name:___________ date:_____ 4) the female dog has black fur. the male dog has black fur. figure out the phenotypes an genotypes of their possible puppies by using a punnett square. % of possible genotypes: ff:_____ ff:_____ ff:_____ % of possible phenotypes: black fur:_____ grey fur:_____

Question

Accepted Answer

To solve this, we assume the black fur is dominant (F) and grey is recessive (f). Since both parents have black fur, let's assume their genotypes (common case: both Ff, but the Punnett square here seems to have F and f from each? Wait, the Punnett square drawn has cells: let's parse the Punnett square. Wait, the Punnett square in the image—let's assume the parents' gametes. Wait, maybe the parents are Ff (since black is dominant). Wait, but the Punnett square in the image: let's look at the cells. The top row and left column are the gametes. Let's assume the female and male are both Ff (so gametes F and f for each). Then the Punnett square would be:

|   | F | f |
|---|---|---|
| F | FF | Ff |
| f | Ff | ff |

But the drawn square in the image—let's check the cells. The first row first cell: maybe FF, first row second: FF? Wait, no, the user's image: the Punnett square has four cells. Let's re-express. Wait, maybe the parents are FF and Ff? No, the problem says both have black fur. Let's proceed with the standard Punnett square for two heterozygotes (Ff x Ff), but maybe the given Punnett square in the image is different. Wait, the user's image: the Punnett square cells are: top left: FF, top right: FF, bottom left: Ff, bottom right: Ff? Wait, no, the writing is unclear, but let's assume the parents are FF (homozygous dominant) and Ff (heterozygous). Wait, no, both black. Let's correct.

Wait, the key is: if both parents are black, and we need to find genotypes. Let's assume the dominant allele is F (black), recessive f (grey).

### Step 1: Determine Parent Genotypes
Since both have black fur, possible genotypes: FF (homozygous dominant) or Ff (heterozygous). Let's assume the Punnett square in the image is for parents with genotypes (let's look at the square). The Punnett square has four cells. Let's suppose the top row (female gametes) and left column (male gametes). Let's say female is Ff (gametes F and f) and male is Ff (gametes F and f), but the drawn square—alternatively, maybe both are FF? No, then all puppies would be FF. But the problem has ff (grey) as a possible genotype, so recessive exists.

Wait, the problem's Punnett square: let's parse the cells. The four cells: top left: FF, top right: FF, bottom left: Ff, bottom right: Ff? Or top left: FF, top right: Ff, bottom left: FF, bottom right: Ff? No, the user's image: the writing in the cells—maybe the parents are FF (female) and Ff (male). Then gametes: female (FF) can only give F, male (Ff) gives F and f. Then Punnett square:

|   | F | f |
|---|---|---|
| F | FF | Ff |
| F | FF | Ff |

Ah, that makes sense. So female is FF (homozygous dominant, black), male is Ff (heterozygous, black). Then:

### Step 2: Calculate Genotype Percentages
- FF: number of FF cells / total cells = 2/4 = 50%
- Ff: number of Ff cells / total cells = 2/4 = 50%
- ff: 0/4 = 0% (since no ff cells)

### Step 3: Calculate Phenotype Percentages
- Black fur: FF and Ff (dominant) → (2 + 2)/4 = 4/4 = 100%
- Grey fur: ff → 0/4 = 0%

But wait, the problem's Punnett square in the image—maybe the parents are Ff and Ff, but the drawn square is different. Wait, the user's image: the % of genotypes: FF, Ff, ff. Let's re-express with the assumed Punnett square (FF x Ff):

#### Genotypes:
- FF: 2 out of 4 cells → 50%
- Ff: 2 out of 4 cells → 50%
- ff: 0% (since no ff)

#### Phenotypes:
- Black fur: all (FF and Ff) → 100%
- Grey fur: 0% (no ff)

But maybe the parents are both FF? Then all puppies are FF (100% FF, 0% Ff, 0% ff; 100% black, 0% grey). But the problem includes ff as a possible genotype, so recessive exists. So the correct approach is:

Assuming parents are Ff (heterozygous, black) x Ff (heterozygous, black):

Punnett square:

|   | F | f |
|---|---|---|
| F | FF | Ff |
| f | Ff | ff |

Then:

- FF: 25% (1/4)
- Ff: 50% (2/4)
- ff: 25% (1/4)

Phenotypes:

- Black fur: FF + Ff = 75% (3/4)
- Grey fur: ff = 25% (1/4)

But the problem's Punnett square in the image—let's check the user's drawing. The Punnett square has four cells. Let's look at the writing: top left: FF, top right: FF, bottom left: Ff, bottom right: Ff? Then parents would be FF (female, gametes F) and Ff (male, gametes F and f). Then:

Gametes from female: F (only, since FF)
Gametes from male: F and f

Punnett square:

|   | F (female) | - |
|---|---|---|
| F (male) | FF | - |
| f (male) | Ff | - |

Wait, no, a Punnett square has two rows and two columns. So if female is FF (gametes F only) and male is Ff (gametes F and f), then the square is 2x2:

|   | F | - |
|---|---|---|
| F | FF | - |
| f | Ff | - |

No, that's not a 2x2. I think the user's image has a 2x2 square, so two gametes from each parent. Let's proceed with the standard case where both parents are Ff (heterozygous), leading to:

### Step 1: Define Alleles
- Dominant (black): F
- Recessive (grey): f

### Step 2: Parent Genotypes
Both parents have black fur, so possible genotypes: FF, Ff. Assume both are Ff (heterozygous, common case for dominant phenotype with recessive allele present).

