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R. A. Emerson crossed two different pure-breeding lines of corn and obtained a phenotypically wild-type \(\mathrm{F}_{1}\) that was heterozygous for three alleles that determine recessive phenotypes: an determines anther; \(b r\) brachytic; and \(f,\) fine. He testcrossed the \(\mathrm{F}_{1}\) with a tester that was homozygous recessive for the three genes and obtained these progeny phenotypes: 355 anther; 339 brachytic, fine; 88 completely wild type; 55 anther, brachytic, fine; 21 fine; 17 anther, brachytic; 2 brachytic; 2 anther, fine. a. What were the genotypes of the parental lines? b. Draw a linkage map for the three genes (include map distances). c. Calculate the interference value.

Short Answer

Expert verified
Parental genotypes: aa, bb, cc and AA, BB, CC. Gene order: an-fine-brachytic. Interference value is calculated using recombination frequencies.

Step by step solution

01

Identify Parental Genotype

First, we determine the genotype of the parental lines based on the recessive and dominant phenotypes. The parental lines must be pure-breeding, meaning they are homozygous. Therefore, assuming wild-type genotype as dominant, one parent could be homozygous recessive for all three traits ("an", "br", "f"), and the other parent could be homozygous dominant (wild type) for all three traits.
02

Define F1 and Tester Genotypes

The 1 generation, resulting from crossing these two parental lines, will be heterozygous for all three traits (AaBbCc). The tester used for testcross is homozygous recessive for all three genes (aabbcc).
03

Analyze Progeny Phenotypes Frequencies

We need to take into account the phenotype frequencies to determine gene order. The most common phenotypes are parental types (355 anther and 339 brachytic, fine), while the least common are double crossover types (2 brachytic and 2 anther, fine). Single crossover phenotypes (88 wild type, 55 anther, brachytic, fine, etc.) will help determine gene order.
04

Determine Gene Order

Comparing single crossover and double crossover phenotypes with parentals, the least frequent ones help infer gene order. Noticing the two least common phenotypes interchanged just one gene, the gene order is determined as: an (A), fine (F), brachytic (B).
05

Calculate Recombination Frequency

To find map distances, calculate recombination frequencies. The recombination frequency between two adjacent genes is calculated as: \( \frac{\text{Number of recombinants between two genes}}{\text{Total number of progeny}} \times 100 \). Use phenotype groups accordingly for each pair.
06

Draw Linkage Map

Add up recombination frequencies between genes to construct a linkage map. If, for example, an-fine is 10cM and fine-brachytic is 20cM, the map will represent gene an, fine, and brachytic in that order, spaced by map distances.
07

Calculate Interference

Interference is calculated as: \( I = 1 - \frac{\text{Observed double crossovers}}{\text{Expected double crossovers}} \). The expected number is the product of individual recombination frequencies between pairs, times the total progeny. Use observed double crossover count for calculation.

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Key Concepts

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

Linkage Map
Linkage maps are an essential tool in genetics, helping to visualize the relative positions of genes on a chromosome. They rely on the concept that genes located physically close to each other on a chromosome are more likely to be inherited together. This is because there's a lower chance of crossover events occurring between them during meiosis.

A linkage map reflects map distances in centimorgans (cM), which represent the frequency of recombination between genes. In a practical sense, 1 cM is equivalent to a 1% chance that a crossover event will separate the linked genes in a resulting gamete. To build a linkage map, first determine the recombination frequencies between each pair of genes. Once that's established, you can arrange the genes in a linear order based on these distances.
  • Recombination frequency data allows calculation of map distances.
  • Genes separated by smaller recombination frequencies are closer together.
  • Linkage maps visually represent gene order and distance, which are crucial for solving genetic inheritance problems.
Recombination Frequency
Recombination frequency is a measure of how often two genes get separated due to crossovers during meiosis. Given as a percentage, it indicates how often recombination produces new allele combinations for two genes. When assessing genetic crosses, this frequency helps determine distances on a linkage map.

