/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 55 An individual heterozygous for f... [FREE SOLUTION] | 91Ó°ÊÓ

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An individual heterozygous for four genes, \(A / a \cdot B / b \cdot C / c \cdot D / d\) is testcrossed with \(a / a \cdot b / b \cdot c / c \cdot d / d,\) and 1000 progeny are classified by the gametic contribution of the heterozygous parent as follows: $$\begin{array}{cc} a \cdot B \cdot C \cdot D & 42 \\ A \cdot b \cdot c \cdot d & 43 \\ A \cdot B \cdot C \cdot d & 140 \\ a \cdot b \cdot c \cdot D & 145 \\ a \cdot B \cdot c \cdot D & 6 \\ A \cdot b \cdot C \cdot d & 9 \\ A \cdot B \cdot c \cdot d & 305 \\ a \cdot b \cdot C \cdot D & 310 \end{array}$$ a. Which genes are linked? b. If two pure-breeding lines had been crossed to produce the heterozygous individual, what would their genotypes have been? c. Draw a linkage map of the linked genes, showing the order and the distances in map units. d. Calculate an interference value, if appropriate.

Short Answer

Expert verified
Genes \(A\) and \(C\), \(B\) and \(D\) are linked. Parent genotypes were \(A \cdot B \cdot c \cdot d\) and \(a \cdot b \cdot C \cdot D\). Interference calculation is appropriate for linked genes.

Step by step solution

01

Analyze Gene Linkage

To determine which genes are linked, identify the non-recombinant and recombinant progeny. Non-recombinants are the most frequent: \(A \cdot B \cdot c \cdot d\) (305) and \(a \cdot b \cdot C \cdot D\) (310). Recombinants are found in lower frequencies. Analyze the progeny data for patterns that indicate linkage.
02

Determine Parent Genotypes

From the non-recombinant types, deduce the parental genotypes. If two pure-breeding lines were crossed to create the heterozygote, the genotypes would be \(A \cdot B \cdot c \cdot d\) (homozygous) and \(a \cdot b \cdot C \cdot D\) (homozygous).
03

Calculate Recombination Frequencies

To calculate the recombination frequencies, use the formula: \( \text{Recombination Frequency} = \frac{\text{Number of Recombinants}}{\text{Total Offspring}} \). Identify recombinant genotypes like \(a \cdot B \cdot C \cdot D\) and \(A \cdot b \cdot C \cdot d\) and compute frequencies between nearby genes.
04

Draw Linkage Map

With recombination frequencies, estimate distances: 1% recombination equals 1 map unit. Arrange genes according to calculated distances to form a linear map.
05

Calculate Interference Value

First, find the expected number of double crossovers using probabilities of single crossovers (product of frequencies). Then, calculate the observed number using given counts. Interference = 1 - (Observed Double Crossovers / Expected Double Crossovers).
06

Verify and Interpret Map

Verify the map by checking if all values are consistent with recombination frequencies. Look for discrepancies that may indicate other factors. The map should be logical and agree with given progeny data.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gene Linkage
Gene linkage occurs when genes are located close to each other on the same chromosome. This proximity tends to prevent the genes from being separated during the formation of reproductive cells, resulting in them being inherited together. In our original exercise, by examining the data from the progeny of a testcross, we can identify which genes are linked by checking which pairs are more frequently passed on together. Non-recombinant types, which appear the most, indicate linkage. For example, in our dataset, the gene combinations \( A \cdot B \cdot c \cdot d \) and \( a \cdot b \cdot C \cdot D \) are found most frequently in the offspring, suggesting these groups of genes are linked. Understanding gene linkage is essential because it affects genetic trait predictions and reveals historical patterns of natural selection.
Recombination Frequency
Recombination frequency is a measure used to estimate the genetic distance between genes on a chromosome. It is calculated by dividing the number of recombinant offspring by the total number of offspring. This frequency provides insight into how often genes recombine during the formation of gametes. For instance, in the exercise at hand, we identify the recombinant types such as \( a \cdot B \cdot C \cdot D \) and compute how often these appear. The recombination frequency translates directly to a map unit measurement, where a 1% recombination frequency corresponds to 1 map unit. These calculations are pivotal in genetic mapping since they help in constructing linkage maps, which show the relative positions of genes on a chromosome.
Testcross
A testcross is a genetic cross between an organism with an unknown genotype and a homozygous recessive organism. This method helps determine the genotype of an individual showing a dominant trait. In the original problem, a testcross is conducted to analyze the parental and recombinant genotypes. The heterozygous individual \( A / a \cdot B / b \cdot C / c \cdot D / d \) is crossed with the homozygous recessive \( a / a \cdot b / b \cdot c / c \cdot d / d \). By examining the progeny, we can infer which alleles were inherited from the heterozygous parent, allowing us to deduce linkage patterns and construct genetic maps. This technique is invaluable because it simplifies the process of analyzing complex genetic interactions by providing clear phenotypic categories for assessment.
Interference
Interference is the phenomenon where one crossover event affects the likelihood of another crossover occurring nearby. It is calculated using the observed number of double crossover events and the expected number based on single crossover probabilities. To determine interference, calculate the expected number of double crossovers by multiplying the probabilities of each crossover event. Then compare this to the observed number by using progeny data. The formula for calculating interference is: \[ \text{Interference} = 1 - \left( \frac{\text{Observed Double Crossovers}}{\text{Expected Double Crossovers}} \right) \]. If the interference value is 0, it means there is no interference (crossovers occur independently), while a positive value indicates one crossover reduces the likelihood of another nearby. Understanding interference helps scientists study how genetic material is inherited in more detail and can explain deviations from expected genetic patterns.

