FLE 8
Which sociological concept is most germane to the findings in Table 1 with respect to heavy housework? A)glass escalator B)glass ceiling C)second shift D)sexual harassment
(C) is the correct answer. Second shift is the tendency of employed women to do a substantial amount of household chores up to a full "second shift" due to the absence, inability or unwillingness of a male partner to contribute to household work. The data suggests that women do more heavy housework than men; this supports the idea that a second shift may be occurring. (A) is incorrect. The glass escalator is the tendency of men in primarily female occupations to advance rapidly in the workplace; this phenomenon is unrelated to the amount of housework performed at home. (B) Glass ceiling is also incorrect for the same reason; it relates to the public work sphere and not the private home sphere. For the same reason, (D) is incorrect.
What temperature is physiological pH at?
98F or 37C (310K)
Suppose a patient suffered from lazy-eye as an infant, which went uncorrected until after the critical development period for the retina. As a result, the patient has severely deficient depth-perception. Which condition describes this patient's inability to use retinal disparity as a cue for depth? A)Stereo blindness B)Achromatopsia C)Binocular rivalry D)Akinetopsia
A is the correct answer, individuals with amblyopia (lazy-eye) which goes uncorrected develop stereo blindness in which they do not develop the ability to use retinal disparity as a cue for seeing depth and must rely instead on pictorial or monocular cues. Achrmoatopsia is a loss of color vision due to damage to the visual cortex, binocular rivalry is a process by which the visual cortex develops ocular dominance columns and results in the development of stereoscopic vision, akinteopsia is loss of motion perception. This question is an example of an MCAT outlier.
Histone acetylation and deacetylation is an important part of gene regulation via the interplay between the euchromatin and heterochromatin forms of chromosomal DNA. Acetylation alters: A) positive charges on the histone and thereby its binding affinity for chromatin. B) the shape of the histone molecule and thereby its binding affinity for chromatin. C) negative charges on the histone and thereby its binding affinity for euchromatin only. D) localization of the histone and thereby its binding affinity for heterochromatin only.
A is the correct answer. Histone acetylation alters the amount of positively charged moieties on histones by acetylating lysine residues, changing that lysine's charge from positive to neutral. This modifies the histone's ability to bind to the negatively charged DNA molecule. Hence B, C and D are incorrect.
Figure 2 demonstrates neutralization of ZEBOV by hybrid MAbs directed against ZEBOV antigens. Another scientist proposes that an alternative approach would be to engineer MAbs which bind the host cell ZEBOV receptor. Is this approach likely to work? A) Yes, because this would block host cell ZEBOV receptors, preventing viral entry into the host cell. B) Yes, because MAbs bound to host cell surface receptors will allow viral entry into the host cell. C) No, because MAbs can only be engineered to bind viral antigens, not host cell receptors. D) No, because MAbs must bind virions directly in order to be effective.
A is the correct answer; Animal viruses gain entrance to cells by binding to receptor proteins on the exterior leaflet of the plasma membrane. Thus, by directing monoclonal antibodies to the ZEBOV receptor on host cells, the antibodies would prevent viral recognition of that receptor; the virus could no longer bind to the cell to gain entrance and infection would be blocked giving the same result as the monoclonal antibodies directed against the viral antigens. This rules out answer D. Answer B is incorrect because if the phenomenon described actually occurred it would not be beneficial—it would cause infection. Answer C is false because antibodies can indeed be designed to bind cell surface receptors and do in many cases.
A chemist examining biological oxidation reactions wishes to reverse a metabolic oxidation step in vitro. Which reagents are appropriate for this purpose? A) NaH or H2S B) Br2 , NaH or H2S C) F2 or HNO3 D) NaH or HNO3
A) Because the chemist is reversing an oxidation, the reaction he is performing will be a reduction. Sodium hydride NaH and H2S are the only two common reducing agents out of the five presented in the answer choices. Choice B is wrong because it includes Br2 which is a weak oxidizing agent. Choice C includes only oxidizing agents, and choice D includes one oxidizing and one reducing agent.
