C/P PT3

Based on the passage, which of the following buffers could be used instead of the citrate buffer in Experiments 1 and 2? A. Acetate (pKa 4.7) B. Phosphate (pKa 6.5) C. Tris (pKa 8.1) D. Carbonate (pKa 9.7)

A. Acetate (pKa 4.7)

Researchers measured the activity of two enzymes that catalyze the same reaction, with results shown below. Which of the enzymes must have multiple active sites? A. Enzyme 1 only B. Enzyme 2 only C. Both Enzyme D. Neither

A. Enzyme 1 only (Choices B and C) Enzyme 2 has a hyperbolic curve, and therefore does not exhibit cooperativity. Although it is possible for a noncooperative enzyme to have multiple active sites, it is not required. (Choice D) Enzyme 1 exhibits positive cooperativity, and therefore must have more than one active site. Educational objective:Enzymes that display positive cooperativity yield sigmoidal curves instead of hyperbolic curves when reaction rate is plotted against substrate concentration. Because cooperativity is a change in the affinity of one active site for its substrate when another active site is bound, any enzyme that exhibits cooperativity must have more than one active site.

In high-performance liquid chromatography (HPLC)

, two types of columns—normal-phase (NP) and reverse-phase (RP)—can be used, depending on the polarities of the compounds being separated. NP-HPLC consists of a polar stationary phase and a nonpolar mobile phase whereas RP-HPLC consists of a nonpolar stationary phase and a polar mobile phase. Molecules with polarity similar to the stationary phase interact with it more and have longer retention times.

(C=O stretch)

1810-1650 cm−1

(sp3 C-H stretch)

3000 cm−1

(sp2 C-H stretch)

3100 cm−1

(sp C-H stretch)

3300 cm−1

(O-H stretch)

3650-3200 cm−1

A 100-g block of iron at 31°C is submerged in 50 g of water at 25°C in a thermally insulated container. If the temperature of the system reaches equilibrium at 26°C, what is the specific heat of iron? (Note: The specific heat of water is 4.2 J/(g⋅°C).) A. 0.42 J/(g⋅°C) B. 1.7 J/(g⋅°C) C. 42 J/(g⋅°C) D. 170 J/(g⋅°C)

A. 0.42 J/(g⋅°C)

The amino acid methionine can be converted to succinyl-CoA to enter the citric acid cycle. How many NADH molecules can be produced by each methionine molecule that enters the citric acid cycle by this pathway? A. 1 B. 2 C. 3 D. 4

A. 1 Choices B, C, and D) To produce 2, 3, or 4 molecules of NADH, a molecule of α-ketoglutarate, isocitrate, or pyruvate, respectively—or one of their precursors—would be needed. However, methionine is metabolized into succinyl-CoA, which enters the cycle after any of these molecules. Furthermore, the oxaloacetate produced at the end of the pathway cannot continue in the citric acid cycle unless additional acetyl-CoA is brought in to provide the energy to produce NADH. Educational objective:Different molecules can enter the citric acid cycle at different points. The cycle oxidizes these molecules to produce the reduced electron carriers NADH and FADH2. The number of electron carriers produced depends on the number of oxidative steps remaining in the cycle from the point of entry.

A black hole emits a jet with a power of 1044 W. If the black hole has a rotational period of 0.1 ms, what is the energy emitted by the jet during one rotation of the black hole? A. 10^40 J B. 10^43 J C. 10^45 J D. 10^48 J

A. 10^40 J

The data in the table above were recorded during an experiment in which a standing wave pattern was produced on a stretched string of length 0.5 m. One end of the string was attached to a fixed wall and the other was connected to an oscillator. The speed of the waves on the string is most nearly Number of antinodes: 2 | 3 | 4 | 5 frequency: 101 Hz | 149 Hz | 201 Hz | 251 Hz A. 50 m/s B. 100 m/s C. 150 m/s D. 200 m/s

A. 50 m/s

Which of the following structures represents the Fischer projection of a β-d-glucose unit found in cellulose?

B. In a Fischer projection, D-sugars always show the second-to-last carbon (carbon 5 in glucose) as having its oxygen atom to the right. (Choice A) This structure is α-d-mannose, a glucose epimer that differs at carbon 2. (Choice C) This structure is β-d-fructose. (Choice D) This is β-l-glucose, the enantiomer of β-d-glucose. Because this is an l-sugar, the β anomer positions the anomeric hydroxyl on the right.

According to the passage, which of the following changes occurs as an astronaut leaves Earth's surface and enters orbit? A. Coefficient of kinetic friction between the astronaut and nearby objects. B. Astronaut center of mass. C. Astronaut bodily inertia. D. Magnitude of the gravitational force between the astronaut and the spacecraft.

B. Astronaut center of mass.

Suppose that citric acid (H3C6H5O7) is titrated with 0.1 M NaOH to form a citrate buffer solution with a pH of 4.5. What is the pH at the first equivalence point? (Note: pKa1 = 3.13, pKa2 = 4.76, pKa3 = 6.40) A. Less than 3.13 B. Between 3.13 and 4.76 C. Equal to 3.13 D. Greater than 4.76

B. Between 3.13 and 4.76

If researchers want to use a coating material that is more soluble in pure water than Mg(OH)2 but less soluble than MgF2, which of the following would be a good option? A. Ca(OH)2 B. CaF2 C. Zn(OH)2 D. Ca3(PO4)2

B. CaF2

The bonds of four salts (MgBr2, NaCl, MgCl2, and NaBr) are evaluated for ionic character based on electronegativity differences between atoms. What is the expected order of these four salts if listed according to bond character from the least ionic character to the greatest ionic character? A. NaBr < NaCl < MgBr2 < MgCl2 B. MgBr2 < MgCl2 < NaBr < NaCl C. NaCl < NaBr < MgCl2 < MgBr2 D. MgBr2 < MgCl2 < NaCl < NaBr

B. MgBr2 < MgCl2 < NaBr < NaCl

If the leucine molecules given to participants were labeled with heavy nitrogen isotopes to monitor the contribution of leucine to urea production, the researchers would in theory have been able to detect the label: A. only in urea. B. first in glutamate and then in urea. C. first in the α-keto acid of leucine and then in urea. D. first in urea and then in citric acid cycle intermediates

B. first in glutamate and then in urea. In the transamination reaction that initiates amino acid oxidation, an amino acid is deaminated to yield an α-keto acid and glutamate. The amino group transferred to glutamate in this reaction can be used to form urea in the reactions of ureagenesis.

Given the in vivo results from the study, the effect of growth hormone on fat oxidation directly results in all of the following EXCEPT: A. increased acetyl-coenzyme A production from triglyceride fatty acids. B. increased glycerol release from triglycerides. C. increased NADH production from fatty acids. D. increased FADH2 production from fatty acids.

B. increased glycerol release from triglycerides. (Choices A, C, and D) Because fat oxidation directly produces acetyl-CoA, NADH, and FADH2, oxidation of fatty acids from any source directly stimulates the production of these three molecules from fatty acids. Oxidation of fatty acids produces acetyl-coenzyme A and the cofactors NADH and FADH2

The experiment was originally performed at physiological pH (7.4). Raising the pH to 12 would: A. promote the corrosion of the magnesium disks. B. inhibit the corrosion of the magnesium disks. C. initially promote the corrosion of the magnesium disks and then begin inhibiting corrosion as the reaction progressed. D. have no effect on the corrosion of the magnesium disks

B. inhibit the corrosion of the magnesium disks. The common ion effect is an application of Le Châtelier's principle, in which a common ion is added to a dissolution reaction, shifting the equilibrium position toward reactants (ie, decreasing solubility).

