1. C
Water has many unique properties that favor life, including (A), a high specific heat, (B), high surface tension and cohesive properties, and (D), high intermolecular forces due to hydrogen bonding. However, water is a very polar molecule and is an excellent solvent, making (C) inaccurate and the correct answer.
2. B
The pH scale is logarithmically based, meaning that each difference of 1 on the log scale is indicative of a tenfold difference in the hydrogen (H+) ion concentration. According to the question, the pH values of the cytoplasm of cells and gastric juices are approximately 7.4 and 1.5, respectively. Therefore, the pH values vary by 6, and the hydrogen (H+) ion concentrations must vary by 6 orders of ten (106), or 1,000,000-fold. Since lower pH values have more hydrogen ions, they are also more acidic.
3. C
All amino acids share a carboxylic acid group, COOH, labeled (B), and an amino-group, NH2, labeled (D). They also share a hydrogen atom bound to the central carbon (A). Differences in amino acids are defined by variations in the fourth position called the R-group, labeled (C).
4. B
The hypothesis that life may have arisen from formation of complex molecules from the primordial “soup” of Earth is not supported by the absence of nucleic acids. All life is DNA-based, yet no nucleic acid molecules were detected. The presence of carbon molecules, amino acids, and sugars, which are common compounds and compose life, supports the hypothesis, so eliminate (A), (C), and (D). Choice (B) is correct.
5. C
The amino acids cysteine and methionine contain the element sulfur (as indicated by the S in the amino acid structure shown). However, no sulfur-based compounds were included in the Miller-Urey experiment, so it was impossible to form these two amino acids under the conditions of the experiment.
6. B
Silica is a mineral form of glass, is not a common component of life-forms, and is largely chemically inert. Since oxygen is already present in several compounds included in the experiment, the addition of this compound does not provide any additional elements or chemical substrates, which would permit generation of additional amino acids or synthesis of nucleic acids. The addition of sulfur compounds and phosphorus is necessary to generate some amino acids and all nucleic acids, which eliminates (A), (C), and (D).
7. D
Hydrolysis adds water to a polymer to break the linkages between monomers. Therefore, (A), free water, will not result because water is being broken down in the reaction. Free water would result from the opposite reaction, for condensation reactions. Choice (B), adenine, results from the hydrolysis of DNA, as it is a monomer of DNA. Choice (C), cholesterol, is a steroid and therefore does not undergo hydrolysis. Dipeptides (D) could result from the hydrolysis of proteins, which are composed of polypeptides.
8. B
Phospholipids have a long fatty acid tail and a polar head group. The fatty acid tails associate in the membrane, and the phospholipid head group associates with water at the boundary of the membrane. Nucleotides, (A), are mostly polar and found in DNA and RNA. Water, (C), is polar and not found in plasma membranes. Amino acids, (D), are zwitterionic, that is, they have both a positive and a negative charge. Free amino acids are not found in the membrane.
9. C
(CH2O)3 = C18H36O18, as in (A), but the dehydration reactions to produce two glycosidic bonds between the three monosaccharides would remove two H2O molecules, or four hydrogens and two oxygens. Therefore, the correct answer is (C).
10. A
All isotopes of hydrogen will contain one proton, by definition. All atoms that have only one proton are identified as hydrogen. Tritium is an isotope that has one proton and two neutrons, for a total of three particles in its nucleus. Choice (A) is correct. Choice (B) is not correct because two protons would mean that the element is helium, not hydrogen. Choice (C) is not correct because the atomic number, one, will never change without changing the identity of the atom. Finally, (D) is not correct because radioactive atoms do not give off electrons.
1. C
Active transport is the movement of substances across a membrane against their concentration gradient through the use of energy (ATP). The sodium-potassium pump is a critical structure which uses ATP hydrolysis to move sodium and potassium ions against their respective concentration gradients. Diffusion of oxygen is an example of simple diffusion and can occur without the need of channels or pores. Water uses aquaporins to travel across the membrane, down its concentration gradient; therefore, it also does not require energy. Movement of sodium ions by a voltage-gated ion channel, (B), is an example of facilitated diffusion. Because the sodium ions are still undergoing diffusion from high to low concentrations without the need of ATP, this does not represent a form of active transport.
