Blueprint FL2
If it was found that individuals in the first study who took Truvada but had a high level of perceived stigma relating to Truvada use were more likely to use condoms than those who took Truvada but had a low level of perceived stigma relating to Truvada use, then which of the following terms would best describe the role of perceived stigma on the relationship between taking Truvada and using condoms? A. Mediating variable Show Explanation B. Moderating variable Show Explanation C. Confounding variable Show Explanation D. Dependent variable
B A moderating variable is one that either decreases or increases the strength of an association. In this case, perceived stigma moderates the relationship between taking Truvada and using condoms. A: A mediating variable is one that provides a causal link between two variables that show a statistical relationship. C:A confounding variable is one that affects two variables that show a relationship. If this was the case for perceived stigma, it would mean that perceived stigma affected BOTH likelihood to use Truvada and condom use. This is not possible in the context of this study since participants were randomized to receive Truvada.
If research continues to indicate that the SAT is biased against students from low socioeconomic backgrounds and does not reflect their knowledge or skills, what does that mean for the test's psychometric properties? A. The test may still have high reliability and validity. Show Explanation B. The test may still have high reliability but low validity. Show Explanation C. The test may have low reliability but still have high validity. Show Explanation D. The test may not have high reliability or validity.
B A test can be reliable but not valid - think of a bathroom scale. If you step on it every day and the weight is the same but wrong, the test has high reliability (consistency) but low validity (accuracy). Students may be getting reliable, consistent scores over time, but if the test isn't a true measure of their ability and skills, then it is not valid.
One theory of high crime prevalence within a certain community is that weakened cultural values have been transmitted throughout the community; another theory is that individuals in the community seek out resources they perceive themselves as lacking. What is the best description of these theories, respectively? A. General strain theory; relative deprivation theory Show Explanation B. Anomie theory; relative deprivation theory Show Explanation C. Relative deprivation theory; general strain theory Show Explanation D. Relative deprivation theory; anomie theory
B Anomie theory states that individuals who experience weakened social values are less likely to behave in ways that are helpful to that society. Relative deprivation theory posits that individuals who perceive themselves as having less resources than others will often act in ways to obtain these resources. C:General strain theory holds that individuals who have experienced negative events feel negative emotions, which lead to negative behaviors. This does not relate to either of the given examples.
Assuming a 95% yield for each coupling step, what would be the final yield for synthesizing a 10 amino-acid length peptide? A. 30% Show Explanation B. 60% Show Explanation C. 80% Show Explanation D. 95%
B Each time we add an amino acid to the chain, we get a 95% pure yield for that step. Here, imagine that each coupling step has a 95% yield and that we are running the cycle 9 times (since no coupling step is needed for the first of the 10 amino acids), with each cycle being independent from the preceding cycle. Thus, a 10-amino acid peptide would be synthesized in a (0.95)9 = 0.63 ≈ 60% final yield. This can be assured because the passage allows us to assume a 100% yield in each unprotection step. There is no need to calculate it all out. A crude estimate of the yield would be that 9 steps, in each of which 5% is lost (corresponding to a 95% yield), would result in 45% being lost. However, the actual yield will be somewhat higher, because each step does not involve exactly a 5% reduction from the starting percentage. 60% is the best approximation.
A circuit is comprised of a 12-volt battery with three light bulbs with the same resistance. If two of the light bulbs are wired in parallel and are in series with the third light bulb, what happens when one of the parallel light bulbs is unscrewed? A. All of the light bulbs remain lit. Show Explanation B. Two of the light bulbs remain lit. Show Explanation C. One of the light bulbs remains lit. Show Explanation D. All of the light bulbs go out.
B Even after unscrewing one of the parallel light bulbs, there is still a circuit for current to flow from the positive terminal to the negative terminal of the battery through the two remaining light bulbs.
Which of Freud's systems of mind are unconscious?
id, ego, and superego. Id is only unconscious; ego and superego are conscious, unconscious, and preconscious
STP
1 atm and 273 K
When 2 moles of hydrofluoric acid are added to 100 mL of water, the resulting solution has a pH equal to 4. What is the percent dissociation of HF? . 0.0005% Show Explanation B. 0.05% Show Explanation C. 0.5% Show Explanation D. 5%
A Remember that HF is a weak acid, meaning that it does not fully dissociate in water. In many weak acid problems, you must use an ICE table and the Ka of the acid to calculate the pH of an acidic solution. However, this question gives us pH outright. Remember, pH = -log[H+], so [H+] = 10-pH. Here, [H+] = 10-4 M. Now, we must understand what is meant by "percent dissociation." The percent dissociation of HF is simply the percent of the original acid concentration that has dissociated into H+ and F- ions. This value is equal to [H+]/[HF] x 100%. Original [HF] = 2 moles / 0.100 L = 20 M [H+]/[HF] = (10-4 M H+) / (20 M HF) = 5 x 10-6 % dissociation = 5 x 10-6 x 100% = 5 x 10-4 % dissociation = 5 x 10-4% = 0.0005% % dissociation is amount of acid dissociated so [HA dissociated]/[HA initial]
A gene found in the upper left corner of Figure 1 is most likely involved in: (Fig1 shows older adults transcript levels on x axis and younger adult transcript levels on y axis so upper L corner is low for older adults and high for younger adults) ("To examine whether AAEL004181 and AAEL 004188 are actually expressed in A. aegypti, the researchers hybridized cDNA derived from female mosquitos to two distinct whole-genome microarrays (see Figure 1).") A. oocyte production. Show Explanation B. spermatocyte maturation. Show Explanation C. mediating Wolbachia's entry into germ line cells. Show Explanation D. the immune response to Wolbachia.
A A careful examination of Figure 1 reveals that a gene found in the upper left corner would be expressed at much higher levels in young adult mosquitos than in old adult mosquitos. Thus, the gene likely has something to do with reproduction, as that is a function that would likely be upregulated in younger as opposed to older mosquitos. The best way to tackle this question is by process of elimination. B: This choice is incorrect, as the passage states that only female mosquitos were used. C and D: The passage states that Wolbachia doesn't infect A. aegypti naturally, and it does not state that infected mosquitos were used for the microarrays. In fact, the use of infected individuals would confound the microarray, whose purpose is to identify whether the transferred genes are expressed in the absence of Wolbachia.
An additional experiment on the bovine extracts using a known ADAMTS inhibitor would serve as: (going along w last Q) A. a positive control. Show Explanation B. a negative control. Show Explanation C. a randomized control. Show Explanation D. a false negative.
A A positive control is a control group that is not exposed to the experimental treatment but that is exposed to some other treatment that is known to produce the expected effect. The scientists wished to test the effect of the analogs on enzyme activity. Using a known inhibitor is an example of comparing the expected effects of this inhibitor to those of the analog treatment. Thus, it is a positive control. Because researchers were trying to test the analogs for inhibition, comparing them to known inhibitors would be a positive control. B: A negative control group is a control group that is not exposed to the experimental treatment or to any other treatment that is expected to have an effect. So a negative control is the control already in the experiment w the "no ADAMTS analog"
The researchers' decision to use their within-subject study design rather than a between-subject design with 100 participants was most likely made in order to: A. allow for all subjects to be compared to the same type of control. Show Explanation B. control for sociodemographic and economic factors, but not genetic ones. Show Explanation C. increase the statistical power of the study. Show Explanation D. decrease the sensitivity of the experiment to detecting the effect of anxiety on decision-making.
