AAMC set
B. the substitution reaction in question serves to replace the hydroxyl group and that hydroxide ion is one of the worst leaving groups in substitution reactions. Under acidic conditions, the hydroxyl group is protonated such that the leaving group is now water, a superior leaving group, rather than hydroxide ion. Thus, B is the best answer.
Alcohols generally require acid catalysis in order to undergo substitution by nucleophiles. The acid catalyst enhances the reaction by: A. increasing the solvent polarity. B. creating a better leaving group. C. neutralizing basic impurities. D. protecting the alcohol group.
C. pH = pKa + log[A-/HA] 10^2 = [H2PO4^2- ] /[H2PO4^-]
Experiment 1 The chemist dissolves 16 mmol CDP in 1 L of an aqueous solution containing PNP and and buffers the solution at pH 8.7. The chemist monitors the reaction by measuring the amount of inorganic phosphate produced. No reaction is detected for 30 min, then the reaction appears to proceed at a constant rate until it stops with a final concentration of 8 mM. The recovered polymer contains 7.5 mmol of cytosine. the pKa for the dissociation of H2PO4^- to H2PO4^2- is 6.7. what is the initial ratio of [H2PO4^2- ] : [H2PO4^-] in the buffer solution of Experiment 1? A. 1:1 B. 2:1 C. 100:1 D. 200: 1
spleen
Filters aged/damaged RBCs Reservoir for blood Immune response (B cell activation site, housing macrophages)
Mesoderm. The heart and blood vessels both differentiate from the mesoderm.
From which germ layer(s) do the tissues of the heart and blood vessels differentiate? Ectoderm Mesoderm Endoderm
C. The question mentions the genes for rRNA. This implies that whatever is needed to rRNA can be passed down from parent to daughter. However, if it is outside the nucleus, this must mean that the gene for rRNA is not in DNA and must be self-replicating for it to be passed down (since what replicates during DNA replication is all in the nucleus).
In the macronucleus, the genes for rRNA are located extrachromosomally. This suggests that the rRNA genes are: Anonlinear. Bnonfunctional. Cself-replicating. Drearranged.
D. E= V/d = 5 / 3 x 10^-2 = 1.6 x10^-2 ~ 166
Ionization detectors, the most common units, contain two parallel electrodes that are typically separated by 3 cm with a 5-V potential difference across them. What is the magnitude of the electric field between the two electrodes in ionization type detectors? A.1.5 N/C B.1.66 N/C C.15 N/C D.166 N/C
A. 60 x 10^6 is the initial amount at time 0. half-life is 430. how many years to get to 3.75 x 10^6 ? 60/2= 30 /2 = 15/2 = 7.5/2 = 3.75. so 4 half-lifes to get to 3.75 x 10^6. 1 half-life = 430 4 half lives (430)= 1600 yrs. Each half-life reduces the number of americium nuclei by half. Because it takes 4 half-lives to go from 60,000,000 nuclei to 3,750,000 nuclei (1/16 of the original amount), the elapsed time is 4(430) = 1720 years.
Most units are initially fueled with 60 million nuclei of radioactive americium 241 (half-life 430 years). When fewer than 3.75 × 10^6 americium nuclei remain, the ionization smoke detector will not operate due to insufficient ionization. How much time will pass before there are this many nuclei remaining? A1720 years B2150 years C4300 years D6880 years
liver cells
Regulation of blood glucose via glycogenesis, glycogenolysis & gluconeogenesis Storage of glycogen, minerals (iron) & vitamins Synthesis of macromolecules such as plasma proteins (clotting factors & albumin), fats, ketone bodies & cholesterol Production & secretion of bile Breakdown/detoxification of numerous drugs & metabolic waste products (bilirubin, ammonia)
B. protein structure and posttranslational modification can both affect the pI of the protein
Researchers purified protein X, experimentally measured its pI value through isoelectric focusing (IEF), and found that it differed significantly from the theoretical pI calculated as described in the passage. The theoretical pI value may have differed from the experimental pI if the theoretical calculation did not consider: I. posttranslational modifications on protein X. II. the tertiary and quaternary structure of protein X. III. the pH in the subcellular localization of protein X. A. I only B. I and II only C. II and III only D. I and III only
A. Sodium. Sodium, aluminum, sulfur, and chlorine are all in the third row of the periodic table. Atomic radius tends to decrease from left to right across a given row of the periodic table. This is because, as one moves across the row, the effective nuclear charge increases, drawing in the outermost electrons. Because sodium is the first element in row three, it is the largest. Answer choice A is the best choice.
Which of the following atoms has the largest atomic radius? A. Sodium B. Aluminum C. Sulfur D. Chlorine
C. A charged particle accelerates in an electric field. The electron starts with a velocity that increases as it approaches the anode through the vacuum. When electrons are ejected from the cathode via the photon beam (photoelectric effect), the anode replaces these "lost" electrons. As a result, this movement of electrons generates a current which flows through the circuit.
Which of the following best describes the movement of an electron after it is ejected from the cathode? A.It is stationary until collisions propel it toward the anode. B.It moves with constant speed toward the anode. C.It accelerates toward the anode. D.It exits through a side of the vacuum photodiode.
C. dialysate concentrations in the table are about isotonic to the normal patient's plasma. If it's not isotonic, water will move. If there is a large osmotic gradient in either direction, then water is being sucked out of or pumped into the patient for the few hours of time that dialysis takes. Pumping water into or out of a patient for that long, probably going to have a bad time.
Which of the following changes in flow rate or in solute concentrations would NOT occur if the blood inflow rate were increased, increasing the pressure in the dialysis chamber? A. The blood volume reaching the outflow tube per unit time would increase. B. The osmotic concentration of proteins in the dialysate fluid would increase. C. The osmotic concentration of proteins in the blood outflow would increase or remain unchanged. D. The filtration rate across the dialysis membrane would increase.
