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Worked Solutions

Module 5: Heredity — Worked Solutions (HSC Biology)

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Worked examples for HSC Biology Module 5: Heredity. Each shows where the marks are awarded, the key idea, and a full model answer explained by your choice of tutor — Stella, Ella or Cassie.

How to use these

Attempt each question first, then check your answer against the model responses. Use the tutor tabs to read the solution in the style that suits you: Stella is direct and challenging, Ella is warm and explains the why, and Cassie is concise and analytical.

Heredity questions reward genetic reasoning. Define your allele symbols, show your Punnett square or pedigree logic, and always link DNA back to the protein it codes for.

Example 1 — Autosomal inheritance and Punnett squares

Standard 4 marks

Question

In pea plants, purple flower colour ($\text{P}$) is dominant over white ($\text{p}$). Two heterozygous purple-flowered plants are crossed. Using a Punnett square, determine the expected phenotypic ratio of the offspring, and explain why a white-flowered offspring can arise from two purple-flowered parents.

Solution

Both parents are heterozygous purple, so $\text{Pp} \times \text{Pp}$.

The Punnett square gives offspring genotypes $\text{PP} : \text{Pp} : \text{pp}$ in a $1 : 2 : 1$ ratio.

Phenotypes: $\text{PP}$ and $\text{Pp}$ are purple (3), $\text{pp}$ is white (1), so the ratio is 3 purple : 1 white.

A white offspring appears because each parent carries one recessive $\text{p}$ allele hidden behind the dominant $\text{P}$. When two $\text{p}$ alleles combine in one offspring ($\text{pp}$), the recessive phenotype is expressed.

Where the marks go

  • 1 mark: Correctly identifies parental genotypes as $\text{Pp} \times \text{Pp}$
  • 1 mark: Completes a correct Punnett square showing genotypes
  • 1 mark: States the correct phenotypic ratio of 3 purple : 1 white
  • 1 mark: Explains that white offspring arise when two recessive alleles combine in a $\text{pp}$ genotype

Key idea

A heterozygous cross ($\text{Pp} \times \text{Pp}$) gives a 3 : 1 phenotypic ratio; recessive phenotypes appear when two hidden recessive alleles combine.

Example 2 — DNA and polypeptide synthesis

Standard 5 marks

Question

Explain how the sequence of nucleotides in DNA is used to synthesise a polypeptide, referring to both transcription and translation.

Solution

Two stages: transcription in the nucleus, translation at the ribosome.

Transcription: the DNA double helix unwinds and one strand acts as a template. RNA polymerase pairs free RNA nucleotides to the template, producing messenger RNA (mRNA). The mRNA carries the base sequence as a complementary copy and leaves the nucleus.

Translation: the mRNA binds to a ribosome. Each group of three bases (a codon) specifies one amino acid. Transfer RNA (tRNA) molecules with complementary anticodons deliver the matching amino acids. The ribosome joins the amino acids with peptide bonds in the order set by the codons, building the polypeptide.

So the nucleotide sequence in DNA determines the codon order in mRNA, which determines the amino acid sequence — and that sequence is the polypeptide.

Where the marks go

  • 1 mark: Identifies transcription and translation as the two stages
  • 1 mark: Describes transcription producing mRNA from a DNA template
  • 1 mark: States that codons (triplets) on mRNA each specify an amino acid
  • 1 mark: Describes the role of tRNA and the ribosome in assembling amino acids
  • 1 mark: Links the DNA nucleotide sequence to the resulting amino acid sequence of the polypeptide

Key idea

DNA is transcribed into mRNA, then translated at the ribosome where codons specify amino acids — so the nucleotide order determines the polypeptide's amino acid sequence.