More Q-Packs.

what makes anything fluoresce?

1. The molecule is a conjugated system AND 2. The molecule just went from an excited state to its ground state

Positive pleural pressure

= expiration because this helps force the air out of the body. Again, things want to reach equilibrium so you want to go from high pressure (in the lungs) to low pressure outside of the body, therefore, allowing the air from the lungs to escape and reach equilibrium

The relationship btw atm pressure and height above earth= inverse the relationship btw atm pressure and depth under water= portional

A reading on a barometer can be obtained by measuring the height of the mercurcy column (in mm), which will be directly proportional to the incident (usually atmospheric) pressure being applied to the mercury. At the top of a mountain, atmospheric pressure is lower, causing the column to fall because mercury flows out of the column under its own weight. Under water, hydrostatic pressure is exerted on the barometer in addition to atmospheric pressure, causing the column to rise because more mercury is forced into the column.

Would an increase in the level of plasma aldosterone be expected to follow ingestion of excessive quantities of NaCl? ANo; aldosterone causes Na+ reabsorption by kidney tubules. BNo; aldosterone causes Na+ secretion by kidney tubules. CYes; aldosterone causes Na+ reabsorption by kidney tubules. DYes; aldosterone causes Na+ secretion by kidney tubules.

A. Aldosterone, which is produced by the adrenal cortex, causes Na+ reabsorption by kidney tubules. Such a mechanism decreases Na+levels in the urine. The steroid aldosterone does not cause Na+ secretion into the urine. Because ingestion of excessive NaCl would trigger Na+secretion into the urine, plasma-aldosterone levels would not increase. Rather, the body would rely on those homeostatic mechanisms that excreted the excess Na+. Thus an increase in plasma aldosterone would not be expected to follow ingestion of large quantities of NaCl. NOT C BC This is describing the function of aldosterone but it doesn't answer the question. They are asking you if aldosterone will increase if you eat NaCl. Aldosterone will actually decrease because is incharge or reabsorbing Na+. The person already has a high Na+ concentration in blood; you wouldn't want to absorb even more Na+

How will W change if the angle of the ramp to the horizontal is increased? A. W will decrease, because the normal force to the surface of the ramp will decrease. B. W will not change, because the coefficient of friction between the box and the ramp will not change. C. W will not change, because the gravitational force is always constant and the length of the ramp is not changed. D. W will increase, because the height from which the box falls increases.

A. FN = mgcos0 as theta approaches 90, cos approaches 0. Therefore it is apparent that FN decreases with increasing angle of the ramp. If the angle decreased cos approaches 1. This is useful knowledge when considering Ffric as well, which also decreases as angle increases However with the parallel component of force, Fpara = mgsin0, which illustrates acceleration a=gsin0, its the opposite, as angle increases acceleration of the block also increases.

In the patient described in the passage, the likely genetic basis of the increased levels of uric acid is a mutation: A. affecting an allosteric site of PRPP synthetase. B. affecting the active site of PRPP synthetase. C. in a promotor gene regulating the rate of transcription of the PRPP synthetase gene. D. in a gene coding for a transcription factor for the PRPP synthetase gene.

A. Generally, active site mutations will decrease enzymatic activity. So cant be B because passage states elevated PRPP synthetase activity. It wouldn't be C or D as the passage specifically says that the patient had normal levels of PRPP synthetase which would indicate that its expression rate is not impacted. The passage indicates that the patient produces the normal amount of PRPP synthetase, but its activity in vivo is 3 times the normal level. However, PRPP synthetase purified from the patient does not show this increased enzymatic activity in vitro. The fact that the enzyme binds the substrate and converts it to product at normal levels in vitro suggests that the active site of the enzyme has not been altered, but rather that an allosteric site on the enzyme has been affected. The mutant PRPP can most likely bind an intracellular molecule at an allosteric site, which changes the shape of the enzyme, enhancing its activity. This activity-enhancing molecule most likely is not present in the in vitro reaction mix. Any given competitive inhibitor concentration can be overcome by increasing the substrate concentration.

Which reaction leads to the formation of DHB whose structure is shown in the passage? ACarboxylation of hydroquinone BOxidation of hydroquinone CReduction of benzoquinone DHydroxylation of benzoquinone

A. Hydroquinone has It has two hydroxyl groups bonded to a benzene ring in a para position. A carboxylation is the addition of a Co2 molecule.

Contraction of the diaphragm results in a: IPP= intrapleural pressure A. more negative IPP and inspiration. B. more negative IPP and expiration. C. more positive IPP and inspiration. D. more positive IPP and expiration.

A. Negative pleural pressure= inspiration because this helps the air from outside the body flow in. Remember things want to flow from high pressure to low pressure in order to maintain equilibrium. The opposite is true for expiration. Now, this expansion notably decreases the intrathoracic pressure relative to the atmospheric pressure further explaining why airflow flows in. Here comes the important concept of Boyle's Law which states that as the volume of the thoracic cavity increases, the pressure within the cavity decreases, and this negative pressure gradient drives air into the lungs for inspiration. this expansion may also be stated as decreasing the pleural pressure to become more negative. while abdominal pressure does not play a significant role during inspiration, the diaphragm, another primary muscle of inspiration contracts and moves downward, slightly increasing the intra-abdominal pressure, assisting the external intercostals creating a more pronounced negative pressure gradient within the thoracic cavity.

The following population pyramid best supports which projected demographic outcome? A. The overall size of the population is likely to decrease. B. The mortality rate of the population is unlikely to change. C. The median age of the population is unlikely to change. D. The birth rate of the population is likely to increase.

A. Population pyramids illustrate the age and sex distribution of a population. The population pyramid in the question shows an aging population, as the size of the population increases with age (until the age groups at the top of the figure, where mortality rates are highest). Because of the "inverse" pyramid shape represented in the figure, with the older age groups being larger than the younger age groups, the overall size of the population is likely to decrease over time.

Kidney failure during severe dehydration is most likely due to: A. inadequate blood volume for effective filtration. B. inability to produce sufficient urine. C. buildup of salts in the distal tubules. D. increased body temperature.

A. Salts (at least NaCl certainly) are actively transported out of the DCT, meaning a "build up" inside isn't all that possible - you would just keep increasing medullary salt concentration, which is probably totally fine. However, glomerular filtration relies on hydrostatic pressure to effectively create filtrate.

Which extraction procedure will completely separate an amide from the by-product of the reaction between an amine and excess carboxylic acid anhydride? A. Add 0.1 M NaOH(aq) to quench unreacted anhydride. Then add diethyl ether and separate the layers. The amide can be obtained from the ether layer by evaporating the solvent. B. Add 0.1 M HCl(aq) to quench unreacted anhydride. Then add diethyl ether and separate the layers. The amide can be obtained from the ether layer by evaporating the solvent. C. Add 0.1 M NaOH(aq) to quench unreacted anhydride. Then add diethyl ether and separate the layers. The amide can be obtained from the aqueous layer by neutralizing with HCl(aq). D. Add 0.1 M HCl(aq) to quench unreacted anhydride. Then add diethyl ether and separate the layers. The amide can be obtained from the aqueous layer by neutralizing with NaOH(aq).

A. The answer to this question is A because the by-product of the reaction will be an acidic carboxylic acid and the excess unreacted starting material will also be acidic. Extraction with aqueous base will hydrolyze and extract both of these into the aqueous layer, leaving the neutral amide in the ether layer When NH2 reacts with the anhydride, it results in an amide and carboxylic acid. The NaOH will not only quench the anhydride into COOH, but will also deprotonate the carboxylic acid and move into the aqueous layer. Then, you have the amide in the organic/ether, and COO- in the aqueous layer and the layers are then separated. "The amide can be obtained from the ether layer by evaporating the solvent." Since the ether layer has an amide and an ether, the ether has a lower boiling point and will thus be boiled out first with the remaining amide left. anything charged goes into the aqeueous layer. so in the answer here, we have NaOH deprotonating the COOH turning it into COO-. theres a charge! so it would go into the aqueous layer. the amide on the other hand doesnt have a charge when you add a base. it stays NH2 uncharged, and nothing happens if you add NaOH to that. so that will stay in the organic diethyl ether layer. molecules with >5 carbons per polar group will dissolve in organic solvents (not water) assuming that they are not charged. Charged molecules no matter how large always dissolve in water.

