Nucleic Acids — IB Biology HL Study Guide (HL Extension)

For: IB Biology HL candidates sitting IB Biology HL.

Covers: IB Topic 7 (HL only) — DNA structure deeper, semiconservative replication mechanism, transcription with promoter and terminator, translation with peptide-bond formation, genetic code features, regulation of gene expression.

You should already know: Molecular biology basics (Topic 2), gene expression intro (Topic 3 SL).

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the IB Biology HL style for educational use. They are not reproductions of past IBO papers.

1. Why Nucleic Acids in HL

Topic 7 is the molecular detail behind SL Topic 2 (Molecular Biology). It covers the mechanism of replication, transcription, and translation — what enzymes do, what cofactors are needed, what regulates each step. About 4-6% of HL Paper 1+2 directly, but it's heavily integrated into Topic 8 (Metabolism) and Topic 11 (Animal Phys: hormones).

2. DNA structure recap (HL detail)

DNA is a double helix with:

• Sugar-phosphate backbone outside.
• Bases (A, T, G, C) hydrogen-bonded inside (A=T 2 bonds; G≡C 3 bonds).
• Antiparallel strands: 5'→3' on one, 3'→5' on the other.

Nucleosomes: DNA wrapped around histone proteins. 8 histones (octamer) + ~146 bp of DNA per nucleosome. Allows compact packaging in the nucleus.

3. DNA replication — semiconservative

1. Helicase unwinds the double helix at the replication fork.
2. DNA gyrase / topoisomerase relieves supercoiling ahead.
3. Each strand serves as template; new nucleotides added by DNA polymerase III following base-pairing rules.
4. Polymerase only synthesises 5'→3', so:
   • Leading strand synthesised continuously toward the fork.
   • Lagging strand synthesised in Okazaki fragments away from the fork, then joined by DNA ligase.
5. DNA polymerase I removes RNA primers (made by primase) and replaces them with DNA.
6. Result: each new double helix has one parent strand + one new strand → "semiconservative".

4. Transcription

DNA → mRNA. Done by RNA polymerase in the nucleus.

1. RNA pol binds promoter sequence (e.g. TATA box) on the DNA template strand.
2. DNA unwinds locally; RNA pol synthesises mRNA 5'→3' using base-pairing (U replaces T).
3. RNA pol stops at the terminator sequence; mRNA is released.
4. Post-transcriptional processing (eukaryotes only): 5' cap added, 3' poly-A tail, splicing removes introns and joins exons.

5. Translation

mRNA → protein. Done by ribosomes in cytoplasm.

1. Initiation: small ribosomal subunit binds 5' cap, scans for AUG (start codon). Initiator tRNA (carrying methionine) base-pairs with AUG. Large subunit joins.
2. Elongation: tRNAs bring amino acids one by one. Each tRNA's anticodon base-pairs with mRNA codon. Ribosome catalyses peptide bond between amino acids on adjacent tRNAs (via peptidyl transferase activity in 23S rRNA).
3. Translocation: ribosome moves one codon along, releases the now-empty tRNA, and exposes the next codon.
4. Termination: stop codon (UAA, UAG, UGA) reached. Release factor binds, polypeptide released, ribosome subunits dissociate.

Multiple ribosomes can translate one mRNA simultaneously — called a polysome.

6. Genetic code features

• Triplet: each codon is 3 nucleotides → 64 possible codons.
• Universal: same code in nearly all organisms.
• Degenerate: 64 codons code for only 20 amino acids → most amino acids have multiple codons (synonyms), often differing only in the 3rd position ("wobble").
• Non-overlapping: codons read sequentially without overlap.
• Comma-less: no separator between codons.
• AUG: start codon (also codes for Met).
• 3 stop codons: UAA, UAG, UGA.

7. Regulation of gene expression

Eukaryotes regulate at multiple levels:

• Transcriptional: promoter accessibility (chromatin remodelling, methylation of DNA, acetylation of histones).
• Post-transcriptional: alternative splicing produces different mRNAs from one gene.
• Translational: regulation of mRNA stability and translation rate.
• Post-translational: protein folding, modification (phosphorylation), degradation.

Operon model (prokaryotes only): clusters of genes controlled by one promoter and operator. Example: lac operon in E. coli — repressor binds operator unless lactose (inducer) is present. Allows efficient response to environmental change.

8. Worked Example

A bacterial gene contains the template DNA strand sequence: 3'-TAC GTA AAG CGC TCG ATT-5'.

(a) Write the mRNA sequence transcribed from this strand. (b) Identify the start codon and the first 5 amino acids encoded. (c) Where is the stop codon?

Solution.

(a) Transcription is 5'→3' on mRNA, complementary to the template (3'→5'). U replaces T. Template: 3'-T A C G T A A A G C G C T C G A T T-5' mRNA: 5'-A U G C A U U U C G C G A G C U A A-3'

(b) Start codon: AUG (first 3 letters). Reading frame: AUG CAU UUC GCG AGC UAA

Amino acids:

• AUG → Met (start)
• CAU → His
• UUC → Phe
• GCG → Ala
• AGC → Ser

(c) UAA at the end is a stop codon — translation terminates here.

9. Common Pitfalls

• 5'→3' direction: DNA polymerase synthesises 5'→3'; this is why the lagging strand needs Okazaki fragments. Get this direction wrong and the whole mechanism flips.
• mRNA = template inverted: mRNA is complementary to the template strand, but identical to the coding strand (with U for T).
• Polysome confusion: multiple ribosomes on ONE mRNA, each making its own protein copy. Not multiple mRNAs.
• Operon vs eukaryotic genes: operons are prokaryote-specific. Eukaryotes regulate genes individually with promoters + transcription factors.

10. Practice Questions

1. Diagram the structure of a single Okazaki fragment, indicating the RNA primer, DNA polymerase III contribution, and where DNA polymerase I and ligase act.
2. Why must AUG be the start codon, and what would happen if a mutation changed an internal AUG to AUA?
3. Explain why the lac operon in E. coli is not transcribed when lactose is absent, but is transcribed when lactose is present.

11. Quick Reference Cheatsheet

• DNA replication: helicase, gyrase, primase, DNA pol III (synthesis), DNA pol I (primer removal), ligase. Semiconservative. Leading + Okazaki on lagging.
• Transcription: RNA pol → mRNA, 5'→3'. Promoter, terminator. Eukaryotes splice introns.
• Translation: small subunit binds AUG → tRNAs deliver amino acids → peptide bond formed → translocation → stop codon → release.
• Code: triplet, universal, degenerate, AUG start, UAA/UAG/UGA stop.
• Regulation eukaryote: chromatin, splicing, mRNA stability, post-translational. Prokaryote: operons.

12. What's Next

Nucleic Acids feeds Topic 8 (Metabolism) for ATP synthesis (mitochondrial DNA, ribosomes inside mitochondria) and Topic 10 (Genetics HL) for inheritance patterns. Use Ollie for any specific transcription/translation problem: "Walk me through what a frameshift mutation does" or "Why is the wobble hypothesis biologically useful?"