interpret and construct genetic diagrams, including Punnett squares, to explain and predict the results of dihybrid crosses that involve autosomal linkage and epistasis (knowledge of the expected ratios for different types of epistasis is not expecte

Cambridge International AS & A Level Biology (9700) – Genetics and Core Cellular Concepts

Learning Objective

Interpret and construct genetic diagrams (including Punnett squares) to explain and predict the results of dihybrid crosses that involve autosomal linkage and epistasis. Integrate this knowledge with the cellular and molecular foundations required by the Cambridge syllabus.

Assessment Objectives (AO)

• AO1 – Knowledge & Understanding: Recall key facts, terminology and processes in genetics, cell structure, biochemistry and molecular biology.
• AO2 – Application: Use knowledge to solve problems, construct genetic diagrams and interpret experimental data.
• AO3 – Analysis & Evaluation: Analyse results, evaluate experimental design and discuss the significance of genetic concepts in health and biotechnology.

1. Core Cellular Foundations (Syllabus Blocks 1‑5)

1.1 Cell Structure & Microscopy

• Prokaryote vs. Eukaryote – nucleus, membrane‑bound organelles, size, DNA organisation (circular vs. linear).
• Key organelles – nucleus (chromatin, nucleolus), mitochondrion, chloroplast, endoplasmic reticulum (rough & smooth), Golgi apparatus, lysosome, vacuole, ribosome, centrosome (centrioles).
• Microscopy techniques – light microscopy (bright‑field, phase‑contrast, fluorescence), electron microscopy (TEM, SEM).
• Practical tip – temporary wet mounts of onion epidermis to visualise cell wall, plasma membrane and nucleus.

1.2 Biological Molecules

1.3 Enzymes

• Action: Lower activation energy; form enzyme–substrate complex.
• Michaelis–Menten kinetics: V = V_max[S] / (K_m + [S])
• Factors affecting activity: Temperature, pH, substrate concentration, competitive & non‑competitive inhibitors.
• Link to later topics: Enzyme regulation in respiration, photosynthesis and DNA replication.

1.4 Cell Membranes & Transport

• Fluid‑mosaic model: Phospholipid bilayer with embedded proteins, cholesterol, glycolipids.
• Passive transport: Diffusion, osmosis, facilitated diffusion.
• Active transport: Primary (ATP‑driven pumps) and secondary (symport/antiport).
• Vesicular transport: Endocytosis, exocytosis.
• Relevance to genetics: Membrane permeability influences uptake of mutagens and delivery of gene‑editing reagents.

1.5 The Mitotic Cell Cycle

• Phases: G₁, S (DNA synthesis), G₂, M (prophase, metaphase, anaphase, telophase) → cytokinesis.
• Key structures: Centrioles, spindle fibres, kinetochores, chromosomes (chromatid pairs), centromere, telomere.
• Control mechanisms: Cyclins, CDKs, checkpoints (G₁/S, G₂/M, spindle).
• Link to genetics: Errors in mitosis → aneuploidy, tumour formation, mutation propagation.

2. Nucleic Acids & Protein Synthesis (Syllabus Block 6)

2.1 DNA Structure & Replication

• Double helix, antiparallel strands, complementary base‑pairing (A–T, G–C).
• Key enzymes:

  • DNA helicase – unwinds the double helix.
  • DNA primase – synthesises short RNA primers.
  • DNA polymerase δ (lagging strand) & DNA polymerase ε (leading strand) – add nucleotides (5’→3’) with 3’→5’ proofreading.
  • DNA ligase – joins Okazaki fragments.
  • Topoisomerase – relieves super‑coiling.

• Semiconservative replication – each daughter DNA contains one parental strand.
• Replication origins (ORI) and replication forks.

2.2 Transcription & RNA Processing (Eukaryotes)

• RNA polymerase II synthesises pre‑mRNA (5’→3’).
• 5’ capping: addition of 7‑methylguanosine; protects mRNA and assists ribosome binding.
• Splicing: removal of introns by the spliceosome; alternative splicing can generate multiple proteins from one gene.
• Poly‑A tail: ~200 A residues added to 3’ end; stabilises mRNA.
• Other RNA types – tRNA (charged by amino‑acyl‑tRNA synthetase), rRNA (ribosomal subunits), miRNA, siRNA (gene regulation).

2.3 Translation & Post‑Translational Modifications

• Ribosome structure: 60S large + 40S small subunit (eukaryotes) forming an 80S ribosome.
• Three stages:

  1. Initiation – mRNA binds to the small subunit, initiator tRNA (Met‑tRNA_i) pairs with the AUG start codon; large subunit joins.
  2. Elongation – amino‑acyl‑tRNAs enter the A‑site, peptide bond forms, ribosome translocates.
  3. Termination – release factors recognise UAA, UAG, UGA; polypeptide is released.

• Post‑translational modifications: phosphorylation, glycosylation, proteolytic cleavage, lipidation, ubiquitination – affect activity, localisation and stability.

2.4 Gene Regulation (Brief Overview)

• Prokaryotes: operons (e.g., lac operon – repressor, inducer, catabolite repression).
• Eukaryotes: transcription factors, enhancers/silencers, epigenetic mechanisms (DNA methylation, histone acetylation/deacetylation), RNA interference.
• Environmental signals can up‑ or down‑regulate gene expression, linking genotype to phenotype.

3. Principles of Genetic Inheritance (Blocks 11 & 17)

3.1 Terminology (AO1 Glossary)

3.2 Mendelian Monohybrid & Dihybrid Crosses

• Single‑gene (monohybrid) cross with complete dominance → 3 : 1 phenotypic ratio in F₂.
• Dihybrid cross with independent assortment (genes on different chromosomes or far apart) → classic 9 : 3 : 3 : 1 phenotypic ratio.
• Incomplete dominance and co‑dominance modify ratios (e.g., 1 : 2 : 1 for incomplete dominance).

