1. D
Because of the cell wall, a plant cell will not lyse when placed in a hypotonic solution. Instead, it will just swell. Both plant and animal cells perform cellular respiration (A) and have transcription factors (B), and both have vacuoles (C).
2. C
If the contents of the cell separated from the cell wall, then water was moving out of the cell. This would cause the space between the cell wall and the cell membrane to expand, so eliminate (B). This also means that the concentration of solutes in the extracellular environment would be hypertonic with respect to the cell’s interior. You can also eliminate (A) because the fluid in the cell was hypotonic to the sugar solution, fluid was moving out of the vacuole, which caused it to become smaller. Choice (C) is the correct answer because if the water was rushing out of the cell, then the water potential inside the vacuole would have been higher than outside the cell. High water potential means the water wants to move. Choice (C) says the water potential would be lower.
3. D
This question tests your ability to associate what happens when enzymes are denatured and what would happen in the synaptic cleft. Acetylcholinesterase is an enzyme that degrades acetylcholine in the synaptic cleft. If acetylcholinesterase is denatured, acetylcholine will still be released from the presynaptic membrane (the nerve before the cleft) and continue to diffuse across the synaptic cleft. Acetylcholine will still bind to the postsynaptic membrane (on the nerve after the cleft) because acetylcholine is not degraded. Therefore, you can eliminate (A), (B), and (C).
4. B
The ratio of purines to pyrimidines should be constant because purines always bind with pyrimidines, no matter which ones they may be.
5. D
Feedback inhibition occurs when something created by a process then inhibits that process. Choices (A) and (B) involve stimulations, so they can be eliminated. Choice (C) doesn’t make sense because it says that “C” is an enzyme and the passage says that the enzymes are the numbers shown between the steps.
6. D
If substance F leads to the inhibition of enzyme 3, then substances D and E and enzymes 3, 4, and 5 will be affected. The activity of enzyme 5 will be decreased, not increased.
7. C
If enzyme 1 were inhibited, then everything that comes after that in the pathway would be reduced. A would not be reduced, but both C and X would. Choice (C) is correct.
8. C
Based on the graph, fetal hemoglobin has a higher affinity for oxygen than maternal hemoglobin. Fetal hemoglobin does not give up oxygen more readily than maternal hemoglobin, so eliminate (A). You can also get rid of (B), as the dissociation curve of fetal hemoglobin is to the left of the maternal hemoglobin. Finally, eliminate (D) because fetal hemoglobin and maternal hemoglobin are different structurally, but you can’t tell this from the graph.
9. C
The passage says that high CO2 and low pH lead to reduced affinity. A right shift curve would represent a reduced affinity. Photosynthesis consumes CO2, so this would not increase the CO2. Mitosis and respiration are not related. Fermentation is done in cells that are doing anaerobic cell respiration, so CO2 should be released, possibly along with lactic acid, which would lower pH.
10. A
Hemoglobin’s affinity for O2 decreases as the concentration of H+ increases (or the pH decreases) and as the concentration of OH– increases (or the pH increases). As the pH decreases, the affinity for oxygen will decrease, and as the pH increases, the affinity for oxygen will increase.
11. D
The CO2-rich environment would decrease the affinity of hemoglobin for oxygen, so the curve should shift toward the right. The regular curve seems to hit 90% saturation around 45–50 pO2, so the CO2-rich curve would take more oxygen to get there.
12. B
Both prokaryotes and eukaryotes possess double-stranded DNA. Choices (A), (C), and (D) are correct differences between prokaryotes and eukaryotes.
13. C
Bb and Bb is the correct set of genotypes that represents a cross that could produce offspring with silver fur from parents that both have brown fur. Complete a Punnett square for this question. In order for the offspring to have silver fur, both parents must have the silver allele.
14. A
Retroviruses have an RNA genome (hereditary material, eliminating (B)) that must be reverse-transcribed into DNA (which is the opposite of the Central Dogma first step) and therefore (A) is correct. The codon sequence is used by all things that do translation (like viruses, eliminating (C)). Retroviruses use transcription as well after their genome has been reverse transcribed, so (D) is also incorrect.
