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Ch. 5 - Chromosome Mapping in Eukaryotes
Klug - Concepts of Genetics 12th Edition
Klug12th EditionConcepts of GeneticsISBN: 9780135564776Non è quello che usi tu?Cambia libro di testo
Capitolo 5, Problema 26

In a certain plant, fruit is either red or yellow, and fruit shape is either oval or long. Red and oval are the dominant traits. Two plants, both heterozygous for these traits, were testcrossed, with the following results.
Table showing progeny counts for two heterozygous plants testcrossed, with phenotypes red/long, yellow/oval, red/oval, and yellow/long.
Determine the location of the genes relative to one another and the genotypes of the two parental plants.

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Step 1: Identify the parental genotypes and phenotypes. Both plants are heterozygous for fruit color (Red is dominant to Yellow) and fruit shape (Oval is dominant to Long). So, the parental genotypes are likely RrOo, where R = red, r = yellow, O = oval, and o = long.
Step 2: Analyze the progeny phenotypes and their counts from the testcross. The testcross involves crossing the heterozygous plants (RrOo) with homozygous recessive plants (rroo). The progeny phenotypes are red/long, yellow/oval, red/oval, and yellow/long with their respective counts for Plant A and Plant B.
Step 3: Determine which phenotypes represent parental (non-recombinant) and recombinant types. The most frequent phenotypes correspond to parental types, while the least frequent correspond to recombinants. For Plant A, red/long and yellow/oval are most frequent, so these are parental; red/oval and yellow/long are recombinant. For Plant B, red/oval and yellow/long are most frequent, so these are parental; red/long and yellow/oval are recombinant.
Step 4: Calculate the recombination frequency (RF) for each plant using the formula: \(\text{RF} = \frac{\text{Number of recombinant progeny}}{\text{Total progeny}} \times 100\). For Plant A, sum the recombinant progeny counts (red/oval + yellow/long) and divide by total progeny (100). Repeat for Plant B.
Step 5: Interpret the recombination frequency to determine gene linkage and relative gene location. A recombination frequency less than 50% indicates that the genes are linked and close on the same chromosome. The parental genotypes can be deduced based on which allele combinations are inherited together (coupling or repulsion phase).

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Testcross and Its Purpose

A testcross involves crossing an individual with a homozygous recessive individual to determine the genotype of the former. It reveals whether the tested individual is heterozygous or homozygous dominant for specific traits by analyzing the phenotypes of the offspring. This method is essential for mapping gene linkage and recombination.
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Purpose of X Inactivation

Gene Linkage and Recombination

Genes located close together on the same chromosome tend to be inherited together, a phenomenon called linkage. Recombination occurs when crossing over during meiosis exchanges genetic material between homologous chromosomes, producing new allele combinations. The frequency of recombinant offspring helps determine the physical distance between genes.
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Chi Square and Linkage

Dominant and Recessive Traits and Phenotypic Ratios

Dominant traits mask the expression of recessive traits in heterozygotes. In this problem, red and oval are dominant, while yellow and long are recessive. Analyzing the phenotypic ratios of progeny from heterozygous parents helps infer parental genotypes and gene linkage, especially when deviations from expected Mendelian ratios occur.
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Traits and Variance
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