SiCl4
Which of the following compounds has the same geometry as methane? (CH4) AH2S BCO2 CXeF4 DSiCl4
D. As molecular weight increases, the rate of filtration is expected to decrease. Membrane 1 has larger pores than Membrane 2, Membrane 1 is expected to allow more molecules to pass at any given molecular weight. D is correct because it shows the trend of decreasing filtration rate with increasing molecular weight, and it shows that Membrane 1 (with larger pores) has a higher filtration rate than Membrane 2 for any molecular weight.
Which of the following figures (A-D) shows expected solute filtration rates (mEq/mL-min) as a function of molecular weight for two dialysis membranes: Membrane 1 with large pores and Membrane 2 with small pores?
The smooth endoplasmic reticulum most resembles the Golgi apparatus in an intact eukaryotic cell when viewed under the microscope. Both organelles appear to be membranes with many folds.
Which of the following organelles most resembles the Golgi apparatus when an intact eukaryotic cell is viewed under the electron microscope? A. Nucleolus B. Mitochondrion C. Plasma membrane D. Smooth endoplasmic reticulum
C. The question asks the examinee to identify the bodily process among the options listed that is LEAST directly influenced by adrenergic drugs. According to the passage, adrenergic drugs mimic activation of the sympathetic nervous system; therefore, the best answer will be the process that is LEAST directly controlled by the sympathetic nervous system. The sympathetic nervous system directly inhibits peristalsis (A) and secretion of digestive enzymes (B). It also increases the blood glucose concentration and causes dilation of the blood vessels that supply the deep muscles and internal organs, which aids nutrient delivery (D) to these tissues. The sympathetic nervous system does not directly affect the activity of digestive enzymes (C) after they have been secreted. Thus, C is the best answer.
Which of the following processes is LEAST directly influenced by adrenergic drugs? A.Peristalsis B.Secretion of digestive enzymes C.Enzymatic breakdown of food molecules D.Nutrient delivery to muscles and organs
A. The problem states that glycine exists "as a dipolar ion in aqueous solution" and asks which of four properties is associated with this fact. Polarity in neutral molecules results from an uneven distribution of electron density, which can arise from separation of unlike charges. This occurs in zwitterions and in ylides. In addition, molecules that contain strongly electron-withdrawing or electron-donating substituents are highly polar and possess correspondingly high dipole moments. Thus, answer choice A is the best answer. D. is tru but this doesn't explain how it's a polar ion in water.
Which of the following properties is associated with the existence of glycine as a dipolar ion in aqueous solution? A High dipole moment B High molecular weight C Low dielectric constant D Low solubility in water
C. In NaCl, chlorine exists as the chloride ion. A chloride ion has 18 electrons with the electron configuration as seen in choice C. When sodium (Na) and chlorine (Cl) are combined, the sodium atoms each lose an electron, forming cations (Na+), and the chlorine atoms each gain an electron to form anions (Cl−). These ions are then attracted to each other in a 1:1 ratio to form sodium chloride (NaCl).
Which of the following shows the electron configuration of chlorine in NaCl? A1s^2 2s^2 2p^6 3s^2 3p^4 B1s^2 2s^2 2p^6 3s^2 3p^5 C1s^2 2s^2 2p^6 3s^2 3p^6 D1s^2 2s^2 2p^6 3s^2 3p^4 4s^2
D. the cellular membrane is made of phospholipids and proteins. sterols are synthesized in the ER. Glycolipids are synthesized in the Golgi apparatus, but prokaryotes do not contain Golgi bc they do not contain mb bound organelles.
Which other cellular components are likely to be located near the lacY6xbs transcript in the cell membrane? A. Proteins and glycolipids B. Glycolipids and sterols C. Sterols and phospholipids D. Phospholipids and proteins
C. Ketone bodies in mitochondrial matrix of liver cells FA synthesis occurs in the cytoplasm of liver cells after acetyl-coA is shuttled as citrate to the cytoplasm Glycogen is produced in the liver (not sure but probably produced in the mitochondrial matrix) Glucose can also be produced in the liver depending on need (i.e high acetyl-coA, high ATP, release of glucagon or epinephrine) The liver is responsible for the production of bile that is stored in the gallbladder and released into the small intestine to digest fats. The liver produces non-essential amino acids. The liver also produces albumin (carrier protein for things like steroids, fatty acids, and thyroid hormones) Lastly, the liver also produces things like clotting factors
he liver synthesizes factors that act cooperatively with platelets to facilitate which physiological process? A. Cholesterol synthesis B. Glucose metabolism C. Blood clotting D. Fat digestion
Heating the distillation flask (i.e., increasing the temperature) at a slower rate will allow both liquids more time in the fractionating column (increase the number of theoretical plates, allowing liquid and vapor to equilibrate); is most likely to improve the degree of separation of the two compounds.
how to improve the degree of separation btw molecules in distillation?
atomic raduis
increases from right to left, and down a group
acetylation
increases gene expression, loosens DNA wrapped around histone.
ionization energy
increases left to right and up a group
period of wave
is just the time it takes to reach the same amplitude point as where you started which is the time it takes for one full cycle. measure crest to crest
Low Rf
most polar, slowest moving.
constructive interference
path difference : whole λ (nλ) 0 degrees/360 degrees phase difference.
destructive interference.
path difference: 1/2 λ (n + 1/2) λ 180 degree phase difference.
macromolecules
proteins, lipids, carbohydrates, and nucleic acids.
turns blue for bases
red litmus paper
Fast glycolytic type 2X fibers.
