NSTest5/MCAT
If a hydrogen NMR spectrum were obtained for a double-stranded amidine dimer, would the amide N-H proton resonances experience any shielding effects? A. Yes, they would exhibit upfield shifts due to shielding effects. B. Yes, they would exhibit downfield shifts due to deshielding effects. C. No, they would not experience any shielding effects. D. Yes, they would exhibit downfield shifts due to shielding effects.
Hydrogen deshielded because nitrogen pulls e- to it B is correct. This question requires you to further extrapolate information from the passage. You can infer from the structure of the helical carboxylic acid dimer shown that the strands of a double-stranded amidine dimer should similarly be held together by hydrogen bonds. If hydrogen bonds are present, the electron density around hydrogen atoms bonded to nitrogen atoms will shrink, as these electron clouds would tend to surround nitrogen more than hydrogen. This deshielding effect will lead to increased chemical (downfield) shifts for N-H proton resonances due to the removal of electron density.
regression to the mean
responses becoming less extreme when participants are assessed again, instead of becoming less moderate or average.
If the tympanic membrane has three times the diameter of the oval window (to amplify sound as it is transmitted through the ossicles) how would it impact the results shown in Figure 2 if the same experiment was performed on a mouse on whom the tympanic membrane had not been bypassed? A. The higher intensity would shift the values shown in Figure 2 to the left. B. The higher intensity would have no impact on the curve shown in Figure 2. C. The higher intensity would increase the wavelength of the sound waves, shifting the values in Figure 2 down and to the right. D. The higher intensity would shift the values shown in Figure 2 up.
"transmitted elastically" - should think spring B is correct. The question asks about how an increase in intensity would change the results shown in Figure 2. The values in Figure 2 were determined by changing the frequency of waves entering the cochlea and measuring the location of maximum amplitude along the basilar membrane. From paragraph 3, it is also clear that the frequency of resonance or maximum amplitude is based on the mass and stiffness of the particular portion of basilar membrane. So, changing the intensity of the incoming sound wave should have no impact on where the maximum amplitude occurs (i.e., what frequencies will vibrate certain locations) but will instead only increase the amplitude of vibration at those locations.
frequency of a spring
(1/2pi)sqrt(k/m) spring constant k in order to have the maximum frequency at the base, the base should have the maximum k and minimum m values while the situation is reversed at the apex. in graph f was highest at base and lowest at apex goes along with place theory, pitch higher at base and lower at apex (pitch = frequency)
latic acid fermentation
-Glycolysis followed by the conversion of pyruvate to to lactate, regenerating NAD+ w/ no release of CO2 in gylcolysis pyruvate being oxidized (reduce NAD+ to NADH) pyruvate in fermentation is reduced (NADH oxidized to NAD+
threshold of hearing
0 dB 1 x 10^-12 W/m^2
A single sports fan is capable of yelling at an intensity level of 80 dB from a given distance. If 10,000 similar fans were all yelling from the same distance, which of the following would be closest to the observed intensity level? A. 84 dB B. 120 dB C. 160 dB D. 320 dB
10,000 = 10^4 B = log 10^4 and dB = 10 log 10^4 so: 10 log 10^4 = 10(4) = 40 dB 80 B is correct. 10,000 fans would be capable of yelling at an intensity 10,000, or 10^4, times greater than the single fan alone. According to the decibel scale, this would correspond to an increase of 40 dB. A: This answer choice recognizes that the decibel scale is logarithmic, but fails to account for the fact that it is defined as 10 times the logarithm of the ratio of I to I0. Thus, each power of 10 in the intensity corresponds to 10 units on the decibel scale, not 1 unit. C, D: These answers stem from miscalculation, and are unreasonably loud. 160 dB is extraordinarily loud (likely to cause eardrums to burst), and 320 dB is much louder than anything that is known to have ever happened on Earth, including hydrogen bomb detonations, the launch of interplanetary rockets, and megavolcano eruptions.
threshold of pain
120 dB 1 W/m^2 other thresholds https://www.youtube.com/watch?v=22b23RYpNNg
What is a possible structure of R shown in Figure 1? A. (CH2)17CH3 B. C6H6 C. (CH2O)8CH3 D. C(O)O(CH2)11CH3
A is correct. According to paragraph 2 and Figure 1, this reaction results in the production of free cholesterol and a fatty acid. The "R" group indicated in Figure 1 refers to the long aliphatic tail of a fatty acid, which is a carboxylic acid with a long, saturated or unsaturated aliphatic tail. Of those choices given, only choice A corresponds to an aliphatic structure.
Is there a correlation or linkage between wing characteristics and eye color? A. No, because each trait is sorted independently from a genetic perspective. B. No, because phenotypes are distributed differently from genotypes. C. Yes, because the loci are identical when alleles appear on homologous chromosomes. D. Yes if homozygosity is present; no if heterozygosity is present.
A is correct. Mendel's Law of Independent Assortment states that separate genes which encode separate traits are passed down from parent to offspring independently of each other. Although Mendel induced this law from empirical observations, without a knowledge of molecular genetics, modern research has clarified the molecular basis for this finding. Independent assortment clearly occurs for genes that are physically located on different chromosomes, because it is random whether an organism inherits the maternal or paternal copy of a given chromosome — that is, during meiosis, the orientation of homologous chromosomes on the metaphase plate is random. Furthermore, even for genes on the same chromosome, recombination during meiosis ensures that they are inherited independently in most cases. However, genes that are physically located close to each other on a chromosome show a tendency to be inherited together. Nonetheless, the data presented in paragraph 2 of the passage conform to expected Mendelian distributions, and we are given no data that would suggest linkage, so it is reasonable to expect that there would be no correlation between the genes for these traits. B: The phenotype/genotype distinction is irrelevant to this issue. C: Homologous chromosomes do not necessarily have identical alleles/loci, and even if they did, this would not result in "linkage" in the technical sense of the term, which refers to a greater than 50% chance of alleles being inherited together. D: Homozygosity versus heterozygosity is irrelevant for linkage, which describes to a pattern where alleles have a >50% chance of being inherited together.
Both narcissistic personality disorder (NPD) and BPD are classified as Cluster B personality disorders in the DSM-5. Carl Rogers would argue from the humanistic perspective that NPD results from incongruence between: A. self-concept and reality. B. self-concept and self-image. C. self-esteem and self-concept. D. ideal self and reality.
A is correct. Narcissistic personality disorder occurs when patients have an inflated sense of their own importance and a deep craving for admiration. The inflated sense of importance implies that patients have a disconnect with reality, so the correct answer must be either choice A or D. Narcissistic personality disorder deals more with who patients think they already are (their self-concept), not who they want to be (their ideal self). B, C: Self-image and self-esteem are components of self-concept. There is no disconnect between these components in individuals with narcissistic personality disorder. D: NPD does not primarily deal with the ideal self. Rogers - real and ideal self Rogers spearheaded the ideas of the real and ideal self. In this framework, the point of psychotherapy is to help the client sort through issues and make positive decisions regarding them, rather than make a diagnosis or provide a concrete solution for the underlying problem.
A pure sample of (R)-limonene has a specific rotation of +125.6. If a mixture of (R)-limonene and (S)-limonene has a specific rotation of +62.8, what are the percentages of the R and S enantiomers in this mixture? A. 75% R, 25% S B. 55% R, 45% S C. 45% R, 55% S D. 25% R, 75% S
A is correct. Since the enantiomers in this case must be present in unequal proportions, a weighted average of the components should be used. In this case, the weighted average is given; therefore, we can set up an equation. If we designate x as the percentage of R and 1 - x as the percentage of S, we can solve for x using this equation: R - x S = 1 - x (125.6)(x) + (-125.6)(1-x) = 62.8 125.6x - 125.6 + 125.6x = 62.8 251.2x = 188.4 x = 0.75 1 - x = 0.25 Therefore, R makes up 75% of the solution, while S comprises the remaining 25%.
