Question

unattached earlobes are dominant to attached earlobes. cleft chin is dominant to no cleft. parents that are heterozygous for both traits are crossed. determine the ratio of offspring with the described trait to the total number of offspring: unattached earlobes and cleft chin \boxed{\quad}:\boxed{\quad} unattached earlobes and no cleft \boxed{\quad}:\boxed{\quad} attached earlobes and cleft chin \boxed{\quad}:\boxed{\quad} attached earlobes and no cleft \boxed{\quad}:\boxed{\quad} done \checkmark \

$$\begin{tabular}{|c|c|c|c|} \\hline & ea & ea & ea & ea \\\\ \\hline ea & eeaa & eeaa & 1 & eeaa \\\\ \\hline ea & 2 & 3 & eeaa & eeaa \\\\ \\hline ea & eeaa & eeaa & 4 & eeaa \\\\ \\hline ea & eeaa & eeaa & eeaa & 5 \\\\ \\hline \\end{tabular}$$

Explanation:

Step1: Fill in the Punnett Square

First, we need to fill in the missing cells (1, 2, 3, 4, 5) in the Punnett square. Let's denote the alleles for earlobes: \( E \) (unattached, dominant), \( e \) (attached, recessive). For chin: \( A \) (cleft, dominant), \( a \) (no cleft, recessive). The parents are heterozygous for both traits, so their genotypes are \( EeAa \), and the gametes are \( EA, Ea, eA, ea \).

- Cell 1 (EA × eA): Genotype is \( EeAA \) (since \( E \times e = Ee \), \( A \times A = AA \))
- Cell 2 (Ea × EA): Genotype is \( EEAa \) ( \( E \times E = EE \), \( a \times A = Aa \))
- Cell 3 (Ea × Ea): Genotype is \( EEaa \) ( \( E \times E = EE \), \( a \times a = aa \))
- Cell 4 (eA × eA): Genotype is \( eeAA \) ( \( e \times e = ee \), \( A \times A = AA \))
- Cell 5 (ea × ea): Genotype is \( eea a \) ( \( e \times e = ee \), \( a \times a = aa \))

Now the complete Punnett square (16 cells) has the following genotypes:

| | EA | Ea | eA | ea |
|-------|----------|----------|----------|----------|
| **EA** | EEAA | EEAa | EeAA (1) | EeAa |
| **Ea** | EEAa (2) | EEaa (3) | EeAa | Eeaa |
| **eA** | EeAA | EeAa | eeAA (4) | eeAa |
| **ea** | EeAa | Eeaa | eeAa | eea a (5) |

Step2: Determine Phenotypes

Now we determine the phenotype for each genotype:

- **Unattached earlobes (E_) and cleft chin (A_)**: Genotypes with at least one \( E \) and at least one \( A \). Let's count these:
- EEAA, EEAa, EeAA, EeAa (from EA row: 4 cells), EEAa, EEaa (no, EEaa has aa), EeAa, Eeaa (no, Eeaa has aa) (wait, no, let's list all 16 cells:
1. EEAA: E_, A_ ✔️
2. EEAa: E_, A_ ✔️
3. EeAA: E_, A_ ✔️
4. EeAa: E_, A_ ✔️
5. EEAa (2): E_, A_ ✔️
6. EEaa (3): E_, aa ❌
7. EeAa: E_, A_ ✔️
8. Eeaa: E_, aa ❌
9. EeAA: E_, A_ ✔️
10. EeAa: E_, A_ ✔️
11. eeAA (4): ee, A_ ❌ (attached earlobes)
12. eeAa: ee, A_ ❌
13. EeAa: E_, A_ ✔️
14. Eeaa: E_, aa ❌
15. eeAa: ee, A_ ❌
16. eea a (5): ee, aa ❌

Wait, let's count properly. Let's list all 16 cells:

1. EEAA: E_, A_ ✔️
2. EEAa: E_, A_ ✔️
3. EeAA: E_, A_ ✔️
4. EeAa: E_, A_ ✔️
5. EEAa: E_, A_ ✔️
6. EEaa: E_, aa ❌
7. EeAa: E_, A_ ✔️
8. Eeaa: E_, aa ❌
9. EeAA: E_, A_ ✔️
10. EeAa: E_, A_ ✔️
11. eeAA: ee, A_ ❌
12. eeAa: ee, A_ ❌
13. EeAa: E_, A_ ✔️
14. Eeaa: E_, aa ❌
15. eeAa: ee, A_ ❌
16. eea a: ee, aa ❌

