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iragen [17]
3 years ago
14

What is the probability of producing a child that will phenotypically resemble either one of the two parents in the following fo

ur cases? How many phenotypically different kinds of progeny could potentially result from each of the crosses? Aa Bb Cc Dd X aa bb cc dd
Biology
1 answer:
Rashid [163]3 years ago
4 0

Complete question:

What is the probability of producing a child that will phenotypically resemble either one of the two parents in the following four cases? How many phenotypically different kinds of progeny could potentially result from each of the crosses?

a. Aa Bb Cc Dd  × aa bb cc dd

b. aa bb cc dd × AA BB CC DD

c. Aa Bb Cc Dd × Aa Bb Cc Dd

d. aa bb cc dd × aa bb cc dd

Answer:

a) the probability of producing a child that will phenotypically resemble either one of the two parents is 1/8.

   There will be 16 possible phenotypes among the progeny

b) the probability of producing a child that will phenotypically resemble either one of the two parents is 1.

   There will be one possible phenotype among the progeny

c) the probability of producing a child that will phenotypically resemble either one of the two parents is 81/256 = 0.316.

   There will be 16 possible phenotypes among the progeny

d) the probability of producing a child that will phenotypically resemble either one of the two parents is 1.

   There will be one possible phenotype among the progeny

Explanation:

Due to technical problems, you will find the complete explanation in the attached files

Download pdf
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Answer:

25%

Explanation:

When looking at a pedigree remember that:

  • squares are males
  • circles are females
  • the solid colored figure represents an individual affected by a disease
  • the empty figure represents a healthy individual

Let us assign the symbol X⁺ to represent the dominant allele linked to the X-chromosome and expressing healthiness, and X⁻ to represent the recessive allele expressing the dissease.

According to this pedigree

  • I1 is a man affected by the disease, YX⁻
  • I2 is a healthy woman X⁺X⁻
  • we can see that among the progeny (generation II) there are two individuals affected (a boy and a girl) and one healthy girl. This means that the mother I2 is heterozygous for the trait.

So, having their genotypes we can know what are the probabilities of getting a son with DMD

Parentals)    YX⁻     x     X⁺X⁻  

Gametes)   Y     X⁻      X⁺     X⁻

Punnett square)

                        X⁺             X⁻

            X⁻      X⁺X⁻         X⁻X⁻    

            Y        X⁺Y           X⁻Y

F1)

  • The probabilities of getting a healthy daughter X⁺X⁻ are 25%
  • The probabilities of getting a healthy son X⁺Y are 25%
  • The probabilities of getting a daughter with DMD X⁻X⁻ are 25%
  • The probabilities of getting a son with DMD X⁻Y are 25%
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