The structure of farnesyl pyrophosphate (FPP) is shown below. In the human body, two FPP molecules are combined in a reaction that directly leads to which class of compounds? A. Cholesterol B. Sphingolipids C. Triacylglycerols D. Purines
A is correct. The structure of FPP is highly suggestive of its combination with a partner and transformation into the fused 4-ring system common to cholesterol. B: Sphingolipids, or glycosylceramides, are a class of lipids containing a backbone of sphingoid bases, a set of aliphatic amino alcohols. There is nothing in their structure to suggest 2 FPP molecules will lead to such a molecule. C: Triacylglycerols are esters derived from glycerol and three fatty acids. D: Purines are heterocyclic aromatic organic compounds. They consist of a pyrimidine ring fused to an imidazole ring (both of which contain nitrogen).
reflection and refraction
A light ray that is incident at 90º will not reflect and will simply travel along the edge of the boundary between two media. B: A light ray that is incident at 0º to the normal will simply pass straight through the boundary without refracting or reflecting. D: The angle of refraction and the angle of incidence can only be the same if the two media have the same index of refraction (in which case there's really no "boundary" between them) or if they're both equal to 0º.
mesolimbic pathway
includes nucleus accumbens, medial forebrain bundle, and ventral tegemental pathway think dopamine (VTA -> nucleus accumbens, amygdala, prefrontal cortex, hippocampus)???
If all of the subjects of the study did so on a voluntary basis, what type of organization would that be? A. A mimetic organization B. A coercive organization C. A normative organization D. A utilitarian organization
Anwer: C Organizations can be divided into three general types: coercive organizations, normative organizations, and utilitarian organizations. Coercive organizations are those for which members do not have a choice about joining. For example, a prison is a coercive organization. Normative organizations are those which members join based on some shared, moral goal. For example, the American Red Cross volunteer division is a normative organization. Utilitarian organizations are those in which members are paid for their efforts. These include businesses and corporations, from small town restaurants all the way up to McDonald's.
Tautomerization (more info)
At room temperature, keto form favored enol form contributes significantly to some reaction mechanisms, and the deprotonated intermediate (known as an enolate ion) is also important for some reactions. Tautomerization also between enamines and imines (the second most important example), lactams and lactims, and amides and imidic acids. tautomers and resonance structures are not the same thing. Resonance - electrons are delocalized, and resonance structures are a somewhat crude way of representing a single underlying, delocalized structure of the molecule. tautomers - two different structures, interconvert by breaking and re-formation of bonds.
In locations with very low dissolved O2 concentrations in drinking water, the observed Ksp of ferrous (II) hydroxide will be: A. slightly smaller. B. the same. C. slightly larger. D. much larger.
B is correct. Only temperature changes Ksp!!!! But, concentration will affect if ppt forms!!! Equilibrium constants, including Ksp, are not affected by concentration. These constants are only altered by changing temperature, as shown below for a variety of salts: A, C, D: Ksp will not change due to concentration differences, although concentration may affect whether a precipitate forms.
Which of the following is the most significant problem with the design of this study? A. The sample size was too small to ensure that the findings were statistically significant. B. The western blots in Figure 4 were performed at week 52, when the body weight of the Nbea+/+ mice and the Nbea+/- mice had become equivalent. C. The dwarf phenotype of the Nbea+/- mice failed to manifest consistently over time. D. The scientists did not study Nbea-/- mice.
B is correct. Because mRNA degrades rapidly in the cytoplasm, analyzing mRNA and protein expression levels typically provides a window into quickly-changing conditions within the cell. Therefore, it is not certain that western blot results from week 52 could provide insight into the development of the dwarf phenotype earlier in the experiment. This reasoning is supported by the fact that the body measurements in Figure 2 and Figure 3 were made at week 16, when the dwarf mice had a significantly different phenotype. A: We are not provided with information about the sample size, but the passage indicates that several of the results reached statistical significance. C: Although it is true that the heterozygotes only showed significantly different phenotypes from 7 to 24 weeks, that may just be a feature of the dwarf phenotype. There is no reason to think that this reflects a flaw in the execution of the experiment. D: In general, it is not a methodological flaw for studies not to evaluate every possible variable. Both practical and conceptual reasons may exist for restricting the scope of a study. Although it is not explicitly discussed in the study, it is possible that Nbea-/- mice may not be viable or may not provide any useful insights into autism, which is the ultimate purpose of this model.
Two separate gel electrophoresis analyses are performed on a sample of purified α enzyme and the following results are obtained: SDS-PAGE: 1 band SDS-PAGE + dithiothreitol: 3 bands Which prediction about the enzyme structure is most strongly supported by these observations? A. Enzyme α contains a high ratio of charged to uncharged residues. B. Enzyme α contains more than one subunit. C. Enzyme α contains a low ratio of charged to uncharged residues. D. Enzyme α is post-translationally methylated.
B is correct. Dithiothreitol (DTT) is a reducing agent often used during SDS-PAGE to further denature proteins by reducing/cleaving disulfide linkages and breaking up the quaternary protein structure (oligomeric subunits). The presence of one band when not exposed to DTT and three bands when exposed to DTT suggests that at least two disulfide linkages are present in the enzyme, and that these linkages hold three separate subunits together.
Some eukaryotic cells are covered with small ciliary projections used for absorption, while others contain larger flagella used for propulsion. These cellular structures are composed of: A. microfilaments. B. microtubules. C. intermediate filaments. D. myosin.
B is correct. Eukaryotic cilia and flagella are composed of bundles of microtubules. Note that this differs from prokaryotic flagella, which are formed from the protein flagellin. A: Microfilaments are composed of actin and are found in the cytoplasm of eukaryotic cells, as well as in muscle. C: Intermediate filaments are less dynamic than actin or microtubular filaments. They are not involved in ciliary or flagellar structure. D: Myosin is present in muscle and aids in the process of contraction.
Under which of the following conditions would IPB grow best? A. 33-40°C, pH 6.6-7.9 B. 17-25°C, pH 6.6-7.9 C. 33-40°C, pH 2.6-3.9 D. 17-25°C, pH 2.6-3.9
B is correct. First, let's consider pH. The third paragraph states that acid can be added to water to kill bacteria, so highly acidic choices like C and D can be eliminated immediately. The passage doesn't directly state anything about the temperature, but paragraph 2 does mention that IPB grow in locations used for potable (drinkable) water delivery. Thus, we can infer that IPB would grow best in the conditions under which tap water is typically found: near-neutral pH and near room temperature. also consider that O2 (needed to be dissolved in water) more available at cooler temp than warmer?
One explanation for the development of personality uses the biological perspective, which emphasizes the influence of genetics and brain biology in determining an individual's behavioral, emotional, and cognitive patterns. Which of the following psychologists' views are most closely aligned with this perspective? A. Abraham Maslow B. Hans Eysenck C. B.F. Skinner D. Gordon Allport
B is correct. Hans Eysenck is noted for the first empirical study he published on genetics of personality in 1951, which investigated the trait of neuroticism in identical (i.e., monozygotic) and fraternal (i.e., dizygotic twins). A: Abraham Maslow is better known for his involvement with the humanistic perspective of personality, where he formed the hierarchy of needs to describe the physiological and psychological needs humans require to be fulfilled C: B.F. Skinner is more closely related to the behaviorist perspective of personality, where he studied the ability of operant conditioning to modify personality over time. D: Gordon Allport is known in connection to the trait perspective of personality. He argued that three key types of traits contribute to personality: cardinal traits, central traits, and secondary traits.
Which of the following is NOT a type of strategy that HM used to try to remember? A. Rehearsal B. Priming C. Chunking D. Dual coding
C is correct. Chunking is not a strategy that HM used. Chunking involves combining individual pieces of information into larger "chunks" so that they may be remembered more easily. A: HM rehearses the numbers he is given when he is adding them together. B: Priming occurs when HM more successfully read the words to which he had previously been exposed than those he hadn't. D: By associating the numbers with different animals, HM was attempting to assign each number to a mental image to be processed at a more connected level. This is the premise of dual coding.
The equivalence point of the titration of tolbutamide with NaOH was reached by adding 50 mL of NaOH. Which of the following correctly describes the solution during this process? A. The solution had a pH less than 7.1 at the equivalence point. B. After addition of 25 mL of NaOH, the pH of the solution was greater than 5. C. The concentration of the charged form of tolbutamide was greater than neutral tolbutamide during the titration. D. The concentration of the charged form of tolbutamide was less than neutral tolbutamide during the titration.
