Punnett Squares Worksheet Collection For Teaching & Learning - Free Printable

Problem Overview:

The worksheet focuses on Mendelian genetics, specifically complete dominance. The task involves understanding key genetic terms and solving two genetic crosses using Punnett squares to determine genotypic and phenotypic ratios.

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Step 1: Define Key Terms

First, let's define the terms in the table:

| Term | Definition | In your own words... | How will you remember? |
|-----------------|-------------------------------------------------------------------------------|-------------------------------------------------------------|--------------------------------------------------|
| Homozygous | Having identical alleles for a particular gene (e.g., AA or aa). | Both copies of the gene are the same. | Think "homo" = same. |
| Heterozygous | Having different alleles for a particular gene (e.g., Aa). | One copy of each allele. | Think "hetero" = different. |
| Phenotype | The observable physical traits of an organism (e.g., black fur, white fur). | What you see. | Think "phenotype" = physical type. |
| Genotype | The genetic makeup of an organism (e.g., AA, Aa, aa). | The actual genes. | Think "genotype" = genetic type. |
| Allele | One of two or more alternative forms of a gene. | Different versions of the same gene. | Think "allele" = alternative. |
| Dominant | An allele that is expressed when present in either homozygous or heterozygous form. | Overpowers the recessive allele. | Think "dominant" = stronger. |
| Recessive | An allele that is only expressed when present in the homozygous state. | Only shows up if no dominant allele is present. | Think "recessive" = weaker. |

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Step 2: Solve Problem 1

#### Problem Statement:
"In rabbits, black fur is dominant over white fur. Show the cross of a heterozygous black male with a homozygous white female."

#### Solution:
1. Identify the alleles:
- Let \( B \) represent the dominant allele for black fur.
- Let \( b \) represent the recessive allele for white fur.

2. Determine the genotypes of the parents:
- Heterozygous black male: \( Bb \)
- Homozygous white female: \( bb \)

3. Set up the Punnett square:
- Parent 1 (male): \( Bb \)
- Parent 2 (female): \( bb \)

```
| b | b |
---|-------|-------|
B | Bb | Bb |
---|-------|-------|
b | bb | bb |
```

4. Determine the genotypes and phenotypes:
- Genotypes: \( Bb, Bb, bb, bb \)
- Phenotypes:
- \( Bb \) → Black fur (dominant phenotype)
- \( bb \) → White fur (recessive phenotype)

5. Calculate the ratios:
- Genotypic ratio: \( 2Bb : 2bb \) → Simplified to \( 1Bb : 1bb \)
- Phenotypic ratio: \( 2 \text{ Black} : 2 \text{ White} \) → Simplified to \( 1 \text{ Black} : 1 \text{ White} \)

6. Fill in the Punnett square and ratios:
- P (parent genotypes): \( Bb \times bb \)
- Genotypic ratio: \( 1 : 1 \)
- Phenotypic ratio: \( 1 : 1 \)

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Step 3: Solve Problem 2

#### Problem Statement:
"Tall is dominant over short in pea plants. Show the cross between a homozygous short plant and a homozygous tall plant."

#### Solution:
1. Identify the alleles:
- Let \( T \) represent the dominant allele for tall plants.
- Let \( t \) represent the recessive allele for short plants.

2. Determine the genotypes of the parents:
- Homozygous short plant: \( tt \)
- Homozygous tall plant: \( TT \)

3. Set up the Punnett square:
- Parent 1 (tall): \( TT \)
- Parent 2 (short): \( tt \)

```
| t | t |
---|-------|-------|
T | Tt | Tt |
---|-------|-------|
T | Tt | Tt |
```

4. Determine the genotypes and phenotypes:
- Genotypes: \( Tt, Tt, Tt, Tt \)
- Phenotypes:
- \( Tt \) → Tall (dominant phenotype)

5. Calculate the ratios:
- Genotypic ratio: \( 4Tt \) → Simplified to \( 1Tt \)
- Phenotypic ratio: \( 4 \text{ Tall} \) → Simplified to \( 1 \text{ Tall} \)

6. Fill in the Punnett square and ratios:
- P (parent genotypes): \( TT \times tt \)
- Genotypic ratio: \( 1 \)
- Phenotypic ratio: \( 1 \)

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Final Answers:

1. Problem 1:
- P (parent genotypes): \( Bb \times bb \)
- Genotypic ratio: \( 1 : 1 \)
- Phenotypic ratio: \( 1 : 1 \)

2. Problem 2:
- P (parent genotypes): \( TT \times tt \)
- Genotypic ratio: \( 1 \)
- Phenotypic ratio: \( 1 \)

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Boxed Final Answer:

\[
\boxed{
\begin{array}{l}
\text{Problem 1:} \\
\text{P (parent genotypes): } Bb \times bb \\
\text{Genotypic ratio: } 1 : 1 \\
\text{Phenotypic ratio: } 1 : 1 \\
\\
\text{Problem 2:} \\
\text{P (parent genotypes): } TT \times tt \\
\text{Genotypic ratio: } 1 \\
\text{Phenotypic ratio: } 1 \\
\end{array}
}
\]