothe cal research project. Independent Variable: The variable that is manipulated or changed by the experimenter to observe its e ect. Dependent Variable: The variable that is measured or observed in response to changes in the independent variable. fi ti ff ti ti ti ti ti ti ti ti ti ti ti ti ti ti ti  3 4. Know the purpose of nega ve and posi ve controls in an experiment. Nega ve Control: To ensure that no e ect is observed when no treatment or interven on is applied. It helps con rm that the results are due to the experimental variable and not external factors. Posi ve Control: To ensure that the experimental setup is capable of producing a known e ect. It helps verify that the experiment can detect a response if one exists. 5. Recognize ways to reduce confounding factors in experiments. - Randomiza on: Randomly assign subjects to di erent groups to ensure that confounding variables are equally distributed. - Control Groups: Use a control group to compare the e ects of the independent variable. - Standardiza on: Keep all other condi ons constant across groups like temperature. 6. Using the Scien c Inquiry research projects and data as an example, interpret hypothe cal data. Your Scien c Inquiry quiz also has examples that will help. Chapter 2 – The Chemical Context of Life 1. List the elements that make up 96% of living ma er. Oxygen (O) Carbon (C) Hydrogen (H) Nitrogen (N) 2. Explain why unpaired electrons in an atom’s outer shell determine the atom’s reac vity. Unpaired Electrons are electrons in the outer shell that are not paired with another electron. Reac vity: Atoms are most reac ve when they have unpaired electrons because they seek to pair with electrons from other atoms to achieve a stable electron con gura on of 8 (usually a full outer shell). So Atoms with unpaired electrons form bonds with other atoms to ll their outer shell, making them more chemically reac ve. ti ti ti ti ti ti ti fi fi ti ti fi ti ti ff ti ti ff tt ff fi fi ti ti ti ff  4 3. Describe the di erence between atoms and their isotopes. Atoms Basic units of ma er made of protons, neutrons, and electrons. De ned by the number of protons in the nucleus (atomic number). Most atoms of an element have a standard number of neutrons. Isotopes Variants of the same element. Same number of protons but di erent numbers of neutrons. Di er in mass number (protons + neutrons). Chemical proper es remain nearly iden cal; physical proper es like stability may vary. 4. Describe the di erence in pH between the cell and the matrix of the mitochondrion. Cell Cytoplasm Typical pH: ~7.0 (neutral). Maintains homeostasis for cellular reac ons outside mitochondria. Mitochondrial Matrix Typical pH: ~7.8 (slightly alkaline). Proton (H⁺) concentra on is lower here due to the ac ve pumping of protons out of the matrix into the intermembrane space during oxida ve phosphoryla on. Key Di erence The intermembrane space between the inner and outer mitochondrial membranes is more acidic (~pH 7.0 or lower), crea ng a proton gradient. This gradient drives ATP synthesis as protons ow back into the matrix via ATP synthase. ff fi ff ff ff ti tt ti ff ti ti ti fl ti ti ti ti  5 5. De ne bu ers and explain why they are important in living things. De ni on: Bu ers are substances that minimize changes in pH by absorbing excess hydrogen ions (H⁺) or releasing hydrogen ions when they are depleted. Importance in Living Things 1. Maintaining pH Homeostasis ◦ Enzymes and biochemical reac ons require speci c pH ranges to func on op mally. ◦ Bu ers help stabilize pH levels in blood, cells, and other uids. 2. Example: Blood Bu er System ◦ Carbonic acid-bicarbonate bu er: Maintains blood pH around 7.4. ▪ H₂CO₃ ⇌ H⁺ + HCO₃⁻ ▪ If pH drops: HCO₃⁻ binds excess H⁺ to form H₂CO₃. ▪ If pH rises: H₂CO₃ dissociates to release H⁺. 3. Protec ng Cellular Func on ◦ Prevents harmful e ects of pH uctua ons, like protein denatura on or metabolic disrup on. 4. Adapta on to Environmental Changes ◦ Helps organisms cope with acidic or alkaline environments (e.g., ocean bu ering systems for marine life). 