2Investigation 10- Energy Dynamics What did we do?
3Investigation 10- Energy Dynamics ConceptsEnergy flow through an ecosystemGross Primary ProductivityNet Primary ProductivityEnergy Audit of an EcosystemEnergy is neither created or destroyed so it must all be accounted for in an energy audit studyLight energy that is not lost as heat or waste by plant forms biomass which is consumed by larvae and that which is not lost as heat or waste forms biomass of the growing insect
4Investigation 10- Energy Dynamics ProcedureGrow Fast Plants from seeds; 6 per containerRemove 10 plants on day 7, wash roots, dry them in incubator, and weigh them.What is Net Primary Productivity?Calculate NPP per plant by (18.27kcal/7 days)/10 plantsContinue growing through day 14, and again remove 10 plants, wash, dry, and weigh
5Day 14 \ Calculate NPP per plant by (40.46kcal/14 days)/10 plants Weigh one large Brussel sprout, cut in half, and place into Brassica BarnWeigh th instar white butterfly larvae and place into Brassica BarnCollect, dry, and mass the frass over the next 3 daysFrass energy =4.76 kcal/gAfter 3 days, mass the larvae and the remaining portion of the Brussel sproutLarvae energy=.40 Wet mass of Larvae * 5.5 kcal/gBrussel sprout=.24 Wet mass of B.S. * 4.35 kcal/g
6Example Calculations: Be able to follow energy flow diagrams
9Investigation 4: Diffusion & Osmosis What did we do?
10Diffusion & Osmosis Description Surface area to cell size dialysis tubing filled with starch-glucose solution in beaker filled with KI solutionEffects of different concentrations of solutions on Elodea leafpotato cores in sucrose solutionsDesign an experiment to determine solute concentration of different colored unknown solutions
11And… you may need to do this with spheres as well Size vs. Surface AreaA cell that is actively metabolizing must stay small by continuously dividing so that it can efficiently move metabolites into and out of the cellAlso, remember that there is only one copy of DNA!
13Diffusion & Osmosis Conclusions water moves from high concentration of water (hypotonic=low solute) to low concentration of water (hypertonic=high solute) or you can say that water moves toward the area of more dissolved solutesolute concentration & size of molecule affect movement through semi-permeable membrane
14Investigation 4: Diffusion and Osmosis In animal cells, the direction of osmosis, in or out of a cell, depends on the concentration of solutes inside and outside of the plasma membrane… How is it different in plants?Water Potential= Pressure Potential + Solute PotentialWater moves from higher potential to lower potentialSolute potential is zero or negative and pressure potential is zero or positiveSolute potential can be calculated using Ψs=-iCRT where i is ionization constant, C is molar concentration, R= (pressure constant), and T is temperature in Kelvin (273 + C)
16Diffusion & OsmosisA 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 using a potato given that the molarity of a potato is 0.28.
18Cell Communication No Distance Short Distance Long Distance Cell to cell contact (APC to Helper T)Short DistanceLocal regulator (Neurotransmitter between two neuronsLong DistanceHormone (Thyroid gland releases T3 and T4)
23Lab 13: Enzyme Catalysis Concepts substrate enzyme product enzyme structureproductdenaturation of proteinexperimental designrate of reactivityreaction with enzyme vs. reaction without enzymeoptimum pH or temperaturetest at various pH or temperature values
26Lab 13: Enzyme CatalysisThe effects of pH and temperature were studied for an enzyme-catalyzed reaction. The following results were obtained.a. How do (1) temperature and (2) pH affect the activity of this enzyme?b. Describe a controlled experiment using a turnip, guaiacol, and hydrogen peroxide that could have produced the data shown for either temperature or pH. Be sure to state the hypothesis that was tested here.
27Investigation 7: Mitosis & Meiosis -What are the similarities and differences?
29Lab 7: Mitosis Description Examine onion root tip slides of group treated with lectins (mitogens) vs. control groupGrow onion root tips in water with lectin vs. water without lectinexam slides of onion root tipscount number of cells in each stage to determine relative time spent in each stage in contol group vs. experimental group
30Number of Cells-Control Number of Cells-Treated TipNumber of Cells-ControlMitosisInterphaseTotal123TipNumber of Cells-TreatedMitosisInterphaseTotal123Run chi-square test by first getting expected by finding percentages of Interphase cells and mitosis cells in the control and then multiply those percentages by total number of cells in the treated group.How many degrees of freedom?If X2 value is greater than critical value we reject the null hypothesis and conclude lectins do effect the cell cycle
31Lab 7: CancerConceptsHela Cells- Taken from Henrietta Lacks cervical tumors in the An immortal cell line that provided cells for many types of research for many decades. These cells were taken without permission of Henrietta or the familyPhiladelphia Chromosome- a segment of chromosome 9 exchanged with a segment of chromosome 22. This translocation resulted in a gene that codes for a protein that greatly accelerates the cell cycle.
