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Lab Review “The Dirty Dozen”. Diffusion & Osmosis.

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Presentation on theme: "Lab Review “The Dirty Dozen”. Diffusion & Osmosis."— Presentation transcript:

1 Lab Review “The Dirty Dozen”

2 Diffusion & Osmosis

3 Description – Dialysis tubing filled with startch-glucose solution in beaker filled with IKI solution – Potato Cores in Sucrose solutions – Determining solute concentration of different solutions

4 Diffusion & Osmosis Concepts – Semi-permeable membrane – Diffusion – Osmosis – Solutions Hypotonic Hypertonic Isotonic – Water Potential

5 Diffusion & Osmosis Conclusions – Water moves from high concentration of water (hypotonic or high water potential) to low concentration of water (hypertonic or low water potential) – Solute concentrations, polarity and size of molecule affect movement through semipermeable membrane

6 Diffusion & Osmosis Essay 1992 A lab assistant prepared solution of.8M,.6M,.4M, and.2M sucrose, but forgot to label them. After realizing the error, the assistant randomly labeled the flasks containing these four unknown solutions as flask A, B, C, and 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.

7 Diffusion & Osmosis Essay 2005 B

8 Try these questions




12 Enzyme Catalysis Description of Honors Biology Lab – Measured factors affecting enzyme activity – H 2 O 2 H 2 O + O 2 – Measured rate of O 2 production Description of AP Biology Lab – Measured factors affecting enzyme activity – Catechol Quinone – Measured rate of light absorbance – Included Hydroquinone and resorsinol (isomers) Catalase Clear Polyphenoloxidase Brown

13 Enzyme Catalysis Concepts – Substrate – Enzyme – Denaturation of protein – Competitive inhibition – Allosteric activation (cooperativity) – Experimental design Reaction rate w/enzyme vs. reaction rate wo/enzyme Optimum pH or temperature

14 Enzyme Catalysis Conclusions – Enzyme reaction rate is affected by: pH Temperature Substrate concentration Enzyme concentration Allosteric reactivity/cooperativity Competitive inhibition

15 Enzyme Catalysis Essay question 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 how 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. Enzyme Activity Temperature pH

16 Try these questions A. Allosteric Inhibition B. Feedback inhibition C. Competitive inhibition D. Noncompetitive inhibition E. Cooperativity 1.Describes inhibition by an enzyme that is capable of either activation or inhibiting a metabolic pathway. 2.A reversible inhibitor that looks similar to the normal substrate and competes for the active site of the enzyme 3.The process by which the binding of the substrate to the enzyme triggers a favorable conformation change which causes similar change in all of the proteins subunits. 4.The process by which a metabolic pathway is shut off by the product it produces 5.Binds to the enzyme at a site other than the active site, causing the enzyme to change shape and be unable to bind substrate.

17 Try these questions 7. The above graph most accurately depicts the energy changes that take place in which of the following types of reaction? A. hypothermic B. hyperthermic C. endergonic D. exergonic E. free range

18 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

19 Mitosis & Meiosis


21 Growing on the plates n Two Hyphae fuse (plasmogamy) Sordaria (Ascomycete fungus) One genetic variability is the color of the ascospores. Wild (dark brown) Tan How the ascospores are arranged in the ascus represents if crossing over has occurred. Karyogamy Spores Meiosis – where crossing over may occur No crossing over: 4:4 Crossing over: 2:4:2 or 2:2:2:2

22 You will look at slides of the plates and calculate the percentage of crossover. This can be used to determine map units.

23 Mitosis & Meiosis Conclusions – Mitosis Cell division – Growth, repair – Making clones – Meiosis Reduction division – Making gametes – Increasing variation Crossing over in Prophase 1

24 Mitosis & Meiosis Essay 1982 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 204 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.

25 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+ DPIP ox = blue (NADP+) DPIP red = clear (NADPH) – Measure light transmittance – Paper chromatography to separate plant pigments

26 Photosynthesis Concepts – Light reaction – Experimental design Control vs. Experimental – Chlorophyll and others Chlorophyll a Chlorophyll b Xanthophylls Carotenoids

27 Photosynthesis Conclusions – Pigments Pigments move at different rates based on solubility and mass in the solvent – Photosynthesis Light and unboiled chloroplasts produced highest rate of photosynthesis Which is the control?

28 Photosynthesis Essay 2004 (part 1) A 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 DPDP, 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 + DPIP Sample 2 – chloroplast suspension surrounded by foil wrap to provide a dark environment + DPIP Sample 3 – chloroplast suspension that has been boiled + DPIP Data are given in the table on the next slide 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.

