Epistasis definition
The genes responsible for expression of the trait in the phenotype. In the case of simple Mendelian inheritance each gene is responsible for the expression of only one phenotypic trait. But, in reality the situation is more complicated. For example, the same gene may act on the expression of multiple traits or the same trait can evolve under the action of multiple genes. Typically, the interaction of genes have a biochemical nature, i.e. it's based on the combined action of proteins, whose synthesis is controlled by certain genes. Interact with each other can both allelic and nonallelic genes. The interaction of several genes that are located in different loci called nonallelic. Nonallelic gene interactions can be divided into two groups: complementary gene interaction and epistatic gene interaction. What is the difference between them? Perhaps, it can be explained in simple words. Complementary genes - is nonallelic genes that act together on trait expression, i.e. the combination of these genes in the genotype give us a new phenotype. And epistasis it is the interaction of nonallelic genes, in which one of them suppress the action of another. A gene that suppress the action of another nonallelic gene called suppressor or epistatic gene, and can be marked by the letters I or S. And the suppressed gene called hypostatic. Both dominant and recessive genes can interact each other. Therefore, there is a dominant and recessive epistasis. When a dominant allele of one gene suppress the action of another gene, it is called dominant epistasis. And when the recessive allele of epistatic gene in the homozygous state suppress the action of another gene, it is called recessive epistasis. Often, the interaction of nonallelic genes can be explained both in terms of complementary interaction of genes and in terms of epistatic interactions of genes. For the complementary interaction of genes the ratio of phenotypes depend on whether or not an independent phenotypic expression in these genes.
9:3:3:1 Epistasis looks like independent assortment
The shape of the chicken's comb.
Let's start with an example where there are nonallelic genes interaction, but the ratio of phenotypes looks like as in the independent assortment. Consider the example of inheritance of shape of the chicken's comb. The shape is controlled by genes that are located in two loci on different chromosomes. As result of interaction of these genes, the chickens can have four different phenotypes. Their comb shape can be Pea, Rose, Walnut and Simple. Wyandotte have Rose comb. Brahmas have Pea comb. Leghorns have Simple comb. When crossed the pure lines of Wyandotte and Brahmas, the all offspring in the first generation had the Walnut comb. That is, the offspring had an absolutely new phenotype not like their parents. From crosses of chickens with Walnut comb in the second generation was obtained the offspring with the phenotypic ratio: 9 Walnut : 3 Rose : 3 Pea : 1 Single. Thus we have a dealing with the case, when each dominant gene has an independent phenotypic expression, combination of two these genes in genotype leads to the development of a new phenotypic trait, and the lack of dominant genes gives another phenotypic trait. Let's mark the genes with letters "P" and "R". If there are dominant allele "P" in genotype and dominant allele "R" is lack, then the chickens have the Pea comb. And if there are dominant allele "R" in genotype and dominant allele "P" is lack, then the chickens have the Rose comb. Chickens with presence of both dominant alleles "P" and "R" in the genotype would have the Walnut comb, and double recessive homozygotes, i.e. with genotype "pprr" would have the Simple comb. Using the rules of making genetic traits files we write our file:
P R:Walnut comb
P:Pea comb
R:Rose comb
*:Simple comb
{
comb:P p
comb:R r
}
Genetic calculation: You can open this file ( Epistasis 1.txt ) and calculate results first for Genotypes and after for Traits phenotypes. It is easy to determine that parents genotypes will be "RRpp" for Wyandotte and "rrPP" for Brahmas. As a result, of these crossing all offspring in first generation will have the genotype "RrPp" with Walnut combs. As a result of crossing of first-generation hybrids, we get the ratio of phenotypes: 9 Walnut : 3 Rose : 3 Pea : 1 Single. From the version 3.3, on the tab "Find" you can choose each phenotype and see from what kind of genotypes it is consist.
Eyes color of fruit flies (Drosophila melanogaster).
