
Human Anatomy & Physiology (11th Edition)
11th Edition
ISBN: 9780134580999
Author: Elaine N. Marieb, Katja N. Hoehn
Publisher: PEARSON
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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}{](https://content.bartleby.com/qna-images/question/1d39eb1c-3154-4bee-b19a-a1be6277f6e8/0cd94ae4-d046-4773-b0ee-36229359716d/8h4j0tl_thumbnail.png)
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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