Archives for February 2007

Question 1: How did natural selection change the morphology of the finch populations on Daphne Island from 1977 to 1978?

In 1977, there was a drought on Daphne Major. Plants began to wilt and shrivel, leaving the finches without much food to eat. The fortis did not breed because of this lack of food. They needed to feed themselves first before worrying about breeding. During the previous year, there had been 10 grams of seed per square meter of lava, but during this year there were only 6 grams of seed per square meter and June and only 3 grams of seed per square meter in December. This dearth of seeds led to a decrease in the selection of seeds available. Most of the easy to eat seeds were already eaten, leaving only the harder to eat seeds. One main seed left to eat was the Tribulus.

In the past, the fortis would almost never attempt to crack open a Tribulus, but now they were forced to try and open them. The smallest fortis could not open Tribulus, and were thus forced to try and eat Chamaesyce seeds. The Chamaesyce plant gave off a milky, sticky latex when its leaves and stems were cut, which got onto the birds’ feathers and helped lead to their deaths. The scientists gathered numbers on the finches and noticed that the populations were dropping, especially the smallest sized finches. The smallest finches were having the most trouble surviving the drought. It was noted that the fortis with the biggest beaks and larger bodies had the highest survival rates.

The next year when the scientists went back to the island, they noticed that the female fortis were very specific with the males they mated with. They only mated with the bigger males. This caused the population of fortis to be bigger than they were in the past. This morphology was cause due to the fact that the bigger fortis had the best chance of survival during the drought. Naturally, the female mated with the males that had the best characteristics to survive during the drought. Eventually, bigger fortis became inferior and the population switched back to smaller fortis.

Question 2: Describe Endler’s “natural” selection experiments with guppies. What did he show?

Endler studied guppies in the wild. He noticed that near the headwaters of the streams where there is only one enemy of the guppy, the guppies for the most part were colorful with large, bright spots of blue. The guppies downstream, however, that had to deal with multiple enemies were for the most part smaller and less bright with spots of black and red. Endler thus concluded that guppy females are attracted to colorful males, but that same trait also attracts predators. Therefore guppies that live in areas with few enemies are more colorful and extravagant because they stand less chance of being eaten and have a better chance of finding a mate. Guppies that live in areas with lots of enemies need to be able to blend in with gravel on the bottom of the stream in order to survive.

In order to test this theory, Endler created artificial streams. He took samples of guppies from every stream and bred them together in an environment free of enemies, creating a heterogeneous mixture of guppies. All the guppies had random attributes of spot size, location, and color. Endler then separated the guppies into different streams. Each stream had different types of gravel at the bottom. Endler introduced predators into some of the streams. In the streams where he introduced enemies, the guppies began to change. They became smaller, less colorful, and their spots changed to match the gravel. In the streams without enemies, the guppies grew bigger, their spots did not change, and they became more colorful. Endler had reproduced what had occurred in the wild. He had proved that the variation in guppies was a direct result of their environment. This showed that natural selection was true.

Question 3: How have biologists reconciled the apparent paradox of rapid evolution in the short term with much slower evolutionary rates measured in the fossil record?

Biologist had once thought that evolution took place over very long periods of time, when in fact, evolution happens quickly. In order to explain this, the analogy of a volcano is used. When close up to a volcano, one can see the violence and intense movement of lava inside and the massive amounts of smoke billowing out. There is a lot of movement and action. When looking from a distance, however, the volcano seems to be passive. Smoke coming out looks to be at a standstill, and one has to watch for a long time to notice much change.

As shown by the finches, evolution happens very fast. In just one year, the size of a finch’s beak can increase by 5 percent. This is a fairly big change. Just as quickly however, the size of the beak can decrease by 5 percent. The finch goes through a back and forth struggle of change to survive. Over the long run, the change in beak size may only be a tenth or hundredth of a percent. This is why the fossil record does not show much change. It takes a very long time for significant changes to become permanent.

