MCAT: questions i missed on exams
RPS4Y was lost in four different species studied. According to Figure 1, what is the minimum number of distinct evolutionary events that must have occurred to cause this result? (see pic on other side)
3. Once a gene is lost, this loss will be mirrored in all future lineages. Therefore, if a common ancestor of two species loses a gene, both of the species will lack it as well. We see that RPS4Y has been lost in opossums, cattle, mice, and rats. If the loss had occurred at a common ancestor of opossums and cattle, then all of the remaining species would exhibit the loss, but this is not the case. Therefore, the opossum must have experienced its own distinct evolutionary event. The diagram below shows the minimum number of events that must have occurred.
Based on Figure 1, adding salt to water causes the boiling point of the water to: A. increase, requiring a greater average kinetic energy of the liquid to produce a vapor pressure equal to the external pressure. B. increase, requiring a greater average kinetic energy of the liquid to produce a vapor pressure that is greater than the external pressure. C. decrease, requiring a lower average kinetic energy of the liquid to produce a vapor pressure equal to the external pressure. D. decrease, requiring a lower average kinetic energy of the liquid to produce a vapor pressure that is less than the external pressure.
A - As stated in the passage, adding salt reduces the vapor pressure of the liquid. Specifically, as the solute concentration is increased, the rate at which water molecules can break through the liquid surface decreases. Remember that boiling point is defined as the temperature at which the vapor pressure of a solution is equal to the atmospheric pressure. A decrease in vapor pressure makes this point more difficult to achieve, resulting in a higher boiling temperature. Boiling does not require that Pvap > Patm; it only requires that Pvap = Patm. Adding solute to water raises the boiling point and lowers the melting point. Be sure to familiarize yourself with these colligative properties.
Is it reasonable to expect that individuals with certain alleles of the gene discussed in the passage might be less likely to develop cancer in the first place? A. Yes, because cancer-causing mutations can be repaired before they can be passed on to daughter cells. B. No, because the error-correcting mechanism only operates when the mutation has already occurred. C. Yes, because metastasis will be prevented. D. No, because these alleles only affect the rate of spread of mutations.
A is correct. Alleles for this gene either include or don't include an effective error-correcting machinery made up of proofreading enzymes that snip and fix mistakes in the DNA-copying process. This has an effect on the spread of cancer and, thereby, life expectancy after diagnosis. However, it makes sense that individuals who do not have cancer will be more or less likely to develop depending on their alleles for the relevant gene. There are two answer choices that say yes, but only A includes the correct reasoning in the answer choice. Metastasis occurs only after cancer has been developed, so it has nothing to do with cancer prevention.
Stimulation of the iris dilator muscle is a result of activation of: A. sympathetic motor neurons. B. parasympathetic motor neurons. C. the fifth cranial nerve. D. sympathetic sensory neurons.
A is correct. Dilation of the pupils is a classic fight-or-flight response. The fight-or-flight response is part of the sympathetic nervous system.
In a leucine zipper motif, every 7 amino acid residues, or 2 full turns of an alpha helix, are leucine resides. Leucine and other amino acids on one face of the helix come together, to an opposite alpha helix that has a similar arrangement of leucine and other amino acids. Which solvent would be LEAST favorable for c-fos/c-jun dimerization? A. Hexane B. Ethanol C. Water D. Phosphate-buffered saline
A is correct. For long questions like this one, begin by summarizing exactly what the question is asking. In short, leucine residues on different parts of a molecule are coming together to form a dimer. Leucine is hydrophobic, since its side chain contains only carbon and hydrogen. If the solvent were also hydrophobic, the face of a leucine zipper could interact just as favorably with the solvent as with the opposite alpha helix. Some leucine residues would likely interact only with the solvent, preventing formation of the dimer entirely. Hexane is the least polar solvent listed.
Boiling chips and vacuum distillation, respectively, are used in distillations to: A. provide nucleation sites that give the liquid a place to start forming bubbles to prevent superheating; lower the boiling points of the substances to be distilled. B. lower the boiling points of the substances to be distilled; work synergistically with the vacuum system to further lower the boiling points. C. lower the boiling points of the substances to be distilled; provide nucleation sites that give the liquid a place to start forming bubbles to prevent superheating. D. provide nucleation sites that give the liquid a place to start forming bubbles to prevent superheating; speed up the distillation process by vacuuming the first distillate out of the apparatus
A is correct. Let's begin with the first part of this question, the function of boiling chips. When transitioning from liquid to gas during boiling, the liquid needs nucleation sites, or places to start forming bubbles. This is typically achieved either by scratching the inside of the flask, or by introducing boiling chips. This eliminates all options except for A and D. Next, remember that boiling occurs when the Pvap of the substance in question equals the Patm. Typically, we boil substances by increasing the temperature, thereby increasing Pvap. Alternatively, however, we can lower boiling point by reducing Patm, which can be accomplished through the introduction of a vacuum. Vacuum distillation is often used when components have very high boiling points and would otherwise be difficult to distill. B: The boiling chip itself does not alter the boiling point of the liquid. D: The vacuum apparatus removes the atmospheric pressure, not the distillate itself. The distillate simply boils more rapidly as a result of the reduced pressure.
