5 Steps to a 5: AP Biology - Mark Anestis 2021
STEP 5 Build Your Test-Taking Confidence
Answer Sheet for AP Biology Practice Exam 1
MULTIPLE-CHOICE QUESTIONS
AP Biology Practice Exam 1: Section I
MULTIPLE-CHOICE QUESTIONS
Time—1 hour and 30 minutes
For the multiple-choice questions that follow, select the best answer and fill in the appropriate letter on the answer sheet.
1. Which of the following characteristics would allow you to distinguish a prokaryotic cell from an animal cell?
A. Ribosomes
B. Cell membrane
C. Chloroplasts
D. Cell wall
2. Which of the following is the source of oxygen produced during photosynthesis?
A. H2O
B. H2O2
C. CO2
D. CO
3. An organism exposed to wild temperature fluctuations shows very little, if any, change in its metabolic rate. This organism is most probably a(n)
A. ectotherm.
B. endotherm.
C. thermophyle.
D. ascospore.
4. Which of the following is a frameshift mutation?
A. CAT HAS HIS → CAT HAS HIT
B. CAT HAS HIS → CAT HSH ISA
C. CAT HAS HIS → CAT HIS HAT
D. CAT HAS HIS → CAT WAS HIT
5. A researcher conducts a survey of a biome and finds 35 percent more species than she has found in any other biome. Which biome is she most likely to be in?
A. Tundra
B. Tiaga
C. Tropical rainforest
D. Temperate deciduous forest
6. On the basis of the following crossover frequencies, determine the relative location of these four genes:
7. A man contracts the same flu strain for the second time in a single winter season. The second time he experiences fewer symptoms and recovers more quickly. Which cells are responsible for this rapid recovery?
A. Helper T cells
B. Cytotoxic T cells
C. Memory cells
D. Plasma cells
8. Which of the following are traits that are affected by more than one gene?
A. Heterozygous traits
B. Pleiotropic traits
C. Polygenic traits
D. Blended alleles
9. A lizard lacking a chemical defense mechanism that is colored in the same way as a lizard that has a defense mechanism is displaying
A. aposometric coloration.
B. cryptic coloration.
C. Batesian mimicry.
D. Müllerian mimicry.
10. Crossover would most likely occur in which situation?
A. Two genes (1 and 2) are located right next to each other on chromosome A.
B. Gene 1 is located on chromosome A, and gene 2 is on chromosome B.
C. Genes 1 and 2 are located near each other on the X chromosome.
D. Gene 1 is located on chromosome A; gene 2 is located far away, but on the same chromosome.
11. Imagine an organism whose 2n = 96. Meiosis would leave this organism’s cells with how many chromosomes?
A. 192
B. 96
C. 48
D. 24
12. A student conducts an experiment to test the efficiency of a certain enzyme. Which of the following protocols would probably not result in a change in the enzyme’s efficiency?
A. Bringing the temperature of the experimental setup from 20°C to 50°C
B. Adding an acidic solution to the setup
C. Adding substrate but not enzyme
D. Placing the substrate and enzyme in a container with double the capacity
13. You observe a species that gives birth to only one offspring at a time and has a relatively long life-span for its body size. Which of the following is probably also true of this organism?
A. It lives in a newly colonized habitat.
B. It is an aquatic organism.
C. It requires relatively high parental care of offspring.
D. The age at which the offspring themselves can give birth is relatively young.
14. In a certain species of plant, the allele to produce green melons (G) is dominant over the allele to produce yellow melons (g). A student performed a cross between a plant that produced green melons and a plant that produced yellow melons. When the student observed the next generation, the 94 seeds that were produced from the cross matured into 53 plants with green melons and 41 plants with yellow melons. Calculate the chi-squared value for the null hypothesis that the green-melon parent was heterozygous for the melon-color gene.
A. 0.76
B. 1.53
C. 0.50
D. 1.26
15. In a certain population of squirrels that is in Hardy-Weinberg equilibrium, black color is a recessive phenotype present in 9 percent of the squirrels, and 91 percent are gray. What percentage of the population is homozygous dominant for this trait?
A. 21 percent
B. 30 percent
C. 49 percent
D. 70 percent
16. Refer to question 15 for details on the squirrel population. Which of the following conditions is required to keep this population in Hardy- Weinberg equilibrium?
A. Random mating
B. Genetic drift
C. Mutation
D. Gene flow
17. A reaction that includes energy as one of its reactants is called a(n)
A. exergonic reaction.
B. hydrolysis reaction.
C. endergonic reaction.
D. redox reaction.
18. To which of the following labeled trophic levels would an herbivore most likely be assigned?
A. A
B. B
C. C
D. D
19. A population undergoes a shift in which those who are really tall and those who are really short decrease in relative frequency compared to those of medium size, due to a change in the availability of resources. This is an example of
A. directional selection.
B. stabilizing selection.
C. disruptive selection.
D. sympatric speciation.
20. Which of the following statements is correct?
A. Water flows from hypertonic to hypotonic.
B. Germinating seeds use less oxygen than do nongerminating seeds.
C. The rate of transpiration decreases with an increase in air movement.
D. Smaller DNA fragments migrate more rapidly than do larger DNA fragments on gel electrophoresis.
21. Which of the following is not a form of interspecies interaction?
A. Commensalism
B. Succession
C. Mutualism
D. Parasitism
22. Sickle cell anemia is a disease caused by the substitution of an incorrect nucleotide into the DNA sequence for a particular gene. The amino acids are still added to the growing protein chain, but the symptoms of sickle cell anemia result. This is an example of a
A. frameshift mutation.
B. missense mutation.
C. nonsense mutation.
D. thymine dimer mutation.
23. In a population of grasshoppers, the allele for tan color is dominant to the allele for green color. A drastic increase in rainfall leads to selection against the tan phenotype. When the rainy season ends, 23 percent of the remaining grasshoppers have the green phenotype. If this population is now in Hardy-Weinberg equilibrium, what will the frequency of the tan allele be in the next generation?
A. 0.52
B. 0.48
C. 0.23
D. 0.071
24. This process couples the production of ATP with the movement of electrons down the electron transport chain by harnessing the driving force created by a proton gradient.
A. Glycolysis
B. Chemiosmosis
C. Fermentation
D. Calvin cycle
25. During prophase 1 of meiosis, homologous chromosomes come together during a synapsis to form a tetrad. What does this lead to?
A. Leads to genetic variation via crossing over
B. Ensures that four daughter cells are produced
C. Creates male and female gametes
D. Separation of sister chromatids
26. The bacteria that cause pimples can be grown in the lab using a suitable nutrient broth, where they will eventually achieve exponential growth. Using the graph that follows, calculate the mean rate of growth, in millions of bacteria per hour, during their exponential phase.
A. 1.46
B. 1.27
C. 0.83
D. 0.89
27. In a large pond consisting of long-finned fish and short-finned fish, a dam is built, splitting the large pond into two separate ponds. Fifty percent of the fish population is randomly separated into each of the smaller ponds leading to two distinct populations of long-finned and short-finned fish. This is an example of:
A. Gene flow
B. Bottleneck
C. Founder effect
D. Allopatric speciation
28. What is the water potation for a solution that is 0.1 M at 20°C? The solution is an open container of
A. -2.4.
B. 2.4.
C. -1.4.
D. 1.4.
For questions 29—30, please refer to the following answers:
29. The specific order of which of the above -molecules is responsible for the primary structure of proteins?
30. Which molecule is the backbone of a structure that is vital to the construction of many cells and is used to produce steroid hormones?
31. A cell, shaped like a cube, contains sides with a measurement of 2.5 cm. Determine the surface- area-to-volume ratio for the cell.
A. 15.6
B. 2.4
C. 37.5
D. 5.2
Questions 32—34: A behavioral endocrinologist captures male individuals of a territorial bird species over the course of a year to measure testosterone (T) levels. In this population, males may play one of two roles: (1) they may stay in their natal group (the group they were born in) and help raise their younger siblings, or (2) they may leave the natal group to establish a new territory. Use this information and the two histograms that follow to answer the following questions.
32. Testosterone level in this population may be an example of
A. adaptive radiation.
B. an adaptation.
C. divergent selection.
D. development.
33. What can you infer about the role of testosterone in reproduction in this species?
A. It is detrimental to breeding.
B. It aids adult males only.
C. It ensures that all males reproduce equally.
D. It aids in breeding.
34. Which of the following is the best explanation of the results presented in the preceding graph, collected from the same population in a different year?
A. The so-called helper males are actually breeding.
B. The population has stopped growing.
C. Females are equally attracted to adult and helper males.
D. Testosterone level is affected by many processes.
Questions 35—38: A researcher grows a population of ferns in her laboratory. She notices, after a few generations, a new variant that has a distinct phenotype. When she tries to breed the original phenotype with the new one, no offspring are produced. When she breeds the new variants, however, offspring that look like the new variant result.
35. What originally caused the change in the variant?
A. Karyotyping
B. Balance polymorphism
C. Mutation
D. Polyploidy
36. What kind of speciation does this example illustrate?
A. Allopatric
B. Sympatric
C. Isolated
D. Polyploidy
37. Which of the following could possibly characterize the new variant?
A. Adaptive radiation
B. Divergent selection
C. Equilibrium
D. Polyploidy
38. Which of the following is likely to exhibit the process described earlier?
A. Fallow deer
B. Fruit flies
C. Grass
D. Spotted toads
For questions 39—41, please refer to the following figure:
39. The DNA placed in this electrophoresis gel separates as a result of what characteristic?
A. pH
B. Charge
C. Size
D. Polarity
40. If this gel were used in a court case as DNA evidence taken from the crime scene, which of the following suspects appears to be guilty?
