/*! 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 38 A Neurospora cross was made betw... [FREE SOLUTION] | 91Ó°ÊÓ

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A Neurospora cross was made between a strain that carried the mating-type allele \(A\) and the mutant allele arg-1 and another strain that carried the matingtype allele \(a\) and the wild-type allele for \(\arg -1(+)\) Four hundred linear octads were isolated, and they fell into the seven classes given in the table below. (For simplicity, they are shown as tetrads.) a. Deduce the linkage arrangement of the mating-type locus and the arg-1 locus. Include the centromere or centromeres on any map that you draw. Label all intervals in map units. b. Diagram the meiotic divisions that led to class 6 Label clearly. $$\begin{array}{ccccccc} \mathbf{1} & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline A \cdot \text {arg} & A \cdot+ & A \cdot \arg & A \cdot \text {arg} & A \cdot \text {arg} & A \cdot+ & A \cdot+ \\ A \cdot \text {arg} & A \cdot+ & A \cdot+ & a \cdot \arg & a \cdot+ & a \cdot \text {arg} & a \cdot \arg \\ a \cdot+ & a \cdot \arg & a \cdot \arg & A \cdot+ & A \cdot \arg & A \cdot+ & A \cdot \arg \\ a \cdot+ & \frac{a \cdot \arg }{127} & \frac{a \cdot+}{100} & \frac{a \cdot+}{36} & \frac{a \cdot+}{2} & \frac{a \cdot \arg }{4} & \frac{a \cdot+}{6} \end{array}$$

Short Answer

Expert verified
The loci arg-1 and mating-type are linked, likely mapping between the centromere and themselves at calculated distances derived from recombinant tetrad frequencies.

Step by step solution

01

Analyze the tetrad types

Examine each given tetrad type and corresponding frequency to understand how alleles are segregating in the octads from this cross. We have the following tetrads: **1:** Non-recombinant for both loci, **2, 3, 5, 6, 7:** indicative of single crossovers between loci or between a locus and the centromere, **4:** possibly a double crossover recombinant. The frequencies provide insight into linkage.
02

Map genetic distances

To determine genetic linkage and distances, apply the formula for genetic map distance. Generally, \( \text{Map Distance} (cM) = \frac{(\text{Recombinant Tetrads})}{400} \times 100 \). Calculate the distances between each gene and the centromere using the class data.
03

Calculate number of recombinants per class

From the frequencies given, calculate the number of recombinants for each class. For instance, classes 2, 3, 5, 6, and 7 involve crossovers, indicating them as recombinant classes. Sum these recombinant numbers: 127 from class 2, 100 from class 3, 36 from class 4, 2 from class 5, 4 from class 6, and 6 from class 7.
04

Determine centromere positions

The map's genetic layout requires you identify which pairings relate to centromere linkage, usually assessed by looking at equal crossover types in the data indicative of gene conversion. Analyze the class 4 data as it potentially reveals double crossovers affecting either end of the map.
05

Construct gene map

Compile your calculations into a genetic linkage map, showing both the arrangement between loci and their distances from centromeres. Ensure correct intervals in centimorgans are marked on the map. Review step calculations to finalize the map with accuracies based on recombination rates.
06

Diagram the meiosis

Illustrate meiosis events leading to class 6 using a diagram. Highlight critical phases such as prophase I (crossing over), metaphase I (alignment), and anaphase I (segregation). Properly label alleles to show crossover events yielding class 6's results.

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Key Concepts

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

Meiosis
Meiosis is a fundamental process in sexual reproduction that reduces the chromosome number by half, resulting in gametes or spores. The process consists of two consecutive divisions: meiosis I and meiosis II.
  • **Meiosis I:** This is known as the reductional division. Homologous chromosomes separate, reducing the chromosome number by half. The key stages include prophase I, where chromosomes condense and crossovers occur, exchange of genetic material between homologous chromosomes happens in the synapsis.
  • **Meiosis II:** This equational division resembles mitosis where sister chromatids are separated. Unlike meiosis I, there is no further reduction in chromosome number; instead, the chromatids separate, and the result is four haploid cells.

