AP Biology · Unit 1 · 8–11% of Exam ⚡ SPRINT MODE

Chemistry of Life

Zero fluff. Maximum exam yield. Every bullet here is either a direct MCQ target, a FRQ component, or a common trap. Master this unit in one focused session.

Exam Weight8–11%

~MCQs5–7 questions

FRQ AppearanceVery Frequent

Sprint Time~90 min

Water Properties CHONPS Dehydration / Hydrolysis Carbohydrates Lipids Nucleic Acids Protein Structure

⚡ Quick Glance — All Topics at a Glance

Topic	Priority	Exam Format	Key Trap
1.1 Water & H-Bonding	★★★	MCQFRQ	Cohesion ≠ Adhesion; ice has more H-bonds than liquid water
1.2 Elements of Life	★★	MCQ	Carbon has 4 bonds, not 2; phosphorus ≠ phosphate
1.3 Dehydration & Hydrolysis	★★★	MCQFRQ	Dehydration RELEASES water (builds); Hydrolysis CONSUMES water (breaks)
1.4 Carbohydrates	★★★	MCQData	Starch vs. cellulose = α vs. β glucose, NOT different monomers
1.5 Lipids	★★★	MCQFRQ	Lipids are NOT polymers; steroids cross membranes freely
1.6 Nucleic Acids	★★★	MCQCalcFRQ	Chargaff (%A=%T, %G=%C) applies to dsDNA only, NOT RNA or ssDNA
1.7 Proteins	★★★	MCQFRQData	Denaturation does NOT break peptide bonds (1° structure stays intact)

MCQ Standalone FRQ Component Data Analysis Calculation

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Topic 1.1

Structure of Water & Hydrogen Bonding

★★★ HIGH PRIORITY MCQ FRQ

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Root Cause

🔬 Why Water is Special

O is highly electronegative → pulls e⁻ away from H
Result: polar covalent bonds → δ− on O, δ+ on H
Bent shape → molecule is polar overall
Polarity → hydrogen bonds between molecules
H-bonds: weak individually, collectively powerful
Each H₂O can form up to 4 H-bonds

Property 1–2

🔗 Cohesion & Adhesion

Cohesion: H₂O ↔ H₂O → surface tension
Enables water striders; paper clip float on water
Adhesion: H₂O ↔ other polar surfaces (xylem walls)
Together → capillary action in xylem
Transpiration pull: tension + cohesion = continuous column
Cohesion ≠ Adhesion: common exam trap!

Property 3–4

🌡 Thermal Properties

High specific heat (4.18 J/g·°C): H-bonds absorb energy before temp rises → stabilizes body temp, moderates climate
High heat of vaporization: evaporation removes lots of heat → sweating cools mammals
Both stem from energy needed to break H-bonds

Property 5–6

🧊 Density & Solvent

Ice less dense than liquid water: crystal lattice = more spacious → ice floats → insulates aquatic life in winter
Ice has MORE organized H-bonds than liquid water (4 each, fixed)
Universal solvent: dissolves polar/ionic solutes via hydration shells
Hydrophobic molecules excluded → drives bilayer formation & protein folding

🎯 Exam Sniper — Where This Shows Up

MCQ: "Why can water striders walk on water?" → cohesion / surface tension (NOT viscosity, NOT adhesion)
MCQ: "Why does sweating cool the body?" → high heat of vaporization
MCQ/FRQ: Xylem water transport always requires both cohesion AND adhesion to be mentioned
FRQ: May show graph of water temp vs. heat added — flat parts = phase changes (H-bonds breaking). Steep parts = kinetic energy increasing
Data Q: Given climate data, connect ocean temperature stability → high specific heat of water

💣 Trap Alert

❌ H-bonds are NOT covalent — they are weak electrostatic attractions
❌ Ice has more H-bonds than liquid water, NOT fewer — "less dense" ≠ "less bonded"
❌ Cohesion (H₂O–H₂O) ≠ Adhesion (H₂O–surface) — both are needed for plant transport
❌ "High specific heat" ≠ "high heat of vaporization" — different properties with different biological roles

MCQTopic 1.1

In a plant on a hot day, water moves from roots to leaves through xylem vessels even without a pump. Which combination of water properties best explains this movement?

