Respuesta :

1. phenotype = black and long hair 2: brown and long hair 2

phenotype ratio = 1:1 and genotype =  BbLL, BbLl,bbLL, bbLl

genotype ratio= 1:1:1:1.

2.  probability of 50% progeny to be normal tailed and funny eyed is there.

3. 3/16 or 18.75% is the probability of long winged black body fly.

Explanation:

1. information given in cat:

long hair = LL, short hair ll, black hair colour BB, Brown hair colour bb

cross between:

homozygous hair length and heterozygous hair colour X heterozygous hait length and homozygous recessive hair colour

genotype = BbLL X bbLl

punnet square:

        BL     bL

bL BbLL   bbLL

bl  BbLl   bbLl

phenotype = black and long hair 2: brown and long hair 2

phenotype ratio = 1:1

genotype =  BbLL, BbLl,bbLL, bbLl

genotype ratio= 1:1:1:1

2. given that in fish:

Normal tail NN (dominant), fancy tail nn (recessive), funny eyes ff (recessive), normal eyes FF (Dominant)

cross between

fancy tail funny eye X Heterozygous normal tail funny eye fish)

genotype = nnff X Nnff

probability of normal tailed funny eye fish=?

cross between in puunet square:

       nf    nf

Nf   Nnff  Nnff

nf   nnff   nnff

phenotype = 2 normal tail funny eye: 2 fancy tail funny eye

there will be probability of 50% progeny to be normal tailed and funny eyed.

3. given that in fly:

gray body GG and long wing LL is dominant

black body gg and short wing is recessive

probability of black body long fly=?

2 heterozygous gray body and long winged bodies crossed

genotype is GgLl

Punnet square:

            Gl       GL       gL       gl

Gl  GGll     GGLl     GgLl    Ggll

GL GGLl   GGLl      GgLL   GgLl

gL GgLl     GgLL    ggLL   ggLl

gl Ggll      GgLl       ggLl    ggll

3/16 is the probability of long winged black body fly.