speed of contraction: fast. Best use: explosive movements (jumping, lifting weights) Easily Fatigue primary use of ATP: anaerobic glycolysis very few mitochondria, capillaries, and myoglobin. white in color
fast oxidative glycolytic type 2A fibers
speed of contraction: medium. Best use: moderate endurance activity susceptibility to Fatigue primary use of ATP: Aerobic respiration & anaerobic glycolysis some mitochondria, capillaries, and myoglobin. Pink in color
Slow oxidative Type 1 muscle fiber.
speed of contraction: slow. Best use: endurance activity Fatigue Resistance primary use of ATP: Aerobic respiration ALOT of mitochondria, capillaries, and myoglobin. Red in color
1. a molecule with a hydrogen bonded to a highly electronegative atom such as oxygen or nitrogen. 2. a hydrogen bond acceptor: an electronegative atom (such as O or N) with an available lone pair of electrons. exmp: H2N-CH2-COOH and CH3OH are able to act both as hydrogen bond donors and acceptors.
the formation of a hydrogen bond requires:
colorless solution
transition metals that have fully filled orbitals (ten d electrons) forms a ______
colored solutions
transition metals that have unfilled d orbitals produce _____
compounds with stronger intermolecular forces, larger masses, and less branching. As hydrogen bonding increases the boiling point of a liquid.
what increases boiling point?
10^-5
which is bigger? 10^-5 or 10^-8
the equatorial position is more favorable rather than a crowded axial position.
which position is it more favorable for large groups to occupy ? the equatorial or axial position?
as harmonics increase, the frequency increases and wavelength decreases and so in this question, since frequency increases with harmonic increase, the period will decrease because frequency and period are inverse.
λ = 2L / n and V = λ f f = nV/2L , where n = harmonics 1st, 2nd, 3rd. With increasing Harmonics there will be increasing Frequency, therefore a decreasing period 1st harmonic is the smallest frequency and it increases as the harmonics increase. 1st harmonic= 50Hz 2nd harmonic = 100Hz and so on
tight junctions
endothelial cells are connected by ________
a. Because there is no relative motion between the jets, there is no frequency shift.
A receiver is in a jet flying alongside another jet that is emitting 2.0 x 106 Hz radio waves. If the jets fly at 268 m/s, what is the change in frequency detected at the receiver? A. 0 Hz B. 0.90 Hz C. 1.79 Hz D. 3.58 Hz
B. Adding a high salt compound which is an ionic compound would attract the cations to cling onto the salt and be removed. The question asks for retaining the ions, so using amount low salt would make sure the cations don't get removed out after they get separated. The mitochondrial matrix is a high-pH environment and has proteins with high isoelectric point (pI) values. High-pI proteins will be positively charged at pH 7, allowing them to bind to a cation-exchange chromatography column at low salt concentrations.
A researcher seeks to study proteins found in the relatively high-pH environment of the mitochondrial matrix. Based on the information in the passage, which of the following procedures will be most successful at causing mitochondrial matrix proteins to bind to an ion-exchange chromatography column? A. Placing the proteins in high-salt buffer at pH 7 and then running it through an anion-exchange column. B. Placing the proteins in low-salt buffer at pH 7 and then running it through a cation-exchange column. C. Placing the proteins in low-salt buffer at pH 7 and then running it through an anion-exchange column. D. Placing the proteins in high-salt buffer at pH 7 and then running it through a cation-exchange column.
C. Electron configurations account for the locations of all electrons within the energy levels (orbital shells and subshells) of an atom or ion. Shells are indicated using the principal quantum number n, which relates to the distance of an orbital from the nucleus. Subshells are labeled by type (s, p, d, f), which relates to the shapes of the orbitals that hold a maximum of 2, 6, 10, and 14 electrons, respectively Electrons typically occupy the lowest energy configuration (ground state), but an electron may achieve an excited state and temporarily jump to a higher energy orbital by absorbing energy equal to the energy difference between the two levels.
A sodium atom in an excited state has an electron configuration of 1s^2 2s^2 2p^6 3p^1. The excited electron within the configuration was excited from the: A. 2s orbital. B. 2p orbital. C. 3s orbital. D. 3p orbital.
A. Melting occurs at a constant temperature because a certain amount of energy, the latent heat of fusion, is needed to convert a substance from its solid to liquid state. The temperature of the metal will not increase above its melting point until all of the metal has melted. The small amount of heat supplied by the bunsen burner is insufficient to melt 100 g of the metal but could melt a small amount of the metal at the constant temperature of the melting point.
A student has a thin copper beaker containing 100 g of a pure metal in the solid state. The metal is at 215°C, its exact melting temperature. If the student lights a Bunsen burner and holds it for a fraction of a second under the beaker, what will happen to the metal? A.A small amount of the metal will turn to liquid, with the temperature remaining the same. B.All the metal will turn to liquid, with the temperature remaining the same. C.The temperature of the metal at the top of the beaker will increase. D.The temperature of the whole mass of the metal will increase slightly.
C. In polymerization, many subunits are connected together leaving you with a single molecule chain. Since it still ends up as a single chain, we need to multiply the HPO4 concentration by 1/n to indicate that.
According to Equation 1, the concentration of the polymer with respect to is [HPO] A. n [HPO] B. n^2 [HPO] C. 1/n [HPO] D. 1/n^2 [HPO]
B. The floating ice cube implies that its weight is balanced by the buoyant force on it Wice = mg = rfluidVsubmergedg Note that both the weight and the buoyant force are proportional to g, making the numerical value of girrelevant to the volume of the ice cube that is submerged. F weight = m g F buoyancy= density of fluid V g = pVg F weight = Fb so gravity cancels out.
