practice problem - hedgehogs!
the allele for mega spikes (m) is dominant over the allele for normal spikes.
the allele for a red nose (r) is dominant over the allele for a yellow nose.
parent 1 is homozygous recessive for spikes and heterozygous for nose color.
parent 2 is heterozygous for spikes and homozygous dominant for nose color.
work out the possible offspring genotypes and phenotypes and give the expected probability of each phenotype.
step 1: work out the parent & the alleles in each ----.
parent 1 genotype:
parent 1 gametes:
parent 2 genotype:
parent 2 gametes:
step 2: write the ---- gametes along the top and ---- of the cross.
step 3: work out offspring ---- by reading across and ----.
step 4: work out the ---- for each genotype. show this by drawing the ---- & coloring in the ----.
step 5: work out the ---- of each offspring phenotype.
mega spikes & red nose = ----/16
normal spikes & red nose = ----/16
mega spikes & yellow nose = ----/16
normal spikes & yellow nose = ----/16

Question

Accepted Answer

To solve this genetics problem with hedgehog traits, we follow these steps:

### Step 1: Determine Parent Genotypes and Gametes
- **Parent 1**: Homozygous recessive for spikes (so genotype $ mm $) and heterozygous for nose color (since $ R $ is dominant for red, $ r $ for yellow; genotype $ Rr $). Thus, Parent 1 genotype: $ mmRr $.
  - Gametes: Parent 1 can produce $ mR $ and $ mr $ (since $ mm $ gives only $ m $, and $ Rr $ gives $ R $ or $ r $).
- **Parent 2**: Heterozygous for spikes (genotype $ Mm $) and homozygous dominant for nose color (genotype $ RR $). Thus, Parent 2 genotype: $ MmRR $.
  - Gametes: Parent 2 can produce $ MR $ and $ mR $ (since $ Mm $ gives $ M $ or $ m $, and $ RR $ gives only $ R $).

### Step 2: Set Up the Punnett Square
Parent 1 gametes: $ mR $, $ mr $ (2 types).
Parent 2 gametes: $ MR $, $ mR $ (2 types). Wait, no—wait, the Punnett square in the image has 4 columns (Parent 2 gametes) and 4 rows (Parent 1 gametes)? Wait, no, let’s recheck. Wait, the problem says Parent 1 is homozygous recessive for spikes ($ mm $) and heterozygous for nose ($ Rr $)—so gametes: $ mR $, $ mr $ (2 gametes? But the Punnett square has 4 rows? Wait, maybe I misread. Wait, the image shows a 4x4 Punnett square (4 rows, 4 columns), so each parent has 4 gametes? Wait, no—maybe Parent 1 is $ mmRr $ (so for spikes: $ mm $ → only $ m $; for nose: $ Rr $ → $ R $ or $ r $)—so 2 gametes: $ mR $, $ mr $. Parent 2 is $ MmRR $ (spikes: $ Mm $ → $ M $ or $ m $; nose: $ RR $ → $ R $)—so 2 gametes: $ MR $, $ mR $. But the Punnett square is 4x4, so maybe there’s a mistake, or maybe Parent 1 is $ mmRr $ (producing 2 gametes) and Parent 2 is $ MmRr $? Wait, no—the problem states: *“Parent 1 is homozygous recessive for spikes and heterozygous for nose color. Parent 2 is heterozygous for spikes and homozygous dominant for nose color.”*

So correct gametes:
- Parent 1 ($ mmRr $): $ mR $, $ mr $ (2 gametes, but Punnett square has 4 rows? Maybe a typo, or maybe Parent 1 is $ mmRr $ (so 2 gametes, but the square is 4x4—maybe the problem expects 4 gametes? Wait, no—let’s proceed with the given Punnett square structure (4x4).

### Step 3: Fill the Punnett Square
Let’s list all combinations:

| Parent 2 \ Parent 1 | $ mR $ | $ mR $ | $ mr $ | $ mr $ |
|---------------------|---------|---------|---------|---------|
| $ MR $            | $ MmRR $ | $ MmRR $ | $ MmRr $ | $ MmRr $ |
| $ MR $            | $ MmRR $ | $ MmRR $ | $ MmRr $ | $ MmRr $ |
| $ mR $            | $ mmRR $ | $ mmRR $ | $ mmRr $ | $ mmRr $ |
| $ mR $            | $ mmRR $ | $ mmRR $ | $ mmRr $ | $ mmRr $ |

Wait, no—this is confusing. Wait, maybe Parent 1 has 4 gametes (e.g., if Parent 1 is $ mmRr $, but that would only produce 2 gametes. Alternatively, maybe Parent 1 is $ mmRr $ (spikes: $ mm $, nose: $ Rr $) and Parent 2 is $ MmRR $ (spikes: $ Mm $, nose: $ RR $), and the Punnett square is 4x4 because we consider independent assortment (but for two genes, a 4x4 square is for two heterozygous genes, but here one gene is homozygous in each parent).

Alternatively, let’s use the traits:
- **Spikes**: $ M $ (mega) dominant over $ m $ (normal).
- **Nose color**: $ R $ (red) dominant over $ r $ (yellow).

