Answer :
Absolutely! Let's break down the genetic cross of two heterozygous fruit flies for body color (Brown, B, is dominant to black, b) and eye color (Red, E, is dominant to brown, e). The genotypes of both parent fruit flies are heterozygous for both traits, i.e., \( BbEe \).
### Step 1: Establish the Genotypes of Gametes
Each parent can produce four types of gametes:
1. BE
2. Be
3. bE
4. be
### Step 2: Create the Punnett Square
We use these gametes to create a Punnett square. Since each parent produces four types of gametes, the Punnett square will have 16 possible combinations (4 from each parent):
[tex]\[ \begin{array}{|c|c|c|c|c|} \hline & BE & Be & bE & be \\ \hline BE & BBEE & BBEe & BbEE & BbEe \\ \hline Be & BBEe & BBee & BbEe & Bbee \\ \hline bE & BbEE & BbEe & bbEE & bbEe \\ \hline be & BbEe & Bbee & bbEe & bbee \\ \hline \end{array} \][/tex]
### Step 3: Determine the Phenotypes
We'll now classify these genotypes into the respective phenotypes:
- \(BB\) or \(Bb\) genotypes will produce brown bodies.
- \(bb\) genotypes will produce black bodies.
- \(EE\) or \(Ee\) genotypes will produce red eyes.
- \(ee\) genotypes will produce brown eyes.
Let's list the phenotypes in each cell of the Punnett square:
1. Brown body and red eyes ( \(B_E_ \) ):
[tex]\[ BBEE, BBEe, BbEE, BbEe, BBEe, BbEE, BbEe, BbEe, BbEe, BbEE, BbEe, Bbee \][/tex]
There are 9 occurrences.
2. Brown body and brown eyes ( \( B_ee \) ):
[tex]\[ BBee, Bbee, Bbee \][/tex]
There are 3 occurrences.
3. Black body and red eyes ( \( bbE_ \) ):
[tex]\[ bbEE, bbEe, bbEe \][/tex]
There are 3 occurrences.
4. Black body and brown eyes ( \( bbee \) ):
[tex]\[ bbee \][/tex]
There is 1 occurrence.
### Step 4: Count and Verify
Counting the phenotypic ratios:
1. Brown body and red eyes: 9 occurrences
2. Brown body and brown eyes: 3 occurrences
3. Black body and red eyes: 3 occurrences
4. Black body and brown eyes: 1 occurrence
So, the ratio of offspring with each phenotype is:
- Brown body and red eyes: 9
- Brown body and brown eyes: 3
- Black body and red eyes: 3
- Black body and brown eyes: 1
These totals sum to 16, confirming the ratios are correct based on the expected 16 offspring combinations from a dihybrid cross.
### Step 1: Establish the Genotypes of Gametes
Each parent can produce four types of gametes:
1. BE
2. Be
3. bE
4. be
### Step 2: Create the Punnett Square
We use these gametes to create a Punnett square. Since each parent produces four types of gametes, the Punnett square will have 16 possible combinations (4 from each parent):
[tex]\[ \begin{array}{|c|c|c|c|c|} \hline & BE & Be & bE & be \\ \hline BE & BBEE & BBEe & BbEE & BbEe \\ \hline Be & BBEe & BBee & BbEe & Bbee \\ \hline bE & BbEE & BbEe & bbEE & bbEe \\ \hline be & BbEe & Bbee & bbEe & bbee \\ \hline \end{array} \][/tex]
### Step 3: Determine the Phenotypes
We'll now classify these genotypes into the respective phenotypes:
- \(BB\) or \(Bb\) genotypes will produce brown bodies.
- \(bb\) genotypes will produce black bodies.
- \(EE\) or \(Ee\) genotypes will produce red eyes.
- \(ee\) genotypes will produce brown eyes.
Let's list the phenotypes in each cell of the Punnett square:
1. Brown body and red eyes ( \(B_E_ \) ):
[tex]\[ BBEE, BBEe, BbEE, BbEe, BBEe, BbEE, BbEe, BbEe, BbEe, BbEE, BbEe, Bbee \][/tex]
There are 9 occurrences.
2. Brown body and brown eyes ( \( B_ee \) ):
[tex]\[ BBee, Bbee, Bbee \][/tex]
There are 3 occurrences.
3. Black body and red eyes ( \( bbE_ \) ):
[tex]\[ bbEE, bbEe, bbEe \][/tex]
There are 3 occurrences.
4. Black body and brown eyes ( \( bbee \) ):
[tex]\[ bbee \][/tex]
There is 1 occurrence.
### Step 4: Count and Verify
Counting the phenotypic ratios:
1. Brown body and red eyes: 9 occurrences
2. Brown body and brown eyes: 3 occurrences
3. Black body and red eyes: 3 occurrences
4. Black body and brown eyes: 1 occurrence
So, the ratio of offspring with each phenotype is:
- Brown body and red eyes: 9
- Brown body and brown eyes: 3
- Black body and red eyes: 3
- Black body and brown eyes: 1
These totals sum to 16, confirming the ratios are correct based on the expected 16 offspring combinations from a dihybrid cross.
