Bio/Biochem Questions

Muscle fibers are composed of small contractile units called sarcomeres. During contraction, which of the following occurs within a sarcomere? I. Myosin filaments shorten. II. Actin filaments shorten. III. Overlap between actin and myosin filaments increases. A. I only B. III only C. II and III only D. I, II, and III

B. Items I and II are false: During contraction, neither myosin nor actin filaments get shorter. The overlap between them increases to make the sarcomere shorter (eliminate choices A, C and D). Item III is true: The overlap between the fibers increases as part of contraction.

Muscle fatigue occurs in humans because: A. there is a buildup of ethyl alcohol. B. of lactic acid accumulation. C. muscle cells have shifted from anaerobic respiration to aerobic respiration. D. there is a shortage of actin and myosin in the cell.

B. Muscle fatigue occurs when lactic acid levels in the muscle have risen due to anaerobic respiration (B is correct). Ethyl alcohol is the end product of fermentation when yeasts reduce pyruvate; in muscles pyruvate is reduced to lactic acid (A is wrong). If muscles had shifted to aerobic respiration they would have more ATP and would not become fatigued (C is wrong), and actin and myosin levels in the cell do not change (D is wrong).

Myasthenia gravis is an autoimmune neuromuscular disorder caused by the production of antibodies against acetylcholine receptors. Which of the following explains why individuals who are affected with this disease experience muscle weakness? A. The antibodies bind to acetylcholine receptors and stimulate a downstream response. B. The antibodies bind to acetylcholine receptors and prevent endogenous acetylcholine from binding. C. The antibodies bind to acetylcholine receptors and cause helper T cells to destroy the receptors. D. The antibodies bind to acetylcholine receptors and directly degrade the receptors

B. Myasthenia gravis manifests as muscle weakness because when the antibodies are bound to ACh receptors, endogenous acetylcholine cannot bind. Thus, muscles cannot contact despite the signal from the motor neuron (choice B is correct). If the antibodies stimulated a downstream response, the muscles would contract rather than feel weak (choice A is wrong). Antibodies can mark antigens for destruction by macrophages and other phagocytes, but not by helper T cells (choice C is wrong), and antibodies are not enzymes; they cannot directly degrade their antigens (choice D is wrong).

When a striated muscle cell metabolizes glucose in the complete absence of O2, which of the following substances is NOT produced in a significant amount? A. Pyruvic acid B. Glucose-6-phosphate C. Lactic acid D. Acetyl-CoA

D. In the absence of oxygen, muscle cells rely on lactic acid fermentation to produce ATP. In this process, glucose is first phosphorylated to glucose-6-P (choice B can be eliminated) and ultimately converted to pyruvic acid (choice A can be eliminated). However instead of being decarboxylated to acetyl-CoA, the pyruvic acid is reduced to lactic acid (choice C can be eliminated and choice D is correct).

Epilepsy may result in motor seizures due to massive synchronous firing of neurons in a small area of the cerebral cortex (the epileptic focus). Excitation spreads from the focus, involving an increasingly larger area of the cortex. A drug for the treatment of epilepsy would be most effective if it caused which of the following changes in the epileptic focus? A. An increase in the neuron-firing threshold B. An increase in extracellular Na+concentration C. A decrease in axon-membrane permeability to negative ions D. A decrease in the length of the depolarization stage

A drug that would best treat epilepsy, as described, should reduce the activity of the neurons in the epileptic focus, which are excitatory neurons based on their effect throughout the cortex. The answer to this question is A because increasing the threshold required to generate an action potential would decrease the chance that individual neurons would fire, thus reducing the overall amount of excitation that spreads from the epileptic focus throughout the cortex.

Gastrin is a peptide hormone that stimulates secretion of gastric acid and acids in gastric motility. Which of the following is a true statement about gastrin? A. It is released by G cells in the antrum of the stomach (the portion of the stomach adjacent the pyloric valve). B. It enters target cells by facilitated diffusion through ATPase carrier channels in the plasma membrane. C. It is a macromolecule that contains peptide, amide, ester and hydrogen bonds. D. It is capable of diffusing in the blood plasma due to its extensive London dispersion forces but minimal van der Waals forces.

A. Choice A is a true statement and the correct answer. Peptide hormones do not enter target cells, and carrier channels that allow facilitated diffusion would not have ATPase activity; this would make them active instead of passive transporters (choice B can be elminated). Peptides do not contain ester bonds (choice C can be eliminated), but do contain the other bonds listed. 'van der Waals forces' is an umbrella term that includes dipole forces, hydrogen bonding, and London forces. If a molecule has extensive London dispersion forces, it cannot have minimal van der Waals forces (choice D can be eliminated).

(+)-Ginkgolide B is a diterpenoid trilactone with six five-membered rings, and is extracted from the root bark and leaves of the Ginkgo biloba tree. This molecule has blood-brain barrier permeability, the molecular formula C20H24O10 and a molecular weight of 424.4 g/mol. This molecule must contain: A. four isoprene units, additional functional groups and is able to perform simple diffusion through endothelial cells connected by tight junctions. B. four isoprene units, no additional functional groups and is able to perform simple diffusion through endothelial cells connected by tight junctions. C. four terpene units, additional functional groups and is able to perform simple diffusion through neurons connected by tight junctions. D. four isoprene units, additional functional groups and is able to perform simple diffusion through endothelial cells connected by gap junctions.

A. Diterpenoids contain four isoprene units (eliminate choice C) and additional functional groups (eliminate choice B). The blood-brain barrier is a highly selective permeability barrier that separates the circulating blood from the central nervous system. It is formed by endothelial cells (another reason to eliminate choice C), which are connected by tight junctions (eliminate choice D). Choice A is correct because it accurately describes the composition of the molecule.

Like α(1→4)-glycosidase, which of the following enzymes acts to cleave α(1→4) linkages between glucose molecules? A. Glycogen phosphorylase B. Glucose-6-phosphatase C. Phosphoglucomutase D. Pyrophosphatase

A. Glycogen phosphorylase catalyzes the rate limiting step of glycogenolysis—the release of glucose-1-phosphate from glycogen via its action on terminal α(1→4) linkages. This makes choice A true and the correct answer. Glucose-6-phosphatase hydrolyzes glucose-6-phosphate, resulting in the release of a phosphate group and free glucose for export from the cell. Choice B is false and an incorrect answer. Phosphoglucomutase is involved in both glycogenolysis and glycogenesis by interconverting glucose 1-phosphate and glucose 6-phosphate, thus choice C is false and the incorrect answer. Pyrophosphatases are members of the acid hydrolase family of enzymes that cleave diphosphate bonds. Choice D is false and an incorrect answer.

During aerobic respiration glycolysis takes place: A. in the cytosol. B. along the inner mitochondrial membrane. C. in the outer mitochondrial membrane. D. within the mitochondrial intermembrane space.

A. Glycolysis always occurs in the cytoplasm (so choice A is correct). Electron transport occurs along the inner mitochondrial membrane (choice B is wrong), neither glycolysis nor the electron transport chain occur in the outer mitochondrial membrane (choice C is wrong), and the proton gradient from electron transport is created in the intermembrane space (choice D is wrong).

Cytochalasins interrupt actin polymerization by binding to the growing end of actin filaments. These drugs would most directly affect which of the following processes? I. Cell crawling movements II. Muscle contraction III. Mitotic spindle formation A. I only B. III only C. I and II only D. I, II, and III

A. Item I is true: Actin filaments are involved in cell contraction functions, like pseudopod formation, cytokinesis, and muscle cell contraction. Pseudopod formation (cell crawling) requires actin polymerization; however during muscle cell contraction actin filaments are not polymerized (eliminate choice B). Item II is false: Existing filaments merely slide over myosin filaments (eliminate choices C and D). Item III is false: Mitotic spindles are made from microtubule polymerization (choice A is correct).

A frameshift mutation involving insertion of one base pair in an exon may result in: I. a protein product of a lower molecular weight. II. a loss of meiotic recombination of the gene. III. premature termination by RNA polymerase II. A. I only B. III only C. I and II only D. II and III only

A. Item I is true: a frameshift mutation will alter the reading frame of all subsequent codons in the gene and can cause a stop codon to be present where none was before, shortening the length of the protein product (choices B and D can be eliminated). Note that since neither of the reamaining choices includes Item III, it must be false and we can focus on Item II. Items II and III are both false: a frameshift mutation only affects translation, not recombination or transcription since the enzymes carrying out these functions do not discriminate codons (choice C can be eliminated and choice A is true).

Which of the following statements is/are true with regard to the mechanics of skeletal muscle contraction? Reduced ATP levels could result in high intracellular calcium and persistent contraction Calcium binds to calmodulin and allows actin and myosin to interact The shorter the sarcomere length, the stronger the contraction A. I only B. I and II only C. II and III only D. I, II, and III

A. Item I is true: in the absence of ATP, not only will the myosin and actin filaments be unable to disconnect, Ca2+ active transporters will not be able to return Ca2+ to the sarcoplasmic reticulum (choice C can be eliminated). Item II is false: calcium binds to troponin (not calmodulin) in skeletal muscle (choices B and D can be eliminated and choice A is correct). Note that Item III is false: the length-tension relationship states that there is an optimum sarcomere length to ensure the most forceful contraction. While it is true that shorter sarcomere length translates to a stronger contraction up to a point, too much overlap causes the actin and myosin filaments to bump together and reduces the strength of contraction.

Chloramphenicol is an antibiotic with an extremely lipid-soluble molecular structure. It binds to the large subunit of 70S ribosomes and prevents peptidyl transferase activity. Chloramphenicol is effective against almost all bacteria; however, it is also considerably toxic to humans. This is most likely because chloramphenicol: A. is able to diffuse into the mitochondria and affect mitochondrial protein synthesis. B. is able to diffuse into the nucleus and block transcription of rRNA. C. affects both human and bacterial ribosomes. D. directly disrupts the cell membrane.

