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AP Biology Practice Exam 1
AP Biology Practice Exam 2
Answer Sheet for AP Biology Practice Exam 1
PART A: MULTIPLE-CHOICE QUESTIONS
PART B: GRID-IN QUESTIONS
AP Biology Practice Exam 1: Section I
PART A: MULTIPLE-CHOICE QUESTIONS
Time—1 hour and 30 minutes (for Parts A and B)
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?
B. Cell membrane
D. Cell wall
2. Which of the following is the source of oxygen produced during photosynthesis?
3. An organism exposed to wild temperature fluctuations shows very little, if any, change in its metabolic rate. This organism is most probably a
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?
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?
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. Which of the following is an example of a detritivore?
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
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 a herbivore most likely be assigned?
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?
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.
For questions 23–26, please refer to the following structures:
23. This represents the backbone of a structure that is vital to the construction of many cells and is used to produce steroid hormones.
24. This structure plays a vital role in energy reactions.
25. This structure is a purine found in DNA.
26. This structure was synthesized in the ribosome.
For questions 27–30, please refer to the following answers:
D. Calvin cycle
27. When oxygen becomes unavailable, this process regenerates NAD+, allowing respiration to continue.
28. This process leads to the net production of two pyruvate, two ATP, and two NADH.
29. 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.
30. This process has as its products NADP+, ADP, and sugar.
For questions 31–34, please refer to the following answers:
D. Deciduous forests
31. This biome has cold winters and is known for its pine forests.
32. This biome is the driest of the land biomes and experiences the greatest daily temperature fluctuations.
33. This biome contains trees that drop their leaves during the winter months.
34. This biome contains plants whose roots cannot go deep due to the presence of a permafrost.
Questions 35–36: 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.
35. Testosterone level in this population may be an example of
A. adaptive radiation.
B. an adaptation.
C. divergent selection.
36. 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.
37. 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 38–41: 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.
38. What originally caused the change in the variant?
B. Balance polymorphism
39. What kind of speciation does this example illustrate?
40. Which of the following could possibly characterize the new variant?
A. Adaptive radiation
B. Divergent selection
41. Which of the following is likely to exhibit the process described earlier?
A. Fallow deer
B. Fruit flies
D. Spotted toads
For questions 42–44, please refer to the following answers:
42. The DNA placed in this electrophoresis gel separates as a result of what characteristic?
43. 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
44. 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 45–48: The frequency of genotypes for a given trait are given in the accompanying graph. Answer the following questions using this information:
45. What is the frequency of the recessive homozygote?
A. 15 percent
B. 19 percent
C. 25 percent
D. 40 percent
46. 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
47. Is this population in Hardy-Weinberg equilibrium?
C. Cannot tell from the information given
D. Maybe, if individuals are migrating
48. 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 49–51: 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.
49. 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 be 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.
50. 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.
51. 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 52–54 all use the following pedigree, but are independent of each other:
52. 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 heterozygous?
53. Imagine that a couple (C and D) go 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?
54. Imagine that a couple (C and D) have 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?
For questions 55–56, please refer to the following diagram:
55. The bold line that point C intersects is known as the
A. biotic potential.
B. carrying capacity.
C. limiting factor.
D. maximum attainable population.
56. 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 57 and 58: 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.
57. 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.
58. 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.
59. 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.
60. 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
61. 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.
62. 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
63. 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?
PART B: GRID-IN QUESTIONS
Calculate the correct answer and enter it on the top line of the grid-in area with each number/symbol in a separate column. Then fill in the correct circle below each number/symbol you entered (only one filled-in circle per column).
1. 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.
2. 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?
3. 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.
4. What is the water potential for a solution that is 0.1 M at 20°C? The solution is in an open container. The equation for water potential is:
5. Determine the surface area–to–volume ratio for a cube that has a side length of 2.5 cm. The volume of a cube = (l)(w)(h). The surface area = 6 × area of a single side.
6. If CiVi = CfVf, where i is the initial solution concentration and f is the final concentration, how many milliliters of a 0.5 M glucose solution would you need in order to make 250 milliliters of a 0.1 M glucose solution?
AP Biology Practice Exam 1: Section II
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 3–8 are short-response questions that should require about 6 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.
A. Describe how a gel electrophoresis experiment works and is set up, why things move the way they do, and why the gel would be able to prove, beyond a shadow of a doubt, that the defendant was indeed guilty as charged.
B. Gel electrophoresis is also used to determine court paternity cases. Describe how a gel could be used to prove whether an individual is the father of a particular baby. Include all the pertinent experimental laboratory procedures in your description.
2. Speciation, the process by which new species are formed, can occur by many mechanisms. Explain how three of the following are involved in the process of species formation.
A. Geographic barriers
C. Balanced polymorphism
D. Reproductive isolation
3. Life on Earth is made possible because of certain unique characteristics of water. Choose two characteristics of water.
A. For each characteristic that you choose, identify and define the property.
B. Describe one example of how the property affects the functioning of living organisms.
4. Evolution is the change in allele frequencies in a population over time. This can occur through a variety of mechanisms, three of which are listed below.
