BIO 1021 (Biology II: Genetic and Evolutionary Biology)

Abstract

This was a study on identifying different mutant E. coli donor and recipient cells and determining the efficiency of conjugation between them. Different dilutions of donor and recipient cells were first plated on L-agar without chloramphenicol, L-agar with chloramphenicol, and MacConkey plates in order to identify them. The donor and recipient cells were then combined and allowed to conjugate. These transconjugate cells were plated on MacConkey plates in order to identify them and determine conjugation efficiency. Lastly, plasmids were isolated from the donor, recipient, and transconjugate cells. The DNA bands were mapped using electrophoresis. It was found that the donor cells could grow on the L-agar without chloramphenicol and on the MacConkey plates. On the MacConkey plates, they produced purple colonies. They could not grow on the L-agar with chloramphenicol. The recipient cells could grow on both L-agar plates, and on the MacConkey plates they yielded white colonies. The transconjugate cells produced both purple and white colonies on the MacConkey plate. The efficiency of conjugation was found to be 8.77%. The photographs of the DNA bands showed that the transconjugate cells had plasmids the same size as the donor cells, and that the recipient cells did not have a plasmid. In conclusion, the recipient and donor cells were both identified, and conjugation successfully produced transconjugates, as shown by the MacConkey plates and DNA bands.

Introduction

There are three ways for bacteria to pass on genetic material: transformation, transduction, and conjugation (Russell, 2003). Conjugation is the main way for the bacteria E. coli to transfer its genetic material (Black, 1999). When transferring genetic material during conjugation, there must be a donor cell and a recipient cell (Hartwell, 2000). The donor cell contains an F plasmid and the recipient cell does not contain one. When the donor cell and the recipient cell physically touch, the F plasmid is copied over to the recipient, sometimes along with other genetic material in the cell (Russell, 2003). The efficiency of this transfer is low, however, due to the fact that it is hard for the cells to stay touching for very long (Hartwell, 2000). After the F plasmid is copied to the recipient, the recipient cell in now considered a transconjugate cell. It can now take the role of donor cell and copy its genetic material to another recipient. These cells can be plated on different cultures to see if they grow (Prescott, 1999). The cells can then be identified and it can be determined if conjugation has happened successfully. In this experiment, mutant E. coli donor and recipient cells were conjugated and plated on different agar and MacConkey plates to identify the cells, determine whether conjugation was successful, and determine the efficiency of conjugation.

Materials and Methods

Teams were first given culture tubes containing either donor or recipient E. coli cells suspended in broth. The donor cells contained the lac operon on their F plasmid. The recipient cells did not contain the lac operon, but instead had a mutation in a chromosomal gene which made it resistant to the antibiotic chloramphenicol. The cultures were then transferred to plates using the sterile technique, which was used throughout all parts of the experiment. 0.1 ml of a 10^-5 and 10^-6 dilution of each culture was transferred to separate L-agar plates without chloramphenicol and separate L-agar plates with chloramphenicol. 0.1 ml of a 10^-5 dilution of each culture was also transferred to separate MacConkey (MAC) agar plates. The culture was spread by placing small glass balls in the plates to roll around and spread the culture evenly. The small glass balls were removed and the plates were all labeled and incubated overnight at 35 ºC. The next day the number of colonies found in each plate was recorded.

For the next part of the experiment, transconjugate cells were plated and counted. The transconjugate cells were made by pipetting 1 ml of prepared donor cells and 1 ml of prepared recipient cells into an empty sterile tube. This culture was mixed by swirling it. The culture was then incubated for two hours at 35 ºC. Next, a 10^-5 and 10^-6 dilution of the donor culture were each transferred to separate MAC agar plates without chloramphenicol. 0.1 ml of a 10^-5 and 10^-6 dilution of the recipient culture were each transferred to separate L-agar plates with chloramphenicol. Finally, 0.1 ml of a 10^-4 and 10^-5 dilution of the transconjugate culture were each transferred to separate MAC agar plates containing chloramphenicol. The cultures were spread evenly on the plates like before using small glass balls. The plates were all labeled and incubated at 30 ºC overnight. The next day the number of colonies in donor and recipient plates that yielded between 30 and 300 colonies were recorded along with the number purple colonies in the transconjugate plates.

