lec 6-Genetics and Molecular Biology: Gene Interaction II

• Understand that most genes do not operate in isolation within a biological system.

• Recognize that modified dihybrid ratios are key indicators of gene interactions, particularly focusing on the concept of epistasis.

• Grasp that a single phenotype is often regulated by a network of interacting genes.

• Learn how to employ the complementation test in model organisms for determining the specific number of genes responsible for a particular characteristic.

Genes Affecting the Same Trait: Independent Operation

• It is rare, but genes that affect the same trait can operate independently of one another.

• Example: Californian Corn Snakes
    - Natural phenotype: A pattern of repeating black and orange colors used for camouflage.
    - Pigment Pathways:
      - Orange pigment is determined by the $o^+$ allele.
      - Black pigment is determined by the $b^+$ allele.
    - Genotypes and Phenotypes:
      - $o^+ - / b^+ -$: Camouflaged (wild type).
      - $o o / b^+ -$: Black (lack of orange pigment).
      - $o^+ - / b b$: Orange (lack of black pigment).
      - $o o / b b$: Albino (absence of both pigments).
    - This scenario leads to typical Mendelian ratios at the phenotypic level ($9:3:3:1$) as these genes operate in different pathways without influencing each other.

Introduction to Epistasis

• Epistasis happens when one gene or gene pair influences or modifies the expression of another gene or gene pair.

• Most phenotypes are governed by several genes that interact within the same biological pathway.

• Epistasis modifies the traditional Mendelian $9:3:3:1$ dihybrid ratio.

1. Complementary Gene Action ($9:7$ Ratio)

• This occurs when genes work together to produce a specific phenotype; both genes must have at least one dominant allele for the phenotype to be expressed.

• Example: Purple Flower Color (Anthocyanin) in Peas
    - Locus P: The dominant allele $P$ produces purple flowers; the recessive allele $p$ results in white flowers (no color).
    - Locus C: The dominant allele $C$ produces color; the recessive allele $c$ prevents color development (leading to white/colorless flowers).   - Homozygous recessive genotypes at either locus ($cc$ or $pp$) will result in a white phenotype.

• Dihybrid Cross ($CcPp imes CcPp$):
    - Phenotype Ratio: $9$ Colored : $7$ White.
    - $9 \, C - / P -$: Colored (both dominant alleles present).
    - $3 \, C - / pp$: White.
    - $3 \, cc / P -$: White.
    - $1 \, cc / pp$: White.

• Biochemical Explanation:
    - There exists a pathway: $ext{Precursor} ightarrow ext{Intermediate X} ightarrow ext{Intermediate Y} ightarrow ext{Pigment (Anthocyanin)}$.
    - Gene $C$ produces an enzyme necessary for one step; Gene $P$ produces an enzyme required for a subsequent step.
    - If the enzyme $C$ is absent, the pathway halts; if enzyme $P$ is absent, the pigment cannot be produced.
    - Allele $C$ is considered epistatic to $P$. A $9:7$ ratio strongly indicates this complementary epistatic relationship.

2. Recessive Epistasis ($9:3:4$ Ratio)

• This occurs when a recessive genotype at one locus masks the expression of a trait at a second locus.

• Example: Coat Colour in Mice
    - Locus B (Pigment): $B-$ (Pigmented), $bb$ (Albino).
    - Locus A (Distribution): $AA$ or $Aa$ (Agouti - grayish with alternating pigment bands), $aa$ (Black).
    - The $bb$ genotype inhibits any pigment synthesis, thus concealing any expression at the $A$ locus.

• Dihybrid Cross ($AaBb imes AaBb$):
    - Phenotype Ratio: $9$ Agouti : $3$ Black : $4$ Albino.
    - $9 \, A - / B -$: Agouti.
    - $3 \, aa / B -$: Black.
    - $3 \, A - / bb$: Albino.
    - $1 \, aa / bb$: Albino.