B is correct. If 50 mL of NaOH represents the volume of titrant required to reach the equivalence point, then 25 mL is the volume added at the half equivalence point. At this point, one half of the original tolbutamide present will have been converted to its conjugate base, and their concentrations will be equal (eliminate choices C and D). According to the Henderson-Hasselbach equation, when the values of protonated acid and conjugate base are equal, pH = pKa + log 1 = pKa + 0 = pKa. At the half-equivalence point, pH of solution equals the pKa of the analyte. According to paragraph 3, tolbutamide's pKa is 5.3. A: The titration of a strong base and a weak acid results in the formation of a basic salt, the hydrolysis of which causes the pH at the equivalence point to be greater than 7. C, D: The relative concentrations of the charged versus neutral forms of tolbutamide changed throughout the titration.
Elena's mom tells her to pay close attention to the new students at school, because first impressions are the most important aspect of determining who your friends should be. What is her mom relying on to make judgments about Elena's potential friends? A. Physical appearance B. Primacy effect C. Recency effect D. Personal constructs
B is correct. In this context, the primacy effect is a phenomenon in which information that is presented first has a disproportionate influence on your perception of a person. Elena's mom heavily relies on first impressions and the primacy effect. A: Elena's mom did not mention physical appearance in particular, just first impressions in general. C: The recency effect is the valuing of information that is presented later, more so than information gained earlier; this does not relate to first impressions. D: Personal constructs are beliefs about which attributes are most important in making judgments about people. Elena's mom did not specify what attributes she was looking for (like kindness or trustworthiness), just that first impressions are important.
In comparison to the cohesive forces between water molecules of the protein solution droplet, how can the strength of interaction between water and oil molecules at an oil-water interface be characterized? A. Weaker, because of the larger surface area of the hydrophobic cap B. Weaker, because they are forces created by induced polarity in nonpolar molecules C. Stronger, because of the predominance of Van der Waals forces D. Stronger, because of sequestration of hydrophobic cap residues from the solvent
B is correct. The Van der Waals forces that exist between molecules of water and of oil are predominately of the induced dipole-dipole type. Here, a small, temporary dipole is induced in molecules of oil by the permanent dipole of water, resulting in a weak attraction between the molecules. This is a weaker interaction than the hydrogen bonding that predominates in the interaction between water molecules. A: In general, as the surface area of molecules increases, so too does the strength of intermolecular forces between molecules. This is a result of the greater area of interaction available per molecule. This, then, would not support a conclusion that oil-water interactions are weaker than those between the water and hydrophobic caps. C, D: The interaction between water and oil molecules is weaker than the bonding which takes place between water molecules and the hydrophilic portions of the protein.
If a signal is transmitted along a segment of nerve axon measuring 5 x 10-4 m, how much time is required for the signal to reach the end of the segment, assuming maximal transmission velocity? A. 2.5 x 10-6 ms B. 2.5 x 10-3 ms C. 1 x 10-1 ms D. 1 x 10-1 s
B is correct. The passage tells us that signals can be transmitted up to 200 m/s, which would be the maximal velocity, and the question tells us that the distance traveled is 5 x 10-4 m. Dividing distance by the speed gives us time: 5 x 10^-4 m / 2 x 10^2 m/s = 2.5 x 10^-6 s Converting to milliseconds, we get 2.5 x 10-3 ms. MILLIseconds!!!!!!
Besides directly killing bacteria, how does pasteurization with hot water treat IPB populations? A. By decreasing the solubility of ferrous (II) hydroxide, preventing further bacterial growth B. By decreasing the solubility of ferric (III) oxide, preventing further bacterial growth C. By decreasing the solubility of oxygen, preventing further bacterial growth D. By increasing the solubility of oxygen, preventing further bacterial growth
C is correct. solubility of gases, all more soluble in cold water than warm This question asks us to think about other ways in which pasteurization might be an effective tool against IPB besides direct bacterial death. The passage states that IPB oxidize iron compounds and require a certain concentration of oxygen to survive. Oxygen, like all gases, is more soluble in cold water than warm water. Thus, the process of heating water (pasteurization) would lower the solubility of oxygen and reduce the amount available to IPB.
A scientist discovers that no hGH was detected in the hippocampus of Nbea+/− mice, the brain region with the highest expression of Nbea. What does this indicate about the expression of hGH? A. The expression of hGH is tissue-specific and is directed by the Nbea promoter, not promoter elements in the hGH transgene itself. B. The expression of hGH is tissue-nonspecific and is directed by the Nbea promoter, not promoter elements in the hGH transgene itself. C. The expression of hGH is tissue-specific and is not directed by the Nbea promoter, but by promoter elements in the hGH transgene itself. D. The expression of hGH is tissue-nonspecific and is not directed by the Nbea promoter, but by promoter elements in the hGH transgene itself.
C is correct. If no hGH was detected in the hippocampus of Nbea+/- mice, the brain region with the highest expression of Nbea, this indicates that the presence of Nbea promoters has no effect on the activation and transcription of hGH. Therefore, the promoter elements must be in the hGH transgene itself and only activated in specific tissues.
George Mead's social theory presents the part of the self which is called the "me" as the: A. autonomous sense of self that reacts to the attitudes taken in from society. B. part of the self that generates moral behavior and is critical of the self's thoughts and actions. C. collection of attitudes taken from society. D. organized, realistic part of the self that mediates between base desires and moral impulses.
C is correct. In Mead's theory of the nature of the self, people are divided into the "I" and the "me." The "me" is the collection of attitudes taken from society, whereas the "I" is the autonomous sense of self that reacts to the "me." A: This is a description of Mead's idea of the "I." B: This describes a Freudian idea, the super-ego. D: The ego, which (according to Freud) mediates between the id and the super-ego, matches this choice, not the "me."
Some people believe that school success should be based on hard work, but in actuality there are a number of extraneous factors that contribute to student success. These are examples of: A. inductive meritocracy and deductive meritocracy, respectively. B. deductive meritocracy and inductive meritocracy, respectively. C. prescriptive meritocracy and descriptive meritocracy, respectively. D. descriptive meritocracy and prescriptive meritocracy, respectively.
C is correct. In sociology, prescriptive refers to what an individual believes should occur, while descriptive refers to what one perceives as actually occurring. A, B: Inductive reasoning extrapolates from individual observations to general principles, while deductive reasoning is the opposite. Neither of these phenomena is occurring in the situations described by the question stem.
Throughout much of its recent existence, Indonesia has had a very strong economy with sharply defined and well-functioning political structures. Its economy and military were some of the strongest in the region, while neighboring Papua New Guinea had much weaker systems and was almost wholly dependent on Indonesia for trade and protection. Indonesia could be described as: A. an undeveloped nation. B. a semi-periphery nation. C. a core nation. D. a periphery nation.
C is correct. In world systems theory, nations are split into core, semi-periphery, and periphery countries on the basis of their wealth, military strength, and government institutions. Core countries are those well-developed, wealthy capitalist nations that control many resources, have powerful militaries, and strong state institutions. This description matches the description of Indonesia in the question.
The HAT mechanism is expected to be: A. spontaneous at low temperatures and nonspontaneous at high temperatures. B. nonspontaneous at both low and high temperatures. C. nonspontaneous at low temperatures and spontaneous at high temperatures. D. spontaneous at both low and high temperatures.
C is correct. Spontaneity can be predicted by the equation ΔG = ΔH - TΔS, where ΔG is the Gibbs free energy and ΔS is change in entropy. Negative values of ΔG indicate spontaneity. Table 1 indicates that ΔH is positive for HAT in all circumstances that were studied, and because the HAT mechanism involves splitting a single molecule into a phenoxyl radical and a hydrogen radical, ΔS must also be positive. Therefore, even without knowing the exact numerical value of ΔS, we can recognize that low temperatures will yield a positive ΔG value, meaning that the reaction will be nonspontaneous, while sufficiently high temperatures will cause the (- TΔS) term to "outweigh" the positive ΔH and make the overall reaction spontaneous (as indicated by a negative ΔG).
