Answer :
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
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