A. Mitochondria have DNA and transcribe and translate a few proteins independently. The mitochondrial system is almost identical to that of the prokaryotes; this is where the theory that mitochondria are ancient bacterial parasites that live symbiotically within eukaryotes comes from. Chloramphenicol is able to diffuse into mitochondria in concentrations high enough to disrupt mitochondrial protein synthesis, ultimately causing the death of the eukaryotic cell (choice A is correct). Human (eukaryotic) ribosomes are 80S ribosomes, so they would not be affected by a drug that inhibits 70S ribosomes (choice C is wrong), and there is no information given to suggest that they disrupt cell membranes or block transcription (choices B and D are wrong).

Retinal isomerase is used to alter the conformation of retinal. Retinal isomerase is most likely manufactured by: A. dehydration synthesis. B. hydrolytic synthesis. C. oxidation. D. glycosidic anabolism.

A. Retinal isomerase, an enzyme, is a protein, and proteins are synthesized by the condensation of amino acids. Condensation involves dehydration.

Which of the following techniques would be LEAST likely to detect a Robertsonian translocation? A. Southern blot B. Routine karyotype C. High-resolution karyotype D. Fluorescent in situ hybridization of chromosomes

A. Southern blots are used to analyze extracted DNA for a potential DNA segment and would not provide detailed information with regard to chromosome structure. In other words, you would know if a gene was present, but would not know if it was on the correct chromosome or not (choice A would be the least likely to detect a translocation and is the correct answer choice). Karyotyping (both routine and high-resolution) is capable of detecting a missing chromosome (choices B and C can be eliminated). Fluorescent in situ hybridization uses fluorescent DNA probes to bind to chromosomes and could detect a translocation by detecting a gene on an unexpected chromosome (choice D can be eliminated).

The amphoteric character of amino acids best explains their ability to: A. form dipolar ions. B. form peptide bonds. C. contribute directly to a protein's secondary structure. D. dissolve in nonpolar solvents.

A. The amphoteric character of amino acids describes their ability to do two things: accept a proton or donate a proton. In other words, amino acids can act as either an acid or a base. When the amino portion of an amino acid deprotonates its own carboxylic acid, a dipolar ion (or zwitterion) forms. This is a direct result of the amphoteric character of the amino acid.

Diabetes insipidus, a disease of ADH deficiency, can be caused by destruction or dysfunction of the supraoptic and paraventricular nuclei of the hypothalamus. One symptom of a person with this condition might be: A. an inability to produce concentrated urine. B. elevated plasma osmolarity and blood pressure. C. an increased heart rate. D. elevated blood glucose levels.

A. The function of ADH is to increase the permeability of the collecting duct of the nephron to water. If ADH is present, water can be reabsorbed from the urine, thereby concentrating the urine and preventing dehydration of the body. An absence of ADH would lead to an inability to concentrate the urine (A is correct). Excreting a dilute urine would certainly increase plasma osmolarity, but this would not increase blood pressure. Plasma volume would tend to decrease due to the excess loss of water, and blood pressure would drop (B is wrong). ADH has no effect on heart rate or blood glucose levels—do not confuse this disorder with diabetes mellitus (C and D are wrong).

The hormones of the adrenal medulla bind to receptors: A. on the cell membrane. B. on the endoplasmic reticulum. C. on the nuclear membrane. D. inside the nucleus.

A. The hormones of the adrenal medulla are epinephrine and norepinephrine, both amino acid derivatives. Therefore, they cannot cross the cell membrane and must bind to a receptor on the cell surface (choice A is correct, and choices B, C, and D are wrong).

would produce the solution with the highest pH? A. Na2CO3 B. NH4Cl C. NaCl D. AlBr3

A. The most basic salt will produce the solution with the highest pH. The carbonate ion from Na2CO3 is the conjugate base of bicarbonate, which is itself weakly basic in aqueous solution; choice A is the most basic salt listed. NaCl is a neutral salt as neither of its ions are reactive with water (eliminate choice C). Both NH4Cl and AlBr3 are acidic salts because both cations will interact with water to increase the [H+] in solution (eliminate choices B and D).

Which of the following can be deduced about a ketotetrose with an absolute configuration of S? A. It is a D sugar. B. It is an L sugar. C. It is levorotatory. D. It is dextrorotatory.

B. A ketotetrose will have a ketone, four carbons, and therefore one chiral center. When drawn in a Fischer projection, the position of the OH group on the chiral carbon determines whether the sugar is D or L. The OH group is on the left side of the molecule in an L sugar, and on the right side of a D sugar. The four substituents of the chiral center are prioritized based on atomic number as indicated below. When connected from 1 to 2 to 3, a clockwise arc generally indicates an R configuration and a counterclockwise arc indicates an S configuration provided the molecule is viewed from the direction that puts the #4 priority group in the back. However, since the fourth priority H is coming out of the page in a Fischer projection of a sugar, the opposite configurations must be assigned. Therefore, the molecule with the S configuration shown is an L sugar since the OH group is on the left side of the molecule (eliminate choice A). The way a compound rotates plane-polarized light cannot be determined solely from its absolute configuration, so without directly measuring this property, we cannot be sure if the molecule is dextrorotatory or levorotatory (eliminate choices C and D).

Spironolactone is an aldosterone antagonist. Which of the following would be expected in someone who was given spironolactone? A. Increased blood [Na+], decreased urine [Na+], increased blood [K+], decreased urine [K+] B. Decreased blood [Na+], increased urine [Na+], increased blood [K+], decreased urine [K+] C. Increased blood [Na+], decreased urine [Na+], decreased blood [K+], increased urine [K+] D. Decreased blood [Na+], increased urine [Na+], decreased blood [K+], increased urine [K+]

B. Because the actions of aldosterone are being inhibited by spironolactone, we would expect the Na+ concentration in the blood to be lower than normal and K+ concentration in the blood to be higher than normal. Since Na+ is not being reabsorbed from the tubular fluid, the concentration of Na+ in the urine should be higher than normal, and K+ concentration in the urine should be lower than normal since it is not being secreted into the fluid.

Which of the following is the best technique for separating 2-butanol from propanoic acid? A. 1H NMR spectroscopy B. Fractional distillation C. Infrared spectroscopy D. Recrystallization

B. Both 2-butanol, an alcohol, and propanoic acid, a carboxylic acid, have low molecular weights, so are likely liquids at room temperature. Since carboxylic acids have stronger hydrogen bonding interactions than alcohols, these compounds will have different boiling points. Therefore, fractional distillation would be the best method for separating these molecules. Choices A and C can be eliminated because they are not separation techniques. Crystallization separates solid compounds based on their solubilities and would not be a good choice as both molecules possess similar solubility characteristics.

ACE inhibitors are a class of drugs frequently prescribed to treat hypertension. Captopril, a compound that is structurally similar to angiotensin I, was developed in 1975 as the first ACE inhibitor. When patients take Captopril, which of the following is true about the kinetics of their ACE? A. Vmax decreases, Km remains the same B. Vmax remains the same, Km increases C. Both Vmax and Km increase D. Both Vmax and Km remain the same

B. From the question text, it can be deduced that Captopril is a competitive inhibitor of ACE (since its structure is similar to that of the substrate, angiotensin I). Competitive inhibitors do not alter an enzyme's Vmax, since the active site can be saturated by increasing the substrate concentration (choices A and C can be eliminated). Competitive inhibitors do increase the apparent Km of the enzyme, because a higher substrate concentration is required to reach ½ Vmax (choice D can be eliminated).

The effect of glucagon was inhibited in the presence of a cAMP antagonist because: A. cAMP acts as a substrate for glucagon. B. glucagon activates a second messenger system once bound to its receptor. C. glucagon binds to intracellular receptors which require cAMP. D. cAMP is used for the actions of steroid hormones.

B. Glucagon is not an enzyme, thus cAMP cannot be a substrate (choice A is wrong). Glucagon is a peptide hormone that binds to cell surface receptors (choice C is wrong) and activates second messenger systems, notably cAMP. The cAMP antagonist would oppose this effect (choice B is correct). cAMP is not used in the actions of steroid hormones; they bind to DNA and modify transcription (choice D is wrong).

It has recently been shown that TBK1 also controls IgA class switching. Activating point mutations in TBK1 would thus most likely impact which of the following processes? A. Immune response to malignant transformation, due to oncogene activation and/or tumor suppressor loss. B. Protection against microbes in the genitourinary, gastrointestinal, and respiratory systems. C. Targeted immune destruction of cells that have been injected by a lysogenic virus. D. MHC I-mediated activation of immune cells that target donor transplant organs.

B. IgA is a type of immunoglobulin and these molecules are important in the humoral immune response. The humoral immune system protects against many toxins, bacteria and free-floating viruses (choice B is correct). Cell-mediated immunity (involving T cells) protects against abnormal self-cells, such as cells that have undergone malignant transformation (choice A is wrong), cells that have been infected by a virus (choice C is wrong) and donor cells (choice D is wrong).

Which one of the following mutations would be most likely to convert a proto-oncogene into an oncogene? A. Silent mutation B. Missense mutation C. Nonsense mutation D. Deletion mutation

B. In order for the mutation to have the described effect, it must modify the protein without completely eliminating it or destroying its effect. A missense mutation converts a codon for one amino acid into a codon for a new amino acid, resulting in a small change within the protein's primary sequence, and an alteration (but usually not a total elimination) of the protein's function (choice B is correct). A silent mutation converts a codon for an amino acid into a new codon for the same amino acid and has no effect on the protein product (eliminate choice A). A nonsense mutation creates a stop codon out of an amino acid codon, resulting in truncation of the protein and (usually) a loss of its function (eliminate choice C). A deletion mutation eliminates one or more base pairs, altering the reading frame and drastically changing the amino acid sequence of the protein (eliminate choice D).

