Complete Dominance Mendelian Genetics Worksheet Answer Key - Fill ... - Free Printable

Problem Analysis:

The task involves solving problems related to Complete Dominance Mendelian Genetics. The goal is to fill out a table with definitions and solve two genetic cross problems using Punnett squares. Let's break this down step by step.

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Step 1: Fill Out the Table

We need to define the terms and provide examples or explanations for each term.

| Term | Definition | In your own words... | How will you remember? |
|-----------------|-------------------------------------------------------------------------------|-------------------------------------------------------------|----------------------------|
| Homozygous | An organism having two identical alleles for a particular gene. | Both alleles are the same (e.g., AA or aa). | "Homo" means "same." |
| Heterozygous| An organism having two different alleles for a particular gene. | Alleles are different (e.g., Aa). | "Hetero" means "different." |
| Phenotype | The observable physical traits of an organism, determined by its genotype. | What the organism looks like (e.g., black fur, white fur). | "Pheno" relates to appearance. |
| Genotype | The genetic makeup of an organism, represented by its alleles. | The combination of alleles (e.g., AA, Aa, aa). | "Geno" relates to genes. |
| Dominant | An allele that expresses its trait even when paired with a recessive allele. | Overpowers the recessive allele (e.g., Black fur in rabbits). | "Dominates" others. |
| Recessive | An allele that is expressed only when paired with another recessive allele. | Only shows up when both alleles are recessive (e.g., White fur in rabbits). | "Recedes" behind dominant. |

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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.
- Parent genotypes: Heterozygous black male (Aa) and homozygous white female (aa).

#### Solution:
1. Identify the alleles:
- Black fur is dominant (represented by A).
- White fur is recessive (represented by a).
- Heterozygous black male: Aa.
- Homozygous white female: aa.

2. Create the Punnett square:
- Place the alleles of the male (Aa) on the top row.
- Place the alleles of the female (aa) on the left column.

```
| A | a
---|----|----
a | Aa | aa
a | Aa | aa
```

3. Determine the genotypes and phenotypes:
- Genotypes: From the Punnett square, the possible genotypes are Aa and aa.
- Aa (heterozygous black)
- aa (homozygous white)
- Phenotypes:
- Aa results in black fur (dominant trait).
- aa results in white fur (recessive trait).

4. Calculate the ratios:
- Genotypic ratio: 2 Aa : 2 aa (or simplified to 1 Aa : 1 aa).
- Phenotypic ratio: 2 black : 2 white (or simplified to 1 black : 1 white).

#### Final Answer for Problem 1:
- P (parent genotypes): Aa × aa
- Genotypic ratio: 1 Aa : 1 aa
- Phenotypic ratio: 1 black : 1 white

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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 tall plant and a homozygous short plant.
- Parent genotypes: Homozygous tall plant (TT) and homozygous short plant (tt).

#### Solution:
1. Identify the alleles:
- Tall is dominant (represented by T).
- Short is recessive (represented by t).
- Homozygous tall plant: TT.
- Homozygous short plant: tt.

2. Create the Punnett square:
- Place the alleles of the tall plant (TT) on the top row.
- Place the alleles of the short plant (tt) on the left column.

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

3. Determine the genotypes and phenotypes:
- Genotypes: From the Punnett square, the possible genotypes are all Tt.
- Phenotypes:
- Tt results in tall plants (dominant trait).

4. Calculate the ratios:
- Genotypic ratio: All Tt (100% Tt).
- Phenotypic ratio: All tall (100% tall).

#### Final Answer for Problem 2:
- P (parent genotypes): TT × tt
- Genotypic ratio: 1 Tt
- Phenotypic ratio: 1 tall

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Final Answers:

1. Problem 1:
- P (parent genotypes): Aa × aa
- Genotypic ratio: 1 Aa : 1 aa
- Phenotypic ratio: 1 black : 1 white

2. Problem 2:
- P (parent genotypes): TT × tt
- Genotypic ratio: 1 Tt
- Phenotypic ratio: 1 tall

Boxed Final Answer:
\[
\boxed{
\begin{array}{l}
\text{Problem 1:} \\
\text{P (parent genotypes): } \text{Aa} \times \text{aa} \\
\text{Genotypic ratio: } 1 \text{ Aa : 1 aa} \\
\text{Phenotypic ratio: } 1 \text{ black : 1 white} \\
\\
\text{Problem 2:} \\
\text{P (parent genotypes): } \text{TT} \times \text{tt} \\
\text{Genotypic ratio: } 1 \text{ Tt} \\
\text{Phenotypic ratio: } 1 \text{ tall}
\end{array}
}
\]