Answer:
Let E and e be the alleles of the genes responsible for earlobes and A and a be the alleles of the gene responsible for a cleft chin.
E is dominant over e and codes for unattached earlobes.
Similarly, A is dominant over a and codes for the cleft chin.
Both the parents are heterozygous for both the traits.
Thus, the genotype of each parent would be EeAa (unattached earlobe and cleft chin).
The gametes produced would be EA, Ea, eA, ea.
The cross would be result in the production of offspring with four possible types of phenotypes; unattached earlobes and cleft chin, unattached earlobes and no cleft, attached earlobes and cleft chin, and attached earlobes and no cleft in the ratio 9:3:3:1.
- unattached earlobes and cleft chin = (EEAA
, EEAa
, EeAA
, EeAa
, EEAa
, EeAa
, EeAa
, EeAA
, EeAa
) = 9/16
- unattached earlobes and no cleft = (EEaa
, Eeaa
, Eeaa
) = 3/16
- attached earlobes and cleft chin = (eeAA
, eeAa
, eeAA
) = 3/16
- attached earlobes and no cleft = (eeaa) = 1/16