When the researcher examined the primary structures of two proteins with very different functions, she found that they had 80% of their precise amino acid sequence in common. The most likely interpretation of this finding is that: A. by coincidence, both proteins share an identical segment of amino acid sequences. B. both proteins evolved from a single common ancestor. C. both proteins will eventually evolve until they have 100% of their precise amino acid sequences in common. D. neither protein was subjected to the process of natural selection.

B. It is highly unlikely that by coincidence alone two proteins with vastly different functions would have 80% similarity between their amino acid sequences (A is wrong). Furthermore, there is no driving force (selection) pushing the proteins to evolve to have identical amino acid sequences and therefore similar functions. In other words, since Function 1 is already being carried out by Protein 1, there is no reason for Protein 2 to evolve to perform Function 1 (C is wrong). All cells and their proteins are continually being subjected to natural selection; in fact, this is most likely the reason they evolved to have different functions and 20% dissimilarity in their primary structure (D is wrong, and B is correct).

Viral infection would lead to an increase in which of the following cellular pathways? Aerobic respiration Pentose phosphate pathway cAMP mediated protein kinase activation A. I only B. I and II only C. II and III only D. I, II, and III

B. Item I is true: viral infected cells are undergoing a fair amount of cellular activity, including DNA replication, RNA transcription, and protein synthesis, all of which contribute to an increased need for ATP, and thus oxidative respiration (choice C can be eliminated). Item II is true: the pentose phosphate pathway produces essential precursors for nucleotide synthesis, necessary for both DNA replication and RNA transcription. It's likely that the activity of this pathway would be increased during viral infection (choice A can be eliminated). Item III is false: cAMP mediated protein kinase activity is increased when G protein coupled receptors are bound and activated; there is no reason to assume this in viral infection (choice D can be eliminated and choice B is correct).

In which of the following people would you expect an increase in PTH activity? A. A person whose parathyroid glands have been removed B. A person with increased osteoblastic activity C. A person with increased osteocytic activity D. A person with higher than normal urinary calcium concentration

B. PTH is secreted in response to low serum levels of calcium. PTH increases osteoclastic activity in order to release calcium into the blood. An increase in osteoblastic activity causes new bone formation as calcium is reabsorbed into the bone, thus reducing calcium serum levels, which in turn causes increased PTH activity (choice B is correct). Because PTH is secreted by the parathyroid glands, if they have been removed, then PTH is not found in the blood (eliminate choice A). An osteocyte is a mature bone cell that does not form or remodel bone, and therefore, osteocytic activity would have no influence on calcium homeostasis (choice C is incorrect). PTH also, as the passage states, causes calcium to be reabsorbed from the distal convoluted tubule. If calcium is being reabsorbed from the distal convoluted tubule, then a lower than normal urinary calcium concentration would be expected (eliminate choice D).

The overall equilibrium of the reaction pathway between pyruvate and glucose lies strongly in the direction of pyruvate. What is the most likely explanation for why the reversible enzymes of gluconeogenesis which bypass irreversible steps in glycolysis can function spontaneously in both glycolysis and gluconeogenesis? A. The stoichiometry of gluconeogenesis from pyruvate as a starting material couples the formation of high-energy phosphate bonds and the reduction of NAD+ with the oxidation of pyruvate. B. Formation of phosphoenolpyruvate, glucose and fructose-6-phosphate shift the reaction equilibrium in favor of gluconeogenesis. C. The activity of the bifunctional gluconeogenic and glycolytic enzymes is reversed upon phosphorylation. D. Specific enzymes present only in gluconeogenesis bypass reversible steps common to gluconeogenesis and glycolysis.

B. Production of PEP, glucose, and fructose-6-phosphate by gluconeogenesis-specific enzymes that bypass irreversible steps of glycolysis (choice D is wrong) push the equilibrium of reversible enzymes that function both in glycolysis and gluconeogenesis in the direction of glucose production (choice B is correct). Gluconeogenesis couples the hydrolysis of high-energy phosphate bonds and the oxidation of NADH with the reduction of pyruvate to glucose (choice A is wrong). While some enzymes that regulate the glycolytic and gluconeogenic pathways such as phosphofructokinase-2 and fructose-2,6-bisphosphatase are bifunctional (the nature of their action depends upon their phosphorylation state) this is not true of the majority of enzymes of glycolysis or gluconeogenesis (choice C is wrong).

Which of the following would be true about cis-oleic acid, a monounsaturated fatty acid with the formula CH3(CH2)7(CH)2(CH2)7COOH? A. It will generate approximately 119 ATP after 9 rounds of β-oxidation, followed by the Krebs cycle and the electron transport chain. B. It will generate approximately 119 ATP after 8 rounds of β-oxidation, followed by the Krebs cycle and the electron transport chain. C. It will generate 90 ATP after 9 rounds of β-oxidation, followed by the Krebs cycle and the electron transport chain. D. It will generate 90 ATP after 8 rounds of β-oxidation, followed by the Krebs cycle and the electron transport chain.

B. Recognize the ability to treat this question as a 2x2 elimination. cis-Oleic acid has 18 carbons and thus will undergo 8 rounds of β-oxidation (eliminate choices A and C). This will generate 9 molecules of acetyl-CoA, 8 molecules of NADH and 7 molecules of FADH2 since it is a monounsaturated fatty acid. Each of the 9 acetyl-CoAs will go through the Krebs cycle and this will generate 27 NADH (which will give 67.5 ATP), 9 FADH2 (which will give 13.5 ATP) and 9 GTPs (9 ATP equivalents). This means the acetyl-CoAs alone generate 90 ATP equivalents. Since the NADH and FADH2 made in β-oxidation will generate even more ATP, the total will exceed 90 (eliminate choice D and choice B is correct). The 8 molecules of NADH made in β-oxidation will lead to 20 ATP and the 7 FADH2will give 10.5. Also remember that fatty acid activation (which must occur before β-oxidation) costs the cell two high energy bonds, or ATP equivalents. This means the electron carriers made in β-oxidation will give a net yield of 28.5 ATP. Overall then, cis-oleic acid will generate 118.5 ATP molecules.

In people suffering from color blindness, which of the following types of cells fails to function normally? A. Rod cells B. Cone cells C. Neurons of the optic nerve D. Corneal cells

B. Rods and cones are the photoreceptor cells of the retina. Rods are the ones more sensitive to light, making them important for night vision. Cones are the cells responsible for detecting color and are the cells responsible for the defect involved in color blindness (choice A is wrong, and choice B is correct). The optic nerve is responsible for all vision, making it unlikely that a defect in the optic nerve would affect only color vision and not signals from rod cells (choice C is wrong). Corneal cells are part of the mechanical structure of the eye but are not photoreceptors and are not likely to specifically affect color vision (choice D is wrong).

In an experiment, various concentations of streptomycin were added to a culture of E. coli. At low concentrations of streptomycin, increased amount of misreading of the mRNA was observed. At high streptomycin concentrations, the initiation of protein translation was inhibited. Which of the following statements is consistent with these results? A. Streptomycin inhibits the large subunit (50S) of the ribosome. B. Streptomycin inhibits the small subunit (30S) of the ribosome. C. Streptomycin binds to RNA polymerase and blocks its function. D. Streptomycin binds to the mRNA and targets it for degradation.

B. Since the errors are observed in translation and not in transcription, it is likely the ribosome that is being affected and not RNA polymerase (choice C is wrong). The role of small subunit (30S) of ribosome is to initiate translation by recognizing the first AUG start site and recruiting the tRNAfmet and the large subunit. It is also important for proofreading and maintaining fidelity during the translation process. Therefore, inhibition of small subunit would result in decreased initiation of protein translation and mRNA misreading (choice A is wrong and choice B is correct). If streptomycin bound to mRNA and caused its degradation, we would likely see less protein being made at low concentration (due to less mRNA available), not misreading of the mRNA (choice D is wrong).

Which of the following is the LEAST characteristic of the Krebs cycle? A. Oxaloacetate is combined with acetyl-coA to produce citrate. B. Most of the energy is produced through substrate-level phosphorylation. C. Most of the energy is produced indirectly through high-energy electron carriers. D. 6 NADH, 2 FADH2, and 2 GTP are produced per glucose.

B. Since the question is looking for the item that is LEAST characteristic of Krebs cycle, eliminate choices that DO characterize it. Choice A is true and can be eliminated: oxaloacetate is combined with acetyl-CoA to form citrate (thus why the cycle is sometimes referred to as the citric acid cycle). Choice C is also true and can be eliminated: most of the energy is produced from the production of high-energy electron carriers (NADH and FADH2). Approximately 18 of the 20 ATP that are associated with Krebs are from these high-energy electron carriers (after they are oxidized in the electron transport chain). Choice D is also true and can be eliminated: six NADH, two FADH2, and two GTP are produced in the Krebs cycle per glucose. Choice B is the only false statement (which is the correct answer for this question). Only two of the 20 ATP made from the Krebs cycle are from substrate-level phosphorylation.

