name: jarret date: feb 0 5 2026
genetic crosses that involve 2 traits - floppy eared bunnies
in rabbits, black hair is dominant to brown hair. also in rabbits, long straight ears are dominant to floppy ears.
these letters represent the genotypes and phenotypes of the rabbits:
bb = black nose ee = long ears
bb = black nose ee = long ears
bb = pink nose ee = floppy ears
1. a male rabbit with the genotype bbee is crossed with a female rabbit with the genotype bbee the square is set up below. fill it out and determine the phenotypes and proportions in the offspring.
be be be be
be
be
be
be
how many out of 16 have:
black noses and long ears? ______
black noses and floppy ears? ______
pink noses and long ears? ______
pink noses and floppy ears? ______

Question

name: jarret date: feb 0 5 2026
genetic crosses that involve 2 traits - floppy eared bunnies
in rabbits, black hair is dominant to brown hair. also in rabbits, long straight ears are dominant to floppy ears.
these letters represent the genotypes and phenotypes of the rabbits:
bb = black nose  ee = long ears
bb = black nose  ee = long ears
bb = pink nose  ee = floppy ears
1. a male rabbit with the genotype bbee is crossed with a female rabbit with the genotype bbee the square is set up below. fill it out and determine the phenotypes and proportions in the offspring.
be  be  be  be
be
be
be
be
how many out of 16 have:
black noses and long ears? ______
black noses and floppy ears? ______
pink noses and long ears? ______
pink noses and floppy ears? ______

Accepted Answer

# Explanation:
## Step1: Determine gametes and fill Punnett square
Male gametes: All $ Be $ (since genotype is $ BBee $, so $ B $ from $ BB $, $ e $ from $ ee $).
Female gametes: From $ bbEe $, gametes are $ bE $ and $ be $ (each twice, as per the Punnett square rows: $ bE, be, bE, be $).

Now fill each cell by combining male ($ Be $) and female gametes:
- $ bE \times Be = BbEe $
- $ be \times Be = Bbee $
- Repeat for all rows/columns.

The filled Punnett square cells (16 total) will be:
- $ BbEe $: 8 times (rows 1,3: 2 rows × 4 columns = 8)
- $ Bbee $: 8 times (rows 2,4: 2 rows × 4 columns = 8)? Wait, no—wait, female gametes are $ bE, be, bE, be $ (4 gametes), male is $ Be, Be, Be, Be $ (4 gametes). So:

Row 1 ($ bE $): $ BbEe, BbEe, BbEe, BbEe $
Row 2 ($ be $): $ Bbee, Bbee, Bbee, Bbee $
Row 3 ($ bE $): $ BbEe, BbEe, BbEe, BbEe $
Row 4 ($ be $): $ Bbee, Bbee, Bbee, Bbee $

Wait, no—wait, female has 4 gametes: $ bE, be, bE, be $ (so two $ bE $, two $ be $). Male has four $ Be $. So total cells: 4×4=16.

Now, analyze phenotypes:
- Black nose ($ B\_ $) and long ears ($ E\_ $): Genotype $ BbEe $ (since $ B $ is dominant, $ E $ is dominant). Count $ BbEe $: rows 1 and 3, 4 cells each, so 8.
- Black nose ($ B\_ $) and floppy ears ($ ee $): Genotype $ Bbee $ ( $ B $ dominant, $ ee $ recessive). Count $ Bbee $: rows 2 and 4, 4 cells each, so 8? Wait, no—wait, $ Ee $ gives long ears, $ ee $ gives floppy. Wait, $ BbEe $: $ Ee $ → long ears. $ Bbee $: $ ee $ → floppy ears. $ bbEe $ would be pink nose, but wait—male is $ BB $, so all offspring have $ Bb $ (black nose). Wait, male is $ BB $, so all offspring get $ B $, so nose color: all $ Bb $ (black nose). Wait, that’s a mistake earlier! Male is $ BB $, so any offspring with $ B $ (from male) and $ b $ (from female, $ bb $) → $ Bb $ (black nose). So all offspring have black nose? Wait, no—female is $ bb $, so offspring must get $ b $ from female, and $ B $ from male → $ Bb $ (black nose). So nose color: all black. Then ear type: $ Ee $ (long) or $ ee $ (floppy).