6. Describe weak interac ons like hydrogen bonding and van der Waals interac ons. Hydrogen Bonding De ni on: A weak interac on between a hydrogen atom covalently bonded to an electronega ve atom (like oxygen or nitrogen) and another electronega ve atom. Key Features: ◦ Direc onal and rela vely strong for weak interac ons (~5–30 kJ/mol). ti ff ff fi fi fi ti ti ti ti ti ff ti ti ff ff ti ti ti ti ff ti fl ti ti fi fl ti ti ti ff ti  6 ◦ Important in stabilizing structures like DNA (between base pairs) and proteins (secondary and ter ary structures). Example: ◦ Water molecules: Hydrogen bonds form between the hydrogen atom of one molecule and the oxygen atom of another. van der Waals Interac ons De ni on: Weak a rac ons between molecules due to temporary dipoles that occur as electrons move around atoms. Key Features: ◦ Non-direc onal and very weak (~0.4–4 kJ/mol). ◦ Depend on close proximity between molecules. ◦ Include two types: ▪ London dispersion forces: Arise from temporary dipoles. ▪ Dipole-dipole interac ons: Occur between permanent dipoles in polar molecules. Example: ◦ Gecko feet s cking to surfaces due to van der Waals forces. 7. Know the di erence between ionic bonds, polar covalent bonds, hydrogen bonds and hydrophobic interac ons. 1. Ionic Bonds ◦ De ni on: A strong electrosta c a rac on between oppositely charged ions (one posi vely charged and one nega vely charged). ◦ Forma on: Formed when one atom donates an electron to another atom, resul ng in ions. ◦ Example: ▪ Sodium chloride (NaCl): Na⁺ and Cl⁻ are held together by an ionic bond. ◦ Characteris cs: ▪ Strong bond but can be broken in water (dissociate into ions). fi fi ti ti ti ti ti ti ti ff ti tt ti ti ti ti ti ti ti tt ti  7 ▪ Occur between metals and non-metals. 2. Polar Covalent Bonds ◦ De ni on: A type of covalent bond where electrons are shared unevenly between two atoms due to a di erence in electronega vity. ◦ Forma on: One atom a racts the shared electrons more strongly, crea ng par al posi ve (δ⁺) and nega ve (δ⁻) charges. ◦ Example: ▪ Water (H₂O): Oxygen is more electronega ve than hydrogen, so electrons are drawn toward oxygen, making it par ally nega ve. ◦ Characteris cs: ▪ Stronger than ionic bonds in a vacuum but weaker than ionic bonds in water. ▪ Molecules with polar covalent bonds are usually polar and can interact with other polar molecules. 3. Hydrogen Bonds ◦ De ni on: A weak a rac on between a hydrogen atom covalently bonded to a highly electronega ve atom (e.g., oxygen, nitrogen) and another electronega ve atom. ◦ Forma on: Occurs when a hydrogen atom bonded to an electronega ve atom (like oxygen) is a racted to another electronega ve atom. ◦ Example: ▪ Water molecules: Hydrogen bonds form between the hydrogen of one water molecule and the oxygen of another. ◦ Characteris cs: ▪ Weak compared to covalent and ionic bonds but important for maintaining the structure of DNA, proteins, and water's high cohesion and adhesion proper es. 4. Hydrophobic Interac ons ◦ De ni on: The tendency of nonpolar molecules or nonpolar regions of molecules to avoid contact with water. fi fi fi ti ti ti ti ti ti ti ti tt ti ti tt ti ti tt ti ff ti ti ti ti ti ti ti ti ti  8 ◦ Forma on: Occurs when nonpolar substances aggregate together in an aqueous environment to minimize contact with water molecules. ◦ Example: ▪ The folding of proteins: Nonpolar amino acid side chains tend to be found in the interior of proteins, away from water. ◦ Characteris cs: ▪ Not a true bond, but a result of the physical proper es of nonpolar molecules. ▪ Important in the forma on of cellular membranes and the structure of proteins. 8. Describe the characteris cs of the water molecule. Polarity Water is a polar molecule, meaning it has a par al posi ve charge on the hydrogen atoms and a par al nega ve charge on the oxygen atom. This occurs because oxygen is more electronega ve than hydrogen, pulling electrons toward itself, crea ng a dipole. Hydrogen Bonding Due to its polarity, water molecules can form hydrogen bonds with each other. The hydrogen atom of one water molecule is a racted to the oxygen atom of a neighboring molecule, crea ng a cohesive network. High Cohesion and Adhesion Cohesion: Water molecules s ck together due to hydrogen bonds, which is why water has high surface tension (e.g., water droplets form beads). Adhesion: Water molecules are a racted to other surfaces (e.g., water climbing up a paper towel by capillary ac on). High Speci c Heat Capacity Water can absorb or release a large amount of heat with only a small change in temperature. This property helps regulate temperature in organisms and the environment. ti fi ti ti ti ti ti ti ti ti ti tt tt ti ti ti ti  9 High Heat of Vaporiza on Water requires a signi cant amount of energy to change from liquid to gas. This helps organisms cool down through processes like swea ng. Universal Solvent Water is an excellent solvent because its polarity allows it to dissolve a wide range of polar and ionic substances. This is crucial for many biological processes, where water dissolves nutrients, ions, and gases. Density of Solid vs. Liquid Water expands as it freezes, making ice less dense than liquid water. This is why ice oats on water and plays an important role in aqua c ecosystems by insula ng the water below. Neutral pH Pure water has a neutral pH of 7, meaning it is neither acidic nor basic. This neutrality is essen al for maintaining the proper pH for biological processes. 