AP Biology Graphing & Calculations Diffusion & Osmosis Students in an AP Biology class explored the diffusion of different molecules through dialysis tubing, a semipermeable membrane. They used glucose test strips to check for the presence of glucose in an aqueous solution and potassium-iodide (IKI) to test for the presence of starch in the same aqueous solution. IKI reacts with starch to give a dark blue, almost black color. When IKI reacts with starch, it becomes part of the starch molecule and is removed from solution. The design of the experiment is shown below and the following data were obtained:

1. What does the data tell you about the sizes of the molecules relative to the pore size of the dialysis tubing?

2. Does this activity account for the diffusion of all the molecules found in the initial set up? If not, what data could have been collected to show the net direction of diffusion of this molecule or molecules?

Students in an AP Biology class designed an experiment to investigate the influence (if any) of solute concentration on the net movement of water molecules through a semipermeable membrane. The solute they used was sucrose (cane or table sugar) in the following molar concentrations: 0.0 M (distilled water); 0.2 M; 0.4 M; 0.6 M; 0.8 M; and 1.0 M. 3. Sketch and label the probable experimental setup used by the students: 4. What would have been an appropriate hypothesis for this experiment? 5. What is the independent variable? 6. What is the dependent variable? 7. What constants should have been used by the students? The students collected their data and graphed it – the following results were obtained:

8. On the basis of the results, write a statement that expresses the relationship of solute concentration and direction of net movement of water molecules in osmosis.

9. In which, if any, of the experimental setups were the solutions in the bag and outside of the bag isotonic to each other?

10. When you drink a glass of water, most of it is absorbed by osmosis through cells lining your small intestine. Drinking seawater can actually dehydrate the body. How?

Enzyme Catalysis In a particular experiment, students observed some characteristics of enzyme action. The specific reaction they investigated was the decomposition of hydrogen peroxide by the enzyme, catalase. At room temperature, hydrogen peroxide very slowly decomposes into water and oxygen. The addition of catalase lowers the activation energy of the reaction until it proceeds swiftly at room temperature. At the end of the reaction, the catalase is unchanged and is available to catalyze the reaction of more hydrogen peroxide. Catalase, like most enzymes, is a protein. Its ability to form a complex with hydrogen peroxide is based on its molecular shape. Any factor that can alter the shape of a protein molecule can be expected to impact the ability of catalase to facilitate the decomposition of hydrogen peroxide. The following graph was produced by the students:

1. What was the independent variable in this experiment? What was the dependent variable?

2. What does the line on the graph represent?

3. What is the rate of the enzyme-catalyzed reaction? What does the graph tell you about the rate of the reaction over time?

Mitosis & Meiosis Environmental factors can affect the rate of mitosis in plant cells. In a recent issue of Biology Weekly, scientists reported that a fungal pathogen may negatively affect the growth of soybeans (Glycine max). Soybean growth decreased during three years of high rainfall, and the soybean roots were poorly developed. Close relatives of the R. anaerobis are plant pathogens and grow in the soil. A lectin-like protein was found in the soil around the soybean roots. This protein may have been secreted by the fungus. Lectins induce mitosis in some root apical meristem tissues. In many instances, rapid cell divisions weaken plant tissues. Suppose you were asked to investigate whither the fungal pahtogen lectin affects the number of cells undergoing mitosis in a different plant, using onion root tips. You hypothesize that treating onion root tip cells with lectin will induce mitosis in the cells, as was the case with the soybean plant. After designing and conducting your experiment, the following data are reported:

Number of Cells

Group Interphase Mitosis Total

Control/Non-Treated 1550 200 1750

Experimental/Treated w/ Lectin 500 1250 1750

1. Are deviations from the expected value due to chance alone? Calculate the chi-square value for this data. Is your hypothesis

supported or refuted? Does lectin induce mitosis in onion root cells? Sordaria fimicola is a common species of ascomycete fungi that grows on the dung of herbivores. Eight spores (ascospores) are produced in an ascus. Many asci are grouped together within a vase-shaped structure called a perithecium. Two nuclei within a developing ascus fuse to produce a diploid (2n) nucleus. This diploid nucleus then undergoes meiosis, followed by a mitotic division to produce eight ascospores in a linear series within the ascus. When the growing filaments of two haploid strains of Sordaria that produce spores of different colors meet, fertilization occurs and zygotes form. The figure below shows spore formation in Sordaria:

The number of map units between two genes or between a gene and the centromere is calculated by determining the percentage of recombinants that result from crossing over. The greater the frequency of crossing over, the greater the map distance. Calculate the percent of crossovers by dividing the number of crossover asci (spores arranged 2:2:2:2 or 2:4:2) by the total number of asci x 100. To calculate map distance, divide the percent of crossover asci by 2. The percent of crossover asci is divided by 2 because only half of the spores in each ascus are the result of crossing over. A group of asci formed from crossing light-spored Sordaria with dark-spored Sordaria produced the following results:

Number of Asci Counted Spore Arrangement

7 4 light / 4 dark spores

8 4 dark / 4 light spores

3 2 light / 2 dark / 2 light / 2 dark spores

4 2 dark / 2 light / 2 dark / 2 light spores

1 2 dark / 4 light / 2 dark spores

2 2 light / 4 dark / 2 light spores

2. How many of these asci contain a spore arrangement that resulted from crossing over?

3. From this small sample, calculate the map distance between the gene and centromere. Plant Pigments & Photosynthesis In the light reactions of photosynthesis, light energy is absorbed by chlorophyll and used to excite electrons. The excited electrons then enter one of two electron transport chains. One chain converts ADP + P to ATP; and the other converts NADP + H to NADPH. In a particular experiment, students added a solution of DPIP (a blue dye) to a suspension of chloroplasts. The DPIP substituted for NADP in the light reactions: DPIP + H → DPIPH. DPIPH is colorless, so as the light reactions took place, the blue color of the solution diminished. The students used this color change as an indication that the light reactions were taking place and they used the rate at which the color change took place as a measure of the rate of the light reactions. The students used a spectrophotometer to measure loss of color by DPIP. They prepared a sample by adding chloroplast suspension, DPIP, and a buffer to water in a tube or vial called a cuvette. The instrument works by shining a light of known intensity into one side of the cuvette. On the opposite side of the test chamber is a photocell. DPIP will absorb some of the light that enters the cuvette, thus, the photocell will “see” less light. As the light reactions take place, there will be less DPIP and the photocell will “see” more light. Since DPIP absorbs light most strongly at orange-red wavelengths, they set the spectrophotometer to read the amount of light transmitted in that part of the spectrum. The following were the content of the cuvettes and data obtained from their experiment:

1. Write a hypothesis that this experiment was designed to test:

2. What was the independent variable? What was the dependent variable?

3. Which cuvette served as a control for this experiment? Explain your answer.

4. What variables are tested in this experiment?

5. What affect did boiling have on the chloroplast suspension?

6. Plot the data from sample table 3 in the space provided below.

Cell Respiration In a particular experiment, students in an AP Biology class used a respirometer to measure the rate of respiration of germinating and nongerminating pea seeds at two different temperatures. The respirometer consists of a vial that contains the peas and a volume of air. The mouth of the vial is sealed with a rubber 1-hole stopper that has a pipet inserted in it. The respirometer is submerged in water. If the peas are respiring, they will use oxygen and release carbon dioxide. Since 1 mole of carbon dioxide is released for each mole of oxygen consumed, there is no change in the volume of gas in the respirometer. (Avogadro’s Law: At constant temperature and pressure, 1 mole of any gas has the same volume as 1 mole of any other gas.) The students altered this equilibrium by placing a solution of potassium hydroxide (KOH) in the vial. Potassium hydroxide reacts with carbon dioxide to form potassium carbonate, which is a solid (CO2 + 2 KOH →K2CO3 + H2O). Since the carbon dioxide produced is removed by reaction with potassium hydroxide, as oxygen is used by cellular respiration the volume of gas in the respirometer will decrease. As the volume of gas decreases, water will move into the pipet. Students used this decrease of volume, as read from the scale printed on the pipet, as a measure of the rate of cellular respiration. The following data were obtained from the experiment:

1. Write 2 hypotheses that this experiment was designed to test.

2. Which respirometer served as the control? Explain.

3. What constants were important in this experiment?

4. Using the graph, summarize your findings, comparing results from respirometers 1 and 2, and results obtained at room temperature vs. results at the colder temperature. Speculate as to the cause(s) of any differences between the treatments.

5. From the graph, calculate the rate of oxygen consumption for each treatment.

6. If you used the same experimental design to compare the rates of respiration of a 25g reptile and a 25g mammal at 10◦C, what results would you expect? Sketch a graph of your expected results and explain your reasoning.

Restriction Enzyme Analysis of DNA Gel electrophoresis is a procedure that separates molecules on the basis of their rate of movement through a gel under the influence of an electrical field. The direction of movement is affected by the charge of the molecules, and the rate of movement is affected by their size and shape, the density of the gel, and the strength of the electrical field. DNA is negatively charged, so it will move toward the positive pole of the gel when current is applied. When DNA has been cut by restriction enzymes, the different-sized fragments will migrate at different rates. Smaller fragments move more quickly, therefore they will migrate the fastest through the gel. The length of each fragment is measured in number of DNA base pairs. The figure below shows the results of electrophoresis performed by a group of AP Biology students. Semilog paper was used to plot the results of the HindiIII digest. Since its fragments sizes are known, this is the standard curve. It was used to determine the other fragement sizes from DNA I and DNA II samples. 1. How many base pairs is the fragment circled in Figure 6.4? Below is a plasmid with restriction sites for BamHI and EcoRI. Several restriction digests were done using these two enzymes either alone or in combination. Use this data to answer the questions below. HINT: Begin by determining the number and size of the fragments produced with each enzyme. “kb” stands for kilobases, or thousands of base pairs. 2. In the samples below, which lane shows a digest with BamHI only? 3. Which lane shows a digest with both BamHI and EcoRI? HINT: Begin by determining the number and size of the fragments

produced with each enzyme. “kb” stands for kilobases, or thousands of base pairs.

Transformation in Bacteria Genetic transformation occurs when a host organism takes in foreign DNA and expresses a foreign gene. Bacterial cells have a single main chromosome and circular DNA molecules called plasmids, which carry additional genetic information. Plasmids are circular pieces of DNA that exist outside the main bacterial chromosome and carry their own genes for specialized functions including resistance to specific drugs. In genetic engineering, plasmids are one means used to introduce foreign genes into a bacterial cell. In a molecular biology laboratory, a student obtained competent E. coli cells and used a common transformation procedure to introduce the uptake of plasmid DNA with a gene for resistance to the antibiotic kanamycin. Competent cells are cells that are most likely to take up extracellular DNA. The obtained results are shown below: If there is no antibiotic in the agar (growth medium for bacteria), then bacteria will cover the plate with so many cells that a “lawn” growth is produced. Only transformed cells can grow on agar with antibiotic. Since only some of the cells exposed to the kanamycin plasmid will actually take them in, only some cells will be transformed, and individual bacterial colonies are produced on the agar plate. If none of the E. coli cells have been transformed, nothing will grow on the agar plate containing antibiotic.

1. On which petri dish do only transformed cells grow?

2. Which of the plates is used as a control to show that nontransformed E. coli will not grow in the presence of kanamycin?

3. If a student wants to verify that transformation has cocurred, what procedure might he/she use?

4. During the course of an E. coli transformation laboratory, a student forgot to mark the culture tube that received the kanamycin-resistant plasmids. The student proceeds with the laboratory because he thinks that he will be able to determine from his results which culture tube contained cells that may have undergone transformation. Sketch and label a plate below that would most likely indicate transformed cells.

Genetics of Organisms – Chi Square Analysis: Consider the following data for F2 grains of corn from a dihybrid cross of corn with grains that are purple and starchy and corn with grains that are yellow and sweet (purple is dominant to yellow and starchy is dominant to sweet). The data are used to determine whether or not corn color and sweetness assort independently or are linked on the same chromosome.

1. What is the correct null hypothesis for this experiment?

2. Calculate χ2 for the actual count above. Show all work and state whether the results support the null hypothesis in favor of the expected ratio or cause you to reject it.

3. If rejected, identify possible sources of error that could cause this.

Population Genetics – Hardy-Weinberg Calculations 1. In the Sample Table 2, are the Generation 5 values

for p and q different from their initial values?

2. Is your answer to 1, above, consistent with the

alleles being at equilibrium? If not, why not? 3. Review the five conditions that must be met for allele frequencies to remain constant. Which, if any, of these conditions might

not have been met in this simulation?

4. In the Sample Table 3, are the Generation 5 values for p and q different from their initial values? Explain why the values did or did not change.

5. Compare the Generation 5 values for p and q calculated in Activity B with the Generation 5 values for p and q that you just

calculated for Activity C. Describe your findings. 6. What would you expect to happen to the frequencies of p and q if you ran the simulation for another five generations? 7. This simulation involved very high selection against the recessive allele. Did selection eliminate the allele from the population?

If not, why did the allele persist?

Transpiration in Plants

1. Graph the class averages from Table 2. Give the graph and axes appropriate titles and labels.

2. Interpret the graph above – explain how each environmental factor affects the transpiration rate in plants:

Graphing Heart Rate of Water Flea 1. Graph the temperature and heart rate data of Daphnia.

Give the graph and axes appropriate titles and labels.

2. Interpret the graph above – explain how temperature affects the heart rate of Daphnia.

3. If this lab had been performed with an endotherm, how might the results have differed? Sketch a graph of your expected results and explain.

Animal Behavior

1. What hypothesis was tested in this experiment?

2. What variables should have been held constant during this experiment?

3. Sketch a graph the data from Table 1 for both the wet side and the dry side of the chamber. Give the graph and axes appropriate titles and labels. Interpret the results of the experiment.

Temperature and Dissolved Oxygen 1. Plot the Class Averages for DO from Table 1. Give the graph and axes

appropriate titles and labels.

2. Interpret the graph above – explain how dissolved oxygen varies according to temperature changes. 3. Predict what would happen to the dissolved oxygen levels at: -10

oC and 40

oC.

Additional Mathematical Calculations