1. Lethal genes are expressed or latent.
2. It causes death of the organism in homozygous condition.
3. Expressed lethal genes cause death of the organism in both homozygous and heterozygous states.
4. Causes death of organisms in the zygote or embryonic stage.
5. Sometimes it manifests with age.

When an organism carrying a homozygous gene dies, it is called a lethal gene. This is called lethal mutation. Organisms may contain one or more lethal genes. In 1905, the French geneticist Cunot was the first to observe the lethal gene in skin color in mice.

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

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

If the manifest feature of the organism does not fully manifest on the latent feature but shows incomplete manifestation, it is called incomplete manifestation. The gene responsible for incomplete expression is called intermediate gene. Scientist Karl Kerns (1903) discovered it.

In many cases the Mendelian 3 : 1 and 9 : 3 : 3 : 1 ratios are not found in genetic tests. This is called non-Mendelian inheritance or Beding inheritance. Due to non-Mendelian inheritance, the ratio of 1 : 2 : 1, 2 : 1, 9 : 7, 13 : 3 etc. is found in genetic tests.

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

Gregor Johann Mendel, the father of genetics, crossed a round-yellow pea (RRYY) with another green-green pea (rryy) to obtain the round-yellow pea (RrYy) in the F1 lineage. A self-crossing of round-yellow beans (RrYy) in the F1 progeny resulted in 9 round-yellow, 3 round-green, 3 shriveled-yellow and 1 shriveled-green beans in the F2 line.

here,

The gene responsible for the spherical trait is R

Genotype RR for spherical trait

 

And the gene Y responsible for the yellow color

Yellow color genotype YY

Consequently, the round-yellow genotype is RRYY

again,

The gene responsible for the stunted trait is r

Genotype rr for stunted traits

 

and the gene responsible for green color y

Genotype in green is yy

As a result, the pale-green genotype is rryy

The first progeny is F1 and the second progeny is F2

Parents : ♂ × ♀

Phenotype: Spherical-yellow shriveled-green

Genotype : RRYY rryy

Gamete : (RY) (ry)

F1 Generation : RrYy (round-yellow)

Self fertilization : F1×F1

Parents : ♂ × ♀

Phenotype : Round-yellow Round-yellow

Genotype : RrYy RrYy

Gamete : (RY) (Ry) (rY) (ry) (RY) (Ry) (rY) (ry)

F2 Generation (checker board)

Remarks: From the checker board, out of 16 beans, 9 are round-yellow, 3 round-green, 3 shriveled-yellow and 1 shriveled-green.

Phenotypic ratio : round-yellow : round-green : shriveled-yellow : shriveled-green = 9 : 3 : 3 : 1

When two or more pairs of organisms with opposite traits are crossed, only the dominant trait is expressed in the first hybrid male (F1), but in subsequent generations, the traits break up and enter different gametes independently. Such a cross is called a dihybrid cross.

Crossing a black (BB) guinea pig with another brown (bb) guinea pig results in a black (Bb) guinea pig in the F1 line. Self-crossing of black (Bb) guinea pigs in the F1 progeny resulted in 3 black and 1 brown guinea pig in the F2 progeny.

here,

The gene responsible for black color is B

Hence, the black genotype is BB

again,

Gene responsible for brown color b

Consequently, the brown genotype is bb

The first progeny is F1 and the second progeny is F2

Parents : ♂ × ♀

Phenotype: Black brown

Genotype : BB bb

Gamit : (B) (b)

1st lineage : Bb (black)

Svanishek : F1×F1

Parents : ♂ × ♀

Phenotype: black black

Genotype: Bb Bb

Gamit : (B) (b) (B) (b)

R2 Vamsa (Checker Board)

Remarks: From the checker board, out of four guinea pigs, 3 are black and 1 is brown.

Phenotypic ratio : Black : Brown = 3:1