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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 14d

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.
Table showing offspring counts for eight phenotypes with combinations of scute, sable, and vermilion traits totaling 1000 individuals.
No determination of sex was made in the data. Calculate the coefficient of coincidence. Does it represent positive or negative interference?

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Step 1: Identify the parental and recombinant phenotypes from the table. The parental phenotypes are the most frequent classes, which are 'sc s v' (314 offspring) and '+++ ' (280 offspring). The double crossover classes are the least frequent, which are 'sc + +' (10 offspring) and '++ v' (14 offspring).
Step 2: Calculate the recombination frequencies between the gene pairs. Use the single crossover classes to find the recombination frequency between sc and s, and between s and v. For example, the recombinants between sc and s are the phenotypes where only one of these genes is recombinant (e.g., '+ s v' and 'sc ++'). Calculate the recombination frequency as the sum of these single crossover offspring divided by the total offspring (1000).
Step 3: Calculate the expected number of double crossovers (DCO) assuming independence. Multiply the recombination frequencies of the two intervals to get the expected DCO frequency, then multiply by the total number of offspring to get the expected DCO count.
Step 4: Calculate the coefficient of coincidence (c) using the formula: \(c = \frac{\text{observed DCO}}{\text{expected DCO}}\). The observed DCO is the sum of the two double crossover classes (10 + 14).
Step 5: Determine the interference (I) using the formula: \(I = 1 - c\). If I is positive, it indicates positive interference (fewer double crossovers than expected). If I is negative, it indicates negative interference (more double crossovers than expected).

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Key Concepts

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

Genetic Linkage and Recombination

Genetic linkage refers to genes located close together on the same chromosome that tend to be inherited together. Recombination occurs during meiosis when crossing over between homologous chromosomes can separate linked genes, producing new allele combinations. The frequency of recombination between genes is used to map their relative positions on chromosomes.
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Coefficient of Coincidence and Interference

The coefficient of coincidence measures the observed double crossover frequency divided by the expected frequency, indicating how often two crossover events occur together. Interference describes how one crossover event affects the likelihood of another nearby; positive interference reduces double crossovers, while negative interference increases them.
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X-linked Inheritance in Drosophila

In Drosophila, X-linked traits are inherited differently in males and females because males have one X chromosome and females have two. Recessive X-linked traits appear in males if they inherit the mutant allele, while females must inherit two copies. This pattern helps interpret crosses involving X-linked genes and their phenotypic ratios.
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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.

Using proper nomenclature, determine the genotypes of the P1 and F1 parents.

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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.

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

776
views
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?

508
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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. No other phenotypes were observed.

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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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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