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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

Respuesta :

coge4eva

Answer: 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

Ver imagen coge4eva
luisvaschetto

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).

  • According to the Mendel principle, the expected phenotypic frequencies for a dihybrid cross between two heterozygous individuals in which the alleles of both genes separate independently into gametes is 9:3:3:1.

  • In this case, the heterozygous genotype of both parents is BbLl, thereby the F1 genotypes from the Punnett square are as follow:

                      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:

https://brainly.com/question/1185199

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