Blueprint Exam 3

During strenuous exercise, lactic acid buildup in cells causes the creation of a hydronium complex known as the Eigen cation (H9O4+). If water molecules then experience hydrogen bond attractions to the Eigen cation, this attractive force: A. results in a semi-stable shell of water molecules around the hydronium. Show Explanation B. results in an inability of hydronium to be neutralized by bases the way normal H+ ions would be. Show Explanation C. results in the ability of muscle cells to reverse both the hydronium creation process and the lactic acid creation process once sufficient oxygen is once again made available. Show Explanation D. results in mobility of hydronium within the environment surpassing the mobility of regular water molecules.

A A process known as "hydration" or "solvation" occurs when the attractive force of an ion molecule causes a thin shell of water molecules to surround it. In the case of hydronium (H3O+), each of the H atoms attracts the O atom in an H2O molecule due to hydrogen bonding. These H2O molecules cause a "shell" of water molecules to surround the hydronium. D is wrong bc: The effects of hydration shells have nothing to do with enhancing the mobility of hydronium.

A woman visits her doctor to receive medical test results. While the patient anxiously waits, the doctor stops in to shake hands before going to analyze the results. The patient mistakes the handshake as a cold dismissal rather than a warm greeting. This is an example of which sociological theory? A. Symbolic interactionism Show Explanation B. Social constructionism Show Explanation C. Exchange-rational Show Explanation D. Feminist theory

A According to symbolic interactionsim, there are 3 core principles to communication: meaning, language and thought. These core principles lead to conclusions about the creation of a person's self and socialization into a larger community. Meaning states that humans act toward people and things according to the meanings that give to those people or things. Symbolic interactionism holds the principal of meaning to be the central aspect of human behavior. Language gives humans a means by which to negotiate meaning through symbols. Humans identify meaning in speech acts with others. Thought modifies each individual's interpretation of symbols. Thought is a mental conversation that requires different points of view. This scenario is an example of symbolic interactionism. C:posits that patterns of behavior in societies reflect the choices made by individuals as they try to maximize their benefits and minimize their costs.

A cell's ATP-to-ADP ratio is most nearly equal to the ratio of its fluorescence emission intensities when Y32 is excited at: "The protonated chromophore, called the "A-state," is favored in the ADP-bound conformation, while the deprotonated chromophore, which is called the "B-state," is favored in the Mg2+-ATP-bound conformation. The fluorescent intensity produced in a cell expressing the sensor is proportional to the number of sensor molecules excited. The difference in fluorescence between the Mg2+-ATP-bound biosensor and the ADP-bound biosensor peaks is seen in their fluorescence excitation spectra (Figure 2). The spectral ratio of the peaks of the excitation spectrum directly reflects the cellular ATP-to-ADP ratio." Graph shows B form peak at 500 nm and A form peak at 420 nm A. 500 nm versus when it is excited at 420 nm. Show Explanation B. 420 nm versus when it is excited at 500 nm. Show Explanation C. 500 nm versus when it is in its ground state at 500 nm. Show Explanation D. 420 nm versus when it is in its ground state at 420 nm.

A According to the passage, the spectral ratio of the peaks of the excitation spectrum of a Y32-expressing cell directly reflects its cellular ATP-to-ADP ratio. The peak absorbance of the B-state, which is favored in the Mg2+-ATP-bound conformation, is near 500 nm, while the peak absorbance of the A-state, which favors the ADP-bound conformation, is near 420 nm. Thus, the ratio of the fluorescence emission intensities when Y32 is excited at 500 nm versus when it is excited at 420 nm serves as a direct indicator of the cellular ATP-to-ADP ratio. *I chose B bc I was not sure about what it was asking, but after looking at the wording in P to Q, just asking for direct reference to what is said in P. C and D are wrong bc comparing to ground state at a certain wavelength isn't going to do anything based on P info.

According to the information presented in the passage, how would the absorbance at 570 nm change over time for enzyme A if a linear starch was used? Table:[Enzyme: Absorbance at 2 mins; avg current] Enzyme A: 0.62; 0.0mA B: 0.36;10mA C: 0.37; 30mA D: 0.79; 0.1mA A. It would remain relatively constant. Show Explanation B. It would decrease. Show Explanation C. It would increase. Show Explanation D. It would initially increase and then decrease as the starch is consumed.

A Based on Figure 3, enzyme A increased absorbance with no change in current. Thus, enzyme A likely breaks down 1,6 linkages, producing more linear polysaccharides, thereby increasing absorbance. However, if the initial starch is an entirely linear polysaccharide, then a debranching enzyme like enzyme A would have no effect on absorbance, and it would remain relatively constant over time.

The AAVrh.10 viral vector was most likely which kind of virus? "Researchers developed an alternative strategy to deliver the coding sequence for an anti-EGFR antibody directly to the CNS via an adeno-associated virus serotype rh.10 (AAVrh.10) gene transfer vector. This would bypass the blood-brain barrier and produce local, sustained therapeutic antibody levels. When administered directly to the CNS in a mouse xenograft model, the AAVrh.10 vector (AAVrh.10CetMab) mediated high levels of protein expression, particularly in neurons." A. A DNA virus Show Explanation B. A positive-sense RNA virus Show Explanation C. A negative-sense RNA virus Show Explanation D. A retrovirus

A Based on the passage, DNA coding for the antibody cetuximab was spliced into the viral vector, resulting in AAVrh.10CetMab. The only way that this splicing process could take place is if the viral genetic material were made of DNA.

The individual thermodynamic contribution of W (RM)chain A was found to increase the interaction energy of the MKR681H dimer. If so, what must be true for chain A? "Interaction energies between chains A and B of both enzymes were calculated in the equilibrated phase according to the equation: W (RM)int = W (RM)dimer - W (RM)chain A - W (RM)chain B Equation 1 where W (RM) is the effective energy of a protein with coordinates RM in solution. W (RM) is given separately by the equation: W (RM) = Hintra + ΔGsolv Equation 2 where Hintra is the intramacromolecular enthalpy, consisting of positive bonded and non-bonded energy terms, and ΔGsolv is the solvation free energy." A. ΔGsolv < 0 Show Explanation B. ΔGsolv = 0 Show Explanation C. W (RM)chain A > 0 Show Explanation D. Hintra < 0

A Consider the following expression: W (RM)int = W (RM)dimer - W (RM)chain A - W (RM)chain B If W (RM)chain A increases the interaction energy of the MKR681H dimer, W (RM)int, W (RM)chain A must be a negative quantity, such that the -W (RM)chain A term in the equation becomes a positive value. Equation 2 also tells us that W (RM) = Hintra + ΔGsolv. If W (RM)chain A < 0, then one or both of the terms Hintra and ΔGsolv must be negative. According to the passage, the intramacromolecular enthalpy term Hintra accounts for positive terms. For this reason, it must be true that ΔGsolv < 0. **basically this is saying chain A increased Wint (at first thought it was Wdimer, but it's not bc it says interaction energies) and A is being subtracted in equation 1 so to increase Wint, you have to subtract a negative. then plug this into Equation 2 and if A is negative here, either Hintra or Gsolv has to be negative and in the Passage it states that Hintra is "positive bonded..." so has to be A.

When an odorous molecule binds to olfactory receptors, the cell transduces the information into an electrical signal that travels to the brain for processing. Which of the following accurately describes the state of the voltage-gated channels on this cell during the relative refractory period? A. Na+ channels are de-inactivated, and K+ channels are activated. Show Explanation B. Na+ channels are inactivated, and K+ channels are activated. Show Explanation C. Na+ channels are de-inactivated, and K+ channels are inactivated. Show Explanation D. Na+ channels are inactivated, and K+ channels are inactivated.

A First, we need to understand the difference between the absolute and relative refractory periods. The absolute refractory period lasts nearly the entire duration of an action potential, during which time a second action potential cannot be generated. During this time, voltage-gated sodium channels are "inactivated." If this term is not familiar, note that voltage-gated sodium channels have two gates that must be open for sodium to flow inward and depolarize the cell membrane. If the "inactivation gate" is closed, the channels are inactivated. If the inactivation gate is open but the activation gate is closed, the channel is "de-inactivated" — it isn't inactivated, but it is not yet open either.This "de-inactivation" occurs once the action potential nears its end and the membrane voltage becomes sufficiently low (generally during the hyperpolarization phase). At this time, the inactivation gate opens and the activation gate closes. Since the channel is not inactivated, a stimulus could theoretically produce an action potential, but since the cell is hyperpolarized, this stimulus would need to be larger than normal. This interval is termed the relative refractory period, which is what this question asks about. At this time, again, sodium channels are de-inactivated; potassium channels are still activated, allowing potassium to continue flowing out of the cell to finish the action potential.