3.3 Sex‑Linked & Polygenic Inheritance

• Genes on the X chromosome show sex‑linked patterns (e.g., colour blindness, haemophilia).
• Polygenic traits (e.g., skin colour, height) involve many genes; phenotypic distribution approximates a normal curve.

3.4 Autosomal Linkage

• Genes on the same chromosome tend to be inherited together; recombination during meiosis I can separate them.
• Recombination frequency (RF):
  

  \$\text{RF (\%)} = \frac{\text{Number of recombinant offspring}}{\text{Total offspring}} \times 100\$

• Interpretation:

  • RF < 5 % – tightly linked.
  • 5 % ≤ RF < 20 % – linked but useful for mapping.
  • ≈ 50 % – genes assort independently (different chromosomes or far apart).

• Gamete proportions from a heterozygote (AB/ab) with known RF:

  • Parental types (AB, ab) = ½ (1 − RF) each.
  • Recombinant types (Ab, aB) = ½ (RF) each.

3.5 Constructing a Linkage Map (Worked Example)

Data (test‑cross of heterozygote AB/ab with aabb):

Steps:

1. Calculate total offspring: 420 + 410 + 80 + 90 = 1000.
2. Recombinant total = 80 + 90 = 170.
3. RF = 170 / 1000 × 100 = 17 % → map distance = 17 cM.
4. Order of genes: the most frequent parental types (AB and ab) indicate that A and B are in coupling phase; the less frequent recombinants (Ab, aB) confirm the order A — B.

Result: A and B are 17 cM apart on the same chromosome.

3.6 Epistasis (Interaction Between Genes)

Epistasis occurs when the phenotype produced by one gene masks or modifies the expression of another gene. The exam does not require memorising the classic ratios, but you must be able to predict the phenotypic classes after the genotype has been worked out.

3.7 Critical‑Thinking Prompt (AO3)

“A measured recombination frequency between two genes is 12 %, but the published genetic map places them 8 cM apart. Discuss possible reasons for this discrepancy and how you would design an experiment to obtain a more accurate estimate.”

• Factors: crossover interference, sample size, scoring errors, presence of double cross‑overs, sex‑specific recombination rates.
• Experimental design: increase progeny number, use molecular markers, perform reciprocal crosses, analyse male vs. female meioses separately.

4. Constructing Genetic Diagrams

4.1 General Procedure for a Punnett Square (AO2)

1. Write the parental genotypes.
2. Determine all possible gametes each parent can produce (consider segregation, independent assortment, linkage, and recombination frequency).
3. Place one parent’s gametes across the top, the other’s down the side.
4. Fill each cell with the combined genotype formed by the intersecting gametes.
5. Convert each genotype to a phenotype using dominance and any epistatic rules.
6. Count the number of each phenotypic class and express as a ratio or percentage.

4.2 Example: Dihybrid Cross with Linked Genes and Recessive Epistasis (RF = 10 %)

Scenario: In pea plants, A (seed colour: A = yellow, a = green) and B (seed shape: B = round, b = wrinkled) are linked on chromosome 5 (RF = 10 %). The allele a is recessive epistatic to B (aa → green seeds irrespective of shape).

1. Parental cross: AABB × aabb → F₁ genotype AaBb (heterozygous for both linked genes).
2. Gamete frequencies from F₁ (using RF = 10 %):

   • Parental: AB = 45 %, ab = 45 %.
   • Recombinant: Ab = 5 %, aB = 5 %.

3. Construct a 4 × 4 Punnett square (scaled to 1000 offspring for clarity). The table shows expected numbers of each genotype and the phenotype after applying the epistatic rule.

Phenotypic percentages (rounded):

4.3 Flowchart for Applying Epistasis

5. Worked Practice Questions

5.1 Question 1 – Independent Assortment & Incomplete Dominance

Cross: CcDd × ccdd where C (purple > white, complete dominance) and D (flower length: DD = long, Dd = intermediate, dd = short; incomplete dominance).

1. List the gametes produced by each parent.
2. Construct a 2 × 4 Punnett square.
3. Give the phenotypic ratio in the F₂ generation.

5.2 Question 2 – Linked Genes, Test‑Cross and Mapping

A plant heterozygous for two linked genes (R r S s) is test‑crossed with a homozygous recessive (rr ss). The offspring numbers are: RS = 310, rs = 300, Rs = 90, rS = 100.

1. Calculate the recombination frequency and map distance.
2. State the most likely order of the genes on the chromosome.
3. Predict the phenotypic classes expected if the heterozygote were crossed with a plant of genotype RrSs (assume the same RF).

5.3 Question 3 – Epistasis with Linked Genes

In corn, kernel colour (Y = yellow, y = white) and kernel texture (R = rough, r = smooth) are linked (RF = 15 %). The allele y is recessive epistatic to texture (yy → white regardless of R/r). A heterozygote YyRr is crossed with a double‑recessive yyrr. Construct the Punnett square, calculate the expected phenotypic percentages, and comment on how the linkage alters the classic 9 : 3 : 4 ratio for recessive epistasis.

6. Linking Genetics to Cellular Foundations (Integration Note)

Accurate DNA replication (Block 6) and faithful chromosome segregation during mitosis (Block 5) are prerequisites for the inheritance patterns described in Section 3. Errors in replication (e.g., polymerase proofreading defects) or in spindle checkpoint control can generate mutations or aneuploidy that alter genotype frequencies in a population. Moreover, the cell membrane’s role in transporting mutagens or gene‑editing reagents provides a direct connection between membrane biology and the manipulation of genetic traits in biotechnology.