• Natural selection
• Genetic drift
A. Define two of the three forces of evolution listed above and give an example.
B. You are 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 is most consistent with this data, and give a hypothetical explanation for the observed changes in phenotypic frequencies in this mouse population.
5. Homeostasis, or the maintenance of a steady-state environment, is a characteristic of all living organisms. For each of the following physiological parameters describe how homeostasis is maintained:
A. Blood glucose levels
B. Body temperature
C. Blood calcium levels
6. Membranes are vital to the transport of substances into and out of cells. Three important forms of cellular transport include:
A. Active transport
B. Endocytosis and exocytosis
C. Facilitated diffusion
For each of the forms listed above, explain how the organization of the cell membranes functions in the movement of specific molecules across membranes and explain the significance of each type of transport to a specific cell.
7. Skin is coated with sebum, an oily substance that slows water loss and inhibits growth of some microorganisms.
A. Briefly explain how sebum is able to perform its function.
B. Explain why lungs are more vulnerable to infection than skin. Include the idea of surface area in your answer.
8. The complete oxidation of a mole of glucose produces 686 kcal of free energy. The oxidation of a mole of glucose in a cell generates a maximum of 38 moles of ATP. Each mole of ATP stores about 7.3 kcal of energy. The efficiency of the ATP energy yield from the complete aerobic respiration of glucose is about 40 percent.
A. Using the laws of thermodynamics, explain what happens to the rest of the energy.
B. How do humans benefit from this energy loss?
C. Why would hibernating animals possess an adaptation to reduce efficiency of cellular respiration even further?
Answers and Explanations for AP Biology Practice Exam 1
PART A: 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 photosynthesis. 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.
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. D—A detritivore is an organism that includes the subcategory of decomposers. Fungi are decomposers.
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 19. 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.
23. A—Cholesterol is one of the lipids that serves as the starting point for the synthesis of sex hormones.
25. D—Purines have a double-ring structure; pyrimidines, a single-ring one.
26. B—The ribosome is the site of protein synthesis.
35. 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.
36. 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.
37. 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.
38. 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).
39. 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.
40. 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.
41. C—Polyploidy is much more common in plants; mutations such as the duplication of whole chromosomes are usually lethal to animals.
42. C—Gel electrophoresis separates DNA on the basis of size. Smaller samples travel a greater distance down the gel compared to larger samples.
43. B—His DNA fingerprint seems to exactly match that of the evidence DNA sample.
44. 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.
45. B—100 − 45 − 36 = 19 percent.
46. C—36 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.
47. B—The expected heterozygous probability does not match up with the actual. This population is not in Hardy–Weinberg equilibrium.
48. B—The homozygous frequency is higher than expected; one explanation for this is that the homozygotes are being selected for.
49. 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.
50. 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.
51. C—Water potential = pressure potential + solute potential. Water passively moves from regions with high water potential toward those with lower water potential.
52. 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.
53. 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.
54. 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.
57. 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).
58. 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).
59. A—Heterozygous individuals carry the recessive gene but are themselves healthy.
60. D—Low numbers of stomata help to reduce water loss, helpful in hot and dry regions.
61. 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.
62. B—Changing envelope proteins are created because of genetic variation in the genes that code for these proteins.
63. D—This cladogram shows a closer relationship between X and Y.
PART B: GRID-IN QUESTIONS
1. 1.56—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.
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.
3. 1.27—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.
4. –2.4—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.
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
(0.5 M)(Vi) = (0.1 M)(250 mL)
Vi = 50 mL
Free-Response Grading Outline
1. Gel electrophoresis question (10 points)
A. Electrophoresis experiment (maximum 5 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)
• 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. Paternity (maximum 5 points)
• Mentioning that DNA samples would need to be taken from the disputed child and the potential parents involved. (1 point)
• Definition of a restriction enzyme as an enzyme that cuts DNA at a particular sequence and creates open fragments of DNA called “sticky ends.” (1 point)
• Mentioning that the DNA from all the different individuals involved must be cut by the same restriction enzyme(s) so that the RFLPs created can be compared with each other. (1 point)
• Mentioning that each sample of DNA must be placed into a different well at the top of the gel plate. (1 point)
• Mentioning that the DNA will migrate from negative charge to positive charge, once the current is applied, to create an RFLP pattern specific for each individual— this is a look at the DNA fingerprint of an individual. (1 point)
• Mentioning that some sort of dye should be added to the DNA samples that will allow for proper viewing of the bands after the current is disconnected. (1 point)
• Mentioning that one of the two DNA cuts from the child’s fingerprint should match up with one of the two DNA cuts from the father’s fingerprint and one from the mother’s fingerprint as well, because the child inherits one chromosome of each homologous pair from the mother and one from the father. (1 point)