Introduction

A chemical reaction usually starts with reactants which react to yield products. Many times the reactants are completely used up to make products. However, the reactants sometimes do not completely turn into products. There is an equilibrium between the concentration of reactants and products. At equilibrium, the reactants turn into product and the products decompose into reactants at the same rate. This ratio of the products to reactants at equilibrium is represented by the equilibrium constant, or K. K is found by taking the concentration and order of the products and dividing by the concentration and order of the reactants. In this experiment, iron(III) ion reacts with thiocyanate ion to produce thiocyanatoiron (III). The reaction is represented by the following equation: Fe³⁺ + SCN^– <—-> FeSCN²⁺. Using a spectrophotometer, the absorbance of FeSCN²⁺ is measured at different concentrations. The absorbance in then put into Beer-Lambert’s law, A = εbc, to find concentration and ultimately the equilibrium constant.

Experimental

First, a clean cuvette was obtained, rinsed, and filled three-fourths full with 0.5 M HNO₃ solution. This was used as a the blank solution for the spectrophotometer, which was set at 447 nm. Next, a 100 mL volumetric flask was obtained and rinsed with distilled water. About 20 mL of 2.00 x 10^-3 M KSCN solution was then dispensed into a clean, dry 50 mL beaker. 10.00 mL of this solution was pipeted into the volumetric flask. The volumetric flask was then filled with distilled water to the line on the neck. The cap was put onto the volumetric flask and it was agitated to ensure consistency of the solution. This solution was then transferred into a clean, dry 250 mL beaker.

Next, about 20 mL of 1.00 x 10^-1 M Fe(NO­₃)₃ solution was dispensed into a clean, dry 50 mL beaker. 1.00 mL of this solution was pipeted into the 250 mL beaker with the KSCN solution. The solution was then mixed with a glass stirring rod. A second clean, dry cuvette was filled three-fourths full with this solution using a disposable Pasteur pipet. The spectrophotometer was blanked with the cuvette filled with 0.5 M HNO­₃ solution and the absorbance of the second cuvette was then measured and recorded.

The KSCN and Fe(NO­₃)₃ solution was then poured back into the 250 mL beaker. Another 1.00 mL of Fe(NO­₃)₃ solution was the pipeted into the 250 mL beaker. The cuvette used for measuring absorbance was filled with this solution using the disposable Pasteur pipet and was rinsed twice. It was finally filled three-fourth full with the solution in the 250 mL beaker. The spectrophotometer was blanked again with the 0.5 M HNO₃ solution and the absorbance of the solution in the other cuvette was recorded again. This process was repeated until 10.00 mL of Fe(NO­₃)₃ solution had been added to the 250 mL beaker.

Results

Average K: 169.13

Calculations

To find [Fe^*], I took the volume of Fe(NO₃)₃ used in liters, multiplied by the molarity of the Fe(NO₃)₃ concentration, 1.00 x 10^-1 M, and divided by the total volume of the solution. For example using the first mixture, (1.00 x 10^-3 L) (1.00 x 10^-1 M) / (0.101 L) = 9.90 x 10^-4 M. I did the same thing to find [SCN^*]. For example using the mixture 1, (10.00 x 10^-3 L) (2.00 x 10^-3 M) / (0.101 L) = 1.98 x 10^-4 M. To find [Fe(SCN)²⁺], I rearranged Beer’s law to c = A / (εb), with b equaling 1 cm and ε equaling 4700 L mol^-1cm^-1. Using the first mixture as an example,

c = 0.092 / (4700 L mol^-1cm^-1 x 1 cm), c = 1.96 x 10^-5 M. To find [Fe³⁺], I used the equation [Fe³⁺] = [Fe^*] – [Fe(SCN)²⁺]. For example, from mixture 1, [Fe³⁺] = 9.90 x 10^-4 M – 1.96 x 10^-5 M = 9.70 x 10^-4 M. To find [SCN-], I used the equation [SCN-] = [SCN^*] – [Fe(SCN)²⁺]. Using the first mixture, [SCN-] = 1.98 x 10^-4 M – 1.96 x 10^-5 M = 1.78 x 10^-4 M. To find K, I took the concentration and order of the products and divided by the concentration and order of the reactants. Using the concentrations from mixture 1, K = [Fe(SCN)²⁺] / ([Fe³⁺] x [SCN-]), K = 1.96 x 10^-5 M / (9.70 x 10^-4 M x 1.78 x 10^-4 M), K = 113.52. Finally, to find the average K, I added the 10 K values found and divided by 10.

Discussion/Conclusions

My results do not seem very accurate. Even during the experiment, my partner and I agreed that our absorbance readings did not seem right. The readings jumped a significant amount between the first and second readings and the fourth and fifth readings. After the seventh reading, the absorbance readings seemed to be evening out, as the difference between readings was getting smaller. When reading some of the absorbances, we took a second reading because a few times the absorbance reading was smaller than the previous. This did not seem right; the absorbance readings should have been going up each reading. Error may be attributed to the fact that a few drops of solution escaped the 250 mL beaker during transfers between the cuvette. Error could have also resulted from the beakers not being totally clean, or the pipetment of solutions could have been inaccurate. This would have caused the volumes and concentrations to be different than they actually were.

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:

Introduction

The kinetics of a chemical equation is determined by its rate. The rate is the speed at which the reactants form into products. The rate is dependent on the concentrations and the orders of the reactants. One way to find the order is by first measuring the concentration of the products as time passes. A spectrophotometer is one tool that can measure relative concentration if the reactants change color as the form products. Graphing absorbance versus time, ln(absorbance) versus time, and 1/absorbance versus time will determine the order depending on which graph produces a straight line. In this experiment, crystal violet and NaOH form a complex that changes from transparent blue to colorless over time. The absorbance is measured using a spectrophotometer, and the rate law is then determined using this information.

Experimental

First, a spectrophotometer was turned on and set at a wavelength of 595 nm. Next, a cuvet was obtained, rinsed, and filled with deionized water. The outside of the cuvet was cleaned Kimwipe to get rid of smudges. The cuvet was then inserted into the spectrophotometer and the spectrophotometer was zeroed. Next, 10.0 mL of 0.010 M NaOH solution was dispensed into one clean 25 mL graduated cylinder and 10.0 mL of 1.50 x 10^-5 M crystal violet solution was dispensed into another clean 25 mL graduated cylinder. The solutions were then simultaneously poured into a clean 50 mL beaker. This mixture was mixed with a glass stirring rod for a few moments to ensure consistency. The cuvet was then rinsed with the mixture two or three times and was then filled with the mixture. The cuvet again cleaned with a Kimwipe and was inserted into the spectrophotometer. The absorbance reading was measured every minute for twenty minutes, starting when the cuvet was first put in. This process was then repeated, replacing the 0.010 M NaOH solution with 0.020 M NaOH solution.

Results

0.010 M NaOH Solution:

0.020 M NaOH Solution:

Discussion/Conclusions

For the graphs using 0.010 M NaOH, the plot of Absorbance vs. Time had the straightest line (R² = 0.9985), but the plot of lnA vs. Time also had a very straight line (R² = 0.9971). The plot of 1/A had the least straight line with R² equaling 0.9906. For the graphs using 0.020 M NaOH, lnA vs. Time had the straightest line (R² = 0.9998). Absorbance vs. Time and 1/A vs. Time were not nearly as straight, with their R² equaling 0.9646 and 0.9695, respectively. I think it is fairly safe to say that the rate equation is first order because lnA vs. Time overall yielded the straightest line in the two runs. The rate equation is rate = k [NaOH].

