Free Printable Sex-linked Traits Worksheets for Students - Free Printable

Sex-Linked Traits Worksheet Solution

#### 1) Albinism (Recessive Autosomal Disorder)

Problem:
Albinism is a recessive autosomal genetic disorder. Fill in the Punnett square and determine the expected genotypic ratios from crossing homozygous recessive and heterozygous dominant parents.

Given:
- Albinism is an autosomal recessive disorder.
- Parents:
- Homozygous recessive parent: `aa`
- Heterozygous dominant parent: `Aa`

Punnett Square Construction:
The alleles for albinism are `A` (dominant, normal pigmentation) and `a` (recessive, albinism). The Punnett square will look like this:

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

Analysis:
- Genotypes:
- `Aa` (heterozygous dominant, normal pigmentation): 2 squares
- `aa` (homozygous recessive, albinism): 2 squares

- Phenotypes:
- Normal pigmentation (`Aa`): 2 squares
- Albinism (`aa`): 2 squares

- Genotypic Ratio:
- `Aa : aa = 1:1`

- % of kids with disorder (albinism):
- `aa` (albinism): 50%

- % of carrier kids (heterozygous dominant):
- `Aa`: 50%

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#### 2) Red-Green Color Blindness (Recessive Sex-Linked Disorder)

Problem:
Red-green color blindness is a recessive sex-linked (X chromosome) genetic disorder. Fill in the Punnett square and determine the expected genotypes and phenotypes from crossing a normal male and a female who is a carrier for colorblindness.

Given:
- Color blindness is an X-linked recessive disorder.
- Alleles:
- `X^H` (normal vision)
- `X^h` (color blindness)
- Parents:
- Normal male: `XY`
- Carrier female: `X^H X^h`

Punnett Square Construction:
The Punnett square will look like this:

| | X^H | X^h |
|-------|-------|-------|
| Y | X^H Y | X^h Y |
| X^H | X^H X^H | X^H X^h |

Analysis:
- Genotypes:
- `X^H Y` (normal male): 1 square
- `X^h Y` (male with colorblindness): 1 square
- `X^H X^H` (normal female): 1 square
- `X^H X^h` (carrier female): 1 square

- Phenotypes:
- Circle all phenotype(s):
- Normal male: `X^H Y`
- Male with colorblindness: `X^h Y`
- Normal female: `X^H X^H`
- Carrier female: `X^H X^h`

- % of kids with disorder (colorblindness):
- Only males can be affected by this X-linked recessive disorder.
- `X^h Y` (male with colorblindness): 25%

- Gender of kids with disorder:
- Male

---

#### 3) Color Blindness (Recessive Sex-Linked Disorder)

Problem:
Color blindness is a recessive sex-linked genetic disorder located on the X chromosome. Fill in the Punnett square for a cross of a male with color blindness and a normal female.

Given:
- Alleles:
- `X^H` (normal vision)
- `X^h` (color blindness)
- Parents:
- Male with color blindness: `X^h Y`
- Normal female: `X^H X^H`

Punnett Square Construction:
The Punnett square will look like this:

| | X^H | X^H |
|-------|-------|-------|
| Y | X^H Y | X^H Y |
| X^h | X^H X^h | X^H X^h |

Analysis:
- Genotypes:
- `X^H Y` (normal male): 2 squares
- `X^H X^h` (carrier female): 2 squares

- Phenotypes:
- Circle all phenotype(s):
- Normal male: `X^H Y`
- Carrier female: `X^H X^h`

- % of kids with disorder (colorblindness):
- None of the offspring will have colorblindness because the father's `X^h` allele is passed only to daughters, and they will be carriers (`X^H X^h`).

- Gender of kids with disorder:
- None

---

Final Answers:

#### 1) Albinism
- Genotypes: `Aa : aa = 1:1`
- Phenotypes: Normal pigmentation: 50%, Albinism: 50%
- % of kids with disorder: 50%
- % of carrier kids: 50%

#### 2) Red-Green Color Blindness
- Genotypes: `X^H Y`, `X^h Y`, `X^H X^H`, `X^H X^h`
- Phenotypes:
- Normal male: `X^H Y`
- Male with colorblindness: `X^h Y`
- Normal female: `X^H X^H`
- Carrier female: `X^H X^h`
- % of kids with disorder: 25%
- Gender of kids with disorder: Male

#### 3) Color Blindness
- Genotypes: `X^H Y`, `X^H X^h`
- Phenotypes:
- Normal male: `X^H Y`
- Carrier female: `X^H X^h`
- % of kids with disorder: 0%
- Gender of kids with disorder: None

---

Summary Table:

| Question | Genotypes | Phenotypes | % of kids with disorder | % of carrier kids | Gender of kids with disorder |
|--------------|-----------------------------------|--------------------------------------------------------------------------------|----------------------------|-----------------------|----------------------------------|
| 1) | `Aa : aa = 1:1` | Normal pigmentation: 50%, Albinism: 50% | 50% | 50% | N/A |
| 2) | `X^H Y`, `X^h Y`, `X^H X^H`, `X^H X^h` | Normal male, Male with colorblindness, Normal female, Carrier female | 25% | N/A | Male |
| 3) | `X^H Y`, `X^H X^h` | Normal male, Carrier female | 0% | N/A | None |

Final Answer: \boxed{\text{See detailed solutions above}}