Gene Interaction and Complementation Test Notes

Gene Interaction II: Departure from Mendelian Ratios

Introduction

Most genes do not operate in isolation.
Gene interaction (epistasis) leads to modified dihybrid ratios.
A phenotype can be controlled by multiple interacting genes.
In model organisms, the complementation test helps determine the number of genes controlling a characteristic.

Genes Operating Independently

Genes affecting the same trait can operate independently, though this is rare.
Example: Color of Californian corn snakes (repeating black and orange for camouflage).
This scenario yields Mendelian ratios at the phenotypic level.
- Natural color is repeating black and orange (camouflage). At least one $o+$ makes orange
- No natural pigment, just blood flowing through (albino)
Example of genotypes:
- $o+$ - / $b+$ -: camouflaged
- $o+$ - / $b b$ : orange
- $o o$ / $b+$ -: black
- $o o$ / $b b$ : albino

Gene Interaction (Epistasis)

Most phenotypes are controlled by various interacting genes (epistasis).
Epistasis modifies Mendelian ratios.
Interacting genes contribute to the same phenotype and often participate in the same pathway.

1. Complementary Gene Action

Genes act in tandem to produce a phenotype.
Example: Purple flower color (anthocyanin) in peas.
Dominant P allele encodes purple color; recessive p gives white color.
- $PP$ and $Pp$ : purple
- $pp$ : white (colorless)
At another locus, C encodes color; recessive c prevents color.
- $CC$ and $Cc$ : color
- $cc$ : white (colorless)
Homozygous recessive genotype at either locus produces the same phenotype.
Wild type C and P are needed to have color.
$C c P p x C c P p$
- 9 colored
- 7 white
A 9:7 ratio (departure from Mendelian ratio) indicates a complementary gene action relationship between two genes.
- 9 $C -$ / $P -$
- 3 $C-$ / $pp$
- 3 $c c$ / $P -$
- 1 $c c$ / $p p$
Genes segregate in Mendelian fashion as phenotypes do not.
Purple-flowered plants appear only when two independent dominant alleles are present together, resulting from their interaction.
Biochemical explanation:
- Precursor --(Enzyme C)--> Intermediate X --(Enzyme P)--> Pigment (anthocyanin)
- If mutant allele C ( $cc$ ), enzyme C doesn’t work.
- A mutant allele at the C locus masks the allele at P.
- C is epistatic to P (a 9:7 ratio suggests an epistatic relationship).

2. Recessive Epistasis

$A-bb$ and $aabb$ individuals have the same phenotype.
One locus, when recessive, affects the other gene (but not vice versa).
Example: Coat color in mice.
- $Bb$ or $BB$ : pigmented
- $bb$ : albino (white) - recessive genotype masks effect of genotype at A locus.
- $AA$ or $Aa$ : agouti
- $aa$ : black (no agouti distribution)
B: dominant, permits production of pigment
A: dominant, agouti (grayish - formed by alternating bands of pigment)
$A a B b x A a B b$
- 9 agouti
- 3 black
- 4 albino
A 9:3:4 ratio indicates a recessive epistasis relationship between two genes.
- 9 $A -$ / $B -$
- 3 $A-$ / $b b$
- 3 $a a$ / $B -$
- 1 $a a$ / $b b$
Biochemical explanation:
- Colourless --(Enzyme B)--> Black --(Enzyme A)--> Agouti
- With B in the presence of A, black pigment is deposited in the agouti pattern.
- In $bb$ mice, no pigment is produced, regardless of A/a.
- When B is present, the black pigment is produced.
- $b$ masks the expression of A (recessive epistasis).
- $a$ does not mask the expression of B.