2. C
Bacterial cells can be visualized using light microscopy. In fact, back in the 17th century, some of the earliest studies using primitive microscopes recorded the shape and organization of bacteria. Choices (A), (B), and (D) are incorrect because virus and cell organelle structures are too small to be observed using light microscopy and require electron microscopy.
3. D
The cell wall is a structure that is present in bacteria but absent in animal cells. Consequently, this structure is targeted by several leading classes of antibiotics and would be an effective target of therapeutics against V. cholerae. Cytoplasm, plasma membrane, and ribosomes—(A), (B), and (C), respectively—are all structures that are present in animal cells.
4. C
Based on the microscopy data, the organism has a cell wall and lacks mitochondria. The presence of a cell wall suggests that the organism is not a protozoan, so eliminate (B). The absence of mitochondria is indicative of prokaryotic structure because they lack organelles, eliminating (A) and (D). The most likely conclusion is that this organism is a bacterial species.
5. B
Although many of the mentioned organelles work closely together, the only choice that can be correct is (B). Ribosomes translate and manufacture proteins, often on the rough endoplasmic reticulum. Those proteins are then packaged in the Golgi bodies into vesicles, which are then fused with the plasma membrane. Choice (A) is incorrect because the nuclear envelope and nucleolus do not interact with vacuoles; they interact with the centrioles only during mitosis. Choice (C) is incorrect because ribosomes don’t often interact with mitochondria, and lysosomes interact only with mitochondria and chloroplasts when they degrade those organelles. Choice (D) is incorrect because the nucleolus doesn’t interact with the smooth endoplasmic reticulum, lysosomes, or centrioles.
6. C
Estrogen is said to be lipid-soluble. This means it can slip through the membrane and bind to an intracellular receptor. A noncompetitive inhibitor would bind to an estrogen receptor, reducing the effectiveness of this binding. Choice (A) is incorrect because estrogen binds to intracellular receptors, not receptors at the plasma membrane. Choice (B) is incorrect because testosterone and estrogen have different effects, but testosterone would not change estrogen effectiveness. Choice (D) is also incorrect because wiping out the ovaries would eliminate, rather than reduce, estrogen levels.
7. B
You would need a cell that makes and secretes a lot of protein or hormones since the ER moves things out of the cell. Bacteria, (C), do not have organelles. Choice (D), blood cells, is not the best answer because leukocytes do not secrete large amounts of hormones or proteins. Choice (A), neurons, is a potential answer, but (B) is the best answer because the pancreas is a gland that secretes copious amounts of protein and hormones, such as insulin.
8. A
Choice (A) is correct because contractile vacuoles expel water that accumulates in cells in a hypotonic environment. The strategy described in (B) is wrong because it would increase water intake in Paramecium. Choice (C) is true for Paramecium but does not help with water gain. Choice (D) would help it find even more dangerously hypotonic environments.
9. B
ATP is consumed by the Na+K+ pump, so (B) is correct. The sodium-potassium pump actively pumps both ions. A cotransporter requires one to be passive and one to be active. Choice (A) is not true. ATP is hydrolyzed and not produced. Choice (C) is not true. Choice (D) is incorrect since ATP is hydrolyzed.
1. B
The Krebs cycle occurs primarily in the matrix of the mitochondria. The inner membrane and the intermembrane space—(A) and (C), respectively—are used in oxidative phosphorylation.
2. D
Inhibitor Y is binding at a site outside the active site and is inducing a conformational change in the enzyme structure. By binding outside the active site, it must be an allosteric inhibitor, eliminating (A) and (C). Because the inhibitor is binding outside the active site, it is not competing with the substrate for binding, so it is considered a noncompetitive inhibitor.
3. B
Enzymes are biological catalysts, which lower the activation energy (the energy threshold that must be met to proceed from reactant to product). The reaction coordinate diagram must reflect a decrease in the activation energy, eliminating (C). Furthermore, the enzyme does not alter the energy of the reactants or products, eliminating (A) and (D).
