2 Science Practices- The student can: -use representations & models to communicate scientific phenomenon & solve scientific problems.-use mathematics appropriately.-engage in scientific questioning to extend thinking or to guide investigations within the context of the AP course.-plan & implement data collection strategies appropriateto a scientific question.5. -perform data analysis & evaluation of evidence.6. -work with scientific explanations & theories.7. -is able to connect & relate knowledge across various scales, concepts & representations in and across domains.
3 The Process of Evolution Drives the Diversity and Unity of Life Big Idea 1The Process of Evolution Drives the Diversity and Unity of LifeChange in the genetic makeup of a population over time is evolution.Organisms are linked by lines of descent from common ancestry.Life continues to evolve within a changing environment.The origin of living systems is explained by natural processes.Hardy-Weinberg EquationsProbabilities
4 Water Potential Gibb’s Free Energy Big Idea 2 Biological Systems Utilize Free Energy and Molecular Building Blocks to Grow, Reproduce and Maintain Dynamic HomeostasisGrowth, reproduction & maintenance of the organization of living systems require free energy & matter.Growth, reproduction & dynamic homeostasis require that cells create & maintain internal environments that are different from their external environments.Organisms use feedback mechanisms to regulate growth & reproduction & to maintain dynamic homeostasis.Growth & dynamic homeostasis f a biological system are influenced by changes in the system’s environment..Many biological processes involved in growth, reproduction & dynamic homeostasis include temporal regulation & coordination.Water PotentialGibb’s Free Energy
5 Chi square Gene Linkage Big Idea 3 Living Systems Store, Retrieve, Transmit and Respond to Information Essential to Life ProcessesHeritable information provides for continuity of life.Expression of genetic information involves cellular & molecular mechanisms.The processing of genetic information is imperfect & is a source of genetic variation.Cells communicate by generating, transmitting& receiving chemical signals.Transmission of information results in changeswithin and between biological systems.Chi squareGene Linkage
6 Population Growth Energy Transfer Primary Productivity Big Idea 4 Biological Systems Interact and These Systems and Their Interactions Possess Complex PropertiesInteractions within biological systems lead to complex properties.Competition & cooperation are important aspects of biological systems.Naturally occurring diversity among & between components within biological systems affect interactions within the systems.Population GrowthEnergy TransferPrimary Productivity
7 100; this is a 3:1 phenotypic ratio In a typical Mendelian monohybrid cross, two heterozygotes produce 400 offspring. How many individuals are expected to have the recessive phenotype?100; this is a 3:1 phenotypic ratio
8 In this genetic cross, Aa x aa, there are 348 offspring In this genetic cross, Aa x aa, there are 348 offspring. How many individuals are expected to have the dominant phenotype?174; this is a 1:1 ratio, so 50% are expected toHave the dominant phenotype
9 This is a 9:3:3:1 ratio, with 9/16 showing both dominant phenotypes. In a dihybrid cross between two heterozygotes, if you have 200 offspring, how many should show both dominant phenotypes?112This is a 9:3:3:1 ratio, with 9/16 showing both dominant phenotypes.9/16 = .56 = 56%56% of 200 is 112
10 The allele for the hair pattern called “widow’s peak” is dominant over the allele for no “widow’s peak”. In a population of 100 individuals, 64 show the dominant phenotype. What is the frequency of the recessive allele?.664 show the dominant phenotype,So 36 show the recessive phenotype.Since this is a population of 100, 36%Show the recessive phenotype.36 = q2.6 = q
11 In a certain population of deer on Fire Island, NY, the allele for a black spot behind the eye is dominant to the allele for no spot. After the hunting season, the percent of deer with no black spot is 15% and the population is in Hardy-Weinberg Equilibrium. What is the frequency for the allele for having no black spot, to the hundredths?15% = .15 = q2q = .39
12 The ability to taste PTC is due to a single dominant allele (A) The ability to taste PTC is due to a single dominant allele (A). You sampled 215 individuals in biology and determined that 150 could taste PTC and 65 could not. How many individuals in this population show the following genotype? AA, Aa, aa65/216 = .3 = q2q = .55p = .45(.45)(.45) = .2 = 20%20% x 215 = 43 AA2(.45)(.55) = .495 = 50%50% x 215 = 107 Aa
