High School Biology (Next Generation Science Standards) • Molecules to Organisms: Structures and Processes

Describe the structure of DNA, including nucleotide composition and base-pairing rules.

Discover how the double helix stores the instructions that build and maintain every living organism on Earth.

High School Biology emblem with a microscope, leaf, and DNA helix
Inside this lesson 10 sections 5 questions 6 figures

SECTION 1

Historical Context & Motivation

The Quest to Understand Heredity's Molecule

For centuries, scientists knew that offspring resembled their parents, but the molecular basis of heredity remained a mystery. By the mid-twentieth century, researchers understood that chromosomes carried genetic information, yet they debated whether proteins or nucleic acids served as the true hereditary molecule. A series of landmark experiments gradually pointed toward deoxyribonucleic acid (DNA) as the molecule responsible for storing and transmitting genetic instructions. Understanding DNA's structure became one of the most important scientific pursuits of the twentieth century.

This lesson traces that discovery and then walks through the molecular architecture of DNA in detail. You will explore the anchoring phenomenon of X-ray diffraction patterns—specifically Photo 51, taken by Rosalind Franklin—which revealed a repeating helical pattern that ultimately guided the construction of a correct structural model for DNA. By investigating this phenomenon, you will use structure–function relationships (a crosscutting concept) to explain how DNA's shape enables it to store and copy genetic information.

1869 • Miescher Isolates 'Nuclein'

Swiss biochemist Friedrich Miescher extracted a phosphorus-rich substance from white blood cells. He named it nuclein, later recognized as nucleic acid.

1928 • Griffith's Transformation Experiment

Frederick Griffith demonstrated that a 'transforming principle' could transfer genetic traits between bacteria, hinting that a specific molecule carried hereditary information.

1944 • Avery, MacLeod & McCarty

Building on Griffith's work, this team identified DNA—not protein—as the transforming principle responsible for heredity.

1952 • Photo 51 — Franklin & Wilkins

Rosalind Franklin's X-ray diffraction image of DNA (Photo 51) revealed a clear helical structure with measurable spacing, providing critical evidence for building a structural model.

1953 • Watson & Crick Propose the Double Helix

James Watson and Francis Crick combined chemical data, Chargaff's rules, and Franklin's X-ray evidence to propose the double-helix model of DNA, one of biology's greatest breakthroughs.

The central question that drove these discoveries was: How does the physical structure of DNA enable it to store vast amounts of genetic information and copy itself accurately? Answering this question requires examining DNA at the molecular level, connecting each structural feature to a biological function—an approach at the heart of the NGSS crosscutting concept of structure and function.

SECTION 2

Core Principles & Definitions

Building Blocks of the Double Helix

DNA is a polymer—a long chain built from repeating units called nucleotides. Each nucleotide consists of three covalently bonded components: a five-carbon sugar, a phosphate group, and a nitrogenous base. The sugar in DNA is deoxyribose, which differs from ribose (the sugar in RNA) by lacking an oxygen atom on its 2′ carbon. Nucleotides link together through phosphodiester bonds to form a long strand with a sugar-phosphate backbone. Two such strands wind around each other in opposite orientations to form the famous double helix.

1

Nucleotide — The Monomer

Each nucleotide has three parts: a phosphate group, a deoxyribose sugar, and a nitrogenous base. These units polymerize to form a single DNA strand.
2

Four Nitrogenous Bases

DNA contains four bases: adenine (A), thymine (T), guanine (G), and cytosine (C). A and G are purines (double ring); T and C are pyrimidines (single ring).
3

Complementary Base Pairing

Adenine pairs with thymine via two hydrogen bonds (A=T). Guanine pairs with cytosine via three hydrogen bonds (G≡C). This specificity is called Chargaff's rules.
4

Antiparallel Strands

The two strands run in opposite directions: one strand reads 5′→3′ while its partner reads 3′→5′. This antiparallel orientation is essential for DNA replication and transcription.
5

Double Helix Geometry

The helix makes one complete turn every 10 base pairs (~3.4 nm). The diameter is approximately 2 nm. The two strands are held together by hydrogen bonds between complementary bases.
✦

KEY TAKEAWAY

Think of DNA like a twisted ladder. The sugar-phosphate backbone forms the two side rails, while the complementary base pairs form the rungs. Just as a zipper's teeth interlock only in one specific way, A only fits with T and G only fits with C. This precise pairing means that if you know one strand's sequence, you automatically know the other—explaining how cells copy DNA so faithfully.