An XsY male is standard: he sires equal numbers of sons and daughters. An XiY male expresses the sex ratio trait: he sires only daughters. Total reproductive output is not affected; XsY males and XiY males sire equal numbers of offspring. A virgin female Drosophila mates and produces 34 daughters and 38 sons. Eighteen of these sons sire only daughters, while the remainder sire approximately equal numbers of daughters and sons. What are the genotypes of the original female and the male with whom she mated? A. XiXs and XsY B. XiXs and XiY C. XiXi and XsY D. XsXs and XiY

A. The male parent must have been XsY or there would only have been daughters in the offspring of the first generation, so options B and D are incorrect. If the female had been XiXi then all its sons would have had only daughters, but twenty of the sons have both sons and daughters, so option C cannot be correct.

What is the electronic configuration of the Co(II) center found in vitamin B12? A. [Ar] 3d7 B. [Ar] 4s2 3d5 C. [Ar] 4s2 3d7 D. [Ar] 4s2 4d5

A. The n+L rule suggests that you take electrons from the lower energy orbital first. 4s=4+0=4, 3d=3+2=5

If none of the Xi-bearing genotypes (XiY, XiXi, or XiXs) is selected against, then the frequency of Xi is expected to increase to 100%, unless other genes act to suppress expression of e and f. If all genotypes are equally fit and if there are no genetic modifiers of the sex ratio trait, what will be the ultimate fate of a population in which 50% of the X chromosomes are currently Xi and 50% are Xs? A. Extinction B. Stable population size, with a predominance of females C. Stable population size, with all individuals producing a 50:50 sex ratio D. Stable population size, with some individuals producing an excess of females and some producing an excess of males

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

When viewing an X ray of the bones of a leg, a doctor can tell if the patient is a growing child, because the X ray shows: A. cartilaginous areas in the long bones. B. bone cells that are actively dividing. C. the presence of haversian cells. D. shorter-than-average bones.

A. The question asks the examinee to identify the characteristic that differentiates growing, developing long bones from adult bones. Long bones grow via endochondral ossification, which requires cartilaginous growth plates at the ends of long bones, that thicken as cartilage and later become ossified (A). Dividing bone cells (B) and haversion canals (C) can be present in fully ossified adult bones. Some adults who are short in stature may have fully ossified long bones that are shorter than those of a developing child (D). Thus, A is the best answer.

For example, family members may enact cultural preferences when asking a provider to withhold aspects of a cancer diagnosis from a patient. If the provider fulfills this request, he or she may struggle to reconcile that action with norms favoring disclosure. Which psychological process is best represented in the hypothetical example at the end of the passage? A. Cognitive dissonance B. A self-fulfilling prophecy C. Confirmation bias D. The fundamental attribution error

A. The question refers to the passage's description of a hypothetical scenario in which family members ask a provider to withhold aspects of a cancer diagnosis from a patient. Negotiating how to handle diagnosis and treatment can take on added complexity when patients and providers come from different cultural backgrounds. As the passage suggests, a provider may feel tension after fulfilling a family's request (to withhold aspects of a diagnosis) that conflicts with professional norms favoring disclosure. The resulting experience of psychological discomfort identifies cognitive dissonance, which can occur when there is a discrepancy among attitudes or an inconsistency between an attitude and a behavior.

What would be the result of complete removal of the parathyroid glands? A. Severe neural and muscular problems due to deficiency of calcium in the plasma B. An increase in calcitonin production to compensate for calcium deficiency in the plasma C. A drastic change in the ratio of mineral to matrix tissue in bones D. Calcification of some organs due to accumulation of calcium in the plasma

A. The removal of PTH gland would result in no PTH production. PTH RIDS bones of Ca2+. So, no PTH means CA2+ is STUCK in bones thus decreasing blood Ca2+. This then would cause increased neuromuscular excitability because of the change in membrane potential, which under normal physiological conditions, is partially kept in balance with extracellular calcium. Typically, the person would eventually die from severe respiratory muscle spasms.

Liposomes were synthesized from 1 mL of various concentrations of Compound 1(0.05-0.20 mM) at pH 7 What mass of Compound 1 (MW = 800 g/mol) is contained in the solution used to prepare liposomes that elute at 20 mL by size-exclusion chromatography? A. 80 µg B. 8 mg C. 80 mg D. 8 g

A. The solution concentration was 0.10 mM for the liposomes that elute at 20 mL. Since the total volume of solution used in their preparation was 1 mL, the mass of lipid can be calculated as: (0.10 × 10^-3 mol/L) × (1 × 10^-3 L) × (800 g/mol) = 8 × 10^-5 g = 80 µg.

In a certain kinetics experiment, the enzymatically catalyzed hydrolysis of ATP proceeds at a constant rate of 2.0 µM•s-1. If the volume of solution is 1.0 mL, what is the total number of ATP molecules that hydrolyzed after 1 min? A. 1.2 × 10^-7 mol B. 3.3 × 10^-6 mol C. 3.3 × 10^-5 mol D. 1.2 × 10^-4 mol

A. The total number of molecules that were hydrolyzed can be calculated by multiplying the rate in µM•s-1 by the time (in seconds) and the volume of the solution (in L): 2.0 × 10^-6mol•L-1•s-1 × 60 s × 1.0 × 10^-3 L = 1.2 × 10^-7 mol.

Sound of a known frequency, wavelength, intensity, and speed travels through air and bounces off an imperfect reflector which is moving toward the source. Which of the following properties of the sound remains the same before and after reflection? A. Speed B. Intensity C. Frequency D. Wavelength

A. Within still air, the speed of sound remains constant. Speed of sound depends on the medium it is in. Hence the answer is speed that remains constant. Intensity = P/A (in W/m^2) and is proportional to amplitude^2 . Also remember A of a sphere is 4pir^2. Intensity could change (attenuation)

Below is a phase diagram for water. As the pressure applied to a sample of water at -0.1°C is increased from 1.0 torr to 200 atm at constant temperature, the: A. vapor will become a solid and then a liquid. B. vapor will become a liquid and then a solid. C. vapor will become and remain a solid. D. solid will become a liquid.

A. at 1 torr and -0.1 C, we are in the vapor phase of water. I think the point of AAMC saying the pressure increases from 1 torr to 200 atm is to just emphasize the process of water vapor at a constant temperature whose pressure is increasing essentially to infinity (1 torr is 1/760 atm, so 200 atm is 152,000 times higher than the original pressure). So essentially staying at the same temperature and following a vertical line to infinity to model the pressure increase shows us that we go from vapor to solid and eventually liquid. 1 atm= 760 torr.

Cancer cells most likely have an abnormality in their: A. DNA. B. rRNA. C. mitochondria. D. lysosomes.

A. cancer caused by mutations that occur in DNA.

Dewlaps that reflect UV light would evolve by natural selection only if: A. individuals with UV-reflective dewlaps produced more offspring than did individuals without them. B. individuals with UV-reflective dewlaps were better able to communicate than individuals without them. C. individuals with UV-reflective dewlaps were less subject to predation than individuals without them. D. individuals with UV-reflective dewlaps mated more frequently than did individuals without them.

A. fitness is generally defined as the number of offspring you produce, or "reproductive success." Natural selection selects for a trait by allowing that trait to be advantageous, but it doesn't end there. After that trait is selected for, more of that individual will be able to give birth and be reproductively fit. NOT B bc relationship between communication and reproductive fitness is not clear

batteries and temperature.

