Biological and Biochemical Section

Which of the following describes the molecular geometry of a carbon dioxide molecule? A. Linear B. Trigonal planar C. Tetrahedral D. Bent

A. Linear VSEPR theory predicts (and experiments have verified) that the carbon dioxide molecule, O=C=O, is linear since the central carbon atom contains no lone-pair electrons (choice A is correct and eliminate choice D) and the two regions of high electron density (the two double bonds) are most stable on opposite sides of the central atom. In order to be trigonal planar or tetrahedral, the central carbon would need to be bonded to three or four atoms, respectively (eliminate choices B and C).

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

Which of the following are products of the pentose phosphate pathway? I.NADH II.NADPH III.Ribose-5-phosphate A. I only B. III only C. II and III only D. I, II, and III

C Item I is false: NADH is an electron-carrier produced in catabolic reactions, like cellular respiration (choices A and D can be eliminated). Note that since both remaining choices include Item III it must be true and we can focus on Item II, which is true: NADPH is produced by the pentose phosphate pathway (choice B can be eliminated and choice C is correct). Item III is in fact true: ribose-5-phosphate is a primary product of the pentose phosphate pathway.

The fundamental difference between the 34-, 17-, and 14-amino acid forms of gastrin can be attributed to: A. primary structure. B. secondary structure. C. tertiary structure. D. quaternary structure.

A These are very small polypeptides. Since they are single polypeptides, they cannot have quaternary structure, which requires interactions between different polypeptides chains (eliminate choice D). Tertiary structure and secondary structure are unlikely to be achieved with such short polypeptides (eliminate choices C and B, respectively). Primary structure is simply the linear amino acid sequence, which certainly will be different since the peptides have different numbers of amino acids (choice A is correct).

A barometer is made by erecting an evacuated glass tube over a dish filled with mercury. Atmospheric pressure at sea level causes the mercury to rise inside the tube. If this barometer were moved from sea level to a mountain several kilometers in altitude, the mercury column would most likely: A. fall due to the decreased atmospheric pressure. B. rise due to the decreased atmospheric pressure. C. fall due to the increased atmospheric pressure. D. rise due to the increased atmospheric pressure.

A Atmospheric pressure on a mountain several kilometers in altitude would be less than at sea level (eliminate choices C and D). Since the mercury in the dish feels less pressure from the atmosphere, it would not be pushed up as high in the tube (eliminate choice B and choice A is correct).

Which of the following statements best describes pyruvate kinase? A. It catalyzes the conversion of phosphoenolpyruvate to pyruvate and uses one ADP molecule. Correct Answer B. Its quaternary protein structure is most similar to hemoglobin and myoglobin. C. It is involved in gluconeogenesis causing the conversion of pyruvate to glucose during high energy states. D. It has several isoforms due to alternative splicing via the spliceosome and ribosome.

A Pruvate kinase catalyzes the last step of glycolysis (eliminate choice C), where phosphoenolpyruvate is converted to pyruvate. Because it is a kinase, pyruvate kinase transfers an inorganic phosphate from phosphoenolpyruvate to ADP to form a molecule of ATP (choice A is correct). Paragraph 3 discusses how PKM2 can exist as a dimer or tetramer, each of which would contain more than one peptide chain and would therefore have quaternary protein structure. Hemoglobin also has quaternary structure (since it contains four peptide chains) but myoglobin does not (eliminate choice B). Since PKM2 is one isoform of pyruvate kinase, there are presumably others. Isoforms are different forms of the same protein and can be due to gene duplication or alternative splicing. However, the ribosome has no function in splicing (eliminate choice D).

RNA polymerase differs from DNA polymerase III in that RNA polymerase: I.does not require a primer. II.lacks exonuclease activity. III.travels in the 5' to 3' direction on the template strand. A. I and II only B. I and III only C. II and III only D. I, II, and III

A RNA polymerase does not require a primer (statement I is true), and it does not have exonuclease activity (statement II is also true). RNA polymerase synthesizes RNA in the 5' to 3' direction, which means it must travel 3' to 5' on the template strand (statement III is false).

A solution is prepared conataining 450 grams of NaI in enough water to create a 2 L solution. What is the molarity of this solution? A. 1.5 M B. 3 M C. 6 M D. 9 M

A Sodium iodide (NaI) has a molar mass of 23 + 127 = 150 g/mol, so 450 g of NaI represents 3 moles. If this is dissolved in enough water to make a 2 L solution, then the concentration of NaI(aq) is [NaI] = 3 mol/2 L = 1.5 M

Blockage of the K+ leak channels in an axon would produce which of the following? A. Depolarization B. More rapid repolarization C. Hyperpolarization D. Increased activity of the Na+/K+ ATPase

A The K+ leak channels are the channels primarily responsible for the negative resting membrane potential. They allow many positive ions to flow out of the cell (K+ moves down its gradient, to the outside of the cell). If these channels were blocked, these positive ions would remain inside the cell, driving the membrane potential in the positive direction. Movement away from rest potential in the positive direction is called depolarization (A is correct, and B and C are wrong). Note that this would not cause an increase in the Na+/K+ ATPase activity, since it is not dependent on ion gradients for its function (D is wrong).

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).