Now count the ✔️ (unattached, cleft): Let's count again. Wait, maybe a better way: For a dihybrid cross (EeAa × EeAa), the phenotypic ratio for dominant-dominant (E_A_), dominant-recessive (E_aa), recessive-dominant (eeA_), recessive-recessive (eeaa) is 9:3:3:1. Wait, that's the standard dihybrid ratio. Let's verify with our Punnett square.

In a dihybrid cross (two traits, both heterozygous parents), the phenotypic ratio is 9 (E_A_): 3 (E_aa): 3 (eeA_): 1 (eeaa). Let's check our Punnett square:

- **E_A_ (unattached, cleft)**: Let's count the number of cells with at least one E and at least one A. From the standard ratio, it's 9. Let's count our filled square:

Looking at each cell:

Row EA: 4 cells (EEAA, EEAa, EeAA, EeAa) – all E_A_ ✔️

Row Ea: First two cells (EEAa, EEAa) – wait no, row Ea: EA column (EEAa) – E_A_ ✔️, Ea column (EEaa) – E_aa ❌, eA column (EeAa) – E_A_ ✔️, ea column (Eeaa) – E_aa ❌. So row Ea: 2 cells (EEAa, EeAa) ✔️

Row eA: EA column (EeAA) – E_A_ ✔️, Ea column (EeAa) – E_A_ ✔️, eA column (eeAA) – eeA_ ❌, ea column (eeAa) – eeA_ ❌. So row eA: 2 cells (EeAA, EeAa) ✔️

Row ea: EA column (EeAa) – E_A_ ✔️, Ea column (Eeaa) – E_aa ❌, eA column (eeAa) – eeA_ ❌, ea column (eeaa) – eeaa ❌. So row ea: 1 cell (EeAa) ✔️

Now total E_A_: 4 (EA row) + 2 (Ea row) + 2 (eA row) + 1 (ea row) = 9. Perfect, matches the 9:3:3:1 ratio.

- **E_aa (unattached, no cleft)**: Genotypes with E_ and aa. Let's count:

Row Ea: Ea column (EEaa) and ea column (Eeaa) – 2 cells

Row ea: Ea column (Eeaa) – 1 cell? Wait no, standard ratio is 3. Let's check:

EEaa (row Ea, Ea column), Eeaa (row Ea, ea column), Eeaa (row ea, Ea column) – that's 3 cells. Yes, 3.

- **eeA_ (attached, cleft)**: Genotypes with ee and A_. Let's count:

Row eA: eA column (eeAA) and ea column (eeAa) – 2 cells

Row ea: eA column (eeAa) – 1 cell? Wait standard ratio is 3. eeAA (row eA, eA column), eeAa (row eA, ea column), eeAa (row ea, eA column) – 3 cells. Correct.

- **eeaa (attached, no cleft)**: Genotype eea a (row ea, ea column) – 1 cell. Correct.

Step3: Calculate Ratios

Now we can determine the ratios:

1. **Unattached earlobes and cleft chin (E_A_)**: Number of offspring with this trait is 9, total offspring is 16. So ratio is \( \frac{9}{16} \), so 9:16.

2. **Unattached earlobes and no cleft (E_aa)**: Number of offspring with this trait is 3, total is 16. Ratio is \( \frac{3}{16} \), so 3:16.

3. **Attached earlobes and cleft chin (eeA_)**: Number of offspring with this trait is 3, total is 16. Ratio is \( \frac{3}{16} \), so 3:16.

4. **Attached earlobes and no cleft (eeaa)**: Number of offspring with this trait is 1, total is 16. Ratio is \( \frac{1}{16} \), so 1:16.

Answer:

• Unattached earlobes and cleft chin: \( \boldsymbol{9:16} \)
• Unattached earlobes and no cleft: \( \boldsymbol{3:16} \)
• Attached earlobes and cleft chin: \( \boldsymbol{3:16} \)
• Attached earlobes and no cleft: \( \boldsymbol{1:16} \)