Final answer:
The genotypes of the parents with both dominant and recessive traits (unattached earlobes and widow's peak) are EeWw for the mother and eeWw for the father. The probability of them having a child with unattached earlobes and a non-widow's peak hairline is 1/8 based on calculating individual probabilities for each trait and then combining them.
Step-by-step explanation:
The genotypes of the parents involve traits for earlobe attachment and hairline shape. The female with unattached earlobes must have one dominant allele (E) and one recessive allele (e) since unattached is dominant and the child has attached earlobes, indicating the mother passed on a recessive allele. Likewise, she has a widow's peak, which is dominant, so her genotype could be Ww or WW. However, since the child has a non-widow's peak, the mother must be heterozygous Ww to pass on the recessive allele. The male has attached earlobes (recessive trait) and a widow's peak (dominant trait). His genotype must be ee for earlobes since attached (ee) is recessive and he exhibits the trait, and for the widow's peak, it can either be WW or Ww, but since the child has a non-widow's peak, the father must also be Ww.
Therefore, the genotypes of the parents are as follows: Female (Mother) - EeWw, Male (Father) - eeWw. To find out the probability of them having a child with unattached earlobes (E) and a non-widow's peak hairline (w), you would construct a Punnett square for each trait and then combine the probabilities. The probability for each trait can be found by multiplying the individual probabilities of receiving an E from the mother (1/2) and an e from the father (1), and a W from the mother (1/2) and a w from the father (1/2). The combined probability of having a child with Ee(Ww or ww) is 1/2 * 1/4 = 1/8.