In Drosophila, the genes for body coloration and eye size are on different chromosomes. Normal-colored bodies are dominant to ebony-colored (very dark) bodies, and normal-sized eyes are dominant to eyelessness. Line A is true breeding for normal bodies and normal eyes, whereas line B is true breeding for ebony bodies and eyelessness. From an F2 cross between lines A and B, what fraction will have exactly one of the recessive traits? 1/16 O 3/16 3/8 9/16 5/8

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**Genetics Problem in Drosophila**

In *Drosophila* (fruit flies), the genes for body coloration and eye size are located on different chromosomes. This implies that these genes assort independently during meiosis.

- **Dominant Traits:**
  - **Body Color:** Normal-colored bodies.
  - **Eye Size:** Normal-sized eyes.
  
- **Recessive Traits:**
  - **Body Color:** Ebony-colored (very dark) bodies.
  - **Eye Size:** Eyelessness (absence of eyes).

**Parental Lines:**
- **Line A:** True breeding for normal bodies and normal eyes.
- **Line B:** True breeding for ebony bodies and eyelessness.

When these lines are crossed (A x B), the F1 generation will have the genotype heterozygous for both traits (normal body and normal eyes, as normal traits are dominant).

In the F2 generation, resulting from the self-cross of F1 individuals, we need to determine what fraction will exhibit exactly one of the recessive traits.

**Choices:**
1. \( \frac{1}{16} \)
2. \( \frac{3}{16} \) (This option is selected.)
3. \( \frac{3}{8} \)
4. \( \frac{9}{16} \)
5. \( \frac{5}{8} \)

**Dominant-Recessive Trait Combinations in the F2 Generation:**
To calculate the proportion displaying exactly one recessive trait, we identify flies with either:
- Normal colored body and eyelessness, or
- Ebony colored body and normal-sized eyes.

Using a Punnett square, we analyze the dihybrid cross:

\[ AaBb \times AaBb \]

Where:
- \( A \) and \( a \) represent the alleles for body color.
- \( B \) and \( b \) represent the alleles for eye size.

From this cross, the genotype ratios for the F2 generation are:
- 9: Normal body and normal eyes (A_B_)
- 3: Normal body and eyeless (A_bb)
- 3: Ebony body and normal eyes (aaB_)
- 1: Ebony body and eyeless (aabb)

Thus, the fraction of flies showing exactly one recessive trait:
\[ \frac{3}{16} + \frac{3}{16} = \frac{6}{
Transcribed Image Text:**Genetics Problem in Drosophila** In *Drosophila* (fruit flies), the genes for body coloration and eye size are located on different chromosomes. This implies that these genes assort independently during meiosis. - **Dominant Traits:** - **Body Color:** Normal-colored bodies. - **Eye Size:** Normal-sized eyes. - **Recessive Traits:** - **Body Color:** Ebony-colored (very dark) bodies. - **Eye Size:** Eyelessness (absence of eyes). **Parental Lines:** - **Line A:** True breeding for normal bodies and normal eyes. - **Line B:** True breeding for ebony bodies and eyelessness. When these lines are crossed (A x B), the F1 generation will have the genotype heterozygous for both traits (normal body and normal eyes, as normal traits are dominant). In the F2 generation, resulting from the self-cross of F1 individuals, we need to determine what fraction will exhibit exactly one of the recessive traits. **Choices:** 1. \( \frac{1}{16} \) 2. \( \frac{3}{16} \) (This option is selected.) 3. \( \frac{3}{8} \) 4. \( \frac{9}{16} \) 5. \( \frac{5}{8} \) **Dominant-Recessive Trait Combinations in the F2 Generation:** To calculate the proportion displaying exactly one recessive trait, we identify flies with either: - Normal colored body and eyelessness, or - Ebony colored body and normal-sized eyes. Using a Punnett square, we analyze the dihybrid cross: \[ AaBb \times AaBb \] Where: - \( A \) and \( a \) represent the alleles for body color. - \( B \) and \( b \) represent the alleles for eye size. From this cross, the genotype ratios for the F2 generation are: - 9: Normal body and normal eyes (A_B_) - 3: Normal body and eyeless (A_bb) - 3: Ebony body and normal eyes (aaB_) - 1: Ebony body and eyeless (aabb) Thus, the fraction of flies showing exactly one recessive trait: \[ \frac{3}{16} + \frac{3}{16} = \frac{6}{
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