AAMC Quiz 4,5,6

lower , bloodstream

An aldosterone deficiency would result in ______Na+ reabsorption into the _________.

Solution: The correct answer is C. (The settling point hypothesis because, with the correct genotype, one's metabolism may allow weight to stabilize at a new level) The question asks the examinee to identify which hypothesis implies that a person can deliberately alter his or her own body maintenance weight. A and B are incorrect because the set point hypothesis suggests that the set point cannot be reset through behavior modification. C correctly states that the settling point hypothesis allows the set point to be reset through behavior modification. D is incorrect because it states that diet and exercise cannot reset the set point, which contradicts the premise of the settling point hypothesis.

Background Info: A healthy weight has been defined as a body mass index (BMI) of 25 or less. Studies suggest that genes account for about 40% of the factors that determine BMI. Stable weight is also believed to be regulated by metabolic feedback loops linking the brain, fat cells, the digestive tract, and muscles. Two hypotheses have been proposed to explain the biological basis of weight control. Set Point Hypothesis: The brain regulates body weight just as a thermostat maintains a constant room temperature. The brain adjusts metabolism and behavior to maintain a predetermined body weight. Genes also influence the set point, which can increase with age—but only to the extent dictated by inheritance. Diet and exercise cannot reset the set point over the long term. Settling Point Hypothesis: Body weight is determined by the interaction of two factors—metabolism and genes—with the environment. Depending on genotype, various metabolic feedback loops may allow weight to be stabilized at a new level. Thus, in an environment where high-calorie food is plentiful, individuals with a genetic predisposition to obesity will tend to become more overweight than those without such a predisposition. Q: Which hypothesis implies that a person can deliberately alter his or her own body maintenance weight? A. The set point hypothesis because a thermostat can be reset B. The set point hypothesis because the set point can change with age C. The settling point hypothesis because, with the correct genotype, one's metabolism may allow weight to stabilize at a new level D. The settling point hypothesis because diet and exercise cannot reset the set point

The correct answer is B(P1 → AB) In Experiment 1, researchers investigated the role of cell-to-cell communication by separating the cells of a two-cell embryo (AB and P1), and culturing them independently. The cultured AB cells produced neurons and skin, but no muscle, whereas the cultured P1 cells gave rise to all of the tissues produced by P1 cells of an intact embryo. The results of Experiment 1 indicate that the direction of signaling between the blastomeres of a two-cell embryo is P1 → AB (since AB's fate differed when isolated from P1, while P1's fate was the same regardless of whether AB was present or not).

Background Info: Experiment 1: To investigate the role of cell‑to-cell communication, researchers separated the cells of a two‑cell embryo and cultured them independently. The cultured AB cells produced neurons and skin, but no muscle, whereas the cultured P1 cells gave rise to all of the tissues produced by P1 cells of an intact embryo. Q: The results of Experiment 1 indicate that the direction of signaling between the blastomeres of a two-cell embryo is: A. AB → P1. B. P1 → AB. C.P1 → P2. D. zygote → AB.

The correct answer is A (The fate of an isolated AB cell differs from that of an AB cell in an intact embryo.) According to the passage, the AB cell from the two-cell stage of the nematode, when kept in contact with the P1 cell, goes on to produce neurons, skin, and muscle. In Experiment 1, researchers investigated the role of cell-to-cell communication by separating the cells of a two-cell embryo (AB and P1), and culturing them independently. The cultured AB cells produced neurons and skin, but no muscle, whereas the cultured P1 cells gave rise to all of the tissues produced by P1 cells of an intact embryo. The observation that the fate of an isolated AB cell is different from that of an AB cell in an intact embryo supports the hypothesis that cell-to-cell communication is involved in the determination of cell fate. C and D is irrelevant to cell fate because that occurs after the blastomere stage, during the gastrulation stage. So they are incorrect.

Background Info: Experiment 1: To investigate the role of cell‑to-cell communication, researchers separated the cells of a two‑cell embryo and cultured them independently. The cultured AB cells produced neurons and skin, but no muscle, whereas the cultured P1 cells gave rise to all of the tissues produced by P1 cells of an intact embryo. Q: Which result of Experiment 1 supports the hypothesis that cell-to-cell communication is involved in the determination of cell fate? A. The fate of an isolated AB cell differs from that of an AB cell in an intact embryo. B. The fate of an isolated P1 cell is indistinguishable from that of a P1 cell in an intact embryo. C. At the two-cell stage, isolated blastomeres can divide and differentiate. D. Several different blastomeres can produce both neurons and muscle tissue.

The correct answer is D(Protein). In Experiment 2, two-cell embryos were incubated in either cycloheximide (an inhibitor of translation) or actinomycin D (an inhibitor of transcription). The AB cells were then isolated and washed to remove inhibitors, and grown in culture. AB cells of embryos treated with cycloheximide (the translation inhibitor, which would have prevented production of proteins at the ribosomes of both AB and P1 cells) produced only neurons and skin, while AB cells of embryos treated with actinomycin D (the transcription inhibitor, which would have prevented production of mRNA) produced neurons, skin and muscle—their normal fate. These results indicate that the signaling interaction (between P1 and AB cells) at the two-cell stage probably involves protein, since proteins of the P1 cells could not have been produced to carry the necessary message(s) to the AB cells prior to isolation.

Background Info: Experiment 2: Two‑cell embryos were incubated in the presence of either cycloheximide, an inhibitor of translation, or actinomycin D, an inhibitor of transcription. The AB cells were then isolated, washed to remove the inhibitors, and grown in culture. AB cells from embryos treated with cycloheximide produced only neurons and skin, whereas those from embryos treated with actinomycin D produced neurons, skin, and muscle. Q: The results of Experiment 2 indicate that the signaling interaction at the two-cell stage probably most involves which class of macromolecules? A. DNA B. Messenger RNA C. Ribosomal RNA D. Protein

The correct answer is C. This item requires knowledge of Snell's law of refraction, which is given by C. The other options are erroneous. Snell's law is...

Background Info: Q: As the light passes from the air into the glass, it makes an angle θa in air and an angle θl in the lens material, relative to the normal at the surface. What equation relates the angles θl and θa? A. θa = θl B. 1/θa = 1/θl C. nasin θa = nlsin θl D. na/sin θa = nl/sin θl ADD BACKGROUND INFO AND ANSWER AFTER GOING OVER OPTICS CHAPTER, physics q pack passage 8

The correct answer is B(Radius of curvature of the mirror). The focal length of the mirror depends only on the radius of the curvature. Because there is no refraction, the index of refraction is irrelevant and the properties of Lens B will not affect the focal length of the mirror

Background Info: Q: Changing which of the following will change the focal length of the convex mirror in Figure 2? A. Index of refraction of the mirror B. Radius of curvature of the mirror C. Position of the lens at B D. Focal length of the lens at B ADD BACKGROUND INFO AND ANSWER AFTER GOING OVER OPTICS CHAPTER

A (is absorbed and re-emitted by the atomic structure of the optically dense medium.) is known to occur and is the reason for the slowing down of light. B (is absorbed and re-emitted by the nucleus of the material in the optically dense medium.) is incorrect because the nucleus is involved. C is incorrect because the motion of the photons is certainly not random.(bounces around randomly inside of the optically dense medium before emerging.) D is true but does not answer the question. (loses amplitude as it passes through the optically dense medium.)

Background Info: Q: Visible light travels more slowly through an optically dense medium than through a vacuum. A possible explanation for this could be that the light: A. is absorbed and re-emitted by the atomic structure of the optically dense medium. B. is absorbed and re-emitted by the nucleus of the material in the optically dense medium. C. bounces around randomly inside of the optically dense medium before emerging. D. loses amplitude as it passes through the optically dense medium. ADD BACKGROUND INFO AND ANSWER AFTER GOING OVER OPTICS CHAPTER , physics q pack passage 8

Conservation of energy implies KEinitial = PEfinal, or mv2/2 = mgh; therefore, h = v2/2g.

Background Info: Q:The mathematical expression for h is: A. mv2/2. B. v2/(2g). C. mg. D. mv. fill in background and reasoning

As the object moves up and slows down, the frequency shift is negative and falls to zero at the peak of the object's flight; as the object falls, the shift becomes increasingly positive. However, to find the magnitude of the frequency shift from 170 Hz, the negative or positive sign of the shift can be ignored and only the absolute value matters. Thus, A (It falls to zero, then increases) is the best answer.

