The NSC Life Sciences exam consists of two 2.5-hour papers (150 marks each). Paper 1 tests Meiosis (11 marks), Reproduction in Vertebrates (6 marks), Human Reproduction (31 marks), Responding to the Environment – Humans: Nervous System & Eye/Ear (40 marks), Endocrine System & Homeostasis (30 marks), and Human Impact on the Environment (32 marks). Paper 2 tests DNA: Code of Life (27 marks), Meiosis review (12 marks), Genetics & Inheritance (45 marks), and Evolution (66 marks).
Worked Example 1: The Mandatory 6-Mark Genetic Cross (Paper 2 · 6 Marks)
DBE markers strictly award marks for correct headings and labels. Missing the labels \( P_1 \), \( F_1 \), Meiosis, or Fertilisation costs marks even if the letters are right:
P_1 Genotype: Bb × bb
-------------------- MEIOSIS --------------------
Gametogenesis: G: B, b × b, b
------------------ FERTILISATION ------------------
Punnett Square:
| B | b |
-----|-------|-------|
b | Bb | bb |
-----|-------|-------|
b | Bb | bb |
F_1 Genotypes: 2 Bb : 2 bb (or 50% Bb : 50% bb)
F_1 Phenotypes: 50% Brown eyes : 50% Blue eyes (1 Brown : 1 Blue)
• Mark 1: \( P_1 \) phenotype and genotype correctly indicated.
• Mark 2: Label "Meiosis" indicated.
• Mark 3: Correct gametes identified.
• Mark 4: Label "Fertilisation" indicated.
• Mark 5: Correct genotypes of offspring (\(Bb\) and \(bb\)).
• Mark 6: Correct phenotype ratio (50% brown, 50% blue).
Worked Example 2: Negative Feedback Loop — Blood Glucose Regulation (Paper 1 · 6 Marks)
- When carbohydrate digestion causes blood glucose levels to rise above normal (above ~5 mmol/L):
- The beta cells in the Islets of Langerhans in the pancreas detect the elevated glucose levels.
- The pancreas secretes the hormone insulin into the bloodstream.
- Insulin stimulates cells (particularly liver and muscle cells) to absorb more glucose from the blood.
- In the liver, excess glucose is converted into glycogen for storage.
- As a result, blood glucose levels decrease back to normal, closing the negative feedback loop.
Worked Example 3: Tracing Sex-Linked Haemophilia in a Pedigree Diagram (Paper 2 · 5 Marks)
(a) State the genotype of the mother (Individual 2).
(b) Explain why females are rarely haemophiliacs compared to males.
1. The affected son (Individual 5) has the genotype \( X^h Y \).
2. He inherits the Y chromosome from his father (who is \( X^H Y \)).
3. Therefore, the son must inherit his \( X^h \) allele from his mother.
4. Since the mother has normal blood clotting, she must carry the dominant allele as well.
Mother's Genotype: \( X^H X^h \) (heterozygous / carrier female).
Males have only one X chromosome (\( XY \)). If a male inherits a single recessive allele (\( X^h \)), it will be expressed because the Y chromosome has no corresponding allele. Females have two X chromosomes (\( XX \)) and would need to inherit two recessive alleles (\( X^h X^h \))—one from a carrier/affected mother AND one from an affected father—to have haemophilia.