15. D
All of the choices are examples of hydrolysis except the conversion of pyruvic acid to glucose. Hydrolysis is the breaking of a covalent bond by adding water. In all of the correct examples, complex compounds are broken down to simpler compounds. The conversion of pyruvic acid to glucose is an example of decarboxylation—a carboxyl group is removed as carbon dioxide and the 2-carbon fragments are oxidized.
16. A
Peroxisomes catalyze reactions that produce hydrogen peroxide, mitochondria contain enzymes involved in cellular respiration, and ribosomes are involved in protein synthesis. Eliminate (B) and (D) because lysosomes are the sites of degradation; they contain hydrolytic enzymes but do not produce hydrogen peroxide. Choice (C) is incorrect, as the Golgi apparatus sorts and packages substances that are destined to be secreted out of the cell.
17. A
The figure shows that water leaves the descending limb and flows by osmotic pressure into the surrounding space and then into the vasa recta. This means that each of those places is more hypertonic than the previous. The most hypotonic is the descending tubule. (This is why the water leaves.)
18. B
When water flows out of the tubule, it is moving by simple osmosis. However, because it is a polar molecule, it must pass through an aquaporin channel, which is facilitated diffusion.
19. D
ADH makes the collecting duct permeable to water. If it is inhibited, then it will be unable to do that, meaning choice (D) is correct. The question doesn’t say that the inhibitor is inhibiting the aquaporins directly. Instead, the inhibitor says it is inhibiting the ADH.
20. C
To make concentrated urine, the medulla must be very osmotic because it must reclaim a lot of water from the collecting duct. To make a very osmotic medulla, the Loop of Henle must be very long. The collecting duct should be very permeable, not impermeable, because concentrated urine means that a lot of water has been reclaimed.
21. C
The carrying capacity is the maximum number of organisms of a given species that can be maintained in a given environment. Once a population reaches its carrying capacity, the number of organisms will fluctuate around it.
22. D
The growth of curve A is exponential, meaning that the bacteria are replicated as fast as possible without any restrictions. It looks like the populations started at the same level, and an inhibitor would not cause high growth as shown in culture A. Not being measured as often would not cause a graph like the one shown for culture B.
23. A
Phototropism is the ability of a plant to move with regard to the sun. A climbing vine uses pressure to sense the trellis. A tree blooming once a year is using photoperiodism.
24. C
Females with Turner’s syndrome lack an X chromosome. If females with this syndrome have a high rate of hemophilia, they must not have the second X to mask the expression of the disease.
25. D
This type of mutation is called frameshift mutation. The insertion of DNA leads to a change in the normal reading frame by one base pair. The other answer choices refer to chromosomal aberrations. Choice (A), duplication, occurs when an extra copy of a chromosome segment is introduced. Choice (B), translocation, occurs when a segment of a chromosome moves to another chromosome. Choice (C), inversion, occurs when a segment of a chromosome is inserted in the reverse orientation.
26. C
The principal inorganic compound found in living things is water. Water is a necessary component for life.
27. D
An extra X chromosome could be created by a problem separating the homologous chromosomes or sister chromatids. When they are lined up, it would not be apparent that they were not separating properly, but once they start to pull apart, one could see if a mistake had occurred.
28. B
This scenario describes natural selection, which eliminates (C) and (D). In this situation, the change in tigers is caused by a long period of stasis followed by a major event: this is punctuated equilibrium (B), and not gradualism (A).
29. A
Metabolic efficiency relates directly to the surface area-to-volume ratio. A higher ratio means higher efficiency. Since the ratio is (2πrh + 2πr2) / πr2h this simplifies to 2/r + 2/h, so if the radius or height is higher, the ratio is smaller. Therefore the smallest radius and height (A) gives the highest ratio of 5.33 and the highest efficiency.
30. C
Helicase is the enzyme that unwinds the helix. The closed helix would exist without helicase.
31. C
The RNA polymerase is not bound, but there is a repressor bound. The transcription must be repressed.
32. D
These four phases must be transcription. Transcription begins with a closed helix, then an enhancer binds, and then RNA polymerase is seen making RNA. Finally, the closed helix and the completed mRNA strand exist.
33. B
Photosynthesis requires light, carbon dioxide, and water and produces oxygen and glucose.
