MCAT The Princeton Review AAMC Practice Exam 2 - Biological and Biochemical Foundations of Living Systems

Passage 5 (Questions 21 - 24)

Calorie restriction (CR) is known to extend lifespan in yeast. CR may exert its effects by reducing the level of reactive oxygen species (ROS) as a result of slowed energy metabolism. Researchers investigated the roles of energy metabolism and histone acetylation level in regulating the lifespan of yeast. Experiment 1 Cells with deletion of the hst3 gene (Δhst3), which encodes a histone deacetylase; deletion of the rtt gene (Δrtt), which encodes an acetyltransferase; or the double deletions of both genes (Δhst3, Δrtt) were prepared, and the lifespan for each mutant was compared with the lifespan of wild-type cells (Figure 1). Experiment 2 Cells with single deletion of the tdh2 gene (Δtdh2), which encodes glyceraldehyde-3-phosphate dehydrogenase (GAPDH), or double deletions of the hst3 and tdh2 genes (Δhst3, Δtdh2) were prepared. The lifespan for each mutant was compared with the lifespan of the Δhst3 mutant and wild-type cells (Figure 2).

Passage 6 (Questions 29 - 32)

Cystic fibrosis (CF) is a common genetic disease in humans. CF is caused by mutations of the cystic fibrosis transmembrane conductance regulator (CFTR) gene. The CFTR gene is found on an autosome. The most common mutation of the CFTR protein (ΔF508) results in a deleted phenylalanine at amino acid residue 508. Individuals homozygous for this allele exhibit CF. In some populations, the frequency of the ΔF508 allele is as high as 0.05, leading researchers to investigate why this harmful allele is so common. One hypothesis is heterozygote superiority. Under this hypothesis, individuals heterozygous for ΔF508 would have increased resistance to typhoid fever. The causative agent of typhoid fever is the bacterium Salmonella typhi. S. typhi enters the body via contaminated food or water and then crosses the epithelial cells of the small intestine. CFTR has been shown to be a major epithelial cell receptor for internalization of S. typhi in the gastrointestinal system. Human epithelial cells expressing ΔF508 CFTR ingest significantly fewer S. typhi than do epithelial cells that express wild-type CFTR. In addition, an antibody specific for the first extracellular domain of human CFTR inhibits the uptake of S. typhi. Therefore, it seems plausible that diminished levels of CFTR in ΔF508 heterozygotes may decrease the heterozygotes' susceptibility to S. typhi infection. If so, the selective advantage of the ΔF508 allele would help explain the high incidence of CF in some populations.

Passage 8 (Questions 39 - 43)

Ebola virus (EboV) causes a rapidly progressing and often fatal hemorrhagic disease that currently has no effective treatment. EboV particles consist of a nucleocapsid encased in a membrane that contains transmembrane viral glycoprotein (EGP). The virus has a single-stranded, negative-sense RNA genome that is used as a template for the mRNAs produced by a virally encoded RNA-dependent RNA polymerase. The entry of an EboV nucleocapsid into the host cell cytoplasm starts with the fusion of the viral membrane with a host cell membrane, a process facilitated by EGP. Recognizing the similarities between how EGP and the glycoproteins of related viruses are processed during viral infection, as well as how inhibitors of endosomal acidification block EboV infection, researchers hypothesized that acid-dependent proteases trigger the conformational changes in EGP that are necessary for fusion. The infectivity of VSV-EGP, vesicular stomatitis virus (VSV) particles engineered to contain EGP instead of VSV glycoprotein in the viral envelope, was reduced more than 99-fold by inhibitors of the mammalian proteases CatB and CatL. In a separate experiment using a strain of EboV, CatB inhibitors reduced the number of EboV particles produced in cell culture 10-fold. Figure 1 shows how the introduction of certain genes into two mutant mouse cell lines affected the subsequent infection of these cells by VSV-EGP.

Passage 3 (Questions 13 - 16)

In myocardial infarction (MI), necrosis of cardiac tissue occurs, typically as a result of the blockage of a coronary artery by a blood clot. Following MI, cardiac repair (or remodeling) begins when necrotic cardiomyocytes trigger an inflammatory response, resulting in the removal of cellular debris. Gradually, cardiac fibroblasts produce new extracellular matrix proteins, and scar tissue is formed. While initially an adaptive process, remodeling becomes progressively adverse as surrounding cardiomyocytes enlarge and the injured area weakens and dilates. Cardiac fibroblasts then begin differentiating into collagen-secreting myofibroblasts, which are not normally found in healthy adult cardiac tissue, and persistent activation of these cells leads to myocardial stiffness and progression to heart failure. Cardiac fibroblast differentiation is thought to be regulated by the Wnt/Frizzled signaling pathway. Wnt proteins are a family of palmitoylated secretory proteins with isoelectric points around 9 that bind and activate the G protein-coupled receptor Frizzled, whose structure includes seven transmembrane α-helical domains. Additional proteins in this pathway include the enzyme casein kinase-1 (CK1), glycogen synthase kinase-3 (GSK3), and β-catenin, which activates expression of Wnt target genes. In the absence of Frizzled activation, CK1 and GSK3 sequentially phosphorylate β-catenin, which targets it for ubiquitination. When Frizzled is activated, CK1 and GSK3 activity is inhibited, and Wnt target genes are transcribed (Figure 1). In healthy adult cardiac tissue, the Wnt signaling pathway is silent; however, it is reactivated by cardiac tissue injury.

Passage 4 (Questions 17 - 20)

Researchers investigated the role of a secreted exotoxin (Protein X) in the pathogenicity of a bacterium. The protein can be detected in the bacterial culture medium only under anaerobic conditions. To analyze the three-dimensional structure of Protein X, its cDNA was cloned and expressed. The purified protein was analyzed by gel electrophoresis under native, denaturing, and denaturing/reducing conditions as shown in Figure 1. The fluorescent molecule 1-anilinonaphthalene-8-sulfonic acid (ANS), which exhibits increased fluorescence upon binding to hydrophobic surface residues of proteins, was used to probe for conformational changes in Protein X in the presence of the detergent dodecyl phosphocholine (DPC) (Figure 2). To investigate the mechanism of Protein X pathogenicity, researchers filled phospholipid-bound liposomes (micelles) with a fluorescent dye. Addition of Protein X to the liposomes caused a gradual release of the fluorescent dye from the liposomes (Figure 3).

Passage 7 (Questions 33 - 38)

The anterior lobe of the pituitary gland produces hormones that regulate the activity of other glands. For example, the anterior pituitary gland (APG) produces luteinizing hormone (LH), which stimulates sex hormone production by the gonads in both sexes and triggers ovulation in females. Hormone secretion from the APG is regulated by specialized neurosecretory neurons located in the hypothalamus. These neurons have terminals that secrete stimulatory or inhibitory factors into a capillary bed. These capillaries empty into venules, which carry the factors to sinusoids in the anterior pituitary. Here the hypothalamic factors act on the hormone-secreting cells. One of these factors, gonadotropin-releasing hormone (GnRH), is a potent stimulator of pituitary LH production and release. In addition to GnRH, there may be other hypothalamic factors that regulate LH secretion. These factors are known as neuromodulators, in contrast to releasing factors, because they do not stimulate LH release on their own, but modulate (enhance or inhibit) the responses of the secretory cells of the anterior pituitary to GnRH. Shown in Table 1 are results from a study in which the pituitary effects of the hypothalamic factor neuropeptide Y (NPY) were examined. In the study, LH levels were measured in female rats treated with saline (control), NPY alone, GnRH alone, or GnRH in combination with NPY.