The synaptonemal complex with which TEX11 interacts serves as a protein-DNA bridge between what nuclear bodies? A. Sister chromatids B. Chromosomal centromeres C. Homologous chromosomes D. Meiotic spindles
C is correct. Synapsis is the pairing of homologous chromosomes during prophase I of meiosis. According to paragraph 2, the synaptonemal complex is "a structure which facilitates interaction between synapsing chromatids." This protein structure bridges together synapsing chromatids from homologous chromosomes during synapsis (and possible recombination). During synapsis, interaction occurs between chromatids from homologous chromosomes, not between sister chromatids. You can remember this because sister chromatids are genetically identical, so crossing over between them would have no effect. B: Centromeres are chromosomal regions linking sister chromatids, not homologous chromosomes. D: In meiosis, the spindle apparatus attaches to chromosomes through the kinetochore, not through the synaptonemal complex Anaphase 1 - Centromeres do not break like in mitosis. Centromeres break when chromatids separate
The azimuthal quantum number corresponds with which of the following? A. The potential energy of the electron B. Approximate radial size of an electron cloud C. Approximate geometric shape of the orbital D. Number of valence electrons orbiting a nucleus
C is correct. The azimuthal quantum number (subshell) for an atomic orbital determines its orbital angular momentum and describes the shape of the orbital. azimuthal, or angular momentum, quantum number (l) describes the subshell of the principal quantum number in which the electron is found, with values ranging from 0 to n − 1, where l = 0 is the s subshell, l = 1 is the p subshell, l = 2 is the d subshell, and l = 3 is the f subshell. max e- in a subshell: (eq 4l + 2) e- in s = 2; p = 6; d = 10; f = 14 A, B: This value is quantified by the principal quantum number, n. first quantum number is the principal quantum number (n). It denotes the energy level of the electron, and can take any integer value (≥1). Higher principal quantum numbers (for example, n = 2 rather than n = 1) have greater energy and are farther from the nucleus. The principal quantum number relates to the row of the periodic table in which the element in question is found. D: This value is not quantified by any quantum number. The number of valence electrons orbiting a nucleus can only be obtained by examining the complete spectroscopic notation of the electron or the location of the atom on the periodic table. **Note that as we progress through the four quantum numbers (n, l, ml, and ms) we get more and more specific about exactly where this electron can be found.
The sequence of a typical HERV is most likely to contain a gene that codes for: A. the production of a DNA-dependent RNA polymerase. B. the production of an RNA-dependent RNA polymerase. C. the production of an RNA-dependent DNA polymerase. D. the production of a helicase enzyme.
C is correct. The passage states that HERVs, or human endogenous retroviruses, arose through the integration of retroviral material into the genome. It also mentions that HERVs are able to perform reverse transcription, a process that requires the enzyme reverse transcriptase. This enzyme catalyzes the production of DNA from an RNA template. Reverse transcriptase thus must have DNA polymerase activity, since it builds a new DNA strand; it is also RNA-dependent, since it reads an RNA template. A: This describes a polymerase that creates RNA from a DNA template. This does happen during regular transcription, but there is no reason for a HERV to code for any transcriptional enzyme. B: This term would describe a polymerase that forms RNA from an RNA template. Such enzymes do not even exist in eukaryotes (although they are found in viruses). D: Helicase is an essential enzyme for DNA replication and would certainly be encoded somewhere in the genome, but there is no reason why it would be found in a HERV.
Based on the passage results, how would the mass and stiffness be expected to vary along the length of the membrane? A. They would both be at a maximum value at the base and decrease along the membrane to a minimum at the apex. B. They would both be at a minimum value at the base and increase along the membrane to a maximum value at the apex. C. The stiffness would be highest at the base, while mass would be highest at the apex. D. The stiffness would be lowest at the base, while mass would be lowest at the apex.
C is correct. This question requires us to examine Figure 2. We need to determine how the mass and stiffness would be expected to vary to produce the results shown in Figure 2, which shows the highest frequency points of resonance at the base and lowest frequency points of resonance at the apex. Paragraph 3 also states that each location can be modeled like an individual spring that will vibrate only at its natural frequency. The equation for the frequency of a spring is f = (1 / 2 π)(√k / m). Thus, in order to have the maximum frequency at the base, the base should have the maximum k and minimum m values while the situation is reversed at the apex.
Which of the following molecules would be expected to have the lowest tissue concentrations in active skeletal muscle deprived of O2? A. Glyceraldehyde 3-phosphate B. Lactate C. Citrate D. Pyruvate
C is correct. Under anaerobic conditions (in the absence of oxygen), pyruvate undergoes fermentation to lactate in the cytoplasm instead of being transported to the mitochondria for conversion to acetyl-CoA. Thus, acetyl-CoA will not be present to enter the Krebs cycle and will not be converted to citrate. A: This is a glycolytic intermediate. Glycolysis does not require oxygen and proceeds similarly under aerobic and anaerobic conditions. B: As a product of anaerobic fermentation, lactate will likely display increased concentrations in the absence of oxygen. D: Pyruvate is formed during glycolysis, which continues regardless of the presence of O2. While pyruvate is later converted to lactate via fermentation, there is no reason to suspect that this would lower its concentrations any more than its conversion to acetyl-CoA under aerobic conditions.
codominance vs. incomplete dominance
Codominance - (tabby cat, black and brown parents) both alleles show up in the phenotype (ex. black and white speckled chickens), Incomplete Dominance - a mixed phenotype is observed (ex. grey chickens, pink flowers from red and white parents) - new phenotype shrunken wings (from wing and wingless parents)
Several Salmonella species are facultative anaerobes. Assuming that other external conditions are controlled for, would the expected growth rate of a Salmonella colony be slower in the presence or absence of O2? A. In the presence of O2, because aerobic respiration produces CO2, a byproduct that is lethal to facultative anaerobes. B. In the presence of O2, because the final product of aerobic respiration contains more energy than the final product of fermentation. C. In the absence of O2, because the bacteria will need to produce pyruvate decarboxylase, an enzyme required for entrance to the Krebs cycle. D. In the absence of O2, because these conditions result in lower production of ATP, which can fuel binary fission.
D is correct. Facultative anaerobes can produce energy in the presence or absence of O2. In the presence of O2, the bacteria undergo aerobic respiration, which produces approximately 19 times as many ATP molecules per molecule of glucose as does anaerobic respiration. (Note that the fermentation step itself does not yield any ATP; it simply regenerates the NAD+ required to continue glycolysis.) ATP is necessary for many cell processes, including reproduction; thus, conditions that lead to less ATP are expected to lead to a lower growth rate. A: CO2 is not toxic to facultative anaerobes. B: The high-energy product formed under both aerobic and anaerobic conditions is ATP; aerobic conditions simply allow more of it to be produced. C: The Krebs cycle is not entered in the absence of O2.
Mr. Smith believes his son John smokes marijuana because he looks very similar to "typical stoners" that Mr. Smith has seen. John sometimes smells a little funny and has been dressing differently because he has started practicing yoga - he has to wear loose clothes at the studio and he smells because his teacher burns incense in class. What cognitive heuristic did Mr. Smith use when judging John? A. Adjustment heuristic B. Anchoring heuristic C. Availability heuristic D. Representativeness heuristic
D is correct. Mr. Smith believed his son's look (and smell) was similar to a category of people (typical teens who smoke marijuana) and inferred that he had all of the attributes of a person who smokes. A, B: Anchoring and adjustment heuristics are the same thing - they mean that we tie impressions to earlier perceptions of people. In this example, Mr. Smith would make a judgment about John's behavior based on what he already knew about him. C: The availability heuristic states that whatever comes to mind first is what we believe to be common and prevalent. This isn't the best option here.