2. Speciation question (here, you can obtain 4 points from a couple of the answers; if 4 points are obtained for an answer, a maximum of 3 points can be obtained from each of the other 2 answers)
A. Geographic barriers (maximum 4 points)
• Mentioning how geographic barriers can lead to reproductive isolation of members from the same species. (½ point)
• Mentioning that if these geographically separated species are moved into regions that have different environments, natural selection might favor different characteristics from the same species in the different environments. (1 point)
• Mentioning that this is an example of allopatric speciation—interbreeding ceases because some sort of barrier separates a single population into two. (1 point)
• Definition of divergent evolution as the evolution of two species farther apart from each other as they are exposed to different environmental challenges. (1 point)
• Mentioning the Galapagos finches as an example of geographic barriers leading to reproductive isolation and divergent evolution. (½ point)
• Mentioning that if after a long period of time, these divergent species come back together and are unable to reproduce, they have become a new species. (1 point)
B. Polyploidy (maximum 4 points)
• Definition of polyploidy as a condition in which an individual has more than the normal number of sets of chromosomes. (1 point)
• Description of how polyploidy initially occurs—an accident during cell division could double the chromosome number in the offspring, producing a tetraploid (4n) organism. (1 point)
• Alternate description of how polyploidy could initially occur—the breeding of two individuals from different species leads to a hybrid that is usually sterile and contains chromosomes that are not able to pair up during meiosis because they are not homologous. (1 point)
• Definition of an autopolyploid—organism with more than two chromosome sets all from the same species. (½ point)
• Definition of an allopolyploid—organism with more than two chromosome sets that come from more than one species. (½ point)
• Mentioning that although an individual may be healthy, it cannot reproduce with nonpolyploidic members of its species. (1 point)
• Mentioning that polyploidic individuals are able to mate only with other individuals who have the same polyploidic chromosomal makeup. (1 point)
C. Balanced polymorphism (maximum 3 points)
• Definition of balanced polymorphism—some characters have two or more phenotypic variants, such as tulip color. (1 point)
• Mention of the fact that if one phenotypic variant leads to increased reproductive success, directional selection will eventually eliminate all other varieties because only those who have the particular phenotypic variant of choice will survive to be able to reproduce, and thus only their genes will be passed along. (1 point)
• Mentioning that this requirement for a particular variant of the trait in order to survive reproductively isolates individuals of the same species from each other, opening the door for sympatric speciation. (1 point)
• Mentioning that if the balanced polymorphism causes the two variants to diverge enough to no longer be able to interbreed, speciation has occurred. (1 point)
• Citing an example of balanced polymorphism. (1 point)
D. Reproductive isolation (maximum 4 points)
• Mentioning that any barrier that prevents two species from producing offspring can be categorized as reproductive isolation. (½ point)
• Definition of prezygotic barriers as reproductive barriers that make the fertilization of the female ovum impossible. (1 point)
• Mentioning, as an example of prezygotic barriers, any of the following (½ point each, up to 1 point total for prezygotic barrier examples): (a) habitat isolation—two species live in different habitats (they just don’t see each other, so they cannot reproduce); (b) temporal isolation—two species mate at either different times of the year or different times of the day (either way, they are isolated from each other because they do not mate at the same time); (c) behavioral isolation—two species have different mating behaviors that do not mix well (members of the other species do not understand the actions of the other as mating signals—a simple communication breakdown ); (d) mechanical isolation—mating may actually be attempted, but the physical sexual structures do not function together properly (they are incompatible).
• Definition of postzygotic barriers as reproductive barriers that prevent a properly formed hybrid between two species from reproducing themselves. (1 point)
• Mentioning, as an example of postzygotic barriers, any of the following (½ point each, up to 1 point total for postzygotic barrier examples): (a) hybrid breakdown— sometimes the first generation of hybrids produced are able to reproduce with each other, but after that the wheels come off and the next generation is infertile; (b) reduced hybrid viability—the two different species are able to mate physically and the hybrid zygote is formed, but problems arise during the development of the hybrid that lead to prenatal death of the individual; (c) reduced hybrid fertility—the two different species are able to mate physically and produce a viable offspring, but the offspring is infertile.
3. Life on earth is made possible because of certain unique characteristics of water. (maximum 4 points for entire question)
A. Properties of water (maximum 2 points)
• Mentioning that hydrogen bonding holds water molecules together (cohesion). (1 point)
• Mentioning that water has high specific heat (water changes its temperature less than most substances when it absorbs or loses a certain amount of heat). (1 point)
• Mentioning that the solid form of water is less dense than its liquid form. (1 point)
• Mentioning water acts as a solvent for so many substances. (1 point)
B. Effects on living organisms (maximum 2 points)
• Mentioning that hydrogen bonding allows for flow of water through plants (xylem). (1 point)
• Mentioning that water’s high specific heat means large bodies of water store heat and leads to milder climates in coastal areas. (1 point)
• Noting that water’s unusual density means ice floats. If it sank, all bodies of water would freeze solid from the bottom up, and sustainable life on earth would not be possible. (1 point)
• Mentioning that water’s ability as a solvent allows for many reactions to occur in the cell. (1 point)
4. Evolution is the change in allele frequencies in a population over time. (maximum 4 points for entire question)
A. Definition and examples (maximum 2 points)
• Defining natural selection. (maximum 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.
