Archives for March 2007

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:

Un giorno, io sono andato al “fort”. “Il fort” e’ un grande magazzino abbandonato. I miei amici sono andati al fort molte volte, ma io non sono mai andato. Loro hanno parlato del fort molto. Allora, io ho voluto andare. Il fort e’ dietro la casa dal mio amico.

Io mi sono svegliato alle otto di mattina. Mi sono lavato e mi sono metto i miei abbiglamenti. Io ho incontrato i miei amici in piazza. Quando tutti eravano la’, noi siamo usciti. Prima, noi siamo andati al “Wawa” e abbiamo comprato il cibo per la colazione. Poi, io sono andato in macchina e i miei amici sono andati in bicicletta a casa del mio amico. Poi, noi siamo andati a piedi alla foresta. Noi siamo arrivati al fort in dieci minuti. Era enorme!

Noi siamo entrati. Io avevo fretta. Io ho fatto molte fotografie. Noi abbiamo fatto la colazione in una camera con un tavolo e alcune sedie. Mi piacevo, poi noi abbiamo sentito qualcosa fuori, ma e’ partito. Dopo noi abbiamo mangiato, noi siamo andati a piedi al fort. Io ho visto tutti dall’edificio. Era interessante. C’ erano i graffiti su tutte le pareti. Era cosi’ da trente anni. La fabbrica era abbandonata. Ora, la gente va al fort per piacere.

Noi siamo andati nello scantinato e io avevo paura. Io non vedevo. Poi io sono andato sul tetto. Io ho potuto vedere molto. Mi piaceva. Infine, noi siamo usciti. E’ stato un bel giorno.

Introduction

Acids are substances that donate hydrogen ions and bases are substances that accept hydrogen ions. Acids and bases react with each other by transferring hydrogen ions. One way to distinguish an acid is by its equivalent mass, which is the number of grams of the acid needed to transfer one mole of hydrogen ion to a base. For a monoprotic acid, which only transfers one hydrogen ion, its equivalent mass equals its molar mass. For a diprotic acid, which transfers two hydrogen ions, its equivalent mass equals half its molar mass. The equivalent mass of a base is simply the number of grams required to accept one mole of hydrogen ion. The equivalent mass of an acid or base is also equal to the mass of the acid or base titrated divided by the number of equivalents of the acid or base.

Acid strength is measured by how much it dissociates. This is determined by the amount of hydronium ion formed in water. Most often, it is expressed by pH, which equals the –log[H₃O⁺]. The equilibrium constant for an acid is represented by K_a = [H₃O⁺] [A^–] / [HA], where HA is the acid and A^– is the dissociated acid. When comparing K_a’s, most commonly pK_a is used, which is equal to –log K_a.

In this experiment, the pK_a of an unknown acid is determined by titrating it with NaOH and graphing its pH levels versus volume of NaOH titrated. The inflection point found by graphing is the equivalence point, and at half that volume is the half-equivalence point. At half equivalence, the [A^–] = [HA], so they cancel out in the equation K_a = [H₃O⁺] [A^–] / [HA], leaving K_a = [H₃O⁺]. With pH being known, K_a is found by [H₃O⁺] equaling 10^-pH, and pK_a = pH.

Experimental

First, about 2 g of an unknown weak acid was obtained. Then about 0.10 to 0.12 g of the acid was weighed out on an analytical scale. This mass was recorded to 4 decimal places. The acid was then transferred to a clean 125 mL Erlenmeyer flask. Distilled water was added to dissolve the acid, and then 3 drops of phenolphthalein indicator was added to the solution. Next, a buret was filled with NaOH solution and its initial reading was recorded to 2 decimal places. The acid solution was titrated with the NaOH solution until the acid solution turned and stayed pink. This final reading was recorded to 2 decimal places and the volume of NaOH used was calculated.