Heart murmurs are extra, abnormal sounds (beyond the normal closure of the valves) produced during the cardiac cycle. They can be caused by stenotic (stiffened) valves, or by valves that do not close properly and allow regurgitation. Murmurs are classified as diastolic or systolic depending on when the additional sound is produced. A heart murmur caused by a failure of the AV valves to close properly would most likely be classified as a: A. diastolic murmur, because this would allow flow from the atria to the ventricles during diastole. B. systolic murmur, because this would allow regurgitation of blood from the ventricles to the atria during systole. C. diastolic murmur, because this would allow regurgitation of blood from the arteries to the ventricles during diastole. D. systolic murmur, because this would allow additional blood to flow from the atria to the ventricles during systole.

B. The AV valves close at the beginning of systole to prevent regurgitation of blood into the atria while the ventricles are contracting. If the AV valves failed to close properly, blood from the high-pressure ventricles would flow back into the low-pressure atria during systole and would produce an abnormal murmur. Flow from the atria to the ventricles during systole would be prevented by the pressure gradient (choice D is wrong). Blood normally flows from the atria to the ventricles during diastole, through the open AV valves; this would not produce a murmur (choice A is wrong), and these valves do not separate the ventricles and the arteries so this would not affect blood flow between those regions (choice C is wrong).

If a saturated ammonia solution were placed under a vacuum, what would be the effect on the ammonia? A. The vapor pressure of NH3 would decrease. B. The boiling point of NH3 would decrease. C. The pKb of NH3 would increase. D. The solubility of NH3 would increase.

B. The boiling point temperature of a substance is the temperature at which its vapor pressure equals the atmospheric pressure. If the atmospheric pressure is lowered, the substance will boil when its vapor pressure is lower, i.e., at a lower temperature, making choice B best. The vapor pressure of a substance is not affected by the external pressure (eliminate choice A). The only way of changing the value of an equilibrium constant, K, and therefore a value like pKb, is to change the temperature (eliminate choice C). The solubility of a gas decreases with decreasing pressure (eliminatie choice D).

Rank the following metals in order of increasing oxidation potential: Sr(s) Cs(s) Fe(s) Ca(s) A. III < II < I < IV B. III < IV < I < II C. II < I < IV < III D. IV < I < III < II

B. The greater an element's oxidation potential, the more readily the atom will lose an electron and the more reactive the element is. This can be approximated by the relative ionization energies of the elements. In general, ionization energies increase across a row and decrease down a column of the periodic table. Elements farthest to the left are most reactive, and reactivity often increases down a metal family, especially for groups I and II. Since this is a ranking question, look for an extreme and eliminate answer choices. Cs is at the bottom of the alkali metal family, so should therefore have the largest oxidation potential. Eliminate choices A and C. Looking at the remaining choices, next determine whether Fe or Ca is more likely to lose electrons. Since transition metals are less reactive than alkaline earth metals eliminate choice D.

Sodium laurel sulfate, a common ingredient in hand soap, functions due to the combination of its hydrophilic head (pKa = 1.9) and hydrophobic tail. This amphipathic nature allows for the removal of hydrophobic substances with water. Which of the following would most likely increase the effectiveness of sodium laurel sulfate? A. Benzene B. Dilute aqueous sodium bicarbonate C. Aqueous phosphoric acid (pH = 1.9) D. Lemon juice

B. The head of sodium laurel sulfate (SLS) is a weak acid. It is most hydrophilic when deprotonated and charged. This head can be protonated in lower pH solutions and become less effective. Therefore, addition of a hydrophilic weak base like sodium bicarbonate (NaHCO3) would favor the deprotonated, active form of SLS. Choices C and D are both acidic and eliminated. Benzene is a hydrophobic compound with essentially non-acidic protons (pKa of about 43), eliminating choice A.

Besides detoxification of drugs such as acetaminophen, the liver is involved in and regulates several different biochemical pathways. Which of the following is NOT a biochemical activity of the liver? A. Regulation of carbohydrate metabolism such as glycogenolysis, glycogenesis, and gluconeogenesis B. Production of lipases and bile for fat digestion C. Deamination of amino acid and conversion of the resulting ammonia to urea D. Lipid metabolism, including cholesterol and lipoprotein synthesis

B. The liver does produce bile, however bile is an emulsifier not a digestive enzyme, and lipases are produced by the pancreas not the liver (choice B is not a function of the liver and is the correct answer choice). The liver plays a large role in the metabolism of virtually all the macromolecules; carbohydrates via glycogenolysis, glycogenesis, and gluconeogenesis (choice A is a function of the liver and can be eliminated), deamination of amino acids and detoxification of ammonia (choice C is a function of the liver and can be eliminated), and lipid metabolism (cholesterol and lipoprotein synthesis, choice D is a liver function and can be eliminated).

Shortly after administration of chlordiazepoxide in a patient suffering from AWS, which of the following steps in propagation of the action potential would be most directly affected? A. Influx of Ca2+ into the pre-synaptic terminal via voltage-dependent calcium channels B. Depolarization of the axon hillock by influx of Na+ C. Fusion of vesicles and release of neurotransmitter into the synaptic cleft D. Binding of neurotransmitter with the receptor on the post-synaptic membrane

B. The passage states that AWS is characterized by increased excitability, and that benzodiazepines can help mediate this effect, i.e., act in an inhibitory manner. The passage also states that chlordiazepoxide has a strong affinity for GABAA receptors, suggesting that treatment with chlordiazepoxide would increase the influx of Cl? into the neuron and inhibit it. A neuron that is inhibited (i.e., hyperpolarized) would be less likely to reach the threshold for voltage-gated Na+ channels to open, and would be less likely to depolarize (choice B is correct). All of the other choices happen after the initial depolarization; influx of calcium is caused by depolarization (choice A is wrong), this leads to fusion of the vesicles and release of neurotransmitter (choice C is wrong), and the final act would be for neurotransmitter to bind to receptors on the post-synaptic membrane (choice D is wrong).

When conducting a hearing test, sounds of varying pitches and loudness are created in order to determine if any aspect of hearing has been compromised. Which of the following best describes how differences in pitch are determined by the brain? A. Differences in pitch are determined by the amplitude of the vibration of the basilar membrane. Larger vibrations cause more frequent action potentials and smaller vibrations cause less frequent action potentials. B. Differences in pitch are determined by which region of the basilar membrane vibrates, stimulating different auditory hair cells. High pitched sounds vibrate the basilar membrane near the oval window and low pitched sounds vibrate the basilar membrane farthest from the oval window. C. Differences in pitch are determined by the amplitude of the vibration of the basilar membrane. Smaller vibrations cause more frequent action potentials and larger vibrations cause less frequent action potentials. D. Differences in pitch are determined by which region of the basilar membrane vibrates, stimulating different auditory hair cells. High pitched sounds vibrate the basilar membrane furthest from the oval window and low pitched sounds vibrate the basilar membrane near the oval window.

B. This is a 2x2 question, where one piece of information can eliminate two answer choices. Differences in pitch are determined by the region of the basilar membrane that vibrates, and loudness is determined by amplitude of the vibration (choices A and C are wrong). High pitched sounds vibrate the region of the basilar membrane near the oval window, and low pitched sounds vibrate the basilar membrane farthest from the oval window (choice B is correct and choice D is wrong).

During a mastectomy, the lymph nodes are often removed in addition to the tumor. This surgical procedure will most likely: A. render the patient less susceptible to viral infections. B. stop the spread of breast cancer to the rest of the body. C. increase the number of circulating lymphocytes. D. remove the oncogenes within the affected tissue.

B. When cancers metastasize, tumor cells leave their original site and enter the blood or lymph vessels to travel to a new part of the body where they can implant and grow. Therefore, removal of nearby lymph nodes during surgery to remove a cancerous tumor is often done to help ensure that the cancer cannot spread (choice B is correct). Removal of the lymph nodes could render the patient more susceptible to infection and decrease the number of circulating lymphocytes (choices A and C are wrong). Finally, to remove the oncogenes within the affected tissue, the surgeons would only need to remove the breast tissue since the lymph nodes are not affected (choice D is wrong).

Which of the following represents the major products of the saponification of linolein, a triacylglycerol in which glycerol is esterified with linoleic acid? A. 1 Glycerol and 1 linoleic acid B. 3 Glycerol and 1 potassium linoleate C. 1 Glycerol and 3 potassium linoleate D. 1 Glycerol and 3 linoleic acid

C. A triacylglycerol is composed of a glycerol backbone and three fatty acids, in this case three linoleic acid molecules. Saponification of linolein yields one glycerol molecule (eliminate choice B) and three linoleic acid salts, so three potassium linoleate equivalents are expected (eliminate choice A). Linoleic acid is not seen, as fatty acids do not remain protonated under basic conditions (eliminate choice D).

During peak cardiac activity, heart muscle utilizes the anaerobic pathway for only 10% of its energy as opposed to up to 85% for maximally contracting skeletal muscle. With respect to skeletal muscle, cardiac muscle would: A. convert more pyruvate to lactate. B. consume 1/8 as much ATP to produce the same contractile force. C. consume less glucose per ATP produced. D. require less oxygen per molecule of ATP produced.

C. Aerobic respiration includes glycolysis, the PDC, the Krebs cycle, and electron transport, to produce a total of 32 ATP per glucose (assuming the malate-aspartate shuttle is used to transfer the electrons from glycolytic NADH into the mitochondria). By contrast, anaerobic respiration involves only glycolysis and fermentation of pyruvate to lactate or alcohol, to produce 2 ATP per glucose. Anaerobic respiration is therefore much less efficient and must consume much more glucose to produce the same amount of ATP. Since cardiac muscle uses mostly aerobic respiration, it will be more efficient than skeletal muscle and uses less glucose to produce an equivalent amount of ATP (choice C is correct). Conversion of pyruvate to lactate occurs during fermentation, which cardiac muscle carries out less of, not more (choice A is wrong). The amount of contractile force generated per ATP is independent of the source of ATP used (choice B is wrong). Cardiac muscle uses more aerobic respiration, so it will use more oxygen per ATP produced, not less (choice D is wrong).

