Solved Punnett Square worksheet Complete the following | Chegg.com - Free Printable

It looks like you've uploaded a Punnett Square worksheet with two dihybrid cross problems. However, the image is not visible to me, so I cannot see the specific traits or alleles involved in each problem.

But based on your description and standard genetics conventions, I can help solve these problems assuming typical Mendelian inheritance patterns, especially for Portuguese Water Dogs (PWD) and plant seed shapes.

Let’s go through both problems step by step using logical assumptions based on common genetic notation.

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🔹 Problem 1: Dihybrid Cross in Portuguese Water Dogs (PWD)

> P generation: Two true-breeding (homozygous) PWDs
> - One has a black, wavy coat
> - One has a brown, curly coat

We assume:
- Coat color: Black (B) is dominant over Brown (b)
- Coat texture: Wavy (W) is dominant over Curly (w)

So:
- Black, wavy = BBWW
- Brown, curly = bbww

#### Step 1: Parental Genotypes
- Parent 1 (black, wavy): BBWW
- Parent 2 (brown, curly): bbww

#### Step 2: Gametes
Each parent produces only one type of gamete because they are homozygous:
- BBWW → BW
- bbww → bw

#### Step 3: F1 Generation (Cross: BW × bw)

All offspring will be: BbWw

| | BW |
|-------|--------|
| bw| BbWw |

✔ All F1 offspring are BbWw — heterozygous for both traits.

#### Resulting F1 Generation:

- Genotypic ratio: All offspring are BbWw → So ratio is 1:0:0:0
But since we're asked for HD:Ht:HR, let's clarify:

Wait! The instructions say:
> For genotypic ratio – it’s HD:Ht:HR
> Where:
> - HD = Homozygous Dominant
> - Ht = Heterozygous
> - HR = Homozygous Recessive

But this applies to each gene separately in a dihybrid cross, but here all F1 are heterozygous for both genes.

So for each gene:
- Color: Bb → Heterozygous
- Texture: Ww → Heterozygous

So for both genes together, the genotype is BbWw, which is heterozygous for both traits.

Thus:
- Genotypic ratio (for each gene):
Since all are Bb and Ww → no homozygous dominant or recessive individuals.
- So: HD : Ht : HR = 0 : 1 : 0 per gene? But wait — that's not quite right.

Actually, the "HD:Ht:HR" likely refers to overall genotype combinations across both loci. But in this case, all F1 are BbWw, so only one genotype.

So:
- Genotypic ratio: Only BbWw → So ratio is 1 (only one type)

But if we follow the instruction format: HD:Ht:HR — we need to define what that means.

Alternatively, perhaps HD = Homozygous Dominant (e.g., BB), Ht = Heterozygous (Bb), HR = Homozygous Recessive (bb). But since we have two genes, we need to consider combinations.

But for F1, all are BbWw, so:

- For coat color: Bb → Heterozygous → Ht
- For coat texture: Ww → Heterozygous → Ht

So overall, genotype is heterozygous for both, so Ht:Ht?

But the instruction says “HD:Ht:HR” — probably meaning three categories for one gene, but we have two genes.

Wait — maybe the HD:Ht:HR is meant for each gene individually, but since it's a dihybrid cross, we should report genotypic ratios as combinations.

But in F1, there is only one genotype: BbWw

So:
- Genotypic ratio: 1 (only BbWw)
- Phenotypic ratio: All have black, wavy coats (since B and W are dominant)

So:
- Phenotypic ratio: 1:0 (all black wavy)

But let's write it properly.

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#### ✔ Final Answer for Problem 1:

Punnett Square (F1 generation):

| | BW |
|-------|--------|
| bw| BbWw |

All offspring: BbWw

- Genotypic ratio: All offspring are BbWw → So only one genotype
But if we interpret HD:Ht:HR as per gene:
- For color: Bb → Ht
- For texture: Ww → Ht
- So overall, no HD or HR, all Ht → So HD:Ht:HR = 0:1:0 per trait?

But the instruction says “genotypic ratio” — so likely expecting genotype frequencies.

But since all are same, genotypic ratio = 1 (only BbWw)

But if we must use HD:Ht:HR, it may refer to combined genotypes in F2, not F1.

Wait — the instruction says:

> For the genotypic ratio - it's HD:Ht:HR
> For the F1 phenotypic ratio - it's Dom:Dom:Dom:Rec... etc.

Ah! It seems HD:Ht:HR is for single-gene ratios, and Dom:Dom:Dom:Rec is for dihybrid phenotypic ratios.

But in F1, we have only one phenotype.

