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

Incomplete dominance explained

Crossing a black (BB) chicken with another white (bb) chicken results in a bluish-grey (Bb) chicken in the F1 line. Self-fertilization of F1 progeny with bluish-brown (Bb) chickens resulted in 1 black, 2 brown and 1 white chicken in the F2 progeny.
here,
The gene responsible for black color is B
Hence, the black genotype is BB
again,
Gene responsible for white color b
As a result, the genotype bb is white
The first progeny is F1 and the second progeny is F2

Parents : ♂ × ♀
Phenotype: Black and white
Genotype : BB bb
Gamit : (B) (b)
Season 1: Bb (Blue Dhusar)

Svanishek : F1×F1
Parents : ♂ × ♀
Phenotype: Bluish Grey
Genotype: Bb Bb
Gamit : (B) (b) (B) (b)
F2 progeny (checker board)
Comment: It can be seen from the checker board, out of 4 chickens, 1 is black, 2 are blue-grey and 1 is white. Chickens are bluish gray rather than black due to incomplete expression of the expression gene B.
Phenotypic Ratio : Black : Blue Gray : White = 1:2:1

Incomplete dominance explained

A cross of sandhyamalati (Mirabilis jalapa) with a red flower (RR) and a white flower (rr) resulted in pink flowers (Rr) in the F1 line. Self-fertilization between the pink (Rr) flowers of the F1 progeny produced 1 red (25%), 2 pink (50%) and 1 white (25%) flowers in the F2 progeny.

here,

The gene responsible for red color is R

Hence, the red genotype is RR

again,

The gene responsible for white color is r

Consequently, the genotype of white color is rr

The first progeny is F1 and the second progeny is F2

 

Parents : ♂ × ♀

Phenotype: Red and white

Genotype : RR rr

Gamit : (R) (r)

F1 Lineage : Rr (Pink)

 

Svanishek : F1×F1

Parents : ♂ × ♀

Phenotype: pinkish pink

Genotype : Rr Rr

Gamit : (R) (r) (R) (r)

F2 progeny (checker board)

Comment: It can be seen from the checker board, out of 4 flowers, 1 is red, 2 are round and 1 is white. Flowers are pink rather than red due to incomplete expression of the expression gene R.

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

Explanation of Mendel’s second law

Crossing a black short haired (BBSS) guinea pig with a brown long haired (bbss) guinea pig results in an F1 breed of black short haired (BbSs) guinea pigs. Inbreeding F1 black shorthair (BbSs) guinea pigs resulted in F2 progeny with 9 black shorthair, 3 black longhair, 3 brown longhair and 1 brown longhair guinea pig.

here,

The gene responsible for black color is B

Genotype BB is black

And the gene S responsible for short fur

Short hair genotype SS

Consequently, the black color and short hair genotype is BBSS

again,

Gene responsible for brown color b

Brown color genotype bb

And the gene responsible for long hair s

The long haired genotype ss

Consequently, the brown color and long hair genotype is bbss

The first progeny is F1 and the second progeny is F2

Parents : ♂ × ♀

Phenotype : Black color – short hair Brown color – long hair

Genotype : BBSS bbss

Gamete : (BS) (bs)

F1 Generation : BbSs

Self fertilization : F1×F1

Parents : ♂ × ♀

Phenotype : Black color Short hair     Black color Short hair

Genotype :        BbSs                             BbSs

Gamit : (BS) (Bs) (bS) (bs) (BS) (Bs) (bS) (bs)

F2 Generation (checker board)

Remarks: From the checker board, out of 16 guinea pigs, 9 are black-short hair, 3 black-long hair, 3 brown-long hair and 1 brown-long hair guinea pig.

Phenotypic Ratio: Black-Short Hair: Black-Long Hair: Brown-Long Hair: Brown-Long Hair = 9:3:3:1