For the last part of the experiment, the plasmids in the donor, recipient, and transconjugate cells were isolated. They were isolated using a QIAprep Spin Miniprep kit and the QIAprep procedure. 1.4 ml of each culture was used for the start of the procedure. The DNA plasmids were each eluted in 50 μl of Tris buffer during the last step. 15 μl of each plasmid DNA was loaded into the gel of the apparatus, and photos of the DNA bands were taken after electrophoresis was finished. The bands were then measured and the points were plotted on semi-logarithmic graph paper.

Results

After one day, the donor cells plated were checked for colonies. The Mac plate with a 10^-5 dilution resulted in 37 purple colonies for my group (Table I). The other donor groups in the class counted 24 purple colonies and 103 purple colonies on their Mac plates (Table I). The L-agar plates with chloramphenicol for my group resulted in no colonies for both the 10^-5 and 10^-6 dilutions (Table I). The other donor groups in the class also found no colonies in their L-agar with chloramphenicol plates (Table I). The L-agar plates free of chloramphenicol for my group resulted with 46 colonies for the 10^-5 dilution and 7 colonies for the 10^-6 dilution (Table I). The other donor groups found 54 and 46 colonies in their 10^-5 dilutions, and 8 and 0 colonies for their 10^-6 dilutions, respectively (Table I).

In regard to the groups that plated recipient cells, their Mac plates resulted in 45 white colonies, 40 white colonies, and 31 white colonies (Table I). For their L-agar plates with chloramphenicol, 65 colonies, 30 colonies, and 25 colonies were found for their 10^-5 dilutions (Table I). For their 10^-6 dilutions, 3 colonies, 4 colonies, and 0 colonies were found, respectively (Table I). Lastly, for the recipient cells plated on L-agar plates without chloramphenicol, 38 colonies, 60 colonies, and 34 colonies were counted from the 10^-5 dilutions (Table I). From the 10^-6 dilution plates, 1 colony, 5 colonies, and 0 colonies were found, respectively (Table I).

The cells from the second plating were also counted after one night. The donor cells plated on Mac plates resulted in 18 colonies for the 10^-5 dilution and 0 colonies for the 10^-6 dilution (Table II). The recipient cells plated on Mac plates resulted in 57 colonies for the 10^-5 dilution and 3 colonies for the 10^-6 dilution (Table II). Lastly, the transconjugate cells plated resulted in 25 purple colonies for the 10^-4 dilution and 9 purple colonies for the 10^-5 dilution (Table II). Only the counts from the stronger dilutions were used in calculations (Table III).

The original concentration of donor cells plated was found to be 1.8 x 10⁷ cells/ml (Table III). The original concentration of recipient cells plated was found to be 5.7 x 10⁷ cells/ml (Table III). The original concentration of transconjugate cells plated was found to be 2.5 x 10⁶ cells/ml (Table III). The efficiency of recipient cells to transconjugate was 8.77% (Table IV). The average plasmid distance on the gel photo was 52.9 mm and the plasmid size was 2.89 kb.

Discussion

The donor and recipient cells both grew on the L-agar plates without chloramphenicol. This is because L-agar plates contain simple chemical medium which E. coli can use as its food source (Miller, 1972). Only the recipients grew on the L-agar plates with chloramphenicol, however. Chloramphenicol is an antibiotic that inhibits bacterial protein synthesis (Davies 1994). The recipient cells contained genes on their bacterial chromosome which made them resistant to chloramphenicol. The donor cells did not carry the genetic information to be resistant to chloramphenicol, so they were not able to grow. In regards to the MacConkey plates, the donor cells plated grew into purple colonies while the recipient cells plated grew into white colonies. This is because the donor cells contained an F plasmid that had the lac operon on it. They were considered lac⁺. The lac operon allows cells to digest lactose, which was on the MacConkey plates. The digestion of lactose produced some acidity. There was an pH indicator in the MacConkey plates that turned the colonies purple in the presence of the acid. This is why the donor cells plated became purple colonies. The recipient cells did not contain an F plasmid, and therefore no lac operon. They were considered lac^–. This meant that they could not digest lactose, and thus did not produce the extra acidity to activate the indicator. This is why the recipient cells plated on the MacConkey plates became white colonies.

