1. B
Adding salt would increase the osmolality (a similar measure to osmolarity) and since Organism 4 is an osmoconformer it would change its intracellular osmolality/osmolarity to mimic the surroundings. This means that it would increase its osmolality and maintain it unless further changes are made to the system.
2. B
The only choice that has direct ties to maintaining osmotic balance is the sodium-potassium pump.
3. D
Since the organisms mimic their surroundings, the only choice that would decrease their osmolalities would be one that decreases the osmolality of their surrounding tank. Choice (A) increases the tank osmolality. Choices (B) and (C) directly increase the organism osmolality. Choice (D) decreases the tank’s osmolality.
4. C
The goal of an osomconformer is to mimic the osmolarity of its surroundings. An isotonic solution has the same concentration as its surroundings. Therefore, an osmoconformer would become isotonic to its tank.
5. A
If Organism 1 decreases osmolality, it means that the tank must have decreased in osmolality. In other words, it gained water or lost solute. Choice (B) increases the tank solute. Choice (D) decreases the water. Choice (C) does not directly relate to the osmolality. Only (A) would cause the tank to gain water as water is removed from the new organism by osmosis.
6. C
RZN45 is the enzyme and it binds to JB-76. Ascorbic acid (vitamin C) is an inhibitor of this reaction and binding to JB-76 is reduced, which suggests the inhibitor binds to the same place as the substrate. The graph confirms that the inhibition is competitive since it shows a decreased reaction rate when ascorbic acid is present. Choice (A) is not true because similarly charged amino acids would repel each other and RZN45 and JB-76 bind to each other. There is nothing to suggest that choice (B) is correct. Choice (C) would make sense since both ascorbic acid and JB-76 bind to the same place so they probably have similar structures. Choice (D) could be true, but it is unlikely. Even things that are the same shape do not necessarily have the same number of amino acids. Proteins are often hundreds of amino acids long, so having the exact same number is like a needle in a haystack.
7. A
The graph shows that at low/medium levels of substrate the reaction rate is low in the presence of ascorbic acid and then at high levels of substrate the reaction rate is high, even with ascorbic acid present. This is because with lots of substrate, the ascorbic acid does not stand a chance at finding a free active site to bind to because the superfluous amount of substrate is likely to be bound at most of the sites.
8. C
Ritzolierre’s Disease is caused by elevated enzyme levels. Only option III would decrease the enzyme activity levels. Options I and II would raise the levels and make the disease worse.
9. C
As shown in the graph, the amount of substrate has reached a saturation point and adding additional substrate will not increase the rate further. The reaction rate will remain at the same high level.
10. D
The islands are separated by ocean. Allopatric speciation occurs when populations are isolated from each other. The finch populations are far away from each other and are likely to mate only with those on the same island. Choice (A) implies that the finches are able to cross between islands, and Choice (B) refers to a natural disaster. Neither is mentioned in the information provided. Choice (C) describes the finches developing similar features, but they developed different types of beaks. Only Choice (D) references the isolated populations associated with allopatric speciation.
11. B
If meiosis occurred as it should, there should be 4 haploid gametes, each with one long and one short chromosome. None of the gametes look correct. There was a nondisjunction at Meiosis I, and the two long homologous chromosomes failed to separate. This caused two gametes to get no copies of the long chromosome and two gametes to get two copies. If the nondisjunction was in Meiosis II, then two of the gametes would look normal and two of the gametes would have incorrect amounts of chromosomes. Barr bodies are small and unviable, but they still have the correct number of chromosomes in them.
12. A
The location appears to be upstream of the transcription start site. This is the promoter region and is likely important for polymerase binding and initiation of transcription. Without it, there would be no transcription and subsequently there could not be any translation.
13. B
Dark fur would be caused by high expression of fursilla. The triggers for that high expression would be an underexpression of DRKFR and high expression of LSFR.
14. D
If something competes with the polymerase, then it is likely an inhibitor and high amounts of it would decrease transcription. DRKFR is the only protein that significantly decreases Fursilla transcript levels when overexpressed.
15. B
Two proteins fitting together would not occur from similarities in their nucleotide sequences, which is referenced in Choices (A) and (C). As proteins, they should not be formed from lipids, so Choice (D) is incorrect. Proteins are formed from amino acids, and complementary conformations would allow them to bind together.
16. A
DRKFR is a transcriptional inhibitor, and it significantly decreases Fursilla transcript levels when overexpressed. If BLD binds to the transcriptional inhibitor, DRKFR, then it would reduce inhibition and increase the transcript levels of Fursilla.