A. Suspect A
B. Suspect B
C. Suspect C
D. Suspect D
41. Which two suspects, while not guilty, could possibly be identical twins?
A. A and B
B. A and C
C. B and C
D. B and D
Questions 42—45: The frequency of genotypes for a given trait are given in the accompanying graph. Answer the following questions using this information:
42. What is the frequency of the recessive homo-zygote?
A. 15 percent
B. 19 percent
C. 25 percent
D. 40 percent
43. What would be the approximate frequency of the heterozygote condition if this population were in Hardy-Weinberg equilibrium?
A. 20 percent
B. 45 percent
C. 48 percent
D. 72 percent
44. Is this population in Hardy-Weinberg equilibrium?
A. Yes
B. No
C. Cannot tell from the information given
D. Maybe, if individuals are migrating
45. Which of the following processes may be occurring in this population, given the allele frequencies?
A. Directional selection
B. Homozygous advantage
C. Hybrid vigor
D. Allopatric speciation
Questions 46—48: An eager AP Biology student interested in studying osmosis and the movement of water in solutions took a dialysis bag containing a 0.5 M solution and placed it into a beaker containing a 0.6 M solution.
46. After the bag has been sitting in the beaker for a while, what would you expect to have happened to the bag?
A. There will have been a net flow of water out of the bag, causing it to decrease in size.
B. There will have been a net flow of water into the bag, causing it to swell in size.
C. The bag will be the exact same size because no water will have moved at all.
D. The solute will have moved out of the dialysis bag into the beaker.
47. If this bag were instead placed into a beaker of distilled water, what would be the expected result?
A. There will be a net flow of water out of the bag, causing it to decrease in size.
B. There will be a net flow of water into the bag, causing it to swell in size.
C. The bag will remain the exact same size because no water will move at all.
D. The solute will flow out of the dialysis bag into the beaker.
48. Which of the following is true about water potential?
A. It drives the movement of water from a region of lower water potential to a region of higher water potential.
B. Solute potential is the only factor that determines the water potential.
C. Pressure potential combines with solute potential to determine the water potential.
D. Water potential always drives water from an area of lower pressure potential to an area of higher pressure potential.
Questions 49—51 all use the following pedigree, but are independent of each other:
49. If the pedigree is studying an autosomal recessive condition for which the alleles are A and a, what was the probability that a child produced by parents A and B would be hetero-zygous?
A. 0.0625
B. 0.1250
C. 0.2500
D. 0.5000
50. Imagine that a couple (C and D) goes to a genetic counselor because they are interested in having children. They tell the counselor that they have a family history of a certain disorder and they want to know the probability of their firstborn having this condition. What is the probability of the child having the autosomal recessive condition?
A. 0.0625
B. 0.1250
C. 0.2500
D. 0.3333
51. Imagine that a couple (C and D) has a child (E) that has the autosomal recessive condition being traced by the pedigree. What is the probability that their second child (F) will have the autosomal recessive condition?
A. 0.0625
B. 0.1250
C. 0.2500
D. 0.5000
For questions 52—53, please refer to the following diagram:
52. The bold line that point C intersects is known as the
A. biotic potential.
B. carrying capacity.
C. limiting factor.
D. maximum attainable population.
53. On the basis of what happens at the end of this chart, what is the most likely explanation for the population decline after point E?
A. The population became too dense and it had to decline.
B. There was a major environmental shift that made survival impossible for many.
C. Food became scarce, leading to a major famine.
D. The population had become too large.
Questions 54 and 55: The solutions in the two arms of this U-tube are separated by a membrane that is permeable to water and sodium chloride, but not to sucrose. Side A is filled with a solution of 0.6 M sucrose and 0.2 M sodium chloride (NaCl), and side B is filled with a solution of 0.2 M sucrose and 0.3 M NaCl. Initially, the volume on both sides is the same.
54. At the beginning of the experiment,
A. side A is hypertonic to side B.
B. side A is hypotonic to side B.
C. side A is isotonic to side B.
D. side A is hypotonic to side B with respect to sucrose.
55. If you examine side A after a couple of days, you will see
A. an increase in the concentration of NaCl and sucrose and an increase in water level.
B. a decrease in the concentration of NaCl, an increase in water level, and no change in the concentration of sucrose.
C. no net change.
D. an increase in the concentration of NaCl and an increase in the water level.
56. Tay-Sachs is a disease caused by a recessive allele. Individuals with the genetic disorder usually do not survive more than a few years, and thus are not able to reproduce and pass on the gene. What would explain how this allele and its associated disease is preserved in the population?
A. Heterozygous individuals will show no symptoms.
B. Spontaneous mutation converts the dominant allele to the recessive form.
C. Occasionally the gene will instead increase the fitness of the individual.
D. Tay-Sachs is asymptomatic in people who are homozygous recessive.
57. A new plant was discovered and determined to have an unusually low number of stomata on the undersides of its leaves. For what environment would this plant most likely be best adapted?
A. Cold and rainy
B. Humid and sunny
C. Hot and humid
D. Hot and dry
58. The first simple cells evolved approximately 3.5 billion years ago, followed by the more complex eukaryotic cells 2.1 billion years ago. Which of the following statements is correct?
A. Eukaryotic organelles helped create separate environments for metabolic reactions, thus increasing their metabolic efficiency.
B. Prokaryotic and eukaryotic cells have no structural similarities.
C. The organelles in larger eukaryotic cells took care of the problems of having a larger surface-area-to-volume ratio.
D. Eukaryotic cells are able to multiply faster based on their more complex structure.
59. Easily produced genetic variation is key to the rapid evolution of viral and microbial populations. Furthermore, pathogens that need to escape the immune system rely on this variation to generate new surface antigens that go unrecognized by the host’s immune system. Which of the following is an example of this antigenic variation?
A. HIV, which can remain integrated into the host genome for many years
B. The flu virus, which changes its envelope proteins
C. MRSA, which has become resistant to many antibiotics
D. Multiple sclerosis, which attacks the cells of the nervous system
60. Two species of hamster (X and Y) are in the genus Cricetulus, whereas a third species (Z) is instead part of genus Mesocricetus. Which of the following phylogenetic trees shows the correct evolutionary relatedness?
AP Biology Practice Exam 1: Section II
FREE-RESPONSE QUESTIONS
Time—1 hour and 30 minutes
(The first 10 minutes is a reading period. Do not begin writing until the 10-minute period has passed.)Questions 1 and 2 are long free-response questions that should require about 20 minutes each. Questions 3—6 are short-response questions that should require approximately 8—10 minutes each.Outline form is not acceptable. Answers should be in essay form.
1. A murder trial court case ended up ruling against the defendant because of DNA evidence found at the crime scene and analyzed in the forensics lab. The following gel was produced after collecting DNA from the crime scene with A = Suspect 1, B = Suspect 2, and C = Murder.
A. Describe how the biotechnology in gel electrophoresis works.
B. Identify the independent and dependent variables used in the gel electrophoresis investigation.
C. Explain who is guilty of the crime? Justify your response.
D. When the forensics lab received the DNA evidence from the crime scene, they noticed that it was not enough to run multiple tests on. Predict what biotechnology the scientist would use in order to ensure they have enough DNA to work with.
2. An experiment was conducted using yeast and hydrogen peroxide to determine the effect of changing environmental conditions on the reaction rate of the catalytic enzyme in yeast (catalase) and hydrogen peroxide.
H2O2 + Catalase → H2O + O2
A. Describe and explain the interaction between the catalase in yeast and hydrogen peroxide in the investigation.
B. Construct an appropriated graph of the data collected in the investigation.
C. Identify and explain the impact that the changing environmental conditions had on the reaction rate of the catalase and hydrogen peroxide.
D. Make a prediction on what would happen to the reaction rate if the catalase-hydrogen peroxide reaction was exposed to a pH of 14? Justify your prediction.
3. Life on Earth is made possible because of certain unique characteristics of water.
A. Describe the characteristics of a water molecule that lead to hydrogen bonding.
B. Describe and explain how the interaction of water molecules results in cohesion, adhesion, and surface tension.
4. Evolution is the change in allele frequencies in a population over time. This can occur through a variety of mechanisms including natural selection, genetic drift, and mutation.
A. Describe the impact on a population for the mechanisms listed.
B. A scientist is studying a population of field mice that includes individuals with light and dark brown coats. Every six months, you perform capture/recapture experiments to census the proportion of light and dark individuals. The following numbers indicate the percentage of dark-coat individuals caught in each successive census over the course of five years: 96, 94, 95, 91, 93, 95, 74, 73, 77, 76. Explain which of the three processes of evolution discussed in part a is most consistent with the data that supports the changes in phenotypic frequencies in the mouse population.
5. Humans maintain homeostasis in regard to blood glucose levels via the actions of insulin and glucagon. The figure below shows the pathway in detail.
