AP Biology Lab Review Based on the PPT

AP Biology Lab Review Based on the PPT from Ms. Foglia

Big Idea 1: Evolution

Lab 1: Artificial Selection • Concepts: • Natural selection = differential reproduction in a population • Populations change over time evolution • Natural Selection vs. Artificial Selection

Lab 1: Artificial Selection • Description: • • Use Wisconsin Fast Plants to perform artificial selection Identify traits and variations in traits Cross-pollinate (top 10%) for selected trait Collect data for 2 generations (P and F 1)

Sample Histogram of a Population

Lab 1: Artificial Selection Analysis & Results: • Calculate mean, median, standard deviation, range • Are the 2 populations before and after selection (P and F 1) actually different? • Are the 2 sub-populations of F 1 (hairy vs. non-hairy) different? • Are the means statistically different? • A T-test could be used to determine if 2 sets of data are statistically different from each other; these are often used in examples where the sample size is small

Lab 2: Mathematical Modeling: Hardy-Weinberg • Concepts: • Evolution = change in frequency of alleles in a population from generation to generation • Hardy-Weinberg Equilibrium • Allele Frequencies (p + q = 1) • Genotypic Frequencies (p 2+2 pq+q 2 = 1) • Conditions: 1. 2. 3. 4. 5. large population random mating no mutations no natural selection no migration

Lab 2: Mathematical Modeling: Hardy-Weinberg Analysis & Results: • Null model: in the absence of random events that affect populations, allele frequencies (p, q) should be the same from generation to generation (H-W equilibrium) • Analyze genetic drift and the effect of selection on a given population • Manipulate parameters in model: • Population size, selection (fitness), mutation, migration, genetic drift

Lab 2: Mathematical Modeling: Hardy-Weinberg • Real-life applications: • Cystic fibrosis, polydactyly • Heterozygote advantage (Sickle-Cell Anemia)

Lab 2: Mathematical Modeling: Hardy-Weinberg ESSAY 1989 Do the following with reference to the Hardy-Weinberg model. a. Indicate the conditions under which allele frequencies (p and q) remain constant from one generation to the next. b. Calculate, showing all work, the frequencies of the alleles and frequencies of the genotypes in a population of 100, 000 rabbits of which 25, 000 are white and 75, 000 are agouti (brownish). (In rabbits the white color is due to a recessive allele, w, and agouti is due to a dominant allele, W. ) c. If the homozygous dominant condition were to become lethal, what would happen to the allelic and genotypic frequencies in the rabbit population after two generations?

a. Indicate the conditions under which allele frequencies (p and q) remain constant from one generation to the next. 1. large population 2. random mating 3. no mutations 4. no natural selection 5. no migration b. Calculate, showing all work, the frequencies of the alleles and frequencies of the genotypes in a population of 100, 000 rabbits of which 25, 000 are white and 75, 000 are agouti (brownish). (In rabbits the white color is due to a recessive allele, w, and agouti is due to a dominant allele, W. ) 25, 000/ 100, 000 =. 25 so. 25 = q 2 (a) q =. 5 and (A) p =. 5 Homozygous Rec =. 25 = 25, 000 Homozygous Dom (agouti) = p 2 =. 25 = 25, 000 Heterozygous (agouti) = 2(p)(q) = 2(. 5) =. 5 = 50, 000 c. If the homozygous dominant condition were to become lethal, what would happen to the allelic and genotypic frequencies in the rabbit population after two generations? The frequency of the ‘A’ allele would decrease and the frequency of the ‘a’ allele would increase.

