Lab Review Based on the PPT from Ms. Foglia

Lab Review Based on the PPT from Ms. Foglia
AP Biology Lab Review Based on the PPT from Ms. Foglia

Big Idea 1: Evolution

Lab 1: Artificial Selection
Concepts: Natural selection = differential reproduction in a population Populations change over time  evolution Natural Selection vs. Artificial Selection

Description: Use Wisconsin Fast Plants to perform artificial selection Identify traits and variations in traits Cross-pollinate (top 10%) for selected trait Collect data for 2 generations (P and F1)

Sample Histogram of a Population

Analysis & Results: Calculate mean, median, standard deviation, range Are the 2 populations before and after selection (P and F1) actually different? Are the 2 sub-populations of F1 (hairy vs. non-hairy) different? Are the means statistically different? A T-test could be used to determine if 2 sets of data are statistically different from each other; these are often used in examples where the sample size is small

Lab 2: 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 (p2+2pq+q2 = 1) Conditions: large population random mating no mutations no natural selection no migration

Analysis & Results: Null model: in the absence of random events that affect populations, allele frequencies (p,q) should be the same from generation to generation (H-W equilibrium) Analyze genetic drift and the effect of selection on a given population Manipulate parameters in model: Population size, selection (fitness), mutation, migration, genetic drift

Real-life applications: Cystic fibrosis, polydactyly Heterozygote advantage (Sickle-Cell Anemia)

ESSAY 1989 Do 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 (brownish). (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?

a. Indicate the conditions under which allele frequencies (p and q) remain constant from one generation to the next. large population random mating no mutations no natural selection no migration 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 (brownish). (In rabbits the white color is due to a recessive allele, w, and agouti is due to a dominant allele, W.) 25,000/ 100,000 = so .25 = q2 (a) q = .5 and (A) p = .5 Homozygous Rec = .25 = 25,000 Homozygous Dom (agouti) = p2 = .25 = 25,000 Heterozygous (agouti) = 2(p)(q) = 2(.5)(.5) = .5 = 50,000 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? The frequency of the ‘A’ allele would decrease and the frequency of the ‘a’ allele would increase.

Lab 3: Comparing DNA Sequences using BLAST  Evolutionary Relationships
Concepts: Bioinformatics: combines statistics, math modeling, computer science to analyze biological data Genomes can be compared to detect genetic similarities and differences BLAST = Basic Local Alignment Search Tool Input gene sequence of interest Search genomic libraries for identical or similar sequences

Description: Use BLAST to compare several genes
Lab 3: Comparing DNA Sequences using BLAST  Evolutionary Relationships Description: Use BLAST to compare several genes Use information to construct a cladogram (phylogenetic tree) Cladogram = visualization of evolutionary relatedness of species

Use this data to construct a cladogram of the major plant groups
Lab 3: Comparing DNA Sequences using BLAST  Evolutionary Relationships Use this data to construct a cladogram of the major plant groups

Fossil specimen in China DNA was extracted from preserved tissue Sequences from 4 genes were analyzed using BLAST

Analysis & Results: BLAST results: the higher the score, the closer the alignment The more similar the genes, the more recent their common ancestor  located closer on the cladogram

Answer is C

Big Idea 2: cellular processes: energy and communication

Lab 4: Diffusion & Osmosis
Concepts: Selectively permeable membrane Diffusion (high  low concentration) Osmosis (aquaporins) Water potential ()  = pressure potential (P) + solute potential (S) Solutions: Hypertonic hypotonic isotonic

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)

Potato Cores in Different Concentrations of Sucrose

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

Answer is A

Answer is B

ESSAY 1992 A laboratory assistant prepared solutions of 0.8 M, 0.6 M, 0.4 M, and 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 description of how you would set up and perform the experiment; the results you would expect from your experiment; and an explanation of those results based on the principles involved. Be sure to clearly state the principles addressed in your discussion.

