AP Biology Assessment 2

Transcription

1 AP Biology Assessment 2 DATE OF ADMINISTRATION: JANUARY 8 12, 2017 TOPICS COVERED: ALL TOPICS THROUGH BASIC GENETICS In the second assessment, the focus will be on all topics through basic Mendelian Genetics. There will be 25 multiple choice questions and 3 grid-in responses (45 minutes) which will account for 50% of the exam score. There will also be 4 free-response questions (45 minutes) accounting for the remaining 50%. Multiple Choice Key and crosswalk Question Key Description Q1 D Big Idea 2, Enduring Understanding 2.A, Essential Knowledge 2.A.3 Q2 C Big Idea 4, EU 4.A, EK 4.A.1 a1,2 SP 7.1 LO 4.1 Q3 A Big Idea 2, EU 2.A, EK 2.A.2g Q4 D Big Idea 4, EU 4.A, EK 4.A.1 a Q5 D Big Idea 4, EU 4.B, EK 4.B.1d Q6 A Big Idea 4, EU 4.B, EK 4.B.1d Q7 B Big Idea 3, EU 3.A, EK 3.A.2b Q8 C Big Idea 4, EU 4.A, EK 4.A.1 Q9 D Big Idea 4, EU 4.A, EK 4.A.2d Q10 A Big Idea 1, EU 1.B, EK 1.B1 Q11 A Big Idea 3, EU 3.A, EK 3.A.2a Q12 D Big Idea 2, EU 2.A, EK 2.A.2f Q13 D Big Idea 2, EU 2.A, EK 2.B.1ab Q14 C Big Idea 2, EU 2.B, EK 2.B.1ab Q15 C Big Idea 1, EU 1.B, EK 1.B.1 Q16 B Big Idea 2, EU 2.A, EK 2.A.2c Q17 D Big Idea 2, EU 2.A, EK 2.A.2c Copyright 2017 National Math + Science Initiative, Dallas, Texas. All rights reserved. Visit us online at 1

2 Question Key Description Q18 A Big Idea 2, EU 2.A, EK 2.A.2c Q19 C Big Idea 3, EU 3.A, EK 3.A.2a2 Q20 A Big Idea 3, EU 3.A, EK 3.A.2c Q21 A Big Idea 1, EU 1.A, EK 1.B.1 Q22 A Big Idea 2, EU 2.A, EK 2.A.3 Q23 C Big Idea 3, EU 3.A, EK 3.A.2c Q24 B Big Idea 2, EU 2.A, EK 2.A.1 Q25 A Big Idea 2, EU 2.A, EK 2.A.1d3 Q26 See Rubric Big Idea 2, EU 2.A, EK 2.A.3b Q27 See Rubric Big Idea 2, EU 2.E, EK 2.E.1b3 Q28 See Rubric Big Idea 3, EU 3.A, EK 3.A.3a Q29 See Rubric Big Idea 2, EU 2.A, EK 2.A.2dg Q30 See Rubric Big Idea 1, EU 1.A, EK 1.A.1, EK 1.A.2 Q31 See Rubric Big Idea 4, EU 4.B, EK 4.B.1d Q32 See Rubric Big Idea 3, EK 3.A.3, SP 2.2, LO 3.14 Copyright 2016 National Math + Science Initiative, Dallas, Texas. All rights reserved. Visit us online at 2

3 GRID-IN ANSWERS (26-28) Question 26 A hypothetical organism is composed of 27 cells, and each cell is in the approximate shape of a cube. The side of each cube is 1 unit long. Determine the surface area-tovolume ratio of the cells contained in this hypothetical organism. When gridding the answer, use a fraction format. For example, a surface area-to-volume ratio may be 5:1 but the fraction format is 5/1 Each cube (cell) has six faces, thus each has a surface area of 6 and a volume of 1 the surface to area ratio is 6:1 or 6 1 Accepted answers: 6/1 Copyright 2017 National Math + Science Initiative, Dallas, Texas. All rights reserved. Visit us online at 3

4 Question 27 Respirometer readings from an experiment measuring oxygen consumption in germinating peas are shown in the table below. An initial reading was recorded at time zero followed by a final reading recorded after the experiment ran for 10 minutes. Calculate the rate of respiration per minute for the peas in Respirometer Number 2. Give your answer to the nearest one thousandth ml. rate = Vol O 2 final Vol O 2 initial final time initial time rate = 0.14 ml 0.01mL 10 min 0 min = ml min Accepted answers: Copyright 2016 National Math + Science Initiative, Dallas, Texas. All rights reserved. Visit us online at 4

5 Question 28 In sheep, eye color is controlled by a single gene with two alleles. When a homozygous brown-eyed sheep is crossed with a homozygous green-eyed sheep, blue-eyed offspring are produced. If the blue-eyed sheep are mated with each other, what percent of their offspring will most likely have brown eyes? Brown Eyes X Green Eyes BB X B B Blue Eyed Offspring = BB Blue Eyes X Blue Eyes BB X BB B B B B BB - Brown BB Blue BB Blue B B - Green 25% are Brown; 50% Blue; 25% Green Accepted answers: 25 1/4 Copyright 2017 National Math + Science Initiative, Dallas, Texas. All rights reserved. Visit us online at 5

6 FREE RESPONSE SCORING GUIDELINES (29-32) QUESTION POINTS Question 29 is a long free-response question. Questions 30 through 32 are short free-response questions. Read each question carefully and write your response. Answers must be written out. Outline form is not acceptable. It is important that you read each question completely before you begin to write. An absorption spectrum indicates the relative amount of light absorbed across a range of wavelengths. The graphs above represent the absorption spectra of individual pigments isolated from two different organisms. One of the pigments is chlorophyll a, commonly found in green plants. The other pigment is bacteriorhodopsin, commonly found in purple photosynthetic bacteria. The table above shows the approximate ranges of wavelengths of different colors in the visible light spectrum. Copyright 2016 National Math + Science Initiative, Dallas, Texas. All rights reserved. Visit us online at 6

7 a. Identify the pigment (chlorophyll a or bacteriorhodopsin) used to generate the absorption spectrum in each of the graphs above. Explain and justify your answer. (3 points maximum) 1 point for each Identify BOTH pigments: Graph 1 = bacteriorhodopsin AND graph 2 = chlorophyll a Explain that an organism containing bacteriorhodopsin appears purple because the pigment absorbs light in the green range of the light spectrum and/or reflects violet or red and blue light. The reflected red and blue light appears purple. Explain that an organism containing chlorophyll a appears green because the pigment absorbs light in the red and blue ranges of the light spectrum and/or reflects green light. b. In an experiment, identical organisms containing the pigment from Graph II as the predominant light- capturing pigment are separated into three groups. The organisms in each group are illuminated with light of a single wavelength (650 nm for the first group, 550 nm for the second group, and 430 nm for the third group). The three light sources are of equal intensity, and all organisms are illuminated for equal lengths of time. Predict the relative rate of photosynthesis in each of the three groups. Justify your predictions. (5 points maximum) Wavelength (Group) 650 nm (1 st Group) Prediction (1 point each) Intermediate rate Justification (1 point each) An intermediate level of absorption occurs at 650 nm (compared to 430 nm and 550 nm); therefore, an intermediate amount of energy is available to drive photosynthesis. 550 nm (2 nd Group) 430 nm (3 rd Group) Lowest rate Highest rate The lowest level of absorption occurs at 550 nm; therefore, the least amount of energy is available to drive photosynthesis. The highest level of absorption occurs at 430 nm; therefore, the greatest amount of energy is available to drive photosynthesis. NOTE: A student who combines two groups (e.g., the 650 nm and 430 nm groups have higher rates of photosynthesis compared to the 550 nm group ) can earn a maximum of 4 points: up to 2 points for the prediction and up to 2 points for the justification. Copyright 2017 National Math + Science Initiative, Dallas, Texas. All rights reserved. Visit us online at 7

8 c. Bacteriorhodopsin has been found in aquatic organisms whose ancestors existed before the ancestors of plants evolved in the same environment. Propose a possible evolutionary history of plants that could have resulted in a predominant photosynthetic system that uses only some of the colors of the visible light spectrum. (2 points maximum; 1 point for proposal including a selective pressure; 1 point for reasoning to support the proposal) Copyright 2016 National Math + Science Initiative, Dallas, Texas. All rights reserved. Visit us online at 8