If a primary immune response confers immunity, why do individuals get the common cold more than once? A.The common cold virus does not trigger the proliferation of T lymphocytes. B.The common cold virus does not have receptors. C.Individuals are not exposed to the same virus. D.There are too few common cold viruses that circulate in the blood.

C. All viruses (and other pathogens) will trigger the production of both B and T lymphocytes ( choice A is wrong). The common cold virus does not have receptors, but receptors are not needed by the virus. Receptors are needed by B and T cells to recognize antigens (choice B is true but does not answer the question, so it is eliminated). The number of viruses circulating in the body has nothing to do with whether a disease can cause immunity or not (choice D is wrong). There are many different viruses that can cause the symptoms we associate with the common cold: runny nose, sore throat, etc. Exposure to one of these viruses will cause the disease and confer immunity to that particular virus, but not to the hundreds of other viruses capable of causing the common cold (choice C is correct).

Which of the following is an accurate statement concerning eukaryotes and prokaryotes? A. In most cases, eukaryotes follow the "one gene, one protein" rule, while prokaryotes are monocistronic. B. Eukaryotes have three DNA polymerases, and prokaryotes have three RNA polymerases. C. In prokaryotes transcription and translation can occur simultaneously, while in eukaryotes, they must remain distinct. D. Eukaryotes modify their primary transcripts by adding a 5' cap and a 3' poly-A tail, but prokaryotes only add a poly-A tail since the Shine-Dalgarno sequence replaces the function of the 5' cap.

C. Because prokaryotes do not have organelles, both transcription and translation occur in the cytoplasm, frequently simultaneously. In eukaryotes, however, transcription occurs in the nucleus, and translation occurs in the cytoplasm. Most eukaryotic RNA codes for a single protein with very few exceptions (alternative mRNA splicing); however, prokaryotic mRNA is often polycistronic and codes for several different proteins, often by utilizing different reading frames (A is wrong). Eukaryotes and prokaryotes both have multiple DNA polymerases (more than three), but eukaryotes have three RNA polymerases while prokaryotes have only a single RNA polymerase (B is wrong). While it is true that the Shine-Dalgarno sequence in prokaryotes essentially replaces the function of the 5' cap in eukaryotes, prokaryotes do not add poly-A tails to their mRNA transcripts (D is wrong).

Case #3 A man's DNA is found at the scene of a crime, but upon investigation it is found that he died far away many years prior to the date of the crime. The man had no criminal record and was known to be a generous benefactor of many scientific charities. Medical records indicate the man was screened at a bone marrow drive and on one occasion was a donor to a child with the blood cancer leukemia that required chemotherapy and total body irradiation to destroy the existing bone marrow followed by bone marrow grafting. What would be the next best step in investigating the crime mentioned in Case #3? A. Locate the bone marrow recipient and obtain a cheek swab for genetic analysis. B. Locate the bone marrow recipient and isolate red blood cells for genetic analysis. C. Locate the bone marrow recipient and obtain a whole blood sample and cheek swab. D. Consider the case closed since the DNA sample is from a deceased suspect.

C. Bone marrow contains the stem cells used to make all the cells found in blood. In a bone marrow transplant, an individual's dysfunctional bone marrow is destroyed, then replaced with a donor's normal bone marrow, thus all the blood cells of the recipient will be derived from the donor's bone marrow stem cells and will contain the genetic material and genotype of the donor. This explains how the DNA of a deceased person could be found at the crime scene (choice D can be eliminated); the perpetrator was most likely the recipient of a bone marrow transplant in the past from the deceased man whose DNA was identified. To test the hypothesis, the recipient's blood cells should be analyzed; a cheek swab alone would be insufficient because only the cells made by the bone marrow could match the DNA found at the crime scene. A cheek swab would reveal cells with the perpetrator's original DNA genotype (choice A can be eliminated). Red blood cells alone are also insufficient, since red blood cells do not contain nuclei and would provide no DNA to analyze (choice B can be eliminated). A whole blood sample and a cheek swab together are the best option. The whole blood sample would contain white blood cells (with nuclei) that would have been produced from the donor's stem cells and thus would match the DNA from the crime scene. A cheek swab is also a good idea to obtain, since it would contain a different sample of DNA that could be compared to any other genetic material that turned up at the crime scene (choice C is correct).

Arachidonic acid is a fatty acid contained in cell membranes. The structure of arachidonic acid is: In which component of the cell membrane would arachidonic acid most likely be found? A. Cholesterol B. Triglycerides C. Phospholipids D. Peptidoglycan layer

C. Cholesterol is an abundant component of animal cell membranes, but it is a steroid, not derived from fatty acids such as arachidonate (eliminate choice A). Triglycerides could contain a fatty acid like arachidonate, but they are not membrane components (eliminate choice B). Peptidoglycans are not found in eukaryotes, only bacteria, and do not consist of fatty acids (eliminate choice D). Phospholipids is the correct answer. Phospholipids are abundant components of the plasma membrane and contain esters of many different fatty acids (choice C is correct).

Conversion of pyruvate into glucose requires enzymes present in: A. the mitochondria only. B. the cytosol only. C. the mitochondria and the cytosol. D. neither the mitochondria nor the cytosol.

C. Conversion of pyruvate to glucose requires its initial conversion into oxaloacetate, in a reaction catalyzed by pyruvate carboxylase in the mitochondria (choices B and D are wrong). Oxaloacetate (OAA) is then decarboxylated and phosphorylated by cytosolic or mitochondrial forms of phosophoenolpyruvate carboxykinase (PEPK). After transport of either OAA (in the form of malate) from the mitochondria, the remainder of gluconeogenesis takes place in the cytosol (choice A is wrong and C is the correct answer).

All of the following processes occur during the conduction of a nerve impulse across the synapse of two neurons EXCEPT: A. the release of vesicles from the presynaptic cell. B. the opening of postsynaptic ion channels. C. a retrograde depolarization of the presynaptic cell. D. the depolarization of the postsynaptic cell.

C. During the conduction of an action potential across a chemical synapse, neurotransmitter is released by the presynaptic cell, it diffuses across the synapse, binds to receptors on the postsynaptic cell, opens ion channels on the postsynaptic cell, and depolarizes (or hyperpolarizes) the postsynaptic cell. Thus, choices A, B, and D are all part of the normal conduction of an action potential across a synapse and can be eliminated. The presynaptic cell does not normally transmit a retrograde action potential, however (choice C is false and the correct response here). This is usually prevented by the refractory period after an action potential passes, causing action potentials to propagate in only one direction along an axon.

Which of the following is true regarding the menstrual cycle and hormones in the female? A. A spike in estrogen directly causes the ovulation of the egg cell from the ovary. B. Progesterone peaks immediately prior to ovulation. C. Follicle-stimulating hormone levels are highest following menses. D. Luteinizing hormone levels peak as a result of ovulation.

C. FSH stimulates follicle development and is released by the anterior pituitary after menses has occurred. This allows for the ovarian follicle to develop prior to ovulation (choice C is correct). The spike in estrogen levels immediately before ovulation is not directly responsible for ovulation (choice A is wrong). The spike in estrogen changes the typical negative feedback system on LH to a positive feedback system, so that the estrogen spike causes a spike in LH. The LH surge is directly responsible for ovulation. After ovulation, the LH levels plummet (choice D is wrong). Progesterone peaks immediately after ovulation (choice B is wrong).

atty acid synthase is most likely suppressed by all of the following hormones EXCEPT: A. glucagon. B. thyroid hormone. C. insulin. D. cortisol.

C. Fatty acid synthesis will only be active if the cells have enough energy and are ready to build storage molecules, especially long term storage molecules. Insulin is released in response to elevated blood glucose and the glucose can be used for cellular respiration and storage (choice C is correct). For a process of elimination approach: glucagon is a hormone that liberates glucose from glycogen, raises blood glucose levels, and induces fat/triglyceride breakdown (eliminate choice A). Thyroid hormone increases the basal metabolic rate of cells. This will increase the ATP consumption of the body and limit production of any long-term storage molecules (eliminate choice B). Cortisol is a stress hormone and will lead to lipolysis to increase ATP availability in stressful situations. It will inhibit lipogenesis (eliminate choice D).

Which of the following are determinants of the G2/M checkpoint? Ensure that genomic replication is complete Take inventory of nucleotide levels Check for mutations or DNA instability A. II only B. I and II only C. I and III only D. I, II, and III

C. Item I is true: Before the cell divides, it must check the integrity of the genome it is passing to the daughter cells. This includes making sure DNA replication is complete. Choice A can be eliminated. Item II is false: Checking the levels of nucleotides in the cell is part of the G1/S checkpoint (the other major checkpoint pathway mentioned in the passage), since these building blocks are required for DNA replication (choices B and D can be eliminated). Item III is true: Before the cell divides, mutations have to be repaired. Choice C is correct.

In which phase of spermatogenesis could a single nondisjunction event occur with the lowest probability of eventually forming a zygote with a Down syndrome karyotype? A. Prophase I B. Anaphase I C. Anaphase II D. Telophase II

C. Nondisjunction involves the failure of homologous chromosomes or sister chromatids to separate; this does not occur in prophase or telophase (choices A and D can be eliminated). If a single nondisjunction event occurred in anaphase II of spermatogenesis, it would only affect two out of the four resulting gametes. The primary spermatocyte would generate two normal secondary spermatocytes. One of the secondary spermatocytes would perform normal anaphase II and would generate two normal spermatids. The secondary spermatocyte that undergoes nondisjunction in anaphase II would produce two abnormal spermatids, one that would fertilize an egg to cause trisomy and another that would fertilize an egg and cause monosomy. Thus, the probability of Down syndrome occurring if nondisjunction occurred in anaphase II of spermatogenesis would be 25% (choice C is correct). If the nondisjunction event occurred in anaphase I, both secondary spermatocytes would be affected and all four spermatids would be affected; two would generate zygotes with trisomy and two would generate zygotes with monosomy. Thus, the probability of Down syndrome occurring if nondisjunction occurs in anaphase I is 50% (choice B is wrong).