The transconjugate cells plated on the MacConkey plates produced white colonies and purple colonies. This is because of conjugation between the donor cells and the recipient cells. When the two cells touch, the F factor from the donor cells is copied to the recipient cells (Black, 1999). This means that the transconjugate cells were recipients that contained the F factor, which had the lac operon on it. Not all of the recipient cells in the transconjugate mixture conjugated however, and this is why both white colonies and purple colonies were found. Some of the cells could digest lactose while other could not. If the transconjugated cells were plated on the L-agar plates with chloramphenicol, I would expect them to grow. The F factor from the donors would have been copied to the recipients, and the recipients would still have the genes coding for resistance to chloramphenicol on their bacterial chromosomes. The transconjugates are able to digest lactose, which was tested, and should be resistant to chloramphenicol. The donor cells would still not be resistant to chloramphenicol even in the presence of transconjugate cells because only cells that contain no F plasmid can be conjugated (Hartwell, 2000). The transconjugate cells could not pass the genetic information on to the donor cells because they already would have an F plasmid.

The efficiency of the donor cells to transconjugate the recipient cells was 8.77% (Table IV). This number should be fairly low because of recipients only receive 3% or less of the donor’s DNA (Hartwell, 2000). It is difficult for the cells stay touching together for very long, so the efficiency of conjugation is not going to be very high. The percent found for my group seems larger than average, but I suppose over time more and more of the recipients will conjugate and receive the F factor. Once a recipient is transconjugated, it could become a donor itself, so that would speed the process.

In Tables II and III, the number of colonies found when reducing the concentration from either a 10^-5 to 10^-6dilution or a 10^-4 to 10^-5 dilution should have be around a tenth of the colonies found from the stronger concentration. This is because the concentration of cells is reduced by a factor of 10%, and thus only 10% of the original number of cells should be in that concentration. The reason why perfect numbers did not show up is because the sample of cells taken from the mixture and plated could have had more or less cells than normal. The mixture may not have been thoroughly agitated. This is the cause for the error there.

In regards to the gel graph, the donors and transconjugates had plasmids nearly the same size. This is because the donors copy their F plasmid to the recipients, so the transconjugates should have plasmids the same size as the donors. If their plasmids were not the same size, then something probably went wrong during the procedure. This was confirmation that the donors truly did copy their F plasmids to the recipients. The recipients in the gel graph did not show an F plasmid. This is also proof that the recipients really had no F plasmid, and that they received the F plasmid from the donors during conjugation.

Literature Cited

Black, Jacquelyn G. 1999. Microbiology: Principles and Explorations (4^th Ed). (Prentice Hall, Upper Saddler River, NJ) 786 p.

Davies, Julian. 1994. Antibiotics and Resistance Genes.

Hartwell, L, L. Hood, M.L. Goldberg, A.E. Reynolds, L.M. Silver, R.C. Veres. 2000. Genetics: From Genes to Genomes. (McGraw Hill, Boston) 813 p.

Prescott, L.M., J.P. Marley, and P.A. Klein. 1999. Microbiology (4^th Ed). (McGraw Hill, Boston) 962 p.

Russell, Peter J. 2003. Essential iGenetics. (Benjamin Cummings, San Francisco).

Tables

Table I:

Table II:

Table III:

Table IV:

Abstract

This was a study on determining the inheritance mechanisms for different physical traits found in Drosophila melanogaster. In order to accomplish this, a reciprocal cross was performed by mating wild type male flies with mutant female flies and then wild type female flies with mutant male flies. After two weeks, the counts and phenotypes of the resulting F1’s were recorded, and then another reciprocal cross was performed by mating the F1’s from each of the parental crosses. Predictions were then made for the phenotypic ratios of the F2’s by following a believed inheritance mechanism. After another two weeks, the actual counts and phenotypes of the F2’s were recorded. The predicted ratios were compared to the actual ratios using a chi-squared test to see if the predicted inheritances mechanisms were correct. The F2 ratios for mutant A were predicted using the autosomal inheritance pattern. Both chi-squared values failed, but this is believed to be because of human error or too small of a sample size. The F2 ratios for mutant D were predicted using the X-linked inheritance pattern. Both chi-squared values these ratios passed, which means that the trait for mutant D is passed on by X-linked inheritance. In conclusion, the inheritance mechanisms were determined, but there is some doubt for the inheritance mechanism of mutant A.

Introduction

The genes that code for specific physical traits found in species are passed on from parent to progeny by different inheritance mechanisms. One mechanism, autosomal inheritance, involves the passing of a gene found on an X chromosome coded for by two alleles. In most cases, there is one dominant allele and one recessive allele, in which trait coded for on the dominant allele shows over the recessive trait. There can however, be more than just two different types of alleles on a gene. Often, the dominant allele codes for the wild type trait of a species. Wild type is a term which refers to the most common physical trait found in a species. This does not mean that the allele coding for this trait is a dominant allele, however this is often the case. When the gene is passed to the progeny, one allele from each parent is contributed to make the gene for the progeny. Physical traits can sometimes be coded for by two different genes, or there can be alleles that are not completely dominant or recessive to each other. This allows room for many different variations and ratios of physical traits.