A patient with back pain visits his doctor. The doctor can choose to give a cursory exam or a thorough one. The patient can choose to follow the advice given, or ignore all advice and get a second opinion. Of the possible outcomes listed below, which one is best in terms of the collective benefit? A) The doctor spends extra time and gives advice; the patient follows his advice. B) The doctor spends extra time and gives advice; the patient seeks a second opinion. C) The doctor spends only 5 minutes; the patient follows his advice. D) The doctor spends only 5 minutes; the patient seeks a second opinion.
Answer A is the correct answer in collective terms. The doctor and patient sacrifice time; the patient's acceptance of the doctor's advice may result in more efficient utilization of the overall system. Answer B is incorrect because it does not contribute to a good collective outcome because the second opinion adds to the collective healthcare costs of the patient's diagnosis even though the doctor sacrificed extra time with the patient. Answer C is incorrect as the doctor has elected not to give sufficient time to the patient's complaint. Answer D is incorrect because it is a suboptimal use of both the doctor's and the patient's time. And, as in answer B, the second opinion adds to the collective healthcare costs of the patient's diagnosis.
IgG antibodies are 150 kDa molecules composed of four peptide chains. What banding pattern would be observed for an IgG molecule separated on an SDS-PAGE gel under reducing conditions? A) 4 bands at 150 kDa B) 2 bands at 25 and 50 kDa C) 3 bands at 25, 50 and 150 kDa D) 4 bands at 25, 50, 75 and 150 kDa
Answer B is correct; The IgG antibody consists of two identical heavy chains and two identical light chains. If run on SDS-PAGE under normal conditions it would produce only one band. The stem tells us, however, that it is under "reducing conditions," which indicates that the antibody will separate into individual peptides. The two identical light chains will be the 25 kDa band and the two identical heavy chains will be the 50 kDa band (adding up to a total of 150 kDa). Answer A would not be correct since the SDS would separate the antibody into individual proteins. One would only get the 150 kDa band under non-denaturing conditions. Answer C is not correct since you would either get the 150 kDa band or the 25 and 50 kDa bands, not both. Answer D is not correct because of the same reason Answer C is incorrect as well as the fact that you would never get a 75 kDa band under either reducing or non-reducing conditions.
In one study, participants were randomly assigned to a no-practice control group and a group which practiced a skill-based sports activity. Both groups were then asked to perform the sports activity and performance was measured and compared. This experimental design is a logical experimental approach to test the impact of: A) implicit memory on recall only. B) explicit memory on recall and performance. C) implicit memory on recall and performance. D) explicit memory on performance only.
Answer C is correct. Implicit memory is unconscious, automatic recall, often of a conditioned response, a task, or a procedure. The experiment described in the passage allows the second group to perform the task using muscle memory, an example of implicit memory. The control group has not performed the task previously, and therefore cannot utilize implicit memory. This eliminates answers B and D, which cite explicit memory—the intentional, conscious recall of information. Answer C is preferred over answer A because the study can be used to examine both recall and the outward behavior, or performance, associated with that recall.
While studying for an upcoming exam, a student memorizes physiology terms by relating them to actual pieces of furniture in her apartment. On the exam the student visualizes the piece of furniture in her apartment and this aides her recall. This memory aide is an example of: A) Spaced repetition B) Peg-word System C) Mnemonic D) Chunking
Answer D is correct. The memory aide in which a person associated the item to be remembered with an actual physical location or object which can be visualized during recall is called the Method of loci. Spaced repetition (A) is the process by which long-term retention is increased through repeated rehearsal and spaced intervals of increasing duration. The peg-word system (B) is the use of pre-memorized list of words or numbers to which one associates new information to be learned. A mnemonic is a memory aide in which an acronym is created for which each letter represents a prominent letter (usually the first) of a word in the phrase to be memorized.