Penicillin is an antibiotic that interrupts the synthesis of bacterial cell walls. Which of the following is most likely to be affected by its action? A. Fatty acids B. Phospholipids C. Peptidoglycans D. Lipoproteins

C. Since bacterial cell walls are made up of proteins and carbohydrates (peptidoglycans), if penicillin affects cell wall synthesis it will most directly affect peptidoglycan (C is correct). Fatty acids and phospholipids are parts of cell membranes (A and B are wrong), and lipoproteins are blood proteins that transport lipids (D is wrong).

The process by which the nitrogen metabolism by-product urea is removed from the blood in the glomerulus is known as: A. tubular secretion. B. reabsorption. C. ultrafiltration. D. osmosis

C. The passage states that filtration occurs at the glomerulus in Bowman's capsule. This allows small substances to travel into the proximal convoluted tubule. Tubular secretion occurs for ions such as K+ in the distal convoluted tubule (A is incorrect). Reabsorption is the process by which substances in the tubular filtrate are brought back into the blood (B is incorrect). Osmosis refers to the movement of water down its concentration gradient (D is incorrect).

Conformity vs Compliance vs Obedience vs Internalization

Conformity is going along with something Obedience is how we follow authority Big distinction is that conformity deals with the thought processes, obedience just deals with the how. Compliance is going along with something because someone tells you to because you either want the reward, or to avoid punishment. Internalization is when you incorporate those beliefs/behaviours/etc., you will do/think these things even if there is no social influence. So compliance is conformity to get reward/avoid punishment, it is discontinued when the reward/punishment is gone. Internalization will persist.

It was subsequently discovered that all the viruses were enveloped. A researcher then hypothesized that this was the reason the viruses could not replicate in insect cells. Is the researcher's hypothesis reasonable? A. Yes; the cell walls of the insect cells prevent an envelope from being acquired. B. Yes; the insect exoskeleton prevents an envelope from being acquired. C. No; insect walls are made of chitin. D. No; insect cells do not possess a cell wall.

D. A viral envelope is acquired as the virus exits its host cell by budding through the plasma membrane and becoming coated in lipid bilayer. Viruses are unable to bud from cells that possess a cell wall (such as bacteria or plants), thus those viruses cannot acquire an envelope. However, insects are members of Kingdom Animalia, and as such their cells do not possess a cell wall, so this could not be the reason for the inability of the viruses to replicate in these cells. (Do not confuse chitinous exoskeleton of insects with a cell wall.)

Which of the following intermolecular attractions will exhibit the greatest strength? A. London dispersion forces B. Induced dipole interactions C. Instantaneous dipole interactions D. Hydrogen bonds

D. Choices A, B, and C are identical, so they can all be eliminated. The answer must be D. The various intermolecular forces, in order of decreasing strength, are the following: Hydrogen bonding > Dipole-Dipole interactions > Dipole- Induced dipole interactions > Induced dipole-Induced dipole interactions (London forces).

Solutions of D-glucose consist of mixtures of α- and β-anomers. The α- and β-anomers individually display +112° and +18.7° optical rotation values, respectively. If the optical rotation of a solution of D-glucose at 40°C is +52.3° in water and +61.6° in DMSO, we can conclude that: A. the optical rotation of DMSO is +9.3°. B. the ratio of α- and β-anomers is greater in water than in DMSO. C. the ratio of α- and β-anomers is the same in water as in DMSO. D. the ratio of α- and β-anomers is lower in water than in DMSO.

D. DMSO is an achiral molecule and therefore has no optical rotation (eliminate choice A). The optical rotation of mixtures is given by the concentration-weighted average of the optical rotations of individual components. (This is why racemic mixtures of enantiomers have no net optical rotation). Since the optical rotations of the two solutions are different, the ratios of α- and β-anomers must also be different (eliminate choice C). The observed (concentration-weighted average) optical rotation is lower in water than in DMSO indicating that the component with the lower optical rotation has greater concentration in water than in DMSO. The β-anomer has the lower optical rotation and therefore is present at higher concentration in water making choice D correct.

The higher than normal frequency of guevedoche in the Dominican Republic is the direct product of consanguineous relationships (relationships between close blood relatives). Which type of inheritance pattern is most consistent with increased phenotypic expression of a rare disease arising as a result of inbreeding within a population? A. X-linked dominant B. Autosomal dominant C. Mitochondrial inheritance D. Autosomal recessive

D. Dominant disorders are typically not rare (they have increased phenotypic expression) because they are caused by a dominant allele. An offspring need only receive one allele from an affected parent to express the disease. For example, suppose a man heterozygous for a dominant disorder (Aa) marries an unaffected woman (aa). The offspring all have a 50% probability of being affected by the disorder themselves. This is true for both autosomal disorders and X-linked disorders (choices A and B are wrong). Mitochondrial disorders are strictly maternally inherited, since the organelles of the zygote come only from the ovum. Thus, expression of this type of disorder does not change if a woman enters into a consanguineous relationship; the probability her offspring will inherit the disease depends only on her (choice C is wrong). However, recessive disorders are typically rare because they require a homozygous recessive genotype to be expressed, with one recessive allele coming from each of the parents. If mating is totally random, the frequency of a recessive disorder in a population is related to the frequency of the recessive allele; for example, if q = .001 (p = .999), then the frequency of affected individuals (qq) is only .000001, or 1 in 1 million. This number is the same from generation to generation, assuming mating stays random. Now consider what happens when mating is consanguineous. We'll use the same allele frequencies and look at two generations. Assume the first-generation parents were not consanguineous; they chose each other randomly. The probability of one of them being a carrier is 2pq. The probability of the father or the mother being a carrier is 2pq + 2pq, or 4pq. The probability that the carrier passes the recessive allele on to a son is 1/2, and the probability that the carrier passes the recessive allele on to a daughter is also 1/2. Now suppose the son and daughter mate and produce offspring. If they are both heterozygous, the probability of them having an affected child is 1/4. So the total probability of an affected child from a consanguineous relationship is: (probability original father is heterozygous + probability original mother is heterozygous)× probability the recessive allele is passed to a son× probability the recessive allele is passed to a daughter× probability son and daughter have an affected offspring= (2pq + 2pq) × 1/2 × 1/2 × 1/4 = 4pq/16 = pq/4 = .000999/4 ≈ 1/4000This is a considerably greater probability than 1 in 1 million from the randomly mating population. Thus, consanguineous relationships (inbreeding) lead to increased expression of rare disorders. This is why this type of relationship is regulated legally, and is taboo in many cultures.

Which of the following represents the correct sequence for embryogenesis? A. Fertilization → gastrulation → blastulation → neural tube formation → somite formation B. Fertilization → gastrulation → blastulation → somite formation → neural tube formation C. Fertilization → blastulation → neural tube formation → gastrulation → somite formation D. Fertilization → blastulation → gastrulation → neural tube formation → somite formation

D. Fertilization is the first step, followed by a series of rapid cell cleavages to form a hollow ball of cells called the blastulam (eliminate choices A and B). Next comes the gastrula, in which cells move into the interior of embryo to form the three germ layers (eliminate choice C). Gastrulation is followed by the formation of the neural tube, which will form the nervous system, followed by the formation of other organs and tissues, such as the somites that will differentiate into bones and muscle. This makes choice D the only possible correct order of events

Which of the following offers the most likely explanation for the substantially higher peak plasma concentration of acetaminophen following IV administration when compared to oral and rectal routes? A. The drug is poorly absorbed through mucous membranes in the oral cavity and rectum. B. A substantial portion of the drug is excreted in the feces before it can be absorbed. C. Much of the drug is broken down in the digestive tract prior to absorption. D. The liver breaks down a large portion of the drug before it reaches the systemic circulation after oral or rectal administration.

D. IV administration allows a drug to bypass the liver (the primary organ responsible for drug metabolism, choice C is wrong) in order to reach the systemic circulation at the highest possible level. While it is true that absorption is poor through the oral and rectal mucosa, the drug is not being aborbed there. Both swallowing the drug (oral administration) and rectal administration introduce acetaminophen into the GI tract, and with a rise in plasma acetaminophen levels only a half hour after oral or rectal administration, acetaminophen is being readily absorbed by intestinal mucosa (choice A is true but does not explain the difference in peak plasma concentrations). With an onset of action of less than a half hour after oral administration, acetaminophen is readily absorbed by the oral mucosa (choice A is wrong). While some of the drug will be excreted unabsorbed, this is not the primary reason for the difference between the peak plasma concentrations (choice B is wrong).

2,4-dinitrophenol (2,4-DNP) is a highly toxic substance which was sold to the public as a weight loss drug in the 1930s. It acts by permeabilizing the inner mitochondrial membrane (IMM) to ions. Which of the following is true of 2,4-DNP's effects on oxidative phosphorylation? A. It causes decreased flux of electrons through ATP synthase. B. It likely leads to a decrease in body temperature. C. It leads to decreased consumption of FADH2and NADH by the electron transport chain proteins. D. It causes a decrease in the electromotive potential built up by the electron transport chain.