Wait, this changes things! Let’s re-express:

Male genotype: $ BBee $ → gametes: $ B e $ (all, since $ BB $ gives $ B $, $ ee $ gives $ e $).
Female genotype: $ bbEe $ → gametes: $ b E $ and $ b e $ (since $ bb $ gives $ b $, $ Ee $ gives $ E $ or $ e $). So female gametes: $ bE $ (50%) and $ be $ (50%), so in 4 gametes: 2 $ bE $, 2 $ be $.

So offspring genotypes:
- $ B e \times b E = Bb Ee $ (black, long ears)
- $ B e \times b e = Bb ee $ (black, floppy ears)

Now, count each:
- $ Bb Ee $: number of times $ bE $ is used. Female has 2 $ bE $ and 2 $ be $ in 4 gametes. Male has 4 $ Be $. So:
  - $ bE $ (2 gametes) × $ Be $ (4) → 2×4=8? Wait, no—Punnett square is 4 (male) × 4 (female) = 16 cells. Female gametes: positions 1: $ bE $, 2: $ be $, 3: $ bE $, 4: $ be $. Male: 1-4: $ Be $.

So cell 1 (row1, col1): $ BbEe $
cell 2 (row1, col2): $ BbEe $
cell 3 (row1, col3): $ BbEe $
cell 4 (row1, col4): $ BbEe $
row2 (be):
cell5: $ Bbee $
cell6: $ Bbee $
cell7: $ Bbee $
cell8: $ Bbee $
row3 (bE):
cell9: $ BbEe $
cell10: $ BbEe $
cell11: $ BbEe $
cell12: $ BbEe $
row4 (be):
cell13: $ Bbee $
cell14: $ Bbee $
cell15: $ Bbee $
cell16: $ Bbee $

Now, count phenotypes:
1. Black nose (all, since $ Bb $) and long ears ($ Ee $): $ BbEe $ cells: rows 1,3 (8 cells: 4+4=8? Wait, row1: 4, row3:4 → 8.
2. Black nose (all) and floppy ears ($ ee $): $ Bbee $ cells: rows 2,4 (4+4=8? Wait, row2:4, row4:4 → 8.
Wait, but the problem asks for “pink noses”—but male is $ BB $, so offspring can only be $ Bb $ (black nose). So pink noses ( $ bb $) are impossible? Wait, that can’t be. Wait, no—wait, male is $ BBee $, female is $ bbEe $. So for nose color: male gives $ B $, female gives $ b $ → all offspring $ Bb $ (black nose). So pink noses ( $ bb $): 0. Long ears: $ Ee $ (since $ EE $ not possible, female is $ Ee $, male is $ ee $ → offspring $ Ee $ or $ ee $). Wait, male is $ ee $, so ear genotype: $ Ee $ (from $ e $ male and $ E $ female) or $ ee $ (from $ e $ male and $ e $ female).

Wait, I made a mistake earlier: male is $ ee $, so ear allele from male is $ e $. Female is $ Ee $, so ear alleles from female: $ E $ or $ e $. So offspring ear genotypes: $ Ee $ (long) or $ ee $ (floppy).

Nose genotype: male $ BB $ → $ B $, female $ bb $ → $ b $ → all $ Bb $ (black nose). So pink noses ( $ bb $): 0. Long ears ( $ Ee $): when female gives $ E $ → $ Ee $. Floppy ears ( $ ee $): when female gives $ e $ → $ ee $.

Female gametes: $ bE $ ( $ b $ nose, $ E $ ear) and $ be $ ( $ b $ nose, $ e $ ear), each 50% (2 of each in 4 gametes).

So:
- $ BbEe $ (black, long): number of $ bE $ gametes × male gametes. Female has 2 $ bE $, male 4 $ Be $ → 2×4=8? Wait, no—Punnett square is 4×4=16. Female gametes: positions 1: $ bE $, 2: $ be $, 3: $ bE $, 4: $ be $. So:

- $ BbEe $: cells 1,2,3,4 (row1: $ bE $) → 4 cells; cells 9,10,11,12 (row3: $ bE $) → 4 cells. Total: 8.
- $ Bbee $: cells 5,6,7,8 (row2: $ be $) → 4 cells; cells 13,14,15,16 (row4: $ be $) → 4 cells. Total: 8.

Now, phenotypes:
- Black nose (all) and long ears ($ Ee $): $ BbEe $ → 8.
- Black nose (all) and floppy ears ($ ee $): $ Bbee $ → 8.
- Pink noses ( $ bb $): 0 (since all $ Bb $).
- Pink noses and long ears: 0.
- Pink noses and floppy ears: 0.