17. D
Solvents dissolve solutes that are similar to them, so polar solutes would no longer dissolve in water if water turned nonpolar. Choice (B) is incorrect, because nonpolar molecules are more likely be found as solids. Less liquid water would be found on Earth due to the lack of strong bonds holding it together, which also ties into the melting and boiling points decreasing instead of increasing. The polarity of water allows it to hydrogen bond and form a less dense solid, so Choice (D) describes what would happen if the polarity changed.
18. C
The wind speed, tree height, and altitude do not tie directly into the function of the jaw and the size of the opening. Only the size of seeds that the mice eat would directly relate to the mouth size.
19. A
It says in the passage that the mice are the same species. This means they should be able to mate with each other. As we do not know anything about the genetics and inheritance of jaw angle, it is not possible to predict how the jaw angle would be affected.
20. A
Since the jaw angle on island B is now large, the smaller jaw angle must have disappeared. The small jaw angle is likely selected against, and only mice with large jaws survive on Island B. The survival on the other islands is not directly relevant.
21. B
Punctuated equilibrium is defined as an isolated episode of rapid development of a population. A slow change, described in Choices (A) and (C), or a fluctuation, described in Choice (D), do not suggest a quick change.
22. D
The percent change in mass will be closest to 0% when the solutions are isotonic. In other words, the potato cells should stay the same mass when they are soaked in a sodium chloride solution with a similar concentration as the cells. The percent change crosses over from a gain to a loss between 0.3M and 0.5M, so the isotonic concentration must occur between these two molarities. Only Choice (D) describes a number in between these concentrations.
23. B
HRIET1 has a signal sequence AND a targeting signal, so it must be part of an organelle that is in the secretory pathway that helps things exit the cell. The Golgi apparatus is the only organelle among these choices that functions in this pathway.
24. D
The receptor must be in the membrane on the surface of the cell. It should have a signal sequence, but it should not have a targeting signal or a localization sequence. The only protein that meets these requirements is K8TE.
25. B
The electron transport chain in humans occurs in mitochondria. Proteins destined for the mitochondria should have a localization sequence. The only two proteins that have a localization sequence are NUH8 and LNACT.
26. B
HAZL2 has none of the sequences or signals represented in the data table. Proteins without these features would remain in the cytosol where it is assembled, so a scientist would have to collect samples that contain cytosol.
27. D
The raccoon population seems steady, with a slight dip in 2008. If raccoons rely on berries as a primary food source, then it is likely that the berry population would have also remained stable.
28. B
The cougar population is shown to be decreasing over time. One possible explanation for the decline may be due to a disappearing food source, so the food source should also show a decrease over time. The only animal that also shows a decreasing population is deer.
29. D
There is no information about what species is a predator of skunks, and the population seems to be stable over time, so this is impossible to determine. Even though the skunks and raccoons are both stable, there is no indication that they are related to each other.
30. A
If the chipmunk is dependent on the acorns, then the acorns affect the carrying capacity. Since the quantity of acorns would affect a large population differently than a small population, this makes it a density-dependent factor.
31. D
None of the populations demonstrate exponential growth because they each have a carrying capacity that sets a limit on their growth. Only populations growing as fast as they can reproduce at a maximal amount will grow at an exponential rate.
32. B
The bases at the 5’ side of the transcript are in the untranslated region because this is generally the site for ribosome binding. Without this region, the ribosome couldn’t bind, and there would be no translation.
33. A
Before the transcripts are completed and leave the nucleus, they undergo certain modifications including the removal of introns and the addition of a 5’ GTP cap and a 3’ poly-A tail. The transcript in the nucleus would not have undergone these modifications yet, so it would contain introns.
34. B
The coding strand is the strand that is identical to the mRNA transcript, except that the mRNA has the nucleotide uracil in place of the nucleotide thymine. It is the partner strand to the coding strand and used as a template during transcription. Therefore, the answer should match the transcript shown in the image with “U’s” in place of any “T’s.”
35. D
A transcription factor binds to DNA before RNA is created during transcription. Since this is already a map of an RNA transcript, transcription has already happened and no transcription factors will bind at any location.
36. D
At each level of the food chain, approximately 10% of energy is used to build biomass for the next level. The primary consumers would have approximately 298.7 g/m2 of biomass, and the secondary consumers would have approximately 29.87 g/m2 of biomass.