Multiple Sclerosis (MS) is a neuro-degenerative disorder wherein the patient's T cells begin attacking the myelin sheath covering the neurons of the central nervous system. Which statement best explains how a T cell contributes to MS? A) T cells secrete antibodies that destroy the myelin. B) T cells non-specifically attack many cells within the body, including neurons. C) T cells engulf the central nervous system neurons via phagocytosis. D) T cells activate macrophages which secrete cytotoxic proteins that destroy the myelin.
Answer choice A can be ruled out since T cells do not secrete antibodies; B cells secrete antibodies. T cells are a component of the adaptive immune response, which is specific (compared to the innate immune system, which is non-specific), making answer choice B incorrect. T cells do not have phagocytic ability, making answer choice C incorrect. Helper T cells do also activate macrophages. Macrophages secrete pro-inflammatory cytokines such as TNF-alpha, which can lead to tissue destruction if they are not regulated. The TNF-mediated destruction of myelin slows the neurotransmission that contributes to motor weakness in MS, which makes answer choice D the correct choice. Specific knowledge of the actual process in MS is not expected for this question; process-of-elimination using basic science allows for selection of the correct answer.
CffDNA testing can be used to test for certain fetal abnormalities by assessing the relative number of each of several types of chromosomal fragments. Relative numbers of chromosomal fragments could indicate: A)trinucleotide repeat expansion B)aneuploidy C) inversions D) tetraploidy
B is the correct answer. Aneuploidy results from a genetic event that produces gametes with an atypical number of chromosomes. If cffDNA were analyzed for the relative number of chromosomal fragments of each type, it should reflect the deviation that results from aneuploidy. CffDNA can be used for many kinds of genetic tests, but none of the other choices lend themselves as well to relative numbers of chromosomal fragments. Trinucleotide repeat expansions are expansion of stable areas on the chromosome that are repeating series of three nucleotides. Some diseases are attributed to an excessive, unstable number of these series. Simple assessment of chromosome contribution to CffDNA would not be the obvious way to detect these; alternative molecular techniques such as sequencing or RFLP analysis may be required. Inversions, Choice C, are regions of chromosomes that are rearranged, and could not be detected by this method since the fragment orientation cannot be determined. Choice D is tetraploidy, or two complete sets of chromosomes. This might be revealed in more total CffDNA, but the relative amounts of each chromosome would be the same.
The extended family of a pregnant black woman living in the rural southern U.S. told her nurse, "We like our greens, soul foods, peas and pork as special pregnancy foods." The nurse included as many of these foods as possible in her prenatal nutrition plan. Which sociological concept is most relevant to this scenario? A)Ethnocentrism B)Cultural relativism C)Racism D)Reverse discrimination
B is the correct answer. By the nurse considering the importance of food culture within the southern African American family and used it to assist with treatment for the pregnant woman. Answer A is wrong because ethnocentrism is when one ignores a group's culture in interpretation or uses one's own culture to explain social behaviors. Answer C is wrong because racism is when a racial group is perceived as inferior to another racial group and is treated as such. Answer D is wrong because reverse discrimination is often brought up by the dominant culture who feel that the non- dominant culture have some advantage because of their race or ethnicity.
The HIV integrase enzyme catalyzes the integration of the HIV genome into the host cell genome and has become one of the primary targets for HIV drug development. This enzyme is an effective drug target because the inhibition of this enzyme prevents: A) fusion of the virus to host cells. B) incorporation of HIV DNA into host DNA. C) budding of the virus and therefore inhibits further infection. D) maturation of the virus after budding and therefore inhibits further infection.
B is the correct answer. Integrase is the enzyme that catalyzes the incorporation of the HIV genome into the genome of infected cells. Answer A is not correct because integrase does not play a role in the initial binding of HIV to target cells. Answer C is not correct because integrase does not play a role in budding of HIV from infected cells. Answer D is not correct because integrase is not involved in the maturation process once HIV is budded as an immature virus.
The passage of the Affordable Care Act of 2010 represented a transition of the U.S. healthcare delivery system from a: A) universal healthcare system to a single-payer system. B) predominantly private system to a hybrid of direct fees and tax subsidies. C) single-payer system to socialized medicine. D) free-market system to universal healthcare.