SECTION 3

Visual Explanation — The Nucleotide and Double Helix

Anatomy of a DNA Nucleotide

Structure of a DNA NucleotidePHOSPHATEPO₄³⁻bondDEOXYRIBOSEC₅H₁₀O₃5′1′3′bondNITROGENOUSBASEA, T, G, or CThe Four Nitrogenous BasesPURINES (double ring)AdenineAGuanineGPYRIMIDINES (single ring)ThymineTCytosineCBase Pairing RulesAT2 H-bondsGC3 H-bondsPurine – PyrimidinePurine – Pyrimidine
A single DNA nucleotide consists of a phosphate group (blue), a deoxyribose sugar (pink), and one of four nitrogenous bases (gold). Purines (A, G) have a double-ring structure, while pyrimidines (T, C) have a single ring. Base-pairing rules dictate that A pairs with T via two hydrogen bonds and G pairs with C via three hydrogen bonds.

The diagram above highlights the modular design of each nucleotide. Notice how the phosphate group connects to the 5′ carbon of deoxyribose, while the nitrogenous base attaches at the 1′ carbon. The 3′ carbon of one sugar connects via a phosphodiester bond to the phosphate of the next nucleotide, forming the continuous backbone. This 5′-to-3′ directionality gives each strand a chemical polarity that is essential for enzymes like DNA polymerase to read and copy the strand correctly. The consistent pairing of a purine with a pyrimidine ensures that the double helix maintains a uniform 2 nm diameter along its entire length.

SECTION 4

How the Double Helix Works — Structure Meets Function

From Sequence to Information Storage

DNA's ability to encode biological information comes from the specific sequence of base pairs along the molecule. Just as the 26 letters of the English alphabet combine to form millions of words, the four bases A, T, G, and C combine in trillions of possible sequences to encode all the instructions an organism needs. Each human cell contains approximately 3.2 billion base pairs of DNA, collectively called the genome. If you stretched all the DNA from a single cell end to end, it would measure about 2 meters—yet it is packed into a nucleus only about 6 micrometers across. This remarkable compaction relies on DNA winding around histone proteins, forming a structure called chromatin.

Chargaff's Rules — A Quantitative Pattern

Before Watson and Crick built their model, biochemist Erwin Chargaff discovered a consistent pattern in DNA from many organisms: the amount of adenine always equals the amount of thymine, and the amount of guanine always equals the amount of cytosine. This can be expressed quantitatively as %A = %T and %G = %C. A direct consequence is that the total percentage of purines always equals the total percentage of pyrimidines, so %A + %G = %T + %C = 50% of all bases. Chargaff's rules are a perfect example of the crosscutting concept of patterns: an observable regularity in data that suggests an underlying mechanism—in this case, complementary base pairing.

CHARGAFF'S FIRST RULE

%A = %T and %G = %C

In any double-stranded DNA molecule, the mole percent of adenine equals that of thymine, and the mole percent of guanine equals that of cytosine. It follows that %A + %G = %C + %T = 50%.

Hydrogen Bonds — The Glue Between Strands

The two strands of the double helix are held together by hydrogen bonds between complementary bases. An A–T pair is stabilized by two hydrogen bonds, while a G–C pair is stabilized by three hydrogen bonds. Individually, hydrogen bonds are relatively weak, but millions of them along a DNA molecule collectively hold the strands firmly together. This balance is crucial: the bonds are strong enough to maintain the double helix during normal cell activity, yet weak enough to be separated when enzymes must access the bases during replication or transcription. This is a clear example of cause and effect at the molecular level—the number and type of bonds cause measurable differences in how easily DNA strands separate.

ℹ

NGSS Connection

DCI LS1.A: All cells contain genetic information in the form of DNA molecules. SEP — Developing and Using Models: You are constructing a mental model of DNA's structure to explain how it stores and transmits information. CCC — Structure and Function: The double-helix structure directly enables DNA's function as a stable, copyable information molecule.

