Dihybrid cross worksheet featuring Punnett squares for genetics problems involving plant height and flower color, and guinea pig fur color and texture.

Explanation:
Let’s solve each part step by step.

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Question 1: Punnett square for DDWW × ddww

We’re crossing two parents:
- Parent 1: DDWW (homozygous dominant for both traits)
- Parent 2: ddww (homozygous recessive for both traits)

Each parent produces only one type of gamete, because they are homozygous at both loci:
- DDWW → gamete = DW
- ddww → gamete = dw

So the Punnett square is 1×1 (but we still draw a 4-box square as usual — all boxes will be the same):

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

All offspring are DdWw.

So the completed Punnett square has DdWw in all 4 boxes.

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Question 2: Using the Punnett square from Q1

All offspring are DdWw.

Recall:
- Tall (D) is dominant over dwarf (d)
- Purple (W) is dominant over white (w)

So genotype DdWw means:
- Dd → tall (since D is dominant)
- Ww → purple (since W is dominant)

Therefore, all offspring are tall with purple flowers.

Now answer each subpart:

a. Probability of tall plants with purple flowers?
→ All 4/4 = 1 or 100%
Possible genotype(s)? → DdWw only

b. Probability of dwarf plants with white flowers?
Dwarf needs dd, white needs ww — but no offspring have dd or ww.
→ 0
Possible genotype(s)? → none

c. Probability of tall plants with white flowers?
Tall = D_ (Dd or DD), white = ww. But all have Ww → not ww.
→ 0
Possible genotype(s)? → none

d. Probability of dwarf plants with purple flowers?
Dwarf = dd, purple = W_. But all have Dd → not dd.
→ 0
Possible genotype(s)? → none

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Question 3: Punnett square for BbRr × BbRr

Wait! The problem says:
> Cross a heterozygous parent (BbRr) with a heterozygous parent (BbRr)

But earlier it says:
> Cross a heterozygous parent (BbRr) with a heterozygous parent (BbRr)
Actually, rereading:
“Cross a heterozygous parent (BbRr) with a heterozygous parent (BbRr)” — yes, both are BbRr.

So this is a classic dihybrid cross: BbRr × BbRr

First, find gametes each parent can produce. Since genes assort independently:

Gametes from BbRr:
- BR, Br, bR, br → 4 types, each 1/4 probability.

Punnett square is 4×4:

Let’s list gametes on top and side:

Top (Parent 1 gametes): BR, Br, bR, br
Side (Parent 2 gametes): BR, Br, bR, br

Now fill in:

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

That’s the full Punnett square.

(We’ll use this for Q4.)

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Question 4: Using the Punnett square from Q3

We need probabilities for phenotypes.

Phenotype rules:
- Black fur = B_ (BB or Bb)
- White fur = bb
- Rough fur = R_ (RR or Rr)
- Smooth fur = rr

Total possible offspring = 16 (4×4 grid)

Let’s count each phenotype:

a. Black, rough → B_ and R_
Look at genotypes where B is not bb AND R is not rr.

From the table:
- BBRR ✔
- BBRr ✔
- BbRR ✔
- BbRr ✔
- BBRr ✔
- BbRr ✔
- BbRR ✔
- BbRr ✔
- BbRr ✔

Let’s count systematically:

Row 1 (BR):
- BBRR, BBRr, BbRR, BbRr → all 4 are B_ R_ → 4

Row 2 (Br):
- BBRr (B_, R_), BBrr (B_, rr) ✘, BbRr (B_, R_), Bbrr (B_, rr) ✘
→ 2 good

Row 3 (bR):
- BbRR (B_, R_), BbRr (B_, R_), bbRR (bb, R_) ✘, bbRr (bb, R_) ✘
→ 2 good

Row 4 (br):
- BbRr (B_, R_), Bbrr (B_, rr) ✘, bbRr (bb, R_) ✘, bbrr (bb, rr) ✘
→ 1 good

Total = 4 + 2 + 2 + 1 = 9

So probability = 9/16
Genotypes: BBRR, BBRr, BbRR, BbRr (all combinations with at least one B and one R)

b. Black, smooth → B_ and rr
rr means genotype ends in rr.

From table:
- BBrr (row2, col2) ✔
- Bbrr (row2, col4) ✔
- Bbrr (row4, col2) ✔

Check:
Row2: BBrr, Bbrr → 2
Row4: Bbrr → 1
Any others?
Row1: none (all R_)
Row3: none (bbRR, bbRr — not B_)
So total = 3

Probability = 3/16
Genotypes: BBrr, Bbrr

c. White, rough → bb and R_
bb and at least one R.

From table:
- bbRR (row3, col3) ✔
- bbRr (row3, col4) ✔
- bbRr (row4, col3) ✔

That’s 3

Probability = 3/16
Genotypes: bbRR, bbRr

d. White, smooth → bb and rr
Only: bbrr (row4, col4) ✔
Only 1

Probability = 1/16
Genotype: bbrr

Double-check totals: 9 + 3 + 3 + 1 = 16 ✔️

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Final Answer:
1. Punnett square: all 4 boxes = DdWw
2. a. Probability = 1; genotype = DdWw
b. Probability = 0; no possible genotype
c. Probability = 0; no possible genotype
d. Probability = 0; no possible genotype
3. Punnett square: 4×4 as filled above (BR, Br, bR, br on both axes; genotypes as listed)
4. a. Probability = 9/16; genotypes = BBRR, BBRr, BbRR, BbRr
b. Probability = 3/16; genotypes = BBrr, Bbrr
c. Probability = 3/16; genotypes = bbRR, bbRr
d. Probability = 1/16; genotype = bbrr