Malonyl-CoA is an intermediate in cytosolic fatty acid biosynthesis. Its inhibition of the β-oxidation of long-chain fatty acids: "In a related experiment, the researchers found that high cytosolic concentrations of malonyl-CoA caused mitochondrial β-oxidation of fatty acids 12 carbons and longer to cease, and the concentration of fatty acyl-CoA decreased." A. prevents the degradation of newly synthesized fatty acids. Show Explanation B. ensures that ample reducing equivalents are available for fatty acid synthetic activity. Show Explanation C. increases the cellular pool of acetyl-CoA available to condense with oxaloacetate. Show Explanation D. is up-regulated in the presence of decreased insulin secretion.

A Malonyl-CoA, an indicator of ongoing fatty acid synthesis, inhibits β-oxidation by preventing the movement of long-chain acyl groups into the mitochondrial matrix, thereby preventing a futile cycle of fatty acid synthesis followed by immediate β-oxidative catabolism of those newly synthesized fatty acids. **malonyl-CoA is an inhibitor of B-oxidation and intermediate of fatty acid synthesis-- read this wrong

Which of the following best explains the tumor reduction capacities of the isotopes observed in the study? "Table 1: shows GKS-Co as Relative Tumor Volume at 1 week post-surgery to be 0.58, 3 months: 0.33, 6 months: 0.12. GKS-X at 1 week: 0.62, 3 months: 0.17, 6 months: .072." A. GKS-X emits more radiation per unit time, causing more cell death. Show Explanation B. GKS-X emits narrower radiation streams that target the tumor. Show Explanation C. GKS-Co anti-tumor emissions decompose much more slowly than GKS-X emissions. Show Explanation D. GKS-Co does not emit sufficient radiation to kill tumor cells.

A Table 1 shows that GKS-X is more effective at reducing tumor size (as shown by its lower RTV values over time). We need to choose a statement that can explain this improved ability of GKS-X to kill cancer cells. We must think about the differences between X and 60Co, chief among them the five fold difference in half-life. Because X has a shorter half-life, it is logical that it would release more radiation in the same period of time because it undergoes decay more quickly. This higher dose of radiation would kill more cells, cancerous and non-cancerous, leading to a larger reduction in tumor size and more undesirable side effects *I chose C bc of the longer half life of GKS-Co (5 yrs) compared to GKS-X with 12 months but if GKS-Co decomposed SLOWER that would mean it would eventually catch up and would be releasing more radiation over time than GKS-X but this is not the case: If GKS-Co involved radioactive material that decayed much more slowly than that in GKS-X, we would expect the GKS-Co tumor-reducing effects to catch up to or surpass GKS-X as time passed. Instead, we can see in Table 1 that the disparity gets larger as time post-surgery increases.

Which of the following processes would NOT be a likely target for anti-IAV therapy? "After receptor binding, IAVs enter cells through receptor-mediated endocytosis. Then, in the low-pH conditions of the late endosome, HA undergoes an irreversible conformational shift in which the N-terminus of the HA2 subunit inserts into the endosomal membrane, causing the viral and endosomal membranes to fuse. The viral M2 protein forms an ion channel, through which proton influx further acidifies the interior of the virus particles. This causes the viral matrix protein (M1) to dissociate from viral ribonucleoprotein (vRNP) complexes, allowing vRNP release into the cytoplasm, from which it is imported into the host nucleus. IAV genomic RNA replication and transcription are catalyzed in the host nucleus by a trimeric viral polymerase complex, composed of PB1, PB2, and PA subunits.The viral protein M2 mediates the release of IAV from the host membrane, and the enzymatic activity of neuraminase (NA) plays an essential role in preventing virion aggregation during budding." A. Proton influx via ion channels in the cell membrane Show Explanation B. Interactions between hemagglutinin and sialic acid-galactose dimers Show Explanation C. Activity of the viral polymerase complex Show Explanation D. Neuraminase activity

A The passage describes proton influx through ion channels in the endosomal membrane, and gives no indication that proton influx via ion channels in the cell membrane is involved in the IAV life cycle. **I assumed proton influx meant increasing the influx which would help, but it's not saying this. So i got it right for wrong reasons.

Which of the following types of electromagnetic radiation would have the shortest wavelength? A. Radiation that ejects an electron from an sp orbital Show Explanation B. Radiation that ejects an electron from an sp2 orbital Show Explanation C. Radiation that ejects an electron from an sp3 orbital Show Explanation D. Radiation that excites but does not eject an electron from an sp3 orbital

A This question asks us about the wavelength of EMR that ejects an electron from an atom. Shorter-wavelength EMR (such as γ rays)carries much more energy than longer-wavelength EMR (such as radio waves). Therefore, we must look for the answer choice that involves the highest-energy EMR. The closer an electron is to the nucleus, the harder it is to eject. Because sp-hybridized orbitals have the most s character of all of the answer choices, they contain the electrons that are hardest to eject **Bc sp orbital has e- closer to nucleus due to higher s%, they are harder to eject and require more energy.

Which of the following is least likely to be observed in a patient experiencing hyperventilation? A. Hypoxia Show Explanation B. Net exhalation of CO2 Show Explanation C. Increased blood pH Show Explanation D. Increased hemoglobin O2 affinity

A This question asks us to determine the effects of hyperventilation. During hyperventilation, there is a loss of CO2 and an increase in O2 in the blood. Hypoxia is another term for oxygen deprivation. D is wrong bc: Loss of CO2 corresponds with increased hemoglobin affinity for O2.

If sodium sulfate was added to the mixture containing silver ions and the yellow precipitate, what might be observed after a significant amount of time elapsed? (Yellow precipitate is AgPO4) A. Little to no silver sulfate formation, because the Ksp of silver sulfate is very large compared to the Ksp of the yellow precipitate. Show Explanation B. Little to no silver sulfate formation, because sodium sulfate cannot be dissolved at all in water. Show Explanation C. Significant silver sulfate formation, because the solubility of sodium sulfate releases sodium ions which catalyze the necessary reactions. Show Explanation D. Significant silver sulfate formation, because the solubility of sodium sulfate releases sulfate ions which catalyze the necessary reactions.

A This question is asking us to find a possible outcome and explanation for the outcome when adding sodium sulfate to the silver/yellow precipitate mix. This means we need to choose the answer that has a valid outcome and a proper explanation for that outcome. The Ksp, or solubility product, of a substance is defined as the product of each of the substance's dissolved ion concentration raised to the power of its stoichiometric coefficient. For silver sulfate, the dissolution reaction is: Ag2SO4 (s) → 2 Ag+ (aq) + SO42- (aq) which means that the Ksp equation is: Ksp =[Ag+]2[SO42-] A large Ksp suggests that a substance is more soluble than a substance with a lower Ksp. It is not certain, given that the coefficients might be different and might affect the power to which the ion concentrations are raised. However, it provides a better possible explanation than the other answer choice options (and the phrasing of the question prompt, using the words "might explain," notes that there is some uncertainty involved). **AgPO4 is insoluble whereas AgSO4 is soluble which means Ksp should be large indicating solubility

The electrical resistance of dry skin is 100 kΩ, but can be lowered to 20 Ω if electrode contact area is large and conducting gel is used on the skin. If Vmax of the defibrillatror 500 V and lasts 0.01 s, what is the maximum possible current delivered to the heart during defibrillation? A. 2.5 × 101 A Show Explanation B. 2.5 × 10-2 A Show Explanation C. 5.0 × 10-2 A Show Explanation D. 5 A

A We are given R, V, and time, and the question asked us to determine current (charge/time, or C/s). We do not need to use the time since we should know Ohm's law, V = IR. If I = V/R and the question asks for the maximum current, we should use the smallest available resistance. I =V/R = (500 V) / (20 Ω) = (50) / (2) = 25 amps = 2.5 × 101 A