SECTION 5

Detailed Breakdown — Antiparallel Strands and the Double Helix

The Antiparallel Arrangement

One of the most important structural details of DNA is that its two strands are antiparallel. This means that one strand runs in the 5′→3′ direction while the complementary strand runs 3′→5′. The 5′ end of a DNA strand has a free phosphate group attached to the 5′ carbon of the terminal sugar, while the 3′ end has a free hydroxyl group (–OH) on the 3′ carbon. This directionality matters because DNA polymerase can only add new nucleotides to the 3′ end of a growing strand. The antiparallel orientation also ensures that the hydrogen-bonding geometry between bases remains consistent along the entire length of the helix.

Antiparallel DNA Double Helix — Ladder Model5′3′Sugar-Phosphate Backbone (Strand 1)A2 H-bondsTT2 H-bondsAG3 H-bondsCC3 H-bondsGA2 H-bondsT3′5′Sugar-Phosphate Backbone (Strand 2)Legend:Adenine (A)Thymine (T)Guanine (G)Cytosine (C)Hydrogen bondStrand 1 backbone (5′→3′)Strand 2 (3′→5′)
This "untwisted ladder" model of DNA shows five base pairs between two antiparallel sugar-phosphate backbones. Strand 1 (cyan) runs 5′→3′ left to right, while Strand 2 (violet) runs 3′→5′ in the same direction. Dashed lines between bases represent hydrogen bonds: two for each A–T pair and three for each G–C pair.

In the diagram above, notice that when you read the sequence of Strand 1 from 5′ to 3′ (A–T–G–C–A), the complementary Strand 2 reads the opposite bases in the 3′-to-5′ direction (T–A–C–G–T). If you instead read Strand 2 from its own 5′ end, you would get T–G–C–A–T—which is different from Strand 1. This directional asymmetry is what makes each strand carry unique information while still being perfectly complementary to its partner.

FeatureMeasurementSignificance
Helix diameter~2.0 nmConsistent width because each rung pairs one purine with one pyrimidine
Rise per base pair0.34 nmRegular spacing allows predictable helix geometry
Base pairs per turn~10 bpOne complete 360° twist of the helix
Pitch (length per turn)~3.4 nm10 bp × 0.34 nm = 3.4 nm per complete turn
Twist directionRight-handed (B-form)The most common form of DNA under physiological conditions

Key dimensions of B-form DNA

SECTION 6

Worked Example — Applying Chargaff's Rules

Predicting Base Composition from Partial Data

Suppose a researcher analyzes a sample of double-stranded DNA and finds that 22% of the bases are adenine. Using Chargaff's rules and complementary base-pairing principles, determine the percentage of thymine, guanine, and cytosine in this sample.

Determining Full Base Composition from %A = 22%

  1. 1

    Step 1 — Apply the A = T Rule

    According to Chargaff's rules, the percentage of adenine equals the percentage of thymine in double-stranded DNA. Since %A = 22%, we know immediately that %T = 22%.
    %T = 22%
  2. 2

    Step 2 — Calculate Total A + T

    Add the A and T percentages together: %A + %T = 22% + 22% = 44%. This means 44% of the total bases are either adenine or thymine.
    %A + %T = 44%
  3. 3

    Step 3 — Determine Remaining Bases

    All four bases must account for 100% of the DNA. Subtract the A + T total from 100%: 100% − 44% = 56%. This remaining 56% is split between guanine and cytosine.
    %G + %C = 56%
  4. 4

    Step 4 — Apply the G = C Rule

    Chargaff's rules tell us that %G = %C. Since %G + %C = 56%, divide by 2: %G = 28% and %C = 28%.
    %G = 28% | %C = 28%
  5. 5

    Step 5 — Verify the Answer

    Check: 22% + 22% + 28% + 28% = 100%. Also confirm: purines (%A + %G) = 22% + 28% = 50%, and pyrimidines (%T + %C) = 22% + 28% = 50%. Both checks pass, confirming the result.
    ✓ A = 22%, T = 22%, G = 28%, C = 28%
💡

WHY THIS WORKS

Chargaff's rules are not just a mathematical trick—they are a direct consequence of the physical structure of DNA. Because every A on one strand must pair with a T on the opposite strand, and every G must pair with a C, the base percentages must be equal in pairs. Knowing one value lets you calculate all four, much like knowing one side of an equation lets you solve for the other side.

