Genetics and Inheritance Worksheets - Free Printable

Problem Overview:

The task involves completing Punnett squares for different genetic crosses and determining the probabilities of specific genotypes or phenotypes. Let's solve each part step by step.

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1. Brown Eyes (Dominant Over Blue Eyes):

- Genotypes:
- Parent 1: \( Bb \) (heterozygous for brown eyes)
- Parent 2: \( bb \) (homozygous recessive for blue eyes)

#### Step 1: Set Up the Punnett Square
- Parent 1 (\( Bb \)): Gametes are \( B \) and \( b \).
- Parent 2 (\( bb \)): Gametes are \( b \) and \( b \).

| | \( b \) | \( b \) |
|-----|---------|---------|
| \( B \) | \( Bb \) | \( Bb \) |
| \( b \) | \( bb \) | \( bb \) |

#### Step 2: Determine Genotype Probabilities
- \( BB \): 0 (not present)
- \( Bb \): 2 out of 4 = \( \frac{2}{4} = \frac{1}{2} \)
- \( bb \): 2 out of 4 = \( \frac{2}{4} = \frac{1}{2} \)

#### Final Answer for Brown Eyes:
- \( BB \): 0
- \( Bb \): \( \frac{1}{2} \)
- \( bb \): \( \frac{1}{2} \)

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2. Curly Hair (Dominant Over Straight Hair):

- Genotypes:
- Parent 1: \( CC \) (homozygous dominant for curly hair)
- Parent 2: \( Cc \) (heterozygous for curly hair)

#### Step 1: Set Up the Punnett Square
- Parent 1 (\( CC \)): Gametes are \( C \) and \( C \).
- Parent 2 (\( Cc \)): Gametes are \( C \) and \( c \).

| | \( C \) | \( c \) |
|-----|---------|---------|
| \( C \) | \( CC \) | \( Cc \) |
| \( C \) | \( CC \) | \( Cc \) |

#### Step 2: Determine Genotype Probabilities
- \( CC \): 2 out of 4 = \( \frac{2}{4} = \frac{1}{2} \)
- \( Cc \): 2 out of 4 = \( \frac{2}{4} = \frac{1}{2} \)
- \( cc \): 0 (not present)

#### Final Answer for Curly Hair:
- \( CC \): \( \frac{1}{2} \)
- \( Cc \): \( \frac{1}{2} \)
- \( cc \): 0

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3. Freckles (Dominant Over No Freckles):

#### Case 1: Homozygous Recessive Freckles (ff) × Homozygous Dominant Freckles (FF)
- Genotypes:
- Parent 1: \( ff \) (homozygous recessive for no freckles)
- Parent 2: \( FF \) (homozygous dominant for freckles)

#### Step 1: Set Up the Punnett Square
- Parent 1 (\( ff \)): Gametes are \( f \) and \( f \).
- Parent 2 (\( FF \)): Gametes are \( F \) and \( F \).

| | \( F \) | \( F \) |
|-----|---------|---------|
| \( f \) | \( Ff \) | \( Ff \) |
| \( f \) | \( Ff \) | \( Ff \) |

#### Step 2: Determine Phenotype Probabilities
- Freckles (\( F_ \)): All offspring have \( Ff \), so 4 out of 4 = \( \frac{4}{4} = 1 \)
- No freckles (\( ff \)): 0 (not present)

#### Final Answer for Case 1:
- Freckles: 1
- No freckles: 0

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#### Case 2: Heterozygous Freckles (Ff) × Homozygous Dominant Freckles (FF)
- Genotypes:
- Parent 1: \( Ff \) (heterozygous for freckles)
- Parent 2: \( FF \) (homozygous dominant for freckles)

#### Step 1: Set Up the Punnett Square
- Parent 1 (\( Ff \)): Gametes are \( F \) and \( f \).
- Parent 2 (\( FF \)): Gametes are \( F \) and \( F \).

| | \( F \) | \( F \) |
|-----|---------|---------|
| \( F \) | \( FF \) | \( FF \) |
| \( f \) | \( Ff \) | \( Ff \) |

