# Nucleic acid structure

> IB Biology SL · Theme B: Form and Function
> Source: https://www.owlsprep.com/study/ib-biology-sl-u2-nucleic-acid-structure/

This sub-topic covers the molecular structure of nucleic acids (DNA and RNA), the building blocks of all genetic information. You will learn monomer structure, base pairing rules, and key features of DNA's double helix, the foundation for all molecular genetics.

**Prerequisites:** [Introduction to biological macromolecules](https://www.owlsprep.com/study/ib-biology-sl-u2-introduction-to-biomolecules/)

## Learning objectives

- Describe the structure of nucleotides as the monomer of nucleic acids
- Distinguish between the structure of DNA and RNA
- Explain the double helix structure of DNA and complementary base pairing rules
- Calculate base composition of double-stranded DNA using base pairing rules

## Nucleotides: The Building Blocks of Nucleic Acids

**Nucleotide** — The universal monomer of all nucleic acid polymers, made of three covalently bonded components: a 5-carbon pentose sugar, a negatively charged phosphate group, and a nitrogenous base.

*Example:* A deoxyadenosine nucleotide is the adenine-containing monomer of DNA.

The key difference between DNA and RNA nucleotides is the pentose sugar. DNA uses **deoxyribose**, which lacks a hydroxyl (-OH) group on the 2' carbon. RNA uses **ribose**, which has a hydroxyl group on the 2' carbon. Nitrogenous bases are divided into two groups: double-ringed purines (adenine, guanine) and single-ringed pyrimidines (cytosine, thymine in DNA, uracil in RNA).

**Worked example:** Draw and label the three components of a DNA nucleotide

1. Draw a pentagon to represent the 5-carbon deoxyribose sugar. Label each carbon 1' through 5' (primes distinguish sugar carbons from base carbons).
2. Add a phosphate group covalently bonded to the 5' carbon of the sugar.
3. Add any nitrogenous base covalently bonded to the 1' carbon of the sugar.
4. Label the hydroxyl group on the 3' carbon, and note that no hydroxyl group is present on the 2' carbon of deoxyribose.

## Nucleic Acid Polymers and Directionality

**Phosphodiester bond** — Covalent bond formed between the 5' phosphate group of one nucleotide and the 3' hydroxyl group of the next nucleotide during a condensation reaction.

All nucleic acid strands have directionality: one end has a free 5' phosphate group (called the 5' end) and the opposite end has a free 3' hydroxyl group (called the 3' end). All biological synthesis of nucleic acids occurs only in the 5' → 3' direction, a rule that is critical for understanding replication and transcription later in the course.

**Worked example:** A strand of RNA has the sequence 5' - A - C - G - 3'. Identify which end has the free phosphate group and which has the free hydroxyl group.

1. Recall that the first nucleotide in the written sequence is the 5' end by convention.
2. The adenine (A) nucleotide at the start of the sequence has a free 5' phosphate group.
3. The phosphodiester bond links the 5' phosphate of each next nucleotide to the 3' hydroxyl of the previous nucleotide.
4. The guanine (G) nucleotide at the end of the sequence has a free 3' hydroxyl group.

> **Exam tip**
>
> Always label 5' and 3' ends when drawing nucleic acid strands. IB exam markers routinely award 1 mark for correct end labeling, and deduct it if you forget.

## DNA Double Helix and Complementary Base Pairing

**Complementary Base Pairing** — Specific hydrogen bonding between nitrogenous bases that holds two DNA strands together. Adenine only pairs with thymine, and guanine only pairs with cytosine.

Watson and Crick used Rosalind Franklin's X-ray crystallography data to propose the double helix structure of DNA in 1953. DNA is made of two antiparallel strands (running in opposite 5'→3' directions) twisted around each other. The sugar-phosphate backbone forms the outside of the helix, and bases are stacked on the inside. Purines always pair with pyrimidines to keep the width of the helix constant: A-T forms 2 hydrogen bonds, G-C forms 3 hydrogen bonds.

**Worked example:** A double-stranded DNA molecule contains 22% adenine. Calculate the percentage of guanine in the molecule.

1. Apply complementary base pairing: the percentage of adenine equals the percentage of thymine, so %T = 22%.
2. Add the percentages of adenine and thymine: 22% + 22% = 44%.
3. The remaining percentage of bases is 100% - 44% = 56%, which is split equally between guanine and cytosine, since %G = %C.
4. Divide 56% by 2 to get %G = 28%.

**Exam command terms**

Common command terms for this topic have specific IB expectations:

- **Distinguish** — State all key differences between two structures *(Distinguish between DNA and RNA structure)*

- **Draw** — Draw the structure and label all key components *(Draw a single DNA nucleotide)*

## Common pitfalls

- **Wrong:** Labeling the 2' carbon of deoxyribose as having a hydroxyl (-OH) group
  - Why it fails: Deoxyribose is missing the oxygen atom on carbon 2, so no hydroxyl group can be present there
  - Correct: Only label the 3' carbon of deoxyribose as having a hydroxyl group
- **Wrong:** Stating that RNA contains thymine instead of uracil
  - Why it fails: Thymine is exclusive to DNA; uracil replaces thymine in all RNA molecules
  - Correct: Memorize: DNA = ATCG, RNA = AUCG
- **Wrong:** Drawing two DNA strands parallel (same direction) instead of antiparallel
  - Why it fails: Parallel strands cannot form the correct complementary hydrogen bonds between bases
  - Correct: Always draw the two strands running in opposite 5'→3' directions
- **Wrong:** Calculating guanine percentage as 50% minus adenine percentage directly
  - Why it fails: This shortcut incorrectly ignores that equal amounts of adenine and thymine are both present
  - Correct: Use the formula: \%G = (100 - 2 \times \%A)/2 for double-stranded DNA

## Cheatsheet

| Feature | DNA | RNA |
| --- | --- | --- |
| Pentose sugar | Deoxyribose (no 2' OH) | Ribose (has 2' OH) |
| Nitrogenous bases | A, T, C, G | A, U, C, G |
| Typical structure | Double-stranded antiparallel double helix | Usually single-stranded |
| 5' end | Free phosphate group | Free phosphate group |
| 3' end | Free hydroxyl group | Free hydroxyl group |
| Base pairs | A-T (2 H bonds), G-C (3 H bonds) | A-U, G-C |

## What's next

Nucleic acid structure is the foundational concept for all molecular genetics in IB Biology SL. The directionality of strands and complementary base pairing rules you learned here directly explain how DNA replicates accurately before cell division, how genetic information is transcribed into RNA, and how translation builds proteins from mRNA instructions. Errors in nucleic acid structure cause mutations, which drive evolution and cause genetic disease. Mastery of drawing and labeling nucleic acid structures is required for almost all exam questions on molecular biology, so practice recalling key features from memory.

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