(A) High specific heat and high heat of vaporization
(B) Cohesion and adhesion
(C) Lower density of ice and universal solvent ability
(D) Hydrogen bonding and high specific heat

Answer: (B) — Cohesion holds the water column together as tension pulls it upward (transpiration pull). Adhesion to xylem cell walls helps support the column against gravity. Neither thermal property explains movement of water in xylem. This is a high-frequency question — always name both cohesion AND adhesion for full credit.

Topic 1.2

Elements of Life

★★ MEDIUM PRIORITY MCQ

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Memorize This

⚛️ CHONPS — The Six Elements

Carbon — backbone of all organic molecules
Hydrogen — in all organic molecules + water
Oxygen — in most macromolecules + water
Nitrogen — in amino acids (amino group) & nucleic acids (bases)
Phosphorus — in DNA/RNA backbone, ATP, phospholipids
Sulfur — in some amino acids (cysteine, methionine) → disulfide bonds

Carbon is Key

🔗 Why Carbon = Life

4 valence electrons → can form 4 covalent bonds
Bonds to C, H, O, N, S, P → enormous molecular diversity
Can form: chains, branches, rings
Functional groups determine reactivity: hydroxyl (–OH), carboxyl (–COOH), amino (–NH₂), phosphate (–PO₄), carbonyl (C=O), methyl (–CH₃)

Functional Groups — Exam Hits

🧪 Key Functional Groups

Carboxyl (–COOH): acidic, in amino acids & fatty acids
Amino (–NH₂): basic, in amino acids
Phosphate (–PO₄): in ATP, DNA, phospholipids; transfers energy
Hydroxyl (–OH): polar, in sugars & alcohols; participates in dehydration reactions
Sulfhydryl (–SH): in cysteine; forms disulfide bonds in proteins

🎯 Exam Sniper

MCQ: Given a molecular diagram, identify which functional group is present → know shapes of –OH, –COOH, –NH₂, –PO₄, –SH
MCQ: "Which element is found in proteins but NOT carbohydrates?" → Nitrogen (and sulfur)
MCQ: "Which element distinguishes nucleic acids from proteins?" → Phosphorus (DNA/RNA backbone) — both have C, H, O, N

Topic 1.3

Introduction to Macromolecules — Dehydration & Hydrolysis

★★★ HIGH PRIORITY MCQ FRQ

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Build vs. Break

🔧 Dehydration Synthesis (Condensation)

Builds polymers from monomers
Releases one H₂O molecule per bond formed
Requires energy input (anabolic)
Forms: glycosidic bonds (carbs), ester bonds (lipids), peptide bonds (proteins), phosphodiester bonds (nucleic acids)
Monomer + Monomer → Polymer + H₂O

Build vs. Break

💧 Hydrolysis

Breaks polymers into monomers
Consumes one H₂O molecule per bond broken
Releases energy (catabolic) — used in digestion
All digestion enzymes catalyze hydrolysis
Polymer + H₂O → Monomer + Monomer

Monomer → Polymer

🧱 The Four Macromolecules

Carbohydrates: glucose → starch/cellulose/glycogen
Proteins: amino acids → polypeptides
Nucleic Acids: nucleotides → DNA/RNA
Lipids: NOT true polymers — no repeating monomers
Triglycerides: glycerol + 3 fatty acids (ester bonds)