Passage 10 (Questions 53 - 56)

The transcription factor Foxp3 is encoded by an X-linked gene in mice and humans. Approximately 60% of female Foxp3sf/+ mice, which carry one inactivating scurfin (sf) allele and one wild-type (+) allele of the Foxp3 gene, spontaneously develop mammary tumors by 2 years of age. The following experimental observations clarify the relationship between Foxp3 and the development of mammary tumors: Wild-type Foxp3 transcripts are easily detected in noncancerous mammary epithelium from female Foxp3sf/+ mice but are undetectable in cancerous mammary epithelium from the same mice. In female Foxp3sf/+ mice, the wild-type Foxp3 allele is carried on the inactivated X chromosome in 15% of noncancerous mammary epithelial cells but in 100% of cancerous mammary epithelial cells. In female Foxp3sf/+ mice, cancerous mammary epithelium has 8- to 12-fold more mRNA for the transmembrane cell surface receptor ErbB2 than does noncancerous mammary epithelium. Foxp3 binds specific sequences in the ErbB2 promoter, and deletion of these Foxp3-binding sites increases expression from the ErbB2 promoter in Foxp3-expressing cells. The growth rate of mammary tumor cells in culture and the ability of these cells to cause death when injected into mice correlate inversely with the level of Foxp3 expression.

Passage 1 (Questions 1 - 4)

Ultraviolet radiation (UVR) causes skin cancer by producing mutagenic cyclobutane pyrimidine dimers (CPDs) and (6-4) photoproducts (PPs) in DNA. In mice and humans, the nucleotide excision repair system (NER) recognizes, removes, and replaces segments of DNA strands that contain CPDs and (6-4) PPs. Figure 1 shows how mouse skin levels of XPA (a damage-recognition and rate-limiting factor in NER) and clock regulatory protein CRY1 fluctuate in accordance with the internal physiological and metabolic time keeper, the circadian clock. Figures 2 and 3 show the circadian rhythms of NER activity and DNA replication in mouse skin, respectively.

Passage 9 (Questions 48 - 52)

ertain cell types, including cytotoxic T lymphocytes (CTLs) and skin melanocytes, use modified lysosomes called secretory lysosomes to secrete cell-specific products in response to appropriate stimuli. Consequently, in mice and humans, mutations that disrupt cellular machinery specific to this secretory pathway can cause immunodeficiency and partial albinism. CTLs use secretory lysosomes called lytic granules to secrete cytolytic proteins. Melanocytes use secretory lysosomes called melanosomes to transport melanin pigments. Melanosomes are transported along microtubules to the cell periphery where the melanosomes transfer to microfilaments prior to the release of melanin from the cell. Rab27a protein on the melanosome membrane facilitates this transfer by interacting with melanophilin protein, which in turn binds myosin Va, a motor protein on microfilaments. People with Griscelli syndrome have hypopigmented skin and silver-gray hair. These pigmentation phenotypes are sometimes but not always associated with an immunodeficiency resulting from a defect in the killing function of CTLs. Characteristics of two mutant mouse lines that are used as models for Griscelli syndrome are shown in Table 1.

Passage 2 (Questions 5 - 8)

Aluminum phosphide (AlP) is a pesticide that can react to produce phosphine, a gas that is highly toxic to humans and animals (Reaction 1). Phosphine toxicity is the result of interference with cellular metabolism. It is thought that phosphine reacts with important sulfhydryl groups in complexes I (NADH dehydrogenase), II (succinate dehydrogenase), and IV (cytochrome oxidase) of the electron transport chain (ETC). Researchers conducted a kinetic analysis of the effect of AlP on cytochrome oxidase (Figure 1). Researchers measured the levels of ATP, the rate of ATP synthesis, and the rate of ATP hydrolysis in rat liver mitochondria. It was found that AlP exposure resulted in a 65% decrease in ATP levels and a 48% decrease in the rate of ATP synthesis. In addition, the effect of AlP on the activity of three ETC complexes was determined (Table 1). The activities of these complexes were determined independently of each other.

38. Which of the following graphs likely illustrates the changes in blood estrogen levels that occur in female rats after treatment with NPY alone?

Answer: A. Reasoning: This is a Biology question that falls under the content category "Structure and functions of the nervous and endocrine systems and ways in which these systems coordinate the organ systems." The answer to this question is A because although LH affects estrogen levels, according to the data in Table 1, NPY treatment alone did not alter LH levels. Therefore, no change in blood estrogen levels is expected. This is a Data-based and Statistical Reasoning question because you are required to interpret the data in Table 1 in order to predict what effect NPY treatment alone would have on estrogen levels.

31. G542X is another CFTR allele. If a female heterozygous for G542X bears a child fathered by a male heterozygous for the ΔF508 allele, what is the probability that the child would be homozygous for the G542X allele, given that neither parent has CF? A. 0.00 B. 0.25 C. 0.75 D. 1.00

Answer: A. 0.00 Reasoning: This is a Biology question that falls under the content category "Transmission of heritable information from generation to generation and the processes that increase genetic diversity." The answer to this question is A because both parents would need to carry the G542X allele in order for a child to be homozygous for the G542X allele. Because only the mother carries the G542X allele, the probability that the child will be homozygous for the G542X allele is 0. It is a Scientific Reasoning and Problem Solving question because it requires you to understand homozygous and heterozygous alleles and calculate genetic patterns of inheritance.

55. In a female mouse born with two wild-type Foxp3 alleles, what is the minimum number of inactivating, recessive mutations that might be sufficient in the Foxp3 alleles in mammary epithelium to significantly increase the likelihood that this mouse will develop mammary tumors? A. 1 B. 2 C. 3 D. 4

Answer: A. 1 Reasoning: This is a Biology question that falls under the content category "Transmission of heritable information from generation to generation and the processes that increase genetic diversity." The answer to this question is A because the passage indicates that approximately 60% of female Foxp3sf/+ mice, which carry one inactivating scurfin (sf) allele and one wild-type (+) allele, spontaneously develop mammary tumors. Therefore, the minimum number of inactivating mutations to increase the likelihood of tumor formation is one. It is a Scientific Reasoning and Problem Solving question because it requires evaluating the information in the passage to make a scientific prediction.