34. B
Avery, MacLeod, and McCarty concluded that DNA is the hereditary material because it was able to transform non-pathogenic bacteria into pathogenic bacteria. Watson, Crick, and Franklin determined the double helix, and Meselson determined that replication is semi-conservative.
35. B
Translation, the synthesis of proteins from mRNA, occurs in the cytoplasm. DNA replication (I) occurs in the nucleus. Transcription (II), the synthesis of RNA from DNA, occurs in the nucleus.
36. C
Crossing-over permits scientists to determine chromosome mapping. Chromosome mapping is a detailed map of all the genes on a chromosome. The frequency of crossing-over between any two alleles is proportional to the distance between them. The farther apart the two linked alleles are on a chromosome, the more often the chromosome will break between them. Crossing-over does not tell us about the chance of variation in zygotes, the rate of mutations, or whether the traits are dominant, recessive, or masked.
37. B
In this scenario, the inside of the cell is hypertonic to the outside environment—this will cause water to move into the cell and can cause the cell to lyse.
38. B
The most likely explanation for this phenomenon is that these birds have different ecological niches. An ecological niche is the position or function of an organism or population in its environment. Eliminate (A) because we do not know if there is a short supply of resources. Choice (C) can also be eliminated because we do not know how long the bird species live together. The breeding patterns of the bird species do not explain the lack of competition, which eliminates (D) as well.
39. B
This relationship is an example of parasitism. Parasitism is a form of symbiosis in which one organism benefits and the other is harmed. Choice (A), commensalism, is a form of symbiosis in which one organism benefits and the other is unaffected. Choice (C), mutualism, is a form of symbiosis in which both organisms benefit. Choice (D), gravitropism, is the growth of a plant toward or away from gravity.
40. C
If the nucleotide sequence of a DNA molecule is 5′-C-A-T-3′, then the transcribed DNA strand (mRNA) would be 3′-G-U-A-5′. The nucleotide sequence of the tRNA codon would be 5′-C-A-U-3′.
41. A
Viruses are considered an exception to the cell theory because they require a cellular host. They can survive only by invading a host. Choice (B) is incorrect, as all viruses have genomes and some have many genes. Choice (C) is also wrong because prokaryotes are cells and do not have nuclei. You can also eliminate (D), as viruses evolve through the same mechanisms that living things do.
42. C
The origin of life refers to the period when the first life began. Life is shown to be present 3.5 billion years ago when no atmospheric oxygen was present yet, so (A) can be eliminated. Also, there was no life at 4 billion years despite having CO2 and no O2; therefore something else must be required, and (B) can be eliminated. There were no proteins when the first life arose so (D) is incorrect. Choice (C) states that self-replicating nucleic acids were necessary, which is why life could only occur when RNA appeared.
43. B
Functional ribosomes would not likely be found before the proteins they make. Therefore, this would be between 3.5 and 3.25 billion years ago.
44. B
The sign that photosynthesis began was when oxygen appeared. Oxygen was absent at 3.25 billion years and present at 3 billion years so photosynthesis must have begun between those times; eliminate (A) and (C). Autotrophs, not heterotrophs, perform photosynthesis. Heterotrophs rely on eating other things for their energy.
45. B
The similarity suggests that humans and chimpanzees are more closely related than humans and dogs. Because these two organisms share similar amino acid sequences, humans must share more recent common ancestors with chimpanzees than with dogs.
46. A
There is no way for these two parents to produce a type O child. The only genotype that will result in type O blood is two “O” alleles, one from each parent, since the allele for O blood is recessive to the alleles for A or B blood. The AB parent has alleles for only A and B blood, so it is impossible for these two individuals to produce a child with type O blood.
47. B
The largest amount of energy is available to producers. Population B is most likely composed of producers because they have the largest biomass.
48. B
An increase in the number of organisms in population C would most likely lead to a decrease in the biomass of B because population B is the food source for population C. Make a pyramid based on the biomasses given. If population C increases, population B will decrease. Eliminate (A) and (C), as we cannot necessarily predict what will happen to the biomass of populations that are above population C. Choice (D) can also be eliminated because the food source available to population C would most likely decrease, not increase.
49. B
Chromosomes replicate during interphase, the S phase. Choices (A) and (C) are incorrect because during G1 and G2, the cell makes protein and performs other metabolic duties.
