A 1.2 Nucleic Acids
Guiding Questions:
- How does the structure of nucleic acids allow hereditary information to be stored?
- How does the structure of DNA facilitate accurate replication?

SL and HL Content
Learning Objectives
A1.2.1—DNA as the Genetic Material
Discuss why DNA is the genetic material in all living organisms.
Explain why some viruses use RNA instead of DNA.
Debate why viruses are not considered living organisms.
DNA as the Genetic Material in Living Organisms
Deoxyribonucleic Acid (DNA) is a molecule composed of two long chains (strands) coiled around each other to form a double helix.
DNA is made up of nucleotides.
Function of DNA:
The sequence of nitrogenous bases (A, T, C, G) acts as instructions for protein synthesis.
Genes (segments of DNA) are transcribed into RNA, then translated into proteins.
DNA carries the genetic instructions used in the growth, development, functioning, and reproduction of all known living organisms and many viruses.
DNA is passed from generation to generation during reproduction.
Universal Genetic Code:
The genetic code is nearly universal, with the same codons coding for the same amino acids in almost all organisms, demonstrating the common ancestry of life.
This supports the common ancestry of life on Earth.
Why RNA Is Used by Some Viruses:
Viruses are infectious agents that replicate only inside the living cells of an organism.
Some viruses use RNA (Ribonucleic Acid) instead of DNA as their genetic material.
RNA viruses can have single-stranded or double-stranded RNA genomes.
Examples of RNA Viruses:
Influenza virus, HIV, and coronaviruses (e.g., SARS-CoV-2)
RNA genomes mutate more quickly, helping viruses adapt rapidly.
Why Viruses Are Not Considered Living
Viruses cannot reproduce on their own; they must infect host cells.
They do not carry out metabolism or respond to stimuli independently.
They are considered biological entities but not fully living organisms by most definitions.
Structure of DNA:
DNA is made up of nucleotides, each consisting of:
a phosphate group
a sugar molecule (deoxyribose)
a nitrogenous base (adenine [A], thymine [T], cytosine [C], or guanine [G])

The sequence of these bases encodes genetic information.
A1.2.2—Components of a Nucleotide
Identify and describe the three components of a nucleotide: phosphates, pentose sugars, and nitrogenous bases.
Discuss the use of circles, pentagons, and rectangles in diagrams to represent these components.
Basic Structure of a Nucleotide:
A nucleotide is the basic building block of nucleic acids (DNA and RNA).