An astronaut on Earth notes that in her soft drink an ice cube floats with 9/10 of its volume submerged. If she were instead in a lunar module parked on the Moon where the gravitation force is 1/6 that of Earth, the ice in the same soft drink would float: A. with more than 9/10 submerged. B. with 9/10 submerged. C. with 6/10 submerged. D. totally submerged.
D. The wavelength of the light detected from the star is smaller than the laboratory value on Earth. This implies a Doppler shift (a blue shift) associated with approaching relative velocity between the star and the Earth. The wavelength is 6.563 x10-7 in the labratory, but we see 6.56186 x10-7 coming from the star. The light from the star has a shorter wavelength than we'd expect(6.561 is smaller than 6.563), so as you said, "Moving towards = observed f will increase (wavelength decreases)" - the star is moving towards us.
An astronomer observes a hydrogen line in the spectrum of a star. The wavelength of hydrogen in the laboratory is 6.563 x 10-7m, but the wavelength in the star's light is measured at 6.56186 x 10-7m. Which of the following explains this discrepancy? A. The star is moving away from Earth. B. The wavelength of light that the star is emitting changes constantly. C. The frequency of light that the star is emitting changes constantly. D. The star is approaching Earth.
D. Mutations which make cells drug resistant are very rare, but the few drug-resistant bacteria that do develop such mutations flourish when the nonresistant cells are killed by antibiotics. Antibiotics are unlikely to have been the source of the mutations; nor do bacteria develop "immune" reactions to antibiotics. Any resistance of the patient's own colon cells to antibiotics is irrelevant to the resistance of bacteria to antibiotic. Thus, answer choice D is the best answer.
At the end of his initial hospital stay, a few E. coli cells remained in the patient's colon, even though he was taking antibiotics. These cells were most likely present because: A.the antibiotics caused drug-resistance mutations to occur in the bacterial DNA. B.the bacteria in the patient developed an immune reaction to the antibiotics. C.the patient's colon cells became increasingly resistant to the antibiotics during his hospitalization. D.chance mutations in a few E. coli before the treatment made these cells and their descendants antibiotic-resistant.
B. Within a sarcomere, the microfilament length remains stable. Because one end of the microfilament is anchored in the Z line, actin monomers are prevented from being added to or subtracted from that end. This rules out the possibility of treadmilling. Therefore, to retain a stable length, both ends of the microfilament must be capped.
Below is a diagram of a muscle sarcomere. Based on the passage, which statement best explains why the microfilament lengths do NOT change when the sarcomere shortens in a muscle contraction? Below is a diagram of a muscle sarcomere. Based on the passage, which statement best explains why the microfilament lengths do NOT change when the sarcomere shortens in a muscle contraction? A.The - ends of the microfilaments are capped by Z lines, and the actin subunit concentration is kept above 1 µM in muscle cells. B.The - ends of the microfilaments are capped by Z lines, and the + ends are capped by another protein. C.The actin subunit concentration is kept above 4 µM in muscle cells. D.The - ends polymerize and the + ends depolymerize at the same rate.
hydrostatic pressure
Blood flowing through capillaries exerts ______ pressure on the vessel walls. As a result, some fluid is forced from vessels into the interstitial space.
A. After introducing the idea of fluid homeostasis, which is obtained by balancing hydrostatic and osmotic pressures, the question asks the examinee to predict the consequences of a blood clot on the venous side of a capillary bed. To answer this, it is necessary to know that blood flows from arteries to capillaries and then to veins. If flow is blocked at the venous side, blood would accumulate in the capillaries. Thus, hydrostatic pressure would build up in the capillaries, causing a net increase in fluid flow into the interstitial spaces. A is therefore correct. The flow gets blocked, but the heart is still pumping, so it keeps trying to jam more fluid into the capillaries. Pressure goes up, and fluid is forced out somewhere. The vein is blocked, and we can't go backwards up the artery, so it goes into the interstitial space.
Capillaries in the kidney and elsewhere in the body maintain fluid homeostasis by balancing hydrostatic and osmotic pressures. Which of the following is the initial effect of a blood clot forming on the venous side of a capillary bed? A. Net fluid flow in the direction of interstitial spaces will increase. B. Net fluid flow in the direction of interstitial spaces will decrease. C. Capillary osmotic pressure will increase. D. Capillary osmotic pressure will decrease.
A. The muscle cells in Culture A use lactic acid fermentation to provide the energy for the contractions that result from electrical stimulation. In this process, NADH reduces pyruvate to produce lactate. Therefore, pyruvate serves as the electron acceptor in production of lactate.
Culture A appeared white when visualized with a microscope. When an electrical current was applied to the culture medium, the concentrations of lactate and H+ ions in that medium increased. The pH of the medium changed from 7 to 5 after 2 minutes of continuous, electrically-stimulated contractions. The terminal electron acceptor in the metabolic pathway responsible for the chemical changes observed when Culture A was electrically stimulated is: A. pyruvate. B. oxygen. C. NAD+. D. water.
Between 12 and 26 hours the heart rate is at 0. When heart rate is at 0 the heart is not pumping and therefore there is no oxygen going into the lungs and therefore no oxygenation of blood, causing only anaerobic metabolism to be the only viable option.
During what time period of the freeze-thaw episode shown in Figure 1 were the frog's tissues relying entirely on anaerobic respiration? Before 0 hours Between 0 and 6 hours Between 0 and 24 hours Between 12 and 26 hours
amplitude of a wave
Height of a wave is the distance from the midline of the wave to either its crest or trough.