D. If the inner mitochondrial membrane became permeable to ions, then hydrogen ions would not need to go through the ATP synthase in order to re-enter the matrix. This would dissipate the proton gradient established by the electron transport chain, and decrease its potential to generate ATP (choice D is correct). FADH2and NADH would still be consumed by the transport chain proteins, since the electron transport chain is not shut down; it's just that the proton gradient would be more easily dissipated (choice C is wrong). The dissipation of the gradient would result in heat production; note that this is similar to what happens in brown fat, used by hibernating animals to stay warm (choice B is wrong). Electrons do not flow through the ATP synthase (choice A is wrong).

An ELISA assay can be used to detect HIV. First, rabbits are inoculated with human IgG. The rabbit immune system forms antibodies to the human antibodies. These rabbit antibodies are purified and conjugated to an enzyme. In the ELISA assay, the plastic is coated with HIV antigen, followed by human plasma (which is washed away), and then conjugated rabbit IgG. If the test is effective, the conjugated antibodies bind directly to: A. HIV particles. B. human white blood cells. C. plastic. D. a variety of human antibodies.

D. In the test described, HIV antigen is bound to the well plate, then human plasma is added. If anti-HIV antibodies (IgGs) are in the plasma (indicating an HIV infection), the anti-HIV antibodies will bind to the HIV antigen on the plate. After the wash, the conjugated antibodies are added, which will bind to the anti-HIV IgG. This is the antigen with which the rabbit was inoculated. Note that the rabbit IgG will recognize a variety of human IgG since the constant domain will be the same across IgG with a wide range of specificity. The only human IgG bound to the plate, however, should be those which are bound to HIV antigens (choice D is correct). The rabbits were immunized with human IgG, and thus the antibodies they make are specific for this, not HIV. This assay detects the patient's immune response to HIV, not the virus itself (choice A is wrong). Although B cells (which are white blood cells) produce antibodies, the rabbit antibody binds only the product, not the synthesizing cell (choice B is wrong). If the antibodies bound to plastic, there would be a very high nonspecific signal (choice C is wrong).

The ion pumps for sodium and chlorine that establish the countercurrent multiplier system in the medulla of a vertebrate kidney are located in the cell membrane of the: A. proximal convoluted tubules. B. distal convoluted tubules. C. descending loops of Henle. D. ascending loops of Henle.

D. Ion pumps for Na+ and Cl- must exist in a part of the nephron that is permeable to those ions. As a result, answer C can be eliminated since the descending loop of Henle is permeable ONLY to water. Answer choices A and B can be eliminated based on knowledge of the anatomy of the nephron and kidney. The convoluted tubules are located in the cortex of the kidney. The question specifically asks "in the medulla of a vertebrate kidney." The loop of Henle is the only part of the nephron, aside from the collecting ducts, that delves into the medulla of the kidney. The deeper into the medulla the loop travels, the greater the countercurrent multiplier concentration that can be established, creating more highly concentrated urine.

When scientists injected a low-molecular-weight yellow dye into a single cell of a 16-cell mollusk embryo, they discovered that the dye was confined to that cell and its progeny. When they injected the dye into a single cell of a 32-cell embryo, they located the dye in that cell as well as in adjacent cells. Based on these findings, it is most reasonable to conclude that communication between cells of a developing mollusk embryo develops at the: A. zygote stage. B. neurula stage. C. gastrula stage. D. blastula stage.

D. The blastula is a fluid-filled ball of cells that arises after cleavage and contains at least 32 cells (choice D is correct). Recall that a morula is a solid ball of cells and contains up to 16 cells. Choice A is incorrect because the zygote is a single cell. Choices B and C are incorrect because these constitute later developmental stages that involve many more cells than the 32-cell embryo examined in the experiment.

Which of the following is a true statement about the diaphragm? A. It contains both skeletal and smooth muscle cells. B. Its effector neurotransmitters are norepinephrine and acetylcholine. C. It is innervated by the phrenic nerve and autonomic nervous system. D. It receives neural signals from the cerebral cortex and the brain stem

D. The diaphragm is purely skeletal muscle (choice A is false) and as such, ACh is the only neurotransmitter used (choice B is false). It's innervated only by the phrenic nerve, not autonomic nerves (choice C is false). The phrenic nerve originates both in the cerebral cortex, for voluntary breathing, and in the brain stem for involuntary control (choice D is correct).

Based on information presented in the passage, which of the following statements is likely true? A. Acute alcohol intoxication results in increased release of inhibitory neurotransmitters due to an increase in intracellular Cl- concentration. B. Acute alcohol intoxication results in increased release of excitatory neurotransmitters due to an increase in intracellular Cl- concentration. C. Acute alcohol intoxication results in decreased release of inhibitory neurotransmitters due to an increase in intracellular Cl- concentration. D. Acute alcohol intoxication results in decreased release of excitatory neurotransmitters due to a nincrease in intracellular Cl- concentration

D. The passage states that acute alcohol intoxication results in an overall sedative effect. This is due to the increased activation of the GABAA receptors, leading to an increase in intracellular Cl-concentration. The increased intracellular Cl- would hyperpolarize the cell, preventing it from releasing neurotransmitter (choices A and B are wrong). Since the overall effect is sedative, the neurotransmitter must be excitatory; the decreased release of an excitatory neurotransmitter would lead to overall sedation (choice C is wrong and choice D is correct).

Which of the following is the most likely sequence of events in the infection of E. coli by phage T1? A. Replication of viral genome, production of viral DNA polymerase, translation of viral lysozyme, assembly of infectious virus B. Production of viral DNA polymerase, replication of viral genome, translation of viral lysozyme, assembly of infectious virus C. Translation of viral lysozyme, production of viral DNA polymerase, replication of viral genome, assembly of infectious virus D. Production of viral DNA polymerase, replication of viral genome, assembly of infectious virus, translation of viral lysozyme

D. The production of viral DNA polymerase must occur prior to replication of the viral genome since the polymerase is needed to do this (choice A is wrong). The last step must be production of lysozyme since this will lyse the cell, bursting it open and stopping any further activities required to produce infectious virus (choices A, B, and C are wrong). The correct events are in choice D: the expression of the viral DNA polymerase must come first to then replicate the genome. Infectious virus can then be assembled, and released by cell lysis induced by viral lysozyme.

PCl5(g)= PCl3(g) + Cl2(g) ΔHo = 92.5 kJ/mol Two identical reaction vessels, X and Y, are charged with equal amounts of gaseous PCl5, and both trials are allowed to reach equilibrium. If flask X is held at 300 K while flask Y is maintained at 500 K, which of the following will be true about the reactions? A. Equilibrium is reached faster in flask Y because higher temperatures shift the reaction toward products. B. Equilibrium is reached faster in flask Y because ΔGYis more negative than ΔGX. C. Equilibrium is established at the same rate in both flasks because the K value of a reaction is a constant. D. Equilibrium is reached faster in flask Y due to the increased collision frequency of reactant molecules.

D. The speed that equilibrium is reached is based on the kinetics of the forward and reverse reactions and not on any thermodynamic parameters. Since both the forward and reverse reactions of the equilibrium will be sped up by increasing temperature, flask Y will reach equilibrium faster (eliminate choice C). The reason must be a kinetic principle, not a thermodynamic principle, so using ΔG or Le Châtelier's Principle, both thermodynamic concepts, as explanations must be wrong (eliminate choices A and B). According to the kinetic molecular theory, temperature is a measure of the average speed (velocity) of the particles. Increasing temperature increases speed, which in turn increases the frequency of collisions between the particles and the vessel.

If size exclusion chromatography (SEC) were used to separate a mixture of ethylene glycol oligomers, with fractions further purified by gas chromatography (GC), what might be expected? A. The use of nonpolar solvents in SEC will lead to good separation and efficient purification by GC. B. The use of ether-type solvents in SEC will lead to poor separation, and subsequently to inefficient purification by GC. C. Oligomers with long retention times on the SEC column have long GC retention times. D. Oligomers with short retention times on the SEC column will have long GC retention times.

D. The use of one solvent or another shouldn't make a large difference in SEC retention times, since it's size, not polarity, that causes separation (eliminate choices A and B). Long retention times on the SEC column mean that the oligomers are small, which also means they are more volatile and have shorter retention times on GC. The converse is also true: short SEC retention times mean the oligomers are large and less volatile, with longer GC retention times, so choice C is false.

Which of the following increases the activity of osteoclasts? A. Thyroid hormone B. Calcitonin C. Epinephrine D. Parathyroid hormone

D. Thyroid hormone and epinephrine do not have their primary effects on calcium metabolism or bone-cell activity (eliminate choices A and C). Osteoclasts are bone-resorbing cells that will release calcium from bone and raise plasma calcium levels. Calcitonin lowers plasma calcium levels, so it must not increase osteoclast activity (eliminate choice B). Parathyroid hormone elevates plasma calcium and acts to stimulate osteoclast activity (choice D is correct).

Which of the following hormones acts on target cells via a second-messenger system? A. Thyroid hormone B. Aldosterone C. Cortisol D. Epinephrine

D. Thyroid hormone, aldosterone, and cortisol are all small hydrophobic molecules that passively diffuse through the plasma membrane to bind to receptors inside the cell in the cytoplasm and nucleus. These receptors then regulate transcription, without the use of second-messenger systems (choices A, B, and C are wrong). A hydrophilic hormone like epinephrine, however, binds to cell-surface receptors and activates adenylate cyclase to make cAMP, a second messenger (choice D is correct). Note that in general, steroid hormones exert their effects by modifying transcription, while protein hormones utilize second-messenger systems. Thyroid hormone, a protein hormone, is an exception.