According to the data in the passage, production of which of the following metabolites was most likely inhibited by infusion of growth hormone? A. α-Keto acids derived from lysine B. α-Keto acids derived from leucine C. Fatty acids derived from triglycerides D. Ketone bodies derived from triglyceride fatty acids

B. α-Keto acids derived from leucine (Choice A) Lysine's concentration, like leucine's, did not change significantly in response to growth hormone infusion. However, unlike leucine, for which oxidation was measured and known to be affected by growth hormone, lysine and its oxidation are not addressed in the passage. Therefore, the effects of growth hormone infusion on the formation of lysine's α-keto acid are less certain. (Choice C) The marked increase in free fatty acid concentration in response to growth hormone infusion suggests an increase in fatty acid release from hydrolyzed triglycerides. (Choice D) The increase in free fatty acid concentration that occurred during growth hormone infusion suggests that, if anything, ketogenesis from free fatty acids would be stimulated rather than inhibited. Likewise, the reduction in leucine oxidation could theoretically enable more use of leucine's α-keto acid for ketogenesis (ie, ketone body production), but to the extent this occurred it too would increase rather than decrease ketogenesis.

The infrared spectrum shows the frequencies of infrared light absorbed by a sample of butyraldehyde that was found to be contaminated. Which of the following could be the contaminant? A. Propene B. Propanone C. 2-Propanol D. Propyl ethanoate

C. 2-Propanol Aldehyde carbonyls (C=O) have a characteristic strong absorption between 1740 and 1720 cm−1. The sample IR spectrum shows a strong absorption at 1730 cm−1, which corresponds to the carbonyl of butyraldehyde. Alcohols have a broad signal between 3650 and 3200 cm−1, so the absorption at 3350 cm−1 most likely corresponds to an O-H stretch. 2-propanol is the only choice that contains an OH group, and therefore it is the most likely contaminant. (Choice A) Propene is an alkene, which has a characteristic sp2 C-H stretch at 3100 cm−1. The sample IR spectrum does not have signals indicative of a compound with a carbon-carbon double bond. (Choices B and D) Propanone contains an internal carbonyl (ketone), which has a strong C=O stretch between 1725 and 1705 cm−1. Propyl ethanoate is an ester, and the ester C=O absorbs IR light between 1750 and 1735 cm−1. The carbonyl stretch in the sample IR spectrum does not correspond to a ketone or an ester but rather to butyraldehyde.

A volunteer with a mass of 80 kg is suspended by his arms on Earth. What is the approximate tension on lumbar disc 5, which is located near the volunteer's center of mass? (Note: The acceleration of gravity is g = 10 m/s2) A. 40 N B. 70 N C. 400 N D. 700 N

C. 400 N Lumbar disc 5 is located close to the volunteer's center of mass. Therefore, approximately half the volunteer's body mass is located below lumbar disc 5. As a result, T on lumbar disc 5 is equal to approximately half the volunteer's W: T=Mg/2=(80 kg)(10 m/s2)/2=400 N (Choice A) Half of the volunteer's total mass is 40 kg, a value that is multiplied by the acceleration of gravity (10 m/s2) to obtain weight. (Choice B) 70 N is obtained by incorrectly calculating 90% of the volunteer's mass, instead of half of the volunteer's weight. (Choice D) 700 N is obtained by incorrectly calculating 90% of the volunteer's weight, instead of half of the volunteer's weight. Educational objective:Newton's second law dictates that an object will accelerate unless the forces acting on the object completely negate one another. For objects suspended in air, the weight of the object is typically countered by a tension force that maintains static equilibrium.

An individual that weighs 700 N in air has an apparent weight of 40 N when submerged in water. What is the volume of the displaced water? Body density can be used to calculate an individual's body fat composition, the percentage of total body mass that is made up of fatty tissue. Human fatty tissue has an average density of 0.9 kg/L, and the density of water is 1 kg/L. The A. 40 L B. 44 L C. 66 L D. 70 L

C. 66 L

The electric force generated by two point charges separated by Distance A is 4 times greater than the electric force generated when separated by Distance B. How does Distance A compare to Distance B? A. Distance A is longer by a factor of 2. B. Distance A is longer by a factor of 4. C. Distance A is shorter by a factor of 2. D. Distance A is shorter by a factor of 4.

C. Distance A is shorter by a factor of 2. Fe=1/r^2

The chemical structures of sulfur compounds H2S, SF6, S2Cl2, and S4N4 each contain only single (sigma) bonds. Based on trends in the atomic radii of the elements, the compound that contains the longest bond between sulfur and an atom of another element is: A. H2S B. SF6 C. S2Cl2 D. S4N4

C. S2Cl2 (Choices A, B, and D) Atoms of H, N, and F each have a smaller atomic radius than Cl because Cl is in Period 3 and has an additional electron shell farther from the nucleus. Therefore, the S-H bonds in H2S, the S-F bonds in SF6, and the S-N bonds in S4N4 are all shorter than the S-Cl bonds in S2Cl2. Educational objective:Covalent sigma bonds are made by sharing electrons through the end-to-end overlap of atomic orbitals. The length of a sigma bond can be estimated as the sum of the atomic radii of the bonded atoms. Atomic radii tend to decrease across a row and increase down a column on the periodic table.

A mirror is placed at the back of the block during the experiment. The laser light from the Doppler sensor reflects off the mirror, as shown in the diagram below. (Note: The figure is NOT drawn to scale.) Which of the following equations is true for any value of the incident angle, θ2? A. θ1 = θ2 B. θ1 = θ3 C. θ2 = θ3 D. θ3 = θ4

C. θ2 = θ3

Which of the following sulfur compounds from Table 1 has a chemical structure that contains both ionic and covalent bonds? A. Hydrogen sulfide B. Sulfuric acid C. Sulfur trioxide D. Potassium sulfate

D. Potassium sulfate The type of bond formed between two atoms depends on the relative difference in electronegativity between the atoms. Atoms with a large difference in electronegativity (usually a metal and a nonmetal) form ionic bonds. Atoms with a small difference in electronegativity (usually two nonmetals) form covalent bonds.

Fatty tissue has a lower electrical conductivity than lean (non-fat) tissue. This property allows body fat content to be estimated by modeling the tissue between two points as a single resistor. For a constant current entering the body, relatively high body fat content would be indicated by: A. an increased current leaving the body. B. a decreased current leaving the body. C. a lower voltage difference across the body. D. a higher voltage difference across the body.

D. a higher voltage difference across the body. Electrical conductivity is the measure of how easily a current can move through a material, and it is inversely proportional to electrical resistance. The question states that fatty tissue is less conductive than lean tissue; in other words, fatty tissue is more resistive. Therefore, an individual with a relatively high body fat content will have a relatively high resistance. Voltage is the difference in electric potential between two points that drives the movement of electric charges. Ohm law describes the voltage differenceV across a resistor, which is the product of the current I and the resistance R: V=IRV=IR Therefore, the voltage difference measured across the resistor is proportional to the resistance, and a high body fat content (high resistance) in an individual will be indicated by a high voltage difference between two points. (Choices A and B) Conservation of electric charge implies that the amount of current exiting the body must be constant because the current entering the body is constant. (Choice C) A relatively high body fat content increases the resistance of the body. When current is constant, higher resistance will increase the voltage difference across the body, not lower the voltage difference. Educational objective:The current entering and exiting a resistor remains the same because electric charge is always conserved. The voltage across a resistor is the product of resistance and current (V = IR). Therefore, tissues with a higher resistance will have a higher voltage across them for the same amount of current.

Developmental protocols for Caco-2 monolayers involve the use of cells with high proliferation potential that can differentiate in a synchronized manner to form homogenous monolayers. This proliferation of Caco-2 cells is primarily achieved through: A. meiosis. B. fission. C. completion of the G2 phase of the cell cycle. D. completion of the M phase of the cell cycle.

D. completion of the M phase of the cell cycle. (Choice A) In eukaryotes, reproductive cells (ie, gametes), not somatic cells, are produced via meiosis, during which a parent cell (2n) divides to produce four genetically distinct daughter cells containing half the original number of chromosomes (1n). (Choice B) The cells of multicellular organisms (eg, humans) cannot divide by binary fission. Binary fission is the process by which single-celled organisms, such as bacteria, reproduce asexually. During binary fission, the parental cell doubles in size and then divides into two identical daughter cells. (Choice C) Cell growth and repair of DNA replication errors occur during the G2 phase of the cell cycle; cell division does not occur at this stage. Educational objective:In multicellular organisms (eg, eukaryotes), nongametic (somatic) cells can divide and multiply via mitosis whereas gametic (ie, reproductive) cells divide and multiply via meiosis. Biological homeostatic mechanisms function to maintain a balanced ratio of proliferating to dying cells in human tissues and organs. In contrast, cell division in single-celled organisms (eg, prokaryotes) occurs via binary fission.