32Control of the Cell Cycle Protein kinases (enzymes that activate other proteins by phosphorylation), Cdks in this case, remain at a constant levelCyclins build upThe two combine to form MPF, maturation promoting factor, which pushes the cell into the M phase
33Stop and Go-Internal and External Signals Nutrient factors in mediumGrowth factors- local regulators ,proteins, that initiate the cell communication that results in cyclins being produced. Cell communication from nearby cells is termed paracrine signalingDensity dependent inhibitionAnchorage dependence
34Lab 7: Meiosis meiosis crossing over meiosis 1 meiosis 2 in prophase 1 separate homologous pairsmeiosis 2separate sister chromatidscrossing overin prophase 1further genes are from each other the greater number of crossovers
35Lab 7: Meiosis Description crossing over in meiosis farther gene is from centromere the greater number of crossoversobserved crossing over in fungus, Sordariaarrangement of ascospores
36distance from centromere Sordaria analysis% crossovertotal crossovertotal offspring=distance from centromere% crossover2=
37Lab 7: Mitosis & MeiosisMeiosis reduces chromosome number and rearranges genetic information.a. Explain how the reduction and rearrangement are accomplished in meiosis.b. Discuss cancer and how a cell might lose contol of the cell cycle.
38What is the I.V. in this experiment? Lab 5: Photosynthesis -How did we use these materials to study photosynthesis?What is the I.V. in this experiment?
39Lab 5: Photosynthesis -The old way Descriptiondetermine rate of photosynthesis under different conditionslight vs. darkboiled vs. unboiled chloroplastschloroplasts vs. no chloroplastsuse DPIP in place of NADP+DPIPox = blueDPIPred = clearmeasure light transmittance
40Lab 5: Photosynthesis Conclusions Photosynthesis light & unboiled chloroplasts produced highest rate of photosynthesisWhich is the control?#2 (DPIP + chloroplasts + light)
42Lab 5: PhotosynthesisA controlled experiment was conducted to analyze the effects of darkness and boiling on the photosynthetic rate of incubated chloroplast suspensions. The dye reduction technique was used. Each chloroplast suspension was mixed with DPIP, an electron acceptor that changes from blue to clear when it is reduced. Each sample was placed individually in a spectrophotometer and the percent transmittance was recorded. The three samples used were prepared as follows.Sample 1 — chloroplast suspension + DPIPSample 2 — chloroplast suspension surrounded by foil wrap to provide a dark environment + DPIPSample 3 — chloroplast suspension that has been boiled + DPIPData are given in the table on the next page.a. Construct and label a graph showing the results for the three samples.b. Identify and explain the control or controls for this experiment.c. The differences in the curves of the graphed data indicate that there were differences in the number of electrons produced in the three samples during the experiment. Discuss how electrons are generated in photosynthesis and why the three samples gave different transmittance results.
45Lab 6: Cellular Respiration Descriptionusing respirometer to measure rate of O2 consumed by pea seedsnon-germinating peasgerminating peaseffect of temperaturecontrol for changes in pressure & temperature in room
46Lab 6: Cellular Respiration Conceptsrespirationexperimental designcontrol vs. experimentalfunction of KOHfunction of vial with only glass beads
48Lab 6: Cellular Respiration ESSAY 1990The results below are measurements of cumulative oxygen consumption by germinating and dry seeds. Gas volume measurements were corrected for changes in temperature and pressure.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°C2. 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.Cumulative Oxygen Consumed (mL)Time (minutes)10203040Germinating seeds 22°C0.08.816.023.732.0Dry Seeds (non-germinating) 22°C0.20.1Germinating Seeds 10°C220.127.116.112.5Dry Seeds (non-germinating) 10°C
49Lab 8: Bacterial Transformation DescriptionTransformationinsert foreign gene in bacteria by using engineered plasmidalso insert ampicillin resistant gene on same plasmid as selectable markerAnd an operator that demands arabinose to be present in the environment for transcription to occur
58Lab 9: Molecular Biology 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).Explain how the principles of gel electrophoresis allow for the separation of DNA fragmentsDescribe 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.DNA digested with only enzyme XDNA digested with only enzyme YDNA digested with enzyme X and enzyme Y combinedUndigested DNA
64Lab 11: TranspirationA group of students designed an experiment to measure transpiration rates in a particular species of herbaceous plant. Plants were divided into four groups and were exposed to the following conditions.Group I: Room conditions (light, low humidity, 20°C, little air movement.)Group II: Room conditions with increased humidity.Group III: Room conditions with increased air movement (fan)Group IV: Room conditions with additional lightThe cumulative water loss due to transpiration of water from each plant was measured at 10-minute intervals for 30 minutes. Water loss was expressed as milliliters of water per square centimeter of leaf surface area. The data for all plants in Group I (room conditions) were averaged. The average cumulative water loss by the plants in Group I is presented in the table below.Construct and label a graph using the data for Group I. Using the same set of axes, draw and label three additional lines representing the results that you would predict for Groups II, III, and IV.Explain how biological and physical processes are responsible for the difference between each of your predictions and the data for Group I.Explain how the concept of water potential is used to account for the movement of water from the plant stem to the atmosphere during transpiration.Average Cumulative Water Loss by the Plants in Group ITime (minutes)Average Cumulative Water Loss (mL H2O/cm2)103.5 x 10-4207.7 x 10-43010.6 x 10-4
65Lab 11: Transpiration How might species A differ from species B If the line for species A leveled off after 10 minutes (Red Line) how would you explain this?
