2. For a particular rodent, black fur (B) is dominant over brown fur (b), and a long tail
(L) is dominant over a short tail (1). Create a Punnett square that shows the genotypes
of all the possible offspring that could result from the breeding of two of the rodents
that are heterozygous for both traits.
Key: Bib

The expected phenotypic frequencies for this dihybrid cross is 9:3:3:1 (i.e., 9 genotype B_L_ : 3 genotype B_ll: 3 genotype bbL_: 1 genotype bbll).

                      1/4 BL             1/4 Bl             1/4 bL           1/4  bl

1/4 BL          1/16 BLBL       1/16 BBLl        1/16 BbLL     1/16 BbLl

1/4 Bl           1/16  BBLl       1/16 BBll         1/16 BbLl      1/16 Bbll

1/4 bL          1/16 BbLL        1/16 BbLl        1/16 bbLL      1/16 bbLl

1/4 bl           1/16 BbLl        1/16 Bbll          1/16 bbLl       1/16 bbll

In consequence, the expected phenotypic frequencies for this dihybrid cross is 9:3:3:1 (i.e., 9 genotype B_L_ : 3 genotype B_ll: 3 genotype bbL_: 1 genotype bbll).

Learn more in:

link

The genotypes of all the possible offspring that could result from the breeding of two rodents that are heterozygous for both traits are BBLL, BBLl, BbLL, BbLl, BBll, Bbll, bbLL, bbLl and bbll. There are 9 possible different genotypes.

Explanation:

The genotype of two rodents that are heterozygous for both traits is BbLl. A dibybrid cross between BbLl and BbLl will produce 16 offsprings with 9 different genotypes.

See the attached punnet square for further illustration of the dibybrid cross

explanation:

okay now i don't know this but for the pints nothing else to say bye