Cambridge AS & A‑Level Biology – Topic 16 Inheritance

Objective

Interpret and construct genetic diagrams (Punnett squares, pedigree charts, linkage maps) to explain and predict the outcomes of monohybrid and dihybrid crosses that involve:

• Complete dominance
• Incomplete (partial) dominance
• Codominance
• Multiple‑allele systems
• Sex‑linked inheritance
• Linkage & recombination (including map‑distance calculation)
• Epistasis (dominant & recessive) and polygenic inheritance
• Statistical validation of expected ratios (χ² test)

1. Syllabus Coverage Checklist (AS 9700 & A‑Level extensions)

2. Fundamental Terminology

• Locus (plural loci) – specific position of a gene on a chromosome.
• Allele – alternative form of a gene at a locus.
• Genotype – the set of alleles an individual carries (e.g., AA, Aa, aa).
• Phenotype – observable trait(s) produced by the genotype together with the environment.
• Homozygous – two identical alleles (AA or aa).
• Heterozygous – two different alleles (Aa).
• Hemizygous – a single allele present because the gene is on a sex chromosome (e.g., males for X‑linked genes).
• Gamete – a haploid reproductive cell (sperm or ovum).
• Recombinant – a gamete that contains a new combination of alleles produced by crossing‑over.

3. Mendelian Laws & Meiosis

3.1 Law of Segregation (First Law)

During meiosis I, homologous chromosomes separate so that each gamete receives only one allele of each autosomal gene.

3.2 Law of Independent Assortment (Second Law)

Alleles of genes that lie on different chromosomes (or are far apart on the same chromosome) are distributed to gametes independently.

3.3 Meiosis – Reduction Division

3.4 Consequences for Genetics

• Segregation of alleles → Law of Segregation.
• Independent orientation of different chromosome pairs → Law of Independent Assortment.
• Crossing‑over creates new allele combinations – the basis of linkage & recombination.

4. Types of Allelic Interaction

• Complete dominance – dominant allele masks recessive allele (AA and Aa have identical phenotype).
• Incomplete (partial) dominance – heterozygote phenotype is intermediate between the two homozygotes.
• Codominance – both alleles are fully expressed in the heterozygote.
• Multiple‑allele systems – more than two alleles exist in the population (e.g., ABO blood groups, sickle‑cell allele Hb^S).
• Sex‑linked inheritance – genes located on the X or Y chromosome; males are hemizygous for X‑linked genes.
• Epistasis – interaction where one gene masks or modifies the expression of another (dominant or recessive).
• Polygenic inheritance – many genes each contribute a small effect; phenotype shows a continuous distribution.

5. Constructing Genetic Diagrams

5.1 General Procedure for a Punnett Square

1. Write the genotype of each parent.
2. List all possible gametes each parent can produce (apply segregation and, where relevant, independent assortment).
3. Place one parent’s gametes across the top and the other’s down the side.
4. Combine the gametes to fill the squares – each square represents a possible genotype of an offspring.
5. Convert genotypes to phenotypes using the appropriate dominance relationship.
6. Count occurrences to obtain genotype and phenotype ratios; express as fractions, percentages or “x : y” ratios.

5.2 Monohybrid Crosses

5.2.1 Complete Dominance (Aa × Aa)

Genotype ratio = 1 AA : 2 Aa : 1 aa (25 % : 50 % : 25 %).

Phenotype ratio (dominant : recessive) = 3 : 1 (75 % dominant, 25 % recessive).

5.2.2 Incomplete Dominance (Rr × Rr)

Genotype ratio = 1 RR : 2 Rr : 1 rr.

Phenotype ratio = 1 red : 2 pink : 1 white.

5.2.3 Codominance (I^AI^B × ii)

Genotype ratio = 2 I^Ai : 2 I^Bi → Phenotype ratio = 1 A : 1 B.

5.3 Dihybrid Cross – Independent Assortment (AaBb × AaBb)

Each parent can produce four gamete types: AB, Ab, aB, ab.

Genotype ratio (simplified) = 9 AB : 3 Abb : 3 aaB : 1 aabb.

Phenotype ratio (both traits dominant) = 9 : 3 : 3 : 1.

Note: The 9:3:3:1 ratio assumes the two genes are on different chromosomes (or far apart). If they are linked, the observed ratio will deviate.

5.4 Sex‑Linked Monohybrid Cross (X‑linked recessive colour blindness)

Cross: carrier female (X^CX^c) × normal male (X^CY).

