Genetics Worksheets and Printables - Free Printable

Monohybrid Cross Worksheet 3 Solutions

#### Problem 1: Mouse Coat Color
A gene encoding mouse coat color has two different alleles:
- G (dominant): encodes gray fur
- g (recessive): encodes white fur

#### Part a: Heterozygous Gray Mouse Mated with a White Mouse
1. P1 Cross:
- Heterozygous gray mouse: Gg
- White mouse: gg
- P1 Cross: Gg × gg

2. Punnett Square:
```
| | g | g |
|---|----|----|
| G | Gg | Gg |
| g | gg | gg |
```

3. F1 Genotypes:
- From the Punnett square, the possible genotypes are: Gg, Gg, gg, gg
- Simplified: Gg, gg

4. F1 Phenotypes:
- Gg: Gray (dominant phenotype)
- gg: White (recessive phenotype)
- Phenotypes: Gray, Gray, White, White
- Simplified: Gray, White

Final Answers for Part a:
- P1 Cross: Gg × gg
- F1 Genotypes: Gg, gg
- F1 Phenotypes: Gray, White

---

#### Part b: Homozygous Gray Mouse Mated with a White Mouse
1. P1 Cross:
- Homozygous gray mouse: GG
- White mouse: gg
- P1 Cross: GG × gg

2. Punnett Square:
```
| | g | g |
|---|----|----|
| G | Gg | Gg |
| G | Gg | Gg |
```

3. F1 Genotypes:
- From the Punnett square, the possible genotypes are: Gg, Gg, Gg, Gg
- Simplified: Gg

4. F1 Phenotypes:
- Gg: Gray (dominant phenotype)
- Phenotypes: Gray, Gray, Gray, Gray
- Simplified: Gray

Final Answers for Part b:
- P1 Cross: GG × gg
- F1 Genotypes: Gg
- F1 Phenotypes: Gray

---

#### Problem 2: Unknown Parental Genotypes
Two mice with unknown coat colors mated and produced the following offspring:
- 8 gray-coated mice
- 4 white-coated mice

#### Part 2a: Determine Parental Genotypes
1. Offspring Ratio:
- Gray:White = 8:4 = 2:1
- This ratio suggests that one of the parents is heterozygous (Gg) and the other is homozygous recessive (gg).

2. Possible Parental Genotypes:
- One parent: Gg (heterozygous gray)
- Other parent: gg (homozygous white)

3. Verification:
- P1 Cross: Gg × gg
- Punnett Square:
```
| | g | g |
|---|----|----|
| G | Gg | Gg |
| g | gg | gg |
```
- Genotypes: Gg, Gg, gg, gg
- Phenotypes: Gray, Gray, White, White
- Ratio: Gray:White = 2:1 (matches the observed ratio)

Final Answer for Part 2a:
- Parental Genotypes: Gg (gray) × gg (white)

---

#### Part 2b: All Offspring Gray
1. Offspring Phenotype:
- All offspring are gray, which means there are no recessive (gg) individuals.

2. Possible Parental Genotypes:
- Both parents must contribute at least one dominant allele (G).
- Possible combinations:
- GG × GG: Both parents are homozygous dominant.
- GG × Gg: One parent is homozygous dominant, and the other is heterozygous.
- Gg × Gg: Both parents are heterozygous.

3. Verification:
- P1 Cross: GG × GG
- All offspring will be GG (gray).
- P1 Cross: GG × Gg
- All offspring will be Gg (gray).
- P1 Cross: Gg × Gg
- Some offspring will be gg (white), which contradicts the condition.

Final Answer for Part 2b:
- Parental Genotypes: GG (gray) × GG (gray) or GG (gray) × Gg (gray)

---

#### Part 2c: 50% Offspring Gray
1. Offspring Ratio:
- 50% gray: 50% white = 1:1
- This ratio suggests a test cross scenario where one parent is heterozygous (Gg) and the other is homozygous recessive (gg).

2. Possible Parental Genotypes:
- One parent: Gg (heterozygous gray)
- Other parent: gg (homozygous white)

3. Verification:
- P1 Cross: Gg × gg
- Punnett Square:
```
| | g | g |
|---|----|----|
| G | Gg | Gg |
| g | gg | gg |
```
- Genotypes: Gg, Gg, gg, gg
- Phenotypes: Gray, Gray, White, White
- Ratio: Gray:White = 1:1 (matches the observed ratio)

Final Answer for Part 2c:
- Parental Genotypes: Gg (gray) × gg (white)

---

#### Problem 3: Orchid Flower Color
A species of orchids has a gene encoding either a dominant pink (P) or recessive white (p) flower color trait.

#### Cross: Heterozygous Pink and White Flower
1. P1 Cross:
- Heterozygous pink flower: Pp
- White flower: pp
- P1 Cross: Pp × pp

2. Punnett Square:
```
| | p | p |
|---|----|----|
| P | Pp | Pp |
| p | pp | pp |
```

3. F1 Genotypes:
- From the Punnett square, the possible genotypes are: Pp, Pp, pp, pp
- Simplified: Pp, pp

4. Probability of White Flowers:
- pp: White (recessive phenotype)
- Number of pp individuals: 2 out of 4
- Probability: \( \frac{2}{4} = \frac{1}{2} = 50\% \)

Final Answer for Problem 3:
- Probability of White Flowers: 50%

---

Final Summary of Answers

1. Part a:
- P1 Cross: Gg × gg
- F1 Genotypes: Gg, gg
- F1 Phenotypes: Gray, White

2. Part b:
- P1 Cross: GG × gg
- F1 Genotypes: Gg
- F1 Phenotypes: Gray

3. Part 2a:
- Parental Genotypes: Gg × gg

4. Part 2b:
- Parental Genotypes: GG × GG or GG × Gg

5. Part 2c:
- Parental Genotypes: Gg × gg

6. Problem 3:
- Probability of White Flowers: 50%

Boxed Final Answer:
\[
\boxed{
\begin{array}{l}
\text{1a: } \text{P1 Cross: } Gg \times gg, \text{ F1 Genotypes: } Gg, gg, \text{ F1 Phenotypes: } \text{Gray, White} \\
\text{1b: } \text{P1 Cross: } GG \times gg, \text{ F1 Genotypes: } Gg, \text{ F1 Phenotypes: } \text{Gray} \\
\text{2a: } \text{Parental Genotypes: } Gg \times gg \\
\text{2b: } \text{Parental Genotypes: } GG \times GG \text{ or } GG \times Gg \\
\text{2c: } \text{Parental Genotypes: } Gg \times gg \\
\text{3: } \text{Probability of White Flowers: } 50\%
\end{array}
}
\]