Salicylic acid is a benzoic acid derivative which can most directly be obtained though the Kolbe-Schmitt carboxylation (shown below). NaOH CO2 ?----> Salicylic acid HCl (Salicylic acid has carboxylic acid on C1 of phenyl ring and OH on C2 of the ring) "C7H6O3 + C4H6O3 → C9H8O4 + C2H4O2" This is most likely to work on: A. a phenol. Show Explanation B. a phenyl aldehyde. Show Explanation C. a phenyl ether. Show Explanation D. a phenyl ester. Show Explanation

A We are told that the reaction in question is a carboxylation of the original reactant, indicating that we are adding a -COO group to form salicylic acid. We can see from Reaction 1 that salicylic acid not only has a carboxylic acid functional group,but also a hydroxyl group attached at what is likely the ortho position. Phenol is a benzene ring with a hydroxyl group attached. **bc Q says it's a carboxylation, take away the CO2 from final structure and pick what should be on molecule prior to this. I was getting too caught up in mechanism

A principal of a high school seeks to establish rules and systems in the school that reflect a meritocracy. Which of the following goals must these systems achieve if the principal is to successfully establish a meritocracy? I. Outcome equality II. Skill equality III. Opportunity equality A. I only Show Explanation B. III only Show Explanation C. II and III only Show Explanation D. I and III only

B A meritocracy is a society of people whose progress within the society is based on ability and talent rather than on class privilege or wealth. This requires that everyone be afforded the same opportunities to advance yet only be rewarded based upon individual outcomes due to their individual talents and/or abilities, which can vary between persons (III). C: wrong bc skill equality is not mandatory for meritocracy since people can vary w regard to their skills

According to French and Raven's bases of power model, when compared to a high motivation high knowledge individual, someone with low motivation and low knowledge is more likely to be influenced by a person with: A. expert power. Show Explanation B. referent power. Show Explanation C. legitimate power. Show Explanation D. coercive power.

B According to the bases of power model, an individual with referent power exerts control by appealing to others' desire to belong to a group. This type of control is most likely to appeal to individuals through external factors, such as appearing desirable or feeling included and not knowledge or logic or evidence. Thus, a low-motivation, low-knowledge individual would most likely be motivated by this type of persuasion. D:Those with coercive power exert control through force or its threat. It is likely that both high and low groups would be persuaded by this.

What is the pH of a 0.10 M aqueous solution of acetylsalicylic acid (pKa=3.5) A. 1.0 Show Explanation B. 2.3 Show Explanation C. 3.5 Show Explanation D. 4.1

B Acetylsalicylic acid is a weak acid, with a pKa of 3.5. Therefore, the pH of this solution must be less than the pKa, because the compound is primarily in its acid form and a pH of 3.5 would mean that the concentration of weak acid and conjugate base were equal (a buffer). Choices C and D can be eliminated. A pH of 1.0 for an acid whose concentration is 0.10 M would require complete dissociation (as in a strong acid), eliminating choice A. Alternatively, the pH can be determined from the equilibrium expression: Ka = [H+][A-]/[HA] 10-3.5 = x2/(0.1-x) Since x will be small, we can approximate 0.1 - x ≈ 0.1, giving: 10-3.5 = x2/(0.1) 10-3.5(0.1) = x 2 10-4.5 = x2 [10-2.25][10-2.25] = x2 10-2.25 = x Since x equals the hydrogen ion concentration, taking the -log(10-2.25) = 2.25. **or can do Ka and then do Henderson-Hasselbach, but this is longer than their example.

That there might be differences between this experiment and actual prison conditions represents a critique of which aspect of the study design? A. Construct validity Show Explanation B. Ecological validity Show Explanation C. Researcher bias Show Explanation D. Reliability

B Ecological validity refers to how findings from an experimental setting can be generalized to the environmental considerations in the real world. A: Construct validity refers to whether measures actually do assess the variables that they are intended to assess. For example, issues with how verbal abuse was defined in this study might pertain to construct validity. D:Reliability refers to the likelihood that results could be replicated.

According to the reported responses, which of the following changed for Student E? "After exams, some of my friends raved about how much it helped them concentrate. I used to think it was a terrible habit, but if study pills helped my grades it must be a good thing!" A. Implicit attitudes towards study pills Show Explanation B. Explicit attitudes toward study pills Show Explanation C. Covert drug use behavior Show Explanation D. Attitude polarization

B Explicit attitudes are conscious attitudes. Student E was aware of his attitudes towards smoking and how they changed over time. He once viewed smoking as a negative habit; now he sees it as a means to an end. A:Implicit attitudes towards smoking are unconscious attitudes.

A defect in the absorption of certain nutrients in the small intestine has been linked to an abnormally-delayed initiation of differentiation. If the levels of HER2, PTEN, and Akt protein can be estimated accurately, what is the earliest point in development at which the defect could be evaluated? (Figure 1: HER2 Receptor>P13K>Akt>mTOR> Cell proliferation and differentiation with PTEN acting as an inhibitor) (Figure 2 shows HER2 is first activated at 14 weeks, Akt is first activated at 16 weeks and PTEN is first activated at 20 weeks) A. 14 weeks Show Explanation B. 16 weeks Show Explanation C. 20 weeks Show Explanation D. 24 weeks

B From Figure 1, it is clear that the initiation in differentiation results from the downstream signaling from Akt, and, thus, differentiation would not occur until the levels of Akt expression increase. From Figure 2, it should be noted that Akt levels are first noticed in the small intestine during week 16; thus, you would not expect differentiation to occur before week 16. **bc Q asks about DIFFERENTIATION you need to look at both Figure 1 and 2 to see that Akt has to be produced for differentiation and that it starts being produced at week 16) I chose C because I looked at when all of the proteins were activated

Efforts to treat lactic acid buildup in muscles were attempted using dissected muscle specimens in the laboratory. One of these experiments involved ammonium formation from dissolved ammonia. Under conditions of excessive lactic acid: A. the final concentration of ammonium will be higher than otherwise due to higher pH. Show Explanation B. the final concentration of ammonium will be higher than otherwise due to lower pH. Show Explanation C. the final concentration of ammonium will be lower than otherwise due to higher pH. Show Explanation D. the final concentration of ammonium will be lower than otherwise due to lower pH.

B In an acidic environment, a base such as ammonia (NH3) will dissolve into its conjugate acid, ammonium (NH4+), to a greater extent than would have been the case in a neutral or a basic environment. An environment with lower pH is an environment that is more acidic.

Four fractions were obtained from the differential centrifugation of a homogenized human liver cell sample: F1: Inner mitochondrial membrane plus matrix F2: Outer mitochondrial membrane F3: Enzymes localized to the intermembrane space F4: Cytosolic enzymes In which of the following fractions was thiokinase most likely isolated? "In order to be β-oxidized, LCFAs must first enter the cell, where they are converted to their CoA derivatives by thiokinase, a membrane-bound enzyme. Because activated fatty acids, including LCFA-CoA derivatives, are impermeable to the inner mitochondrial membrane, they must be transported into the mitochondrial matrix by carnitine, a specialized carrier protein." A. F1 Show Explanation B. F2 Show Explanation C. F3 Show Explanation D. F4

B Paragraph 2 states that LCFAs must first enter the cell, where they are converted to their CoA derivatives by thiokinase, a membrane-bound enzyme. Also according to the passage, this conversion occurs prior to their transport into the the mitochondrial matrix by carnitine. This indicates that thiokinase must act after an activated fatty acid enters a cell, but before its transport into the mitochondrial matrix. Thiokinase must then be bound to the cytosolic face of the outer mitochondrial membrane, which makes this choice correct. C is wrong bc: The second paragraph states that thiokinase is a "membrane-bound enzyme." These fractions are not associated with a membrane.

In a third test, patients are given sentences that contain content that relate to their lives (such information having been gained through prior interviews with family members). Moreover, these sentences state facts that are either wrong (e.g. "Your niece is named Juliana" when in fact her name is "Julia") or distorted somehow. Using such sentences would likely show: A. increased recall but significantly lower recognition. Show Explanation B. proactive interference in the normal patients but not in the Korsakoff's patients. Show Explanation C. proactive interference in the Korsakoff's patients but not in the normal patients. Show Explanation D. proactive interference in both groups in the experiment.