Lab 3: Comparing DNA Sequences using BLAST Evolutionary Relationships • Concepts: • Bioinformatics: combines statistics, math modeling, computer science to analyze biological data • Genomes can be compared to detect genetic similarities and differences • BLAST = Basic Local Alignment Search Tool • Input gene sequence of interest • Search genomic libraries for identical or similar sequences

Lab 3: Comparing DNA Sequences using BLAST Evolutionary Relationships • Description: • Use BLAST to compare several genes • Use information to construct a cladogram (phylogenetic tree) • Cladogram = visualization of evolutionary relatedness of species

Lab 3: Comparing DNA Sequences using BLAST Evolutionary Relationships

Lab 3: Comparing DNA Sequences using BLAST Evolutionary Relationships • Use this data to construct a cladogram of the major plant groups

Lab 3: Comparing DNA Sequences using BLAST Evolutionary Relationships • Fossil specimen in China • DNA was extracted from preserved tissue • Sequences from 4 genes were analyzed using BLAST

Lab 3: Comparing DNA Sequences using BLAST Evolutionary Relationships • Analysis & Results: • BLAST results: the higher the score, the closer the alignment • The more similar the genes, the more recent their common ancestor located closer on the cladogram

Lab 3: Comparing DNA Sequences using BLAST Evolutionary Relationships

Answer is C

Big Idea 2: cellular processes: energy and communication

Lab 4: Diffusion & Osmosis • Concepts: • • Selectively permeable membrane Diffusion (high low concentration) Osmosis (aquaporins) Water potential ( ) • = pressure potential ( P) + solute potential ( S) • Solutions: • Hypertonic • hypotonic • isotonic

Lab 4: Diffusion & Osmosis

Lab 4: Diffusion & Osmosis • Description: • Surface area and cell size vs. rate of diffusion • Cell modeling: dialysis tubing + various solutions (distilled water, sucrose, salt, glucose, protein) • Identify concentrations of sucrose solution and solute concentration of potato cores • Observe osmosis in onion cells (effect of salt water)

Lab 4: Diffusion & Osmosis

Potato Cores in Different Concentrations of Sucrose

Lab 4: Diffusion & Osmosis • Conclusions • Water moves from high water potential ( ) (hypotonic=low solute) to low water potential ( ) (hypertonic=high solute) • Solute concentration & size of molecule affect movement across selectively permeable membrane

Answer is A

Answer is B

Lab 4: Diffusion & Osmosis ESSAY 1992 A laboratory assistant prepared solutions of 0. 8 M, 0. 6 M, 0. 4 M, and 0. 2 M sucrose, but forgot to label them. After realizing the error, the assistant randomly labeled the flasks containing these four unknown solutions as flask A, flask B, flask C, and flask D. Design an experiment, based on the principles of diffusion and osmosis, that the assistant could use to determine which of the flasks contains each of the four unknown solutions. Include in your answer: a. a description of how you would set up and perform the experiment; b. the results you would expect from your experiment; and c. an explanation of those results based on the principles involved. Be sure to clearly state the principles addressed in your discussion.

A laboratory assistant prepared solutions of 0. 8 M, 0. 6 M, 0. 4 M, and 0. 2 M sucrose, but forgot to label them. After realizing the error, the assistant randomly labeled the flasks containing these four unknown solutions as flask A, flask B, flask C, and flask D. Design an experiment, based on the principles of diffusion and osmosis, that the assistant could use to determine which of the flasks contains each of the four unknown solutions. Include in your answer: a. a description of how you would set up and perform the experiment; b. the results you would expect from your experiment; and c. an explanation of those results based on the principles involved. Be sure to clearly state the principles addressed in your discussion. Think about our lab that we did…. Talk about putting a cell (potato) in each of the solutions – mass before and then mass after and see in which solution did it gain/lose the most weight. The principle involved is simple osmosis. The potato will lose weight in hypertonic solutions and gain weight in the hypotonic solutions. Based on how much weight is gained/lost in each of the solutions, you would be able to put them in order. 0. 8 M – potato will lose the most mass 0. 6 M – potato will lose the second most mass 0. 4 M – potato will remain mostly the same with losing just a bit 0. 2 M – potato will gain mass *potatos are about 0. 3 molarity

Lab 5: Photosynthesis • Concepts: • Photosynthesis • 6 H 2 O + 6 CO 2 + Light C 6 H 12 O 6 + 6 O 2 • Ways to measure the rate of photosynthesis: • Production of oxygen (O 2) • Consumption of carbon dioxide (CO 2)