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. 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. Homeostasis, or the maintenance of a steady-state environment, is a characteristic of all living organisms. (maximum 4 points for entire question)
A. Blood glucose levels (maximum 1½ points)
• Mentioning insulin/glucagon is released from the pancreas. (½ point)
• Mentioning insulin stimulates uptake of glucose from the blood to the liver. (½ point)
• Mentioning insulin causes glucose to be stored as glycogen in the liver. (½ point)
• Mentioning insulin signals body cells to take up glucose for energy use. (½ point)
• Mentioning glucagon stimulates the liver to release glucose into the bloodstream. (½ point)
B. Body temperature (maximum 1½ points)
• Describing how body insulation (hair, fat, feathers) reduces heat loss. (½ point)
• Mentioning that vasodilation results in increased blood flow and increased heat loss. (½ point)
• Mentioning that vasoconstriction results in decreased blood flow and decreased heat loss. (½ point)
• Mentioning that sweating results in evaporative cooling. (½ point)
• Mentioning that shivering generates energy. (½ point)
C. Blood calcium levels (maximum 1½ points)
• Mentioning that parathyroid hormone (PTH) is released by the parathyroid gland. (½ point)
• Mentioning that PTH increases the amount of calcium in circulation. (½ point)
• Mentioning that PTH causes release of calcium from bones. (½ point)
• Mentioning that PTH leads to increased absorption of calcium by the intestines and kidneys. (½ point)
• Mentioning that calcitonin is released by the thyroid gland. (½ point)
• Mentioning that calcitonin decreases the amount of calcium in circulation. (½ point)
• Mentioning that calcitonin promotes reabsorption of calcium by the bones. (½ point)
• Mentioning that calcitonin leads to decreased absorption of calcium by the kidneys and intestines. (½ point)
6. Membranes are vital to the transport of substances into and out of cells. (maximum 4 points for entire question)
A. Active transport (maximum 1½ points)
• Describing active transport as the movement of a particle against its concentration gradient (from low concentration to high concentration), which requires input of energy. (½ point)
• Mentioning that it allows the cell to concentrate substances within the cell membrane. (½ point)
• Mentioning that it is performed by protein pumps in the membrane. (½ point)
• Possible example: sodium-potassium pump moves potassium into the cell and sodium out of the cell. (½ point)
B. Endocytosis and exocytosis (maximum 2 points)
• Mentioning that endocytosis brings a substance into a cell by enclosing it within a membrane-created vesicle. (½ point)
• Mentioning that the vesicle then fuses with lysosome (contains hydrolytic enzymes). (½ point)
• Possible example: phagocytes (white blood cells) of the immune system engulf foreign invaders. (½ point)
• Mentioning that receptor-mediated endocytosis uses proteins embedded in a membrane that contains receptors for specific molecules. (½ point)
• Mentioning that exocytosis expels waste substances for export by enclosing these substances in a vesicle that fuses with the membrane. (½ point)
• Possible examples: cells expelling waste or a pancreatic cell exporting insulin protein into the bloodstream. (½ point)
C. Facilitated diffusion (maximum 1½ points)
• Describing facilitated diffusion as the diffusion of particles with the assistance of membrane transport proteins. (½ point)
• Mentioning that transport proteins are specific and have a binding site for the specific molecule of interest. (½ point)
• Mentioning that it does not require energy. (½ point)
• Mentioning that osmosis occurs from a hypotonic solution to a hypertonic solution. (½ point)
7. Skin is coated with sebum, an oily substance that slows water loss and inhibits growth of some microorganisms. (maximum 4 points for entire question)
A. How sebum works (maximum 2 points)
• Mentioning that sebum (or any oily substance) is composed of lipids. (1 point)
• Mentioning that lipids are hydrophobic (repel water). (1 point)
• Mentioning that sebum creates a hydrophobic barrier on skin that slows water loss. (1 point)
B. Why lungs are more vulnerable to infection (maximum 2 points)
• Mentioning that lungs provide a moist environment, ideal for microbial growth. (1 point)
• Mentioning that lungs have a highly folded surface (high surface area). (1 point)
• Mentioning that a higher surface area provides more opportunities for invasion. (1 point)
8. The complete oxidation of a mole of glucose produces 686 kcal of free energy. (maximum 4 points for entire question)
A. Loss of energy (maximum 3 points)
• Mentioning that energy cannot be created or destroyed but only transformed from one form to another (first law of thermodynamics). (1 point)
• Mentioning that every energy transformation results in some energy loss as heat (second law of thermodynamics). (1 point)
• Mentioning that the remaining 60 percent energy from glucose oxidation is lost as heat. (1 point)
B. Benefits from loss of energy (maximum 1 point)
• Mentioning that humans use some of the heat to maintain body temperature. (1 point)
C. Adaptation of hibernating animals (maximum 2 points)
• Mentioning that hibernating animals don’t need a lot of ATP because they are inactive. (1 point)
• Mentioning that they must still maintain internal body heat. (1 point)
• Mentioning that lower efficiency means heat generation without much ATP being produced. (1 point)
Scoring and Interpretation
AP BIOLOGY PRACTICE EXAM 1
1. ____ / 10
2. ____ / 10
3. ____ / 4
4. ____ / 4
5. ____ / 4
6. ____ / 4
7. ____ / 4
8. ____ / 4
Add up the total points accumulated in the eight questions and multiply the sum by 1.57 to obtain the free-response raw score:
CALCULATE YOUR SCORE
Now combine the raw scores from the multiple-choice and free-response sections to obtain your new raw score for the entire practice exam. Use the ranges listed below to determine your grade for this exam. Don’t worry about how we arrived at the following ranges, and remember that they are rough estimates on questions that are not actual AP exam questions . . . do not read too much into them.