B Proactive interference refers to the fact that currently existing long-term memories can interfere with the process of forming new long-term memories. This is unlikely to affect patients with Korsakoff's syndrome, as they are already unable to form new long-term memories. C:This choice may be tempting, since we might expect the Korsakoff's patients to be more likely to display a memory problem (here, proactive interference) than the normal patients. However, normal individuals are absolutely subject to proactive interference. In contrast, Korsakoff's patients already have such impaired memory formation (as evidenced by the results in Figure 1) that they are unlikely to display additional impairment as a result of previously formed memories. After all, if memory formation is already so impaired that it is nonexistent or almost nonexistent, there is little room for more impairment due to proactive interference. **PROACTIVE interference is forgetting the new info

ssuming that oligomycin has a high affinity for ATP synthase, which of the following is the best explanation for the findings presented in Table 1 regarding the amount of ATP produced with oligomycin pre-treatment? "Researchers studying the energy yield of fatty acid metabolism estimated the amount of ATP produced in the mitochondria of cultured human cells by the β-oxidation of saturated, straight-chain fatty acids of various lengths in both the presence and the absence of oligomycin, an inhibitor of ATP synthase (Table 1)." (Table shows fatty acids # of ATP produced being high in the 100s and # of ATP produced w oligomycin pre-treatment being in the 50s) A. Oligomycin binds directly to acetyl-CoA roughly 33% of the time. Show Explanation B. An insufficient concentration of oligomycin was used. Show Explanation C. Despite the inhibition of oxidative phosphorylation, glycolysis can still produce a smaller baseline amount of ATP. Show Explanation D. Substrate-level phosphorylation, instead of oxidative phosphorylation, accounts for roughly 33% of the ATP produced through the electron transport chain.

B Table 1 shows that oligomycin pre-treatment reduced the ATP production from fatty acids by about 67%. The acetyl-CoA generated from fatty acids is passed through the citric acid cycle, culminating in oxidative phosphorylation. Oxidative phosphorylation, which is powered by the electron transport chain, produces ATP by using ATP synthase. Thus, the information in the passage and question stem indicating that oligomycin effectively blocks ATP synthase would lead us to expect that ATP production from fatty acids would completely, or almost completely, cease in response to oligomycin pre-treatment. However, we don't see anything close to a complete reduction. Keeping in mind that for oligomycin to completely block ATP synthase, there must be enough oligomycin molecules in a location where they can physically interact with ATP synthase molecules, the simplest explanation for this finding is that the oligomycin concentration might not be high enough for complete effectiveness. C is wrong bc:Baseline ATP production through glycolysis would be a plausible explanation on the level of the cell, but the table provides the amount of ATP produced in the mitochondria by fatty acid metabolism, which generates acetyl-CoA, which cannot enter glycolysis.

Besides aspirin, what is the other product of Reaction 1? C7H6O3 + C4H6O3 → C9H8O4 + C2H4O2 A. Ethanol Show Explanation B. Ethanoic acid Show Explanation C. Ethanal Show Explanation D. Methyl formate

B The other product of Reaction 1 has a formula of C2H4O2. Ethanoic acid, a carboxylic acid, has this formula. The reaction between acetic anhydride will involve the acylation of the alcohol functional group of the salicylic acid, with the by-product being acetic acid, also known as ethanoic acid. D: Methyl formate is an ester and does have the correct formula of C2H4O2.However, Reaction 1 yields a by-product of ethanoic acid,which preserves the bond between the methyl group and the carbonyl carbon that was present in the acetic anhydride. For methyl formate to form instead, the methyl group would have to rearrange somehow (magic?) and bond to an oxygen atom to form the ester

A 60-kg runner raises his center of mass approximately 0.5 m with each step. Although his leg muscles act as a spring, recapturing the energy each time his feet touch down, there's an average 10% loss with each compression. What must the runner's additional power output be to account for just this loss, if he averages 0.8 s per stride? A. 0 W Show Explanation B. 37 W Show Explanation C. 46 W Show Explanation D. 331 W

B This is a multi-step question, though each step is relatively straightforward. The gravitational potential energy at the runner's height is: PE =(60 kg)(10 m/s2)(0.5 m) = 300 J Most of this energy is conserved as the runner hits the ground and his muscles capture the energy as spring potential energy, so the question only asks about the lost energy, amounting to 10%, or 30 J. Don't worry about the mechanics of which foot lands first and how much of the kinetic energy each one absorbs. The question stem doesn't give that kind of information, so there's nothing for it anyway. So, an additional 30 J is needed per stride, and a stride occurs every 0.8 s. Thus: P =(30 J)/(0.8 s) = 40 W This is between answer choices B and C. So is it really 40 W? Well, no, 8 goes into 30 less than 4 times, so the answer should be lower. Also, g is closer to 9.8 than 10, so that also caused the value of energy, used in the power calculation, to be overestimated. 37 W, then, must be the match.

Which of the following would likely decrease an individual's self-efficacy in regard to a particular task? A. Seeing a task performed successfully Show Explanation B. Feeling that past failures were due to other individuals Show Explanation C. Being offered positive encouragement by others D. Learning stress-reduction techniques in regard to the task

B This is an example of an external locus of control, which might increase an individual's self-esteem, but not self-efficacy.

Which of the following discoveries would most support the statement that p53 is the most important protein regulator of DNA repair to have been discovered? A. When mutated, p53 continues to function to repair damaged DNA. Show Explanation B. More human tumors can be traced to a mutation in the p53 protein gene than in any other protein. Show Explanation C. Most cancerous cells lack the p53 protein. Show Explanation D. p53 levels are always elevated in patients with cancer.

B This question is asking us to determine the best support for a claim made in the passage. This choice directly relates p53 to the incidence of tumors in humans. More importantly, it supports the claim that p53 is the "most important" protein that may have a role in DNA repair, since p53 mutations are implicated in more tumors than are mutations in any other gene. I chose C but this is wrong bc: The simple fact that cancerous cells lack a protein do not mean that protein has anything to do with the incidence of that cancer (or with DNA repair). It could be that another protein deleted this protein or turned it off and something else is controlling for DNA repair

Which of the following best explains why arginine is more basic than lysine? A. The electron-donating groups around the basic nitrogen on arginine make its conjugate base less stable. Show Explanation B. The electron-donating groups around the basic nitrogen on arginine make its conjugate acid more stable. Show Explanation C. The lack of electron-donating groups on lysine make its conjugate acid more stable. Show Explanation D. The lack of electron-withdrawing groups on lysine make its conjugate base more stable.

B This question is asking us to determine why arginine is more basic than lysine. The reason must be related to how arginine is better able to handle being protonated, as this is the essence of being a base. Since, in its protonated form, arginine has electron-donating groups via resonance with other nitrogens, it is a more stable conjugate acid. The resonance structures of arginine at this position are shown below. Note that the backbone amino and carboxylic acid groups are deprotonated, meaning that this is the structure of arginine at relatively high pH (albeit not high enough to deprotonate the side chain).

Conn's syndrome, also known as primary hyperaldosteronism, is most likely to cause which symptom? A. High renin concentration Show Explanation B. Low blood potassium Show Explanation C. Low blood sodium Show Explanation D. Hypotension

B This question is asking you to recall the effects of aldosterone and how it achieves those effects. Aldosterone increases H2O and Na+ reabsorption from the kidney while exchanging K+ ions for Na+ ions. The triggers for and results of aldosterone secretion are shown below.

A student determined that her yield of aspirin was 3.9 g. What was her percent yield? "Aspirin (180 g/mol; pKa = 3.5) is easily prepared by reacting salicylic acid (138 g/mol; pKa = 2.97) with acetic anhydride (102 g/mol) as shown in Reaction 1. C7H6O3 + C4H6O3 → C9H8O4 + C2H4O2 Reaction 1 In a laboratory experiment, the following procedure was used to prepare aspirin. 1. Weigh out 5 g of salicylic acid (C7H6O3) and place it in a 125-ml Erlenmeyer flask. 2. Using a graduated cylinder, add 7 ml of acetic anhydride (d = 1.08 g/cm3) to the Erlenmeyer flask." A. 45% Show Explanation B. 60% Show Explanation C. 78% Show Explanation D. 92%

B We first need to determine if there is a limiting reagent and then determine the theoretical yield. The moles of acetic anhydride can be determined by using the volume and density information in step 2 of the procedure and the molecular weight given in the passage. 7 mL x 1.08 g/mL x 1 mol/102 g ≈ 7 x 10-2 mol We can also calculate the moles of salicylic acid from the number of grams used instep 1 and the molecular weight given in the passage. 5.0 g x 1mol/138 g ≈ (5/1.4) x 10-2 ≈ 3.5 x 10-2 mol Since the stoichiometry is 1:1 from Reaction 1, the acetic anhydride is in excess and the limiting reagent is the salicylic acid. The stoichiometric ratio between salicylic acid and the acetyl salicylic acid is also 1:1; therefore, the theoretical yield can be calculated using the molecular weight from the passage. 3.5 x 10-2 mol x 180 g/mol ≈ 3.5 x 2 x 10-2 x 102 ≈ 7 g Since the question indicates that the actual yield was 3.9 g, which is about 4 g, the percent yield is (4/7) x 100 ≈ 60%.

Which of the following contradicts Weber's law? A. A weightlifter who doesn't notice a 5-pound difference in weight added in 1 pound increments, but who notices the difference in weight added in one 5-pound increment Show Explanation B. A non-linear relationship between the intensity of a stimulus and an individual's ability to detect it Show Explanation C. The existence of a threshold, above which stimuli are detectable and below which stimuli are not detectable Show Explanation D. Hearing a whisper in a quiet room but not hearing a shout in a noisy room

B Weber's law postulates that there is a linear relationship, not a non-linear relationship, between the intensity of a stimulus and its detection.

When does a runner output the most additional energy to keep the ground reaction forces most nearly vertical and through her body's center of mass? "The pitch moment is the torque on a runner resulting from a horizontal force in line with the direction of motion either above or below the center of mass. To put it another way, it's the tendency to fall forward, and must be constantly counter-acted while running. Pitch can be minimized if the body applies counter-acting forces to ensure ground reaction forces are approximately vertical through the center of mass. In this case, the force and energy required to maintain such a stance are: Fx=Fztan(φ) Wx=Fztan(φ)Δz where F x is the horizontal force required, Fz is the ground reaction force, Δz is the vertical displacement of a single step, and φ is the angle of ground reaction forces with the vertical, related to the angle at which the body is leaning forward (i.e., the tendency to pitch is greater at higher angles, requiring higher counter-acting forces)." A. When she takes high, bouncing strides and keeps her spine fairly vertical Show Explanation B. When she takes long, low strides and keeps her spine fairly vertical Show Explanation C. When she takes high, bouncing strides and leans her upper half into her run Show Explanation D. When she takes long, low strides and leans her upper half into her run

C Both the vertical displacement of the runner's steps and the angle of her body from the vertical increase the energy required to realign the ground reaction forces. Recall that tanθ = sinθ/cosθ, so when θ is close to zero, so is tanθ. When θ is close to 90°, tanθ becomes arbitrarily large. The more the runner leans into the run, the greater (or closer to 90°) tanθ is, and the greater the work expended by the runner must be, according to the work equation given in the passage (Wx = Fz tan(θ) Δz). Also, in this equation, the passage states that Δz represents the vertical displacement of a single step. Thus, the higher the vertical displacement ("high, bouncing strides..."), the greater the energy expenditure.

Y32 is pH-sensitive. This can be a problem when employing it to monitor changes in the energy balance of a cell because: A. pH can differ between cells. Show Explanation B. the ATP-to-ADP ratio of a cell depends on pH. Show Explanation C. pH may change over time in the same cell. Show Explanation D. decreased pH favors the CPV B-state.

C If the biosensor is pH-sensitive, then changes in cellular pH over the course of a measurement could confuse changes in the energy balance of the cell, as reflected by the cell's ATP-to-ADP ratio B is wrong bc: While the ATP-to-ADP ratio in most cells does to some extent depend on pH, this does not address the question of how pH changes influence the utility of the biosensor in measuring a cell's ATP-to-ADP ratio. **ANSWER the Q they ask!!

Which of the following would be LEAST useful in cellular movement? A. Flagella Show Explanation B. Actin polymerization Show Explanation C. Microtubule depolymerization Show Explanation D. Cilia

C In order for cells to travel to the site of injury, they need to migrate. Microtubule de-polymerization is responsible for separating chromosomes during anaphase of mitosis or meiosis I or II. It does not contribute to overall cell migration. B is wrong bc: Rapid actin polymerization near the edge of the cellular membrane is responsible for cellular motility in complex eukaryotic cells.

The passage of IgG antibodies from mother to fetus illustrates: A. natural immunity. Show Explanation B. cell-mediated immunity. Show Explanation C. passive immunity. Show Explanation D. nonspecific immunity.

C Passive immunity is the transfer of active humoral immunity in the form of ready-made antibodies, from one individual to another. **I picked C but thought it could be A also since antibodies through breast milk were described in anatomy as natural passive, but I guess the passage of the antibodies is more describing a passive process

How many liters of carbon dioxide at STP are produced by reacting 100 g of calcium carbonate with an excess of hydrochloric acid? A. 0.0 L Show Explanation B. 11.2 L Show Explanation C. 22.4 L Show Explanation D. 44.8 L

C The balanced chemical reaction is: CaCO3 (s) + 2 HCl (aq) → CO2 (g) + H2O (l) + CaCl2 (aq) From the periodic table, the formula weight of calcium carbonate is 40 + 12 + 3(16) = 100 g/mol. 100 g of calcium carbonate therefore represents one mole, and based on the reaction, this will produce one mole of carbon dioxide gas. Remember, one mole of any gas at STP (standard temperature and pressure) has a volume of 22.4 L! Thus, the volume of gas produced will be 22.4 L at STP. 100 g CaCO3 x 1 mol/100 g x 1 CO2/1 CaCO3 x 22.4 L/mol = 22.4 L CO2

A second student repeated the experiment using glucose and the equivalent enzymes of glycolysis instead of starches. How would his results compare to those shown in Figure 3? "To test enzymes that could potentially digest starch in a battery, the student created four solutions by adding 100 µmol of starch, electrolytes, cofactors, AQDS molecules, and electrodes to 50 ml of buffered saline solution. Five µmol of phosphoglucomutase (PGM) and enzyme X were then added to each solution along with several drops of iodine. Iodine selectively binds linear chain polysaccharides, and when bound, it changes the color of the solution from brown to purple with a characteristic peak absorption wavelength of 570 nm corresponding to the purple color. The student initially observed no current and measured the initial absorbance at 570 nm for the solution using UV-Visible spectroscopy. Finally, the student tested the ability of four enzymes (A, B, C, and D) to break down starch by adding 5 µmol of each enzyme to separate solutions. The average current over 2 minutes was measured, and after 2 minutes, the absorbance at 570 nm was measured again for each solution. The results are shown in Figure 3." A. Purplish color with higher initial current Show Explanation B. Purplish color but no current because of incomplete glycolysis Show Explanation C. Brownish color with higher initial current Show Explanation D. Brownish color with lower initial current

C The first difference between the first student and the second is the use of glucose, a monosaccharide (shown below), as the fuel. Paragraph 2 states that iodine binds selectively to linear-chain polysaccharides. As a result, the iodine will NOT bind the glucose, causing the solution to remain brown. We can eliminate choices A and B. Next, we must determine the effect of using glucose and the glycolytic enzymes on the current generated by the battery. From Figure 1, we can infer that the amount of current generated should be directly related to the production of NADH. Unlike in the biostarch battery, glycolysis does not require the breakdown of starch. Thus, NADH is generated directly and more quickly via the conversion of glyceraldehyde 3-phosphate (GAP) to 1,3-bisphosphoglycerate (1,3-BPG). The enzymes should produce 1 NADH per GAP molecule, or 2 NADH per glucose molecule (since each unit of glucose breaks down into two GAP molecules). The original experiment only produced 1 NADH per glucose conversion to phosphogluconate, so we can conclude that the second student should observe a higher initial current than the first. **also it says "initial" current so you can assume that since glucose is entering the cycle right away, it would a higher current faster rather than starch which takes time to be broken down.

If a 3-kg rabbit's leg muscles act as imperfectly elastic springs, how much energy will they hold if the rabbit lands from a height of 0.5 m and its legs are compressed by 0.2 m? A. -0.6 J Show Explanation B. 0 J Show Explanation C. 13 J Show Explanation D. 14.7 J

C This is a tricky question, because the passage states that nature, i.e., biology, has no perfectly elastic springs, which means energy cannot be completely conserved. This does not, however, mean that all the energy is lost, nor does it make sense for the potential energy stored in a spring to be negative. So, if energy were completely conserved, what would it be? Potential energy of the rabbit at the peak of its height is PE = (3)(10)(0.5) = 15 J, or, if we're more exact and use 9.81 for g, slightly less than that. 14.7 J seems about right. 13 J is clearly a little on the low side, but this is exactly what's needed for this question, since the question stem is dropping a big hint that some energy is lost. **bc it states it's a IMPERFECTLY elastic spring, it can't be equal to GPE

The passage data regarding the thermal stability and enzyme activity of MKR681H is most consistent with what conclusion regarding the role of Arg681 in CCT? "Finally, the thermal stability of MKR681H was characterized and its inactivation temperature (Ti), the temperature at which 75% of its activity is lost, determined." (Figure 2 shows Relative enzyme activity % on y and incubation temperature on x with MK(WT) being higher and having higher Ti than MK(R681H) being lower with lower Ti) A. Arg681 replacement reduces the CTP binding affinity of MKR681H. Show Explanation B. Arg681 provokes increased heat sensitivity in the dimer. Show Explanation C. Arg681 is engaged in the catalytic function of the enzyme. Show Explanation D. Arg681 absence promotes dissociation of the CCT dimer.

C This question is NOT asking about the mutant. R681H denotes that amino acid 681 (arginine, R) is changed to a histidine (H). The question is asking what we can conclude about the wild-type enzyme given the data presented in the passage. Examining Figure 1 for the enzyme activity data plotted for MKR681H and MKWT shows that substitution of arginine by histidine decreases maximum enzyme activity. Enzyme activity depends principally on an enzyme's intrinsic catalytic efficiency, its concentration, the initial substrate concentration, the presence of inhibitors or allosteric activators, temperature, and pH. In the experiment performed in the passage, all of these factors were controlled for between the enzyme variants other than their intrinsic activity. Of the choices given, this most strongly suggests that Arg681 is involved in the catalytic function of the normal enzyme. B is wrong bc: This is a tempting answer but it is actually an opposite answer trap. The question is asking about what we can say about the normal enzyme form (the one with R). In Figure 2, the Ti for MKR681H is lower than the Ti of the wild-type enzyme. This means that the mutant loses 25% of its enzymatic activity at a lower temperature than the wild-type enzyme. We can also tell from the definition of Tithat Ti is inversely related to heat sensitivity - a lower Ti means the protein is more heat sensitive aka less heat stable. Since the mutant has a lower Ti (when the R is replaced by H) this tells us that the mutant is less heat stable (i.e. more heat sensitive) compared to the wild-type enzyme with the Arg681 in it. Thus, the arginine confers more heat stability aka decreased heat sensitivity than the histidine in the mutant. **picked B because I just saw "heat" and didn't distinguish b/t heat sensitivity vs STABILITY.

A hospital purchases brand-new GKS-Co and GKS-X machines. Five years after installation, what is the expected ratio of the total atomic mass of material in the Co machine to that in the X machine, assuming both machines start with the same mass of radioactive material? A. 1:16 Show Explanation B. 1:5 Show Explanation C. 1:1 Show Explanation D. 5:1

C When reading questions, be careful not to read too quickly. In this case, fast but inefficient reading will lead us to assume that it is asking about the percentage of a certain isotope that is left after radioactive decay. However, the question is asking about atomic mass. While β-decay does cause a nuclear transmutation of protons to neutrons (β-plus) or neutrons to protons (β-minus), the atomic mass lost in these processes is negligible. This means that whether after one (Co) or five (X) half-lives, the atomic mass will be the same in both samples. D would be right if it were asking about decayed material, but it's asking about total atomic mass: This choice involves the differences in half-lives between 60Co (5 years) and X (1 year). However, since the question is asking about undecayed and decayed material, this distinction does not matter.

A balloon has a volume of 3.0 L at 25°C. What is the approximate volume of the balloon at 50°C? A. 1.5 L Show Explanation B. 2.0 L Show Explanation C. 3.3 L Show Explanation D. 6.0 L

Charles' law states that there is a direct relationship between the volume of an ideal gas and its temperature, when pressure is constant. Note that the temperature must be in Kelvin! We can approximate the initial temperature as 300 K and the final temperature as 320 K. (3.0 L)/(300 K) = V2/(320 K) [(3.0)(320)] / (300) = (3.0)(1.1) = 3.3 L

creativity requires

Creativity requires several characteristics, including flexibility, openness to experience, new ideas, an internal locus of evaluation, an ability to toy with elements and concepts, perceiving freshly, concern with outside and inside worlds, ability to defer closure and judgment, and skilled performance of the traditional arts, among others.

A student theorizes that the differences in ATP-to-ADP ratios between cells detected in an experiment could be due to variations in Y32 expression rather than differences in the cells' metabolic conditions. Does information presented in the passage support this possibility? "The fluorescent intensity produced in a cell expressing the sensor is proportional to the number of sensor molecules excited. The difference in fluorescence between the Mg2+-ATP-bound biosensor and the ADP-bound biosensor peaks is seen in their fluorescence excitation spectra (Figure 2). The spectral ratio of the peaks of the excitation spectrum directly reflects the cellular ATP-to-ADP ratio." A. Yes, Y32 fluorescence intensity is proportional to the number of sensor molecules excited. Show Explanation B. Yes, Y32 expression levels can influence the cellular ATP-to-ADP ratio. Show Explanation C. No, the shape of the fluorescence excitation spectrum of Y32 is influenced by the bound state. Show Explanation D. No, the spectral ratio intrinsically normalizes for the amount of biosensor.

D According to the passage, the spectral ratio of the peaks of the excitation spectrum of a Y32-expressing cell directly reflects its cellular ATP-to-ADP ratio. The peak absorbance of the B-state, which is favored in the Mg2+-ATP bound conformation, is near 500 nm, while the peak absorbance of the A-state, which favors the ADP-bound conformation, is near 420 nm. Thus, the ratio of the fluorescence emission intensities when Y32 is excited at 500 nm versus when it is excited at 420 nm serves as a direct indicator of the cellular ATP-to-ADP ratio. If, as stated in the passage, the fluorescence intensity is proportional to the number of sensor molecules excited, then excitation at both 500 nm and 420 nm would change proportionally with changes in biosensor expression, and would have no impact on determining the ratio in which ATP and ADP are present in the cell. *I was confused when first read this Q but since Y32 is a biosensor, it wouldn't make sense that it would impact the concentration of ATP/ADP so A is wrong: While A is true, it does not take into account the proportional increased fluorescence corresponding to both the ATP- and ADP-sensitive peaks. The ratio of ATP to ADP in the cell will not be affected. C is wrong bc: While this is true, it does not provide a way to distinguish between the two possibilities described in the question stem (i.e. variations across cells, rather than variations within a cell's metabolic activity).

Which of the following represents a limitation to the design of this study? "Overgeneralization of conditioned fear may be relevant to anxiety disorders like obsessive compulsive disorder (OCD).... A study was run to determine the degree to which individuals with obsessive-compulsive (OC) traits generalize conditioned fear when compared to healthy participants." A. The age of the participants Show Explanation B. Lack of knowledge about OCD Show Explanation C. Ethical implications of delivering shocks to participants Show Explanation D. Using a non-clinical patient sample

D Any findings during this study would be more valid if the researchers could have gathered patients who had been diagnosed with OCD, instead of having participants that simply had obsessive-compulsive traits. An important note here is that this limitation specifically pertains to the goal of the study, which was conducted to investigate overgeneralization. In turn, paragraph 1 tells us that overgeneralization is linked to OCD. Therefore, conducting a study among participants with OCD, perhaps in addition to those with only obsessive-compulsive traits, would have strengthened the findings of the study. C is wrong bc: conducting research ethically does not necessarily avoiding all techniques that might be uncomfortable and delivering shocks is fairly common.

If a flat Petri dish containing a single layer of cells suspended in viscous culture medium is tapped, some cells will collide into each other. If cell 1 collides into cell 2 on such a plate, which of the following describes what happens to cell 2 after the collision, assuming that it undergoes an elastic collision and experiences drag from the medium? I. Cell 2 continuously accelerates. II. Cell 2 moves with decreasing speed. III. Cell 2 moves with constant speed. IV. Cell 2 decelerates until it reaches a velocity of 0 m/s. A. I only Show Explanation B. II only Show Explanation C. III only D. II and IV only

D Cell 2 can only accelerate while it is being pushed/in contact with cell 1, so statement I is false. Since the cells experience drag from the medium, cell 2 will decelerate after the collision, making statement II correct and statement III incorrect. The drag will cause cell 2 to continuously decelerate until it comes to rest.

Which of the following is NOT a strategy to induce compliance in a target group, as described in the passage? "In the Stanford Prison Experiment, researchers assigned a number of college-aged participants to portray either prisoners or guards in a mock prison. The researchers were interested in assessing the effect this contrived situation had on the attitudes and behavior of both the prisoners and guards. The experiment began with the prisoners being arrested at their residences and then processed in a manner similar to what might occur in a genuine prison, including having the prisoners stripped of clothes, sprayed with a cleaning spray, given a uniform, and given a number, which is what they were called. Guards also wore uniforms. There was a typical prison schedule that the prisoners and guards followed. There was little instruction from the researchers about how the guards and prisoners should behave." A. A dress code at an organization Show Explanation B. Referring to military recruits by their military number Show Explanation C. Hazing rituals in a club involving humiliation Show Explanation D. Pointing out differences between one group of students' performance and another group's performance

D Compliance is induced in groups that view themselves as similar. The other strategies were specified in the passage as ways to induce compliance. *compliance is b/t groups that are similar and A, B, and C all show groups who are similar whereas D shows 2 diff groups

All of the following are potential limitations of the research discussed in the passage EXCEPT: "Researchers sought to address the question of whether monthly welfare check distribution and accompanying bias affects police interactions with mental health patients, and any impact this has on ED activity." A. Seattle is a busy urban center, with some core regions of the city having a high concentration of inhabitants with mental illness and addictions. Show Explanation B. the impact of MHAs were not correlated to ED or police workload or flow. Show Explanation C. the increase in MHAs following income assistance payments may be related to other unknown and undetermined confounders. Show Explanation D. sociological activity, such as welfare check distribution, is known to be causally related to increased police or ED activity.

D If welfare check distribution were known to be an established causal element to the relationship observed between MHAs and welfare distribution, it would strengthen the implications of the study. A is wrong bc: This is a limitation of the study which make the results potentially not generalizable to entire regions. If Seattle has more people w mental illness then this can't be generalized to rest of US.

The Rorschach inkblot test is often used to identify distinct personalities in those with DID. This test is best categorized as a: I. subjective personality assessment. II. objective personality assessment. III. projective personality assessment. A. I only Show Explanation B. II only Show Explanation C. II and III only Show Explanation D. I and III only

D In subjective assessments, patients project their own subjective feelings, perceptions, and thoughts onto the assessment stimuli, yielding results that are open for inaccuracy (I). For example, physicians may reach a different conclusion despite seeing the same patient who says the same thing. Projective personality assessments require the participant to respond, and then their response is assessed for meaning (III).

A pregnant woman has her fetus genotyped, suspecting a numerical abnormality in the sex chromosomes. Restriction enzymes cut the DNA at common, X chromosome-specific site(s), and the product is subjected to SDS-PAGE. The mother's and father's DNA are also run, with the results summarized below. (Gel shows: Mom and fetus have band at top of gel of same appearance; mom and dad have band at bottom of gel of same appearance; fetus has very dark and smudged band at bottom of gel) Which of the following is the correct genotype for the child? A. Euploid female Show Explanation B. Euploid male Show Explanation C. Trisomy X with complete XCI Show Explanation D. Trisomy X with incomplete XCI

D Restriction enzymes cut near or at their recognition sequences. However, if the gene is methylated, this cutting cannot occur. The question states that they are using a common X-specific site, not the XIST site described in the passage. The mother's result shows one heavy strip at the top of the gel, indicating that it is bulky and did not travel the length of the gel, as it shows large pieces. We can infer that this is the inactivated and methylated X that was not cut in as many places. Both the mother and the father have a smear at the bottom of the gel, indicating smaller fragments, or the active X that was cut in many places. The fetus shows a band at the top, indicating an inactive X; however, it also has a much darker and broader smear, indicating more active X chromosomes. Therefore, this child has a genotype of XaXaXi, or Trisomy X with incomplete XCI.

Which of the following types of proteins would be most efficiently fractionated by an SEC column? A. Two proteins with very high molecular weights Show Explanation B. Two proteins with similar molecular weights Show Explanation C. Two proteins with very low molecular weights Show Explanation D. Two proteins with substantially different molecular weights

D SEC separates on the basis of differences in molecular weight, the technique's resolution should increase if the pair of proteins differ in their molecular weights to a greater extent. Too small or too large would mean they are both eluting together and those w similar molecular weights would also elute very similarly

Which of the following is NOT correct regarding the results of the FISH analysis? (Table shows Signals upon FISH analysis:Percent of unique chromosomal signals observed using XIST (X inactivation)-specific probe (note: numbers in parentheses represent the number of studies observed: [Signals: 0; 1; 2] Euploid female: 0.6(2); 99.4(321);0(0) Euploid male: 95.0(385);5(20);0(0) Triple X: 0(0); 47.6(59); 52.4(65) Klinefelter: .6(1); 62.1(100); 37.3(60) A. The incidence of incomplete XCI is less prevalent in patients with Klinefelter syndrome than those with Trisomy X syndrome. Show Explanation B. There is more statistical power for normal phenotypes than abnormal phenotypes. Show Explanation C. Euploid females may experience hypomethylation of their X chromosome. Show Explanation D. The incidence of incomplete XCI is more prevalent in patients with Klinefelter than those with Triple X syndrome.

D The FISH analysis results are found in Table 2, which shows that in the case of Triple X syndrome, about 47.6% show the XaXaXi genotype (incomplete XCI) and 52.4% show the XaXiXi genotype. For Klinefelter syndrome, 62.1% of patients show a XaXiY genotype, 37.3% show a XiXiY genotype, and 0.6% show the XaXaY genotype (incomplete XCI). Thus incomplete XCI is much more prevalent in Triple X syndrome (47.6%) than in Klinefelter (0.6%). This contradicts choice D, making it the correct answer to this "NOT" question. incomplete X inactivation happens when 2 Xs are unmethylated so Klinefelter's is incompletely X inactivated when two Xs are still there so XaXaY and triple X is incompletely inactivated when XaXaXi so for Klinefelter's look at the 0 column and for Triple X look at the 1 column

Examining the results in Figure 2, the common ion effect accounts for: I. the increased yellow precipitate observed in Figure 2 when AgNO3 is added. II. the increased yellow precipitate observed in Figure 2 when K3PO4 is added. III. the decreased yellow precipitate observed in Figure 2 when Ba(OH)2 is added. ^^ these are all true of the table "Attempts to replicate this were conducted at a macroscopic scale, and mass spectrometry showed the yellow precipitate to contain two absorption peaks, correlating to both a phosphate group and a silver component." A. I only B. II only C. III only Show Explanation D. I and II only

D The common ion effect occurs when a solution contains a salt that partially dissolves into ions, and an extra quantity of one of these ions is added. This results in the combination of this ion with existing dissolved ions, causing an additional quantity of the salt to precipitate out of solution. In this case,the partially-dissolved salt is Ag3PO4. Adding AgNO3 causes dissolution of extra Ag+ions, which then combine with already-present PO43- ions and precipitate out of solution into the yellow Ag3PO4 precipitate (identified with the mass spectrometer used in Figure 1). Similarly, adding K3PO4 results in dissolution of extra PO43- ions, which then combine with already-present Ag+ ions and precipitate into yellow Ag3PO4 (I and II).

