Dihybrid cross worksheet with Punnett squares for genetic inheritance of traits in plants and guinea pigs.

Let's solve each part of this Dihybrid Cross Worksheet step by step.

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Question 1: Set up a Punnett square

Given:
- Dominant allele for tall plants = D
- Recessive allele for dwarf plants = d
- Dominant allele for purple flowers = W
- Recessive allele for white flowers = w
- Cross: Homozygous dominant parent (DDWW) × Homozygous recessive parent (ddww)

#### Step 1: Determine gametes
Each parent produces gametes with one allele for each trait.

- DDWW → Gametes: DW
- ddww → Gametes: dw

Since both parents are homozygous, they produce only one type of gamete each.

So the cross is:
- DW × dw

We can set up a 2×2 Punnett square (but since both parents produce only one type of gamete, it will be simple):

| | dw |
|-------|-------|
| DW | DdWw |

But to make it look like a proper dihybrid square (as expected), we write:

```
dw dw dw dw
---------------------------
DW | DdWw DdWw DdWw DdWw
```

Wait — actually, since each parent produces only one kind of gamete, all offspring are identical.

But in standard form, even though there's only one type per parent, we still represent it as:

| | dw |
|-------|-------|
| DW | DdWw |

So all offspring are DdWw.

✔ So the Punnett square looks like:

| | dw |
|-------|-------|
| DW | DdWw |

But typically, we use a 4-box grid. Since both parents are homozygous, and produce only one gamete type, the full Punnett square is:

| | dw |
|-------|-------|
| DW | DdWw |

But to follow convention, let’s expand it properly with four boxes, even if they're all the same:

```
dw dw dw dw
--------------------------------
DW | DdWw DdWw DdWw DdWw
```

Actually, that’s not correct — each parent produces only one type of gamete, so we don’t need multiple copies.

Standard format: One gamete from each parent.

So:

| | dw |
|-------|-------|
| DW | DdWw |

All offspring are DdWw.

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Question 2: Using the Punnett square from #1

From above, all offspring are DdWw.

Now answer the questions:

#### a. Probability of producing tall plants with purple flowers

- Tall = D_ (D dominant)
- Purple = W_ (W dominant)
- Genotype DdWw → Dd (tall) and Ww (purple)

→ All offspring are DdWw, so 100% are tall and have purple flowers.

✔ Probability = 1 (or 100%)

Possible genotype(s): DdWw

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#### b. Probability of producing dwarf plants with white flowers

- Dwarf = dd
- White = ww
- But all offspring are DdWw → not dd or ww

So none have these traits.

✔ Probability = 0

Possible genotype(s): None (or no such offspring)

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#### c. Probability of producing tall plants with white flowers

- Tall = D_
- White = ww
- But all offspring are Ww → not ww, so cannot have white flowers

So no white-flowered offspring

✔ Probability = 0

Possible genotype(s): None

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#### d. Probability of producing dwarf plants with purple flowers

- Dwarf = dd
- Purple = W_
- But all offspring are Dd → not dd, so not dwarf

✔ Probability = 0

Possible genotype(s): None

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> ✔ Summary for Question 2:
>
> a. 100%, genotype: DdWw
> b. 0%, no genotype
> c. 0%, no genotype
> d. 0%, no genotype

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Question 3: Set up a Punnett square

Given:
- Black fur = B (dominant)
- White fur = b (recessive)
- Rough fur = R (dominant)
- Smooth fur = r (recessive)
- Cross: Heterozygous parent BbRr × Heterozygous parent BbRr

This is a classic dihybrid cross.

Each parent produces gametes: BR, Br, bR, br

So we make a 4×4 Punnett square.

Gametes from Parent 1: BR, Br, bR, br
Gametes from Parent 2: BR, Br, bR, br

Set up the square:

| | BR | Br | bR | br |
|-------|--------|--------|--------|--------|
| BR| BBRR | BBRr | BbRR | BbRr |
| Br| BBRr | BBrr | BbRr | Bbrr |
| bR| BbRR | BbRr | bbRR | bbRr |
| br| BbRr | Bbrr | bbRr | bbrr |

This gives all possible combinations.

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Question 4: Using the Punnett square from #3

Now analyze the probabilities.

#### a. Probability of black, rough fur

- Black = B_ (BB or Bb)
- Rough = R_ (RR or Rr)

We count how many offspring have B_ and R_

Look at the 16 boxes:

Let’s list genotypes that are B_ R_ (i.e., not bb and not rr):

- BBRR → 1
- BBRr → 2 (top row: BBRr appears twice)
- BbRR → 2 (left column: BbRR appears twice)
- BbRr → 4 (appears in four positions)

Total:
- BBRR: 1
- BBRr: 2
- BbRR: 2
- BbRr: 4
→ Total = 9

✔ Probability = 9/16

Possible genotypes: BBRR, BBRr, BbRR, BbRr

---

#### b. Probability of black, smooth fur

- Black = B_
- Smooth = rr

So: B_ rr

Genotypes: BBrr, Bbrr

Count them:

- BBrr: 1 (middle of top row)
- Bbrr: 2 (bottom two cells in middle columns)

Wait: check the square:

- BBrr → one box (Br × Br)
- Bbrr → two boxes: (Br × br) and (br × Br)

So total: 1 + 2 = 3

✔ Probability = 3/16

Possible genotypes: BBrr, Bbrr

---

#### c. Probability of white, rough fur

- White = bb
- Rough = R_

So: bb R_

Genotypes: bbRR, bbRr

Count:

- bbRR → 1 (bR × bR)
- bbRr → 2 (bR × br and br × bR)

Total = 1 + 2 = 3

✔ Probability = 3/16

Possible genotypes: bbRR, bbRr

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#### d. Probability of white, smooth fur

- White = bb
- Smooth = rr

So: bbrr

Only one box: br × br

✔ Probability = 1/16

Possible genotype: bbrr

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✔ Final Answers:

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Question 1: Punnett Square

| | dw |
|-------|-------|
| DW | DdWw |

All offspring: DdWw

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Question 2: Probabilities

a. Tall & purple: 100%, genotype: DdWw
b. Dwarf & white: 0%, no genotype
c. Tall & white: 0%, no genotype
d. Dwarf & purple: 0%, no genotype

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Question 3: Punnett Square (BbRr × BbRr)

| | BR | Br | bR | br |
|-------|--------|--------|--------|--------|
| BR| BBRR | BBRr | BbRR | BbRr |
| Br| BBRr | BBrr | BbRr | Bbrr |
| bR| BbRR | BbRr | bbRR | bbRr |
| br| BbRr | Bbrr | bbRr | bbrr |

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Question 4: Probabilities

a. Black, rough fur: 9/16, genotypes: BBRR, BBRr, BbRR, BbRr
b. Black, smooth fur: 3/16, genotypes: BBrr, Bbrr
c. White, rough fur: 3/16, genotypes: bbRR, bbRr
d. White, smooth fur: 1/16, genotype: bbrr

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