Each nucleotide consists of three components:
Phosphate Group: Contains one phosphorus atom bonded to four oxygen atoms
Pentose Sugar: A five-carbon sugar.
In DNA, the sugar is deoxyribose
In RNA, the sugar is ribose
Nitrogenous Base: A nitrogen-containing molecule that is one of the four bases:
In DNA - adenine [A], thymine [T], cytosine [C], guanine [G]
In RNA - adenine [A], uracil [U], cytosine [C], guanine [G]
Diagrammatic Representation:
Circle: Represents the phosphate group.
Pentagon: Represents the pentose sugar (deoxyribose in DNA or ribose in RNA).
Rectangle: Represents the nitrogenous base.
Relative Positions:
The phosphate group is attached to the 5’ carbon of the pentose sugar.
The nitrogenous base is attached to the 1’ carbon of the pentose sugar.
A1.2.3—Sugar–Phosphate Bonding
Explain how sugar-phosphate bonds form the backbone of DNA and RNA.
Discuss the significance of the continuous chain of covalently bonded atoms in nucleotides.
Sugar–phosphate bonding makes a continuous chain of covalently bonded atoms in each strand of DNA or RNA nucleotides, which forms a strong “backbone” in the molecule.
Sugar–Phosphate Bonding:
Covalent Bonds:
In DNA and RNA, nucleotides are linked together by covalent bonds between the phosphate group of one nucleotide and the hydroxyl group on the 3’ carbon of the sugar of another nucleotide.
These bonds are called phosphodiester bonds.
Formation of Phosphodiester Bonds:
A phosphodiester bond forms between the 3’ hydroxyl group of the sugar of one nucleotide and the 5’ phosphate group of the next nucleotide.
This creates a continuous chain of covalently bonded atoms, resulting in a strong and stable structure.
Sugar–Phosphate “Backbone”:
Structure:
The repeating pattern of sugar and phosphate groups forms the “backbone” of DNA and RNA strands.
In diagrams, the backbone is often depicted as a series of alternating pentagons (sugars) and circles (phosphates).
Directional Nature:
DNA and RNA strands have directionality, indicated by the 5’ (five-prime) and 3’ (three-prime) ends.
The 5’ end has a free phosphate group attached to the 5’ carbon of the sugar.
The 3’ end has a free hydroxyl group attached to the 3’ carbon of the sugar.
Stability:
The covalent phosphodiester bonds in the sugar-phosphate backbone provide significant stability to the DNA and RNA molecules.
This stability is crucial for the preservation of genetic information and the integrity of the nucleic acid structure.
Differences Between DNA and RNA Backbones:
Sugar Component:
DNA: Deoxyribose sugar (lacking an oxygen atom at the 2’ position).
RNA: Ribose sugar (containing an oxygen atom at the 2’ position).
Single vs. Double Stranded:
DNA typically exists as a double-stranded molecule with two complementary strands running in opposite directions (antiparallel).
RNA is usually single-stranded, but it can form complex secondary structures through intramolecular base pairing.
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A1.2.4—Nitrogenous Bases
List the names of the nitrogenous bases in nucleic acids.
Discuss the role of these bases in the genetic code.
A1.2.5—RNA Structure
Describe how RNA is formed by the condensation of nucleotide monomers.
Draw and recognize diagrams of single nucleotides and RNA polymers.
A1.2.6—DNA Structure
Explain the structure of DNA as a double helix with antiparallel strands.
Discuss hydrogen bonding between complementary base pairs: adenine (A) with thymine (T) and guanine (G) with cytosine (C).
A1.2.7—Differences between DNA and RNA
Compare and contrast DNA and RNA in terms of strand number, types of nitrogenous bases, and type of pentose sugar.
Sketch the differences between ribose and deoxyribose sugars.
A1.2.8—Complementary Base Pairing
Discuss the role of complementary base pairing in DNA replication and expression.
Explain how hydrogen bonding is crucial for complementarity.
A1.2.9—DNA’s Capacity for Storing Information
Explain the diversity of possible DNA base sequences and their implications.
Emphasize the enormous capacity of DNA for storing genetic information.
A1.2.10—Universal Genetic Code
Discuss the conservation of the genetic code across all life forms as evidence of common ancestry.
Additional Higher Level
A1.2.11—Directionality of RNA and DNA
Explain the significance of 5’ to 3’ linkages in the sugar-phosphate backbone.
Discuss the importance of directionality for replication, transcription, and translation.
A1.2.12—Purine-to-Pyrimidine Bonding
Discuss how purine-to-pyrimidine bonding contributes to DNA helix stability.
Explain why adenine–thymine (A–T) and cytosine–guanine (C–G) pairs result in a consistent three-dimensional structure.
A1.2.13—Structure of a Nucleosome
Describe the structure of a nucleosome: DNA wrapped around histone proteins.
Use molecular visualization software to study the association between proteins and DNA within a nucleosome.
A1.2.14—Hershey–Chase Experiment
Explain the significance of the Hershey–Chase experiment in identifying DNA as the genetic material.
Discuss how technological advancements enable new experimental possibilities.
A1.2.15—Chargaff’s Data
Analyze Chargaff’s data on the relative amounts of pyrimidine and purine bases across different life forms.
Discuss how the data falsified the tetranucleotide hypothesis, addressing the “problem of induction” through “certainty of falsification”.