Question

can someone help me step by step!

4. In cats, black color is dominant to a special, temperature-sensitive albino gene which produces cats with dark legs, faces and tails (Siamese cats). A short haired (dominant) Siamese colored female is bred to a long-haired recessive) black male. They have eight kittens: 2 black, short-haired; 2 black, long-haired; 2 Siamese, short-haired; and 2 Siamese, long-haired. What were the genotypes of the two parents?

Answer 1

Given that black (B) colour is dominant over albino or Siamese (b). And short hair (S) is dominant over long hair (s).

A short haired (dominant) Siamese or albino coloured female is crossed with long haired (recessive) black male is crossed and they produce 8 kittens as : 2 black, short hair ; 2 black, long hair; 2 Siamese, short hair and 2 Siamese,long hair.

They have phenotypic ratio as 2:2:2:2 or 1:1:1:1.

To find genotypes of parents we need to perform a test cross. Test cross involves the cross between the Homozygous recessive to dominant ( can be Homozygous or heterozygous).

1) with Homozygous dominant { SSBB}

SSBB - short hair and black coloured. -----dominant

ssbb - Recessive; long hair and Siamese

Gametes are a) SB and b) sb. Perform the cross between a and b

	SB	SB	SB	SB
sb	SsBb	SsBb	SsBb	SsBb

In this case we get all progenies as heterozygous dominant so the parents can't be Homozygous dominant.

2) With heterozygous dominant { SsBb}

SsBb - short haired and Siamese -------dominant

ssbb - long haired and black --------recessive

Gametes are a) SB b)Sb c)sB d)sb and e)sb. Perform the cross between a-d with e

	SB	Sb	sB	sb
sb	SsBb	Ssbb	ssBb	ssbb
sb	SsBb	Ssbb	ssBb	ssbb

Here sb is taken 2 times as kittens produced are 8, we can get the results by crossing a-d with e once also but then we have to caculatce the number of each type of kittens by using the ratio and total progenies.

So, in heterozygous parent cross, cross 2 we observe that we get the ratio as 1:1:1:1 similar to the expected. It shows that the parents are heterozygous.

As given P1 = Short hair and Siamese female so the genotype will be Ssbb.

P2 = long hair and black colour male so the genotype will be ssBb.

On crossing these parents we get

	sB	sb	sB	sb
Sb	SsBb	Ssbb	SsBb	Ssbb
sb	ssBb	ssbb	ssBb	ssbb

SsBb - Short hair and black cat ------- 2

Ssbb - Short hair and Siamese cat ------ 2

ssBb - Long hair and black cat ------- 2

ssbb - Long hair and Siamese cat ------ 2

Ratio = 2:2:2:2 or 1:1:1:1.

Parent genotypes are : 1) male --- ssBb

2) female ----- Ssbb

Thank you....????