Genotypic analysis of another colorectal cancer cell line, CRC200, indicates that the cancerous cells are expressing a single nonsense mutation in one of their copies of p53. Given that information, which of the following would most likely represent the number of colonies observed after an identical transfection assay? A. pC27-53: 16, pC27-53X 56 B. pC27-53: 67, pC27-53X 713 C. pC27-53: 712, pC27-53X 56 D. pC27-53: 647, pC27-53X 748
D is correct. Paragraph 2 states that the point of this experiment was to determine whether introducing a copy of wild-type p53 would affect growth. Since we are told that p53 is a tumor suppressor gene, we can infer that it must experience loss-of-function mutations in cancer cells. Therefore, since human cells are diploid, to remove the function of a tumor suppressor protein such as p53, both copies of the allele must be mutated. (This is in contrast to mutations in oncogenes, where a single gain of function mutation can lead to an overactive protein product). The question states that CRC200 cells express only a single mutation in one copy of p53. This implies that they have another functioning copy of p53, and are cancerous nonetheless. (Loss of p53 function may be one mechanism underlying oncogenesis, but it is not the case that all cancer cells express the same suite of mutations). Therefore, since the CRC200 cells are already expressing wild-type p53 proteins, transfection with an additional copy of wild-type p53 would not significantly alter cell growth. Of the options, only D fits this description.
Which of the following solvents would lead to the fastest SN1 reaction? A. n-hexane B. Benzene C. Tetrachloromethane D. Propanol
D is correct. SN1 reactions proceed faster in protic solvents. SN2 reactions prefer polar aprotic solvents. The only protic solvent here is choice D. A, B, C: These are all nonpolar solvents. Note that another name for C is carbon tetrachloride. pay attention!!! asking for solvent not reactant!!!
According to the passage, the factors associated with teen marijuana use are examples of: I. dispositional attribution. II. external attribution. III. situational attribution. A. I only B. II only C. I and II only D. I, II, and III
D is correct. The passage lists a variety of factors that contribute to marijuana use, including internal (or dispositional) processes like beliefs about marijuana safety, and external (or situational) processes like peer pressure or parental beliefs about use. Each factor is attributed to a different cause, but all three are covered in the passage.
What is the molar solubility of ferrous (II) hydroxide in water at 25°C? A. 2.1 x 10-8 B. 8.2 x 10-8 C. 2.8 x 10-6 D. 2.1 x 10-5
D is correct. The passage states that the Ksp of ferrous (II) hydroxide is 3.2 x 10-14. When Ksp is known, we can determine molar solubility from the dissociation reaction. Fe(OH)2 (aq) + H2O (l) → Fe2+(aq) + 2 OH- (aq) Note that Fe(OH)2 dissociates into three ions (one Fe2+ and 2 OH-). Given this 2:1 ratio, Ksp = [Fe2+][OH-]2 = [x][2x]2 = 4x^3, where x represents the molar solubility. Next, we must divide Ksp by 4, then take its cube root to solve for x. Dividing 3.2 by 4 is more difficult than dividing 32 by 4, so we can manipulate scientific notation and rewrite Ksp in an easier format. Ksp = 4x3 = 3.2 x 10^-14 = 32 x 10-15 8 x 10^-15 = x^3 2 x 10^-5 = x
Which of the following statements describe an obstacle researchers would face in implementing a similar transduction procedure to treat cancerous growth within patients? I. Gene incompatibility between patients with different endogenous copies of p53 II. Patient immunological response to the transduction virus III. Effective virus delivery into the entirety of the tumor A. I only B. II only C. I and III only D. II and III only
D is correct. There are many obstacles facing successful gene therapy treatments for cancer, but gene incompatibility is not one of them. Critical genes are often conserved across species, and a wild-type copy of human p53 would certainly be genetically compatible with another human patient. (In fact, many genes are so highly conserved that human homologues can be used interchangeably with their bacterial counterparts.) For this reason, statement I is incorrect. Since statement II is in both remaining choices, we know that it must be correct. Transduction is mediated by viruses, which produce an immunological response when encountered by human cells. Finally, we turn to statement III. In order to arrest the growth of a tumor, nearly all (if not all) of the tumorous cells have must be successfully altered with the gene in question. Simply introducing the new gene to a small number of cells will be ineffective, as the remaining cancerous cells will repopulate the tumor and continue to grow uncontrollably. Thus, statement III is accurate as well.
defense mechanisms
Defense mechanisms include a wide range of psychological dynamics through which an individual deals with undesired thoughts or feelings (conceptualized as conflict between the superego and id in Freudian psychology). Repression is the process the ego uses to push undesired or unacceptable thoughts and urges down into the unconscious. The conscious, deliberate form of this is known as suppression or denial, and is typically used to willfully forget an emotionally painful experience or event. Regression is the unconscious process of reverting back to behaviors that are less sophisticated and often associated with children (sucking one's thumb, wetting the bed). Reaction formation is the process of repressing a feeling by outwardly expressing the exact opposite of it. For example, if you really hate a person, you would pretend to really like them. Projection is the process of attributing one's own undesired thoughts or feelings onto another person. For example, if you have a serious problem with your roommate, you will believe that your roommate has a serious problem with you. Displacement is the process of redirecting violent, sexual, or otherwise unseemly impulses from being directed at one person or thing to another. For example, if a teacher having trouble with a problem student begins to feel aggressive urges towards the student, the teacher might displace those aggressive feelings towards their spouse when arriving at home. Rationalization is the process of justifying one's behaviors, which might be socially unacceptable and impulsive, with intellectual explanations that are more acceptable. For example, you might explain your vandalism of a public wall with spray paint by saying, "Well, so many other people do it, look how many walls have graffiti on them." Finally, sublimation is the process of transferring unacceptable urges or impulses into acceptable and perhaps laudatory behaviors, such as creative work.
IR spectra
Diatomic molecules (e.g. O2, N2, and Br2) do not return IR signals because no net change in the dipole moment occurs. Molecules respond to the influx of energy by either stretching or bending. stretching - changing distance bending - changing angle
alcohols (RC-OH), aldehydes (RC(=O)H), ketones (RC(=O)R'), and carboxylic acids (R(C=O)OH).
Due to hydrogen bonding, alcohols and carboxylic acids have higher melting/boiling points than aldehydes and ketones, and can function as organic weak acids
tautomerization
Even if unfamiliar with the Diels-Alder reaction, don't panic! Familiarity with tautomerization, or the keto and enol forms of a structure, can help us answer this question. The keto form of a molecule has a typical C=O bond (as in a ketone). To tautomerize into the enol form, an alpha hydrogen must first be abstracted. The resulting free electrons move to form a double bond between the alpha carbon and the carbonyl carbon, and the C=O bond becomes C-O and gains a proton. The result is the enol form, an alcohol sitting adjacent to a C=C bond. For any of this to occur, an alpha hydrogen must first be abstracted. The marked carbonyl carbon has no α-hydrogens.
NMR spectroscopy
In 1H NMR, each hydrogen atom (or set of equivalent hydrogen atoms) produces a single peak on the spectrum. Equivalent hydrogen atoms are protons that exist in the same magnetic environment. Such protons do not differ in any measurable way, so they correspond to only one peak on the spectrum; however, if a peak corresponds to multiple equivalent hydrogen atoms, the area under that peak will be proportionally greater than the area under a peak that corresponds to a single hydrogen atom. In other words, the peak area (or integration) directly correlates to the number of protons represented by that peak. Peaks on a proton NMR spectrum can range from 0-12 ppm. The location depends on the extent of shielding or deshielding experienced by 1H nuclei, which varies by functional group. In a phenomenon known as splitting, each signal is affected by protons on atoms adjacent to the carbon to which the proton is attached. Splitting patterns are predicted the n + 1 rule, which states that any peak will be split into a number of smaller peaks equal to the number of adjacent hydrogen atoms plus one. For example, if a hydrogen atom is positioned on a terminal carbon adjacent to a carbon bound to one additional hydrogen atom, the peak that represents the first hydrogen atom will be split into a doublet (1 adjacent hydrogen atom + 1 = 2).
Which of the following rankings (from most to least favored) best describes the thermodynamic favorability of the antioxidant mechanisms in benzene? A. SET-PT > SPLET > HAT B. HAT > SPLET = SET-PT C. HAT > SPLET > SET-PT D. SET-PT = SPLET > HAT
Just look at first step of reaction, has to do that first, so only that enthalpy counts and remember more positive, less thermodynamically favored C is correct. The enthalpic changes (ΔH) in Table 1 indicate the enthalpic barrier that must be overcome for a reaction to proceed. As such, mechanisms with a lower ΔH are favored thermodynamically. For all three compounds, HAT shows markedly lower ΔH values in benzene than the other two mechanisms. To distinguish between SPLET and SET-PET, we must observe that although their total ΔH is comparable, SET-PT has a first step with a particularly high ΔH, meaning that it is especially thermodynamically unfavorable. Therefore, the correct ranking is HAT > SPLET > SET-PT.