47. In humans, the characteristic tissue of which of the following organs is NOT derived from mesoderm? A. Brain B. Heart C. Kidney D. Skeletal muscle

Answer: A. Brain Reasoning: This is a Biology question that falls under the content category "Processes of cell division, differentiation, and specialization." The answer to this question is A because the brain is part of the central nervous system, which is derived from ectoderm. Heart, kidney, and skeletal muscle are derived from mesoderm. It is a Knowledge of Scientific Concepts and Principles question because it involves major structures that arise from primary germ layers.

51. Melanosomes most likely move along microtubules that originate in and radiate from the: A. centrosome. B. kinetochores. C. Golgi apparatus. D. microfilaments under the plasma membrane.

Answer: A. Centrosome Reasoning: This is a Biology question that falls under the content category "Assemblies of molecules, cells, and groups of cells within single cellular and multicellular organisms." The answer to this question is A because microtubules are cellular structures that originate from centrosomes. It is a Knowledge of Scientific Concepts and Principles question because you must recall the organization of microtubules and their role in cellular transport.

25. Vasopressin regulates the insertion of aquaporins into the apical membranes of the epithelial cells of which renal structure? A. Collecting duct B. Proximal tubule C. Bowman's capsule D. Ascending loop of Henle

Answer: A. Collecting duct Reasoning: This is a Biology question that falls under the content category "Structure and functions of the nervous and endocrine systems and ways in which these systems coordinate the organ systems." The answer to this question is A because vasopressin regulates the fusion of aquaporins with the apical membranes of the collecting duct epithelial cells. It is a Knowledge of Scientific Concepts and Principles question because it requires knowledge of hormone function in the excretory system.

7. When researchers determined the total cellular concentration of ATP in AlP-exposed rat liver cells, they found the concentration to be equal to the control value. Which conclusion about the metabolic state of the cell is best supported by these data? A. Glycolytic flux is increased after AlP treatment. B. Glycolytic flux is decreased after AlP treatment. C. Citric acid cycle flux is increased after AlP treatment. D. Citric acid cycle flux is decreased after AlP treatment.

Answer: A. Glycolytic flux is increased after AIP treatment Reasoning: This is a Biochemistry question that falls under the content category "Principles of bioenergetics and fuel molecule metabolism." The answer to this question is A because ATP production is the same in both control and AlP-exposed cells, and the data in the passage show that mitochondrial ATP production is decreased. This indicates that the flux through glycolysis is increased, because this would be the major pathway for ATP production once the electron transport chain is shut down. It is a Scientific Reasoning and Problem Solving question because you must reason about how changes in metabolic flux affect the amount of products from a metabolic pathway.

35. Scientists are hopeful that NPY can be used in combination with GnRH to treat certain cases of female infertility. Individuals with a deficiency in what receptor system would be most likely to benefit from such a treatment? A. GnRH B. LH C. NPY D. LH and NPY

Answer: A. GnRH Reasoning: This is a Biology question that falls under the content category "Structure and integrative functions of the main organ system." The answer to this question is A because according to the data in Table 1, NPY enhances the normal function of GnRH in stimulating blood LH levels. It is a Scientific Reasoning and Problem Solving question because it requires you to make a prediction based on the data presented in Table 1.

16. The middle region, as opposed to either end, of each of the seven α-helical domains of Frizzled is most likely to contain a high proportion of which type of amino acid residue? A. Nonpolar B. Polar uncharged C. Positively charged D. Negatively charged

Answer: A. Nonpolar Reasoning: This is a Biochemistry question that falls under the content category "Structure and function of proteins and their constituent amino acids." The answer to this question is A because according to the passage, the seven α-helical domains are transmembrane domains and, thus, cross the phospholipid bilayer. As a result, these domains are most likely to have a high proportion of hydrophobic residues. Hydrophobic residues are nonpolar. It is a Scientific Reasoning and Problem Solving question because it requires you to identify the type of amino acid residues contained within the middle region of Frizzled, based on the information presented in the passage.

23. What is the best experimental method to analyze the effect of tdh2 gene deletion on the rate of histone acetylation? Comparing histone acetylation in wild-type and Δtdh2 cells by: A. Western blot B. Southern blot C. Northern blot D. RT-PCR

Answer: A. Western blot Reasoning: This is a Biology question that falls under the content category "Transmission of genetic information from the gene to the protein." The answer to this question is A because posttranslational modification of proteins such as histone acetylation is analyzed by Western blotting. It is a Reasoning about the Design and Execution of Research question because you must reason about the appropriateness of tools that are used to assess posttranslational modification of proteins.

36. Scientists have hypothesized that NPY is necessary for the generation of the preovulatory LH surge, a hormonal event that triggers ovulation. Which of the following findings best supports this hypothesis? A. When the actions of NPY are blocked in female rats, the LH surge and ovulation do not occur. B. NPY release from the hypothalamus increases just prior to the preovulatory LH surge. C. NPY can enhance GnRH-stimulated LH secretion in female rats during the preovulatory period. D. NPY has no effect on GnRH-stimulated LH secretion in male rats.

Answer: A. When the actions of NPY are blocked in female rats, the LH surge and ovulation do not occur Reasoning: This is a Biology question that falls under the content category "Structure and integrative functions of the main organ system." The answer to this question is A because when the function of only NPY is blocked, the LH surge and ovulation no longer occur. This provides strong evidence that NPY function is required for these events. B is incorrect because NPY release could occur just prior to the preovulatory LH surge, but these could be completely unrelated events. C is incorrect because in this case, NPY is enhancing the LH secretion, but it does not prove that NPY is necessary for the generation of the preovulatory LH surge. D is incorrect because ovulation occurs in females. It is a Scientific Reasoning and Problem Solving question because it requires you to choose the observation that would best support the hypothesis.

43. The precursor of EGP is translated from a transcript that has had one nontemplated nucleotide added to the open reading frame. This change does not create or eliminate a stop codon. Compared with the protein sGP, which is produced from the unedited transcript, EGP most likely has the same primary: A. amino-terminal sequence as sGP, but a different primary carboxy-terminal sequence. B. carboxy-terminal sequence as sGP, but a different primary amino-terminal sequence. C. sequence as sGP except that EGP has one additional amino acid. D. sequence as sGP except that EGP has one less amino acid.

Answer: A. amino-terminal sequence as sGP, but a different primary carboxy-terminal sequence Reasoning: This is a Biology question that falls under the content category "Transmission of genetic information from the gene to the protein." The answer to this question is A because the addition of one nucleotide to the open reading frame of EGP results in a frameshift mutation and an aberrant carboxy-terminal domain. It is a Scientific Reasoning and Problem Solving question because it requires you to evaluate the effect of a nucleotide addition to the open reading frame of EGP in its protein sequence.

15. In the absence of Frizzled activation, β-catenin is covalently modified and: A. bound by a proteasome to initiate degradation into short peptides. B. translocated into the Golgi body for secretion through exocytosis. C. engulfed by a lysosome where it is hydrolyzed by proteases. D. stored in vesicles until the signaling pathway is activated.