Another inheritance mechanism is X-linked inheritance. In this type of inheritance, the gene for the physical trait is encoded for on the X sex chromosome. The gene can also have different variations of alleles, just as in autosomal inheritance, but in this case the male will only have one allele. This is because males only have one X sex chromosome. Females have two X sex chromosomes, so they will have two alleles that work much like they do in autosomal inheritance. Because the males only have one X chromosome, this means they only pass one allele on to their female progeny and none to their male progeny (Goodnenough, 1984). Females have two alleles to pass on, so they pass one allele to the male progeny and one to the female progeny.

A way to find the inheritance mechanism for a certain trait is by crossing a male wild type with a female that has the trait, and then doing the reciprocal cross, female wild type with male with the trait (Greenspan, 1997). After performing this, the F1’s can then be used for another reciprocal cross. By this time, it should be possible to analyze the counts of the progeny and map out the inheritance pattern. In this experiment, the inheritance patterns of different fly traits found in Drosophila melanogaster were determined though reciprocal crosses of parental flies and their F1 progeny.

Materials and Methods

Six different variations of drosophila melanogaster were distributed to teams in vials containing about 10 adult flies each. The six variations of drosophila were labeled wild type, mutant A, mutant B, mutant C, mutant D, and mutant E. Each species of fly was observed under a microscope. In order to view the flies under the microscope without them flying away, a cooler with ice was first obtained and an empty vial was placed into the ice. The flies in the first vial were knocked down from the top by tapping the vial on the table. While the flies were still down, the foam cork was removed and the vial was quickly flipped upside-down on top of the empty vial in the cooler. In a few moments, the flies fell asleep from the coldness of the ice and fell into the empty vial. The dormant flies were then transferred from the new vial to a metal outlet cover placed on a plate of ice. The flies were now observed under a 10x lensed light microscope. Fine paint brushes and toothpicks were used to manipulate the flies to better view them. The sex was determined by looking for sex combs, which are only on males, or by looking for darker tails, which are also found only on males. The phenotype was determined by eye color, eye size, or wing size. Once observations were finished being recorded, the flies were discarded into the fly morgue, which was a can with about one inch of vegetable oil in the bottom.

After observing all six species of drosophila, teams then picked a mutant to mate with the wild type. A reciprocal cross was performed, mating the wild type male with the mutant female and the wild type female with the mutant male. First, two empty vials were filled about a third of the way up with instant blue drosophila food and 0.5% propionic acid, which was needed to feed the larvae. Six to eight grains of yeast were also put in each vial to feed the adults. Then, vials containing wild type males and females and mutant males and females were distributed to each team. The flies were “knocked out” in the same manner as they were before, and were observed under the microscope in order to determine phenotypes and sex them. Once teams became better at discerning fly characteristics, the flies could be viewed simply with the naked eye. These figures were recorded, then about six mutant females and wild type females were put into one of the new vials with food, and about six mutant males and six wild type females were put into the other new vial. These vials were stored in a 20º C lab room.

A week later, the adult flies were cleared (put into the morgue) and observations were recorded about the appearance of the vial. After another week, some F1 progeny had emerged. These flies were knocked out and observed. Their sex and phenotypes were recorded. During the next two weeks, the F1’s were observed and cleared four more times. These counts were recorded on a sheet containing all of the class’s information. The F1’s were then crossed in the same manner the parentals were crossed. About six female and six male progeny from the mutant males by wild type female cross were put in a vial together, and about six female and six male progeny from the mutant female by wild type male cross were put in a vial together. These flies were observed and their F2 progeny counts were recorded after two weeks. This data was also recorded on a sheet along with the rest of the class’s data. All of the class’s data was recorded into a notebook to take home and analyze.

Results

When observing our mutant fly (mutant D), we noticed it had eyes which were the same size as the wild type flies, but they were white in color (or this could be described as being colorless). Other than this trait, it appeared the same as the wild type. It seemed to be the same width and length, have the same sized wings, and have the same beige body color. The eye color is what distinguished it as being different than the wild type.