13 is 0.2, and the population is in Hardy-Weinberg equilibrium? In geckos, spots are dominant to the solid color. If the frequency In a population of 700 geckos, what percentage of the geckos would have spots, if the frequency of the recessive alleleis 0.2, and the population is in Hardy-Weinberg equilibrium?96%q= .2p= .8Homozygous dominant (p2) = .64 (64%)Heterozygotes (2pq) = .32 (32%)
14 Number of Cells Observed Number of Cells Expected A Cellular Biologist wants to double check that statement that cells spend 90 percent of their time in Interphase as compared to the various stages of Mitosis. She grows some Allium in her laboratory. She then takes one of the plants, cuts off the root tips, stains the DNA in the cells so as to be able to see the stages of the cell cycle. Her hypothesis states “If cells spend 90 percent of their time in Interphase, then she should be able to calculate the relative time existing between Interphase and Mitosis based upon the cells counted in her specimen.” She counted 1000 cells from her preserved specimen under the microscope. Her data are shown below. Calculate the X2 to the nearest hundredth.Stage of the Cell CycleNumber of Cells ObservedNumber of Cells ExpectedInterphase872900Mitosis128100
16 Forty flies were put into a choice chamber with two chambers Forty flies were put into a choice chamber with two chambers. In one chamber there was a cotton ball soak with vinegar. The other chamber had nothing. After 20 minutes the number of flies were counted in both chambers. This was repeated four more times. Perform a chi-square analysis to determine if the difference between in the number of flies found in the two chambers is significant.
20 The formula is easy: it is the square root of the Variance The formula is easy: it is the square root of the Variance. So now you ask, "What is the Variance?“The Variance is defined as: The average of the squared differences from the Mean.To calculate the variance follow these steps:Work out the Mean (the simple average of the numbers)Then for each number: subtract the Mean and square the result(the squared difference).3. Then work out the average of those squared differences.
24 To measure the population density of monarch butterflies occupying a particular park, 100 butterflies are captured, marked with a small dot on a wing and then released. The next day, another 100 butterfliesare captured, including the recapture of 20 marked butterflies. Onewould estimate the population to be:a. 200b. 500c. 1,000d. 10,000(100 x 100)/ 20 = 500
25 Use the Station 1 data to calculate the Primary Productivity of a water sample. Report your answer in units of mg Carbon fixed/LiterStation 14.2 mg O2/L = 2.9 mL O2/L2.9 mL O2/L 0.536= 1.6 mg Carbon fixed/L
26 1st Law of Thermodynamics- energy cannot be created or destroyed, but it can change form. 18,000 energy accumulated as biomass; 12,000 going to the tree layer; 4,400 going to the shrub layer; 1,600 left, to go to the grass layer. 1,600 is 9% of 18,000 (1,600/18,000 x 100)
28 (the average partial pressure of oxygen at sea level is 21%) Atmospheric pressure is the combined partial pressures of all of the gases that make up the atmosphere. At the summit of a high mountain, the atmospheric pressure is 380mm/Hg. The partial pressure of oxygen is 69mm/Hg. What percentageOf the atmosphere is made up of oxygen at this altitude?18%69/380 = .18 = 18%(the average partial pressure of oxygen at sea level is 21%)
30 Sometimes we concentrate on the complicated math formulas, and we have students whocan’t calculatepercentages.PhaseNumberPercent spent in each phaseInterphaseProphase/prometaphaseMetaphaseAnaphaseTelophase
31 The purpose of a particular investigation was to see the effects of varying salt concentrations of nutrient agar and its effect on colony formation. Below are the results Determine the mean for each treatment and graph the results.TrialNo Treatment1% salt3% salt5% salt7% salt9% salt1474125282452464232232163341845744173927
33 What is the mean rate of growth per day between day 5 and day 25 What is the mean rate of growth per day between day 5 and day 25? Record your answer to the nearest hundredth of a cm.If this same rate of growth continues, how tall will the plant be on day 50? Record your answer to the nearest hundredthof a cm.18-3 = 1515/20 days = .75 cm.75 x 25 = 18.75= 36.75
34 What is the water potential of a cell with a solute potential of -0 What is the water potential of a cell with a solute potential of MPa and a pressure potential of 0.43 MPa?-.24MPa
35 You measure the total water potential of a cell and find it to be -0 You measure the total water potential of a cell and find it to be MPa. If the pressure potential of the same cell is 0.46 MPa, what is the solute potential of that cell?Since water potential is equal to the solute potential + the pressure potential,-0.24 MPa = 0.46 MPa + X. Solve for x= -0.7