Let's consider at another example in more detail. In this example, the ratio of phenotypes also looks as in the independent assortment. Fruit fly ( Drosophila melanogaster ) is a popular subject of many genetic experiments. In the fruit fly has a large number of forms that differ in the eyes color. The wild-type flies, which is prevalent in nature, have a brick-red eyes. There are flies with bright-red eyes. This trait is determined by a recessive allele of the gene, called the "scarlet" and marked as "st". The dominant allele of this gene is marked as "st+". Thus, flies with the genotype "stst" will have bright-red eyes, and flies with the genotype "st+st+" and "st+st" will have the eyes of wild-type - brick-red. There are also flies with brown eyes. It is also a recessive trait, which is determined by the gene "brown". Recessive allele of this gene ismarked as "br", a dominant gene as "br+". Thus, flies with the genotype "brbr" will have brown eyes, and flies with the genotype "br+br+" and "br+br" will have the eyes of wild-type - brick-red. If we cross flies with bright-red eyes and flies with brown eyes, then all the offspring will have the eyes of wild-type ( brick-red ), and from crossing of the first generation hybrids in the second-generation was obtained the offspring with the ratio of phenotypes: 9 brick-red : 3 bright-red : 3 brown : 1 white. We can understand that there is an interaction of the two nonallelic genes "st" and "br". The mechanism of this interaction has been studied well enough. Eye color of a fruit fly is determined by the synthesis of the two pigments - red and brown. Recessive allele "br" in the homozygous state blocks the synthesis of red pigment, so in the eyes of these flies will be only brown pigment. Recessive allele "st" in the homozygous state blocks the synthesis of brown pigment, so in the eyes of these flies will be only red pigment. That is, we have described this interaction in terms of the epistatic action of the recessive genes. Using the rules of making genetic traits files, let's write our file (you can mark the wild-type with symbol "+" instead of the symbols "st+" and "br+" :
<st> <st> <br> <br>:white eyes
<st> <st>:bright-red eyes
<br> <br>:brown eyes
<+>:brick-red eyes
{
eyes color:<+> <st>
eyes color:<+> <br>
}
Genetic calculation: You can open this file ( Epistasis 2.txt ) and calculate results first for Genotypes and after for Traits phenotypes. To obtain the first generation of offspring we must cross fruit flies with genotypes <st><st><+><+> ( bright-red eyes ), and <+><+><br><br> ( brown eyes ). As a result of these cross all offspring in first generation will have the genotype <+><st><+><br> ( brick-red eyes ). As result of crossing of the first generation hybrids in the second-generation we get the offspring with the ratio of phenotypes: 9 brick-red : 3 bright-red : 3 brown : 1 white. On the tab "Find" you can choose each phenotype and see from what kind of genotypes it is consist.
Color of budgerigars feathering.
Let's consider the example of color of budgerigars feathering. For example, gene "A" responsible for the forming of blue color of budgerigars feathering , a gene "B" - of yellow color of feathering, budgerigars with genotype "A_B_" - have a green color, and with the genotype "aabb" - white color. This description corresponds to the complementary interaction of genes. Using the rules of making genetic traits files, let's write our file:
A B:green budgerigars
A:blue budgerigars
B:yellow budgerigars
*:white budgerigars
{
color:A a
color:B b
}
However, we can consider this case in terms of epistatic interactions of genes. Then, a description will look different. Recessive allele "b" in the homozygous state blocks the effect of the dominant allele "A", and the budgerigars will have blue color. Recessive allele "a" in the homozygous state blocks the effect of the dominant allele "B", and the budgerigars will have yellow color. The double recessive homozygote with the genotype "aabb" will have white color. And the budgerigars will have a green color in all other cases. Using the rules of making genetic traits files, let's write our file:
a a b b:white budgerigars
b b:blue budgerigars
a a:yellow budgerigars
*:green budgerigars
{
color:A a
color:B b
}
Variant, which describe this interaction of genes in terms of epistatic interactions, looks more complicated, and in this case, variant with complementary interaction will be more correct. You can use both variants. In any case, the result will be the same.