For the cross between mutant A males and wild type females, the resulting F1 progeny were 87 wild type males and 103 wild type females (Table I). The reciprocal of this cross, mutant A female by wild type male, yielded 75 wild type males and 88 wild type females (Table I). For the crosses between these F1’s, the progeny of the mutant A males by wild type female cross produced 40 mutant A males, 35 mutant A females, 131 wild type males, and 177 wild type females (Table II). These F2’s were compared to an expected ratio of 1 mutant A male to 1 mutant A female to 3 wild type males to 3 wild type females (Table III). The resulting chi-squared value of 13.62 was compared the 95% confidence value of 7.82 for 3 degrees of freedom (Table III, IX). Chi-squared was greater than this value, so that meant the proposed ratio had to be rejected. The progeny of the mutant A female by wild type male cross yielded 7 mutant A males, 16 mutant A females, 53 wild type males, and 71 wild type females (Table II). These F2’s were also compared to an expected ratio of 1 mutant A male to 1 mutant A female to 3 wild type males to 3 wild type females (Table IV). The resulting chi-squared value of 12.01 was compared the 95% confidence value of 7.82 for 3 degrees of freedom (Table IV, IX). Chi-squared was greater than this value, so that meant the proposed ratio had to be rejected.

For the cross between mutant D males and wild type females, the resulting F1 progeny were 25 wild type males and 25 wild type females (Table V). The reciprocal of this cross, mutant D female by wild type male, yielded 78 mutant D males and 87 wild type females (Table V). For the crosses between these F1’s, the progeny of the mutant D males by wild type female cross produced 58 mutant D males, 66 wild type males, and 159 wild type females (Table VI). These F2’s were compared to an expected ratio of 1 mutant D male to 1 wild type male to 2 wild type females (Table VII). The resulting chi-squared value of 4.78 was compared the 95% confidence value of 5.99 for 2 degrees of freedom (Table VII, IX). Chi-squared was less than this value, so that meant the proposed ratio could not be rejected. The progeny of the mutant D female by wild type male cross yielded 26 mutant D males, 31 mutant D females, 27 wild type males, and 36 wild type females (Table VI). These F2’s were compared to an expected ratio of 1 mutant D male to 1 mutant A female to 1 wild type male to 1 wild type female (Table VIII). The resulting chi-squared value of 2.06 was compared the 95% confidence value of 5.99 for 2 degrees of freedom (Table VIII, IX). Chi-squared was less than this value, so that meant the proposed ratio could not be rejected.

Discussion

After analyzing the counts of the F1’s, I was able to determine the mechanism by which the specified trait was passed on. For mutant D, it appeared that the trait for colorless eyes is an X-linked recessive trait. This is the only model of inheritance that seemed to fit with the resulting progeny. I denoted the wild type gene as X⁺ and the mutant gene as X^D. The first parental cross was mutant D male by wild type female (X^D/Y x X⁺/ X⁺), which hypothetically should have resulted in ½ wild type females (X⁺/ X^D) and ½ wild type males (X⁺/Y). The second parental cross was wild type male by mutant D female (X⁺/Y x X^D/ X^D), which hypothetically should have resulted in ½ wild type females (X⁺/ X^D) and ½ mutant D males (X^D/Y). The F1 data seemed to follow these numbers. Following the X-linked recessive inheritance pattern, I predicted what the F2’s should be. With the use of a Punnett square, I found that the cross between F1 wild type males and F1 wild type females (X⁺/Y x X⁺/ X^D) should result in ½ wild type females, ¼ wild type males, and ¼ mutant males. The genotypes of these progeny would be ¼ X⁺/ X⁺, ¼ X⁺/ X^D, ¼ X⁺/Y, and ¼ X^D/Y. In order to determine whether my predicted phenotypic ratio matched the observed ratio, I used a chi squared test.

A chi-squared test takes a null hypothesis, which says that there is no difference between observed data and predicted data, and determines whether the null hypothesis is valid (Russell, 2003). The observed data is compared to expected data using a mathematical formula to produce a chi-squared value. That value is then compared to a probability value from a table, which is obtained depending on the degrees of freedom. Degrees of freedom is found by taking the total number of classes and subtracting one. The most common probability value is 95% confidence. If the chi-squared value is greater than the probability value, the null hypothesis is rejected, but if it is less than the probability value, the null hypothesis cannot be rejected.