36 A population of ground squirrels has an annual per capita birth rate of 0.06 and an annual per capital death rate of Estimate thenumber of individuals added to (or lost from) a population of1,000 individuals in one year.120 individuals added40 individuals added20 individuals added20 individuals lostdN/dt = B-DChange in population size/time = Birth rate – Death rate0.06 – 0.02 = 0.04 x 1000 = 40 individuals added per year
37 A small population of white-footed mice has the same intrinsic rate of increase (r) as a large population. If everything else is equal,a. the large population will add more individuals per unit timeb. the small population will add more individuals per unit timec. the two populations will add equal number of individuals per unit timed. the J-shaped growth curves will look identical
38 In 2005, the United States had a population of approximately 295,000,000 people. If the birth rate was 13 births for every1,000 people, approximately how many births occurred in theUnited States in 2005?a. 3, 800b. 38,000c. 380,000d. 3,800,000295,000,000/1,000 = 295,000295,000 x 13 = 3,835,000
39 As N approaches K for a certain population, which of the following is predicted by the logistic equation?a. the growth rate will not changeb. the growth rate will approach zeroc. the population will increase exponentiallyd. the carrying capacity of the environment will increase
40 Which of the following might be expected in the logistic model of population growth?a. as N approaches K, b increasesb. as N approaches K, r increasesc. as N approaches K, d increasesd. both a and b are true
41 According to the logistic growth equation: a. the number of individuals added per unit time is greatest when N is close to zerob. the per capita growth rate (r) increases as N approaches Kc. population growth is zero when N equals Kd. the population grows exponentially when K is small
42 An experiment determined that when a protein unfolds to its denatured (D) state from the original folded (F) state, the change in Enthalpy is ΔH = H(D) – H(F) = 46,000 joules/mol. Also the change in Entropy is ΔS = S(D) – S(F) = 178 joules/mol. At a temperature of 20⁰C, calculate the change in Free Energy ΔG, in j/mol, when the protein unfolds from its folded state.The correct answer is joules/mol.ΔG = ΔH – TΔSΔG = (46,000 joules/mol) – 293 K (178 joules /mol)ΔG = 46,000 joules/mol – 52,154 joules/mol = joules/mol
43 14,000 (a) (c)(b) (d)You are starting with 87,400 kJ and simply subtracting to get the answers.
44 Geneticists working in an agriculture lab wanted to develop a crop that combines the disease resistance of rye grain with the high crop yielding capacity of wheat grain. Rye grain has a diploid chromosome number (2n) of 14 and wheat grain has a diploid chromosome number of 42. The resulting grain is called triticale and is an alloploidy plant. How many chromosomes are found in the pollen grain of triticale?Alloploidy results when two different plant species combine their diploid genome to make new and unique species. That would mean that this particular species would have 56 chromosomes. The cells in a pollen grain of would be haploid so the resulting number is 28.
45 A study was conducted on the island of Daphne Major in the Galapagos Islands by Peter and Rosemary Grant. This study lasted over 20 year s. The study investigated how the type of seeds available to the finches impacted the depth of their beaks. In years when rain and water were plentiful, the available seeds were smaller and easy to crack. In years experiencing drought, fewer seeds were produced, and the finches had to eat the larger, leftover seeds produced from previous years. During years of drought, birds with a greater beak depth had a selective advantage.Use the data above to determine the increase in the mean of the depth of the beak between the wet and dry years. Give your answer to the nearest hundredth of a millimeter.
46 How many unique gametes could be produced through independent assortment by an individual with the genotype AaBbCCDdEE?a. 4b. 8c. 16d. 32
47 Given the parents AABBCc x AabbCc, assume simple dominance and independent assortment. What proportion of the progeny will be expected to phenotypically resemble the first parent?a. 1/4b. 1/8c. 3/4d. 3/8
48 Consider a field plot containing 200 kg of plant material Consider a field plot containing 200 kg of plant material. Approximately how many kg of carnivore production can be supported?a. 200b. 100c. 20d. 2
49 Cytosine makes up 38% of the nucleotides in a sample of DNA from an organism. Approximately, what percentage of the nucleotides in this sample will be thymine?a. 12b. 24c. 31d. 38