B is the correct answer. Since 2010, the Affordable Care Act has created more health insurance and health care systems benefits for many citizens, but those citizens are still required to pay insurance fees and the system is subsidized by state and federal taxes, making it a hybrid of direct-fees and tax-subsidies. Answer A is wrong because a universal health care system is fully paid for by tax subsidies and requires no direct fees to individuals, which was not the case before or after this change. Answer C is wrong because a single-payer national health care system is the same as a universal system in which tax subsidies pay the costs of health care for all individuals enrolled with no out-of-pocket costs. Answer D is wrong because prior to the change the U.S. system was not a fully free-market system. Medicare and Medicaid were socialized aspects within that system covering some but not all individuals. Further, the new system is not truly universal healthcare, making D false on both accounts.
A compound that blocks the α-subunits of the ATP synthase complex must enter which region of the mitochondria to function? A)Cristae B)Matrix C)Lumen D)Vesicle
B is the correct answer; the matrix is the inner most compartment of the mitochondria. A compound that blocks the F1 portion of ATP synthase has to enter this part of the mitochondria, since this is where the F1 component of the ATPase is located. The cristae are the folds of the mitochondria and the compound would have to cross these membranes to function (Answer A). The lumen is the intermembrane space and this compound would have to pass through this region to act. Finally, a vesicle is a membrane enclosed region of the cell for transport of items and is not related to the mitochondria
If a student immediately identified a solution to the case study that turned out to be inaccurate, yet the student persisted to espouse this solution even when verified errors were pointed out to him, the medical student may be experiencing: A) confirmation bias. B) the Gambler's fallacy. C) functional fixedness. D) overconfidence.
Belief perseverance occurs when an individual continues to hold a given belief even in the face of evidence to the contrary. B is incorrect. Gambler's fallacy is the erroneous belief that the future probabilities are affected by past events. C is incorrect. Functional fixedness refers to a situation in which an individual cannot imagine using an object in any way other than its traditional use. D is incorrect. Overconfidence is the tendency to overestimate our ability to make good decisions.
Research suggests that men make up 66 percent of today's physician workforce, whereas 91 percent of registered nurses are women. Which sociological concept best explains this phenomenon? A)Glass escalator B)Glass ceiling C)Gender inequality D)Economic inequality
C is the correct answer. Gender inequality captures the issue here that women and men are valued differently in the health professions which leads to gender role divisions in labor. In addition, women are often seen in society as better suited for "nurturing" professions and have been historically restricted from participating in male-dominated professions like physicians. Answer A is wrong because the glass escalator is a sociological term that explains the quick advancement of men in women-dominated fields. Answer B is wrong because the glass ceiling concept explains the lack of advancement for women in many professions in comparison to men. While the concept does have some applicability in specific cases when a women is not allowed to become a doctor, the much larger issue is that men and women are valued differently within the profession. Answer D is wrong because this a result of gender inequality in which women find themselves being paid less due to the gender division of labor that leads to economic inequality.
Which digestive enzyme functions in vivo in the absence of the bile salts mentioned in the passage? A) Chymotrypsin B) Trypsin C) Pepsin D) Pancreatic amylase
C is the correct answer. Pepsin is the only one of the listed enzymes that functions in the stomach. Answers A, B and D are all small intestine enzymes that would be functioning in the presence of bile, which arrives to the duodenum via the pancreatic duct.
Which feature is most likely to be absent from cells of B. pseudomallei? A)Plasma membrane B)Cell surface polysaccharides C)Cell wall D) Peptidoglycans
D is correct. According to the passage, B. pseudomallei are Gram-negative bacteria. The Gram stain recognizes peptidoglycans coating the cell wall and Gram-positive bacteria have a cell wall containing a high concentration of this molecule. Bacteria that are Gram-negative contain relatively little peptidoglycan and will appear unstained next to Gram-positive bacteria. For this reason, D is the correct answer. All bacteria, whether they stain positive or negative with the Gram stain have a cell wall, polysaccharides on the extracellular surface, and an inner plasma membrane; therefore, answers A through C are incorrect.