Following a meal with a high phenylalanine content, the tyrosine levels in a PKU patient would be expected to be: A. lower than in unaffected individuals. B. the same as in unaffected individuals. C. higher than in unaffected individuals. D. reduced due to absence of the conversion enzyme

A The passage states that in PKU individuals, the enzyme that converts phenylalanine to tyrosine is missing or mutated, thus phenylalanine levels would be expected to be higher than normal, and tyrosine levels lower than normal (B and C are wrong). D is a poor choice since it insists the reduction in tyrosine levels be due to a missing enzyme, when the enzyme could merely be mutated. A is a better choice.

A soluble protein with a specific three-dimensional conformation is gently heated until it no longer displays activity. If at this point it is then allowed to cool, the protein resumes its function. The best explanation for these observations is that: A. the bonds that produce the three-dimensional configuration of a protein, once broken, are able to reform under certain conditions. B. the bonds that produce the three-dimensional configuration of a protein are impervious to changes in temperature. C. when a protein undergoes denaturation, the denaturation is always reversible. D. extremes in temperature will always produce reversible denaturation of a protein.

A The three-dimensional structures of proteins are very important to their functions. Destroying the three dimensional structure by heating or otherwise denaturing the protein can thus destroy the protein's function. In some cases however, such as mild heating, the bonds are able to reform once the protein has cooled, and the protein function is restored (A is correct, and B is wrong). C and D are poor answer choices because of the word "always." Denaturation is not always reversible (C is wrong), and extremes in temperature often produce irreversible destruction (D is wrong). Also in the case described in the question, the protein was only mildly heated, implying that extremes in temperature were not reached.

If the measurements taken in Step 4 of Experiments 1 and 2 had shown no change in bacterial lactose concentration, then it is likely that within the medium: A. hydrogen ion concentration had remained unchanged in Experiments 1 and 2. B. hydrogen ion concentration had increased in Experiments 1 and 2. C. potassium ion concentration had remained unchanged in Experiments 1 and 2. D. potassium ion concentration had increased in Experiments 1 and 2.

A These experiments show that a gradient of either potassium or protons can drive the transport of lactose into the cell against its gradient. If there is no lactose transport, it is safe to assume that the concentration in the media of the driving ion will remain the same (choice A is correct). There is no reason to assume that hydrogen ion concentration would increase (regardless if lactose transport had occurred or not) since this is not the case in any of the experiments (eliminate choice B). Step 4 involves a proton gradient, not a potassium gradient (eliminate choices C and D).

Aldosterone's mechanism of action on its target cell is most similar to that of which of the following hormones? A. Testosterone B. Glucagon C. Insulin D. ACTH

A Two general classes of hormones are those that are small hydrophobic molecules like steroid hormones and those that are peptides. The steroid hormones, which include aldosterone and testosterone, diffuse through the plasma membrane to bind to a receptor which enters the nucleus to regulate transcription of a specific set of genes (choice A is correct). Peptide hormones, such as glucagon, insulin, and ACTH, cannot diffuse into the cell since they are large and hydrophilic, so they bind to cell-surface receptors to transduce a signal into cells (choices B, C, and D are wrong).

As described in the passage, which of the following best describes how PrPSc causes disease formation? A. PrPSc induces changes in the secondary structure of PrPc proteins by causing alpha helices to convert to beta sheets. B. PrPSc induces changes in the primary structure of PrPc proteins by causing changes in the amino acid sequence that alter protein folding. C. PrPSc induces changes in the primary structure of PrPc proteins by causing alpha helices to convert to beta sheets. D. PrPSc induces changes in the secondary structure of PrPc protein by causing the formation of mutations in chaperone proteins, preventing proper protein folding.

A. While a defect in chaperone proteins could plausibly cause disease, the question asks for the method by which PrPSc causes disease as described in the passage, and chaperone proteins were not discussed (eliminate choice D). The passage does, however, explain how PrPSc causes changes in the structure of normal proteins. The passage states that the amino acid sequence (the primary structure) does not change (choices B and C can be eliminated, and choice A is correct). Changes in prion proteins take place at the alpha-helix and beta-sheet levels of protein structure; this is the secondary structure of a protein.

During times of abundant energy, ATP acts as a: A. positive feedback regulator by stimulating the activity of PFK. B. negative feedback regulator by inhibiting the activity of PFK. C. competitive inhibitor by inhibiting the activity of PFK. D. noncompetitive inhibitor by stimulating the activity of PFK

B

Certain digestive enzymes secreted by pancreatic acinar cells are synthesized as longer polypeptide chains than they actually possess when carrying out their specific functions. The most reasonable explanation for this observation is that these enzymes: A. are targeted by lysosomal hydrolases for protein degradation. B. are zymogens. C. have not gone through mRNA splicing. D. have been inactivated through cleavage by proteases.

B Because the digestive enzymes secreted by the pancreatic acinar cells would digest the pancreas itself if secreted in an active form, the enzymes are secreted in an inactive precursor form called a zymogen. Activation of the zymogen usually involves removal of a portion of the protein precursor, therefore the zymogen can be considerably longer than the active form of the enzyme (B is correct). The proteins are to be secreted, not digested by the lysosome (A is wrong), and if they had been cleaved by proteases they would be shorter, not longer (D is wrong). Finally, enzymes do not go through mRNA splicing, mRNA does (C is wrong).

