4/20 All section Questions, ****

​​​​​​​ . (Choice C) Mucus—a thick, sticky, water-based substance containing salts, lipids, cellular debris, and protein—is produced by specialized epithelial cells in many areas of the body, including throughout the respiratory tract. Specifically, mucus coats the inside of the nose, pharynx, larynx, and trachea as well as the bronchi leading to the lungs. This sticky mucus can trap foreign particles (eg, pathogens) to prevent them from reaching the lungs. . (Choice D) Stomach acid, which is secreted by parietal cells, contains hydrochloric acid that helps to break down food and also maintains the low pH environment in the stomach that is necessary for activating the digestive enzyme pepsin. In addition, many pathogens cannot survive in the presence of stomach acid.

Researchers hypothesize that the hormone adiponectin significantly increases the activity of sebaceous glands in humans. Which of the following observations in patients deficient in adiponectin would best support this hypothesis? . A. Decreased production of bile by the liver B. Decreased production of oils by the skin C. Decreased production of mucus by the trachea D. Decreased production of acid by the stomach . . . The skin contains several structures, including sweat (sudoriferous) glands and sebaceous glands, both of which are exocrine glands located in the dermis: . Sweat glands secrete sweat, a hypotonic solution (ie, a solution with a low solute concentration) that contributes to thermoregulation. In warm environments, the evaporation of water from sweat on the skin leads to heat loss and cooling of the body. In addition, antimicrobial substances secreted by sweat glands can contribute to the innate immune response. . Sebaceous glands secrete sebum, an oil-containing (hydrophobic) substance that helps prevent water loss from the skin and hair. Sebum also contains certain substances (eg, fatty acids) that inhibit the survival of pathogens, contributing to innate immunity. . Therefore, observing decreased production of oils in the skin of these patients would best support this hypothesis. . (Choice A) Bile, which is synthesized and released by liver cells, is a nonenzymatic solution composed of bile salts, bile pigments (eg, bilirubin), and cholesterol. Bile salts have both a hydrophobic region that that can associate with lipid surfaces and a hydrophilic region that associates with water. Consequently, bile salts can act as detergents, breaking down large lipid globules into smaller droplets (micelles) via emulsification.

Number of AAs in a single LPL subunit × 3 nucleotides per codon = number of nucleotides in mRNA sequence . 450 AAs in LPL subunit × 3 nucleotides = 1,350 nucleotides . Because the mRNA must be translated twice to produce the two subunits of the LPL homodimer, the total number of mRNA nucleotides translated by ribosomes to form the homodimer can then be calculated: . Number of nucleotides translated to form a single subunit × 2 = total number of nucleotides translated to form homodimer . 1,350 nucleotides × 2 = 2,700 total nucleotides . Therefore, ribosomes must translate approximately 2,700 total nucleotides to form one 100-kDa LPL homodimer.

A scientist studying the sequence of the LPL homodimer would most likely expect that: . " Generally, active LPL is a homodimer consisting of two individual subunits that are each approximately 450 residues in length and weigh 50-kDa." . A. ribosomes must translate a total of 1,350 mRNA nucleotides into amino acids to form one 100-kDa LPL homodimer. B. ribosomes must translate a total of 2,700 mRNA nucleotides into amino acids to form one 100-kDa LPL homodimer. C. ribosomes must translate a total of 1,350 mRNA nucleotides into amino acids to form one 50-kDa LPL homodimer. D. ribosomes must translate a total of 2,700 mRNA nucleotides into amino acids to form one 50-kDa LPL homodimer. . . A dimer is a protein generally composed of two noncovalently bound polypeptide chains (also called subunits, or monomers). A homodimer is a protein in which the polypeptide chains of the two monomers have the same sequence (number, order, and type) of amino acids (AAs). . According to the passage, active LPL is generally a homodimer consisting of two individual subunits, which each are approximately 450 amino acid residues in length and weigh 50-kDa. Accordingly, the LPL homodimer (consisting of two subunits) is made up of approximately 900 total amino acid residues and weighs around 100-kDa. . In this scenario, the scientist is calculating the number of mRNA nucleotides that ribosomes must translate into amino acids to form the 100-kDa LPL homodimer. Specifically, LPL mRNA must be translated twice to form each subunit. . First, the total number of mRNA nucleotides that must be translated by ribosomes to form one LPL subunit can be calculated as follows:

Hence, L is directly proportional to L1. Therefore, L must decrease because the torque due to the 25-kg child decreases. . (Choice B) Each child's weight is constant, but L must decrease as the 25-kg child moves to the right. . (Choice C) Torque depends on the distance from the fulcrum. Hence, the torque deceases as the child's distance from the fulcrum decreases. . (Choice D) As the 25-kg child moves to the right, the torque produced on the seesaw will decrease, not increase. . Educational objective:The net torque exerted on an object in static equilibrium is zero. If the torque on one side of a fulcrum decreases, the torque on the other side of the fulcrum must also decrease for the system to remain in equilibrium.

A seesaw is in equilibrium when a 25-kg child sits 3 m to the left of the fulcrum and a 38-kg child sits 2 m to the right of the fulcrum. If the 25-kg child moves to the right, how must the 38-kg child's distance L from the fulcrum change for the seesaw to remain in equilibrium? . A. L must decrease, because the torque produced by the 25-kg child decreases. B. L must not change, because each child's weight is constant. C. L must not change, because the torque produced by the 25-kg child is constant. D. L must increase, because the torque produced by the 25-kg increases.

​​​​​​​ (Choice B) Ex-convict is a status that is attained through one's actions, so it would be considered an achieved status, not an ascribed status. . (Choice C) Socioeconomic status defines one's standing in society based on income (wealth), education, and occupation. A status that dominates all of one's other statuses, like that of ex-convict, is best defined as a master status, not a socioeconomic status. . (Choice D) Generational status refers to whether individuals were born in the country in which they reside. For example, an individual who was born in the United States but whose parents were born in Mexico has a generational status of "first-generation American." Ex-convict is not considered a generational status.

According to the experts cited in the third paragraph, "ex-convict" is best described as what type of status? .​​​​​​ ​​​Experts suggest that, once convicted of a felony, individuals find that their status as a former felon (or "ex-convict") dominates all their other statuses, making it difficult to reintegrate into society." . A. Master B. Ascribed C. Socioeconomic D. Generational . . . Relative to others in society, individuals hold social status positions, which are ascribed or achieved. An ascribed status is one that is socially assigned, such as race. An achieved status is one that is attained, such as one's occupation (eg, doctor), notable achievements (eg, Olympic athlete), or failures (eg, college dropout). . Individuals may hold many ascribed and achieved statuses, but their master status is the one that dominates in social situations. A master status can be either ascribed or achieved, may be context-dependent, and tends to influence most aspects of one's social life.. . The experts cited in the third paragraph suggest that, once convicted of a felony, individuals find that their status as a former felon (or "ex-convict") dominates all their other statuses, which describes a master status.

Because the question specifies an enveloped virus, viral surface proteins must also be trafficked to the plasma membrane through the host endomembrane system. . Therefore, inhibiting lysosomal trafficking would have the least impact on the assembly of a new virus, because lysosomes function primarily in the degradation of macromolecules in the cell, not protein synthesis or trafficking to the cell membrane. Lysosomes do have a role in viral entry by endocytosis, but the virus in question has already entered the host cell. The other choices would affect the assembly of a new enveloped virus in an infected cell. . (Choice A) Viral mRNA is translated by host ribosomes in the cytosol to create viral proteins needed to assemble viruses. Therefore, inhibiting protein synthesis in the cytosol would limit viral assembly.