During prophase I, crossing over introduces genetic variation by exchanging segments between homologous chromosomes. Metaphase I aligns the homologous pairs at the cell's equator, whereas anaphase I separates them to opposite poles. The crossover and segregation events are pivotal in generating genetic diversity. This explanation helps in understanding the creation of recombinant combinations, such as those seen in recombination frequency during genetic mapping analysis.
Genetic mapping
Genetic mapping involves determining the relative positions of genes on a chromosome, highlighting their genetic linkage. Linkage occurs when genes are located close to each other on the same chromosome and tend to be inherited together.

**Key Principles of Genetic Mapping:**
  • **Linkage and Recombination:** Genes closer together have a lower chance of recombination compared to those further apart. Recombination frequency is used to infer distances on genetic maps.
  • **Centimorgan (cM):** This is the unit of map distance, where 1 cM corresponds to a 1% chance of recombination occurring between two genes.

In genetic mapping, the distances between gene pairs are calculated using the formula: \[ \text{Map Distance} \,(\text{cM}) = \frac{\text{(Recombinant Tetrads)}}{\text{Total Tetrads}} \times 100 \]
By analyzing tetrad data from crosses such as in the Neurospora example, scientists are able to position gene loci relative to each other and the centromere, providing insights into genetic traits and inheritance patterns.
Recombination frequency
Recombination frequency is a measure of how often crossing over happens between two loci during meiosis. It's a direct estimate of the physical distance separating two genetic markers on a chromosome.

**Understanding Recombination Frequency:**
  • **Calculation**: It is calculated by taking the number of recombinant progeny and dividing it by the total number of progeny observed, then multiplying by 100 to get a percentage.
  • **Expression in cM**: 1% recombination frequency approximates to 1 map unit or 1 centimorgan on a genetic map. However, distances above 50% typically indicate independent assortment and are not useful for map calculations.

When crossovers occur between genes, they generate recombinant chromosomes. In studies, like those involving Neurospora, different classes of recombinants are observed, each indicating particular crossover events. These frequencies provide essential data for constructing genetic linkage maps and understanding genome architecture.

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Most popular questions from this chapter

The recessive alleles \(k\) (kidney-shaped eyes instead of wild-type round), \(c\) (cardinal-colored eyes instead of wild-type red), and \(e\) (ebony body instead of wildtype gray) identify three genes on chromosome 3 of Drosophila. Females with kidney-shaped, cardinalcolored eyes were mated with ebony males. The \(\mathrm{F}_{1}\) was wild type. When \(\mathrm{F}_{1}\) females were testcrossed with \(k k\) \(c c\) ee males, the following progeny phenotypes were obtained: $$\begin{array}{cccr}k & c & e & 3 \\\k & c & \+ & 876 \\\k & \+ & e & 67 \\\k & \+ & \+ & 49 \\ \+ & c & e & 44 \\\\+ & c & \+ & 58 \\\\+ & \+ & e & 899 \\\\+ & \+ & \+ & 4 \\\\\text { Total } & & & 2000\end{array}$$ a. Determine the order of the genes and the map distances between them. b. Draw the chromosomes of the parents and the \(\mathrm{F}_{1}\) c. Calculate interference and say what you think of its significance

In the plant Arabidopsis, the loci for pod length \((L,\) long; \(l,\) short) and fruit hairs (H, hairy; \(h,\) smooth) are linked 16 map units apart on the same chromosome. The following crosses were made: (i) \(L H / L H \times l h / l h \rightarrow \mathrm{F}_{1}\) (ii) \(L h / L h \times l H / l H \rightarrow \mathrm{F}_{1}\) If the \(F_{1}\) 's from cross i and cross ii are crossed, a. what proportion of the progeny are expected to be \(l h / l h ?\) b. what proportion of the progeny are expected to be \(L h / l h ?\)

From parents of genotypes \(A / A \cdot B / B\) and \(a / a \cdot b / b, a\) dihybrid was produced. In a testcross of the dihybrid, the following seven progeny were obtained: \\[\begin{array}{c}A / a \cdot B / b, a / a \cdot b / b, A / a \cdot B / b, A / a \cdot b / b \\ a / a \cdot b / b, A / a \cdot B / b, \text { and } a / a \cdot B / b\end{array}\\] Do these results provide convincing evidence of linkage?

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?

A fruit fly of genotype \(B R / b r\) is testcrossed with \(b r /\) \(b r .\) In 84 percent of the meioses, there are no chiasmata between the linked genes; in 16 percent of the meioses, there is one chiasma between the genes. What proportion of the progeny will be \(B r / b r ?\)

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