A drug is discovered that reduces smooth muscle contraction by inhibiting the muscle's response to acetylcholine. This is likely to: A. reduce the transcription of Na+/K+ ATPase. B. reduce the hypertensive effect of angiotensin II. C. increase the strength of cardiac muscle contractions. D. decrease renal blood flow.

B In the absence of acetylcholine signaling, the overall tone of smooth muscle in artery walls will be reduced, reducing blood pressure (choice B is correct). Transcription is not related to smooth muscle contraction (eliminate choice A), and the drug acts on smooth muscle, not cardiac muscle (eliminate choice C). If the smooth muscle in blood vessels fails to contract, then blood vessels would dilate. This would increase, not decrease, renal blood flow (eliminate choice D)

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.

Which one of the following is true for all liquids? A. Liquids are always more stable than gases or solids. B. A higher vapor pressure indicates a lower heat of vaporization. C. As the pressure on a liquid increases, the temperature of that liquid decreases. D. Fluid pressure is equal at all points in a liquid.

B Statements A, C, and D are all false, so the answer must be B. The vapor pressure of a liquid is the pressure at which vaporization and condensation are at equilibrium. Higher vapor pressure implies weaker intermolecular forces, and, consequently, a lower heat of vaporization.

An activated aldosterone receptor most directly regulates activity of which of the following enzymes? A. DNA polymerase B. RNA polymerase II C. Na+/K+ ATPase D. Renin

B Aldosterone binds to aldosterone receptors to regulate transcription of a specific set of genes. The enzyme that synthesizes mRNA is RNA polymerase II, so it is this enzyme that would be most directly affected by the activated aldosterone receptor (choice B is correct). DNA polymerase is used in replication, not transcription (choice A is incorrect), and while aldosterone may ultimately affect the activity of the Na+/K+ ATPase and renin, these would be indirect effects (choices C and D are incorrect).

The OCA2 gene overlaps with another gene called HERC2, which has two alleles: the wild type allele, and a recessive allele A1 which has been linked to a genetic predisposition to Crohn's disease. A woman with an OCA2305R : HERC2wt chromosome and an OCA2305W : HERC2A1 chromosome mates with a homozygous OCA2305R / 305R : HERC2A1 / A1 man. Their children will most likely be: A. 25% normal with blue eyes, 25% normal with brown eyes, 25% blue eyed with a risk of Crohn's disease and 25% brown eyed with a risk of Crohn's disease. B. 50% normal with blue eyes and 50% brown eyed with a risk of Crohn's disease. C. 50% blue eyed with a risk of Crohn's disease and 50% normal with brown eyes. D. 100% blue eyed with a risk of Crohn's disease.

B Based on information in the passage, alleles of OCA2 are one of the major determinants of eye color. The woman in the question stem will have brown eyes because she is heterozygous for OCA2 and has an allele that is associated with blue eyes (305R) and an allele associated with brown eyes (305W). Since the question stem says that the OCA2 gene and the HERC2 gene overlap, they must be linked (they are 0 map units apart). This means that crossing over will not occur between these two genes and they will be inherited as a unit. The father in this question is acting like a testcross; that is, he is homozygous recessive for both genes. Therefore, the offspring produced will either get the OCA2305R : HERC2wt chromosome or the OCA2305W : HERC2A1 chromosome from the mother and an OCA2305R : HERC2A1 chromosome from the father. The children will be 50% OCA2305R / 305R : HERC2wt / A1 (blue eyed with no increased risk of Crohn's disease; note that the A1 allele of HERC2 is recessive to the wild type allele) and 50% OCA2305W / 305R : HERC2A1 / A1 (brown eyes with an increased risk of Crohn's disease). Thus, choice B is correct (eliminate choices A, C, and D). Note than choice A is what would occur if the two genes were not linked.

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 portion of the central nervous system confers balance by coordinating the activity of various motor units? A. Cerebrum B. Cerebellum C. Medulla D. Hippocampus

B Coordination of motor skills is one of the primary functions of the cerebellum (choice B is correct). The cerebrum triggers skeletal muscle contraction, but the cerebellum coordinates it (eliminate choice A). The medulla regulates many homeostatic functions (eliminate choice C) and the hippocampus is a component of the limbic system (eliminate choice D).

Gastrin secretion is inhibited by low pH in the stomach. This is an example of: A. competitive inhibition. B. negative feedback. C. acid hydrolysis of proteins. D. exocrine secretion of a hormone

B Gastrin stimulates acid production, and its release is inhibited by stomach acid. This is a negative-feedback loop designed to maintain acid within a certain pH range (choice B is correct). Competitive inhibition does not apply in this case, because there is no enzyme to be inhibited (eliminate choice A). Since gastrin, a hormone, is not excreted into the stomach lumen, it will not be degraded by stomach acid (eliminate choice C). A hormone cannot be secreted in an exocrine manner (eliminate choice D).

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. Your Answer 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 been suggested that one reason tumor cells have novel metabolism is to generate the additional biomolecules required to increase biomass, which is needed to support high rate of proliferation. Which is the best explanation of how this could occur? A. Fructose-6-phosphate is shuttled from glycolysis to the pentose phosphate pathway to support both nucleotide and fatty acid biosynthesis. B. Glutamate can be used to produce Krebs cycle intermediates such as oxaloacetate and citrate and these molecules can be used to generate lipids, amino acids and nucleotides. C. High glycolysis rates can support the pentose phosphate pathway, which generates ribose-2-phosphate for nucleotide catabolism. D. Glutamate is converted into glutamine to provide the cell with an amino acid precursor, thus powering translation.