All batteries are influenced by temperature changes. For instance, in lead-acid batteries in cars, like most galvanic cells, tend to fail the most often in cold weather.

aerobic metabolism

Assuming no inhibition occurs, protons are pumped into the intermembrane space, thereby increasing the proton-motive force. The proton-motive force is directly proportional to the energy stored in the concentration gradient; therefore, the larger the proton-motive force is, the more energy available for generating ATP. Because NADH donates electrons to the ETC at Complex I and FADH2 donates electrons to the ETC at Complex II, this explains why NADH yields more ATP than FADH2. (2.5 vs. 1.5 ATP)

Colchicine most likely relieves gout symptoms through what mechanism? A. Prevention of uric acid diffusion through cell membranes B. Inhibition of leukocyte phagocytosis of uric acid crystals C. Inhibition of uric acid crystal formation D. Maintenance of the pH optimum for PRPP synthetase

B.

Which of the following statements correctly describes the distinction between the exocrine and endocrine portions of the testis? A. The exocrine portion secretes only peptides; the endocrine portion secretes only steroids. B. The exocrine portion releases its products into ducts; the endocrine portion releases its products into the blood. C. The exocrine portion secretes only cellular elements; the endocrine portion secretes only chemical substances. D. The exocrine portion is the target tissue for the products of the endocrine portion.

B.

The sequence of events in the human menstrual cycle involves close interaction among which organs? A. Hypothalamus-thyroid-ovary B. Hypothalamus-pituitary-ovary C. Pituitary-thyroid-ovary D. Pituitary-adrenal glands-ovary

B. Hypothalamus releases releasing factors in menstrual cycle. The pituitary gland releases FSH and LH. Ovaries release estrogen and progesterone

Which of the following conclusions about dewlap reflectance is supported by information in the passage? A. Lizard habitat is determined by dewlap reflectance for each species. B. High UV dewlap reflectance is most important in brightly lit habitats. C. High dewlap reflectance is most important in dimly lit habitats. D. Dewlap reflectance is highest at the blue end of the visible spectrum.

B. Option A is reversed. The dewlap reflectance for each species is determined by the lizard habitat.

According to the passage and the data in Table I, what feature must distinguish laser type A from laser type C in order to be suitable for the cleavage of the bonds specified? Laser type A must: A. be better focused laser type C. B. have a higher frequency than laser type C. C. have a longer wavelength than laser type C. D. emit fewer photons per unit time than laser type C.

B. in the passage it says, "Because the photon energy is designed to be larger than the energy needed to break the bond, the excess energy is absorbed by the solution, causing a local temperature increase. " The laser uses excess energy to break the bond. Therefore, stronger bond dissociation enthalpy's warrant a higher energy laser and because E=hf and h is constant, we would have to increase the frequency for more energy. laser A breaks an OH bond and laser C breaks an SH bond. O has a smaller atomic radius than S; therefore, it will be more difficult to remove an H from it. This would require more energy which is directly proportional to frequency (E = hf). You can also use process of elimination with the same logic. (A) increase focus is out of scope of the question. Nowhere in the passage is changing focus mentioned. (C) wavelength -> is inversely related to Energy (E=ch/lambda) (D) Emitting fewer photons would results in less energy.

Embryonic mouse cells divide every 10 hours at 37C. How many cells would be produced from an egg after three days? A. Fewer than 50 B. Between 50 and 500 C. Between 500 and 5000 D. More than 5000

B. So 3 days= 72 hours. 0 hour = 1 cell 10 hours = 2 cells 20 hours = 4 cells 30 hours = 8 cells 40 hours = 16 cells 50 hours = 32 cells 60 hours = 64 cells 70 hours = 128 cells 80 hours = 256 cells Now, since only 72 hours have past, the number of cells must be between 128 and 256 cells, more closer to the 128 side. The only answer choice that fits in this range is B.

When a downward force is applied at a point 0.60 m to the left of a fulcrum, equilibrium is obtained by placing a mass of 10^-7 kg at a point 0.40 m to the right of the fulcrum. What is the magnitude of the downward force? A. 1.5 × 10^-7 N B. 6.5 × 10^-7 N C. 9.8 × 10^-7 N D. 1.5 × 10^-6 N

B. Torque (left)= Torque (right) rF = rF (0.60m) F= (0.40m)(9.8 x 10^-7) solve for F= 6.5 x 10^-7 The system will be in equilibrium only if the right and left forces exert torques that cancel each other. One force's torque will attempt to twist the "see-saw" supported by the fulcrum clockwise, while the other force provides a counterclockwise torque. The torque of a force Fapplied at a distance L from the fulcrum is given by FLsinθ, with θ is the angle between the beam and the force. At equilibrium the left and right forces point vertically downward with the beam horizontal, hence θ = 90° and sin90° = 1. At equilibrium the torque magnitudes are equal so: Fleft Lleft = Fright Lright. The force at Lright = 0.40 m is the weight of the given mass, mg = (10-7 kg)(9.8 m/s2) = 9.8 × 10-7 N. The unknown force is at Lleft = 0.60 m, so Fleft (0.60 m) = (9.8 × 10-7 N)(0.40 m) which may be solved to yield Fleft = 6.5 × 10-7 m.

The biomolecules M are adducted by free protons from DHB and form [M+H]+ and [M+2H]2+ ions. The ions leave the MALDI plate due to the effect of the 4.5 kV electric voltage between the MALDI plate and the MS detector, and travel a distance of 0.5 m to the MS detector within the uniform electric field E region. The velocity of the ions is inversely proportional to their mass-to-charge ratio (m/z). Proteins can be "fingerprinted" using MALDI if they are subjected to proteolytic cleavage before analysis. Which experimental feature of the MALDI-MS technique allows the separation of ions formed after the adduction of tissue molecules? A. Distance travelled by ions depends on the ion charge. B. Velocity of ions depends on the ion mass-to-charge ratio. C. Time of travel is inversely proportional to the ion mass-to-charge ratio. D. Electric field between the MALDI plate and the MS analyzer is uniform.

B. "Translational motion, forces, work, energy and equilibrium in living systems." The answer to this question is B because the passage states that all ions travel the same distance of 0.5 m to the MS detector within the uniform electric field E region, and that the velocity of the ions is inversely proportional to their mass-to-charge ratio (m/z). Thus, the fastest ions are those with smallest m/z ratio, and these ions arrive first at the MS detector, being separated from the slower ones. I chose C. this is wrong bc the passage states "the velocity of the ions is inversely proportional to their mass-to-charge ratio". velocity = d /t. thus, C does not make sense. this is the opposite of what the passage is saying.

People who are born without sweat glands are likely to die of heat stroke in the tropics. This indicates that, under tropical conditions, the human body may: A. gain, rather than lose, heat by evaporation. B. gain, rather than lose, heat by radiation. C. need to use different mechanisms than in temperate zones to maintain body temperature. D. be better able to regulate body temperature than under temperate conditions.

B. Here they're talking about how we're retaining solar radiation, gaining heat from the sun via radiation of heat.So normally we would couple radiation of heat (achieved by dilation of our capillaries so that even more heat can diffuse out through our skin) and couple that with evaporation of sweat, released from our sweat glands.Without sweat glands, we can't do any heat loss via evaporation. While we still try to do the vasodilation, it clearly isn't enough to compensate for the lack of sweat glands, since these people suffer from heat stroke more often. Thus, rather than losing heat due to the radiation our skin is trying to accomplish, we're gaining heat from radiation of solar energy. A is wrong because evaporation simply isn't happening without sweat glands, and evaporation never makes you gain heat. C is wrong because the passage only every discusses evaporation and radiation as methods to manage heat. I think it's a little bit unreasonable to think that humans have developed an undefined skin adaptation to survive in the tropics. I know the passage didn't bring up any alternative, so without any suggestion, I would only look for overtly stated concepts in the passage. D is wrong because dying of heat stroke implies that these people are not compensation at all for body temperature regulation.