9. Describe several reasons why the hydrogen bonding of water supports life. High Cohesion and Surface Tension The hydrogen bonds between water molecules create cohesion, causing water to s ck to itself. This results in high surface tension, which allows water to form droplets, move up plant roots (capillary ac on), and maintain its liquid form even in small volumes, suppor ng cellular processes. Moderates Temperature Hydrogen bonding gives water a high speci c heat capacity, meaning water can absorb a lot of heat before its temperature rises signi cantly. This property helps organisms regulate their internal temperature, maintaining stable condi ons despite environmental uctua ons (e.g., ocean temperatures remaining rela vely constant). E ec ve Solvent for Biochemical Reac ons ff ti ti ti ti fl ti ti fi ti fl ti ti fi fi ti ti ti ti  10 Water's polarity, due to hydrogen bonds, allows it to dissolve many polar and ionic compounds. This makes water an excellent solvent, allowing for the e cient transport of nutrients, gases, and waste products in biological systems (e.g., blood and cytoplasm). Forma on of Water’s Unique Proper es High Heat of Vaporiza on: Hydrogen bonding means water requires signi cant energy to evaporate, making it e ec ve for cooling mechanisms like swea ng or transpira on in plants, helping regulate temperature. Density Anomaly: Hydrogen bonds cause water to be less dense as a solid than as a liquid. This causes ice to oat, insula ng bodies of water and allowing life to survive beneath the frozen surface during cold temperatures. Supports Cellular Structure and Func on The hydrogen bonds between water molecules allow water to maintain a liquid state at room temperature, which is crucial for biological processes. Water's cohesive and adhesive proper es help maintain the shape of cells, assist in nutrient transport, and facilitate chemical reac ons in the cytoplasm. Regulates pH Water has a neutral pH of 7, and its ability to form hydrogen bonds helps maintain the pH balance in organisms. This regula on ensures that biochemical processes can occur within the op mal pH range for enzymes and other molecules to func on correctly. Chapter 3 – Carbon and the Molecular Diversity of Life 1. Recognize the func onal groups of amino acids. Amino group (-NH₂) and carboxyl group (-COOH). 2. Describe how the structure of phospholipids a ects cell membrane permeability. The phospholipid bilayer's hydrophobic tails restrict polar and large molecules, while its hydrophilic heads allow selec ve passage of water and small nonpolar molecules. ti ti ti ti ff fl ti ti ti ti ti ti ff ti ti ffi ti fi ti ti  11 3. Describe the monomers that make up the polymers of nucleic acids, carbohydrates, lipids and proteins. Nucleic Acids: Nucleo des Carbohydrates: Monosaccharides Lipids: Fa y acids and glycerol (not true monomers but building blocks) Proteins: Amino acids 4. Explain the di erences between the primary, secondary, ter ary and quaternary structures of protein. Primary: Sequence of amino acids in a polypep de chain. Secondary: Folding into α-helices or β-pleated sheets, stabilized by hydrogen bonds. Ter ary: 3D shape formed by interac ons among R-groups (e.g., hydrogen bonds, ionic bonds, hydrophobic interac ons). Quaternary: Assembly of mul ple polypep de chains into a func onal protein. 5. Predict how amino acids would interact in the quaternary structure of a protein, based on their side chains (ionic bonds, hydrophobic interac ons, Van der Waals, disul de bridges and hydrogen bonding) Ionic bonds: Form between posi vely and nega vely charged side chains (e.g., lysine and glutamate). Hydrophobic interac ons: Nonpolar side chains cluster together to avoid water. Van der Waals forces: Weak a rac ons between closely posi oned nonpolar side chains. Disul de bridges: Covalent bonds between cysteine residues stabilize the structure. Hydrogen bonding: Occurs between polar side chains or backbone atoms, adding stability. 