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Enzyme Activity Lab Ap Biology Essay

Presentation on theme: "AP Biology Lab Review AP Biology Lab 1: Enzyme Activity  Concepts:  Enzyme  Structure (active site, allosteric site)  Lower activation energy, speed."— Presentation transcript:


2 AP Biology Lab Review

3 AP Biology Lab 1: Enzyme Activity  Concepts:  Enzyme  Structure (active site, allosteric site)  Lower activation energy, speed up rate  Substrate  product  Proteins denature (structure/binding site changes)

4 AP Biology Lab 1: Enzyme Activity  Description:  Determine which factors affect the rate of enzyme reaction  H 2 O 2  H 2 O + O 2  Measure rate of O 2 production catalase

5 AP Biology Lab 1: Enzyme Activity  Conclusions:  Enzyme reaction rate affected by:  pH (acids, bases)  Temperature  Substrate concentration  Enzyme concentration  Ionic conditions Calculate Rate of Reaction

6 AP Biology

7 ESSAY 2000 The 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? In your answer, include a discussion of the relationship between the structure and the function of this enzyme, as well as a discussion of ho structure and function of enzymes are affected by temperature and pH. b. Describe a controlled experiment that could have produced the data shown for either temperature or pH. Be sure to state the hypothesis that was tested here. Lab 1: Enzyme Activity

8 AP Biology Lab 2: Diffusion & Osmosis (Lab Two)  Concepts:  Selectively permeable membrane (cell membrane!)  Diffusion (high  low concentration)  Osmosis (aquaporins)  Water potential (  )   = pressure potential (  P ) + solute potential (  S )  Solutions:  Hypertonic  hypotonic  isotonic

9 AP Biology Lab 2: Diffusion & Osmosis

10 AP Biology Lab 2: 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)

11 AP Biology Lab 2: Diffusion & Osmosis

12 AP Biology Potato Cores in Different Concentrations of Sucrose

13 AP Biology Lab 2: 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


15 AP Biology

16 Lab 2: 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 explanation of those results based on the principles involved. Be sure to clearly state the principles addressed in your discussion.

17 AP Biology Lab 3: Mitosis  Concepts:  Cell Cycle (G1  S  G2  M)  Control of cell cycle (checkpoints)  Cyclins & cyclin-dependent kinases (CDKs)  1 cell forms two identical (clone) daughter cells  Phases:  interphase  prophase  metaphase  anaphase  telophase

18 AP Biology Lab 3: Mitosis

19 AP Biology Lab 3: Mitosis Count # cells in interphase, stages of mitosis

20 AP Biology Lab 3: Mitosis  Conclusions  Mitosis (PMAT!)  longest phase = interphase

21 AP Biology Lab 3: Mitosis

22 AP Biology Lab 3: Mitosis  Description:  How environment affects mitosis of plant roots  Effect of caffeine on root growth

23 AP Biology Labs 3 & 6: Mitosis vs Meiosis

24 AP Biology Lab 4: Cellular Respiration

25 AP Biology Lab 4: Cellular Respiration

26 AP Biology Lab 4: Cellular Respiration  Concepts:  Respiration  Measure rate of respiration by:  O 2 consumption  CO 2 production

27 AP Biology Lab 4: 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

28 AP Biology

29 Lab 4: Cellular Respiration  Conclusions:   temp =  respiration   germination =  respiration  Animal respiration > plant respiration   surface area =  respiration Calculate Rate

30 AP Biology Lab 4: Cellular Respiration

31 AP Biology



34 Lab 4: 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. 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. Cumulative Oxygen Consumed (mL) Time (minutes)010203040 Germinating seeds 22°C0.08.816.023.732.0 Dry Seeds (non-germinating) 22°C0. Germinating Seeds 10°C0. Dry Seeds (non-germinating) 10°C0.0

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

36 AP Biology 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  leaf disk rises  Measure rate of photosynthesis by O 2 production  Factors tested: types of plants, light intensity, colors of leaves, pH of solutions

37 AP Biology Plant Pigments & Chromatography

38 Floating Disk Technique


40 AP Biology 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

41 AP Biology


43 2004 - 2005 Lab 6: Meiosis  Concepts  meiosis  meiosis 1  meiosis 2  crossing over  tetrad in prophase 1  Conclusions:  4:4 arrangement in ascospores  no crossover  any other arrangement  crossover  2:2:2:2 or 2:4:2

44 AP Biology Lab 6-Meiosis: Crossing over in Prophase I

45 AP Biology 2004 - 2005 Lab 6:Meiosis  crossing over in meiosis  further the gene is from centromere the greater number of crossovers  Counted spore crossing over in fungus, Sordaria  arrangement of ascospores-2/4/2, 4/4, or 2/2/2/2

46 AP Biology Lab 6: Meiosis  Observed crossing over in fungus (Sordaria)  Arrangement of ascospores

47 AP Biology Sordaria Analysis % crossover total crossover total offspring = distance from centromere % crossover 2 =

48 AP Biology Abnormal karyotype = Cancer

49 AP Biology Labs 3,6: 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.

50 AP Biology Lab 7: Genetics (Fly Lab)

51 AP Biology Lab 7: Genetics (Fly Lab)  Description  Cross flies of known genotype for wing shape to observe inheritance of the trait  Conduct chi square analysis on results