Background Info: Q:What is the magnitude of the detected sound frequency shift from 170 Hz during the projectile flight described in the passage? A. It falls to zero, then increases. B. It is constant throughout the flight. C. It rises continuously. D. It falls continuously. PUT IN BACKGROUND INFO

The passage states that rickets is caused by insufficient vitamin D activity. Insufficient vitamin D activity would reduce the ability of the body to absorb ingested calcium from the small intestine. To maintain calcium levels in the blood plasma, parathyroid hormone would promote the breakdown of bone tissue, causing the bones to become weak. If there were a metabolic deficiency of parathyroid hormone, the body would be unable to break down bone tissue (option I), causing a higher than normal ratio of mineral to organic matter in the bones instead of a lower than normal ratio. However, if the body were unable to convert vitamin D to its active form, or if vitamin D were unable to act on its target tissue, overall vitamin D activity would be impaired (options II and III), which can lead to rickets. Thus, D is the best answer.

Background Info: Bone tissue contains about 99% of the body's calcium. Throughout life, bone is continuously resorbed and reformed—a process intimately related to the maintenance of an adequate level of calcium in the blood plasma. Some important agents that affect this process are: Vitamin D, which in its activated form functions like a hormone. This nonpolar compound acts on the small intestine to stimulate absorption of calcium and also on bone tissue to enhance the effect of parathyroid hormone. It can be obtained from the diet or by the action of ultraviolet light on the skin. Common disease that affects bone metabolism: Rickets affects children, causing inadequate mineralization of new bone matrix. The ratio of mineral to organic matter is lower than normal. This abnormality can result in distortion of the bones, especially long bones, leading to bowed legs. It is caused by insufficient vitamin D activity. Q: Which of the following conditions could produce rickets? I. Metabolic deficiency of parathyroid hormone II. Impairment of conversion of vitamin D to its active form III. Inability of the active form of vitamin D to act on its target tissue A. I only B. I and II only C. I and III only D. II and III only

The correct answer is C.(copies of the plasmid containing the genes were not equally distributed to the new daughter cells.) The passage states that the plasmids are not attached to the cell membrane and thus may not be equally distributed among daughter cells in asexual reproduction. The passage also informs a reader that the building of a cytoplasmic bridge (sex pilus) is directed by plasmid genes rather than by chromosomal genes. Therefore neither chromosomal replication nor chromosomal segregation is directly pertinent to sex-pilus construction. Bacteria generally lack membrane enclosed organelles such as lysosomes. Which is why D (one copy of the plasmid containing the genes was digested by a bacterial lysosome) is incorrect. The best explanation is that in some of the daughter cells plasmids were absent due to chance distribution.

Background Info: E. coli undergo asexual reproduction. Copies of the bacterial chromosome, which are attached to the cell membrane, move to opposite ends of the cell as the cell divides. Plasmids (small, circular DNA molecules within the bacterial cell) contain various nonessential genes-including genes that confer resistance to specific antibiotics. Because plasmids are not attached to the cell membrane, they may not be equally distributed among daughter cells. Bacteria also carry out a form of "sexual" reproduction called conjugation. Conjugation occurs when the plasmid genes of one bacterium direct the building of a cytoplasmic bridge, or sexpilus, between that organism and a bacterium lacking such genes. Following plasmid DNA replication, a copy of the plasmid DNA moves unidirectionally through the pilus, bestowing new genes upon the recipient cell. Conjugation may occur between members of the same, or different, bacterial species. Q: A bacterium containing genes for sex pilus construction gave rise to a daughter cell lacking these genes. This most probably occurred because: A. the bacterial chromosome was not completely replicated prior to cell division. B. the cell membrane failed to move the replicated chromosomes apart. C. copies of the plasmid containing the genes were not equally distributed to the new daughter cells. D. one copy of the plasmid containing the genes was digested by a bacterial lysosome.

The correct answer is C ( E. coli entering the abdominal cavity from the appendix.) The passage indicates that the ruptured appendix preceded infection with M. tuberculosis, so the infective agent would have had to be the E. coli bacteria, a species common to the colon. Therefore we can eliminate choice A. This item further relies on anatomical knowledge of the gastrointestinal tract for an answer. The passage states that E. coli are found within the colon, where they function in digestion and vitamin production. The appendix is continuous with the colon so that bacteria can move between these two structures; a ruptured appendix would allow E. coli into the abdominal cavity, which is not normal. therefore C is correct. B and D are incorrect because the passage said E.coli is present in the colon, which is connected to the appendix, so having E coli go from the colon to the appendix in either direction is normal.

Background Info: Escherichia coli, a bacterial species that inhabits the colon, benefits humans by breaking down foods for which we lack digestive enzymes and by producing vitamins such as B12 and K. A man was hospitalized with a ruptured appendix and received massive doses of the antibiotics ampicillin and kanamycin over a period of weeks. Shortly after leaving the hospital, he contracted Mycobacterium tuberculosis, the pathogen that causes tuberculosis, from a coworker. He was again hospitalized and treated with large doses of first ampicillin and then kanamycin, with no effect. Recovery was rapid when a third antibiotic, streptomycin, was administered. Q:The patient's ruptured appendix required treatment with antibiotics because he had a bacterial infection caused by: A. M. tuberculosis. B. E. coli entering the colon from the appendix. C. E. coli entering the abdominal cavity from the appendix. D. E. coli entering the appendix from the colon.

The correct answer is D (chance mutations in a few E. coli before the treatment made these cells and their descendants antibiotic-resistant) Mutations which make cells drug resistant are very rare, but the few drug-resistant bacteria that do develop such mutations flourish when the nonresistant cells are killed by antibiotics. Antibiotics are unlikely to have been the source of the mutations; Bacteria develop "immune" reactions to antibiotics. Immune reactions may be caused by certain bacteria. Faulty definitons. So the answer cannot be choice B. Any resistance of the patient's own colon cells to antibiotics is irrelevant to the resistance of bacteria to antibiotic. So we can eliminate choice C.

Background Info: Escherichia coli, a bacterial species that inhabits the colon, benefits humans by breaking down foods for which we lack digestive enzymes and by producing vitamins such as B12 and K. About every 20 min, E. coli undergo asexual reproduction. Copies of the bacterial chromosome, which are attached to the cell membrane, move to opposite ends of the cell as the cell divides. Plasmids (small, circular DNA molecules within the bacterial cell) contain various nonessential genes-including genes that confer resistance to specific antibiotics. Because plasmids are not attached to the cell membrane, they may not be equally distributed among daughter cells. A man was hospitalized with a ruptured appendix and received massive doses of the antibiotics ampicillin and kanamycin over a period of weeks. Shortly after leaving the hospital, he contracted Mycobacterium tuberculosis, the pathogen that causes tuberculosis, from a coworker. He was again hospitalized and treated with large doses of first ampicillin and then kanamycin, with no effect. Recovery was rapid when a third antibiotic, streptomycin, was administered. Q:At the end of his initial hospital stay, a few E. coli cells remained in the patient's colon, even though he was taking antibiotics. These cells were most likely present because: A. the antibiotics caused drug-resistance mutations to occur in the bacterial DNA. B. the bacteria in the patient developed an immune reaction to the antibiotics. C. the patient's colon cells became increasingly resistant to the antibiotics during his hospitalization. D. chance mutations in a few E. coli before the treatment made these cells and their descendants antibiotic-resistant.

The strain of M. tuberculosis in the coworkers was killed by both ampicillin and kanamycin, indicating that this strain did not carry a plasmid gene that made it resistant to these two antibiotics. However, once the M. tuberculosis coexisted in the patient with other antibiotic-resistant bacteria, the M. tuberculosis could survive despite treatment with large doses of these two antibiotics. The antibiotic does not induce mutations; the few cells that are already resistant flourish in the absence of the nonresistant bacteria. Therefore answer choice B is incorrect. The M. tuberculosis did not adapt to its new environment by modifying its metabolism; rather, there was a strain with a metabolic capability that was not compromised by the antibiotic. Therefore answer choice D is incorrect. The passage indicates that conjugation may occur between members of different bacterial species. Therefore, it is most likely that these M. tuberculosis bacteria underwent conjugation with resistant cells, answer choice A.