6. Compare and contrast hydrolysis and dehydra on reac ons. Dehydra on Reac on: Joins monomers by removing water, forming a covalent bond. Hydrolysis Reac on: Breaks polymers into monomers by adding water, breaking a covalent bond. Similarity: Both involve water and are crucial for building or breaking biological molecules. ti fi ti tt ff ti ti ti ti ti ti tt ti ti ti ti ti ti ti ti ti ti ti ti fi  12 Di erence: Dehydra on Reac on: Water is removed to form a bond. Hydrolysis Reac on: Water is added to break a bond. 7. Describe why carbs, proteins and nucleic acids are polymers but lipids are not. Carbs, Proteins, Nucleic Acids: Made of repea ng monomers (e.g., monosaccharides, amino acids, nucleo des) linked by covalent bonds. Lipids: Composed of glycerol and fa y acids but lack a repea ng monomer structure, so they are not true polymers. 8. Explain why humans can digest starch but not cellulose. Starch: Humans can digest starch because it is made of α-glucose monomers linked by α(1→4) and α(1→6) glycosidic bonds. The enzyme amylase can break these bonds. Cellulose: Humans cannot digest cellulose because it is made of β-glucose monomers linked by β(1→4) glycosidic bonds. The human diges ve system lacks the enzyme cellulase, which is required to break these bonds. Key Di erence: The di erence lies in the type of glycosidic bonds and the speci c enzymes available in the human diges ve system. 9. Know the type of bond a pep de bond is, the product of a pep de bond and when this occurs in gene expression. Type of Bond: Covalent bond. Product: A dipep de or polypep de and water (via dehydra on synthesis). Occurs During: Transla on, when ribosomes synthesize polypep des from mRNA. Chapter 4 – A Tour of the Cell 1. Know the func ons of the smooth and rough ER, lysosomes, and bound ribosomes. Smooth ER: Synthesizes lipids, detoxi es drugs, and stores calcium. Rough ER: Synthesizes and processes proteins for secre on or membrane inser on. ff ff ti ti ti ti ti ti ff ti ti ti ti tt fi ti ti ti ti ti ti ti fi ti  13 Lysosomes: Digest macromolecules and recycle cellular components. Bound Ribosomes: Synthesize proteins des ned for secre on or organelles. Chapter 17 – Viruses 1. Describe the ly c and lysogenic cycle and know which virus can alternate between the two. Ly c Cycle: The virus replicates, lyses the host cell, and releases new virions. Lysogenic Cycle: The viral DNA integrates into the host genome and replicates passively with it. Virus: Temperate phages (e.g., Lambda phage) alternate between the two cycles. 2. Describe how viruses evolve resistance to drugs. Viruses evolve resistance through muta ons in their gene c material, which can alter drug targets, evade immune responses, or enhance replica on e ciency. High muta on rates and rapid replica on accelerate this process. 3. Speci cally describe how HIV has evolved resistance to compe ve inhibitors of reverse transcriptase. HIV evolves resistance by muta ng the reverse transcriptase gene, altering its ac ve site to reduce the binding a nity of compe ve inhibitors while maintaining its enzyma c func on. 4. List the reasons that viruses are not considered living. Lack cellular structure. Cannot carry out metabolism independently. Require a host cell for reproduc on. Do not grow or respond to s muli. Do not maintain homeostasis. ti fi ti ti ffi ti ti ti ti ti ti ti ti ti ti ffi ti ti ti ti ti ti  14 5. Describe the best ways to decrease the severity of viral infec ons. Vaccina on to boost immunity. An viral medica ons to inhibit viral replica on. Good hygiene prac ces (e.g., handwashing). Isola on to prevent spread. Strengthening the immune system through a healthy lifestyle. 6. Compare and contrast eukaryo c cells and viruses. Which have proteins, nucleic acids, lipids and/or carbs? Eukaryo c Cells: ◦ Proteins: Yes ◦ Nucleic Acids: Yes (DNA and RNA) ◦ Lipids: Yes (cell membranes) ◦ Carbs: Yes (glycogen, glycoproteins) Viruses: ◦ Proteins: Yes (capsid, enzymes) ◦ Nucleic Acids: Yes (DNA or RNA) ◦ Lipids: Some mes (envelope) ◦ Carbs: Rare (on the viral envelope, if present) Comparison: Eukaryo c cells are complete, independent life forms with all essen al components for metabolism and reproduc on. Viruses are non-living en es that require a host cell to replicate and do not have the necessary machinery for independent life. 7. Describe methods of reproduc on in viruses, eukaryo c cells including plants and prokaryo c cells. ti ti ti ti ti ti ti ti ti ti ti ti ti ti ti ti ti ti  15 Viruses: Reproduce by infec ng a host cell and using the host's machinery to replicate and assemble new virions. Two cycles: Ly c (cell lysis) and Lysogenic (integra on into host genome). Eukaryo c Cells (including plants): Mitosis (asexual): Produces iden cal daughter cells. Meiosis (sexual): Reduces chromosome number for gametes (sperm and egg). Prokaryo c Cells: Binary Fission: Asexual reproduc on where the cell divides into two iden cal daughter cells. 