A According to the first paragraph, the goal of the study was to investigate the effects of anticipatory anxiety on individual decision-making and underlying neural activity.Within-subjects design controls for individual variations in a measurement by comparing the scores of a subject in one condition to the scores of the same subject in other conditions. Thus, each subject serves as his or her own control. In this experiment, all subjects were compared to the same type of control - themselves.
Which of the following best explains why a Robertsonian carrier may have no health problems due to his or her chromosome rearrangement? A. Genes on the lost p arms can be found elsewhere in the genome. Show Explanation B. There are no genes found on any of the p arms of acrocentric chromosomes. Show Explanation C. All acrocentric p arms contain the same base sequence. Show Explanation D. The long q arm chromosome copies the genetic information from the p arms before they are lost.
A In the figures, we can see that the short chromosomal arms are lost. However, if all of the genes on a short arm are available on the short arms of other acrocentric chromosomes, a Robertsonian translocation carrier will have no health problems due to his or her chromosome rearrangement. B: This would imply that loss of the p arms will never result in genotypic or phenotypic abnormalities, which we know is incorrect. C: The 5 acrocentric chromosomes carry a diverse set of genes, and it is unlikely (in fact untrue) that they would all carry the same base sequence on their p arms.
Researchers noted that when a lysine residue on a histone is acetylated, its side chain becomes neutral in charge. Combined with the information in the passage, which of the following conclusions will the researchers most likely reach? A. Histone deacetylation generally decreases gene expression. Show Explanation B. Histone acetylation tends to promote a more condensed chromatin structure. Show Explanation C. Histone deacetylation reduces the extent to which the gene in question will be replicated. Show Explanation D. Histone acetylation favors the formation of heterochromatin rather than euchromatin.
A Lysine is a basic amino acid and typically has a positively-charged side chain at physiological pH. When lysine is acetylated, this charge becomes neutral, as shown in the figure below. Since DNA is negatively charged due to its phosphate backbone, the charge on lysine allows for tight histone-DNA interactions thanks to electrostatic attraction between the charged atoms on each molecule. Acetylation of lysine makes the residue neutral, lessening these interactions and promoting a looser structure. Loose chromatin structure is typically associated with euchromatin, the less dense, transcriptionally active chromatin structure that appears light under a microscope. In contrast, histone deacetylation will restore the positive charge to the residue, allowing the electrostatic attractions to return. Therefore, deacetylation of lysine residues on histones should lead to a denser chromatin structure and lowered transcription/gene expression. C: Gene expression and histone modifications impact transcriptional activity, not replication. A gene should be replicated during each S phase regardless of the post-translational modifications on its histones.
What is the mechanical advantage of using the ramp, as opposed to lifting the gurney straight up? (Ignore any effects of friction.) ("In a second experiment, the gurney brakes are unlocked and the gurney is pushed from rest to the top of a ramp, then brought to a stop. The ramp makes an angle of 30° with the horizontal, and the top of the ramp is 5 meters above ground level. Once at the top of the ramp, the gurney is released and allowed to roll downward.") Gurney is 200kg A. 2 Show Explanation B. 1.5 Show Explanation C. 1 Show Explanation D. 0.5
A One form of mechanical advantage represents an ability to move a load (do work) using less force than simply lifting the load directly. For an inclined plane, this advantage is represented using the formula: MA = hypotenuse / height From trigonometry, we know this as sin-1. We're told the ramp makes a 30º angle with the ground. This means it is a 30º-60º-90º right triangle, with side ratio of x : x√3 : 2x. Thus: MA = 2x / x = 2 That is, by putting in some measure of force (e.g. 1000 N) you're able to lift an object that would otherwise take twice as much force (e.g. 2000 N) to lift straight up. Also: MA=Fout/Fin Fout= (200kg)(10)=2000N Fin= 2000sin30=1000 MA=2000/1000=2
The acceleration of the gurney during the initial 6 seconds of Experiment 1 is: ("A medical gurney rests on four wheels with its brakes locked to prevent rotation. A dummy is placed on the gurney to replicate the weight of a typical patient, with the gurney and dummy together totaling a mass of 200 kg. A medical student is asked to push the gurney horizontally. During the experiment, the force exerted on the gurney is increased until the tires begin to slide along the floor without rotating. Once the gurney begins to slide, the pushing force is decreased to the lowest force needed to maintain a fixed speed. Figure 1 shows data from a force sensor attached to the gurney, which records the student's applied force over time. (The floor has a coefficient of static friction, µ.)") (Graph shows Fapp increasing until 8 seconds, a slight decrease from 8 to 9 seconds and then plateau at 9 to 16) A. zero. Show Explanation B. nonzero and in the direction of Fapplied. Show Explanation C. constant and greater than 0. Show Explanation D. rising as Fapplied increases.
A The passage states that the Fapplied is increased until the gurney moves. This means that while Fapplied is less than its maximum (t = 0-6s), the gurney is not moving. Here, Fnet = 0 N. According to Newton's second law, Fnet = ma; thus, a = Fnet /m = 0
Given the procedure outlined in the passage, what must be true about SP peptide synthesis? ("During the procedure, small, porous beads of resin are treated with functional units called linkers, on which peptide chains can be built. The carboxyl end of the new peptide remains immobilized by its covalent attachment to the bead until cleaved by a reagent. This allows the peptide to be retained during the filtration process while liquid reagents and waste products are washed away. Synthesis consists of repeated cycles of bind-wash-unprotect-wash. The free N-terminal amine of a solid-phase attached peptide is bound to a single N-protected amino acid unit. Then, the protective unit is removed, revealing a new N-terminal amine to which the next amino acid is attached. Efficiency is increased by performing a wash cycle after each reaction, removing excess reagent with the growing peptide remaining attached to the resin. Physical properties of the solid support vary with the material from which the support is constructed, the amount of cross-linking, and the linker being used. Commonly accepted characteristics of an efficient support are listed.") A. The protein is synthesized from C-terminal to N-terminal. Show Explanation B. The protein is synthesized from N-terminal to C-terminal. Show Explanation C. The protein can be extended at either terminus. Show Explanation D. The peptides created contain only L-amino acids.
A Unlike ribosome protein synthesis, solid-phase peptide synthesis proceeds in a C-terminal to N-terminal fashion because the C-terminus is bound to the linker throughout the synthetic process. In the absence of a permanent support at one end — as in SP synthesis — amino acids could be added to either terminus. B: It is the C terminal, not the N terminal, that is bound to the resin bead during SP synthesis.
When a patient is on the operating table for heart surgery, surgeons need to clamp major nearby blood vessels before cutting them because, according to Bernoulli's equation, a volume of pressurized blood opened to the atmosphere: A. will convert its pressure to kinetic energy, rapidly exiting the body and causing shock. Show Explanation B. will convert its pressure to potential energy, pooling at the high places of the body and causing a potential brain aneurysm or stroke. Show Explanation C. will match outside pressure by decreasing fluid speed, starving the body of oxygen. Show Explanation D. is at risk of over-oxidizing, dangerously increasing pressure.