🎯 Exam Sniper

MCQ: "A cell is digesting stored glycogen. Which reaction type occurs?" → Hydrolysis
MCQ: "During translation, amino acids are joined. Which reaction type occurs and what is released?" → Dehydration synthesis; water is released
FRQ: May ask you to compare anabolic vs. catabolic reactions — always tie dehydration = anabolic, hydrolysis = catabolic
FRQ: "How many water molecules are released when a 50-amino acid polypeptide forms?" → 49 (one per peptide bond = n−1)

💣 Trap Alert

❌ Dehydration does NOT add water — it removes water; Hydrolysis does NOT remove water — it uses water
❌ The number of water molecules released = n − 1 (where n = number of monomers)
❌ Lipids are NOT polymers — do not say "lipid monomers"

Topic 1.4

Carbohydrates

★★★ HIGH PRIORITY MCQ Data Analysis

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The Critical Comparison: α vs. β Glucose

Feature	α-Glucose	β-Glucose
–OH on C1	Points down	Points up
Bond formed	α-1,4-glycosidic bond	β-1,4-glycosidic bond
Polysaccharides	Starch (amylose/amylopectin), Glycogen	Cellulose, Chitin
3D shape	Coiled helix → energy storage	Straight, stacked sheets → structural
Humans digest?	✓ Yes (amylase breaks α bonds)	✗ No (no enzyme for β bonds)
Function	Short-term energy storage	Cell walls (plants), exoskeletons (chitin in insects/fungi)

Know These Cold

🍞 Polysaccharide Summary

Starch: α-glucose; plants store energy; amylose (unbranched) + amylopectin (branched)
Glycogen: α-glucose; animals/fungi store glucose; highly branched → quick mobilization
Cellulose: β-glucose; plant cell walls; hydrogen bonds between chains = incredible strength
Chitin: β-glucose + nitrogen; insect/crustacean exoskeletons & fungal cell walls

Monosaccharides

🔬 Simple Sugars

Glucose (C₆H₁₂O₆): most common; cellular respiration substrate
Fructose: found in fruits; isomer of glucose (same formula, different structure)
Ribose (C₅H₁₀O₅): in RNA; Deoxyribose: in DNA (missing –OH at C2)
Disaccharides: sucrose = glucose + fructose; lactose = glucose + galactose; maltose = glucose + glucose

🎯 Exam Sniper

MCQ (top hit): "Why can cows digest grass but humans cannot?" → Cows have bacteria with β-glucosidase that break β-1,4 bonds in cellulose; humans lack this enzyme
MCQ: "Starch and cellulose are both made of glucose. Why do they have different functions?" → Different bond types (α vs. β) → different 3D shapes → different properties
Data Analysis: Iodine test — starch turns blue-black (iodine intercalates into helix). Does NOT react with cellulose, glucose, or other sugars
MCQ: Animals store excess glucose as glycogen (NOT starch) — mainly in liver and muscle

💣 Trap Alert

❌ Starch and cellulose differ by bond type, NOT by monomer — both are 100% glucose
❌ Glycogen is the animal storage polysaccharide, NOT starch (plants make starch)
❌ Chitin contains nitrogen (it's N-acetylglucosamine) — do not describe it as "pure glucose"

MCQTopic 1.4HIGH FREQUENCY

A researcher compares the digestibility of starch and cellulose in human subjects. Starch is completely broken down, but cellulose passes through undigested. Which molecular difference between starch and cellulose best explains this result?

(A) Starch is made of glucose while cellulose is made of galactose
(B) Starch contains α-glycosidic bonds while cellulose contains β-glycosidic bonds
(C) Starch is a disaccharide while cellulose is a polysaccharide
(D) Cellulose monomers are linked by hydrogen bonds rather than covalent bonds

Answer: (B) — Both starch and cellulose are made entirely of glucose (eliminating A). Human amylase is specifically shaped to break α-1,4 bonds; it cannot break β-1,4 bonds. The difference is bond geometry, not monomer identity. This makes cellulose indigestible (structural) and starch digestible (energy storage).