Background Info: Escherichia coli, a bacterial species that inhabits the colon, benefits humans by breaking down foods for which we lack digestive enzymes and by producing vitamins such as B12 and K. About every 20 min, E. coli undergo asexual reproduction. Copies of the bacterial chromosome, which are attached to the cell membrane, move to opposite ends of the cell as the cell divides. Plasmids (small, circular DNA molecules within the bacterial cell) contain various nonessential genes-including genes that confer resistance to specific antibiotics. Because plasmids are not attached to the cell membrane, they may not be equally distributed among daughter cells. Bacteria also carry out a form of "sexual" reproduction called conjugation. Conjugation occurs when the plasmid genes of one bacterium direct the building of a cytoplasmic bridge, or sexpilus, between that organism and a bacterium lacking such genes. Following plasmid DNA replication, a copy of the plasmid DNA moves unidirectionally through the pilus, bestowing new genes upon the recipient cell. Conjugation may occur between members of the same, or different, bacterial species. A man was hospitalized with a ruptured appendix and received massive doses of the antibiotics ampicillin and kanamycin over a period of weeks. Shortly after leaving the hospital, he contracted Mycobacterium tuberculosis, the pathogen that causes tuberculosis, from a coworker. He was again hospitalized and treated with large doses of first ampicillin and then kanamycin, with no effect. Recovery was rapid when a third antibiotic, streptomycin, was administered. Q: Tests performed on the M. tuberculosis strain infecting the patient's coworker indicated that the strain was susceptible to both ampicillin and kanamycin, and the coworker was successfully treated. The M. tuberculosis most likely survived in the patient because it had: A. undergone conjugation with cells of resistant E. coli. B. undergone an antibiotic-induced mutation that conferred antibiotic resistance. C. reproduced more rapidly than the strain in the coworker. D. adapted to its new environment by modifying its metabolism.

The correct answer is psychophysical testing methods (such as the Method of Limits) directly assess our perception of stimuli in relation to their true physical properties. The nature of the illusions discussed in the passage suggests that individuals would be prone to either over- or under-state the size of items like those in Panel A.

Background Info: Humans experience a variety of visual illusions, highlighting that their perception of the world can differ from how the world really is. For example, the introduction of other stimuli around the perimeter of two central dots, as shown in Panel A, can lead to the misperception of the central dot in the left half of the figure as being smaller than the central dot in the right half of the figure. Q:Which methodology will best show how perceptual illusions impact our judgments of the nature of stimuli such as those in Panel A? A. Partial report technique B. Word association testing C. Psychophysical discrimination testing D. Operational span testing see if definitions of these are in book and then fill in answer explanation w definitions of each answer choice

The correct answer is A (Amoeboid movement stops upon exposure to cytochalasins) The passage proposes that force is generated as a microfilament elongates and pushes against a structure such as the plasma membrane. This is representative of how an amoeba moves. Cytochalasins are drugs that inhibit the growth of microfilaments. Therefore, if amoeboid movement stops upon exposure to cytochalasins, microfilaments and their ability to elongate are both implicated as being necessary to generate the force for movement in an amoeba.

Background Info: Microfilaments were first identified as the actin-containing thin filaments of muscle cells. All eukaryotic cells are thought to contain microfilaments. Researchers suspect that microfilaments can generate force, even in the absence of myosin, by elongating and pushing against a structure such as the plasma membrane. A microfilament has two ends, each of which can either gain or lose actin subunits. During microfilament growth, the plus (+) end of the microfilament grows faster than the minus (-) end Researchers believe that regulation of microfilament growth helps determine the shape of cells and the stability of their microfilaments. Within the cell, the addition or loss of subunits at each end of a microfilament can be controlled by capping proteins, which bind selectively to one of the ends. Some natural poisons also affect microfilament metabolism: the cytochalasins bind to the + end of a microfilament and prevent the addition of actin subunits to that end. Q:The theory of force generation proposed in the passage is best supported by which of the following observations about Amoeba locomotion? A. Amoeboid movement stops upon exposure to cytochalasins. B. Amoeboid movement cannot occur if mitosis is blocked. C. Moving Amoeba cells produce more troponin than do stationary ones. D. The rate of movement is inversely proportional to the viscosity of the medium in which the Amoeba moves.

The answer to this question is B (context effects), because the phenomena discussed in the passage reflect how both the context in which stimuli are presented and the processes of perceptual organization contribute to how people perceive those stimuli (and also that the context can establish the way in which stimuli are organized).

Background Info: Psychologists have investigated how humans organize their perceptual experiences in ways that allow them to quickly understand the meaning of those experiences. However, in many cases, people's perceptions of objects, scenes, and events in the world differ from the stimulus that is registered by the sensory receptors. Humans experience a variety of visual illusions, highlighting that their perception of the world can differ from how the world really is. For example, the introduction of other stimuli around the perimeter of two central dots, as shown in Panel A, can lead to the misperception of the central dot in the left half of the figure as being smaller than the central dot in the right half of the figure. Q: The types of perceptual experiences, illustrated in Panel A, that are not relevant to a person's judgment and decision making processes, but can still have a biasing impact on those processes, are best described as: A. recall cues. B. context effects. C. feature detectors. D. practice effects. fill in proper answer, and add in definitions after looking in book/NC

Solution: The correct answer is A(Extinction). The passage states that if none of the Xi genotypes are selected against, then the Xi chromosome will increase to 100%. If all the males are XiY, then only females will be produced and the population will become extinct. Answer A is correct. The population will not be stable, but will increasingly become female, so options B and C are incorrect. Option D is also incorrect, not only because it assumes a stable population size, but also because it supposes that some individuals will produce an excess of males.

Background Info: Several species of Drosophila have X-linked e and f genes that affect the sex ratios of individuals' offspring. However, the genes only affect sex ratios if they are brought close together by an inversion of one arm of the X chromosome. An XsY male is standard: he sires equal numbers of sons and daughters. An XiY male expresses the sex ratio trait: he sires only daughters. Total reproductive output is not affected; XsY males and XiY males sire equal numbers of offspring. If none of the Xi-bearing genotypes (XiY, XiXi, or XiXs) is selected against, then the frequency of Xi is expected to increase to 100%, unless other genes act to suppress expression of e and f. Q: If all genotypes are equally fit and if there are no genetic modifiers of the sex ratio trait, what will be the ultimate fate of a population in which 50% of the X chromosomes are currently Xi and 50% are Xs? A. Extinction B. Stable population size, with a predominance of females C. Stable population size, with all individuals producing a 50:50 sex ratio D. Stable population size, with some individuals producing an excess of females and some producing an excess of males

When the level of calcium in the blood plasma is low, the body responds by mobilizing stores of calcium from the bones via the activity of parathyroid hormone. Parathyroid hormone will increase the number of osteoclasts, which break down bone cells. Therefore, one would expect an increase in both parathyroid hormone and osteoclast activity in order to increase the level of calcium in the blood plasma (options I and II). However, vitamin C (option III) promotes bone formation, a process that would further lower the calcium level in the plasma. Thus, B (osteoclast activity and parathyroid hormone) is the best answer.

Background Info: Parathyroid hormone, which acts on bone tissue to encourage the formation and activity of osteoclasts (which break down bone cells) and to impair new bone formation. Vitamin C, which is required for the synthesis of bone matrix and is therefore needed for bone formation . Q: A low level of calcium in the plasma will trigger an increase of: I. osteoclast activity. II. parathyroid hormone. III. vitamin C. A. I only B. I and II only C. I and III only D. II and III only

The correct answer is B (the resistance R must be continually decreased). As the capacitor discharges the voltage across it falls, thus to maintain a constant current, R must be proportionately reduced. This is so from Ohm's law, I = V/R. To keep I fixed, R must fall with V.

Background Info: A capacitor is a device that stores charge. The voltage V across a capacitor and the charge q on the capacitor are related by q = CV, where C is the capacitance measured in farads, F (1.0 F = 1.0 coulomb per volt). Q: To keep the current constant during the discharge cycle: A. the resistance R must be continually increased. B. the resistance R must be continually decreased. C. the resistance r must be continually increased. D. the resistance r must equal R. ADD BETTER ANSWER ONCE YOU UNDERSTAND/GO OVER CHAPTER MORE

The answer to this question is D (Verbal inputs to the left ear, which first go to the auditory cortex in the right hemisphere, must be processed by the language areas of the left hemisphere) because, in all but a relatively small number of people, the brain areas that process linguistic information (both production and comprehension) are lateralized in the left hemisphere.