A When the surgeons cut into the artery, blood will obviously shoot out of the body as a result of blood pressure. That pressure is converted to kinetic energy, causing the blood to move rapidly as it exits the body. B: wrong bc the pt is horizontal on the bed so there wouldn't be a potential energy change. C: wrong bc pressure in vessels is higher than atm pressure so having decreased pressure would cause increased speed D: doesn't have anything to do w Bernoulli's.
imine
A double bond between a carbon and a nitrogen. R=NR
Electrical synapses lack
gain: signal in postsynaptic neuron is same or smaller than that in originating neuron
The reactions catalyzed by xanthine oxidase in Figure 2 are reversible. If Keq for the conversion of hypoxanthine, water, and oxygen to xanthine and hydrogen peroxide is approximately 1, which change in an equilibrium mixture of the five compounds will NOT necessarily increase the ratio of xanthine to hypoxanthine? A. Adding water Show Explanation B. Increasing temperature Show Explanation C. Adding O2 Show Explanation D. Removing H2O2
B For the reaction described, Keq is approximately 1. As we do not know whether this is an endothermic or exothermic reaction, changing the temperature may drive the reaction forward, or it may drive it backwards. In that case, a change in temperature might not necessarily lead to a change in concentration of reactants and products. So increasing temperature may increase this ratio, but it may decrease it as well! **I didn't read water in the Q stem and instead only looked at figure.
Which of the following statements accurately describes a major assumption of the functionalist theory of social stratification? I. Social stratification affects what people believe about themselves and others and influences their daily life, lifestyle choices, and interactions with others. II. Social stratification is unnecessary and results from both lack of opportunity and discrimination against the poor and people of color. III. Social stratification is necessary and results from the need for those with special intelligence, knowledge, and skills to be a part of the most important professions and occupations. A. I only Show Explanation B. III only Show Explanation C. I and II only Show Explanation D. I and III only
B Functionalist theories assert that aspects of culture are necessary and need-based (III). D: This statement describes symbolic interactionism in relation to social stratification (I).
Fluid pressure changes with depth are assumed to be linear. Which statement best explains why this does not hold true for atmospheric pressure changes? A. At very high temperatures, air behaves less ideally. B. The volume of a mass of air is not constant. C. The majority of molecules in air are nonpolar. D. Air is not of a uniform composition.
B Hydrostatic pressure for liquids is linear because as depth changes, the density of the liquid remains constant. Gases, however, have densities that change according to the forces applied to them. Gases are compressible, while liquids and solids are not. Because a force can be applied to a gas and therefore compress the gas, with more/less forces, the density of the gas would change aka mass/volume=density. A: at high temps, air behaves more ideally C: this is true but doesn't explain why it's nonlinear D: gas mixtures don't act differently from pure gases according to molecular kinetic theory of gases and even though it isn't just one molecule, this wouldn't explain why it's not a linear depth.
A 50-kg child is riding on a playground merry-go-round. If the radius of the circular path of the merry-go-round is 5.0 m and the frequency is 0.1 hertz, what is the force required to keep the child on the ride? A. 25 N Show Explanation B. 100 N Show Explanation C. 1000 N Show Explanation D. 2500 N
B If the frequency of the merry-go-round is 0.1 Hz, then the period is 10 s. We can calculate the speed of the merry-go-round from V = 2πr/T = 2(3.14)(5.0 m)/(10 s) = 3.14 m/s. Using this in the centripetal force equation: Fc = mv2/r = (50 kg)(3.14 m/s)2/(5.0 m) Approximate the square of the speed at 10 m2/s2. Fc = (50)(10)/(5) = 100 N ** OR: v=wr w=2(pi)f --> v=2(pi)fr and plug this into centripetal force equation
Arachidonic acid, released during AEA hydrolysis, is NOT a precursor for the synthesis of what class of molecules? A. Prostaglandins Show Explanation B. Catecholamines Show Explanation C. Thromboxanes Show Explanation D. Phospholipids
B In the first paragraph of the passage, it's written that AEA is degraded to ethanolamine and arachidonic acid. Ethanolamine is included as a polar head group in the phospholipid phosphatidylethanolamine, while arachidonic acid is important as a precursor for the biosynthesis of the eicosanoid signaling molecules: prostaglandins, thromboxanes, and leukotrienes. Neither molecule contributes to the synthesis of catecholamines, which are a class of molecules derived from tyrosine. D:Ethanolamine is found in the phospholipid phosphatidylethanolamine; arachidonic acid is a polyunsaturated fat found in phospholipids, especially phosphatidylinositol, phosphatidylcholine, and phosphatidylethanolamine.
How could the studies be changed to assess operant conditioning? A. Present a reward while participants view the stimuli and then have them rate the stimuli. Show Explanation B. Present a reward after participants rate the stimuli and then have them rate the stimuli again. Show Explanation C. Present a reward after participants view the stimuli and then have them rate the stimuli. Show Explanation D. Present a reward before participants view the stimuli and then have them rate the stimuli.
B Operant conditioning involves shaping behavior. This is the only condition where the participants are given a reward after a behavior (rating the stimuli) and then asked to perform that same behavior again. The fact that the participants are then asked to rate the stimuli again implies that we are testing the impact of the reward on their behavior. C:The reward serves no clear operant conditioning function here. In typical operant conditioning protocols, a reward is presented upon performance of a desired behavior. Here, the participants are asked to rate the stimuli, which is a behavior, but the reward is given beforehand and not in conjunction with the desired action.
In order to find the concentration of metal ions needed to determine the distribution coefficient of copper (II), what calculation could the researchers have performed in addition to the procedure described in the passage? ("Researchers performed a series of extractions of a known concentration of copper (II) nitrate in aqueous solution containing HL into six organic solvents of differing polarities. During the trials, the aqueous phase pH was varied. After the two phases were separated completely, concentrations of the metal remaining in the aqueous phase were determined spectrophotometrically at 820 nm. The effect of solvent choice and pH on the distribution coefficient, D, defined as the ratio between the concentration of metal in organic and aqueous phase following extraction, is shown in Figure 2.") A. Average the concentrations of metal ions in the aqueous and organic phases after extraction. Show Explanation B. Subtract the concentration of metal ions in the aqueous phase after extraction from the concentration before extraction. Show Explanation C. Subtract the concentration of metal ions in the organic phase after extraction from the concentration before extraction. Show Explanation D. Average the concentrations of metal ions in the organic phase before and after extraction.
B The passage defines the distribution coefficient as "the ratio between the concentration of metal in organic and aqueous phase following extraction." In order to determine D, the researchers must know the concentrations of metal ions in the organic phase and of the metal ions remaining in the aqueous phase following the extraction. According to the description provided by the passage of the researchers' procedure "after the two phases were separated completely, concentrations of the metal remaining in the aqueous phase were determined spectrophotometrically at 820 nm." This measurement provides the concentration of metal ions remaining in the aqueous phase directly. The concentration of metal in the organic phase after the extraction is not directly known. However, the passage also indicates that the concentration of metal contained in the aqueous solution prior to extraction is known—"researchers performed a series of extractions of a known concentration of copper (II) nitrate in aqueous solution." In order to determine the concentration of metal ions in organic solvent after extraction, the researchers could use the known concentrations of metal in aqueous solution before and after extraction—the difference, found by subtracting the value, should equal the concentration of metal ions in the organic phase after extraction.
