Respuesta :
Answer:
d. BbSs
Explanation:
As given;
Black coat color (B) is dominant on brown coat color (b);
Striped fur pattern (S) is dominant on a marbled fur pattern (s);
Cross of black, striped cat with brown marbled cat.
As given, brown and marbled is recessive, so brown marbled cat is bbss.
And according to the given offspring, the other parent must be heterozygous.
So black, striped cat is BbSs
cross is between BbSs and bbss
gametes are: parent-1: BS, Bs, bS, bb; Parent-2; bs
offspring is BbSs, Bbss, bbSs, bbss.
d) is the correct option
Explanation:
B= black coat color (dominant)
b= brown coat color (recessive)
S= striped fur pattern (dominant)
s= marbled fur pattern (recessive)
A cross is made between black striped cat(genotype unknown) and brown marbled cat (bbss genotype)
(here in this brown marbled cat is homozygous recessive)
Resulting offsprings are:
3 brown marbled= both coat and fur color recessive
2 brown striped= coat color recessive fur color dominant
2 black marbled= coat color dominant fur color recessive
3 black striped= both coat color and fur color dominant
Crossing brown marbled cat with the genotypes given in options, the correct genotype of rescued cat comes out to be: BbSs
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