4. B
Based on the pathway provided, consumption of one glucose and two ATP results in production of four ATP. In other words, each glucose results in a net gain of two ATP. Therefore, two glucose molecules would result in a net gain of four ATP.
5. B
In fermentation, pyruvic acid is converted into either ethanol or lactic acid. During this process, NADH is recycled into NAD+.
6. B
Glycolysis results in the production of ATP (energy), so it is considered an exergonic process.
7. B
Thermophilic bacteria live in hot environments. Also, a DNA polymerase replicates DNA. Therefore, in the PCR technique, the stage in which DNA is elongated by a DNA polymerase is part three of the cycle, at 72°C. Therefore, bacteria growing in hydrothermal vents between 70–75°C is the answer. The conditions described in (A) and (D) do not reflect the data, which focuses on temperature. Choice (C) describes hot springs, but the temperature does not reflect the enzyme activity most likely for Taq polymerase.
8. D
The free energy does not change between catalyzed and uncatalyzed reactions; therefore, (A) is incorrect. Choice (B) is a correct statement, but it does not answer the question. Choice (C) is incorrect because enzymes catalyze reactions.
9. B
ATP production will increase in the treated mitochondria because the low pH provides more H+ ions in the solution. Also, oxygen provides a terminal electron acceptor for oxidative phosphorylation.
10. D
Choices (A) and (B) are clearly wrong, as they defy physics. Choice (C) is not correct because the anabolic and catabolic reactions are not necessarily equal nor are they direct opposites. Choice (D) is the correct answer because it explains the increase in entropy. Even though organisms build and develop as ordered systems, heat is lost continuously. Additionally, organisms exhale gases and produce waste products that balance the effect of order.
1. C
During anaphase, the chromatids are separated by shortening of the spindle fibers. Chemically blocking the shortening of these fibers would arrest the cell in metaphase. The cells are arrested in metaphase as indicated by the alignment of the chromosomes in the center of the cell and their attachment to spindle fibers, eliminating (A) and (D). The chromosomes still seem to be attached to the fibers, so there doesn’t appear to be dissociation of the fibers, eliminating (B).
2. C
The synthesis, or S phase, of the cell cycle represents the step in which the genetic material is duplicated. The only phase labeled in the experiment that represents an increase is phase B. Based on the time scale on the x-axis, this phase lasts approximately 30 minutes.
3. D
Anaphase represents the cell division stage of the cell cycle and would be the phase that occurs right before the amount of genetic material should decrease. Phase D is the phase right before the genetic material would drop, so (D) is the correct answer.
4. D
Choices (A) and (B) do not occur, but if they did, they would not impact the gametes. Choice (C) is not likely to occur because the mitotic spindles attach only to kinetochore protein complexes at the centromere. Therefore, (D) is the answer.
5. B
Both (A) and (B) have sperm or ovum as the first type of cell. Because these are both gametes, they will both have half of the DNA of other types of cells. Choices (C) and (D) can thus be eliminated. Muscles and neurons are both terminally differentiated in G0 arrest. However, liver and taste buds potentially differ. Although liver cells can divide, taste buds divide at a higher rate, so they are most likely to be in G2 phase. Therefore, the answer is (B).
6. A
Nondisjunction results in major changes to the genome, so (B) and (C) can be eliminated. Deletion of an enhancer, (D), could affect the gene expression and then phenotype. However, in (A), translocation may not produce any effect, as the same information exists in the genome.
1. D
The father and mother are both AB blood type. Since neither parent has a recessive allele, it is impossible for their child to be O blood type.
2. B
When the phenotype associated with two traits is mixed, this is considered an example of incomplete dominance. In this case, neither red nor blue is dominant, and the resulting progeny exhibited a mixture of the traits (purple).
3. B
Crossing the pea plant that is heterozygous for both traits (TtGg) with a plant that is recessive for both traits (ttgg) results in the following possible combinations, each of which should occur 25 percent of the time: TtGg (tall and green), Ttgg (tall and yellow), ttGg (short and green), and ttgg (short and yellow). Using the rules of probability, there is a 1/2 likelihood of it being tall and also a 1/2 likelihood of it being yellow. Multiply them together to get 1/4.