Answer Sheet for AP Biology Practice Exam 2
ANSWER SHEET FOR MULTIPLE-CHOICE QUESTIONS
PART B: GRID-IN QUESTIONS
AP Biology Practice Exam 2: Section I
PART A: MULTIPLE-CHOICE QUESTIONS
Time–1 hour and 30 minutes (for Parts A and B)
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
B. fixed-action pattern.
C. agonistic behavior.
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
B. active transport.
C. facilitated diffusion.
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?
C. Specialized transduction
7. Chemiosmosis occurs in
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?
B. Protein kinases
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. Which of the following cells is most closely associated with phagocytosis?
B. Plasma cells
C. B cells
D. Memory cells
13. Which of the following statements about photosynthesis 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?
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?
18. Which of the following tends to be highest on the trophic pyramid?
A. Primary consumers
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
20. The transfer of DNA between two bacterial cells connected by sex pili is known as
A. specialized transduction.
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?
For questions 24–27, please use the following answers:
A. Aposomatic coloration
B. Batesian mimicry
C. Deceptive markings
D. Cryptic coloration
24. Those being hunted adopt a coloring scheme that allows them to blend in to the colors of the environment.
25. An animal that is harmless copies the appearance of an animal that is dangerous as a defense mechanism to make predators think twice about attacking.
26. Warning coloration adopted by animals that possess a chemical defense mechanism.
27. Some animals have patterns that can cause a predator to think twice before attacking.
Questions 28–31 refer to the following choices:
28. Short sequence by promoter that assists transcription by interacting with regulatory proteins.
29. Protein that prevents the binding of RNA polymerase to the promoter site.
30. Transcription-affecting DNA region that may be located thousands of basepairs away from the promoter.
31. Basepair sequence that signals the start site for gene transcription.
Questions 32–35 refer to the following choices:
A. Divergent evolution
B. Convergent evolution
C. Parallel evolution
32. Two unrelated species evolve in a way that makes them more similar.
33. Similar evolutionary changes occurring in two species that can be related or unrelated.
34. The tandem back-and-forth evolution of closely related species, which is exemplified by predator-prey relationships.
35. Two related species evolve in a way that makes them less similar.
Questions 36–39 refer to the preceding pedigree.
36. 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
37. What is the probability that couple C and D will produce a child that has the condition?
38. Which of the following conditions could show the same kind of pedigree results?
A. Cri-du-chat syndrome
B. Turner syndrome
39. If child E does in fact have the condition, what is the probability that child F will also have it?
Questions 40–42: 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.
40. From the data presented above, one can conclude that these genes are
41. What is the crossover frequency of these genes?
A. 10 percent
B. 20 percent
C. 30 percent
D. 35 percent
42. 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 43–45: A laboratory procedure involving plants presents you with the data found in the following 2 charts:
Transpiration Rate → 1.0 = Control Rate (All Leaves Have the Same Surface Area)
43. 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 ↑
44. According to the Rf values given in the preceding smaller table, which pigment would migrate the fastest on chromatography paper?
B. Chlorophyll a
C. Chlorophyll b
D. Beta carotene
45. From the transpiration rate data presented in the preceding 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 46–48: 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.
46. 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.
47. 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.
48. 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 49–52: 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:
49. What is this lab setup called?
A. A gel sheet
B. A choice chamber
C. A potometer
D. An incubation chamber
50. 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.
51. After 20 minutes, 20 slugs are in the high-temperature, low-salinity quadrant. What kind of animal behavior has this experiment displayed?
52. 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 2 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.
53. 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
54. 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 55–57 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:
55. Which line represents the bag that contained a solution isotonic to the 0.5 M solution?
56. Which line represents that bag with the highest initial concentration of glucose?
57. Which line or lines represent bags that contain a solution that is hypertonic at 50 minutes?
A. A and B
D. D and E
58. 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 59 and 60:
59. 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
60. 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
61. A single gene from five related species of leafhopper 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 62 and 63 based on the following cladogram:
62. What is the common ancestor for B and E?
63. Which two species are most closely related?
A. A and E
B. A and B
C. B and C
D. D and E
PART B: GRID-IN QUESTIONS
Calculate the correct answer and enter it on the top line of the grid-in area with each number or symbol in a separate column. Then fill in the correct circle below each number or symbol you entered (only one filled-in circle per column).
1. 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?
2. 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.
3. If the pH of a solution is calculated using the equation pH = −log[H+], what is the pH of a solution with a hydrogen ion concentration of 1.33 × 10−8?
4. 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.
