1. D
The resting potential depends on active transport (the Na+K+-ATPase pump) and the selective permeability of the axon membrane to K+ than to Na+, which leads to a differential distribution of ions across the axonal membrane.
2. A
The Krebs cycle releases carbon dioxide as the carbon molecules are broken down and electron carriers are generated. Glucose is at the start of glycolysis and pyruvate is at the end of glycolysis. Lactic acid is generated through fermentation.
3. B
A heterotroph obtains its energy from organic molecules. An autotroph obtains energy from sunlight utilizing pigments such as chlorophyll and uses CO2 and water to make organic molecules. Therefore, (A) and (C) can be eliminated. Heterotrophs can obtain their energy from ingesting autotrophs, but they can also consume other heterotrophs. So you can eliminate (D), leaving (B) as your answer.
4. B
In meiosis, the sister chromatids separate during the second metaphase of meiosis (Meiosis II), whereas the sister chromatids separate during metaphase of mitosis. Choice (A) is incorrect because in meiosis there are two rounds of cell division, whereas in mitosis there is only one round of cell division. Chromosomes are replicated during interphase in both meiosis and mitosis, so (C) is incorrect. Choice (D) can also be eliminated because spindle fibers never form during interphase.
5. C
Amino acids are organic molecules that contain carbon, hydrogen, oxygen, and nitrogen, so eliminate (A), (B), and (D). Don’t forget to associate amino acids with nitrogen because of the amino group (NH2).
6. B
Unlike eukaryotes, prokaryotes (which include bacteria) do not contain membrane-bound organelles. Bacteria contain circular double-stranded DNA, ribosomes, and a cell wall, so (A) and (D) are incorrect. Also eliminate (C) because bacterial cell membranes are made up of a bilipid layer with proteins interspersed.
7. D
Populations can be described as the evolutionary unit because changes in the genetic makeup of populations can be measured over time. Eliminate (A), as genetic changes occur only at the individual level. Only under Hardy-Weinberg equilibrium does the gene pool remain fixed over time in a population. However, this statement does not explain why the population is the evolving unit, so (B) is incorrect. Choice (C) is true but does not address the question.
8. A
Positive feedback occurs when a stimulus causes an increased response. Choices (B), (C), and (D) are examples of negative feedback.
9. B
In order to determine the genotype of the parents, use the ratio of the offspring given in the question and work backward. The ratio of black-haired to white-haired guinea pigs is 3:1. In order to get a white-haired offspring, each parent must have been able to contribute a b allele. However, since the parents were both black in color, they must have each been Bb.
10. D
The mean weight of the offspring in the next generation will be heavier than the mean weight of the original population because all the lighter horses in the original population died off. The normal distribution for weight will therefore shift to the heavier end (to the right of the graph). You can therefore eliminate (C) because the mean weight should increase. The mean weight of the offspring could be heavier or lighter than their parents, so you can also eliminate (A) and (B).
11. C
Apoptosis (programmed cell death), hox and homeotic genes (genes that control differentiation), and differentiation itself play a role in morphogenesis. Operons are sets of multiple genes regulated by a single regulatory unit in bacteria.
12. A
The primitive atmosphere lacked oxygen (O2). It contained methane (CH4), ammonia (NH3), and hydrogen (H2).
13. A
Villi and microvilli are fingerlike projections present in the small intestine which dramatically increase the surface area available for nutrient absorption.
14. C
The pancreas produces digestive enzymes, and enzymes are sensitive to pH. The inability to neutralize stomach acid would disrupt the function of the enzymes. Food moving too quickly would decrease digestion/absorption rates but the function of the enzyme would be intact. Removal of water should not greatly affect enzyme function.
15. D
Ribosomes are the site of protein synthesis. Therefore, the correct answer should start with ribosome. So eliminate (A) and (C). The polypeptide then moves through the rough ER to the Golgi apparatus, where it is modified and packaged into a vesicle. The vesicle then floats to the plasma membrane and is secreted. Choice (D) is your answer.
16. D
Choice (D) is false because an individual with two identical alleles is said to be homozygous, not heterozygous, with respect to that gene. Alleles are different forms of the same gene found on corresponding positions of homologous chromosomes, so (A) and (B) are incorrect. More than two alleles can exist for a gene, but a person can have only two alleles for each trait.
17. C
A transformation is the uptake of DNA by a bacterium. DNA uptake by a virus (A) would be a transduction, although this requires a virus to gain the DNA via infection of a bacterium.