An epithelial cell in the upper respiratory tract was infected with an enveloped virus, such as influenza virus. Upon entering the host cell, the viral life cycle was initiated and the host cell machinery was used to synthesize new viral components. Given this information, which of the following statements is LEAST likely to be true? . A. Inhibition of protein synthesis in the cytosol would limit the assembly of new viruses. B. Disruption of trafficking between the Golgi apparatus and cell surface would result in the production of fewer viral glycoproteins. C. Inhibition of lysosomal trafficking would prevent new viruses from being released. D. Disruption of protein synthesis in the rough endoplasmic reticulum would inhibit the release of new viruses. . . Viruses are intracellular parasites that require host cell machinery to replicate. Once a virus enters a host cell, the viral genome replicates and viral proteins must be synthesized to assemble new viruses. The virus uses the host cell's protein synthesis and trafficking machinery to synthesize and correctly position viral proteins for assembly. . Viral assembly begins once all viral components are synthesized and properly located. Enveloped viruses acquire a plasma membrane-derived envelope containing viral proteins while exiting the host cell. These viral proteins are synthesized and trafficked to the plasma membrane using the host's endomembrane system (ie, endoplasmic reticulum [ER], Golgi apparatus, transport vesicles). . The question asks which component of host cell machinery is least involved with the assembly of an enveloped virus. The virus must use host machinery to synthesize viral proteins in the cytosol.

​​​​​​​ (Choices A, C, and D) A police officer who believes black people commit more crimes (prejudice) and who spends the majority of his time patrolling predominantly black neighborhoods (discrimination) is more likely to witness black people than white people committing crimes, which confirms his original belief (confirmation bias). . Educational objective:Prejudice is having a preconceived belief about individuals or groups based on their group membership. Discrimination is treating people unequally on the basis of group membership. Confirmation bias occurs when individuals accept evidence supporting their beliefs and dismiss evidence refuting their beliefs.

As described in the second paragraph, disparities in prison demographics may be explained by all of the following, EXCEPT: . A. prejudice. B. stereotype boost. C. discrimination. D. confirmation bias. . . ​​​​​​​ ​​​​​​​ Stereotype boost (also known as stereotype lift) occurs when positive stereotypes about social groups cause improved performance. For example, Asians reminded of the stereotype "Asians are good at math" before completing a math test tend to outperform Asians who are not first reminded of this stereotype. . This passage does not discuss improved performance as a result of a positive stereotype. . Prejudice refers to preconceived ideas and beliefs about people or groups based on their group membership. Discrimination is the unequal treatment of individuals or groups on the basis of group membership. . Confirmation bias is a type of cognitive bias (common error in thinking) in which individuals tend to embrace evidence supporting their beliefs, dismiss or ignore evidence refuting their beliefs, and interpret ambiguous evidence as support. Prejudice may cause an individual to act in a discriminatory fashion, creating the conditions for confirmation bias.

(Choice A) The nucleotide here contains an adenine base, which is complementary to thymine (not cytosine). In DNA, adenine always base pairs with thymine via two hydrogen bonds. In RNA, adenine base pairs with uracil (U). . (Choices C and D) These nucleotides contain a ribose sugar with a hydroxyl group (-OH) on the 2′ and 3′ carbon atoms. Therefore, these are ribonucleotides found in RNA (not deoxyribonucleotides found in DNA). . Deoxyribonucleotides are composed of a phosphate, a deoxyribose sugar, and a nitrogenous base (A, T, C, or G). Cytosine and thymine are pyrimidines (single-ringed bases) whereas guanine and adenine are purines (double-ringed bases). In double-stranded DNA, cytosine always aligns with guanine (and vice versa), and adenine always aligns with thymine (and vice versa) via complementary base pairing.

Based on Figure 1, substitution of cytosine for which of the following nucleotides would cause the T291-1 loss-of-function mutation? . T291-1 ( † ) (Note: * indicates mutation resulting in a truncated protein; † indicates substitution of a cytosine for its complementary base.) . Deoxyribonucleotides, the building blocks of DNA, are each composed of a phosphate, a deoxyribose sugar, and one of four nitrogenous bases (A, T, C, or G). Individual nucleotides can be identified by analyzing the structure of their attached base: . Cytosine (C) and thymine (T) are one-ringed bases known as pyrimidines. Thymine contains a methyl group (-CH3) attached to the carbon atom in the fifth position, a feature that helps distinguish it from cytosine. . Guanine (G) and adenine (A) are two-ringed bases known as purines. Guanine has a carbonyl group (-C=O) at the sixth position that helps differentiate it from adenine. . In double-stranded genetic material, complementary base pairing is the bonding of a particular nucleotide with only one other type of nucleotide. Such base pairing is determined by the alignment of the hydrogen bond donors and acceptors of each nucleotide's nitrogenous base. Cytosine and guanine always base pair via three hydrogen bonds. The caption for Figure 1 states that the T291-1 mutation in ANGPTL4 results after substituting a cytosine with its complementary base (guanine), which is depicted in Choice B.

(Number I) LDL cholesterol has a p-value of 0.06, which is not considered statistically significant. . (Number II) HDL cholesterol has a p-value of 0.087, which is not considered statistically significant. . Educational objective:The p-value is the probability of observing a result due to chance alone, assuming that the null hypothesis is true. A value of p ≤ 0.05 is generally considered statistically significant whereas p > 0.05 is not considered statistically significant.

Based on Figure 2, ANGPTL4 mutations were found to decrease CAD risk by significantly altering which of the following? . LDL cholesterol levels HDL cholesterol levels Triglyceride levels . A. I only B. III only C. I and II only D. II and III only . . The p-value is widely used to determine the significance of experimental results. A p-value is formally defined as the probability of observing a given result due to chance alone, assuming that the null hypothesis (status quo assumption that there is no actual difference between the experimental and control groups) is true. . The p-value ranges from 0 to 1 and is generally interpreted as follows: . p > 0.05 signifies a greater than 5% probability that the observed result is due to chance alone, not due to an actual association between the variables under study. Values in this range are generally not considered statistically significant. p ≤ 0.05 signifies a 5% or lower probability that the observed result is due to chance alone. Values in this range are typically considered statistically significant. . The plasma triglyceride level in mutation carriers is significantly lower than the plasma triglyceride level in mutation noncarriers (p = 0.023) (Number III).

Therefore, the reaction that was run for 0.75 h in acetone with 3.0 eq IBX gave the greatest yield for the oxidation of Compound 3. . (Choice A) Compared to 0.75 h in acetone, these conditions yield a smaller spot for Compound 4 as well as a spot for Compound 5, indicating some of the product was overoxidized. . (Choice D) These conditions yield only Compound 5, indicating that all the Compound 4 that formed was overoxidized. . The relative amounts of a product formed in different trials of the same reaction can be assessed by comparing the size of the spots on a thin-layer chromatography plate if the reactants all have the same initial concentration.

Based on the TLC plate shown in Figure 2, the best yield of Compound 4 results from which reaction conditions? . A. 48 h in DMF with 2.0 eq of IBX B. 0.75 h in acetone with 3.0 eq IBX C. 24 h in acetone with 1.5 eq IBX D. 72 h in ethyl acetate with 3.0 eq IBX . . The passage states that the reaction conditions for the oxidation of Compound 3 are optimized and monitored by thin-layer chromatography (TLC) to determine the best conditions for formation of Compound 4 (the product). The optimal conditions do not further oxidize Compound 4 into the side-product, Compound 5. . The TLC plate shows that the reactions in acetone are the only ones in which Compound 4 is not overoxidized. Because the reactions used the same initial concentrations of Compound 3, the relative amounts of Compound 4 formed and Compound 3 remaining can be qualitatively assessed by comparing the size of the corresponding spots. . For the reaction that was run for 0.75 h, the Compound 4 spot is larger and the Compound 3 spot is smaller compared to the corresponding spots for the reaction that was run for 24 h (Choice C).