Lab 5: Photosynthesis • Description: • Paper chromatography to identify pigments • Floating disk technique • Leaf disks float in water • Gases can be drawn from out from leaf using syringe leaf sinks • Photosynthesis O 2 produced bubbles form on leaf disk rises • Measure rate of photosynthesis by O 2 production • Factors tested: types of plants, light intensity, colors of leaves, p. H of solutions

Plant Pigments & Chromatography

Floating Disk Technique

To make comparisons between experiments, a standard point of reference is needed. Repeated testing of this procedure has shown that the point at which 50% of the disks are floating (the median or ET 50) is a reliable and repeatable point of reference. In this case, the disks floating are counted at the end of each time interval. The median is chosen over the mean as the summary statistic. The median will generally provide a better estimate of the central tendency of the data because, on occasion, a disk fails to rise or takes a very long time to do so. A term coined by G. L Steucek and R. J Hill (1985) for this relationship is ET 50, the estimated time for 50% of the disks to rise. That is, rate is a change in a variable over time. The time required for 50% of the leaf disks to float is represented as Effective Time = ET 50.

Lab 5: Photosynthesis • Concepts: • photosynthesis • Photosystems II, I • H 2 O split, ATP, NADPH • chlorophylls & other plant pigments • • chlorophyll a chlorophyll b xanthophylls carotenoids • experimental design • control vs. experimental

Answer is A

Answer is D

Lab 6: Cellular Respiration • Concepts: • Respiration • Measure rate of respiration by: • O 2 consumption • CO 2 production

Lab 6: Cellular Respiration • Description: • Use respirometer • Measure rate of respiration (O 2 consumption) in various seeds • Factors tested: • • • Non-germinating seeds Germinating seeds Effect of temperature Surface area of seeds Types of seeds Plants vs. animals

Lab 6: Cellular Respiration

Lab 6: Cellular Respiration

Lab 6: • Cellular Respiration Conclusions: • • temp = respiration germination = respiration Animal respiration > plant respiration surface area = respiration Calculate Rate

Lab 6: Cellular Respiration

Answer is A

Answer is A

Answer is C

Lab 6: Cellular Respiration ESSAY 1990 The results below are measurements of cumulative oxygen consumption by germinating and dry seeds. Gas volume measurements were corrected for changes in temperature and pressure. Cumulative Oxygen Consumed (m. L) Time (minutes) 0 10 20 30 40 Germinating seeds 22°C 0. 0 8. 8 16. 0 23. 7 32. 0 Dry Seeds (non-germinating) 22°C 0. 0 0. 2 0. 1 0. 0 0. 1 Germinating Seeds 10°C 0. 0 2. 9 6. 2 9. 4 12. 5 Dry Seeds (non-germinating) 10°C 0. 0 0. 2 0. 1 0. 2 a. Plot the results for the germinating seeds at 22°C and 10°C. b. Calculate the rate of oxygen consumption for the germinating seeds at 22°C, using the time interval between 10 and 20 minutes. c. Account for the differences in oxygen consumption observed between: 1. germinating seeds at 22°C and at 10°C 2. germinating seeds and dry seeds. d. Describe the essential features of an experimental apparatus that could be used to measure oxygen consumption by a small organism. Explain why each of these features is necessary.

Lab 6: Cellular Respiration Cumulative Oxygen Consumed (m. L) Time (minutes) 0 10 20 30 40 Germinating seeds 22°C 0. 0 8. 8 16. 0 23. 7 32. 0 Dry Seeds (non-germinating) 22°C 0. 0 0. 2 0. 1 0. 0 0. 1 Germinating Seeds 10°C 0. 0 2. 9 6. 2 9. 4 12. 5 Dry Seeds (non-germinating) 10°C 0. 0 0. 2 0. 1 0. 2 a. Plot the results for the germinating seeds at 22°C and 10°C. I would say m. L of O 2 consumed instead of “corrected difference” and this graph shows all 4 lines when we only needed 2 (this was all I could find online) **make sure graph has a title, axis are labeled, lines labeled