B Glucose-6-phosphate (not fructose-6-phosphate) is shuttled from glycolysis to the pentose phosphate pathway (choice A is wrong), although this does support nucleotide and fatty acid biosynthesis. Paragraph 2 describes how glutamine is converted into the Krebs cycle intermediate α-ketoglutarate, which can then be used to generate other Krebs cycle intermediates to power biomolecule synthesis (choice B is correct). The pentose phosphate pathway generates ribose-5-phosphate (not ribose-2-phosphate) and this allows nucleotide anabolism (not catabolism, choice C is wrong). Paragraph 2 describes how glutamine is first converted into glutamate, not the other way around (choice D is wrong).

A highly proliferating cell would most likely: A. overexpress hexokinase and fructose-1,6-bisphosphatase. B. express high levels of the lactate transporter and the glutamine transporter. C. power cell growth by running the electron transport chain and oxidative phosphorylation. D. upregulate pyruvate dehydrogenase kinase and downregulate phosphofructokinase.

B Highly proliferating cells would express large amounts of hexokinase (a key glycolysis enzyme) but would not also overexpress fructose-1,6-bisphosphatase because this is an enzyme involved in gluconeogenesis. Even highly proliferative cells will avoid concurrently running reciprocally regulated pathways (eliminate choice A). The passage says that rapidly dividing tumor cells produce and export large amounts of lactate, and import large amounts of the amino acid glutamine (choice B is correct). The focus of this passage is how highly proliferative tumor cells power growth via glycolysis and fermentation, making choice C an unlikely correct answer (eliminate choice C). Based on information in the last paragraph, pyruvate dehydrogenase kinase activity is high in some rapidly growing cells, but a high rate of glycolysis will lead to high PFK expression (eliminate choice D).

During DNA replication, which of the following enzymes is necessary for completion of the lagging strand? A. Helicase B. DNA ligase C. Topoisomerase II D. Primase

B In DNA replication, the lagging strand is a result of the discontinuous 5' → 3' synthesis on the template strand of DNA. These Okazaki fragments need to be joined together to make one continuous strand of DNA. DNA ligase is the enzyme that will connect the Okazaki fragments (choice B is correct). Helicase is used to unzip the double-helix structure of DNA to allow for DNA polymerase to enter. Topoisomerase II (DNA gyrase) is the enzyme that relieves supercoiling of DNA that typically occurs with circular DNA, not linear DNA. It relieves the supercoiling by making nicks in both strands of DNA to allow uncoiling and then reconnecting those strands. Primase is the enzyme that is necessary to generate an RNA primer upon which DNA polymerase can latch to begin DNA polymerization.

According to the scheme depicted in Figure 1, if the carbon dioxide concentration increases within the cell: A. more Cl- will be secreted into the interstitium. B. more HCO3- will be secreted into the interstitium. C. less Cl- will be excreted into the lumen. D. less H+ will be excreted into the lumen

B In Figure 1, chloride moves inward from the interstitium (eliminate choice A). The more CO2, the more bicarbonate which is formed and transported into the interstitium in exchange for chloride (choice B is correct). If more bicarbonate is produced and excreted into the interstitium, more chloride will enter the cell and be secreted into the lumen (eliminate choice C). The more CO2 in the cell, the more H+ will be produced and driven into the lumen (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).

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).

Retroviruses, which are a subclass of RNA viruses, are unique in that they contain: A. DNA-dependent DNA polymerase. B. RNA-dependent DNA polymerase. C. DNA-dependent RNA polymerase. D. RNA-dependent RNA replicase. Your Answer

B Retroviruses have RNA genomes but undergo the lysogenic cycle in hosts with double-stranded DNA genomes. Therefore, the virus must be able to reverse transcribe its genome into DNA so that it can successfully integrate into the host-cell genome. Enzymes that are able to create a strand of DNA by reading a strand of RNA are called RNA-dependent DNA polymerases (B is correct). DNA-dependent DNA polymerases make a strand of DNA by reading a strand of DNA and are what the host cell normally uses for replicating its own genome (A is wrong). DNA-dependent RNA polymerases can make a strand of RNA by reading a strand of DNA and are what the host cell uses to transcribe its genome (C is wrong), and there is no such thing as an RNA-dependent RNA replicase (D is wrong).

Among the following routes, the one most likely taken in the formation of Protein X and its incorporation in a lysosome by a normal (non-Hurler's) cell is: A. synthesis in Golgi apparatus, passage to endoplasmic reticulum, secretion to extracellular space, uptake by same or different cells, and incorporation into lysosome. B. synthesis in endoplasmic reticulum, passage to Golgi apparatus, secretion to extracellular space, uptake by same or different cells, and incorporation into lysosome. C. synthesis in Golgi apparatus, passage to endoplasmic reticulum, and budding off endoplasmic reticulum to produce lysosome containing Protein X. D. synthesis in endoplasmic reticulum and direct passage to lysosome containing Protein X.

B Secreted, membrane-bound, and organelle proteins are synthesized by ribosomes on the rough ER (choices A and C are wrong). From the rough ER they are directed to the Golgi for processing and packaging before traveling to the cell surface or to another organelle (choice B is correct, and choice D is wrong).

