# Crossing of sordaria strains promotes genetic exchange

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The purpose of this study is to determine what color the ascospores of the fungus Sordaria fimicola will yield during meiosis, between the crossover of wild type and mutant alleles. A black spore (g+ and t+) the normal allele, and a tan spore (g+ and t-), the mutant, are the ones being crossed in this hybrid. My hypothesis is that all tan alleles will appear in the final results because it is the mutant allele that is raised in the final phase of meiosis. The experiment was done by retrieving a Petri dish containing a mycelia mass of hyphae, wild-type black and tan cultures, cut into small squares. Two squares from each mass were removed using the aseptic technique, and then placed into a sterile Petri dish. When collected, the squares were flipped upside down on the surface of the correspondent crossing agar, then left to incubate in the dark, at 22 to 24 degrees Celsius for seven days. Approximately a week after, the Sordaria fimicola are matured and hybrids have formed. The spores are removed from the Petri dish and placed under the microscope at 10x objective for observation. Out of the twenty-five Asci observed in MI + MII, eleven were MI and fourteen were MII. Nine of the fourteen were in the formation of two plus two plus two plus two crossover, and the remaining five were in the two plus four plus two formation. As a result, the frequency (percentage) of Asci showing crossover is fifty-six percent.

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The objective of studying Sordaria is to examine the genetic exchange between the black and tan alleles. The ascomycete fungus Sordaria fimicola, family of Sordariaceae, is a haploid organism, which spends most of its life cycle in the haploid condition (Helms). Under this condition, the body is made up of mostly long filamentous haploid (n) cells, hyphae, which aggregate to form a mass known as Mycelium. Two spherical, hyphal cells from two different mycelia combine, making one of the haploid nuclei move to the other, resulting in a nucleus containing diploid zygotes (2n). This is when meiosis occurs, forming four separate haploid cells, then the organism returns back in its original haploid condition. The haploid nuclei then undergo a mitotic division, yielding eight haploid cells. The cells thicken and form resistant cell walls, they are called ascospores, thus a sac is called an ascus (Helms). The ascocarp is where the asci linearly assemble to form hyphae, and when they come apart, discharge through the perithecium (round shape at the base with a long end upwards, similar to a flask). The spores of the Sordaria are usually black (g+ and t-), however mutations in alleles can form resulting in gray (g- and t+), tan (g+ and t-), or colorless spores (g- and t-). Dr. Lindsay S. Olive of the University of North wouCarolina at Chapel Hill was the first to study ascospore color mutants in Sordaria fimicola. She found that when two such cultures are paired distantly apart on an agar plate, a distinct line of crossed, mutant and/or wild-type perithecia will form. Crosses containing both color mutant and wild-type spores will produce perithecia that has both heterozygous and homozygous asci (Olive, 1956). This study needs to be done because Sordaria is a commonly known fungus, that can be grown on organic decay and manipulated in a laboratory to demonstrate the results of crossing over of alleles in meiosis, however the organism does have a mitotic stage attributed to it. This experiment operates using simple techniques and equipments to provide the opportunity for others to learn and have a firm understanding of fungal genetics (Mertens, 1968). For my hypothesis, I expect that all the yielding alleles from the crossover will be the tan, since the organism is haploid, the probability of a mutation resulting in another gene increases because the mycelia will be fused together.

## Materials and Methods

This experiment occurred in the University of Houston biology lab, room number 210, and during the course several instruments were used. Two types of Petri dishes were used, one containing the agar covered with mycelia hyphae, and the other one sterile where the crossing (labeled with + and -) will take place. The mycelia mass of hyphae were cut into squares on the agar plate, so it'll be easier to do the crossing. Before the exuding the next procedure of crossing the cultures, the instrument, spatula, has to be made sterile, using the aseptic technique. This was done by dipping the end of the spatula into a 150mL beaker of ninety-five percent ethanol, then carefully holding it above the alcohol lamp's flame. Afterwards, the spatula was removed, to allow cooling for fifteen seconds before proceeding on. The lid containing the wild-type culture was slowly opened midway, to prevent any contaminants from entering, and a brief test of heat was made by touching the side of the dish (avoiding the agar) to make sure the ends were cool enough for collection. When satisfaction was met, a square of mycelium was removed and placed upside down so the surface contacts the crossing agar marked with a plus (+) sign. These procedures were repeated again with another wild-type square strain, and twice more with a mutant tan square strain. Now, there should be a total of four squares: two black wild-type strains add two mutant tan strains inside of the crossing agar, approximately two centimeters apart. The plates should be allowed to incubate in a dark setting at twenty-two to twenty-four degrees Celsius for seven days. The dark is a necessary factor because the Sordaria can release its ascospores in the presence of light, without having a chance to mature. After a week has passed, the perithecia inside of the Petri dish should now appear as fused fruiting bodies, similar to the appearance of scattered seeds or dots. Also, a dark color 'X' should be visible in the place where the initial divisions of the cultures were located. For observation under the microscope, the Sordaria had to be moved onto a clean slide, so a single toothpick was used to gently graze the perithecia in an outward motion, without piercing the agar. When collected, it was gently smeared on the slide, adding a drop of water, and covered with a coverslip. Little pressure was applied with the end of a pencil eraser to press the coverslip so the perithecia would rupture in-between the two surfaces to allow viewing of the asci. Under the 10x objective magnification, eight ascospores lined up in each ascus were accounted for.

## Results

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The non-hybrid asci were not accounted for or recorded, to only emphasize the asci that went through reproduction and minimize error. The percentage of asci showing crossover can be calculated by taking the total amount of MII asci, 14, dividing by the overall total of MI plus MII, 25, then multiplying by 100, 56. This number is also known as the Frequency. Map Units of the asci is found by taking the frequency and dividing it by two, 0.28, then multiplying that by 100, yields 28 Map Units.

