뒤로AP Biology Unit 1: Macromolecules & Structure-Function Relationships – Guided Study
스터디 가이드 - 스마트 노트
자료에 맞춘 맞춤형 노트, 핵심 정의, 예시, 맥락을 확장해 제공합니다.
Q1. Differentiate between a dehydration synthesis reaction and a hydrolysis reaction in terms of water involvement, polymer assembly/breakdown, and bond formation.
Background
Topic: Macromolecule Assembly & Breakdown
This question tests your understanding of how biological polymers are built and broken down, focusing on the role of water in these processes.
Key Terms:
Dehydration synthesis (condensation reaction): A chemical reaction that joins monomers to form polymers by removing a water molecule.
Hydrolysis: A chemical reaction that breaks polymers into monomers by adding a water molecule.
Polymer: A large molecule made up of repeating subunits (monomers).
Covalent bond: The type of bond formed or broken during these reactions.
Step-by-Step Guidance
Describe what happens to water in each reaction: Is it added or removed?
Explain whether the reaction results in the assembly (building) or breakdown of a polymer.
Identify the type of bond formed or broken during each process (e.g., peptide bond, glycosidic bond).
Summarize the key differences between the two reactions in a comparative statement.
Try solving on your own before revealing the answer!
Final Answer:
Dehydration synthesis removes a water molecule to join two monomers, forming a covalent bond and assembling a polymer. Hydrolysis adds a water molecule to break a covalent bond, resulting in the breakdown of a polymer into monomers. In dehydration synthesis, water is produced and a bond is formed; in hydrolysis, water is consumed and a bond is broken.
Q2. Describe how a molecule of water is removed during the synthesis of a dipeptide from two amino acids.
Background
Topic: Protein Synthesis (Peptide Bond Formation)
This question focuses on the process of forming a peptide bond between two amino acids, specifically how water is involved in this reaction.
Key Terms and Concepts:
Amino acid: The monomer unit of proteins, containing an amino group (-NH2) and a carboxyl group (-COOH).
Dipeptide: A molecule consisting of two amino acids joined by a peptide bond.
Peptide bond: The covalent bond formed between the carboxyl group of one amino acid and the amino group of another.
Dehydration synthesis: The reaction mechanism that removes water to form the bond.
Step-by-Step Guidance
Identify the functional groups involved in the reaction: the carboxyl group of one amino acid and the amino group of the other.
Describe how the -OH from the carboxyl group and an -H from the amino group combine to form water ().
Explain that the removal of water allows the formation of a peptide bond between the two amino acids.
Optionally, sketch or describe the resulting dipeptide structure, indicating the location of the new bond.
Try solving on your own before revealing the answer!
Final Answer:
During dipeptide synthesis, the carboxyl group (-COOH) of one amino acid reacts with the amino group (-NH2) of another. The -OH from the carboxyl and an -H from the amino group combine to form a water molecule (), which is removed. This allows a peptide bond to form between the two amino acids, resulting in a dipeptide.
Q3. Explain why humans can digest starch but cannot extract energy from cellulose, despite both being polymers composed entirely of glucose monomers.
Background
Topic: Carbohydrate Structure and Enzyme Specificity
This question examines the structural differences between starch and cellulose and how these differences affect human digestion.
Key Terms:
Starch: A polysaccharide made of α-glucose monomers, digestible by humans.
Cellulose: A polysaccharide made of β-glucose monomers, indigestible by humans.
Enzyme specificity: The ability of enzymes to recognize and break specific bonds.
Glycosidic linkage: The type of bond connecting glucose monomers (α-1,4 in starch; β-1,4 in cellulose).
Step-by-Step Guidance
Describe the type of glycosidic bonds in starch (α-1,4) and cellulose (β-1,4).
Explain how human digestive enzymes (like amylase) can hydrolyze α-linkages but not β-linkages.
Discuss why the inability to break β-linkages prevents humans from extracting energy from cellulose.
Relate this to the concept of enzyme-substrate specificity.
Try solving on your own before revealing the answer!
Final Answer:
Humans can digest starch because our enzymes can break the α-1,4 glycosidic bonds between glucose monomers. However, cellulose has β-1,4 glycosidic bonds, which our digestive enzymes cannot hydrolyze. As a result, we cannot extract energy from cellulose, even though both are made of glucose.
Q4. What physical property makes "insoluble fiber" pass through the human digestive tract unabsorbed? Name the polymer involved.
Background
Topic: Dietary Fiber and Carbohydrate Structure
This question focuses on the properties of dietary fiber and why it is not digested or absorbed by humans.
Key Terms:
Insoluble fiber: Plant material that cannot be dissolved in water and is not digested by human enzymes.
Cellulose: The main polymer that makes up insoluble fiber.
Hydrophilic vs. hydrophobic: Relates to solubility in water.