Specific rotation [α] describes the direction (+ or −) and magnitude (degrees) by which a chiral molecule rotates plane-polarized light. The observed rotation αobs is dependent upon sample concentration (c in g/mL) and path length (l in dm) of the cell used; temperature T (°C) and wavelength λ (nm) affect a molecule's rotation and are reported with [α]. Specific rotation [α] is standardized to account for c and l: . [α]λ𝑇 = αobs/𝑐×𝑙

Based on the specific rotation stated in the passage, what would be the observed rotation of NNG3 if a 2.0 × 10−3 g/mL sample in a 0.50 dm cell were used to take the measurement? . ."After its synthesis, a 1 g/mL sample of NNG3 in a 1 dm sample cell was found to have a specific rotation of −350° at 20 °C using 589 nm light." . . A. −7.0° B. −1.4° C. −0.70° D. −0.35° .

(Choice B) Platelets are non-nucleated cellular fragments that adhere to injured blood vessel walls, forming a plug at the injury site. Platelets generally aggregate and adhere to injury sites but do not form the interior lining of veins. Therefore, red blood cells would not adhere to platelets in patients with CRVO. . (Choice C) Leukocytes (white blood cells) are cells of the immune system that protect the body from infectious agents and foreign antigens. These leukocytes can adhere to the endothelial lining and migrate between endothelial cells to exit the bloodstream and travel to affected tissue during infection or injury. However, because leukocytes do not make up the interior lining of the central retinal vein, erythrocytes do not likely adhere to leukocytes in patients with CRVO.

Central retinal vein occlusion (CRVO), a disorder that leads to blindness, is thought to be caused by erythrocytes attaching to the interior lining of the central retinal vein. Erythrocytes most likely adhere to which structures in patients with CRVO? . A. Smooth muscle cells B. Platelets C. Leukocytes D. Endothelial cells . . Endothelial cells are a type of specialized epithelial cell. They form the endothelium, or interior lining of the cardiovascular system (heart and blood vessels). The endothelium provides a smooth surface that reduces the friction of blood moving through the cardiovascular system. However, substances such as erythrocytes may become attached to the endothelium, disrupting the normal flow of blood and causing vascular occlusion (ie, blockage). . In this question, central retinal vein occlusion (CRVO) is described as a disorder that leads to blindness. It is thought to be caused by erythrocytes (red blood cells) attaching to the interior lining of the central retinal vein. The interiors of all cardiovascular components (heart, blood vessels) are lined with endothelial cells. Therefore, in a patient with CRVO, erythrocytes most likely adhere to the endothelial cells lining the central retinal vein, causing vein occlusion that leads to blindness. . (Choice A) The endothelium of arteries and veins is surrounded by a layer of smooth muscle cells. These cells can contract, causing vasoconstriction and increased blood pressure, or they can relax, causing vasodilation and decreased blood pressure. Because smooth muscle cells surround the endothelium, they do not make up the interior lining of the central retinal vein.

Because antibodies mark cellular materials for destruction by cells of the immune system, production of antibodies against intestinal proteins would likely cause the destruction of these proteins, damaging the intestinal tract and decreasing nutrient absorption (Choice C). . Intestinal nutrient transporters facilitate nutrient absorption from the intestinal tract into the body. Therefore, increased nutrient transporter activity would increase (not decrease) nutrient absorption. . Nutrient absorption in the gastrointestinal tract is affected by the diversity of intestinal bacteria (gut flora), the surface area of the small intestine, and the functions of intestinal proteins (digestive enzymes, nutrient transporters, structural proteins).

During a nutrition study, a participant consumes five times the physician-recommended amount of food but is unable to sufficiently absorb nutrients. Given this information, this participant would be LEAST likely to exhibit which of the following conditions? . A. Loss in diversity of bacteria that produce fatty acids in the large intestine B. Decreased surface area in the small intestine C. Heightened production of serum antibodies against intestinal proteins D. Increased activity of nutrient transporter proteins in the small intestine . . . The presence and diversity of fatty acid-producing bacteria in the large intestine. These bacteria metabolize undigested carbohydrates from the small intestine into short-chain fatty acids, which are absorbed and used as additional energy sources. A decrease in these bacterial species would lead to decreased absorption of additional fatty acids as nutrients (Choice A). . The small intestine's surface area. Large folds in the intestine's epithelial lining slow the movement of chyme through the intestinal tract. In addition to these large folds, villi (fingerlike projections extending from the lining) and microvilli (smaller fingerlike extensions of individual absorptive cells) maximize the time and surface area available for nutrient absorption. Malabsorption is sometimes caused by structural damage (eg, villi destruction) that decreases small intestinal surface area, which thereby impairs nutrient absorption (Choice B). . The functions of intestinal proteins. Structural proteins, digestive enzymes, and nutrient transporters work synergistically to break down and absorb nutrients within the gastrointestinal tract.

(Choice A) Figure 2 shows that cell senescence is increased (not decreased) by the presence of cGAS. . (Choice C) Figure 1 (not Figure 2) provides data on STING. Figure 2 demonstrates senescence data only in cells lacking cGAS, not STING. . (Choice D) Cell senescence is significantly lower (not higher) at cell culture cycle 6 than at cycle 11 in WT cells, according to Figure 2. . Educational objective:Senescence is an aging process that occurs when cell cycle progression no longer occurs.

From the data in Figure 2, researchers can conclude that cell senescence is: . "The role of cGAS in MEF cell senescence was tested by staining cGas-/- and WT cells with SA-β-Gal, a commonly used method to detect senescent cells. The percentage of SA-β-Gal positive cells during 20 cycles of cell culture was measured in both cell types (Figure 2)." . A. decreased by the presence of cGAS. B. increased by the presence of cGAS. C. increased at least 2 times by the absence of STING. D. significantly higher at culture cycle 6 than at cycle 11 in WT cells. . . Senescence is a natural aging process during which cells and tissues acquire damage (eg, mutations), causing diminished function over time. In cells, senescence occurs when cell cycle progression no longer occurs. . In Experiment 2, researchers explored the role of cGAS in senescence phenotypes in MEFs during their spontaneous immortalization (extended proliferation). Cells lacking cGAS (cGas-/-) and wild-type (WT) cells with SA-β-Gal were stained, a commonly used method to detect whether cells are senescent. Figure 2 shows that when SA-β-Gal positive cells are detected, there is a significantly higher percentage of these cells in WT than in cGas-/- cells. . These results suggest that when cGAS is not present, there are fewer senescent cells. Likewise, when cGAS is present (ie, in WT cells), there are more senescent cells. Therefore, researchers can conclude that cell senescence is increased by the presence of cGAS.

In this question, the temperature of the liquid equals its exact boiling point. As a result, adding a small amount of heat will cause some of the liquid to transition into gas, but the temperature will not change. . (Choices A and C) The temperature will remain constant until the heat of vaporization is exceeded and all the liquid has turned to gas. . (Choice B) A small amount of heat will convert a small amount of the liquid to gas, not all the liquid. . Educational objective:During phase transitions, the temperature of a substance remains constant. A liquid at the exact temperature of its boiling point is transitioning from a liquid to a gas. It must gain an amount of heat equal to its heat of vaporization before all the liquid turns to gas and the temperature increases.

If a liquid is at its exact boiling point, what will happen to the liquid when a small amount of heat is added to it? . A. None of the liquid will turn to gas, and the temperature will increase. B. All the liquid will turn to gas, and the temperature will remain the same. C. A small amount of the liquid will turn to gas, and the temperature will increase. D. A small amount of the liquid will turn to gas, and the temperature will remain the same.