B. hemodialysis is described as a procedure that filters blood for a patient whose kidneys are unable to do so. Before the procedure, the patient's kidneys do not effectively filter blood, which would lead to high blood volume and high Capillary hydrostatic pressure, ultimately causing a net exit of fluids from capillaries. The question states that hemodialysis removes relatively large amounts of solutes and water in a short period of time. As hemodialysis occurs and blood is filtered, the removal of solutes and fluid from the patient's body by the dialysis machine reduces blood volume, which would in turn decrease CHP. Therefore, the net direction of fluid flow near the end of the procedure is into the capillaries because of reduced capillary hydrostatic pressure.
Hemodialysis is a procedure that filters blood for a patient whose kidneys are unable to do so. This procedure removes relatively large amounts of solutes and water from the patient's bloodstream in a short period of time. If a patient is undergoing hemodialysis, in which net direction would fluid flow across capillary walls near the end of the procedure? A. Into the capillaries, because the procedure decreased capillary solute concentration B. Into the capillaries, because the procedure reduced capillary hydrostatic pressure C. Into the interstitial fluid, because interstitial fluid osmolarity is increased D. Into the interstitial fluid, because overall blood volume is reduced
A. Olestra is not metabolized because digestive enzymes are blocked from cleavage sites. If a molecule is not metabolized, it can provide no energy and thus no calories. Choice A, 0 Cal, is the correct answer.
How many dietary calories does a 1-g sample of Olestra contribute to a human consumer? A. 0 Cal B. 4 Cal C. 5 Cal D. 9 Cal
B. Ksp= [Na+] [CHO-] 34.9 = x^2 x= 6 Ksp = square root of 34.9= 6 The solubility product constant Ksp represents the limit of solubility for a compound, and it is defined as the product of the molar concentrations of the dissolved species each raised to the power of their respective balanced equation coefficients. This relationship can be used to find molar concentrations of ions in saturated solutions.
If the solubility product constant Ksp for NaC9H7O4 is estimated to be 34.9, what is the approximate acetylsalicylate ion concentration in the saturated NaC9H7O4 solution used for the reaction? A. 2.95 M B. 5.91 M C. 17.4 M D. 34.9 M
linked genes will have low recombination progeny in F1 and F1 will resemble the parental phenotype.
If two genes are located close together on a chromosome (measured in centimorgans, or map units), they are relatively unlikely to be separated by a recombination event. This is because there is less distance between the two genes in which a recombination can occur. Therefore, substantially fewer progeny will have recombinant combinations than parental combinations Homologous chromosomes can exchange genetic material by recombining. Genes that are located close together on a chromosome have a lower probability of being separated by recombination than those that are far apart. Fewer progeny from a cross will have recombinant genotypes than will have parental genotypes.
C. The formula for the number of possible peptides that contain one each of n amino acids is n! (n factorial). For n = 3 (a tripeptide), n! = 3! = 3 × 2 × 1 = 6 = Answer C. Alternatively, for a tripeptide ABC, the following combinations are possible: ABC, ACB, BAC, BCA, CAB, and CBA, or six = Answer C.
Ignoring stereochemistry, how many different tripeptides may exist that contain the same three amino acids as the molecule shown below? A. 1 B. 3 C. 6 D. 9
Photoelectric effect
KE= hf-work function; the photons are a measure of energy in light. we know the energy of a photon is represented by "E=hf"; Planks Contant times frequency. We know that the photons must hold enough energy to liberate an electron from the metal and then have some left over to accelerate that electron-- KE(1/2mv2)= hf(of the photon)- hf(work function, intrinsic to the metal). To increase the velocity we need to either increase the energy of the photon we are shooting at the metal or the work constant would somehow have to decrease; either way, we're trying to increase the difference between the incident photon and the work function, so a greater amount of energy will go toward accelerating the electron. By increasing the frequency of the incident photon we have increased KE [KE= hf(of the incident photon)- hf(work function)], as KE increases so does the velocity, since KE=1/2mv2 (the answer the second question) However, we are NOT shooting more electrons at the metal, just ones with greater energy (due to the increase in frequency). the _________is saying that 1 photon in =1 electron out. If we increase the number of photons shot at the metal, we liberate more electrons, because of this one to one thing. and because they didnt increase the energy of the photons (question 1) the velocity of those ejected electrons did NOT increase. So, increasing frequency increases velocity of the ejected electron and increasing the amount of photons increases the amount electrons ejected. If we wanted to increase speed and amount we would need to increase frequency AND the amount of incident photons
A. S An element which can act as an electron acceptor should have a high electron affinity
Like oxygen atoms in methanogens, which of the following elements can act as an electron acceptor? AS BHe CH2 DFe
B. The first step in working stoichiometry problems is to balance the equation. Na2CO3 (aq) + 2 HCl (aq) → CO2 (g) + H2O (l) + 2 NaCl (aq) The problem seeks the volume of 2 M Na2CO3solution which, when reacted with HCl solution, will produce 11.2 L of CO2 gas at STP (standard temperature and pressure). It is an important fact that one mole of an ideal gas, at STP, will occupy a volume of 22.4 L. Therefore, 11.2 L of CO2 gas, at STP, must represent 0.5 mole of CO2. (Virtually all gases can be approximated as ideal gases at common temperatures and pressures.) According to the balanced chemical equation, one mole of CO2 is produced when one mole of Na2CO3 reacts. Therefore, the amount of Na2CO3 required to produce 0.5 mole of CO2 gas must also be 0.5 moles. We must find the volume of 2 M Na2CO3 solution which contains the required 0.5 moles of reactant.