Lab 6: Cellular Respiration Cumulative Oxygen Consumed (m. L) Time (minutes) 0 10 20 30 40 Germinating seeds 22°C 0. 0 8. 8 16. 0 23. 7 32. 0 Dry Seeds (non-germinating) 22°C 0. 0 0. 2 0. 1 0. 0 0. 1 Germinating Seeds 10°C 0. 0 2. 9 6. 2 9. 4 12. 5 Dry Seeds (non-germinating) 10°C 0. 0 0. 2 0. 1 0. 2 b. Calculate the rate of oxygen consumption for the germinating seeds at 22°C, using the time interval between 10 and 20 minutes. Rate = ∆Y/ ∆X (16. 0 – 8. 8) = 7. 2 =. 72 m. L/ min (20 -10) 10 **make sure to show work and make sure to include units!

Lab 6: Cellular Respiration Cumulative Oxygen Consumed (m. L) Time (minutes) 0 10 20 30 40 Germinating seeds 22°C 0. 0 8. 8 16. 0 23. 7 32. 0 Dry Seeds (non-germinating) 22°C 0. 0 0. 2 0. 1 0. 0 0. 1 Germinating Seeds 10°C 0. 0 2. 9 6. 2 9. 4 12. 5 Dry Seeds (non-germinating) 10°C 0. 0 0. 2 0. 1 0. 2 c. Account for the differences in oxygen consumption observed between: 1. germinating seeds at 22°C and at 10°C increased temp causes increased respiration/ metabolism so they use more O 2 at a higher temp 2. germinating seeds and dry seeds germinating seeds are active and thus will use more O 2, dry seeds are dormant and will respire very little d. Describe the essential features of an experimental apparatus that could be used to measure oxygen consumption by a small organism. Explain why each of these features is necessary. Needs to be a closed system that separates oxygen consumption and carbon dioxide release (like KOH that turns CO 2 into a solid). Must be able to control temp (water bath), control volume (glass beads), have a timer, equal numbers of organisms in each setup, etc.

Big Idea 3: genetics and information transfer

Lab 7: Mitosis & Meiosis • Concepts: • Cell Cycle (G 1 S G 2 M) • Control of cell cycle (checkpoints) • Cyclins & cyclin-dependent kinases (CDKs) • Mitosis vs. Meiosis • Crossing over genetic diversity

Lab 7: Mitosis & Meiosis

Lab 7: Mitosis & Meiosis

Lab 7: Mitosis & Meiosis • Description: • Model mitosis & meiosis (pipecleaners, beads) • How environment affects mitosis of plant roots • Lectin - proteins secreted by fungus • Root stimulating powder • Count # cells in interphase, mitosis • Observe karyotypes (cancer, mutations) • Meiosis & crossing over in Sordaria (fungus)

Lab 7: Mitosis & Meiosis

Lab 7: Mitosis & Meiosis

Abnormal karyotype = Cancer

Meiosis: Crossing over in Prophase I

Lab 7: Mitosis & Meiosis • Observed crossing over in fungus (Sordaria) • Arrangement of ascospores

Sordaria Analysis % crossover = distance from = centromere total crossover total offspring % crossover 2

Lab 7: Mitosis & Meiosis ESSAY 1987 Discuss the process of cell division in animals. Include a description of mitosis and cytokinesis, and of the other phases of the cell cycle. Do not include meiosis. ESSAY 2004 Meiosis reduces chromosome number and rearranges genetic information. a. Explain how the reduction and rearrangement are accomplished in meiosis. b. Several human disorders occur as a result of defects in the meiotic process. Identify ONE such chromosomal abnormality; what effects does it have on the phenotype of people with the disorder? Describe how this abnormality could result from a defect in meiosis. c. Production of offspring by parthenogenesis or cloning bypasses the typical meiotic process. Describe either parthenogenesis or cloning and compare the genomes of the offspring with those of the parents.