Introduction

The ideal gas law is used to define how gasses typically act. It is not precise, as gasses do not usually act ideally, but it works for most laboratory conditions. The ideal gas law is defined by PV = nRT (pressure in atm * volume in L = moles * constant * temperature in K). There is also a rule that states the total pressure of a system is equal to the sum of its partial pressures. Using this information, a small piece of metal magnesium is reacted with hydrochloric acid. The reaction is represented by the equation Mg (s) + 2HCl (aq) –> MgCl₂ (aq) + H₂ (g). Using the sum of partial pressures rule and the ideal gas law to find moles, the molar mass of Mg (s) can be calculated only knowing the pressure of H₂ (g).

Experimental

First, a strip of magnesium metal was obtained and weighed on an analytical balance. This weight was recorded to three significant figures. Next, the magnesium strip was wrapped around the tip of a copper wire, and was then encaged by the same copper wire by wrapping the wire around the magnesium. Following this, the tip of a buret was calibrated and this volume was recorded. Next, about 3 mL of concentrated HCl was put in the empty buret with the stopcock closed, and distilled water was then added until the buret was almost full. The copper wire with the magnesium was then lowered into the buret and held in place with a one-holed rubber stopper to make sure the wire would stay in place and would not fall to the bottom. Distilled water was squirted into the one-holed rubber stopped to make sure the buret was completely filled. The buret was then quickly inverted and clamped into a 600 mL beaker half filled with tap water. The water level on the buret was recorded after production of gas ceased. Finally, the distance between the height of water in the beaker and water in the buret was also recorded. The process was repeated for one more trial.

Results

Average percent error: 0.4525

Calculations

In order to find the volume of hydrogen, I took the total volume of the buret (50 mL plus the calibrated part) and subtracted my final reading. Using the results from my first trial, 50.0 mL + 1.20 mL – 9.40 mL = 41.80 mL. To find the pressure difference of water levels, I took the distance in difference of water levels, multiplied by the density of water, and divided by the density of Hg. From the first trial, 135 mm (1.0 g/mL) / 13.5 g/mL = 10.0 mm. The aqueous vapor pressure was found using a chart. At 23 ºC, the aqueous vapor pressure is 21.2 torr (or 21.1 mm Hg). Using the equation P_atm = P_{H2O (l)} + P_{H20 (g)} + P_{H2 (g)}, I could solve for the pressure of H₂ (g). Subbing in the numbers from the first trial, 762.76 torr = 21.2 torr + 10.0 torr + P_{H2 (g)}. P_{H2 (g)} = 731.6 torr. To find atmospheres, I multiplied by 1 atm/760 torr. 731.6 torr (1 atm/760 torr) = 0.9626 atm. I then rearranged the ideal gas law to find moles of hydrogen produced. n = PV/RT, n = 0.9626 atm * 0.04180 L / (0.0821 L atm mole^-1 K^-1 * 296 K), n = 0.001656 moles. In the equation for the reaction, there are an equal number of moles of hydrogen and magnesium used. To find the molar mass of magnesium, I took the mass of the magnesium used divided by the number of moles used. From trial 1, 0.0406 g / 0.001656 moles = 24.52 g/mol. To find percent error, I took the absolute value of the true value minus my calculated value divided by the true value and multiplied by 100%. Using trial 1, ( | 24.31 – 24.52 | ) / 24.31 * 100% = 0.8638%. Finally, to find the average percent error, I added the two percents and divided by two: (0.8638% + 0.04113%) / 2 = 0.4525%.

Discussion/Conclusions

My results are very good. An average percent error of 0.4525% is extremely small, so I feel that I performed the experiment well and had some good luck. I felt that when performing the experiment, I could have gotten some error from my temperature readings. The temperature seemed to fluctuate a couple degrees during the experiment and I thought that would negatively affect my final results. I also thought I could have gotten error from my difference in water levels measurement because it was hard to hold the ruler steady and read the ruler at the same time. Lastly, since the ideal gas law is not precise, error could have come from that. Other than those questionable factors, the experiment went smoothly.