4. C
Because the woman is a carrier, she must have one normal copy of the X chromosome and one diseased copy. Since the boy will receive an X chromosome from his mother, there is a 50 percent chance that he will receive a diseased copy. Because he doesn’t have a second X chromosome, he must have the disease if he receives the diseased X chromosome.
5. D
They both must have a normal copy of the X chromosome. It is possible that the woman may be affected with hemophilia; however, that scenario is extremely unlikely because such a case would require two diseased copies of the X chromosome.
6. D
Essentially, transmission of hemophilia to a girl born with Turner syndrome would be very similar to the conditions by which a boy would receive the disease. Both boys and girls with Turner syndrome have only one copy of the X chromosome. Therefore, they both would have a 50 percent chance of receiving the diseased copy of the X chromosome.
7. B
Choice (A) is true, but it doesn’t explain why the parents are normal. Choice (B) is a better explanation. Carriers of recessive genetic diseases frequently do not have a recognizable phenotype because one normal allele provides enough functioning protein to avoid the ill effects of the disease allele. Choice (C) is also possible, but (B) is still the better answer. Finally, (D) is not correct because CLN3 is an autosomal gene.
8. C
Choice (C) describes a testcross that will actually let the breeder know if the male is heterozygous. If any white-spotted pups result from that cross, then the male would have contributed a spotted allele. Choice (B) describes a male that is homozygous for the spot allele, so it’s incorrect. Choice (A), stop breeding Speckle, is a way to remove spots from the line, but it will not help you identify males with the spot allele. Choice (D) also will not help you determine which black Labrador males have a spot allele.
9. A
Choice (B) does not apply to this question. Choices (C) and (D) refer to a gene not following Mendel’s Law of Dominance. Choice (A)—genes are linked on the same chromosome—describes a normal reason that certain traits would not follow Mendel’s Law of Independent Assortment.
10. A
Choices (B) and (C) are unlikely to be true and can be eliminated. Choice (D) is true but does not answer the question. Choice (A) is the easiest and most probable explanation.
1. B
Okazaki fragments are generated during DNA replication when the DNA polymerase must create short DNA segments due to its requirement for 5’ to 3’ polymerization. Since the newly discovered yeast cell has 3’ to 5’ activity, there would be no lagging strand and likely no Okazaki fragments.
2. C
Since the gene is much shorter than expected, a stop codon must have been introduced by mutagenesis. This is an example of a nonsense mutation.
3. D
The order for DNA replication is helicase, RNA primase, DNA polymerase, and ligase.
4. C
If an mRNA codon is UAC, the complementary segment on a tRNA anticodon is AUG.
5. D
During post-translational modification, the polypeptide undergoes a conformational change. Choices (A), intron excision, and (B), a poly(A) addition, are examples of post-transcriptional modifications. Formation of peptide bonds (C) occurs during translation, not afterward.
6. B
If 21 nucleotides compose a sequence and 3 nucleotides compose each codon, there would be 7 codons and thus a maximum of 7 amino acids.
7. A
Choice (B) is incorrect because bacteria make membrane proteins that reside on the plasma membrane. Choice (C) is also incorrect because bacteria use transcription factors. Choice (D) is possible, but (A) is the best answer.
8. C
Choice (A) is likely incorrect. Choice (B) would be deleterious for the cell. Choice (D) is also incorrect, as DNA replication occurs only during or preceding binary fission.
9. B
Viruses do not have their own ribosomes, flagella, or independent metabolism, so (A), (C), and (D) are incorrect. The only possible answer is (B).
10. C
Transformation occurs when bacteria take up DNA from their surroundings. The pathogenic bacteria did not come alive again, as suggested by (A). Protein was not the transforming agent, so (B) is incorrect. Finally, (D) does not make sense, as genes cannot turn into other genes simply by being in a different cell context.