Lab 7: Mitosis & Meiosis ESSAY 1987 Discuss the process of cell division in animals. Include a description of mitosis and cytokinesis, and of the other phases of the cell cycle. Do not include meiosis. Simply explain cell cycle (G 1, S, G 2, Mitosis, cytokinesis) see notes on topic ESSAY 2004 Meiosis reduces chromosome number and rearranges genetic information. a. Explain how the reduction and rearrangement are accomplished in meiosis. Chromosomes are paired in tetrads first, they get split in half so the cells after the first division of meiosis have ½ the number of chromosomes as the parent cell. Rearrangement is due to independent assortment, where maternal and paternal chromosomes line up randomly in the tetrads, and crossing over which occurs in Prophase I. Crossing over is when a piece of one chromosomes breaks off and switches places with a piece on the homologous chromosome. b. Several human disorders occur as a result of defects in the meiotic process. Identify ONE such chromosomal abnormality; what effects does it have on the phenotype of people with the disorder? Describe how this abnormality could result from a defect in meiosis. c. Production of offspring by parthenogenesis or cloning bypasses the typical meiotic process. Describe either parthenogenesis or cloning and compare the genomes of the offspring with those of the parents.

Lab 7: Mitosis & Meiosis b. Several human disorders occur as a result of defects in the meiotic process. Identify ONE such chromosomal abnormality; what effects does it have on the phenotype of people with the disorder? Describe how this abnormality could result from a defect in meiosis. Down Syndrome is trisomy 21 (an extra copy of chromosome #21). This is characterized by low muscle tone, small stature, an upward slant to the eyes, a single deep crease across the center of the palm, and some mental deficiencies. Down Syndrome is caused by nondisjunction – chromosomes not separating correctly during meiosis. This can happen in either meiosis I or meiosis II. c. Production of offspring by parthenogenesis or cloning bypasses the typical meiotic process. Describe either parthenogenesis or cloning and compare the genomes of the offspring with those of the parents. Parthenogenesis is an example of asexual reproduction where growth and development of embryos occur without fertilization. In animals, parthenogenesis means development of an embryo from an unfertilized egg cell. This would produce clones and all offspring would be genetically identical to the parent cell.

Lab 8: Bacterial Transformation Concepts: • Transformation: uptake of foreign DNA from surroundings • Plasmid = small ring of DNA with a few genes • Replicates separately from bacteria DNA • Can carry genes for antibiotic resistance • Genetic engineering: recombinant DNA = p. GLO plasmid

Lab 8: Bacterial Transformation

Lab 8: Bacterial Transformation

Lab 8: Bacterial Transformation • Conclusions: • Foreign DNA inserted using vector (plasmid) • Ampicillin = Selecting agent • No transformation = no growth on amp+ plate • Regulate genes by transcription factors (ara. C protein)

Plates that have only ampicillinresistant bacteria growing include which of the following? A. I only B. III only C. IV only D. I and II

Plates that have only ampicillinresistant bacteria growing include which of the following? A. I only B. III only C. IV only D. I and II Answer is C

Answer is A

Answer is A

Answer is B

Answer is C

Lab 9: Restriction Enzyme Analysis of DNA • Concepts: • Restriction Enzymes • Cut DNA at specific locations • Gel Electrophoresis • DNA is negatively charged • Smaller fragments travel faster

Lab 9: Restriction Enzyme Analysis of DNA • Description

Lab 9: Restriction Enzyme Analysis of DNA • Determine DNA fragment sizes

Lab 9: Restriction Enzyme Analysis of DNA • Conclusions: • • Restriction enzymes cut at specific locations (restriction sites) DNA is negatively charged Smaller DNA fragments travel faster than larger fragments Relative size of DNA fragments can be determined by distance travelled • Use standard curve to calculate size (remember we made this from the marker when we KNEW how many bases were in each band)