Based upon the researchers' analysis of the extraction equilibria, the organic phase extractant forms what complex or complexes in low-polarity organic solvents and high-polarity organic solvents, respectively? ("Copper (II) can be extracted from an aqueous medium into an organic solvent using a lauric acid (HL; molecular formula C11H23COOH) extractant. In the organic phase, the metal can be recovered as part of a complex with the extractant. Depending upon the solvent used, one of two complexes can form, labeled as complexes A and B in Figure 1, involving dimeric HL and its conjugate base laurate (L), or monomeric HL, coordinated to copper (II)." Complex A is 2 dimers and Complex B is a 2 monomers. "Cu2+ + (HL)2, organic --> CuL2(HL2), organic, +2H+ for low polarity organic solvents Cu2+ +2(HL), organic--> CuL2, organic, +2H+ for high polarity organic solvents.") A. Complex A; complex A Show Explanation B. Complex A; complex B Show Explanation C. Complex B; complex A Show Explanation D. Complex B; complex B
B The two chemical equations shown in the passage indicate the structure of the organic phase extractant complex formed in both low-polarity and high-polarity organic solvents. In low-polarity organic solvents, copper (II) is coordinated with the organic phase extractant as part of a complex consisting of two dimers. Both dimers contain one molecule of lauric acid (HL) and another of its conjugate base, laurate (L). This is consistent with the structure of complex A shown in Figure 1. In high-polarity organic solvents, copper (II) is coordinated with the organic phase extractant as part of a complex consisting of two monomers of laurate. This is consistent with the structure of complex B shown in Figure 1. B should be high polarity bc it has 2 monomers and A should be low polarity bc it has 2 dimers.
According to the passage, which of the following relationships must hold? (Note: 1 m3 = 1000 L.) A. 100 kPa is equivalent to 100 kJ/L. Show Explanation B. 100 kPa is equivalent to 100 J/L. Show Explanation C. 100 Pa is equivalent to 100 J. Show Explanation D. 100 kPa is equivalent to 100 J.
B This is a unit conversion problem: 100 kPa = 100,000 Pa = 100,000 N/m2 = 105 N/m2 100 J/L x (1000 L/m3) = 105 J/m3 = 105 Nm/m3 = 105 N/m2
Which of the following is NOT a correct description of a cognitive bias that adolescents may experience when making health-related decisions? A. Availability heuristic; judging an event to not be risky because friends have not suffered consequences Show Explanation B. Negativity bias; only focusing on the good aspects of alcohol use and ignoring the bad This is an incorrect pairing of bias and description. Negativity bias is when the negative aspects of a situation are focused upon, not the good aspects. C. Social desirability bias; peers report using more substances than they really did, leading others to follow suit This is an accurate description of social desirability bias, in which people answer questions in a way they think is socially beneficial to them. D. Self-serving bias; interpreting that one's positive experiences with substances are due to positive personality traits and dismissing negative experiences with substances as due to external factors
B This is an incorrect pairing of bias and description. Negativity bias is when the negative aspects of a situation are focused upon, not the good aspects C:This is an accurate description of social desirability bias, in which people answer questions in a way they think is socially beneficial to them.
If pharmacologists wished to convert morphine into a form available to brain tissue, which of the following changes could be made to its molecular structure to allow for the best chance to use the drug as a direct brain treatment? A. Attach a glucose molecule to the nitrogen atom Show Explanation B. Replace the alcoholic protons with acetyl groups Show Explanation C. Lyse the molecule to make it smaller Show Explanation D. Remove all double bonds from the molecule
B This question asks us to consider the structure of morphine given in the passage. What needs to change for morphine to enter the brain while maintaining its function? We are told in the passage that lipid-soluble molecules are able to pass through, while hydrophilic molecules are not. Morphine is hydrophilic in part because of its hydroxyl groups. So, replacing them with acetyl groups would allow the drug to become more lipophilic; although the C=O bond in the acetylated molecule is polar, it cannot undergo hydrogen bonding, so the acetylated structure is effectively much more lipophilic than the original structure with -COOH groups. Interestingly, diacetylated morphine is heroin. The effects and pharmacology of heroin are somewhat distinct from those of morphine, although they belong to the same general class of opiates, and heroin does indeed more readily cross the BBB. C:This would destroy the molecule's structure, making it unlikely to maintain its therapeutic function.
Which of the following is a kinship of affinity? A. Grandparents rearing their grandchildren Show Explanation B. Extended families living in the same household Show Explanation C. A husband and wife merging their children from a previous marriage Show Explanation D. Divorced parents sharing custody of their children
C A kinship of affinity is one in which individuals are related by choice, such as through marriage, rather than through blood, such as the other choices.
Which of the following was NOT a finding by Harry Harlow in his experiments with rhesus monkeys and their pairing with a simulated mother made from wire (Wire Mother) or one made from soft cloth (Cloth Mother)? A. Monkeys paired with Wire Mothers drank similar amounts of milk and grew at a rate comprable to monkeys paired with Cloth Mothers. Show Explanation B. Monkeys paired with a Wire Mother sought soothing from their "mothers," but at a rate less than monkeys paired with Cloth Mothers. Show Explanation C. After early neglectful conditions, monkeys' abnormal behavior could be corrected by pairing them with the appropriate simulated mother. Show Explanation D. When given a choice, monkeys spent more time holding Cloth Mothers than Wire Mothers.
C In Harlow's experiment, the infant monkeys preferred spending their time clinging to the cloth mother. Even when only the wire mother could provide food, the monkeys visited her just to eat. Harlow concluded that there was much more to the mother/infant relationship than milk and that this "contact comfort" was crucial to the psychological development and health of infants. After monkeys were paired with a Wire Mother, they showed abnormal behavior, which could not be corrected later in life by pairing them with a Cloth Mother.
In a follow-up experiment, two identical gurneys are placed side-by-side on a ramp with their wheels locked to eliminate spinning. Gurney 1 has a dummy placed on it to give it a total mass of 200 kg, while Gurney 2 is loaded with a dummy that makes it only 50 kg overall. If the ramp has a coefficient of friction of μs, which gurney is more likely to slide down the ramp? A. Gurney 1, due to the increased force of gravity Show Explanation B. Gurney 2, due to the reduced force of static friction Show Explanation C. Gurney 1 and Gurney 2 are equally likely to slide. Show Explanation D. Neither gurney can slide unless the wheels are unlocked.
C In order for the gurney to slide down the ramp, the force pulling it downward (mgsinƟ) must be greater than the static frictional force (μsFN = μsmgcosƟ). The net force on each gurney is thus Fnet = (mgsinƟ) - (μsmgcosƟ). Since net force is equal to the product of mass and acceleration, we can rewrite this equation as ma = (mgsinƟ) - (μsmgcosƟ), where an acceleration greater than 0 means the gurney will slide down the ramp. The mass of the gurney is present in all terms and can be canceled, meaning that it is not a factor in whether the gurney slides. Thus, both gurneys have an equal likelihood of slipping down the ramp, regardless of the fact that they have different total masses.