Genetic calculation: You can open file ( ( Epistasis 3A.txt or Epistasis 3B.txt ) and calculate results for Traits phenotypes. Crossing the green budgerigars with genotype "AaBb" we get the ratio of phenotypes: 9 green : 3 blue : 3 yellow : 1 white. On the tab "Find" you can choose each phenotype and see from what kind of genotypes it is consist.
In considering interactions of nonallelic genes changing not the ratio of phenotypes, but their character, i.e. the interaction led to the emergence of absolutely new traits. But in most cases of the interaction of nonallelic genes, we can see the reduction in the number of phenotypic classes, which is the result of combining of some of them. In any case, the base of the interaction of the two nonallelic genes is the ratio of phenotypes, which is typical for independent assortment - it is 9:3:3:1. In studying of the interaction of genes it's necessary to understand, what phenotypic classes is merged and then identify the type of interaction. However, for many morphological traits the biochemical mechanisms of the genes expression is unknown, and therefore, often we have only the formal scheme of their genetic inheritance.
9:6:1 Duplicate interaction
The shape of pumpkins.
As an example, consider the interaction of genes, that determine the shape of pumpkin. The mechanism of their interaction was not been studied for now. However, we can note, that the interaction of genes is very similar to the interactions, that was described above. As a result of the interaction of two genes, the pumpkin can have three different types of shapes.The pumpkins can be disk-shaped, ball-shaped, and long-shaped. From the crossing of pure lines of ball-shaped pumpkins with genotypes "AAbb" and "aaBB", all pumpkins in the offspring of first generation was disk-shaped. It was hybrids with the genotype "AaBb". From crosses of these hybrids, in the second generation was obtained offspring with phenotypic ratio : 9 disk-shaped : 6 ball-shaped : 1 long-shaped. The interaction of these genes can be described in terms of the complementary interaction of genes. The combination of dominant alleles of two genes form the first trait, the combination of recessive alleles of these genes - the second trait, and presence in the genotype only one from dominant genes - the third trait. For example, a pumpkin with genotype "A_B_" is disk-shaped, with the genotype "aabb" - long-shaped, and with genotypes "A_ bb" and "aaB_" - ball-shaped. Using the rules of making genetic traits files, let's write our file:
A B:disk-shaped pumpkin
A:ball-shaped pumpkin
B:ball-shaped pumpkin
*:long-shaped pumpkin
{
shape:A a
shape:B b
}
Genetic calculation: You can open file ( Epistasis 4.txt )and calculate results first for Genotypes and after for Traits phenotypes. In this case, the phenotypic classes "Ab" and "aB" is merged, and will have the same phenotype. As result of crossing of fist generation hybrids we get phenotypic ratio : 9 disk-shaped : 6 ball-shaped : 1 long-shaped. On the tab "Find" you can choose each phenotype and see from what kind of genotypes it is consist.
9:7 Duplicate recessive epistasis
Flower color in sweet peas.
Sweet peas - is the classic object of genetic research of traits inheritance. When scientists studied the color of flowers of this plant, they discover, that in some crosses the ratio of phenotype was 9:7. That is, from the crossing of two pure lines of sweet peas with white flowers, all plants in first-generation had the purple flowers. And from crossing of these hybrids was obtained the offspring with the phenotypic ratio : 9 purple: 7 white. If the ratio was 3:1, we could suppose, that we are dealing with a simple monohybrid crossing, but in this case the ratio is differ. So obviously, the genetic trait of color of sweet peas flowers is controlled by two nonallelic genes, which interact each other. For convenience, we mark these genes as "C" and "P". Thus, the pure lines will have genotypes "CCpp" and "ccPP", and first generation hybrids respectively genotype "CcPp". The mechanism of genetic interaction of these genes has become known recently. It was found that the color of sweet pea flowers depends on the synthesis of pigments - anthocyanins. The synthesis of anthocyanins occurs in two stages. Gene "C" is responsible for the first stage of the synthesis, while the gene "P" for second, and even if only one dominant allele from these genes is lack, then synthesis of anthocyanins is not occurs. That is to say, that the presence of a recessive allele in the homozygous state in the genotype of either of the two genes is blocks the development of trait of flowers coloration. Such interaction of nonallelic genes is called a double recessive epistasis. We can describe this interaction like this - recessive allele of each gene in the homozygous state suppresses the expression of a dominant allele of another ( "cc" suppresses "P", "pp" suppresses "C" ). Genetic traits file for this case should be like this :