Topic 1.5

Lipids

★★★ HIGH PRIORITY MCQ FRQ

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Critical Fact

⚠️ Lipids are NOT Polymers

No repeating monomer unit — unified by being hydrophobic / nonpolar
Poorly soluble in water because they lack polar functional groups
Four main types: triglycerides, phospholipids, waxes, steroids

Type 1

🛢 Triglycerides (Fats & Oils)

Structure: 1 glycerol + 3 fatty acids → linked by ester bonds (dehydration synthesis)
Saturated fat: no double bonds; straight chains; pack tightly; solid at room temp (animal fats)
Unsaturated fat: one or more C=C double bonds; kinked chains; can't pack tightly; liquid at room temp (plant oils)
Function: long-term energy storage (2× more energy per gram than carbs)

Type 2 — Most Tested

🫧 Phospholipids

Structure: glycerol + 2 fatty acids + 1 phosphate group
Amphipathic: hydrophilic phosphate "head" + hydrophobic fatty acid "tails"
In water → spontaneously form bilayer (tails face inward, heads face out)
Foundation of ALL cell membranes
Fluidity affected by: unsaturation (↑ fluidity), cholesterol (buffers fluidity), temperature

Type 3

🔬 Steroids

4 fused carbon rings — very different structure from other lipids
Cholesterol: in animal cell membranes → regulates fluidity; precursor for other steroids
Steroid hormones: testosterone, estrogen, cortisol — all derived from cholesterol
Steroid hormones are lipid-soluble → cross membranes directly → bind intracellular receptors (NOT membrane receptors)

🎯 Exam Sniper

MCQ (top hit): "Why do phospholipids form a bilayer in water?" → amphipathic nature (hydrophilic heads + hydrophobic tails) — thermodynamically favorable to minimize hydrophobic exposure
MCQ: "Which of the following is NOT a polymer?" → Lipid / Triglyceride (proteins, carbs, nucleic acids are polymers)
MCQ: A steroid hormone can enter a cell without a membrane receptor because it is nonpolar/lipid-soluble → diffuses through the phospholipid bilayer
FRQ (Unit 2 connection): Membrane fluidity — unsaturated fatty acids ↑ fluidity; cholesterol acts as a buffer (↑ fluidity when cold, ↓ when hot); connects to Unit 2
MCQ: Butter (saturated fat) is solid at room temperature; olive oil (unsaturated) is liquid — explain based on molecular packing

💣 Trap Alert

❌ Lipids are NOT polymers — never say "lipid monomers"
❌ Steroid hormones do NOT need membrane receptors — they enter cells directly. Only hydrophilic (protein/peptide) hormones need surface receptors
❌ "Saturated" refers to hydrogen saturation of C–C bonds, NOT to the fat being absorbed by the body
❌ Waxes are ester-linked lipids — they are also NOT polymers and are highly hydrophobic (leaf cuticle, ear canal)

Topic 1.6

Nucleic Acids

★★★ HIGH PRIORITY MCQ Calculation FRQ

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DNA vs. RNA — The Must-Know Comparison

Feature	DNA	RNA
Sugar	Deoxyribose (no –OH at C2)	Ribose (–OH at C2)
Bases	A, T, G, C	A, U (uracil, no thymine), G, C
Strands	Double-stranded (helix)	Single-stranded (usually)
Location	Nucleus (+ mitochondria, chloroplasts)	Nucleus → cytoplasm
Function	Long-term information storage	mRNA (message), tRNA (translator), rRNA (ribosome)
Stability	More stable (no 2'–OH)	Less stable

Calculation Target

🧬 Chargaff's Rules (dsDNA only)

A pairs with T (2 hydrogen bonds)
G pairs with C (3 hydrogen bonds) → stronger
Therefore: %A = %T and %G = %C
Also: %A + %G = %T + %C = 50% (purines = pyrimidines)
Applies to double-stranded DNA only! NOT ssDNA, NOT RNA
More G≡C pairs → higher melting temperature (3 H-bonds harder to break)