Step-by-Step Guidance
Identify the main polymer that constitutes insoluble fiber in plants.
Describe the structural property (e.g., β-glycosidic bonds) that makes it resistant to digestion.
Explain why this property causes the fiber to pass through the digestive tract unabsorbed.
Try solving on your own before revealing the answer!
Final Answer:
Insoluble fiber is primarily cellulose, which contains β-1,4 glycosidic bonds that human digestive enzymes cannot break. This makes cellulose insoluble and indigestible, so it passes through the digestive tract unabsorbed.
Q5. A biochemist isolates a pure lipid sample with double bonds in the carbon chain tails. Is this fatty acid saturated or unsaturated? Will it be solid or liquid at room temperature? Explain at the molecular level.
Background
Topic: Lipid Structure and Physical Properties
This question tests your understanding of the difference between saturated and unsaturated fatty acids and how their structure affects their physical state.
Key Terms:
Saturated fatty acid: No double bonds; straight chains; usually solid at room temperature.
Unsaturated fatty acid: One or more double bonds; kinked chains; usually liquid at room temperature.
Double bond: A chemical bond that introduces a bend (kink) in the fatty acid chain.
Step-by-Step Guidance
Determine whether the presence of double bonds indicates saturation or unsaturation.
Explain how double bonds affect the shape of the fatty acid chains.
Discuss how the shape influences the ability of the molecules to pack together and their state at room temperature.
Try solving on your own before revealing the answer!
Final Answer:
The presence of double bonds means the fatty acid is unsaturated. Double bonds create kinks in the carbon chains, preventing tight packing, so unsaturated fats are usually liquid at room temperature.
Q6. Draw a simplified phospholipid molecule and label the hydrophilic head and hydrophobic tail regions. Explain how this amphipathic nature causes phospholipids to automatically form bilayers in aqueous environments.
Background
Topic: Membrane Structure and Amphipathic Molecules
This question explores the structure of phospholipids and how their properties lead to the formation of biological membranes.
Key Terms:
Phospholipid: A lipid with a hydrophilic (water-loving) head and two hydrophobic (water-fearing) tails.
Amphipathic: Having both hydrophilic and hydrophobic regions.
Bilayer: A double layer of phospholipids that forms the basis of cell membranes.
Step-by-Step Guidance
Sketch or visualize a phospholipid: a circle (head) with two lines (tails).
Label the head as hydrophilic (contains phosphate group) and the tails as hydrophobic (fatty acid chains).
Explain how, in water, the hydrophilic heads face outward toward water, while the hydrophobic tails face inward, away from water.
Describe how this arrangement leads to the spontaneous formation of a bilayer.
Try solving on your own before revealing the answer!
Final Answer:
A phospholipid has a hydrophilic (polar) head and two hydrophobic (nonpolar) tails. In water, phospholipids arrange themselves so that the heads face the water and the tails face each other, forming a bilayer. This amphipathic property drives the self-assembly of cell membranes.
Q7. Identify the four major structural levels of proteins (1°, 2°, 3°, 4°) and name the specific types of bonds/interactions responsible for maintaining each level.
Background
Topic: Protein Structure
This question tests your knowledge of the hierarchy of protein structure and the chemical forces that stabilize each level.
Key Terms:
Primary structure (1°): Sequence of amino acids.
Secondary structure (2°): Local folding (α-helix, β-sheet).
Tertiary structure (3°): Overall 3D shape of a single polypeptide.
Quaternary structure (4°): Association of multiple polypeptide chains.
Bonds/interactions: Peptide bonds, hydrogen bonds, disulfide bridges, ionic bonds, hydrophobic interactions, van der Waals forces.
Step-by-Step Guidance
List each structural level and briefly describe it.
For ea0ch level, identify the main type(s) of bond or interaction that stabilize it (e.g., peptide bonds for primary, hydrogen bonds for secondary).
Include examples of interactions for tertiary and quaternary structures.
Try solving on your own before revealing the answer!
Final Answer:
Primary structure: peptide (covalent) bonds. Secondary: hydrogen bonds. Tertiary: disulfide bridges, ionic bonds, hydrogen bonds, hydrophobic interactions, van der Waals forces. Quaternary: same as tertiary, but between different polypeptide chains.
Q8. If a protein undergoes thermal denaturation, which level(s) of protein structure remain intact, and why?
Background
Topic: Protein Denaturation
This question examines the effects of heat on protein structure and which bonds are disrupted or preserved.
Key Terms:
Denaturation: Loss of protein structure due to disruption of non-covalent interactions.
Primary structure: Sequence of amino acids held by peptide bonds.
Secondary, tertiary, quaternary structures: Maintained by weaker interactions (hydrogen bonds, ionic bonds, etc.).