8. Describe the organelles that viruses, eukaryo c cells including plants and prokaryo c cells have and are lacking. Viruses: Have: None (lack organelles). Lack: All organelles; they rely on the host cell machinery for replica on. Eukaryo c Cells (including plants): Have: Nucleus, mitochondria, endoplasmic re culum, Golgi apparatus, lysosomes, ribosomes, chloroplasts (in plants), vacuoles, and cytoskeleton. Lack: Cell walls (in animal cells), agella (in most plant cells). Prokaryo c Cells: Have: Ribosomes, plasma membrane, cell wall, cytoplasm, nucleoid region. Lack: Nucleus, membrane-bound organelles (e.g., mitochondria, Golgi apparatus). 9. Describe which kinds of cells/viruses have mitochondria, chloroplasts, both or none. Eukaryo c Animal Cells: Have: Mitochondria. Lack: Chloroplasts. ti ti ti ti ti ti ti ti fl ti ti ti ti ti ti ti  16 Eukaryo c Plant Cells: Have: Mitochondria and chloroplasts. Prokaryo c Cells: Have: None (lack mitochondria and chloroplasts). Viruses: Have: None (lack mitochondria and chloroplasts). Chapter 5 – Transport and Signaling 1. Describe how the structure of the phospholipid bilayer creates a selec vely permeable membrane and be able to predict which molecules can cross the membrane and which cannot. Structure: The phospholipid bilayer has hydrophilic (water-a rac ng) heads facing outward and hydrophobic (water-repelling) tails facing inward, crea ng a semi-permeable barrier. Molecules that can cross: Small, nonpolar molecules (e.g., oxygen, carbon dioxide) Lipid-soluble molecules (e.g., steroids) Molecules that cannot cross: Large, polar molecules (e.g., glucose) Ions (e.g., Na+, K+) Charged molecules (e.g., amino acids) 2. De ne and describe di usion as passive transport. De ni on: Di usion is the movement of molecules from an area of higher concentra on to an area of lower concentra on. Passive Transport: Di usion does not require energy (ATP) and relies on the natural tendency of molecules to spread out evenly. Example: Oxygen and carbon dioxide move across the cell membrane by di usion. fi fi ti ti ti ff ff ff ti ti tt ti ti ff ti  17 3. Know general de ni ons used to describe osmosis such as hypertonic, hypotonic, isotonic. Hypertonic: Solu on with a higher concentra on of solutes compared to another solu on. Water moves out of the cell, causing it to shrink. Hypotonic: Solu on with a lower concentra on of solutes compared to another solu on. Water moves into the cell, causing it to swell or burst. Isotonic: Solu on with the same concentra on of solutes as another solu on. No net movement of water occurs, and the cell maintains its shape. 4. Describe the adapta ons of animals and plants that increase membrane uidity in freezing temperatures. Animals: Increase produc on of unsaturated fa y acids, which have kinks that prevent ght packing, maintaining membrane uidity. Produce cholesterol to stabilize and maintain uidity at lower temperatures. Plants: Adjust the fa y acid composi on of phospholipids by increasing unsaturated fats to maintain membrane uidity. Some plants produce an freeze proteins to protect membranes from freezing damage. 5. Solve varying osmosis problems as described on the slide “Solving Osmosis Problems.” 1. Iden fy the solu on types: Determine if the solu on is hypertonic, hypotonic, or isotonic in rela on to the cell. 2. Determine water movement: Hypertonic: Water moves out of the cell (cell shrinks). Hypotonic: Water moves into the cell (cell swells). Isotonic: No net movement of water (cell maintains shape). 3. Assess nal state: Based on the direc on of water movement, predict the nal state of the cell (shrinking, swelling, or unchanged). ti ti fi tt ti ti ti ti fi ti ti ti fl ti ti fl tt ti ti ti ti fl ti fl ti fi ti ti ti  18 6. Know the di erences between di usion, facilitated di usion, ac ve transport, passive transport, phagocytosis and pinocytosis. Di usion: Movement of molecules from high to low concentra on, no energy required (passive). Facilitated Di usion: Movement of molecules from high to low concentra on through a membrane protein, no energy required (passive). Ac ve Transport: Movement of molecules against their concentra on gradient, requires energy (ATP). Passive Transport: Movement of molecules down their concentra on gradient, no energy required (includes di usion and facilitated di usion). Phagocytosis: Type of endocytosis where the cell engulfs large par cles or microorganisms (cell ea ng). Pinocytosis: Type of endocytosis where the cell engulfs extracellular uid and dissolved solutes (cell drinking). 