(Choices A and B) Osteoblasts are bone-forming cells that secrete the bone matrix to which calcium phosphate crystals bind, and osteoclasts are cells that resorb (ie, break down) bone. Both osteoblasts and osteoclasts are active in bone remodeling throughout life. . (Choice C) Bone marrow cells are responsible for hematopoiesis (ie, blood cell production) in adults and remain active throughout life. Although bone marrow cells in long bones become less active with age, a residual amount of bone marrow cell activity is typically preserved into adulthood. . Long bone growth occurs when chondrocytes in the growth plate divide and produce collagen, to which calcium phosphate crystals attach to form hardened bone. When chondrocytes in bone stop dividing, bone growth is complete.

In adults, which of the following cell types would be LEAST active in healthy long bones? . A. Osteoblasts B. Osteoclasts C. Bone marrow cells D. Chondrocytes . . Long bone grows in length by a process where chondrocytes (cartilage-producing cells) divide and produce collagen to which calcium phosphate attaches to form hardened bone. The chondrocytes are located at the interface between the long shaft (diaphysis) and the widened ends (epiphyses) of long bones, and this cartilaginous region is known as the epiphyseal (growth) plate. The cell division and collagen added to this region force the epiphyses and diaphysis farther apart, thereby increasing the length of the bone. . The question asks for the cell type that would be least active in healthy long bones of adults. The activity of the chondrocytes producing the cartilage of the growth plate stops when linear growth of long bones is complete (except during fracture healing). Therefore, of the cell types listed chondrocytes would be the least active in healthy long bones of adults.

To improve the accuracy of the experimental results, the researchers could include gene expression datasets relating to other types of cancers. This would allow analysis of the role of cGAS in different types of cancer cells, rather than only lung adenocarcinoma cancer cells. Having this comparison data may reveal previously unrecognized patterns in cGAS expression in cancer cells, or it may provide evidence that there is no correlation between cGAS expression and survival across multiple cancer types. . The other choices would not increase the accuracy of the experimental results.

In an additional experiment, researchers seek to determine whether cGAS expression in human cancer patients is correlated with decreased patient survival. As one step in this experiment, the researchers analyzed gene expression data from a publicly available dataset relating to adult lung adenocarcinoma patients. Which change to this protocol would increase the accuracy of results? . A. Limit the portion of the dataset analyzed to individuals of one gender. B. Include survival data pertaining to children with lung adenocarcinoma. C. Include gene expression datasets relating to other types of cancers. D. Limit the portion of the dataset analyzed to individuals living in the same location. . . Tumor suppressor genes regulate DNA repair by pausing or repressing the cell cycle. This regulation helps ensure that only normal cells finish cell division. Programmed cell death (apoptosis) is induced in cells that cannot be repaired. In cancer cells, tumor suppressor genes are inhibited, allowing uncontrolled progression of the cell cycle and tumor formation and growth. . In this question, researchers seek to determine whether cGAS expression in human cancer patients is correlated with decreased patient survival. These researchers analyzed gene expression data from a dataset relating to adult lung adenocarcinoma patients to assist in elucidating the role of cGAS expression in cancer patient survival.

(Choice A) The sp2 hybridized carbons in an alkene can be stereocenters if the alkene can be classified as either cis/trans or E/Z. Carbon 1 is a stereocenter as it is part of a Z alkene, but its configuration cannot be assigned as R or S. . (Choices B and D) Carbon 4 is a chiral carbon because it has four different substituents. The substituents with priority 1, 2, and 3 are in a counterclockwise arrangement with the lowest-priority group pointed behind the plane; therefore, carbon 4 has an S configuration.

NNG1 is shown below with certain carbon atoms labeled. Which of the following carbon atoms are stereocenters with an R configuration? . A. Carbon 1 only B. Carbon 4 only C. Carbons 2 and 3 only D. Carbons 2, 3, and 4 only . . . Carbons 2 and 3 are chiral centers because each has four different substituents. The substituents with priority 1, 2, and 3 on carbons 2 and 3 are arranged in a counterclockwise configuration, and the lowest-priority group (H) is pointed in front of the plane (wedged) on carbons 2 and 3. Therefore, carbons 2 and 3 have only an R configuration. . To determine the R/S configuration of a chiral carbon, the priority of the substituents and their circular arrangement must be examined. When the lowest-priority group is pointed behind the plane (dashed), a clockwise arrangement of priorities 1, 2, and 3 is R whereas a counterclockwise arrangement is S. When the lowest-priority group is pointed in front of the plane (wedged), a clockwise arrangement is S whereas a counterclockwise arrangement is R.

(Choice B) Cell cycle checkpoints are critical for the health of an organism. Therefore, inactivation of cGAS to prevent cell cycle checkpoints from taking place would not be beneficial and could lead to cancer. . (Choice C) G0 is a state of cell cycle arrest, in which a cell is not actively preparing to divide but is performing other essential functions. During the G0 phase, the nuclear envelope is intact; therefore, cytosolic cGAS would not encounter nuclear DNA. . (Choice D) Gamete formation is the result of meiosis, not mitosis.

Researchers have noted that cGAS associates with a cyclin-dependent kinase during mitosis, causing cGAS to temporarily become inactive. The best explanation for this temporary inactivation would be: . A. to prevent cGAS from sensing nuclear DNA. B. to prevent cell cycle checkpoints from taking place. C. to prevent cGAS from associating with nuclear DNA during the G0 phase. D. to prevent nondisjunction during gamete formation. . . . In this question, researchers have noted that cGAS associates with a cyclin-dependent kinase during mitosis, causing cGAS to temporarily become inactive. The passage states that cGAS is a cytosolic DNA sensor. This means that when the cell is performing its normal duties, cGAS would not sense nuclear DNA. . However, during mitosis, the nuclear envelope disintegrates, exposing nuclear DNA that is not normally encountered by cGAS. Therefore, the association of cGAS with a cyclin-dependent kinase and the subsequent temporary inactivation of cGAS likely prevents cGAS from sensing nuclear DNA during mitosis.

Therefore, substituting the equilibrium partial pressures into the Kp expression gives: . Kp=x/(Pi,W−x)(Pi,X2−x) . (Choice A) The equilibrium constant is defined as the ratio of products to reactants, not the ratio of reactants to products. . (Choice B) The reaction proceeds toward products, so W(g) and X2(g) must decrease (−x) and WX2(g) must increase (+x). . (Choice C) This expression incorrectly uses the initial partial pressures rather than the equilibrium partial pressures. Additionally, the ratio is inverted (ie, reactants to products instead of products to reactants).

Suppose W(g) has an initial partial pressure of Pi,WPi,W and X2(g) has an initial partial pressure of Pi,X2Pi,X2 near the walls of the bulb. Which of the following expressions is equivalent to Kp for Reaction 1 if the equilibrium partial pressure of WX2(g) is represented by x? . W(g) + X2(g) ⇄ WX2(g) Reaction 1 . A. Kp = (Pi,W−x)(Pi,X2−x) /x B. Kp= −x / (Pi,W+x)(Pi,X2+x) C. Kp= (Pi,W)(Pi,X2) / x D. Kp= x / (Pi,W−x)(Pi,X2−x)

G≡C base pairs form more hydrogen bonds and have stronger stacking interactions than A=T base pairs. The higher the GC content, the greater the melting temperature (Tm) and stability of double-stranded DNA.