## Asci in Meiosis I and Meiosis II patterns that incubated at 22 to 24Ù’Ù’Ù’C.

## Number of MI Asci showing no crossover (4+4)

## Number of Asci showing crossover (2+2+2+2) or (2+4+2)

## Total MI + MII Asci

## Asci Percentage showing crossover

## (also known as Frequency)

## Map Units

## - - - - + + + +: 7 + + + + - - - -: 4

## - - + + - - + + : 5

## + + - - + + - - : 4

## - - + + + + - - :2

## + + - - - - + + : 3

## 25

## (11 MI +14 MII)

## 56%

## 28

Table 1: This figure shows the number of asci that occurred during meiosis I that showed no crossover, and the asci hybrid crossover that occurred in meiosis II. The '-' sign represents the mutant tan allele, and '+' sign represents the wild-type black allele.

## Class Data

## Total MI + MII asci counted

## 155

## Total MI asci

## 74

## Total MII asci

## 81

## Asci Percentage showing crossover

## (Also known as Frequency)

## 52%

## (81/155 Ã- 100)

## Map Units

## 26

## ((81/155)÷2 Ã- 100)

Table 2: This figure represents the total amount of asci that occurred in meiosis I and II for the class.

## Chi-square Test

The Chi-square test is used to determine the significant differences, if found, between my data, the observed, and the hypothetical valves, the expected (Helms).

Ï‡Â² = Î£ (81 - 80.6) Â² ÷ 80.6 + (74 - 74.4) Â² ÷ 74.4

Ï‡Â² = 0.16/80.6 + 0.16/74.4

Ï‡Â² = 0.004

Ï‡ = 0.06325 The formula for the Chi-square is calculated by: Ï‡Â² = Î£ (observed - expected) Â² ÷ expected

Chart 1 My Data Class Data

Ï‡Â² = Î£ (81 - 80.6) Â² ÷ 80.6 + (74 - 74.4) Â² ÷ 74.4

Ï‡Â² = 0.16/80.6 + 0.16/74.4

Ï‡Â² = 0.002 + 0.002 â‰ˆ 0.004

Ï‡ = 0.063

The percentage of asci crossover shown by the class was fifty-two percent, yet in order to get the number of crossovers to use in the formula is done by dividing 52 by 100 to yield 0.52, then that number is multiplied by 155, the total MI + MII asci by the class to yield 80.6 (the expected value). To figure out the number of non-crossovers is to take the number of crossovers subtracted from the MI + MII asci total, which yields 74.4 (the expected value). The expected values are subtracted from the observed values (MI and MII values found in table 2, the class data). Now the numbers can be plugged in for formula use.

## Critical Values of the Chi-Square Test

Degrees of Freedom(df)

Probability(p) 0.9 (9 in 10)

Probability(p) 0.5 (1 in 2)

Probability(p) 0.2 (1 in 5)

Probability(p)

0.05 (1 in 20)

Probability(p) 0.001 (1 in 100)

## 1

## 0.016

## 0.46

## 1.64

## 3.84

## 6.64

Table 3: These are the critical values for the Chi-square test.

This statistical test analyzes my data to the effect of how well the map distance value of the mutant tan allele matches the data obtained from the class, twenty-six map units, as well to test if my null hypothesis is valid. If the chi-square value calculated is greater than or equal to the critical value, then the null hypothesis is rejected. If the null hypothesis is rejected then the alternative hypothesis is accepted. This means that the expected value is too large to be accounted for resulting in a difference. If the null hypothesis is accepted, that means the chi-square value is less than the critical value, meaning the expected value is smaller or adequate enough that there is no difference. This is initially determined by the degrees of freedom (df) which is (n-1). In this case n=2 since crossover and non-crossovers are the ones being studied. The probability value (p) is the minimum "level of rejection" for the null hypothesis; for example a probability of 0.2 means that there is a 20% chance of wrong doing (Helms).

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Examples of our workMy null hypothesis for this investigation is twenty-six map units, and I accept it since the chi-square value 0.063 (Chart 1) for the class data is smaller than critical value 3.84 (Table 3). The class data fit the expected value very well. My experiment data for the distance of the mutant tan gene from the centromere on the Sordaria was a bit higher than the expected value of twenty-six map units, because it was twenty-eight map units.

## Discussion

The results represent in this study that both the wild-type black allele, and the mutant tan allele appeared when viewing with the microscope. This also shows that my hypothesis was slightly incorrect. Production of an ascus containing eight ascospores derived from a diploid zygote undergoes four nuclear divisions. The first two are meiotic, resulting in four nuclei and the last two are mitotic, with the third division consisting of four haploid cells giving rise to eight and the fourth division is where the spores mature (Mertens, 1968). The ascospores have a four plus four sequence, when no crossing over occurs in meiosis I, referred to as the MI pattern (Helms). When crossing over does occur, however, it is known as the MII pattern, having a sequence of two plus two plus two plus two, or two plus four plus two; this is when the chromatids exchange, resulting in genetic exchange. Map units are the frequency of crossing over is determined by the distance of where the gene and centromere cross in the second division of meiosis (Helms). The percentage of asci showing crossover can be used to figure what the map unit is, by taking the percentage number and dividing that by two. The reason for this is because the map unit distance is equal to the one percent frequency (Helms) so taking one half of that distance, gives you the correct amount of units. Asci in MII experience crossing over at the tan locus and the centromere. Crossover involves two of the four chromatids present, because it only crosses at one region, locus and the centromere, so half of the strands are recombinant and half are parental (Mertens, 1968).