A. Describe how insulin and glucagon functions in the homeostasis.
B. Explain the signal transduction pathway that takes places in body cells of humans in terms of blood glucose levels.
C. Predict what occurs if there is a change in the pathway. Justify your prediction.
Figure is for Question 5 (Reproduced with permission from Raven P, Johnson G, Mason K, Losos J, Duncan T; Biology, 12th ed. New York: McGraw Hill; 2020)
6. An herbivorous grasshopper feeds on a leaf of an autotrophic plant to maintain homeostasis. For all of the energy that the grasshopper consumes 17 percent is for growth, 33 percent for cellular respiration, and 50 percent as feces.
A. Describe the allocation of energy by the grasshopper for homeostasis. Explain why all the energy consumed by the grasshopper is not available for the next trophic level.
B. A long-term drought causes a decrease in the number of autotrophic plants in the area that the grasshopper feeds. Make a prediction to how the grasshopper energy allocation would be different. Justify your prediction.
Figure is for Question 6 (Reproduced with permission from Raven P, Johnson G, Mason K, Losos J, Duncan T; Biology, 12th ed. New York: McGraw Hill; 2020)
Answers and Explanations for AP Biology Practice Exam 1
MULTIPLE-CHOICE QUESTIONS
1. D—Cell walls are present in prokaryotes but not eukaryotic animal cells. Ribosomes and cell membranes are present in both of them. Chloroplasts and large central vacuoles are not seen in either of them. Animal cells have small vacuoles.
2. A—The oxygen released by plants is produced during the light reactions of photo-synthesis. The main inputs to the light reactions are water and light. Water is the source of the oxygen.
3. B—Endotherms are organisms whose metabolic rates do not respond to shifts in environmental temperature.
4. B—A frameshift mutation is one in which the reading frame for the protein construction machinery is shifted. It is a deletion or addition of nucleotides in a number that is not a multiple of 3. Often this can lead to premature stop codons, which lead to nonfunctional proteins.
5. C
6. C—We can see from the data that m and f have the highest crossover frequency. They must therefore be farthest apart of any pair along the chromosome. This leaves only answer choice C.
7. C—Memory B cells are able to recognize foreign invaders if they come back into our systems and lead to a more rapid and efficient attack on the invader.
8. C—Polygenic traits are traits that require the input of multiple genes to determine the phenotype. Skin color is a classic example of a polygenic trait; three genes combine to provide the various shades of skin tone seen in humans.
9. C—This is a classic example of Batesian mimicry.
10. D—Crossover is most likely to occur between two genes that are located far away from each other on the same chromosome.
11. C—Meiosis reduces the number of chromosomes in an individual by half: 96 ÷ 2 = 48.
12. D—The volume of the container is not a major factor that affects enzyme efficiency.
13. C—The original question describes an organism that can be classified as a K-selected population. Individuals of this class tend to have fairly constant size, low reproductive rates, and offspring that require extensive care.
14. B—If the green-melon parents were Gg, you would expect a cross with a yellow-melon plant (gg) to produce 50 percent Gg and 50 percent gg offspring. What you actually observed was 53 green and 41 yellow. Based on a total number of 94 offspring, your expected half-and-half ratios would be 47 of each color.
The chi-square value is 1.53 (less than the critical value of 3.84), so the null hypothesis is accepted.
15. C—If 9 percent of the population is homozygous recessive, this means that q2 = 0.09, and that the square root of q2 = 0.30 = q. This means that p = 0.70 since p + q = 1. Thus, the percentage of the population that is homozygous dominant: p2 = (0.7)2 = 0.49 or 49 percent.
16. A—All the other answer choices are violations of the Hardy-Weinberg equilibrium.
17. C—Exergonic reactions give off energy, and hydrolysis reactions are reactions that use water to break apart a compound. Redox reactions are reactions that involve the movement of electrons.
18. D—Herbivores tend to be the primary consumers of trophic pyramids, and thus would take up the first level up from the bottom.
19. B—Stabilizing selection tends to eliminate the extremes of a population, directional selection is a shift toward one of the extremes, and disruptive selection is the camel-hump selection in which the two extremes are favored over the middle. Sympatric speciation is the formation of new species due to an inability to reproduce that is not caused by geographic separation.
20. D—This is a lab experiment question based on the material in Chapter 13. We threw it in here just to remind you that you should not ignore the concepts of this very important chapter. You will be asked about these concepts on the exam.
21. B—Succession is an ecological process in which landforms evolve over time in response to the environmental conditions. Commensalism is when one organism benefits while the other is unaffected. Mutualism is when both organisms reap benefits from the interaction. Parasitism is when one organism benefits at the other’s expense.
22. B
23. A
tan = p; green = q
green phenotype = q2 = 0.23; frequency of green allele = √0.23 = 0.48
Since p + q = 1, the tan allele (p) = 1 - 0.48 = 0.52
24. B
25. A—Crossing over occurs when the homologous pairs match up during prophase I of meiosis. Complementary pieces from the two homologous chromosomes wrap around each other and are exchanged between the chromosomes leading to genetic variation in a population.
26. B—Logarithmic growth takes place during the time where the slope is the greatest, approximately between 12 and 30 hours. During that time (18 hours), the bacterial population started at 10 million and increased to 33 million (a difference of 23 million). Therefore, 23 million divided by 18 hours gives a rate of 1.27 million bacteria per hour.
27. C—Founder effect takes place when a new small population is created from members of a larger population resulting in reduced genetic variation for the new population.
28. A—The solute potential is —(1) × (0.1 M) × (0.00831 MPa/mole K) × (293 K) = —0.24 MPa. The pressure potential is zero because the solution is in an open container. Therefore, (—0.24) + 0 = —0.24 MPa.
29. B—The primary structure of a protein consists of a specific order of amino acids held together by peptide bonds.
30. A—Cholesterol is one of the lipids that serves as the starting point for the synthesis of sex hormones.
31. B
SA = 6 × (2.5 cm × 2.5 cm) = 37.5 cm2
V = (2.5 cm)3 = 15.6 cm3
SA/V = 37.5/15.6 = 2.4
32. B—Testosterone level is an adaptive trait in this population, one that has been molded by natural selection (or possibly sexual selection; we cannot determine this from the question) to aid in reproduction. Adaptive radiation is a process by which many speciation events occur in a newly exploited environment and does not apply here. This is not an example of divergent selection because both breeding and nonbreeding males have low testosterone levels during at least one part of the year; if the two male types always differed in testosterone level, this population could eventually split into two populations. Development and sperm production may be related to testosterone but are not addressed in this experiment.
33. D—Since testosterone levels are increased only during the breeding season, we can infer that testosterone has some role in breeding. Since reproductive males express higher testosterone levels only during the breeding season, we hypothesize that testosterone is beneficial, as opposed to detrimental, to breeding.
34. A—Since testosterone seems to be linked with reproduction, we infer from the new data that the “nonbreeding” males are actually breeding and therefore have elevated testosterone levels. Females, population growth, and number of offspring produced are not considered in this example. Finally, although testosterone does affect many physiological processes, none of these are discussed or illustrated in this example.
35. C—Although several processes can affect the frequency of a new phenotype or genotype, once it is in place, the original genetic change must have been the result of a mutation (probably a chromosomal aberration).
36. B—No physical barrier separated the two populations; this is therefore an example of sympatric, not allopatric speciation. The other answer choices are not types of speciation.
37. D—Polyploidy is the only answer that can describe an individual. All the others are processes or states that describe population events. Polyploidy is the duplication of whole chromosomes that leads to speciation because the new variety can no longer breed with the original.
38. C—Polyploidy is much more common in plants; mutations such as the duplication of whole chromosomes are usually lethal to animals.
39. C—Gel electrophoresis separates DNA on the basis of size. Smaller samples travel a greater distance down the gel compared to larger samples.
40. B—His DNA fingerprint seems to exactly match that of the evidence DNA sample.
41. B—A and C seem to share the exact same restriction fragment cut of their DNA. Perhaps they are messing with our heads and added the DNA from the same individual twice.
42. B—100 — 45 — 36 = 19 percent.
43. C—Thirty-six percent of the population is AA. Taking the square root of 0.36, we find the frequency of the A allele to be 0.6. This means that the a allele’s frequency must be 1 — 0.6, or 0.4. From these numbers, we can calculate the expected Hardy-Weinberg heterozygous frequency is 2pq = 2(A)(a) = 2(0.6)(0.4) = 48.0, or 48 percent.
44. B—The expected heterozygous probability does not match up with the actual. This population is not in Hardy-Weinberg equilibrium.
45. B—The homozygous frequency is higher than expected; one explanation for this is that the homozygotes are being selected for.
46. A—Water will flow out of the bag because the solute concentration of the beaker is hypertonic compared to the dialysis bag. Osmosis passively drives water from a hypotonic region to a hypertonic region.
47. B—Water would now flow into the bag because the solute gradient has been reversed. Now the beaker is hypotonic compared to the dialysis bag. Water thus moves into the bag.
48. C—Water potential = pressure potential + solute potential. Water passively moves from regions with high water potential toward those with lower water potential.
49. D—The mother (person B) must be heterozygous Aa because she and her husband (aa) have produced children that have the double recessive condition. This means that person B (the mother) must have contributed an a and that the cross is Aa × aa—and the probability is ½.