Another sample of unknown acid was weighed out using an analytical scale requiring about 25 mL of standardized NaOH solution. This mass was recorded to 4 decimal places. The acid was transferred into a clean 150 mL beaker, and was again dissolved using distilled water. A pH meter and electrode was then obtained. The electrode was rinsed with distilled water and put into the acid solution. Standardized NaOH solution was again used to titrate the acid. The initial buret reading was recorded to 2 decimal places and NaOH was added 0.5 mL at a time. After every portion added, the buret reading and pH meter reading were recorded to 2 decimal places. This was done until the pH spiked and stopped increasing significantly.

Results

Calculations

In order to find the volume of NaOH used, I subtracted the initial buret reading from the final reading (14.37 mL – 0.75 mL = 13.62 mL). To find the mass of unknown acid needed using 25 mL of NaOH solution, I used a proportion. The proportion was 0.1167 g / 13.62 mL = x / 25.00 mL, with x equaling the mass of unknown acid needed. To find the half-equivalence point, I divided the equivalence point volume by 2. I then just looked at the graph to find the pH at the half-equivalence point. The pK_a is equal to the pH, which was 4.62. The K_a is equal to the concentration of [H₃O⁺]. [H₃O⁺] is equal to 10^-pH, so using my numbers, 10^-4.62 = 2.40 x 10^-5. To find percent error, I took the theoretical pK_a – my found pK_a, divided by the theoretical pK_a­, and multiplied by 100%. Using my numbers, (4.69 – 4.62) / 4.69 x 100% = 1.49%. In order to find the number of equivalents of NaOH, I took the volume at the equivalence point (25.75 x 10^-3 L) and multiplied by the molarity (1.00 x 10^-1 M). The number of equivalents of acid titrated was equal to the equivalents of NaOH. Finally, to find the equivalent mass of the acid, I divided the mass of the acid used by the number of equivalents of acid. Using my numbers, 0.2152 g / 2.58 x 10^-3 equiv. = 83.41 g/equiv.

Discussion/Conclusions

Looking at the table in the lab book, the acid with the closest pK_a to my calculated pK_a is trans-crotonic. It has a pK_a of 4.69 while my calculated pK_a was 4.62. I feel that these numbers are very close, as the percent error is only 1.49%. The percent error for my calculated equivalent mass of the acid is also small at 3.11%. Error could have come from reading the graph, as I had to estimate the equivalence point. I should have taken measurements at smaller intervals when approaching the inflection point. This would have allowed me to read the graph more accurately. Error could have also come from the pH meter, which may not have been calibrated precisely. Lastly, error could have resulted if I inaccurately read the buret while recording the readings.

Choose a song from your personal library of music which best and most thoroughly represents the emotions and conflict in Shakespeare’s Othello. The range of emotions is impressive—Betrayal, Trust, Jealous Rage, pathological lying, reality vs. appearance etc.

Please print out a full set of lyrics for the song and analyze the song in terms of Othello.

• Begin the paper with an introduction of the song, its writer, performer and any background if available.
• Begin by analyzing the song stanza by stanza and how it may directly relate to Othello. Find quotes from the character in the play to support your theory on why this song is the best fit.

• Some portion of the song, perhaps the bridge or chorus must be analyzed line by line. Again, corresponding action and quotes must appear in the play in order to maintain that your song is most appropriate.

The challenge is to be able to dually analyze the song in light of the play and vice versa. Back up your analysis with direct quotes from the play as well as detailed explanation!

I feel that the song “K05-0564” by Kid Dynamite effectively portrays the emotions of envy, betrayal, deceit, and regret strewn throughout Shakespeare’s Othello. It is a frantic, fast paced song lasting under 2 minutes, which is much like the pace of Othello. Othello regresses from a stable, powerful man into a fickle, vulnerable lunatic over the course of only a few days. The song is about the singer not being able to trust his girlfriend and putting an end to their relationship. This basically mirrors the feelings and actions of Othello, though Othello literally puts an “end” to his relationship with Desdemona. Specifically, the lyrics of this song echo how Othello feels about Desdemona as he is being brainwashed by Iago. The lyrics also describe regret, which Othello feels when he is “forced” to take Desdemona’s life. Finally, the song also talks about being controlled and manipulated, which is what Iago does to all the characters in the play.

