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4. A woman who is heterozygous for freckles and a carrier for PKU disease marries a
man who is also heterozygous for freckles and a carrier for PKU disease. List the
probability for each of the four possible phenotypes (not genotypes) in their offspring.

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Answer:

Freckles, no PKU- 9/16

Freckles, PKU- 3/16

No freckles, no PKU- 3/16

No freckles, PKU- 1/16

Explanation:

This question involve two distinct genes; one coding for possession or not of freckles and the other for PKU disease in humans. Allele for freckles (F) and no PKU (P) are dominant over the alleles for no freckles (f) and PKU (p) respectively.

N.B: Possession of freckles is a dominant trait i.e. FF and Ff possess freckles

If a woman who is heterozygous for freckles and a carrier for PKU disease (FfPp) marries a man who is also heterozygous for freckles and a carrier for PKU disease (FfPp), they produce the following gametes: FP, Fp, fP, and fp

Using these gametes in a punnet square (see attached image), the following offsprings will be produced:

Freckles, no PKU (F_P_) - 9/16

Freckles, PKU (F_pp) - 3/16

No freckles, no PKU (ffP_) - 3/16

No freckles, PKU (ffpp) - 1/16

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Oseni

The possible phenotypes of the offspring with their probabilities are:

  1. Freckles and PKU = 3/4 x 1/4 = 3/16
  2. Freckles and no PKU = 3/4 x 3/4 = 9/16
  3. No freckles and PKU = 1/4 x 1/4 = 1/16
  4. No freckles and no PKU = 1/4 x 3/4 = 3/16

Each of the traits would be independently inherited:

Freckles are autosomal dominant traits. Assuming the allele for freckles is A, a heterozygous individual will have the genotype Aa.

                      Aa    x    Aa

                    AA   Aa   Aa   aa

Probability of having freckles = 3/4

Probability of not having freckles = 1/4

PKU is an autosomal recessive trait. Assuming the allele to be b, heterozygous individuals will have the genotype Bb.

                                  Bb    x    Bb

                                BB   Bb  Bb   bb

Probability of having PKU = 1/4

Probability of not having PKU = 3/4

The four possible phenotypes with their probabilities are:

  1. Freckles and PKU = 3/4 x 1/4 = 3/16
  2. Freckles and no PKU = 3/4 x 3/4 = 9/16
  3. No freckles and PKU = 1/4 x 1/4 = 1/16
  4. No freckles and no PKU = 1/4 x 3/4 = 3/16

More on dihybrid crosses can be found here: https://brainly.com/question/1185199?

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