The equation for solute potential is Ψs = −i CRT where
i = ionization constant (assume that it is 1)
C = molar concentration
R = pressure constant (R = 0.0831 liter MPa/mole K)
T = temperature in kelvins (room temperature is 293 K)
5. Treatment of tomato plants with a growth hormone yielded the following weights of tomatoes: 100 g, 86 g, 123 g, 98 g, 104 g, 71 g. What is the average weight of a tomato after treatment?
6. After seven days of growth, a plant’s weight was 14.3 grams. The percent biomass of that plant was determined to be 23.1 percent. What amount of energy (in kcal) is stored in the plant, if the amount of stored energy = (g biomass) × 4.35 kcal?
AP Biology Practice Exam 2: Section II
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 3–8 are short-response questions that should require about 6 minutes each. Outline form is not acceptable. Answers should be in essay form.
1. The immune system is the body’s defense against foreign invaders and is divided into specific and nonspecific immunity, and humoral and cell-mediated immunity. Answer three of the following four questions:
A. Describe the primary immune response and how an invading antigen is met, dealt with, and eliminated. Describe the cells involved and how they are created.
B. Describe the mechanism by which the immune system deals with viruses, invaders that make it inside our cells.
C. Define nonspecific immunity, and list three examples of nonspecific defense mechanisms in humans.
D. Define the term vaccination, and describe how a vaccination works.
2. You just started working at a local laboratory and are usually given the grunt-work lab assignments to perform. Design and describe how you would do the following experiments:
A. Describe how you would design an experiment to prove the theory that photosynthesis requires both light and chloroplasts. Describe what equipment you would use, what your control would be, and how your expected outcome would support your hypothesis.
B. You are told that you need to determine how the following factors affect the rate of transpiration in plants: temperature, humidity, light intensity, and air movement. Describe how you would perform an experiment that could accomplish that task, and give your prediction of the expected results. Be sure to describe your experimental setup.
3. 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. Based on the data presented here, determine the P-generation genotypes. Provide Punnett squares that support your answer.
B. The white-scaled female in the F1 generation resulted from a mutational change. Explain what a mutation is and discuss a type of mutation that might have produced the white-scaled female in the F1 generation.
4. The idea of surface area is an important concept in biology. Explain how surface area plays a critical role in the digestive system.
5. The following table includes data from scan samples conducted on a fictional mammal called a googabear every 10 minutes over the course of 42 hours. At each scan, it was noted whether the googabear was active or inactive. The percentage of active (feeding, moving, engaging in social behavior) and inactive (resting or sleeping) scans recorded for each time period are shown in the table. Describe the pattern of activity for the googabear and discuss possible reasons for this pattern.
6. In earth’s early history, the evolution of photosynthesis in simple cells occurred before the evolution of more complex cells. Briefly describe the significance of photosynthesis being present first.
7. What evidence supports the theory that chloroplasts and mitochondria are evolved from prokaryotic cells?
8. You are asked to estimate if a certain species of plant could live in a salt marsh. You collect the following data:
• The overall Ψ of the soil (Ψsoil): −2.2 MPa
• Solute concentration of plant cell contents: 0.08 M (assume i = 1, and 12°C
• Pressure potential of the plant cells: −1.2 MPa
• R = 0.00831 liter MPa/mole K
Do you think the plant could grow in this environment? Why or why not? Show your work.
Answers and Explanations for Practice Exam 2
PART A: 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 19, 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.
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—Neutrophils are phagocytic cells of the immune system. They roam the body looking for rubbish to clear.
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.
36. 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.
37. 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.
38. C—Albinism is the only autosomal recessive condition on this list.
39. 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.
40. 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.
41. 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 = 1200). This results in 120/1200 or 10 percent.
42. B—One map unit is equal to a 1 percent recombination frequency.
43. B—The data in the table show you that this answer is the correct choice.
44. 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.
45. 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.
46. C—The average enamel thickness started at 10, increased to 12, and then increased to 15. It is therefore increasing overall.
47. 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).
48. B—Because thicker enamel in this species indicates foods that are more difficult to process, the answer is B.
49. B—Experimental setups where individuals are given a choice as to where to move are called “choice chambers.”
50. 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.
51. 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.
52. 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.
53. C—New genes are introduced into the bacterium through viral transduction.
54. 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.
55. 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.
56. A—The most water would diffuse into the most hypertonic solution; line A shows the biggest increase in weight.
57. B—Line B still shows an increase in weight at 50 minutes, whereas line A has leveled out and is isotonic at 50 minutes.
58. 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.
59. 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.
60. B—The promoter would be located upstream from where transcription would begin.
61. 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.
62. A—Both B and E branches originate from point 1.
63. C—Species B and C reside the closest to one another.
PART B: GRID-IN QUESTIONS
1. .0025—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 = .0025. In other words, 25 out of 10,000 people (0.25 percent) will have cystic fibrosis.
2. 13.33—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.
3. 7.88—pH = −log(1.33 × 10−8) = 7.88
4. –19.5—Ψs = −(1)(0.8 M)(0.0831 L bars/mole K)(294 K) = –19.5 MPa
5. 97—Add the weight of all tomatoes and divide by the number of tomatoes. The average is 97 grams.