9 QUESTION 30 4 POINTS Many aquatic animals can hold their breath and dive underwater for long periods of time. During long dives, lactic acid builds up in tissues of the organism. The graph above shows blood lactate levels following dives of different duration for three species. Lactate is produced when lactic acid is dissolved in the blood and becomes ionized. a. Based on the data, propose a hypothesis to explain the change in blood lactate levels in Weddell seals for dives lasting longer than 20 minutes. (1 point) Proposal: Increase in lactate levels are due to fermentation/anaerobic metabolism b. Describe the most likely shape of the curve if blood oxygen levels of Weddell seals were plotted rather than blood lactate levels. Include in your description the likely shape of the curve between 0 and 20 minutes and the shape of the curve after 20 minutes. (2 points maximum; points may be earned from only one row) 0-20 min (1 point) After 20 min (1 point) Oxygen levels start high and decline steadily Oxygen levels start and remain high Oxygen levels decline more slowly or remain flat Oxygen levels decline Copyright 2017 National Math + Science Initiative, Dallas, Texas. All rights reserved. Visit us online at 9

10 c. The data suggest that Baikal seals can sustain much longer dives than Emperor penguins. Propose a hypothesis that could explain the evolution of different dive responses in Emperor penguins and Baikal seals. (1 point) The (genetic/heritable) capacity to sustain dives for longer periods of time provides selective/reproductive advantages (access to food/avoidance of predators) for seals but not for penguins. Copyright 2016 National Math + Science Initiative, Dallas, Texas. All rights reserved. Visit us online at 10

11 QUESTION 31 3 POINTS. The graph above shows the initial rate of an enzyme-catalyzed reaction at different substrate concentrations in the presence of a constant concentration of enzyme. a. Connect the primary structure of the enzyme to its overall shape. (1 point) The amino acid sequence determines the overall shape (of the polypeptide/protein/enzyme) R-groups interact and stabilize the structure b. Predict the effect of adding a noncompetitive inhibitor to the reaction mixture on the rate of reaction at a high substrate concentration. Support your prediction by describing how a noncompetitive inhibitor affects the structure and function of an enzyme. (2 points maximum; 1 point for prediction, 1 point for support) Noncompetitive/allosteric inhibitor will decrease the (initial) rate of reaction (at any substrate concentration) Inhibitor binding to site other than active site (allosteric site) changes shape of enzyme, which alters the interaction of substrate with active site Copyright 2017 National Math + Science Initiative, Dallas, Texas. All rights reserved. Visit us online at 11

12 QUESTION 32 3 POINTS Fruit flies (Drosophila melanogaster) with a wild-type phenotype have gray bodies and red eyes. Certain mutations can cause changes to these traits. Mutant flies may have a black body and/or cinnabar eyes. To study the genetics of these traits, a researcher crossed a true-breeding wild-type male fly (with gray body and red eyes) with a true-breeding female fly with a black body and cinnabar eyes. All of the F1 progeny displayed a wild-type phenotype. Female flies from the F1 generation were crossed with true-breeding male flies with black bodies and cinnabar eyes. The table below represents the predicted outcome and the data obtained from the cross. Explain three (3) differences between the expected data and the actual numbers observed. F2 Generation Phenotypes Body Color Eye Color Number Predicted Number Observed Gray Red Black Cinnabar Gray Cinnabar Black Red (3 points maximum; 1 point for each difference, only the first 3 differences are accepted.) Student explanations could include the following: Prediction of a 1:1:1:1 ratio with correct phenotypes based on independent assortment. Support for prediction with a diagram of the cross of BbEe x bbee. Correct application of chi-square analysis to show that observed results do not conform to expected Mendelian frequencies. Identification of body color and eye color as linked genes/loci. Use of ratios to show linkage and independent assortment of wing type versus linked traits. Identification of the bottom two phenotypes as products of crossing over (recombinant chromosome). Mentioning that crossover rate is approximately 9 10 percent. Copyright 2016 National Math + Science Initiative, Dallas, Texas. All rights reserved. Visit us online at 12