SECTION 7

Comparing DNA and RNA Structure

Two Nucleic Acids, Two Roles

DNA and RNA are both nucleic acids, but structural differences give them distinct roles. Understanding these differences clarifies why DNA serves as the long-term information archive while RNA typically acts as a short-term messenger or functional molecule. Comparing the two highlights the crosscutting concept of structure and function: each structural feature correlates with a specific biological role.

FeatureDNARNA
SugarDeoxyribose (missing –OH at 2′ carbon)Ribose (has –OH at 2′ carbon)
StrandsDouble-stranded (usually)Single-stranded (usually)
BasesA, T, G, CA, U, G, C (uracil replaces thymine)
StabilityVery stable; suited for long-term storageLess stable; suited for temporary tasks
Primary functionStores and transmits genetic informationCarries messages (mRNA), builds proteins (rRNA, tRNA)
Location (eukaryotes)Primarily in nucleus; also in mitochondria/chloroplastsMade in nucleus; functions mainly in cytoplasm

Key structural and functional differences between DNA and RNA

✦

KEY TAKEAWAY

Think of DNA as the master blueprint kept safely in an architect's vault, while RNA is a working photocopy carried to the construction site. The double-stranded structure of DNA protects the original plans, and the missing 2′ hydroxyl group on deoxyribose makes the molecule more chemically resistant to degradation. RNA's single-stranded flexibility, by contrast, allows it to fold into functional shapes and act as a disposable messenger.

SECTION 8

Connections to Advanced Topics

From Structure to Replication, Expression, and Biotechnology

Understanding DNA's structure is a gateway to many advanced topics in biology. The complementary base-pairing rules you have learned directly explain how DNA is copied during semiconservative replication: each strand serves as a template for building a new partner strand. This same complementarity also underlies transcription, where an RNA copy of a gene is produced. In biotechnology, knowledge of DNA structure enables powerful tools like polymerase chain reaction (PCR) and CRISPR gene editing, both of which exploit base-pairing specificity.

This LessonAdvanced TopicConnection
Complementary base pairingDNA ReplicationEach strand serves as a template; new nucleotides are added following A–T and G–C rules
Antiparallel orientationLeading & Lagging StrandsDNA polymerase reads 3′→5′ but synthesizes 5′→3′, creating asymmetric replication
Hydrogen bond strength (A–T vs. G–C)Melting Temperature (Tₘ)DNA with more G–C pairs requires higher temperatures to separate strands
Nucleotide sequenceGene Expression & MutationsChanges in base sequence (mutations) can alter protein structure and organismal traits
Base-pairing specificityPCR & CRISPRPrimers and guide RNAs use complementary pairing to target specific DNA sequences

How DNA structure concepts connect to advanced biology

As you continue in biology, you will see that nearly every topic in genetics, evolution, and biotechnology traces back to the structural principles covered in this lesson. The double-helix model is not just an iconic image—it is the foundation for understanding how life stores, reads, and modifies its own instructions. Future lessons will build on these structural details to explain how genetic information flows from DNA to RNA to protein, a pathway often called the central dogma of molecular biology.

SECTION 9

Practice Problems

Test Your Understanding

Question 1: CONCEPTUAL

Which of the following correctly describes the composition of a single DNA nucleotide? A) A ribose sugar, a phosphate group, and a nitrogenous base B) A deoxyribose sugar, a phosphate group, and a nitrogenous base C) A deoxyribose sugar, an amino acid, and a nitrogenous base D) A ribose sugar, a phosphate group, and an amino acid
Show answer
Correct Answer: B. A DNA nucleotide consists of three parts: a deoxyribose sugar (not ribose, which is found in RNA), a phosphate group, and one of four nitrogenous bases (A, T, G, or C). Choice A is incorrect because ribose is the sugar in RNA, not DNA. Choice C replaces the phosphate group with an amino acid, which is a component of proteins, not nucleic acids. Choice D combines two errors—ribose sugar and amino acid—making it doubly incorrect for DNA.

Question 2: BASIC CALCULATION

A double-stranded DNA molecule is found to contain 18% cytosine. What percentage of the molecule is adenine? A) 18% B) 32% C) 36% D) 64%
Show answer
Correct Answer: B. By Chargaff's rules, %C = %G, so %G = 18%. Together, G + C = 36%. Since all bases total 100%, A + T = 100% − 36% = 64%. Because %A = %T, each equals 64% ÷ 2 = 32%. Choice A incorrectly assumes adenine equals cytosine (they are not complementary partners). Choice C represents the combined G + C content, not adenine alone. Choice D represents the combined A + T content before dividing by 2.