What is a likely explanation for the difference in column affinity exhibited by the perpromethate and the neodymate? A. Perpromethate nuclei have greater atomic masses and are therefore unable to form tight bonds to the column. Show Explanation B. Neodymate atoms have a greater mass defect and are unable to form covalent bonds to the column. Show Explanation C. Perpromethate ions carry less charge than the neodymate ions, and their electrostatic attraction to the column will be less. Show Explanation D. The potassium ions in the solution are able to surround the neodymate ions more efficiently than the perpromethate ions.
C Paragraph 2 states that the perpromethate ion "is less tightly bound to the gallium." You need to examine the structure of the two molecules to identify any difference between them that could impact their ability to bind to the gallium. Looking at the structures, neodymate carries a -2 charge while perpromethate carries a -1 charge. Thus, it follows that this difference in charge can explain the varying affinity for the gallium column. The more highly charged neodymate will have a greater affinity for the column, while the less charged perpromethate will be less able to form electrostatic attractions to the gallium. A: the atomic mass of both ND and Pm is the same bc they are both listed as having 145 protons + neutrons. B: The bonds between the metal and the oxygen atoms are ionic, not covalent. In addition, the passage states that perpromethate has the weaker attractions to gallium. Finally, the mass defect, or the mass of the atom that is converted to energy when a nucleus undergoes fusion, would have no bearing on its attraction to the column. D: If the potassium ions were able to surround the neodymate ions more efficiently, then it should follow that the neodymate ions are more easily separated from the gallium during the wash.
Pentane is a straight-chain hydrocarbon with the molecular formula C5H12. How many additional structural isomers can be constructed using this molecular formula? A. 0 Show Explanation B. 1 Show Explanation C. 2 Show Explanation D. 3
C Since the molecular formula follows the CnH2n+2 formula, we can only build alkanes. This makes the job of finding isomers much easier. We can make a tertiary carbon, 2-methylbutane. We can make a quaternary carbon, 2,2-dimethylpropane. And that's it. DIDN'T READ "ADDITIONAL"
Based on the information in the passage, compared to the T11 residue, the T3 residue is: ("Among other results, it was indicated that phosphorylation of the T11 residue on histone H3 fully prevents demethylation of the trimethylated K9 residue by either KDM4A or KDM4C. However, when residue T3 was phosphorylated instead of residue T11, demethylation at the trimethylated K9 position was observed. Figure 1 shows the results of the kinetics analysis performed on this reaction, where [S] represents the concentration of T3-phosphorylated, K9-trimethylated H3 histone protein.") A. closer to the C-terminus, and that terminus is likely to be positively charged in vivo. Show Explanation B. closer to the C-terminus, and that terminus is likely to be negatively charged in vivo. Show Explanation C. closer to the N-terminus, and that terminus is likely to be positively charged in vivo. Show Explanation D. closer to the N-terminus, and that terminus is likely to be negatively charged in vivo.
C The T11 (threonine at position 11) residue was introduced in the beginning of the third paragraph with T3 being discussed later in the same paragraph. The amino acid structure of a protein is conventionally written from the N-terminus to the C-terminus, meaning that the T3 residue (which is the third residue in the protein) must be closer to the N-terminus than T11. In vivo means "in the living" which indicates we are talking about the conditions found inside the living human body. Physiological pH ranges from 7.2 to 7.4, and at this pH, the N-terminus tends to exist in the form of a positively charged -NH3+ group. D: It is the carboxylic acid, or C-terminus, of a protein that is typically negatively charged at physiological pH, since it takes the form of -COO-. The N-terminus should never be negatively charged; it can only be positively charged (-NH3+) or neutral (-NH2).
Given the data discussed in the passage, if a hospital requires at least 250 g of 145Nd to complete a standard cardiac image, how long will it be before it must replace a 2000 g sample? A. 25 minutes Show Explanation B. 50 minutes Show Explanation C. 75 minutes Show Explanation D. 100 minutes
C The decay of a substance can be calculated by knowing its half-life, the time it takes for half of the original amount to decay. The final paragraph states that about 93% of the decay occurs in the first 100 minutes. This leaves about 7% of the sample left, which would indicate that close to 4 half-lives (0.54 = 0.0625 = 6.25% remaining) have passed at the 100-minute mark. Thus, 1 half-life = 100/4 = 25 minutes. The hospital sample is 2000 g and must be replaced once it reaches 250/2000, or 1/8 of its original radioactivity. Using the shortcut that the amount remaining = (1/2)n, where n represents the number of half-lives that have passed, we can calculate 1/8 = (1/2)3. Therefore, 3 half-lives will pass before the sample drops below 250 g. At 25 minutes apiece, 25 min (3) = 75 minutes.
A physician suspects their patient's schizophrenia is caused by an overactive hippocampus. According to the passage what technique would most effectively eliminate the activity? ("Biopsychological studies tend to impair or enhance neuronal function in specific key areas of the brain to determine how they affect behavior. Scientists disable or decrease neural function, through surgical or chemical lesions. Alternatively some utilize transcranial magnetic stimulation to disturb electrical activity in small areas, functioning as a very local and temporary lesion. Scientists also have the ability to enhance neural functioning through chemical receptor agonists, electric stimulation, or optogenetics, a technique that uses light to control neurons that have been genetically modified to express light-sensitive ion channels.") A. Hippocampal receptor agonists Show Explanation B. Transcranial magnetic stimulation Show Explanation C. Surgical lesions Show Explanation D. Optogenetics
C The question asks for a permanent solution to the overactive hippocampal activity. The final paragraph states that lesions can be used to disable or reduce neural activity. Surgical lesions are most likely to be permanent, while chemical lesions could be permanent as well. Therefore, surgical or chemical lesions impairing brain regions is likely what the physician would recommend. B:The final paragraph informs us that transcranial magnetic stimulation is able to effect a local and temporary lesion on a brain region. However, the question asks for a permanent fix.
According to the data in Figure 1, what is the probability that a male Robertsonian translocation carrier who mates with a normal female will produce a viable offspring? (Fig1: normal gamete, a balanced translocation (all genetic info is present; this gamete would give rise to another ROB carrier), Trisomy 21, Monosomy 14, Monosomy 21, and Trisomy 14.) A. 1/6 Show Explanation B. 1/3 Show Explanation C. 1/2 Show Explanation D. 5/6
C There are no viable autosomal monosomies, and the only viable autosomal trisomies you should know are Trisomy 21, 18, and 13 (though Trisomy 8, 9, and 22 can also survive to term). Therefore, of the 6 gametes produced by the man, 3 of them contain genetic information that would produce a viable offspring (two normal, one with Down Syndrome). 3/6 = 1/2.