Can young White Leghorn female chicks respond to estrogenic stimulation? A. Yes, because their weight normally increases during the gestational period. B. Yes, because their oviduct tissue enlarges in response to stilbestrol injection. C. No, because post-pubertal females have estrogen-insensitive oviduct tissue. D. No, because endogenous estrogen is directed toward respiratory, digestive, and other nonreproductive functions.

B The animals used in the experiment shown in Figure 1 were young White Leghorn chicks. These animals responded to stilbestrol with a large increase in oviduct weight, indicating that these animals can respond to estrogenic stimulation such as that of stilbestrol (choice B is correct). It is certainly true that chick weight will increase during gestation, but the relation of this to estrogen is not clear (eliminate choice A). After puberty, estrogen will play a key role in reproduction, and it is likely that oviducts will be regulated by estrogen (eliminate choice C). Stilbestrol is only a synthetic compound used in this experiment to mimic the effects that endogenous estrogen would normally produce, of which reproductive functions are particularly important (eliminate choice D).

A tall plant with curly leaves (TT/CC ) was crossed with a dwarf plant with straight leaves (tt/cc). Two of the resulting F1s were crossed with each other and produced F2s as follows: 34 tall plants with curly leaves 4 dwarf plants with straight leaves 11 tall plants with straight leaves 12 dwarf plants with curly leaves. Based on these results, are the genes for plant height and leaf shape linked? A. No, since recombination occurred. B. No, since the phenotype ratio is close to the expected ratio for this cross. C. Yes, since the recombination frequency is less than 50%. D. Yes, since there are considerably fewer recombinant plants than expected.

B The cross between the parental plants produced an F1 generation that all have the genotype Tt/Cc (double heterozygotes). These plants were then crossed to produce the F2 generation. The expected unlinked phenotype ratio from a cross between two double heterozygotes is 9:3:3:1, with 9/16 of the offspring double-dominant, 3/16 dominant/recessive, 3/16 recessive/dominant, and 1/16 double-recessive. Based on the numbers given in the question, the actual ratio (34:11:12:4) is very close to this, so these genes are not linked (choice B is correct). Recombination can occur between linked genes; it just happens less frequently (eliminate choice A). The recombination frequency (RF, the number of recombinants divided by the total number of offspring) does not have to be 50% for the genes to be unlinked; in the unlinked 9:3:3:1 ratio, the RF is only 37.5% (eliminate choice C). Based on the total number of F2's produced (34 + 4 + 11 + 12 = 61), the expected number of recombinants was (37.5%)(61) = (3/8)(61) ≈ 23, and 11 + 12 = 23 were produced (eliminate choice D).

Based on the test results in Figure 1 and Table 1, Patient 3 is most likely suffering from: A. fulminant hepatitis caused by HBV and HDV coinfection. B. prion disease. C. both prion disease and fulminant hepatitis. D. another condition causing neurological symptoms, because this patient is negative for both prion diseases and hepatitis.

B. This question requires the use of Figure 1 and Table 1 to determine whether Patient 3 is suffering from a prion disease, hepatitis, or both. According to Figure 1, an analysis of Patient 3's blood using SOFIA showed high levels of light emittance. As described in the passage, this means that Patient 3 had high levels of PrPSc, which is consistent with a prion disease (eliminate choices A and D). According to Table 1, Patient 3 tested negative for any antibodies associated with Hepatitis B or Hepatitis D (eliminate choice C). Patient 3 is suffering from prion disease, and not hepatitis (choice B is correct).

If pirenzepine is administered to a Zollinger-Ellison syndrome patient, which of the following will occur? A. Luminal secretion of H+ will decrease, because high intracellular Ca2+ will inhibit protein kinases. B. Luminal H+ secretion will increase, because low intracellular cAMP will increase protein kinase activity. C. Luminal H+ secretion will decrease, because low intracellular Ca2+ will lead to lowered protein kinase activity. D. Luminal H+ secretion will increase, because intracellular pH will rise.

C If acetylcholine is blocked, calcium levels will be low, protein kinase activity low, and H+ secretion low (choice C is correct). Calcium will be low (eliminate choice A) and H+ secretion will decrease, not increase (eliminate choices B and D).

A bacteriologist initiated an E. coli culture from one E. coli cell and hypothesized that some of the progeny in the culture would be genetically different from the original parent cell. Is this hypothesis true? A. Yes; bacteria are capable of undergoing genetic recombination through a variety of mechanisms. B. Yes; bacteria reproduce sexually, and the progeny of any one cell are genetically distinct from the parent cell. C. No; bacteria are asexual organisms, and in the absence of mutation, all progeny of any one cell are genetically identical to the parent cell. D. No; bacteria can reproduce only by meiosis, which ensures preservation of the genome

C Bacteria reproduce asexually, by one cell replicating its genome and then splitting into two cells that are genetically identical to the original cell. The only potential sources of genetic variation in bacteria are mutation and the transfer of genetic information through conjugation, transduction, or transformation, none of which are linked to reproduction. In the absence of mutation, all progeny of a cell will be identical to the original cell (choice C is correct). Bacteria only perform recombination under special circumstances such as through the presence of Hfr plasmids that replicate a portion of the bacterial genome to make it transiently diploid (eliminate choice A). There is no indication of a role for Hfr in this case and in a clonal cell line, it could not play a role. Bacteria do not perform the recombination, independent assortment and independent segregation that create genetic diversity in eukaryotes that reproduce sexually (eliminate choice B). They also do not perform meiosis (eliminate choice D).