Answer: A. bound by a proteasome to initiate degredation into short peptides Reasoning: This is a Biology question that falls under the content category "Transmission of genetic information from the gene to the protein." The answer to this question is A because according to the passage and Figure 1, in the absence of Frizzled activation, β-catenin is phosphorylated and ubiquitinated. Ubiquitination targets a protein for degradation by a proteasome. It is a Knowledge of Scientific Concepts and Principles question because it requires knowledge of a posttranslational process.

53. Based on the passage, Foxp3 affects ErbB2 expression in noncancerous mammary epithelium most likely by directly: A. inhibiting synthesis of ErbB2 mRNA. B. stimulating synthesis of ErbB2 mRNA. C. inhibiting synthesis of ErbB2 protein. D. stimulating synthesis of ErbB2 protein.

Answer: A. inhibiting synthesis of ErbB2 mRNA Reasoning: This is a Biology question that falls under the content category "Transmission of genetic information from the gene to the protein." The answer to this question is A because the passage states that cancerous mammary epithelium in female Foxp3sf/+ mice has 8- to 12-fold more ErbB2 mRNA than noncancerous mammary epithelium. Because the passage states that Foxp3 binds the ErbB2 promoter, Foxp3 regulation of ErbB2 expression occurs at the transcriptional, not translational level. Noncancerous epithelium has reduced ErbB2 mRNA expression compared with cancerous epithelium, which suggests that in noncancerous epithelium, Foxp3 inhibits synthesis of ErbB2 mRNA. It is a Scientific Reasoning and Problem Solving question because the question involves determining which cellular process would reduce ErbB2 mRNA levels.

5. AlP exposed to an aqueous solution in which pH range will result in the largest amount of phosphine production? A. pH < 4 B. 4 < pH < 7 C. 7 < pH < 10 D. pH > 10

Answer: A. pH < 4 Reasoning: This is a General Chemistry question that falls under the content category "Principles of bioenergetics and fuel molecule metabolism." The answer to this question is A because H+ is a reactant, and the increase in the concentration of H+ at low pH will favor product formation. It is a Scientific Reasoning and Problem Solving question because it is necessary to take the information from the chemical equation in the passage and reason about how changing solution conditions will affect that reaction.

52. Lytic granules are generally released from CTLs when the T-cell receptors on these cells bind specifically to: A. viral antigens presented on the surface of virus-infected cells. B. growth factors secreted by helper T lymphocytes. C. B-cell receptors on activated B lymphocytes. D. constant regions of secreted antibodies.

Answer: A. viral antigens presented on the surface of virus-infected cells Reasoning: This is a Biology question that falls under the content category "Structure and integrative functions of the main organ system." The answer to this question is A because cytotoxic T lymphocytes target virus-infected cells by recognizing the viral antigen presented on the cell surface. It is a Knowledge of Scientific Concepts and Principles question because you must recall the function of cytotoxic T lymphocytes.

27. Osmotic pressure Π is given by the relation: Π = iMRT where i is the van't Hoff factor, M is the concentration of solute, R is the gas constant, and T is the temperature. The osmotic pressure of sea water is approximately 24 atm at 25°C. What is the approximate concentration of salt in sea water (approximated by NaCl with i = 2)? (Note: Use R = 0.08 L•atm/mol•K.) A. 0.25 M B. 0.50 M C. 0.75 M D. 1.0 M

Answer: B. 0.50 M Reasoning: This is a General Chemistry question that falls under the content category "Assemblies of molecules, cells, and groups of cells within single cellular and multicellular organisms." The answer to this question is B since algebraic manipulation of the relation gives M = Π/iRT = 24 atm / (2 × 0.08 L•atm/mol•K × 300 K) = 24 / (2 × 24) mol/L = 0.5 M. It is a Scientific Reasoning and Problem Solving question since you must manipulate a scientific formula algebraically to solve a problem.

37. NPY amplified pituitary responses to GnRH by: A. 27%. B. 48%. C. 67%. D. 83%.

Answer: B. 48% Reasoning: This is a Biology question that falls under the content category "Structure and functions of the nervous and endocrine systems and ways in which these systems coordinate the organ systems." The answer to this question is B. From Table 1, GnRH + NPY = 7.03 and GnRH alone = 4.74. 7.03 / 4.74 = 1.48, so NPY amplified pituitary responses to GnRH by 48%. This is a Data-based and Statistical Reasoning question because you must determine the percentage increase in LH levels from the data presented in Table 1.

45. Under anaerobic conditions, how many net molecules of ATP are produced by the consumption of 5 moles of glucose? A. 3 × 10^24 B. 6 × 10^24 C. 9 × 10^24 D. 1.2 × 10^25

Answer: B. 6 x 10^24 Reasoning: Under anaerobic conditions, 2 moles of ATP are produced from each mole of glucose. Thus, 10 moles of ATP would be generated from 5 moles of glucose. Since there are 6 × 10^23 molecules per mole, 10 moles of ATP is equal to 6 × 10^24 molecules. It is a Scientific Reasoning and Problem Solving question because you must use knowledge about the production of ATP under anaerobic conditions and Avogadro's number relating molecules to moles to scale the answer to the correct value and units.

49. Based on the passage, myosin Va most likely directly binds: A. tubulin. B. actin. C. melanin. D. Rab27a.

Answer: B. Actin Reasoning: This is a Biology question that falls under the content category "Assemblies of molecules, cells, and groups of cells within single cellular and multicellular organisms." The answer to this question is B because based on the passage, myosin Va is a motor protein. Motor proteins such as myosin Va move along microfilaments through interaction with actin. It is a Scientific Reasoning and Problem Solving question because it requires you to evaluate which cellular protein would directly bind to myosin Va.

59. A homodimeric protein was found to migrate through SDS polyacrylamide gel electrophoresis (SDS-PAGE) with a mobility that matched that of a 45-kDa standard. What change in the experiment would increase the chances of observing the mobility expected for the 22.5-kDa monomer? A. Increasing the gel running time B. Adding a reducing agent C. Using a higher voltage D. Removing the SDS

Answer: B. Adding a reducing agent Reasoning: This is a Biochemistry question that falls under the content category "Structure and function of proteins and their constituent amino acids." The answer to this question is B because adding a reducing agent would eliminate any disulfide bridges and allow the monomers to run separately-thus leading to a migration expected for the 22.5-kDa protein. It is a Reasoning about the Design and Execution of Research question because you must use knowledge about the experimental procedure to determine an outcome that would result if this procedure was changed.

6. Based on the passage, which amino acid will most likely react with phosphine? A. Met B. Cys C. Ser D. Thr

Answer: B. Cys Reasoning: This is an Organic Chemistry question that falls under the content category "Structure and function of proteins and their constituent amino acids." The answer to this question is B because the passage states that phosphine reacts with sulfhydryl groups, and the cysteine side chain contains a sulfhydryl group. It is a Knowledge of Scientific Concepts and Principles question because you must recall the structure of the cysteine side chain and apply it to the reactivity described in the passage.