In severe diabetic hyperglycemia (high blood sugar), insulin cannot effectively induce the uptake of glucose by cells. Chronic hyperglycemia directly leads to the presence of which of these molecules in the urine? I. Proteins II. Glucose III. Ketone bodies A. I only Show Explanation B. I and II only Show Explanation C. II and III only Show Explanation D. I, II, and III
C This question requires outside knowledge about glucose metabolism. If cells cannot take up glucose, it will remain in the blood and eventually be excreted in the urine when it builds up to the point that it cannot be reabsorbed by the nephron (II). In a state of extended hyperglycemia, the body relies on fat metabolism to generate energy, which produces ketone bodies that are also excreted in the urine (III).
Uracil is usually found in: I. tRNA. II. ribosomes. III. single-stranded DNA. A. I only B. II only C. I and II only Show Explanation D. I, II, and III
C Uracil is found in any structure made of RNA. Both tRNA and ribosomes are made of RNA (I and II). Single-stranded DNA has thymine rather than uracil (III). **at first picked tRNA only, but remembered that the ribosome is made of RNA that is a ribozyme.
Which statement best explains why the lung volume in Figure 1 never drops below 15 ml/kg? ( 15ml/kg is the residual volume) A. Since PV is constant, pressures below that value will cause ruptures in the lung tissue. Show Explanation B. At this volume, maximum contraction of the diaphragm has forced out as much air as possible from the lungs. Show Explanation C. The residual volume is such that the lungs are at their minimum volume under maximum intrapleural pressure. Show Explanation D. Below this volume, spontaneous lung collapse will occur, and the body has evolved safeguards.
C this 15 mL/kg of air is the residual volume, which is the minimum volume the lungs can have, and which is attained at maximum intrapleural pressure. A: Remember, work = PΔV, so PV would not be constant in a case where work is being done on or by a mass of air. Additionally, you may be thinking of the ideal gas law, where for a certain number of moles of an ideal gas at constant temperature, PV is constant. During breathing, however, the molar amount of air in the lungs changes over time. B: contraction of the diaphragm flattens the muscle and increases thoracic volume. This facilitates inspiration, not expiration (as would be expected to explain minimum lung volumes). Forced exhalation beyond the norm uses not the diaphragm muscles (which simply relax during expiration), but the intercostal muscles. D: This is true because if air was forcibly removed from the lungs, below a certain point, the lungs would collapse under intrapleural pressure. However, this will not occur spontaneously from exhaling with too much force.
Cupric
Cu+2
Which components of cells are physically connected by a gap junction?
Cytoskeleton of one to cytoskeleton of the other
What characteristics of society are associated with high levels of anomie? I. Rapid changes in society II. Low levels of income III. High heterogeneity A. I and II only B. I and III only C. II and III only D. I, II, and III
D Anomie refers to society feeling fragmented and lacking cohesiveness. All of the characteristics listed are associated with anomie (I, II, III).
Two known competitive inhibitors were studied to analyze their effects on the reaction rate catalyzed by the enzyme lysozyme. Equimolar amounts of inhibitor A (hydronium ion) and inhibitor B (hydroxide ion) were mixed in a beaker before being added into the reaction mixture containing the substrates. Then, lysozyme was also added into the reaction mixture. Based on Figure 1, where would the resulting enzyme kinetics curve for this experiment fall? A. Above line 1 Show Explanation B. Below line 1 but above line 3 Show Explanation C. Below line 3 Show Explanation D. The same as line 1
D Be sure to read the question stem carefully! Inhibitor A is a strong acid (H3O+) and inhibitor B is a strong base (OH-). Since these inhibitors are mixed together before being added to the reaction mixture, they would simply neutralize each other and become water before having any effect at all. Thus, the enzyme kinetics would not be affected, and the curve would fall along line 1. B: This choice represents an inhibited reaction, with a rate lower than that of the uninhibited reaction. In reality, however, the two inhibitors would neutralize each other before being added to the substrate-containing mixture, and no inhibition will occur.
Many elderly individuals struggle with the concept of death and dying. The Kübler-Ross model provides a series of stages that most people experience when facing the end of their life. Which of the following is the correct sequence of emotional stages from this model? A. Anger, denial, depression, acceptance, bargaining Show Explanation B. Denial, depression, bargaining, anger, acceptance Show Explanation C. Bargaining, anger, depression, denial, acceptance Show Explanation D. Denial, anger, bargaining, depression, acceptance
D For these stages, you can remember the acronym Death Always Brings Definite Acceptance. The stages are sequenced in the following order: denial, anger, bargaining, depression, and acceptance.
Which of the following is NOT an example of a strategy people use in impression management? A. Complimenting coworkers on their clothing Show Explanation B. Bragging about one's financial success while on a date Show Explanation C. Matching the behavior of a boss Show Explanation D. Revealing a friend's personal information while on a date
D Impression management involves the control of information about oneself and is characterized by flattery, boasting, and ingratiation. In contrast, in this choice, an individual is attempting to control information about someone else.
Which of the following most closely approximates the pKa of phenolphthalein? ("The acidity of each broth sample was evaluated by acid-base titration with a standardized solution of 0.1 N sodium hydroxide, using phenolphthalein as an indicator. The acetic acid content of the fermentation products was measured 24, 48, and 72 hours after inoculation and reported as titratable acidity (%TA) in Table 2.") A. 2.4 Show Explanation B. 4.8 Show Explanation C. 6 Show Explanation D. 9.3
D Indicators function in acid-base titrations to identify, via color change or a similar mechanism, that an expected pH, and thus the titration endpoint, has been reached. In order to function in this way, an indicator must undergo a color change near the desired pH. This typically occurs because of a reversible change in the protonation state of the indicator. It is desirable then that the pKa of a chosen indicator be within ±1 unit of the target pH. For the titrations performed in the study of acetic acid, a weak acid, and sodium hydroxide, a strong base, the endpoint of the titration will occur at a pH greater than 7. Of the possible pKa values given as answer choices, only D could function well as an indicator for a pH in this range.
An outspoken skeptic and opponent of automatic writing attends multiple séances where a medium appears to successfully contact someone who has recently died. If the skeptic's views shift over time to view the practice as harmless entertainment, this evolution is best described by: A. subjective norms. Show Explanation B. a locus of control. Show Explanation C. cognitive dissonance. Show Explanation D. reciprocal determinism.
D Reciprocal determinism is the theory set forth by Albert Bandura that a person's behavior both influences and is influenced by personal factors and the environment. In this scenario, the skeptic does not believe in the supernatural power of a medium, but attends multiple séances where they may see evidence that could soften their views. It appears this behavior of attending séances, in conjunction with the social environment of others who attend the séances, caused the skeptic's views on supernatural phenomena to shift. C:Cognitive dissonance is found when an individual has inconsistent beliefs or attitudes, especially when making behavioral decisions and during changes in attitude. For example, conflicts from inconsistencies may cause discomfort, leading to an alteration in one of the attitudes, beliefs or behaviors the individual holds in order to reduce the discomfort and restore balance. From the question stem evidence, it does not appear that the skeptic has changed their mind on the validity of séances, only that they now believe séances are harmless.
