Answer :
Let's determine the offspring's genetic makeup using the Punnett square for a cross involving pea plants with the alleles for pod seed type. In this case, the allele "I" is dominant (inflated pod seed), and the allele "i" is recessive (constricted pod seed).
We can represent the cross as follows:
- Parent 1 (homozygous dominant): \( II \)
- Parent 2 (homozygous recessive): \( ii \)
Now, let's create the Punnett square step-by-step.
Step 1: Write the alleles of each parent along the sides of the square.
- Parent 1 will have alleles \( I \) and \( I \) along the top.
- Parent 2 will have alleles \( i \) and \( i \) along the side.
```
I I
+---+---+
i | | |
+---+---+
i | | |
+---+---+
```
Step 2: Fill in the squares by combining the alleles from each parent.
Each square combines one allele from the horizontal row (parent 2) and one from the vertical column (parent 1):
- Top-left cell: Combine \( I \) from the top with \( i \) from the side, resulting in \( Ii \).
- Top-right cell: Combine \( I \) from the top with \( i \) from the side, resulting in \( Ii \).
- Bottom-left cell: Combine \( I \) from the top with \( i \) from the side, resulting in \( Ii \).
- Bottom-right cell: Combine \( I \) from the top with \( i \) from the side, resulting in \( Ii \).
The completed Punnett square is:
```
I I
+---+---+
i | Ii| Ii|
+---+---+
i | Ii| Ii|
+---+---+
```
Step 3: Determine the genotype of the offspring.
Based on the Punnett square above:
- All offspring have the genotype \( Ii \).
Step 4: Count the number of homozygous dominant offspring.
- Homozygous dominant means having two dominant alleles (II).
- In this Punnett square, none of the cells have the genotype \( II \).
Therefore, there are 0 homozygous dominant (\( II \)) offspring in this cross. All the offspring will have an \( Ii \) genotype.
Summary:
- Genotypes of offspring: \( Ii \) (heterozygous)
- Number of homozygous dominant offspring: 0
We can represent the cross as follows:
- Parent 1 (homozygous dominant): \( II \)
- Parent 2 (homozygous recessive): \( ii \)
Now, let's create the Punnett square step-by-step.
Step 1: Write the alleles of each parent along the sides of the square.
- Parent 1 will have alleles \( I \) and \( I \) along the top.
- Parent 2 will have alleles \( i \) and \( i \) along the side.
```
I I
+---+---+
i | | |
+---+---+
i | | |
+---+---+
```
Step 2: Fill in the squares by combining the alleles from each parent.
Each square combines one allele from the horizontal row (parent 2) and one from the vertical column (parent 1):
- Top-left cell: Combine \( I \) from the top with \( i \) from the side, resulting in \( Ii \).
- Top-right cell: Combine \( I \) from the top with \( i \) from the side, resulting in \( Ii \).
- Bottom-left cell: Combine \( I \) from the top with \( i \) from the side, resulting in \( Ii \).
- Bottom-right cell: Combine \( I \) from the top with \( i \) from the side, resulting in \( Ii \).
The completed Punnett square is:
```
I I
+---+---+
i | Ii| Ii|
+---+---+
i | Ii| Ii|
+---+---+
```
Step 3: Determine the genotype of the offspring.
Based on the Punnett square above:
- All offspring have the genotype \( Ii \).
Step 4: Count the number of homozygous dominant offspring.
- Homozygous dominant means having two dominant alleles (II).
- In this Punnett square, none of the cells have the genotype \( II \).
Therefore, there are 0 homozygous dominant (\( II \)) offspring in this cross. All the offspring will have an \( Ii \) genotype.
Summary:
- Genotypes of offspring: \( Ii \) (heterozygous)
- Number of homozygous dominant offspring: 0