I compared my predicted phenotypic ratio to the actual F2 counts and my chi-squared value of 4.78 was less than the 95% confidence value of 5.99 for 2 degrees of freedom, which means my predicted ratio could not be rejected (Table VII). I also used a Punnett square to find the hypothetical F2’s between the F1 mutant D males and F1 wild type females (X^D/Y x X⁺/ X^D). The Punnett square predicted ¼ wild type females, ¼ wild type males, ¼ mutant D females, and ¼ mutant D males. The genotypes of these progeny would be ¼ X⁺/ X^D, ¼ X⁺/Y, ¼ X^D/ X^D, and ¼ X^D/Y. I also compared the phenotypic ratio to the actual F2 counts and my chi-squared value of 2.06 was less than the 95% confidence value of 7.82 for 3 degrees of freedom, which means my predicted ratio could not be rejected (Table VIII). This means that I am most likely correct in my assumption of colorless eyes being an X-linked recessive trait. Also, whenever the reciprocal crosses give significantly different F1 and F2 phenotypic ratios, the trait is usually X-linked, which was the case here (Goodenough, 1984). This also supports the hypothesis of X-linked inheritance for white eyes.

In concern to mutant A, after analyzing the results of the F1’s, it seemed that having small eyes is an autosomal recessive trait. I denoted the wild type gene as X⁺ and the mutant gene as X^A. The first parental cross was mutant A male by wild type female (X^A/X^A x X⁺/ X⁺), which hypothetically should have resulted in ½ wild type females (X⁺/ X^A) and ½ wild type males (X⁺/ X^A). The second parental cross was wild type male by mutant A female (X⁺/ X⁺ x X^A/ X^A), which also hypothetically should have resulted in ½ wild type females (X⁺/ X^A) and ½ wild type males (X⁺/ X^A). The F1 data seemed to come close to these numbers. Using a Punnett square, I predicted what the F2 progeny should be after crossing the F1’s. I found that the cross between F1 wild type males and F1 wild type females (X⁺/ X^A x X⁺/ X^A)should result in 1/8 mutant A males, 1/8 mutant A females, 3/8 wild type males, and 3/8 wild type females. The genotypes of these progeny would be ¼ X⁺/ X⁺, ½ X⁺/ X^A, and ¼ X^A/ X^A. I compared the phenotypic ratio to the actual F2 counts and my chi-squared value of 13.62 was greater than the 95% confidence value of 7.82 for 3 degrees of freedom, which means my predicted ratio had to be rejected (Table III). I also used a Punnett square to find the hypothetical F2’s between the other cross for mutant A, which was also F1 wild type males and F1 wild type females (X⁺/ X^A x X⁺/ X^A). The Punnett square again predicted 1/8 mutant A males, 1/8 mutant A females, 3/8 wild type males, and 3/8 wild type females. The genotypes were the same as before, being ¼ X⁺/ X⁺, ½ X⁺/ X^A, and ¼ X^A/ X^A. I also compared the phenotypic ratio to the actual F2 counts and my chi-squared value of 12.01 was greater than the 95% confidence value of 7.82 for 3 degrees of freedom, which means my predicted ratio could had to be rejected (Table IV).

Even though my proposed ratios were rejected for mutant A, I still believe that the trait for small eyes is an autosomal recessive trait. There is no other method of inheritance that would fit with the phenotypes found in the F1’s and F2’s. If more flies were counted, I feel that the chi-squared value may be lower, and thus my proposed ratios would be accepted. However, there were less mutant D flies counted than mutant A flies and my proposed ratios for mutant D passed the chi-squared test. This could mean that there was error in counting the mutant A flies. The trait of colorless eyes is easier to discern than small eyes, so I feel the results for mutant D are more accurate than the results for mutant A.

The genes for white eyes (colorless eyes) and bar eyes (small eyes) are both found on chromosome 1 of drosophila melanogaster (Russell, 1994). This is an X chromosome, which rules out either of the traits being passed on by Y-linked inheritance. Only genes on the Y chromosome as passed on by Y-linked inheritance. This proves that my assumptions of autosomal inheritance for mutant A and X-linked inheritance for mutant D could be correct. Those are the only two inheritance mechanisms that involve the X chromosome. The traits for white eyes and bar eyes are also recessive (Russell, 1994). This goes along with my prediction of both traits being recessive. The only thing I do not know for certain is if the specific traits are either autosomal or X-linked.

Literature Cited

Goodenough, Ursula. 1984. Genetics (3^rd edition). (Saunders College Publishing, Philadelphia PA).

Greenspan, Ralph J. 1997. Fly Pushing: The Theory and Practice of Drosophila Genetics. (Cold Spring Harbor Laboratory Press, Plainview NY).

Russell, Peter. 1994. Fundamentals of Genetics. (HarperCollins, New York NY).