37. A
The question states that the change affects the place where the substrate binds, which is defined as the active site. Position A is identified as the active site.
38. D
Position B has nonpolar residues and is not the active site, since Position A is the active site. Therefore, Choice (A) is not correct since the substrate does not bind at Position B. It doesn’t look like it is where the homodimer is, so Choice (C) is incorrect. There is no information to indicate Choice (B). Nonpolar residues are known to be helpful for inserting into the lipid bilayer, which makes (D) a strong answer.
39. D
A homodimer is formed when more than one polypeptide subunit come together. This is defined as a quaternary structure, (D).
40. C
The protein is known to be a homodimer, so it must have two polypeptides in it. Each polypeptide has one N-terminus, so this dimer protein would have two. They would not occur at the active site, so position A should not be in the answer.
41. B
If fats are insoluble in water, figuring out what type of mixture it forms when mixed with water would help target this property of lipids. The only option that discusses mixing with water is Choice (B).
42. A
While Choice (A) is true, the question is asking about stored energy. Choice (C) describes glycerol, which is a component of cell membranes, and glucagon’s function as a hormone will not affect energy storage.
43. B
Organic compounds are defined as molecules that contain carbon and are found in living organisms. Choice (B) is the only option that references carbon.
44. A
Figure 1 is adenine, which is a nucleotide. Nucleotides are the monomers of nucleic acids, so the image represents Choices (B) and (C). According to the information in the passage, nucleic acids relay genetic information, so Choice (A) is the only answer that incorrectly describes the figure.
45. A
A true-breeding white nettle would be homozygous dominant (WW). A true breeding pink nettle would be homozygous recessive (ww). All offspring would inherit one dominant allele and one recessive allele, so they would all be heterozygous (Ww). Heterozygous nettles would display the dominant white phenotype.
46. C
The F1 generation refers to the first generation of offspring. The previous question determined that all of the offspring from the first cross would be heterozygous. If the heterozygote offspring was taken and crossed with a true-breeding pink nettle, which would be homozygous recessive (ww), half of the offspring would be Ww and half would be ww. The ww offspring would have pink flowers.
47. B
The heterozygote (WwSs) would be crossed with a nettle recessive for both traits (wwss). If the genes were not linked, the flowers should be equally white/ruffled, white/slotted, pink/ruffled, and pink/slotted. If the genes are linked, then some combinations of color/texture should be more common. Since a pattern is found in the offspring, and only some of those combinations are present, the traits are likely linked.
48. B
The equations of Hardy-Weinberg (p + q = 1 and p2 + 2pq + q2 = 1) should be used. There are 5,300 plants, and 400 of them are pink plants (homozygous recessive). Therefore, the frequency of homozygous recessive alleles = 400/5,300 = q2 = 0.075 and thus q = 0.27.
49. B
If you correctly solved the previous problem, you’re halfway there, as you’ve already found that q = 0.27. However, here’s that information again. The equations of Hardy-Weinberg (p + q = 1 and p2 + 2pq + q2 = 1) should be used. There are 5,300 plants, and 400 of them are pink plants (homozygous recessive). Therefore, the frequency of homozygous recessive alleles = 400/5,300 = q2 = 0.075 and thus q = 0.27. If p + q = 1, the equation becomes p + 0.27 = 1, or p = 0.73. The frequency of heterozygotes is 2pq, which becomes 2(0.73) (0.27) = 0.3942 = 39%.
50. D
G-protein coupled receptors (GPCRs) are necessary to bring a signal from outside the cell to inside the cell. If the ligand is a small nonpolar molecule, then it can go through the membrane without trouble and does not need an extracellular receptor.
51. B
According to the figure, the first step in the activation pathway was the binding of epinephrine, the external ligand.
52. C
The G-protein coupled receptor (GPCR) allows a message to be transmitted inside the cell without the ligand actually entering the cell. This is most like a drive-through window where the order is communicated to someone inside a restaurant, but the patron does not actually enter the building.
53. B
The conformational change allows a message to be passed from the outside of the cell to the inside. The conformational change is a sign that the ligand is bound. The ligand does not bind to the membrane, so Choice (A) is incorrect. The ligand triggers binding of the GTP-protein, so Choice (D) is incorrect. A conformational change would not make an attachment more solid, which rules out Choice (C).
54. C
According to the table, the average carrot mass was stable around 60 grams, but the carrot length decreased by almost 4 cm over time.