The construction of the first study seems designed to: ("To study the so-called "Mozart Effect," in which listening to 15 minutes of Mozart supposedly increases performance on cognitive tasks, researchers conducted a study (Study 1) in which grade-school children were given a frustrating cognitive task (solving an unsolvable puzzle). Prior to attempting the task, the children were either told to sit quietly in a room for 15 minutes (Group 1), to sit quietly while listening to Mozart for 15 minutes (Group 2), or to sit quietly in a room while listening to "metal" rock music featuring aggressive lyrics (Group 3). Each group consisted of 100 children. Figure 1 shows the percent of the group that was still attempting to work on the problem as a function of time.") A. confirm the theory that mood affects IQ. Show Explanation B. investigate whether children with shorter attention spans prefer aggressive music. Show Explanation C. test the hypothesis that frustration increases aggressive behavior. Show Explanation D. test the hypothesis that aggression increases frustration.
D The children were presented with aggressive or soothing music and then given an impossible puzzle to solve. Thus, the experiment seems to be testing whether the aggressive feelings created by the music increased the children's frustration with the impossible puzzle. A:The first study is not assessing IQ. Note that the puzzle the children are asked to solve is unsolvable, so the researchers cannot possibly be assessing the children's intelligence. Instead, they are monitoring the length of time the children persisted at the frustrating task.
If trifluoroacetic acid is the reagent used in SP synthesis, which of the following would be the most effective for creating an effective support system? ("Commonly accepted characteristics of an efficient support are listed. 1. It must be physically stable and permit the rapid filtration of liquids. 2. It must be inert to all reagents and solvents used during SP synthesis. 3. It must swell significantly in the solvents used to allow for penetration of the reagents. 4. It must allow for the binding of the first amino acid.") A. NH3 Show Explanation B. NaOH Show Explanation C. p-bromophenol Show Explanation D. Benzene
D The list of characteristics in the passages states that an ideal support "...must be inert to all reagents and solvents used during SP synthesis." The question tells you that we are using a weak acid in the synthesis, so we want a molecule that will be inert to the actions of an acid. Benzene, the aromatic hydrocarbon shown below, is very stable (inert) in the presence of acids and bases and will not react with the reagents.
The findings of the experiment support which of the following conclusions? ("Aggrecan, a large aggregating proteoglycan, is a major constituent of the cartilage matrix. Increased proteolytic processing of aggrecan is related to the pathology of OA. Two proteinases (ADAMTS-4 and ADAMTS-5) are thought to be responsible for the deleterious cleavage that occurs within the inter-globular domain of aggrecan (Figure 1). It is still debated which of the two proteinases is mainly responsible for the aggrecan degradation in human OA. Researchers tested the effect of two suspected ADAMTS substrate analogs on cartilage matrix concentration in bovine cartilage extracts (17 mg). Extracts were washed three times in phosphate-buffered saline and cultured in well plates for four days before baseline measurements were taken. On day 5, the cultures were divided into three groups as follows: 1) media without analog treatment 2) media treated with an ADAMTS-5 analog and 3) media treated with ADAMTS-4 and -5 analogs. Treatment consisted of 3 μM of the analog. Media were changed and a new analog dose was added every 2 days, during which the cartilage concentration was measured (Figure 2). The enzymatic reactions were stopped on day 15." Graph shows: No ADAMTS analog being lowest for cartilage composition, then ADAMTS-5 analog is second highest, with ADAMTS-4 and -5 analog being highest) A. ADAMTS-5 is mainly responsible for the aggrecan degradation during human OA. Show Explanation B. ADAMTS-5 has a higher specificity constant than ADAMTS-4. Show Explanation C. The ADAMTS-5 analog acts as an agonist. Show Explanation D. Both analogs act as ADAMTS inhibitors.
D The passage states that ADAMTS-4 and -5 promote degradation (breakdown) of cartilage, specifically of aggrecan. Thus, we would expect the addition of ADAMTS-4 and -5 themselves to decrease the mass of cartilage. However, Figure 2 shows that the ADAMTS-4 and -5 analogs do the opposite; their addition increases the cartilage concentration. As such, we can deduce that the ADAMTS-4 analog actually inhibits ADAMTS-4, while the ADAMTS-5 analog inhibits ADAMTS-5. This is also supported by the description of these compounds as "substrate analogs." This means that the ADAMTS-4 analog mimics the substrate of ADAMTS-4, causing ADAMTS-4 to act on it instead of acting on aggrecan (and the same is true of the ADAMTS-5 analog). I chose C because I was confused why the analog was going against what the researchers had originally said, but since it is a substrate ANALOG and not the actual enzyme, it makes sense that the binding would promote inhibition of the enzyme and increase cartilage. B: specificity constant is kcat/km and we don't have an experiment on only -4 analog alone so we cannot say for certain which enzyme is more efficient. A: w/o testing effect of -4 alone, again no conclusions can be drawn.
Following extraction into chloroform, the spectrophotometric absorbance at 820 nm of copper in aqueous solution will be greatest at which pH? (Graph shows chloroform to have lowest pH point at 5.15 and greatest pH point at 5.45) A. 5.45 Show Explanation B. 5.35 Show Explanation C. 5.25 Show Explanation D. 5.15
D The passage states that spectrophotometric absorbance is used to measure the concentration of metal ion remaining in aqueous solution following extraction. This value should be maximized when the extent of the extraction is minimized. In Figure 2, this should correspond to the aqueous solution pH in which log D is smallest for the extraction in chloroform. Of those pH values given as answer choices, log D is smallest at pH 5.15.
Based upon the information provided in the passage, which of the following could be a cultural factor which impacts adolescent risky behavior? I. Culturally-based expectations II. Cultural response to onset of puberty III. Cultural taboos regarding alcohol use ("Puberty impacts the way adolescents see themselves. Girls who believe they are overweight as they begin puberty might develop an eating disorder. Early onset of puberty in adolescence is associated with increased risky behaviors, such as substance abuse and unprotected sex....While friends and peer groups become more influential during adolescence, parent influence remains a salient factor. The presence of a romantic relationship has been found to increase some risky behaviors, but decrease others. It is unclear how peer group might influence adolescent health behavior, as health behavior tends to be a defining characteristic of peer group status and adolescents may select peer groups based on individual characteristics. Peer groups then reinforce these behaviors, leading to further behavior.") A. I only B. I and II only C. I and III only D. I, II, and III
D The second paragraph mentions that how a girl views herself during puberty can impact risk behavior (II). Cultural response can contribute to these views. Paragraphs four and five describes how peer groups and parental influence can impact behavior. Both of these factors can be reflections of cultural taboos (III).
The hemoglobin (Hb) dissociation curve at high altitude has a distinct shape from that at atmospheric pressure. Which of the following best explains this shape? A. Dissolved oxygen in the blood serves to increase pH. Show Explanation B. The primary structure of Hb is changed by oxygen binding. Show Explanation C. Dissolved oxygen forms a chain of atoms that is easier for Hb to bind. Show Explanation D. Homotropic regulation by oxygen occurs.