Question 3: INTERMEDIATE

A segment of one DNA strand reads 5′–ATCGGA–3′. What is the sequence of the complementary strand, written in the standard 5′-to-3′ direction? A) 5′–TAGCCT–3′ B) 5′–TCCGAT–3′ C) 5′–ATCGGA–3′ D) 5′–AGCCTA–3′
Show answer
Correct Answer: B. First, write the complement of each base in the given strand: A→T, T→A, C→G, G→C, G→C, A→T, giving 3′–TAGCCT–5′ for the complementary strand. However, convention requires writing sequences 5′→3′, so reverse the order: 5′–TCCGAT–3′. Choice A writes the complement but fails to reverse the direction. Choice C simply copies the original strand. Choice D incorrectly pairs some bases.

Question 4: APPLIED

A forensic scientist isolates a DNA sample and determines that it has an unusually high melting temperature (the temperature at which the two strands separate). Which base composition would best explain this observation? A) 35% adenine, 35% thymine, 15% guanine, 15% cytosine B) 15% adenine, 15% thymine, 35% guanine, 35% cytosine C) 25% adenine, 25% thymine, 25% guanine, 25% cytosine D) 40% adenine, 10% thymine, 40% guanine, 10% cytosine
Show answer
Correct Answer: B. G–C base pairs are held together by three hydrogen bonds, while A–T pairs have only two. DNA with a higher proportion of G–C pairs requires more energy (higher temperature) to denature. Choice B has 70% G+C content, which would produce the highest melting temperature among the options. Choice A has only 30% G+C, resulting in a lower melting temperature. Choice C is 50% G+C—average, not unusually high. Choice D violates Chargaff's rules (A ≠ T and G ≠ C), so it cannot represent normal double-stranded DNA.

Question 5: CRITICAL THINKING

A student analyzes a nucleic acid sample and finds the following base percentages: A = 30%, U = 30%, G = 20%, C = 20%. Another sample shows: A = 30%, T = 30%, G = 20%, C = 20%. Which statement best explains the difference between these two samples? A) Both samples are double-stranded DNA; the presence of U is a measurement error. B) The first sample is single-stranded RNA; the second is double-stranded DNA. Both obey Chargaff's rules. C) The first sample is double-stranded RNA; the second is single-stranded DNA. D) The first sample is single-stranded RNA that happens to have equal A and U; the second is double-stranded DNA where Chargaff's rules require A = T and G = C.
Show answer
Correct Answer: D. The key clue is the presence of uracil (U) in the first sample, which identifies it as RNA. RNA is typically single-stranded, so Chargaff's base-pairing rules do not necessarily apply—the equal amounts of A and U could be coincidental. The second sample contains thymine (T) instead of uracil, identifying it as DNA, and its %A = %T and %G = %C values are consistent with double-stranded DNA following Chargaff's rules. Choice B incorrectly states that both samples obey Chargaff's rules; Chargaff's rules apply strictly to double-stranded DNA. Choice A ignores the biological significance of uracil. Choice C misidentifies the strand counts for both molecules.

SUMMARY

Lesson Summary

Summary

DNA is a double-stranded helix composed of repeating units called nucleotides. Each nucleotide contains three parts: a deoxyribose sugar, a phosphate group, and one of four nitrogenous bases: adenine (A), thymine (T), guanine (G), or cytosine (C). The sugar and phosphate groups alternate to form the sugar-phosphate backbone, while the bases project inward and pair with complementary bases on the opposite strand. Adenine always pairs with thymine via two hydrogen bonds, and guanine always pairs with cytosine via three hydrogen bonds.

The two strands are antiparallel, running in opposite 5′-to-3′ directions. Chargaff's rules (%A = %T and %G = %C) are a direct mathematical consequence of complementary base pairing. The B-form double helix has a diameter of approximately 2 nm, with 10 base pairs per turn and a pitch of 3.4 nm. This precise structure–function relationship enables DNA to store genetic information in its base sequence, replicate faithfully through complementary pairing, and serve as the molecular foundation for all of genetics and modern biotechnology.

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