Which of the following correctly describes the orbital hybridization of XeF4 and NH3, respectively? A. sp3d2, sp3 B. sp3, sp3 C. sp3, sp2 D. sp3d2, sp2
A is correct. One quick way to determine the orbital hybridization around the central atom is to simply count up the number of bonds and lone pairs. For example, ammonia has three bonds and one lone pair around its central nitrogen atom, for a total of four regions of electron density. This orbital hybrid therefore needs four orbitals to hybridize: s, p, p, and p. Thus, ammonia is an sp3 hybrid. We can narrow it down to choices A and B. Next, XeF4 has two lone pairs and four bonds around the central xenon atom. So we need six hybrid orbitals to hold those electrons: s, p, p, p, d, and d. That's an sp3d2 hybrid.
Over millions of years, recombination between sex chromosomes proved to be harmful, and population genetics resulted in the loss of nearly all of the genes on the Y chromosome. Which form of genetic change best explains the loss of the majority of genes from the Y chromosome? A. Genetic drift B. Founder effect C. Gene flow D. Low mutation rate
A is correct. Paragraph 1 tells us that the Y genes were lost partially due to population genetics. Genetic drift encompasses the fact that an adult male who has only female offspring will never pass on his Y chromosome. In other words, even if the male has a well-adapted Y chromosome, it may not be passed to the next generation. The repeated random loss of well-adapted Y chromosomes can result in the degeneration of the chromosome as a whole.
Which of the following are products of the decay of I-131 if it underwent beta minus decay? A. Ionizing radiation, Xe-131, and an electron B. Ionizing radiation, Te-131, and an electron C. Non-ionizing radiation, Xe-131, and a positron D. Non-ionizing radiation, Te-131, and a positron
A is correct. Since each answer has multiple parts, we should divide and conquer, eliminating wrong choices as we go. Paragraph 3 mentions that the nuclei emit gamma rays, a form of electromagnetic radiation that consists of high-energy photons. Gamma rays represent ionizing radiation, which makes them biologically hazardous (eliminate C and D). The passage also states that I-131 undergoes β-minus decay. In β-minus decay, a neutron is converted to a proton as an electron is emitted. Therefore, iodine must be converted to an element with one additional proton, which can only be Xe. (Te-131 would be formed if I-131 underwent positron decay or electron capture)
The researchers participating in the experiments described above were assigned to conduct new experimentation, also on Leigh syndrome-positive mouse specimens. As a precursor step, it was decided to create a new population of Leigh syndrome-positive mice. What would be the option most likely to be successful in creating this population? Note, Leigh syndrome = passed mother -> child A. Induction of intentional mutations in the mitochondria of pre-fertilization ova B. Induction of intentional mutations in the mitochondria of pre-fertilization sperm cells C. Induction of intentional mutations in the mitochondria of either type of gamete D. Induction of intentional mutations in the mitochondria of synaptonemal complexes
A is correct. The information in the second paragraph of the prompt establishes that Leigh syndrome is passed on from mother to child through cytoplasmic components. Mitochondria are the most likely culprits, as they have their own DNA and they been implicated in a number of genetic diseases that are passed on from mother to child. This narrows the choices to A, B and C. Here, (A) is correct because in a fertilized egg, the mitochondria of ova predominate due to their much larger numbers compared to the mitochondria of sperm B, C: For the above reasons, mitochondrial diseases are nearly always the result of the child inheriting mitochondria from ova (the female gamete) rather than from sperm (the male gamete). D: Synaptonemal complexes are complexes made out of chromosomes and have nothing to do with mitochondria, which are organelles in the cellular cytoplasm.
the cytosolic CK M subunit is enriched in asparagine and lysine. CK M subunits are slowly converted into two modified subunits, M1 and M2, by plasma carboxypeptidases via successive cleavage of C-terminal amino acid residues. CK isoforms containing M1 and M2 subunits migrate nearer the anode than do CK isoforms containing the unmodified M subunit. The region near the C-terminus of the unmodified M subunit is enriched in which of the following amino acids? A. Lys, Asn, Arg B. Leu, Cys, Pro C. Lys, Ala, Asp D. Gln, Asp, Glu
A is correct. The passage states that once released from tissue, the CK M subunits are converted into modified subunits M1 and M2 by successive cleavage of C-terminal amino acid residues. If the modified subunits migrate nearer the positive electrode than do unmodified subunits, then modified subunits possess a lower pI than unmodified subunits and should be enriched in lower-pI amino acids via the removal of higher-pI residues. Only choice A—lysine, asparagine, arginine—contains three higher-pI, basic residues and is the correct answer. Choices B, C, and D all contain some combination of non-basic amino acids and are incorrect.
Which hormone upregulates water reuptake via aquaporin channels?