Why is the Ames test for mutagens used to test for carcinogens? A. Salmonella transform mutagens into carcinogens. B. Most mutagens are also carcinogens. C. Salmonella contain oncogenes. D. Salmonella's RNA distinguishes between carcinogens and mutagens.

B. In the Ames test, the chemicals that cause mutations in Salmonella test strains are possibly carcinogens, due to the fact that they mutate DNA and DNA mutations can cause cancer (B). The mutagens are transforming the Salmonella, and not vice versa (A). While Salmonella may contain genes that are oncogenes, the statement doesn't explain the relationship between mutagens and carcinogens that allows the Ames test to be used to detect carcinogens (C). Salmonella's RNA is not able to distinguish between carcinogens and mutagens (D).

Why did the liposomes fluoresce during size-exclusion chromatography? A. The macromolecule had extensive conjugation. B. Fluorescent dye was trapped inside. C. Intermolecular interactions lower the energy of the excited state. D. Light reflects from the surface of the sphere.

B. Liposomes can be difficult to detect since they do not absorb visible light and many molecules absorb UV light. The experimental design allowed fluorescent dye to be trapped inside during liposome formation, which allowed their detection by fluorescence spectroscopy. It must be inferred that liposomes cannot absorb light. Why? They're micelle like structures, composed of fatty acids that lack any extensive conjugation necessary for absorbing light and subsequently light emission. what makes anything fluoresce? The molecule is a conjugated system AND The molecule just went from an excited state to its ground state A) The macromolecule had extensive conjugation -- NO, not a conjugated system. C) Intermolecular interactions lower the energy of the excited state -- Okay, assuming the molecule can fluoresce, when a molecule is excited, there are steps that can be taken to lower its energy level. Typically when the energy level is lowered and it does not fluoresce, then this means it was a non-radiative transition - the excitation energy is dispersed as vibrations or heat. The photon is NOT emitted in this case. Close, but not quite the answer. D) Light reflects the surface of a sphere -- Nope, this does not match the basic concept of fluorescent. So we eliminate this answer. B) Fluorescent dye was trapped inside -- BINGO, if the molecule could not do it (because of the reasons above) then something else has to be causing the fluorescence. Ultrasonication and agitation was used to essentially open up the liposome and stuff it with fluorescent dye. From there they were able to tell the relative quantity of each size of liposome that was formed.

DDT would most likely initiate cancer or cause a mutation if which of the following structures is damaged? A. Nuclear envelope B. Chromosome C. Ribosome D. Histone

B. Mutations are heritable changes in the sequence of the nucleic acid component of chromosomes (B), and mutations that lead to unregulated cellular growth can lead to cancer. The other options listed do not mention chromosomes. the passage talks about how DDT can readily cross biological mb's and uncouple oxidation metabolism but the answer is still B. also a nuclear envelope is "The nuclear envelope is the phospholipid bilayer that covers the nucleus. It does not contain any genetic information."

Which technique was most likely used to measure increased neural activation of specific brain regions when viewing the images of food? A. MRI B. fMRI C. CT D. PET

B. The answer to this question is B because an fMRI is an imaging technique that measures brain activity by detecting associated changes in blood flow.

What does the behavior of liposomes prepared from compounds 1 and 2 upon mixing indicate about the energetics of their transformations? Liposomes prepared from: A. both Compound 1 and Compound 2 are under kinetic control. B. Compound 1 are under kinetic control, but those prepared from Compound 2 are under thermodynamic control. C. Compound 1 are under thermodynamic control, but those prepared from Compound 2 are under kinetic control. D. both Compound 1 and Compound 2 are under thermodynamic control.

B. The idea of kinetic control means that the kinetic product will be product which will not be as stable as the thermodynamic product since they're usually run at lower temperatures, irreversible conditions, so the fast product w/ the lowest activation energy forms. Thermodynamic control has reversible conditions, and what usually happens is the kinetic product has enough energy to GO BACK and take the higher activation energy route to form the more stable, thermodynamic product. This is bc they're run at higher temps. Not sure what the prev redditor meant but when I first saw "stable" for compound 1, I automatically assumed that Compound 1 was the thermodynamic product since it was "stable". But in context, stable to mixing was supposed to indicate irreversible conditions and the fact that Compound 2 was able to go back and turn into a different liposome indicates reversible conditions. I think we were all assuming thermodynamic/kinetic control to be the same as thermodynamic/kinetic product. Basically it says kinetic control is irreversible and thermodynamic control is reversible conditions bc of equilibrium attained

If the magnitude of a positive charge is tripled, what is the ratio of the original value of the electric field at a point to the new value of the electric field at that same point? A. 1:2 B. 1:3 C. 1:6 D. 1:9

B. The magnitude of the electric field E of a point charge is given by E=kq/r^2 If q is tripled, E also will be tripled. Thus, B is the best answer.

If the e and f genes are expressed, the Xi chromosome will be prevented from reaching 100% frequency if selection pressures cause which of the following to be true? A. XsXs flies have the lowest fitness of any genotype. B. XsXs flies have the highest fitness of any genotype. C. XiY flies and XsY flies have equal fitness. D. XiXs flies and XsXs flies have equal fitness.

B. The question is basically asking which of these conditions (Choice A,B,C,D) would prevent the Xi chromosome from reaching 100% frequency. The only choice that makes sense would be B because XsXs flies having the highest fitness would mean an increase in flies with XsXs genotypes. This is directly selecting AGAINST the Xi gene, which would prevent it from reaching 100%. The third paragraph says the frequency of Xi is expected to increase to 100% unless other genes act to suppress it. In this case, increased XsXs flies would suppress the frequency of Xi and prevent it from reaching 100%.

The outer layers of human skin are composed of dead cells impregnated with keratin and oil, which make the epidermis relatively impermeable to water, yet humans sweat freely in hot temperatures. This occurs because: A. the salt in sweat allows it to diffuse through the skin. B. sweat glands have special channels through the skin. C. an osmotic gradient in sweat moves it through the skin. D. sweating occurs in only those areas of the body where the skin is water permeable.

B. The sweat glands secrete onto the surface of the skin through channels continuous with the most superficial layer of the skin, the epidermis. These channels prevent water loss by isolating the water-permeable, sweat-secreting cells from dry surface air. The openings of the sweat glands on to the surface of the epidermis are pores. The correct answer is B. All the other answers require some movement of water through the epidermis itself, which is relatively impermeable.

A ray of light in air strikes the flat surface of a liquid, resulting in a reflected ray and a refracted ray. If the angle of reflection is known, what additional information is needed in order to determine the relative refractive index of the liquid compared to air? A. Angle of incidence B. Angle of refraction C. Refractive index of air D. Wavelength of the light

B. To find the relative refractive index to air one needs both the incident and refracted angles. Since the angle of incidence is equal to the angle of reflection we will know the angle of reflection. We still must know the angle of refraction and this is answer B. None of the other answers will allow one to know the angle of refraction. Snell's law. n1sinθ1 = n2sinθ2 so n2/n1 = sinθ1/sinθ2, and since θ1 can be calculated from the angle of reflection, you only need θ2 the angle of refraction to determine n2/n1.

What is the volume flow rate of blood that moves at 0.20 m/s through an artery with a diameter of 1.0 × 10^-2 m? A. 5.0 × 10^-6 m3/s B. 5π × 10^-6 m3/s C.π × 10^-5 m3/s D. 2π × 10^-5 m3/s

B. Volume flow rate is the product of blood speed and artery cross-sectional area: (0.20 m/s)·(π/4)·(1.0 × 10-2 m)2 = 5π × 10-6 m3/s. Therefore, answer choice B is the best answer.