But the problem has blanks for pink noses, so maybe I misread the male genotype? Wait, the male is $ BBee $, female $ bbEe $. Wait, maybe the male is $ Bbee $? No, the problem says $ BBee $. Wait, let’s check again:

“In rabbits, black hair is dominant to brown hair. Also in rabbits, long straight ears are dominant to floppy ears. These letters represent the genotypes and phenotypes of the rabbits: BB = black nose, Bb = black nose, bb = pink nose; EE = long ears, Ee = long ears, ee = floppy ears.”

Ah! Wait, “black hair” but the letters are for nose? Wait, typo? Maybe “black hair” is $ BB $, “brown hair” $ bb $, but the letters are for nose? Or maybe it’s a typo, and “black hair” is $ BB $, “brown hair” $ bb $, but the table uses “black nose” for $ BB $, etc. Maybe it’s a typo, and “hair” is “nose”? Or maybe the problem has a mistake. But assuming the table is correct: $ BB $, $ Bb $ = black nose; $ bb $ = pink nose. So male is $ BBee $ (black nose, floppy ears), female $ bbEe $ (pink nose, long ears).

So for nose color: male $ BB $ → $ B $, female $ bb $ → $ b $ → offspring $ Bb $ (black nose). So pink nose ( $ bb $): 0.

Ear type: male $ ee $ → $ e $, female $ Ee $ → $ E $ or $ e $. So offspring ear: $ Ee $ (long) or $ ee $ (floppy).

So:
- Black nose ( $ Bb $) and long ears ( $ Ee $): $ BbEe $. How many? Female gametes: $ bE $ (2) and $ be $ (2). Male: $ Be $ (4). So $ BbEe $ is when female gives $ E $ ( $ bE $): 2 (female $ bE $) × 4 (male $ Be $) = 8? Wait, no—Punnett square is 4×4=16. Female gametes: 4 ( $ bE, be, bE, be $ ), male: 4 ( $ Be, Be, Be, Be $ ). So:

Each $ bE $ (2 gametes) × $ Be $ (4) → 2×4=8 cells of $ BbEe $.
Each $ be $ (2 gametes) × $ Be $ (4) → 2×4=8 cells of $ Bbee $.

Thus:
- Black nose and long ears ( $ BbEe $): 8
- Black nose and floppy ears ( $ Bbee $): 8
- Pink nose ( $ bb $): 0, so pink nose and long ears: 0; pink nose and floppy ears: 0

But the problem asks for “how many out of 16”, so:

Black noses and long ears: 8
Black noses and floppy ears: 8
Pink noses and long ears: 0
Pink noses and floppy ears: 0

But this seems odd (pink nose offspring impossible). Maybe the male is $ Bbee $ instead of $ BBee $? Let’s check the problem again: “A male rabbit with the genotype BBee is crossed with a female rabbit with the genotype bbEe”. So as per problem, male is $ BBee $, female $ bbEe $.

Alternatively, maybe I messed up the gametes. Let’s list all 16 cells:

Row 1 ( $ bE $):
1. $ BbEe $
2. $ BbEe $
3. $ BbEe $
4. $ BbEe $

Row 2 ( $ be $):
5. $ Bbee $
6. $ Bbee $
7. $ Bbee $
8. $ Bbee $

Row 3 ( $ bE $):
9. $ BbEe $
10. $ BbEe $
11. $ BbEe $
12. $ BbEe $

Row 4 ( $ be $):
13. $ Bbee $
14. $ Bbee $
15. $ Bbee $
16. $ Bbee $

Now, count each phenotype:
- Black nose ( $ Bb $) and long ears ( $ Ee $): cells 1-4, 9-12 → 8 cells.
- Black nose ( $ Bb $) and floppy ears ( $ ee $): cells 5-8, 13-16 → 8 cells.
- Pink nose ( $ bb $): 0 (no cells with $ bb $).
- So pink nose and long ears: 0; pink nose and floppy ears: 0.

# Answer:
Black noses and long ears: 8
Black noses and floppy ears: 8
Pink noses and long ears: 0
Pink noses and floppy ears: 0

(But note: If there’s a typo and male is $ Bbee $, the answer would differ. But based on given genotypes, this is the result.)

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QUESTION IMAGE

Question

Explanation:

Step1: Determine gametes and fill Punnett square

Answer: