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BU-102 / L22 · lecture

Nucleotides and Nucleic Acid Structure

~6 min read · Tier toggle changes depth (T1 plain · T2 standard · T3 rigour)

BEFORE YOU START

You should be familiar with the basic structure of carbohydrates and lipids, and understand how covalent bonds and non-covalent interactions determine molecular shape.

Why this matters

Nucleic acids are the physical carriers of genetic information, and their specific chemical structure dictates how that information is stored, copied, and read.

Unlike proteins, which are diverse in sequence and function, nucleic acids rely on a highly regular structural framework to ensure the fidelity of genetic transmission. Understanding the components of nucleotides and the rules governing their assembly is essential for grasping how biological systems maintain hereditary information. This lecture focuses on the monomeric units and the primary structural features that precede the formation of the double helix.

The Building Blocks: Nucleotides

A nucleotide is the fundamental monomer of nucleic acids. It consists of three distinct chemical groups covalently bonded together. The first is a five-carbon sugar, known as a pentose. In deoxyribonucleic acid (DNA), this sugar is deoxyribose, which lacks an oxygen atom at the 2' carbon position. In ribonucleic acid (RNA), the sugar is ribose, which possesses a hydroxyl group at the 2' position. The second component is a phosphate group, which is attached to the 5' carbon of the sugar. The third component is a nitrogenous base, which is attached to the 1' carbon of the sugar. These bases are classified into two families: purines, which have a double-ring structure (adenine and guanine), and pyrimidines, which have a single-ring structure (cytosine, thymine, and uracil). The specific combination of sugar, phosphate, and base defines the identity of each nucleotide.

Phosphodiester Linkages and Polarity

Nucleotides are linked together to form polynucleotide chains via covalent bonds known as phosphodiester bonds. This linkage forms between the 3' hydroxyl group of one sugar and the 5' phosphate group of the next nucleotide. This reaction releases a molecule of water and creates a stable backbone of alternating sugar and phosphate residues. Because the sugar is chiral and the linkage is directional, the resulting chain has a distinct polarity. One end of the chain terminates with a free 5' phosphate group, while the other end terminates with a free 3' hydroxyl group. This directionality is critical for biological processes, as enzymes that synthesise or degrade nucleic acids operate in a specific 5' to 3' direction.

\[ \text{5'-P-Sugar-Base-3'} \quad \xrightarrow{\text{phosphodiester bond}} \quad \text{5'-P-Sugar-Base-P-Sugar-Base-3'} \]

Base Pairing and Hydrogen Bonding

The nitrogenous bases project inward from the sugar-phosphate backbone. In double-stranded nucleic acids, these bases interact with complementary bases on the opposing strand through hydrogen bonds. Adenine (A) pairs specifically with Thymine (T) in DNA, or Uracil (U) in RNA, forming two hydrogen bonds. Guanine (G) pairs specifically with Cytosine (C), forming three hydrogen bonds. This specificity is known as Watson-Crick base pairing. The geometry of these pairs is nearly rectangular, which allows the two strands to maintain a consistent distance from each other. The strength of the interaction depends on the number of hydrogen bonds; G-C pairs are thermodynamically more stable than A-T pairs due to the presence of three hydrogen bonds rather than two.

\[ \mathrm{A} \dots \mathrm{T} \quad (2 \text{ H-bonds}) \qquad \mathrm{G} \dots \mathrm{C} \quad (3 \text{ H-bonds}) \]

Antiparallel Orientation

When two polynucleotide strands associate to form a double helix, they must be oriented in opposite directions. This is described as an antiparallel arrangement. If one strand runs in the 5' to 3' direction, the complementary strand must run in the 3' to 5' direction. This orientation is a direct consequence of the geometry of the phosphodiester backbone and the requirement for optimal hydrogen bonding between the bases. The antiparallel nature ensures that the sugar-phosphate backbones are equidistant from the central axis of the helix, creating a uniform width. This structural regularity is a prerequisite for the stability of the double-stranded molecule.

The concept

◆ CONCEPT · T-009 · full treatment on its own page

The DNA double helix

Complementary base pairing on two antiparallel strands.

\[ \text{Two antiparallel, complementary strands (A-T, G-C), wound into a regular right-handed double helix.} \]
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Tier 3 · watch out

A common misconception is that the hydrogen bonds between bases are the primary force holding the double helix together. In reality, the hydrophobic effect, which drives the stacking of the non-polar bases in the interior of the helix away from the aqueous environment, provides the majority of the thermodynamic stability. Hydrogen bonds provide specificity, ensuring the correct pairing, but base stacking interactions are energetically dominant. C

Worked example

Problem. A single-stranded RNA molecule contains 40% adenine (A), 30% guanine (G), 20% cytosine (C), and 10% uracil (U). If this RNA strand were to hybridize with a complementary DNA strand, determine the percentage of thymine (T) in the DNA strand and calculate the total number of hydrogen bonds formed per 1000 nucleotides in the double-stranded region.

  1. Determine the base composition of the complementary DNA strand. By the base-pairing rules, A in RNA pairs with T in DNA, and G in RNA pairs with C in DNA. Therefore, the DNA strand will contain 40% T (paired with A in RNA) and 30% C (paired with G in RNA). The remaining bases in DNA will be 20% G (paired with C in RNA) and 10% A (paired with U in RNA).
  2. Calculate the number of each base pair type in 1000 nucleotides. The RNA strand has 400 A, 300 G, 200 C, and 100 U. The complementary DNA strand will have 400 T, 300 C, 200 G, and 100 A.
  3. Count the hydrogen bonds for each pair type. A-T pairs have 2 hydrogen bonds. G-C pairs have 3 hydrogen bonds. The A·T-type pairs number 400 (A in RNA with T in DNA) plus 100 (U in RNA with A in DNA), a total of 500. The number of G-C pairs is 300 (from G in RNA and C in DNA) plus 200 (from C in RNA and G in DNA), totaling 500 G-C pairs.
  4. Sum the total hydrogen bonds. Total H-bonds = (500 pairs × 2) + (500 pairs × 3) = 1000 + 1500 = 2500.
\[ 2500 \text{ hydrogen bonds} \]

Units check. The result is a dimensionless count of bonds, which is consistent with the input of 1000 nucleotides.

CHECK YOURSELF

Three quick tests

  1. What is the key structural difference between the sugar in DNA and the sugar in RNA?
    AnswerDNA contains deoxyribose, which lacks an oxygen atom at the 2' carbon, whereas RNA contains ribose, which has a hydroxyl group at the 2' carbon.
  2. Why is the antiparallel orientation of the two strands in a double helix necessary?
    AnswerThe antiparallel orientation allows the sugar-phosphate backbones to maintain a consistent distance from the central axis, ensuring uniform helix width and optimal hydrogen bonding geometry between the bases.
  3. If a double-stranded DNA molecule is 30% guanine, what percentage is cytosine?
    AnswerIt must contain 30% cytosine, as guanine and cytosine pair specifically with each other in double-stranded DNA.
UNIT

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NOTES

Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.

Concepts cited: T-009 · Threads: structure, energy