Step-by-Step Guidance
Recall which bonds are affected by heat (non-covalent vs. covalent).
Identify which level of structure is held together by covalent peptide bonds.
Explain why this level remains intact while others are disrupted.
Try solving on your own before revealing the answer!
Final Answer:
Thermal denaturation disrupts secondary, tertiary, and quaternary structures, but the primary structure (amino acid sequence) remains intact because peptide bonds are not broken by heat.
Q9. How do the chemical properties of an amino acid's R-group (side chain) dictate its location in a folded, water-soluble globular protein?
Background
Topic: Protein Folding and Amino Acid Properties
This question tests your understanding of how hydrophobic and hydrophilic side chains influence protein structure in aqueous environments.
Key Terms:
R-group (side chain): The variable group on an amino acid that determines its properties.
Hydrophobic: Water-fearing; nonpolar side chains.
Hydrophilic: Water-loving; polar or charged side chains.
Globular protein: A compact, water-soluble protein.
Step-by-Step Guidance
Classify amino acid side chains as hydrophobic or hydrophilic based on their chemical properties.
Explain how hydrophobic side chains tend to be buried inside the protein, away from water.
Describe how hydrophilic side chains are usually found on the protein's surface, interacting with water.
Try solving on your own before revealing the answer!
Final Answer:
Hydrophobic (nonpolar) R-groups are typically found in the protein's interior, away from water, while hydrophilic (polar or charged) R-groups are exposed on the surface, interacting with the aqueous environment.
Q10. Name the three component parts that combine to make a single nucleotide monomer.
Background
Topic: Nucleic Acid Structure
This question tests your knowledge of the basic building blocks of DNA and RNA.
Key Terms:
Nucleotide: The monomer of nucleic acids.
Phosphate group, pentose sugar, nitrogenous base: The three components of a nucleotide.
Step-by-Step Guidance
List the three parts that make up a nucleotide.
Briefly describe the role of each component in the structure of nucleic acids.
Try solving on your own before revealing the answer!
Final Answer:
A nucleotide consists of a phosphate group, a five-carbon (pentose) sugar, and a nitrogenous base.
Q11. Compare and contrast DNA and RNA using the following criteria: sugar type, nitrogenous base differences, and structural strand count.
Background
Topic: Nucleic Acid Structure and Function
This question asks you to distinguish between DNA and RNA based on their chemical and structural features.
Key Terms:
Deoxyribose vs. ribose: The sugars in DNA and RNA, respectively.
Thymine vs. uracil: The nitrogenous base found in DNA but not RNA, and vice versa.
Double-stranded vs. single-stranded: The typical structure of DNA and RNA.
Step-by-Step Guidance
Identify the sugar present in DNA and RNA.
List the nitrogenous bases found in each (noting the difference between thymine and uracil).
State whether each molecule is typically single- or double-stranded.
Try solving on your own before revealing the answer!
Final Answer:
DNA contains deoxyribose sugar, uses thymine, and is double-stranded. RNA contains ribose sugar, uses uracil instead of thymine, and is usually single-stranded.
Q12. State the Central Dogma of molecular biology, outlining the direction of genetic information flow within a cell.
Background
Topic: Information Flow in Cells
This question tests your understanding of how genetic information is transferred from DNA to functional proteins.
Key Terms:
Central Dogma: The process by which genetic information flows from DNA to RNA to protein.
Transcription: DNA to RNA.
Translation: RNA to protein.
Step-by-Step Guidance
State the sequence of information flow (DNA → RNA → protein).
Briefly describe the processes of transcription and translation.
Try solving on your own before revealing the answer!
Final Answer:
The Central Dogma states that genetic information flows from DNA to RNA (transcription), then from RNA to protein (translation).
Q13. Which of the following best describes the chemical reaction that occurs when a glycosidic linkage forms between two monosaccharides?
Background
Topic: Carbohydrate Synthesis
This multiple-choice question tests your knowledge of how disaccharides are formed from monosaccharides.
Key Terms:
Glycosidic linkage: The covalent bond formed between two monosaccharides.
Dehydration synthesis: The reaction that removes water to form the bond.
Step-by-Step Guidance
Recall what happens during the formation of a glycosidic bond (is water added or removed?).
Eliminate answer choices that do not match the process of dehydration synthesis.
Focus on the answer that describes water removal and covalent bond formation.
Try solving on your own before revealing the answer!
Final Answer:
B) A water molecule is removed to form a covalent bond.
Q14. Hemoglobin is a protein composed of four polypeptide chains (two α and two β chains) working as a single functional complex. A single point mutation in the gene encoding the β chain causes valine (nonpolar) to replace glutamic acid (charged/polar). Which of the following best predicts the immediate structural consequence?
Background
Topic: Protein Structure and Mutations
This question tests your understanding of how changes in amino acid sequence affect protein structure and function.