56. Human breast cancer patients whose tumors overexpress HER2, the human homolog of ErbB2, may be treated with trastuzumab, an antibody that was developed to be highly specific for the extracellular domain of HER2. Given this, which of the regions of trastuzumab shown in the figure is(are) most likely highly specific for the extracellular domain of HER2? A. Regions 1 and 2 only B. Regions 1 and 3 only C. Regions 3 and 4 only D. Region 4 only

Answer: B. Regions 1 and 3 only Reasoning: This is a Biology question that falls under the content category "Structure and integrative functions of the main organ system." The answer to this question is B because regions 1 and 3 correspond to the variable portion of the light and heavy chains, respectively, of an antibody. The variable region of an antibody will enable recognition of a particular antigen, such as HER2, which typically has elevated expression in breast cancer tumors. It is a Scientific Reasoning and Problem Solving question because it involves reasoning about a scientific model of antigen-antibody binding.

13. Based on the passage, which statement describes Wnt proteins? A. They are composed of multiple subunits. B. They have a positive charge. C. They are synthesized in the smooth endoplasmic reticulum. D. They fold into their tertiary structure in the cytoplasm.

Answer: B. They have a positive charge Reasoning: This is a Biochemistry question that falls under the content category "Structure and function of proteins and their constituent amino acids." The answer to this question is B because based on the passage, Wnt proteins are a family of secretory proteins with isoelectric points around 9, implying that they are positively charged at physiological pH. A is incorrect because there is no information in the passage to support this response. C is incorrect because secretory proteins are synthesized in the rough endoplasmic reticulum. D is incorrect because folding of secretory proteins occurs in the rough endoplasmic reticulum. It is a Scientific Reasoning and Problem Solving question because it requires you to estimate the charge of Wnt proteins, based on the value of their isoelectric points as described in the passage.

46. In humans, eggs and sperm are most similar with respect to: A. cell size. B. genome size. C. the time required for development. D. the numbers produced by a single individual.

Answer: B. genome size Reasoning: This is a Biology question that falls under the content category "Processes of cell division, differentiation, and specialization." The answer to this question is B because both eggs and sperm contain a haploid number of chromosomes and therefore they are most similar with respect to their genome size. It is a Knowledge of Scientific Concepts and Principles question because it requires knowledge of gametogenesis and meiosis.

21. The enzyme encoded by the tdh2 gene catalyzes the reversible conversion of: A. 3-phosphoglycerate to 1,3-bisphosphoglycerate. B. glyceraldehyde-3-phosphate to 1,3-bisphosphoglycerate. C. fructose-1,6-bisphosphate to glyceraldehyde-3-phosphate. D. 2-phosphoglycerate to 3-phosphoglycerate.

Answer: B. glyceraldehyde-3-phosphate to 1,3-bisphosphoglycerate Reasoning: This is a Biochemistry question that falls under the content category "Principles of bioenergetics and fuel molecule metabolism." The answer to this question is B because GAPDH, the enzyme encoded by the tdh2 gene, is a glycolytic enzyme that catalyzes the reversible conversion of glyceraldehyde-3-phosphate to 1,3-bisphosphoglycerate. It is a Knowledge of Scientific Concepts and Principles question because it requires knowledge of glycolytic enzymes and their substrates and products.

48. Given that secretory lysosomes form normally in the melanocytes and CTLs of ashen and dilute mice, the data in Table 1 best support the conclusion that melanosome secretion and lytic granule secretion differ in that the secretion of lytic granules does NOT require: A. Rab27a. B. myosin Va. C. microtubules. D. fusion of these secretory lysosomes with the plasma membrane.

Answer: B. myosin Va Reasoning: This is a Biology question that falls under the content category "Transmission of genetic information from the gene to the protein." The answer to this question is B because based on the data in Table 1, lytic granule secretion (as indicated by killing by CTLs) remains unaffected by a mutation that inactivates myosin Va. Therefore, it can be concluded that in contrast to melanosome secretion, the secretion of lytic granules does not require myosin Va. It is a Data-based and Statistical Reasoning question because it requires you to draw conclusions about how the inactivation of different genes affects secretory pathways in different cells.

34. NPY is not classified as a releasing factor because it has: A. no effect on LH secretion. B. no effect on LH secretion without GnRH. C. no effect on LH secretion in combination with GnRH. D. the same effect on LH secretion as GnRH.

Answer: B. no effect on LH secretion without GnRH Reasoning: This is a Biology question that falls under the content category "Structure and functions of the nervous and endocrine systems and ways in which these systems coordinate the organ systems." The answer to the question is B because LH levels in the control and NPY alone treatment conditions were similar, and GnRH had to be present for LH levels to be significantly different from control levels. According to the passage, releasing factors stimulate LH release on their own. Therefore, NPY cannot be classified as a releasing factor because it had no effect on LH secretion without GnRH. In this case, NPY functioned as a neuromodulator. This is a Data-based and Statistical Reasoning question because it requires you to draw conclusions about how the different treatment conditions affect LH levels.

17. Subunits of Protein X are linked covalently by bonds between the: A. thiol groups of methionine residues. B. thiol groups of cysteine residues. C. hydroxyl groups of serine residues. D. hydroxyl groups of threonine residues.

Answer: B. thiol groups of cysteine residues Reasoning: This is a Biochemistry question that falls under the content category "Structure and function of proteins and their constituent amino acids." The answer to this question is B because Figure 1 shows that reducing agents separate subunits of Protein X. This indicates that subunits of Protein X are linked together by disulfide bonds, which implicate the thiol groups of cysteine residues. It is a Knowledge of Scientific Concepts and Principles question because it requires knowledge of the involvement of the thiol groups of cysteine residues in the formation of disulfide bonds in proteins.

4. Which image shows an example of a cyclobutane pyrimidine dimer?

Answer: C. Reasoning: This is a Biochemistry question that falls under the content category "Transmission of genetic information from the gene to the protein." The answer to this question is C because it is the only image that displays a dimer of two pyrimidine residues. It is a Knowledge of Scientific Concepts and Principles question because it requires knowledge of purine and pyrimidine structures.

54. Given that the tumorigenicity of a certain mouse mammary tumor cell line is dependent on ErbB2-mediated intracellular signaling, mice injected with variants of this cell line that have which of the following modifications would be most likely to survive the longest? A. The Foxp3 genes deleted B. One of the ErbB2 genes amplified C. A Foxp3-expressing plasmid introduced D. The Foxp3 binding sites deleted in the promoter of one of the ErbB2 genes

Answer: C. A Foxp3-expressing plasmid introduced Reasoning: This is a Biology question that falls under the content category "Transmission of genetic information from the gene to the protein." The answer to this question is C because according to the passage, cancerous tumors are associated with the inactivating scurfin allele, not the wild-type Foxp3 allele. Because Foxp3 regulates ErbB2 transcription and high levels of ErbB2 mRNA are found in cancerous epithelium, the best modification to a cell line would be Foxp3-expressing plasmid, to repress transcription of ErbB2. It is a Scientific Reasoning and Problem Solving question because it requires you to evaluate which cellular modification would enhance survival of the mice.