6. 14.4—The computation is shown below:
(14.3 g)(0.231) = 3.30 g biomass
(3.30 g biomass)(4.35 kcal) = 14.4 kcal
Free-Response Grading Outline
1. Immune system question (here, the student can obtain 4 points from two of the answers; if 4 points are awarded for an answer, a maximum of 3 points can be obtained for each of the remaining answers)
A. (maximum 4 points)
• Definition of an antigen as a molecule foreign to the body. (½ point)
• Mentioning that the primary immune response is an example of humoral immunity. (½ point)
• Description of a B cell and how each B cell has a specific antigen recognition site on its surface that will match up with only one antigen. (1 point)
• When B cells meet and attach to the appropriate antigen, they become activated and undergo mitosis and differentiation into two types of cell. (1 point)
• The two types of cell are the memory cells and plasma cells. (½ point)
• Definition of plasma cells as the cells that produce the specific antibodies. (½ point)
• Definition of memory cells as the cells that head up the secondary immune response. (½ point)
• Description of how an antibody recognizes a particular antigen, including the fact that antibodies have two functional regions: Fab, which binds to the antigen; and Fc, which binds to the effector cells, and later comes in and cleans up the trash left behind. (1 point)
• Mentioning that complement is the one that binds to the antigen–antibody complex and aids in the quicker removal of the complex from the body. (½ point)
B. (maximum 3 points)
• Mentioning that this portion of the immune system is known as cell-mediated immunity. (½ point)
• Mentioning that the major player involved here is the cytotoxic T cell. (1 point)
• Mentioning that the cells infected by a virus are forced to produce viral antigens, some of which show up on the surface of the cell, and that it is these antigens that cytotoxic T cells recognize and attack. (1 point)
• Mentioning that all cells of the human body (except red blood cells) have class I histocompatibility antigens (MHC I) on their surfaces. (1 point)
• Further discussion of MHC—mentioning that MHC I antigens are slightly different for each person and the immune system accepts any cell that has the identical MHC I as friendly, and any cell that has a different form of MHC on its surface as an enemy. (1 point)
C. (maximum 4 points)
• Definition of nonspecific immunity as the nonspecific prevention of the entrance of invaders into the human body. (1 point)
• Examples (each example is worth 1 point)
a. Lysozyme in the saliva can kill germs before they have the chance to take hold.
b. The skin covering the body is a major form of nonspecific protection against invasion.
c. The mucous membrane lining the trachea and lungs prevent bacteria from entering cells and actually assist in the expulsion of bacteria by ushering them up and out with a cough.
d. The low pH of the stomach (acidity) is a nonspecific defense mechanism because it is able to kill a lot of bacteria that enter the body that cannot handle such an acidic environment.
D. (maximum 4 points)
• Definition of a vaccination as something given to an individual in an effort to prime the immune system to be prepared to fight a specific sickness if confronted again in the future. (1 point)
• Recognition that a vaccination is the injection of an antigen into the system (human body). (½ point)
• Description of how the reception of an antigen by a B cell causes B-cell differentiation into memory and plasma cells. (½ point)
• Mentioning that at the time of the vaccination, the plasma cells will produce antibodies to wipe out the small dose of antigen presented during the vaccination, and that the memory cells will remember the antigen and be ready to react later if necessary. (1 point)
• Definition of a secondary immune response. Memory cells are stored instructions on how to handle a particular invader. When the invader returns to the body, the memory cells recognize it and produce antibodies in rapid fashion. (1 point)
• Mentioning that the secondary immune response is faster and more efficient than the primary immune response. (½ point)
• Mentioning that the principle of a successful vaccination rests on the belief that the secondary immune response will succeed and wipe out the sickness if the individual is exposed in the future. (½ point)
2. Plant laboratory question
A. (maximum 5 points)
• Mentioning that the products of the light reactions of photosynthesis are ATP, NADPH, and oxygen. (½ point)
• Mentioning that in this experiment, the NADP+ would be replaced by a compound known as DPIP. (½ point)
• Mentioning that normally this compound DPIP has a nice blue color, but when reduced, it changes to a colorless solution. (½ point)
• Mentioning that a machine called a spectrophotometer will be used to measure the amount of light that can pass through various samples. (½ point)
• Description of the experiment.
a. Set aside three beakers—one with boiled chloroplasts, two with unboiled chloroplasts. (1 point)
b. Take initial reading on spectrophotometer to determine how much light passes through the unboiled chloroplasts before the experiment begins. (½ point)
c. Take one sample (unboiled chloroplasts) and measure how much photosynthesis occurs while it sits in a dark environment. After a certain amount of time, use the spectrophotometer to measure how much light can pass through the solution. (1 point)
d. Take a second sample (unboiled chloroplasts) and measure how much photosynthesis occurs when it is exposed to light. After a certain amount of time, use the spectrophotometer to measure how much light can pass through the solution. (1 point)
• Mentioning that they would now compare the two samples to see the effect of light on photosynthesis. (1 point)
• Take a third sample (boiled chloroplasts) and expose it to light, and after a certain period of time, measure how much light can pass through the solution. (1 point)
• Mentioning that they would now compare the third sample and the second sample to see the effect of the presence or absence of chloroplasts on photosynthesis. (1 point)
B. (maximum 5 points)
• Definition of transpiration as the evaporative water loss from plants. (1 point)
• Mentioning that they will use a potometer to measure the amount of water loss from plants. (1 point)
• Mentioning that the surface area of a leaf is important to the measurement of transpiration rate in an experiment of this nature. (½ point)
• Description of experiment.