52 AP Biology Lab 7: Genetics (Fly Lab)  Concepts  phenotype vs. genotype  dominant vs. recessive  P, F1, F2 generations  sex-linked  monohybrid cross  dihybrid cross  test cross  chi square

53 AP Biology Lab 7: Genetics (Fly Lab)  Conclusions: Can you solve this? Case 1 Mode of inheritance=________

54 AP Biology Now solve this one! Case 2 Mode of inheritance=________

55 AP Biology Lab 7: Genetics (Fly 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 offspring The 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 Male Wild-Type Female White-eyed Male White-Eyed Female Brown-Eyed Female F-1 0455501 Wild-Type Male Wild-Type Female White-eyed Male White-Eyed Female Brown-Eyed Female F-2 233122240

56 AP Biology Lab 7: Genetics (Fly Lab) ESSAY 2003 (part 2) The formula for Chi-square is: Probability (p) Degrees of Freedom (df) 12345.053.845.997.829.4911.1 22 =  (observed – expected) 2 expected

57 AP Biology 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 = pGLO plasmid

58 AP Biology Lab 8: Bacterial Transformation

59 AP Biology 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 (araC protein)

60 AP Biology





65 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

66 AP Biology Lab 9: Restriction Enzyme Analysis of DNA  Description

67 AP Biology Lab 9: Restriction Enzyme Analysis of DNA  Determine DNA fragment sizes

68 AP Biology 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

69 AP Biology Labs 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.

70 AP Biology Labs 8-9: Biotechnology ESSAY 2002 The human genome illustrates both continuity and change. a.Describe the essential features of two of the procedures/techniques below. For each of the procedures/techniques you describe, explain how its application contributes to understanding genetics.  The use of a bacterial plasmid to clone and sequence a human gene  Polymerase chain reaction (PCR)  Restriction fragment polymorphism (RFLP analysis) b.All humans are nearly identical genetically in coding sequences and have many proteins that are identical in structure and function. Nevertheless, each human has a unique DNA fingerprint. Explain this apparent contradiction.

71 AP Biology Lab 10: Population Genetics

72 AP Biology Lab 10: 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 +2pq+q 2 = 1)  Conditions: 1. large population 2. random mating 3. no mutations 4. no natural selection 5. no migration


Patrick McCrystal

Enzymes: Natural Catalysts

Enzymes are catalytic proteins, meaning they speed up chemical reactions without beingused up or altered permanently in the process. Although various enzymes use different methods,all accomplish catalysis by lowering the activation energy for the reaction, thus allowing it tooccur more easily. Enzymes have very specific shapes (conformations). Part of the conformationis the active site of the enzyme, where the actual catalysis occurs. The specific molecule or closely related molecules on which an enzyme functions is known as its substrate. Shape playssuch an important role in enzymatic catalysis that often even isomers of a substrate will berejected. Once the substrate enters the active site, it may begin a process known as induced fit inwhich the enzyme perfectly conforms to the molecule to allow for more efficient catalysis.Changes in environment can severely impact enzyme catalysis in both negative and positiveways. Each enzyme has specific ranges at which it optimally functions; in general, increasing thetemperature will help the reaction along, until the point at which the protein degrades anddenatures. Denatured proteins will often return to their original state, after the removal of thedenaturing agent, except when they are degraded multiple levels.


1.Peel a fresh potato tuber and cut the tissue into small cubes.2.Weigh out 50 grams of tissue.3.Place the tissue, 50 mL of cold distilled water, and a small amount of crushed ice in a prechilled blender.4.Homogenize for 30 seconds at high speed.5.Filter the potato extract using cheesecloth.6.Pour the filtrate into a 100 mL graduated bylinder and add cold distilled water to bring upthe final volume to 100 mL.


Label eight 50 mL beakers as follows: 100 units/mL, 80 units/mL, 75 units/mL, 60units/mL, 50 units/mL, 25 units/mL, 10 units/mL, 0 units/mL.


Prepare 40 mL of enzyme for each of the above concentrations in the following ratio of enzyme:distilled water – 40:0, 32:8, 30:10, 24:16, 20:10, 10:30, 4:36, and 0:40.9.Using forceps, immerse a 2.1cm filter paper disc into the prepared catalase solution for 5seconds.10.Remove the disc and drain for 10 seconds on a paper towel.11.Place the disc at the bottom of the first substrate solution. The oxygen produced from the breakdown of the hydrogen peroxide by catalase becomes trapped in the fibers of thedisc, thereby causing the disc to float to the surface of the solution12.Measure (using a stopwatch) the reaction time for the amount of time from when the discwas placed at the bottom of the beaker until the disc floats on top of the solution. (ratewill be measured in seconds). The rate (R) of the reaction is calculated as R = 1/t.13.Repeat this procedure twice for each enzyme concentration and average the results.