11. The graph shows the average relative concentration of ions in pond water and in the cytoplasm of green algae cells. Which process moves chlorine ions into the cells of the green algae? A. Osmosis B. Diffusion C. Active transport D. Facilitated diffusion

Answer: C. Active Transport Reasoning: This is a Biology question that falls under the content category "Assemblies of molecules, cells, and groups of cells within single cellular and multicellular organisms." The answer to this question is C because in order to maintain a higher concentration of chlorine ions inside the cell, the ions must be moved into the cell against their concentration gradient, which requires energy. This process is active transport. In the other processes, ions would move along their concentration gradient, either with or without the help of transport proteins. It is a Knowledge of Scientific Concepts and Principles question because it addresses assumed knowledge of membrane transport processes.

18. Based on the results in Figure 2, what effect does DPC have on the hydrophobic amino acids in Protein X? A. DPC phosphorylates these amino acids. B. DPC hydrolyzes these amino acids. C. DPC exposes these amino acids. D. DPC suppresses these amino acids.

Answer: C. DPC exposes these amino acids Reasoning: This is a Biochemistry question that falls under the content category "Structure and function of proteins and their constituent amino acids." The answer to this question is C because Figure 2 shows that in the presence of DPC, Protein X has increased fluorescence, indicating that it has adopted a conformation that exposes hydrophobic residues on its surface. It is a Scientific Reasoning and Problem Solving question because it requires you to evaluate the information presented in Figure 2 to explain the scientific basis of an observation.

39. Based on the passage, CatB and CatL most likely act on EGP in which of the following cellular compartments to facilitate membrane fusion? A. Endoplasmic reticulum B. Golgi apparatus C. Endosomes D. Cytosol

Answer: C. Endosomes Reasoning: This is a Biology question that falls under the content category "Assemblies of molecules, cells, and groups of cells within single cellular and multicellular organisms." The answer to this question is C because based on the passage, the entry of the virus into the host cell requires CatB and CatL proteases and involves endocytosis through the fusion of the viral membrane with the host cell membrane. Internalization of viral particles through endocytosis is mediated by endosomes. It is a Knowledge of Scientific Concepts and Principles question because it requires knowledge of cellular compartments and their functions within eukaryotic cells.

57. Where in the human male reproductive system do the gametes become motile and capable of fertilization? A. Testis B. Urethra C. Epididymis D. Prostate gland

Answer: C. Epididymis Reasoning: This is a Biology question that falls under the content category "Structure and integrative functions of the main organ system." The answer to this question is C because sperm, produced in the seminiferous tubules of the testes, completes maturation and becomes motile in the epididymis. It is a Knowledge of Scientific Concepts and Principles question because it asks you to recall aspects of human male reproductive anatomy.

10. Inhibition of phosphofructokinase-1 by ATP is an example of: I. allosteric regulation. II. feedback inhibition. III. competitive inhibition. A. I only B. III only C. I and II only D. II and III only

Answer: C. I and II only Reasoning: This is a Biochemistry question that falls under the content category "Structure and function of proteins and their constituent amino acids." The answer to this question is C because ATP, the end product of glycolysis, downregulates through feedback inhibition the activity of phosphofructokinase-1 by binding to a regulatory site other than the active site of the enzyme (allosteric regulation). In contrast, competitive inhibition involves competition for binding to the active site. It is a Scientific Reasoning and Problem Solving question because you need to reason about how the product of glycolysis (ATP) inhibits a regulatory enzyme of glycolysis (phosphofructokinase-1).

12. Dendrotoxin from the mamba snake blocks voltage-gated potassium channels in somatic motor neurons that regulate skeletal muscle contraction. In what way would initial exposure to dendrotoxin affect the ability of a somatic motor neuron to propagate an electrical signal in response to a stimulus? A. It would inhibit the initiation of an action potential. B. It would shorten the refractory period. C. It would prolong the action potential. D. It would prevent depolarization.

Answer: C. It would prolong the action potential Reasoning: This is a Biology question that falls under the content category "Structure and function of the nervous and endocrine systems and ways in which these systems coordinate the organ systems." The answer to this question is C because if potassium ion channels are blocked, the membrane would fail to repolarize, extending the length of the action potential and simulating excessive muscle contractions. It is a Scientific Reasoning and Problem Solving question because it requires you to predict which outcome would occur.

9. A large carbohydrate is tagged with a fluorescent marker and placed in the extracellular environment around a macrophage. The macrophage ingests the carbohydrate via phagocytosis. Which cellular structure is most likely to be fluorescently labeled upon viewing with a light microscope soon after phagocytosis? A. Nucleus B. Golgi apparatus C. Lysosome D. Endoplasmic reticulum

Answer: C. Lysosome Reasoning: This is a Biology question that falls under the content category "Assemblies of molecules, cells, and groups of cells within single cellular and multicellular organisms." The answer to this question is C because when a macrophage ingests foreign material, the material initially becomes trapped in a phagosome. The phagosome then fuses with a lysosome to form a phagolysosome. Inside the phagolysosome, enzymes digest the foreign object. Of the cell structures listed, the labeled carbohydrate is most likely to be microscopically visualized within a lysosome (phagolysosome). It is a Scientific Reasoning and Problem Solving question because the question makes a scientific prediction.

2. Which cells harvested from adult mice were most likely used as the highly proliferative benchmark in the experiment that generated the data shown in Figure 3? A. Adipocytes B. Cardiac muscle cells C. Gastrointestinal epithelial cells D. Neurons

Answer: C. Neurons Reasoning: This is a Biology question that falls under the content category "Structure and integrative functions of the main organ system." The answer to this question is C because the epithelial cells that line the gastrointestinal tract are typically highly proliferative. It is a Reasoning about the Design and Execution of Research question because you must apply your knowledge of the different cell types to predict which cell type would best suit the experiment.

40. Based on the passage, is CatL expression sufficient for VSV-EGP infection of the mouse cell lines presented in Figure 1? (Note: In these experiments, assume that values of <3% of total cells infected are too low to be measured accurately.) A. Yes, because VSV-EGP infects cells expressing CatL better than it infects cells not expressing CatL B. Yes, because VSV-EGP infects cells expressing CatB better than it infects cells not expressing CatB C. No, because VSV-EGP does not infect CatB-/- cells expressing CatL better than it infects CatB-/- cells not expressing CatL D. No, because VSV-EGP infects cells expressing both CatB and CatL better than it infects cells expressing CatB but not CatL

Answer: C. No, because VSV-EGP does not infect CatB^-/- cells expressing CatL better than it infects CatB^-/- cells not expressing CatL Reasoning: This is a Biology question that falls under the content category "Transmission of genetic information from the gene to the protein." The answer to this question is C because the data presented in Figure 1 indicate that compared with the control mouse cell line (bacterial; column 5), introduction of only the CatL gene into the CatB−/−, CatL−/− cell line (column 7) does not affect cell infectivity and therefore is not sufficient for VSV-EGP infection of the mouse cell lines. It is a Data-based and Statistical Reasoning question because it requires you to draw conclusions about how the expression of different genes affects cell infectivity levels.