ADH =It's a hormone made by the hypothalamus in the brain and stored in the posterior pituitary gland. It tells your kidneys how much water to conserve. a peptide hormone that increases water permeability of the kidney's collecting duct and distal convoluted tubule by inducing translocation of aquaporin-CD water channels in the plasma membrane of collecting duct cells
Analogous structure
Analogous structures are those structures that evolved independently to carry out the same function. Basically: Different evolution, same function
Increasing plasma concentration of aldosterone is most likely to be followed by which of the following? A. Increased water reabsorption through increased aquaporin channels in the collecting duct B. Increased sodium reabsorption in the distal tubule C. Decreased water reabsorption in the collecting duct D. Decreased plasma calcium concentration
B is correct. Aldosterone is released from the adrenal cortex in response to low blood pressure. Its primary function is to increase sodium reabsorption in the distal tubule and collecting duct. Aldosterone upregulates the sodium-potassium pumps along the lining of the nephron, pumping three sodium ions OUT of the nephron lining (and toward the blood) for every two potassium ions it pumps IN (toward the nephron and away from the blood). Since we have a net solute movement out of the nephron, aldosterone also increases the gradient that favors water reabsorption.
Which of the following correctly lists a pair of analogous structures and a pair of homologous structures, respectively? A. The wing of a bee and the wing of a bird; the wing of a bird and the leg of a bird B. The wing of a bee and the wing of a bird; the arm of a human and the flipper of a walrus C. The arm of a human and the wing of a bat; the wing of a bird and the wing of a bat D. The wing of a bird and the wing of a bat; the wing of a bee and the wing of a bat
B is correct. Analogous structures are those structures that evolved independently to carry out the same function. Thus, the wing of a bee and the wing of a bird are analogous structures. Homologous structures are those that have a similar evolutionary history, arising from the same source, even if they now have different functions. The forelimbs of mammals (human arm, walrus flipper, bat wing) would all be homologous despite their different functions. Thus, this choice correctly indicates a pair of analogous structures and then a pair of homologous structures. A: The first set of structures listed here are analogous, but the second set (which come from the same type of organism) are neither analogous nor homologous. C: These are homologous and analogous structures, respectively. D: Both of these sets of structures are analogous
Cancer cells use fermentation more frequently than healthy cells; why could researchers use measurements of intracellular lactate levels (ILL) in cancer cells to assess efficacy of cancer drugs? A. High ILL would indicate that glycolysis is significantly inhibited. B. Low ILL would indicate that glycolysis is significantly inhibited. C. High ILL would indicate that the pentose phosphate pathway is significantly inhibited. D. Low ILL would indicate that the pentose phosphate pathway is significantly inhibited.
B is correct. Lactate is a product of fermentation, and the first paragraph states that cancerous cells rely heavily on fermentation for energy production. Inhibiting glycolysis or fermentation would reduce ILL and indicate that the metabolism of the cancer cell is being effectively inhibited by the drug. A: High ILL would indicate that the cell is metabolizing effectively, so this answer is the opposite. C, D: ILL have nothing to do with the pentose phosphate pathway. Avoid these distracting answers.
A book rests horizontally on a table. The book experiences a gravitational force of mg due to the Earth's gravity. According to Newton's third law: A. the book experiences a normal force of mg pushing up due to the table. B. the earth experiences a gravitational force of mg from the book. C. the table exerts a gravitational force of mg on the earth. D. the earth exerts a normal force up on the table equal to mg plus the weight of the table.
B is correct. Newton's third law can be expressed as: FA on B = - FB on A Here, the force of the earth pulling on the book is equal to and opposite of the book pulling up on the earth. Thus, choice B is correct. A: Remember that Newton's third law applies very narrowly to the interactions between two objects. While the table does exert a normal force upwards on the book, that normal force is not what Newton's third law refers to. Notice that in the equation above, the reaction force of A on B is strictly limited to B on A.
While the blood is buffered primarily through the equilibrium between carbon dioxide and carbonic acid, coupled with hemoglobin, the blood may also be buffered through other plasma proteins. Which of the following is true? A. A shift in the pH can alter the tertiary or quaternary structure of the protein, allowing it to buffer the pH by precipitating out of plasma in response to pH shifts. B. The amino acid residues that make up the protein may act as Brønsted acids or bases, reducing any shifts in pH. C. Plasma soluble proteins have enzymatic function allowing them to sequester hydronium ions from the blood inside membrane-bound organelles in the podocytes lining the capillaries. D. In the presence of altered pH, any plasma-soluble proteins will undergo either acid- or base-catalyzed cleavage, thus depleting the acid or base causing the disruption to blood pH.
B is correct. The amino acids that make up a protein may include many acidic or basic side chain groups. Those side chains can either release or absorb protons, allowing them to help buffer the blood through action as a Brønsted-Lowry acid or base. A: While changes in pH can denature proteins, this choice is wrong in reference to precipitating out of the plasma. That would create serious biological harm by potentially blocking circulatory vessels. Also, the buffering happens through controlling the pH, not through precipitation as the choice says. C, D: These choices are factually false.
Eosinophils play a major role in the promotion of the inflammatory response, which is marked by increased vasodilation, blood vessel permeability, and pain. The inflammatory response is classified as: A. a function of the adaptive immune system. B. a function of the innate immune system. C. a function of passive immunity. D. a function of antibody-mediated immunity.