50. D—To answer this question, we must first determine the probability that person D is heterozygous. We know she is not aa because she does not have the condition. Since we know that the father has the condition, we know for certain that his genotype is aa. Both of mother D’s parents must be heterozygous since neither of them have the condition, but they have produced a child with the condition. The probability that mother D is heterozygous Aa is ⅔. The probability that a couple with the genotypes Aa × aa have a double recessive child is ½. The probability that these two will have a child with the condition is ½ × ⅔ = ⅓ = 0.3333.
51. D—If the couple has a child (person E) with the recessive condition, then we know for certain that mother D must be heterozygous. It is definitely an aa × Aa cross, leaving a 50 percent chance that their child will be aa.
52. B
53. B
54. A—The total solute potential for side A is 1.0 MPa (remember that for NaCl, i = 2), and the total solute potential for side B is 0.8 MPa. Therefore, side A has a higher concentration of solute (hypertonic).
55. D—Water will move from a hypotonic solution (side B) toward a hypertonic solution (side A). Sodium will diffuse from a region of more sodium (side B) to a region of less sodium (side A).
56. A—Heterozygous individuals carry the recessive gene but are themselves healthy.
57. D—Low numbers of stomata help to reduce water loss, helpful in hot and dry regions.
58. A—Prokaryotic and eukaryotic cells do have similar structures, the organelles in eukaryotic cells took care of having a smaller surface-area-to-volume ratio, and eukaryotic cells are not able to multiply faster.
59. B—Changing envelope proteins are created because of genetic variation in the genes that code for these proteins.
60. D—This cladogram shows a closer relationship between X and Y.
Free-Response Grading Outline
1.
A. Describe how the biotechnology in gel electrophoresis works. (maximum 4 points)
✵ Mentioning that smaller particles travel faster. (1 point)
✵ Mentioning that the fragments of DNA are placed into wells at the head of the gel to begin their migration to the other side. (1 point)
✵ Mentioning that the DNA migrates only as electric current is passed through the gel. (1 point)
✵ Mentioning that the DNA migrates from negative charge to positive charge. (1 point)
✵ Mentioning that when DNA samples from different individuals are cut with restriction enzymes, they show variations in the band patterns on gel electrophoresis known as restriction fragment length polymorphisms (RFLPs). (1 point)
✵ Mentioning that DNA is specific to each individual, and when it is mixed with restriction enzymes, different combinations of RFLPs will be obtained from person to person. (1 point)
✵ Providing a definition of a DNA fingerprint as the combination of an individual’s RFLPs inherited from each parent. (1 point)
✵ Mentioning that if an individual’s electrophoresis pattern identically matches that of the crime scene evidence, the DNA has spoken and shown the individual to be the perpetrator, since the probability of two people having an identical set of RFLPs is virtually non-existent. (1 point)
B. Identify the independent and dependent variables used in the gel electrophoresis investigation.
✵ Independent variable = Samples of DNA (1 point)
✵ Dependent variable = Number of base pairs based on bands in gel (1 point)
C. Explain who is guilty of the crime. Justify your response.
✵ Suspect 2 committed the crime. (1 point)
✵ Sample A (Suspect 1) and Sample C (Murder) contain the same bands on the gel. This confirms that Suspect 1 is the murder and Suspect 2 is innocent due to the fact that two people having identical sets of RFLPs is virtually non-existent. (1 point)
D. When the forensics lab received the DNA evidence from the crime scene, they noticed that it was not enough to run multiple tests on. Predict what biotechnology the scientist would use in order to ensure they have enough DNA to work with.
✵ Scientist would use PCR, polymerase chain reaction, to make copies of the DNA. PCR is a technology that uses original DNA sequences to amplify the original DNA sequences. (1 point)
2.
A. Describe and explain the interaction between the catalase in yeast and hydrogen peroxide in the investigation.
✵ The enzyme (catalase)—substrate (hydrogen peroxide) interaction is specific between the catalase and hydrogen -peroxide. (1 point)
✵ The enzyme’s active site has a unique confirmation (shape) that only the specific substrate is able to fit into like a lock and key. (1 point)
B. Construct an appropriated graph of the data collected in the investigation.
✵ Independent and dependent variable correctly plotted (1 point)
✵ Correctly labeled (1 point)
✵ Correctly scaled axis (1 point)
C. Identify and explain the impact that the changing environmental conditions had on the reaction rate of the catalase and hydrogen peroxide.
✵ Enzyme 2 functions most efficiently at 20°C, Enzyme 1 at 35°C, and Enzyme 3 at 10°C. (1 point)
✵ Enzyme 3 functions most efficiently at a pH of 6, Enzyme 1 and 2 at pH of 2. (1 point)
✵ Enzymes have optimum temperatures and pH that they function at. Enzymes are proteins which have a unique confirmation created by interactions of their R groups. These interactions are interrupted when their temperature or pH is outside their optimum range, causing them to denature. (1 point)
D. Make a prediction on what would happen to the reaction rate if the catalase hydrogen peroxide reaction was exposed to a pH of 14. Justify your prediction.
✵ The three enzymes would denature in a pH of 14. (1 point)
✵ The protein structure of each enzyme would denature, causing them to lose their shape and function. (1 point)
3.
A. Describe the characteristics of a water molecule that lead to hydrogen bonding.
✵ Water is a covalently bonded molecule. Each hydrogen molecule is covalently bonded to the oxygen molecule. (1 point)
✵ Water is a polar molecule. (1 point)
✵ The oxygen atom in water is slightly negatively charged and the hydrogen atoms are slightly positively charged due to the electrons from the hydrogen atoms being closer to the oxygen atom due to increased attraction. (1 point)
B. Describe and explain how the interaction of water molecules results in cohesion, adhesion, and surface tension.
✵ Water molecules form hydrogen bonds, which is an attraction between the positively charged hydrogen atoms of one water molecule and a nearby negatively charged oxygen atom of a water molecule. (1 point)
✵ Cohesion is when two water molecules are attracted to each other via hydrogen bonds. Adhesion is when a water molecule is attracted to a different molecule. Surface tension is when water molecules interact with each other instead of the air molecules, causing a strong tension in the water molecules at the surface of a body of water. (1 point)
4.
A. Describe the impact on a population for the mechanisms listed.
Definition and examples (maximum 2 points)
✵ Defining natural selection. (1 point)
a. Mentioning it is the process by which certain alleles increase in frequency in a population because of the survival or reproduction benefit they give to those individuals who possess them. (½ point)
b. Possible example: sickle cell allele persists in populations where malaria is present (having sickle-shaped red blood cells makes you less likely to contract malaria). (½ point)
c. Defining genetic drift. (maximum 1 point)
d. Describing how random processes can change allele frequencies. (½ point)
e. Possible example: allele frequencies in a new population are dependent on which alleles are present in the founders of that population (founder effect). (½ point)
✵ Defining mutation. (1 point)
a. Mentioning that changes in DNA create genetic variation and new alleles. (½ point)
b. Mentioning that evolution by “neutral mutations” can occur even if the new alleles are not acted on by natural selection. (½ point)
c. Possible example: eye color gene mutates to a different color without any change in vision or behavior as a result of the mutation. (½ point)
B. A scientist is studying a population of field mice that includes individuals with light and dark brown coats. Every six months you perform capture/recapture experiments to census the proportion of light and dark individuals. The following numbers indicate the percentage of dark-coat individuals caught in each successive census over the course of five years: 96, 94, 95, 91, 93, 95, 74, 73, 77, 76. Explain which of the three processes of evolution discussed in part a is most consistent with the data that supports the changes in phenotypic frequencies in the mouse population.
Explanation of data (maximum 2 points)
✵ Mentioning the changes could not be the gradual process of natural selection because they occurred rapidly between two censuses. (1 point)
✵ Indicating that the changes could be caused by genetic drift. (1 point)
✵ Indicating that the changes could be caused by some environmental event (flood, fire) that randomly killed many dark-coated mice. (1 point)
5.
A. Describe how insulin and glucagon functions in the homeostasis.
✵ Insulin and glucagon function as ligands or signals for blood sugar. (1 point)
B. Explain the signal transduction pathway that takes place in body cells of humans in terms of blood glucose levels.
✵ Insulin and glucagon are hormones produced in the pancreas that work to control the level of glucose in the blood. Insulin is released from the pancreatic beta cells when glucose levels are higher in the blood—insulin stimulates cellular uptake of glucose for utilization or glycogen production. On the other hand, glucagon is released from the pancreatic alpha cells when glucose levels are lower in the blood. Glucagon stimulates the release of glycogen from the liver to raise glucose levels in the blood. The opposing effects of insulin and glucagon maintain the homeostasis of blood glucose. (2 points)
C. Predict what occurs if there is a change in the pathway. Justify your prediction.
✵ Diabetes occurs when the insulin/glucagon pathway is interrupted, causing the body’s cells to not receive the correct amount of blood sugar. (1 point)
6.
A. Describe the allocation of energy by the grasshopper for homeostasis. Explain why all the energy consumed by the grasshopper is not available for the next trophic level.