29 Photosynthesis )Time (min)Light, Unboided % Transmittance Sample 1 Dark, Unboiled % Transmittance Sample 2 Light, Boiled %Transmittance Sample 3 028.829.228.8 548.730.129.2 1057.831.229.4 1562.532.428.7 2066.731.828.5

30 Cellular Respiration

31 Description – Using a CO 2 probe to measure the rate of respiration in crickets at different temperatures Concepts – Respiration – Endothermic vs. Exothermic – Experimental design Conclusions – As temp. decreases, respiration decreases – As germination (of peas) increases, respiration increases

32 Cellular Respiration Germination pea experiment – Set up – Purpose of KOH

33 Cellular Respiration Essay 1990 The results below are measuremnts of culumlative oxygen consumption by germination and dry seeds. Gas volume measurements were corrected for changes in temperature and pressure a.Plot the results for the germinating seeds at 22⁰C and 10⁰C. b.Calculate the rate of oxygen consumption for the germinating seeds at 22⁰C, using the time interval between 10 and 20 minutes. c.Account for the differences in oxygen consumption observed between: a.Germinating seeds at 22⁰C and at 10⁰C b.Geminating 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 (min)010203040 Germinating seeds 22⁰C0.08.816.023.732.0 Dry seeds (non germinating) 22⁰C0. Germinating Seeds 10⁰C0. Dry seeds (non germinating) 10⁰C0.0

34 Molecular Biology

35 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

36 Molecular Biology Concepts – Transformation – Plasmid – Selectable marker Ampicillin resistance – Restriction Enzyme – Gel Electrophoresis DNA is negatively charged Smaller fragments travel faster Correlate distance traveled with size of fragment

37 Molecular Biology Conclusions – Can insert foreign DNA using a vector – Ampicillin becomes selecting agent No transformation = no growth on amp+ plate

38 Molecular Biology


40 Try these questions







47 Genetics Fly Lab Description – Given a fly of unknown genotype use crosses to determine the mode of inheritance of a trait.

48 Genetics Fly Lab Concepts – Phenotype vs. genotype – Dominant vs. recessive – P, F1, F2 generations – Sex-linked – Monohybrid cross – Dihybrid cross – Test cross – Chi-square

49 Genetics Fly Lab Conclusions: Can you solve these? Case 1 Case 2

50 Genetics Fly Lab


52 Population Genetics Size of population & gene poolRandom vs. Non-random mating

53 Population Genetics Description – Simulations were used to study effects of different parameters on frequency of alleles in a population Selection Heterozygous advantage Genetic drift

54 Population Genetics Concepts – Hardy Weinberg equilibrium P + q = 1 P 2 + 2pq + q 2 = 1 Required conditions – Large population – Random mating – No mutations – No natural selection – No migration – Gene pool – Heterozygous advantage – Genetic drift Founder effect Bottleneck

55 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! Allele frequencies…solve for p and q Genotypic frequencies…use p 2, 2pq, or q 2

56 Populations Genetics Essay 1989 Do the following with reference to the Hardy-Weinberg model. a.Indicate the conditions under which allele frequencies 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?

57 Transpiration

58 Description – Test the effects of environmental factors on rate of transpiration Temperature Humidity Air flow (wind) Light intensity OR plants wrapped in plastic wrap, put in the above varying conditions and taking mass over time. Rate at which plant loses mass is equivalent to rate of water loss due to transpiration.

59 Transpiration Concepts – Transpiation – Stomates – Guard cells – Xylem Adhesion Cohesion – H bonding

60 Transpiration Conclusions – Transpiration – Increase in wind – Increase in light – Transpiration – Increase in humidity

61 Transpiration

62 Circulatory Physiology

63 Description – Study factors that affect heart rate Body position Level of activity – Determine whether an organism in an endotherm or an ectotherm by measuring change in pulse rate as temperature changes Daphnia Goldfish

64 Circulatory Physiology Concepts – Themoregulation – Endotherm – Ectotherm – Q 10 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

65 Circulatory Physiology Conclusions – Activity increases heart rate In a fit individual pulse and blood pressure are lower and 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

66 Circulatory Physiology

67 Animal Behavior Fruit Fly Behavior Sow bugs

68 Animal Behavior Description – Set up an experiment to study behavior in an organism Betta fish agonistic behavior Drosophila mating behavior Pillbug/Sowbug kinesis

69 Animal Behavior Concepts – Innate vs. learned behavior – Experimental design Control vs. experimental Hypothesis Number of trials/subjects – Choice chamber Temperature Humidity Salinity Other factors

70 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 place on both sides of a wet/dry choice chamber and allowed to move about freely for 10 minutes, then most will be found on the wet side.

71 Animal Behavior Experimental design

72 Animal Behavior


74 Dissolved Oxygen Dissolved O 2 availability

75 Dissolved Oxygen

76 Description – Measure primary productivity by measuring O 2 production – Factors that affect amount of dissolved O 2 Temperature – As water temp. increases it’s ability to hold O 2 decreases Photosynthetic activity – In bright light, aquatic plants produce more O 2 Decomposition activity – As organic matter decays, microbial respiration consumes O 2 Mixing and Turbulence – Wave action, waterfalls and rapids aerate water and O 2 increases Salinity – As water becomes more salty, its ability to hold O 2 decreases

77 Dissolved Oxygen Concepts – Dissolved Oxygen – Primary productivity Measured in 3 ways – Amount of CO 2 used – Rate of sugar (biomass) formation – Rate of O 2 production – Net Productivity vs. Gross productivity – Respiration

78 Dissolved Oxygen Conclusions

79 Dissolved Oxygen

80 Essay 2008


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