The GC content is the percentage of guanine and cytosine bases in DNA, and the higher the GC content, the greater the stability and melting temperature (Tm) of double-stranded DNA. DNA double helix stability is primarily maintained by stacking interactions: the interactions between neighboring G≡C base pairs are stronger than those between adjacent A=T base pairs. It also takes more energy (heat) to break the three hydrogen bonds in G≡C base pairs than the two hydrogen bonds in A=T base pairs. As a result, the Tm for GC-rich DNA molecules is higher as additional energy is required to separate the more stable helix. . Figure 1 shows that SW-480 cells have a G→T base substitution at codon 12 of the K-ras gene, depicting a loss in GC content. Assuming there are additional similar mutations in the mutant K-ras gene, the mutant DNA will have lower GC content than the WT DNA and therefore be less stable and denature faster (ie, exhibit lower Tm).

According to the passage, as the tungsten filament in the bulb is heated (Tfilament ≈ 3300 K), a temperature gradient forms between the hot filament and the cooler glass bulb wall (Tbulb wall ≈ 1000 K). . Near the bulb wall, W(g) reacts with X2(g) to form WX2(g) (Reaction 1). When WX2(g) diffuses toward the hot filament, it decomposes back into W(g) and X2(g). . Therefore, the equilibrium favors the formation of WX2(g) at lower temperatures and favors the formation of W(g) and X2(g) at higher temperatures, indicating that the reaction is exothermic and ∆H is negative.

The enthalpy of reaction ∆H for Reaction 1 is best described by which of the following? . In a typical incandescent bulb, a solid tungsten (W) filament, surrounded by a sealed glass bulb filled with an inert gas, is heated to a high temperature (Tfilament ≈ 3300 K) via an electric curren. " W(g) + X2(g) ⇄ WX2(g) Reaction 1 As W(g) is formed, it diffuses toward the cooler bulb wall (Tbulb wall ≈ 1000 K). . A. ∆H is positive because the formation of WX2(g) is favored with decreasing temperature. B. ∆H is positive because the formation of WX2(g) is favored with increasing temperature. C. ∆H is negative because the formation of WX2(g) is favored with decreasing temperature. D. ∆H is negative because the formation of WX2(g) is favored with increasing temperature. . . . The amount of heat released or absorbed by a chemical reaction measured under constant pressure is equal to the change in enthalpy of the reaction (∆H). Exothermic reactions release heat (ie, heat is a product), resulting in a negative ∆H, whereas endothermic reactions absorb heat (ie, heat is a reactant), resulting in a positive ∆H. For an exothermic reaction at equilibrium, a temperature decrease causes the reaction to shift toward products to compensate for the heat loss in accordance with Le Châtelier's principle. Conversely, for an endothermic reaction at equilibrium, a temperature decrease causes the reaction to shift toward reactants.

(Choice A) −245 kJ/mol is the sum of the cumulative free energy changes after each of the reactions of the energy investment phase of glycolysis. Summing would be the correct approach if the graph depicted the individual free energy changes associated with each reaction. However, the graph depicts the cumulative net change in free energy relative to glucose, so each bar already incorporates the individual free energy changes associated with all previous reactions. . (Choice B) −95 kJ/mol is the approximate free energy change after the payoff phase of glycolysis (ie, after the tenth reaction). . (Choice D) −39 kJ/mol is the approximate free energy change occurring during the payoff phase alone (ie, the free energy of the fifth reaction products subtracted from the free energy of the tenth reaction products).

The graph above shows the free energy of the products of each step of glycolysis relative to the free energy of glucose under typical cellular conditions (free energy of glucose is set to zero). Based on the data shown, the total free energy change during the energy investment reactions of glycolysis, in which two molecules of glyceraldehyde-3-phosphate are formed, is closest to which of the following? . A. −245 kJ/mol B. −95 kJ/mol C. −57 kJ/mol D. −39 kJ/mol . . Glycolysis is the series of ten cytosolic reactions by which a glucose molecule is metabolized into two pyruvate molecules, generating ATP and the reduced electron carrier NADH in the process. Glycolysis consists of successive energy investment (ie, the first five reactions) and energy payoff (ie, the last five reactions) phases during which two ATP molecules are invested and four ATP molecules are synthesized, respectively. . Each reaction of glycolysis is accompanied by a free energy change, which determines whether the reaction is spontaneous or not. Reactions with small free energy changes are considered reversible whereas those with larger free energy changes are considered irreversible (ie, a considerable amount of energy would need to be added to drive the reaction in the opposite direction). . The graph in the question depicts the free energy changes that occur during the reactions of glycolysis, depicted as the net free energy change relative to the original free energy of glucose. The question asks for the cumulative free energy change during the energy investment reactions of glycolysis, in which glyceraldehyde-3-phosphate is formed (ie, the first five reactions). Correct reading of the graph indicates that the cumulative free energy change after the fifth reaction of glycolysis is about −57 kJ/mol.

(Choices A, B, and D) A knockout model suggests that ANGPTL4 gene expression and consequently its mRNA are absent; this condition eliminates Profile 1 because loss of ANGPTL4 expression leads to increased (not decreased) LPL activity. In the constitutively active mouse model, ANGPTL4 gene mRNA levels are increased due to constant ANGPTL4 expression; this leads to decreased (not increased) activity of LPL, eliminating Profile 2 as an option. . Educational objective:A constitutively active gene is transcribed at a relatively constant rate regardless of current cell conditions. In a knockout model, an existing gene is replaced or disrupted, which leads to an absence of the protein product.

The knockout mouse model and the constitutively active mouse model of ANGPTL4 are best represented by: . " Specifically, angiopoietin-like 4 inhibits the activity of the enzyme lipoprotein lipase (LPL). . A. Profile 1 and Profile 2, respectively. B. Profile 1 and Profile 4, respectively. C. Profile 3 and Profile 4, respectively. D. Profile 3 and Profile 2, respectively. . Per the passage, ANGPTL4 codes for angiopoietin-like 4, a protein that inhibits the action of LPL. The passage also states that LPL catalyzes the degradation of triglycerides (lipid molecules) in the blood. . Consequently, expression of the wild-type ANGPTL4 gene, whose protein product inhibits LPL activity, leads to higher levels of triglycerides in the blood and therefore increases the risk for CAD. . Based on Table 1, Profile 3 represents the knockout mouse model in which the gene of interest (ANGPTL4) has been inactivated. This results in absentANGPTL4 mRNA and angiopoietin-like 4 protein activity. Consequently, LPL activity is increased due to the lack of inhibition, which causes decreased levels of blood triglycerides (as these are degraded by LPL). . Profile 4 represents the constitutively active mouse model in which ANGPTL4 is transcribed at a relatively constant rate regardless of current cell conditions. This results in increasedANGPTL4 mRNA and angiopoietin-like 4 protein activity. In turn, the angiopoietin-like 4 protein inhibits LPL activity, which leads to increased triglycerides in the blood.