7. Describe the e ects of hypertonic, hypotonic and isotonic solu ons on plant and animal cells. Hypertonic Solu on: Animal Cells: Shrink (crenate) as water leaves the cell. Plant Cells: Lose turgor pressure, become plasmolyzed (shrivel inside the cell wall). Hypotonic Solu on: Animal Cells: Swell and may burst (lyse) as water enters the cell. Plant Cells: Become turgid ( rm), but generally do not burst due to the cell wall's rigidity. Isotonic Solu on: Animal Cells: Maintain shape as water moves in and out at equal rates. Plant Cells: Maintain shape with some turgor pressure, but are less rm than in a hypotonic solu on. 8. Generally, describe the three stages of cell signaling. ff ti ti ff ti ff ff ti ti ti ff fi ff ff ff ti ti ti ti ti ti fi fl ti  19 1. Recep on: A signaling molecule (ligand) binds to a receptor on the target cell's surface or inside the cell. 2. Transduc on: The receptor ac vates a signal transduc on pathway, o en involving a series of proteins or second messengers that amplify and relay the signal. 3. Response: The signal triggers a cellular response, such as gene expression, enzyme ac va on, or changes in cell behavior. 9. Explain why mul -step signal transduc on pathways are advantageous to cells. Ampli ca on: Each step can amplify the signal, leading to a stronger response. Regula on: Allows for more control at di erent points, ensuring appropriate responses. Diversity of Responses: Mul ple pathways can be ac vated, allowing cells to integrate di erent signals and produce varied responses. Coordina on: Enables ne-tuned regula on of cellular processes, ensuring e ciency and adaptability. Chapter 9, 16.3 - Cell Cycle, Mitosis, and Cancer 1. Know the de ni ons of benign tumor, metastasis, passive tumor, and malignant tumor. Benign Tumor: A non-cancerous growth that does not spread to other parts of the body. Metastasis: The spread of cancer cells from the original site to other parts of the body. Passive Tumor: Not a standard medical term; likely a misunderstanding or incorrect term. Malignant Tumor: A cancerous growth that invades nearby ssues and can spread to other parts of the body. 2. Explain the reason cancer incidence generally increases with age. Cancer incidence increases with age because DNA damage accumulates over me, and the body's ability to repair gene c errors and suppress abnormal cell growth declines. ff fi ti ti ti ti ti fi ti ti fi ti ti ti ti ti ff ti ti ti ft ffi ti ti ti  20 3. Describe the reason that proto-oncogenes are not selected against and are therefore s ll occurring in the popula on. Proto-oncogenes are not selected against because they play essen al roles in normal cell growth and division; they only become harmful when mutated into oncogenes. 4. De ne tumor suppressor genes. Tumor Suppressor Genes: Genes that regulate cell growth and division, preven ng uncontrolled cell prolifera on. 5. Explain why early-onset cancer arises at a younger age and why those individuals also have rela ves with early-onset cancer. Early-onset cancer arises due to inherited muta ons in cri cal genes, increasing cancer risk. Rela ves may share the same gene c muta on, leading to familial cases of early-onset cancer. 6. List the phases of mitosis and describe the events characteris c of each phase in animal cells. Prophase: Chromosomes condense, spindle bers form, and the nuclear envelope breaks down. Metaphase: Chromosomes align at the cell's equatorial plane. Anaphase: Sister chroma ds separate and move to opposite poles. Telophase: Nuclear envelopes reform, and chromosomes decondense. Cytokinesis (not mitosis but occurs a er): The cell divides into two daughter cells. 7. Describe the phase that causes stem cells to divide infrequently. G0 Phase: A res ng phase where cells exit the cell cycle, remaining inac ve and dividing infrequently. Chapter 6 - Energy and ATP 8. Describe the First Law of Thermodynamics. First Law of Thermodynamics: Energy cannot be created or destroyed, only transferred or transformed. fi ti ti ti ti ti ti ti ft ti fi ti ti ti ti ti ti ti  21 9. De ne metabolism. Metabolism is the set of chemical reac ons in a living organism that convert food into energy and build or break down molecules to sustain life. 10. Describe how enzymes catalyze reac ons through the lowering of ac va on energy. Enzymes catalyze reac ons by lowering the ac va on energy, which makes it easier for reactants to form products. They do this by stabilizing the transi on state and providing an op mal environment for the reac on. 11. Describe how compe ve inhibi on works. Compe ve inhibi on occurs when a molecule similar to the substrate binds to the enzyme's ac ve site, blocking the actual substrate from binding and slowing down the reac on. 12. Explain the rela onship between catabolism, anabolism, exergonic and endergonic reac ons. Catabolism: Breaks down molecules, releasing energy (exergonic). Anabolism: Builds molecules, requiring energy input (endergonic). Exergonic reac ons fuel endergonic reac ons in metabolism. 