B is correct. The innate immune system includes nonspecific immune responses. In other words, its responses provide general protection, rather than protection against specific pathogens that have been previously encountered and "remembered." Inflammation is a function of the innate immune system, as it serves as a general response to infection or injury. A: Adaptive immunity is directed against particular pathogens to which the body has been previously exposed. These pathogens are "remembered" by specialized cells, leading to a quicker response in the case of a second exposure. Inflammation is a not a specific protective mechanism, so this choice does not fit. C: Passive immunity involves the transfer of antibodies from one individual to another. This is not relevant to the question. D: Inflammation does not directly relate to antibodies, which are specific proteins produced by B lymphocytes.
A typical eosinophil would be expected to differ from an erythrocyte in that the eosinophil: A. would not contain a membrane-bound nucleus. B. would contain a membrane-bound nucleus. C. would not be expected to be found in human blood. D. would be expected to be found in human blood.
B is correct. Unlike almost all cells in the human body, erythrocytes (red blood cells) do not contain DNA or a membrane-bound nucleus. This allows these cells to have more space for oxygen-carrying hemoglobin molecules. Thus, eosinophils differ from erythrocytes in that eosinophils do contain nuclei and DNA. A: This option is backwards. Be sure to read the question stem carefully! C, D: Both eosinophils (a type of white blood cell) and erythrocytes (red blood cells) would typically be found in human blood.
What aspects separate single-crossover events from double-crossover events? A. Single-crossover events result in one-way displacement of chromosomal content from one chromosome to another, while double-crossover events always reverse this one-way displacement, resulting in chromosomes identical to the pre-crossover chromosomes. B. Single-crossover events occur during mitosis when a cell splits into two cells, while double-crossover events can only occur during meiosis when a cell splits into four cells. C. Single-crossover events affect only the ends of chromosome arms, while double-crossover events can affect segments in the middle of chromosome arms. D. Single-crossover events only affect one arm of each chromosome, while double-crossover events affect two arms of each chromosome.
C is correct. A double-crossover event is one in which chromosomal arms of homologous chromosomes cross over in two different places along the arm. This results in a section in the middle of each chromosome being exchanged. A: A double-crossover event occurs in two different places along the chromosome arm, resulting in a segment exchange, not necessarily a reversal of the original crossover. D: Both single- and double-crossover events will only affect one arm of each chromosome.
Drinking ocean water is ultimately fatal to a human because: A. the water has a very low solute concentration relative to the body's cells, resulting in cell swelling and death. B. the kidney must work very hard to excrete the excessive levels of bivalent ions in the ocean water, causing kidney failure. C. the water has a very high solute concentration relative to the body's cells, resulting in cell shrinkage and death. D. ocean water contains toxic levels of environmental pollutants that can damage organs or cause cancer, leading to death.
C is correct. Ocean water contains very high levels of sodium and chloride ions. This creates an osmotic gradient that pulls water out of the body's cells, leading to dehydration. A: Water would flow into the cells if the ocean water had a low solute concentration. B: The ions in the ocean are primarily monovalent (Na+ and Cl-), not bivalent. D: Nothing in the question suggests that the person is specifically drinking from a part of the ocean that is polluted.
Nitrogen primarily exists in the atmosphere as a diatomic gas. Which of the following is true about this form of nitrogen? A. The presence of a lone pair of electrons on each nitrogen atom in the molecule allows it to act as a strong Lewis base. B. The triple bond of electrons creates a region of high electron density that allows N2 to be very reactive as a nucleophile. C. Diatomic nitrogen gas is relatively inert and can be used as the atmosphere in laboratory conditions to prevent unwanted side reactions. D. Atmospheric nitrogen reacts spontaneously with carbon dioxide, which keeps atmospheric CO2 levels at a relatively low 0.04% (on a molar basis) of the atmosphere.
C is correct. The MCAT will expect you to be familiar with N2 as a very inert gas. It makes up approximately 80% of the air you breathe, yet has no significant chemical reactions with your lungs - or with anything other than nitrogen-fixing plants. This information implies that nitrogen is very inert (unreactive). As such, it would serve as a good artificial atmosphere when working with reagents that might react with oxygen or other gases. A, B: Although N2 does have relatively high electron density, the strength and stability of the triple bond means it doesn't react strongly as many other compounds. Nitrogen and carbon dioxide do not react readily with each other.
A follow-up study by the researchers showed that the Y chromosome experiences a much higher mutation rate than the X chromosome. This is likely due to the fact that: A. the X chromosome encodes more proteins responsible for genetic repair than the Y chromosome. B. the Y chromosome is much smaller and therefore more susceptible to mutation. C. the Y chromosome goes through many more divisions than the X chromosome in a typical population. D. the Y chromosome has significantly fewer genes, making mutations more evident.