The structure in Choice B correctly shows the inversion of stereochemistry at the electrophilic carbon and the retention of stereochemistry of the leaving group. . (Choice A) This structure does not show the inversion of stereochemistry at the electrophilic carbon. . (Choice C) This structure incorrectly shows the inversion of stereochemistry at both epoxide carbons. Only the carbon attacked by the nucleophile undergoes inversion. . (Choice D) This structure shows the outcome of an SN2 reaction if the nucleophile attacked the other carbon (not marked by an asterisk) in the epoxide. . SN2 reactions are concerted substitution reactions in which a nucleophile attacks an electrophile and the bond between the electrophile and the leaving group simultaneously breaks. The nucleophile does a backside attack, and if the electrophilic carbon is chiral, the stereochemistry of the chiral carbon is inverted

The structure of the epoxide formed from Compound 2 is shown below. . Based on the reaction described in the passage, what is the structure of the product when hydroxide reacts with the carbon marked by an asterisk, followed by protonation using HCl? . . . An SN2 reaction is a concerted substitution reaction. A nucleophile attacks an electrophile on the opposite side of the leaving group (ie, the nucleophile does a backside attack), and if the electrophilic center is a chiral center, the stereochemistry of the chiral center is inverted. . The passage states that the epoxide shown in this question undergoes an SN2 reaction. In this reaction, hydroxide (-OH) is the nucleophile that attacks the carbon of the epoxide marked with the asterisk (ie, the electrophile). At the same time, the bond between the electrophile and the leaving group (ie, the oxygen of the epoxide) breaks, causing the epoxide to open and form an alkoxide. . Because the leaving group is pointed in front of the plane (wedged), the nucleophile approaches the electrophile from behind the plane (ie, opposite side of leaving group) and forms a bond with the electrophile that points behind the plane (dashed). Therefore, the electrophilic carbon undergoes inversion of stereochemistry (S to R, in this case). The carbon atom bonded to the alkoxide retains the stereochemistry of the epoxide (wedged) because it was not attacked by the nucleophile. . After nucleophilic attack, the alkoxide is protonated by an acid.

(Choice B) The voltage across the battery is not less than the voltage across the resistors, and it is equal to the sum of the voltage drop across each resistor. . (Choice C) R1 is larger than R2. Therefore, IR1 is greater than IR2. . (Choice D) All three resistance values are not equal, so the voltage across each resistor will not be equal, and the battery will have the largest voltage. . Educational objective:The sum of all voltages in any closed loop within a circuit is zero. Voltage is the product of resistance and current.

Three resistors are connected in series with a battery with voltage V. If R1 > R2 > R3, which of the following expressions correctly shows the relationship between the battery voltage and the voltages across the resistors? . A. V=IR1+IR2+IR3 B. IR1>IR2>IR3>V C. IR3<IR1<IR2<V D. V=IR1=IR2=IR3 . . . Moreover, Ohm's Law states that voltage is equal to the product of resistance R and current: . ΔV=IR . In this question, the current is the same through all resistors because they are in series. Also, the voltage across the battery V is positive and potential differences across resistors are negative. Therefore, applying Kirchhoff's loop rule and Ohm's law: . ∆Vi=V-IR1-IR2-IR3=0 . Therefore, the potential difference across the battery is related to the potential differences across the resistors as: . V=IR1+IR2+IR3

(Choice A) A cell undergoing mitosis displays two centrosomes at each phase. Consequently, the presence of two centrosomes would not in itself signify that a cell is just prior to metaphase. . (Choice B) After sister chromatids split in anaphase and prior to cytokinesis, the chromosome number is double the diploid number for replicating cells. However, in cells just prior to metaphase, the chromosomes have replicated (ie, during the S phase of interphase) but have not yet split; therefore, the chromosome number is still equal to the diploid number for these cells. . (Choice D) The cleavage furrow forms in cells at the beginning of cytokinesis, which typically overlaps partially with telophase. This occurs after the cells have gone through metaphase.

To further elucidate the role of IRF3, researchers developed a protocol to synchronize replicating cells just prior to metaphase, allowing direct comparison of cellular IRF3 levels during mitosis. Which characteristic best describes the cells used in the experiment? Each cell displays: . A. two centrosomes. B. a chromosome number that is double the diploid number for these cells. C. migration of chromosomes toward alignment at the equator of the dividing cells. D. the beginning of cleavage furrow formation . . . . . The passage states that activated cGAS produces the second messenger cGAMP, which activates STING. This leads to inflammation and cytokine release via recruitment of IRF3. In this question, researchers developed a protocol to synchronize replicating cells just prior to metaphase to further elucidate the role of IRF3. . This allows the researchers to directly compare cellular IRF3 levels during mitosis. Because the experimental cells were synchronized just prior to metaphase, they would be characterized by migration of chromosomes toward alignment at the equator (ie, metaphase plate) of the dividing cells.

Correct Answer is B. .​​​​​​​ Translational kinetic energy is the energy of motion whereas gravitational potential energy describes the energy of position. Due to the conservation of energy, in the absence of friction, the total energy of an object along a path remains constant, but contributions from potential energy may convert to kinetic energy (or vice versa).

Two identical balls of equal mass m1 and m2 are placed at rest at the top of separate hills. . How do the velocities v1 and v2 of the balls compare, measured after each has rolled down to the bottom of its respective hill? (Note: Assume the absence of friction and negligible rotational kinetic energy.) . .A. v1= 1/2⋅v2 .B. v1= (√2 /2)⋅v2 .C. v1= 2⋅v2 .D. v1= 4⋅v2

According to the passage, the K-ras gene normally encodes a protein responsible for activating the signaling pathways that promote DNA replication (cell cycle S-phase) along with cell growth and survival. Mutated K-ras in SW-480 cells is most likely constantly inducing signaling pathways responsible for the characteristic uncontrolled growth, division, and survival of cancer cells. Therefore, K-ras is most likely a proto-oncogene with an activating mutation in at least one allele that turned it into an oncogene. . Oncogenes are mutated forms of genes involved in cell cycle progression (proto-oncogenes) that transform normal cells into cancer cells. Tumor suppressor genes regulate, pause, and inhibit cycle cell progression to ensure damaged DNA is repaired. In cancerous cells, mutations inactivate tumor suppressor genes and activate oncogenes.

Which of the following is most likely to be identified in SW-480 cells? . "The DNA of SW-480 cells, a line originating from a colon adenocarcinoma, is the standard positive control used to identify codon 12 mutations in K-ras testing." . A. An activating mutation in at least one K-ras allele B. Inactivating mutations in both K-ras alleles C. Decreased expression of K-ras protein D. Increased expression of hypoactive K-ras protein . . Cancer cells are able to grow and divide excessively due to loss of cell cycle control. Various types of cancers result from oncogene activation, tumor suppressor gene inactivation, or both. . Proto-oncogenes are wild-type (WT) genes that are normally involved in cell cycle progression (eg, cell division stimulation, apoptosis prevention). However, when proto-oncogenes are mutated into oncogenes via activating mutations in one or two alleles, cells become cancerous. As a result, oncogenes lead to either an overexpressed protein, a hyperactive protein, or both, promoting uncontrolled cell proliferation. . Tumor suppressor genes (or antioncogenes) regulate DNA repair by repressing or pausing the cell cycle to ensure that only normal cells proceed to the division (mitosis) stage. These genes also prevent accumulation of mutations in cancer cells by either repairing the mutations or inducing programmed cell death (apoptosis) if repair fails. In the setting of cancer, tumor suppressor genes become inactivated by loss of function mutations in both alleles and lose the ability to prevent abnormal growth and division of damaged cells.

(Choice A) Regions I and II both involve hydrogen bonding. Neither has a charged species that can participate in an electrostatic interaction necessary to form a salt bridge. . (Choice C) Region II does have hydrogen bonding, but region IV contains only London dispersion forces between nonpolar amino acid side chains (ie, no ionic bonding). . (Choice D) Region III contains an electrostatic ion-ion interaction (ie, a salt bridge), and region IV involves only London dispersion interactions. . Educational objective:Hydrogen bonding is a special type of dipole-dipole interaction in which at least one of the dipoles involves a hydrogen atom covalently bonded to one of the highly electronegative elements oxygen, nitrogen, or fluorine. Salt bridges form from strong electrostatic attractions (eg, ion-ion interactions) between two oppositely charged side chains in a protein.