18. D
The least likely explanation for why mutations are low is that mutations produce variability in a gene pool. Any gene is bound to mutate. This produces a constant input of new genetic information into a gene pool. Choice (D) is therefore correct, because it doesn’t give us any additional information about the rate of mutations. Some mutations are subtle and cause only a slight decrease in reproductive output, so eliminate (A). Some mutations are harmful and decrease the productive success of the individual, so (B) is incorrect. Some mutations are deleterious and lead to total reproductive failure. The zygote fails to develop. Therefore, (C) can also be eliminated.
19. B
The independent variable is the condition the scientist sets up. The dependent variable is the thing that is the measured outcome of the experiment. The scientist should set different temperatures and then measure the germination rate. The outdoor temperature is interesting, but it is too variable to be a good control for the groups with a set temperature.
20. D
Sympatric speciation occurs when there is no geographic barrier separating the two species (as is the case here). Allopatric speciation requires a geographic barrier.
21. B
Every cell in an organism has the same set of genes. Differences in the timing and expression levels of these genes lead to structural and functional differences.
22. C
Deoxygenated blood flows through all chambers before going to the lungs. This means the correct answer is chambers 1, 2, 3, and 4.
23. C
The action potentials cause the heart to contract. If chamber 3 is full of more blood, then that would indicate that the preceding chamber, chamber 2 (right atrium), had suddenly contracted a larger amount.
24. B
Normal cells can become cancerous when a virus invades the cell and takes over the replicative machinery. This would occur if the mammalian genome is altered, such as if the viral genome inserted itself and disrupted mammalian gene expression. A pilus forms between two bacteria, so (A) is wrong. Also eliminate (C) because the host’s genome is not converted to the viral genome. Choice (D) is incorrect because spores are released by fungi, not viruses.
25. C
Crossing-over and synapsis occur during meiosis, which produces haploid gametes. Separation of homologous chromosomes occurs during meiosis I, while separation of sister chromatids does not occur until meiosis II.
26. D
If potential mates have different alleles, this doesn’t keep them from mating. Having different alleles of a gene is just a normal part of genetic variation. Use common sense to eliminate the other answer choices. If the organisms don’t meet, they won’t reproduce; eliminate (A). Also eliminate (B) and (C): if potential mates do not share the same behaviors (such as courtship rituals), they may not mate.
27. D
There is no union of gametes in mitosis. Choices (A) and (C) are incorrect: asexual reproduction involves the production of two new cells with the same number of chromosomes as the parent cell. If the parent cell is diploid, then the daughter cells will be diploid. The daughter cells are identical to the parent cell, so eliminate (B).
28. D
Cohesion, adhesion, and capillary action are special properties of water that are necessary for transpiration to occur. Water is a hydrophilic, polar molecule.
29. B
The two species differ in their preference for water volume. Species A prefers 80 mL, and Species B prefers 40 mL. The different watering times do not seem to play a role in growth as there is no obvious pattern and all of the times had similar growth. Both plants prefer pH 7.
30. A
The top growth conditions for Species B are pH 7, 40 mL, and any time of day. In those conditions, the Species B plants grew the tallest.
31. D
The plants in the watering time experiment seemed to be around 60 and 20 cm in height. This corresponds to a pH of 7 rather than 4 and a watering volume of 80 mL rather than 40 mL.
32. D
Choices (A), (B), and (C) would all give interesting information about the experiment, but the only one that would make the results more statistically significant is to increase the number of plants in each group. Two plants is not enough to be sure about the results of the experiment.
33. D
Photoperiodism is a response of plants to the changing length of day/night. It is a way that plants respond to a signal from the environment. Animal circadian rhythms are also a response to the environment (D). Viral infection (A), increased appetite (B), and meiotic division (C) are not responses to the environment.
34. B
The gross primary productivity is the total amount of energy stored, but the plant has energy needs of its own that are fulfilled with cellular respiration. Choice (A) is true but not relevant to this question. Choice (C) is not true. Heterotrophs must get energy from autotrophs and (D) is not true because most biomass is stored at the producer level.
35. C
Choices (A) and (B) are incorrect: these organisms exhibit the same behavior because they were subjected to the same environmental conditions and similar habitats. This is an example of convergent evolution. However, they are not genetically similar, so (C) is the answer. (One is an insect, and the other a bird.) They are analogous, so (D) is also incorrect. They exhibit the same function but are structurally different.