50. A
The first sign of prophase in mammalian cells is the appearance of chromosomes, which are usually invisible. They are condensed at the start of mitosis.
51. C
To induce cell cycle progression, an inactive CDK (cyclin-dependent kinase) binds a regulatory cyclin. Once together, the complex is activated, can affect many proteins in the cell, and causes the cell cycle to continue. To do this, the CDK transfers phosphate groups onto other molecules. If there are inadequate phosphate groups, CDK would not function properly and the cell cycle would not progress, which rules out (A). Tumor suppressor proteins inhibit cell cycle progression and can trigger apoptosis, which rules out (B). To inhibit cell cycle progression, CDKs and cyclins are kept separate. Together, they promote the cell cycle. If a cyclin is unable to release from its cyclin dependent kinase, the cell cycle would be promoted, not stopped, making (C) correct. Choice (D) could’ve been elimianted because an inhibitor of a cyclin gene would prevent expression of the cyclin gene. With lower levels of the cyclin protein, cell cycle progression would be inhibited.
52. B
Because neurons are not capable of dividing, it is reasonable to conclude that these cells will stay in the G1 phase (specifically in an extension of G1 called G0). phase. This is a reading comprehension question. The passage states that cells that do not divide are arrested at the G1 phase. Choice (A) is incorrect because these cells will not be committed to go through cell division. You can also eliminate (C) and (D), as the cells will not enter the M or S phase.
53. D
Since Snippeiq has the most differences with the other organisms it must be at position S. This means that Gerdellen cannot be there, which makes (D) the best answer. (Position Q or R is the best position for Gerdellen, though positions O and P could possibly be correct as well.)
54. C
Position III, (C), contains the oldest common ancestor because it is the node for all the current species designated by the lettered positions.
55. A
Snippeiq, (A), is the outgroup because it is the least related to the other species. The table shows that Snippeiq has the most differences from all of the other species.
56. D
Snorflak and Fixxels, (D), would have the most shared derived characters because they have the fewest amino acid differences. Snippeiq and Sqellert would be expected to have the least since they are so different, which rules out (A). Sqellert and Gerdellen, (B), were close with only 4 differences but still not as low as Snorflak and Fixxels. Snorflak and Gerdellen, (C), must be an intermediate with 9 differences observed in the table.
57. B
The two most closely related organisms are the two with the most shared derived characteristics.
58. D
Shared derived characteristics are newly evolved traits that are shared with every group on a phylogenic tree except for one. Vertebral columns are present in every group except for the sea anemone, so it must have evolved first. Walking legs are found only in the salamander, indicating that it most likely evolved most recently.
59. C
Pre-zygotic barriers to reproduction are those that prevent fertilization, so you can eliminate (A), (B), and (D). Choice (C) is an example of a post-zygotic barrier to reproduction.
60. B
Convergent evolution occurs when two organisms that are not closely related independently evolve similar traits, such as the wings of insects and birds. Divergent evolution occurs when two closely related individuals become different over time and can lead to speciation.
Short student-style responses have been provided for each of the questions. These samples indicate an answer that would get full credit, so if you’re checking your own response, make sure that the actual answers to each part of the question are similar to your own. The structure surrounding them is less important, although we’ve modeled it as a way to help organize your own thoughts and to make sure that you actually respond to the entire question.
Note that the rubrics used for scoring periodically change based on the College Board’s analysis of the previous year’s test takers. This is especially true as of the most recent Fall 2019 changes to the AP Biology exam! We’ve done our best to approximate their structure, based on our institutional knowledge of how past exams have been scored and on the information released by the test makers. However, the 2020 exam’s free-response questions will be the first of their kind.