c c:white pea flower
p p:white pea flower
*:purple pea flower
{
color:C c
color:P p
}
The interaction of these genes we can also describe in terms of complementary interaction of genes. For this case, it will be like this - dominant and recessive alleles of complementary genes don't have an independent phenotypic expression, and only a combination of dominant alleles of two genes provides the coloration of flowers. And then the genetic traits files can be formulated differently:
C P:purple pea flower
*:white pea flower
{
color:C c
color:P p
}
Genetic calculation: You can open one from this traits files ( Epistasis 5A.txt or Epistasis 5B.txt ) and calculate results first for Genotypes and after for Traits phenotypes. In this case, the phenotypic classes "Cp" , "cP" and "cp" is merged, and will have the same phenotype. As result of crossing of fist generation hybrids we get phenotypic ratio : 9 purple : 7 white. On the tab "Find" you can choose each phenotype and see from what kind of genotypes it is consist.
Eyes color of fruit flies (Drosophila melanogaster).
Let's look at another example of a double recessive epistasis. Let's return again to the inheritance of eye color in the fruit fly, as you may remember, there are large number of forms that differ in the eyes color. In the previous example, we had to deal with fruit flies with bright-red eyes. f there are recessive allele of gene "st" ("scarlet") in the homozygous state in genotype of the fruit fly, then these fruit flies have bright-red eyes. The dominant allele of this gene is marked as "st+". Thus, flies with the genotype "stst" will have bright-red eyes, and flies with the genotype "st+st+" and "st+st" will have the eyes of wild-type - brick-red. It is known, that besides of recessive alleles "st", the recessive alleles of the gene "purple" is also blocking the synthesis of brown pigment . We mark the dominant allele of gene "purple" as "pr +", and recessive as "pr". Flies with the genotype "prpr" will have bright-red eyes, and flies with the genotype "pr+pr+" and "pr+pr" will have the eyes of wild-type - brick-red. If you cross two pure lines of fruit flies with bright red eyes, all first generation offspring will have the eyes of wild-type - brick-red. From the crossing of these hybrids was obtained the offspring with ratio : 9 brick-red : 7 bright-red. Obviously, the parents with bright-red eyes have genotypes "<st+><st+><pr><pr>" and "<st><st><pr+><pr+>", and hybrids in first-generation have genotype "<st +><st><pr+><pr>". To marking the wild-type in the genetic traits file you can use the symbol "<+>" instead of the "<st+>" and "<pr+>" :
<st> <st>:bright-red eye
<pr> <pr>:bright-red eye
<+>:brick-red eye
{
eye color:<+> <st>
eye color:<+> <pr>
}
And as in the previous case, we can describe this interaction in terms of complementary interaction of genes. Then our trait file would look even simpler:
<+>:brick red eye
*:bright red eye
{
eye color:<+> <st>
eye color:<+> <pr>
}
Genetic calculation: You can open one from this traits files ( Epistasis 6A.txt or Epistasis 6B.txt ) and calculate results first for Genotypes and after for Traits phenotypes. To obtain the first generation of offspring we must cross fruit flies with genotypes <st><st><+><+> ( bright-red eyes ), and <+><+><pr><pr> ( bright-red eyes ). As a result of these cross all offspring in first generation will have the genotype <+><st><+><pr> ( brick-red eyes ). In this case, the phenotypic classes "<+><pr>", "<st><+>" and "<st><pr>" is merged, and will have the same phenotype. As result of crossing of the first generation hybrids in the second-generation we get the offspring with the ratio of phenotypes: 9 brick-red : 7 bright-red. On the tab "Find" you can choose each phenotype and see from what kind of genotypes it is consist.
9:3:4 Recessive epistasis
Color of rabbit fur.
