Question

cross a parent that is homozygous recessive for both traits (bbee) with a dihybrid (bbee). first, draw a punnett square in your notes. for each offspring in the punnett square, determine the phenotype. in other words, what are the fur color and eye color? finally, use that information to fill in the predicted fraction of each phenotype in the data table below. data table with phenotypes: black fur and black eyes, black fur and red eyes, white fur and black eyes, white fur and red eyes, and fraction columns with some pre - filled fractions like /16, /16, /16, /16

Explanation:

Step1: Identify Parent Genotypes

One parent is homozygous recessive for both traits: \( bbee \) (so it can only produce \( be \) gametes). The other parent is \( BbEe \), which can produce gametes: \( BE \), \( Be \), \( bE \), \( be \).

Step2: Create Punnett Square

| | \( BE \) | \( Be \) | \( bE \) | \( be \) |
|-------|---------|---------|---------|---------|
| \( be \) | \( BbEe \) | \( Bbee \) | \( bbEe \) | \( bbee \) |
| \( be \) | \( BbEe \) | \( Bbee \) | \( bbEe \) | \( bbee \) |
| \( be \) | \( BbEe \) | \( Bbee \) | \( bbEe \) | \( bbee \) |
| \( be \) | \( BbEe \) | \( Bbee \) | \( bbEe \) | \( bbee \) |

Wait, actually, for a \( BbEe \times bbee \) cross, the Punnett square has 4 rows (from \( bbee \): all \( be \)) and 4 columns (from \( BbEe \): \( BE, Be, bE, be \)). So 16 cells.

Step3: Determine Phenotypes

- \( BbEe \): Black fur, Black eyes (since \( B \) is dominant for fur, \( E \) for eyes)
- \( Bbee \): Black fur, Red eyes ( \( B \) dominant, \( e \) recessive for eyes)
- \( bbEe \): White fur, Black eyes ( \( b \) recessive for fur, \( E \) dominant)
- \( bbee \): White fur, Red eyes ( \( b \) and \( e \) recessive)

Step4: Count Each Phenotype

- Black fur, Black eyes (\( BbEe \)): 4 cells (since \( BbEe \) occurs when \( BE \times be \), and there are 4 \( be \) gametes from \( bbee \), so 4 times? Wait, no: \( BbEe \) is 1/4 of \( BbEe \)'s gametes, times 4 (from \( bbee \)): 4.
- Black fur, Red eyes (\( Bbee \)): 4 cells ( \( Be \times be \): 4 times)
- White fur, Black eyes (\( bbEe \)): 4 cells ( \( bE \times be \): 4 times)
- White fur, Red eyes (\( bbee \)): 4 cells ( \( be \times be \): 4 times) Wait, no, that can't be. Wait, \( BbEe \) produces 4 gametes: \( BE, Be, bE, be \) (each with 25% chance). \( bbee \) produces only \( be \). So each combination is 4 times (since 4 rows of \( be \)). So:

- \( BbEe \): Black fur, Black eyes: 4/16 = 1/4? Wait, no, the Punnett square is 4x4=16 cells. Let's list each cell:

1. \( BE \times be = BbEe \) (Black, Black)
2. \( Be \times be = Bbee \) (Black, Red)
3. \( bE \times be = bbEe \) (White, Black)
4. \( be \times be = bbee \) (White, Red)

Since there are 4 rows (each \( be \)), each of these 4 genotypes appears 4 times. So:

- Black fur, Black eyes: 4/16 = 1/4 (4 cells)
- Black fur, Red eyes: 4/16 = 1/4 (4 cells)
- White fur, Black eyes: 4/16 = 1/4 (4 cells)
- White fur, Red eyes: 4/16 = 1/4 (4 cells)

But the table in the problem has fractions like /16. Let's check the table:

First row: Black Fur and Black Eyes: Fraction? Let's see the correct counts. Wait, maybe I made a mistake. Wait, \( BbEe \) is \( B \) (black fur), \( E \) (black eyes). \( Bbee \) is \( B \) (black fur), \( e \) (red eyes). \( bbEe \) is \( b \) (white fur), \( E \) (black eyes). \( bbee \) is \( b \) (white fur), \( e \) (red eyes). So each phenotype:

- Black, Black: \( BbEe \): 4 cells (since \( BbEe \) has 4 gametes, each with \( be \) gives 4)
- Black, Red: \( Bbee \): 4 cells
- White, Black: \( bbEe \): 4 cells
- White, Red: \( bbee \): 4 cells

So each fraction is 4/16. Wait, but the table in the problem has:

- Black Fur and Black Eyes: /16 (should be 4)
- Black Fur and Red Eyes: /16 (4)
- White Fur and Black Eyes: /16 (4)
- White Fur and Red Eyes: /16 (4)

Wait, maybe the parent is \( BbEe \) (dihybrid) and \( bbee \) (homozygous recessive). So the Punnett square is 4x4=16. Each of the four genotypes ( \( BbEe, Bbee, bbEe, bbee \)) occurs 4 times. So the fractions are 4/16 for each.

So filling the table:

- Black Fur and Black Eyes: 4/16
- Black Fur and Red Eyes: 4/16
- White Fur and Black Eyes: 4/16
- White Fur and Red Eyes: 4/16

Answer:

For the data table:

• Black Fur and Black Eyes: \( \frac{4}{16} \)
• Black Fur and Red Eyes: \( \frac{4}{16} \)
• White Fur and Black Eyes: \( \frac{4}{16} \)
• White Fur and Red Eyes: \( \frac{4}{16} \)