A laboratory assistant prepared solutions of 0. 8 M, 0. 6 M, 0
A laboratory assistant prepared solutions of 0.8 M, 0.6 M, 0.4 M, and 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 description of how you would set up and perform the experiment; the results you would expect from your experiment; and an explanation of those results based on the principles involved. Be sure to clearly state the principles addressed in your discussion. Think about our lab that we did…. Talk about putting a cell (potato) in each of the solutions – mass before and then mass after and see in which solution did it gain/lose the most weight. The principle involved is simple osmosis. The potato will lose weight in hypertonic solutions and gain weight in the hypotonic solutions. Based on how much weight is gained/lost in each of the solutions, you would be able to put them in order. 0.8 M – potato will lose the most mass 0.6 M – potato will lose the second most mass 0.4 M – potato will remain mostly the same with losing just a bit 0.2 M – potato will gain mass *potatos are about 0.3 molarity

Lab 5: Photosynthesis Concepts: 6H2O + 6CO2 + Light  C6H12O6 + 6O2
Ways to measure the rate of photosynthesis: Production of oxygen (O2) Consumption of carbon dioxide (CO2)

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  O2 produced  bubbles form on leaf  leaf disk rises Measure rate of photosynthesis by O2 production Factors tested: types of plants, light intensity, colors of leaves, pH of solutions

Plant Pigments & Chromatography

Floating Disk Technique

To make comparisons between experiments, a standard point of reference is needed. Repeated testing of this procedure has shown that the point at which 50% of the disks are floating (the median or ET50) is a reliable and repeatable point of reference. In this case, the disks floating are counted at the end of each time interval. The median is chosen over the mean as the summary statistic. The median will generally provide a better estimate of the central tendency of the data because, on occasion, a disk fails to rise or takes a very long time to do so. A term coined by G. L Steucek and R. J Hill (1985) for this relationship is ET50, the estimated time for 50% of the disks to rise. That is, rate is a change in a variable over time. The time required for 50% of the leaf disks to float is represented as Effective Time = ET50.

Lab 5: Photosynthesis Concepts: photosynthesis Photosystems II, I
H2O split, ATP, NADPH chlorophylls & other plant pigments chlorophyll a chlorophyll b xanthophylls carotenoids experimental design control vs. experimental

Answer is A

Answer is D

Lab 6: Cellular Respiration
Concepts: Respiration Measure rate of respiration by: O2 consumption CO2 production

Description: Use respirometer Measure rate of respiration (O2 consumption) in various seeds Factors tested: Non-germinating seeds Germinating seeds Effect of temperature Surface area of seeds Types of seeds Plants vs. animals

Conclusions: temp = respiration germination = respiration Animal respiration > plant respiration  surface area =  respiration Calculate Rate

Answer is A

Answer is C

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) 10 20 30 40 Germinating seeds 22°C 0.0 8.8 16.0 23.7 32.0 Dry Seeds (non-germinating) 22°C 0.2 0.1 Germinating Seeds 10°C 2.9 6.2 9.4 12.5 Dry Seeds (non-germinating) 10°C

a. Plot the results for the germinating seeds at 22°C and 10°C. Cumulative Oxygen Consumed (mL) Time (minutes) 10 20 30 40 Germinating seeds 22°C 0.0 8.8 16.0 23.7 32.0 Dry Seeds (non-germinating) 22°C 0.2 0.1 Germinating Seeds 10°C 2.9 6.2 9.4 12.5 Dry Seeds (non-germinating) 10°C I would say mL of O2 consumed instead of “corrected difference” and this graph shows all 4 lines when we only needed 2 (this was all I could find online) **make sure graph has a title, axis are labeled, lines labeled

Cumulative Oxygen Consumed (mL) Time (minutes) 10 20 30 40 Germinating seeds 22°C 0.0 8.8 16.0 23.7 32.0 Dry Seeds (non-germinating) 22°C 0.2 0.1 Germinating Seeds 10°C 2.9 6.2 9.4 12.5 Dry Seeds (non-germinating) 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. Rate = ∆Y/ ∆X (16.0 – 8.8) = = .72 mL/ min (20-10) 10 **make sure to show work and make sure to include units!

c. Account for the differences in oxygen consumption observed between: 1. germinating seeds at 22°C and at 10°C  increased temp causes increased respiration/ metabolism so they use more O2 at a higher temp 2. germinating seeds and dry seeds  germinating seeds are active and thus will use more O2, dry seeds are dormant and will respire very little 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. Needs to be a closed system that separates oxygen consumption and carbon dioxide release (like KOH that turns CO2 into a solid). Must be able to control temp (water bath), control volume (glass beads), have a timer, equal numbers of organisms in each setup, etc. Cumulative Oxygen Consumed (mL) Time (minutes) 10 20 30 40 Germinating seeds 22°C 0.0 8.8 16.0 23.7 32.0 Dry Seeds (non-germinating) 22°C 0.2 0.1 Germinating Seeds 10°C 2.9 6.2 9.4 12.5 Dry Seeds (non-germinating) 10°C

Big Idea 3: genetics and information transfer

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

Lab 7: Mitosis & Meiosis

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)

Lab 7: Mitosis & Meiosis

Abnormal karyotype = Cancer

Meiosis: Crossing over in Prophase I

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

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

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.