In the previous examples we considered the cases of double recessive epistasis, when recessive allele of each gene in the homozygous state suppresses the expression of a dominant allele of another. Let's now consider the case of a single recessive epistasis. In the case of single recessive epistasis the recessive allele of one gene in the homozygous state suppresses the expression of a dominant allele of another ( for example, "bb" suppress "A" ). For example, the color in rabbits is determined by two genes: "P" - presence of pigment, "p" - lack of pigment, and "B" - black color, "b" - blue color. So in this example we have two traits in definiton - pigment and color. If the rabbit have the dominant allele of gene "B" in genotype, then he will have a black fur, and otherwise the rabbit will have a blue fur. But if there are the recessive allele of a gene "p" in the homozygous state ("pp") in genotype, then the rabbit will be albino, even if there are the dominant allele "B" in genotype. In other words, homozygous recessive allele "p" suppresses the expression of gene of coloration "B". So we can write trait file like this:
p p:albino rabbit
B:black rabbit
b:blue rabbit
{
pigment:P p
color:B b
}
We also can describe this interaction in terms of complementary interaction of genes, when the dominant and recessive alleles of complementary genes have the independent phenotypic expression. If there are two dominants allels of genes "P" and "B" in genotype, then rabbit is black and if we have only one dominant allel "P" then rabbit is blue. Otherwise rabbit will be albino. So we can form trait file like this:
P B:black rabbit
P:blue rabbit
*:albino rabbit
{
pigment:P p
color:B b
}
Genetic calculation: You can open one from this traits files ( Epistasis 7A.txt or Epistasis 7B.txt ) and calculate results for Traits phenotypes. In this case, the phenotypic classes "pB" and "pb" is merged, and will have the same phenotype. As result of crossing of the hybrids with genotype "PpBb" in the second-generation we get the offspring with the ratio of phenotypes: 9 black : 3 blue : 4 albino. On the tab "Find" you can choose each phenotype and see from what kind of genotypes it is consist.
13:3 Dominant and recessive epistasis
Primula Petal Color.
So we considered the cases of recessive epistasis, but often in the nature, we also can find the examples of dominant epistasis. Dominant epistasis is called the suppression of the effect of one gene by the dominant allele of another gene. Let's consider this on the example of petal color inheritance of Primula plant. In the presence of a substance called malvidin, the flowers are colored in blue. Synthesis of malvidin is due to the presence in the genotype of a dominant allele of the gene "K". If this gene in the genotype is presented as the recessive allele in the homozygous state "kk", then malvidin is not produced and the flowers will be uncolored (white). But there are also another gene, which is located in a completely different locus, and which affects the formation of color. We mark this gene as "D". It turned out, that if there are the dominant allele of gene "D" in genotype, then regardless of the presence or absence of the dominant allele of the gene "K" in the genotype, the flowers of this plant will be also uncolored. We can say, that the dominant allele of the gene "D" suppresses the effect of a gene "K", i.e. inhibits the synthesis of malvidin. If we cross uncolored heterozygous plants with the genotype "DdKk", then we get the offspring with ratio of phenotypes : 13 white: 3 blue. Genetic traits files for this case should be:
D:white flower
K:blue flower
k:white flower
{
pigment:D d
color:K k
}
Genetic calculation: You can this trait file ( Epistasis 8.txt ) and calculate results for Traits phenotypes. In this case, the phenotypic classes "DK", "Dk" and "dk" is merged, and will have the same phenotype. As result of crossing of the hybrids with genotype "DdKk" in the second-generation we get the offspring with the ratio of phenotypes: 13 white: 3 blue. On the tab "Find" you can choose each phenotype and see from what kind of genotypes it is consist.
12:3:1 Dominant epistasis
Coloration of dog's fur.
