Biomolecules · Lesson 10 of 12
Nucleic Acids: Nucleotides, DNA and RNA
“Build nucleic-acid components and directional strands, explain complementarity and connect DNA information with RNA-mediated protein synthesis.”
• Distinguish bases, sugars, nucleosides and nucleotides. • Identify DNA/RNA sugar and base differences. • Read 3′–5′ backbone connectivity and antiparallel complementary sequences. • Explain DNA replication and the roles of mRNA, tRNA and rRNA. • Interpret DNA profiling without assuming absolute individual uniqueness.
Chemical information requires an ordered polymer
Organisms resemble their parents partly because they inherit molecular information. DNA, deoxyribonucleic acid, and RNA, ribonucleic acid, are polynucleotides: chains of nucleotide units. In eukaryotic nuclei, DNA associates with proteins in chromosomes. DNA also occurs outside nuclei, and bacteria lack a membrane-bound nucleus, so heredity should not be restricted to one cellular compartment.
A chain can carry information because its constituent bases occur in a specific sequence. Knowing the total number of four letters does not specify their order, just as amino-acid composition does not specify a protein sequence. DNA sequence is the primary chemical information reservoir in cellular organisms, while RNAs perform information-transfer, structural, regulatory and catalytic roles. Some viruses use RNA genomes, so DNA should not be claimed as the only possible hereditary material in all biological systems.
Three components of a nucleotide
Complete chemical breakdown of nucleic acids yields pentose sugar, phosphate-related material and nitrogenous bases. RNA uses D-ribose; DNA uses 2-deoxy-D-ribose. In their nucleotides these sugars occur as β-D-furanose derivatives. The important difference is at C2′: RNA ribose has OH, while DNA deoxyribose has H. Deoxy means loss of this particular oxygen relative to ribose, not a sugar with no oxygen.
Sugar carbons are numbered 1′, 2′, 3′, 4′ and 5′ to distinguish them from the atom numbers of the base. The ring contains C1′ through C4′ and an oxygen; C5′ lies outside the ring as a CH₂ group. C1′ bears the base. C3′ bears an OH available for backbone connection, and the common mononucleotide drawn here has phosphate on the C5′ oxygen.
| Feature | DNA | RNA |
|---|---|---|
| Sugar | 2-Deoxyribose: C2′ has H | Ribose: C2′ has OH |
| Bases | Adenine, guanine, cytosine, thymine | Adenine, guanine, cytosine, uracil |
| Typical organisation | Antiparallel double helix | Usually single strand with folded paired regions |
| Major roles | Long-term storage and replication of cellular genetic information | Information transfer, translation, structure, catalysis and regulation |
Bases have distinct ring structures
Adenine and guanine are purines with a fused two-ring framework. Cytosine, thymine and uracil are pyrimidines with a single six-membered ring. Thymine differs from uracil by a methyl substituent at the corresponding ring position. Amino and carbonyl groups on these rings provide particular hydrogen-bond donors and acceptors, so pairing depends on chemical structure as well as size.
Nucleoside versus nucleotide
A nucleoside consists of a nitrogenous base attached to a sugar at C1′ through a β-N-glycosidic bond. It contains no phosphate as part of the definition. For purines the bond uses base N9; for pyrimidines it uses N1. Add a phosphate ester, commonly at C5′ in the introductory structure, and the molecule is a nucleotide. Examples of base–sugar names include adenosine and deoxyadenosine; adding phosphate gives the corresponding nucleotide.
Hydrolysis must be specified. Cleaving a nucleotide’s phosphate ester yields a nucleoside and phosphate. Complete hydrolysis also cleaves the base–sugar bond, giving free base and sugar. A DNA nucleotide containing thymine therefore ultimately yields thymine, 2-deoxyribose and phosphate material; ribose would be the wrong sugar. In solution, phosphate protonation depends on pH, so “phosphoric acid” is a conventional description of the complete hydrolysis component rather than its sole ionic form in every medium.
Phosphodiester bonds give strand direction
A nucleic-acid strand joins the C3′ oxygen of one sugar to the C5′ oxygen of the next through a phosphate group. Because that phosphate forms two ester connections, the bridge is a phosphodiester. The backbone sequence is sugar–phosphate–sugar, and the bases project from C1′. Hydrogen bonds between bases do not hold the covalent backbone together.
The different ends give a strand direction. Sequences are normally written 5′ to 3′. A chain with n nucleotide residues has n−1 internal phosphodiester connections, although terminal phosphate groups can vary. Biological synthesis uses activated nucleotide triphosphates; the simple linkage description does not mean polymerisation is just free nucleotides spontaneously losing water in a cell.
Complementary DNA strands run in opposite directions
In the usual DNA double helix, A pairs with T through two hydrogen bonds, and G with C through three. A purine–pyrimidine pair helps maintain a consistent helix width. The strands are complementary rather than identical: knowing one base tells us its partner. They are antiparallel, so a top strand running 5′→3′ faces a partner running 3′→5′. Complementary base pairing and other interactions, especially base stacking, stabilise the helix.