Helicobacter pylori, a bacterium found in the stomach and duodenum, has been implicated in the formation of peptic ulcers (lesions due to inflammation and low pH). Which of the following would be the most effective treatment to eliminate the infection and allow for healing? A. Solid antacids to neutralize the stomach acid B. Parietal cell agonist to kill the H. pylori C. Oral antibiotic and solid antacids D. Oral antibiotic and parietal cell antagonist

D. To treat peptic ulcers we must both eliminate the infection (so that the ulcer does not recur) and neutralize the stomach acid (to reduce the inflammation and allow healing). Treatment with the appropriate antibiotic will eliminate the infection (choices A and B are wrong) and blocking release of HCl by parietal cells with an antagonist will provide a sustained higher stomach pH than will solid antacids (either alone or in combination, choice D is better than choice C). Note that choice B might be particularly harmful; an agonist would increase acid secretion, the exact opposite of what we are trying to accomplish.

Summary of DNA replication in E. coli

Helicase opens up the DNA at the replication fork. Single-strand binding proteins coat the DNA around the replication fork to prevent rewinding of the DNA. Topoisomerase works at the region ahead of the replication fork to prevent supercoiling. Primase synthesizes RNA primers complementary to the DNA strand. DNA polymerase III extends the primers, adding on to the 3' end, to make the bulk of the new DNA. RNA primers are removed and replaced with DNA by DNA polymerase I. The gaps between DNA fragments are sealed by DNA ligase.

Fructose-2,6-bisphosphate

It exerts reciprocal control on glycolysis and gluconeogenesis by stimulating phosphofructokinase and inhibiting fructose-1,6-bisphosphatase.

Structural proteins Motor proteins Cell Adhesion Molecules (CAM)

Structural proteins: Make up the cytoskeleton, anchoring proteins, & extracellular matrix. Include collagen, elastin, keratin, & tubulin. Motor proteins: Have 1 or more heads that can generate force through a conformational change. They have catalytic activity, acting as ATPases to generate movement like muscle contraction, vesicle movement, & cell motility. Include myosin, kinesin, & dynein. Cell Adhesion Molecules (CAM): Allow cells to bind to other cells or surfaces. Include cahedrins (hold similar cells together), integrins (help cells stick to proteins), & selectins(help sells to stick to carbohydrates)

What is the difference between substrate-level phosphorylation and oxidative phosphorylation?

Substrate-level phosphorylation is directly phosphorylating ADP with a phosphate and energy provided from a coupled reaction. SLP will only occur if there is a reaction that releases sufficient energy to allow the direct phosphorylation of ADP. Oxidative phosphorylation is when ATP is generated from the oxidation of NADH and FADH2 and the subsequent transfer of electrons and pumping of protons. That process generates an electrochemical gradient, which is required to power the ATP synthase.

In humans, the lining of which structure is NOT primarily derived from the endoderm? A.Mouth B.Bronchi C.Bladder D.Stomach

The answer to this question is A because most of the lining of the mouth is derived from an invagination of ectoderm. Most of the epithelial tissues inside the body (B, C, D) are derived from endoderm.

The initial filtration step in the glomerulus of the mammalian kidney occurs primarily by: A. passive flow due to a pressure difference. B. passive flow resulting from a countercurrent exchange system. C. active transport of water, followed by movement of electrolytes along a resulting concentration gradient. D. active transport of electrolytes, followed by passive flow of water along the resulting osmolarity gradient.

The answer to this question is A because the initial filtration in the glomerulus occurs as blood pressure forces the fluid from the glomerulus into the lumen of Bowman's capsule.

ntroduction of which amino acid substitution would result in the largest decrease in the entropic penalty associated with a protein folding into its native conformation? A. Ile to Asp substitution at a buried site B. Leu to Thr substitution at a surface-exposed site C. Gly to Pro substitution in a flexible loop D. Arg to Tyr substitution at a surface-exposed site

The answer to this question is B because changing a surface-exposed hydrophobic residue for a more hydrophilic residue eliminates the entropic penalty associated with ordered water molecules around hydrophobic groups. "Entropic penalty" refers to the loss of disorder when water molecules must arrange themselves in an orderly manner around a hydrophobic group "Entropic penalty" refers to the loss of disorder when water molecules must arrange themselves in an orderly manner around a hydrophobic group. Substituting leucine (a nonpolar amino acid) for threonine (a polar amino acid) at the surface of a protein reduces this entropic penalty as the hydrophobic leucine would be gone and the water molecules would be less orderly (choice B is correct). Substituting isoleucine (nonpolar) for aspartic acid (polar) at a buried site would not lead to as big a drop, as the interiors of proteins are typically hydrophobic and there isn't as much water; water molecules are typically found near the surface as proteins usually exist in aqueous environments (choice A is wrong). Glycine and proline are both hydrophobic, and arginine and tyrosine are both polar; substituting one for the other (regardless of location) would likely not change the entropic penalty (choices C and D are wrong).

If a patient with Addison's disease is given too high a replacement dose of glucocorticoids, the effect over time will be an increase in: A. muscle mass. B. muscle weakness. C. red blood cell count. D. heart rate

The answer to this question is B because glucocorticoids act on skeletal muscle causing the breakdown of muscle proteins. Therefore, if a patient is given too high a replacement dose of glucocorticoids, this will result in muscle weakness.

Glucose transporter proteins in the liver do not require the presence of insulin to facilitate the uptake of glucose. However, insulin does stimulate the first step in the glycolytic pathway within the liver. Therefore, in liver cells, insulin most likely: A. hinders glucose uptake by increasing the cellular concentration of glucose. B. aids glucose uptake by decreasing the cellular concentration of glucose. C. hinders glucose uptake by using the ATP needed by the glucose transporter proteins. D. aids glucose uptake by providing the ATP needed by the glucose transporter proteins.

The answer to this question is B because the stem states that insulin stimulates the first step in the glycolytic pathway in the liver, thus decreasing the cellular concentration of glucose. This results in increased uptake of glucose to maintain the cellular concentration of glucose. Insulin--->Increased Glycolysis in the Liver Cells---> Decreased concentration of glucose in the liver cells ----> Increased re-uptake of Glucose in the liver in order to equilibrate glucose concentration in liver cells.

What is the total number of fused rings present in a steroid? A. 1 B. 2 C. 4 D. 6

The answer to this question is C because a steroid is defined by its fused 4-ring structure.

The aldosterone deficiency associated with Addison's disease will cause a decrease in the serum levels of all of the following ions EXCEPT: A. Na+ ions. B. Cl- ions. C. K+ ions. D. HCO3- ions.

The answer to this question is C because the passage states that aldosterone causes excretion of K+. Therefore, aldosterone deficiency will not lead to a decrease in serum levels of K+ ions.

Which primer is most suitable for PCR? A. 5-ATTACGTTAACATGAAG-3 B. 5-ATATCGTTAACAAATTG-3 C. 5-GCTATAAAGATTGCAAA-3 D. 5-GCATAGAAGCATTCCGC-3

The answer to this question is D because suitable primers have a high GC content and have G or C base pairs at the 5 and 3 ends.

Enzymes alter the rate of chemical reactions by all of the following methods EXCEPT: A. co-localizing substrates. B. altering local pH. C. altering substrate shape. D. altering substrate primary structure.

The answer to this question is D. The primary structure of a protein substrate is the amino acid sequence of the protein. Enzymes cannot alter primary structures of protein, but can colocalize substrates, alter local pH, and alter substrate shape.

Bisphosphonates are a class of medications that function to inhibit bone resorption. They are frequently used by physicians to prevent or reduce the bone loss that occurs during osteoporosis. Which of the following is a potential mechanism of action for bisphosphonates? A) Increase calcium absorption in the small intestine B) Trigger apoptosis, or programmed cell death, in osteoclasts C) Stimulate osteoblasts D) Prevent renal losses of calcium into the urine

The correct answer is B. The question stem states that bisphosphonates function to "inhibit bone resorption." By directly targeting the osteoclasts (the predominant cell type involved in bone resorption) for apoptosis, bone loss should be reduced (choice B is correct). Increasing calcium absorption from the intestines and preventing its loss in the urine would increase serum calcium concentrations and potentially improve bone formation. However, promoting bone formation is not equivalent to preventing bone loss, and the question text specifies that bisphosphonates act to inhibit bone resorption (choices A and D are wrong). Similarly, stimulating osteoblasts would target bone formation, not resorption (choice C is wrong).

medulla oblongata cerebellum hypothalamus midbrain hippocampus thalamus

The medulla oblongata is responsible for autonomic functions, such as breathing rate and heart rate regulation. Dysfunction of the medulla oblongata could result in problems with breathing rhythm. The cerebellum is responsible for coordination and balance. The hypothalamus regulates the fight-or-flight response (activation of the sympathetic nervous system), sex drive, thirst, and hunger, temperature, endocrine function. The midbrain is the center for auditory and visual signal relay to the cortex. The hippocampus functions in the retention of memories.The thalamus is primarily responsible for integrating and coordinating sensory input to the brain.

Lung capillaries are so narrow that RBCs must pass through them in single file. This feature aids respiration by: A. increasing the production of CO2 in the RBCs. B. allowing RBCs to have direct contact with alveoli. C. giving maximum exposure of each RBC to diffusing gases. D. making Hb available for CO2 but not O2 to bind.

c. If the RBCs must pass through the capillaries in single file, more of their surface area will be exposed to the diffusing gases, plus they will have to proceed a little more slowly and orderly. This will increase the rate of diffusion to more effectively move the gases in and out of the cell (choice C is correct). There is no reason to assume this would cause an increase in CO2 production (choice A is wrong) and also no reason to assume that Hb would only be available to CO2 and not O2 (choice D is wrong). The RBCs never have direct contact with the alveoli; the alveolar walls and the capillary walls separate the alveoli from the RBCs (choice B is wrong).