Although ELISA can accurately detect the concentration of an analyte in a sample, a scientist claims ELISA can generate faulty results under certain experimental settings. Inaccurate quantification of an analyte such as HIF-1α using ELISA would most likely result from any of the following EXCEPT: A. supplementing the sample with a denaturing agent during the initial sample preparation. B. adding an amount of substrate disproportionate to the amount of analyte. C. quantifying the color change in the presence of unbound antibodies. D. identifying a color change proportional to analyte concentration after adding the substrate.

D. identifying a color change proportional to analyte concentration after adding the substrate. Choice A) Because analytes such as HIF-1α are proteins, a denaturing agent (eg, sodium dodecyl sulfate) would denature (unfold) the native protein into a linear arrangement of amino acids. This result could disassemble the antibody binding site on the protein and render ELISA ineffective. (Choice B) If the amount of substrate added is disproportionate to the amount of analyte, either too much or too little substrate was added. Too much substrate can lead to excessive color change and an inaccurate report of higher protein levels whereas too little substrate can lead to insufficient color change and an inaccurate report of lower protein levels. (Choice C) If unbound antibodies remain in the sample during the measurement of the color change, the ELISA result is faulty because the enzyme linked to the unbound antibody would have also created colored product. The ensuing color change would not be representative of actual protein levels. Educational objective:Enzyme-linked immunosorbent assay (ELISA) can detect and quantify proteins. Initially, a primary antibody (linked to a "reporter" enzyme) is added, which binds the antigen (protein). The samples are washed to remove unbound proteins, and the reporter enzyme substrate is added. The enzyme-substrate reaction creates a product that results in a quantifiable/detectable signal.

O-linked D-mannose can form a glycosidic bond with any sugar that has a free anomeric carbon. Therefore it could theoretically form a glycosidic bond with all of the following sugars EXCEPT: A. galactose. B. ribose. C. lactose. D. sucrose.

D. sucrose. (Choices A and B) Galactose and ribose are monosaccharides (single carbohydrate units). All free monosaccharides are reducing sugars and can form glycosidic bonds. (Choice C) Lactose is a disaccharide formed by a β-1,4 linkage between galactose and glucose. The glucose unit has a free anomeric carbon, and therefore is a reducing sugar. Educational objective:Glycosidic bond formation requires a sugar with a free anomeric carbon, called a reducing sugar. All free monosaccharides are reducing sugars whereas disaccharides are reducing only if one anomeric carbon does not participate in a glycosidic bond. Sucrose is the most common nonreducing sugar.

A pitot tube is used to measure the velocity of air flow delivered to a patient from a mechanical ventilator. Opening A is at the front of the tube and opening B is on the side of the tube. The fluid between these two openings blocks the air flow, causing the air velocity at opening A vA to equal zero. Which of the following expressions is proportional to the height h of the fluid inside the pitot tube? (Note: Assume the air behaves like an ideal fluid; vA and vB are the air velocities at openings A and B, respectively.) A. 1/vB B. √vB) C. vB D. vB^2

D. vB^2

When one chemical process provides the energy for another, the processes are said to be coupled.

Decoupling two otherwise coupled processes causes the energy-providing step to continue without driving the dependent process

Pluripotent stem cells

can give rise to only fetal cells (ie, all cell lineages from the three germ layers).

The adaptive (specific) immune system

comprises cells that mount responses against specific pathogens via recognition of specific antigens (ie, foreign molecules). Adaptive immune cells include both B cells and T cells (eg, cytotoxic T cells, helper T cells). Upon initial exposure to a specific pathogen, some of these B and T cells divide and differentiate into specialized cells called memory cells. Memory B and T cells are stored in various tissues (eg, lymphoid tissues) and can more rapidly recognize and respond to future infections by the same pathogen.

The innate (nonspecific) immune system

comprises physical and chemical barriers (eg, skin, mucous membranes, stomach acid) that act to prevent pathogens from entering the body, as well as innate immune cells (eg, neutrophils, macrophages, natural killer cells) that can destroy pathogens and regulate the responses of other immune cells (ie, via the release of signaling molecules). Innate immune responses are directed at many types of pathogens (ie, are nonspecific) and can occur immediately upon initial contact with a pathogen.

Endoderm (innermost layer)

gives rise to accessory digestive organs (eg, liver, pancreas) as well as to the lining (epithelium) of the digestive and respiratory tracts. alveolar cells islet beta cells

Mesoderm (middle layer)

gives rise to the circulatory system, muscles, bones, and parts of the urinary and reproductive systems

Ectoderm (outermost layer)

gives rise to the nervous system (neurulation) and develops into the integumentary system, which includes hair, skin, nails, and the lining of the mouth, nostrils, and anus melanocytes

nonreducing sugars are

sucrose, stachyose, verabscose

Anomers

are a special kind of epimer that differ at the anomeric carbon and are designated as the α- and β-forms of the same sugar. This difference can only occur in the cyclical form of the carbohydrate because the anomeric carbon is not a stereocenter in the linear form.

Multipotent cells

are able to differentiate only into the specialized cells of certain tissues; these cells are also found in adults

Totipotent stem cells

are the least specialized cells and can give rise to both placental and fetal cells.

Magnesium hydroxide begins to precipitate onto the magnesium metal surface when which of the following expressions is true? A. [Mg2+][OH−]^2 > 8.9 × 10−12 B. [Mg2+][OH−]^2 < 8.9 × 10−12 C. [OH−] > 8.9 × 10−12 D. [OH−] < [Mg2+]

A. [Mg2+][OH−]^2 > 8.9 × 10−12

Matter falling in the gravitational field of a central mass can form a hot, charged fluid. The fluid matter orbiting some stars radiates infrared light, but the matter orbiting black holes emits x-rays. Which of the following explains this discrepancy? The matter orbiting black holes must: A. be hotter than the matter orbiting stars. B. be colder than the matter orbiting stars. C. have lower kinetic energy than the matter orbiting stars. D. exhibit higher viscosity than the matter orbiting stars.

A. be hotter than the matter orbiting stars. The electromagnetic (EM) spectrum classifies EM waves by frequency or wavelength. Objects with high energy have a higher temperature and emit EM waves with relatively higher frequencies and energy.

Which of the following experimental changes to the initially prepared oxytocin acetate solutions will decrease the percent of oxytocin acetate recovered in Experiment 1 after the 4-week storage period? A. Adding more sodium citrate after the NaOH has been added B. Decreasing the initial concentration of oxytocin acetate C. Storing the solutions at 0 °C rather than 5 °C D. Increasing the citrate buffer concentration

A. Adding more sodium citrate after the NaOH has been added In Experiment 1, the citrate buffer is made of citric acid [HA] and sodium citrate [A−]. Adding more sodium citrate to the initially prepared solutions (pH of 4.5) will increase the pH of the solution (pH > 4.5). The passage states that oxytocin is more stable at a pH of 4.5, and Figure 1 shows that at the higher pH of 6.8 (water), less oxytocin was recovered. Therefore, adding more sodium citrate will decrease the percent recovery of oxytocin acetate. (Choice B) The experiment measures the percent recovery of oxytocin. Decreasing the initial amount of oxytocin acetate dissolved in solution will lower the total amount present in solution but will not change the percentage recovered. (Choice C) Storing the solutions at 0 °C rather than 5 °C during the experiment will not decrease the percent recovery because the results in Experiments 1 and 2 show that oxytocin acetate is more stable at cooler temperatures. (Choice D) Increasing the buffer concentration will increase buffer capacity to resist changes in pH and assist in stabilizing the oxytocin acetate. Based on the results in Experiment 1, this will not decrease oxytocin acetate recovery. Educational objective:A buffered solution consists of a weak acid and a salt of its conjugate base or a weak base and its conjugate acid. Buffers resist changes in pH by neutralizing added H+ and OH− ions. The pH of a buffered solution depends on the ratio of weak acid and its conjugate base.