Problem
A DNA strand is 5′-AGCTTA-3′. Write its partner directly opposite, then write that partner in the usual 5′→3′ direction.
- 1.Pair A with T and G with C while keeping aligned positions: 3′-TCGAAT-5′.
- 2.To express that partner 5′→3′, reverse the written order: 5′-TAAGCT-3′.
- 3.The complement and reverse complement are different written strings because direction matters. A correct set of letters with incorrect end labels is not a complete answer.
For an ordinary complete double-stranded DNA sample with canonical pairing, total A equals T and total G equals C. A single isolated strand need not satisfy these equalities internally. Many RNA molecules contain unpaired regions and are not paired end-to-end with a complementary second strand. Their hydrolysis products therefore need not have equal A/U or G/C amounts. Unequal amounts suggest absence of complete canonical complementary pairing, not absence of all local RNA helices.
Replication and expression are different uses of information
During DNA replication, each parental strand guides construction of a complementary new strand, producing daughter duplexes. Replication requires enzymes and activated substrates; DNA is not an independent self-copying chemical machine. Copying is highly accurate but not absolutely error-free, and mutation and recombination contribute to variation. DNA’s hereditary role does not mean species remain chemically unchanged over millions of years.
During gene expression, information from suitable DNA regions is transcribed into RNA. Messenger RNA carries coding information to the translation machinery. Transfer RNA supplies amino acids through its adaptor function. Ribosomal RNA forms a central structural and catalytic component of ribosomes, including peptide-bond formation. These three major RNA classes cooperate in protein synthesis; many other RNA classes also exist.
Evidence, history and DNA profiling
The double-helix model emerged from chemical, diffraction and genetic evidence. Watson and Crick proposed a model in 1953 drawing on important experimental work by Rosalind Franklin, Maurice Wilkins and others. Later work by researchers including Har Gobind Khorana, Marshall Nirenberg and Robert Holley helped connect nucleotide information with protein synthesis and decipher the genetic code. The important chemical lesson is that base sequence can specify a different sequence of molecular building blocks.
DNA profiling compares variable regions to assess relationships or the likely source of a sample. Applications include forensic identification, parentage analysis and identification of remains. A profile is not an absolute guarantee of one-person uniqueness: identical twins usually share standard profiles, samples can be mixed or degraded, and most but not all nucleated cells in an individual share the same genome. Mature red blood cells lack nuclear DNA. Population ancestry studies require careful interpretation and do not justify treating human groups as fixed biological races.
Problem
A canonical double-stranded DNA sample has 30% adenine. Find the other base percentages.
- 1.A = T gives T = 30%. Together they account for 60%.
- 2.The remaining 40% is G + C. G = C gives G = 20% and C = 20%.
- 3.The calculation assumes the entire double-stranded sample with canonical pairing. It would not be justified from an arbitrary single DNA strand or a general RNA sample.
Quick Quiz
Quiz
What additional component distinguishes a nucleotide from a nucleoside?
Explanation: Nucleoside is base plus sugar; nucleotide includes phosphate.
Which structural sugar difference distinguishes RNA from DNA?
Explanation: Ribose versus 2-deoxyribose differs at the 2′ substituent.
Which bond joins successive sugars in a nucleic-acid backbone?
Explanation: Phosphate bridges sugar O3′ and O5′ positions.
What partner lies directly opposite 5′-AGTC-3′?
Explanation: Complementarity requires both correct bases and antiparallel direction.
Which RNA is central to ribosome structure and peptide-bond catalysis?
Explanation: Ribosomal RNA has structural and catalytic roles in translation.
Practice Problems
- Compare the complete hydrolysis products of thymine-containing DNA nucleotide and uracil-containing RNA nucleotide.
- Distinguish nucleoside and nucleotide and identify common attachment sites.
- Give the antiparallel partner of 5′-ATGCCA-3′ and then its 5′→3′ sequence.
- A double-stranded DNA sample has 18% guanine. Calculate the remaining percentages under canonical pairing.
- Explain two principal functions of nucleic acids and the three major RNA roles.
- Why can RNA hydrolysis give unequal amounts of its bases?
- Distinguish a backbone bond from the interactions between DNA strands.
- Why is a standard DNA profile not an absolute claim of unique identity in every case?
The DNA nucleotide yields thymine, 2-deoxyribose and phosphate material on complete cleavage. The RNA nucleotide yields uracil, ribose and phosphate material. Cleavage of only the phosphate ester stops at a nucleoside; it is not yet complete hydrolysis of base–sugar connectivity.
Key Takeaways
• Nucleic acids are ordered nucleotide polymers. • A nucleoside is base plus sugar; a nucleotide includes phosphate. • DNA contains 2-deoxyribose and thymine; RNA contains ribose and uracil. • Purines are A and G; pyrimidines are C, T and U. • The backbone uses 3′–5′ phosphodiester connectivity. • Canonical DNA pairs A–T and G–C in antiparallel strands. • mRNA, tRNA and rRNA cooperate in translation. • DNA profiling is powerful evidence with specific limitations.