The formula to calculate recombination frequency is: \[ \text{Recombination Frequency} = \frac{\text{Number of Recombinant Offspring}}{\text{Total Number of Offspring}} \times 100 \]For instance, if you observe 88 recombinant offspring in a total of 888 progenies, the recombination frequency would be \( \frac{88}{888} \times 100 = 9.91\% \).
  • This percentage can be directly interpreted as the number of centimorgans between two genes.
  • Lower recombination frequencies signify genes are closer.
  • Essential for accurately constructing linkage maps and determining gene order.
Gene Order
Gene order determines the sequence of genes along a chromosome. It's vital for understanding the genetic layout and potential crossover events between them. To determine this order, geneticists typically analyze the frequency of recombinant offspring in genetic tests.

Using single and double crossover data, the most likely order can be inferred. For instance, when observing a genetic cross, the least frequent progeny are usually double crossover types, revealing crucial changes between parentals. By comparing single crossovers to parental types, the gene shifting in rare phenotypes can highlight the order.
  • Single crossover events help suggest relative positions between gene pairs.
  • Double crossovers confirm gene order by indicating which gene lies intermediately.
  • This understanding is crucial for accurate genetic mapping and studying genetic linkages.
Interference Calculation
Interference, in genetics, describes the phenomenon where the occurrence of one crossover event on a chromosome segment can inhibit additional crossover events in its vicinity. Calculating this factor is key to understanding the genetic landscape and crossover events' reliability.

The interference is calculated as follows: \[ I = 1 - \frac{\text{Observed Double Crossovers}}{\text{Expected Double Crossovers}} \]To determine the expected double crossovers, multiply the recombination frequencies of the adjacent gene pairs and then by the total number of offspring. For instance, if observed double crossovers are 2, while you expected 5, then \( I = 1 - \frac{2}{5} = 0.6 \), indicating 60% interference.
  • High interference suggests fewer crossovers than expected.
  • Useful for predicting crossover events' frequency and distribution.
  • Helps refine linkage maps and improve understanding of chromosome behavior during meiosis.

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Most popular questions from this chapter

An RFLP analysis of two pure lines \(A / A \cdot B / B\) and \(a / a \cdot b / b\) showed that the former was homozygous for a long RFLP allele ( \(l\) ) and the latter for a short allele (s). The two were crossed to form an \(\mathrm{F}_{1}\), which was then backcrossed to the second pure line. A thousand progeny were scored as follows: $$\begin{array}{lc} \text {Aa Bb ss} & 9 \\ \text {Aa Bb ls} & 362 \\ \text {aa bb ls} & 11 \\ \text {aa bb ss} & 358 \\ \text {Aa bb ss} & 43 \\ \text {Aa bb ls} & 93 \\ \text {aa Bb ls} & 37 \\ \text {aa Bb ss} & 87 \end{array}$$ a. What do these results tell us about linkage? b. Draw a map if appropriate. c. Incorporate the RFLP fragments into your map.

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.)

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?

The father of Mr. Spock, first officer of the starship Enterprise, came from planet Vulcan; Spock's mother came from Earth. A Vulcan has pointed ears (determined by allele \(P\) ), adrenals absent (determined by \(A\) ), and a right-sided heart (determined by \(R\) ). All these alleles are dominant to normal Earth alleles. The three loci are autosomal, and they are linked as shown in this linkage map: If Mr. Spock marries an Earth woman and there is no (genetic) interference, what proportion of their children will have a. Vulcan phenotypes for all three characters? b. Earth phenotypes for all three characters? c. Vulcan ears and heart but Earth adrenals? d. Vulcan ears but Earth heart and adrenals? (Problem 60 is from D. Harrison, Problems in Genetics. Addison-Wesley, 1970.)

In a haploid organism, the \(C\) and \(D\) loci are 8 m.u. apart. From a cross \(C d \times c D,\) give the proportion of each of the following progeny classes: (a) \(C D ;\) (b) \(c d ;(\mathbf{c}) C d\) (d) all recombinants.

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