#### Step 2: Determine Phenotype Probabilities
- Freckles (\( F_ \)): All offspring have either \( FF \) or \( Ff \), so 4 out of 4 = \( \frac{4}{4} = 1 \)
- No freckles (\( ff \)): 0 (not present)

#### Final Answer for Case 2:
- Freckles: 1
- No freckles: 0

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#### Case 3: Heterozygous Freckles (Ff) × Heterozygous Freckles (Ff)
- Genotypes:
- Parent 1: \( Ff \) (heterozygous for freckles)
- Parent 2: \( Ff \) (heterozygous for freckles)

#### Step 1: Set Up the Punnett Square
- Parent 1 (\( Ff \)): Gametes are \( F \) and \( f \).
- Parent 2 (\( Ff \)): Gametes are \( F \) and \( f \).

| | \( F \) | \( f \) |
|-----|---------|---------|
| \( F \) | \( FF \) | \( Ff \) |
| \( f \) | \( Ff \) | \( ff \) |

#### Step 2: Determine Phenotype Probabilities
- Freckles (\( F_ \)): Offspring with \( FF \) or \( Ff \): 3 out of 4 = \( \frac{3}{4} \)
- No freckles (\( ff \)): 1 out of 4 = \( \frac{1}{4} \)

#### Final Answer for Case 3:
- Freckles: \( \frac{3}{4} \)
- No freckles: \( \frac{1}{4} \)

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#### Case 4: Heterozygous Freckles (Ff) × Homozygous Recessive Freckles (ff)
- Genotypes:
- Parent 1: \( Ff \) (heterozygous for freckles)
- Parent 2: \( ff \) (homozygous recessive for no freckles)

#### Step 1: Set Up the Punnett Square
- Parent 1 (\( Ff \)): Gametes are \( F \) and \( f \).
- Parent 2 (\( ff \)): Gametes are \( f \) and \( f \).

| | \( f \) | \( f \) |
|-----|---------|---------|
| \( F \) | \( Ff \) | \( Ff \) |
| \( f \) | \( ff \) | \( ff \) |

#### Step 2: Determine Phenotype Probabilities
- Freckles (\( F_ \)): Offspring with \( Ff \): 2 out of 4 = \( \frac{2}{4} = \frac{1}{2} \)
- No freckles (\( ff \)): 2 out of 4 = \( \frac{2}{4} = \frac{1}{2} \)

#### Final Answer for Case 4:
- Freckles: \( \frac{1}{2} \)
- No freckles: \( \frac{1}{2} \)

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Final Summary of Answers:

1. Brown Eyes:
- \( BB \): 0
- \( Bb \): \( \frac{1}{2} \)
- \( bb \): \( \frac{1}{2} \)

2. Curly Hair:
- \( CC \): \( \frac{1}{2} \)
- \( Cc \): \( \frac{1}{2} \)
- \( cc \): 0

3. Freckles:
- Case 1: Freckles: 1, No freckles: 0
- Case 2: Freckles: 1, No freckles: 0
- Case 3: Freckles: \( \frac{3}{4} \), No freckles: \( \frac{1}{4} \)
- Case 4: Freckles: \( \frac{1}{2} \), No freckles: \( \frac{1}{2} \)

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Final Boxed Answers:

\[
\boxed{
\begin{array}{l}
\text{Brown Eyes: } BB = 0, Bb = \frac{1}{2}, bb = \frac{1}{2} \\
\text{Curly Hair: } CC = \frac{1}{2}, Cc = \frac{1}{2}, cc = 0 \\
\text{Freckles: } \\
\text{Case 1: Freckles = 1, No freckles = 0} \\
\text{Case 2: Freckles = 1, No freckles = 0} \\
\text{Case 3: Freckles = } \frac{3}{4}, \text{No freckles = } \frac{1}{4} \\
\text{Case 4: Freckles = } \frac{1}{2}, \text{No freckles = } \frac{1}{2}
\end{array}
}
\]