11. C
If there are DNA/RNA fragments, then helicase, (A), is present and has opened the strands. RNA primase, (D), is also present because RNA primers have been added. DNA polymerase, (B), must be present because there are small chunks and long chunks. If there was not polymerase, then there wouldn’t be small chunks because nothing new would be made. It is likely that DNA ligase (C) is missing and couldn’t attach the short Okazaki fragments and the longer leading strand fragments.
1. D
Mammals and cephalopods developed similar eye structures independently due to similar selective pressures. This is an example of convergent evolution.
2. B
The data provided show a transition toward one extreme (black) and away from another (white). This is an example of directional selection.
3. B
Based on the data, the number of white-bodied pepper moths decreased between 1802 and 1902, and the number of black-bodied pepper moths increased during the same period.
4. B
Longtail moths were included in the experiment as a control to compare the effects that are not associated with color.
5. B
If the color of ash or soot produced by the Industrial Revolution were white or light gray, this would likely reverse the trend observed, applying additional selection against the black moths.
6. B
The mole rats live in the same location, which means there is not a geographic barrier so (A) is incorrect. They do not attempt to breed, so (C) and (D) are incorrect. They have formed two separate species in the same geographic area, so (B) is correct.
7. D
Choice (A) is incorrect because this is a small population, so genetic drift is likely. Choice (B) is incorrect because this situation does not describe convergent evolution, and the second part of the answer choice does not accurately describe the change of the population. Choice (D) is correct because the number of people that are homozygous recessives will likely become fewer as the redhead allele is mixed with the more dominant hair colors.
8. D
Choice (D) is the answer because it is the only choice that does not predict evolution occurring on the island, thus supporting a claim for Hardy-Weinberg equilibrium. Choices (A), (B), and (C) all prevent a Hardy-Weinberg equilibrium and predict evolution. The population is small, mutations are inevitable, and humans usually do not mate randomly. This population will definitely evolve or undergo genetic drift.
9. A
Choice (A) is correct because the trait is being selected based on female mating preferences. Longer, bigger tails indicate reproductive fitness. Peahens choose males who are healthy and strong enough to grow these big tails and will hopefully produce the best offspring. Choices (B), (C), and (D) are incorrect because they do not describe the female mating preferences.
1. D
The behavior displayed by the chimpanzee represents insight because the chimpanzee has figured out how to solve the problem without external influence or learning.
2. A
Viruses would display reproductive strategies most similar to r-strategists because they aim to reproduce as fast as possible and create as many progeny as possible in order to increase their odds of transmission to other hosts.
3. C
They would be considered secondary consumers because they consume the primary consumers (plankton), which consume algae (the producers).
4. C
Since brook trout can tolerate pH values as low as 4.9 and do not appear to diminish, the pH of the creek must exceed 4.9, so we can eliminate (D). Since crayfish cannot tolerate pH levels lower than 5.4, the pH of the stream must have dropped lower than this value. Only (C) falls in this range.
5. C
Only acid rain would directly explain why the pH would drop in the creek over the time period.
6. D
Based on Table 1, the pH must exceed 6.1 for snails to be able to return to the creek ecosystem.
7. D
Choices (A) and (B) refer to light and gravity responses of plants. Choice (C) is not an actual tropism—it is a made-up word. Thigmotropism is the term for the way in which plants respond to touch. The correct answer is (D).
8. D
If (D) is a correct statement, then simply cutting out beef and lamb from your diet can dramatically decrease your carbon footprint—even if you do not decide to eat vegan. Choices (A), (B), and (C) are all true statements with respect to reducing the carbon footprint of food.
9. B
A community that has been destroyed and then rebuilds is a result of secondary succession. Choice (A) would be the answer only if there were no life present before the new growth. Choice (C) is true only of a mature forest. Choice (D) is incorrect, as the species taking over the field is native to the rainforests on the periphery of the field. Nothing in the question suggests invasive species. Therefore, (B) is correct.
10. A
If the intestines are occupied with beneficial bacteria, there will not be any room available for pathogenic bacteria to grow by density-dependent limitations on the population. Choice (B) is not accurate because pathogenic bacteria do not require a niche prepared for them by probiotic bacteria. Choice (C) is true, but it does not answer the question. Choice (D) is also incorrect, leaving (A) as your answer.