Arachidonic acid is a fatty acid contained in cell membranes. The structure of arachidonic acid is: A linear hydrocarbon chain of 20 carbon atoms containing 1 carboxyl group on the first carbon and double bonds between 5th&6th carbon atoms, 8th&9th carbon atoms, 11th&12th carbon atoms and 14th&15th carbon atoms. 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).

A FISH probe should be made of: A. an antibody that binds to a DNA epitope, covalently linked to a fluorescent segment of DNA. B. a segment of double-stranded DNA which is complementary to the gene being studied, covalently linked to a fluorescent protein. C. a segment of single-stranded DNA which is complementary to the gene being studied, covalently linked to a fluorochrome. D. a single-stranded piece of RNA.

C The passage states that during FISH, fluorescently labeled probes are hybridized to chromosome clusters. This is similar to how probes are used in Southern and Northern blots. The probe must be single-stranded DNA or RNA if it is going to hybridize or bind to the chromosomes (which are denatured first, also similar to the process in Southern or Northern blotting), and should be covalently linked to the fluorescent molecule (choice C is correct). An antibody is not typically used to bind DNA, and in any case, there is no easy way to generate a "fluorescent segment of DNA." It is much easier to use fluorescent molecules or dyes (choices A and D are wrong). A double-stranded probe will not bind the chromosomes because it will be complementary to itself (choice B is wrong).

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).

2H2O2 --> 2 H2O +O2 The substrate in the experiment is: A. catalase. B. potassium permanganate. C. hydrogen peroxide. D. water.

C Catalase is an enzyme that acts on hydrogen peroxide, reducing it to water and oxygen (choice A is wrong and choice C is correct). In an enzymatic reaction, the substrate is that on which the enzyme acts. Potassium permanganate serves in the titration but is not the substrate (choice B is wrong). Choice D refers to water, a product.

Metformin is used in the treatment of diabetic patients. One of the possible effects of metformin on hepatocytes will be to: A. stimulate gluconeogenesis to decrease blood glucose levels. B. stimulate glycolysis to increase blood glucose levels. C. inhibit gluconeogenesis to decrease blood glucose levels. D. inhibit glycolysis to decrease blood glucose levels.

C Diabetics have increased glucose levels in their blood. Therefore, it is expected that the actions of metformin will be to lower the levels of glucose in the blood. Stimulating gluconeogenesis would increase blood glucose levels (choice A is wrong). Stimulating glycolysis will result in decreased blood glucose levels, while inhibiting it will result in increased blood glucose levels, not the other way around; and in any case, diabetics are unable to take up glucose and thus could not run glycolysis at all (choices B and D are wrong). Inhibiting gluconeogenesis will decrease the glucose production by the liver and hence decrease the blood glucose levels, making C the correct answer choice.

If a spontaneous mutation occurred which eliminated a restriction site on the DNA sample from Individual #1, one would expect the electrophoresis results to show: A. one smaller fragment. B. two smaller fragments. C. one larger fragment. D. two larger fragments.

C If a restriction site is eliminated, the DNA can no longer be cut at that site. Thus the results would show the presence of one larger fragment, which should be equal in size to the combination of the two smaller fragments.

Tryptophan, an essential amino acid found in banana, turkey, and milk proteins, can induce sleep in some people. Warm milk causes greater sleepiness than cold milk because heating the milk: A. reduces the solubility of tryptophan in the milk. B. causes hydrolysis of lactose, releasing tryptophan. C. releases free tryptophan from proteins, causing more rapid intestinal absorption. D. increases the rate of absorption of tryptophan by the stomach.

C If heating the milk reduced tryptophan solubility, this would decrease, not increase the sleep-inducing properties of milk (eliminate choice A). Lactose is a disaccharide, not a protein, and its hydrolysis cannot release an amino acid (eliminate choice B). Amino acids, like most nutrients, are absorbed in the small intestine mostly, not the stomach (eliminate choice D). Although heating the milk does not create more tryptophan, it might help to hydrolyze some of the milk proteins and release tryptophan so it can be absorbed more rapidly after ingestion and cause greater sleepiness (choice C is correct).

Which of the following could explain the Warburg effect? I. etabolic differences allow cancer cells to adapt to hypoxic (oxygen-deficient) conditions inside solid tumors. II. ells with low proliferation rates often have a high ratio of glycolysis to mitochondrial respiration. III. ome oncogenic changes shut down the mitochondria because these organelles are involved in apoptosis, which would result in cell death. A. I only B. III only C. I and III only D. I, II, and III

C Item I is true: tumors can grow quite quickly and often the inside of tumors don't have sufficient blood supply. This can lead to hypoxic conditions inside the tumor. Since glycolysis and fermentation don't require oxygen, they could facilitate cell growth in anaerobic conditions (choice B can be eliminated). Item II is false: the passage is about highly proliferative cells that use glycolysis and fermentation to power growth so a statement about slow growth does not explain the Warburg effect discussed in the passage (choice D can be eliminated). Item III is true: if a tumor cell is relying on glycolysis and fermentation instead of mitochondrial cell respiration, it will have less use for the mitochondria in general. Since the mitochondria can initiate apoptosis, less reliance on this organelle could lead to apoptosis resistance, conferring a survival advantage to the cancerous cell (choice A can be eliminated and choice C is correct).