36. C
A mutualistic relationship is a relationship among two organisms in which both benefit. Choice (A) describes parasitism, and (B) describes commensalism. Choice (D) is an example of altruistic behavior.
37. D
Beta cells secrete insulin. Binding of antibodies to the beta cells in the pancreas will halt the production of insulin. Therefore, eliminate (A), (B), and (C). This will lead to an increase in blood glucose levels.
38. C
The graph does show a drought, but it says nothing about genetic variation. It is true that the rainfall could affect evolution of the population, but the graph doesn’t address evolution. It shows only the decrease in biomass. Choice (C) is the best answer since it addresses how the environmental effects ripple through the levels of the ecosystem. Invasive species do not need an unoccupied niche, and that is not shown in the graph.
39. A
The biomass seems to correlate fairly well with the rainfall. Therefore, a rainfall higher than any recorded would give a biomass higher than any recorded. The axis on the right shows the biomass. With a rainfall of 120, the biomass should be greater than 150.
40. A
The secondary consumers’ biomass should always be less than that of the primary consumers, so the only option is (A).
41. C
This is an example of mutualism. Both organisms benefit. Choice (A), parasitism, is a type of symbiotic relationship in which one organism benefits and the other is harmed. Choice (B), commensalism, occurs when one organism benefits and the other is unaffected. Choice (D), endosymbiosis, is the idea that some organelles originated as symbiotic prokaryotes that live inside larger cells.
42. A
Both bacteria and fungi contain genetic material (DNA), a plasma membrane, and a cell wall. Unlike fungi, bacteria lack a definite nucleus. Therefore, eliminate (B). Bacteria are unicellular, whereas fungi are both unicellular and multicellular. Therefore, eliminate (C) and (D).
43. B
This question tests your understanding of what stage of gene expression utilizes RNA polymerase. Since the presence of RNA polymerase is required for transcription, the regulation is pre-transcriptional.
44. C
Trypsin is an enzyme. Enzymes are proteins and organic catalysts that speed up reactions without altering them. They are not consumed in the process. Therefore, you can eliminate (A) and (B). The rate of reaction can be affected by the concentration of the substrate up to a point, so (D) can be eliminated.
45. C
DNA polymerase, not RNA polymerase, is the enzyme that causes the DNA strands to elongate. DNA helicase unwinds the double helix, so (A) is true and therefore incorrect. Choice (B), which states that DNA ligase seals the discontinuous Okazaki fragments, is also true. Eliminate it. In the presence of DNA polymerase, DNA strands always grow in the 5′ to 3′ direction as complementary bases attach. Therefore, (D) is also incorrect.
46. A
Ecosystems high in variation are more resilient than those with low biodiversity (so (B) is out). Meiosis provides genetic variation but mitosis should provide almost none, so eliminate (C). Since crossing-over is a source of genetic variation it can be a source of speciation (making (D) incorrect). Choice (A) is the only correct answer.
47. C
Electrons passed down along the electron transport chain from one carrier to another lose energy and provide energy for making ATP. Glucose is decomposed during glycolysis, but this process is not associated with energy given up by electrons; eliminate (A). Glucose is made during photosynthesis, so eliminate (B). NADH is an energy-rich molecule, which accepts electrons during the Krebs cycle. Therefore, (D) is incorrect as well.
48. A
The oxygen released during the light reaction comes from splitting water. (Review the reaction for photosynthesis.) Therefore, water must have originally contained the radioactive oxygen. Carbon dioxide is involved in the dark reaction and produces glucose, so eliminate (B). Glucose is the final product and would not be radioactive unless carbon dioxide was the radioactive material, so (C) is incorrect. Finally, eliminate (D), because nitrogen is not part of photosynthesis.
49. D
A ligand binding to a receptor and causing a cellular response is signal transduction. In this case, the ligand is the pheromone and the response is increased expression of transcription factors. All other answers do not make sense in this context.
50. D
Homologous structures are organisms with the same structure but different functions. The forelegs of an insect and the forelimbs of a dog are not structurally similar. (One is an invertebrate, and the other a vertebrate.) They do not share a common ancestor. However, both structures are used for movement. All of the other examples are vertebrates that are structurally similar.
51. B
Speciation occurred in the Galapagos finches as a result of the different environments on the islands. This is an example of divergent evolution. The finches were geographically isolated. Choice (A), convergent evolution, is the evolution of similar structures in distantly related organisms. Choice (C), disruptive selection, is selection that favors both extremes at the expense of the intermediates in a population. Choice (D), stabilizing selection, is selection that favors the intermediates at the expense of the extreme phenotypes in a population.