Which of the following regions from the protein shown in Figure 2 depict noncovalent hydrogen bonding and a salt bridge, respectively? . A. I and II B. II and III C. II and IV D. III and IV . . Hydrogen bonding is a special type of dipole-dipole interaction in which at least one of the dipoles involves a hydrogen atom that is covalently bonded to an atom of oxygen, nitrogen, or fluorine. . In the protein shown in Figure 2, regions I and II both depict attractions associated with hydrogen bonding. Region I has hydrogen bonding between the N−H bonds (ie, the hydrogen bond donors) and the C=O bonds (ie, the hydrogen bond acceptors) from the peptide backbone. . Region II has a hydrogen bond between the oxygen atom of a C=O bond in the peptide chain and the O−H bond in the tyrosine side chain. . When two oppositely charged side chains are close enough to each other, they experience a strong, mutual electrostatic attraction and form an ion-ion interaction known as a salt bridge. . The only depiction of a salt bridge in Figure 2 is in region III, where the side chain of lysine (−CH2CH2CH2CH2NH3+) interacts with the side chain of glutamic acid (−CH2CH2COO−). . Therefore, regions II and III depict noncovalent hydrogen bonding and a salt bridge, respectively.

Stereoisomers stereocenters chiral centers Diastereomers Epimers . (Choice A) This structure differs in orientation from NNG3 at all the stereocenters and is the enantiomer of NNG3. . (Choice B) This is the structure of NNG1; it contains the same molecular formula as NNG3 but different atom connectivity (ie, the CH3 group on the ring is bonded to a different carbon atom). Therefore, this structure is a constitutional isomer (or structural isomer) of NNG3 rather than an epimer. . (Choice D) This structure differs in orientation from NNG3 at two stereocenters rather than one stereocenter. It is a diastereomer of NNG3 but not an epimer. . Educational objective:Stereoisomers are compounds with the same molecular formula and atom connectivity but with bonds oriented differently in space. Epimers are a type of diastereomer that differ in spatial orientation at only one stereocenter.

Which of the following structures is an epimer of NNG3? . . . _______________________ are compounds with the same molecular formula and atom connectivity but with bonds oriented differently in space. . Stereoisomers have one or more _________________, also known as ____________________, that consist of an atom bonded to four different substituents. . ___________________ are a type of stereoisomer where at least one (but not all) of the stereocenters differ in orientation. . _________________ are a type of diastereomer that differ in spatial orientation at only one stereocenter. . NNG3 has three stereocenters. Therefore, an epimer of NNG3 must have the same atom connectivity and number of stereocenters but differ in orientation at only one of the stereocenters. Choice C shows the only structure that fits these characteristics.

(Choices B and D) A surge in LH and follicular rupture from the ovary are necessary steps for ovulation to occur. LH provides the necessary stimulus for follicular rupture and the release of a secondary oocyte into the fallopian tube. Because these steps must happen before fertilization can occur, they would not be the cause of embryonic attachment outside the uterus. . (Choice C) Gastrulation is a post-implantation process by which a developing embryo progresses from the blastula stage to the tri-layered gastrula stage. Because this process occurs post-implantation, abnormal gastrulation would not cause implantation outside the uterus. . After fertilization, fallopian cilia help propel the fertilized oocyte toward the uterus for implantation. An inadequate number of cilia in the fallopian tube can cause implantation of the fertilized egg outside the uterus.

Which of the following would most likely cause an embryo to implant in a location other than the uterine lining? . A. Reduced number of fallopian cilia B. Surge in luteinizing hormone prior to ovulation C. Incomplete gastrulation D. Rupture of follicle from the ovary . Oogenesis is the process by which females produce sex cells (gametes) called eggs. All of a woman's developing eggs (oocytes) are produced during fetal development. At birth, their maturation is arrested in prophase I; these arrested eggs are known as primary oocytes. At puberty, the menstrual cycle begins, and each month a single primary oocyte develops into a secondary oocyte by continuing meiosis up to metaphase II. . During the follicular phase of the cycle, the follicle (oocyte plus supporting granulosa cells) matures until it bulges at the outer wall of the ovary. Maturation and growth of the follicle is driven by the release of follicle-stimulating hormone (FSH) from the anterior pituitary. A subsequent release of luteinizing hormone (LH) stimulates follicle rupture and the release of the secondary oocyte near the opening of the fallopian tube (ovulation). . Fertilization (joining of male and female gametes into a zygote) becomes possible after the released secondary oocyte has been drawn into the fallopian tube. It is normal for fertilization to occur in the fallopian tube prior to implantation within the uterus. Typically, small hair-like structures within the fallopian tube called fallopian cilia help propel the fertilized oocyte toward the uterus for implantation. However, with a reduced number of fallopian cilia, improper implantation of the fertilized egg can occur outside the uterus.

The following parameters influence the thermodynamic stability of the DNA duplex: . DNA length: Longer DNA molecules take more time to both melt and reanneal. . pH: Extreme changes in pH outside the physiological range lead to loss of hydrogen bonding and destabilize the DNA helix. . Salt concentration (ionic strength): High salt concentration of the solution increases double helix stability, but decreased salt concentration decreases stability.

Which sequence from Table 1 would have the fastest reannealing time in this experiment? . . . Melted DNA strands will reanneal to form the double helix once temperature decreases, but the time it takes for complementary strands to reanneal depends on the following: DNA length: Longer DNA molecules have more hydrogen bonds and will take more time to both melt and reanneal. pH: The physiological pH range (7.3-7.4) allows maximal hydrogen bonding between nitrogenous bases of DNA. . At low pH, hydrogen bond acceptor atoms in the bases become protonated, and these protonated acceptors cannot form hydrogen bonds, causing double helix separation. In contrast, high pH causes deprotonation of hydrogen bond donors in the bases; the loss of protons results in the loss of hydrogen bonds and a destabilized double helix. . Salt concentration (ionic strength): The electrostatic repulsion between negatively charged phosphate groups on the sugar-phosphate backbone destabilizes the double helix. However, this repulsion is neutralized and shielded by the binding of positively charged species in solution (eg, Na+ and Mg2+ cations). High salt concentration of the solution increases double helix stability, but low salt concentration decreases stability. . The four short tandem repeats sequences shown in Table 1 were denatured during melting curve analysis and cooled in environments with different pH and NaCl concentrations. Sequence 3 will have the fastest reannealing time as it is the shortest sequence and was allowed to cool in the solution closest to physiological pH with a high salt concentration (>0.15 M).

(Choice A) Steroid hormones are produced primarily in the adrenal cortex and the gonads from cholesterol and other precursors. They are not required in the diet and do not generally accumulate in adipose tissue. . (Choice C) Peptide hormones are readily synthesized by the body, and are not required in the diet. They are generally soluble in water, and are not stored in adipose tissue. . Educational objective:Certain biomolecules such as vitamins and the essential amino and fatty acids cannot be synthesized in the human body and must be obtained through the diet. Water-soluble vitamins (B series, C) are excreted in the urine whereas fat-soluble vitamins (A, D, E, and K) are stored in adipose and other fatty tissues.

Which type of molecule, primarily obtained from the diet, would most likely accumulate in adipose tissue? . A. Steroid hormones B. Water-soluble vitamins C. Peptide hormones D. Fat-soluble vitamins . . . The human body is capable of using precursor molecules to synthesize many of the biomolecules needed for proper function. However, humans lack the enzymes necessary to synthesize certain molecules such as the essential amino and fatty acids, and the class of molecules known as vitamins. These molecules must be obtained from the diet. . Vitamins are classified as either water-soluble (B series, C) or fat-soluble (A, D, E, K). Water-soluble vitamins contain several hydrophilic groups and are easily dissolved in aqueous environments. They are not stored in the body, and the excess is excreted in the urine (Choice B). For this reason, water-soluble vitamins must be frequently replenished. . In contrast, fat-soluble vitamins are highly hydrophobic. They dissolve in lipid rather than aqueous environments and cannot be easily excreted in urine. Instead, the excess is stored and accumulates in adipose tissue, which consists mainly of fat cells.