If the majority of the baseball's kinetic energy comes from power generation in the legs and hips, approximately how much energy do the lower extremities produce in the pitch? A. 65 J B. 70 J C. 140 J D. 810 J
KEinitial + PEinitial = KEfinal + PEfinal Etotal - Enonconservative = Efinal. get KE from ball (final energy output) find energy lost (multiply efficiencies) remember PE = KE - energy lost so add energy lost back which was about 50 percent to find original energy given C is correct. The kinetic energy of the baseball can be calculated using the equation 1/2 mv2. Paragraph 2 tells us that the average velocity of the ball is 30 m/s and its mass is 150 g, which is equivalent to 0.15 kg. KE = (½)(0.15 kg)(30 m/s)^2 KE = (½)(0.15)(900) KE = (0.15)(450) = 67.5 J The ball ends up with 67.5 J of KE, but the question asked for how much energy the lower extremities generated (eliminate choice A). Table 1 shows us the overall efficiency of each energy transfer. Using these, we can calculate how much energy was needed to end up with 67.5 J of KE in the ball. The overall efficiency of the kinetic chain can be calculated by multiplying the efficiencies of each step, and we can round the numbers to make our calculations easier. Efficiency = (0.9)(0.9)(0.7)(0.8) = 9*9*7*8 x 10-4 = (81)(56) x 10-4 ≈ (80)(60) x 10^-4 = 4800 x 10^-4 = 0.48 ≈ 0.5 Thus, the total energy generated in the lower extremities in order to ensure that 67.5 J makes it to the ball is 67.5 J / 0.5 = 135 J (eliminate choice D). If you are uncomfortable choosing a number that is not exactly what was calculated, you only need to remember that we estimated by rounding up to 0.5. Thus, our denominator in the final calculation, 67.5/0.5, is a bit larger than it should be, making our final value smaller than it would be had we not rounded or used a calculator (exact value of 67.5/ 0.48 = 140 J). Thus, we round up and choose choice C.
lipids
Lipids are hydrophobic molecules with a diverse range of structural functions, including energy storage, structure, and signaling. For the MCAT, you should be aware of four main classes of lipids: (1) fatty acids and their derivatives; (2) cholesterol and its derivatives, such as steroid hormones; (3) eicosanoids, including prostaglandins; and (4) terpenes and terpenoids. A fatty acid (R-COOH) is a carboxylic acid attached to a nonpolar hydrocarbon tail. Fatty acids form part of larger dietary lipids termed triacylglycerols, which have three fatty acid chains attached to a glycerol backbone via ester linkages. A diverse range of lipids can be generated by replacing one of the fatty acid chains with something else. In phospholipids, this "something else" is a polar phosphate group, which itself can be modified by the addition of other organic substituents. Phospholipids are most well-known for being the major components of the lipid bilayer of the eukaryotic plasma membrane. Cholesterol is a lipid with several crucial cellular functions. Namely, it contributes to the fluidity of the plasma membrane and serves as the precursor for steroid hormones and other biomolecules, such as vitamin D. Cholesterol and its derivatives can be identified by their structures, which include four fused hydrocarbon rings. A separate broad family of lipids is the eicosanoids, which include signaling molecules such as prostaglandins. Prostaglandins are synthesized from arachidonic acid, contain 20 total carbons including one 5-carbon ring, and help regulate inflammation. Finally, the MCAT expects you to be aware of terpenes and terpenoids. Terpenes are hydrocarbons composed of repeating isoprene (C5H8) units, and terpenoids are terpenes that are modified with other organic substituents. Squalene, a 30-carbon terpene, is a precursor in cholesterol synthesis. Last but not least, waxes are lipids that are common in plants and that do not fit into the above categories. In nature, waxes appear as naturally-occurring mixtures of lipids that include fatty acids, long-chain alcohols, aromatic compounds, and other functional groups.
Content Foundations: Power
Power is defined as work divided by time (P = W/t). Its units are watts (W), and 1 W = 1 J/s. The idea of power is essentially that a given amount of work could be expended either quickly or slowly, and that a system capable of doing so quickly is more "powerful" than a system in which the energy represented by a given amount of work is dissipated slowly. Rearranging the units of J/s indicates that power can also be expressed as a constant force multiplied by a constant velocity (P = Fv). Another important definition of power occurs in the context of circuits, where power is defined as current multiplied by voltage (P = IV). This is also derived from the general concept of power being work divided by time, but it is best memorized separately for the MCAT. From a problem-solving point of view, it is useful to remember that work can be expressed in multiple ways. A common definition is W = F⋅d⋅cos(θ), but other ways to calculate work exist. For instance, the work-energy theorem states that the work performed on or by an object is equal to the change in its kinetic energy: W = KEfinal - KEinitial. It is therefore important to be alert to which parameters are given in specific MCAT problems in order to approach power effectively.
While evaluating the timing of pitches, the researchers noticed a couple of high-performing athletes who were able to generate higher velocities by increasing the length of their acceleration phase. How did the power production for these athletes compare to those originally mentioned? A. These high-performing athletes had an increased power output due to the higher velocities. B. These high-performing athletes had a decreased power output due to the increased time spent in the acceleration phase. C. These high-performing athletes had an increased power output due to the increased time spent in the acceleration phase. D. There may have been no change in power output between the two groups.
Remember: P = Fv P = W/t also: F = ma (if velocity increases, acc increases) W = Fd Power increased b/c of increased velocity Power decreased b/c of increased time D is correct. The power output is given by the equation P = W / t, where W is the work done in joules (or energy output) and t is the time period over which the work was performed. If the velocity increased, then the work also increased, but the time spent in the acceleration phase increased as well (and this increased time of acceleration is the only reason given for the increased velocity, meaning that both v and t increased to the same degree). Thus, there should be no significant difference in power output between the two groups. A: Higher velocities would mean that this group did more work, but this does not necessarily require an increased power output, as power takes time into account. B: If there was no change in work, then the increased time would result in a decreased power output. However, the work also increased, so there should be minimal change to the power generated. C: The increased time could not result in an increased power output, as they are inversely related.
Meiosis II
Results in the separation of sister chromatids -similar to mitosis in that sister chromatids are separated from one another -> no change in ploidy -Not separation of homologues
saturated and unsaturated fatty acids
Saturated fatty acids have higher melting/freezing points, no kinks **generally solid at room temp. need higher temp to melt freeze at higher temp Two ways to look at this: - There are no kinks, so interactions between the fatty acid chains are maximized. More London forces = more heat necessary to melt it = higher melting point. - There is no kinks to disrupt formation of the 3D structure (solid form) as you lower temperature. Therefore, you don't need to cool it down as much to get it into a solid form = higher freezing point. Unsaturated fatty acids have lower melting/freezing points, kinks **generally liquid at room temp. melt at lower temp freeze at lower temp - Kinks in the structure disrupt interactions between the fatty acid chains. Less Van der Waals forces = less heat necessary to melt it = lower melting point. - Kinks in the structure disrupt formation of the 3D structure (solid form) as you lower temperature. We need to cool it down more (lower temperature) to get it to solid form.
misattribution of arousal
The subjects who were uninformed about the potential effects of the adrenaline injection incorrectly thought that they were emotionally responding to the arousing situation. This perfectly constitutes misattribution of arousal, as the subjects attributed their arousal to the wrong cause.
polygenic traits
Traits controlled by two or more genes Polygenic - "many genes." For example, at least three genes are involved in making the reddish-brown pigment in the eyes of fruit flies. Polygenic traits often show a wide range of phenotypes. The broad variety of skin color in humans comes about partly because at least four different genes probably control this trait.