The true weight of the p53 protein, as calculated by summing the weight of the amino acids encoded by its gene, is 43.7 kDa. Which of the following possible mechanisms would best explain this discrepancy from its estimated weight, as mentioned in the passage? "Perhaps the most important protein to have been discovered with regards to regulation of DNA repair is called p53, named so for the weight of the protein as estimated by SDS-PAGE." A. The protein migrated a greater distance because it contains many positively charged residues. Show Explanation B. The protein migrated a smaller distance because it contains many negatively charged residues. Show Explanation C. The protein migrated a greater distance because it contains many negatively charged residues. Show Explanation D. The protein migrated a smaller distance because it contains many positively charged residues.

D The final paragraph of the passage mentions that p53 is so named for its estimated molecular weight using SDS-PAGE. This question is asking us to determine a plausible reason for the 9.3-kDa discrepancy. SDS-PAGE is used to grant a uniform negative charge to all proteins in an assay. However, if the protein has enough charged residues on its own, it causes the measurements to be less accurate. If there are enough positive charges, the negative charge of the protein will not be as great as anticipated and the molecule will travel a smaller distance, which, to the individual running the assay, usually means the mass of the protein is greater and would lead to the discrepancy described. B is wrong bc: In SDS-PAGE, proteins migrate toward a positively-charged pole. As such, containing many negatively-charged residues would make a protein migrate a larger distance, not a smaller one.

Which pair of enzymes, if used simultaneously, will produce the greatest amount of glucose if the experiment is repeated? "Batteries that run on starch must break down the linear 1,4-D-glucose linkages and the 1,6-D-glucose linkages found at branch points (Figure 2), which is a complex, multi-step process.... Iodine selectively binds linear chain polysaccharides, and when bound, it changes the color of the solution from brown to purple with a characteristic peak absorption wavelength of 570 nm corresponding to the purple color. The student initially observed no current and measured the initial absorbance at 570 nm for the solution using UV-Visible spectroscopy. Finally, the student tested the ability of four enzymes (A, B, C, and D) to break down starch by adding 5 µmol of each enzyme to separate solutions." Table:[Enzyme: Absorbance at 2 mins; avg current] Enzyme A: 0.62; 0.0mA B: 0.36;10mA C: 0.37; 30mA D: 0.79; 0.1mA A. A and B Show Explanation B. A and C Show Explanation C. B and C Show Explanation D. C and D Show Explanation

D The greatest glucose production per mass of starch will occur when there is the greatest rate of starch breakdown. For complete breakdown, this requires both enzymes that break the 1,4 linkages in linear sequences and those that break the 1,6 linkages at branch points. The enzymes that are most active at the 1,4 sites will cause the largest amount of current, as they will produce the most glucose 6-phosphate from the available linear portions of the starch.The enzymes most active at the 1,6 linkages will result in the largest absorbance of light at the peak wavelength (570 nm). Based on paragraph 3, iodine binds linear polysaccharides, so as the 1,6 linkages are broken, more linear polysaccharides are formed; this will lead to more iodine-starch interactions and an increased absorbance. Thus, the combination that maximizes total breakdown should have the maximum current and maximum absorbance at 570 nm, which matches to enzymes C and D, respectively.

When Y32 is expressed within a normal cell, what is true of its nucleotide binding site? "The biosensor can bind both Mg2+-ATP and ADP with very high affinity (Km ∼ 1 μM). In the cytosol of a normal cell, the concentrations of ADP and Mg2+-ATP range in the hundreds of μM and approximately 1 mM, respectively. As a result, there is competition between these two ligands for binding to the nucleotide binding site of the sensing domain, allowing the biosensor to function as a detector of the cytosolic ratio of Mg2+-ATP to ADP." A. It is most likely to be occupied by ADP. Show Explanation B. It is unlikely to be occupied by Mg2+-ATP. Show Explanation C. It is unlikely to be occupied by Mg2+-ATP or ADP. Show Explanation D. It is effectively always occupied by Mg2+-ATP or ADP.

D The passage indicates that the biosensor binds Mg2+-ATP and ADP with very high affinity (with a Km ∼ 1 μM). This greatly exceeds the normal cytosolic concentrations of ADP and Mg2+-ATP (hundreds of μM and about 1 mM,respectively). Given this, it is quite likely that the nucleotide binding site of a Y32 molecule would be occupied by Mg2+-ATP or ADP under such conditions.

The high-energy radiation produced by the γ rays has sufficient energy to: I. generate free radicals. II. excite electrons to higher energy levels. III. eject electrons from molecular orbitals. "The radiation is strong enough to cause molecular electronic transitions by exciting electrons to higher energy levels in molecular orbitals. In some molecules the beam passes through, the radiation is sufficiently strong to produce molecules with unpaired valence electrons by way of electron ejection." A. I only B. I and II only C. II and III only D. I, II, and III

D The second paragraph tells us that the radiation is strong enough to cause molecular electronic transitions by exciting electrons to higher energy levels in molecular orbitals. This indicates that the radiation can either excite or eject electrons (depending on its energy) and can create free radicals (atoms with unpaired valence electrons) (I, II, III).

Viruses are distinct from most biological organisms in that they: A. are able to reproduce inside a host. Show Explanation B. only use RNA as genetic material. Show Explanation C. are obligate parasites. Show Explanation D. are pathogenic entities which have been described as not being "living."

D Viruses are unique in that they occupy a gray area between living and non-living. They have been described as non-living. C is wrong bc although true, other organisms like tapeworms and lots of bacteria species are obligate parasites and they are alive but we are looking for what makes viruses different from living things

An artificial leg designed for use by runners is spring-based, to mimic the compression required of a muscle during hard running. For safety reasons, it was determined that the leg should be able to absorb as much as 125 J of kinetic energy without compressing more than 10 cm, or the runner would be likely to stumble. What should the spring constant be? A. 250 N/m Show Explanation B. 2,500 N/m Show Explanation C. 12,500 N/m Show Explanation D. 25,000 N/m

D When the leg "absorbs" kinetic energy, it is converted to elastic potential energy. While this process involves the loss of some energy as heat, we can assume that it is a perfectly elastic process here for the sake of simplicity. Thus, the leg needs to hold up to 125 J of elastic PE. The formula for potential energy contained in a spring is PE = (1/2) kx2. Since PE and KE are interconverted as the individual runs, we can write this formula as KE = (1/2) kx2, which can be rearranged to yield k = 2 KE/x2. k = (2)(125 J) / (0.10 m)2 k = (250 J) / (.01 m2) k = 25,000 N/m

emission from n=4 to n=3 is

IR light

expectation bias

expectations influence attitudes or behavior

Photon emission from n=2 to n=1 is

UV light

Common ion effect

a decrease in the solubility of an ionic compound caused by the addition of a common ion, which causes more precipitation and reduced solubility

compliance is shown to be severely weakened by

just one person disagrees with the group

are the ETC carrier molecules mobile and hydrophobic

no: In the ETC, carriers travel inside the inner mitochondrial membrane, passing electrons from one to another and pumping protons across the inner mitochondrial membrane. Therefore, they are mobile. In order to travel inside the hydrophobic interior of the membrane, we would expect them to all be hydrophobic. However, cytochrome c is a highly water-soluble protein, unlike other cytochromes.

framing bias/effect

one of many factors which affect a person's decisions. Just like how a picture may be framed in many different ways to change the viewer's impression, the way an option or decision is presented to a person will change how they feel about it and influence their likelihood to make a particular choice. An example from everyday life could be the way that two different people frame their beliefs about the same used car. One person might say, "oh, it's a real reliable car", giving off a positive impression, and the other might say, "it's old and worn", spinning a negative impression.

how many electrons can electron carriers of ETC carry at one time?

one or two at max

moral hypocrisy

situation in which a person appears to be a moral person but doesn't actually try to pursue moral behavior.

status quo bias

tendency to avoid situations or actions that may produce change, instead preferring to choose action that will keep normalcy, or the status quo

automation bias

tendency to excessively depend on automated systems, which can lead to erroneous automated information overriding correct decisions.

egocentric bias

tendency to overstress changes between the past and present in order to make oneself appear more worthy or competent than one actually is. According to the results from several conducted studies, individuals are also more likely to favor circumstances that are beneficial to themselves compared to those that favor the people around them.

Emission from n=3 to n-2 is

visible light