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. Simply explain cell cycle (G1, S, G2, Mitosis, cytokinesis)  see notes on topic ESSAY 2004 Meiosis reduces chromosome number and rearranges genetic information. a. Explain how the reduction and rearrangement are accomplished in meiosis. Chromosomes are paired in tetrads first, they get split in half so the cells after the first division of meiosis have ½ the number of chromosomes as the parent cell. Rearrangement is due to independent assortment, where maternal and paternal chromosomes line up randomly in the tetrads, and crossing over which occurs in Prophase I. Crossing over is when a piece of one chromosomes breaks off and switches places with a piece on the homologous chromosome. 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.

Lab 7: Mitosis & 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. Down Syndrome is trisomy 21 (an extra copy of chromosome #21). This is characterized by low muscle tone, small stature, an upward slant to the eyes, a single deep crease across the center of the palm, and some mental deficiencies. Down Syndrome is caused by nondisjunction – chromosomes not separating correctly during meiosis. This can happen in either meiosis I or meiosis II. 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. Parthenogenesis is an example of asexual reproduction where growth and development of embryos occur without fertilization. In animals, parthenogenesis means development of an embryo from an unfertilized egg cell. This would produce clones and all offspring would be genetically identical to the parent cell.

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

Conclusions: Foreign DNA inserted using vector (plasmid) Ampicillin = Selecting agent No transformation = no growth on amp+ plate Regulate genes by transcription factors (araC protein)

Plates that have only ampicillin-resistant bacteria growing include which of the following?
I only III only IV only I and II

Plates that have only ampicillin-resistant bacteria growing include which of the following?
I only III only IV only I and II Answer is C

Answer is A

Answer is B

Answer is C

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

Description

Determine DNA fragment sizes

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 (remember we made this from the marker when we KNEW how many bases were in each band)

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.

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. Gel electrophoresis separates segments of DNA based on their size. The DNA is put into the gel at the negative end and it runs to the positive. The electric current that moves through the buffer and the gel allow the negatively charged DNA to be attracted to the positive end of the chamber. The smaller pieces move through the gel faster than the larger pieces.

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). 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. DNA digested with only enzyme X  4 pieces; sizes in base pairs are: 400, 1700, 1300, 1500 DNA digested with only enzyme Y  2 pieces; sizes in base pairs are: 900, 4000 DNA digested with enzyme X and enzyme Y combined  5 bands; sizes in base pairs are: 400, 500, 1200, 1300, 1500 Undigested DNA  one huge band

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). c. Explain both of the following: The mechanism of action of restriction enzymes  Restriction enzymes identify specific splice sites in DNA and cut there. Restriction sites are palindromes which read the same forward as backward on the opposite strands of the DNA molecule. Restriction enzymes can either create blunt ends or sticky ends. The different results you would expect if a mutation occurred at the recognition site for enzyme Y. I would expect the Y enzyme not to cut the piece of DNA, so if you digested the whole piece of DNA with both the X and Y enzymes, it would only be cut at the X restriction sites.

Big Idea 4: interactions

Lab 10: Energy Dynamics Concepts:
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

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

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

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

Lab 10: Energy Dynamics

Lab 10: Energy Dynamics Conclusions:

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

Answer: 125 – (60 + 5) = 60 g/m2

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

Lab 11: Transpiration

Lab 11: Transpiration Description: Nail polish  stomatal peels
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

Analysis of Stomata

Rates of Transpiration

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

Answer:

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

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

Lab 12: Animal Behavior

Lab 12: Animal Behavior Hypothesis Development
Keep this in mind if you have to write a hypothesis on the AP exam… what makes a good hypothesis? 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.

Lab 12: Animal Behavior Experimental Design sample size

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)

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

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.

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

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

Turnip peroxidase  Color change (O2 produced)

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

Answer is B

Lab Review Based on the PPT from Ms. Foglia

Similar presentations

Presentation on theme: "Lab Review Based on the PPT from Ms. Foglia"— Presentation transcript:

Similar presentations

About project

Feedback

Log in

Auth with social network:

Lab Review Based on the PPT from Ms. Foglia

Similar presentations

Presentation on theme: "Lab Review Based on the PPT from Ms. Foglia"— Presentation transcript:

Similar presentations

About project

Feedback