Object between f and lens
VUL virtual, upright, and larger
vitamin functions B1, D, E, K
Vitamin B1, or thiamine, is a coenzyme in metabolic processes involving amino acids and carbohydrates. Vitamin D is a lipid-soluble molecule primarily involved in calcium metabolism. Vit. E - antioxidant Vitamin K refers to a group of closely-related molecules that contribute to blood coagulation and clotting.
rotation think:
acl rotation/(concentration x length)
independent assortment
also applies to genes on the same chromosome due to crossing over during prophase I of meiosis.
Amines (R-NH2, R-NHR', or R-NR'R"), imines (R=NH or R=NR'), and enamines (C=C-NH2, C=C-NHR, or C=C-NRR')
are nitrogen-containing compounds with medium melting/boiling points that can act as weak bases. Sulfur-containing functional groups contain the root "thio" and generally act similarly to the corresponding oxygen-containing groups.
Intermolecular forces
attractive forces between molecules that are notably weaker than intramolecular bonds. For this reason, it is typically fairly easy to disrupt these attractions using common lab techniques, such as heating. There are three main categories of intermolecular forces to know for the MCAT: hydrogen bonding, dipole-dipole interactions, and van der Waals forces. Hydrogen bonding (as in water and nitrogenous bases in DNA) attractive force between an electronegative atom (oxygen, nitrogen, or fluorine) and a hydrogen atom in a nearby molecule that is covalently bonded to an electronegative atom. For hydrogen bonding to take place, one of the molecules present must contain one or more O-H, N-H, or F-H bonds. The hydrogen bonds in question are not the same as the O-H, N-H, or F-H bonds themselves, all of which are covalent and thus intramolecular. Instead, the hydrogen "bond" is the attraction between the positive dipole of that covalent bond and the O, N, or F position on a different molecule. Examples of hydrogen bonding include between H2O molecules in water and between nitrogenous bases in complementary DNA strands. Of the three main types of intermolecular attractions, hydrogen bonds are the strongest, because of the strength of the O-H, N-H, or F-H dipole. Dipole-dipole interactions between polar molecules that do not contain O-H, N-H, or F-H bonds. These attractive forces draw the negative end of one dipole closer to the positive end of another. London dispersion forces (van der Waals) are the weakest (in nonpolar and covalent molecules) all molecules experience these attractions but mainly nonpolar b/c polar molecules with H bonding and dipole-dipole so much stronger that van der Waals not significant ...stem from instantaneous dipoles, which are transient dipoles created by the random movement of electrons within a molecule. Even nonpolar molecules display these instantaneous dipoles, so even they will experience London dispersion forces. However, polar molecules that display hydrogen bonding or dipole-dipole attraction will not be affected by London dispersion forces to a significant extent, since those other attractions are much stronger, which is why London dispersion forces are often associated with nonpolar molecules.
Allport's cardinal, central, secondary
cardinal - obvious, whole life revolves around it central - can infer it from their actions secondary - infrequent, can happen in social settings (shyness)
Chromaffin cells
catecholamine-secreting neuroendocrine cells of the adrenal medulla. Remember, the catecholamines include epinephrine and norepinephrine.
L1 and L2 have the structures shown below. What kind of isomers are L1 and L2? A. Enantiomers B. Diastereomers C. Geometric isomers D. Constitutional isomers
constitutional, structural - not same connectivity conformational - stereoisomers just need to rotate, same connectivity stereoisomers - same connectivity (ie, enantiomers, diastereomers) geometric - cis and trans (diastereomers) D is correct. Constitutional isomers have the same molecular formula but differ in their connectivity, as exhibited by L1 and L2. A: Enantiomers are chiral mirror images of each other. They have the same atomic connectivity. B: Diastereomers, or stereoisomers that are not enantiomes, also have the same connectivity. C: Geometric isomers are another form of stereoisomer. A classic example is the "cis" and "trans" configurations around a double bond, which does not apply here.
If the average pitcher is releasing the ball from a height of 1.8 m above the ground, and the pitcher's mound is 0.2 m higher than the rest of the baseball field, at what height would the catcher need to hold his glove to catch the pitched ball? (Note: neglect air resistance, estimate the acceleration due to gravity as 10 m/s2, and assume the pitcher is only throwing the ball horizontally.) A. 2.0 m above the ground B. 1.8 m above the ground C. 0.5 m above the ground D. 0.2 m above the ground
d = rt for time x = v(initial)t + 1/2at^2 (gravity for acceleration) to get distance ball falls then subtract from initial height to get place catcher's glove needs to be D is correct. The passage tells us that pitchers throw the ball at an average of 30 m/s and that the ball travels 18 m horizontally. This means the ball's flight time is: (18 m) / (30 m/s) = (18/30) s = 3/5 s = 0.6 s The ball is released from a position 2 m off the ground (0.2 m from the pitcher's mound and 1.8 m from the pitcher). To calculate the distance the ball falls during 0.6 s, we can use the equation d = v0t + 1/2at2: d = (0 m/s)(0.6 s) + 1/2(10 m/s2)(0.6 s)^2 d = 1/2(10)(0.36) = 1/2(3.6) = 1.8 The ball has fallen 1.8 m from an initial height of 2.0 m. Thus, the catcher must hold his glove 0.2 m above the ground to catch the pitch.
exothermic enthalpy and endergonic delta G
deltaH < 0 but DeltaG > 0 which would mean non-spontaneous and with oil and water makes sense b/c deltaS is greatly reduced by mixing oil with water micelles formed to lesson entropy penalty with hydrophobic inside and hydrophilic outside and rememeber hydrophobic effect reduces entropy penalty
Retroviruses
distinct class of single-stranded RNA viruses, including HIV, that use an enzyme known as reverse transcriptase to synthesize DNA from their RNA genome.
D DUV
diverging lens is always diminished, upright, virtual
if have 120 dB, think 120 = 10 log x (whatever you are looking for I/Io if sound intensity)
divide by 10 then raise both sides by 10 to get rid of log 10^12 = x
Ten moles of the monoprotic, weakly acidic medication aspirin were added to water to make one liter of solution. If the pH of the resulting solution was 5.9, what is the approximate Kb for the non-diffusible form of aspirin? A. 0.1 B. 0.01 C. 0.001 D. 1
find Ka Ka = [H+][A-]/[HA] pH = around 6 so [H+] = 10^-6 H and A separate with same concentration so: Ka = (10^-6 x 10^-6)/10^1 10^1 b/c M = 10 moles/1 L Ka = 10^-13 Ka x Kb = -14 so: Kb = -1
Single-stranded RNA viruses
further subdivided into positive-sense and negative-sense viruses. Positive-sense RNA viruses contain mRNA that can immediately be translated by the cell. negative-sense RNA viruses contain RNA that is complementary to mRNA, meaning that mRNA must be synthesized by an enzyme known as RNA replicase that is carried in the virion.
Cardiac glycosides like ouabain have long been used in the treatment of heart failure. The above reaction of ouabain would most likely result in: A. the formation of 2 ketones. B. the formation of 1 aldehyde and 1 ketone. C. the formation of 1 carboxylic acid and 1 ketone. D. no reaction.
had 3 oxygen as reactant (O3) in reductive conditions (DMS) break alkene; secondary C --> ketone primary C --> aldehyde with H2O2 primary aldehyde instead becomes COOH B is correct. The reaction shown is ozonolysis. Ozonolysis is the cleavage of an alkene or alkyne with ozone (O3) that results in the multiple carbon-carbon bond being replaced by a double bond to oxygen (carbonyl). There is only 1 alkene on ouabain and its cleavage will result in the formation of 1 aldehyde (the alkene C on the right) and 1 ketone (the alkene C on the left). B is correct. The reaction shown is ozonolysis. Ozonolysis is the cleavage of an alkene or alkyne with ozone (O3) that results in the multiple carbon-carbon bond being replaced by a double bond to oxygen (carbonyl). There is only 1 alkene on ouabain and its cleavage will result in the formation of 1 aldehyde (the alkene C on the right) and 1 ketone (the alkene C on the left). A: The alkene C on the right side will form an aldehyde when the molecule is cleaved. C: Ozonolysis will not result in the formation of a carboxylic acid in this case. D: Ozonolysis requires a C=C bond, so a reaction will take place.
order of reactivity
highest to lowest: acid halide, anhydride, ester, amide The -OH group of carboxylic acids can be replaced by other functional groups to form carboxylic acid derivatives, the most notable are amides (R(C=O)NR'R''), esters (R(C=O)OR'), acid anhydrides (R(C=O)O(C=O)R'), and acid halides (R(C=O)X), in increasing order of reactivity.
MCAT Biology: Meiosis
https://quizlet.com/293147266/mcat-biology-meiosis-flash-cards/
solubility in gases vs. liquids
inc. T, inc. ionic substance solubility inc. T, dec. gas solubility inc. P, inc. gas solubility generally, solubility of ionic substances in water increases with temperature, while the opposite pattern is observed for gases. This is because higher temperatures provide gases with more kinetic energy that they can use to escape the solution. Additionally, pressure favors the solubility of gases.
Le Chatelier's principle with moles of gases and temp.
increasing the reactant concentration or decreasing the product concentration will shift the reaction towards the product side, and vice versa. with gases, increasing the volume (decreasing the pressure) will shift the equilibrium to the side with more moles of gas (and vice versa). For a reaction where ∆H > 0, increasing the temperature will shift the reaction toward the products, while decreasing it will shift the reaction toward the reactants; the opposite pattern is found if ∆H < 0.
Fermentation/Lactate
is the process by which our muscle cells deal with pyruvate during anaerobic respiration. When our cells need energy, they break down simple molecules like glucose. ... The cells turn pyruvate, the products of glycolysis, into lactic acid.
Gases
lack a fixed shape and volume b/c vol can change, density is not constant The lack of a fixed volume—or compressibility—is especially important because it means that the density of a given gas is not constant. Rather, if the gas is forced into a smaller container, its density will increase as its particles pack more closely together.
magnetic quantum number and spin quantum number
magnetic quantum number (ml) (range from -l to +l) maximum e- in s = 2; p = 6; d = 10; f = 14 the spatial orientation of the orbital in question within its subshell. Potential values of ml range from −l to +l. Since each orbital can hold a maximum of two electrons, this means that an s subshell can contain up to two electrons, a p subshell can hold up to six electrons, a d subshell can contain up to 10, and an f subshell can hold up to 14. spin quantum number (ms) spin orientation of the electron, which relates to its angular momentum. The two possible spin orientations are ms = −1⁄2 and ms = +1/2. Two electrons in the same orbital (and thus with the same ml value) are said to be paired and must have opposite spin.
facultative anaerobe
makes ATP by aerobic respiration if oxygen is present, but is capable of switching to fermentation if oxygen is absent
asters
microtubules known as asters extend from the centrosome to anchor it to the cell membrane.
Which of the following compound-solvent combinations would be expected to show the highest antioxidant activity? A. Compound 1 in benzene B. Compound 2 in water C. Compound 2 in benzene D. Compound 1 in water
notice which one is meta, ortho, para donating - better base w/d - better acid Also, table provided here All the enthalpic (ΔH) changes presented in Table 1 are positive, indicating that all three mechanisms are endothermic for all three compounds in both solvents. In other words, all three mechanisms must overcome an enthalpic barrier to move forward. The lower the enthalpic barrier, the more thermodynamically favorable the reaction is. Thus, lower ΔH values are associated with higher antioxidant activity. Compound 1 generally shows lower ΔH values than compound 2, and the mechanisms have lower ΔH values in water than in benzene, making choice D correct.
According to the results of the study, we would predict that females from low-income families would be most likely to: A. major in biology and view their grades as being based on effort. B. major in biology and view their grades as being based on external factors. C. major in literature and view their grades as being based on effort. D. major in literature and view their grades as being based on external factors.
null hypothesis not disproven according to p-value C is correct. According to the study, females majored in arts and literature at a greater rate than males. In addition, people from low-SES groups had statistically equivalent views of meritocracy to those of high-SES individuals. A, B: Figure 1 clearly tells us that females were more strongly represented in arts and literature. D: Be careful not to fall for the apparent discrepancy in Figure 2. The final paragraph tells us that the p-value for this trend was 0.15, which is greater than the threshold of p<0.05 usually used for statistical significance. Therefore, we cannot consider the null hypothesis to have been disproven.
fatty acid
one of the major components of a triglyceride triglyceride - ester derived from glycerol and three (tri) fatty acids (from tri-and glyceride)
nucleus accumbens (NAC)
pleasure center, reward pathway a nucleus of the basal forebrain near the septum; receives dopamine-secreting terminal buttons from neurons of the ventral tegmental area and is thought to be involved in reinforcement and attention
maginfication RI-N and DU-P
real, inverted and will have negative magnification greater than 1 magnification (absolute value) is enlarged less than 1 (ab. value, so decimal) is reduced object at focal point, no image
reductional division vs equational division
reductional - diploid to haploid (meiosis I) equational - sister chromatids separated (meiosis II)
Cultural capital
relates to non-financial things of value associated with higher social class (certain accent, high education, style of dress) that promote social mobility symbolic and interactional resources
molar solubility and Ksp
solubility or solubility product given Ksp includes # of ions molar solubility = x (just solve for x) Ksp = [Mg2+][Cl−]2. Ksp is similar to, but distinct from, molar solubility, which refers to the moles of a substance that will dissolve in a solvent, because Ksp takes into account the relevant ions from any source.
London disperson forces (van der Waals)
temporary induced dipole (causing temporary polarity) mainly found in nonpolar but exists in all molecules just in polar to insignificant extent
Note that ∆H, entropy, and temperature
typically given in units: ∆H in kJ entropy in units of J/K temperature in units of K (Kelvin)
A patient presents in the emergency department having ingested a large quantity of tolbutamide. Intravenous administration of which of the following compounds is most likely to increase the rate of urinary excretion of the drug? A. KCl B. NaHCO3 C. NH4ClO4 D. NaCl
weak acid is conj of strong base, so need to deprotonate tolbutamide (defined as weak acid in passage) to give charge B is correct. In order to increase the percentage of drug excreted in the urine, it is necessary to decrease the fraction of tolbutamide capable of reabsorption, or diffusion out of the lumen of the nephron. Paragraph 3 says that for weakly acidic drugs, the uncharged state is capable of diffusion through membranes much more than the charged form. Thus, to prevent reabsorption, we must maximize the charged form of tolbutamide. We can do this to a weak acid by deprotonating the drug (rendering it negatively charged) via administration of a base. This can will increase blood and urinary pH (i.e. decreasing [H+]) and increase the fraction of ionized drug present. Urinary alkalization can be accomplished by administration of a basic salt, like NaHCO3 (the salt of the conjugate base of carbonic acid). A, D: NaCl and KCl are salts that are very close to neutral (neither acidic nor basic). Because Cl- is the conjugate base of a strong acid (HCl), it will only be able to act as a base when pH is extremely low. C: ClO4- is also the conjugate base of a strong acid (HClO4), so it will also fail to act as a base under biological conditions. Additionally, NH4+ is acidic. Because of this, NH4ClO4 is acidic overall, not basic.
Intermolecular Forces vs. Intramolecular
weaker than intramolecular intramolecular: covalent (remember OH is covalent) intermolecular include: (and these can be disrupted easily as in heat) hydrogen bonding (make sandwich - H in middle, can even be "heter" as in N-H-O) dipole-dipole interactions van der Waals forces (think London dispersion)
ethanol fermentation (alcoholic fermentation)
yeast cells convert pyruvate (from glucose, sucrose, fructose) into ethanol and carbon dioxide. anaerobic, redox reaction (reduce pyruvate) products and byproducts: 2 ATP, 2 pyruvate, 2 CO2, 2 acetaldehydes, then 2 ethanols, 2 NADH then convert back to 2 NAD+ In ethanol fermentation, (1) one glucose molecule breaks down into two pyruvates. The energy from this exothermic reaction is used to bind the inorganic phosphates to ADP and convert NAD+ to NADH. (2) The two pyruvates are then broken down into two acetaldehydes and give off two CO2 as a by-product. (3) The two acetaldehydes are then converted to two ethanol by using the H- ions from NADH, converting NADH back into NAD+.