The first stanza of “K05-0564” talks about not being able to depend on someone when that person should be counted on. When Othello marries Desdemona, he expects that they are to be faithful to each other, as should any husband and wife. Also considering Othello’s power and stature, it is makes it even more unlikely Desdemona would cheat on him. This is why he does not believe Iago when he first tries to plant the seeds of envy into Othello’s head. Othello says, “I do not think but Desdemona’s honest” (3.3.225). But later on, Othello is fully convinced by Iago that Desdemona betrayed him. Lines 3 and 4, “I feel like a fool because I believed in you. / I compared the likes of you to the things I do,” echo how Othello feels betrayed. He thought very highly of Desdemona, but once Iago gets the thought of Cassio and Desdemona together, there is no changing his thinking. Othello says, “Ay, let her rot, and perish, and be damned tonight, for she shall not live. No, my heart is turned to stone; I strike it, and it hurts my hand. O, the world hath not a sweeter creature! She might lie by an emperor’s side and command him tasks” (4.1.183-187). This shows how Othello still does consider Desdemona a beautiful person and how he puts her on his level, but that she cannot be trusted.

Stanza two of “K05-0564” relates to the betrayal felt by Othello. Line 5, “Then the bother builds, I go through it at times,” could be the way Othello feels as Iago builds jealousy. Othello gets only slightly worried when Iago begins to set his plan in motion, as he is a strong man and is not easily swayed by his emotions. When Iago first tries to talk to Othello about Desdemona and Cassio, Othello says, “No, Iago; I’ll see before I doubt; when I doubt, prove; and on the proof there is no but this; away at once with love or jealousy” (3.3.189-192). Othello wants proof before he believes anything. Over time however, Iago’s words get to Othello. The following line in the song, “You’d think I’d be used to it, but I don’t have the mind”, could show how Othello has become vulnerable to Iago’s tactics. Iago’s consistent efforts to create jealousy and envy in Othello have succeeded. This is the point where Othello has fallen into Iago’s trap and is now under his control. Othello is overtaken by emotion at one point and falls into a trance. Iago says, “Work on. My med’cine works! Thus credulous fools are caught, and many worthy and chaste dames even thus, all guiltless, meet reproach” (4.1.46-49). This is where Iago knows he has control of Othello and his ability to reason is gone. This is very out of character for Othello to lose control.

Lines 7 through 9 of the song, “…to deal with your deceit / or wallow at the feet / of empty promises or its royalty,” could describe Othello not being able to trust Desdemona anymore. It does not matter what she says to him at one point, Othello will not listen to her. Othello confronts Desdemona and says, “Come, swear it, damn thyself; lest, being like one of heaven, the devils themselves should fear to seize thee. Therefore be double-damned: swear thou art honest” (4.2.34-37). Desdemona responds saying she is truthful, but Othello does not believe her. It too late for him to change his mind about her. The lines in the song could also be interpreted as relating to the deceit of Iago. They are all simply pawns in his master plan, especially Roderigo and Cassio. They are given “empty promises” by Iago, and these words are along the lines of what they might say to him if they were to realize they were being used.