Structure

🔬 Nucleotide Structure

Monomer = nucleotide = phosphate + sugar + nitrogenous base
Nucleotides linked by phosphodiester bonds (3'–5' direction)
DNA strands are antiparallel: 5'→3' on one strand, 3'→5' on the other
Purines (2 rings): Adenine, Guanine — A & G
Pyrimidines (1 ring): Cytosine, Thymine, Uracil — C, T, U
ATP: adenine + ribose + 3 phosphates — energy currency (nucleotide derivative)

🎯 Exam Sniper

Calculation MCQ: "DNA sample has 30% adenine. What is the percentage of cytosine?" → %T = 30%, so %G+%C = 40%, thus %C = 20%
MCQ: "Which base pairs with guanine in DNA? In RNA?" → DNA: C (3 H-bonds); RNA: C (same, G-C pairing is universal)
MCQ: "Two DNA sequences are compared. Sequence A has 60% G+C. Sequence B has 40% G+C. Which has a higher melting temperature?" → Sequence A (more G≡C = 3 H-bonds each = harder to denature)
FRQ: Often integrated into gene expression questions in Unit 6 — Unit 1 nucleic acid structure is foundational
MCQ: "Where is DNA found in eukaryotes?" → Nucleus, mitochondria, chloroplasts (endosymbiosis connection!)

💣 Trap Alert

❌ Chargaff's rules apply to double-stranded DNA only — in single-stranded DNA or RNA, %A ≠ %T
❌ RNA uses Uracil (U) not Thymine — RNA has no thymine
❌ DNA backbone: phosphate–deoxyribose (NOT ribose). RNA: phosphate–ribose
❌ Purines are the larger bases (A and G) — A is a purine (pairs with T, a pyrimidine)

Calculation MCQTopic 1.6HIGH FREQUENCY

Analysis of a double-stranded DNA molecule reveals that 22% of the nucleotides contain adenine. Which of the following correctly describes the nucleotide composition of this molecule?

(A) 22% thymine, 28% guanine, 28% cytosine
(B) 22% thymine, 28% guanine, 28% cytosine
(C) 22% thymine, 56% guanine, 0% cytosine
(D) 44% thymine, 6% guanine, 6% cytosine

Answer: (B) — By Chargaff's Rules: %A = %T = 22%. Since all bases must sum to 100%: %G + %C = 100% − 22% − 22% = 56%. Since %G = %C: each is 28%. So composition: A=22%, T=22%, G=28%, C=28%. This calculation type appears almost every year — memorize the formula: %G = %C = (100 − 2×%A) ÷ 2.

Topic 1.7

Proteins

★★★ HIGH PRIORITY MCQ FRQ Data Analysis

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The Four Levels of Protein Structure — CRITICAL

Level	What it is	Bond/Interaction	Exam Key Point
1° Primary	Sequence of amino acids (AA)	Peptide bonds (covalent)	Determined by DNA. Dictates all other levels. NOT broken by denaturation
2° Secondary	α-helix or β-pleated sheet	Hydrogen bonds (between backbone NH and C=O)	Broken by heat/pH changes. Results from H-bonding of backbone
3° Tertiary	Overall 3D shape of polypeptide	H-bonds, ionic bonds, hydrophobic interactions, disulfide bonds (covalent)	Shape = function. Disulfide bonds most stable; broken by reducing agents
4° Quaternary	Multiple polypeptide chains assembled together	Same as 3° (between subunits)	Only proteins with multiple subunits (e.g., hemoglobin = 4 subunits; collagen)

Key Concept

🔴 Denaturation

Loss of 3D shape = loss of function
Caused by: high temperature, extreme pH, heavy metals, certain chemicals
Breaks: H-bonds, ionic bonds, hydrophobic interactions (2°, 3°, 4° structure)
Does NOT break peptide bonds (1° structure intact)
Usually irreversible (cooked egg white)
Some proteins can refold (chaperones assist)