✵ The grasshopper is using 100 percent of the energy it consumes to survive with 33 percent being used for cellular respiration in order to maintain homeostasis. Fifty percent of the energy is lost through waste in the form feces and heat. Only 17 percent is left over for the grasshopper to grow in terms of muscle, bones, and other tissue. (1 point)
✵ The grasshopper must first maintain homeostasis, which includes regulating internal and external processes. This takes an average up to 90 percent of all energy consumed by the grasshopper through cellular respiration and waste. Thus, on average, only 10 percent is left over for the grasshopper to grow, which is the energy available for the next trophic level to consume. (1 point)
B. A long-term drought causes a decrease in the number of autotrophic plants in the area that the grasshopper feeds. Make a prediction to how the grasshopper energy allocation would be different. Justify your prediction.
✵ The grasshopper would have to spend more energy looking for food. (1 point)
✵ The increased expenditure of energy looking for food would cause less energy to be available for growth, leading to less energy for the next trophic level. (1 point)
Scoring and Interpretation
AP BIOLOGY PRACTICE EXAM 1
AP BIOLOGY
SCORE CONVERSION CHART
Answer Sheet for AP Biology Practice Exam 2
ANSWER SHEET FOR MULTIPLE-CHOICE QUESTIONS
AP Biology Practice Exam 2: Section I
MULTIPLE-CHOICE QUESTIONS
Time—1 hour and 30 minutes
For the multiple-choice questions to follow, select the best answer and fill in the appropriate letter on the answer sheet.
1. A baby duck runs for cover when a large object is tossed over its head. After this object is repeatedly passed overhead, the duck learns there is no danger and stops running for cover when the same object appears again. This is an example of
A. imprinting.
B. fixed-action pattern.
C. agonistic behavior.
D. habituation.
2. In a population of giraffes, an environmental change occurs that favors individuals that are tallest. As a result, more of the taller individuals are able to obtain nutrients and survive to pass along their genetic information. This is an example of
A. directional selection.
B. stabilizing selection.
C. sexual selection.
D. disruptive selection.
3. The relatives of a group of pelicans from the same species that separated from each other because of an unsuccessful migration are reunited 150 years later and find that they are unable to produce offspring. This is an example of
A. allopatric speciation.
B. sympatric speciation.
C. genetic drift.
D. gene flow.
4. A cell is placed into a hypertonic environment and its cytoplasm shrivels up. This demonstrates the principle of
A. diffusion.
B. active transport.
C. facilitated diffusion.
D. plasmolysis.
5. Which of the following is a biotic factor that could affect the growth rate of a population?
A. Volcanic eruption
B. Glacier melting
C. Destruction of the ozone layer
D. Sudden reduction in the animal food resource
6. Which of the following is not a way to form recombinant DNA?
A. Translation
B. Conjugation
C. Specialized transduction
D. Transformation
7. Chemiosmosis occurs in
I. Mitochondria
II. Nuclei
III. Chloroplasts
A. I only
B. II only
C. III only
D. I and III
8. Which of the following theories is based on the notion that mitochondria and chloroplasts evolved from prokaryotic cells?
A. Fluid mosaic model
B. Endosymbiotic model
C. Taxonomic model
D. Respiration feedback model
9. Which of the following is not known to be involved in the control of cell division?
A. Cyclins
B. Protein kinases
C. Checkpoints
D. Fibroblast cells
10. Which of the following statements about posttranscriptional modification is incorrect?
A. A poly-A tail is added to the 3′ end of the mRNA.
B. A guanine cap is added to the 5′ end of the mRNA.
C. Introns are removed from the mRNA.
D. Posttranscriptional modification occurs in the cytoplasm.
11. In a certain pond, there are long-finned fish and short-finned fish. A horrific summer thunderstorm leads to the death of a disproportionate number of long-finned fish to the point where the relative frequency of the two forms has drastically shifted. This is an example of
A. gene flow.
B. natural selection.
C. genetic drift.
D. stabilizing selection.
12. During the central dogma, transcription occurs in the nucleus of eukaryotic cells by transcribing the cells’ DNA. Located thousands of base pairs away from the promoter region is the transfer-affecting region on the DNA. What is that region called?
A. Enhancer
B. Repressor
C. Operator
D. Promoter
13. Which of the following statements about photo-synthesis is incorrect?
A. H2O is an input to the light-dependent reactions.
B. CO2 is an input to the Calvin cycle.
C. Photosystems I and II both play a role in the cyclic light reactions.
D. O2 is a product of the light-dependent reactions.
14. If a couple has had three sons and the woman is pregnant with their fourth child, what is the probability that child 4 will also be male?
A. ½
B. ¼
C. ⅛
D. 1⁄16
15. Which of the following is an incorrect statement about gel electrophoresis?
A. DNA migrates from positive charge to negative charge.
B. Smaller DNA travels faster.
C. The DNA migrates only when the current is running.
D. The longer the current is running, the farther the DNA will travel.
16. You are told that in a population of guinea pigs, 4 percent are black (recessive) and 96 percent are brown. Which of the following is the frequency of the heterozygous condition?
A. 16 percent
B. 32 percent
C. 40 percent
D. 48 percent
17. Which of the following is known to be involved in the photoperiodic flowering response of angiosperms?
A. Auxin
B. Cytochrome
C. Phytochrome
D. Gibberellins
18. Which of the following tends to be highest on the trophic pyramid?
A. Primary consumers
B. Herbivores
C. Primary carnivores
D. Primary producers
19. A form of species interaction in which one of the species benefits while the other is unaffected is called
A. parasitism.
B. mutualism.
C. commensalism.
D. symbiosis.
20. The transfer of DNA between two bacterial cells connected by sex pili is known as
A. specialized transduction.
B. conjugation.
C. transformation.
D. generalized transduction.
For questions 21—22, please use the preceding diagram:
21. If inhibitor 1 is able to bind to the active site and block the attachment of the substrate to the enzyme, this is an example of
A. noncompetitive inhibition.
B. competitive inhibition.
C. a cofactor.
D. a coenzyme.
22. Which of the following is not a change that would affect the efficiency of the enzyme shown above?
A. Change in temperature
B. Change in pH
C. Change in salinity
D. Increase in the concentration of the enzyme
23. Which of the following points on the preceding energy chart represents the activation energy of the reaction involving the enzyme?
A. A
B. B
C. C
D. D
24. When an animal is harmless but has a similar appearance of a more dangerous animal, that animal is exhibiting what type of defense mechanism?
A. Aposomatic coloration
B. Batesian mimicry
C. Deceptive markings
D. Cryptic coloration
25. Twenty people decide to start a new population, totally isolated from anyone else. Two of the individuals are heterozygous for a recessive allele, which in homozygotes causes cystic fibrosis. Assuming this population is in Hardy-Weinberg equilibrium, what fraction (expressed as a decimal) of people in this new population will have cystic fibrosis?
A. 0.05
B. 0.005
C. 0.0025
D. 0.25
26. During meiosis, trisomy 21 can result from what chromosome abnormality?
A. An extra chromosome 21 from a mutation
B. An extra chromosome 21 from nondisjunction
C. An autosomal dominant disorder
D. A missing chromosome 21 from nondisjunction
27. The semiconservative model of DNA produces what complementary strand to the sequence 5′ — TTAACGAACG — 3′ during DNA replication?
A. 5′ — UUAAGCUUGC—3′
B. 5′ — AATTGCTTGC—3′
C. 3′ — AATTGCTTGC 5′
D. 3′ — UUAAGCUUGC5′
28. A local scientist has been re-creating Mendel’s experiments and recently produced an equal number of true breeding yellow (Y), smooth (R) pea plants, and true breeding green (y), wrinkled (r) pea plants. What is/are the expected F2 genotype(s)?
A. YYrr
B. YyRr; YYRr; yyrr
C. YyRr
D. YYRR; YyRr; yyRR; yyrr
29. Describe the outcome when a carrot with an osmolarity of 0.25 M is placed in a solution of 0.45 M saltwater.
A. The saltwater will enter the cell until equilibrium is reached.
B. The saltwater and carrot are isotonic, resulting in no movement.
C. Water will enter the cell until equilibrium is reached.
D. Water will leave the cell until equilibrium is reached.
30. A certain mutation found in fruit flies (Drosophila melanogaster) is hypothesized to be autosomal recessive. The experimenter crossed two Drosophila flies that were heterozygous for the trait. The next generation produced 70 wild-type males, 65 wild-type females, 36 males with the mutation, and 40 mutant females. Calculate the chi-squared value for the null hypothesis that the mutation is autosomal recessive.
A. 3.35
B. 9.98
C. 13.33
D. 6.64
31. A recent levy has been breached, causing flooding in a local community with saltwater. A local farmer is trying to determine if his crops can survive the flooding. The saltwater has a water potential of −8.73 MPa. The farmer’s crops have a molar concentration of 0.25 M with a pressure potential of −1.0 MPa at 27°C. What will happen to the crops?
A. The crops will maintain turgid pressure due to the crop’s water potential being −7.23 MPa and the saltwater being −8.73 MPa.
B. The crops will become flaccid and eventually die due to the crop’s water potential being −7.23 MPa and the saltwater being −8.73 MPa.
C. The pressure potential of the plant will cause the crops to gain water from the flooding saltwater.
D. The pressure potential of the plant will push the sugar out of the crops into the flooding saltwater.
32. A cell is in equilibrium with its surroundings. The molarity of the surrounding solution at 20°C is 0.8 M. Calculate the solute potential of the surrounding solution.
A. -3.90
B. -39.0
C. -19.5
D. -1.95
Questions 33—36 refer to the following pedigree.