C is correct. The Y chromosome is inherited through sperm, which undergo many divisions during gametogenesis. Each cellular division makes mutation more likely. Additionally, sperm develop in the testes, which exist under unusually oxidative conditions; this also promotes mutation. The X chromosome undergoes fewer total replications in the population due to the limited number of replicative rounds during oogenesis. Additionally, females have crossing-over events to restore function to damaged X chromosomes. A: While this may be true, proteins responsible for gene repair are not exclusive to their chromosome of origin. B: The size of the Y chromosome does not contribute to its susceptibility to mutation. D: While the Y chromosome has a very limited number of genes, the majority of the chromosome is "junk" DNA; therefore, mutation of the Y chromosome is actually relatively unlikely to have a phenotypic effect.
aneuploidy in Alzheimer's disease may result from disruption of the mitotic spindle. This apparatus is composed of what structures? A. Microfilaments B. Intermediate filaments C. Microtubules D. Myosin filaments
C is correct. The mitotic spindle is composed of microtubules, which are cytoskeletal components made of tubulin.
Another possible method of separating 2-methylundecanal and 2-methylundecanoic acid could be based on: A. their differences in the rotation of plane-polarized light. B. a mass spectrometry analysis. C. an extraction based on their differing solubilities. D. the very different scent profiles of each molecule.
C is correct. This question asks about separating an acidic compound from a fairly neutral one. The carboxylic acid group, if deprotonated, will be much more soluble in water than the aldehyde group. Thus, a careful extraction with a dilute weak base (such as a solution of sodium bicarbonate) would separate these two molecules. Remember, the more polar a compound is, the higher its solubility in aqueous solution! A, B, D: These all represent ways to analyze, or gather information about, a substance. Analytic techniques are distinct from separatory techniques, which aim to isolate substances from each other. Note also that the compounds discussed in the question are the compounds mentioned in paragraph 2, not the enantiomers depicted in Figure 1.
If a 65-kg man undergoes a turning acceleration of 5 m/s2 during a running turn, what is the magnitude of force experienced by the foot due to the ground? A. 325 N B. 650 N C. 750 N D. 1075 N
C is correct. To turn while running, the foot must push off the ground, applying a shearing force while simultaneously supporting the weight of the body. When faced with problems like this, always consider the forces involved. Here, we must account for the normal force exerted by the ground on the foot; this is a vertical force which occurs as a result of the runner's weight. We also must consider the acceleration force, which (since the person is turning) is horizontal. These two force vectors are perpendicular and will form a right triangle. We are looking for the overall force experienced, so we must find the hypotenuse. Specifically, we need to find the hypotenuse of a triangle with legs of Fnormal = mg = (65 kg)(10 m/s2) = 650 N and Fturning = (65 kg)(5 m/s2) = 325 N. The combined vector will be bigger than either component alone, so eliminate choices A and B. This calculation can be approximated as: √(3002 + 7002) = √(90000 + 490000) = √(580000) = √(58 x 104) = (√58) x 102 The square root of 58 falls between 7 and 8, so the overall value of our answer falls between 700 and 800, meaning that choice C must be correct (the actual value is 761).
Troponin isoenzymes are used as an alternative biomarker in the diagnosis of heart attacks. In which of the following muscle types does the troponin complex function in contraction? I. Skeletal muscle II. Smooth muscle III. Cardiac muscle A. I only B. I and II only C. I and III only D. I, II, and III
C is correct. Troponin is a complex of three proteins (troponin I, troponin C, and troponin T) required for muscle contraction in skeletal muscle and cardiac muscle, but not smooth muscle. I and III are true; II is not. Choice C is then correct.
In a population of Amish people, the frequency of the recessive autosomal allele for polydactyly is 1.2%. What percent of the population are carriers for this gene? A. 0.0144% B. 1.19% C. 2.37% D. 97.6%
C is correct. We can use the Hardy-Weinberg equation to solve this question. Remember that the total number of alleles in the population has to add up to 1: A + a = 1 And the total number of genotypes in the population must also add up to 1: AA + 2Aa + aa = 1 We're told that a = 0.012. By the first equation above, A = 0.988. The CARRIERS are the heterozygotes with the genotype Aa. Their frequency is: 2Aa = 2 x 0.988 x 0.012 = 0.988 x 0.024 At this point we've got what looks like a tough calculation to do, so we should probably back up and start estimating. Our calculation is telling us to take 98.8% of 0.024, so our answer is going to be really close to 0.024 (since 100% of any number is just that number itself [e.g. 100% of 56.7 is 56.7]). If we look at the answer choices, the only one that's remotely close is 0.0237, or 2.37%. A: This is the percent of the population with genotype aa, or 0.0122. B: This is half the right answer, and a trap for folks who forget that the heterozygote population is calculated as 2Aa. D: This is the percent of the population with genotype AA, or 0.9882.
acetate
C2H3O2-
oxalate
C2O42-
Cyanide
CN-
Cyanate
CNO-
carbonate
CO32-
Would Fluorine be stabilizing or destabilizing as a group on a carbocation?
