Question

3. Predict the possible appearance of the children of Bob and Sue Everyperson, a young couple who now live in

Centralia. We are tracking two genes: hair line and earlobes. These genes are located on non-homologous

chromosomes. The allele for widows peak (W) is dominant to the allele for a smooth hairline (w). The allele for free

ear lobes (F) is dominant to the allele for attached ear lobes (f). Bob’s genotype is WwFf. Sue’s genotype is wwff.

3a. What type of hair line and earlobes does Bob have? What type of hair line and earlobes does Sue have?

3b. How many different gametes can Bob produce? List these gametes? How many different gametes can Sue produce? List these gametes.

3c. Using a Punnett square, show the genotype and phenotype of all possible combinations between the

gametes from Bob and Sue. What is the likelihood that their first child will have a widow’s peak and attached

ear lobes?

Answer 1

a) Bob's genotype is WwFf, means he is heterozygous. He contains dominant alleles of hairline and earlobes, and they are being expressed. W means he has widow peak and F means free ear lobes.

Therefore we can easily conclude that his hairline type is widow peak and his earlobes are of free type.

Sue's genotype is wwff, means she is homozygous. She contains recessive alleles of hairline and earlobes, and they are being expressed. w means she has smooth hairline and f means attached earlobes.

Therefore we can easily conclude that her hairline type is smooth and her earlobes are attached type.

b)

Gametes of Bob :

He can produce 4 different combinations of gametes, namely WF, Wf, wF and wf.

Gametes of Sue :

She can produce 4 copies of only a single type of gamete wf.

c)

Punnett Square is given as :

No. of progenies possible with Widow peak and attached earlobes = 4 ( genotype of Wwff)

Total number of progenies possible = 16

Probability = 4/16

= 0.25

= 25%

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