How will W change if the initial speed of the box at Point A is increased by a factor of 2? A. W will decrease by 50%. B. W will not change. C. W will increase by 50%. D. W will increase by 100%.

B. Work is equivalent to Force x Distance. The force of friction is only dependent on the normal force of the box, which does not change with increasing speed since the weight due to gravity remains the same. Next, the distance also does not change since the length of the ramp is the same no matter how fast the box is traveling. Thus, since the force nor distance changes, the Work done by friction should not change either. Know that Work is dependent on Fdcostheta NOT velocity. Hence doubling velocity wont have an effect on work. Also know the equation Fk = ukFn = ukmg is the force of friction Wk = ukFnd. velocity isnt in this equation has no effect

The ATP-dependent phosphorylation of a protein target is catalyzed by which class of enzyme? A. Oxidoreductase B. Transferase C. Hydrolase D. Ligase

B. kinases catalyze the transfer phosphate groups from ATP to target proteins and are classified as transferases.

A drug that binds to tubulin molecules of plant cells and prevents the cells from assembling spindle microtubules would most likely cause the resulting plants or plant cells to have: A. greater genetic variability than the parent plants. B. more than two sets of chromosomes. C. a stronger cell wall because of excess tubulin. D. independent movement because of excess tubulin.

B. the microtubules are disturbed, there will likely be changes in ploidy due to uneven separation during anaphase. The fibers that attach to the chromosomes in mitosis and form the mitotic spindle (the spindle fibers) are microtubules. They are assemblies of tubulin proteins. Disruption of the mitotic spindle by drugs prevents the proper segregation of chromosomes into the daughter cells and usually results in unequal numbers being distributed to the two daughter cells. Many of the resulting plant cells would have more than two sets of chromosomes, thus making answer choice B the correct answer. While a plant might differ from its parent plants in the number of chromosomes, it would contain the same genes. Only in a limited sense would it have more genetic variability. Thus, answer choice A is incorrect. Answer choice C is incorrect because tubulin is not found in the cell wall. Answer choice D is incorrect because tubulin has only a minor role in motility, thus making independent movement unlikely.

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

C.

Which of the following will occur when the magnet used in the flowmeter discussed in the passage is replaced with a stronger magnet? A. The electric field will reverse polarity. B. The electric field will decrease. C. The voltage will increase. D. Blood will flow faster.

C. A stronger B field increases the magnetic force, Fm = q v B^. The electric force must also increase to achieve equilibrium. This implies a larger electric field in the artery and a larger voltage across the artery. if: magnet strength increases ^ then: magnetic field (B) increases ^ since F = qvB increasing B means F increases since F = qE increasing F means E increases since E = ∆V/d increasing E means V increases

Preliminary studies suggest that tobacco cessation is more successful when individuals are motivated by culturally consistent values that support their efforts to quit smoking. Which psychological theory best explains the role of community values in the preliminary studies of tobacco cessation? A. Drive theory B. Humanistic theory C. Incentive theory D. Psychoanalytic theory

C. As described in the passage, the preliminary studies suggest that smoking cessation can be successful when individuals are motivated by culturally consistent values that support quitting. The incentive theory of motivation calls attention to how factors outside of individuals, including community values and other aspects of culture, can motivate behavior. The other options would be more likely to look for motivational factors within the individual.

The knowledge, skills, and education required to practice medicine are associated with high social status in the United States. Patient-provider interactions are partially structured by status differences that are also based in culture. Which sociological concept is suggested by the passage's discussion of the knowledge, skills, and education required for practicing medicine? A. The social capital of physicians B. The hidden curriculum of medicine C. The cultural capital of physicians D. The ascribed status of medicine

C. Cultural capital refers to knowledge, skills, education, and similar characteristics that are used to make social distinctions and that are associated with differences in social status. This description is referenced in the passage, which suggests that status differences based in cultural capital are involved in structuring patient-provider interactions.

In a laboratory population of Drosophila, all the males are XsY. Among the females, 15% are XiXi, 50% are XiXs, and 35% are XsXs. Assuming random mating, what proportion of male flies in the next generation will be XiY? A. 12% B. 30% C. 40% D. 65%

C. For the XiXi mothers: the probability of donating an Xi allele is 100%. The probability of the male mating with this mother is 15%. These are INDEPENDENT EVENTS, so we multiply the two probabilities. 100% x 15% = 15% For the XiXs mothers: the probability of donating an Xi allele is 50%. The probability of the male mating with this mother is 50%. Again, these are INDEPENDENT EVENTS, so we multiply the two probabilities. 50% x 50% = 25% For the XsXs mothers: the probability of donating an Xi allele is 0%. The probability of the male mating with this mother is 35%. Again, these are INDEPENDENT EVENTS, so we multiply the two probabilities. 0% x 35% = 0% Now for a given offspring, it can only result from a mating pair of the father with ONE mother. So each of the options above are MUTUALLY EXCLUSIVE. You cannot have an individual offspring result from two mothers. So we ADD the probabilities of each of the three types of mothers. 15% + 25% + 0% = 40%

The energy meter, based on the photoelectric effect, uses a detector with a work function of 3.4 eV. What is the kinetic energy of a photoelectron produced in the energy meter of the PAC device when the frequency of an incident photon that is NOT absorbed in the solution is f = 5.0 × 10^15 Hz? (Note: Use h = 4.1 × 10^-15 eV•s.) A. 23.9 eV B. 20.5 eV C. 17.1 eV D. 12.3 eV

C. KE = hf - work (20)- 3 = 17 I accidentally did 20 + 3 = 23 this is WRONG.

The liver is different from many other organs in that it can at least partially regenerate following illness or damage. This regeneration is accomplished primarily through: A. fission. B. meiosis. C. mitosis. D. cell growth.

C. Mitosis is the process that somatic cells reproduce

Which of the following best explains why bacterial colonies formed on Plate IV in Figure 1? A. The air contained mutagens. B. The agar contained mutagens. C. Spontaneous mutations occurred. D. The DNA repair system became activated.

C. One can assume, based on information provided in the passage, that the air (A) and media (B) to which the bacteria are exposed are pure and free of mutagens, and thus not causing any mutations in the test strains. If DNA repair systems were activated any back-mutations that occurred would be corrected and no colonies would form on the plates (D). Thus, C is the best answer.

How many ppm is 1%? A. 100 B. 1,000 C. 10,000 D. 100,000

C. Ppm means "parts per million" and is often used as a term in dilutions. 1% of a million would be 10,000.

Which of the following changes would NOT interfere with the repeated transmission of an impulse at the vertebrate neuromuscular junction? A. Addition of a cholinesterase blocker B. Addition of a toxin that blocks the release of acetylcholine C. An increase in acetylcholine receptor sites on the motor end plate D. Addition of a substance that binds to acetylcholine receptor sites

C. Question asks "what would not interfere with repeated transmission". Increasing acetylcholine sites on the end plate would not interfere with action potential transmission. A) blocking cholinesterase = increased AcCH = continued stimulation and therefore this would interfere with a "repeated impulse". B) blocking AcCH release would prevent repeated transmission. D) A substance which blocks AcCH sites would prevent repeated transmission as AcCH won't bind to its receptor and post-synaptic ion channels remain closed

Which cation is most likely to be found in place of Fe(II) in the square planar binding domain of hemoglobin? A. Mg2+ B. Li+ C. Co2+ D. Na+

C. The answer to this question is C because Co2+ is closely related to Fe2+ as a transition metal and can support a square planar coordination environment.