Background Info: A modern air force wanted a standardized test to identify which flight cadets were most likely to acquire proficiency in flying high-performance aircraft. Candidates who failed were either unable to appropriately divide attention among concurrent activities or concurrent signals, or too slow to redirect attention to crucial signals arriving on unattended channels. Therefore, researchers needed a good measure of how quickly and effectively a person can redirect attention. Researchers used a task in which participants were required to focus attention on one of two different messages, presented simultaneously, one to each ear. All stimuli were presented on earphones at a subjectively comfortable loudness level. The study consisted of two phases. In Phase 1, each participant was instructed by a 2500-Hz tone to attend to the right ear or by a 250-Hz tone to attend to the left ear. The indicator tone was presented to the relevant ear. Then either two or four digits were presented to the relevant ear, interspersed in a stream of bisyllabic words. Six digits were presented to the irrelevant ear. Digits were never presented simultaneously in Phase 1. Participants were instructed to repeat immediately any digits they heard in the attended ear. Most errors in Phase 1 were omissions of relevant items, and there was a greater likelihood of omissions when four digits had been presented than with only two. There was no clear difference between the left and right ears in omission errors. Most intrusions of digits from the unattended ear occurred when participants were instructed to attend to the left ear. Q:What is the best explanation for the difference in the number of intrusion errors in Phase 1? A. Verbal inputs to the right ear, which first go to the auditory cortex in the right hemisphere, must be processed by the language areas of the left hemisphere. B. Verbal inputs to the right ear, which first go to the auditory cortex in the left hemisphere, must be processed by the language areas of the right hemisphere. C. Verbal inputs to the left ear, which first go to the auditory cortex in the left hemisphere, must be processed by the language areas of the right hemisphere. D. Verbal inputs to the left ear, which first go to the auditory cortex in the right hemisphere, must be processed by the language areas of the left hemisphere GO OVER PSYCH HEARING TO REALLY HAVE GOOD ANSWER

The answer to this question is A (Yes, because cadets who completed training on more advanced equipment had performed significantly better on the task than those who dropped out of the training earlier)because the main requirement was for a test of how quickly a flight cadet can redirect attention, and because the selective listening task fulfilled that requirement and allowed the "modern air force" to test this ability quickly and inexpensively.

Background Info: A modern air force wanted a standardized test to identify which flight cadets were most likely to acquire proficiency in flying high-performance aircraft. Candidates who failed were either unable to appropriately divide attention among concurrent activities or concurrent signals, or too slow to redirect attention to crucial signals arriving on unattended channels. Therefore, researchers needed a good measure of how quickly and effectively a person can redirect attention. Researchers used a task in which participants were required to focus attention on one of two different messages, presented simultaneously, one to each ear. All stimuli were presented on earphones at a subjectively comfortable loudness level. The study consisted of two phases. There was a strong negative correlation between the number of errors in Phase 2 and the successful completion of pilot training. Redirecting attention from the left ear to the right ear was easier than vice versa. The number of errors in Phase 2 had low correlations with all other tests used for pilot selection. Q: Did the task used in the study fulfill the original need described in the first paragraph? A. Yes, because cadets who completed training on more advanced equipment had performed significantly better on the task than those who dropped out of the training earlier B. No, because performance in Phase 2 of the task did not predict performance on other tests used in pilot selection C. Yes, because cadets' ability to divide attention between simultaneous inputs was predicted by their performance in Phase 1 D. No, because cadets' performance in Phase 1 did not predict their ability to redirect attention in Phase 2 FILL IN MORE BACKGROUND, NOT ENOUGH INFO

Activated vitamin D acts on the small intestine to stimulate the absorption of calcium into the bloodstream. The inclusion of vitamin D in calcium supplements would ensure that vitamin D is present in the body to help promote this absorption. Thus, B (The activated form of vitamin D stimulates the absorption of calcium into the blood) is the best answer.

Background Info: Vitamin D, which in its activated form functions like a hormone. This nonpolar compound acts on the small intestine to stimulate absorption of calcium and also on bone tissue to enhance the effect of parathyroid hormone. It can be obtained from the diet or by the action of ultraviolet light on the skin. Q: Why do calcium supplements often include vitamin D? A. Vitamin D is needed to prevent rickets. B. The activated form of vitamin D stimulates the absorption of calcium into the blood. C. The activated form of vitamin D enhances the action of calcitonin. D. The activated form of vitamin D enhances the uptake of calcium by bone tissue.

In a one-dimensional problem such as this, work is force times distance. W=FdcosΘ Hence, W=(19.6 N)(0.5 m)(1) = 9.8 J

Background Info: A student performs an experiment to determine the distance moved and velocity of a projectile as a function of time. A 1.00-kg object is initially at rest. The student applies a force of 19.6 N through a distance of 0.50 m to propel the object straight upward. It has an initial speed v and reaches a peak height h above the launch point. Q:How much work is done in launching the object? A. 9.80 kg m/s2 B. 19.6 J C. 9.8 J D. 19.6 N

B (the study's measure of attention capabilities reduces the predictive validity of the pilot selection test battery.) is correct. because the flight candidates who had successfully completed two years of training (and had thus reached the point where they were flying more advanced aircraft) have attention deployment capabilities that predict their success in meeting with increasingly complex information processing requirements. Thus, the negative correlation means that the cadets with fewer errors in the selective listening task scored higher on all the other selection criteria

Background Info: Add Q: The observed correlations mentioned in the passage suggest all of the following EXCEPT: A. the study's measure of attention capabilities enhances the predictive validity of the pilot selection test battery. B. the study's measure of attention capabilities reduces the predictive validity of the pilot selection test battery. C. flight cadets who perform better on the task are more likely to complete training on more advanced equipment than those whose performance is poorer. D. flight cadets who make more errors in Phase 2 are less likely to complete training on more advanced equipment. Add background Info and sean's reasoning after session 4/23

The correct answer is C(F⁻). According to the passage, the ionic compound in Solution A was completely soluble in water. The information in Table 1 shows that the only Ag+ salt studied that is soluble in water is AgF. Thus, C is the best answer.

Background Info: Aqueous Solutions A, B, and C each contained a different ionic compound. The three ionic compounds were completely soluble in water. The cation in Solutions A, B, and C was either Ag+, Ca 2+, Cu 2+, or Fe 3+. The anion in Solutions A, B, and C was either F−, Cl−, CrO 2−4, or S 2−. Solution A was colorless, Solution B was yellow, and Solution C was blue. It was determined that Solution B contained CaCrO 4 and that Solution C contained CuCl 2. Q: If Solution A contains Ag+, the anion component must be: A. CrO₄²⁻ B. Cl⁻. C. F⁻. D. S²⁻.

Assuming that the lizards use the UV-reflectivity of the dewlap primarily as a means of intraspecies communication, species E would most likely be least affected by a mutation that eliminated UV photoreceptors. Its dewlaps are the least UV-reflective of the five lizard species, which indicates that species E is least likely to rely heavily upon this form of communication in the first place. Thus D (Species E) is the correct answer.

Background Info: During a study on visual communication in five species of lizards in the genus Anolis, investigators discovered ultraviolet (UV) photoreceptors in the eyes of all five species. Additional studies were done to determine what role, if any, the ability to detect UV light plays in intraspecific communication. The five species are closely related and live in Puerto Rico. Three species (A, B, and C) live in open unshaded fields, and the other two species (D and E) live in the understory of a closed canopy forest. Male lizards have a dewlap, a large fold of skin under the throat that they can fan out like a flag. Flashing the dewlap plays an important role in lizard communication such as territorial displays, warning signals, and courtship. Q: If these lizards use UV light in communication, a mutation that eliminated UV photoreceptors would probably cause the LEAST disadvantage to: A. species A. B. species B. C. species D. D. species E.

Based on the information presented, the gene encoding UV-reflectance pigment could be on a sex chromosome or an autosome. The fact that the pigment is expressed in the dewlap, a structure found only in males, is not sufficient to eliminate any chromosome as the location of this gene. Thus, D (could be on a sex chromosome or on an autosome.) is the best answer.

Background Info: During a study on visual communication in five species of lizards in the genus Anolis. These lizards have a dewlap, a large fold of skin under the throat that they can fan out like a flag. Flashing the dewlap plays an important role in lizard communication such as territorial displays, warning signals, and courtship. Q: If Anolis lizards have X-Y chromosomal sex determination, the locus of a gene for the UV reflectance pigment: A. must be on the X chromosome. B. must be on the Y chromosome. C. must be on an autosome. D. could be on a sex chromosome or on an autosome.

Solution: The correct answer is A. According to Equation 2, the fertilizer ammonium monohydrogen phosphate, (NH4)2HPO4, in the presence of soil moisture, will dissolve and ionize to release hydroxide ions (OH-). The effect of increasing the amount of the reactant water should be to force the equilibrium to the right, an example of LeChatelier's Principle, with an increase in the amount of OH- released. This would represent an increased degree of ionization. Only response A includes both a release of more OH- and an increase in the degree of ionization. Thus, answer choice A is the best answer

Background Info: Fertilizers are substances added to the soil to improve plant growth and productivity. Commonly used fertilizers contain ionic salts of N, P, and K. Because of the complexity of soil chemistry, it is often difficult to predict how soil pH will be affected by the addition of a fertilizer. Q:After adding the fertilizer shown in Equation 2 to the soil, what will be the most likely effect of excessively moist soil conditions? A. The degree of ionization will be greater, releasing more OH-. B. The degree of ionization will be greater, consuming more OH-. C. The degree of ionization will be reduced, releasing more OH-. D. The degree of ionization will be reduced, consuming more OH-.

The answer to this question is D because the proximal stimulus is the stimulus registered by the sensory receptors (e.g., the pattern of light falling on the retina), which is referred to in the first paragraph.

Background Info: Psychologists have investigated how humans organize their perceptual experiences in ways that allow them to quickly understand the meaning of those experiences. However, in many cases, people's perceptions of objects, scenes, and events in the world differ from the stimulus that is registered by the sensory receptors. Q: The technical term for the type of stimulus registered by the sensory receptors, mentioned in the first paragraph, is: A. incentive stimulus. B. sensory stimulus. C. distal stimulus. D. proximal stimulus.

Solution: The correct answer is A. Figure 1 shows that at free actin concentrations less than 1µM, both the + and the - ends of the microfilament experience a net loss of actin subunit

Background Info:A microfilament has two ends, each of which can either gain or lose actin subunits. During microfilament growth, the plus (+) end of the microfilament grows faster than the minus (-) end. At a particular concentration of actin subunits, the rate of subunit addition (polymerization) is exactly balanced by the rate of subunit loss (depolymerization). This critical concentration is different for the + and - ends of the microfilament (Figure 1) Q:Based on Figure 1, at what free actin subunit concentration (or range of concentrations) will both the + and - ends of the microfilament experience a net loss of subunits? A. At any concentration below B. 1 µM Exactly at 1 µM C. At any concentration above 1 µM D. Only between 1 µM and 4 µM

Figure 1 shows that at a free actin concentration of 1.5µM, the rate at which actin subunits are added to the + end of the microfilament is equal to the rate at which actin filaments are removed from the - end. This fits the definition of treadmilling. (Meaning there is the same amount above/below 0 point of addition/loss)

Background Info:A microfilament has two ends, each of which can either gain or lose actin subunits. During microfilament growth, the plus (+) end of the microfilament grows faster than the minus (-) end. At a particular concentration of actin subunits, the rate of subunit addition (polymerization) is exactly balanced by the rate of subunit loss (depolymerization). This critical concentration is different for the + and - ends of the microfilament (Figure 1). When the rate of subunit addition at the + end equals the rate of subunit loss at the - end, the microfilament is undergoing a process called treadmilling. Q: Based on Figure 1, at what free actin subunit concentration (or range of concentrations) will the microfilament treadmill? A. 0.25 µM B. 1.0 µM C. 1.5 µM D. Any concentration between 1.0 µM and 4.0 µM

The correct answer is C(At any concentration greater than 1 µM). Figure 1 shows that at free actin concentrations greater than 1µM, actin is added to the + end of a microfilament. At concentrations lower than 1µM, the + end loses actin subunits. The - end of the microfilament does not begin to add actin until the free actin concentration is greater than 5.5µM. The rate at which actin is added to the + end is greater than the rate it is added to the - end, implying that at any concentration greater than 1µM, the + end of the microfilament grows faster than the - end.

Background Info:Microfilaments were first identified as the actin-containing thin filaments of muscle cells. All eukaryotic cells are thought to contain microfilaments. Researchers suspect that microfilaments can generate force, even in the absence of myosin, by elongating and pushing against a structure such as the plasma membrane. A microfilament has two ends, each of which can either gain or lose actin subunits. During microfilament growth, the plus (+) end of the microfilament grows faster than the minus (-) end. At a particular concentration of actin subunits, the rate of subunit addition (polymerization) is exactly balanced by the rate of subunit loss (depolymerization). This critical concentration is different for the + and - ends of the microfilament (Figure 1). When the rate of subunit addition at the + end equals the rate of subunit loss at the - end, the microfilament is undergoing a process called treadmilling. Q:At what concentration of free actin will the + end of the microfilament grow faster than the - end? A. Exactly at 1 µM B. Only between 1 µM and 4 µM C. At any concentration greater than 1 µM D. At any concentration

The answer to this question is A (Glucose metabolism in the fetal brain increased.)because cocaine is a stimulant it would have a physiological effect similar to stress and hence glucose metabolism is expected to increase. B (Imaging studies showed that cocaine entered the fetal circulation.) is incorrect because transport of cocaine to the fetal circulation does not indicate whether the drug is pharmacologically active in the fetus. C (The fetus had an increase in tolerance to pain )is incorrect because the fetus may not be mature enough to feel pain. In any case, pain relief is associated with heroin, not cocaine. D(Imaging studies showed increased internalization of dopamine receptors.) is incorrect because, if the receptors were internalized, dopamine could not be pharmacologically active. Neurotransmitters bind to receptors on the outside, not inside, of neurons.

Background Info:Research using animal models indicates that cocaine exposure during pregnancy: (1) reduces oxygen levels and blood flow to the uterus, (2) interferes with neural proliferation and synaptogenesis, (3) alters functioning of three neurotransmitters (dopamine, serotonin, and norepinephrine) in the offspring during gestation, and thereafter, (4) impairs post-weaning discrimination learning and attention skills. Q: Researchers hypothesized that cocaine is pharmacologically active in the primate fetal brain when pregnant primates are administered cocaine at doses typically used by human drug users. Which experimental observation best supports their hypothesis? A. Glucose metabolism in the fetal brain increased. B. Imaging studies showed that cocaine entered the fetal circulation. C. The fetus had an increase in tolerance to pain. D. Imaging studies showed increased internalization of dopamine receptors.

The answer to this question is C because the hypothalamus is concerned largely with the maintenance of homeostatic equilibrium. Damage restricted to the hypothalamus would be unlikely to interfere with discrimination learning or attention skills A. is incorrect because The frontal lobe is involved in tasks of paragraph 1. B. (Hippocampus) is incorrect because it is involved in learning and memory, which is listed as (4) in paragraph 1 D. (Thalamus) is incorrect because it is involved in..

Background Info:Research using animal models indicates that cocaine exposure during pregnancy: (1) reduces oxygen levels and blood flow to the uterus, (2) interferes with neural proliferation and synaptogenesis, (3) alters functioning of three neurotransmitters (dopamine, serotonin, and norepinephrine) in the offspring during gestation, and thereafter, (4) impairs post-weaning discrimination learning and attention skills. Q:In mammals, which brain area is LEAST involved in the abilities mentioned in the first paragraph? A. Frontal lobe B. Hippocampus C. Hypothalamus D. Thalamus finish explanation after reviewing psych ch 1.

Because the projectile has stopped moving, there is no Doppler shift. The wavelength is λ = vsound/f λ= 340/170 = 2 m. (C)

Background Info:object is initially at rest. The student applies a force of 19.6 N through a distance of 0.50 m to propel the object straight upward. It has an initial speed v and reaches a peak height h above the launch point. The projectile contains a small speaker that emits a sound at a frequency of 170 Hz. The sound is detected by a microphone and recording device located on the ground directly beneath the vertical trajectory of the projectile. The recorded data are used to compute the velocity of the object using the Doppler effect. (speed of sound is 340 m/s) Q:The wavelength of the detected sound when the projectile is at h is: A. 0.50 m. B. 1.00 m. C. 2.00 m. D. 170 m.

Firstly it is important to note that in the question, it says that glycerol reacts with 3 DIFFERENT f.a.'s, so when triacyglycerol is formed, it has 3 different ester R groups. It is NOT the triacylglycerol shown in the image. Also if glycerol is reacting When glycerol reacts with three different fatty acids, only carbon 2 in the resulting triacylglycerol is attached to four different groups. The other two carbons have 2 H groups making it achiral. Thus, B is the best answer.

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correct answer is A( Eq/m). The force on the charge is qE and force is also ma. Setting qE = ma and solving for acceleration a yields: a=Eq/m

Q: A charged particle with a mass of m and a charge of q is injected midway between the plates of a capacitor that has a uniform electric field of E. What is the acceleration of this particle due to the electric field? A. Eq/m B. Em/q C. mq/E D. Emq

The correct answer is B (It accelerates toward Q) We know that a charge is going to move towards a positive charged particle (Q), so we can automatically eliminate A and C. The Coulomb force between the negative and positive charges is attractive. That force accelerates the negatively-charged particle toward the positively charged particle. -It doesn't have to move with constant speed, eliminating D.

Q: A small negatively charged particle is placed near a fixed positively charged particle (Q). Which of the following describes the motion of the negatively charged particle? A. It accelerates away from Q. B. It accelerates toward Q. C. It moves with constant speed away from Q. D. It moves with constant speed toward Q.

The floating ice cube implies that its weight is balanced by the buoyant force on it W(ice)= mg = r(fluid)*V(submerged)*g we can thus remove g as a variable from the equation, so we don't have to consider that g might be different on the moon. (because g isn't a variable we consider!!) Note that both the weight and the buoyant force are proportional to g, making the numerical value of g irrelevant to the volume of the ice cube that is submerged. Thus, B is the best answer.

Q: An astronaut on Earth notes that in her soft drink an ice cube floats with 9/10 of its volume submerged. If she were instead in a lunar module parked on the Moon where the gravitation force is 1/6 that of Earth, the ice in the same soft drink would float: A. with more than 9/10 submerged. B. with 9/10 submerged. C. with 6/10 submerged. D. totally submerged.

The correct answer is B (It will be approximately equal.). The near equality of the photon energy and the work function means that little initial kinetic energy will be left for the electron. This initial kinetic energy is small compared to the 50 eV it will gain from the potential difference between electrodes.

Q: An electron is ejected from the cathode by a photon with an energy slightly greater than the work function of the cathode. How will the final kinetic energy of the electron upon reaching the anode compare to its initial potential energy immediately after it has been ejected? A. It will be 2 times as large. B. It will be approximately equal. C. It will be 1/4 as large. D. It will be 0. ask sean if he has content needed for this flashcard

The question asks about the effect of an increase in the level of albumin, one of the major plasma proteins. Because albumin has nothing to do with the immune response, answer choice A is incorrect. The plasma proteins can not cross the walls of blood vessels, but water molecules can. The wall of the artery acts as a semipermeable membrane setting up the conditions needed for osmosis to occur. An increase in plasma albumin will upset the osmotic balance because the blood will become hypertonic with respect to the tissue. Water will have to flow into the bloodstream to reestablish equilibrium. One of the causes of edema, increased fluid in body tissues, is a decrease in the plasma protein level. This occurs, for instance, in starvation when the body is forced to use its albumin as an energy source. An increase in the plasma protein level would have the opposite effect: fluid would enter the bloodstream (answer D-influx of tissue fluid to the bloodstream.).

Q: An intravenous infusion causes a sharp rise in the serum level of albumin (the major osmoregulatory protein in the blood). This will most likely cause an: A. increase in the immune response. B. increase in tissue albumin levels. C. outflow of blood fluid to the tissues. D. influx of tissue fluid to the bloodstream.

When in a series, capicators C₁ and C₂ are added by the inverse rule: 1/Ctot= 1/C₁ +1/C₂ Therefore, if C₁=C₂= C, then, 1/Ceq= 1/C+ 1/C = 2/C, 1/Ceq= 2/C, Ceq=C/2 Therefore, A (1/2 has great) is the correct answer.

Q: Another capacitor, identical to the original, is added in series to the circuit described in the passage. Compared to the original circuit, the equivalent capacitance of the new circuit is: A. 1/2 as great. B. the same. C. 2 times as great. D. 4 times as great.

Based on the passage, there was no reaction at 100K. The reaction rate at 300K proceeded faster than of that at 200K. If the reaction is performed at 400K, the rate will most likely be faster than of that at 300K. Rates of chemical reactions always increase with increasing temperature, since a greater number of molecules possess enough energy to overcome the energetic barrier to reaction - the activation energy - at higher temperature. Thus, A (greater) is the best answer.

Q: Compared to the rate of the reaction between Reactants X and Y at 300 K, the rate of the reaction at 400 K would be: A. greater. B. the same. C. less. D. unrelated to the temperature. PUT BACKGROUND ON THIS

Calcium undergoes the following reaction with water: Ca(s) + 2H2O(l)→ Ca²⁺(aq) + 2OH⁻(aq) + H2(g). The gas produced was H2. Therefore we can eliminate C and D. If 0.40 g calcium reacted, then the number of moles of calcium reacted was equal to (0.40 g)/(40.1 g/mol) = 0.01 mol. The amount of H2 formed was also 0.01 mol. (B/c same coefficients) PV=nRT, (note 27°C= 300L) V= [(.01mol)(.08)(300K)]/1 atm =.246 L= ~250 mL At 1.0 atm and 27°C, the volume of 0.01 mol H2 = [(0.01 mol)(0.0821 L×atm/mol×K)(300 K)]/1.0 atm = 0.246 L = 246 mL. Thus, A is the best answer. practice this math

Q: Experiment 1 was repeated with 0.40 g of calcium, and the gas that evolved was collected. The identity of the gas, and its approximate volume at 1.0 atm and 27°C were: (Note: R = 0.0821 L•atm/mol•K) A. H2, 250 mL. B. H2, 500 mL. C.O2, 250 mL. D. O2, 500 mL.

Electrical force depends on the particle's charge and the strength of the electric field experienced by the particle, not on the particle's speed. Thus, B (remain the same) is the best answer.

Q: If the speed of the charged particle described in the passage is increased by a factor of 2, the electrical force on the particle will: A.decrease by a factor of 2. B. remain the same. C.increase by a factor of 2. D.increase by a factor of 4.

HPO₄²⁻ is the conjugate base of H₂PO₄⁻(answer D). An acid differs from its conjugate base only by the presence of an additional H+. (Therefore Answers A and B cannot be true (NH4+ is a completely different molecule). A base differs from its conjugate acid only in the lack of a single H+. Acids donate H+, it's conjugate base will have one less H+. Since HPO₄²⁻ has one fewer hydrogen atom, and one more negative charge, than H₂PO₄⁻, then HPO₄²⁻must be the conjugate base of H₂PO₄⁻. Thus, answer choice D is the best answer.

Q: In Equation 1, HPO₄²⁻ is the conjugate: A. acid of NH4+. B. base of NH4+. C. acid of H₂PO₄⁻ . D. base of H₂PO₄⁻.

C. move to orbits of lower energy. In the Bohr model of an atom, energy is emitted only when an electron falls from a higher to a lower energy level. Thus, answer choice C is the best answer.

Q: In the Bohr model of the atom, radiation is emitted whenever electrons: A. change orbits. B. undergo acceleration. C. move to orbits of lower energy. D. move to orbits of larger radius.

Solution: The correct answer is D (increase the speed of the ejected electrons.) The only effect the photon frequency has on the ejected electron is on its kinetic energy. Photon energy equals cathode work function plus electron kinetic energy. The number of electrons ejected (the current) depends on the number of incident photons.

Q: Increasing the frequency of each photon that is directed at the cathode will: A. decrease the number of photons ejected. B. increase the number of photons ejected. C. decrease the speed of the ejected electrons. D. increase the speed of the ejected electrons. PUT A BETTER EXPLANATION.

Capicitance(C)= Q/V Capacitance C depends on geometric factors only. A. I assume would have the opposite effect to the capacitance. B. Is incorrect bc chancing the voltage from 10 to 20V. would make C= 1/10 to 1/20, making the capicitance smaller C. (Increasing the separation of the capacitor plates) is incorrect bc separation of the capacitor plates is inversely proportional to the separation distance of the plates D.( Increasing the area of the capacitor plates) is correct bc Capicitance C is proportional to the plate area.

Q: Making which of the following changes to a circuit element will increase the capacitance of the capacitor described in the passage? A. Replacing the 500-Ω resistor with a 250-Ω resistor B. Replacing the 10-V battery with a 20-V battery C. Increasing the separation of the capacitor plates D. Increasing the area of the capacitor plates

he question asks the examinee to identify a characteristic common to most bacterial and human cells. Of the options listed, only A, the ability to produce ATP via ATP synthase is common to both bacterial and human cells. Both types of cells possess a membrane-embedded electron transport chain capable of generating a H+ gradient, which drives synthesis of ATP via ATP synthase. This ATP synthesis takes place on the plasma membrane of bacteria and on the inner mitochondrial membrane in human cells. B. (the chemical composition of their ribosomes) is incorrect, because Eukaryotic and prokaryotic have different sized ribosomes. The chemical composition of prokaryotic and eukaryotic ribosomes (B), although similar, is distinct enough that several types of antibiotics are able to preferentially target prokaryotic ribosomes over eukaryotic ribosome Of human and bacterial cells, only bacterial cells have cell walls (C) (incorrect). D (the shape of the self-replicating structures that carry their DNA) is incorrect. Most bacterial chromosomes are circular whereas human chromosomes are linear

Q: Most bacterial cells and human cells are alike in: A. the ability to produce ATP via ATP synthase. B. the chemical composition of their ribosomes. C. their enclosure within cell walls. D.the shape of the self-replicating structures that carry their DNA.

The question asks the examinee to identify the tissue that is NOT of mesodermal origin. Mesoderms is middle embryonic layer, and is means of getting around, inside body and sexually;) Meaning Circulatory, excretory, musculoskeletal, gonads. muscular and connective tissue layers of the digestive tract. Of the tissues listed, the only one that does not arise from the mesoderm during embryonic development is nervous tissue (D). Nervous tissue arises developmentally from ectoderm, not mesoderm.

Q: Of the following tissues, which is NOT derived from embryonic mesoderm? A. Circulatory B. Bone C. Dermal D. Nerve

Any disruption of mitochondria is likely to decrease ATP production since they are a major cellular source of that molecule. Answer choices C and D can not be right because they propose an increase in ATP production. There is no information in the question to suggest that valinomycin will cause K+ to compete with H+ for an active site on ATP synthetase. ALSO: Since K+ is in the intermembrane space, it creates a positive charge int he intermembrane space, so H's wont need to cross ETC/ETC won't occur so ATP synthase will decrease. So B is correct. A is incorrect because protons power the ATP synthse not other molecules.

Q: The chemical valinomycin inserts into membranes and causes the movement of K+ into the mitochondria. Based on Figure 1, if mitochondria are treated with valinomycin, the rate of ATP synthesis in the mitochondria will most likely: A. decrease, because the K+ will compete with protons at the active site on ATP synthetase. B. decrease, because movement of K+ into the mitochondrial compartments will disrupt proton movement into the intermembrane space. C. increase, because the net positive charge in the mitochondria will cause increased movement of protons into the intermembrane space. D. increase, because the additional positive charge will further activate ATP synthetase.

We are solving for I (Threshold of intensity) Because the intensity level is 120 dB = 10 log (I/Io), 120/10 = log(I/Io) 12= log(I) - log (Io), where Io= 10⁻¹² 12= log(I) - log (1*10⁻¹²) 12= log(I) +12 log(I)=0 (BNE) 10⁰= I log (I/Io) must equal 12 and I/Io must equal 10¹². Therefore I = 10° W/m2, because I0 = 10-12 W/m2.

Q: The intensity level in decibels is defined as 10 log10(I/I0), where I0 is a reference intensity equal to the human threshold of hearing, 10-12 W/m2. What is the intensity of the threshold of pain, 120 decibels? A. 10⁻¹² W/m2 B. 10⁰ W/m2 C. 10⁻² W/m2 D. 10⁻¹² W/m2

Two sounds A and B would have intensity levels: βa = 10Log(Ia/I₀) and βb = 10Log(Ib/I₀) which may be subtracted to yield: βb- βa = 10Log(Ib/Ia). βb is 20 db larger than βa, therefore βb- βa = 20dB, which can be inserted into the equation earlier: 20dB = 10Log(Ib/Ia). log(Ib/Ia)= 2 10²= (Ib/Ia)= 100 (Answer D)

Q: The intensity level of Sound B is 20 dB greater than the intensity level of Sound A. How many times greater is the intensity level of Sound B than the intensity level of Sound A? A. 2 B. 10 C. 20 D. 100

The answer to this question relies both upon the molar ratios of Equation 1a and on the fact that, at STP, one mole of a gas will occupy 22.4 L. Equation 1a shows that mixing 2 moles of R₂NBCl₂ with 5 moles of LiPH₂ will produce 3 moles of PH₃. Coefficients are impactful!! mol *coefficient is used to figure out L.R. .002 R₂NBCl₂ *2(coefficient)= .004 (L.R.) .005 *5(coefficient)= .005 always compares moles for L.R. to product given the trend above, we can assume there is .003 moles of PH₃ will be produced. At 0°C and 1 atm (STP), this will occupy 0.003 mol × 22.4 L/mol or 67.2 mL.

Q: What is the maximum volume of PH₃(g) that a chemist can obtain from the reaction shown by Equation 1a, if 0.005 mol LiPH₂ reacts with 0.002 mol R₂NBCl₂ at 0°C and 1 atm? A. 0.672 mL B. 6.72 mL C. 67.2 mL D. 67.2 L

The correct answer is C. (lower in frequency than it actually is) The question requires knowledge of the Doppler-shift phenomenon. A sound source moving away from an observer appears to have its source frequency shifted to a lower frequency when detected

Q: When a sound source moves away from an observer, the observer has the impression that the sound source is: A. rotating. B. louder than it actually is. C. lower in frequency than it actually is. D. higher in frequency than it actually is.

Solution: The correct answer is A. The polarity of each of the bromomethanes is determined by the vector sum of all of the individual bond polarities within a given molecule of the compound. Carbon is sp3 hybridized and tetrahedral in all of the bromomethanes. Therefore, the vector sum of the bond polarities in CBr4 is zero, whereas the vector sum is greater than zero in the other compounds because the 4 groups on the C are not the same, changing the vector sum of bond polarities (because they are not all equal). Thus, answer choice A is the only possible answer.

Q: Which of the bromomethanes is LEAST polar? A. CBr4 B. CHBr3 C. CH2Br2 D. CH3Br

The correct answer is C (It accelerates toward the positive plate.) Opposites attract, so the negatively charged particle will move toward the positive plate. Because there is a constant force qE on the particle, it will accelerate toward the positive plate.

Q: Which of the following best describes the motion of a negatively charged particle after it has been injected between the plates of a charged, parallel-plate capacitor? (Note: Assume that the area between the plates is a vacuum.) A. It moves with constant speed toward the positive plate. B. It moves with constant speed toward the negative plate. C. It accelerates toward the positive plate. D. It accelerates toward the negative plate.

The correct answer is C (It accelerates toward the anode.). A charged particle accelerates in an electric field. The electron starts with a velocity that increases as it approaches the anode through the vacuum.

Q: Which of the following best describes the movement of an electron after it is ejected from the cathode? A. It is stationary until collisions propel it toward the anode. B. It moves with constant speed toward the anode. C. It accelerates toward the anode. D. It exits through a side of the vacuum photodiode.

The correct answer is C. (Current flows through the circuit.) The electrons ejected from the cathode are replaced by electrons from the battery and anode. The only effect on the apparatus of electron ejection from the cathode is the current in the circuit.

Q: Which of the following occurs when electrons are ejected from the cathode? A. The voltage across the electrodes reverses polarity. B. The voltage difference between the electrodes increases. C. Current flows through the circuit. D. The total resistance of the circuit increases.

Solution: The correct answer is B.(Ionization Potential) In experiment 1 the RQ is if different metals would react with deionized water. Electronegativity and e-affinity are technically the same thing= is how much a species wants an electron. Since they have the same definition we can eliminate answers A and C. Since we are looking at metals, which are not polarized (polarized is used for organic compounds). So we can eliminate choice D. Metals are ions--> so it will likely be ionized in water (ionization potential), so Answer B is correct. Kaplan Answer: The metals that reacted in Experiment 1 underwent oxidation. The energy required to remove an electron from an atom is the ionization potential. The reactivity of a metal depends on its ionization potential. Among the metals listed in Table 1, potassium has the smallest ionization potential and magnesium has the largest ionization potential. Therefore, potassium reacted the most vigorously and there was no obvious reaction with magnesium. Thus, B is the best answer.

Q: Which of the following properties is most useful in explaining the trend in the reactivities in Experiment 1? A. Electronegativity B. Ionization potential C. Electron affinity D. Polarizability

Aldosterone, which is produced by the adrenal cortex, causes Na+ reabsorption by kidney tubules. Such a mechanism decreases Na+ levels in the urine. The steroid aldosterone does not cause Na+ secretion into the urine. Because ingestion of excessive NaCl would trigger Na+ secretion into the urine, plasma-aldosterone levels would not increase. Rather, the body would rely on those homeostatic mechanisms that excreted the excess Na+. Thus an increase in plasma aldosterone would not be expected to follow ingestion of large quantities of NaCl. The best answer choice is A (No; aldosterone causes Na+ reabsorption by kidney tubules.).

Q: Would an increase in the level of plasma aldosterone be expected to follow ingestion of excessive quantities of NaCl? A. No; aldosterone causes Na+ reabsorption by kidney tubules. B. No; aldosterone causes Na+ secretion by kidney tubules. C. Yes; aldosterone causes Na+ reabsorption by kidney tubules. D. Yes; aldosterone causes Na+ secretion by kidney tubules.

The correct answer is B ( EMS to E.). Put simply: If you look at figure one and follow P1 down to gut, you follow this exact path. If the zygote contains all unique cell contents that are necessary for gut differentiation, segregation of these substances during cell division would occur in the sequence of zygote to P1 to EMS to E. This information comes directly from the flow chart in Figure 1, which shows that gut cells are derived from the following source: zygote → P1 → EMS → E.

Q:If the zygote contains unique cell contents that are necessary for gut differentiation, segregation of these substances during cell divisions would occur in the sequence of zygote to P1 to: A. AB. B. EMS to E. C. P2 to EMS to E. D. both P2 and EMS.

If precipitate is formed, that means it was not soluble. (over saturated) According to Table 1, Ag+ had one soluble compound, Ca2+ had three soluble compounds, and Cu2+ and Fe3+ each had two soluble compounds. Ca2+ had the greatest number of soluble compounds. Thus, B is the best answer.

Q:In Table 1, which cation allowed for the greatest number of soluble compounds? A. Ag+ B. Ca2+ C. Cu2+ D. Fe3+

C. (depriving the subject of some desirable stimulus item for a period of time.) is correct.

Q:In operant conditioning studies, the subject's motivational state is most typically operationally defined by: A. observing the subject's behavior over a long period of time. B. using a type of reinforcement that the experimenter knows the subject usually likes. C. depriving the subject of some desirable stimulus item for a period of time. D. using a novel stimulus that the subject is sure to like. make sure to fill in answer after sean session 4/24

The Q is really why do we exclude liquids (in this case H20) from Keq equations. We exclude liquids because they are usually solvents, and the concentration will stay constant. So we can eliminate answer choices B and D. So next we need to look at how water acts in the equation. H20 is creating OH-s so we can assume basicity. We know its weakly basic because its creating OH in this ionic reaction.

Q:In the equilibrium constant expression for Equation 2, [H2O] is omitted because the salt is: A. only weakly basic, and [H2O] is nearly constant. B . strongly basic, and [H2O] is nearly zero. C. only weakly acidic, and [H2O] is nearly constant. D. strongly acidic, and [H2O] is nearly zero.

The answer to this question is A because, based on the standardization system used to score IQ, the Wechsler Scales of Intelligence (WISC) scores are "normalized" to a mean of 100 and a standard deviation of 15. So 68% of the scores will be between 85 and 115.

Q:Interpretation of intelligence test scores is based on the assumption that the scores are normally distributed within a population such that: A. more than two-thirds of children will score between 85 and 115. B. a quarter of all children will perform poorly on least one component of the exam. C. significantly more children will score at the extremes of the scoring distribution than in the middle. D. only exceptionally gifted children would have IQ scores above 85 by age 6; perfect test scores are rare during early childhood. FIX AFTER SEAN SESSION

The correct answer is C. (decrease, because the proton gradient will rapidly reach equilibrium.) Hydrogen ions (H+) are protons. The provision of a channel for proton flow across the membrane would allow hydrogen ions to flow across the membrane until equilibrium had been achieved between the concentrations on each side of the membrane. Because ATP synthesis is driven by a flow of hydrogen ions down a concentration gradient, ATP production will decrease and eventually stop as equilibrium is established because H's wont be used to power the ETC to eventually reach the ATP synthase if there is no gradient. (not increase as suggested in answer choice A or remain unchanged as suggested in answer choice D). The decrease has nothing to do with the rate of hydrogen ion donation by NADH, answer choice B is incorrect.

Q:The chemical gramicidin inserts into membranes and creates an artificial pathway for proton movement. Based on Figure 1, if mitochondria are treated with gramicidin, the rate of ATP synthesis will most likely: A. increase, because of increased proton movement back into the mitochondria. B. decrease, because of a decreased rate of hydrogen-atom donation by NADH. C. decrease, because the proton gradient will rapidly reach equilibrium. D. not be altered, because sufficient protons will remain between the membranes to generate ATP.

The answer to this question is B because Gestalt psychology was the theoretical approach that emphasized the idea that the ways in which people's perceptual experience is organized result from how human brains are organized. Put simply, Gestalt principles are all the visual illusion concepts

Q:The principles of perceptual organization described in the first paragraph are most closely related to which psychological theory? a. Behaviorist B. Gestalt C. Humanistic D. Cognitive PUT DEFINITIONS OF WRONG ANSWERS IN

Since 1 equivalent of NaHCO3 provides 1 equivalent of Na+, the molar concentration of Na+(aq) in 0.010 M NaHCO3(aq) solution is also 0.010 M = 0.010 mole/L. [.01M]*1 eq of Na=.01 mol/L The molar concentration of Na+(aq) in 0.010 M Na₂CO3(aq) solution is 0.020 mol/L since 1 equivalent of Na2CO3 provides 2 equivalents of Na+. [.01M]*2 eq of Na=.02 mol/L When equal volumes of these two solutions are mixed, the resulting molar concentration is equal to their average, (0.010 mol/L + 0.020 mol/L)/2 = [.01M]*1 eq of Na=.01 mol/L Thus, B ([.01M]*1 eq of Na=.01 mol/L) is the correct answer.

Q:What is the molar concentration of Na+(aq) in a solution that is prepared by mixing 10 mL of a 0.010 M NaHCO3(aq) solution with 10 mL of a 0.010 M Na₂CO3(aq) solution? A. 0.010 mole/L B. 0.015 mole/L C. 0.020 mole/L D. 0.030 mole/L

Answer A is incorrect (Seeing a portion of food served on a large plate as being smaller than the same portion of food served on a small plate), because it has a similar principle of perceptual organization as ones described in the passage. This example is analagous to Panel A shown in the figure and described in the passage. Answer B is incorrect (Seeing birds flying in the same direction as being part of an integrated flock), because it has a similar principle of perceptual organization as ones described in the passage. It is similar to panel D, in which we group similar objects together. Answer C is incorrect (Seeing cars lined up in a parking lot as being in long rows, rather than in pairs facing each other with a gap between each pair), because it has a similar principle of perceptual organization as ones described in the passage. It is similar to panel D, in which we group similar objects together. The answer to this question is D because it is the only option that describes a perceptual experience that results from a form of top down processing (i.e., context effects), but is neither a perceptual illusion nor the result of one of the Gestalt principles of perceptual grouping

Q:Which of the following experiences is NOT related to any of the principles of perceptual organization that are described in the passage? A. Seeing a portion of food served on a large plate as being smaller than the same portion of food served on a small plate B. Seeing birds flying in the same direction as being part of an integrated flock C. Seeing cars lined up in a parking lot as being in long rows, rather than in pairs facing each other with a gap between each pair D. Seeing a word with a missing letter and being able to identify the word, based on the sentence in which it is contained

Molecular nitrogen is normally a very unreactive molecule, because of the very strong triple bond holding the nitrogen atoms together, as stated in answer choice D(N₂ is very unreactive because of the great strength of the N≡N triple bond, which is why plants CANNOT utilize N from the atomosphere). Since plants require nitrogen in a reduced form, this would be the main reason for their inability to utilize nitrogen directly from the air. The other responses are readily eliminated. Nitrogen is the most abundant element in the air, so A (N₂ is present in very low concentrations in the atmosphere.) is incorrect. Because the two nitrogen atoms in N2 are identical, and have the same electronegativity, the molecule is not polar at all, and B (N₂ is too polar.) is incorrect. Finally, N2 is not classified with the noble gases, which are in group 8 of the periodic table. Thus, answer C (N₂ is very unreactive because it is a noble gas) is incorrect as well.

Q:Which of the following is the most likely reason that plants CANNOT utilize nitrogen from the atmosphere? A. N₂ is present in very low concentrations in the atmosphere. B. N₂ is too polar. C. N₂ is very unreactive because it is a noble gas. D. N₂ is very unreactive because of the great strength of the N≡N triple bond.

Removal of the parathyroid gland would lead to... hypocalcemia, a condition of low blood calcium, resulting from the lack of parathyroid hormone. This would cause increased neuromuscular excitability because of the change in membrane potential, which under normal physiological conditions, is partially kept in balance with extracellular calcium. Typically, the person would eventually die from severe respiratory muscle spasms

Removal of the parathyroid gland would lead to...

current

The number of electrons ejected is the

kinetic energy

The only effect the photon frequency has on the ejected electron is on its_____.

kinetic energy, cathode work function plus electron kinetic energy, (the current), number of incident photons.

The only effect the photon frequency has on the ejected electron is on its_____. Photon energy equals________ . The number of electrons ejected (____) depends on the ________.

1*10^neg number (much less than 1)

the dissociation constant of the weak acid is