26. What are the primary myelin-forming cells in the peripheral nervous system? A. Microglia B. Astrocytes C. Schwann cells D. Oligodendrocytes

Answer: C. Schwann cells Reasoning: This is a Biology question that falls under the content category "Structure and functions of the nervous and endocrine systems and ways in which these systems coordinate the organ systems." The answer to this question is C because Schwann cells are the myelin-forming cells in the peripheral nervous system. It is a Knowledge of Scientific Concepts and Principles question because it assesses knowledge of the different types of glia within the nervous system.

20. What event will most likely occur if Protein X is inserted into the inner membrane of mitochondria? A. The citric acid cycle will cease to function. B. The electron transport chain will cease to function. C. The proton gradient across the inner membrane will dissipate. D. The pH of the intermembrane space will decrease.

Answer: C. The proton gradient across the inner membrane will dissipate Reasoning: This is a Biochemistry question that falls under the content category "Principles of bioenergetics and fuel molecule metabolism." The answer to this question is C because based on information from the passage, Protein X forms membrane-spanning channels that alter the permeability of the inner membrane, thereby dissipating the proton gradient across the inner membrane. It is a Knowledge of Scientific Concepts and Principles question because it requires the knowledge of proton motive force across the inner mitochondrial membrane.

14. Based on the passage, β-catenin most likely has: A. multiple subunits. B. very few disulfide bonds. C. a nuclear localization sequence. D. a high proportion of surface-exposed nonpolar residues.

Answer: C. a nuclear localization sequence Reasoning: This is a Biology question that falls under the content category "Transmission of genetic information from the gene to the protein." The answer to this question is C because according to Figure 1, β-catenin activates transcription factors for Wnt target genes. As transcription factors are found in the nucleus, β-catenin must enter the nucleus. Proteins that are translocated into the nucleus usually contain a nuclear localization sequence. It is a Scientific Reasoning and Problem Solving question because it requires you to bring together assumed knowledge and passage-based observations to make a conclusion.

30. Researchers have noted that chloramphenicol (a commonly used antibiotic) is becoming less effective in treating typhoid fever. The best explanation for this observation would be selection: A. against chloramphenicol in ΔF508 heterozygotes. B. against chloramphenicol in wild-type homozygotes. C. for chloramphenicol resistance in populations of S. typhi. D. for ΔF508 CFTR, which cannot bind chloramphenicol.

Answer: C. for chloramphenicol resistance in populations of S. typhi Reasoning: This is a Biology question that falls under the content category "Transmission of heritable information from generation to generation and the processes that increase genetic diversity." The answer to this question is C because according to the passage, S. typhi is the causative agent of typhoid fever. As S. typhi becomes resistant to chloramphenicol, this antibiotic will become less effective in treating typhoid fever. It is a Scientific Reasoning and Problem Solving question because it requires you to explain the scientific basis of an observation.

28. A prion is best described as an infectious: A. prokaryote. B. transposon. C. protein. D. virus.

Answer: C. protein Reasoning: This is a Biology question that falls under the content category "The structure, growth, physiology, and genetics of prokaryotes and viruses." The answer to this question is C because a prion is an abnormally folded protein that induces a normally folded version of the protein to also adopt the abnormal structure, which is often deleterious. It is a Knowledge of Scientific Concepts and Principles question because it assesses knowledge of different types of pathogens.

41. ased on the passage, does optimal VSV-EGP infection in vitro require CatB, CatL, or both? Optimal infection: A. requires CatB only. B. requires CatL only. C. requires both CatB and CatL. D. does not require CatB or CatL.

Answer: C. requires both CatB and CatL Reasoning: This is a Biology question that falls under the content category "Transmission of genetic information from the gene to the protein." The answer to this question is C because the data presented in Figure 1 indicate that CatB―/―, CatL−/− mouse cells expressing either CatB or CatL have low levels of viral infection. In contrast CatB−/−, CatL−/− mouse cells expressing both CatB and CatL exhibit several folds higher viral infection levels. It is a Data-based and Statistical Reasoning question because it requires you to draw conclusions about how the expression of different genes affect cell infectivity levels.

33. The purpose of the study described in the passage was: A. to determine whether LH can modulate GnRH or NPY secretion. B. to determine whether GnRH can modulate NPY secretion. C. to determine whether NPY can modulate LH secretion. D. to determine whether NPY can modulate GnRH secretion.

Answer: C. to determine whether NPY can modulate LH secretion Reasoning: This is a Biology question that falls under the content category "Structure and functions of the nervous and endocrine systems and ways in which these systems coordinate the organ systems." The answer to the question is C because the dependent variable of the experiment is blood LH levels. According to the passage, GnRH is a potent stimulator of LH production, but there may be other hypothalamic factors that regulate LH secretion. Therefore, because the treatment conditions were control, NPY alone, GnRH alone, and GnRH + NPY, the researchers were testing whether NPY can modulate LH secretion. It is a Reasoning about the Design and Execution of Research question because it requires you to determine the purpose of the study from the information in the passage and the data presented in Table 1.

42. Based on the passage, CatB or CatL or both would be expected to have which of the following effects, if any, on EGP? A. No effect B. Reduction of enzyme activity C. Formation of protein dimers D. Digestion into smaller protein fragments

Answer: D. Digestion into smaller protein fragments Reasoning: This is a Biology question that falls under the content category "Structure and function of proteins and their constituent amino acids." The answer to this question is D because the passage states that CatB and CatL are proteases. Proteases function to digest proteins into smaller fragments. It is a Knowledge of Scientific Concepts and Principles question because it requires knowledge of enzyme functions in catalyzing different biological reactions.

19. Which interpretation(s) is(are) consistent with the observations in the passage? I. Surface amino acids of Protein X are mostly hydrophilic in aqueous solution. II. Surface amino acids of Protein X are mostly hydrophilic in presence of DPC. III. Surface amino acids of Protein X are mostly hydrophobic in presence of DPC. A. I only B. II only C. III only D. I and III only

Answer: D. I and III only Reasoning: This is a Biochemistry question that falls under the content category "Structure and function of proteins and their constituent amino acids." The answer to this question is D because Figure 2 indicates that Protein X has few exposed hydrophobic surfaces in aqueous solution. In the presence of DPC, hydrophobic amino acid residues are exposed on the surface, as evidenced by ANS binding and increased fluorescence levels. It is a Scientific Reasoning and Problem Solving question because it requires you to evaluate the information presented in Figure 2 to explain the scientific basis of an observation.

24. Which experimental approach(es) can be used to analyze the effect of ROS on the lifespan of yeast? Comparing the lifespans of: I. wild-type yeast versus yeast lacking antioxidant enzymes II. wild-type yeast versus yeast overexpressing antioxidant enzymes III. yeasts growing in the presence or absence of hydrogen peroxide A. I only B. II only C. II and III only D. I, II, and III

Answer: D. I, II, and III Reasoning: This is a Biochemistry question that falls under the content category "Principles of bioenergetics and fuel molecule metabolism." The answer to this question is D because all listed options influence ROS levels in yeast and can be used to analyze the role of ROS in regulating the lifespan of yeast. It is a Reasoning about the Design and Execution of Research question because you must identify the appropriate research studies that test the stated hypothesis.