33. What kind of inheritable condition does this pedigree appear to show?
A. Autosomal dominant
B. Autosomal recessive
C. Sex-linked dominant
D. Sex-linked recessive
34. What is the probability that couple C and D will produce a child that has the condition?
A. 0
B. 0.125
C. 0.250
D. 0.333
35. Which of the following conditions could show the same kind of pedigree results?
A. Cri-du-chat syndrome
B. Turner syndrome
C. Albinism
D. Hemophilia
36. If child E does in fact have the condition, what is the probability that child F will also have it?
A. 0
B. 0.250
C. 0.500
D. 0.750
Questions 37—39: An experiment involving fruit flies produced the following results:
Vestigial wings are wild type; crumpled wings are mutant.
Gray body is dominant; black body is mutant.
37. From the data presented above, one can conclude that these genes are
A. sex-linked.
B. epistatic.
C. holandric.
D. linked.
38. What is the crossover frequency of these genes?
A. 10 percent
B. 20 percent
C. 30 percent
D. 35 percent
39. How many map units apart would these genes be on a linkage map?
A. 5 map units
B. 10 map units
C. 20 map units
D. 30 map units
Questions 40—42: A laboratory procedure -involving plants presents you with the data found in the following two charts:
40. From the transpiration rate data, it appears that transpiration rate rises as
A. temperature ↑, wind speed ↓, humidity ↓
B. temperature ↑, wind speed ↑, humidity ↓
C. temperature ↑, wind speed ↑, humidity ↑
D. temperature ↓, wind speed ↑, humidity ↑
41. According to the Rf values given in the preceding smaller table, which pigment would migrate the fastest on chromatography paper?
A. Xanthophyll
B. Chlorophyll a
C. Chlorophyll b
D. Beta-carotene
42. From the transpiration rate data presented in the larger table, which of the following plants appears to be most resistant to transpiration?
A. Plant A
B. Plant B
C. Plant C
D. Plants B and C are similarly resistant.
Questions 43—45: A population of rodents is studied over the course of 100 generations to examine changes in dental enamel thickness. Species that are adapted to eat food resources that require high levels of processing have thicker enamel than do those that eat softer, more easily processed foods. Answer the following questions using this information and the curves that follow.
43. How is average enamel thickness changing in this population?
A. There is no real change.
B. The color and size are changing.
C. It is increasing.
D. It is decreasing.
44. You randomly pick one data point from all three sets of data (all three generations), and the individual’s enamel thickness score is 15. Which of the following can be inferred?
A. The individual comes from generation 1.
B. The individual comes from generation 50.
C. The individual comes from generation 100.
D. The individual could be from any of these generations.
45. What inference can you make about this species’ diet?
A. Its food resources are getting softer and easier to process.
B. Its food resources are getting harder and more difficult to process.
C. The population is growing.
D. The population is shrinking.
Questions 46—49: A student sets up a lab experiment to study the behavior of slugs. She sets up a large tray filled with soil that measures 1 square meter and has four sets of conditions, one in each quadrant:
She places 20 slugs in the tray, 5 in each quadrant. Use this information to answer the following questions:
46. What is this lab setup called?
A. A gel sheet
B. A choice chamber
C. A potometer
D. An incubation chamber
47. After 5 minutes, there are 5 slugs in each quadrant. Which of the following is not a viable explanation for this finding?
A. The slugs haven’t had time to move yet.
B. The slugs have no preference for temperature or salinity conditions.
C. The slugs can’t move from one area of the tray to another.
D. The slugs do not like to live in high-temperature areas.
48. After 20 minutes, 20 slugs are in the high- temperature, low-salinity quadrant. What kind of animal behavior has this experiment displayed?
A. Kinesis
B. Taxis
C. Survival
D. Feeding
49. A classmate has set up a similar experiment in the following manner:
Of the 20 slugs that she puts in her tray, 18 move to the high-salinity, high-temperature section within one hour, while the other 2 move to the low-salinity, low-temperature section. She concludes that slugs prefer conditions of high salinity and temperature. What is wrong with this conclusion?
A. She didn’t specify what the two temperatures or salinities were.
B. The slugs may not have been able to move where they wanted.
C. Crowding may have affected the behavior of the slugs, causing the two others to move to the other section.
D. She is measuring two variables at once with no control, and therefore can’t conclude anything about slug tastes.
50. Viral transduction is the process by which viruses carry bacterial DNA from one bacterial cell to another. In what way does this process play a role in bacterial evolution?
A. By making the bacterial cell more resistant to predators
B. By directly creating new species of bacteria
C. By increasing genetic variation of the bacteria
D. By selecting for viruses better able to infect bacteria
51. ADH is a hormone secreted by the kidneys that reduces the amount of water excreted in the urine. ADH is released in times of dehydration. This is an example of
A. innate behavior.
B. maintaining homeostasis.
C. failure to respond to the environment.
D. positive feedback.
For questions 52—54, refer to the information and graph that follows.
Five dialysis bags, made from a semipermeable membrane that is impermeable to glucose, were filled with various concentrations of glucose and placed in separate beakers containing 0.5 M glucose solution. The bags were weighed every 10 minutes and the percent change in mass for each bag was graphed:
52. Which line represents the bag that contained a solution isotonic to the 0.5 M solution?
A. A
B. B
C. C
D. D
53. Which line represents the bag with the highest initial concentration of glucose?
A. A
B. B
C. C
D. D
54. Which line or lines represent bags that contain a solution that is hypertonic at 50 minutes?
A. A and B
B. B
C. C
D. D and E
55. A mutation in a bacterial enzyme changed a previously polar amino acid into a nonpolar amino acid. This amino acid was located at a site distant from the enzyme’s active site. How might this mutation alter the enzyme’s substrate specificity?
A. By changing the enzyme’s pH optimum
B. By changing the enzyme’s location in the cell
C. By changing the shape of the protein
D. An amino acid change away from the active site cannot alter the enzyme’s substrate specificity.
Use the following picture of DNA to answer questions 56 and 57:
template strand 5′ ___________ 3′
complementary strand 3′ ___________ 5′
56. Based on the preceding picture, which direction would RNA polymerase move?
A. 3′ → 5′ along the template strand
B. 3′ → 5′ along the complementary strand
C. 5′ → 3′ along the template strand
D. 5′ → 3′ along the complementary strand
57. If the DNA segment is a transcriptional unit, where would the promoter be located?
A. To the left of the complementary strand
B. To the right of the template strand
C. To the left of the template strand
D. To the right of the complementary strand
58. A single gene from five related species of -leafhoppers was compared, and the nucleotide differences between the genes are as shown in the table:
Which of the following phylogenetic trees best shows the correct evolutionary relationship between the leafhoppers?
Answer questions 59 and 60 based on the following cladogram:
59. What is the common ancestor for B and E?
A. 1
B. 2
C. 3
D. 4
60. Which two species are most closely related?
A. A and E
B. A and B
C. B and C
D. D and E
AP Biology Practice Exam 2: Section II
FREE-RESPONSE QUESTIONS
Time—1 hour and 30 minutes
(The first 10 minutes is a reading period. Do not begin writing until the 10-minute period has passed.)
Questions 1 and 2 are long free-response questions that should require about 20 minutes each. Questions 3—6 are short-response questions that should require approximately 8—10 minutes each. Outline form is not acceptable. Answers should be in essay form.
1. Photosynthesis occurs in photosynthetic autotrophs in which carbon dioxide convert light energy from the sun into chemical energy in the form of sugar such as glucose. In an investigation, leaf disks are used to assay the net rate of photosynthesis under light and dark conditions. Leaf disks normally float when placed in water but when the air spaces are infiltrated with carbon dioxide, the density of the leaf disk increases, and the leaf disks sinks. Using sodium bicarbonate dissolved in water as a carbon source for photosynthesis, which causes the leaf disks to sink, photosynthetic activity will lead to oxygen production, and the leaf disks will become less dense and begin to float. The chart at the right shows the results of the investigation with leaf disks being placed in the light and in the dark.
A. Explain how carbon fixation by the leaf disks creates the oxygen needed to float when placed in light.
B. Identify the independent and dependent variable used in this experiment.
C. Construct an appropriately labeled graph.
D. Make a prediction if the leaf disks were placed under more intense light source. Justify your prediction.
2. Gila monsters are the only venomous lizard native to the United States. While they are slow and heavy, their venom is created in their salivary glands. The phenotype for scale color in gila monsters is determined by a specific locus. The dominant allele (black) is represented by G and the recessive allele (brown) is represented by g. The cross between a male gila monster with black scales and a female gila monster with brown scales produced the following F1 generation:
✵ Black-scaled gila monsters: 52
✵ Brown-scaled gila monsters: 55
✵ White-scaled gila monsters: 1
The black-scaled females and brown-scaled males from the F1 generation were then crossed to produce the following F2 generation:
✵ Black-scaled gila monsters: 53
✵ Brown-scaled gila monsters: 54
✵ White-scaled gila monsters: 0
A. Calculate the P-generation genotypes. Justify your response.
B. Describe and explain the white-scaled female that was present in the F1 generation.
C. The phenotype for scale color is claimed to be autosomal. Support the claim by calculating the chi-square value. Provide reasoning to justify the claim.