Amino Acid Basics

🧪 Amino Acid Structure

Central carbon bonded to: NH₂ (amino), COOH (carboxyl), H, and R-group
R-group determines identity and properties
R-group polarity determines folding: nonpolar R-groups → hydrophobic core; polar/charged R-groups → outside, facing water
20 different amino acids → ~unlimited protein diversity
Peptide bond: between carboxyl of one AA + amino of next (dehydration)

Shape = Function

⚙️ Protein Functions

Enzymes: catalysts; shape of active site = substrate specificity
Structural: collagen (connective tissue), keratin (hair/nails), actin/myosin (muscle)
Transport: hemoglobin (O₂), channel/carrier proteins (membrane)
Defense: antibodies (immunoglobulins)
Signaling: some hormones (insulin, glucagon); receptors
Motor: myosin, dynein, kinesin

🎯 Exam Sniper

MCQ (top hit): "A mutation changes one amino acid in a protein. How might this affect protein function?" → Could alter 3D shape (if R-group change affects folding) → loss of function. If charge/polarity changes in active site → loss of enzyme function. If conservative substitution → possibly no effect
FRQ: "Explain why an enzyme stops working at high temperatures" → Must mention: heat breaks hydrogen bonds and other non-covalent interactions holding 3° structure → active site loses shape → substrate cannot bind → enzyme non-functional (denatured). Note: peptide bonds remain intact
Data Analysis: Graph of enzyme activity vs. pH or temperature — identify optimal conditions, explain loss of function at extremes using structure knowledge
MCQ: "Hemoglobin has four subunits. This represents _____ structure." → Quaternary
FRQ: Connecting protein structure to evolution — mutation in DNA → change in AA sequence → change in protein shape → change in function → selection pressure

💣 Trap Alert

❌ Denaturation does NOT break peptide bonds — primary structure (AA sequence) is always preserved. This is the #1 protein error on the exam
❌ All proteins have 1°, 2°, and 3° structure — only some have 4° (only multi-subunit proteins)
❌ Disulfide bonds (S–S) are covalent — they are the one covalent bond in 3° structure and are NOT broken by temperature alone
❌ "R-group" is not always nonpolar — R-groups can be polar, nonpolar, acidic, or basic; this determines protein folding

FRQ-style MCQTopic 1.7HIGH FREQUENCY

A researcher heats an enzyme solution from 37°C to 90°C and observes a complete loss of enzymatic activity. When the solution cools back to 37°C, activity does not return. Which of the following best explains why activity was lost and did not recover?

(A) The peptide bonds holding the amino acid chain together were hydrolyzed by the heat
(B) The substrate concentration decreased as temperature increased
(C) High temperature disrupted non-covalent interactions, causing the protein to unfold and lose its active site shape irreversibly
(D) Heat caused the DNA encoding the enzyme to denature, preventing new enzyme production

Answer: (C) — Heat disrupts hydrogen bonds, hydrophobic interactions, and ionic bonds that maintain 2° and 3° structure. This causes the protein to unfold (denature), destroying the active site's shape. The substrate can no longer bind. The enzyme cannot recover because re-folding into the exact correct shape is thermodynamically unlikely without assistance. Note: peptide bonds (A) are covalent and are NOT broken by heat — this is a classic trap.

Practice

Sprint Practice — Mixed Questions

MCQCross-Topic

Which of the following correctly pairs a macromolecule with its monomer AND the type of bond linking monomers together?

(A) Starch — fructose — glycosidic bond
(B) Protein — nucleotide — hydrogen bond
(C) DNA — deoxyribonucleotide — phosphodiester bond
(D) Triglyceride — amino acid — ester bond

Answer: (C) — DNA monomers are deoxyribonucleotides, linked by phosphodiester bonds (between the 3' OH of one nucleotide and the 5' phosphate of the next). (A) Starch is made of glucose, not fructose. (B) Proteins are made of amino acids, linked by peptide bonds (not H-bonds). (D) Triglycerides are made of glycerol + fatty acids (not amino acids), linked by ester bonds — the bond is correct but the monomer is wrong.