Carbocations are typically unstable due to their high concentration of positive charge. If a substituent can contribute some electron density to the carbocation, this positive charge will be "balanced" and the species stabilized. In contrast, electron-withdrawing substituents pull even more electron density away from the carbocation, decreasing its stability further. Since fluorine is destabilizing, and since it is highly electronegative, we can deduce that it must be electron-withdrawing.
hypochlorite
ClO-
chlorite
ClO2-
chlorate
ClO3-
perchlorate
ClO4-
dichromate
Cr2O72-
chromate
CrO42-
when does crossing over occur?
Crossover events only occur during meiosis, not during mitosis.
If the students perform an enzyme inhibition assay using captopril, a competitive inhibitor, which of the following changes in the kinetic parameters of ACE should be expected? A. Vmax decreased; Km unchanged B. Vmax decreased; Km increased C. Vmax unchanged; Km decreased D. Vmax unchanged; Km increased
D is correct. According to the passage, captopril is a "competitive inhibitor of ACE." Competitive inhibitors increase the Km of their associated enzymatic reactions without altering the Vmax value. A: These changes are characteristic of noncompetitive inhibition. B: These changes are characteristic of certain mixed inhibitors.
On average, every gamete has one or two mutations that causes its DNA to differ from the cells of its parent. In other words, every person has, on average, three mistakes dating from their initial zygote self, counting both parents. Based on the passage, how many nucleotides does this suggest are found in a complete diploid set of human chromosomes? A. One billion B. One and a half billion C. Two billion D. Three billion
D is correct. Given a total of three mistakes for the complete diploid set, with one mistake per billion copies (according to paragraph two of the passage), that must mean that three billion nucleotides were copied. Indeed, this is approximately correct as an estimate of the size of a complete set of human DNA. A, B, and C are a result of improper calculations.
he liquid remaining in the round-bottom flask at the end of the procedure was most likely: B. 2-methylundecanal. D. 2-methylundecanoic acid.
D is correct. In distillation procedures, the component with the lower boiling point boils off first, leaving the remaining components in the original flask. Thus, this question is basically asking which component has the higher boiling point. The hydrogen bonding created by the carboxylic acid functional group in 2-methylundecanoic acid would lead to a vastly higher boiling point than the aldehyde (over 300ºC compared to 170ºC).
Several samples are analyzed for nucleotide composition. Which of the following compositions most likely represents a single-stranded piece of DNA? A. 17% A, 17% T, 33% G, 33% C B. 29% A, 14% U, 11% G, 46% C C. 4% A, 4% U, 46% G, 46% C D. 12% A, 12% T, 30% G, 46% C
D is correct. In double-stranded DNA, all of the A's are paired with T's and all of the G's are paired with C's. Only choice D breaks this rule and so must be single-stranded instead of double-stranded. A: This most likely represents double-stranded DNA and can be eliminated. B, C: U is only found in RNA.
Vinblastine is a microtubule-disrupting drug that inhibits tubulin polymerization. Which of the following processes would be directly inhibited upon vinblastine treatment? I. Phagosome transport to the lysosome II. Mitosis III. Meiosis IV. Electron transport You can choose more than one!!!
D is correct. Microtubules are used in vesicle transport and in both mitosis and meiosis. However, they have no direct role in electron transport.
A child is rolling his toy car (with mass = m) down a ramp. The coefficient of static friction between the car and the ramp is 0.25. When the car is halfway down the ramp, the child pushes down on the car, halting it. What is the minimum force the child must apply to keep the car from starting to roll down the ramp? A. mg sin θ B. 0.25 mg cos θ + mg sin θ C. mg sin θ - 0.25 mg cos θ D. [(mg sin θ) / 0.25] - mg cos θ
D is correct. The gravitational force pulling the car down the ramp is mg sin θ. (This is the typical value for the gravitational force that acts on an object to drag it down an incline.) To stop the car from sliding down the ramp, we must have an equal and opposite frictional force. Remember, frictional force is equal to the product of the appropriate coefficient of friction and the normal force. Ff = mg sin θ Ff = μF,N, Ff = 0.25 x F,N, = mg sin θ The car itself has a mass m and thus generates a normal force of F,N, = mg cos θ. Again, this is the standard value for the normal force on an object positioned on an inclined plane. From this information alone, we may be tempted to pick C, which is the difference between the gravitational force and the frictional force. If the child were pushing the car upwards along the plane in a parallel fashion, this choice would be correct. However, the child is actually pushing down on the car, perpendicular to the plane. Thus, the force exerted by the child (Fa) will add to the force created by the mass of the car itself and alter the value for the normal force. Our total FN = mg cos θ + Fa. Substituting, we get: Ff = 0.25 x (mg cos θ + Fa) = mg sin θ mg cos θ + Fa = (mg sin θ) / 0.25 Thus, the force with which the child must push down on the car is Fa = [(mg sin θ) / 0.25] - mg cos θ.
It is found that small to moderate doses of I-131 are more likely to cause thyroid cancer than extremely large doses. Which of the following provides the best explanation for this observation? A. Small doses of I-131 are not absorbed by the body. B. Extremely large doses of I-131 are not absorbed by the body. C. The amount of radiation released by large doses of I-131 is enough to damage DNA without killing cells. D. The amount of radiation released by small doses of I-131 is enough to damage DNA without killing cells.