When Gd3+ binds to DOTA, Gd3+ is acting as a(n): A. Lewis acid. B. Lewis base. C. Arrhenius acid. D. Arrhenius base.

A. Lewis acid.

If the pathological effects observed in patients with the L156P mutation are caused by decreased Complex V activity, the effects could be best counteracted by a drug that increases which of the following? A. Malate-aspartate shuttle activity to transfer cytosolic NADH to the mitochondria B. Potassium concentration in the mitochondrial matrix to decrease the charge gradient C. The pH of the intermembrane space to equalize it with the pH of the matrix D. Electron transfer from NADH to FAD via reduction of glycerol 3-phosphate to DHAP

A. Malate-aspartate shuttle activity to transfer cytosolic NADH to the mitochondria (Choice B) The energy released when protons flow through Complex V is partially due to the difference in charge across the inner membrane. Decreasing the charge difference by adding potassium to the matrix would cause a reduction in ATP synthesis. (Choice C) ATP synthase is driven by protons flowing from a high concentration in the intermembrane space toward a low concentration in the matrix. Equalizing the pH of the intermembrane space and the mitochondria by raising the pH of the intermembrane space would destroy the proton gradient and stop ATP synthesis. (Choice D) The glycerol 3-phosphate shuttle effectively converts cytosolic NADH to mitochondrial FADH2 by oxidizing glycerol 3-phosphate to DHAP, not by reducing it. In addition, FADH2 pumps fewer protons than NADH does, so this system is less efficient than malate translocation at stimulating Complex V activity. Educational objective:Complex V activity depends on proton availability. Proton concentration in the intermembrane space can be increased by increasing the number of NADH molecules that enter the electron transport chain. NADH from glycolysis can indirectly enter the mitochondrial matrix by first passing its electrons to oxaloacetate, forming malate.

To determine whether Gd-daa3 is selective for Zn2+, researchers compared relaxivity measurements of Gd-daa3 in the presence of different cations. Varying amounts of XCl2 (where X is Zn2+, Ca2+, or Mg2+) were added to a buffered 0.1 mM solution of Gd-daa3 at pH 7.4. What is the most appropriate control for this experiment? A. Relaxivity measurements without XCl2 B. Relaxivity measurements with ZnCl2 C. Relaxivity measurements with BeCl2 D. Relaxivity measurements without Gd-daa3

A. Relaxivity measurements without XCl2 For these testing conditions, the independent variables include the amount of XCl2 and the identity of X, and the dependent variable is relaxivity. The most appropriate experimental control includes baseline readings of the relaxivity of the Gd-daa3 solution in the absence of XCl2. Therefore, the control solution contains only Gd-daa3 and buffer at pH 7.4. All instrumental settings should be the same for all experimental trials. (Choice B) Because Zn2+ is one of the ions tested in this experiment (ie, an independent variable), ZnCl2 would not be part of the control. (Choice C) Even though BeCl2 is not one of the ions tested in this experiment, it would not be included in the control. Be2+ ions are not present in all testing samples, and therefore could skew baseline relaxivity readings. (Choice D) A control without Gd-daa3 would not provide baseline relaxivity measurements in the absence of the independent variable (ie, the amount of XCl2 for each X). Educational objective:The experimental control is a measurement taken with all experimental conditions in place except the independent variable of interest. Controls provide baseline measurements so that the effects of independent variables on dependent variables can be accurately determined.

An object with the same density as water is stationary and suspended in a container filled with water. A downward force is momentarily applied to the object. Assuming the water density remains constant and ignoring the effects of fluid friction, what will happen to the object immediately after this downward force is removed? A. The object will sink at a constant velocity. B. The object's downward velocity will increase with time. C. The object will quickly stop sinking. D. The object will sink before returning to its initial position.

A. The object will sink at a constant velocity. (Choice B) Immediately after the force is removed, the velocity of the object will remain constant, not increase with time. (Choice C) An object that experiences a net force of zero will remain at its current velocity; it does not imply that the object will stop moving. (Choice D) A net force pointing upward would be required to return the object to its initial position. After the applied force is removed, the net force returns to zero because neither FB nor W depend on depth. Educational objective:When a mass experiences a net force, the mass will accelerate in the direction of the force (F = ma). When no net force acts on an object, its velocity remains unchanged. Therefore, an object will accelerate to a velocity when a force is applied and will continue at the same velocity after the force is removed.

The laser Doppler sensor is positioned at the top of the ramp and emits light at 500 nm toward a block sliding down the ramp. The sensor records the data shown in Figure 2. During the interval from 0.3 s to 0.7 s, the wavelength of the light reflected back to the sensor is: A. greater than the emitted light by a constant value. B. less than the emitted light by a constant value. C. continuously increasing with time. D. continuously decreasing with time

A. greater than the emitted light by a constant value.

All of the following could confirm apoptosis due to increased oxidative stress in cells with an L156P mutation in Complex V EXCEPT: A. inhibition of caspase activity. B. increased cytosolic cytochrome C. C. activation of proteolysis. D. increased mitochondrial superoxide concentration.

A. inhibition of caspase activity. Apoptosis is the controlled, programmed death of cells. It occurs in response to a variety of circumstances, including oxidative stress as well as DNA damage and certain developmental events. Apoptosis is initiated when the mitochondrial membrane is permeabilized in response to these environmental cues. This allows the release of cytochrome C from the mitochondria, resulting in increased cytosolic cytochrome C (Choice B). Cytochrome C induces proteolysis (Choice C) and other degradative pathways by activating caspase proteases (inactive caspase is found in healthy cells). Educational objective:Apoptosis (programmed cell death) can be caused by certain developmental events, DNA damage, or reactive oxygen species. It is induced when cytochrome C is allowed to leave the mitochondria and enter the cytosol, where it activates caspase. Caspase in turn activates several degradative pathways such as proteolysis.

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.

A. to prevent cGAS from sensing nuclear DNA. (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. Educational objective:The cell cycle is controlled by various checkpoints, which are regulated by cyclins and cyclin-dependent kinases, within the process to ensure that the steps of the cell cycle are executed in the proper manner

Laser light from the Doppler sensor travels through air with a refractive index of n1 and into a glass block at an incident angle of θ1. The refractive index of the glass block is n2, and the angle of refraction is θ2. What is the ratio of n1 to n2? A. θ2/θ1 B. sinθ2/sinθ1 C. θ1/θ2 D. sinθ1/sinθ2

B n1sinθ1=n2sinθ2 In this question, the laser light is refracted as it passes from air into the glass block. The ratios of the refractive indexes of the two materials are calculated by rearranging Snell's law: n1=n2sinθ2/sinθ1 n1/n2=sinθ2/sinθ1 Therefore, the ratio of n1 to n2 equals sinθ2/sinθ1

A star orbiting a supermassive black hole makes two approximately circular revolutions around the black hole every 20 hours. What is the approximate angular velocity of the star? A. 5.0 × 10-3 degrees/second B. 1.0 × 10-2 degrees/second C. 3.0 × 10-1 degrees/second D. 6.0 × 10-1 degrees/second

B. 1.0 × 10-2 degrees/second

Compound 1 is a common component of plasma membranes. How many unique types of soap molecules will be generated through the complete hydrolysis of Compound 1 using excess NaOH? A. 1 B. 2 C. 3 D. 5

B. 2 The structural features of Compound 1 indicates that it is a phospholipid variant, as it contains: Two fatty acyl groups A glycerol core An inorganic phosphate A small, charged organic group (choline) A total of four esters and phosphoesters linking the components Treatment with excess NaOH will hydrolyze all ester and phosphoesters in Compound 1, as well as deprotonate any acidic protons. Since the two fatty acyl groups contain a different number of carbon atoms, basic hydrolysis will lead to the formation of two different sodium carboxylate salts (soaps). (Choice A) One unique soap molecule would be produced if the fatty acyl groups in Compound 1 had identical structures. However, the fatty acyl groups in Compound 1 have different chain lengths. (Choice C) Three soap molecules would be generated if Compound 1 was a triglyceride containing three different fatty acyl groups. However, Compound 1 is a phospholipid. (Choice D) Although the complete basic hydrolysis of Compound 1 generates five total molecular fragments, only two of these materials are classified as unique sodium carboxylate salts (soaps). Educational objective:A soap is a sodium carboxylate salt of a fatty acid. Soaps can be produced through basic hydrolysis of fatty acyl-containing lipids like triacylglycerols or phospholipids. The number of unique soap molecules that can be generated is dependent on the number of unique fatty acyl groups in the source lipid.

When pyruvate enters the mitochondria, it is converted to acetyl-CoA and further oxidized in a series of reactions that produces four NADH molecules and one FADH2 molecule. If all the NADH and FADH2 produced by 2 moles of pyruvate digestion in the mitochondria enter the ETC, how many moles of ATP can wild-type Complex V produce? Four protons are needed to synthesize an ATP molecule, and thus 12 protons produce 3 ATP molecules each time the γ subunit makes a complete revolution. Every NADH molecule that enters the electron transport chain (ETC) pumps 10 protons into the intermembrane space for use by Complex V whereas only 6 protons are pumped per FADH2 molecule. A. 18 B. 23 C. 28 D. 32

B. 23 B - According to the passage, 10 protons are pumped into the intermembrane space per NADH molecule that enters the ETC, and 6 protons are pumped per FADH2 molecule. Therefore, each pyruvate can cause the pumping of 46 protons(40 from the four NADH molecules and six from FADH2). The passage also states that Complex V produces three ATP for every twelve protons (or one ATP per four protons) that flow through it; therefore, it produces 46/4 = 11.5 ATP molecules for every pyruvate molecule, or 11.5 moles of ATP for every mole of pyruvate. Because the question asks for the ATP yield of 2 moles of pyruvate, this result must be doubled, giving 23 moles of ATP for 2 moles of mitochondrial pyruvate.

A marble rolls down a slope at a velocity of 10 m/s. If the kinetic energy of the marble is 2 J, what is the mass of the marble? A. 20 g B. 40 g C. 60 g D. 80 g

B. 40 g

A 0.3-kg object with half the density of human fatty tissue is placed in water. Ignoring the atmospheric pressure, what percentage of the mass will be submerged? Body density can be used to calculate an individual's body fat composition, the percentage of total body mass that is made up of fatty tissue. Human fatty tissue has an average density of 0.9 kg/L, and the density of water is 1 kg/L. The acceleration due to gravity is 10 m/s2. A. 30% B. 45% C. 90% D. 100%

B. 45% fraction submerged=ρobject/ρfluid

Which of the following accurately describes the magnetic properties of Gd3+ ions? A. Gd3+ is diamagnetic because of unpaired electrons. B. Gd3+ is paramagnetic, and its electron spins align parallel to the applied magnetic field. C. Gd3+ is diamagnetic because ions are attracted to the magnetic field due to their positive charge. D. Gd3+ is paramagnetic, and therefore repels magnetic field lines.

B. Gd3+ is paramagnetic, and its electron spins align parallel to the applied magnetic field.

If researchers wanted to replicate physiological conditions by increasing the SBF temperature to 37°C, how would this change affect the solubility product constant and equilibrium pH of Mg(OH)2 dissolution? A. Ksp will decrease and equilibrium pH will increase. B. Ksp and equilibrium pH will both increase. C. Ksp will stay the same and equilibrium pH will increase. D. Ksp will stay the same and equilibrium pH will decrease.

B. Ksp and equilibrium pH will both increase. The solubility product constant Ksp is defined as the product of the molar concentrations of the product ions, each raised to the power of their balanced stoichiometric coefficients. When temperature increases, Ksp decreases for exothermic reactions and increases for endothermic reactions.

Which of the following statements explains the discrepancy between the microgravity data and the laboratory data for the experiment described in the passage? A. In microgravity, gravitational forces were converted to tension. B. In the laboratory setting, the weight of the body still acted on spinal cartilage. C. The mass of the laboratory volunteers did not decrease. D. Laboratory volunteers were not suspended upside-down to account for PFS.

B. In the laboratory setting, the weight of the body still acted on spinal cartilage. In contrast to compression, tension acts to elongate objects. As a result, the extra spinal column length seen in Figure 1 of the laboratory data may be explained by tension due to body weight. Therefore, it cannot be said that the effects of gravity were eliminated; gravity still acted on the body but was experienced as tension rather than compression. (Choice A) The effect of tension was not observed in the microgravity condition because little gravity was present in space. Greatly reduced gravity leads to greatly reduced weight. (Choice C) Mass is an intrinsic property of an object that depends only on the characteristics of the object and not on external factors (eg, local gravity). (Choice D) Although hanging a subject upside-down may reproduce the redistribution of bodily fluids experienced in microgravity, fluid weight pooling at the head of an upside-down subject suspended by the feet would contribute to further spinal elongation. Educational objective:Weight can act as a compressive force or as a tensile (pulling) force. Tensile forces may lead to the elongation of objects under tension.

Astronauts on the Moon decide to turn on a loudspeaker outside their lunar module to broadcast Earth communications during a moonwalk. Will the astronauts hear the sound coming from the speaker while they are outside the lunar module? A. No, because radio waves require a medium and there is no air on the moon B. No, because sound is a mechanical wave requiring a medium and there is no air on the moon C. Yes, because sound waves, like radio waves, can travel through a vacuum D. Yes, if the speaker's power is turned up to a large enough value

B. No, because sound is a mechanical wave requiring a medium and there is no air on the moon

The germ layer that gives rise to the notochord also gives rise to which of the following cell types? A. Alveolar cells B. Osteocytes C. Melanocytes D. Islet beta cells

B. Osteocytes

Undifferentiated gastrula cells presumed to give rise to neural tissue were transplanted from a donor frog embryo into a host frog embryo. The cells were transplanted among undifferentiated host gastrula cells presumed to give rise to epidermal tissue. Both host and donor embryos were allowed to develop following the transplant and gave rise to two normal embryos. What was the purpose of this experiment? A. To determine which cells in the gastrula ultimately give rise to epidermal tissue B. To determine if cell specialization is influenced by communication between cells C. To determine how developmental fates of cells are determined during blastula formation D. To determine whether cell differentiation could be reversed based on embryonic location

B. To determine if cell specialization is influenced by communication between cells

Why did the investigators include ADP in the incubation solution for the experiments measuring the effect of growth hormone infusion on mitochondrial ATP synthesis? A. To provide a source of energy to ATP synthase B. To provide a substrate for the reaction catalyzed by ATP synthase C. To ensure continued substrate-level phosphorylation essential for acetyl-coenzyme A production D. To act as an allosteric inhibitor of glycolytic ATP production

B. To provide a substrate for the reaction catalyzed by ATP synthase (Choice A) The energy for ATP synthesis is provided by the proton gradient across the inner mitochondrial membrane (not ADP). (Choice C) Substrate-level phosphorylation occurs in the cytosol when acetyl-coenzyme A is produced through glycolysis. However, the investigators measured ATP synthesis in isolated mitochondria (ie, in the absence of the cytosol), so the ADP would be used for ATP synthesis by oxidative phosphorylation. (Choice D) Because the measurements were conducted in isolated mitochondria, glycolysis (and glycolytic ATP production) would not occur. Educational objective:Mitochondrial ATP synthesis is carried out by ATP synthase as protons pass through it. This passage of protons from the mitochondrial intermembrane space to the matrix causes movement within ATP synthase that facilitates phosphorylation of ADP to form ATP.

Based on the passage, Caco-2 cells originate from a segment of the gut that mainly functions to: A. absorb nutrients. B. absorb water. C. produce proteolytic enzymes. D. digest nutrients.

B. absorb water. The large intestine is composed of three subdivisions: the cecum, colon, and rectum. The colon functions to absorb electrolytes (eg, sodium, chloride) and additional water from the mass of undigested material. As water is absorbed by the colon, the undigested material concentrates into feces (waste matter), which is stored in the rectum for subsequent excretion. The passage states that Caco-2 cells are a cell line originally derived from colon cells. Therefore, these cells originate from a section of the gut that mainly functions to absorb water (ie, the colon). (Choices A and D) The brush border of the small intestine is a microvilli-covered epithelial surface where digestion and absorption of nutrients occur. The small intestine is composed of the duodenum, jejunum, and ileum; the duodenum is involved in additional digestion of nutrients, and the jejunum and ileum are involved in absorption of these nutrients. Although the passage states that Caco-2 cells were cultured under specific conditions to mimic the functional and morphological phenotype of wild-type enterocytes lining the small intestine, Caco-2 cells originate in the colon, not the small intestine. (Choice C) Proteolytic enzymes are able to digest proteins (polypeptides) but the large intestine does not function to produce these enzymes. However, enzymes secreted by the stomach (pepsin), pancreas (trypsinogen, chymotrypsinogen), and the microvilli-covered brush border of the small intestine (peptidases) function in protein digestion. Educational objective:The colonic segment of the large intestine functions to absorb water and electrolytes from undigested waste material that is left over from digestion and absorption in the small intestine.

The Heisenberg uncertainty principle impacts the study of nuclei and small particles such as electrons. Which of the following is a consequence of this principle? A. Two electrons in the same orbital cannot both have parallel alignment with the magnetic field. B. Neutrons and protons are affected by the nuclear force almost identically. C. Electrons in coordination bonds can be described only as probability distributions. D. Particles release energy when transitioning from an excited state to a ground state.

C. Electrons in coordination bonds can be described only as probability distributions.

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

C. Fertilization → morula formation → blastulation → gastrulation → neurulation

Which of the following statements about HIF-1 is most likely true? At low oxygen levels, HIF-1 is nonfunctional. At low oxygen levels, HIF-1 is functional. At atmospheric oxygen levels, HIF-1 concentration is decreased. At atmospheric oxygen levels, HIF-1 concentration is increased. A. I and III only B. I and IV only C. II and III only D. II and IV only

C. II and III only

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.

C. Include gene expression datasets relating to other types of cancers.

Ignoring any attachments between the spinal discs and vertebrae, how would the maximum force of static friction between these anatomical structures compare for a person standing on the surface of the Moon to a person standing on the surface of the Earth? A. It would be 6 times as great. B. It would be approximately equal. C. It would be 1/6 as great. D. It would be 0.

C. It would be 1/6 as great. Specifically, the maximum value of Ffr on the surface of the Moon will be 1/6 the maximum value of Ffr on the surface of the Earth because g on the Moon is 1/6 the magnitude of g on the Earth. (Choices A and B) Because the force of gravity on the Moon is 1/6 the force of gravity on the Earth, the maximum static friction force on the Moon will be 1/6 as great the maximum force on the Earth, not 6 times as great or approximately equal. (Choice D) The passage explains that the Moon is a reduced-gravity environment, not a microgravity environment. On the Moon, the effects of gravity are not eliminated. Educational objective:Friction is the force that resists sliding between two surfaces. Maximal static friction is proportional to the normal force exerted by the surfaces. Reducing gravitational acceleration also reduces the normal force, which in turn reduces the static friction by the same degree.

An object is launched with a one-time burst of propulsion away from the surface of the Moon. After the burst, which of the following best describes the changes that occur as the object moves away from the Moon's surface? A. The object mass is dissipated as heat. B. Potential energy is converted into kinetic energy. C. Kinetic energy is converted into potential energy. D. Total mechanical energy is not conserved.

C. Kinetic energy is converted into potential energy. Conservation of energy states that any change in kinetic energy is accompanied by an equal and opposite change in potential energy such that the total mechanical energy remains constant. The kinetic energy of an object launched upward in gravity is converted into gravitational potential energy.

Magnesium corrodes quickly in the body because Mg2+ combines with anions other than OH− to form soluble salts. Based on the solubility data given in Table 1, which of the following soluble salts, if present in the body, will cause the magnesium implant to corrode most readily? A. (NH4)3PO4 B. NaOH C. NaCl D. KF

C. NaCl The rate of metal corrosion increases in the presence of anions with which the metal forms soluble salts. Solubility tables provide information on the relative solubilities of potential salts to be formed. Greater solubility corresponds to a faster corrosion rate.

Neural crest cells migrate away from the neural tube to give rise to most of the peripheral nervous system. Given this information, would neural crest cells be involved in the pathology of myelomeningocele? A. Yes; neural crest cells secrete signals that directly inhibit neural tube closure. B. Yes; neural crest cells fail to differentiate into the posterior spinal cord. C. No; neural crest cells continue to differentiate into glial cells of the brain. D. No; neural crest cells continue to act normally as temporary migratory cells that give rise to other cell types.

C. No; neural crest cells continue to differentiate into glial cells of the brain. (Choice A) Neural crest cells have no substantial inductive or inhibitory influence on neural tube formation. (Choice B) Neural crest cells are not associated with myelomeningocele pathology (neural tube closure) as they do not contribute to the formation of the brain or spinal cord (ie, CNS). (Choice C) Nervous system cell types are neurons (ie, conduct electrical signals) and glia (ie, support neural function). Astrocytes and oligodendrocytes are CNS glial cells that arise from the neural tube. Oligodendrocytes form an insulating covering (myelin sheath) around neurons, and astrocytes provide neurons with physical support and anchor them to nutrient supply sources. In contrast, neural crest cells give rise to PNS glia: Schwann cells and satellite cells, which are functionally similar to oligodendrocytes and astrocytes, respectively. Educational objective:Cell migration in embryogenesis is the movement of cells into their final positions within the embryo. The migratory action of neural crest cells during neurulation, the formation of the nervous system, gives rise to many peripheral nervous system structures. In contrast, the central nervous system is derived from the neural tube.

The interval between successive heart beats is modulated by the balance between sympathetic and parasympathetic input to the heart. Given this, an increase in parasympathetic input that increases variability in heart rate will most likely have which of the following effects? A. The mean heart rate and the standard deviation of the interval between beats will both increase. B. The mean heart rate and the standard deviation of the interval between beats will both decrease. C. The mean heart rate will decrease and the standard deviation of the interval between beats will increase. D. The mean heart rate will increase and the standard deviation of the interval between beats will decrease.

C. The mean heart rate will decrease and the standard deviation of the interval between beats will increase.

The researchers made blocks of ice by placing square containers of water in a freezer and monitoring the block's temperature with a thermometer, as shown in the plot below. Why was the block's temperature a constant 0 °C from 15 min to 20 min in the plot? A. The densities of water and ice are different. B. The temperature of the freezer was 0 °C. C. The water must overcome the heat of fusion. D. The water must overcome the heat of vaporization.

C. The water must overcome the heat of fusion. In this question, water changes phase from liquid to solid as it is cooled. During this phase change, heat is removed from the water but the temperature remains constant. Therefore, the liquid water is transforming into solid ice during the time interval from 15 min to 20 min in the plot, and the temperature remains constant until the heat of fusion is overcome. (Choice A) The densities of water and ice are different, but that fact does not cause the temperature to remain constant when water undergoes a change from one phase to another. (Choice B) The temperature of the block continued to decrease to about −5 °C. Therefore, the temperature of the freezer was not 0 °C. (Choice D) The heat of vaporization is required when changing phases between liquid and gas, not between liquid and solid. Educational objective:The heat of fusion is the amount of heat energy required to transform a substance between its liquid phase and solid phase. The material remains at its freezing temperature during this phase change until it is all in solid form.

Assume that neural stem cells are multipotent. Which of the following best describes this classification of stem cells? A. They are able to differentiate into all fetal structures but not placental structures. B. They are able to differentiate into both placental and fetal structures. C. They are able to differentiate into specialized cell types of a particular tissue. D. They are able to differentiate in the zygote but not in the adult organism.

C. They are able to differentiate into specialized cell types of a particular tissue.

An astronaut drifts through space along the surface of her spaceship. She is perpendicular to the spaceship with her feet toward it. The astronaut moves as a rigid body with constant velocity and no rotation. Which of the following occurs when the astronaut momentarily strikes a protruding portion of the spaceship with her foot? The astronaut: A. slows down and continues along her original path with no rotation. B. turns upside down and slowly falls to the ship's surface. C. continues along her original path while spinning about her center of mass. D. stops, spins about the protrusion, and falls to the ship.

C. continues along her original path while spinning about her center of mass. The astronaut's translational (ie, non-rotational) motion does not change significantly because gravitational forces are negated in microgravity. However, the astronaut continues along her original path with reduced velocity because a portion of the impact force acts perpendicular to her original path, causing negative acceleration. (Choice A) The impact between the astronaut's foot and the protrusion on the ship generates a torque, making her spin about her center of mass. (Choices B and D) The astronaut does not fall to the ship because the effects of the gravitational force are minimal in microgravity. The impact with the protrusion induces rotation but the motion of her center of mass is mostly unchanged. Educational objective:Torque is a rotational force that makes objects rotate about a pivot point. For an unrestrained object, the pivot point is the object's center of mass.

POMT can use activated D-mannose, but not D-glucose, as a substrate. D-glucose and D-mannose are: A. constitutional isomers. B. enantiomers. C. epimers. D. anomers.

C. epimers.

Compared to WT cells, cells with deletion of cGAS are likely to exhibit: A. increased apoptosis. B. decreased cell immortalization. C. increased telomerase activity. D. decreased telomere length.

C. increased telomerase activity.

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.

C. migration of chromosomes toward alignment at the equator of the dividing cells. (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.

When Caco-2 cells are cultured in hypoxic conditions, HIF-1 is most likely located in the: A. tight junctions. B. cytoplasm. C. nucleus. D. lysosomes.

C. nucleus. (Choice A) HIF-1 regulates the expression of genes in the nucleus. This transcription factor does not interact directly with tight junctions, or seals between adjacent cells. (Choice B) The cytoplasm is the thick fluid that fills a cell and includes all materials (eg, organelles, enzymes) within the cell but outside the nucleus. Although transcription factors are translated in the cytoplasm, they function in the nucleus. Therefore, in hypoxic conditions, most HIF-1 is located in the nucleus, not in the cytoplasm. (Choice D) Lysosomes are membrane-bound organelles that degrade many macromolecules (eg, proteins, nucleic acids, carbohydrates, lipids). As a transcription factor, HIF-1 is located in the nucleus, not in lysosomes. Educational objective:Transcription factors are translated in the cytoplasm but act in the nucleus to control gene expression. They co

A star orbiting a black hole in the clockwise direction begins to slowly spiral inward due to a counterclockwise drag force. When the star's radial distance from the black hole is 1.0 × 106 km, a drag force of 250 N acts perpendicular to the orbit radius. What is the torque on the star? A. 0 N∙m B. 2.5 × 10^9 N∙m in the counterclockwise direction C. 2.5 × 10^11 N∙m in the clockwise direction D. 2.5 × 10^11 N∙m in the counterclockwise direction

D. 2.5 × 10^11 N∙m in the counterclockwise direction

The fundamental frequency of a glass tube is measured to be f when the tube is capped at one end, but open at the other end. What is the fundamental frequency of the tube when the end cap is removed and both ends are open? A. f/4 B. f/2 C. f D. 2f

D. 2f v=f(wavelength)

What volume of the solutions tested would contain 0.080 g of oxytocin acetate? 0.25 mg/mL A. 0.020 mL B. 0.32 mL C. 20. mL D. 320 mL

D. 320 mL

Suppose the 10 mM MgCl2 solution used in Experiment 2 is replaced with a 5 mM NaCl solution. Compared to the Cl− concentration in the 10 mM MgCl2 solution, the concentration of Cl− in the 5 mM NaCl solution would be: A. greater. B. the same. C. 2 times less. D. 4 times less.

D. 4 times less.

Gadolinium becomes ionized to Gd3+ when it loses electrons from which orbital(s)? A. 4f, 4d, and 5s B. 4f C. 4f, 5d, and 6s D. 5d and 6s

D. 5d and 6s

Provided that the initial velocity in both circumstances is identical, how much longer will an astronaut's vertical jumps last on the surface of the Moon compared to on the surface of the Earth? In theory, spinal elongation would also occur in reduced-gravity environments like the surface of the Moon, where the acceleration of gravity is one-sixth that observed on the Earth's surface. A. 1/6 as long B. 1/3 as long C. 3 times longer D. 6 times longer

D. 6 times longer

A gram-negative bacterium moves along a flat surface. What is the maximum displacement of the bacterium during a 1-hour period if the bacterium moves at a constant speed of 25 µm/s? A. 1.5 × 10−3 m B. 9.0 × 10−3 m C. 1.5 × 10−2 m D. 9.0 × 10−2 m

D. 9.0 × 10−2 m

Alpha mannosidase cleaves the link between D-mannose and the hydroxyl group of serine or threonine. If researchers treated the radiolabeled Dg from each autoradiogram with alpha mannosidase, where would the radiolabels be found? A. Both 32P and 3H-ribitol would be attached to the Dg protein. B. The 32P would be attached to Dg and the 3H-ribitol to the cleaved polysaccharide. C. The 3H-ribitol would be attached to Dg and the 32P label to the cleaved polysaccharide. D. Both 32P and 3H-ribitol would be found in the cleaved polysaccharide.

D. Both 32P and 3H-ribitol would be found in the cleaved polysaccharide. According to the passage, D-mannose is the first sugar in the glycan chain and links to Dg via a glycosidic bond through the anomeric carbon of D-mannose. Mannosidase is in a class of enzymes called glycosidases that break glycosidic bonds. In this case, the broken bond is the bond that links the glycan chain to Dg through D-mannose, so the entire glycan chain will be removed from Dg as a single polysaccharide. The 32P is attached to carbon 6 of mannose, and the 3H ribitol 5-phosphate is within the cleaved polysaccharide. Therefore, both radiolabels will be removed from the Dg protein and remain on the cleaved polysaccharide. (Choice A) The polysaccharide would have to be cleaved between the ribitol 5-phosphate added by fukutin and the end of the glycan chain for both radiolabels to remain associated with Dg. Instead, mannosidase cleaves the bond that links D-mannose to Dg and removes all radiolables from Dg. (Choice B) If the polysaccharide were cleaved between mannose and ribitol 5-phosphate, this would be the correct answer. Because the bond linking the polysaccharide to Dg is cleaved, both radiolabels are removed from Dg. (Choice C) For this option to be correct, the polysaccharide would have to be cleaved in such a way as to remove mannose from Dg without removing ribitol 5-phosphate. Because mannose is the only sugar directly linked to Dg, this is not possible. Educational objective:Glycosidases break glycosidic bonds to produce monosaccharides, disaccharides, and polysaccharides. The length and composition of the newly formed saccharides depends on which glycosidic bond the glycosidase breaks.

The results presented in Table 1 suggest that the effect of growth hormone on ATP synthesis in skeletal muscle mitochondria primarily involved which metabolic pathways? A. The pyruvate dehydrogenase reaction B. Citric acid cycle reactions that produce FADH2 but not NADH C. Citric acid cycle reactions that produce NADH but not FADH2 D. Entry into the citric acid cycle from nonglycolytic pathways

D. Entry into the citric acid cycle from nonglycolytic pathways (Choice A) ATP synthesis was not significantly different in growth hormone-treated and control mitochondria when pyruvate was used as a substrate. This finding suggests that growth hormone did not affect the activity of pyruvate dehydrogenase. (Choices B and C) Table 1 shows that growth hormone affected the ATP synthesis rate when NADH entry into complex I was inhibited by rotenone and when FADH2 entry into complex II was inhibited by malate. These data suggest that levels of both NADH and FADH2 produced in citric acid cycle reactions were increased in response to growth hormone treatment. Educational objective:Different molecules can enter the citric acid cycle at different points. The cycle oxidizes these molecules to produce the electron carriers NADH and FADH2. Different entry points into the citric acid cycle may be inhibited by different molecules.

Two MRI images of the pancreas were taken at different times (Time A and Time B) in a mouse injected with Gd-daa3. Greater relaxivity was detected at Time B, suggesting all the following EXCEPT: A. Time B was soon after a meal when insulin is released. B. Fewer diaminoacetate arms were bound to Gd3+ ions at Time B than at Time A. C. The relaxivity of water molecule nuclei was increased at Time B. D. Gd3+ ions are bound to Zn2+ ions at Time B.

D. Gd3+ ions are bound to Zn2+ ions at Time B. The passage describes the mechanism by which Gd-daa3 acts as a contrast agent in the presence of zinc ions. Zinc preferentially binds to diaminoacetate (daa), not Gd3+, allowing Gd3+ to bind with water molecules and increase relaxivity. An increase in relaxivity at Time B suggests an increase in the release of zinc ions. Because Zn2+ is co-released with insulin, the image at Time B was likely taken shortly after a meal, when insulin is released from pancreatic beta cells (Choice A). When Zn2+ binds to the diaminoacetate arms on Gd-daa3, gadolinium is released. Consequently, fewer diaminoacetate arms are bound to Gd3+ at Time B because more Zn2+ ions are present at Time B than at Time A (Choice B). The released Gd3+ ions are then free to coordinate with water molecules, which increases the relaxivity of nuclei in water molecules (Choice C). This is the mechanism underlying the enhanced MRI signal. Therefore, the greater relaxivity detected at Time B suggests the conclusions in all the answer choices EXCEPT (Choice D) because Gd3+ ions bind to water molecules, not Zn2+ ions. Educational objective:Contrast agents increase MRI signals by increasing the relaxivity of water molecules. These agents can be used to monitor biological processes if their mechanisms involve biochemical changes associated with those processes.

Compared to newborns without NTDs, a newborn diagnosed with myelomeningocele would most likely have exhibited which of the following during gestation? A. Lower frequency of cell damage at critical developmental periods B. Lower cellular concentrations of free radicals C. Higher levels of anti-apoptotic transcription factors D. Higher incidence of apoptosis

D. Higher incidence of apoptosis According to the passage, NTDs can be caused by increased embryonic oxidative stress during crucial developmental periods; therefore, a child born with myelomeningocele, a severe NTD, most likely experienced oxidative stress in utero. Compared to a healthy newborn, a newborn with myelomeningocele likely had a higher incidence of apoptosis induced by oxidative stress during gestation, as well as: Higher frequency of cell damage at critical developmental periods during gestation (Choice A), because exposure to amniotic fluid can trigger the damage to developing cells that the passage states is associated with NTDs Higher cellular concentrations of free radicals during gestation (Choice B), as free radicals are the molecular species that cause the oxidative stress thought to contribute to NTDs Lower levels of anti-apoptotic transcription factors during gestation (Choice C), because such transcription factors, if present, might translocate to the nucleus and elicit responses suppressing apoptosis Educational objective:Apoptosis (programmed cell death) is crucial for normal embryonic patterning and development. Oxidative stress occurs when concentrations of reactive oxygen species are excessive, resulting in cell damage. This damage can result in abnormal apoptosis and congenital malformations.

Based on the in vitro data presented in Table 1, growth hormone infusion affects mitochondrial ATP synthesis fueled by molecules derived from which of the following precursors? I. Proteins II. Carbohydrates III. Fats A. II only B. III only C. I and II only D. I and III only

D. I and III only The question asks for the fuel molecules influenced by growth hormone based on the measurements of in vitro mitochondrial ATP synthesis. Table 1 shows that ATP synthesis during incubation with either palmitoylcarnitine (derived from fat) or glutamate (an amino acid derived from proteins) plus malate was significantly affected by growth hormone treatment. Therefore, the in vitro data suggest that the use of molecules derived from protein (Number I) and fat (Number III) for mitochondrial ATP synthesis was affected by growth hormone infusion. (Number II) Carbohydrates are converted to pyruvate before they are used for the citric acid cycle and ATP synthesis in mitochondria. Table 1 shows that pyruvate plus malate results in approximately the same level of ATP synthesis in the presence and absence of growth hormone. The difference between the two scenarios is statistically insignificant (p > 0.05). Therefore, growth hormone infusion does not affect mitochondrial metabolism of molecules derived from carbohydrates. Note that changes in the use of the citric acid cycle intermediates succinate and α-ketoglutarate for ATP synthesis could occur as part of growth hormone-induced changes in noncarbohydrate (ie, fat or protein) metabolism. Carbohydrates, proteins, and fats can all be metabolized to contribute to mitochondrial ATP synthesis. Many hormones influence metabolic processes such as mitochondrial ATP synthesis

Assume that a patient who was originally infected with a certain strain of bacteria becomes reinfected with the same strain. Of the following cell types, which is LEAST likely to produce memory cells during the initial infection that would rapidly respond to the new infection? A. B cell B. Cytotoxic T cell C. Helper T cell D. Neutrophil

D. Neutrophil Neutrophils are important components of the innate (nonspecific) immune system. Neutrophils perform many functions, including the destruction of pathogens via phagocytosis, the release of granules containing antimicrobial compounds, and the release of neutrophil extracellular traps containing DNA. Neutrophils also release cytokines (chemical signals) that regulate different immune responses.

A block made of glass and a block made of wood have the same mass. Each block is pulled by a rope at constant speed the same distance along a horizontal steel surface. Would the same amount of work be required to pull the glass block as the wood block? A. Yes, because the masses of the blocks are the same B. Yes, because the blocks are pulled along a horizontal surface C. No, because the coefficients of static friction for the block materials on steel are different D. No, because the coefficients of kinetic friction for the block materials on steel are different

D. No, because the coefficients of kinetic friction for the block materials on steel are different (Choices A and B) The amount of work required to pull the blocks is not the same. Although the blocks have the same mass and are pulled along the same horizontal surface, pulling the glass block requires more work because its coefficient of kinetic friction on steel is greater. (Choice C) The coefficients of static friction are not relevant because the blocks are sliding along a surface. Educational objective:The work required to pull an object at constant speed along a surface equals the product of the coefficient of kinetic friction, the object's weight, and the distance over which the object is pulled.

Throughout the experiment, researchers monitored the amount of magnesium that dissolved from the disks in each beaker at various time points. Which of the following is NOT a valid test for the rate of magnesium corrosion? A. Periodically measuring the volume of H2 gas evolved from the reaction. B. Periodically measuring the pH of the solution. C. Periodically measuring Mg2+ concentration in the SBF using sensitive elemental analysis techniques. D. Periodically measuring the amount of OH− produced by titrating HCl directly into each beaker.

D. Periodically measuring the amount of OH− produced by titrating HCl directly into each beaker. (Choice A) Hydrogen gas is evolved when magnesium metal corrodes into Mg(OH)2 and H2 gas. Therefore, measuring the amount of H2 gas evolved over time would give a good estimate of the corrosion rate of the magnesium disks. (Choice B) Hydroxide ions (OH−) form as the magnesium metal dissolves and corrodes, so direct measurement of the solution pH over time would be a good way to determine the rate of corrosion. (Choice C) Magnesium ions form as the metal dissolves, so direct measurement of the concentration of Mg2+ in solution over time would be a good way to determine the rate of corrosion. Educational objective:In a good experimental design, the method of measurement must not alter the experimental outcome.

Given that non-small cell lung cancer (NSCLC) increases the uptake and use of plasma branched chain amino acids (BCAAs) for synthesis of tumor proteins, which structure below represents a BCAA that would be transported at a greater rate into NSCLC cells than into healthy lung cells? A. Arginine B. Leucine C.Phenylalanine D. Threonine

b. Leucine Branched chain amino acids have branched alkyl side chains.