Na2CO3 + HCl → CO2 + H2O + NaCl Consider the above unbalanced equation. For this reaction, how many mL of a 2 M solution of Na2CO3 are required to produce 11.2 L of CO2 at STP? A. 125 mL B. 250 mL C. 375 mL D. 500 mL
High Rf
Non-polar, least polar, fastest moving
D. Light with a wavelength of 2.0 × 10^-7 m has a frequency of f = c/λ = (3.0 ×10^8 m/s)/(2.0 × 10^-7 m) = 1.5 × 10^15 Hz and light with a wavelength of 6.0 × 10^-7 m has a frequency of f = c/λ = (3.0 × 10^8 m/s)/(6.0 × 10^-7 m) = 5.0 × 10^14 Hz. The frequency difference is 10.0 × 10^14 Hz, or 1.0 × 10^15 Hz.
Photoelectric detectors, by contrast, contain a light-emitting diode that sends a beam of unpolarized light across a small chamber. The light beam usually has a wavelength of 6.0 × 10^-7 m and has an intensity of 1.0 × 10^-3 W. When smoke particles enter the chamber, the light scatters in all directions. A photocell then senses either the increase in the scattered light or the reduced intensity of the light beam and sets off the alarm. The speed of light in air is 3.0 × 10^8 m/s. Advanced photodiode detectors have a second light-emitting diode, operating at a wavelength of 2.0 × 10^-7 m, to detect even smaller smoke particles from smoldering flames. What is the frequency difference between the two light beams? A.12.0 × 10^15 Hz B. 3.0 × 10^15 Hz C. 2.0 × 10^15 Hz D. 1.0 × 10^15 Hz
C. meiosis. meiosis and mitosis do not occur in prokaryotes (which includes bacteria) The answer here is C because as it is mentioned, after initial conjugation, the bacteria now divide through binary fission instead of meiosis..
Some of the DNA sequences that are eliminated during macronuclear differentiation (Figure 1, Step 6) may be sequences involved in: A. transcription. B. translation. C. meiosis. D. ribosome production.
N1 > N2 theta 1 < theta 2
TIR?
A. The strain of M. tuberculosis in the coworkers was killed by both ampicillin and kanamycin, indicating that this strain did not carry a plasmid gene that made it resistant to these two antibiotics. However, once the M. tuberculosis coexisted in the patient with other antibiotic-resistant bacteria, the M. tuberculosis could survive despite treatment with large doses of these two antibiotics. The antibiotic does not induce mutations; the few cells that are already resistant flourish in the absence of the nonresistant bacteria. The M. tuberculosis did not adapt to its new environment by modifying its metabolism; rather, there was a strain with a metabolic capability that was not compromised by the antibiotic. The passage indicates that conjugation may occur between members of different bacterial species. Therefore, it is most likely that these M. tuberculosis bacteria underwent conjugation with resistant cells.
Tests performed on the M. tuberculosis strain infecting the patient's coworker indicated that the strain was susceptible to both ampicillin and kanamycin, and the coworker was successfully treated. The M. tuberculosismost likely survived in the patient because it had: A.undergone conjugation with cells of resistant E. coli. B.undergone an antibiotic-induced mutation that conferred antibiotic resistance. C.reproduced more rapidly than the strain in the coworker. D.adapted to its new environment by modifying its metabolism.
C. prokaryotes contain cytoskeletal filaments. and moving the transcript via active transport is much faster than diffusion. question states "short half-life" which indicates that the speed of transport is important. Diffusion across the cytoplasm: this is quite slow, if the transcript has a short half life it may degrade before reaching its destination. Transport from the endoplasmic reticulum in vesicles: this is incorrect bc prokaryotes do not contain mb bound organelles.
The bglF transcript is known to have a short half-life within the cytosol. What mechanism is most likely responsible for transport of this transcript to the cytoplasmic membrane once it is synthesized? A. Diffusion across the cytoplasm B. Transport via attachment to the mitotic spindle C.Active transport along cytoskeletal filaments D. Transport from the endoplasmic reticulum in vesicles
B. "germ-line" micronucleus. germ-line = only thing passed down, so you cross only the micronuclei which is the Hh x Hh = so 1/4 hh
The ciliate protozoan Tetrahymena contains two nuclei: a diploid, germ-line micronucleus and a 45-ploid macronucleus that is the site of gene expression during the vegetative state. Sexual reproduction in Tetrahymena occurs by the process of conjugation. In a mating of two Tetrahymena strains that are homozygous in their macronuclei and heterozygous in their micronuclei for a recessive gene, what percentage of the F1 generation will express the recessive phenotype? A. 0% B. 25% C. 50% D. 100%
D. Meiosis II then proceeds similarly to mitosis but differs in that each cell begins with a haploid, rather than a diploid, genome. During anaphase II of meiosis, sister chromatids are separated to opposite poles of the cell. Assuming no recombination, the identical e alleles on one set of sister chromatids will be pulled to opposite poles by the mitotic spindle during anaphase II.
The diagram above represents the immature gamete of a male fly that is heterozygous for Ena. If the gamete undergoes meiotic division and Ena alleles do not undergo recombination, during which phase of meiosis will the sister chromatids containing the e allele be pulled to opposite poles of the cell? A. Metaphase I B. Anaphase I C. Metaphase II D. Anaphase II
A. The chemical behavior depends on electron interactions and bond formation. The number and configuration of electrons depends on the element which is solely determined by the atomic, or proton, number.
The explanation for the fact that radioactive isotopes of an element exhibit the same chemical behavior as the stable isotopes of the element is that each has the same: Aatomic number. Bnumber of neutrons. Cmass number. Datomic weight.
c. A and B: disruption doesn't match up well with cancer since cancer is hyperactive cell growth. Blood vessels tend to dilate during inflammation. Inflammation is the accumulation of WBC @ the site of injury.