52. D
Mutations produce genetic variability. All of the other answer choices are forms of asexual reproduction.
53. C
Secondary consumers feed on primary consumers. If you set up a pyramid of numbers, you’ll see that the herrings belong to the third trophic level.
54. A
The biomass is the total bulk of a particular living organism. The phytoplankton population has both the largest biomass and the most energy.
55. C
A decrease in the herring population will lead to an increase in the number of crustaceans and a decrease in the phytoplankton population. Reorder the organisms according to their trophic levels and determine which populations will increase and decrease accordingly.
56. C
Relatedness is shown on a phylogenetic tree by how far back you have to go to reach the common ancestor of the two species (shown by a branch node). Since the recent species are on the right, traveling to the left goes back in time. Because you have to go to the oldest common ancestor to connect B. musculus and B. brydei, they are the least closely related. Even though they are near each other on the tree, clades can rotate around a node so this does not matter.
57. D
Because molecular biology demonstrates where the actual mutations occured (rather than just what phenotype is observed) it is the most reliable, because it allows us to detect changes that are not phenotypically observable.
58. D
Because the brain is destroyed, it is not associated with the movement of the leg. Choice (A) is incorrect, as reflex actions are automatic. Choices (B) and (C) can also be eliminated; both of these statements are true but are not supported by the experiment.
59. C
The DNA template strand is complementary to the mRNA strand. Using the mRNA strand, work backward to establish the sequence of the DNA strand. Don’t forget that DNA strands do not contain uracil, so eliminate (B) and (D).
60. B
Use the amino acid chart to determine the sequence after uracil is deleted. The deletion of uracil creates a frameshift.
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. Describe the carrying capacity in a population. Can a brood size have a carrying capacity? Justify your choice. (2 points)
A carrying capacity is the maximum amount of individuals in a population that an environment can support with the biotic and abiotic factors available. There just isn’t enough space, food, etc., for more individuals to survive well. A brood could probably have a carrying capacity because the mother/father might be limited in resources that the babies will need. For instance, as the number of babies increases, it becomes more difficult to protect them from predators, or to gather enough food. In this data, it looks like the carrying capacity is around 9.
b. This study involved observing birds in their natural habitat. Identify the independent variable in this brood size vs. percent survival study. Describe what makes an observational study different from a study inside a lab environment. (3 points)
The independent variable would be the number of offspring in a brood; the scientist would count how many survive from each brood size. In an observational study, it is difficult to control anything other than the independent variable. Maybe the nests being studied are in slightly different environments. Maybe the parents of different broods are better than others. There is a lot that might be different between groups in an observational study. In a lab study, variables are more easily controlled, but this is an artificial setup that doesn’t quite mimic the natural environment.
c. Describe the brood size(s) in which 100% of the offspring survive. Identify which brood size has the highest percent of mortality. (2 points)
For a brood size of 3 it shows that 3 offspring survive, so 100%. For a brood size of 13, only 8 survive. 8/13 is the least surviving percentage so a brood size of 13 has the highest mortality percentage.
d. Over many more years the number of breeding adults was measured, and it was found that over a long time the number of breeding adults averaged around 100. Describe how this would change information in Figure 1. Justify your answer. (3 points)
It would not change Figure 1, which measures only the percentage of surviving offspring per brood size. If the number of surviving offspring were counted, then the information could be put on Figure 2, but just the number of mating adults is not enough for Figure 1 or Figure 2.
a. Explain how the citric acid cycle and the electron transport chain are related and how they are affected by exercise. (2 points)
The citric acid cycle (Krebs cycle) is an important part of cell respiration. The reduced electron carriers created during the citric acid cycle bring their electrons to the electron transport chain, which is used to generate ATP. During exercise the muscle cells will need more energy. The process of cell respiration will speed up. When muscle cells have exhausted their stores of oxygen, then the electron transport chain will cease and fermentation will take over.
b. Construct a graph of the data indicating citric acid cycle activity. Be sure to use error bars. (4 points)
1 point—proper axes
1 point—proper labels/data organization
1 point—proper size of bars
1 point—graphing error bars
c. Identify which group did not show an increase in cardiolipin after exercise was done. Justify your choice with data. (2 points)
The obese group did not show an increase after exercise. Although the numbers 72 and 83 seem different, when the std error is considered, the values overlap. 72 + 7.9 = 79.9 and 83 – 4.8 = 78.2. The lean and T2DM participants each saw an increase in cardiolipin with exercise.
d. If another measure of the citric acid cycle were added in a future experiment, explain what results you would expect. (2 points)
If another measure were included, I would expect an increase for all groups after exercise.
a. In mitochondrial inheritance the offspring have the genotype and phenotype of the mother. Under Mendelian inheritance it is possible for the offspring to have a different genotype but the same phenotype as the mother. Describe how this is possible.
