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

Roles of Genes in Determining the Phenotype

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 (knowledge of the expected ratios for different types of epistasis is not required).

1. Basic Concepts

• Gene: A segment of DNA that encodes a functional product, usually a protein.
• Allele: Alternative forms of a gene that arise by mutation and are found at the same locus on homologous chromosomes.
• Genotype: The genetic constitution of an individual (e.g., AA, Aa, aa).
• Phenotype: The observable characteristics produced by a genotype interacting with the environment.
• Dominance relationships:

  • Complete dominance – one allele masks the other completely.
  • Incomplete dominance – heterozygote shows an intermediate phenotype.
  • Co‑dominance – both alleles are expressed simultaneously.

2. From Genotype to Phenotype

The pathway can be summarised as:

\$\text{Genotype} \;\xrightarrow{\text{Transcription}} \;\text{mRNA} \;\xrightarrow{\text{Translation}} \;\text{Protein} \;\xrightarrow{\text{Function}} \;\text{Phenotype}\$

Regulatory genes, epigenetic modifications and environmental factors can modify any step, leading to variation in the phenotype even among individuals with the same genotype.

3. Genetic Diagrams

Genetic diagrams are visual tools that help to track the segregation and assortment of alleles during meiosis and fertilisation. The most common diagrams are:

1. Punnett squares – for monohybrid and dihybrid crosses.
2. Pedigrees – for tracking inheritance through families.
3. Linkage maps – for locating genes on chromosomes.

4. Constructing a Punnett Square

Steps:

1. Write the parental genotypes.
2. Determine the possible gametes each parent can produce (considering segregation and independent assortment).
3. Place one parent’s gametes across the top and the other’s down the side.
4. Fill each cell with the genotype formed by the combination of the intersecting gametes.

5. Dihybrid Crosses

A dihybrid cross involves two genes, each with two alleles. For genes that assort independently, the classic phenotypic ratio in the F₂ generation is \$9:3:3:1\$.

When the two genes are on different chromosomes, the gametes from the heterozygous F₁ parent are \$AB\$, \$Ab\$, \$aB\$, \$ab\$ – giving a \$4 \times 4\$ Punnett square.

6. Autosomal Linkage

Genes that lie on the same chromosome may be inherited together. The degree of linkage depends on the distance between the loci.

• Recombination frequency (RF): \$RF = \dfrac{\text{Number of recombinant offspring}}{\text{Total offspring}} \times 100\%\$.
• Genes with \$RF < 5\%\$ are considered tightly linked; those with \$RF > 50\%\$ behave as if they assort independently.

When constructing a dihybrid cross with linked genes, the initial gamete set is reduced (e.g., mostly parental types \$AB\$ and \$ab\$) and a smaller proportion of recombinant types \$Ab\$ and \$aB\$ must be added according to the known recombination frequency.

7. Epistasis

Epistasis occurs when the expression of one gene (the epistatic gene) masks or modifies the effect of another gene (the hypostatic gene). Common types include:

• Recessive epistasis: Two copies of the epistatic allele are required to mask the other gene (e.g., \$aa\$ blocks colour production).
• Dominant epistasis: A single dominant epistatic allele overrides the other gene (e.g., \$A\$ produces a fruit shape regardless of \$B\$).
• Duplicate recessive epistasis: Either gene can produce the same phenotype; both must be homozygous recessive to lose the trait.

When drawing Punnett squares for epistatic crosses, first determine the combined genotype of the two loci, then apply the epistatic rule to convert genotypes into phenotypes.

8. Worked Example – Dihybrid Cross with Autosomal Linkage and Recessive Epistasis

Problem: In pea plants, gene A (seed colour) is dominant for yellow (\$A\$) and recessive for green (\$a\$). Gene B (seed shape) is dominant for round (\$B\$) and recessive for wrinkled (\$b\$). The two genes are linked on chromosome 5 with a recombination frequency of \$10\%\$. Additionally, \$aa\$ is epistatic to \$B\$ – if a plant is homozygous recessive at A, seed colour is green regardless of the \$B\$ locus.

1. Parental cross: \$AABB \times aabb\$ → F₁ genotype \$AaBb\$.
2. Gametes from F₁ (considering linkage):

   • Parental types: \$AB\$ and \$ab\$ (each \$45\%\$ of gametes).
   • Recombinant types: \$Ab\$ and \$aB\$ (each \$5\%\$ of gametes).

3. Construct a \$4 \times 4\$ Punnett square using the above frequencies (numbers can be scaled to a total of 1000 offspring for clarity).
4. Apply the epistatic rule: any genotype containing \$aa\$ gives green seeds, irrespective of \$B\$.

The resulting phenotypic classes (rounded to the nearest whole number) are:

9. Practice Questions

1. Two genes, C (flower colour) and D (petal length), are on different autosomes. C shows complete dominance (purple \$C\$ over white \$c\$). D shows incomplete dominance (long \$DD\$, intermediate \$Dd\$, short \$dd\$). A \$C c Dd \times c c dd\$ cross is performed. Construct the Punnett square and state the phenotypic ratio.
2. Gene E (seed coat) is dominant for brown (\$E\$) and recessive for white (\$e\$). Gene F (seed texture) is dominant for smooth (\$F\$) and recessive for rough (\$f\$). The two genes are 20 % linked. If the parental genotypes are \$EeFf \times eeff\$, calculate the expected proportion of smooth‑brown seeds in the F₂ generation, assuming no epistasis.
3. In a plant, gene G (height) is dominant for tall (\$G\$) and recessive for dwarf (\$g\$). Gene H (leaf colour) is dominant for green (\$H\$) and recessive for yellow (\$h\$). \$g h\$ is epistatic to \$G\$ – any plant homozygous recessive at both loci is dwarf and yellow regardless of \$G\$. Predict the phenotypic classes from a \$GgHh \times GgHh\$ dihybrid cross.

10. Summary Checklist

• Identify alleles and their dominance relationships.
• Determine possible gametes for each parent, accounting for linkage and recombination frequency.
• Construct Punnett squares (monohybrid or dihybrid) using the appropriate gamete set.
• Apply epistatic rules after the genotype has been established.
• Convert genotypes to phenotypes and calculate expected percentages or ratios.