FRQ-StyleData Analysis

A DNA sample from a bacterium is analyzed and found to contain 15% adenine. A student predicts that the DNA of an extremophile living in hot springs would have a higher percentage of G+C pairs than the bacterium. Provide TWO pieces of reasoning that support this prediction.

(A) G-C pairs have 2 hydrogen bonds; more G-C = lower melting point = more stable in heat
(B) G-C pairs have 3 hydrogen bonds; more energy is needed to break G-C pairs; high G+C content raises the DNA melting temperature, making the genome more stable in high-heat environments
(C) Extremophiles use RNA instead of DNA, which has more uracil to stabilize at high temperatures
(D) High G+C content shortens the DNA strands, making them more compact and heat-resistant

Answer: (B) — G≡C pairs form 3 hydrogen bonds (vs. A=T which form only 2). Therefore, DNA with a higher G+C content requires more thermal energy to separate the strands. Extremophiles in hot springs are under selection pressure to maintain genome integrity at high temperatures. Natural selection would favor organisms with higher G+C content, as their DNA would be more thermally stable. The bacterium has %A=15%, so %T=15%, %G+%C=70% — already fairly high; an extremophile would be even higher.

MCQReasoning

A student argues that lipids should not be classified with carbohydrates, proteins, and nucleic acids as "macromolecules" because lipids lack a defining characteristic shared by the other three groups. Which characteristic is the student referring to?

(A) Lipids do not contain carbon atoms
(B) Lipids are not polymers composed of repeating monomer subunits
(C) Lipids cannot form hydrogen bonds with water
(D) Lipids are not involved in any cellular processes

Answer: (B) — Carbohydrates, proteins, and nucleic acids are all polymers made of repeating monomers (monosaccharides, amino acids, nucleotides respectively) linked by dehydration synthesis. Lipids have no such repeating monomer unit — triglycerides are glycerol + 3 fatty acids; steroids are 4 fused rings. This is why lipids are often called "the macromolecule that isn't a polymer." (A) is wrong — lipids do contain carbon. (C) — technically true but not the "defining characteristic" asked about.

⚠ Trap Alert

Unit 1 High-Frequency Exam Traps

💧
Cohesion ≠ Adhesion (and both are needed for xylem transport)Cohesion = water-to-water (surface tension, maintains water column). Adhesion = water-to-other-surface (xylem walls). Saying only "cohesion" for xylem transport loses points — you must mention BOTH.
🧊
Ice has MORE H-bonds than liquid water, not fewerIn ice, each water molecule forms exactly 4 H-bonds in a rigid lattice. In liquid water, H-bonds constantly break and reform. Ice floats because the lattice is more spacious (lower density), NOT because it has fewer bonds.
🍞
Starch vs. cellulose: different bonds, SAME monomerBoth are 100% glucose. The difference is the bond angle: α-1,4 (starch, coiled, digestible) vs. β-1,4 (cellulose, straight, indigestible). If you say "different monomers" you will lose points.
🫧
Lipids are NOT polymers — never write "lipid monomers"This is a heavily tested distinction. Carbohydrates, proteins, and nucleic acids are all polymers with repeating monomers. Lipids are not. Triglycerides have a glycerol backbone + 3 fatty acids, but this is NOT described as monomer–polymer relationship.
🔥
Denaturation does NOT break peptide bonds (1° structure intact)Heat and pH changes break non-covalent bonds (H-bonds, ionic, hydrophobic interactions) holding 2°, 3°, 4° structure. Peptide bonds are covalent and require much more energy (hydrolysis by proteases, not heat) to break. The amino acid sequence remains after denaturation.
🧬
Chargaff's rules apply ONLY to double-stranded DNAIn dsDNA: %A=%T and %G=%C. In single-stranded DNA or RNA, this relationship does NOT hold because there is no complementary strand. A common trick question gives you an RNA strand and asks you to apply Chargaff's rules — DO NOT.
💊
Steroid hormones cross membranes directly — no receptor on membrane surfaceSteroids are lipid-soluble (nonpolar) and diffuse through the phospholipid bilayer. They bind intracellular receptors (cytoplasm or nucleus). Only hydrophilic hormones (peptide hormones, e.g., insulin) cannot cross the membrane and require surface receptors.
⚛️
"Which element is unique to proteins/nucleic acids?"Nitrogen: in both amino groups (proteins) and nitrogenous bases (nucleic acids). Phosphorus: unique to nucleic acids (backbone) and phospholipids — NOT found in proteins or carbohydrates. Sulfur: found in some amino acids (cysteine), NOT in carbohydrates or nucleic acids.

