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Animations and Videos Bozeman - AP BIO Labs Review.

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1 Animations and Videos Bozeman - AP BIO Labs Review

2 Lab 1: Diffusion & Osmosis

3 Lab 1: Diffusion & Osmosis
Description dialysis tubing filled with starch-glucose solution in beaker filled with KI solution potato cores in sucrose solutions

4 Lab 1: Diffusion & Osmosis
Concepts semi-permeable membrane diffusion osmosis solutions hypotonic hypertonic isotonic water potential

5 Lab 1: Diffusion & Osmosis
Conclusions water moves from high concentration of water (hypotonic=low solute) to low concentration of water (hypertonic=high solute) solute concentration & size of molecule affect movement through semi-permeable membrane

6 When a solution such as that inside dialysis tubing is separated from pure water by a selectively-permeable membrane water will move by osmosis from the surrounding area where the water potential is higher into the cell where water potential is lower due to the presence of solute. The movement of water into the cell causes the cell to swell and the cell membrane pushes against the cell wall to produce an increase in pressure (turgor). This process will continue until the water potential of the cell equals the water potential of the pure water outside the cell. At this point, a dynamic equilibrium is reached and net water movement will cease.

7 Animations and Videos Bozeman - AP BIO Lab 1 - Diffusion and Osmosis
AP LAB 1 - Diffusion and Osmosis AP WEB LAB - Osmosis and Diffusion

8 Lab 2: Enzyme Catalysis Description
measured factors affecting enzyme activity H2O2  H2O + O2 measured rate of O2 production catalase

9 Lab 2: Enzyme Catalysis Concepts substrate enzyme product
enzyme structure product denaturation of protein experimental design rate of reactivity reaction with enzyme vs. reaction without enzyme optimum pH or temperature test at various pH or temperature values

10 Lab 2: Enzyme Catalysis Conclusions
enzyme reaction rate is affected by: pH temperature substrate concentration enzyme concentration calculate rate?

11 In the first few minutes of an enzymatic reaction, the number of substrate molecules is usually so large compared to the number of enzyme molecules that changing the substrate concentration does not (for a short period at least) affect the number of successful collisions between substrate and enzyme. During this early period, the enzyme is acting on substrate molecules at a constant rate. The slope of the graph line during this early period is called the initial velocity of the reaction. The initial velocity, or rate, of any enzyme catalyzed reaction is determined by the characteristics of the enzyme molecule. It is always the same for an enzyme and its substrate as long as temperature and pH are constant and substrate is present in excess. Also, in this experiment the disappearance of the substrate is essential in this reaction. Eventually, the rate of the reaction levels off.

12 Animations and Videos Bozeman - AP BIO Lab 2 - Enzyme Catalyst
AP LAB 2 - Enzyme Catalyst AP WEB LAB - Enzyme Catalyst

13 Lab 3: Mitosis & Meiosis

14 Lab 3: Mitosis & Meiosis Description cell stages of mitosis
exam slide of onion root tip count number of cells in each stage to determine relative time spent in each stage crossing over in meiosis farther gene is from centromere the greater number of crossovers observed crossing over in fungus, Sordaria arrangement of ascospores

15 Lab 3: Mitosis & Meiosis Concepts I P M A T mitosis meiosis
interphase prophase metaphase anaphase telophase meiosis meiosis 1 meiosis 2 crossing over tetrad in prophase 1 I P M A T

16 Lab 3: Mitosis & Meiosis Conclusions Mitosis Meiosis
longest phase = interphase each subsequent phase is shorter in duration Meiosis 4:4 arrangement in ascospores no crossover any other arrangement crossover 2:2:2:2 or 2:4:2

17 distance from centromere
Sordaria analysis % crossover total crossover total offspring = distance from centromere % crossover 2 =

18 The relative lengths of the mitotic stages are: 53. 4% prophase, 17
The relative lengths of the mitotic stages are: 53.4% prophase, 17.4% metaphase, 16.8% anaphase and 12.4% telophase. Meiosis is important for sexual reproduction because it reduces the chromosome number by half and it also results in new combinations of genes through independent assortment and crossing over, followed by the random fertilization of eggs by sperm.

19 Animations and Videos Bozeman - AP BIO Lab 3 - Mitosis and Meiosis
AP LAB 3 - Mitosis and Meiosis Bozeman - The Sodaria Cross AP WEB LAB – Mitosis AP WEB LAB - Meiosis

20 Lab 4: Photosynthesis

21 Lab 4: Photosynthesis Description
determine rate of photosynthesis under different conditions light vs. dark boiled vs. unboiled chloroplasts chloroplasts vs. no chloroplasts use DPIP in place of NADP+ DPIPox = blue DPIPred = clear measure light transmittance paper chromatography to separate plant pigments

22 Lab 4: Photosynthesis Concepts photosynthesis Photosystem 1
NADPH chlorophylls & other plant pigments chlorophyll a chlorophyll b xanthophylls carotenoids experimental design control vs. experimental

23 Lab 4: Photosynthesis Conclusions Pigments Photosynthesis
pigments move at different rates based on solubility in solvent Photosynthesis light & unboiled chloroplasts produced highest rate of photosynthesis

24 Lab 4: Photosynthesis Time (min) Light, Unboiled % transmittance
Sample 1 Dark, Unboiled % transmittance Sample 2 Light, Boiled % transmittance Sample 3 28.8 29.2 5 48.7 30.1 10 57.8 31.2 29.4 15 62.5 32.4 28.7 20 66.7 31.8 28.5

25 Animations and Videos Bozeman - AP BIO Lab 4 - Plant Pigments and Photosynthesis AP LAB 4 - Plant Pigments and Photosynthesis Using a Spectrophotometer AP WEB LAB – Pigments AP WEB LAB - Photosynthesis

26 The solvent moves up the paper by capillary action, which occurs as a result of the attraction of solvent molecules to the paper and the attraction of solvent molecules to one another. As the solvent moves up the paper, it carries along any substances dissolved in it, in this case pigments. The pigments are carried along at different rates because they are not equally soluble in the solvent and because they are attracted, to different degrees, to the cellulose in the paper through the formation of hydrogen bonds. Also, as the DPIP is reduced and becomes colorless, the resultant increase in light transmittance is measured over a time course using a spectrophotometer. 

27 Lab 5: Cellular Respiration

28 Lab 5: Cellular Respiration
Description using respirometer to measure rate of O2 production by pea seeds non-germinating peas germinating peas effect of temperature control for changes in pressure & temperature in room

29 Lab 5: Cellular Respiration
Concepts respiration experimental design control vs. experimental function of KOH function of vial with only glass beads

30 Lab 5: Cellular Respiration
Conclusions temp = respiration germination = respiration calculate rate?

31 Germinating peas respire and need to consume oxygen in order to continue the growing process. Pea seeds are non-germinating and do not respire actively. These seeds are no longer the site of growth and thus do not need oxygen for growth. In consideration to temperature, at higher temperatures more oxygen is consumed which means more respiration is occurring. 686 kilocalories are released during respiration. When temperature decreases molecular motion slows down and respiration decreases because less energy is made available.

32 Animations and Videos Bozeman - Lab 5 - Cell Respiration
Bozeman - AP BIO Lab 5 - Cell Respiration AP LAB 5- Cell Respiration AP WEB LAB - Cell Respiration

33 Lab 6: Molecular Biology

34 Lab 6: Molecular Biology
Description Transformation insert foreign gene in bacteria by using engineered plasmid also insert ampicillin resistant gene on same plasmid as selectable marker Gel electrophoresis cut DNA with restriction enzyme fragments separate on gel based on size

35 Lab 6: Molecular Biology
Concepts transformation plasmid selectable marker ampicillin resistance restriction enzyme gel electrophoresis DNA is negatively charged smaller fragments travel faster

36 Lab 6: Transformation Conclusions can insert foreign DNA using vector
ampicillin becomes selecting agent no transformation = no growth on amp+ plate

37 Lab 6: Gel Electrophoresis
Conclusions DNA = negatively charged correlate distance to size smaller fragments travel faster & therefore farther

38 Bacterial Transformation-Ampicillin Resistance: In this exercise, we will introduce competent E. Coli cells to take up the plasmid pAMP, which contains a gene for ampicillin resistance. Normally, E. Coli cells are destroyed by the antibiotic ampicillin, but E. Coli cells that have been transformed will be able to grow on agar plates containing ampicillin. Thus, we can select for transformants; those cells that are not transformed will be killed by ampicillin; those that have been transformed will survive.

39 Restriction Enzyme Cleavage of DNA: Restriction endonuclease recognizes specific DNA sequences in double-stranded DNA and digests the DNA at these sites. The result is the production of fragments of DNA of various lengths corresponding to the distance between identical DNA sequences within the chromosome. By taking DNA fragments and systematically reinserting the fragments into an organism with minimal genetic material, it is possible to determine the function of particular gene sequences

40 Electrophoresis: Fragments of DNA can be separated by gel electrophoresis when any molecule enters the electrical field, the mobility or speed at which it will move is influenced by the charge (negative charges travel to positive/top pole of gel), the density of the molecule, (the smaller the molecule, the faster it travels), the strength of the electrical field, and the density of the medium (gel) which it is migrating.

41 Animations and Videos Bozeman - AP BIO Lab 6 - Molecular Biology
AP LAB 6 - Molecular Biology AP WEB LAB - Molecular Biology

42 Lab 7: Genetics (Fly Lab)

43 Lab 7: Genetics (Fly Lab)
Description given fly of unknown genotype use crosses to determine mode of inheritance of trait

44 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

45 Lab 7: Genetics (Fly Lab)
Conclusions: Can you solve these? Case 1 Case 2

46 From this lab, you will be able to find genotypes and phenotypic expression within a fruit fly. Also, recessive genes and mutations will be revealed as the student crosses a variety of Drosophila alleles. For example, if a female carrier for an x-linked, recessive trait, was crossed with a male without the recessive trait the results would be: ½ males with x-linked trait ½ males without ½ female carriers ½ females without 0 females express sex linked traits

47 Animations and Videos Bozeman - AP BIO Lab 7 -Genetics of Organisms
AP LAB 7 - Genetics of Organisms AP WEB LAB - Genetics of Organisms Bozeman - Chi-squared Test

48 Lab 8: Population Genetics
Description simulations were used to study effects of different parameters on frequency of alleles in a population selection heterozygous advantage genetic drift

49 Lab 8: Population Genetics
Concepts Hardy-Weinberg equilibrium p + q = 1 p2 + 2pq + q2 = 1 required conditions large population random mating no mutations no natural selection no migration gene pool heterozygous advantage genetic drift founder effect bottleneck

50 Lab 8: Population Genetics
Conclusions recessive alleles remain hidden in the pool of heterozygotes even lethal recessive alleles are not completely removed from population know how to solve H-W problems! to calculate allele frequencies, use p + q = 1 to calculate genotype frequencies or how many individuals, use, p2 + 2pq + q2 = 1

51 Assuming that Hardy-Weinberg equilibrium is maintained allele and genotype frequencies should remain constant from generation to generation. For this to happen the five following situations must all occur: 1. Population is very large. The effects of chance on changes in allele frequencies is thereby greatly reduced. 2. Individuals show no mating preference, i.e. random mating. 3.There is no mutation of alleles. 4. No differential migration occurs, (no immigration or emigration). 5. All genotypes have an equal chance of surviving and reproducing, i.e. there is no natural selection.

52 In humans, several genetic diseases have been well characterized
In humans, several genetic diseases have been well characterized. Some of these diseases are controlled by a single allele where the homozygous recessive genotype has a high probability of not reaching reproductive maturity. If this were to occur both the homozygous dominant and heterozygous individuals will survive while the homozygous recessive will become extinct.

53 Animations and Videos Bozeman - Lab 8 - Population Genetics and Evolution AP LAB 8 -Population Genetics and Evolution AP WEB LAB - Population Genetics and Evolution

54 Lab 9: Transpiration

55 Lab 9: Transpiration Description
test the effects of environmental factors on rate of transpiration temperature humidity air flow (wind) light intensity

56 Lab 9: Transpiration Concepts transpiration stomates guard cells xylem
adhesion cohesion H bonding

57 Lab 9: Transpiration Conclusions transpiration transpiration  wind
 light transpiration  humidity

58 Conditions that cause a decreased rate of water loss from leaves result in a decreased water potential gradient from stem to leaf and therefore in a decreased rate of water movement up the stem to the leaves. Conditions that cause an increased rate of water loss from leaves result in an increase in the water potential gradient from stem to leaf and therefore in an increase in the rate of water movement up the stem to the leaves.

59 Normal Room Conditions (CONTROL)
When you expose a plant to room conditions nothing is supposed to happen. The reasoning for this is room conditions don’t cause drastic changes in the plants environment for major transpiration or even water gain to occur. The plant under room conditions is considered to be your control.

60 Floodlight When light is absorbed by the leaf, some of the light energy is converted to heat and remember that transpiration rate increases with temperature. We learned in Unit One of the Campbell’s edition that when the temperature of liquid water rises, kinetic energy of the water molecules increases. As a result, the rate at which liquid water is converted to water vapor increases. When the water is turned into water vapor, it easily passes out through stomata out into the outer atmosphere. The floodlight is an example of a plant near the sun (which is why the plant is one meter away from the light). Due to the aforementioned properties of plants you should see a loss of water.

61 Fan (WIND) An increase in wind speed results in an increase in the rate of leaf water loss because increased wind decreases the boundary layer of still air at the leaf surface. This boundary layer acts to slow leaf water loss. Increased wind also causes the rapid removal of evaporating water molecules from the leaf surface. This results in a low water potential in the air immediately and the water level should drop.

62 Mist (HIGH HUMIDITY) Increased humidity in the air surrounding the leaf decreases the water potential gradient between the saturated air in the leaf air spaces and the air surrounding the leaf, resulting in a decreased rate of leaf water loss. However, when the humidity of the air surrounding the leaf if very low, the water potential of the air is low. Therefore, the water potential gradient between the air spaces of the leaf and the surrounding air is high, and the rate of leaf water loss increases.

63 When there is a great amount of humidity, transpiration decreases because of water potential. When the humidity is at a low or normal, the mesophyll cells in the plant are much higher in water potential than the relatively drier surrounding air. Due to the properties of water potential, which states that water tends to evaporate from the leaf surface moving from an area of higher water potential to an area of lower water potential, transpiration occurs. But, because of the high humidity, the surrounding air has a higher water potential than the mesophyll cells and water loss is at a minimum. 

64 Adaptations to reduce leaf water loss include a reduced number of stomates, an increase in the thickness of the leaf cuticle, a decrease in leaf surface area, and adaptations that decrease air movements around stomates, such as dense hairs and sunken stomates. Because leaves are all different in size, reporting the water loss without considering a unit area would provide non-comparable data.

65 Animations and Videos Bozeman - AP BIO Lab 9 – Transpiration
AP LAB 9 – Transpiration AP WEB LAB - Transpiration

66 Lab 10: Circulatory Physiology

67 Lab 10: Circulatory Physiology
Description study factors that affect heart rate body position level of activity determine whether an organism is an endotherm or an ectotherm by measuring change in pulse rate as temperature changes Daphnia

68 Lab 10: Circulatory Physiology
Concepts thermoregulation endotherm ectotherm Q10 measures increase in metabolic activity resulting from increase in body temperature Daphnia can adjust their temperature to the environment, as temperature in environment increases, their body temperature also increases which increases their heart rate

69 Lab 10: Circulatory Physiology
Conclusions Activity increase heart rate in a fit individual pulse & blood pressure are lower & will return more quickly to resting condition after exercise than in a less fit individual Pulse rate changes in an ectotherm as external temperature changes

70 The sphygmomanometer measures the blood pressure
The sphygmomanometer measures the blood pressure. The blood pressure cuff is inflated so that blood flow stops to through the brachial artery in the upper arm. A stethoscope is used to listen to blood flow entering the brachial artery. When blood first enters the artery, snapping sounds called the sounds of Korotkoff are generated. A. Blood pressure and heart rate increase when you move from a reclining to a standing position counteracting gravitational pull on the blood B. Elevated arterial blood pressure indicates increased arterial resistance to blood flow C. Fit individuals can pump a larger volume of blood with each contraction and deliver more oxygen to muscle tissue than the hearts of unfit individuals. As a result, blood pressure and heart rate increases are smaller for fit individuals, and the time required to return to normal conditions is shorter for fit individuals than unfit individuals. D. For the Daphnia, remember that ectothermic animals use behavior to regulate their body temperatures and that Q10 cannot be determined for endothermic animals because body temperatures remain constant regardless of environmental temperatures.

71 Animations and Videos Bozeman - AP BIO Lab 10 - Circulatory Physiology
AP LAB 10 - Circulatory Physiology

72 Lab 11: Animal Behavior

73 Lab 11: Animal Behavior Description
set up an experiment to study behavior in an organism Betta fish agonistic behavior Drosophila mating behavior pillbug kinesis

74 Lab 11: Animal Behavior Concepts innate vs. learned behavior
experimental design control vs. experimental hypothesis choice chamber temperature humidity light intensity salinity other factors

75 Lab 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.

76 Lab 11: Animal Behavior Experimental design sample size

77 When conducting this experiment, a couple of things should be understood.
A. Three variables are tested: light, temperature and pH. The control is exposed to room light, room temperature, and neutral pH is also prepared. For each of the variables, a gradient is established providing a continuous variation from weak to strong intensities. B. The histograms are prepared showing the number of isopods in each of the four intensities. A histogram is prepared for each of the three variables and the control. From the data in the histograms, conclusions cab be made describing the habitat preferences of the isopod.

78 Animations and Videos Bozeman - AP BIO Lab 11 - Animal Behavior
AP LAB 11 - Animal Behavior Bozeman - Q10: The Temperature Coefficient AP WEB LAB - Animal Behavior

79 Lab 12: Dissolved Oxygen Dissolved O2 availability

80 Lab 12: Dissolved Oxygen

81 Lab 12: Dissolved Oxygen Description
measure primary productivity by measuring O2 production factors that affect amount of dissolved O2 temperature as water temperature, its ability to hold O2 decreases photosynthetic activity in bright light, aquatic plants produce more O2 decomposition activity as organic matter decays, microbial respiration consumes O2 mixing & turbulence wave action, waterfalls & rapids aerate H2O & O2 salinity as water becomes more salty, its ability to hold O2 decreases

82 Lab 12: Dissolved Oxygen Concepts dissolved O2 primary productivity
measured in 3 ways: amount of CO2 used rate of sugar (biomass) formation rate of O2 production net productivity vs. gross productivity respiration

83 Lab 12: Dissolved Oxygen Conclusions temperature = dissolved O2
light = photosynthesis = O2 production O2 loss from respiration respiration = dissolved O2 (consumption of O2)

84 The amount of oxygen dissolved in natural water samples is measured and analyzed to determine the primary productivity of the sample. The amount of dissolved oxygen is dependent upon many factors. A. Temperature B. Salinity C. Photosynthesis D. Respiration Primary productivity is a measure of the amount of biomass produced by autotrophs through photosynthesis per unit time. It can be examined by the following factors A. Gross Primary Productivity B. Net Primary Productivity C. Respiratory Rate

85 Animations and Videos Bozeman - Lab 12 - Dissolved Oxygen and Aquatic Primary Productivity AP LAB 12 - Dissolved Oxygen and Aquatic Primary Productivity AP WEB LAB - Dissolved Oxygen AP WEB LAB - Aquatic Primary Productivity


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