### Step 4: Analyze Phenotypes
We need to find the fraction of offspring with each phenotype:

1. **Mega spikes & red nose**: Genotype has $ M $ (for spikes) and $ R $ (for nose).
   - From Punnett square, offspring with $ M $ (spikes) and $ R $ (nose): Let’s count.
   - Parent 1: $ mmRr $ (spikes: $ mm $ → normal; nose: $ Rr $ → red/yellow).
   - Parent 2: $ MmRR $ (spikes: $ Mm $ → mega/normal; nose: $ RR $ → red).
   - Offspring genotypes:
     - $ MmRR $ (mega spikes, red nose)
     - $ MmRr $ (mega spikes, red nose)
     - $ mmRR $ (normal spikes, red nose)
     - $ mmRr $ (normal spikes, red nose)
   Wait, no—wait, Parent 1 is $ mm $ (spikes: normal), Parent 2 is $ Mm $ (spikes: mega/normal). So:
   - Spikes: $ Mm $ (mega) or $ mm $ (normal).
   - Nose: $ RR $ or $ Rr $ (both red, since $ R $ is dominant; $ rr $ would be yellow, but Parent 2 is $ RR $, so all offspring have $ R $ from Parent 2, so nose color is red (since $ R $ is dominant). Wait, Parent 1 is $ Rr $ (nose: red/yellow), Parent 2 is $ RR $ (nose: red). So offspring nose genotype: $ RR $ or $ Rr $ (both red). So **all offspring have red noses** (since $ R $ is dominant, and Parent 2 is $ RR $, so no $ rr $ possible).

Now spikes:
   - Parent 1: $ mm $ (normal) → gives $ m $.
   - Parent 2: $ Mm $ (mega/normal) → gives $ M $ or $ m $.
   So offspring spikes genotype: $ Mm $ (mega) or $ mm $ (normal), each with 50% chance.

But the Punnett square in the image has 4x4=16 cells (since it’s 4 rows, 4 columns). Wait, maybe the problem has a typo, and Parent 1 is $ mmRr $ (producing 2 gametes) and Parent 2 is $ MmRr $ (producing 2 gametes), but the square is 4x4. Alternatively, let’s assume the Punnett square has 16 cells (4x4), so each parent has 4 gametes (e.g., Parent 1: $ mR, mR, mr, mr $; Parent 2: $ MR, MR, mR, mR $)—so repeated gametes to make 4 each.

### Step 5: Calculate Phenotype Ratios
Since all offspring have red noses (because Parent 2 is $ RR $, so no $ rr $ possible), the nose color is always red. Now spikes:

- **Mega spikes ($ Mm $)**: From Parent 2’s $ M $ and Parent 1’s $ m $ → $ Mm $.
- **Normal spikes ($ mm $)**: From Parent 2’s $ m $ and Parent 1’s $ m $ → $ mm $.

With the Punnett square (16 cells):

- Mega spikes ( $ Mm $): How many? Parent 2’s $ M $ (2 gametes) × Parent 1’s $ m $ (2 gametes) → 4 cells? Wait, no—let’s count:

If Parent 1 gametes: $ mR, mR, mr, mr $ (4 gametes: 2 $ mR $, 2 $ mr $).
Parent 2 gametes: $ MR, MR, mR, mR $ (4 gametes: 2 $ MR $, 2 $ mR $).

Punnett square cells:
1. $ mR $ (P1) × $ MR $ (P2): $ MmRR $ (mega, red)
2. $ mR $ (P1) × $ MR $ (P2): $ MmRR $ (mega, red)
3. $ mr $ (P1) × $ MR $ (P2): $ MmRr $ (mega, red)
4. $ mr $ (P1) × $ MR $ (P2): $ MmRr $ (mega, red)
5. $ mR $ (P1) × $ mR $ (P2): $ mmRR $ (normal, red)
6. $ mR $ (P1) × $ mR $ (P2): $ mmRR $ (normal, red)
7. $ mr $ (P1) × $ mR $ (P2): $ mmRr $ (normal, red)
8. $ mr $ (P1) × $ mR $ (P2): $ mmRr $ (normal, red)
Wait, no—this is 8 cells? But the Punnett square is 4x4 (16 cells). Oh, I see—each parent has 4 gametes (2 types, each repeated twice), so 4×4=16 cells. So:

- Mega spikes ( $ Mm $): Cells 1–4 (4 cells? No, 1–4: $ MmRR, MmRR, MmRr, MmRr $) → 4 cells? Wait, no, 1–4: 4 cells (mega spikes). Then cells 5–8: $ mmRR, mmRR, mmRr, mmRr $ (normal spikes). Wait, but that’s 8 cells? No, the Punnett square in the image has 4 rows and 4 columns (16 cells), so maybe each parent has 4 gametes (e.g., Parent 1: $ mR, mR, mr, mr $ (4 gametes), Parent 2: $ MR, MR, mR, mR $ (4 gametes)). So:

- **Mega spikes & red nose**: Cells with $ Mm $ (spikes) and $ R $ (nose). From the 16 cells:
  - $ MmRR $: 4 cells (2 $ mR $ × 2 $ MR $)? Wait, no—let’s list all 16 cells:

Row 1 (P1: $ mR $) × Column 1 (P2: $ MR $): $ MmRR $
Row 1 × Column 2 (P2: $ MR $): $ MmRR $
Row 1 × Column 3 (P2: $ mR $): $ mmRR $
Row 1 × Column 4 (P2: $ mR $): $ mmRR $

Row 2 (P1: $ mR $) × Column 1: $ MmRR $
Row 2 × Column 2: $ MmRR $
Row 2 × Column 3: $ mmRR $
Row 2 × Column 4: $ mmRR $

Row 3 (P1: $ mr $) × Column 1: $ MmRr $
Row 3 × Column 2: $ MmRr $
Row 3 × Column 3: $ mmRr $
Row 3 × Column 4: $ mmRr $

Row 4 (P1: $ mr $) × Column 1: $ MmRr $
Row 4 × Column 2: $ MmRr $
Row 4 × Column 3: $ mmRr $
Row 4 × Column 4: $ mmRr $

Now, count phenotypes:

1. **Mega spikes & red nose**: Genotypes $ MmRR $, $ MmRr $.
   - $ MmRR $: Rows 1,2 × Columns 1,2 → 4 cells.
   - $ MmRr $: Rows 3,4 × Columns 1,2 → 4 cells.
   - Total: $ 4 + 4 = 8 $ cells.

2. **Normal spikes & red nose**: Genotypes $ mmRR $, $ mmRr $.
   - $ mmRR $: Rows 1,2 × Columns 3,4 → 4 cells.
   - $ mmRr $: Rows 3,4 × Columns 3,4 → 4 cells.
   - Total: $ 4 + 4 = 8 $ cells.

Wait, but the problem asks for:
- Mega spikes & red nose = ____/16
- Mega spikes & yellow nose = ____/16 (but Parent 2 is $ RR $, so no $ rr $ → 0/16)
- Normal spikes & red nose = ____/16
- Normal spikes & yellow nose = ____/16 (0/16)

But the problem’s left side says:
“Mega spikes & red nose = ___/16
Mega spikes & yellow nose = ___/16
Normal spikes & red nose = ___/16
Normal spikes & yellow nose = ___/16”

Wait, but Parent 2 is $ RR $ (homozygous dominant for nose), so all offspring have $ R $ (red nose). Thus, yellow nose ($ rr $) is impossible (0/16 for yellow nose phenotypes).

For mega spikes ($ Mm $): From Parent 2’s $ M $ and Parent 1’s $ m $ → $ Mm $ (mega).
For normal spikes ($ mm $): From Parent 2’s $ m $ and Parent 1’s $ m $ → $ mm $ (normal).

From the 16-cell Punnett square:
- Mega spikes ($ Mm $): 8 cells (as above).
- Normal spikes ($ mm $): 8 cells.

Since all have red noses,

- **Mega spikes & red nose**: 8/16
- **Mega spikes & yellow nose**: 0/16
- **Normal spikes & red nose**: 8/16
- **Normal spikes & yellow nose**: 0/16

### Final Answers (Assuming the Punnett Square Has 16 Cells)
- Mega spikes & red nose = $ \boldsymbol{8/16} $
- Mega spikes & yellow nose = $ \boldsymbol{0/16} $
- Normal spikes & red nose = $ \boldsymbol{8/16} $
- Normal spikes & yellow nose = $ \boldsymbol{0/16} $

(Note: If the Punnett square is 4x4 but with different gametes, adjust accordingly. The key is that Parent 2 is $ RR $, so no yellow noses, and spikes are $ Mm $ (mega) or $ mm $ (normal) in a 1:1 ratio, leading to 8/16 for each spike phenotype with red nose.)

Parent 2 \ Parent 1	\( mR \)	\( mR \)	\( mr \)	\( mr \)
\( MR \)	\( MmRR \)	\( MmRR \)	\( MmRr \)	\( MmRr \)
\( mR \)	\( mmRR \)	\( mmRR \)	\( mmRr \)	\( mmRr \)
\( mR \)	\( mmRR \)	\( mmRR \)	\( mmRr \)	\( mmRr \)

Parent 2 \ Parent 1	\( mR \)	\( mR \)	\( mr \)	\( mr \)
\( MR \)	\( MmRR \)	\( MmRR \)	\( MmRr \)	\( MmRr \)
\( mR \)	\( mmRR \)	\( mmRR \)	\( mmRr \)	\( mmRr \)
\( mR \)	\( mmRR \)	\( mmRR \)	\( mmRr \)	\( mmRr \)

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Step 1: Determine Parent Genotypes and Gametes

Step 2: Set Up the Punnett Square

Step 3: Fill the Punnett Square

Step 4: Analyze Phenotypes

Answer:

Step 1: Determine Parent Genotypes and Gametes

Step 2: Set Up the Punnett Square

Step 3: Fill the Punnett Square

Step 4: Analyze Phenotypes

Step 5: Calculate Phenotype Ratios

Final Answers (Assuming the Punnett Square Has 16 Cells)