1. A
In order to answer this question, you must first calculate how much weight each subject lost and then divide by the number of subjects (in this case, five).
Note that subject 3 gained four pounds. Total weight lost is 60 pounds (remember to subtract 4 pounds for subject 3, not add), divided by 5 subjects is 12 pounds. The average weight lost is 12 pounds ((A) is correct).
2. C
Plant 6 is 68 inches. All the answer choices list median values smaller than this, so the answer must start with “Yes” (eliminate (B) and (D)). In order to determine the median height for all six plants, their heights must first be organized in ascending order: 61, 66, 67, 68, 70, 72. The middle two numbers are 67 and 68; when averaged, this produces a median of 67.5 inches ((C) is correct). Notice that if all the data points are whole numbers, the median value must end in either .0 or .5, so you can quickly eliminate (A) and (B) in this question. In fact, (A) and (B) are giving the value for mean, not median.
3. D
The test scores ordered from smallest to largest are:
32, 65, 66, 67, 68, 68, 69, 70, 71, 72, 73, 75, 75, 75, 78, 82
The most frequently recurring number in the set above is 75, so this is the mode (eliminate (A) and (B)). The smallest number is 32 and the largest is 82. The range is the difference between the two, or 50 ((D) is correct).
4. D
This question is testing the product rule.
P(AABbCC) = P(AA) × P(Bb) × P(CC)
P = 
× 
×
P = 
5. A
This question is testing the sum rule, but you also need to use the product rule.
P(Aabb or aaBb) = P(Aabb) + P(aaBb)
P = [P(Aa) × P(bb)] + [P(aa) × P(Bb)]
P = 
+
= +
P = 
=
6. C
In chi-square tests, a calculated χ2 value is compared to a critical value (CV) from a chi-square table, like the one on the AP Biology Equations and Formulas sheet. If χ2 < CV, you accept the null hypothesis (Ho). If χ2 > CV, you reject Ho. Our Ho is that the observed and expected data match and that the experimental plants have a 3:1 ratio of yellow to green plants. Based on this, the only possible answer choices are (B) and (C); the other choices mix up how χ2 values and critical values are compared (eliminate (A) and (D)).
One hundred plants were studied, and we expect three-fourths of them to be yellow (75 plants) and one-fourth of them to be green (25 plants). Next, we compare expected (E) and observed (O) data and calculate an χ2 value.
Because two possibilities are being compared (green and yellow), the degrees of freedom in this test = 2 – 1 = 1. Using p = 0.05, you can determine the critical value to be 3.84 (which you don’t need to look up because it is listed in all answer choices). Since χ2 > CV, you reject H0. The observed data does not match the expected data, and the correct answer is (C).
7. B
The first thing you need to do is sort out the alleles for each gene. Since the F1 generation had long phenotypes, you know these must be dominant to the short phenotypes. Let’s use:
L = long leaves
l = short leaves
S = long shoots
s = stubby shoots
The parental cross must have been LLSS × llss, and all F1 plants were LlSs. Our null hypothesis (Ho) is that the two genes are segregating independently. This means the expected ratio of F2 phenotypes will be 9LS:3Ls:3lS:1ls. Since 1,000 F2 plants were generated, we expect:
P(long leaves, long shoots) = P(LS phenotype) = 
× 1,000 = 562.5
P(long leaves, stubby shoots) = P(Ls phenotype) = 
× 1,000 = 187.5
P(short leaves, long shoots) = P(lS phenotype) = × 1,000 = 187.5
P(short leaves, stubby shoots) = P(ls phenotype) = 
× 1,000 = 62.5
Next, generate a chart:
χ2 = 1,911.4 and degrees of freedom = # of possibilities – 1 = 4 – 1 = 3 (eliminate (C) and (D)). Using p = 0.05, we get a critical value of 7.82. Since χ2 > CV, we reject Ho. These two genes are not segregating independently (eliminate (A); (B) is correct).