✓ Last-Min Checklist

Pre-Exam 10-Minute Checklist

Click each item to mark as confirmed. Review any you can't check off.

Water (1.1)

I can explain WHY water is polar (electronegativity of O → δ charges → H-bonds)
I can name all 4 water properties and their biological significance
I know cohesion ≠ adhesion, and both are needed for xylem water transport
I know ice is less dense because of its MORE ordered H-bond lattice (NOT fewer bonds)

Elements & Reactions (1.2–1.3)

I can name CHONPS and what each element is found in
I know phosphorus is in nucleic acids, ATP, and phospholipids — NOT proteins or carbs
Dehydration synthesis = builds polymers + releases H₂O
Hydrolysis = breaks polymers + consumes H₂O
n monomers joined = (n−1) water molecules released

Carbohydrates (1.4)

α-glucose → starch (plants) & glycogen (animals) — energy storage, digestible
β-glucose → cellulose (plants) & chitin (fungi/insects) — structural, indigestible
Starch vs. cellulose = SAME monomer, DIFFERENT bond type (α vs. β)
Glycogen is more branched than starch → faster glucose mobilization in animals

Lipids (1.5)

Lipids are NOT polymers — no repeating monomer unit
Phospholipids are amphipathic → form bilayers in water
Unsaturated fats = kinked chains = more fluid at room temperature
Steroid hormones are lipid-soluble → cross membranes freely → intracellular receptors

Nucleic Acids (1.6)

DNA: deoxyribose, T (no U), double-stranded
RNA: ribose, U (no T), single-stranded
A=T (2 H-bonds), G≡C (3 H-bonds) — Chargaff's rules for dsDNA ONLY
I can calculate %G and %C if given %A in a dsDNA sample
Higher G+C content → higher melting temperature

Proteins (1.7)

4 levels of protein structure: 1° (peptide bonds), 2° (H-bonds, helix/sheet), 3° (H-bonds, ionic, hydrophobic, disulfide), 4° (multi-subunit)
Denaturation breaks NON-covalent bonds (2°, 3°, 4°) — does NOT break peptide bonds
Shape determines function — a mutation changing AA sequence can alter shape → loss of function
I can explain why an enzyme stops working at high temp or extreme pH

⚡ Final Sprint Strategy for Unit 1

Highest yield topics (do these first): Protein denaturation, α vs. β glucose, Chargaff's rule calculations, phospholipid bilayer formation, water properties + xylem
FRQ danger zones: Always explain HOW a property causes a biological outcome (not just name it). "High specific heat" alone is wrong — must say "stabilizes body/ocean temperature because it resists temperature change"
Data questions: Iodine = starch only; Biuret reagent = proteins; Benedict's = reducing sugars. Know what each test detects
Connection forward: Unit 1 appears in Unit 2 (membrane structure), Unit 3 (ATP = nucleotide), Unit 6 (DNA structure), Unit 7 (heredity). Everything connects back here