Gene Interaction, Linkage and Mapping

Learning Goals:
- Design genetic crosses to provide information about genes, alleles, and gene functions.
- Interpret experimental results comparing phenotypes from single mutations in two different genes with the phenotype of the double mutant, contrasting epistatic interactions.
- Solve dihybrid/trihybrid crosses where gene interactions occur.

Complementation tests determine if mutations are in the same gene.
Consider two separate recessive mutations, $h1$ and $h2$ , both resulting in the same "S" phenotype.
- If both mutations are present in a trans configuration (i.e., on different chromosomes)
  - and the S phenotype is observed, the mutations do not complement each other and are alleles of the same gene.
  - but, if the trans configuration results in the wild-type phenotype, the mutations do complement each other and are alleles of different genes.

Gene interaction occurs when two or more genes affect the same phenotype by influencing a common pathway, also known as Epistatic interactions.
Example: Harebell color.
- If two genes interact, there are fewer phenotypes than in the offspring of a dihybrid cross.
- Consider genes A and B involved in anthocyanin production
  - Pathway: Precursor 1 (white) → Precursor 2 (white) → Anthocyanin (blue).
  - Gene A encodes Enzyme A, and Gene B encodes Enzyme B
- F2 Generation Phenotype Ratio: 9:7 ratio
  - A–B–: Wild-type blue (9/16)
  - A–bb: White (3/16)
  - aaB–: White (3/16)
  - aabb: White (1/16)

The 9:7 ratio indicates two genes interacting in the same pathway.
Mutations in either gene disrupt the pathway, causing the mutant phenotype.
- Mutation in gene A (aa B–) blocks the conversion of Precursor 1 to Precursor 2.
- Mutation in gene B (A– bb) blocks the conversion of Precursor 2 to Anthocyanin.

The genes in a redundant system have duplicate gene action; they encode the same product or products that have the same effect in a pathway or compensatory pathways.
Example: Bean flower color. Having either the P or R allele will produce a purple phenotype.
- P-R-: Purple (9/16)
- P-rr: Purple (3/16)
- ppR-: Purple (3/16)
- pprr: White (1/16)

Plants with one or two dominant alleles for just one of either of the genes will have round fruit (disk or sphere), and those with only recessive alleles of both genes will have long fruit.
Example: Squash fruit shape.
- A-B-: Disk (9/16)
- A-bb: Sphere (3/16)
- aaB-: Sphere (3/16)
- aabb: Long (1/16)

In recessive epistasis, homozygosity for the recessive alleles at one locus will mask the phenotypic expression of the alleles at a second locus.
Example: Labrador retriever coat color.
- B-E-: Black (9/16)
- bbE-: Chocolate (3/16)
- B-ee: Golden (3/16)
- bbee: Golden (1/16)

In dominant epistasis, a dominant allele of one gene masks or reduces the expression of either allele of the other gene.
Example: Summer squash coloration.
- W-Y-: White (9/16)
- W-yy: White (3/16)
- wwY-: Yellow (3/16)
- wwyy: Green (1/16)

Dominant suppression occurs when the dominant allele of one gene suppresses the expression of a dominant allele of a second gene.
Example: Chicken feather color.
- C-I-: White (9/16)
- C-ii: Colored (3/16)
- ccI-: White (3/16)
- ccii: White (1/16)

Find two mutants affecting the same phenotype.
Do a complementation test to find out if two genes are involved.
If so, perform a dihybrid cross and examine the offspring ratios to infer the pathways governed by the two genes.

Wild-type blue-eyed mary has blue flowers.
Two genes control the pathway that makes the blue pigment:
- Gene W turns a white precursor into magenta pigment.
- Gene M turns the magenta pigment into blue pigment.
Each gene has a recessive loss-of-function allele: $w$ and $m$ , respectively.
A double heterozygote (WwMm) is self-pollinated. What proportion of offspring will be magenta?
In other words: What proportion of $WwMm \times WwMm$ offspring will be $W-mm$ ? i.e., 3/16
- Normal pathway: White Pigment → Magenta Pigment → Blue Pigment
- W catalyzes White to Magenta; M catalyzes Magenta to Blue

In mice, wild-type coat color is agouti, produced by a combination of yellow and black pigments along each hair.
A dominant allele of agouti, $A^Y$ , causes yellow pigment to be deposited along the entire hair, resulting in a yellow coat.
The $A^Y$ allele is recessive embryonic lethal, so all yellow mice are heterozygous ( $A^YA$ = lethal).
agouti x agouti results in all agouti (AA).
agouti x yellow results in 1/2 agouti (AA), 1/2 yellow ( $A^YA$ ).
yellow x yellow results in 1/3 agouti (AA), 2/3 yellow ( $A^YA$ ). The last cross produces 1/4 with the $A^YA^Y$ = lethal phenotype; thus, fractions are counted based on survivors only.

One allele, many phenotypes. e.g., the agouti allele affects both coat color and viability