VCE Biology Unit 3 AoS 1: Nucleic Acids & Proteins — Flashcards & Quiz

Q1: Describe the structure of DNA.

DNA is a double-stranded helical molecule made of nucleotides. Each nucleotide consists of a deoxyribose sugar, a phosphate group and a nitrogenous base (adenine, thymine, guanine or cytosine). The two strands are antiparallel and held together by hydrogen bonds between complementary base pairs: A-T (2 H-bonds) and G-C (3 H-bonds). The sugar-phosphate backbone runs 5′ to 3′.

Q2: Outline the process of semi-conservative DNA replication.

Steps: 1) Helicase unwinds and separates the double helix at the replication fork. 2) Each original strand serves as a template. 3) DNA polymerase III adds complementary nucleotides in the 5′→3′ direction. 4) Leading strand is synthesised continuously; lagging strand is synthesised in Okazaki fragments. 5) DNA ligase joins fragments. Result: two identical DNA molecules, each with one original and one new strand.

Q3: What are the key features of the genetic code?

The genetic code is: 1) Triplet — three bases (codon) code for one amino acid. 2) Degenerate (redundant) — most amino acids are coded by more than one codon. 3) Universal — the same codons code for the same amino acids in almost all organisms. 4) Non-overlapping — codons are read sequentially without overlap. 5) Has start codon (AUG = methionine) and stop codons (UAA, UAG, UGA).

Q4: Describe the process of transcription.

Transcription copies DNA into mRNA in the nucleus. Steps: 1) RNA polymerase binds to the promoter region on the template (antisense) strand. 2) RNA polymerase reads the template strand 3′→5′ and synthesises mRNA 5′→3′, using complementary RNA bases (U pairs with A). 3) The mRNA strand grows until RNA polymerase reaches a terminator sequence. 4) Pre-mRNA undergoes processing before leaving the nucleus.

Q5: What is RNA processing and why is it necessary in eukaryotes?

Pre-mRNA undergoes three modifications before translation: 1) 5′ capping — a modified guanine cap is added to protect from degradation and aid ribosome binding. 2) 3′ polyadenylation — a poly-A tail is added for stability and export from nucleus. 3) Splicing — introns (non-coding sequences) are removed by spliceosomes; exons (coding sequences) are joined together. Alternative splicing allows one gene to produce multiple proteins.

Q6: Describe the process of translation.

Translation converts mRNA into a polypeptide at the ribosome. Steps: 1) Initiation — ribosome binds to mRNA at the start codon (AUG); initiator tRNA (carrying methionine) binds to P site. 2) Elongation — tRNAs bring amino acids to the A site; anticodon pairs with codon; peptide bond forms between amino acids; ribosome translocates along mRNA. 3) Termination — ribosome reaches a stop codon (UAA, UAG, UGA); release factor binds; polypeptide is released.

Q7: Describe the four levels of protein structure.

Primary structure: linear sequence of amino acids (determined by gene). Secondary structure: local folding into alpha helices and beta pleated sheets (H-bonds between backbone atoms). Tertiary structure: overall 3D shape of a single polypeptide (interactions between R groups: H-bonds, disulfide bridges, ionic bonds, hydrophobic interactions). Quaternary structure: arrangement of multiple polypeptide subunits into a functional protein.

Q8: How do enzymes catalyse biochemical reactions?

Enzymes are biological catalysts (mostly proteins) that lower activation energy. The induced-fit model: the substrate binds to the active site, causing a conformational change in the enzyme that stabilises the transition state. Factors affecting enzyme activity: temperature (increases rate until denaturation), pH (optimal range), substrate concentration (saturation kinetics), and enzyme concentration.

Q1: DNA strands run in the same direction (parallel) from 5′ to 3′.

Answer: FALSE

DNA strands are ANTIPARALLEL — one runs 5′→3′ and the complementary strand runs 3′→5′. This is essential for replication and transcription.

Q2: DNA replication is described as semi-conservative because each new molecule contains one original strand and one new strand.

Answer: TRUE

Semi-conservative means each daughter DNA molecule consists of one parental (conserved) strand and one newly synthesised strand, as demonstrated by Meselson and Stahl.

Q3: The genetic code is overlapping, meaning the same base can be part of two different codons during translation.

Answer: FALSE

The genetic code is NON-OVERLAPPING — each base belongs to only one codon. Codons are read sequentially in groups of three without sharing bases.

Q4: RNA polymerase reads the template strand in the 3′ to 5′ direction during transcription.

Answer: TRUE

RNA polymerase reads the template (antisense) strand 3′→5′ and synthesises the mRNA strand in the 5′→3′ direction, following complementary base pairing rules.

Q5: Introns are the coding sequences of a gene that are retained in the mature mRNA.

Answer: FALSE

INTRONS are non-coding sequences that are removed (spliced out) during RNA processing. EXONS are the coding sequences that are retained in mature mRNA and translated into protein.

DNA Structure and Replication

DNA's double helix structure with complementary base pairing (A-T, G-C) enables semi-conservative replication. Understanding the roles of helicase, DNA polymerase, and ligase is critical. You should be able to explain why replication is semi-conservative and how the leading and lagging strands differ in their synthesis mechanisms.

Transcription and mRNA Processing

Transcription converts a DNA template strand into pre-mRNA using RNA polymerase. In eukaryotes, the pre-mRNA undergoes processing including 5' capping, polyadenylation, and splicing to remove introns. Understanding these modifications explains how one gene can produce multiple protein variants through alternative splicing.

Translation and Protein Folding

Ribosomes read mRNA codons and assemble amino acid chains using tRNA molecules. The genetic code is degenerate, meaning multiple codons can specify the same amino acid. After translation, proteins fold into specific three-dimensional shapes determined by their amino acid sequence, which dictates their biological function.

Gene Expression Overview

Gene expression is the process by which information from a gene directs protein synthesis. Not all genes are expressed in every cell — differential gene expression allows cells with identical DNA to have specialised functions. Understanding the central dogma (DNA to RNA to protein) provides a framework for analysing genetic phenomena.

What topics are covered in VCE Biology Unit 3 AoS 1?

Unit 3 AoS 1 covers nucleic acid structure (DNA and RNA), DNA replication, the genetic code, transcription and translation, RNA processing (splicing, capping, polyadenylation), protein structure and function, enzyme action, gene mutations, and gene technologies including PCR, gel electrophoresis, CRISPR-Cas9 and recombinant DNA technology.

How does PCR amplify DNA for VCE Biology?

PCR (Polymerase Chain Reaction) amplifies specific DNA sequences through repeated thermal cycles: denaturation at 95°C separates strands, annealing at 55-65°C allows primers to bind, and extension at 72°C lets Taq polymerase synthesise new strands. Each cycle doubles the target DNA, producing millions of copies from a tiny sample.

What is the difference between transcription and translation in VCE Biology?

Transcription occurs in the nucleus where RNA polymerase copies the DNA template strand into mRNA. Translation occurs at ribosomes in the cytoplasm where the mRNA codons are read and tRNA molecules deliver corresponding amino acids to build a polypeptide chain.