D The sigmoidal shape of the curve implies that as each oxygen molecule binds to Hb, the affinity of Hb for oxygen goes up. Homotropic regulation is when a molecule serves as a substrate for its target enzyme, as well as a regulatory molecule of the enzyme's activity. O2 is a homotropic allosteric modulator of hemoglobin. The four subunits of hemoglobin actually bind to oxygen cooperatively, meaning the binding of oxygen to one of the four subunits will increase the likelihood that the remaining sites will bind with oxygen as well. This is the cause of the sigmoidal curve shown in the figure. Cooperative binding in hemoglobin is also depicted below. C: there is no chain formation of oxygen. B: the primary structure is the AA sequence which isn't changed by anything other than AA sequence break down by enzymes or acid/base. A: oxygen doesn't really affect pH, CO2 does. And it doesn't explain the sigmoidal shape, and only allows oxygen to bind more easily. I was focused on the first part of the sentence instead of it asking specifically about SHAPE, so this Q took me longer. Remember to focus on what Q is ACTUALLY asking.
An artificial heart valve was tested for its ability to function under extreme conditions, to a maximum flow rate of 4.00 x 10-4 m3/s. What speed would this correspond to for an average red blood cell within a blood vessel of cross-sectional area 5.00 x 10-6 m2? A. 2.00 x 10-9 m/s Show Explanation B. 1.25 x 10-2 m/s Show Explanation C. 4.05 x 10-4 m/s Show Explanation D. 8.00 x 101 m/s
D Watch your units! We're given one number in m3/s and another in m2. To get an answer in the units of m/s, just divide the two numbers in the question: Speed = (4 x 10-4 m3/s) / (5 x 10-6 m2) = (4/5) x (10-4/10-6) = 0.8 x 102 m/s = 8 x 101 m/s This can also be solved using the equation flow rate = cross-sectional area x velocity.
While the composition of oxygen and nitrogen in air does not change with altitude, the decreasing temperature at high altitude does change the percent of air that is composed of H2O. Assuming constant relative humidity, which of the following can be asserted about the total grams of H2O in a given volume of air at 3000 m above sea level versus at sea level? A. Assuming constant relative humidity means that air has roughly the same mass of H2O per unit volume at 3000 m above sea level. Show Explanation B. Whether air at very high altitude has more or less mass of H2O per unit volume than it does at sea level depends on the temperature at high altitude. Show Explanation C. Air has significantly more mass of H2O per unit volume at 3000 m above sea level. Show Explanation D. Air has significantly less mass of H2O per unit volume at 3000 m above sea level.
D With decreasing temperature, air is able to hold less H2O. Since colder air can hold less total H2O, this means the same relative humidity would result in less total water in the air. For example, if warm air could hold 100 g of water in a given volume, then 50% relative humidity would be 50 g of water in the air. And if cold air could only hold 40 g of water in the same given volume, then 50% relative humidity would be 20 g of water in the air. So with the same relative humidity, the cold high-altitude air has less mass of water. I chose C because I was thinking if the same amount of water were present in the sample, then there's less air at a higher altitude so the mass composition of water in a sample of air would be higher at a higher altitude than a sample at a lower altitude BUT it SAYS DIRECTLY IN THE QUESTION STEM "While the comp of oxygen and nitrogen in air does not change with altitude, the decreasing temp... does change the percent of air that is composed of H2O." SO ANSWER THE QUESTION ASKED.
If the DNA of a representative species from each of the major kingdoms was examined, the sequences coding for which of following would be expected to be most similar? A. Photosynthesis Show Explanation B. Cholesterol synthesis Show Explanation C. Protein modification Show Explanation D. DNA synthesis
D. DNA sequences that are common among different species, phyla, or even kingdoms are called conserved sequences. Conserved sequences tend to remain that way due to the fact that they code for a vital function that is common among disparate species.
imide
O=C-NH-C=O (NH replaces O in middle on anhydride)
face validity
describes the extent to which a study appears to assess what it is intended to assess—that is, more or less, the degree to which it "seems right" to participants and researchers.
disulfide
R-S-S-R
thiol
R-SH
thioketone
R2C=S. (sulfur replaces oxygen as ketone)
sulfoxide
R2S=O
house money effect
Thaler and Johnson have found that individual risk-taking behavior is affected by prior gains and losses. They found that after a prior gain, people become more open to assuming risk since the new money is not treated as one's own. This is known as the house money effect.
Viable trisomies
Trisomy 21, 18, 13
At what pH will a solution of 2.2 M cysteine be isoelectric? (pKa of COOH is 1.96, of HS is 8.18 and of NH2 is 10.12) A. 1.96 Show Explanation B. 5.07 Show Explanation C. 8.18 Show Explanation D. 9.15
We need to determine the pH at which Cys will be isoelectric, which means having no net electric charge. There are two ways to solve this question. First, we can use the given pKa values to determine the charge that Cys will carry in each solution. Alternatively, we can recall that an amino acid will be in its zwitterion (i.e. isoelectric) form when the pH of the environment is equal to the pKa of the amino acid. This means we need to take the average of the two most relevant pKa's on cysteine. While Cys is not commonly listed as an acidic amino acid, its side chain includes a thiol (-SH) group, which can be deprotonated/made negative. A way to figure this out is to note that the structure of cysteine at pH = 7 shows that the side group is protonated. So it must be that even though the pKa is 8.33, the thiol is able to act as an acid (reminder, a pKa > 7 does NOT mean the group is basic). For amino acids with acidic side chains, the isoelectric point is the average of the two most acidic (lowest) pKa's. Alternatively, we can evaluate each choice and calculate the overall charge on the Cys residue. At the pH of choice B (the correct answer), the COOH group will be deprotonated (-1 charge), the amino group will be protonated (+1 charge), and the side chain will be protonated (neutral charge). As such, the overall net charge on the molecule will be 0.
An object with a mass of 10 kg is rolled down a frictionless ramp from a height of 3 meters. If a factory worker at the bottom of the ramp slows the object until it comes to a stop, how much work must the factory worker have done? (Assume g = 10 m/s2.) A. 30 N Show Explanation B. 300 N Show Explanation C. 30 J Show Explanation D. 300 J
You can solve this question using conservation of energy and the work-energy theorem. The object has 300 J of gravitational potential energy at the top of the ramp (U = mgh = 10 x 10 x 3). That energy is converted into kinetic energy at the bottom of the ramp. The work-energy theorem tells us that Wnet = ΔKE. Since the object is coming to rest, its ΔKE = 300 J. Thus, the work done must also equal 300 J. I chose answer B while reminding myself not to go for the trap answer of wrong units and picked the one w the wrong units instead of the one with the correct units...
Stronger acids have
a more stable conjugate base
primary aging
aging of biological factors and the physical body
Case control design
compare information about individuals with a disease against people without the disease
gambler's fallacy
mistaken belief that, if something happens more frequently than normal during some period, it will happen less frequently in the future
Deoxyadenosine is a
nucleoside; has a deoxyribose so is present in DNA
Adenosine is a
nucleoside; has a ribose sugar so is present in RNA
thinning in operant conditioning
reducing the frequency of rewards for a given action
content validity
refers to whether a study comprehensively accounts for all the relevant facets of the phenomenon it is intended to investigate.
prisoner's dilemma
standard example from psychological game theory that shows why two completely "rational" individuals might not cooperate, even if it appears that it is in their best interests to do so.