What physical quantity is represented by the area of quadrilateral OABC in Figure 2? A. Average speed of the box B. Average acceleration of the box C. Distance traveled by the box D. Work done on the box

C. The area of a v/t graph is distance traveled, while the slope of a v/t graph is acceleration.

Researchers investigated the effect of pH and Compound 1 concentration on liposome formation. Liposomes are synthetic spherical lipid bilayers that mimic a cell membrane. Compound 1 is a member of which class of lipid molecules? A. Triacylglycerols B. Pyrophosphates C. Phosphatides D. Phosphonic acids

C. The article said its mimics the cell membrane which is a Phospholipid. It can't be the acid and it cant be a pyrophosphates (neither has nothing to do with cell membranes) and since its only 2 chains, it isn't a tri anything. (remember phospholipids only have 2 tails) You don't need to know the structures if you remember that cell membranes are made of phospholipids. another name for phospholipid is phosphatides. phospholipid: a glycerol with 2 FA chains and a phosphate + nitrogenous base

What is the conjugate base of the bisulfate ion (HSO4-)? A. H+ B. OH- C. SO42- D. H2SO4

C. The conjugate base of HSO4-, formed by the loss of one proton, is SO42-.

In comparison with the wall of the right ventricle of the heart, the left ventricular wall is: A. thinner and generates a higher pressure when it contracts. B. thinner and generates a lower pressure when it contracts. C. thicker and generates a higher pressure when it contracts. D. thicker and generates a lower pressure when it contracts.

C. The difference in size between the thin wall of the right ventricle and the thick wall of the left ventricle is dramatic when viewed in cross section. The peak pressure in the left ventricle is 120 mmHg while the peak pressure in the right ventricle is only 25 mmHg. The right ventricle pumps blood through the lungs. The left ventricle pumps blood through the entire rest of the body. The difference in the size of the lung compared to the rest of the body suggests that the left ventricle develops more pressure and has a thicker wall than the right. The organs through which the left ventricle pumps blood are farther away from the heart than the lungs and resistance in a tube is inversely proportional to the length of the tube. This would also suggest a thicker wall for the left ventricle and greater pressure. The correct answer is choice C, that the left ventricle has a thicker wall and develops more pressure than the right.

Because this unique type of binding activates approximately 20% of the T lymphocytes, as opposed to 1 in 100,000 T cells activated by conventional antigenic stimulation According to the passage, superantigens increase the number of activated T cells over activation levels observed with conventional antigens by a factor of: A. 20. B. 5,000. C. 20,000. D. 100,000.

C. The passage states that 20 percent of T lymphocytes are activated by superantigens, while one in 100,000 is activated by conventional antigens. Taking 20 percent of 100,000 yields 20,000, so the correct response is answer choice C.

The time dependence of the potassium current through a cell membrane channel subject to a constant 80-mV depolarization voltage is shown. What is the minimum electrical resistance of the ion channel during the time interval shown? A. 120 kΩ B. 160 kΩ C. 200 MΩ D. 240 MΩ

C. The questions asks for the minimum resistance. Since voltage is constant and current and resistance are inversely related, resistance is at its minimum when current is at its maximum. According to the graph, current is max at 400 x 10^-12 A. V=IR R=V/I = (80 )/ (400)= 0.2 *10^9= 200 M Ω

To test this hypothesis, they proposed follow-up studies using twins who were born in SB states but who were adopted into different families. Which finding from the follow-up studies proposed by the researchers in Study 1 would best support the hypothesis about genetic predispositions to stroke? A. Monozygotic twins who were raised in the same state outside of the SB both have decreased risk of stroke in adulthood. B. Dizygotic twins who were raised in the same state inside of the SB both have excess risk of stroke in adulthood. C. Monozygotic twins who were raised in different states, one twin inside the SB and one twin outside, both have excess risk of stroke in adulthood. D. Dizygotic twins who were raised in different states, one twin inside the SB and one twin outside, both have excess risk of stroke in adulthood.

C. The researchers also considered the hypothesis that genetic predispositions to stroke may be more likely among people living in the SB. To test this hypothesis, they proposed follow-up studies using twins who were born in SB states but who were adopted into different families.

An artery is constricted at one location to 1/2 its normal cross-sectional area. How does the speed of blood past the constriction compare to the speed of blood flow in the rest of the artery? (Note: Assume ideal fluid flow.) A. It is 1/4 as fast. B. It is 1/2 as fast. C. It is 2 times as fast. D. It is 4 times as fast.

C. if they give you area you don't need to square anything.

The concentration of the protein cyclin rises and falls during the cell cycle as shown in Figure 1. Changes in the concentration of cyclin during phases of the cell cycle What mechanism could account for this oscillation of cyclin protein concentration? A. Replication of the cyclin gene during S phase of interphase B. Segregation of chromosomes carrying the cyclin genes during mitosis C. Translation of cyclin mRNA in interphase and proteolysis of cyclin protein in mitosis D. Translation of cyclin mRNA in mitosis and proteolysis of cyclin protein in interphase

C. look at graph. Incorrectly thought answer was A bc DNA synthesis does occur during interphase however this does NOT answer the question. The graph shown in the question indicates that the concentration of cyclin rises and falls in a regular manner throughout the cell cycle, reaching a peak just at the beginning of mitosis, gradually declining during mitosis, reaching a minimum at the end of mitosis, and gradually increasing during interphase. The mechanism that can best account for this oscillation in the concentration of cyclin is translation of cyclin mRNA (creating the protein from mRNA template) followed by proteolysis (destruction) of cyclin protein during mitosis.

What is the net volume of fresh air that enters the alveoli each minute, assuming that the breathing rate is 10 breaths/min, the tidal volume is 800 mL/breath, and the nonalveolar respiratory system volume (dead space) is 150 mL? A. 65 mL B. 95 mL C. 6500 mL D. 7850 mL

C. tidal volume means how much air is going into the lungs. In this case, its 800 ml. The question is how much air enters the alveoli. This is not the same thing as tidal volume. Remember alveoli is at the bottom of the lungs. The deadspace is the air thats occupying the trachea or your windpipe. The air here does not reach the alveoli. You have to subtract this 150 from 800 because this air will not reach the alveoli. When you exhale, it will be the first air to be pushed out. So 650 ml/breath reaches the alveoli * 10 breaths = 6500 ml

The investigators used specially enhanced cameras to measure the degree to which the dewlaps of the five species reflected UV light (defined as wavelengths around 360 nm). If these lizards use UV light in communication, a mutation that eliminated UV photoreceptors would probably cause the LEAST disadvantage to: A. species A. B. species B. C. species D. D. species E.

D. in the passage, it says that the uv spectrum is designated to be 360 nanometers so at that part of the graph E has the lowest wavelength

If DDT accumulates in the liver, all of the following bodily functions may be significantly impaired EXCEPT: A. absorption of fats in the small intestine. B. production of bile. C. detoxification of poisons. D. regulation of blood pressure.

D. NOT A bc if the liver is not able to produce bile, then we won't be able to emulsify fats so it would be harder for the small intestine to absorb the fat through the lacteals

What is the pH of a .001 M NaOH solution? A. .001 B. 3 C. 7 D. 11

D. pOH= 3 -14 pH= 11.

Would methane gas (CH4) be a candidate for determination by the method described in the passage? A. Yes, because carbon, like oxygen, is a nonmetal B. Yes, because carbon, like oxygen, is in the 2nd period of the periodic table C. No, because hydrogen is already at its lowest oxidation state in methane D. No, because carbon is already at its lowest oxidation state in methane

D. H oxidation state:+1 C oxidation state: -4 CH4 oxidation state: 0 (stable) Carbon is more reduced with the more hydrogens bonded to it and more oxidized with the more oxygens bounded to it. So, methane cannot be reduced further because you cannot bind anymore hydrogens to it.