Russell, Peter J. 2003. Essential iGenetics. (Benjamin Cummings, San Francisco).

Tables

Table I:

Table II:

Table III:

Table IV:

Table V:

Table VI:

Table VII:

Table VIII:

Table IX:

Abstract

This study was on the Mendelian phenotype ratio of corn and sorghum. Second generation corn and sorghum seeds were planted, monitored, and observed for three weeks. The phenotypes of the plants grown were recorded. Using a chi-squared test, the observed phenotypic ratios were compared to Mendelian ratios to see if the ratios matched. The first generation of corn agreed with a 3:1 ratio of tall plants to short plants. The second generation of corn agreed to a 9:3:3:1 ratio of tall green plants to tall white plants to short green plants to short white plants. The sorghum agreed with a 9:4:3 ratio of green plants to white plants to white with green plants. In conclusion, the plants all matched Mendelian ratios, and therefore their genotypes and their parent’s genotypes were able to be determined.

Introduction

Gregor Mendel first came up with the concept of genetics in the 1860’s (Dewitt, 2003). He discovered that parents pass genes on to their offspring coding for different physical characteristics (phenotypes). Genes can come in different variations, called alleles. Alleles for different phenotypes can be dominant or recessive when paired with another allele, which means that only one allele will be expressed physically. Mendel found that when organisms with different allele combinations (genotypes) were mated, they produced specific ratios of offspring with specific phenotypes (Griffiths, 2000). When organisms with one dominant allele and one recessive allele for a gene were crossed, a 3 to 1 ratio in the dominant phenotype to recessive phenotype resulted. Depending on how many genes accounted for a phenotype and if the alleles were completely dominant to each other or incompletely dominant among other factors, different ratios of phenotypes resulted, including 9:3:3:1, 1:2:1, and 9:4:3. In this experiment, observed ratios of phenotypes of plants were compared to Mendelian ratios in order to figure out the possible genotypes of the plants.

Materials and Methods

A 1.5 inch plastic flat was obtained containing seemingly equal amounts of potting soil, perlite, and sphagnum. These three parts were mixed evenly and the mixture was then slightly packaged down. About 50 second generation (F₂)corn seeds were obtained and inserted into the soil. The seeds were pushed about 0.5 inches into the soil about an inch away from each other in rows containing about 8 seeds each. Rows were about 2 inches away from each other. Once all seeds were planted, the flat was then watered with tap water until the soil was moist. A second and third flat were prepared in the same manner, only containing a different type of F₂ corn seeds and F₂ sorghum seeds. The flats were kept in a 20º C classroom with lights on for three weeks. The flats were watered as necessary, and plant growth was monitored and recorded during the three weeks.

Results

For the first F₂ generation of corn observed, 38 tall plants and 11 short plants were counted (Table I). The tall plants had skinny long leaves while the short plants had shorter wider leaves. The observed ratio was 3.45 tall plants for every 1 short plant (Table I). Comparing this ratio to a Mendelian ratio of 3 tall plants for every 1 short plant resulted in a chi-squared value of 0.169 (Table II). This number was less than the 95% confidence value of 3.84 for 1 degree of freedom, so this Mendelian ratio was unable to be rejected (Table II).

For the other F₂ generation of corn observed, 22 green and tall plants, 9 green and short plants, 9 white and tall plants, and 4 white and short plants were counted (Table III). The tall phenotype meant the plants had long skinny leaves and the short phenotype meant the plants had short wide leaves. The green phenotype meant the plant was green and the white phenotype meant the plant was a white color. The observed ratio was 5.5 green tall plants to 2.25 green short plants to 2.25 white tall plants to 1 white short plant (Table III). Comparing this ratio to a Mendelian ratio of 9 green tall plants to 3 green short plants to 3 white tall plants to 1 white short plant resulted in a chi-squared value of 1.016 (Table IV). This number was less than the 95% confidence value of 7.82 for 3 degree of freedom, so this Mendelian ratio was unable to be rejected (Table IV).

For the F₂ generation of sorghum observed, 29 green plants, 8 white with green plants, and 9 white plants were counted (Table V). The green plants were green in color, the white with green plants were white in color with spots of green, and the white plants were white in color. The observed ratio was 3.68 green plants to 1.38 white plants to 1 white with green plant (Table V). Comparing this ratio to a Mendelian ratio of 9 green plants to 4 white plants to 3 white with green plants resulted in a chi-squared value of 0.36 (Table VI). This number was less than the 95% confidence value of 5.99 for 2 degrees of freedom, so this Mendelian ratio was unable to be rejected (Table VI).