Lab 8 -9: Biotechnology ESSAY 1995 The diagram below shows a segment of DNA with a total length of 4, 900 base pairs. The arrows indicate reaction sites for two restriction enzymes (enzyme X and enzyme Y). a. Explain how the principles of gel electrophoresis allow for the separation of DNA fragments b. Describe the results you would expect from electrophoretic separation of fragments from the following treatments of the DNA segment above. Assume that the digestion occurred under appropriate conditions and went to completion. I. DNA digested with only enzyme X II. DNA digested with only enzyme Y III. DNA digested with enzyme X and enzyme Y combined IV. Undigested DNA c. Explain both of the following: 1. The mechanism of action of restriction enzymes 2. The different results you would expect if a mutation occurred at the recognition site for enzyme Y.

Lab 8 -9: Biotechnology ESSAY 1995 The diagram below shows a segment of DNA with a total length of 4, 900 base pairs. The arrows indicate reaction sites for two restriction enzymes (enzyme X and enzyme Y). a. Explain how the principles of gel electrophoresis allow for the separation of DNA fragments. Gel electrophoresis separates segments of DNA based on their size. The DNA is put into the gel at the negative end and it runs to the positive. The electric current that moves through the buffer and the gel allow the negatively charged DNA to be attracted to the positive end of the chamber. The smaller pieces move through the gel faster than the larger pieces.

Lab 8 -9: Biotechnology ESSAY 1995 The diagram below shows a segment of DNA with a total length of 4, 900 base pairs. The arrows indicate reaction sites for two restriction enzymes (enzyme X and enzyme Y). b. Describe the results you would expect from electrophoretic separation of fragments from the following treatments of the DNA segment above. Assume that the digestion occurred under appropriate conditions and went to completion. I. DNA digested with only enzyme X 4 pieces; sizes in base pairs are: 400, 1700, 1300, 1500 II. DNA digested with only enzyme Y 2 pieces; sizes in base pairs are: 900, 4000 III. DNA digested with enzyme X and enzyme Y combined 5 bands; sizes in base pairs are: 400, 500, 1200, 1300, 1500 IV. Undigested DNA one huge band

Lab 8 -9: Biotechnology ESSAY 1995 The diagram below shows a segment of DNA with a total length of 4, 900 base pairs. The arrows indicate reaction sites for two restriction enzymes (enzyme X and enzyme Y). c. Explain both of the following: 1. The mechanism of action of restriction enzymes Restriction enzymes identify specific splice sites in DNA and cut there. Restriction sites are palindromes which read the same forward as backward on the opposite strands of the DNA molecule. Restriction enzymes can either create blunt ends or sticky ends. 2. The different results you would expect if a mutation occurred at the recognition site for enzyme Y. I would expect the Y enzyme not to cut the piece of DNA, so if you digested the whole piece of DNA with both the X and Y enzymes, it would only be cut at the X restriction sites.

Big Idea 4: interactions

Lab 10: Energy Dynamics • Concepts: • Energy from sunlight drives photosynthesis (store E in organic compounds) • Gross Productivity (GPP) = energy captured • But some energy is used for respiration (R) • Net primary productivity (NPP) = GPP – R • Energy flows! (but matter cycles) • Producers consumers • Biomass = mass of dry weight

Lab 10: Energy Dynamics Pyramid of Energy Pyramid of Biomass Pyramid of Numbers

Lab 10: Energy Dynamics • Description: • Brassica (cabbage) cabbage white butterfly larvae (caterpillars)

Lab 10: Energy Dynamics • Measuring Biomass: • Cabbage mass lost • Caterpillar mass gained • Caterpillar frass (poop) dry mass

Lab 10: Energy Dynamics

Lab 10: Energy Dynamics • Conclusions:

Lab 10: Energy Dynamics • Conclusions: • Energy is lost (respiration, waste) • Conservation of Mass • Input = Output

Answer: 125 – (60 + 5) = 60 g/m 2

Lab 11: Transpiration • Concepts: • Transpiration • Xylem • Water potential • Cohesion-tension hypothesis • Stomata & Guard cells • Leaf surface area & # stomata vs. rate of transpiration