• Biochemical Explanation:
    - The pathway follows: $ext{Colourless} ightarrow ext{(Enzyme B)} ightarrow ext{Black} ightarrow ext{(Enzyme A)} ightarrow ext{Agouti}$.
    - If $bb$ is present, then Enzyme $B$ is missing, and the pathway stops at colorless.
    - If $B$ is present but $aa$ is present, the pathway stops at black.
    - Recessive $b$ masks the expression of $A$; conversely, recessive $a$ does not mask the expression of $B$.

3. Dominant Epistasis ($12:3:1$ Ratio)

• This occurs when a dominant allele at one locus obscures the genotype at a second locus.

• Example: Colour in Summer Squashes
    - Phenotypes include White, Yellow, and Green.
    - Dominant allele $B$ is epistatic to the $A$ locus.

• Dihybrid Cross ($AaBb imes AaBb$):
    - Phenotype Ratio: $12$ White : $3$ Yellow : $1$ Green.
    - $9 \, A - / B -$: White.
    - $3 \, aa / B -$: White.
    - $3 \, A - / bb$: Yellow.
    - $1 \, aa / bb$: Green.

• Biochemical Logic:
    - The presence of dominant $B$ results in a white phenotype regardless of the alleles at the $A$ locus (thus yielding $12/16$ white).
    - If the genotype is $bb$, the $A$ locus then determines color ($A-$ is yellow, $aa$ is green).

4. Duplicate Gene Action ($15:1$ Ratio)

• This happens when two or more genes carry out the same redundant function.

• Example: Seed Shape in Wheat
    - At least one dominant allele from either locus ($T$ or $V$) results in a triangular seed shape.
    - Only the double recessive genotype ($ttvv$) produces an ovate seed shape.

• Dihybrid Cross ($TtVv imes TtVv$):
    - Phenotype Ratio: $15$ Triangular : $1$ Ovate.

• Biochemical Explanation:
    - Enzyme $T$ (from Gene $T$) and Enzyme $V$ (from Gene $V$) have redundant functions.
    - Both enzymes convert $ext{Intermediate X}$ into $ext{Intermediate Y}$. Only one active enzyme is necessary to achieve the triangular shape.

Summary of Epistatic Dihybrid Phenotypic Ratios

Complex Gene Control and Mutation Study

• Traits are often governed by extensive networks of genes:
    - For example, coat colour in mice involves at least $5$ genes ($A, B, C, D, S$).
    - Locus C (Albino): The $cc$ genotype is epistatic to all other mouse coat color genes.
    - Locus D: Influences the intensity of the pigment.
    - Locus S: Controls the presence of spots (affecting melanocyte migration).
    - In Drosophila, around $100$ genes contribute to eye pigmentation.

• Identifying Genes:
    - Researchers induce mutations (e.g., through X-ray irradiation) to create mutants for a specific trait.
    - The aim is to determine whether new mutations are allelic (same gene) or occur in different genes.

Use of Model Organisms and the Complementation Test

• The Complementation Test: A genetic test applied to ascertain whether mutations are allelic and to find out the total number of genes involved in a trait.

• Step-by-Step Procedure:
    1. Induce Mutations: Create mutant organisms (e.g., harebell flowers) to find mutants (e.g., white-petaled instead of blue wild-type).
    2. Cross with Wild-Type: If the $F_1$ generation shows the wild-type phenotype and the $F_2$ generation segregates at a $3:1$ ratio, the mutation represents a single recessive allele at one gene.
    3. Cross Mutants: Conduct a cross between two different homozygous recessive mutants for testing.

• Interpreting Results:
    - No Complementation (Allelic Mutations): If offspring exhibit the mutant phenotype (white), the mutations belong to the same gene; both parents provided a defective version of the same gene.
    - Complementation (Different Genes): If the offspring appear wild-type (blue), the mutations are located on different genes, with each parent contributing the wild-type allele that the other lacks ($aaBB imes AAbb ightarrow AaBb$).

• Complementation Groups:
    - A complementation group represents a single gene.
    - The total number of complementation groups identified signifies the total number of genes mutated in the study.