Result:

• Females: 50 % normal, 50 % carriers.
• Males: 50 % normal, 50 % colour‑blind.

5.5 Linked Genes, Recombination & Map Distance

When two loci are close on the same chromosome they tend to be inherited together (parental types). Crossing‑over during prophase I can produce recombinant gametes.

Example – Genes A and B 10 cM apart

Calculating Map Distance

Map distance (cM) = (Number of recombinant offspring ÷ Total offspring) × 100.

Worked example – Test cross AB ab × ab ab yields 180 offspring: 162 parental (AB or ab) and 18 recombinant (Ab or aB).

• Recombination frequency = 18 / 180 = 0.10 → 10 cM.
• Thus A and B are 10 map units apart.

Constructing a Simple Linkage Map

1. Perform a test cross and record numbers of each gamete type.
2. Calculate recombination frequencies for each pair of genes.
3. Arrange genes on a line so that the sum of adjacent distances equals the total distance between the most‑distant genes.

6. Statistical Validation – χ² Test

The χ² test determines whether observed offspring ratios fit the expected Mendelian ratios.

Step‑by‑Step Procedure

1. State the null hypothesis (H₀): observed ratios do not differ from expected ratios.
2. Calculate expected numbers (E) for each class: E = (total offspring) × (expected proportion).
3. Compute χ² using χ² = Σ[(O – E)² / E], where O = observed number.
4. Determine degrees of freedom (df = number of classes – 1 – number of parameters estimated).
5. Compare the calculated χ² with the critical value from the χ² table (df, p = 0.05).
   

   • χ² ≤ critical → accept H₀ (fit).
   

   • χ² > critical → reject H₀ (significant deviation).

Example – Monohybrid cross Aa × Aa (observed 78 dominant, 22 recessive; total = 100)

df = 2 – 1 = 1. Critical χ² (0.05, df = 1) ≈ 3.84.

Since 0.48 < 3.84, we accept H₀; the data fit the 3 : 1 ratio.

7. Epistasis – Gene Interactions

7.1 Recessive Epistasis

One recessive allele at a second locus masks the expression of the first gene.

• Classic example: mouse coat colour (A = agouti, B = black, a = non‑agouti, b = non‑black).
  

  Genotype aa → albino (no colour), regardless of B locus.

Expected phenotypic ratio = 9 : 3 : 4.

7.2 Dominant Epistasis

A dominant allele at one locus suppresses the expression of a second gene.

• Example: squash colour – A (dominant) suppresses B locus.

Expected phenotypic ratio = 12 : 3 : 1.

7.3 Worked Example – Recessive Epistasis

Cross: AaBb × AaBb (both genes on different chromosomes). Expected phenotypes:

Phenotypic ratio = 9 : 3 : 4.

8. Polygenic Inheritance

Traits controlled by many loci each with a small additive effect produce a continuous distribution of phenotypes.

8.1 Example – Human Skin Colour

• Assume 5 loci (A, B, C, D, E) each with two alleles (dark‑producing D and light‑producing d). The more D alleles an individual carries, the darker the skin.
• Possible genotype scores range from 0 (ddddd) to 10 (DDDDD).

8.2 Visualising the Distribution

A histogram of a large population typically shows a bell‑shaped (normal) curve.

• Mean (μ) – average phenotype score.
• Standard deviation (σ) – spread of the distribution.
• Environmental factors (e.g., sun exposure) shift the mean without altering σ.

8.3 Quick Exercise

Given the following frequencies for a 5‑locus skin‑colour trait (score = number of dark alleles):

Plot these values on a histogram; the shape should approximate a normal distribution, illustrating polygenic inheritance.

9. Summary Checklist for Revision

• Define locus, allele, genotype, phenotype, homo‑/heterozygous, hemizygous.
• State Mendel’s two laws and link them to meiosis I & II.
• Construct Punnett squares for:

  • Complete dominance
  • Incomplete dominance
  • Codominance
  • Multiple‑allele systems (e.g., ABO)
  • Sex‑linked traits (X‑linked recessive & dominant)

• Perform dihybrid crosses assuming independent assortment; recognise when linkage will alter the 9:3:3:1 ratio.
• Calculate recombination frequency and map distance; draw a simple linkage map.
• Apply the χ² test to check observed vs. expected ratios (step‑by‑step).
• Identify epistatic relationships and write the expected phenotypic ratios (9:3:4, 12:3:1, etc.).
• Describe polygenic traits, sketch a normal distribution, and explain environmental effects.
• Practice with past‑paper style questions covering each of the points above.