Discussion

A chi-squared test takes a null hypothesis, which says that there is no difference between observed data and predicted data, and determines whether the null hypothesis is valid (Russell, 2003). The observed data is compared to expected data using a mathematical formula to produce a chi-squared value. That value is then compared to a probability value from a table, which is obtained depending on the degrees of freedom. Degrees of freedom is found by taking the total number of classes and subtracting one. The most common probability value is 95% confidence. If the chi-squared value is greater than the probability value, the null hypothesis is rejected, but if it is less than the probability value, the null hypothesis cannot be rejected.

For the first generation of corn observed, the observed proportion of 3.45 tall plants to every 1 short plant was compared to the Mendelian ratio of 3 tall plants to every 1 short plant (Table I). This resulted in a chi-squared value of 0.169, which is less than the 95% confidence value of 3.84 for 1 degree of freedom (Table II). This means the proposed ratio could not be rejected and that it is a possibility for the actual proportion of tall plants to short plants. The genotypes would have to be ¼ TT, ½ Tt, and ¼ tt, with the tall allele (T) being completely dominant to the short allele (t). Both plants with TT and Tt genotype would be tall, resulting in ¾ tall plants and ¼ short plants, matching the 3:1 Mendelian ratio. This means that the parents of this generation must have been both heterozygous in order to produce the 1:2:1 genotype ratio.

For the second generation of corn observed, the observed proportion of 5.5 green tall plants to 2.25 green short plants to 2.25 white tall plants to 1 white short plant was compared to the Mendelian ratio of 9 green tall plants to 3 green short plants to 3 white tall plants to 1 white short plant (Table III). This resulted in a chi-squared value of 1.016, which is less than the 95% confidence value of 7.82 for 3 degrees of freedom (Table IV). This means the proposed ratio could not be rejected and that it is a possibility for the actual proportion. The genotypes would have to be 9/16 G- T-, 3/16 G- tt, 3/16 gg T-, and 1/16 gg tt, with the green allele (G) being completely dominant to the white allele (g), and the tall allele (T) being completely dominant to the short allele (t). The parents for this generation must have been both heterozygous for both traits in order to produce a 9:3:3:1 phenotype ratio.

Lastly, for the generation of sorghum observed, the observed proportion of 3.68 green plants to 1.38 white plants to 1 white with green plant was compared to the Mendelian ratio of 9 green plants to 4 white plants to 3 white with green plants (Table V). This resulted in a chi-squared value of 0.36, which is less than the 95% confidence value of 5.99 for 2 degrees of freedom (Table VI). This means the proposed ration could not be rejected and that it is a possibility for the actual proportion. The genotype for this plant is dependent on two genes, green (G) being completely dominant to white (g), and normal pigmentation (N) is completely dominant to a variation in pigmentation (n). The recessive allele for variation in pigmentation causes a plant with a dominant green allele to have white patches (a lack of pigmentation). A white plant with the recessive pigmentation genotype will have no visible effect in its phenotype because it does not have any pigmentation. This is an example of epitasis, in which one gene is dependent on another (Griffiths, 2000).

The green plants consists of the genotype G- N-, which adds up to 9/16 of the proportion (1/16 GG NN, 1/8 GG Nn, 1/8 Gg NN, and 1/4 Gg Nn). The white phenotype consists of the genotypes gg NN, gg Nn, and gg nn, which adds up to 4/16 of the proportion (1/16 gg NN, 1/8 gg Nn, and 1/16 gg nn). Finally, the white with green phenotype includes the genotypes Gg nn and GG nn, which adds up to 3/16 of the proportion (1/8 Gg nn and 1/16 GG nn). The parents for this generation had to have been heterozygous for both genes in order to produce a 9:4:3 phenotypic ratio.

Literature Cited

DeWitt, Stetten Jr. 2003. The Genetic Basics: What Are Genes and What Do They Do?. (National Institute of Health). http://www.history.nih.gov/exhibits/genetics/sect1f.htm.

Griffiths, Anthony J.F., Miller, Jeffrey H., Suzuki, David T., Lewontin, Richard C., and William

M. Gelbart. 2000. An Introduction to Genetic Analysis. W.H. Freeman and Company.

Russell, Peter J. 2003. Essential iGenetics. Benjamin Cummings, San Francisco.

Tables

Table I:

Table II:

Table III:

Table IV:

Table V:

Table VI: