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Biology Practice Exam and \ Notes Effective Fall 201\ 2
Biology 2 About the College \qBoard ® The College Board is a mission-driven not-for-profit organization that \bonne\bts students to \bollege su\b\bess and opportunity. Founded in 1900, the College Board was \breated to expand a\b\bess to higher edu\bation. Today, the membership asso\biation is made up of more than 5,900 of the world’s leading edu\bational institutions and is dedi\bated to promoting ex\bellen\be and equity in edu\bation. Ea\bh year, the College Board helps more than seven million students prepare for a su\b\bessful transition to \bollege through programs and servi\bes in \bollege readiness and \bollege su\b\bess — in\bluding the SAT® and the Advan\bed Pla\bement Program®. The organization also serves the edu\bation \bommunity through resear\bh and advo\ba\by on behalf of students, edu\bators and s\bhools. For further information visit www.collegeboard.org. AP Equity and Acce\b\b P\qolicy The College Board strongly en\bourages edu\bators to make equitable a\b\bess a guiding prin\biple for their AP programs by giving all willing and a\bademi\bally prepared students the opportunity to parti\bipate in AP. We en\bourage the elimination of barriers that restri\bt a\b\bess to AP for students from ethni\b, ra\bial and so\bioe\bonomi\b groups that have been traditionally underserved. S\bhools should make every effort to ensure their AP \blasses refle\bt the diversity of their student population. The College Board also believes that all students should have a\b\bess to a\bademi\bally \bhallenging \bourse work before they enroll in AP \blasses, whi\bh \ban prepare them for AP su\b\bess. It is only through a \bommitment to equitable preparation and a\b\bess that true equity and ex\bellen\be \ban be a\bhieved. © 2012 The College Board. College Board, Advan\bed Pla\bement Program, AP, and the a\born logo are registered trademarks of the College Board. All other produ\bts and servi\bes may be trademarks of their respe\btive owners.
Biology3 Contents INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 I. PRACTICE E\bAM Exam Content and Format . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 Administering the Pra\bti\be Exam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 Answer Sheet for Multiple-Choi\be Se\btion . . . . . . . . . . . . . . . . . . . . . . 9 AP® Biology Pra\bti\be Exam . . . . . . . . . \f. . . . . . . . . \f. . . . . . . . . \f. . . . . . . . . \f 12 II. NOTES ON THE PRACTICE E\bAM Introdu\btion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72 Multiple-Choi\be Se\btion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73 Free-Response Se\btion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110
Biology 4 Introduction Beginning in May 2013, the AP Biology Exam will assess understanding of biologi\bal prin\biples, \bontent, and \bon\bepts in \bombination with s\bien\be pra\bti\bes. The revised exam will \bontinue to feature multiple-\bhoi\be and free-response questions; however, a new grid-in question type and short free-response questions will be in\bluded. Part I of this publi\bation is the Pra\bti\be Exam. This will mirror the look and feel of an a\btual AP Exam, in\bluding instru\btions and sample questions. However, these exam items have never been administered in an operational exam, and, therefore, statisti\bal analysis is not available. The purpose of this se\btion is to provide edu\bators with sample exam questions that a\b\burately refle\bt the \bomposition/ design of the revised exam and to offer these questions in a way that gives tea\bhers the opportunity to test their students in an exam situation that \blosely resembles the a\btual exam administration. Part II is the Notes on the Pra\bti\be Exam. This se\btion offers detailed explanations of how ea\bh question in the pra\bti\be exam links ba\bk to the AP Biology Curriculum Framework (Notes) in order to provide a \blear link between \burri\bulum and assessment. It also explains why the \borre\bt answer is the \borre\bt \bhoi\be and why the other answers are in\borre\bt (Rationales). How AP Courses and Exams Are Developed AP \bourses and exams are designed by \bommittees of \bollege fa\bulty and AP tea\bhers who ensure that ea\bh AP \bourse and exam refle\bts and assesses \bollege- level expe\btations. These \bommittees define the s\bope and expe\btations of the \bourse, arti\bulating through a \burri\bulum framework what students should know and be able to do upon \bompletion of the AP \bourse. Their work is informed by data \bolle\bted from a range of \bolleges and universities to ensure that AP \bourse work refle\bts \burrent s\bholarship and advan\bes in the dis\bipline. These same \bommittees are also responsible for designing and approving exam spe\bifi\bations and exam questions that \blearly \bonne\bt to the \burri\bulum framework. The AP Exam development pro\bess is a multi-year endeavor; all AP Exams undergo extensive review, revision, piloting, and analysis to ensure that questions are high quality and fair and that the questions \bomprise an appropriate range of diffi\bulty. Throughout AP \bourse and exam development, the College Board gathers feedba\bk from se\bondary and post-se\bondary edu\bators. This feedba\bk is \barefully \bonsidered to ensure that AP \bourses and exams provide students with a \bollege- level learning experien\be and the opportunity to demonstrate their qualifi\bations for advan\bed pla\bement and \bollege \bredit upon \bollege entran\be.
Biology5 Cour\be Development Ea\bh \bommittee first arti\bulates its dis\bipline’s high-level goals and then identifies the \bourse’s spe\bifi\b learning obje\btives. This approa\bh is \bonsistent with “ba\bkward design,” the pra\bti\be of developing \burri\bula, instru\btion, and assessments with the end goal in mind. The learning obje\btives des\bribe what students should know and be able to do, thereby providing \blear instru\btional goals as well as targets of measurement for the exam. Exam Development Exam development begins with the \bommittee making de\bisions about the overall nature of the exam. How long will it be? How many multiple-\bhoi\be questions? How many free-response questions? How mu\bh time will be devoted to ea\bh se\btion? How will the \bourse \bontent and skills be distributed a\bross ea\bh se\btion of the exam? Answers to these questions be\bome part of the exam spe\bifi\bations. With the exam spe\bifi\bations set, test developers design questions that \bonform to these spe\bifi\bations. The \bommittee reviews every exam question for alignment with the \burri\bulum framework, \bontent a\b\bura\by, and a number of other \briteria that ensure the integrity of the exam. Exam questions are then piloted in AP \blassrooms to determine their statisti\bal properties. Questions that have been approved by the \bommittee and piloted su\b\bessfully are in\bluded in an exam. When an exam is assembled, the \bommittee \bondu\bts a final review of the exam to ensure overall \bonformity with the spe\bifi\bations. How AP Exams Are Scored The exam s\boring pro\bess, like the \bourse and exam development pro\bess, relies on the expertise of both AP tea\bhers and \bollege fa\bulty. While multiple-\bhoi\be questions are s\bored by ma\bhine, the free-response questions are s\bored by thousands of \bollege fa\bulty and expert AP tea\bhers at the annual AP Reading. AP Exam Readers are thoroughly trained, and their work is monitored throughout the Reading for fairness and \bonsisten\by. In ea\bh subje\bt, a highly respe\bted \bollege fa\bulty member fills the role of Chief Reader, who, with the help of AP Readers in leadership positions, maintains the a\b\bura\by of the s\boring standards. S\bores on the free-response questions are weighted and \bombined with the results of the \bomputer-s\bored multiple-\bhoi\be questions, and this raw s\bore is summed to give a \bomposite AP s\bore of 5, 4, 3, 2, or 1. The s\bore-setting pro\bess is both pre\bise and labor intensive, involving numerous psy\bhometri\b analyses of the results of a spe\bifi\b AP Exam in a spe\bifi\b year and of the parti\bular group of students who took that exam. Additionally, to ensure alignment with \bollege-level standards, part of the s\bore-setting pro\bess involves \bomparing the performan\be of AP students with the performan\be of students enrolled in \bomparable \bourses in \bolleges throughout the United States. In general, the AP \bomposite s\bore points are set so that the lowest raw s\bore needed to earn an AP s\bore of 5 is equivalent to the average s\bore among \bollege students earning grades of A in the \bollege \bourse. Similarly, AP Exam s\bores of 4 are equivalent to \bollege grades of A–, B+, and B. AP Exam s\bores of 3 are equivalent to \bollege grades of B–, C+, and C.
Biology 6 \bsing and Interpreting AP Scores The extensive work done by \bollege fa\bulty and AP tea\bhers in the development of the \bourse and the exam and throughout the s\boring pro\bess ensures that AP Exam s\bores a\b\burately represent students’ a\bhievement in the equivalent \bollege \bourse. While \bolleges and universities are responsible for setting their own \bredit and pla\bement poli\bies, AP s\bores signify how qualified students are to re\beive \bollege \bredit and pla\bement:AP Score Qualification5 Extremely well qualified 4 Well qualified 3 Qualified 2 Possibly qualified 1 No re\bommendation Additional Resources Visit ap\bentral.\bollegeboard.org for more information about the AP Program.
Biology7 Practice Exam Exam Content and For\qmat The AP Biology Exam is approximately 3 hours in length. There are two se\btions. • Se\btion I is 90 minutes and \bonsists of 63 multiple-\bhoi\be questions and 6 grid-in questions a\b\bounting for 50 per\bent of the final s\bore. • Se\btion II is 90 minutes and \bonsists of 2 long free-response questions and 6 short free-response questions a\b\bounting for 50 per\bent of the final s\bore. It begins with a 10-minute reading period to read the questions and plan your answers. The remaining 1 hour and 20 minutes is for writing. The 2 long free-response questions should require about 20 minutes ea\bh to answer. Questions 3 through 8 are short free-response questions and should require about 6 minutes ea\bh to answer. Admini\btering the Pra\qctice Exam This se\btion \bontains instru\btions for administering the AP Biology Pra\bti\be Exam. You may wish to use these instru\btions to \breate an exam situation that resembles an a\btual administration. If so, read the indented, boldfa\be dire\btions to the students; all other instru\btions are for administering the exam and need not be read aloud. Before beginning testing, have all exam materials ready for distribution. These in\blude test booklets, answer sheets, and the AP Biology formula list. SECTION I: Multiple-Choice Questions When you are ready to begin Se\btion I, say: Section I is the multiple-choice portion of the exam. This section includes traditional multiple-choice questions as well as grid-in questions. For the grid-in questions, you will write in your final answers. Mark all of your responses on your answer sheet, one response per question. If you need to erase, do so carefully and completely. Your score on the multiple-choice section will be based solely on the number of questions answered correctly. Four-function calculators (with square root) are allowed. Are there any questions? You have 1 hour and 30 minutes for this section. Open your Section I booklet and begin. Note Start Time here______ . Note Stop Time here______ . Che\bk that students are marking their answers in pen\bil on their answer sheets, and that they are not looking at their Se\btion II booklets. After 1 hour and 30 minutes, say:Stop working. I will now collect your Section I booklet. There is a 10-minute break between Se\btions I and II. When all Se\btion I materials have been \bolle\bted and a\b\bounted for and you are ready for the break, say:Please listen carefully to these instructions before we take a 10-minute break. Everything you placed under your chair at the beginning of the exam must stay there. Leave your Section II packet on your desk during the break. Are there any questions?
Biology 8 You may begin your break. Testing will resume at _______. SECTION II: Free-Response Questions After the break, say: Section II begins with a 10-minute reading period. During the reading period, you will read the questions and plan your answers to the questions. You may use the blank page of this booklet to organize your answers and for scratch work, but you must write your answers on the lined pages provided for each question. Answers must be written in ink. Are there any questions? You may now open the Section II booklet and begin the 10-minute reading period. Note Start Time here______ . Note Stop Time here______ . After 10 minutes, say: Stop. The reading period is over. You have 1 hour and 20 minutes to answer the questions. You are responsible for pacing yourself, and may proceed freely from one question to the next. Write your answers on the lined pages provided for each question. If you need more paper during the exam, raise your hand. At the top of each extra piece of paper you use, be sure to write your name and the number of the question you are working on. Are there any questions? Begin Section II. Note Start Time here______ . Note Stop Time here______ . Che\bk that students are using pens and that they are writing their answers in their exam booklets and not in their inserts. After 1 hour and 10 minutes, say: There are 10 minutes remaining. After 10 minutes, say: Stop working and close your exam booklet. Put your exam booklet on your desk, face up. Remain in your seat, without talking, while the exam materials are collected. If any students used extra paper for the free-response se\btion, have those students staple the extra sheet/s to the first page \borresponding to that question in their exam booklets. Colle\bt a Se\btion II booklet from ea\bh student and \bhe\bk that ea\bh student wrote answers on the lined pages \borresponding to ea\bh question. Then say: You are now dismissed.
Biology9 Name: AP ® Biology Student Answer Shee\wt for Multi\ble-Choice \wSection No. An\bwer No.An\bwer Grid-In Section 1 33 No.An\bwer 2 34 1 3 35 4 36 2 5 37 6 38 3 7 39 8 40 4 9 41 10 42 5 11 43 12 44 6 13 45 14 46 15 47 16 48 17 49 18 50 19 51 20 52 21 53 22 54 23 55 24 56 25 57 26 58 27 59 28 60 29 61 30 62 31 63 32
Biology 10 �ADVANCED�PLACEMENT�BIOLOGY�EQUATIONS�AND�FORMULAS� ss� � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � Statistical Analysis and Probability Standard Error � � � Mean � x SE n s �� � 1 1 n i n x x i � Standard Deviation Chi\bSquare 2 ( ) 1 n ix x s � � � 2 2 o e e c Chi\bSquare Table Degrees of Freedom p 1 2 3 4 5 6 7 8 0.05 3.84 5.9 .9 7.82 9.49 11.07 12.59 14.07 15.51 1 0.01 6.64 9.32 11.34 13.28 15.09 16.81 18.48 20.09 Laws of Probability \bf A and B are mutually exclusive, then P (A or B) = P(A) + P(B) \bf A and B are independent, then P (A and B) = P(A) x P(B) Hardy Weinberg Equations p2 + 2pq + q 2 = 1 p + q = 1 � s = sample standard deviation (i.e., the sample based estimate of the standard deviation of the population) x p = frequency of the dominant allele in a population q = frequency of the recessive allele in a population = mean n = size of the sam\7ple o = observed individuals with observed genotype e = expected individuals with observed genotype Degrees of freedom equals the number of distinct possible outcomes minus one. Mode = value that occurs most frequently in a data set Median = middle value that separates the greater and lesser halves of a data set Mean = sum of all data points divided by number of data points Range = value obtained by subtracting the smallest observation ( sample minimum ) from the greatest (sample maximum )� Metric Prefixes Factor Prefix Symbol 10 9 giga G 106 mega M 103 kilo k 10-2 centi c 10-3 milli m 10-6 micro 10 -9 nano n 10-12 pico p � �
Biology11 � � � uuu� Rate and Growth Rate dY/dt Population Growth dN/dt=B-D Exponential Growth max dN r N dt Logistic Growth dN K N r N dt K max = \bemperature Coefficient Q 10 2 1 10 2 10 1 Q t t k k Primary Productivity Calculation mg O 2/L x 0.698 = mL O 2/L mL O 2/L x 0.\b36 = mg carbon fixed/L dY= amount of change t = time B = birth rate D = death rate N = population size K = carrying capacity r max = maximum per capita growth rate of population Surface Area and Volume Volume of Sphere V = 4/3 r3 Volume of a cube (or square column) V = l w h Volume of a column V = r2 h Surface area of a sphere A = 4 r2 Surface area of a cube A = 6 a Surface area of a rectangular solid A = (surface area of each side ) r = radius l = length h = height w = width A = surface area V = volume =Sum of all a = surface area of one side of the cube Water Potential ( �) � = �p + �s �p = pressure potential �s = solute potential The water potential will be equal to the solute potential of a solution in an open container, since the pressure potential of the solution in an open container is zero. \bhe Solute Potential of the Solution �s = – iCRT i = ionization constant (For sucrose this is 1.0 because sucrose does not ionize in water) C= molar concentration R= pressure constant (R = 0.0831 liter bars/mole K) T = temperature in Kelvin (273 + ºC) � Dilution - used to create a dilute solution from a concentrated stock solution CiVi = C fVf i=initial (starting) C = concentration of solute f=final (desired) V = volume of solution t2 = higher temperature t1 = lower temperature k2 = metabolic rate at t2 k1 = metabolic rate at t1 Q 10 = the factor by which the reaction rate increases when the temperature is raised by �ten degrees pH = – log [H +] Gibbs Free Energy �G = � H – T�S �G = change in Gibbs free energy �S = change in entropy �H = change in enthalpy T= absolute tem perature (in Kelvin )
Biology 12 At a Glance Total Time1 hour, 30 minutes Number of Questions69 Percent of Total Score50% Writing InstrumentPencil required Electronic \bevice\bour-function calculator allowed Instructions Section I of this exam contains 63 multiple-choice questions and 6 grid -in questions . Indicate all of \bour answers to the Section I questions on the answer sheet .No credit will be given for an\bthing written in this exam booklet ,but \bou ma\b use the booklet for notes or scratch work .When answering the multiple -choice items ,write on the answer sheet the letter corresponding to the answer \bou think is best .Give onl\b one answer to each question .For the grid -in items ,write in the appropriate number ,using decimals and other s\bmbols as appropriate .Units are not required for grid -in items . Use \bour time effectivel\b ,working as quickl\b as \bou can without losing accurac\b .Do not spend too much time on an\b one question .Go on to other questions and come back to the ones \bou have not answered if \bou have time .It is not expected that ever\bone will know the answers to all of the questions . Your total score on Section I is based onl\b on the number of questions answered correctl\b . Points are not deducted for incorrect answers or unanswered questions . \bO NOT OPEN THIS BOOKLET UNTIL YOU ARE TOL\b TO \bO SO. AP ® Biology Practice Exam SECTION I: \bultiple Choice Minimum 20% post-c\aonsumer waste
Biology13 GO ON TO THE NEXT PAGE. BIOLOGY SECTION I Part A: 63 Mult\bple-Cho\bce quest\bons Part B: 6 Gr\bd-In Quest\bons T\bme----- -90 M\bnutes Part A D\brect\bons: Each of the questions or incomplete statements below is followed by four suggested answers or completions\b Select the answer that is best in each case and enter the appropriate letter in the corresponding space on the answer sheet\b When you have completed pa rt A, you should continue on to part B\b 1\b Membrane-bound organelles have been an important component in the evolution of complex, multicellular organisms\b Which of the following best summarizes an advantage of eukaryotic cells having internal membranes? (A) Eukaryotic cells are able to reproduce faster because of the presence of organelles\b (B) Some organelles, such as mitochondria and chloroplasts, are similar to prokaryotic cells in structure\b (C) Organelles isolate specific reactions, increasing metabolic efficiency\b (D) Compartmentalization leads to a higher mutation rate in DNA, which leads to more new species\b 2\b By discharging electric sparks into a laboratory chamber atmosphere that consisted of water vapor, hydrogen gas, methane, and ammonia, Stanley Miller obtained data that showed that a number of organic molecules, including many amino acids, could be synthesized\b Miller was attempting to model early Earth conditions as understood in the 1950s\b The results of Miller’s experiments best support which of the following hypotheses? (A) The molecules essential to life today did not exist at the time Earth was first formed\b (B) The molecules essential to life today could not have been carried to the primordial Earth by a comet or meteorite\b (C) The molecules essential to life today could have formed under early Earth conditions\b (D) The molecules essential to life today were initially self-replicating proteins that were synthesized approximately four billion years ago\b 3\b Simple cuboidal epithelial cells line the ducts of certain human exocrine gl ands\b Various materials are transported into or out of the cells by diffusion\b (The formula for the surface area of a cube is 6 ¥ S 2, and the formula for the volume of a cube is S 3, where S = the length of a side of the cube\b) Which of the following cube-shaped cells would be most efficient in removing waste by diffusion? (A) (B) (C) (D)
Biology 14 GO ON TO THE NEXT PAGE. 4. The area covered by tropical rain forest is reduced by millions of hectares per year due to a\briculture and lo\b\bin\b. Which of the followin\b best describes a likely result of tropical rain forest deforestation? (A) Populations of plants and animals will decrease as more rain forest disappears, leadin\b to a decrease in biodiversity. (B) An increase of soil moisture will lead to a rapid increase in new ve\betation covera\be. (C) An increase in atmospheric carbon dioxide will lead to hi\bher levels of ultraviolet radiation reachin\b the surface of Earth. (D) More oxy\ben will be available to other or\banisms as plant numbers decrease. 5. Which of the followin\b statements best supports the claim that or\banisms share fundamental processes as a result of evolution? (A) All or\banisms that are introduced into new environments have the capacity to fill vacant ecolo\bical roles. (B) All or\banisms have the ability to utilize oxy\ben to harness ener\by from the chemical breakdown of or\banic compounds. (C) All or\banisms share a \benetic code or\banized into triplet codons, makin\b it possible for one or\banism to express a \bene from another or\banism. (D) All or\banisms possess structures such as chloroplasts and mitochondria within their cells that reflect past symbiotic relationships between prokaryotic precursors. 6. When DNA replicates, each strand of the ori\binal DNA molecule is used as a template for the synthesis of a second, complementary strand. Which of the followin\b fi\bures most accurately illustrates enzyme-mediated synthesis of new DNA at a replication fork? (A) (B) (C) (D)
Biology15 GO ON TO THE NEXT PAGE. Questions 7-10 A biologist spent many years researching the rate of evolutionary change in the finch populations of a group of islands. \bt was determined that the average beak size (both length and mass) of finches in a certain population increased dramatically during an intense drought between 1981 and 1987. During the drought, there was a reduction in the number of plants producing thin-walled seeds. 7. Which of the following procedures was most likely followed to determine the change in beak size? (A) A few finches were trapped in 1981 and again in 1987, and their beak sizes were compared. (B) The beak size in fifteen finches was measured in 1987, and the beak size in the original finches was determined by estimation. (C) The beak size in a large number of finches was measured every year from 1981 to 1987. (D) Finches were captured and bred in 1981, and the beak size of the offspring was measured. 8. Which of the following statements might best explain the increase in average beak size in the finch population during the drought? (A) Finches with bigger beaks are better able to crack thick-walled seeds and produce more surviving offspring. (B) Finches with bigger beaks can attack and kill finches with smaller beaks. (C) Finches with bigger beaks possess more- powerful flight muscles and are able to find more food. (D) Finches that crack large seeds develop larger beaks over time. 9. Which of the following best describes the mechanism behind the change in beak size in the finch population? (A) The formation of two new finch species from a single parent species (B) A change in gene frequencies in the finch population due to selective pressure from the environmental change (C) A new allele appearing in the finch population as a result of mutation (D) The achievement of dynamic equilibrium in the finch population as a result of homeostasis 10. The biologist discovered that from 1988 to 1993, the average beak size declined to pre-1981 levels. The reversal in beak size from 1988 to 1993 was most likely related to which of the following events? (A) A loss of food supply for the finches (B) The end of the drought (C) An increase in drought conditions (D) An increase in predators consuming finches
Biology 16 GO ON TO THE NEXT PAGE. 11. Which of the following correctly illustrates a dipeptid e and an amino acid in the optimal position to form a tripeptide\b (A) (B) (C) (D)
Biology17 GO ON TO THE NEXT PAGE. 12. The following is a food web for a meadow habitat that occu\bies 25.6 km 2. The \brimary \broducers’ biomass is uniformly distributed throughout the habitat and totals 1,500 kg/km 2. Develo\bers have a\b\broved a \broject that will \berman ently reduce the \brimary \broducers’ biomass by 50 \bercent and remove all rabbits and deer. Which of the following is the most likel y result at the com\bletion of the \broject? (A) The biomass of coyotes will be 6 kg, and the biomass of hawks will be 0.5 kg. (B) The biomass of coyotes will be dramatically reduced. (C) The coyotes will switch \brey \breferences and outcom\bete the hawks. (D) There will be 50 \bercent fewer voles and 90 \bercent fewer hawks.
Biology 18 GO ON TO THE NEXT PAGE. Segment 1: 5′ - ATATGAGTAGT - 3 ′ Segment 2: 5′ - GCGCAGACGAC - 3′ 3′ - TATACTCATCA - 5′ 3′ - CGCGTCTGCTG - 5′ 13. The sequences for two short fragments of \bNA are s hown above. Which of the following is one way in which these two segments would differ? (A) Segment 1 would not code for mRNA because bot h strands have T, a base not found in RNA. (B) Segment 1 would be more soluble in water than segment 2 because it has more phosphate groups. (C) Segment 1 would become denatured at a lower temp erature than would segment 2 because A-T base pairs have two hydrogen bonds whereas G-C base pairs have three. (\b) Segment 1 must be from a prokaryote becau se it has predominantly A-T base pairs.
Biology19 GO ON TO THE NEXT PAGE. 14. A pathogenic bacterium has been engulfed b\b a phagoc\b tic cell as part of the nonspecific (innate) immune response. Which of the following illust rations best represents the response? (A) (B) (C) (D)
Biology 20 GO ON TO THE NEXT PAGE. 15. The diagram above shows the progression of eco\bogica\b ev ents after a fire in a particu\bar ecosystem. Based on the diagram, which of the fo\b\bowing best exp\bains why the oak trees are \bater rep\baced by other trees? (A) Eventua\b\by the other trees grow ta\b\ber than the oak trees and form a dense canopy that shades the understory. (B) Oak trees a\bter the pH of the soi\b, making the forest better suited for shrubs and other trees. (C) Roots of shrubs pro\biferate in the soi\b of the fo rest and prevent the oak trees from obtaining water. (D) Oak trees succumb to environmenta\b po\b\butants more readi\by than do either the shrubs or the other trees.
Biology21 GO ON TO THE NEXT PAGE. 16. Lactose digestion in E. coli begins with its hydrolysis by the enzyme b\bgalactosidase. The gene encoding b\bgalactosidase, lacZ , is part of a coordinately regulated operon containing other genes required for lactose utilization. Which of the following figures correctly depicts the interactions at the lac operon when lactose is NOT being utilized? (The legend below defines the shapes of the molecules illustrated in the options.) (A) (B) (C) (D)
Biology 22 GO ON TO THE NEXT PAGE. Questions 17-19 An experiment to measure the rate of respiration in crickets and mice at 10∞C and 2\b ∞C was performed using a respirometer, an apparatus that measures changes in gas volume. Respiration was measured in mL of O 2 consumed per gram of organism over seve ral five-minute trials and the following data were obtained. Organism Temperature (∞C) Average respiration (mL O 2/g/min) Mouse 10 0.0\b18 Mouse 2\b 0.0321 Cricket 10 0.0013 Cricket 2\b 0.0038 17. During aerobic cellular respiration, oxygen gas is consumed at the same rate as carbon dioxide gas is produced. In order to provide accurate volumetric measurements of oxygen gas consumption, the experimental setup should include which of the following? (A) A substance that removes carbon dioxide gas (B) A plant to produce oxygen (C) A glucose reserve (D) A valve to release excess water 18. According to the data, the mice at 10 ∞C demonstrated greater oxygen consumption per gram of tissue than did the mice at 2\b ∞C. This is most likely explained by which of the following statements? (A) The mice at 10 ∞C had a higher rate of ATP production than the mice at 2\b ∞C. (B) The mice at 10 ∞C had a lower metabolic rate than the mice at 2\b ∞C. (C) The mice at 2\b ∞C weighed less than the mice at 10 ∞C. (D) The mice at 2\b ∞C were more active than the mice at 10 ∞C. 19. According to the data, the crickets at 2\b ∞C have greater oxygen consumption per gram of tissue than do the crickets at 10∞C. This trend in oxygen consumption is the opposite of that in the mice. The difference in trends in oxygen consumption among crickets and mice is due to their (A) relative size (B) mode of nutrition (C) mode of internal temperature regulation (D) mode of ATP production
Biology23 GO ON TO THE NEXT PAGE. 20. The FtsZ protein is present in prokaryotes and in chloroplasts. The protein is structurally and \bunctionally similar to tubulin proteins o\b eukaryotic cells. Which o\b the \bollowing is a likely conclusion to draw \brom this in\bormation? (A) FtsZ and tubulin proteins were both present in a common ancestor. (B) Microtubules are involved in the mechanics o\b photosynthesis. (C) Tubulin genes are evolutionarily derived \brom the gene that codes \bor the FtsZ protein. (D) The sequences o\b the genes encoding the FtsZ and tubulin proteins are identical. 21. In most \breshwater \bish, nitrogenous waste is primarily excreted as ammonia, which is highly soluble in water and is toxic at low concentrations. In terrestrial mammals, ammonia is converted to urea be\bore it is excreted. Urea is also highly soluble in water but is less toxic than ammonia at low concentrations. Which o\b the \bollowing best explains why \breshwater \bish do not convert ammonia to urea \bor excretion? (A) The metabolic pathways o\b \bish do not normally involve nitrogen consumption. (B) The dilution o\b ammonia by direct excretion into \breshwater conserves energy. (C) Ammonia is concentrated in tissues, where it is stored prior to excretion. (D) The nitrogen in ammonia is recycled \bor use in protein and nucleotide synthesis. 22. During the in\bection cycle \bor a typical retrovirus, such as HIV, which uses RNA as genetic material, the genetic variation in the resulting population o\b new virus particles is very high because o\b (A) damage to the virus particle \brom envelope loss during in\bection (B) errors introduced in the DNA molecule through reverse transcription (C) errors in the protein molecules produced in translation (D) recombination o\b the genomes o\b \bree virus particles
Biology 24 GO ON TO THE NEXT PAGE. 23. A sample of human blood was placed in a test t ube containing a ph\bsiological saline solution (0.9% sodium chloride). This t\bpe of solution is often used intrav enousl\b to quickl\b reh\bdrate patients. A drop of the blood from the test tube was placed on a slide and red blood cells (RBCs) were observed under a microscope. Three possible outcomes are diagrammed below. Which of the following best predicts which diagrammed microscope view the laborator\b worker would see and best explains wh\b? (A) View 1 because RBC membranes are freel\b permeable to water (B) View 2 because the RBCs use energ\b to allow sodium entr\b and to pump water out (C) View 2 because the rate of water movement into th e RBCs equals the rate of water movement out of the cells (D) View 3 because the sodium-potassium pumps in the RBC membranes use energ\b to keep the sodium out but allow water to freel\b flow into the cells
Biology25 GO ON TO THE NEXT PAGE. 24. Which of the following statements most directly su\b\borts the claim that different s\becies of organisms use different metabolic strategies to meet their energy requirements for growth, re\broduction, and homeostasis? (A) During cold \beriods \bond-dwelling animals can increase the number of unsaturated fatty acids in their cell membranes while some \blants make antifreeze \broteins to \brevent ice crystal formation in tissues. (B) Bacteria lack introns while many eukaryotic genes contain many of these intervening sequences. (C) Carnivores have more teeth that are s\becialized for ri\b\bing food while herbivores have more teeth that are s\becialized for grinding food. (D) Plants generally use starch molecules for storage while animals use glycogen and fats for storage. 25. Paramecia are unicellular \brotists that have contractile vacuoles to remove excess intracellular water. In an ex\berimental investigation, \baramecia were \blaced in salt solutions of increasing osmolarity. The rate at which the contractile vacuole contracted to \bum\b out excess water was determined and \blotted against osmolarity of the solutions, as shown in the gra\bh. Which of the following is the correct ex\blanation for the data? (A) At higher osmolarity, lower rates of contraction are required because more salt diffuses into the \baramecia. (B) The contraction rate increases as the osmolarity decreases because the amount of water entering the \baramecia by osmosis increases. (C) The contractile vacuole is less efficient in solutions of high osmolarity because of the reduced amount of ATP \broduced from cellular res\biration. (D) In an isosmotic salt solution, there is no diffusion of water into or out of the \baramecia, so the contraction rate is zero.
Biology 26 GO ON TO THE NEXT PAGE. Questions 26-29 A student placed 20 tobacco seeds of the same species on mois t paper towels in each of two petri dishes. \bish A was wrapped completely in an opaque co ver to exclude all light. \bish B was not wrapped. The dishes were placed equidistant from a light source set to a cycle of 14 hours of light and 10 hours of dark. All other conditions were the same for both dishes. The dishes were examined after 7 days and the opaque cover was permanently removed from dish A. Both dishes were returned to the light and examined again at 14 days. The following data were obtained. Dish A Dish B \bay 7 Covered \bay 14 Uncovered \bay 7 Uncovered \bay 14 Uncovered Germinated seeds 12 20 20 20 Green-leaved seedlings 0 14 15 15 Yellow-leaved seedlings 12 6 5 5 Mean stem length below first set of leaves 8 mm 9 mm 3 mm 3 mm 26. According to the results of this experiment, germination of tobacco seeds during the first week is (A) increased by exposure to light (B) unaffected by light intensity (C) prevented by paper towels (\b) accelerated in green-leaved seedlings 27. The most probable cause for the difference in mean stem length between plants in dish A and plants in dish B is which of the following? (A) Shortening of cells in the stem in response to the lack of light (B) Elongation of seedlings in response to the lack of light (C) Enhancement of stem elongation by light (\b) Genetic differences between the seeds 28. Which of the following best supports the hypothesis that the difference in leaf color is genetically controlled? (A) The number of yellow-leaved seedlings in dish A on day 7 (B) The number of germinated seeds in dish A on days 7 and 14 (C) The death of all the yellow-leaved seedlings (\b) The existence of yellow-leaved seedlings as well as green-leaved ones on day 14 in dish B 29. Additional observations were made on day 21, and no yellow-leaved seedlings were found alive in either dish. This is most likely because (A) yellow-leaved seedlings were unable to absorb water from the paper towels (B) taller green-leaved seedlings blocked the light and prevented photosynthesis (C) yellow-leaved seedlings were unable to convert light energy to chemical energy (\b) a higher rate of respiration in yellow-leaved seedlings depleted their stored nutrients
Biology27 GO ON TO THE NEXT PAGE. 30. Arctic foxes typically have a white coat in the winter. In s\bmmer, when there is no snow on the gro\bnd, the foxes typically have a darker coat. Which of the following is most likely responsible for the seasonal change in coat color? (A) The decrease in the amo\bnt of daylight in winter ca\bses a change in gene expression, which res\blts in the foxes growing a lighter- appearing coat. (B) The diet of the foxes in s\bmmer lacks a partic\blar n\btrient, which ca\bses the foxes to lose their white coat and grow a darker- colored coat. (C) Competition for mates in the spring ca\bses each fox to increase its camo\bflage with the environment by prod\bcing a darker- appearing coat. (D) The lower temperat\bres in winter denat\bre the pigment molec\bles in the arctic fox coat, ca\bsing the coat to become lighter in color. 31. The endocrine system incorporates feedback mechanisms that maintain homeostasis. Which of the following demonstrates negative feedback by the endocrine system? (A) D\bring labor, the fet\bs exerts press\bre on the \bterine wall, ind\bcing the prod\bction of oxytocin, which stim\blates \bterine wall contraction. The contractions ca\bse the fet\bs to f\brther p\bsh on the wall, increasing the prod\bction of oxytocin. (B) After a meal, blood gl\bcose levels become elevated, stim\blating beta cells of the pancreas to release in s\blin into the blood. Excess gl\bcose is then converted to glycogen in the liver, red\bcing blood gl\bcose levels. (C) At high elevation, atmospheric oxygen is more scarce. In response to signals that oxygen is low, the brain decreases an individ\bal’s rate of respiration to compensate for the difference. (D) A transcription factor binds to the reg\blatory region of a gene, blocking the binding of another transcription factor req\bired for expression.
Biology 28 GO ON TO THE NEXT PAGE. 32. Five new species of bacteria were discovered in Antarctic ice core samples. \bhe nucleotide (base) sequences of rRNA subunits were determined for the new species. \bhe table below shows the number of nucleotide differences between the species. NUCLEOTIDE DIFFERENCES Species 1 2 3 4 5 1 - 3 19 18 27 2 - 19 18 26 3 - 1 27 4 - 27 5 - Which of the following phylogenetic trees is most consistent with the data? (A) (B) (C) (D) 33. \bhe mechanism of action of many common medications involves interfering with the normal pathways that cells us e to respond to hormone signals. Which of the following best describes a drug interaction that directly interferes with a signal transduction pathway? (A) A medication causes the cell to absorb more of a particular mineral, eventually poisoning the cell. (B) A medication enters the target cell and inhibits an enzyme that normally synthesizes a second messenger. (C) A medication enters the target cell’s nucleus and acts as a mutagen. (D) A medication interrupts the transcription of ribosomal RNA genes.
Biology29 GO ON TO THE NEXT PAGE. 34. The first diagram below shows the levels of mRN\b from two different genes ( bicoid and caudal) at different positions along the anterior-posterior axis of a Drosophila egg immediately before fertilizati on. The second diagram shows the levels of the two corresponding proteins along the anterior-posterior axis shortly after fertilization. Which of the following conc lusions is best supported by the data? (\b) Bicoid protein inhibits translation of caudal mRN\b. (B) Bicoid protein stabilizes caudal mRN\b. (C) Translation of bicoid mRN\b produces caudal protein. (D) Caudal protein stimulates development of anterior structures.
Biology 30 GO ON TO THE NEXT PAGE. 35. Living cells typically ha ve biosynthetic pathways to synthesize at least some o\b the amino acids used in making proteins. Some strains o\b E. coli, a prokaryote, can synthesize the amino acid tryptophan, while other E. coli strains cannot. Similarly, some strains o\b the yeast S. cerevisiae, a eukaryote, can synthesize tryptophan, while other S. cerevisiae strains cannot. Which o\b the \bollowing describes the most likely source o\b genetic variation \bound in the tryptophan synthesis pathways o\b both species? (A) Exchange o\b genetic in\bormation occurs through crossing over. (B) Viral transmission o\b genetic in\bormation required to synthesize tryptophan occurs. (C) Random assortment o\b chromosomes leads to genetic variation. (D) Errors in DNA replication lead to genetic variation. 36. Sickle-cell anemia results \brom a point mutation in the HBB gene. The mutation results in the replacement o\b an amino acid that has a hydrophilic R-group with an amino acid that has a hydrophobic R-group on the exterior o\b the hemoglobin protein. Such a mutation would most likely result in altered (A) properties o\b the molecule as a result o\b abnormal interactions between adjacent hemoglobin molecules (B) DNA structure as a result o\b abnormal hydrogen bonding between nitrogenous bases (C) \batty acid structure as a result o\b changes in ionic interactions between adjacent \batty acid chains (D) protein secondary structure as a result o\b abnormal hydrophobic interactions between R-groups in the backbone o\b the protein
Biology31 GO ON TO THE NEXT PAGE. 37. In a hypothetical population of beetles, there is a wi\be variety of color, matching the range of coloration of the tree trunks on which the beetles hi\be from pre\bators. The graphs below illustrate four possible changes to the beetle population as a result of a change in the envi ronment \bue to pollution that \barkene\b the tree trunks. Which of the following inclu\bes the most likely change in the coloration of the beetle population after pollution an\b a correct rationale for the change? (A) The coloration range shifte\b towar\b more light-colore\b beetles, as in \biagram I. The pollution helpe\b the pre\bators fin\b the \barkene\b tree trunks. (B) The coloration in the population split into two extremes , as in \biagram II. Both the lighter-colore\b an\b the \barker-colore\b beetles were able to hi\be on the \barker tree trunks. (C) The coloration range became narrower, as in \biagram III. The pre\bators selecte\b beetles at the color extremes. (D) The coloration in the population shifte\b towar\b more \barker-colore\b beetles, as in \biagram IV. The lighter- colore\b beetles were foun\b more easily by the pre\b ators than were the \barker-colore\b beetles.
Biology 32 GO ON TO THE NEXT PAGE. 38. Testosterone oxido-reductase is a liver enzyme t\b at regulates testosterone levels in alligators. One study compared testosterone oxido-reductase activity between male and female alligators from Lake Woodruff, a relatively pristine environment, and fro m Lake Apopka, an area t\bat \bas suffered severe contamination. T\be grap\b above depicts t\be findings of t\bat study. T\be data in t\be grap\b best support w\bic\b of t\be following claims? (A) Environmental contamination elevates total te stosterone oxido-reductase activity in females. (B) Environmental contamination reduces total te stosterone oxido-reductase activity in females. (C) Environmental contamination elevates total testosterone oxido-reductase activity in males. (D) Environmental contamination reduces total te stosterone oxido-reductase activity in males.
Biology33 GO ON TO THE NEXT PAGE. 39. An individual’s humoral response to a particular antigen di\b\bers depending on whether or not the individual has been previously exposed to that antigen. Which o\b the \bollowing graphs properly represents the humoral immune response when an individual is exposed to the same antigen more than once? (A) (B) (C) (D)
Biology 34 GO ON TO THE NEXT PAGE. Questions 40-44 In a transformation experiment, a sample of E. coli bacteria was mixed with a plasmid containing the gene for resistance to the antibiotic ampicillin ( amp r\b. Plasmid was not added to a second sample. Samples were plated on nutrient agar plates, some of which were supplemen ted with the antibiotic ampicillin. The results of E. coli growth are summarized below. The shaded area represents extens ive growth of bacteria; dots represent individual colonies of bacteria. 40. Plates that have only ampicillin-resistant bacteria growing include which of the following? (A\b I only (B\b III only (C\b IV only (D\b I and II 41. Which of the following best explains why there is no growth on plate II? (A\b The initial E. coli culture was not ampicillin- resistant. (B\b The transformation procedure killed the bacteria. (C\b Nutrient agar inhibits E. coli growth. (D\b The bacteria on the plate were transformed. 42. Plates I and III were included in the experimental design in order to (A\b demonstrate that the E. coli cultures were viable (B\b demonstrate that the plasmid can lose its amp r gene (C\b demonstrate that the plasmid is needed for E. coli growth (D\b prepare the E. coli for transformation 43. Which of the following statements best explains why there are fewer colonies on plate IV than on plate III? (A\b Plate IV is the positive control. (B\b Not all E. coli cells are successfully transformed. (C\b The bacteria on plate III did not mutate. (D\b The plasmid inhibits E. coli growth. 44. In a second experiment, the plasmid contained the gene for human insulin as well as the amp r gene. Which of the following plates would have the highest percentage of bacteria that are expected to produce insulin? (A\b I only (B\b III only (C\b IV only (D\b I and III
Biology35 GO ON TO THE NEXT PAGE. 45. Experimental evidence shows that the process of \blycolysis is present and virtually identical in or\banisms from all three domains, Archaea, Bacteria, and Eukarya. Which of the followin\b hypotheses could be be st supported by this evidence? (A) All or\banisms carry out \blycolysis in mitochondria. (B) Glycolysis is a universal ener\by-releasin\b process and therefore su\b\bests a common ancestor for all forms of life. (C) Across the three domains, all or\banisms depend solely on the process of anaerobic respiration for ATP production. (D) The presence of \blycolysis as an ener\by- releasin\b process in all or\banisms su\b\bests that conver\bent evolution occurred. 46. The last part of the metamorphosis of a tadpole to an adult fro\b results in the disappearance of the tail. This sta\be of development most likely occurs by (A) cells of the tail dyin\b and the nutrients bein\b absorbed and reused by the body (B) sheddin\b of the tail so ener\by is not spent on maintenance of an unneeded part (C) bilateral division of the tail and fusion with the developin\b hind limbs (D) individual cells of the tail mi\bratin\b to the developin\b \bonads
Biology 36 GO ON TO THE NEXT PAGE. 47. MRSA is the acronym for methicillin-resistant Staphylococcus aureus . Many of the strains of the common \bacterium are also resistant to other anti\biotics in use today. The resistance is linked to a collection of genes carried on plasmids that are passed from one \bacterium to another \by conjugation. Suppose a newly discovered, chemically different anti\biotic is used in place of methicillin. Which of the following would \be the most likely effect on Staphylococcus aureus anti\biotic resistance? (A) The gene for methicillin resistance, no longer needed, would disappear entirely from Staphylococcus aureus populations within a few generations. (B) Transmission of the methicillin-resistance plasmid \by conjugation would increase among the Staphylococcus aureus population as the genes would confer resistance to the new anti\biotic. (C) Transmission of the methicillin-resistance plasmid would gradually decrease \but the plasmid would not entirely disappear from the Staphylococcus aureus population. (D) Transmission of the methicillin-resistance plasmid \by conjugation would increase among the Staphylococcus aureus population due to destruction of \bacteria without the plasmid through use of the new anti\biotic. 48. A human kidney filters a\bout 200 liters of \blood each day. Approximately two liters of liquid and nutrient waste are excreted as urine. The remaining fluid and dissolved su\bstances are rea\bsor\bed and continue to circulate throughout the \body. Antidiuretic hormone (ADH) is secreted in response to reduced plasma volume. ADH targets the collecting ducts in the kidney, stimulating the insertion of aquaporins into their plasma mem\branes and an increased rea\bsorption of water. If ADH secretion is inhi\bited, which of the following would initially result? (A) The num\ber of aquaporins would increase in response to the inhi\bition of ADH. (B) The person would decrease oral water intake to compensate for the inhi\bition of ADH. (C) Blood filtration would increase to compensate for the lack of aquaporins. (D) The person would produce greater amounts of dilute urine.
Biology37 GO ON TO THE NEXT PAGE. 49. The diagram above shows a developing worm embryo at the \bour-cell stage. Experiments have shown that when cell 3 divides, the anterior daughter cell gives rise to muscle and gonads and the posterior daughter cell gives rise to the intestine. However, i\b the cells o\b the embryo are separated \brom one another early during the \bour-cell stage, no intestine will \borm. Other experiments have shown that i\b cell 3 and cell 4 are recombined a\bter the initial separation, the posterior daughter cell o\b cell 3 will once again give rise to normal intestine. Which o\b the \bollowing is the most plausible explanation \bor these \bindings? (A) A cell sur\bace protein on cell 4 signals cell 3 to induce \bormation o\b the worm’s intestine. (B) The plasma membrane o\b cell 4 interacts with the plasma membrane o\b the posterior portion o\b cell 3, causing invaginations that become microvilli. (C) Cell 3 passes an electrical signal to cell 4, which induces di\b\berentiation in cell 4. (D) Cell 4 trans\bers genetic material to cell 3, which directs the development o\b intestinal cells. 50. The tiny blue-eyed Mary \blower is o\bten one o\b the \birst \blowers seen in the spring in some regions o\b the United States. The \blower is normally blue, but sometimes a white or pink \blower variation is \bound. The \bollowing data were obtained a\bter several crosses. Parents F 1 F 2 Blue ¥ white Blue 196 blue, 63 white Blue ¥ pink Blue 149 blue, 52 pink Pink ¥ white Blue 226 blue, 98 white, 77 pink Which o\b the \bollowing statements best explains the data? (A) The appearance o\b blue in the F 1 generation o\b the pink and white cross demonstrates that \blower color is not an inherited trait but is determined by the environment. (B) Flower color depends on stages o\b \blower development, and young \blowers are white, advancing to pink and then blue. (C) Since the F 1 and F 2 phenotypes o\b the pink and white cross do not \bit the expected genotypic and phenotypic ratios, blue-eyed Mary must reproduce by vegetative propagation. (D) Flower color is an inherited trait, and the F 1 and F 2 phenotypes o\b the \blowers arising \brom the pink and white cross can best be explained by another gene product that in\bluences the phenotypic expression.
Biology 38 GO ON TO THE NEXT PAGE. Questions 51-53 Both myoglobin and hemoglobin are proteins that bind reversibly with molecular oxygen\b The graph below shows the oxygen-binding saturation of each protein at different concentrations of oxygen\b 51\b Which of the following statements is correct? (A) At 10 mm Hg partial pressure, hemoglobin binds oxygen but myoglobin does not\b (B) At 20 mm Hg partial pressure, myoglobin and hemoglobin bind oxygen in equal amounts\b (C) At 40 mm Hg partial pressure, myoglobin has a greater affinity for oxygen than hemoglobin has\b (D) At 80 mm Hg partial pressure, myoglobin binds twice as much oxygen as hemoglobin binds\b 52\b Strenuous exercise lowers the blood pH, causing the curves for both hemoglobin and myoglobin to shift to the right\b This shift results in (A) an unloading of O 2 at higher partial pressures (B) an increase in the number of O 2-binding sites (C) the capture of more O 2 by hemoglobin (D) the capture of more O 2 by myoglobin 53\b Which of the following best describes the physiological significance of the different oxygen-binding capabilities of hemoglobin and myoglobin? (A) They prevent muscles from depleting oxygen levels in the blood\b (B) They cause muscles to become anaerobic\b (C) They prevent glycogen depletion in muscles\b (D) They enhance movement of oxygen from the blood into the muscles\b
Biology39 GO ON TO THE NEXT PAGE. 54. The regulatory sequences of the ope ron controlling arabinose \betabolis\b (ara operon) were studied to deter\bine whether bacteria can respond to changes in nutrien t availability. It is predicted that if those regulatory sequences are functioning properly, the bacteria will produce the enzy\bes involved in arabinose \betabolis\b (structural genes B, A, and D) in the presence of arabinose. If a gene that encodes a green fluorescent protein (GFP) is substituted for the structural genes of the operon, activation of the regulatory sequences can be assayed by GFP expression. A culture of E. coli cells underwent a transfor\bation procedure with a plas\bid c ontaining the regulatory sequences of the ara operon directly upstrea\b of the gene encoding the GFP. The plas\bid also confer s a\bpicillin resistance to bacteria. Sa\bples were then plated on different types of culture \bedia. (Note: The GFP fluoresces only under UV light, not under white light.) The table below shows the results. Transformation Results Type of Culture Media Agar Ampicillin Ara\binose Color of Colonies Under White Light Color of Colonies Under UV Light + - - All white All white + + - All white All white + - + All white Mostly white, so\be green + + + All white All green + Indicates the presence of the indicat ed substance in the culture \bedia - Indicates the absence of the indicated substance in the culture \bedia Which of the following can best be used to justify why the GFP is expressed by E. coli cells after transfor\bation with the plas\bid? (A) The presence of arabinose in the nutrient agar activ ated the expression of the genes located downstrea\b of the ara operon regulatory sequences. (B) The co\bbination of a\bpicillin and arabinose in the nutrient agar inhibited the expression of certain gene products, resulting in the increased expression of the GFP. (C) The nutrient agar without arabinose but with a\bpi cillin activated the expression of the genes located downstrea\b of the ara operon regulatory sequences. (D) Both arabinose and a\bpicillin were required in the nutrient agar to activate the expression of genes located downstrea\b of the ara operon regulatory sequences.
Biology 40 GO ON TO THE NEXT PAGE. 55. The chemical reaction for photosynthesis is 6 CO 2 \b 12 H 2O \b light energy → C 6H12O6 \b 6 O 2 \b 6 H 2O If the input water is labeled with a radioactive isotope of oxygen, 18O, then the oxygen gas released as the reaction proceeds is also labeled with 18O. Which of the following is the most likely explanation? (A) During the light reactions of photosynthesis, water is split, the hydrogen atoms combine with the CO 2, and oxygen gas is released. (B) During the light reactions of photosynthesis, water is split, removing electrons and protons, and oxygen gas is released. (C) During the Calvin cycle, water is split, regenerating NADPH from NADP \b, and oxygen gas is released. (D) During the Calvin cycle, water is split, the hydrogen atoms are added to intermedia tes of sugar synthesis, and oxygen gas is released.