Assuming Hardy-Weinberg equilibrium conditions, what percent of Europeans are homozygous for RhD? (15% don't have RhD) A) 85 % B) 70 % C) 15 % D) 61 %
D is correct. From the passage, it can be determined that the frequency of homozygous recessive individuals is 0.15 (or 15%, since 85 % of Europeans are either homozygous dominant or heterozygous). This is value is q2 in the Hardy-Weinburg equation, so its square root or 0.39 is q, or the frequency of the recessive allele. The sum of the two frequencies, p and q, must equal 1, so p = 0.61, or 61 %, Choice D. Choice A is the percentage of individuals with the dominant phenotype, but not necessarily the homozygous for the dominant allele, and Choice C is the percentage of homozygous recessive individuals—information that is needed in the calculation, but not the answer to the question. Choice B can only be eliminated by correctly performing the calculation.
Mitochondrial function and host cell viability are affected by impaired lipid transport to the mitochondria from the ER. Which of the following would NOT explain this phenomenon? A) Loss of mitochondrial membrane integrity B) Depletion of the proton motive force C) Inactivity of the electron transport chain D) Impairment of the glycolysis pathway
D is correct. Impaired lipid import from the ER would compromise mitochondrial membrane integrity, a structure critical to the production of adequate amounts of chemical energy. Answer A is therefore a good explanation for the phenomenon and thus incorrect. Glycolysis occurs in the cytosol, so the ER-mitochondrial import of lipids would have little impact on this process. Answer D is therefore not a good explanation for this phenomenon and is the correct answer. The pyruvate generated by glycolysis, however, is actively imported into the mitochondrial matrix where it combines with acetyl CoA. This ATP-driven process would begin to fail quickly, as would the citric acid cycle. The electron carriers NADH and FADH2 would not be generated in sufficient amounts to support the electron transport chain; therefore, answer C is a good explanation and thus incorrect. The failure of the electron transport chain would allow the proton gradient maintained in mitochondrial cristae to be depleted, in part because of failing membrane integrity; therefore, answer B is a good explanation for this phenomenon and thus incorrect. As a result, ATP synthase would cease to function and the primary source of chemical energy used by the host cell would be lost. Cell viability would be impacted negatively.
Figure 1 shows examples of a fundamental frequency with three overtones. What is the frequency ratio of the fundamental frequency to the frequency of the third overtone? A) 1:1 B) 2:1 C) 3:1 D) 4:1
D is the correct answer. The frequencies of the harmonics are equal to the frequencies of the fundamental, times the harmonic number. Choice C might be tempting, but remember that the third overtone is the fourth harmonic. Thus the frequency will be 4× the fundamental frequency, which is choice D
Figure 1 shows examples of a fundamental frequency with three overtones. What is the frequency ratio of the fundamental frequency to the frequency of the third overtone? A) 1:1 B) 2:1 C) 3:1 D) 4:1
D is the correct answer. The frequencies of the harmonics are equal to the frequencies of the fundamental, times the harmonic number. Choice C might be tempting, but remember that the third overtone is the fourth harmonic. Thus the frequency will be 4× the fundamental frequency, which is choice D. Another way of looking at this is to consider the wavelengths. As given above and below the pictures of the standing waves, for the first harmonic the length L = ½ λ and for the fourth harmonic the length L = 2λ. Since the lengths are the same, the wavelength for the fourth harmonic is ¼ the wavelength of the first harmonic (this can also be seen directly from the figure by comparing sections of the waves). And since frequency is inversely related to wavelength (assuming v is constant, which it is), that means that the frequency for the fourth harmonic is 4× the frequency of the fundamental wave. Either way you look at it, you end up with choice D as the correct answer.
Which fact best accounts for the observation that ionic compounds containing gold are uncommon in nature? A) Elemental gold has a high electronegativity compared to most metallic elements. B) Elemental gold has a low electronegativity compared to most metallic elements. C) Elemental gold is easily oxidized by O2. D) Elemental gold requires strong reducing agents to form compounds.
Gold has an electronegativy of 2.4, which is higher than any other metal. This number need not be known or memorized. The question simply requires a logical evaluation of the most likely reason gold compounds are rare. The higher the electronegativity, the less prone the atom is to lose electrons. Like all metals, gold almost exclusively exists in compounds in cationic form, but because of its relatively high electronegativity, it has less of a tendency to do so than do other metals. Choice B is a false statement which can be deduced for the general trend of electronegativity to increase towards fluorine. Choice C is a false statement; gold in jewelry is not oxidized to any visible extent by atmospheric oxygen. Choice D is a false statement; gold only exists in compounds in cationic form, so oxidizing agents, not reducing agents are required.
Is the cytosol hydrophobic or hydrophilic?
Hydrophilic
The theory of endosymbiosis suggests endocytosis of an aerobic prokaryote as the origin for mitochondria. What similarities between mitochondria and prokaryotes support this theory? I. Circular DNA genome II. Aerobic respiration III. Anaerobic respiration IV. Cytokinesis
I, II; Both mitochondria and aerobic prokaryotes have a circular DNA genome (I) and use oxygen as the electron acceptor to drive ATP synthesis (II). Anaerobic respiration occurs in some prokaryotes in the absence of oxygen, but not in mitochondria. The endosymbiont theory suggests that aerobic prokaryotes formed a symbiotic relationship with anaerobic cells. This eliminates choice III. Both mitochondria and prokaryotic cells reproduce by binary fission, not mitosis, eliminating choice IV Only answer A provides the right combination of variables to be the correct answer.
Triglycerides undergo base-catalyzed transesterification with methanol or ethanol, which involves: A) attack of the carbonyl carbon by the proton from an alcohol. B) attack of the carbonyl carbon by the methoxide or ethoxide ion. C) the release of ethylene glycol. D) the formation of three fatty acid esters, carbon dioxide, and water.
In transesterification reactions, the base extracts a proton from the alcohol to generate the RO- ion. This is a nucleophile that attacks the carbonyl carbon. Answer A can be eliminated. The product of the transesterification is three fatty acid esters and glycerol, making Answers C and D incorrect. Answer B is therefore the best answer.
Metabolic steps that are IRREVERSIBLE have what type of equilibrium constant and free energy change?
Large negative free energy changes make a step irreversible as this is a driving force for a reaction in the forward direction. Answer C is the best choice
Metabolic steps that are reversible have what type of equilibrium constants and free energy changes?
Metabolic steps that are reversible have equilibrium constants near 1 and free energy changes close to zero.
Given the predominate state of most metals in nature, purifying a metal from one of its natural source compounds will involve which reaction type? A) Combustion B) Oxidation C) Reduction D) Synthesis
Metals almost exclusively exist in nature (when combined) as cations. Therefore, the predominant process to recover the pure metals from the natural compounds would be reduction. One example is purifying Fe from Fe2O3; iron must be reduced for the 3+ to 0 oxidation state. This is a consequence of the fundamental fact that metals tend to form cations and nonmetals form anions. Choice A is wrong because combustion adds oxygen molecules to what is burned which is oxidation. Choice B is wrong because oxidation causes an increase in oxidation number; the opposite of what is required here. Choice D cannot be correct because synthesis reactions take elements and combine them with other elements to form compounds.
Ion channels specific to the fluoride ion are inhibited by N3- ions.N3- is also a known inhibitor of the electron transport chain. The chemical behavior of these two anions may be explained by a difference in size resulting from the fact that the: A) N3- ion has two more valence electrons than the F- ion. B) N3- ion has the ability to form sp3 hybridized orbitals while the fluorine ion does not. C) additional electrons in the N3- ion all occupy a higher-energy shell than the one extra electron in the fluorine ion. D) two ions have the same number of shielding and valence electrons, but the fluorine ion has a greater effective nuclear charge.
N3- and F- are isoelectronic, each has 8 valence electrons and 2 inner electrons. The inner electrons shield the outer electrons from the pull of the nucleus which is greater in the case of fluoride with 9 protons compared to 7 protons in the nucleus of nitride. Fluoride is significantly smaller than nitride because it has a greater nuclear charge acting on the same number of valence electrons with the same amount of shielding in between. Choice A is a false statement; the two ions have the same number of valence electrons. Choice B is true in theory, but does not explain anything about ionic radius. Choice C is a false statement; the "added" electrons in both ions go into the n = 2 shell.
According to the First Law of Thermodynamics, a body's change of internal energy is equal to the: A) difference between the heat absorbed by the body and the work done by the body. B) heat energy of the body plus the work done by the body. C) body's internal temperature divided by its total mass. D) total heat absorbed by the body plus the work done by the body.
The First Law of Thermodynamics states that Q = ΔU+W, which means that the heat added to or taken away from a system is equal to the change in the internal energy of the system plus the work done by the system. Rearranging the formula, we see that the heat going into a system minus the work done by the system equals the internal energy of the system (DU = Q - W). Hence, A is the best answer. B is not the best answer because it is unclear which entity "heat energy" refers to, and adding heat to work done will not give total internal energy. C is not the best answer because mass is not addressed by the First Law of Thermodynamics. Choice D may seem reasonable at first glance because it accounts for both heat and work, but work done by the body must be subtracted from its total internal energy, not added to it.
If an individual continues to take a drug in order to avoid the aversive properties of drug withdrawal, the drug serves as : A) negative punishment for drug taking behavior. B) negative reinforcement for drug taking behavior. C) positive punishment for drug taking behavior. D) positive reinforcement for drug taking behavior.
The correct answer is B. Anything that makes a behavior more likely to occur in the future through the avoidance (or removal) of some aversive stimulus is known as negative reinforcement. A is incorrect. Negative punishment involves making a behavior less likely to occur because of the removal of some appetitive stimulus. C is incorrect. Positive punishment involves making a behavior less likely to occur because of the presentation of some aversive stimulus. D is incorrect. Positive reinforcement involves presenting some appetitive stimulus to make a behavior more likely to occur.
In the cell, tau protein is soluble at approximately one micromole per liter. Given that the molar mass of tau protein is 55,000 Da and the volume of a cell is approximately 100 μL, how much soluble tau protein could be found in one cell? A)0.55 μg B)5.5 μg C)0.55 mg D)5.5 mg
The correct answer is B. To answer this question, you need to do a simple units conversion. The simplest way to do it is: which is the molar mass times the soluble concentration times the volume of the cell. Using this, you can cancel out every unit except grams, which will give you the mass of tau protein in the cell that will remain soluble. With a calculation like this, you don't need to do the math, you can just move the decimal. 55,000 x 0.000001 = 0.055 (because you move the decimal over 6 places). 0.055 x 0.1 = 0.0055 (moved the decimal over 1 place). 0.0055 / 1000 = 0.0000055 (move the decimal over 3 places). And 0.0000055 g = 5.5 μg. A, C, and D are incorrect because each one is off by a factor of ten or more. When doing equations like this, make sure that the units make sense, make sure you can cancel out any unit you don't need in the final answer, and watch your decimals carefully.
Research has demonstrated that addiction can be divided into two distinct components: drug liking and drug wanting. Often addicts will report that their drug liking was pronounced early on, but that it has faded over time even as they increase the dosage of the drug. Based on this information, one might argue that addicts are experiencing: A) discrimination for drug liking. B) generalization for drug liking. C) sensitization for drug liking. D) tolerance for drug liking.
The correct answer is D. Tolerance refers to a decrease in response to a given drug as a function of prior experience with the drug. In this case, it would appear that addicts actually develop a tolerance for the hedonic properties of the drug. A is incorrect. Discrimination occurs in a classical conditioning context when an individual learns to discriminate among like stimuli and only show a conditioned response to a very specific conditioned stimulus. B is incorrect. Generalization occurs when an individual learns to respond to stimuli that are similar to the original conditioned stimulus. C is incorrect. Sensitization involves an enhanced response to a given stimulus. In the context of drug taking behavior, sensitization can be conceived as "reverse tolerance" to a given drug.
What is a null hypothesis?
The null hypothesis is the hypothesis a scientist would want to disprove in their experiment. It typically takes the form of, "there is no relationship between X and Y" when one wants to demonstrate a relationship between X and Y. Disproving the null hypothesis by experiment shows there is in fact a relationship between X and Y.
A chemotherapeutic technique has been proposed in which micromolar amounts of a mixture of a strong acid and a strong base are direct-injected into growing tumors. One scientist expressed concern that even small amounts of the acids or bases may enter the bloodstream and form a dangerous precipitate. Which strong acid-strong base pair is of most concern? A) KOH and HNO3 B) CH3COOH and HCl C) Ba(OH)2 and HCl D) Ba(OH)2 and H2SO4
The only cation-anion pair out of all the compounds listed that together comprise an insoluble compound are Ba2+ and SO42- so the only chance of precipitation exists if solutions of the compounds listed in D are mixed together. Choice A is a strong acid-strong base pair, but when the cations are exchanged with each other KNO3 and H2O (HOH) would result, neither of which are insoluble solids in water. Choice B lists two acids so it doesn't meet the requirement given. Furthermore, each has H+ as the cation; no insoluble cation-anion combinations are present. Choice C lists a strong acid-strong base pair, but when the cations are exchanged with each other BaCl2 and H2O (HOH) would result, neither of which are insoluble solids in water.
The posterior pituitary and the adrenal medulla: Ainteract with the nervous system directly, through synapses. A) is the correct answer. B)synthesize hormones. C)release steroid hormones. D)are regulated by the hypothalamus.
The posterior pituitary (neurohypophysis) and the adrenal medulla are both composed of nervous tissue and synapse directly with neurons. This is Choice A. The posterior pituitary receives hormones from the hypothalamus but does not synthesize them, so Answer B can be eliminated. Answer C can be eliminated because only the adrenal medulla releases steroids—the steroids released by the posterior pituitary are peptide hormones. Answer D can be eliminated as only the posterior pituitary is directly regulated by the hypothalamus.
Mercury (II) ions bound most strongly to the [15aneN4S] macrocycle, forming Hg[15aneN4S]+2. How many valence d electrons are found in this complex? A) 4 B) 6 C) 8 D) 10
This question is slightly tricky. The valence configuration of Mercury is 1s22s22p63s22p63s22p6 3d10 4s23p64d10 f145s24p65d106s2. Many people refer to elements such as Mercury as post-transition metals, meaning that in their valence shell they are using 6s orbital electrons. So when Mercury loses two electrons to bind in the +2 oxidation state, there are still 10 d valence electrons. If mercury is in the +2 oxidation state, whether it is coordinating with the macrocycle or not, it has 10 d valence electrons.
Following proteosomal degradation of a virion into viral peptides, the peptides are transported to the endoplasmic reticulum via the TAP protein. Eventually, these peptides are presented on the MHC-I complex on the cell's surface. Where in the cell are the viral proteins most likely to encounter and be loaded onto MHC-I? A)The nucleus B) The endoplasmic reticulum C) The Golgi apparatus D)The cell surface
This question may be over-thought by some students. The MHC is itself a complex of proteins, and proteins bound for the cell surface or excretion are synthesized in ribosomes on the Rough ER and folded in the endoplasmic reticulum. If the peptides are being transported to the ER by TAP, it is logical that they will be loaded into the peptide binding grove of the MHC in the ER, which is in fact the case. Answer choice B is correct. The MHC does not even exist as a protein in the nucleus (the MHC-I mRNA has not been translated yet), making answer choice A incorrect. Although the complex will travel through the Golgi (Answer choice C) to the cell membrane (Answer choice D), the peptide cannot travel through this network unless it is attached to something (e.g. the MHC).