13. Describe the regenera on of ATP. ATP is regenerated through cellular respira on, where ADP and inorganic phosphate (Pi) are combined using energy from catabolic processes like glycolysis, the Krebs cycle, and oxida ve phosphoryla on. 14. Recognize the parts of the ATP molecule and how the molecule changes when it stores and releases energy. Parts: Adenine (nitrogenous base), ribose (sugar), and three phosphate groups. Energy Storage: Energy is stored in the high-energy bonds between phosphate groups. Energy Release: Energy is released when ATP loses a phosphate group, becoming ADP (adenosine diphosphate). ti ti fi ti ti ti ti ti ti ti ti ti ti ti ti ti ti ti ti ti ti ti ti ti ti ti ti 22 15. Describe how ATP is used to drive endergonic reac ons through coupling. ATP drives endergonic reac ons by transferring a phosphate group to a reactant, crea ng a phosphorylated intermediate. This makes the reac on more exergonic, allowing it to proceed. 16. Explain why exergonic reac ons spontaneous and why endergonic reac ons are not spontaneous. Exergonic Reac ons: Spontaneous because they release energy and have a nega ve Gibbs free energy (-ΔG). Endergonic Reac ons: Not spontaneous because they require energy input and have a posi ve Gibbs free energy (+ΔG). 17. Describe why the enzyme cellulase can hydrolyze glycosidic linkages of starch but not cellulose. Cellulase is speci c to β-glycosidic linkages in cellulose, but starch has α-glycosidic linkages, which cellulase cannot recognize or break down. Chapter 7 - Cellular Respira on and Fermenta on You do not have to know the names of speci c intermediate molecules unless stated in this document. 1. Explain the three stages of respira on in general terms. What are they called? What goes in? What comes out? Where do they occur? Three Stages of Respira on 1. Glycolysis ◦ What goes in? Glucose ◦ What comes out? Pyruvate, ATP, NADH ◦ Where? Cytoplasm 2. Krebs Cycle (Citric Acid Cycle) ◦ What goes in? Acetyl-CoA ◦ What comes out? CO₂, ATP, NADH, FADH₂ ti ti fi ti ti ti ti ti fi ti ti ti ti ti ti ti  23 ◦ Where? Mitochondrial matrix 3. Electron Transport Chain (ETC) ◦ What goes in? NADH, FADH₂, Oxygen ◦ What comes out? ATP, Water ◦ Where? Inner mitochondrial membrane 2. Compare the reactants, products, and purpose of anaerobic alcoholic and lac c acid fermenta on. Alcoholic Fermenta on Reactants: Glucose Products: Ethanol, CO₂, ATP Purpose: Energy produc on in yeast and some microorganisms under anaerobic condi ons Lac c Acid Fermenta on Reactants: Glucose Products: Lac c acid, ATP Purpose: Energy produc on in animal cells and some bacteria under anaerobic condi ons 3. Explain the func on or purpose of respira on. Func on/Purpose: To produce ATP, the energy currency of the cell, by breaking down glucose or other molecules. Why? ATP powers cellular processes like growth, repair, and maintenance. 4. Describe why oxygen is the last electron acceptor in the ETC and where it comes from. Why is it the last electron acceptor? Oxygen has a high a nity for electrons, allowing it to accept electrons at the end of the Electron Transport Chain. What does it form? Combines with electrons and hydrogen ions to produce water. Where does it come from? From the air we breathe. ti ti ti ti ti ti ti ti ti ti ti ti ffi ti  24 5. Trace the ow of electrons through the stages of respira on. Glycolysis: Electrons from glucose are transferred to NAD⁺, forming NADH. Krebs Cycle: Electrons from Acetyl-CoA are transferred to NAD⁺ and FAD, forming NADH and FADH₂. Electron Transport Chain (ETC): Electrons from NADH and FADH₂ are passed through protein complexes, crea ng a proton gradient. Electrons end at oxygen, forming water. 6. Describe how energy from the electron transport chain is used to create a proton- mo ve force (chemiosmosis). Energy Source: Electrons moving through the ETC release energy. What Happens? This energy pumps protons (H⁺) from the mitochondrial matrix into the intermembrane space. Result: Creates a proton gradient (proton-mo ve force) with high H⁺ concentra on in the intermembrane space. Purpose: Drives ATP synthesis as protons ow back into the matrix through ATP synthase. Chapter 8- Photosynthesis You do not have to know the names of speci c intermediate molecules unless stated in this document. 