Biology41 GO ON TO THE NEXT PAGE. 56. A culture of Spirogyra (an autotrophic alga) is maintaine\b in a water solution containing \bissolve\b carbon \bioxi\be an\b a source of phosphates but lacking nitrogen compoun\bs. A researcher \betermines the rates of synthesis of several orga nic compoun\bs foun\b in the Spirogyra before an\b after several weeks in the water solution. Which of the following graphs best illustrates a likely result of the experiment? (A) (B) (C) (D)
Biology 42 GO ON TO THE NEXT PAGE. 57. The illustration above depicts a neurom uscular junction o\b a patient with an autoimmune disorder. Acetylcholine is a stimulatory neurotransmitter. Which o\b the \bollo wing would be the most likely result o\b the continued presence o\b the antibody? (A) An increase in action potentials in th e motor neuron and constant nerve pain (B) A decrease in action potentials in the muscle, causing muscle weakness and \batigue (C) A decrease in the opening o\b sodium-gated channels in the muscle, causing less sodium to be released \brom the muscle (D) An increase in the opening o\b sodium-gated channe ls in the motor neuron because o\b the accumulation o\b acetylcholine in the junction
Biology43 GO ON TO THE NEXT PAGE. 58. A group of students summarized inform ation on five great extinction events. \bass Extinction Time of Extinction Organisms Greatly Reduced or \bade Extinct End of the Ordovician period 443 million years ago Trilobites, brachiopods, echinoderms, and corals End of the Devonian period 354 million years ago \barine families on tropical reefs, corals, brachiopods, and bivalves End of the Permian period 248 million years ago Trilobites, mollusks, brachiopods, and many vertebrates End of the Triassic period 206 million years ago \bollusks, sponges, marine vertebrates, and large amphibians End of the Cretaceous period 65 million y ears ago Ammonites, dinosaurs, brachiopods, bivalves, and echinoderms The students are sampling a site in search of fossils from the Devonian period. Based on the chart, which of the following would be the most reasonable plan for the students to follow? (A) Searching horizontal rock layers in any class of rock and try to find those that contain the greatest number of fossils (B) Collecting fossils from rock layers deposited prior to the Permian period that contain some early vertebrate bones (C) Looking in sedimentary layers next to bodies of water in order to find marine fossils of bivalves and trilobites (D) Using relative dating techniques to determine the geological ages of the fossils found so they can calculate the rate of speciation of early organisms
Biology 44 GO ON TO THE NEXT PAGE. 59. Anabaena is a simple multicellular photosynthetic cyanobacter ium. In the absence of fixe\b nitrogen, certain newly \beveloping cells along a filament express genes that co\be for nitrogen-fixing enzymes an\b become non- photosynthetic heterocysts. The specialization is a\bvant ageous because some nitrogen-fixing enzymes function best in the absence of oxygen. Heterocysts \bo not carry out photosynthesis but instea\b provi\be a\bjacent cells with fixe\b nitrogen, in exchange receivi ng fixe\b carbon an\b re\buce\b energy carriers. As shown in the \biagram above, when there is low fixe\b nitrogen in the environment, an increase in the concentration of free calcium ions an\b 2-oxyglutarate stimulates the expression of genes that pro\buce two transcription factors (NtcA an\b HetR) that promot e the expression of genes responsible for heterocyst \bevelopment. HetR also causes pro\buction of a signal, Pa tS, that prevents a\bjacent cells from \beveloping as heterocysts. Base\b on your un\berstan\bing of the wa ys in which signal transmission me\biates cell function, which of the following pre\bictions is most consiste nt with the information given above? (A) In an environment with low fixe\b nitrogen, treating the Anabaena cells with a calcium-bin\bing compoun\b shoul\b prevent heterocyst \bifferentiation. (B) A strain that overexpresses the patS gene shoul\b \bevelop many more hete rocysts in a low fixe\b nitrogen environment. (C) In an environment with abun\bant fixe\b nitrogen, free calcium levels shoul\b be high in all cells so that no heterocysts \bevelop. (D) In environments with abun\ba nt fixe\b nitrogen, loss of the hetR gene shoul\b in\buce heterocyst \bevelopment.
Biology45 GO ON TO THE NEXT PAGE. 60. The figure above shows several steps in the process of bacteriophage trans\buction in bacteria. Which of the following explains how genetic variation in a popul ation of bacteria results from this process? (A) Bacterial proteins transferre\b from the \bonor bacter ium by the phage to the recipient bacterium recombine with genes on the recipient’s chromosome. (B) The recipient bacterium incorporat es the trans\buce\b genetic material co\bing for phage proteins into its chromosome an\b synthesizes the correspon\bing proteins. (C) The phage infection of the recipient bacterium an\b the intro\buction of DNA carrie\b by the phage cause increase\b ran\bom point mutations of the bacterial chromosome. (D) DNA of the recipient bacterial chromosome un\ber goes recombination with DNA intro\buce\b by the phage from the \bonor bacterium, lea\bing to a change in the recipient’s genotype.
Biology 46 GO ON TO THE NEXT PAGE. 61. Figure I shows the growth of an algal species in a flask of sterili\bed pond water. If phosphate is added as indicated, the growth curve changes as shown in Figure II. Which of the following is the best prediction of the algal growth if nitrate is added instead of phosphate? (A) (B) (C) (D)
Biology47 GO ON TO THE NEXT PAGE. 62. The figures below illustrate the similarities betw een ATP s\bnthesis in mitochondria and chloroplasts. The figures can best assist in answering which of the following questions? (A) Do electron transport chains create a gradient so that ATP s\bnthase can generate ATP molecules? (B) What are the sources of energ\b that drive mito chondrial and chloroplast electron transport s\bstems? (C) What is the optimal temperature at which ATP s\bnthase chemicall\b converts ADP and a phosphate group into one molecule of ATP? (D) What is the evolutionar\b relationship between the ATP s\bnthase in mitochondria and the ATP s\bnthase in chloroplasts?
Biology 48 GO ON TO THE NEXT PAGE. 63. The Hedgehog protein (Hh) plays a critical role d\bring a certain period of embryo development, b\bt it normally has no role in ad\blts except for the maintenance of ad\blt stem cells. However, th e Hedgehog protein has been detected in 70 percent of pancreatic cancer cell samp les. As ill\bstrated in the fig\bres below, the Hedgehog protein binds to an integral membrane protein receptor known as Patched (Ptc), th\bs initiating a pathway of gene expression. When Hedgehog is absent, Pt c inhibits another protein known as Smoothened (Smo), which, in t\brn, blocks the activation of a gro\bp of proteins collectiv ely known as the Hedgehog signaling complex (HSC). The inactivation is the res\blt of proteolytic cleavage of one component of the HSC complex, a transcription factor known as C\bbit\bs interr\bpt\bs (Ci). When Hedgehog is present, it binds to Ptc, which prevents the inhibition of Smo by Ptc. The res\blt is that Ci remains intact and can enter the n\bcle\bs, where it binds to and activates certain genes. One approach to treating patients with pancreatic cancer and other cancers in which the Hedgehog protein is detected is to modify the Hedgehog si gnaling pathway. Which of the following is the most \bsef\bl approach? (A) Treating patients with a molec\ble that is str\bct\bra lly similar to Hedgehog and that will bind to and interact with Ptc in the same fashion as Hedgehog (B) Injecting patients with embryonic cells so that Hedgehog will bind to those cells instead of the cancer cells (C) Treating patients with a membrane-sol\bble com po\bnd that can bind to Smo and block its activity (D) Injecting patients with a preparation of p\brified me mbrane-sol\bble Ci that will enter the n\bclei of the cancer cells and ind\bce gene transcription END OF PART A DO NOT STOP PLEASE CONTINUE TO PART B
Biology49 GO ON TO THE NEXT PAGE. P art B D irections: Part B consists of six questions requiring numeric answers. Ca\bcu\bate the correct answer for each question, and enter on the \bine provided on the answer sheet.
Biology 50 GO ON TO THE NEXT PAGE. 1. Use the graph above to calculate the mean rate of population growth \bindividuals per day) between day 3 and day 5. Give your answer to the nearest whole number. 2. In a certain species of flowering plant, the purple allele P is dominant to the yellow allele p. A student performed a cross between a purple- flowered plant and a yellow-flowered plant. When planted, the 146 seeds that were produced from the cross matured into 87 plants with purple flowers and 59 plants with yellow flowers. Calculate the chi-squared value for the null hypothesis that the purple-flowered parent was heterozygous for the flowe r-color gene. Give your answer to the nearest tenth.
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