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Most popular questions from this chapter

A rice breeder obtained a triple heterozygote carrying the three recessive alleles for albino flowers \((a l),\) brown awns \((b),\) and fuzzy leaves \((f u),\) all paired with their normal wild-type alleles. This triple heterozygote was testcrossed. The progeny phenotypes were \(170 \quad\) wild type \(150 \quad\) albino, brown, fuzzy \(5 \quad\) brown 3 albino, fuzzy \(710 \quad\) albino \(698 \quad\) brown, fuzzy \(42 \quad\) fuzzy \(38 \quad\) albino, brown a. Are any of the genes linked? If so, draw a map labeled with map distances. (Don't bother with a correction for multiple crossovers.) b. The triple heterozygote was originally made by crossing two pure lines. What were their genotypes?

The Neurospora cross al- \(2^{+} \times a l-2\) is made. A linear tetrad analysis reveals that the second-division segregation frequency is 8 percent a. Draw two examples of second-division segregation patterns in this cross. b. What can be calculated by using the 8 percent value?

In corn, a triple heterozygote was obtained carrying the mutant alleles \(s\) (shrunken), \(w\) (white aleurone), and \(y\) (waxy endosperm), all paired with their normal wild-type alleles. This triple heterozygote was testcrossed, and the progeny contained 116 shrunken, white; 4 fully wild type; 2538 shrunken; 601 shrunken, waxy; 626 white; 2708 white, waxy; 2 shrunken, white, waxy; and 113 waxy. a. Determine if any of these three loci are linked and, if so, show map distances. b. Show the allele arrangement on the chromosomes of the triple heterozygote used in the testcross. c. Calculate interference, if appropriate.

In a tetrad analysis, the linkage arrangement of the \(p\) and \(q\) loci is as follows: Assume that in region i, there is no crossover in 88 percent of meioses and there is a single crossover in 12 percent of meioses; in region ii, there is no crossover in 80 percent of meioses and there is a single crossover in 20 percent of meioses; and there is no interference (in other words, the situation in one region does not affect what is going on in the other region) What proportions of tetrads will be of the following types? (a) \(\mathrm{M}_{\mathrm{f}} \mathrm{M}_{\mathrm{I}}, \mathrm{PD}\) \((\mathbf{b}) \mathrm{M}_{1} \mathrm{M}_{\mathrm{l}}, \mathrm{NPD}\) \((\mathbf{c}) M_{1} M_{11}, T ;\) (d) \(\mathrm{M}_{\mathrm{D}} \mathrm{M}_{\mathrm{l}}, \mathrm{T}\) \((\mathbf{e}) M_{1} M_{\Perp}, P D\) \((\mathbf{f}) M_{11} M_{11}, N P D ;\) (g) \(\mathrm{M}_{\mathrm{II}} \mathrm{M}_{\mathrm{U}}\), T. (Note: Here the M pattern written first is the one that pertains to the \(p\) locus.) Hint: The easiest way to do this problem is to start by calculating the frequencies of asci with crossovers in both regions, region i, region ii, and neither region. Then determine what \(\mathrm{M}_{1}\) and \(\mathrm{M}_{\mathrm{II}}\) patterns result.

An autosomal allele \(N\) in humans causes abnormalities in nails and patellae (kneecaps) called the nail-patella syndrome. Consider marriages in which one partner has the nail-patella syndrome and blood type A and the other partner has normal nails and patellae and blood type O. These marriages produce some children who have both the nail-patella syndrome and blood type A. Assume that unrelated children from this phenotypic group mature, intermarry, and have children. Four phenotypes are observed in the following percentages in this second generation: nail-patella syndrome, blood type A \(66 \%\) normal nails and patellae, blood type 0 if \(16 \%\) normal nails and patellae, blood type A \(9 \%\) nail-patella syndrome, blood type \(\mathrm{O}\), \(9 \%\) Fully analyze these data, explaining the relative frequencies of the four phenotypes. (See pages \(214-215\) for the genetic basis of these blood types.)

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