There is a relationship between H. pylori infection and cancer. Infected individuals have a two-fold increased risk of gastric cancer, although >75% of patients with active infections do not develop cancer. Genetic studies of H. pylori have identified genes that are expressed in different strains of this bacterium. One gene, vacA, encodes a toxin. Expression of another gene, cagA, leads to inflammation and may be related to the genesis of gastric cancer. Although many individuals develop antibodies against H. pylori antigens, these antibodies rarely eradicate the infection; evidently, this pathogen has developed effective ways to elude host defenses. According to the passage, the cagA gene product will cause: A. the disruption of host cell enzymatic activity. B. the disruption of host cell protein synthesis. C. the movement of leukocytes into mucosal tissue. D. the vasoconstriction of arterioles in the mucosal layer.
A. The waveform in Figure 1c begins to repeat at the zero displacement point near the end of the time axis. This is the same time period as the first harmonic as seen in Figure 1a.
The period of the waveform shown in Figure 1c is the: A. same as the period of the first harmonic. B. same as the period of the second harmonic. C. same as the period of the third harmonic. D. sum of the periods of the first, second, and third harmonics.
C. The equation shows the migration of a hydrogen atom from an α-carbon atom to an oxygen atom. The hybridization of the α-carbon atom and oxygen atom change during the process. This is a description of tautomerism, answer choice C; the other three answer choices are terms for distinctly different processes. Thus, answer choice C is the best answer.
The phenomenon exemplified in the equilibrium equation below is called: A. reduction. B. resonance. C. tautomerism. D. mutarotation.
C. So pressure increases linearly with depth, from the data given, it seems like the formula is P=40D+50, which shows why 0 depth has a pressure of 50. So then for the doubled density: P=80D+50, which plugging in 10 gives 850. Gauge pressure = hydrostatic pressure (pgh) + atmospheric pressure (ie pressure measured at 0 depth) Might make intuitive sense if you imagine using a pressure gauge widget to measure underwater pressure at some point. The gauge will show whatever the starting (atmospheric) pressure is + hydrostatic pressure from being dunked. Then you can use system of equations with the table to figure out what the atmospheric pressure is and then calculate 2pgh + p_atm.
The table above gives pressure measured at various depths below the surface of a liquid in a container. A second liquid, whose density is twice that of the first liquid, is poured into a second container. Similar pressure measurements are taken for the second liquid at various depths below the surface of the second liquid. What is the pressure at a depth of 10 cm for the second liquid? A.250 N/m2 B.450 N/m2 C.850 N/m2 D.1650 N/m2
A. "Some natural poisons also affect microfilament metabolism: the cytochalasins bind to the + end of a microfilament and prevent the addition of actin subunits to that end, and phalloidin blocks subunit loss from either end." So by that logic, if cytochalasins are added and the amoeba does not move then that means the amoeba was using microfilaments to move since cytochalasins stop microfilaments by binding to the ends.
The theory of force generation proposed in the passage is best supported by which of the following observations about Amoeba locomotion? A.Amoeboid movement stops upon exposure to cytochalasins. B.Amoeboid movement cannot occur if mitosis is blocked. C.Moving Amoeba cells produce more troponin than do stationary ones. D.The rate of movement is inversely proportional to the viscosity of the medium in which the Amoeba moves.
When two gene products function in the same biological pathway, altering an upstream product will have an effect on the activity of downstream targets or products. Altering a downstream product in a way that mimics the function of the upstream product can compensate for the loss of an upstream product.
Upstream gene products in a biological pathway act on downstream gene products. Failure of an upstream product to act can be compensated for by alterations to downstream targets that mimic the effect of the upstream product.
C. (1 kg) × (1000 g/kg) × (1 cal/g oC) × (1 kcal/1000 cal) × (50oC) = 50 kcal which gives the energy provided by the combustion of ten peanuts. One peanut therefore provides 5 kcal.
What is the energy content in kcal of one peanut, if the temperature of 1 kg of water in a calorimeter increases by 50oC upon the combustion of 10 peanuts? A. 0.5 kcal B. 1 kcal C. 5 kcal D. 10 kcal
C. The number of protons in the nucleus of an element is given by the atomic number. In a neutral atom, the number of electrons is equal to the number of protons. The number of neutrons can be found by subtracting the atomic number from the mass number. The atomic number of strontium (Sr) is 38; so the number of neutrons is 90 - 38 = 52. The sum of protons, neutrons, and electrons in strontium is 38 + 38 + 52 = 128. This value is given as answer choice C.
What is the sum of the protons, neutrons, and electrons in strontium-90? A. 90 B. 126 C. 128 D. 218
D. 60% % yield= actual/theoretical. find limiting reagent 2.0x10^-4 mol XeF6. 2 x 3/4 = 1.5 mol Theoretical % = 9.0 /1.5 = 60% The percent yield for a reaction is calculated as the ratio of the actual yield (the amount obtained) to the theoretical yield (the amount possible) expressed as a percentage. This value assesses the extent to which the starting materials are converted into products during a reaction.
When 2.0 × 10^−4 mol of XeF6(s) is placed in 0.020 L of 0.30 M NaOH(aq), the following disproportionation reaction occurs: 4 XeF6(s) + 36 NaOH(aq) → 3 Na4XeO6(s) +Xe(g) + 24 NaF(aq) + 18 H2O(l) What is the percent yield of Na4XeO6(s) for the reaction if 9.0 × 10^−5 mol Na4XeO6(s) is produced? A. 18% B. 33% C. 45% D. 60%
strongly influence the amount of NE released by the sympathetic nerve terminal.
When an action potential reaches a nerve terminal it triggers the opening of Ca+ channels in the neuronal membrane. Because the extracellular concentration of Ca+ is greater than the intracellular Ca+ concentration, Ca+ flows into the nerve terminal. This triggers a series of events that cause the vesicles containing NE to fuse with the plasma membrane and release NE into the synapse.