D is correct. The passage states that radionuclides can damage DNA and kill cells. Presumably, it would take a larger dose to kill a cell than to damage its DNA in a non-lethal manner. Therefore, smaller doses may be able to cause cancer in tissues whose cells would be killed by larger amounts.
Assuming that ECP (Eosinophil Cationic Protein) was named for its electrical charge at physiological pH, which of the following must be true? A. The primary structure of ECP contains more acidic residues than uncharged residues. B. The primary structure of ECP contains more basic residues than uncharged residues. C. The primary structure of ECP contains more acidic residues than basic residues. D. The primary structure of ECP contains more basic residues than acidic residues.
D is correct. The passage tells us that ECP is the abbreviation for eosinophil cationic protein. From the question stem, we can assume that ECP must be cationic at physiological pH (7.4). At this pH, a typical protein will have a -1 charge on its deprotonated carboxylic acid terminal and a +1 charge on its protonated amino terminal, for a total of 0 net charge from its termini. If ECP is cationic, then, its positive charge must be due to its side chains. Only basic amino acids have the potential to become positively charged; in contrast, acidic amino acids have the ability to become negatively charged. In fact, virtually any acidic residue should have a -1 charge at a pH of 7.4, so for ECP to have a positive net charge, it must have more basic residues than acidic ones.
formal charge equation
Formal charge = # of valence electrons - bonds - dots so for oxygen w/ 3 bonds and a lone pair it would be: 6(valence electrons) - 3 (bonds) - 2(dots/electrons) = +1
Founder effect?
Founder effect explains the loss of genetic diversity when a new population is founded with a small population. While it may explain why there are fewer alleles for each gene, it doesn't explain a loss of genes within a chromosome.
Gene flow?
Gene flow, or the movement of alleles between local populations, adds to genetic diversity and would not result in the loss of genetic information.
dihydrogen phosphate
H2PO4-
sulfuric acid
H2SO4
hydronium
H3O+
phosphoric acid
H3PO4
hydrobromic acid
HBr
bromic acid
HBrO3
acetic acid
HC2H3O2
hydrogen carbonate
HCO3-
hydrochloric acid
HCl
chlorous acid
HClO2
Chloric acid
HClO3
nitric acid
HNO3
hydrogen phosphate
HPO42-
hydrogen sulfate
HSO4-
Homologous structure
Homologous structures are those that have a similar evolutionary history, arising from the same source, even if they now have different functions. Basically: Same evolution, Different function
Which of the following do NOT have proteins with a nuclear localization signal? I. E. coli II. Homo sapiens III. Fungi IV. Archaea can choose more than 1
I and IV E. coli, a bacteria, and archaea do not have nuclei, and thus do not have a need for nuclear localization signal on their proteins. Homo sapiens and fungi are eukaryotes with nuclei.
A ray of white light strikes the surface of water in a beaker. The index of refraction of the water is 1.33 and the angle of incidence is 30º. All of the following are true EXCEPT: I. the angle of reflection is 30º. II. the angle of refraction is 30º. III. total internal reflection will result, depending on the critical angle.
I is true; the angle of incidence always equals the angle of reflection II: This is false, as light entering a more dense medium will bend towards the normal. With an angle of incidence of 30º, the angle of refraction must be less than 30º. III: This is false because total internal reflection can only result when a ray of light begins in a higher-index material and reaches a boundary with a lower-index one (e.g. starting in water and moving towards air). Here, the light ray started in air (n = 1) and moved into water (n ∼ 1.3), making total internal reflection impossible. SO: answer = II and III
talk about redox potential of the Electron transport chain... (can unstar this if it doesn't make any sense, but should look it up, is section 1 question 40, free practice test on next step mcat)
In the electron transport chain, electrons are passed from species with less positive reduction potential to those with more positive reduction potential. O2 serves as the final electron acceptor of the electron transport chain and must possess a standard reduction potential that is more positive than any other acceptor in the chain. Of the standard reduction potentials mentioned in the passage, the greatest is that of Fe3+/Fe2+ in cytochrome c, for which E° = +0.22 V. Only choice A exceeds this value.
role of microfilaments and intermediate filaments?
Microfilaments (actin filaments) are smaller than either microtubules or intermediate filaments; they play an important role in muscle contraction, but they are not found in the mitotic spindle. B: Intermediate filaments play a role in cell adhesion; they do not make up the mitotic spindle.
permanganate
MnO4-
ammonia
NH3
ammonium
NH4+
nitrite
NO2-
nitrate
NO3-
Which, if any, of the following represent differences between growth of cell cultures containing cytoplasmic organelle disease versus genetic transmission of nuclear DNA from parent to offspring? I. Meiosis maintains the genetic integrity of growing cell cultures, while mitosis arranges nuclear DNA into a form which is transmissible to offspring. II. The process of arranging nuclear DNA in preparation for transmission to offspring is a cyclical process, while the growth of cell cultures is carried out by a noncyclical process. III. Growth of cell cultures requires transformation of diploid cells into haploid cells, while genetic transmission of nuclear DNA to offspring requires transformation of haploid cells into diploid cells.