D. Coyotes are natural predators of gila monsters. During a recent drought, the coyote’s population experienced a bottleneck effect. Predict the impact this would have on the gila monsters. Justify your prediction.
3. A fictional mammal called a googabear was scanned for activity every 10 minutes for 42 hours. The percentage of active behavior including feeding, moving and other social behavior, as well as in active behavior including resting or sleeping was recorded for the googabear.
A. Based on the graph below, describe and explain the pattern of activity for the googabear.
B. Winter is quickly approaching, and the scientist claims that the googabear population is a species that hibernates, requiring adaptations to reduce efficiency of cellular respiration. Predict the physiological and behavioral adaptations that googabear will have. Provide evidence that supports the scientist’s claim.
Figure is for Question 3
4. The Florida Everglades is a wet sawgrass prairie that extends from Lake Okeechobee in Central Florida to the tidal salt marshes and mangroves along the coast of southern Florida. It is made of several different types of wetlands, starting with freshwater marshes in the north moving into deep water swamps as it move south until it reaches the tidal salt marshes and coastal mangrove along the coasts of the Gulf of Mexico and the Atlantic Ocean. A scientist claims that a new plant species could thrive in the tidal salt marsh and provides the following data to help convince the community:
✵ Overall water potential of the soil is Ψsoil= —2.2 MPa
✵ Plant cell contents: 0.08 M and 12°C (assume i = 1)
✵ Pressure potential of the plant cell is Ψpressure = —1.2 MPa
A. Predict the ability of the new plant species’ ability to survive in the tidal salt marshes of the Everglades. Provide evidence to support your prediction and justify your response.
B. Burmese pythons, native to the tropics of Southeast Asia, which started out as a popular pet species in south Florida, is now threatening the Florida Everglades as an invasive species when pet owners release them into the Everglades’ ecosystem. Explain how the Burmese python has affected the ecosystem of the Everglades.
5. Earth’s early atmosphere was full of gases such as NH3, CH4, H2O(g), and H2 from numerous volcanic activities. However, there was no O2. Energy was abundant in the form of UV light, lightning, heat, and radioactivity. It was these characteristics that Stanley Miller and Harold Urey used to simulate Earth’s primordial environment.
A. Describe the outcome of the Miller-Urey experiment.
B. Explain the significance of photosynthesis being present in simple cells before the evolution of more complex cells. Justify your response.
C. Identify the evidence that supports the evolution of the eukaryotic cell.
Figure is for Question 5 (Reproduced with permission from Raven P, Johnson G, Mason K, Losos J, Duncan T; Biology, 12th ed. New York: McGraw Hill; 2020)
6. Sea otters, sea urchins, kelp, orcas, and other organisms live in the ecosystem along the West Coast of the United States. In the ecosystem, the orcas feed on sea otters as one of their sources of food. Graph A represents when the orcas feed mainly on sea otters, while Graph B represents when the orcas feed on a wide variety of mammals in the ecosystem.
A. Explain the impact of sea otters on the diversity of the ecosystem.
B. Scientists are consistently concerned by the potential loss of species due to human activity. Identify and explain one human impact that could affect the ecosystem of sea otters, sea urchins, kelp, orcas, and other organisms along the West Coast of the United States.
Figures are for Question 6 (Reproduced with permission from Raven P, Johnson G, Mason K, Losos J, Duncan T; Biology, 12th ed. New York: McGraw Hill; 2020)
Answers and Explanations for AP Biology Practice Exam 1
MULTIPLE-CHOICE QUESTIONS
1. D—Habituation is the loss of responsiveness to unimportant stimuli or stimuli that do not provide appropriate feedback. This is a prime example of habituation.
2. A—Directional selection occurs when members of a population at one end of a spectrum are selected against, while those at the other end are selected for. Taller giraffes are being selected for; shorter giraffes are being selected against.
3. A—When interbreeding ceases because some sort of barrier separates a single population into two (an area with no food, a mountain, etc.), the two populations evolve independently, and if they change enough, then, even if the barrier is removed, they cannot interbreed. This is allopatric speciation.
4. D—Chapter 13, despite being last, is a very important chapter. The experiments are very well represented on the AP Biology exam, and you should read this chapter carefully and learn how to design and interpret experiments.
5. C
6. A
7. D
8. B—The endosymbiotic theory proposes that mitochondria and chloroplasts evolved through the symbiotic relationship between prokaryotic organisms.
9. D—Fibroblast growth factor is said to be involved, but fibroblast cells are not.
10. D—Posttranscriptional modification actually occurs in the nucleus.
11. C—Genetic drift is a change in allele frequencies that is due to chance events. When drift dramatically reduces population size, it is called a “bottleneck.”
12. A
13. C—Only photosystem I is involved in the cyclic reactions. Photosystem II is not.
14. A—Genetics has no memory. It will be ½ forever.
15. A—DNA migrates from a negative charge to a positive charge. The rest are true.
16. B—0.04 = q2. Therefore, the square root of 0.04 = q = 0.20 and p + q = 1. So p + 0.20 = 1. Therefore, p = 0.80, and 2pq is the frequency of the heterozygote condition: 2(0.20)(0.80) = 0.320 = 32 percent.
17. C—Phytochrome is an important pigment to the process of flowering. Of its two forms, the active form, Pfr, is responsible for the production of the hormone florigen, which is thought to assist in the blooming of flowers.
18. C—Primary carnivores > primary consumers = herbivores > primary producers.
19. C—The example to know is the cattle egrets that feast on insects aroused into flight by cattle grazing in the insects’ habitat. The birds benefit because they get food, but the cattle do not appear to benefit at all.
20. B—Conjugation is the sexual reproduction of bacteria.
21. B—In competitive inhibition, an inhibitor molecule resembling the substrate binds to the active site and physically blocks the substrate from attaching.
22. C—This is the only factor that is not a major factor affecting enzyme efficiency.
23. B—The activation energy of a reaction is the amount of energy needed for the reaction to occur. Notice that the activation energy for the enzymatic reaction is much lower than the nonenzymatic reaction.
24. B
25. C—For this specific gene in this specific population, there are a total of 40 alleles, two of which are the recessive cf allele (2/40 = 0.05 = q).
Since you need to be homozygous recessive to have cystic fibrosis, (q) × (q ) = q2 = (0.05)2 = 0.0025. In other words, 25 out of 10,000 people (0.25 percent) will have cystic fibrosis.
26. B—Nondisjunction occurs during meiosis, resulting in an extra chromosome in one gamete and a missing chromosome in another gamete. trisomy 21 (Down syndrome) is a result of an extra chromosome 21 from nondisjunction.
27. C—During DNA replication, the semiconservative model of DNA is produced where the complementary strand is anti-parallel.
28. C—One hundred percent of the offspring in F2 generation would YrRr.
29. D—When the carrot is immersed in the saltwater with a concentration of 0.45 M (a hypertonic solution), the carrot will lose water through osmosis.
30. C—If both parents were heterozygous and if this trait is indeed recessive, you would expect the next generation to show 75 percent normal-looking flies and 25 percent of the flies with the recessive trait. Based on a total of 211 flies, that would mean you would expect 158 normal flies and 53 recessive flies. Your observed numbers were, instead, 135 normal flies and 76 recessive flies.
Since your chi-squared value (13.33) is higher than the critical value of 6.64 (based on 1 degree of freedom), you have to reject your hypothesis. Something other than an autosomal recessive trait is going on.
31. B—Water potential (Ψ) = −7.23 MPa (−6.23 MPa + −1 MPa)
Solute potential (Ψs) = (0.35 M)(1.0)(0.0831 liter bars/mole) (300 K) = −6.23 MPa
32. C—Ys = —(1)(0.8 M)(0.0831 L bars/mole K) (294 K) = —19.5 MPa
33. B—It is not autosomal dominant because in order for the second generation on the left to have those two individuals with the condition, one parent would need to display the condition as well. It is probably not sex-linked because it seems to appear as often in females as in males. Autosomal recessive seems to be the best fit for this disease.
34. D—One first needs to determine the probability that person C is heterozygous (Bb). We know that person D is double recessive because she has the condition. We know that the parents for person C must be Bb and Bb because neither of them has the condition, but they produced children with the condition. The probability of person C being heterozygous is ⅔, because a monohybrid cross of his parents (Bb × Bb) gives the following Punnett square:
Since you know that he doesn’t have the condition, he cannot be bb. This leaves just three possible outcomes, two of which are Bb. A cross must then be done between the father (person C) Bb and the mother (person D) bb. The chance of their child being bb is 50 percent, or ½. This means that the chance of these two having a child with the condition is ⅔ × ½, or ⅓.
35. C—Albinism is the only autosomal recessive condition on this list.
36. C—It is ½, because finding out that one of their children has the condition lets us know that the father (person C) is definitely Bb. This changes the probability of ⅔ to 1, meaning that the probability of the two having another child with this condition is simply the result of the Punnett square of Bb × bb, or ½.
37. D—When you see a ratio like the one in this problem—7:7:1:1 (approximately)—the genes are probably linked. The reason the crumpled, gray, and vestigial black flies exist at all is because crossover must have occurred.
38. A—To determine the crossover frequency in a problem like this, simply add up the total number of crossovers (75 + 45 = 120) and divide that sum by the total number of offspring (120 + 555 + 525 = 1,200). This results in 120/1,200, or 10 percent.