A. If such a burn patient is treated with continuous intravenous infusions of albumin during the healing process, the capillary solute concentration will increase, which will ultimately increase the capillary oncotic pressure. This pressure will promote water movement into the capillaries via osmosis. Compared to a control patient who did not receive albumin infusions, the treated patient will likely display increased capillary oncotic pressure due to a higher albumin (ie, solute) concentration in the plasma
When body surface area is affected by a burn, capillaries leak fluid containing serum proteins such as albumin. A burn patient is treated with continuous intravenous infusions of albumin during the healing process. Compared to a burn patient control (no albumin infusions), the treated patient's capillary oncotic pressure will most likely be: A. increased. B. decreased. C. the same. D. increased, but fluid leakage will also increase over time.
A. The number of incident photons affects only the number of electrons, not their energies. The electron energies depend on photon energy, the cathode work function, and the potential difference between the cathode and anode.
When the number of photons incident on the cathode with energies above the value of the work function increases, which of the following quantities also increases? A. Number of electrons ejected B. Potential energy of each ejected electron C. Magnitude of the electric field between the electrodes D. Speed of electrons at the anode
P= CO x VR
blood (hydrostatic) pressure, vascular resistance (VR, force opposing blood flow through a vessel), and cardiac output (CO, blood volume expelled from the ventricles per unit time).
vasoconstriction
blood vessel diameter decreases blood flow decreases blood pressure increases
C. Dissociation of the myosin head from the actin filament requires the binding of ATP (I). Attachment of the myosin head to the actin filament requires calcium and a troponin/tropomyosin shift (II). The conformational changes that move actin and myosin relative to one another require that ATP be hydrolyzed, for these changes occur upon release of the products of hydrolysis (ADP and Pi) by the myosin head (III). Binding of troponin to actin does not require the hydrolysis of ATP (IV). Release of calcium from the sarcoplasmic reticulum also does not require ATP hydrolysis. This release occurs when calcium ions move via voltage-gated ion channels down their concentration gradient (V). The reuptake of calcium into the sarcoplasmic reticulum occurs via an ATP-hydrolyzing pump that moves calcium against its concentration gradient (VI). Thus, C is the correct response.
Which steps involved in the contraction of a skeletal muscle require binding and/or hydrolysis of ATP? I. Dissociation of myosin head from actin filament II. Attachment of myosin head to actin filament III. Conformational change that moves actin and myosin filaments relative to one another IV. Binding of troponin to actin filament V. Release of calcium from the sarcoplasmic reticulum VI. Reuptake of calcium into the sarcoplasm A. I, II, and III only B. II, III, and IV only C. I, III, and VI only D. III, IV, and VI only
A. The Doppler equation for frequency is Δf/f = -v/c for a given relative velocity vbetween source and detector. Thus, the frequency shift Δf depends inversely on the speed of the wave in the medium in which it propagates, c. The velocity of sound is much smaller than that of electromagnetic radiation, so for the same relative velocity the frequency and wavelength shifts are much greater for sound than for radio waves. Sound is faster in solids, slower in everything else
Why are the percentages of the change in frequency and wavelength much greater when sound waves are used instead of radio waves in these experiments? A. Sound waves travel more slowly. B. Sound waves have a much higher frequency. C. Sound waves have a much shorter wavelength. D. Interference in the atmosphere affects sound waves much more.
vasodilation
blood vessel diameter increases blood flow increases blood pressure decreases
turns red for acids.
blue litmus paper
osmotic pressure
circulating proteins in the bloodstream cause blood ______ pressure to be higher than that of the interstitial fluid, drawing fluid from the interstitial space into the capillaries.
boiling point The normal boiling point is measured at 1 atm pressure. The vapor pressure of a liquid increases with increasing temperature. Hence, the boiling point of a liquid decreases as the pressure on the surface of the liquid is decreased. If a leak develops in the apparatus, the surface pressure will increase, as will the boiling points of both liquids.
______ of a liquid is the temperature at which the vapor pressure of the liquid equals the surface pressure.
Coulomb's Law
_______________ states that the electrostatic force between two charged objects is directly proportional to the product of charge magnitudes. Increasing the charge on one object increases the electrostatic force on the other object.
Thromboplastin
________is released when damage occurs to a tissue. Prothrombin and fibrinogen are both present in the blood. When _________ interacts with the blood, it converts prothrombin to its active state, thrombin. Thrombin then cleaves fibrinogen, converting it to its active state, fibrin. Fibrin is exceptional at forming cross-linked mesh-like interactions, and in doing so traps passing blood cells and plasma proteins. This blockage and accumulation of blood elements becomes the clot.
Ka= [A-][H+]/[HA] Strong acids ionize essentially 100% in aqueous solutions, resulting in a large acid dissociation constant (Ka > 1), whereas weak acids ionize only to a small extent, resulting in a small acid dissociation constant (Ka< 1).
acid dissociation constant?
ebulliator/boiling chip
added to introduce small air bubbles into the system. The air bubbles break the surface tension of the liquid being heated and prevent superheating and bumping.
protease
are enzymes that facilitate the breaking of peptide bonds in a hydrolysis reaction. does not cleave disulfide bonds.
autosomal traits
are generally expressed in the same proportions in males as in females.
Sex-linked traits
arise due to the expression of alleles present on an organism's sex chromosomes. Because the X chromosome contains significantly more genes than the Y chromosome, the majority of sex-linked traits are linked to the X chromosome. Due to sex determination, male offspring (XY) receive their X chromosome from their mother (XX) and their Y chromosome from their father (XY). Because males have only one X chromosome and cannot compensate with another, males in a population generally express recessive X-linked traits in higher proportions than their female counterpart.
as harmonics increase, the frequency increases and wavelength decreases
as harmonics(n) ______, the frequency ______ and wavelength _________