Aldosterone stimulates Na+ reabsorption by the kidneys. What changes in blood volume and pressure would be expected as a result of aldosterone deficiency? A. Increased volume and increased pressure B. Increased volume and decreased pressure C. Decreased volume and increased pressure D. Decreased volume and decreased pressure

D. An aldosterone deficiency consequently would cause a decrease in water and salt reabsorption as a result you would have a lowNa+ reabsorption low blood volume and low blood pressure due to low water reabsorption. the result of an aldosterone deficiency on blood volume and blood pressure, given that aldosterone stimulates Na+ reabsorption in the kidney. An aldosterone deficiency would result in lower Na+ reabsorption into the bloodstream. Because H2O passively follows Na+ during reabsorption in the kidney, less Na+reabsorption would result in less H2O reabsorption into the bloodstream. This would result in decreased blood volume. Blood volume would also be affected by lower blood Na+ levels because there would be less of this ion to osmotically hold water in the extracellular fluid. Decreased blood volume would result in decreased blood pressure as well I said B but the arterial system is not an ideal system. Blood slows down due to friction with the walls, pressure is lost due to the elasticity of the walls of vessels, etc. Boyles law (PV are inverse does NOT apply to blood. So for arteries or capillaries if volume increases, pressure increases.

The cell type in the male reproductive system that is most analogous to the female ovum is the: A. spermatogonium. B. primary spermatocyte. C. spermatid. D. spermatozoon.

D. Both an ovum and a spermatozoa are the mature form. They can both interact via fertilization.

Glucose is labeled with 14C and followed as it is broken down to produce CO2, H2O, and ATP in a mammalian liver cell. In theory, during this process the label will be detectable: A. in the mitochondria only. B. first in the nucleus, then in the mitochondria. C. first in the mitochondria, then on the ribosomes. D. first in the cytoplasm, then in the mitochondria.

D. Breakdown of glucose proceeds first by glycolysis, then by oxidation in the citric acid (Krebs or tricarboxylic acid) cycle. The enzymes for the former process are located in the cytoplasm and those for the latter are in the matrix of mitochondria. The 14C label, therefore, would first appear in the cytoplasm, then in the mitochondria.

The diameter d versus elution volume V for liposomes using identical chromatographic conditions is shown. Using the ultrasonication and agitation method of liposome preparation, what concentration of lipid is necessary to prepare liposomes with a diameter of 250 nm? A. 0.10 mM B. 0.15 mM C. 0.20 mM D. 0.30 mM

D. First, you should notice that as micelle diameter increases, elution volume decreases. If we turn to Figure 1, we see that at 15 mL elution volume, we had a 0.15 mM lipid concentration. For the 10 mL elution volume, we had a lipid concentration of 0.20 mM. The trend is now this, as lipid concentration increases, elution volume decreases. If we want a 250 nm micelle diameter, we will need to decrease the elution volume. If we need to decrease the elution volume, we need to increase the lipid concentration to a concentration >0.20 mM. The only concentration listed in the answer that is >0.20 mM is D (0.30 mM), and is therefore the correct answer

What is the reading of the energy meter in Figure 1 when an appropriate laser is used in PAC to dissociate a particular chemical bond? A. Em B. ΔHu C. ΔHnr D. 0

D. Looking at the diagram, you can see that the basic flow of energy goes from laser (photon energy) --> cell excitation on the lens (releases heat, causing sound waves) --> microphone --> and is ultimately picked up by the energy meter in some form. When the laser/photon energy is just right, that means no excess heat is released from bond breakages inside the cell, and therefore no sound waves are produced. Thus the energy meter will read 0

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

D. Nervous tissue arises developmentally from ectoderm. Dermis = mesoderm Epidermis = ectoderm

The enzyme pepsin, which catalyzes the hydrolysis of proteins in the stomach, has a pH optimum of 1.5. Under conditions of excess stomach acidity (pH of 1.0 or less), pepsin catalysis occurs very slowly. The most likely reason for this is that below a pH of 1.0: A. pepsin is feedback-inhibited. B. pepsin synthesis is reduced. C. the peptide bonds in pepsin are more stable. D. the three-dimensional structure of pepsin is changed.

D. Remember: Structure determines Function. this is the best asnswer as it is referring to a change in an enzyme itself because enzymes need very specific conditions on where they can function (pH, temp, etc). for a protein to function properly, it must have a very specific three-dimensional structure. This three-dimensional structure of a protein is stabilized by covalent bonds and noncovalent interactions between different regions of the linear peptide. This three-dimensional structure can be disrupted by heating or by changing the pH. The disorganization of proteins by such agents is called denaturation. An example of denaturation is the hardening of an egg during cooking. Enzymes are proteins that act as organic catalysts, speeding chemical reactions but not being consumed in them. Their function is highly dependent on a precise three dimensional structure, especially at the site of catalysis, known as the active site. The lowering of pH described in the question is likely to have caused the enzyme pepsin to lose its three-dimensional shape and thus its catalytic activity.

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

D. Since albumin levels increases, causing an increase in osmotic pressure, or solute content in blood water would consequently flow in to the blood to reach equilibrium. This water would flow in from the tissues to the bloodstream as stated in the answer. The plasma proteins can not cross the walls of blood vessels, but water molecules can. The wall of the artery acts as a semipermeable membrane setting up the conditions needed for osmosis to occur. An increase in plasma albumin will upset the osmotic balance because the blood will become hypertonic with respect to the tissue. Water will have to flow into the bloodstream to reestablish equilibrium. One of the causes of edema, increased fluid in body tissues, is a decrease in the plasma protein level. This occurs, for instance, in starvation when the body is forced to use its albumin as an energy source. An increase in the plasma protein level would have the opposite effect: fluid would enter the bloodstream. water follows solute. (everything that is not water is a solute). An increase in blood volume= an increase in blood pressure.

Which alcohol will most likely undergo substitution by an SN1 mechanism in acidic conditions? A. Primary Alcohol B. secondary alcohol C. 2- methyl-1- propanol D. teriarty alcohol

D. The answer to this question is D because this alcohol produces the most stable carbocation (tertiary) and consequently will most easily lose a water molecule upon protonation of the hydroxyl group in acidic media. Sn1 3>2>1

The pancreas produces which of the following substances for the digestive system? A. Bile salts B. Emulsifier C. Gastric juices D. Proteolytic enzymes

D. The pancreas produces several proteolytic enzymes, which are released into the small intestine where they are converted to their active forms of trypsin, chymotrypsin, and carboxypeptidase. proteolytic enzymes produced by pancreas which are secreted by acinar cells of pancreas. (C is produced by the stomach.) & Bile is produced by the liver and stored in the gall bladder.

Which statement below most accurately describes the roles of the proteins actin and myosin during muscular contraction? A. Both actin and myosin shorten, causing the muscle tissue to which they are attached to contract. B. Both actin and myosin catalyze the reactions that result in muscle contraction. C. Actin molecules are disassembled by myosin, leading to a shortening of muscle sarcomeres. D. Bridges between actin and myosin form, break, and re-form, leading to a shortening of muscle sarcomeres.

D. The sliding filament model describes the interaction of actin and myosin during muscle contraction. According to this model, neuronal impulses cause the release of calcium from the sarcoplasmic reticulum within muscle cells. The calcium then binds to troponin, a molecule that along with tropomyosin, blocks the binding sites for myosin on actin molecules. Calcium binding to troponin causes a shift in the troponin/tropomyosin complex, revealing the binding site for myosin. Myosin then binds to actin, causing a conformational change in myosin that "cocks" the head of the myosin molecule and slides the actin filament relative to myosin. ATP binds to myosin, causing it to detach from actin and "recharge" (rebend again). If another binding site is available on actin, myosin will bind again, and slide the actin filament even further. Thus, D is the best answer.

