Solved Monohybrid Cross Worksheet Name Period Part A: | Chegg.com - Free Printable

Problem Analysis and Solution

#### Part A: Guinea Pig Hair Length (Short vs. Long)
We are given that short hair (`S`) is dominant over long hair (`s`). Let's solve the Punnett squares step by step.

---

#### 16a. One guinea pig is `Ss` and one is `ss`

- Genotypes of parents:
- Parent 1: `Ss` (heterozygous for short hair)
- Parent 2: `ss` (homozygous recessive for long hair)

- Punnett Square Construction:
- Parent 1 contributes alleles: `S` or `s`
- Parent 2 contributes alleles: `s` only
- Fill in the Punnett square:

| | s | s |
|---|----|----|
| S | Ss | Ss |
| s | ss | ss |

- Phenotypes and Genotypes of Offspring:
- `Ss`: Short hair (dominant phenotype)
- `ss`: Long hair (recessive phenotype)

- Expected Number of Offspring:
- Short hair (`Ss`): 2 out of 4 (50%)
- Long hair (`ss`): 2 out of 4 (50%)

Answer for 16a:
- Expected number of offspring:
- Short hair (`SS` or `Ss`): 2
- Long hair (`ss`): 2

---

#### 16b. Both guinea pigs are heterozygous for short hair

- Genotypes of parents:
- Parent 1: `Ss` (heterozygous for short hair)
- Parent 2: `Ss` (heterozygous for short hair)

- Punnett Square Construction:
- Parent 1 contributes alleles: `S` or `s`
- Parent 2 contributes alleles: `S` or `s`
- Fill in the Punnett square:

| | S | s |
|---|----|----|
| S | SS | Ss |
| s | Ss | ss |

- Phenotypes and Genotypes of Offspring:
- `SS`: Short hair (dominant phenotype)
- `Ss`: Short hair (dominant phenotype)
- `ss`: Long hair (recessive phenotype)

- Expected Number of Offspring:
- Short hair (`SS` or `Ss`): 3 out of 4 (75%)
- Long hair (`ss`): 1 out of 4 (25%)

Answer for 16b:
- Expected number of offspring:
- Short hair: 3
- Long hair: 1

---

#### Part C: Monohybrid Cross Problems

---

#### 17. Hornless (H) in cattle is dominant over horned (h)

- Given:
- Homozygous hornless bull: `HH`
- Homozygous horned cow: `hh`

- Cross: `HH × hh`

- Punnett Square Construction:
- Parent 1 (`HH`) contributes alleles: `H`
- Parent 2 (`hh`) contributes alleles: `h`
- Fill in the Punnett square:

| | h | h |
|---|----|----|
| H | Hh | Hh |
| H | Hh | Hh |

- Offspring Genotypes and Phenotypes:
- All offspring will have the genotype `Hh`.
- Since `H` (hornless) is dominant, all offspring will have the phenotype hornless.

Answer for 17:
- Genotype of first generation: Hh
- Phenotype of first generation: Hornless

---

#### 18. Tongue rolling (R) is dominant over non-rolling (r)

- Given:
- Father: Non-roller (`rr`)
- Mother: Heterozygous for tongue rolling (`Rr`)

- Cross: `rr × Rr`

- Punnett Square Construction:
- Father (`rr`) contributes alleles: `r`
- Mother (`Rr`) contributes alleles: `R` or `r`
- Fill in the Punnett square:

| | R | r |
|---|----|----|
| r | Rr | rr |
| r | Rr | rr |

- Offspring Genotypes and Phenotypes:
- `Rr`: Tongue roller (dominant phenotype)
- `rr`: Non-roller (recessive phenotype)

- Probability of having a tongue roller:
- Out of 4 possible offspring, 2 will be `Rr` (tongue rollers).
- Probability = \( \frac{2}{4} = \frac{1}{2} = 50\% \)

Answer for 18:
- Father’s phenotype: Non-roller
- Mother’s phenotype: Tongue roller
- Father’s genotype: rr
- Mother’s genotype: Rr
- Probability of having a child who is a tongue roller: 50%

---

Final Answers:

16a.
- Expected number of offspring:
- Short hair: 2
- Long hair: 2

16b.
- Expected number of offspring:
- Short hair: 3
- Long hair: 1

17.
- Genotype of first generation: Hh
- Phenotype of first generation: Hornless

18.
- Father’s phenotype: Non-roller
- Mother’s phenotype: Tongue roller
- Father’s genotype: rr
- Mother’s genotype: Rr
- Probability of having a child who is a tongue roller: 50%

Boxed Final Answer:
\[
\boxed{
\begin{array}{l}
\text{16a: Short hair: 2, Long hair: 2} \\
\text{16b: Short hair: 3, Long hair: 1} \\
\text{17: Genotype: Hh, Phenotype: Hornless} \\
\text{18: Father's phenotype: Non-roller, Mother's phenotype: Tongue roller} \\
\text{Father's genotype: rr, Mother's genotype: Rr, Probability: 50\%}
\end{array}
}
\]