Solved Dihybrid Cross Worksheet 2. Using the punnett square ... - Free Printable

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

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Question 1: Set up a Punnett square for cross: FFDD × ffdd

- Parent 1 (homozygous dominant): genotype = FFDD
→ Gametes: only FD (since both alleles are same for each gene)

- Parent 2 (homozygous recessive): genotype = ffdd
→ Gametes: only fd

So the Punnett square is 1×1 in terms of unique gametes, but since it's a dihybrid, we still draw a 4×4 grid (because each parent produces 4 gamete combinations *if heterozygous*, but here they’re homozygous — so all gametes are identical).

Actually, for a dihybrid cross, each parent contributes one allele for each gene. Since FFDD has only F and D, all gametes = FD. Similarly, ffdd → all gametes = fd.

So the Punnett square will have:
- Top row (gametes from FFDD): FD, FD, FD, FD
- Left column (gametes from ffdd): fd, fd, fd, fd

Each box = FD + fd = FfDd

So the full 4×4 square is filled with FfDd in all 16 boxes.

✔ So all offspring are FfDd — heterozygous for both traits.

Now answer Question 2 using this.

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Question 2a: Probability of tall plants with purple flowers?
- Tall = dominant D → need at least one D → D_ (DD or Dd)
- Purple = dominant F → need at least one F → F_ (FF or Ff)

All offspring are FfDd, so:
- Ff → has F → purple ✔
- Dd → has D → tall ✔

So all 16/16 = 100% = 1

Possible genotype(s): FfDd only.

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Question 2b: Short plants with pink flowers?
- Short = dd (recessive)
- Pink = ff (recessive)

Offspring are all FfDd, so none are ff or dd.
Probability = 0
Possible genotype(s): none

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Question 2c: Short plants with purple flowers?
- Short = dd
- Purple = F_

We need genotype: F_ dd
But all offspring are Dd → no dd. So impossible.
Probability = 0
Possible genotype(s): none

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Question 2d: Tall plants with pink flowers?
- Tall = D_
- Pink = ff

Need: ff D_
But all offspring are Ff → no ff. So impossible.
Probability = 0
Possible genotype(s): none

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Now Question 3: Cross BbRr × BbRr
Both parents heterozygous for two traits:
- B = black (dominant), b = white (recessive)
- R = rough (dominant), r = smooth (recessive)

Each parent produces 4 types of gametes:
BR, Br, bR, br
(All combinations of one allele from each gene)

So Punnett square is 4×4:

Gametes (top): BR, Br, bR, br
Gametes (left): BR, Br, bR, br

Now fill in each box by combining alleles:

Let me list all 16 genotypes:

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

Now count phenotypes:

- Black, rough: B_ R_
Genotypes: BBRR, BBRr, BbRR, BbRr (multiple copies)

Let’s count how many out of 16:

From table:
- Row 1: BBRR (1), BBRr (1), BbRR (1), BbRr (1) → 4
- Row 2: BBRr (1), BBrr (no — rr = smooth), BbRr (1), Bbrr (no) → 2
- Row 3: BbRR (1), BbRr (1), bbRR (no — bb = white), bbRr (no) → 2
- Row 4: BbRr (1), Bbrr (no), bbRr (no), bbrr (no) → 1

Wait — better to use standard dihybrid ratio: for BbRr × BbRr, expected phenotypic ratio is 9:3:3:1
Where:
- 9 = B_ R_ → black, rough
- 3 = B_ rr → black, smooth
- 3 = bb R_ → white, rough
- 1 = bb rr → white, smooth

So total 16 squares.

Thus:
- Black, rough = 9/16
- Black, smooth = 3/16
- White, rough = 3/16
- White, smooth = 1/16

Now answer Question 4:

4a: Probability of black, rough fur?
= 9/16
Possible genotypes: Any with at least one B and one R:
BBRR, BBRr, BbRR, BbRr
(These appear in the table — 9 combos total)

4b: Black, smooth = B_ rr
= 3/16
Genotypes: BBrr, Bbrr

4c: White, rough = bb R_
= 3/16
Genotypes: bbRR, bbRr

4d: White, smooth = bb rr
= 1/16
Genotype: bbrr

Let me double-check counts from the table:

List all 16:

1. BBRR
2. BBRr
3. BbRR
4. BbRr
5. BBRr
6. BBrr
7. BbRr
8. Bbrr
9. BbRR
10. BbRr
11. bbRR
12. bbRr
13. BbRr
14. Bbrr
15. bbRr
16. bbrr

Now classify:

- Black, rough (B_ R_): need B≠bb and R≠rr
→ #1 (BBRR), #2 (BBRr), #3 (BbRR), #4 (BbRr), #5 (BBRr), #7 (BbRr), #9 (BbRR), #10 (BbRr), #13 (BbRr) → that’s 9 ✔

- Black, smooth (B_ rr): BBrr (#6), Bbrr (#8), Bbrr (#14) → 3 ✔

- White, rough (bb R_): bbRR (#11), bbRr (#12), bbRr (#15) → 3 ✔

- White, smooth: bbrr (#16) → 1 ✔

Perfect.

Now compile final answers.

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