66Lab 2: Modeling Evolution Descriptionsimulations are used to study effects of different parameters on frequency of alleles in a populationComputer Spreadsheet SimulationYarn Worm Simulation
68Lab 2: Modeling Evolution Conclusionsrecessive alleles remain hidden in the pool of heterozygoteseven lethal recessive alleles are not completely removed from populationknow how to solve H-W problems!to calculate allele frequencies, use p + q = 1to calculate genotype frequencies or how many individuals, use, p2 + 2pq + q2 = 1
69Lab 2: Modeling Evolution ONLINEPopulation genetics simulation program: Bob Sheely from Radford University has created a simulation and documentation in the form of a Web application. It is available for free at .
70Lab 2: Modeling Evolution ESSAY 1989Do 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. (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?
71Lab 12: Animal Behavior -What did we do? 2=(observed – expected)2expectedProbability (p)Degrees of Freedom (df)12345.053.845.997.829.4911.1
72Fruit Fly Life CycleA molt must occur between each of the three instars… this is where the fly grows out of it’s old exoskeletonThe pupa is immobileThe speed of the cycle is dependent upon the external temperature and ranges between 1 and 2 weeks
73Male vs. Female Sex? How can you tell? Males have a dark posterior abdomenMales have tiny claspers on their front legs known as sex combsMales are typically a little smaller than females
74Lab 11: Animal Behavior Description Concepts set up an experiment to study behavior in an organismConceptsexperimental designcontrol vs. experimentalI.V. vs. D.V.hypothesischoice chambertemperaturehumiditylight intensity
75Degrees of Freedom (df) Lab 11: Animal BehaviorFruit Fly BehaviorKinesis vs. TaxisGeotaxis (aka gravotaxis)ChemotaxisPreference for rotting fruit as opposed to fresh.Preference for vinegar and alcoholStatistical Analysis of DataChi-Square AnalysisExpected is 50% of total amount of flies to be on one side of choice chamberObserved is the counted number of flies to one sideDegrees of freedom =1Greater than 3.84 reject null or less than 3.84 accept nullProbability (p)Degrees of Freedom (df)12345.053.845.997.829.4911.1
77Lab 11: Animal Behavior Pill Bug Behavior (the old lab) Same ideas with different organism
78Lab 11: Animal Behavior Hypothesis development Poor: I think pillbugs will move toward the wet side of a choice chamber.Better: If pillbugs prefer a moist environment, then when they are randomly placed on both sides of a wet/dry choice chamber and allowed to move about freely for 10 minutes, most will be found on the wet side.
80Lab 11: Animal 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.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.time of day12 mid4am8am12 noon4pm8pmaverage distance between individuals8.08.944.8174.0350.560.5
81Lab 11: Animal 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.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.temperaturefood availabilitypresence of predatorssocial behaviorPropose 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.
82Lab 3- Comparing DNA Sequences to Understand Evolutionary Relationships with BLAST What did we do?
83Lab 3- Comparing DNA Sequences to Understand Evolutionary Relationships with BLAST DescriptionUse BLAST to compare several genes, and then use information to construct a cladogramCladogram, also called a phylogenetic tree, is a visualization of the evolutionary relatedness of species
84Lab 3- Comparing DNA Sequences to Understand Evolutionary Relationships with BLAST ConclusionFossil Specimen is related to birds based on 3 of the 4 genes we uploaded into BLAST1 of the 4 genes, Fruit Fly, must be the specimens lunch!
85Lab 3- Comparing DNA Sequences to Understand Evolutionary Relationships with BLAST Use the following data to construct a cladogram of the major plant groups
86Investigation 1- Artificial Selection What did we Do?