1. Explain the two stages of photosynthesis in general terms. What is the func on of each? What are they called? What goes in? What comes out? Where do they occur? 1. Light-Dependent Reac ons Func on: Capture light energy to produce ATP and NADPH. What goes in? Light, Water, ADP, NADP⁺ ti ti fl ti fl fi ti ti ti ti ti  25 What comes out? Oxygen, ATP, NADPH Where? Thylakoid membranes 2. Calvin Cycle (Light-Independent Reac ons) Func on: Use ATP and NADPH to synthesize glucose. What goes in? CO₂, ATP, NADPH What comes out? Glucose, ADP, NADP⁺ Where? Stroma of the chloroplast 2. Describe the rela onship between photosynthesis and respira on in terms of similari es, di erences, overall func ons and loca ons in which each occurs. Photosynthesis vs. Respira on 1. Similari es 1. Both involve energy conversion processes. 2. Both use electron transport chains and ATP produc on. 3. Both involve the movement of electrons and crea on of a proton gradient. 2. Di erences 1. Photosynthesis: Converts light energy into chemical energy (glucose). 2. Respira on: Breaks down glucose to release stored energy as ATP. 3. Inputs: Photosynthesis uses CO₂ and water; respira on uses glucose and oxygen. 4. Outputs: Photosynthesis produces glucose and oxygen; respira on produces CO₂ and water. 3. Overall Func ons 1. Photosynthesis: Stores energy in organic molecules. 2. Respira on: Releases energy for cellular ac vi es. 4. Loca ons 1. Photosynthesis: Chloroplasts in plants and some microorganisms. 2. Respira on: Mitochondria in nearly all eukaryo c cells. ff ti ti ti ti ti ti ti ti ff ti ti ti ti ti ti ti ti ti ti ti ti ti  26 3. Explain why there are two loca ons for ATP synthase pumps in plants and where the loca ons are. ATP Synthase Loca ons in Plants Why Two Loca ons? ◦ Dual Energy Processes: Plants perform both photosynthesis and respira on, each requiring ATP produc on. ◦ Separate ATP Needs: Chloroplasts generate ATP for the Calvin cycle, while mitochondria produce ATP for overall cellular func ons. Where Are They Located? ◦ Chloroplasts ▪ Func on: Produce ATP during the light-dependent reac ons of photosynthesis. ◦ Mitochondria ▪ Func on: Generate ATP through cellular respira on to power various cellular ac vi es. Chapter 13 – DNA Replica on 1. Know the phase of the cell cycle in which DNA replica on occurs. DNA replica on occurs in the S phase (Synthesis phase) of the cell cycle. 2. Know the func ons of each of the proteins and enzymes at the replica on fork. Helicase: Unwinds and separates the DNA strands. Primase: Synthesizes RNA primers to start DNA synthesis. DNA polymerase: Synthesizes the new DNA strand by adding nucleo des. Single-strand binding proteins (SSBs): Prevent the separated strands from re-annealing. Ligase: Joins Okazaki fragments on the lagging strand. Topoisomerase: Relieves tension ahead of the replica on fork. ti ti ti ti ti ti ti ti ti ti ti ti ti ti ti ti ti ti ti ti  27 3. Know the direc on (5’ to 3’ or 3’ to 5’) of the leading strand, lagging strand and the RNA primer at the replica on fork in DNA replica on. Leading strand: 5’ to 3’ Lagging strand: 3’ to 5’ RNA primer: 5’ to 3’ 4. Describe the direc onality of DNA polymerase III. DNA polymerase III synthesizes DNA in the 5’ to 3’ direc on. 5. De ne gene. Gene: A segment of DNA that contains the instruc ons for the synthesis of a speci c protein or RNA molecule. 6. Be able to write the complementary sequence of a DNA sequence including the direc on (5’ to 3’ or 3’ to 5’) Given a DNA sequence, the complementary sequence in the 5’ to 3’ direc on is wri en by pairing each nucleo de with its complementary base: For example, the complementary sequence to 5’-A T G C C T A-3’ would be 3’-T A C G G A T-5’. 7. Explain how a muta on in DNA can change a phenotype. A muta on in DNA can change a phenotype by altering the sequence of nucleo des, which may lead to changes in the protein produced. This can a ect the protein's func on, structure, or regula on, ul mately impac ng the traits or characteris cs (phenotype) of an organism. 8. Explain why muta ons in gametes can cause gene c diseases in o spring. Muta ons in gametes can cause gene c diseases in o spring because these muta ons can be passed on to the next genera on. If a gamete (sperm or egg) carrying a muta on fuses with a normal gamete during fer liza on, the resul ng o spring inherits the mutated gene, which can lead to a gene c disease. fi ti ti ti ti ti ti ti ti ti ti ti ti ti ti ti ti ti ti ti ti ti ff ff ff ti ti ff ti ti ti ti ti fi tt  28 9. Explain epigene cs and how it leads to changes in gene expression.