55. C
If a weevil lived at 17 cm, the carrots that are longer than 17 cm would not survive. The shorter carrots would be untouched by the weevils and have more evolutionary fitness. The shorter carrots would be naturally selected for over time and the short trait would become more prevalent in the population.
56. B
By removing the weevil population, any long carrots would have the chance to thrive. Additionally, by crossing the longest carrots, the genotypes that promote carrot length will be selected for. Since all of the carrots are longer than 10 cm, a predator at 10 cm would still be able to prey on the carrots.
57. B
If the carrot became shorter, the weevil population would still survive as long as the radish supply was sufficient. Without the radishes, the weevils that lived at those depths would die, and only the weevils that can live at depths where some food is found will survive, (B). Carrots grow to a depth of 17 cm, so the population of weevils might be selected to live at 17 cm. If only a small percentage of weevils died from a virus, this shouldn’t change the preferred depth of the entire population.
58. B
The table shows that the weevil predation is a significant selective pressure on carrot length. Length determines whether or not a carrot is able to prosper in its environment and display a high level of fitness.
59. D
Sexual selection is defined as natural selection specifically for a mate of the opposite sex. The dance would increase mate attraction, but the predator would also decrease the health of the same mates. The sexual selection and the predation are selective pressures acting in opposing ways at the same time, and both are important.
60. B
Bacteria reproduce through asexual reproduction, so they don’t have “parents” as mentioned in Choice (A). Other species are not mentioned as in Choice (C), and the bacteria still fill the same niche before and after the resistance trait is developed as mentioned in Choice (D).
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) Bacterial communities containing different species of bacteria are no different than other communities with multiple species. Explain what factors contribute to a carrying capacity. (2 points)
Carrying capacity is determined by the available resources including food/nutrients, safe, healthy space to live in, water, sunlight, etc. Anything that is required but is in short supply contributes to the carrying capacity. When populations are competing for resources, there is a finite population that the environment can sustain, and that is the carrying capacity.
(b) Identify the two important controls that are used in this experiment. In addition, the experiment includes a 7-day waiting period between when the bacterial transplant takes place. Justify why this is necessary. (3 points)
One important control is the use of a clean swab. Another control is to deposit the interdigital swab onto another interdigital space. These are important to confirm that the process of applying the swab to the new area is not a tainted or contaminated process. The 7-day waiting period is important to allow the new bacteria to compete a little with the existing bacteria. Then we can better judge which bacteria is going to stick around in the new site.
(c) Interpret the effect of the transplants between interdigital web space and the buttock, external auditory canal, and plantar heel areas. (3 points)
When deposited on the buttock, the bacteria community already on the buttock did not get changed much. The existing bacteria still made up the community. This means the interdigital bacteria didn’t survive well there or could not compete with the existing bacteria.
When deposited on the external auditory canal and the plantar heel areas it looked like the new bacteria made an impact and were competing with the existing bacteria. The plantar heel area, in particular, seemed to be recolonized by the bacteria or the interdigital web space.
(d) If each deposition site is pretreated with known antibiotics, explain how this would likely affect the results. (2 points)
Antibiotics destroy bacteria. If the new sites had been treated with antibiotics this would have wiped out the existing bacteria. If the antibiotics were still present, then they would kill the newly deposited bacteria as well. Either way, the skin sites would be clear of the existing bacteria and ready to receive the newly deposited bacteria, like a clean slate. This would make the results show a population more like the interdigital than the existing bacteria at those sites.
(a) Describe how CO2 is a measure of respiration. Explain why temperature might affect the rate of respiration. (2 points)
CO2 is a byproduct of cellular respiration, which is why we need to get rid of it when we exhale. Living things, like crickets, make more CO2 when they are performing more cellular respiration (with the processes glycolysis and the Krebs cycle and the electron transport chain). CO2 is a byproduct of the Krebs cycle.
Temperature might affect respiration because the crickets change their behavior levels with the temperature. This makes them use more oxygen and energy and need more respiration when they are more active. Also, certain enzymes work better at different temperatures, so this might increase the rate of respiration or decrease the rate of respiration depending on how the temperature impacts enzyme efficiency.
(b) On the axes provided, construct an appropriately labeled graph to demonstrate the CO2 levels with changing temperature. Please include all temperatures, timepoints, and error bars. (4 points)
(c) Identify which temperature led to having the highest respiration rate in the crickets. (1 point)
The highest respiration rate was found at 35°C as that showed the greatest increase over time.
