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In Labrador retrievers, black coat (B) is dominant to brown coat (b) and normal vision (N) is dominant to blindness (n). The genes for coat color and blindness are on separate chromosomes. You cross a dog that is heterozygous for coat color and vision with a blind brown dog. Which statements are TRUE about this scenario

a.The probability of getting a brown blind puppy is 1/16
b.The probability of getting a brown puppy with normal vision is ¼
c.The probability of getting a black blind puppy is 1/2.
d.The probability of getting a brown puppy is ½
e.Because blindness is recessive, the probability of getting a blind puppy is ¼
f.The brown blind perent will make ell bn gametes
g.A black blind puppy that results from this cross could have the genotype BBnn
h.A black puppy with normal vision that results from this cross will have the genotype BbNn
i.Because two traits are being followed, the phenotypic ratio of offspring will be 9:3:31

Answer :

Answer:

Correct options are

B, D, F, H,  

Explanation:

Given -

Black coat (B) is dominant to brown coat (b)

and

Normal vision (N) is dominant to blindness (n)

Genotype of dog  heterozygous for both coat color and normal vision - BbNn

Genotype of blind brown dog - bbnn

The cross is between BbNn * bbnn

The genotype of the offspring would be

BbNn * bbnn

BN        Bn        bN       bn

bn BbNn Bbnn bbNn bbnn

bn BbNn Bbnn bbNn bbnn

bn BbNn Bbnn bbNn bbnn

bn BbNn Bbnn bbNn bbnn

Genotype of offspring  

BbNn– 4 in number

Bbnn– 4 in number

bbNn– 4 in number

bbnn – 4 in number  

Phenotype and probability of offspring  

BbNn– Black and normal vision   [tex]\frac{4}{16} = \frac{1}{4}[/tex]

Bbnn– Black and blind    [tex]\frac{4}{16} = \frac{1}{4}[/tex]

bbNn– Brown and normal vision    [tex]\frac{4}{16} = \frac{1}{4}[/tex]

bbnn – Brown and blind     [tex]\frac{4}{16} = \frac{1}{4}[/tex]

Correct options are

B, D, F, H,  

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