So:

✔ Correct Interpretation:

For F1 generation:
- All offspring are BbWw
- Phenotype: Black, wavy (since both dominant)

So:
- Genotypic ratio: All are BbWw → So only one genotype → Ratio is 1
But if forced into HD:Ht:HR, then:
- HD = BB or WW? No — HD would be homozygous dominant, but none exist.
- So: HD : Ht : HR = 0 : 1 : 0 (for each gene)

But since it's dihybrid, perhaps they want genotypic ratio as BbWw only, so just 1

But let’s assume the format is HD:Ht:HR meaning:
- HD = Homozygous Dominant (e.g., BB or WW)
- Ht = Heterozygous (Bb or Ww)
- HR = Homozygous Recessive (bb or ww)

But again, in F1, all are BbWw, so:
- For color: Ht
- For texture: Ht

So genotypic ratio is Ht:Ht → but the format expects HD:Ht:HR

So perhaps it's 0:1:0 for each gene.

But better to say:

> Genotypic ratio: All offspring are BbWw → So only one genotype → Ratio is 1:0:0 if we consider HD:Ht:HR as total count? Not really.

Actually, HD:Ht:HR likely refers to genotypic classes for a single gene, but here we have two genes.

Given confusion, perhaps the genotypic ratio is simply 1 (BbWw)

And phenotypic ratio is 1 (all black, wavy)

But the instruction says: “For the F1 phenotypic ratio - it's Dom:Dom:Dom:Rec”

Wait — that suggests four phenotypes — but F1 has only one phenotype.

So likely, “Dom:Dom:Dom:Rec” is for F2, not F1.

Let me re-read:

> For the F1 phenotypic ratio - it's Dom:Dom:Dom:Rec

That doesn’t make sense — F1 from two homozygous parents should have one phenotype.

Unless there's a typo.

Wait — the instruction says:

> For the F1 phenotypic ratio - it's Dom:Dom:Dom:Rec
> For the F2 phenotypic ratio - it's DomDom:DomRec:RecDom:RecRec

Ah! So:
- F1 phenotypic ratio uses Dom:Dom:Dom:Rec — but that’s four parts — but F1 has only one phenotype.

So likely, this is a mistake — or perhaps Dom:Dom:Dom:Rec means Dominant for both, Dominant for first and recessive for second, etc.

But in F1, all are dominant for both, so:

- Phenotypic ratio: 1:0:0:0 → But they want Dom:Dom:Dom:Rec

Wait — maybe they mean:
- DomDom = dominant for both
- DomRec = dominant first, recessive second
- RecDom = recessive first, dominant second
- RecRec = recessive for both

So:
- F1: all DomDom → so ratio: 1:0:0:0

But the instruction says “For the F1 phenotypic ratio - it's Dom:Dom:Dom:Rec” — that’s confusing.

Perhaps it's a typo and should be:

> For the F2 phenotypic ratio: DomDom:DomRec:RecDom:RecRec

And for F1, just 1:0:0:0 in that format.

But the worksheet says:

> For the F1 phenotypic ratio - it's Dom:Dom:Dom:Rec

That’s strange.

Alternatively, maybe Dom:Dom:Dom:Rec is meant to be DomDom:DomRec:RecDom:RecRec, and they wrote it poorly.

Let’s proceed.

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✔ Final Answer for Problem 1:

P generation:
- Parent 1: BBWW (black, wavy)
- Parent 2: bbww (brown, curly)

Gametes:
- BW × bw

F1 offspring: All BbWw

Punnett Square:

| | BW |
|-------|--------|
| bw| BbWw |

All 4 squares (if expanded) would be BbWw — but since only one gamete type from each, it's a 1×1 square.

So:

- Genotypic ratio: All are BbWw → So only one genotype → Ratio: 1
But if we must use HD:Ht:HR, and interpret as for each gene, then:
- For color: Bb → Ht → so Ht
- For texture: Ww → Ht → so Ht
- So HD:Ht:HR = 0:1:0 for each gene

But since it's dihybrid, perhaps genotypic ratio is just 1 (BbWw)

- Phenotypic ratio: All have black, wavy coats → So 1:0:0:0 in DomDom:DomRec:RecDom:RecRec format

So:
- Genotypic ratio: 0:1:0 (HD:Ht:HR) — but this is ambiguous

Better to write:

> Genotypic ratio: All offspring are BbWw → So only one genotype
> Phenotypic ratio: All are black, wavy → So 1:0:0:0 (DomDom:DomRec:RecDom:RecRec)

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🔹 Problem 2: F1 generation – Hybrid (heterozygous) smooth seed shape plants are crossed.

This sounds like a monohybrid cross for seed shape, but the term "smooth seed shape" suggests a trait.

But it says "Hybrid (heterozygous) smooth seed shape plants are crossed"

So:
- Smooth seeds are dominant (S)
- Wrinkled seeds are recessive (s)
- Heterozygous = Ss

So cross: Ss × Ss

This is a monohybrid cross, not dihybrid.

But the Punnett square is 4×4 — suggesting dihybrid?

Wait — the title says "dihybrid cross", but problem 2 says "smooth seed shape" — only one trait.

Possibly a typo.

But let’s assume it's monohybrid for seed shape.

But the Punnett square is 4×4 — so likely dihybrid.

Wait — maybe "smooth seed shape" is one trait, and another trait is implied?

But the problem says: "Hybrid (heterozygous) smooth seed shape plants are crossed"

So only one trait mentioned.