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).

In normal individuals phenylalanine is considered to be an "essential" amino acid, because it cannot be synthesized by the body. However, tyrosine is not considered "essential," because it can be synthesized from phenylalanine. Based on this knowledge and the information in the passage, which of the following statements is true concerning individuals suffering from PKU? I.Tyrosine is considered an essential amino acid. II.Tyrosine should be supplemented but is not considered essential. III.Phenylalanine is considered an essential amino acid. A. I only B. II only C. I and III only D. II and III only

C The definition of an essential amino acid is one that cannot be synthesized by the body. Since PKU sufferers lack (or have a mutated version of) the enzyme that converts phenylalanine to tyrosine, they are unable to synthesize tyrosine, and in them, tyrosine is "essential" (statement I is true). Tyrosine should be supplemented in these individuals because it is essential (statement II is false). This disease has no effect on the body's ability to synthesize phenylalanine; this is still not possible in PKU sufferers, so phenylalanine is still considered essential (statement III is true, it must simply be consumed in a much lower quantity).

Which of the following represents the correct sequence for lytic DNA bacteriophage replication? A. Genome transcription, infection of the host cell, mRNA translation, progeny assembly B. Infection of the host cell, mRNA translation, genome transcription, progeny assembly, genome replication C. Infection of the host cell, genome transcription, mRNA translation, progeny assembly D. Infection of the host cell, genome replication, mRNA translation, genome transcription, progeny assembly

C The first step in any viral life cycle is infection (A is wrong). Since the question asks about a DNA virus, the next step must be transcription of the viral genome into mRNA (B and D are wrong, and C is correct).

Eye color, skin color and hair color in humans are examples of: A. monogenism, where the trait is determined by a single gene with dominant and recessive alleles. B. codominance, where two alleles of a single gene are expressed simultaneously. C. polygenism, where the trait is determined by several genes with several alleles. D. incomplete dominance, where alleles of one gene blend together to give intermediate phenotypes.

C The passage discusses many proteins and genes that control pigmentation in humans (MC1R, MSH, ASIP, TYR and OCA2) and mentions that the OCA2 gene alone has 58 alleles that have so far been documented. This information best matches choice C, since several genes are involved in pigmentation phenotypes. All other answer choices indicate that only a single gene is involved (choices A, B, and D are wrong). Note that while examples in genetics often use eye color to demonstrate monogenism, based on information in the passage, this would be an oversimplification (choice A 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).

In a population of 18,000 Caucasians, how many are expected to be carriers of cystic fibrosis? A. 50 B. 295 C. 590 D. 1180

C The passage states that the frequency of the autosomal recessive condition cystic fibrosis, q2, is 1 in 3600. The frequency of the recessive allele, q, then is 1 in 60. The frequency of the dominant non-disease producing allele, p, is 59 in 60. The carriers of a population are determined by the expression 2pq. In the given population, the number of carriers would be (2)(1/60)(59/60)(18000) or 590. Thus, choice C is correct and choices A, B, and D are eliminated.

If an individual suffering from symptoms identical to PKU is found to have normal phenylalanine hydroxylase activity, what is the most likely cause of the symptoms? A. A spontaneous mutation in the phenylalanine hydroxylase gene B. Adrenal hyperplasia C. An excess of phenylalanine in the diet D. Peripheral nervous system dysfunction

C The question states that phenylalanine hydroxylase activity is normal, so the enzyme cannot have been mutated (A is wrong). The adrenal glands have nothing to do with this disorder (B is wrong), and peripheral nervous system dysfunction cannot cause the mental retardation seen in PKU individuals (D is wrong). The only logical choice is C. If the levels of phenylalanine in the diet exceed the amount of phenylalanine hydroxylase available to convert it to tyrosine, then some unmetabolized phenylalanine would remain and could cause symptoms similar to those of PKU.

If radiolabeled stilbestrol were administered to the experimental chicks, given its effect on the oviduct, the stilbestrol would be found most heavily concentrated: A. at the cell membrane of oviduct tissue. B. in the cytoplasm of oviduct tissue. C. in the nuclei of oviduct tissue. D. in the mitochondria of oviduct tissue

C When stimulated by the addition of a ligand such as stilbestrol, estrogen receptor will localize within the nucleus, where it regulates genes by binding to enhancers and promoters. Radiolabeled stilbestrol would localize with estrogen receptor in the nucleus (choice C is correct). There is no estrogen receptor in the plasma membrane or mitochondria (eliminate choices A and D). Some estrogen receptor may be located in the cytoplasm, particularly in the absence of ligand, but it will localize mostly in the nucleus when it has ligand bound (eliminate choice B).

Based on the results of Experiments 1 and 2, it is reasonable to assume that all components of the mechanism necessary for concentrating lactose intracellularly within E. coli are located in the: A. rough endoplasmic reticulum. B. genome. C. cell membrane. D. Golgi bodies.

C The key experiment is Experiment 2, in which vesicles formed from the cell membrane had the same ability to concentrate lactose as whole cells (choice C is correct). The genome will encode protein components, but the active components are not themselves physically located in the genome (eliminate choice B). The ER and Golgi are not found in bacteria (eliminate choices A and D).

