Apparent epistasis explained (13 : 3)

Crossing a White Leghorn (CCII) chicken with another White Wyndot (ccii) chicken produces a white chicken in the F1 line. Self-fertilization of white (CcIi) chickens in the F1 progeny resulted in 13 white and 3 colored chickens in the F2 progeny.
here,
White Leghorn chicken genotype CCII
White Wyndot chicken genotype ccii
The first progeny is F1 and the second progeny is F2
Parents : ×
Phenotype: White Leghorn White Wyndot
Genotype: CCII ccii
Gamit : (CI) (ci)
R1 lineage: CcIi (White)

Svanishek : F1×F1
Parents : ×
Phenotype: White White
Genotype : CcIi CcIi
Gamit : (CI)(Ci)(cI)(ci) (CI)(Ci)(cI)(ci)
F2 progeny (checker board)
Comment: It can be seen from the checker board, out of 16 chickens, 13 are white and 3 are colored chickens. In this case the expressed gene I non-allele inhibits expression of the expressed gene C.
Phenotypic ratio : White : Colored = 13 : 3

Apparent epistasis definition

Expressed epistasis occurs when an expressed gene at one locus on a chromosome interferes with the expression of a non-allelic expressed gene at another locus. Leghorn, Wyndot, Plymouth Rocks, etc., feather colors are the result of pronounced epistasis. Bateson I Punnett (1908) noted this.

Due to pronounced epistasis, the ratio of Mendel’s second law of inheritance is 13 : 3 instead of 9 : 3 : 3 : 1.

Complementary gene explanation

Crossing a sweet pea with a white flower (AAbb) with another white flower (aaBB) resulted in an F1 line with purple flowers (AaBb). Self-fertilization between purple (AaBb) flowers in the F1 progeny resulted in 9 purple and 7 white flowers in the F2 progeny.

here,

The first white flower genotype is AAbb

The second white flower genotype is aaBB

The first progeny is F1 and the second progeny is F2

Parents : ♂ × ♀

Phenotype: White flowers White flowers

Genotype : AAbb aaBB

Gamit : (Ab) (aB)

F1 Lineage : AaBb (Purple)

 

Self fertilization : F1×F1

Parents : ♂ × ♀

Phenotype: Purple flowers Purple flowers

Genotype : AaBb AaBb

Gamit : (AB)(Ab)(aB)(ab) (AB)(Ab)(aB) (ab)

F2 progeny (checker board)

Comment: From the checker board, out of 16 flowers, 9 are purple and 7 are white. The purple color of the flower is due to the combined effect of expressed genes A and B.

Phenotypic ratio : Purple : White = 9 : 7

Explanation of Codominance

A cross of a red (AA) cow with another white (BB) cow results in an F1 breed of rowan (AB) cows. Self-insemination of F1 rowan (AB) cows resulted in 1 red, 2 rowan and 1 white cow in R2.

here,

The gene responsible for red color is A

Genotype AA in red

again,

The gene responsible for white color is B

Genotype BB is white

The first progeny is F1 and the second progeny is F2

 

Parents : ♂ × ♀

Phenotype: Red and white

Genotype : AA BB

Gamit : (A) (B)

F1 Breed : AB (Rowan)

 

Self fertilization : F1×F1

Parents : ♂ × ♀

Phenotype: Rowan Rowan

Genotype : AB AB

Gamit : (A) (B) (A) (B)

F2 progeny (checker board)

Comment: It can be seen from the checker board, out of 4 cows, 1 is red, 2 is rowan and 1 is white.

Phenotypic ratio : Red : Rowan : White = 1:2:1

Explanation of Lethal gene

In the third stage, a yellow (Aya) mouse was crossed with another yellow (Aya) mouse, resulting in 1 dead (AyAy), 2 yellow (Aya) and 1 meto (aa) mice.

here,

The genotype of yellow mice is Aya

Parents : ♂ × ♀

Phenotype: Yellow Yellow

Genotype : Aya Aya

Gamit : (Ay) (a) (Ay) (a)

F2 progeny (checker board)

Remarks : From the checker board, out of 4 rats, 1 is dead, 2 are yellow and 1 is meto colored. The mouse died because the expressed gene Ay was homozygous (AyAy).

Phenotypic ratio : Yellow : Meto = 2 : 1