1. AP Biology
Lab Review

2. Big Idea 3: genetics and information transfer

3. Lab 7: Mitosis & Meiosis Concepts: Cell Cycle (G1  S  G2  M)
Control of cell cycle (checkpoints)
Cyclins & cyclin-dependent kinases (CDKs)
Mitosis vs. Meiosis
Crossing over  genetic diversity

4. Lab 7: Mitosis & Meiosis

5. Lab 7: Mitosis & Meiosis

6. 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)

7. Lab 7: Mitosis & Meiosis

8. Lab 7: Mitosis & Meiosis

9. Abnormal karyotype = Cancer

10. Meiosis: Crossing over in Prophase I

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

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

13. 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.

14. 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

15. Lab 8: Bacterial Transformation

16. Lab 8: Bacterial Transformation

17. 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)

23. 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

24. Lab 9: Restriction Enzyme Analysis of DNA
Description

25. Lab 9: Restriction Enzyme Analysis of DNA
Determine DNA fragment sizes

26. 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

27. 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).
Explain how the principles of gel electrophoresis allow for the separation of DNA fragments
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.
DNA digested with only enzyme X
DNA digested with only enzyme Y
DNA digested with enzyme X and enzyme Y combined
Undigested DNA
Explain both of the following:
The mechanism of action of restriction enzymes
The different results you would expect if a mutation occurred at the recognition site for enzyme Y.

28. Lab 8-9: Biotechnology ESSAY 2002
The human genome illustrates both continuity and change.
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)
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.

29. Big Idea 4: interactions

30. Lab 10: Energy Dynamics Concepts: Biomass = mass of dry weight
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

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

32. Lab 10: Energy Dynamics Description:
Brassica (cabbage)  cabbage white butterfly larvae (caterpillars)

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

34. Lab 10: Energy Dynamics

35. Lab 10: Energy Dynamics
Conclusions:

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

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

40. Lab 11: Transpiration

41. 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

42. Analysis of Stomata

43. Rates of Transpiration

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

49. 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

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

51. Lab 12: Animal Behavior

52. Lab 12: Animal Behavior Hypothesis Development
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.

53. Lab 12: Animal Behavior
Experimental Design
sample size

54. 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 X2 < critical value  accept null hypothesis (any differences between observed and expected due to CHANCE)

55. 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.
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 day
12 mid
4am
8am
12 noon
4pm
8pm
average distance between individuals
8.0
8.9
44.8
174.0
350.5
60.5

56. 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.
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
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.

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

58. Lab 13: Enzyme Activity Description:
Determine which factors affecting rate of enzyme reaction
H2O2  H2O + O2
Measure rate of O2 production
catalase

59. Turnip peroxidase  Color change (O2 produced)

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

62. Lab 13: Enzyme Activity 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.

63. Any Questions??