a. Begin by measuring the amount of water that evaporates from the surface of a plant over a certain amount of time under normal conditions. Use this as the control. (1 point)
b. Change the temperature, humidity, light intensity, and air movement that the plant is exposed to by 5-degree increments, and measure the amount of transpiration that occurs at the various temperatures. (1 point for each variable mentioned up to a maximum of 2 points)
c. Mention that these values will be compared to the control to determine the effect of temperature, humidity, light intensity, and air movement. (½ point for each variable mentioned up to a maximum of 1 point)
• Mentioning that transpiration increases with an increase in temperature, decrease in humidity, increase in light intensity, and increased air movement. (1 point for each, up to a maximum of 2 points)
3. The phenotype for scale color in gila monsters is determined by a specific locus. (maximum 4 points for entire question)
A. P-generation genotypes (maximum 2 points)
• Creating the correct Punnett square, as shown here (1 point):
• Mentioning that genotype of P-generation = Gg (heterozygous) crossed with gg (homozygous recessive). (1 point)
B. Explanation for mutation (maximum 3 points)
• Mentioning that mutation is a random event that causes change in allele frequencies. (1 point)
• Possible explanation: scale-color gene is tied to another gene that controls pigment distribution (gene at one locus alters phenotypic expression of a gene at another locus). (1 point)
• Using the term epistasis correctly. (1 point)
• Possible explanation: point mutation occurred as DNA responsible for the production of protein that determines scale color was undergoing replication. (1 point)
4. The idea of surface area is an important concept in biology. Explain how surface area plays a critical role in the digestive system. (maximum 4 points for entire question)
• Mentioning that majority of absorption occurs in small intestine. (1 point)
• Mentioning that numerous folds and ridges increase surface area. (1 point)
• Mentioning that brush border and microvilli increase surface area. (1 point)
• Mentioning that large surface area leads to greater absorption of nutrients. (1 point)
• Mentioning that chewing (mastication) breaks up food into smaller pieces. (1 point)
• Mentioning that higher surface area gives greater access to salivary amylase. (1 point)
5. The following table includes data from scan samples conducted on a fictional mammal called a googabear every 10 minutes over the course of 42 hours. (maximum 4 points for entire question)
• Mentioning that googabears are more active from 6 a.m. until 6 p.m. (during daylight hours). (1 point)
• Mentioning that they decrease activity from 6 p.m. to 6 a.m. (when it is dark). (1 point)
• Mentioning that googabears’ food source is available during daylight. (1 point)
• Mentioning that googabear predators are nocturnal (out at night), so it is safest for googabears to remain hidden at night. (1 point)
• Mentioning that googabears rely on collective body heat at night (huddling); activity of huddled group is low. (1 point)
6. In earth’s early history, the evolution of photosynthesis in simple cells occurred before the evolution of more complex cells. (maximum 4 points for entire question)
• Mentioning that photosynthesis releases oxygen as a by-product. (1 point)
• Mentioning that photosynthesis led to an increase in atmospheric oxygen. (1 point)
• Mentioning that the presence of more oxygen in the atmosphere allowed the evolution of cellular respiration. (1 point)
• Mentioning that oxygen allowed cells to generate more energy and grow larger and more complex. (1 point)
• Mentioning that the first photosynthetic cells were prokaryotic. (1 point)
• Mentioning that eukaryotic cells could not evolve until there was a higher level of atmospheric oxygen. (1 point)
7. What evidence supports the theory that chloroplasts and mitochondria are evolved from prokaryotic cells? (maximum 4 points for entire question)
• Mentioning that chloroplasts and mitochondria have their own DNA. (1 point)
• Mentioning that chloroplast and mitochondrial DNA consist of a single, circular molecule (like bacterial DNA). (1 point)
• Mentioning that chloroplast and mitochondrial DNA are not associated with histones (like bacterial DNA). (1 point)
• Mentioning that chloroplasts and mitochondria replicate by a process similar to prokaryotes. (1 point)
• Mentioning that the inner membranes of both organelles have enzymes homologous to those found in prokaryotes. (1 point)
8. You are asked to estimate if a certain species of plant could live in a salt marsh. (maximum 4 points for entire question)
• Calculate Ψplant cell:
Ψ = pressure potential + solute potential solute potential of plant cell = Ψs = –iCRT = –(1)(0.08 M)(.00831)(273 + 12) = –0.189 MPa (1 point)
Ψ = –1.2 MPa (pressure potential) + –0.189 MPa (solute potential) = –1.39 MPa (1 point)
• Mentioning that the plant cell’s water potential (–1.39 MPa) is higher than that of the soil (–2.2 MPa). (1 point)