But the Punnett square is large — 4×4 — so perhaps it's a dihybrid cross, and "smooth seed shape" is one of two traits.

But no second trait is mentioned.

Alternatively, perhaps "smooth seed shape" is the trait, and "hybrid" means heterozygous, so Ss × Ss

Then:

Punnett Square (Ss × Ss):

| | S | s |
|-------|--------|--------|
| S | SS | Ss |
| s | Ss | ss |

Offspring:
- Genotypes: 1 SS : 2 Ss : 1 ss
- Phenotypes: 3 smooth : 1 wrinkled

But the instruction says:

> For the F2 phenotypic ratio - it's DomDom:DomRec:RecDom:RecRec

Again, that’s for two traits.

So likely, problem 2 is a dihybrid cross, but only one trait is mentioned.

Wait — perhaps the seed shape is one trait, and another trait is implied — like seed color?

But it’s not stated.

Alternatively, maybe "smooth seed shape" is being used to describe a dihybrid, but it's not clear.

Wait — perhaps the F1 generation is heterozygous for two traits, and we’re crossing them.

But the problem says: "Hybrid (heterozygous) smooth seed shape plants are crossed"

So only one trait is specified.

But the Punnett square is 4×4 — so likely dihybrid cross.

So perhaps the full context is missing.

But let’s assume it's a standard dihybrid cross between two heterozygous individuals for two traits.

For example:
- Seed shape: Smooth (S) dominant, Wrinkled (s) recessive
- Seed color: Yellow (Y) dominant, Green (y) recessive

Then F1: SsYy × SsYy

This is a classic dihybrid cross.

Then:

Gametes: SY, Sy, sY, sy

Punnett Square (4×4):

| | SY | Sy | sY | sy |
|-------|--------|--------|--------|--------|
| SY| SSYY | SSYy | SsYY | SsYy |
| Sy| SSYy | SSyy | SsYy | Ssyy |
| sY| SsYY | SsYy | ssYY | ssYy |
| sy| SsYy | Ssyy | ssYy | ssyy |

Now count:

Genotypic ratio:
- SSYY: 1
- SSYy: 2
- SsYY: 2
- SsYy: 4
- SSyy: 1
- Ssyy: 2
- ssYY: 1
- ssYy: 2
- ssyy: 1

But the instruction wants HD:Ht:HR — likely for each gene.

But for phenotypic ratio, we use:

- DomDom: Smooth, Yellow → S_Y_ → 9
- DomRec: Smooth, Green → S_yy → 3
- RecDom: Wrinkled, Yellow → ssY_ → 3
- RecRec: Wrinkled, Green → ssyy → 1

So:
- Phenotypic ratio: 9:3:3:1

Which matches DomDom:DomRec:RecDom:RecRec = 9:3:3:1

So likely, Problem 2 is a dihybrid cross between two heterozygous plants for two traits, and "smooth seed shape" is just one of them.

Even though it's not fully specified, we assume standard dihybrid cross.

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✔ Final Answer for Problem 2:

F1 cross: SsYy × SsYy (heterozygous for seed shape and color)

Punnett Square (4×4):

| | SY | Sy | sY | sy |
|-------|--------|--------|--------|--------|
| SY| SSYY | SSYy | SsYY | SsYy |
| Sy| SSYy | SSyy | SsYy | Ssyy |
| sY| SsYY | SsYy | ssYY | ssYy |
| sy| SsYy | Ssyy | ssYy | ssyy |

Resulting F2 generation:

- Phenotypic ratio: 9:3:3:1
→ DomDom : DomRec : RecDom : RecRec = 9 : 3 : 3 : 1

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🟩 Final Answers:

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1. P generation – Two true-breeding PWDs

- Parent 1: BBWW (black, wavy)
- Parent 2: bbww (brown, curly)
- Gametes: BW × bw
- F1 offspring: All BbWw

Punnett Square:

| | BW |
|-------|--------|
| bw| BbWw |

(Only one cell, but typically shown as 1×1)

- Genotypic ratio: All BbWw → So only one genotype
If forced into HD:Ht:HR: 0:1:0 (but this is per gene; better to say all are heterozygous)
- Phenotypic ratio: All black, wavy → So 1:0:0:0 (DomDom:DomRec:RecDom:RecRec)

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2. F1 generation – Heterozygous hybrid plants crossed

Assume dihybrid cross: SsYy × SsYy

Punnett Square (4×4) as above

- Phenotypic ratio: 9:3:3:1
→ DomDom : DomRec : RecDom : RecRec = 9 : 3 : 3 : 1

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✔ Summary:

#### Problem 1:
- Genotypic ratio: 1 (all BbWw)
- Phenotypic ratio: 1:0:0:0 (DomDom:DomRec:RecDom:RecRec)

#### Problem 2:
- Phenotypic ratio: 9:3:3:1

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If you can provide the actual image or clarify the traits, I can adjust accordingly. But based on standard genetics, this is the most likely interpretation.