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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 genotype of the following offspring.

| | EA | Ea | eA | ea |
|-----|-----|-----|-----|-----|
| EA | EEAA | EEAa | EeAA | EeAa |
| Ea | EEAa | EeAa | EeAa | Eeaa |
| eA | EeAA | EeAa | eeAA | eeAa |
| ea | EeAa | Eeaa | eeAa | eeaa |

Offspring genotypes:

1. EeAa
2. EEAa
3. EeAa
4. EeAA
5. eeAa

Sagot :

GinnyAnswer
To determine the genotypes of the offspring from the given Punnett square, we first need to fill in the blanks in the table by following the Mendelian inheritance rules for a dihybrid cross (where both parents are heterozygous for both traits).

The specific parents' genotypes are:
- Both parents have the genotype "EeAa".

The possible gametes formed by these parents are:
- EA, Ea, eA, ea.

Let’s fill in the table based on the cross of these gametes:

Here’s the detailed step-by-step solution:

1. EA x EA = EEAA (already provided in table)
2. EA x Ea = EEAa (already provided in table)
3. EA x eA = EeAA (already provided in table)
4. EA x ea = EeAa (already provided in table)

Moving on to the next row:

5. Ea x EA = EEAa (already provided in table)
6. Ea x Ea = EEaa
7. Ea x eA = EeAa (already provided in table)
8. Ea x ea = Eeaa (already provided in table)

Continuing to the next row:

9. eA x EA = EeAA (already provided in table)
10. eA x Ea = EeAa (already provided in table)
11. eA x eA = eeAA
12. eA x ea = eeAa (already provided in table)

For the last row:

13. ea x EA = EeAa (already provided in table)
14. ea x Ea = Eeaa (already provided in table)
15. ea x eA = eeAa (already provided in table)
16. ea x ea = eeaa (already provided in table)

Now, we’ll extract the genotypes of the specific positions indicated in your question.

1. EeAa (Already given)
2. From first row, second column: EEAa (EA x Ea)
3. From first column, second row: EeAA (Ea x EA)
4. From second column, third row: EeAa (eA x Ea)
5. From last row, last column: eeaa (ea x ea)

So the genotypes of the offspring in the specified positions are:

1. EeAa
2. EEAa
3. EeAA
4. EeAa
5. eeaa

All positions are now defined as:
[tex]\[ \begin{tabular}{|c|c|c|c|} \hline EA & Ea & eA & ea \\ \hline EEAA & EEAa & EeAA & EeAa \\ \hline EEAa & EEaa & EeAa & Eeaa \\ \hline EeAA & EeAa & eeAA & eeAa \\ \hline EeAa & Eeaa & eeAa & eeaa \\ \hline \end{tabular} \][/tex]

Note that genotypes corresponding to position numbers are:

1. EeAa [tex]\(\sim\)[/tex]
2. EEAa
3. EeAA
4. EeAa
5. eeaa
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