None are correct! Here, (I) is inaccurate; it is mitosis, not meiosis, which is responsible for cell culture growth and for maintaining genetic integrity between parent cells and daughter cells within a cell culture. Meiosis does not maintain genetic integrity, as one of its functions is genetic recombination in order to create genetic diversity in offspring. (II) is inaccurate because arranging nuclear DNA in preparation for transmission to offspring - in other words, the creation of gametes through meiosis - is a noncyclical process. It happens once and terminates in gamete creation (either sperm cells or ova). Growth of cell cultures, though, occurs through mitosis, which is a cyclical process. (III) is also inaccurate; mitosis, which causes growth of cell cultures, always results in cells with a diploid number of chromosomes. Also, during meiosis, which allows genetic transmission of nuclear DNA to offspring, the number of chromosomes is reduced from diploid to haploid, not increased from haploid to diploid. Hence, none of the three statements are correct.
peroxide
O22-
Hydroxide
OH-
phosphite
PO33-
phosphate
PO43-
O2 and Co2 exchange in lungs is active or passive?
Passive! O2 and CO2 are not actively transported by lung tissue
thiosulfate
S2O32-
thiocyanate
SCN-
Sulfite
SO32-
Sulfate
SO42-
what do enzymes change (ie equilibrium constant, activation energy, spontenaeity, etc). what don't they change?
The catalytic action of enzymes changes the activation energy, and thus the reaction rate, of reactions in which enzymes are involved. The presence of various ATP-hydrolyzing enzymes in the body, and the absence of such enzymes in the laboratory, could account for the observed difference in reaction rates. Enzymes do not alter the equilibrium constant for a given reaction, nor do they alter equilibrium concentrations of reactants and products. -Because enzymes are unable to alter the equilibrium of a reaction, they also cannot change the spontaneity of a reaction, which is determined by the free energy change associated with the reaction.
3 results of PPP?
The generation of reducing equivalents, in the form of NADPH, used in reductive biosynthesis reactions within cells (e.g. fatty acid synthesis). Production of ribose 5-phosphate (R5P), used in the synthesis of nucleotides and nucleic acids. Production of erythrose 4-phosphate (E4P) used in the synthesis of aromatic amino acids.
Pentose phosphate pathway
a metabolic pathway parallel to glycolysis. It generates NADPH and pentoses (5-carbon sugars) as well as ribose 5-phosphate, the last one a precursor for the synthesis of nucleotides. While it does involve oxidation of glucose, its primary role is anabolic rather than catabolic.
IR spectroscopy ranges for -OH and C=O?
approx. 3300 cm-1 for -OH (alcohol functional group) Approx. 1700 cm-1 for C=O (carbonyl functional group)
If a rxn is spontaneous, is the dG positive or negative? Is the Keq less than 1 or more than one?
dG = negative (system is losing energy; products have less energy than the reactants) Keq = MORE than one; (Keq is roughly the ratio of products over reactants; if the rxn is spontaneous, there will theoretically be more products than reactants SO: products/reactants > 1;;; ALSO: dG = -ln(Keq); so if dG<0, than Keq>1)
Given the structure of the molecule on the other side, is it hydrophilic or hydrophobic?
hydrophobic; is a nonpolar compound with lots of electron delocalization, meaning that it is poorly soluble in polar (hydrophilic) solvents like water
Besides the mother having a lot of mitochondria; why else does the sperm not contribute that many mitochondria
sperm mitochondria's association with ubiquitin protein which causes them to be targeted for destruction post-fertilization.
Passive immunity is what?
the short-term immunity that results from the introduction of antibodies from another person or animal. Passive immunity is the transfer of active humoral immunity in the form of ready-made antibodies. Passive immunity can occur naturally, when maternal antibodies are transferred to the fetus through the placenta, and it can also be induced artificially, when high levels of antibodies specific to a pathogen or toxin (obtained from humans, horses, or other animals) are transferred to non-immune persons through blood products that contain antibodies, such as in immunoglobulin therapy or antiserum therapy
When does total internal reflection occur?
total internal reflection can only result when a ray of light begins in a higher-index material and reaches a boundary with a lower-index one (e.g. starting in water and moving towards air)
2 phases of the PPP and where they occur
two distinct phases in the pathway. The first is the oxidative phase, in which NADPH is generated, and the second is the non-oxidative synthesis of 5-carbon sugars. For most organisms, the pentose phosphate pathway takes place in the cytosol;
What does Hess' law state?
△Hreaction = ∑ Hf products - ∑ Hf reactants the sum of standard enthalpy of formation for products - the sum of standard enthalpy of formation for reactants = the change in heat of reaction (BASICALLY: Products - reactants) -NOTE: The standard enthalpy of formation of any pure element in its standard state (e.g. C (graphite), N2 (g), O2 (g), etc....) is zero so like: H2 + 1/2O2 => H2O; you only have to consider H2O (i think?)