39. B—One map unit is equal to a 1 percent recombination frequency.
40. B—The data in the table show you that this answer is the correct choice.
41. D—The larger the value of Rf for a bunch of pigments dissolved in a particular chromatography solvent, the faster the pigments will migrate. Beta-carotene has the highest Rf value.
42. B—Across the board it seems to have the lowest rate of transpiration. You can make this leap because, as mentioned on top of the larger chart, all the leaves have the same surface area, allowing you to compare their transpiration values.
43. C—The average enamel thickness started at 10, increased to 12, and then increased to 15. It is therefore increasing overall.
44. D—The average enamel thickness does not describe the range of possible values; an individual with a thickness of 15 could reasonably come from any of the three generations (if we took into account probability, we could say that the individual most likely came from the 100th generation because this population has the highest frequency of individuals with this thickness; however, the question does not ask for probabilities).
45. B—Because thicker enamel in this species indicates foods that are more difficult to process, the answer is B.
46. B—Experimental setups where individuals are given a choice as to where to move are called “choice chambers.”
47. D—All the answers except D are possible, and are important things to consider when setting up an experiment. For example, it is important to allow your study animals enough time to move and/or get used to their new surroundings and conditions before drawing conclusions about their behavior. D is not a good answer because half of the slugs started in a high-temperature area and haven’t moved.
48. A—Kinesis is the movement of animals in response to current conditions; animals tend to move until they find a favorable environment, at which point their movement slows.
49. D—It is important to try to measure only one variable at once. The 18 slugs may have moved to the higher-temperature, higher-salinity conditions because they need high temperatures to survive, even if they dislike high salinity, and vice versa. The original experiment circumvents this problem by giving a choice for all the possible combinations of variables.
50. C—New genes are introduced into the bacterium through viral transduction.
51. B—When the body has too little water, ADH works to increase the amount of water available. This drive to maintain a stable condition is an example of homeostasis.
52. C—Line C showed no net change in weight, indicating the concentration of the solution inside the bag was the same (isotonic) as the solution in the beaker.
53. A—The most water would diffuse into the most hypertonic solution; line A shows the biggest increase in weight.
54. B—Line B still shows an increase in weight at 50 minutes, whereas line A has leveled out and is isotonic at 50 minutes.
55. C—Even though an amino acid doesn’t have direct contact with the substrate, it still plays a role in the overall shape of the enzyme.
56. A—As RNA polymerase adds new nucleotides to the 3′ end of the new strand, it is moving toward the 5′ end of the (antiparallel) template strand.
57. B—The promoter would be located upstream from where transcription would begin.
58. C—There are few nucleotide differences between species 1 and 2, indicating they would reside close to one another on the cladogram. The same holds true for species 3 and 5. There are large numbers of differences between species 4 and all others, indicating it would be positioned on its own branch.
59. A—Both B and E branches originate from point 1.
60. C—Species B and C reside the closest to one another.
Free-Response Grading Outline
1.
A. Explain how carbon fixation by the leaf disks creates the oxygen needed to float when placed in light.
✵ Located in the chloroplast. (1 point)
✵ Light reaction transfers light energy to chemical energy via ATP and NADPH. (1 point)
✵ Uses CO2 from environment, ATP, and NADPH from light reaction to create sugar, which is stored chemical energy. (1 point)
✵ Oxygen is the by-product of splitting of H2O during light reaction. (1 point)
B. Identify the independent and dependent variables used in this experiment.
✵ Independent variable — exposure to light. (1 point)
✵ Dependent variable — number of floating leaf disks. (1 point)
C. Construct an appropriately labeled graph.
✵ Independent and dependent variables correctly plotted. (1 point)
✵ Correctly labeled. (1 point)
✵ Correctly scaled axis. (1 point)
D. Make a prediction if the leaf disks were placed under more intense light source. Justify your prediction.
✵ Predict that the leaf disk would float quicker due to an increase of photosynthetic activity in the leaves. (1 point)
✵ Increase in light would cause an increase of light energy being absorbed, and more oxygen would be released as a by-product of the light reaction of photosynthesis. (1 point)
2.
A. Calculate the P-generation genotypes. Justify your response.
✵ The P-generation is Gg × gg. (1 point)
✵ The P-generation results in a F1 generation that consists of Gg (black-scaled) and gg (brown-scaled) gila monsters. Then a cross of the F1 generation results in 50 percent black-scaled monsters and 50 percent brown-scaled monsters in the F2 generation. (1 point)
B. Describe and explain the white-scaled female that was present in the F1 generation.
✵ The white-scaled female is caused by a point mutation, which is a random event, that can lead to changes in allele frequencies. (2 points)
✵ The point occurred as DNA responsible for the production of protein that determines scale color was undergoing replication. (2 points)
C. The phenotype for scale color is claimed to be autosomal. Support the claim by calculating the chi-square value. Provide reasoning to justify the claim.
✵ x2 = (53 — 53.5)2 + (54 — 53.5)2/2 — 1 = 0.5 (1 point)
✵ Null hypothesis—The phenotype for scale color is autosomal. (1 point)
✵ Scale color for gila monsters is, in fact, autosomal. A chi-square value of 0.5 on the F2 generation data resulted in a failure to reject the null hypothesis. (1 point)
D. Coyotes are natural predators of gila monsters. During a recent drought, the coyote’s population experienced a bottleneck effect. Predict the impact this would have on the gila monsters. Justify your prediction.
✵ The population of gila monsters would randomly decrease. (1 point)
✵ During a bottleneck effect, a population is randomly reduced, causing a decrease in the population and genetic diversity, leading to a genetic drift. (1 point)
3.
A. Based on the graph, describe and explain the pattern of activity for the googabear.
✵ Googabears are more active from 6 a.m. until 6 p.m. (during daylight hours). (1 point)
✵ Decrease activity from 6 p.m. to 6 a.m. (when it is dark). (1 point)
✵ Googabears’ food source is available during daylight. (1 point)
✵ Googabear predators are nocturnal (out at night), so it is safest for googabears to remain hidden at night. (1 point)
✵ Googabears rely on collective body heat at night (huddling); activity of huddled group is low. (1 point)
B. Winter is quickly approaching, and a scientist claims that the googabear population is a species that hibernates, requiring adaptations to reduce efficiency of cellular respiration. Predict the physiological and behavioral adaptations that googabears will have. Provide evidence that supports the scientist’s claim.
(2 Points maximum)
✵ Googabears are endothermic—Must maintain body temperature through increase fat storage and thick fur than when active.
✵ Googabears need a constant supply of ATP, which will come from a large supply of stored energy in the form of fat.
✵ Googabears are inactive during hibernation, during which cells require less ATP on a daily basis.
4.
A. Predict the ability of the new plant species’ ability to survive in the tidal salt marshes of the Everglades. Provide evidence to support your prediction and justify your response.
✵ The plant cell’s water potential (—1.39 MPa) is higher than that of the soil (—2.2 MPa). (1 point)
✵ Water would flow out of the plant cell (hypotonic) into soil (hypertonic). (1 point)
✵ The plant cell would not survive. (1 point)
B. Burmese pythons, native to the tropics of Southeast Asia, which started out as a popular pet species in south Florida, is now threatening the Florida Everglades as an invasive species when pet owners release them into the Everglades ecosystem. Explain how the Burmese python has affected the ecosystem of the Everglades.
✵ Introduction of an invasive species allows for the species to exploit a new niche free of predators and/or competitors, allowing them to outcompete other organisms for resources. (1 point)
5.
A. Describe the outcome of the Miller-Urey experiment.
✵ Organic molecules could have originated in the early atmosphere. (1 point)
B. Explain the significance of photosynthesis being present in simple cells before the evolution of more complex cells. Justify your response.
✵ With the first prokaryotic cells being photosynthetic, they release oxygen as a by-product, leading to an increase in atmospheric oxygen and allowing for the evolution of cellular respiration. Increased oxygen allowed cells to generate more energy and grow larger and more complex. (1 point)
C. Identify the evidence that supports the evolution of the eukaryotic cell.
✵ Chloroplasts and mitochondria have their own DNA that consists of a single, circular molecule. They replicate by a process similar to prokaryotes and have enzymes homologous to those found in prokaryotes. (2 points)
6.
A. Explain the impact of sea otters on the diversity of the ecosystem.
✵ The sea otter is the keystone species that keeps the sea urchin population in check, allowing kelp and other organisms to thrive. (1 point)
✵ Keystone species are a species that holds a habitat together by playing a pivotal role in how the ecosystem functions. It may not be the largest or most abundant species in an ecological community, but its removal sets off a chain of events that can lead to dramatic structural and diversity changes to the ecosystem. (1 point)
B. Scientists are consistently concerned by the potential loss of species due to human activity. Identify and explain one human impact that could affect the ecosystem of sea otters, sea urchins, kelp, orcas, and other organisms along the West Coast of the United States.
✵ Pollution, habitat degradation, or overfishing are a few examples of human impact. (1 point)
✵ These human impacts on the ecosystem accelerate changes to the local ecosystem’s and eventually the global ecosystem’s structure and dynamics. (1 point)
Scoring and Interpretation
AP BIOLOGY PRACTICE EXAM 2
AP BIOLOGY
SCORE CONVERSION CHART