The chorus of “K05-0564” could describe Othello’s feelings after killing Desdemona. Lines 11 and 12, “I only needed you to be there for me. / I just wanted you to stop taking advantage of me,” can show Othello’s sorrow while and after taking Desdemona’s life. These lines makes it sound like he was forced to kill Desdemona, which is also shown in the play. Before Othello kills Desdemona, he says, “O balmy breath, that dost almost persuade justice to break her sword. One more, one more! Be thus when thou art dead, and I will kill thee, and love thee after. One more, and that’s the last. So sweet was ne’er so fatal. I must weep, but they are cruel tears. This sorrow’s heavenly; it strikes where it doth love” (5.2.16-22). These lines show how Othello still loves Desdemona, but he is going to kill her anyway. He acts almost if he has to kill her, like he has no choice. Lines 13 and 14 in the song, “Now, I’m not coming around anymore. / You can call it fucked up if you want,” could show how far Othello has fallen. These lines show that he knows it was wrong to kill Desdemona, but he was so jealous he had to do the deed. Othello was so caught up in envy that he lost his ability to reason.

The next stanza could be used to describe Iago’s feelings toward the whole situation. Lines 16 and 17, “I don’t care, I don’t need to be the better man. / I’m sorry if it’s not the decent thing to do,” accurately shows how Iago feels during about the whole situation. When devising his plan, Iago says, “Two things are to be done: my wife must move for Cassio to her mistress; I’ll set her on; myself awhile to draw the Moor apart and bring him jump when he may Cassio find soliciting his wife. Ay, that’s the way! Dull not device by coldness and delay” (2.3.383-388). This shows how he is only there to help himself. He does not care if other people are hurt, as long as he gets what he wants. The last line in this stanza, “Talk about it, maybe someday you’ll see the truth,” could be Iago hinting to everyone that if they were less like him, they could have avoided their fates. All the characters in Othello look out mainly for themselves, just like Iago, except he is craftier than them. The jealousy and envy Iago puts in everyone’s mind causes them to be cautious of each other, and therefore not figure out Iago was controlling each one of them.

The last stanza could be interpreted as Othello giving advice after everything is over. Lines 19 to 21, “There’s one thing that I know. / Friends, they come and go. / A lesson learned in life and I have you to owe,” can be interpreted as Othello regretting killing Desdemona because he knows that he cannot trust most people, but he Desdemona was one person he could depend on. She was one of the only people in the play that was truthful. When Cassio tells him he did not cheat with Desdemona, Othello says, “…speak of one that loved not wisely, but too well; of one not easily jealous, but, being wrought, perplexed the extreme; of one whose hand, like the base Judean, threw a pearl away richer than all his tribe…” (5.2.339-344). Othello kills himself after realizing how badly he messed up. He knows he was lucky to have Desdemona and he should have cherished and trusted her. Line 22 of the song, “I’m growing everyday and fools get in the way,” describes how Iago got in his way with his relationship with Desdemona. Without Iago in the picture, Othello and Desdemona most likely would have had a happy relationship. However, their relationship is ruined by Iago and his pawns Roderigo and Cassio. The last line, “If I whistle loud will you come and play?,” could be perceived as one last desperation by Othello to say he is sorry to Desdemona. He knows she is gone and his emotions got the best of him.

“K05-0564” – Kid Dynamite

When you say you will it really means you might.
When you don’t come through I shut up, it may start a fight.
I feel like such a fool cause I believed in you.
I compared the likes of you to the things I do.

Then the bother builds, I go through it at times. ₅
You’d think I’d be used to it, but I don’t have the mind
to deal with your deceit
or wallow at the feet
of empty promises or its royalty.

I never asked you to change. ₁₀
I only needed you to be there for me.
I just wanted you to stop taking advantage of me.
Now, I’m not coming around anymore.
You can call it fucked up if you want.

Smile if you will, a mile, if you can. ₁₅
I don’t care, I don’t need to be the better man.
I’m sorry if it’s not the decent thing to do.
Talk about it, maybe someday you’ll see the truth.

There’s one thing that I know.
Friends, they come and go. ₂₀
A lesson learned in life and I have you to owe.
I’m growing everyday and fools get in the way.
If I whistle loud will you come and play?

I never asked you to change.
I only needed you to be there for me. ₂₅
I just wanted you to stop taking advantage of me.
Now, I’m not coming around anymore.
You can call it fucked up if you want.