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In cocker spaniels, solid coat color is dominant (S) over spotted coat (s). Suppose a true-breeding...

In cocker spaniels, solid coat color is dominant (S) over spotted coat (s). Suppose a true-breeding solid-colored dog is crossed with a spotted dog, and the F1 dogs are interbred.

What is the probability that the first puppy born will have a solid coat?

What is the probability that the solid-coated puppy in the previous question is homozygous for the solid coat allele?

What is the probability that if four puppies are born, all of them will have a spotted coat?

What is the probability that if six puppies are born, 2 of them would have a spotted coat and 4 of them would have a solid coat? Please round your answer to the nearest hundredth or leave as a fraction.

The following two individuals are crossed: Aa Bb Cc Dd x Aa bb Cc Dd What proportion of the progeny will have the following genotype? Express as either a fraction or as a decimal] Aa Bb CC dd p=

What is the probability that one of the progeny will have all dominant OR all recessive phenotype? p=

Expert Solution

Solid colored dog= SS

Spotted dog= ss

F1= Ss

When F1 is interbred, Ss X Ss

We get, SS(solid]), Ss(solid), Ss(solid), ss(spotted)

1)The probability of firstborn puppy will have solid coat is 3/4 or 0.75. As the solid coat is dominant, the heterozygous genotype will also show a solid phenotype.

2) The probability of the solid-coated puppy in the previous question is homozygous for the solid coat allele is 1/2 or 0.5

3) The probability that if four puppies are born, all of them will have a spotted coat is 1/256 or 0.0039

As the probability of a puppy having a spotted coat is 1/4. so if four puppies are born, then probability of all of them having spotted coat becomes 1/4 X 1/4 X 1/4 X 1/4

4) The probability that if six puppies are born, 2 of them would have a spotted coat and 4 of them would have a solid coat is 81/4096 or 0.019

As the probability of a puppy having a solid coat is 3/4. so if four puppies are born, then the probability of them having spotted coat becomes 3/4 X 3/4 X 3/4 X 3/4

As the probability of a puppy having a spotted coat is 1/4. so if two puppies are born, then the probability of them having spotted coat becomes 1/4 X 1/4.

Probability of six puppies, where 2 of them would have a spotted coat and 4 of them would have a solid coat becomes 3/4 X 3/4 X 3/4 X 3/4 X 1/4 X 1/4 = 81/4096

5) Aa Bb Cc Dd x Aa bb Cc Dd

To calculate the probability of genotype. perform a monohybrid cross of each gamete and then write the probability of gamete according to the monohybrid cross to find out the probability of genotype asked.

the first monohybrid cross, Aa X Aa = AA Aa Aa aa. So the probability of Aa becomes 1/2

the second monohybrid cross, Bb X bb= Bb Bb bb bb. So the probability of Bb becomes 1/2

the third monohybrid cross, Cc X Cc= CC Cc Cc cc. So the probability of CC becomes 1/4

the fourth monohybrid cross, Dd X Dd= DD Dd Dd dd. So the probability of dd becomes 1/4

the probability of Aa Bb CC dd genotype = 1/2 X 1/2 X 1/4 X 1/4 = 1/64

the first monohybrid cross, Aa X Aa = AA Aa Aa aa. So the probability of aa becomes 1/4

the second monohybrid cross, Bb X bb= Bb Bb bb bb. So the probability of bb becomes 1/2

the third monohybrid cross, Cc X Cc= CC Cc Cc cc. So the probability of cc becomes 1/4

the fourth monohybrid cross, Dd X Dd= DD Dd Dd dd. So the probability of dd becomes 1/4

Probability of all recessive phenotype aa bb cc dd = 1/4 X 1/2 X 1/4 X 1/4 = 1/128

gladiator answered 7 months ago

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