. (Choice C) An Rf value is the ratio of the distance a compound migrates up the TLC plate to the distance traveled by the mobile phase. Relative Rf values of two compounds can be compared without running TLC because the relative strengths and numbers of interactions two compounds have with silica in the stationary phase are known previously. A compound that has a stronger interaction with the stationary phase will have a lower Rf than a compound that has a weaker interaction with the stationary phase. . (Choice D) Optimal reaction conditions should provide the greatest amount of the desired product (Compound 4) and the least amount of the side-product (Compound 5).

Why was the optimization for the oxidation of Compound 3 monitored by thin-layer chromatography? The resulting TLC plate given in the passage is used to determine: . "​​​​​​​ Compound 3 is then oxidized using the oxidizing agent IBX to give a mixture of the desired product (Compound 4) and an overoxidized product (Compound 5)." . A. which reaction conditions provide the greatest amount of Compound 4 and the least amount of Compound 5. B. the percent yield of the reaction. C. whether Compound 4 or Compound 5 has a greater Rf value. D. which solvent and reaction time provide the greatest amount of Compound 5 and the least amount of Compound 4. . . . ​​​​​​​ . The passage states that the oxidation of Compound 3 yielded a mixture of Compound 4 (the desired product) and Compound 5 (the overoxidized side-product). . TLC was then performed using a nonpolar mobile phase and a polar silica stationary phase. Compounds 3-5 have different strengths and numbers of interactions with the stationary phase: Compound 5 has the fewest and weakest interactions and migrates the fastest and farthest up the TLC plate whereas Compound 3 migrates the slowest. . The given TLC plate shows that different solvents and reaction times yield different relative amounts of Compounds 3-5. This information can be used to determine the optimal conditions—those that yield the greatest amount of Compound 4 (the product) and the least amount of Compound 5 (the side-product). . ​​​​​​​ (Choice B) Although TLC can show which conditions produce more or less of a compound, it is not a quantitative technique and does not allow determination of exact amounts of a compound present in a mixture. Therefore, the percent yield of the reaction cannot be calculated from the TLC plate.

Differential association

__________________ theory suggests that deviance is learned through interaction. For example, someone goes to prison for drug possession and learns from incarcerated sellers how to efficiently sell drugs. . . Labeling theory suggests that deviance lies not in the act but in the social response of applying a label to individuals. Labeling individuals as deviant has consequences, such as stigmatization, that lead to further deviance.

​​​​​​​Reaction optimization

_________________________ is necessary to determine under what conditions the desired product readily forms while preventing or minimizing side-product formation. . Researchers studied different reaction conditions for the oxidation of Compound 3 by monitoring the reaction with thin-layer chromatography (TLC) on a silica plate with a nonpolar solvent. . The rate at which each compound travels up the TLC plate depends on the strength of the competing intermolecular interactions between the compound and the stationary (polar) and mobile phases (nonpolar) . . Thin-layer chromatography (TLC) is a separation technique commonly used to visually monitor reaction progress when the starting material and products have different relative polarities. As such, TLC can be used to determine whether starting material is still present and whether desired products and/or side-products have formed. . ​​​​​​​Reaction optimization is necessary to determine under what conditions the product readily forms while preventing or minimizing side-product formation. Thin-layer chromatography is a separation technique commonly used to monitor reactions because it allows visualization of the reaction's progress.

​​​​​​​ (Choice A) Informal sanctions are individual responses not codified into laws, such as disapproving looks or comments. The "war on drugs" resulted in more aggressive laws against drug use and distribution, which describes more formal (not informal) sanctions. . (Choice C) Folkways are less formal social norms that, when violated (eg, wearing clothes backward), typically result in informal sanctions (eg, raised eyebrows). The "war on drugs" did result in drug-related behaviors being viewed as less of a folkway (more taboo), but it also increased formal (not informal) sanctions. . (Choice D) The passage explains that drug use and distribution came to be considered a higher level of deviance than it was prior to the "war on drugs," suggesting that drug-related behaviors became more taboo, not more of a folkway.

​​​​​​​ According to the passage, the "war on drugs" made drug use and distribution: . A. less taboo through the implementation of more informal sanctions. B. more taboo through the implementation of more formal sanctions. C. less of a folkway through the implementation of more informal sanctions. D. more of a folkway through the implementation of more formal sanctions. . . . Norms exist in every culture and serve to disseminate information about appropriate conduct in a variety of situations. Deviance is any behavior that violates culturally established norms. Deviant behavior can be viewed as simply odd or off-putting (eg, nose-picking in public) or can be viewed as egregious or criminal (eg, murder). . ​​​​​​​Taboos are the most serious form of deviance—acts considered illegal and/or reprehensible by society. These forms of deviance meet with the most serious types of disapproval in the form of institutionalized, or formal, sanctions, such as arrest and incarceration. . Taboos are deviant behaviors considered morally reprehensible by society. Committing a taboo often results in a formal sanction, the application of an institutional response, such as arrest or incarceration. Since the "war on drugs" beginning in the 1980s, the use and distribution of drugs has become increasingly treated as not only deviant, but reprehensible and criminal in the U.S., and was increasingly met with formal sanctions (eg, jail time).

(Choice B) Occupational prestige is a measure of the respect and esteem (prestige) of a given occupation. Occupations requiring the most education are generally ranked highest. However, the passage does not mention occupational status or prestige. (Choice C) Social capital refers to one's social network and the value of those connections. Although one may have connections in the criminal justice system who can impact outcomes, the example given here addresses only sex/gender disparities. Racial disparities are more emphasized in the passage, and both racial and sex/gender disparities are accounted for with intersectionality.

​​​​​​​ Which of the following helps explain the demographic disparities of the prison population described in the passage? . A. Social reproduction, because a child whose parent has been incarcerated is more likely to stay away from criminal activities. B. Occupational prestige, because individuals with higher-paying, professional jobs are not motivated to commit crime. C. Social capital, because females are more likely to have contacts in the criminal justice system who can get them released without a conviction. D. Intersectionality, because white males are more likely to be incarcerated than black females, but less likely than black males. . . . . ​​​​​​​ The intersectionality approach describes how all individuals hold multiple, interconnected identities that simultaneously impact their lives and perspectives. For example, being female is not the only identity that affects how a woman sees, experiences, or is treated in the world. Her experiences and perspectives are also a product of her other identities (eg, race/ethnicity, class, sexual orientation). . With respect to incarceration, the U.S. prison population is both disproportionately black and disproportionately male, placing individuals who are black and male at the intersection of statuses most represented in prisons. Although white males are more likely to be incarcerated than black females, black males are more likely to be imprisoned than white males. . (Choice A) Social reproduction refers to the transmission of society's values, norms, and practices, including social inequality, from one generation to the next. Children with a parent who has been incarcerated are more (not less) likely to be incarcerated themselves. The passage does not mention generational effects on prison demographics.

Accordingly, the correct order in which a human embryo develops is fertilization → morula formation → blastulation → gastrulation → neurulation. . Educational objective:Embryogenesis is the process by which an embryo develops from a single-celled zygote. It includes the following stages in order: fertilization, morula formation, blastulation, gastrulation, neurulation.

​​​​​​​ Which series shows the order in which a human embryo develops? . A. Fertilization → blastulation → morula formation → gastrulation → neurulation B. Fertilization → morula formation → blastulation → neurulation → gastrulation C. Fertilization → morula formation → blastulation → gastrulation → neurulation D. Fertilization → neurulation→ morula formation → blastulation → gastrulation