3. Dominant Epistasis

$A-B-$ displays the same phenotype as $aaB-$ .
Dominant B is epistatic to A (B masks effect of A).
Phenotypic ratio in offspring from a dihybrid cross is 12:3:1.
Example: Color in Summer squashes.
- $AABB$ : white x $aabb$ : green
- F1: $AaBb$ white
Three color phenotypes: white, green, yellow.
A dominant allele of one gene masks the effect of the genotype at another locus.
$A a B b x A a B b$
- 12 white
- 3 yellow
- 1 green
A 12:3:1 ratio in the F2 indicates a dominant epistasis relationship between two genes.
- 9 $A -$ / $B -$
- 3 $A-$ / $b b$
- 3 $a a$ / $B -$
- 1 $a a$ / $b b$
COLOURLESS --(B)--> white --(A)-->yellow --(a)-->green

4. Duplicate Gene Action

Both genes perform the same function (biological redundancy).
Example: Shape of seeds in wheat.
T: dominant, triangular seed
V: dominant, triangular seed
T and V are different loci
F1: $TtVv$ All triangular
$TTVV$ Triangular x $ttvv$ Ovate
A 15:1 ratio in the F2 indicates duplicate gene action
$TtVv x TtVv$
- 15 triangular
- 1 ovate
Biochemical explanation:
- Enzyme T and Enzyme V are redundant.
- Both make Y, so they duplicate each other.
- Precursor --> Intermediate X --(Enzyme T or Enzyme V)--> Intermediate Y

Summary of Epistatic Interactions

Table of Modified dihybrid-phenotypic ratios due to gene interaction:
- None: 9:3:3:1
- Recessive epistasis: 9:3:4
- Dominant epistasis: 12:3:1
- Complementary gene action : 9:7
- Duplicate recessive epistasis: 9:6:1
- Duplicate interaction : 9:6:1
- Duplicate gene action: 15:1
- Dominant and recessive epistasis: 13:3

Complex Traits

Most genes do not operate in isolation.
Coat color in mice is controlled by at least 5 genes:
- A: distribution of pigment
- B: pigment itself (B black)
- C: expression of color; $cc$ - albino (epistatic to all other genes)
- D: intensity of pigment
- S: spots (controls migration of melanocytes)
In Drosophila, approximately 100 genes contribute to the pigmentation of the compound eye.

Identifying Genes Controlling a Trait

As a geneticist, you may be interested in a particular characteristic.
To identify genetic determinants, introduce mutations in the organism of interest (e.g., irradiation with X rays).
This will lead to a number of organisms which are mutant for the trait.
Questions to address:
- How many genes have I mutated?
- Are these mutations in the same gene (allelic) or in a series of different genes?
To answer these questions use model organisms & the complementation test

Complementation Test

The genetic test used to determine both allelism among a group of mutations and the number of different genes is the complementation test.
Example: Harebell flower color (wildtype is blue), induction of white-petaled mutants (homozygous & pure breeding)
- Make mutants that affect the phenotype you want to investigate
- Mutate (after radiation will get lots of mutants (eg diff petal size, stem length, etc), but ur only interested in plants that fail to make blue pigment so u only collect those ones
- Only work with plants that are white FROM mutating at one locus (because if mutated in diff loci cant further analyse, will mix up later on)
Make sure mutant is due to change in one gene
- If F1 all blue & F2 segregate at 3:1 (blue to white) indicates mutant is in one gene only..
- Typical Mendelian segregation where each mutant is determined by a recessive allele at a single gene.
Cross the mutants with each other - the complementation test
Is the wild type recovered from the progeny?
- No (produce white flowers): mutations represent alleles of the same gene (no complementation)
- Yes (produce blue): mutations represent alleles of different genes (has complementation)
If recessive mutations represent alleles of the same gene, they will never complement (show the wild type blue), because they both represent loss of function of the same gene.
- $1 a/a X 2 a/a$
- $F1 All white$
If recessive mutations are in different genes, a cross will give blue progeny; because the mutations are in different genes, each parent provides what the other one lacks.
- $1 a/a X 3 b/b$
- $F1 All blue$
- $a/A B/b$
This type of test leads to the establishment of complementation groups
- No. of complementation groups is always equal to the number of genes that have been mutated.

Key Concepts

Most traits are controlled by a number of genes which interact with each other (epistasis).
Genes that control a trait can be identified by mutating a model organism.
Gene interaction leads to a departure from Mendelian ratios (at the phenotypic level).
The number of genes mutated can be determined by the complementation test.