Lab 11: Transpiration

Lab 11: Transpiration • Description: • Determine relationship between leaf surface area, # stomata, rate of transpiration • Nail polish stomatal peels • Effects of environmental factors on rate of transpiration • Temperature, humidity, air flow (wind), light intensity

Analysis of Stomata

Rates of Transpiration

Lab 11: Transpiration • Conclusions: • • transpiration: wind, light transpiration: humidity Density of stomata vs. transpiration Leaf surface area vs. transpiration

Answer:

Answer:

Answer:

Answer:

Lab 12: Animal Behavior • Concepts: • Experimental design • IV, DV, control, constants • Control vs. Experimental • Hypothesis • innate vs. learned behavior • choice chambers • • • temperature humidity light intensity salinity other factors

Lab 12: Animal Behavior • Description: • Investigate relationship between environmental factors vs. behavior • Betta fish agonistic behavior • Drosophila (fruit fly) behavior • Pillbug kinesis

Lab 12: Animal Behavior

Lab 12: Animal Behavior • Hypothesis Development Keep this in mind if you have to write a hypothesis on the AP exam… what makes a good hypothesis? • Poor: I think pillbugs will move toward the wet side of a choice chamber. • Better: If pillbugs are randomly placed on two sides of a wet/dry choice chamber and allowed to move about freely for 10 minutes, then more pillbugs will be found on the wet side because they prefer moist environments.

Lab 12: Animal Behavior • Experimental Design sample size

Lab 12: Animal Behavior • Data Analysis: • • Chi-Square Test Null hypothesis: there is no difference between the conditions Degrees of Freedom = n-1 At p=0. 05, if X 2 < critical value accept null hypothesis (any differences between observed and expected due to CHANCE)

Lab 12: Animal (Fruit Fly or Pillbug) Behavior ESSAY 1997 A scientist working with Bursatella leachii, a sea slug that lives in an intertidal habitat in the coastal waters of Puerto Rico, gathered the following information about the distribution of the sea slugs within a ten-meter square plot over a 10 -day period. time of day average distance between individuals 12 mid 4 am 8 am 12 noon 4 pm 8 pm 12 mid 8. 0 8. 9 44. 8 174. 0 350. 5 60. 5 8. 0 a. For the data above, provide information on each of the following: • Summarize the pattern. • Identify three physiological or environmental variables that could cause the slugs to vary their distance from each other. • Explain how each variable could bring about the observed pattern of distribution. b. Choose one of the variables that you identified and design a controlled experiment to test your hypothetical explanation. Describe results that would support or refute your hypothesis.

Lab 12: Animal (Fruit Fly or Pillbug) Behavior ESSAY 2002 The activities of organisms change at regular time intervals. These changes are called biological rhythms. The graph depicts the activity cycle over a 48 -hour period for a fictional group of mammals called pointy-eared bombats, found on an isolated island in the temperate zone. a. Describe the cycle of activity for the bombats. Discuss how three of the following factors might affect the physiology and/or behavior of the bombats to result in this pattern of activity. • temperature • food availability • presence of predators • social behavior b. Propose a hypothesis regarding the effect of light on the cycle of activity in bombats. Describe a controlled experiment that could be performed to test this hypothesis, and the results you would expect.

Lab 13: Enzyme Activity • Concepts: • Enzyme • Structure (active site, allosteric site) • Lower activation energy • Substrate product • Proteins denature (structure/binding site changes)

Lab 13: Enzyme Activity • Description: • Determine which factors affecting rate of enzyme reaction • H 2 O 2 H 2 O + O 2 catalase • Measure rate of O 2 production

Turnip peroxidase Color change (O 2 produced)

Lab 13: Enzyme Activity • Conclusions: • Enzyme reaction rate affected by: • • p. H (acids, bases) Temperature Substrate concentration Enzyme concentration Calculate Rate of Reaction

Answer is B