Variations to Mendel's First Law
Epistasis
Gene Interactions
The genes of an individual do not operate isolated from one another, but obviously are functioning in a common cellular environment. Thus, it is expected interactions between genes would occur. Bateson and Punnett performed a classical experiment that demonstrated genetic interactions. They analyzed the three comb types of chicken known to exist at that time:
| Chicken Varieties | Phenotype |
|---|---|
| Wyandotte | Rose Comb |
| Brahmas | Pea Comb |
| Leghorns | Single Comb |
Rose 
Pea
Single 
Walnut
Result: The F1 differed from both parents and two new phenotypes not seen in the parents appeared in the F2. How can this result be explained? The first clue is the F2 ratio. We have seen this ratio before when the F1 from a dihybrid cross is selfed (or intermated). This observation suggests that two genes may control the phenotype of the comb. The gene interactions and genotypes were determined by performing the appropriate testcrosses.
A series of experiments demonstrated that the genotypes controlling the various comb phenotypes are as follows.
| Phenotypes | Genotypes | Frequency |
|---|---|---|
| Walnut | R_P_ | 9/16 |
| Rose | R_pp | 3/16 |
| Pea | rrP_ | 3/16 |
| Single | rrpp | 1/16 |
It was later shown that the genotypes of the initial parents were:
Rose = RRpp
Pea = rrPP
Therefore, genotypically the cross was:
The development of any individual is obviously the expression of all the genes that are a part of its genetic makeup. Therefore, it is not an unexpected conclusion that more than one gene could be responsible for the expression of a single phenotype. We will now discuss this situation. First let's give a definition.
Epistasis - the interaction between two or more genes to control a single phenotype
The interactions of the two genes which control comb type was revealed because we could identify and recognize the 9:3:3:1. Other genetic interactions were identified because the results of crossing two dihybrids produced a modified Mendelian ratio. All of the results are modifications of the 9:3:3:1 ratio.
Example 1: 15:1 Ratio
Phenotypes: Kernel Color in Wheat
For this type of pathway a functional enzyme A or B can produce a product from a common precursor. The product gives color to the wheat kernel. Therefore, only one dominant allele at either of the two loci is required to generate the product.
Thus, if a pure line wheat plant with a colored kernel (genotype = AABB) is crossed to plant with white kernels (genotype = aabb) and the resulting F1 plants are selfed, a modification of the dihybrid 9:3:3:1 ratio will be produced. The following table provides a biochemical explanation for the 15:1 ratio.
| Genotype | Kernel Phenotype | Enzymatic Activities |
|---|---|---|
| 9 A_B_ | colored kernels | functional enzymes from both genes |
| 3 A_bb | colored kernels | functional enzyme from the A gene pair |
| 3 aaB_ | colored kernels | functional enzyme from the B gene pair |
| 1 aabb | colorless kernels | non-functional enzymes produced at both genes |
If we sum the three different genotypes that will produce a colored kernel we can see that we can achieve a 15:1 ratio. Because either of the genes can provide the wild type phenotype, this interaction is called duplicate gene action.
Example 2: 9:7 Ratio
Example: Flower color in sweet pea
If two genes are involved in a specific pathway and functional products from both are required for expression, then one recessive allelic pair at either allelic pair would result in the mutant phenotype. This is graphically shown in the following diagram.
If a pure line pea plant with colored flowers (genotype = CCPP) is crossed to pure line, homozygous recessive plant with white flowers, the F1 plant will have colored flowers and a CcPp genotype. The normal ratio from selfing dihybrid is 9:3:3:1, but epistatic interactions of the C and P genes will give a modified 9:7 ratio. The following table describes the interactions for each genotype and how the ratio occurs.
| Genotype | Flower Color | Enzyme Activities/TH> |
|---|---|---|
| 9 C_P_ | Flowers colored; anthocyanin produced |
Functional enzymes from both genes |
| 3 C_pp | Flowers white; no anthocyanin produced |
p enzyme non-functional |
| 3 ccP_ | Flowers white; no anthocyanin produced |
c enzyme non-functional |
| 1 ccpp | Flowers white; no anthocyanin produced |
c and p enzymes non-functional |
Because both genes are required for the correct phenotype, this epistatic interaction is called complementary gene action.
Example 3: 12:3:1 Ratio
Phenotype: Fruit Color in Squash
With this interaction, color is recessive to no color at one allelic pair. This recessive allele must be expressed before the specific color allele at a second locus is expressed. At the first gene white colored squash is dominant to colored squash, and the gene symbols are W=white and w=colored. At the second gene yellow is dominant to green, and the symbols used are G=yellow, g=green. If the dihybrid is selfed, three phenotypes are produced in a 12:3:1 ratio. The following table explains how this ratio is obtained.
Shapes of Squash Fruit
| Genotype | Fruit Color | Gene Actions |
|---|---|---|
| 9 W_G_ | White | Dominant white allele negates effect of G allele |
| 3 W_gg | White | Dominant white allele negates effect of G allele |
| 3 wwG_ | Yellow | Recessive color allele allows yellow allele expression |
| 1 wwgg | Green | Recessive color allele allows green allele expression |
Because the presence of the dominant W allele masks the effects of either the G or g allele, this type of interaction is called dominant epistasis.
Example 4: 13:3 ratio
Phenotype: Malvidin production in Primula
Certain genes have the ability to suppress the expression of a gene at a second locus. The production of the chemical malvidin in the plant Primula is an example. Both the synthesis of the chemical (controlled by the K gene) and the suppression of synthesis at the K gene (controlled by the D gene) are dominant traits. The F1 plant with the genotype KkDd will not produce malvidin because of the presence of the dominant D allele. What will be the distribution of the F2 phenotypes after the F1 was crossed?
| Genotype | Phenotype and genetic explanation |
|---|---|
| 9 K_D_ | no malvidin because dominant D allele is present |
| 3 K_dd | malvidin productions because dominant K allele present |
| 3 kkD_ | no malvidin because recessive k and dominant D alleles present |
| 1 kkdd | no malvidin because recessive k allele present |
The ratio from the above table is 13 no malvidin production to 3 malvidin production. Because the action of the dominant D allele masks the genes at the K locus, this interaction is termed dominant suppression epistasis.
Suppressor - a genetic factor that prevents the expression of alleles at a second locus; this is an example of epistatic interaction
Remember that epistasis is the interaction between different genes. If one allele or allelic pair masks the expression of an allele at the second gene, that allele or allelic pair is epistatic to the second gene. Therefore, the following table summarizes the four epistatic interactions discussed above.
| Example | Allelic Interactions | Type of Epistasis |
|---|---|---|
| Wheat kernel color | A epistatic to B, b B epistatic to A, a | Duplicate genes |
| Sweet pea flower color | cc epistatic to P, p pp epistatic to C, c | Complementary gene action |
| Squash Fruit Color | W epistatic to G, g | Dominant epistasis |
| Primula malvidin production | D epistatic to K, k | Dominant suppression |
Copyright © 2000. Phillip McClean