### Step 3: Gametes
Each parent produces gametes: F and f (since Ff → F, f).

### Step 4: Punnett Square
|       | F (female) | f (female) |
|-------|------------|------------|
| F (male) | FF         | Ff         |
| f (male) | Ff         | ff         |

### Step 5: Genotype Percentages
- FF: 1/4 = 25%
- Ff: 2/4 = 50%
- ff: 1/4 = 25%

### Step 6: Phenotype Percentages
- Black fur (FF, Ff): 3/4 = 75% (since FF and Ff show black)
- Grey fur (ff): 1/4 = 25% (recessive, shows grey)

But the problem's Punnett square in the image—if the drawn square is different (e.g., parents are FF and Ff), let's check:

If female is FF (gametes F) and male is Ff (gametes F, f):

|       | F (female) | - |
|-------|------------|-----|
| F (male) | FF         | - |
| f (male) | Ff         | - |

No, that's not 2x2. I think the user's image has a 2x2 square with cells: FF, FF, Ff, Ff. So parents are FF (female) and Ff (male)? No, FF x Ff would give FF, FF, Ff, Ff? Wait, FF (female, gametes F) and Ff (male, gametes F, f):

Punnett square:

|   | F | - |
|---|---|---|
| F | FF | - |
| f | Ff | - |

No, that's 2x2 with empty cells. I think the intended problem is parents with genotypes Ff x Ff, leading to the standard 75% black, 25% grey, and genotypes 25% FF, 50% Ff, 25% ff.

But the problem's Punnett square in the image—let's look at the cells again. The user's image: the four cells are:

Top left: FF

Top right: FF

Bottom left: Ff

Bottom right: Ff

So that would mean the female is FF (gametes F only) and male is Ff (gametes F, f). Then:

Gametes from female: F (only)

Gametes from male: F, f

Punnett square:

|   | F | - |
|---|---|---|
| F | FF | - |
| f | Ff | - |

No, that's not a 2x2. I think there's a misdrawing, and the intended parents are both Ff, leading to the standard ratios.

### Final Answers (Assuming Ff x Ff):
- % Genotypes:
  - FF: 25%
  - Ff: 50%
  - ff: 25%
- % Phenotypes:
  - Black fur: 75% (FF + Ff)
  - Grey fur: 25% (ff)

If the parents are FF x FF (homozygous dominant), then:

- Genotypes: 100% FF
- Phenotypes: 100% black, 0% grey

But the presence of ff as a possible genotype (grey fur) implies the recessive allele is present, so parents must be Ff (or one FF and one Ff, but then ff can't occur). So the most probable case is Ff x Ff, leading to the above ratios.

### Step-by-Step (Correcting for Ff x Ff):

## Step 1: Identify Alleles
- Dominant (black): $ F $
- Recessive (grey): $ f $

## Step 2: Parent Genotypes
Both parents have black fur (dominant phenotype), so assume $ Ff $ (heterozygous, carries recessive allele $ f $).

## Step 3: Gametes Produced
Each parent ($ Ff $) produces two types of gametes: $ F $ and $ f $ (1:1 ratio).

## Step 4: Construct Punnett Square
|       | $ F $ (Female) | $ f $ (Female) |
|-------|-----------------|-----------------|
| $ F $ (Male) | $ FF $         | $ Ff $         |
| $ f $ (Male) | $ Ff $         | $ ff $         |

## Step 5: Calculate Genotype Percentages
- $ FF $: $ \frac{1}{4} = 25\% $
- $ Ff $: $ \frac{2}{4} = 50\% $
- $ ff $: $ \frac{1}{4} = 25\% $

## Step 6: Calculate Phenotype Percentages
- Black fur ($ FF, Ff $): $ \frac{3}{4} = 75\% $ (dominant phenotype)
- Grey fur ($ ff $): $ \frac{1}{4} = 25\% $ (recessive phenotype)

# Answer:
- % of possible Genotypes:
  FF: $ 25\% $
  Ff: $ 50\% $
  ff: $ 25\% $

- % of possible Phenotypes:
  Black fur: $ 75\% $
  Grey fur: $ 25\% $

(Note: If the Punnett square in the image differs, adjust based on the actual gametes/genotypes, but the above is the standard solution for heterozygous parents with dominant black fur.)

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QUESTION IMAGE

Question

Explanation:

Step 1: Determine Parent Genotypes

Step 2: Calculate Genotype Percentages

Step 3: Calculate Phenotype Percentages

Genotypes:

Phenotypes:

Answer: