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Ch. 7 - Linkage and Chromosome Mapping in Eukaryotes
Klug - Essentials of Genetics 10th Edition
Klug10th EditionEssentials of GeneticsISBN: 9780135588789Not the one you use?Change textbook
Chapter 7, Problem 15a

Drosophila melanogaster has one pair of sex chromosomes (XX or XY) and three pairs of autosomes, referred to as chromosomes II, III, and IV. A genetics student discovered a male fly with very short (sh) legs. Using this male, the student was able to establish a pure breeding stock of this mutant and found that it was recessive. She then incorporated the mutant into a stock containing the recessive gene black (b, body color located on chromosome II) and the recessive gene pink (p, eye color located on chromosome III). A female from the homozygous black, pink, short stock was then mated to a wild-type male. The F1 males of this cross were all wild type and were then backcrossed to the homozygous b, p, sh females. The F2 results appeared as shown in the following table. No other phenotypes were observed.
Table showing F2 phenotypic counts of Drosophila by sex for wild, pink, black short, and black pink short traits.
Based on these results, the student was able to assign short to a linkage group (a chromosome). Which one was it? Include your step-by-step reasoning.

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Step 1: Understand the genetic background and the traits involved. The traits are black body color (b) on chromosome II, pink eye color (p) on chromosome III, and short legs (sh) of unknown chromosomal location. The mutant alleles are recessive, and the female used in the backcross is homozygous recessive for all three traits (b, p, sh). The male parent in the backcross is wild type for all traits.
Step 2: Analyze the F2 phenotypic classes and their counts. The phenotypes are Wild type, Pink only, Black and Short together, and Black, Pink, and Short together. Since the female parent is homozygous recessive for all three traits, the phenotypes in the progeny reflect the alleles contributed by the male parent and the recombination events between the genes.
Step 3: Determine linkage by comparing expected independent assortment ratios to observed data. If short legs (sh) were on a different chromosome from black (b) or pink (p), the phenotypes involving sh would assort independently, and the phenotypic ratios would reflect independent assortment. However, if sh is linked to either b or p, the phenotypes involving those traits will show non-independent assortment, with certain combinations appearing more frequently.
Step 4: Compare the phenotypic counts of 'Black, Short' and 'Black, Pink, Short' classes. If short legs (sh) is linked to black (b) on chromosome II, these two classes should be more frequent and appear together more often than expected by chance. Conversely, if sh is linked to pink (p) on chromosome III, the 'Pink' and 'Black, Pink, Short' classes would show linkage patterns.
Step 5: Conclude the linkage group of short legs (sh) by identifying which chromosome's recessive trait co-segregates with sh. The chromosome whose recessive trait appears together with sh more frequently than expected indicates linkage. Therefore, based on the phenotypic distribution, assign short legs (sh) to the chromosome that shows this pattern of linkage.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Linkage and Chromosome Mapping

Linkage refers to genes located close together on the same chromosome that tend to be inherited together. By analyzing offspring phenotypes and their frequencies, geneticists can determine whether genes are linked and assign them to specific chromosomes. This concept is essential for interpreting the inheritance patterns in the F2 generation and identifying the chromosome carrying the short (sh) gene.
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Recessive Traits and Pure Breeding

A recessive trait only appears phenotypically when an organism has two copies of the recessive allele (homozygous recessive). Pure breeding stocks are homozygous for the trait, ensuring consistent expression in offspring. Understanding this helps explain the initial crosses and the phenotypic ratios observed in the progeny.
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Sex Chromosomes vs. Autosomes in Drosophila

Drosophila melanogaster has one pair of sex chromosomes (XX or XY) and three pairs of autosomes (II, III, IV). Knowing the chromosomal location of genes (e.g., black on chromosome II, pink on III) allows geneticists to deduce linkage groups by comparing expected and observed phenotypic ratios, especially when incorporating a new gene like short.
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Related Practice
Textbook Question

In Drosophila, a cross was made between females—all expressing the three X-linked recessive traits scute bristles (sc), sable body (s), and vermilion eyes (v)—and wild-type males. In the F1, all females were wild type, while all males expressed all three mutant traits. The cross was carried to the F2 generation, and 1000 offspring were counted, with the results shown in the following table.

No determination of sex was made in the data.

Determine the sequence of the three genes and the map distances between them.

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Textbook Question

In Drosophila, a cross was made between females—all expressing the three X-linked recessive traits scute bristles (sc), sable body (s), and vermilion eyes (v)—and wild-type males. In the F1, all females were wild type, while all males expressed all three mutant traits. The cross was carried to the F2 generation, and 1000 offspring were counted, with the results shown in the following table.

No determination of sex was made in the data.

Are there more or fewer double crossovers than expected?

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Textbook Question

In Drosophila, a cross was made between females, all expressing the three X-linked recessive traits scute bristles (sc), sable body (s), and vermilion eyes (v)—and wild-type males. In the F₁, all females were wild type, while all males expressed all three mutant traits. The cross was carried to the F₂ generation, and 1000 offspring were counted, with the results shown in the following table.

No determination of sex was made in the data. Calculate the coefficient of coincidence. Does it represent positive or negative interference?

908
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Textbook Question

Drosophila melanogaster has one pair of sex chromosomes (XX or XY) and three pairs of autosomes, referred to as chromosomes II, III, and IV. A genetics student discovered a male fly with very short (sh) legs. Using this male, the student was able to establish a pure breeding stock of this mutant and found that it was recessive. She then incorporated the mutant into a stock containing the recessive gene black (b, body color located on chromosome II) and the recessive gene pink (p, eye color located on chromosome III). A female from the homozygous black, pink, short stock was then mated to a wild-type male. The F₁ males of this cross were all wild type and were then backcrossed to the homozygous b, p, sh females. The F₂ results appeared as shown in the following table.

The student repeated the experiment, making the reciprocal cross, with F₁ females backcrossed to homozygous b, p, sh males. She observed that 85 percent of the offspring fell into the given classes, but that 15 percent of the offspring were equally divided among b + p, b + +, + sh p, and + sh + phenotypic males and females. How can these results be explained, and what information can be derived from the data?

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Textbook Question

Drosophila females homozygous for the third chromosomal genes pink and ebony (the same genes from Problem 16) were crossed with males homozygous for the second chromosomal gene dumpy. Because these genes are recessive, all offspring were wild type (normal). F1 females were testcrossed to triply recessive males. If we assume that the two linked genes, pink and ebony, are 20 mu apart, predict the results of this cross. If the reciprocal cross were made (F1 males—where no crossing over occurs—with triply recessive females), how would the results vary, if at all?

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Textbook Question

In Drosophila, two mutations, Stubble (Sb) and curled (cu), are linked on chromosome III. Stubble is a dominant gene that is lethal in a homozygous state, and curled is a recessive gene. If a female of the genotype

is to be mated to detect recombinants among her offspring, what male genotype would you choose as a mate?

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