87Investigation 1- Artificial Selection Can extreme selection change the expression of a quantitative trait in a population in one generation?DescriptionGrow a random sampling of Wisconsin Fast Plant seedsSelect a trait that can be quantified… we chose Trichomes. What are trichomes?Decide on- how many days after planting (12 days) and a uniform counting method to count trichomesCollect data and graph into a histogramAllow only the top 10% hairiest plants to flowerCross pollinate the hairiest plants and then collect seedsReplant seeds and recount trichomes of second generation on day 12Compare the mean number of trichomes for each populationConstruct error barsRun a two tailed t-test to find out if there is a significant difference between the means of the two populations
88Investigation 1- Artificial Selection Directional SelectionThe mean number of trichomes per plant in the population was influenced by extreme selection
91Lab 12: Dissolved Oxygen (Old Lab) Descriptionmeasure primary productivity by measuring O2 productionfactors that affect amount of dissolved O2temperatureas water temperature, its ability to hold O2 decreasesphotosynthetic activityin bright light, aquatic plants produce more O2decomposition activityas organic matter decays, microbial respiration consumes O2mixing & turbulencewave action, waterfalls & rapids aerate H2O & O2salinityas water becomes more salty, its ability to hold O2 decreases
92Lab 12: Dissolved Oxygen (Old Lab) Conceptsdissolved O2primary productivitymeasured in 3 ways:amount of CO2 usedrate of sugar (biomass) formationrate of O2 productionnet productivity vs. gross productivityrespiration
94Lab 12: Dissolved Oxygen (Old Lab) ESSAY 2001A biologist measured dissolved oxygen in the top 30 centimeters of a moderately eutrophic (mesotrophic) lake in the temperate zone. The day was bright and sunny and the wind was calm. The results of the observation are presented below.Using the graph paper provided, plot the results that were obtained. Then, using the same set of axes, draw and label an additional line/curve representing the results that you would predict had the day been heavily overcast.Explain the biological processes that are operating in the lake to produce the observed data. Explain also how these processes would account for your prediction of results for a heavily overcast day.Describe how the introduction of high levels of nutrients such as nitrates and phosphates into the lake would affect subsequent observations. Explain your predictions.hour6am8am10amnoon2pm4pm6pm8pm10pmmid[O2] mg/L0.91.73.14.18.104.22.168.24.02.4
95Lab 12: Dissolved Oxygen (Old Lab) ESSAY 2004BIn most aquatic environments, primary production is affected by light available to the community of organisms.Using measurements of dissolved oxygen concentration to determine primary productivity, design a controlled experiment to test the hypothesis that primary productivity is affected by either the intensity of light or the wavelength of light. In your answer, be sure to include the following.A statement of the specific hypothesis that you are testingA description of your experimental design (Be sure to include a description of what data you would collect and how you would present and analyze the data using a graph.)A description of results that would support your hypothesis
96Lab 7: Genetics (Old Lab) Descriptiongiven fly of unknown genotype use crosses to determine mode of inheritance of trait
97Lab 7: Genetics (Old Lab) Conceptsphenotype vs. genotypedominant vs. recessiveP, F1, F2 generationssex-linkedmonohybrid crossdihybrid crosstest crosschi square
99Lab 7: Genetics (Old Lab) Conclusions: Can you solve these?Case 1Case 2
100Lab 7: Genetics (Old Lab) ESSAY 2003 (part 1)In fruit flies, the phenotype for eye color is determined by a certain locus. E indicates the dominant allele and e indicates the recessive allele. The cross between a male wild type fruit fly and a female white eyed fruit fly produced the following offspringThe wild-type and white-eyed individuals from the F1 generation were then crossed to produce the following offspring.a. Determine the genotypes of the original parents (P generation) and explain your reasoning. You may use Punnett squares to enhance your description, but the results from the Punnett squares must be discussed in your answer.b. Use a Chi-squared test on the F2 generation data to analyze your prediction of the parental genotypes. Show all your work and explain the importance of your final answer.c. The brown-eyed female of the F1 generation resulted from a mutational change. Explain what a mutation is, and discuss two types of mutations that might have produced the brown-eyed female in the F1 generation.Wild-Type MaleWild-Type FemaleWhite-eyed MaleWhite-Eyed FemaleBrown-Eyed FemaleF-145551Wild-Type MaleWild-Type FemaleWhite-eyed MaleWhite-Eyed FemaleBrown-Eyed FemaleF-223312224
101Lab 7: Genetics (Old Lab) ESSAY 2003 (part 2)The formula for Chi-squared is:Probability (p)Degrees of Freedom (df)12345.053.845.997.829.4911.12=(observed – expected)2expected