29. Assume that S. typhi immediately enters the bloodstream from the small intestine. Of the following, which would be the first major organ that bloodborne S. typhi would encounter? A. Stomach B. Pancreas C. Large intestine D. Liver

Answer: D. Liver Reasoning: This is a Biology question that falls under the content category "Structure and integrative functions of the main organ system." The answer to this question is D because blood from the small intestine is transported first to the liver, which regulates nutrient distribution and removes toxins from the blood. It is a Knowledge of Scientific Concepts and Principles question because it requires knowledge of how organ systems function in the body and their relationship to one another.

32. Which of the following best describes the phenotype of an individual who is heterozygous with one ΔF508 and one wild-type CFTR allele? A. More susceptible to typhoid fever than wild-type homozygotes and has CF B. More susceptible to typhoid fever than ΔF508 homozygotes and does not have CF C. More resistant to typhoid fever than ΔF508 homozygotes and has CF D. More resistant to typhoid fever than wild-type homozygotes and does not have CF

Answer: D. More resistant to typhoid fever than wild-type homozygotes and does not have CF Reasoning: This is a Biology question that falls under the content category "Transmission of heritable information from generation to generation and the processes that increase genetic diversity." The answer to this question is D because the passage states that individuals homozygous for the ΔF508 allele have CF. Therefore, an individual who is heterozygous for the ΔF508 allele does not have CF. However, this individual is more resistant to typhoid fever because the passage states that human epithelial cells expressing ΔF508 ingest significantly fewer S. typhi than do epithelial cells that express wild-type CFTR. It is a Scientific Reasoning and Problem Solving question because it requires you to predict the effect that a particular genotype would have on disease outcomes.

3. After a section of a DNA strand containing a UVR-induced lesion is removed and resynthesized, the newly synthesized strand is rejoined to the remainder of the DNA strand by what type of bond? A. Disulfide B. Hydrogen C. Peptide D. Phosphodiester

Answer: D. Phosphodiester Reasoning: This is a Biochemistry question that falls under the content category "Transmission of genetic information from the gene to the protein." The answer to this question is D because phosphodiester bonds link the 3ʹ carbon atom of one deoxyribose and the 5ʹ carbon atom of another deoxyribose within the DNA molecules. It is a Knowledge of Scientific Concepts and Principles question because it requires knowledge of DNA strand synthesis through the polymerization of nucleotides.

44. Which molecule is NOT formed during the citric acid cycle? A. Malate B. Succinate C. α-Ketoglutarate D. Phosphoenolpyruvate

Answer: D. Phosphoenolpyruvate Reasoning: This is a Biochemistry question that falls under the content category "Principles of bioenergetics and fuel molecule metabolism." The answer to this question is D because phosphoenolpyruvate is a product of glycolysis, not the citric acid cycle. It is a Knowledge of Scientific Concepts and Principles question because you must recall a key concept about a metabolic cycle.

22. Based on Figure 1, how do the deletions of the hst3 or the rtt genes affect the lifespan of yeast? A. The deletion of the hst3 gene compensates for the ∆rtt-induced decrease in lifespan. B. The deletion of the hst3 gene has no effect on the ∆rtt-induced increase in lifespan. C. The deletion of the rtt gene compensates for the ∆hst3-induced decrease in lifespan. D. The deletion of the rtt gene has no effect on the ∆hst3-induced decrease in lifespan.

Answer: D. The deletion of the rtt gene has no effect on the ∆hst-3 induced decrease in lifespan Reasoning: This is a Biology question that falls under the content category "Processes of cell division, differentiation, and specialization." The answer to this question is D because according to data shown in Figure 1, compared to wild type cells, the ∆hst3 cells exhibit a decrease in lifespan. Furthermore, the deletion of rtt gene in ∆hst3 cells has no effect on the cells lifespan. It is a Scientific Reasoning and Problem Solving question because you need to evaluate data shown in Figure 1 and reason about how deletion of genes affects the phenotypes in yeast.

58. According to the cross-bridge model of muscle contraction, the muscles stiffen after death because ATP is unavailable to bind and directly release: A. ADP from the actin head. B. ADP from the myosin head. C. the actin head from the myosin filament. D. the myosin head from the actin filament.

Answer: D. The myosin head from the actin filament Reasoning: This is a Biology question that falls under the content category "Structure and integrative functions of the main organ system." The answer to this question is D because during normal muscle contraction, ATP is required to break the bonds between the actin filament and the myosin head. After death, no new ATP is generated, so the myosin head cannot be released from the actin filament, resulting in stiffening of muscles. It is a Knowledge of Scientific Concepts and Principles question because it requires knowledge of muscle structure and control of contraction.

8. Why was it necessary for the researchers to determine the activity of the complexes independent of one another? A. Complex stability is lost if the complexes are able to interact structurally. B. The complexes have different cellular locations, and it is not feasible to isolate them together. C. The complexes all use the same substrates, so their use must be monitored separately. D. The reactions catalyzed by the complexes are coupled to one another.

Answer: D. The reactions catalyzed by the complexes are coupled to one another Reasoning: This is a Biochemistry question that falls under the content category "Principles of bioenergetics and fuel molecule metabolism." The answer to this question is D because studying the complexes all together would lead to erroneous results because inhibition of complexes I and II affects the activity of Complex III, which affects the activity of Complex IV. It is a Scientific Reasoning and Problem Solving question because you must use your knowledge of the electron transport chain to reason about the best way to study individual components of it.

50. Secretory lysosomes are classified as lysosomes because secretory lysosomes have some functional components in common with conventional lysosomes. Given this, secretory lysosomes most likely contain: A. ribosomes. B. Krebs cycle enzymes. C. RNA and DNA polymerases. D. degradative enzymes that function at low pH.

Answer: D. degradative enzymes that function at low pH Reasoning: This is a Biology question that falls under the content category "Assemblies of molecules, cells, and groups of cells within single cellular and multicellular organisms." The answer to this question is D because lysosomes are defined as membrane-bound organelles that contain hydrolytic enzymes activated by a low pH. These enzymes are capable of degrading many kinds of biomolecules. It is a Knowledge of Scientific Concepts and Principles question because you must recall the characteristics of lysosomes.

1. The information in the passage suggests that in mice CRY1 most likely affects XPA by: A. activating XPA protein activity. B. activating translation of XPA-encoding transcripts. C. repressing replication of the XPA-encoding gene. D. repressing transcription of the XPA-encoding gene.

Answer: D. repressing transcription of the XPA-encoding gene Reasoning: This is a Biology question that falls under the content category "Transmission of genetic information from the gene to the protein." The answer to this question is D because Figure 1 shows that XPA levels decrease as CRY1 levels increase. It is a Scientific Reasoning and Problem Solving question because the question involves determining which cellular process would lead to a reduction in XPA levels.