This example is very similar to the previous one. But in this case, such phenotypic ratio is observed, when individual, which homozygous by recessive trait, has a special phenotype. Consider this on example of inheritance of fur color in dogs. For fur color is responsible gene "B". With dominant allel "B" in genotype a dog will be black, otherwise we have brown dog. On the fur color affect also another nonallelic gene "I". And in any case, if we have dominant allel "I" in the genotype, then dog will be only white. Thus, pups with the genotype "I_bb" have a white color, with genotype "iiB_" - black, and with the genotype "iibb" - brown. Genetic traits files for this case should be:
I:white dog
B:black dog
b:brown dog
{
pigment:I i
color:B b
}
Genetic calculation: You can open this traits file ( Epistasis 9.txt ) and calculate results for Traits phenotypes. In this case, the phenotypic classes "IB" and "Ib" is merged, and will have the same phenotype. As result of crossing of two diheterozygous dogs with genotype "DdKk", we get the offspring with the ratio of phenotypes: 12 white : 3 black : 1 brown. On the tab "Find" you can choose each phenotype and see from what kind of genotypes it is consist.
15:1 Duplicate dominant epistasis
Wheat Kernel Color.
Phenotypic ratio 15:1 is typical for the double dominant epistasis. The two previous examples we have considered only in terms of epistatic interactions of genes. And it would be difficult to describe in terms of complementary interaction of genes these two examples. But in this case it is more suitable the description in the terms of the complementary interaction of genes, because it is more corresponds to the biochemical mechanism of formation of this trait. Wheat kernel can be colored and uncolored. Coloration of wheat kernel is depends on the biochemical reaction, which resulted in a substance of precursor is transformed into a pigment. This reaction occurs under the action of products, which are synthesized by two genes, "A" and "B". Moreover, this biochemical reaction is possible both in the presence in the genotype the dominant alleles of both genes, and in the presence of only one dominant allele of these genes. Thus wheat with genotypes "A_B_", "A_bb" and "aaB_" will have colored kernel, and only plants with the genotype "aabb" will have uncolored kernel. Genetic traits files for this case should be:
A:colored kernel
B:colored kernel
*:noncolored kernel
{
color:A a
color:B b
}
Genetic calculation: You can open this traits file ( Epistasis 10.txt ) and calculate results for Traits phenotypes. In this case, the phenotypic classes "AB", "Ab" and "aB" is merged, and will have the same phenotype. As result of crossing of hybrids with genotype "AaBb", we get the offspring with the ratio of phenotypes: 15 colored : 1 noncolored. On the tab "Find" you can choose each phenotype and see from what kind of genotypes it is consist.
Sometimes when we are dealing with recessive epistasis, genetic traits files looks pretty complicated. For example, in the case of color of budgerigars feathering :
a a b b:white budgerigars
b b:blue budgerigars
a a:yellow budgerigars
A:green budgerigars
{
color:A a
color:B b
}
When we work with recessive epistatic genes, we should mark them as two recessive alleles. This is necessary to obtain correct results. If you mark the epistatic gene as only one recessive allele, then you will get an incorrect ratio of Traits phenotypes.
a b:white budgerigars
b:blue budgerigars
a:yellow budgerigars
A:green budgerigars
{
color:A a
color:B b
}
The fact that in the program the ratio of Traits phenotypes generated by the ratio of genotypes and if you will use such traits file, then the some of the genotypes will be interpreted not correctly. For example, only the budgerigars with genotype "aabb" should have the white color, but individuals with genotypes "AaBb", "aaBb", "Aabb" will also be identified as white budgerigars. Is it possible to use this type traits files? Yes, of course. If you choose Phenotypes for type of results, then you will see, that each gene in the phenotype is represented as a single allele - "AB", "Ab", "aB", "ab". Thus, if Traits phenotypes will be generated by the ratio of phenotypes, the results will be correct. To do this you need check on the checkbox "From Phenotypes". You can open this traits file ( Epistasis 3C.txt ) and calculate results for Traits phenotypes.
If you want to use this feature, you must clearly understand how and what you are doing. If you are dealing with both allelic and nonallelic interaction of genes, then you should not use this ability, because in this case Traits phenotypes must be generated from only the ratio of genotypes. If you do it right, you get the correct results. Do not be afraid to experiment and you will succeed.
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