Key Terms:
Primary structure: The sequence of amino acids.
R-group properties: Hydrophobic vs. hydrophilic side chains.
Quaternary structure: The arrangement of multiple polypeptide chains.
Step-by-Step Guidance
Identify the type of mutation (amino acid substitution) and its effect on primary structure.
Consider how replacing a charged/polar amino acid with a nonpolar one affects R-group interactions.
Predict how this change could alter the folding and aggregation of the protein complex.
Match your reasoning to the answer choices provided.
Try solving on your own before revealing the answer!
Final Answer:
B) Primary structure changes, which alters hydrophobic R-group interactions, leading to abnormal quaternary aggregation.
Q15. A research group analyzes an unknown cellular macromolecule and finds its stoichiometric formula is CnH2nOn. Which class of biomolecules does this sample belong to?
Background
Topic: Biomolecule Identification
This question tests your ability to recognize the general formula for carbohydrates.
Key Terms:
Stoichiometric formula: The ratio of elements in a molecule.
Carbohydrate: General formula is CnH2nOn.
Step-by-Step Guidance
Recall the general formula for carbohydrates.
Compare the given formula to those of other biomolecule classes (lipids, proteins, nucleic acids).
Identify which class matches the formula CnH2nOn.
Try solving on your own before revealing the answer!
Final Answer:
C) Carbohydrate
Q16. Pyrimidines and purines are two classes of nitrogenous bases found in nucleic acids. Which pair contains only pyrimidine bases found in nucleic acids?
Background
Topic: Nucleic Acid Chemistry
This question tests your knowledge of the classification of nitrogenous bases.
Key Terms:
Pyrimidines: Single-ring bases (cytosine, thymine, uracil).
Purines: Double-ring bases (adenine, guanine).
Step-by-Step Guidance
Recall which bases are pyrimidines and which are purines.
Eliminate answer choices that include purines or non-base molecules.
Select the pair that includes only pyrimidines found in DNA and/or RNA.
Try solving on your own before revealing the answer!
Final Answer:
A) Cytosine and Uracil
Q17. (FRQ) An enzyme isolated from a soil bacterium catalyzes the breakdown of starch into maltose.
(a) Describe the primary structure of this enzyme and identify the type of covalent bond that holds this primary sequence together.
Background
Topic: Protein Structure
This part asks you to define the primary structure of a protein and the bond type involved.
Key Terms:
Primary structure: Linear sequence of amino acids.
Peptide bond: Covalent bond linking amino acids.
Step-by-Step Guidance
Define what is meant by the primary structure of a protein.
Identify the specific covalent bond that connects amino acids in this sequence.
Try solving on your own before revealing the answer!
Final Answer:
The primary structure is the unique linear sequence of amino acids in the enzyme, held together by covalent peptide bonds.
(b) Explain how heating this enzyme to 80℃ will affect its tertiary structure and its catalytic functionality.
Background
Topic: Protein Denaturation and Enzyme Function
This part asks you to connect the effects of heat on protein structure and enzyme activity.
Key Terms:
Tertiary structure: 3D folding of a polypeptide.
Denaturation: Loss of structure and function due to disruption of non-covalent interactions.
Active site: The region of the enzyme where catalysis occurs.
Step-by-Step Guidance
Explain how high temperature affects the non-covalent interactions stabilizing tertiary structure.
Describe how loss of tertiary structure alters the shape of the active site.
Connect these changes to the enzyme's ability to catalyze reactions.
Try solving on your own before revealing the answer!
Final Answer:
Heating to 80℃ disrupts the tertiary structure by breaking non-covalent interactions, causing the enzyme to denature. This changes the shape of the active site, so the enzyme loses its catalytic function.
(c) Predict what would happen to the primary structure of the protein if it is placed in an 80℃ environment versus if it is treated with a digestive protease enzyme. Justify your prediction.
Background
Topic: Protein Stability and Enzymatic Digestion
This part asks you to compare the effects of heat and protease treatment on the primary structure of a protein.
Key Terms:
Primary structure: Sequence of amino acids.
Protease: Enzyme that hydrolyzes peptide bonds.
Denaturation vs. hydrolysis: Denaturation affects higher-order structure; hydrolysis breaks peptide bonds.
Step-by-Step Guidance
Explain what happens to the primary structure when exposed to high heat (denaturation).
Describe what happens when a protease enzyme is added (hydrolysis of peptide bonds).
Justify why the outcomes are different for each treatment.
Try solving on your own before revealing the answer!
Final Answer:
At 80℃, the primary structure (amino acid sequence) remains intact because peptide bonds are not broken by heat. However, treatment with a digestive protease breaks peptide bonds, so the primary structure is destroyed and the protein is broken into smaller peptides or amino acids.