To solve this genetics problem, we need to consider the traits given and their inheritance patterns. We will use a Punnett square to represent the genetic cross. Since both parents are heterozygous for both traits (unattached earlobes and cleft chin), their genotypes are EeAa.
### Step-by-Step Solution:
1. Identify the genes and alleles:
- Unattached earlobes (E) are dominant to attached earlobes (e).
- Cleft chin (A) is dominant to no cleft chin (a).
2. Parents' genotypes: EeAa (heterozygous for both traits)
3. Determine possible gametes from each parent:
- Each parent can produce four types of gametes: EA, Ea, eA, and ea.
4. Create a Punnett square:
| | EA | Ea | eA | ea |
|------|--------|--------|--------|--------|
| EA | EEA A | EEA a | EeA A | EeA a |
| Ea | EEA a | EEA a | EeA a | Eeaa |
| eA | EeA A | EeA a | eeA A | eeA a |
| ea | EeA a | EeA a | eeA a | eeaa |
5. List possible genotypes from the Punnett square:
- EEA A
- EEA a
- EeA A
- EeA a
- eeA A
- eeA a
- Eeaa
- eeaa
6. Identify phenotypes:
- EEAA, EEAa, EeAA, and EeAa will have unattached earlobes and a cleft chin.
- EEaa and Eea will have unattached earlobes and no cleft chin.
- eeAA and eeAa will have attached earlobes and a cleft chin.
- eeaa will have attached earlobes and no cleft chin.
7. Determine the ratio of offspring phenotypes:
- Unattached earlobes and cleft chin: 9/16
- Unattached earlobes and no cleft chin: 3/16
- Attached earlobes and cleft chin: 3/16
- Attached earlobes and no cleft chin: 1/16
8. Punnett Square Analysis:
| Phenotype | Count | Ratio |
|------------------------------------|-------|---------|
| Unattached earlobes and cleft chin | 9 | 9/16 |
| Unattached earlobes and no cleft | 3 | 3/16 |
| Attached earlobes and cleft chin | 3 | 3/16 |
| Attached earlobes and no cleft | 1 | 1/16 |
Thus, the ratio of offspring with attached earlobes and no cleft to the total number of offspring is 1:16.