It is possible for a mother and offspring to have a different genotype and the same phenotype under classic Mendelian inheritance because there are more than one genotype that have the same phenotype. For example, a homozygous dominant and a heterozygous genotype will both have the dominant phenotype.
b. Explain what important control(s) the scientists should do to account for the procedure in Step 2.
It would be important to make sure the injection process itself is not causing a problem. There should be a control group of wild-type cells that is injected with saline to account for the injection and make sure it is the actual mitochondria that is causing the mutant phenotype.
c. If the nuclear DNA was injected into the wild-type Neurospora cells instead of the mitochondria, explain whether the resulting fungi would have the poky phenotype.
They should NOT have the poky phenotype.
d. Justify your prediction.
The resulting fungi should not have the poky phenotype because the trait is not carried on a nuclear genome. It is part of the mitochondrial genome, and giving the nuclear genome is not going to provide the gene for the poky phenotype.
a. Finches are incapable of digging burrows yet approximately 25% of the finches found on the island live in underground dwellings. Describe the source of the finch dwellings.
The source of the burrows must come from some other animal on the island. Small rodents could dig the burrows which the finches live in opportunistically either simultaneously or after the rodents have abandoned the burrows.
b. Describe the relationship between the finches and rodents on the island.
The finches and the rodents have a symbiotic relationship. The rodents may provide shelters for the birds and the birds spread seeds which will eventually develop into a food source for the rodents as well. This describes a mutualistic behavior, or the relationship could be commensalistic where the birds benefit but the rodents do not benefit from the birds.
c. Protein HITB8 is found to be present in high levels in medium-sized beak finches and less so in other finch types. Predict what will happen to the frequency of this protein product in the finch population after two generations if the drought continues.
The protein HITB8 will decrease in abundance.
d. Justify your answer to (c).
A drought will make surviving on the island more difficult due to the birds’ decreased ability to find the food with their beak type. As such, the frequency of the gene encoding the HITB8 protein will be less plentiful and the protein will be expressed in lower numbers.
a. Describe the importance of being the sole occupant of a niche in an ecosystem.
A niche is helpful because it is a set of resources that an organism is well-suited to utilize. Having no other competitors in the niche means that there is a set of resources (physical space, food, water, etc.) that the organism will have exclusive access to.
b. Describe the impact of an invasive species that competes with phytoplankton and zooplankton
If an invasive species competed with phytoplankton and zooplankton they would possibly disappear. Invasive species often have no natural enemies and they can steal resources from the other occupants in their niche that still do have to worry about predators. The consumers of the zooplankton and phytoplankton might also disappear when their food source disappears and an invasive species can cause trouble throughout the food web.
c. Identify the number of species with no natural predator on this food web.
There are 7 species that have no natural predator (they have no lines pointing away from them).
d. Explain why toxins are more of a concern for tertiary consumers than they are for primary consumers.
Toxins are a concern higher up the food web because they can often become concentrated. For example, it is bad for a toxin to be absorbed by a plant, but then a fish eats 10 toxin-affected plants and a bigger fish eats 10 of the little fish, which is like 100 toxin-affected plants. Higher up the food web toxins can concentrate and the animals can begin to be harmed by the level of toxin even though the primary consumers were not.
a. Identify the null hypothesis of this experiment Squirrels do not prefer one side of the road over another.
b. Complete the table with the average data and values necessary to test the hypothesis. (All white empty blanks should be filled.)
c. Use statistics to determine whether the null hypothesis should be rejected.
Since our degree of freedom is 1 and we are using p = 0.05, our critical value is 3.841. Our χ2 value is less than our critical value so we fail to reject our null hypothesis.
d. Explain how the results of the experiment support the theory that animal behavior changes as their environment changes.
According to our data, the squirrels do not prefer one side of the road over the other. As the road was constructed, what was likely a continuous forest became separated. As both pine trees and maples provide food and shelter resources, the squirrels have adapted to the change in their habitat by migrating to both sides of the road frequently.
(Any answer properly linking the experimental data to changes in the environment and the squirrel behavior earns this point.)