(d) Predict what would happen to the CO2 levels in the crickets with an additional 5°C increase above the highest temperature. Justify your prediction with evidence from the data. (2 points)
With a 5°C increase in temperature, the respiration rate would increase to approximately 2100-2200ppm. This is justified because the rate increased by approximately 500ppm when the temperature increased by 15°C. This means that a 5°C increase in temp would increase the respiration rate by about 166ppm.
(a) Explain how changing a single nucleotide can change the sequence of the protein.
Changing a nucleotide will change the codon, which is the group of three nucleotides that gets read by the ribosome when the protein is made. If the codon changes, it might change the amino acid that gets added to the protein. It could also make the chain shorter if it changes to a stop codon.
(b) Given what is known about the neuraminidase enzyme, describe what is measured to provide the results in Figure 2 and Figure 3.
The neuraminidase enzyme cleaves a tether that holds the viral particle to the membrane. To measure the “neuraminidase activity” for figure 2 and figure 3, the cleavage action of the enzyme must be measured.
(c) Will an NAH virus with an N234Y substitution be a rapid spreading or slow spreading virus?
It will be a slow spreading virus.
(d) Justify your prediction.
The N234Y causes the enzyme activity to decrease, and this would mean that the viral particles do not get released. This means that the virus should not spread quickly. NAH has high enzyme activity and spreads quickly, but the N234Y version should not spread quickly.
(a) Identify which cellular process thylakoids are associated with.
Thylakoids are associated with photosynthesis.
(b) Explain why ions in the hydroponic tank cannot pass into the roots by simple diffusion.
The roots, like all cells, have a phospholipid bilayer membrane. This membrane is not permeable to polar or charged things. The inner hydrophobic space prevents things like ions from diffusing through. Ions must enter by facilitated diffusion or sometimes active transport using a pump.
(c) Predict the immediate effect this would have on the cell’s ability to produce reactants necessary for carbon fixation.
The reactants for the carbon fixation in the Calvin cycle would not be produced.
(d) Justify your prediction.
The permeability of the thylakoid membrane would make the hydrogen ions able to flow freely from one side to the other. Without the ability to create a concentration gradient with hydrogen ions, the electron transport chain would fail. This would prevent production of ATP and NADPH, both of which are required for carbon fixation in the Calvin cycle.
(a) Describe what is represented by a node in a phylogenetic tree.
A node represents a common ancestor. This means that the species that branch from that point had a common ancestor and were once the same species. That species got split and each group evolved differently.
(b) Identify two of the viruses that are LEAST similar to GGE.
Two viruses with the least similarity to GGE are YF and SRE. Actually, YF and TYU and SRE are all distantly related to GGE.
(c) In 50 years, new data have identified that TBEwest(Hypr) is no longer in circulation, but it seems to have evolved in two directions: TBEwest1 and TBEwest2. Using the empty tree below, correctly indicate TBEwest(neu), TBEwest(Kuml), TBEwest1m and TBEwest2.
(d) This phylogenetic tree was created using nucleotide sequences. If protein sequences were used instead, explain why this might increase the similarity between the viruses?
If protein sequences were used, the sequences might show more similarity because there is more than one codon for some amino acids. This means that the nucleotide sequence might vary, but the amino acid sequence would be the same. These are called silent mutations.
(a) Identify which color moth was the most plentiful between 1969 and 1995?
Although the two moth colors take turns being the most plentiful, most of the years showed the beige colored moth being the most plentiful.
(b) Explain why the lines of the graph never exceed 100?
The graph is showing the percentages of moths of each color. 100% would represent the entire population of moths. It cannot be higher than 100%. The two colors of moths should always add up to 100% since they are the only possible colors of moth.
(c) Predict during which years there was a drought?
There was likely a drought from approximately 1976-1988. The beginning of the drought seemed to happen quickly in 1976 when the green moths disappeared and the beige moths rose in numbers. It is hard to know exactly when the drought ended, but seems like maybe the green moths are making a comeback starting around 1988, and by the end of the graph the numbers of each were approaching 50% again, suggesting the drought conditions might be disappearing.
(d) Explain how natural selection led to the rise of the beige moth in the late 1970s?
Natural selection caused the beige moths to become more plentiful because they survive better when the grass turns brown from the drought. This is probably because the birds cannot find them to eat them as well as the green moths, which are not camouflaged against the grass because the grass is dry and brown and not green. As the green moths get eaten, they do not reproduce, and the population becomes filled with a higher percentage of beige moths.