Our advice is to over-prepare. Find a comfortable structure that works for you, and really make sure that you’re providing all of the details required for each question. Also, continue to check the College Board’s website, as they may release additional information as the test approaches. For some additional help, especially if you’re worried that you’re not being objective in scoring your own work, ask a teacher or classmate to help you out. Good luck!
a. Explain why the ability to respond to a visual stimulus would be naturally selected. Would a rod receptor with a resting membrane potential of –65mV be selected for? (2 points)
If an organism could use visual cues, this could give it an advantage since it might be able to navigate the world more easily. It might access alternative ecological niches or visualize predators or other dangers in the world. Increasing survival will increase evolutionary fitness and having this trait will be selected for. A rod with a low resting membrane potential of –65mV would not be selected for because when the cell is hyperpolarized to such a large degree the passage says it does not respond to light stimuli anymore. This means that the visual cues will be absent and the trait won’t be selected for.
b. Explain how the control in Experiment 1 would differ from the control in Experiment 2. (3 points)
The control in Experiment 1 would be the rod cell in the vitreous humor that is not stimulated by light. Experiment 2 would have a similar control except that it would be in vitreous humor devoid of sodium and then without stimulation by light.
c. Describe the relationship between a stimulus and hyperpolarization. Use evidence from the data. (3 points)
When the stimulus was a low light intensity, it caused a small hyperpolarization shown by the small downward bump in Figure 2. However, when the stimulus was a high light intensity, the membrane potential dipped very low, all the way to –65mV. The intensity of the light stimulus and the level of hyperpolarization have a direct relationship.
d. If the rods were bathed in fluid with a high sodium content, explain how the resting membrane potential would change. How does the sodium change the resting membrane potential? (1 point)
If the sodium content were high in the fluid, it would make the resting membrane potential more positive/less negative, perhaps –20mV. This is the opposite of what happened when the sodium was removed from the vitreous humor. The cell would depolarize at rest because sodium is always flowing through the sodium channel and going into the cell and the positive sodium ions make it less negative. When a light stimulus occurs, the channel closes and the cell hyperpolarizes.
a. Explain how biotic and abiotic factors affect where plants grow. (2 points)
Biotic factors are things that are alive, like plants and fungi and animals and bacteria. Some of these things compete with the plants, but other things like certain types of fungi and bacteria can help the plant acquire nutrients. Abiotic things are nonliving, like sunlight and soil, nutrients, water, and atmosphere composition. These things are also important, especially since plants need appropriate water and sunlight for photosynthesis.
b. Construct a graph of averages and standard error for the western and eastern flowers. (4 points)
1 point—proper axes
1 point—proper size of bars showing mean
1 point—correct labels
1 point—graphing error bars
c. Perform a chi-squared analysis using the average data for both flower types. (3 points)
Using the null hypothesis that the flowers do not prefer the Western side, we would have expected values of equal numbers living in each place. There were 2,196 flowers total in the average. So, we can expect that there would be 1,098 flowers in each location for the null hypothesis. For the western flowers observed – expected = 1,303 – 1,098 = 205 and for the eastern side observed – expected = 892 – 1,098 = –206. When we square each of them, it is 42,025 and 42,436. Divide each by the expected = 42,025/1,098 and 42, 436/1,098 = 38.27 and 38.64. Those get added together = 76.91 for the chi-squared value.
d. Using your chi-squared analysis data, explain if you can reject the null hypothesis that flowers do not prefer the western side. (1 point)
Yes, the null hypothesis can be rejected. The flowers appear to grow significantly more on the Western side.
a. Describe how a mutation in DNA can lead to a protein that no longer functions correctly.
Mutations in DNA are a mistake in the recipe used to make a protein. When the DNA gets expressed, it gets turned into an RNA during transcription. This causes an error in the mRNA. Then, the mRNA gets translated and the codons get read and amino acids get added to build the protein. If there was a mistake in the RNA, then the wrong amino acid would be added or maybe a Stop codon would be added. This can cause the protein to be the wrong shape when it folds up. Protein shape is important for protein function and if it is the wrong shape, it probably won’t work correctly.
b. Explain the impact of the radiolabeled dye on the cell cycle.
The radiolabeled dye does not impact the cell cycle. It is only being used to help track the phases of the cell cycle and visualize them. We know the dye has no effect because it is used in the wild-type cells and it doesn’t cause them to have an arrested cell cycle.
c. Predict what would happen if the cdc mutants were initially incubated at 37°C and then switched to 25°C.
If the mutants were started at 37°C, they would not grow. This is what happened in Experiment 1. If they were switched to 25°C, then they would be able to start growing.
d. Justify your prediction.