A force F is used to raise a 4-kg mass M from the ground to a height of 5 m. What is the work done by the force F? (Note: sin 60° = 0.87; cos 60° = 0.50. Ignore friction and the weights of the pulleys.) A. 50 J B. 100 J C. 174 J D. 200 J

D. Work is the product of force and distance. The easiest way to calculate the work in this pulley problem is to multiply the net force on the weight mg by the distance it is raised: 4 kg × 10 m/s2 × 5 m = 200 J. Since the mass is also being displaced vertically, the angle between the displacement and the force is 0. So, W = Fdcos(theta) = Fdcos(0) = Fd Also, the F acting on the mass is going to have twice the magnitude of the F being applied to the rope because of mechanical advantage (or d is doubled where the force is applied on the rope). So now it's W = 2Fd = 10F But notice they dont give us anything else, so we can't plug anything in to find F or W. We get stuck using this method. Instead we can just use E1 + W = E2. We can assume E1 = 0 so, W = E2. And E2= PE + KE KE = 0 (it's static at h=5m) PE = mgh = (4)(10)(5) = 200J = E2 = W. *we can also now plug 200 into W = 10F to get F = 20N Pulley's reduce the force needed to move an object by increasing the distance needed to move that object. So now I want you to think about distance in terms of time. Imagine if you didn't have the pulley and you were trying to move that weight up, it would require more forceto move that object in that moment, but you will probably move it faster. However, if you have the pulley, you will need less force, but it will take you longer to move the object up (think about pulling a rope in the picture above where F is pointing). In both instances, you are doing the same amount of work, whether it takes more or less time because W = Force x Distance(or time). So less Force needed = more distance. Less distance needed = more force. Either way Work done on the object stays the same.

The researchers mixed liposomes of different sizes and observed that those formed from Compound 1 were stable to mixing, but mixing those from Compound 2 formed new liposomes with an average size expected for the effective final lipid concentration. Liposomes derived from Compound 2 are prepared at pH 8.5 from two different solution concentrations (0.10 mM and 0.20 mM) as described in the passage. What is the expected appearance of the size-exclusion chromatograph of the liposomal suspension that results after mixing equal volumes of these?

D. mixing liposomes of various sizes created from Compound 2 is unstable and creates new liposomes with an average size (from the last sentence). average size means one peak and then average concentration is .15mM (0.10 + 0.20= 0.30/2 = 0.15 (new average)).

Accumulation of DDT in the testes may cause reduced fertility in males because the uncoupling of oxidative metabolism from ATP production may reduce: A. glucose concentration of semen. B. testosterone concentration of semen. C. blood circulation in the testes. D. sperm motility.

D. sperm motility requires large amounts of ATP as evidenced by the high concentration of mitochondria in the sperm midpiece. Therefore, of the choices given, reduced sperm motility (D) is most likely to result from reduced ATP levels.

Which of the following figures best illustrates demographic transition theory?

Demographic transition theory addresses changes in the birth rate and the death rate that are associated with economic development (specifically, related to industrialization). The typical pattern begins with a drop in the death rate, leading to population growth, followed by a drop in the birth rate, leading to population stabilization. This demographic pattern was seen in the United States and several European countries during the 19th century, with many other countries following this pattern during the 20th century.

concentration of (volatile) solute added to a solvent and vapor pressure of the solvent are inverse.

PA = XA PºA where PA is the vapor pressure of solvent A when solutes are present, XA is the mole fraction of the solvent A in the solution, and PºA is the vapor pressure of solvent A in its pure state. As more solute is added to the solvent, the mole fraction of solvent is decreased, causing the vapor pressure of the solvent to decrease proportionately. Raoult's Law explains the phenomenon of Vapor Pressure Depression.

reductional potential

The species in a reaction that will be oxidized or reduced can be determined from the reduction potential of each species, defined as the tendency of a species to gain electrons and to be reduced. Each species has its own intrinsic reduction potential; the more positive the potential, the greater the tendency to be reduced. Since reduction potential is defined as the tendency to be reduced, a less positive Eºred means a greater tendency for oxidation to occur because oxidation and reduction are opposite processes.

One characteristic common to arteries, veins, and capillaries is the: A. presence of a layer of endothelial cells. B. presence of numerous valves that prevent the backflow of blood. C. ability to actively dilate or constrict in regulating blood flow. D. ability to supply surrounding tissues with nutrients by filtration and diffusion.

a. A is correct because all three types of vessels possess an inner layer of endothelial cells. C. is wrong bc only certain types of arteries can dilate and constrict. D is wrong bc: only capillaries do exchange.

Electrolytes

are solutes that enable solutions to carry currents. The electrical conducitivity of aqueous solutions is governed by the presence and concentration of ions in solution; a solute is considered a strong electrolyte if it dissociates completely into its constituent ions. Examples of strong electrolytes include certain ionic compounds, such as NaCl and KI, and molecular compounds with highly polar covalent bonds that dissociate into ions when dissolved, such as HCl in water.

If an artery that supplies blood to a lung lobe was blocked but ventilation to the lobe was unaffected, how would alveolar gas partial pressures change? A. Both PO2 and PCO2 would increase. B. Both PO2 and PCO2 would decrease. C. PO2 would increase and PCO2 would decrease. D. PO2 would decrease and PCO2 would increase.

c. If the blood flow to an alveolus were blocked there would be no flow of hemoglobin-rich red blood cells to take away O2 and no influx of CO2 from the blood. As a result the air in the alveolus would become more like that of the atmosphere. It would acquire a higher PO2 and a lower PCO2. no blood going to lungs means no CO2 exchange. ventilation means oxygen is still getting in. so O2 would increase but CO2 would decrease Ventilation is normal, thus you could imagine an individual alveoli receiving PO2 and PCO2 similar to that of the atmosphere (PO2 ~ 150 mm Hg and PCO2 is small enough to be neglected ~ 0 mm Hg). The gases then equilibrate with the deoxygenated blood perfusing that particular alveoli -- NORMALLY. In this case, there is no blood perfusion to the tissue and you would expect the partial pressure to match that of the atmosphere (since it does not equilibrate with the blood). So, relative to the normal, PO2 is much higher and PCO2 is much lower.

Skeletal and cardiac muscles

contain striated muscle fibers. Smooth muscles do not

DNA polymerase catalyzes the replication of chromosomal DNA in bacteria as shown below. A double-stranded DNA molecule contains bases with a ratio of (A + T)/(G + C) = 3:1. This molecule is replicated with DNA polymerase in the presence of the four deoxynucleoside triphosphates with a molar ratio of (A + T)/(G + C) = 1:1. What is the expected ratio of (A + T)/(G + C) in the double-stranded daughter DNA molecule? A. 1:3 B. 1:1 C. 2:1 D. 3:1

d. they're just testing whether or not you know that once duplicated, you're still going to have the same ratio. The double stranded daughter DNA molecule would be an exact duplicate of the parent molecule. It would have the same (A + T)/(G + C) ratio. The correct answer is therefore 3:1. dNTPs are just DNA building blocks, does not change the ratio of the product. DNA replication is semi-conservative.

symbolic interactionism

examines small scale (or micro level) social interactions, focusing attention on how shared meaning is established among individuals or small groups.

Social epidemiology

focuses on the contribution of social and cultural factors to disease patterns in populations. It is also well positioned to supplement the biomedical approach because social epidemiology is a sub-field of epidemiology.

spectroscopy

in a wavelength vs. absorption plot, the area under the curve represents the amount of UV light a sunscreen can absorb. For a given sunscreen, more area under the curve = more effective sunscreen

mechanical advantage

the ratio of the output force to the input force Machines allow you to increase the distance over which work acts but ultimately decreases the force required. (same amount of work). W=Fd is expressed in ideal terms, no loss of energy to heat for friction. based on one of the equations of mechanical Work: W = F • d Using simple machines to provide mechanical advantage (reducing the force needed to accomplish a given amount of work) does have a cost associated with it; the distance through which the smaller force must be applied in order to do the work must be increased. This is illustrated in inclined planes. For example, pushing a 100 N block up an incline over a distance of 20 m to a height of 10 m requires half as much force than trying to raise the block vertically 10 m. This is because the incline distributes the workload over a greater distance to perform the same amount of work.