I predicted that based on what occurred in Experiment 2. In that experiment they started at the lower temperature which allowed the cell cycle to proceed. Then at the higher temperature it caused them to stop the cell cycle. I predict if they begin at the high temp, they won’t have a cell cycle and then when they shift to the lower temperature, it will begin to grow and the cell cycle will continue.
a. Describe what happened to the biodiversity of the ecosystem before and after the fire.
The biodiversity of the ecosystem increased following the fire.
b. Explain your description in (a) using the Simpson’s Diversity Index.
Before the fire:
Insect
n
n/N
(n/N)2
Hazel-hued bugs
3
3/44 = 0.068
0.004624
Hairy bear moths
14
14/44 = 0.318
0.101124
Telluride tiger beetles
6
6/44 = 0.136
0.018496
Golden-winged butterflies
21
21/44 = 0.477
0.227529
D = 0.004624 + 0.101124 + 0.018496 + 0.227529 = 0.351773
Simpson’s Diversity Index = 1 – D = 0.648227
After the fire:
Insect
n
n/N
(n/N)2
Hazel-hued bugs
10
10/44 = 0.227
0.05129
Hairy bear moths
9
9/44 = 0.205
0.042025
Telluride tiger beetles
15
15/44 = 0.341
0.098596
Golden-winged butterflies
10
10/44 = 0.227
0.05129
D = 0.05129+ 0.042025 + 0.098596+ 0.05129 = 0.24301
Simpson’s Diversity Index = 1 – D = 0.756799
The Simpson’s Diversity Index increased, meaning that biodiversity of the ecosystem increased.
c. Predict what would happen to the biodiversity if there was an invasive insect species introduced into this ecosystem.
The addition of an invasive species is likely to decrease the biodiversity of the ecosystem.
d. Justify your prediction.
Since invasive species are unlikely to have a natural predator, they will likely increase in number and consume a large portion of the resources leading to other organisms being unable to consume those numbers and decreasing in number. Therefore a greater percentage of the organisms would be made up of the invasive species while fewer would be made up of the original species decreasing the biodiversity.
a. Explain how it is possible for a person to have two rare unlinked genetic diseases at the same time.
Everybody has thousands of genes, and it is possible for a person to have mutations in more than one gene. This causes them to have a phenotype that is the combination of all their genes. If a person has the bad luck to have more than one mutated gene, then they could have more than one genetic disease at the same time.
b. Calculate the ratios of affected males and affected females for Disease 2.
There are 21 males and 4 are affected = 4/21
There are 20 females and 0 are affected = 0/21
c. Identify the likely inheritance patterns of Disease 1 and Disease 2.
Disease 1 is autosomal dominant
Disease 2 is X-linked recessive
d. Describe how inheritance patterns are shaped by the chromosomes upon which the genes are located.
If a gene is located on an X or a Y chromosome, then it will be inherited differently in males and female since males have XY and females have XX. This is sex-linked inheritance. In this pedigree with X-linked recessive, the affected males never have affected fathers. They get the mutated X from their mothers, but the mothers are never affected with the disease themselves. Things on autosomes get inherited the same in males and females because they each have the same number of autosomes.
a. Describe how you can determine that tortoises raised at 34°C had the highest variability in enzyme reaction rates.
Russian tortoises raised at 34°C had the largest error bars and therefore the largest variability.
b. Reptiles in the study needed to have a gertimtonase reaction rate exceeding 10 mmol/s for an observed improvement in water retention. Describe the range of temperatures that Russian tortoises could be raised in to achieve this benefit.
Russian tortoises could be raised between 37°C and 30°C for gertimtonase to have the water retention effect.
c. Desert tortoises live in environments where the ground temperature can exceed 60°C. Using the data above, predict and justify how the graph would look for desert tortoise gertimtonase.
The graph’s peak would shift to the higher temperatures. Because the tortoise has lived in the desert for many years, its enzymes you would expect to be optimized for a higher temperature. Therefore, we would expect to see the peak around 60°C with reaction rates dropping significantly at lower and higher temperatures.
d. Explain at a molecular level how the temperature change might affect gertimtonase.
Raising temperatures for Russian tortoises to those exceeding 37°C would likely denature the enzyme. Denaturation of the enzyme greatly limits its function which would lower the reaction rate.