Based on the data collected from the medical students' study, what percentage of the isolates were methicillin-resistant Staphylococcus aureus? A. 115/120 B. 115/160 C. 115/345 D. 115/785

C The table lists the number of cultures of each type of organism. The number of positive isolates for methicillin-resistant Staphylococcus aureus is 115. The total number of cultures reported is the number of Gram-(+) plus the number of Gram-(-) plus the fungal cultures: 160 + 180 + 5 = 345. Therefore, choice C is correct (and choices A, B, and D are eliminated).

Which of the following best describes the role of fructose-2,6-bisphosphate? A. It exerts reciprocal control on glycolysis and gluconeogenesis by stimulating fructose-1,6-bisphosphatase and inhibiting phosphofructokinase. B. It exerts reciprocal control on glycolysis and gluconeogenesis by stimulating phosphofructokinase and inhibiting hexokinase. C. It exerts reciprocal control on glycolysis and gluconeogenesis by stimulating phosphofructokinase and inhibiting fructose-1,6-bisphosphatase. D. It exerts reciprocal control on glycolysis and gluconeogenesis by stimulating hexokinase and inhibiting fructose-1,6-bisphosphatase

C. Fructose-2,6-bisphosphate exerts reciprocal control on glycolysis and gluconeogenesis through phosphofructokinase and fructose-1,6-bisphosphatase, and has no impact on hexokinase activity (eliminate choices B and D). Fructose-2,6-bisphosphate stimulates phosphofructokinase, which is used in glycolysis, and inhibits fructose-1,6-bisphosphatase, which is used in gluconeogenesis (choice C is correct and eliminate choice D).

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).

The Gram-staining procedure used in the laboratory enables the inspector to: A. identify bacterial species present in the incubate. B. distinguish between aerobic and anaerobic organisms. C. differentiate pathogenic from nonpathogenic colonies. D. distinguish bacteria whose peptidoglycan layer lies outside the cell membrane from those whose peptidoglycan layer lies within the periplasmic space.

D Gram-positive bacteria have a thick peptidoglycan layer outside of the cell membrane. It is this outermost layer of peptidoglycan that takes up the Gram dye and gives a positive Gram-stain result. Gram-negative bacteria, on the other hand, have a considerably thinner layer of peptidoglycan, which is located within the periplasmic space and is isolated from the outside environment by an outer phospholipid membrane. Both of these factors cause the Gram dye to be removed when alcohol is applied to these latter bacteria as part of the staining process, thus making them "Gram-negative" (choice D is correct). Choice A is incorrect as Gram-staining does not allow specific identification of bacterial species; it only narrows the options by specifying whether the bacteria are Gram-positive or Gram-negative. Choices B and C are wrong because Gram staining does not elucidate metabolic processes or pathogenicity.

The diaphragm plays an important role in respiration. During inspiration, the diaphragm: A. relaxes, causing alveolar pressure to drop below atmospheric pressure. B. contracts, causing alveolar pressure to rise above atmospheric pressure. C. relaxes, causing alveolar pressure to rise above atmospheric pressure. D. contracts, causing alveolar pressure to drop below atmospheric pressure.

D Recognize theh opportunity to treat this question as a 2 X 2 elimination. Inspiration is the drawing of air into the lungs. The diaphragm contracts and flattens during inspiration (eliminate choices A and C), expanding the chest cavity; the lungs (which are stuck to the inside wall of the chest cavity) expand as well. The expansion of the lungs decreases the pressure in the alveoli, causing air to move into the lungs from the exterior (eliminate choice B and choice D is correct).

Some vitamins are essential to humans because they act as precursors of: A. auxins. B. glucose. C. enzymes. D. coenzymes.

D . Auxins are plant hormones not found in humans (eliminate choice A). Glucose is not made from vitamins (eliminate choice B). Enzymes are proteins made from amino acids (eliminate choice C). Many coenzymes (which are not themselves amino acids and so cannot be made into enzymes) are required for enzyme activity and are derived from water-soluble vitamins, such as thiamine, biotin, folate, and niacin (choice D is correct).

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 is a true statement? A. If anticancer treatment is working well on a patient, more FDG will be detected on a positron emission tomography scan. B. Expression of telomerase, a ribonucleoprotein complex, is typically higher in normal somatic cells than in tumor cells. C. Glutamine is a hydrophobic amino acid with one chiral center, a basic amino group and an acidic carboxyl group. D. Tumor cells can use cap-independent translation to make proteins when growth conditions are not ideal.

D Cap-independent translation uses internal ribosomal entry sites (IRESs) and allows a cell to translate proteins during sub-optimal growth conditions (because less regulation is necessary). Since tumors typically grow quickly, they are often short of oxygen and nutrients and must deal with acidic and CO2-rich growth conditions. Activating cap-independent translation would allow the tumor cells to continue proliferating even under these less-than-optimal conditions (choice D is correct). Based on information in the last paragraph, tumors uptake large amounts of FDG, which can be imaged via PET. This means that if an anticancer treatment is working, less FDG should be detected (choice A is wrong). Normal somatic cells do not express telomerase. This enzyme is only expressed in the germ line, by some white blood cells, and in some tumor cells (choice B is wrong). Glutamine is a hydrophilic polar amino acid (choice C is wrong). Note that explicit knowledge of cap-independent translation was not necessary, only the ability to recognize the other three statements as false.

The HCO3-/Cl- exchange is an example of: A. exocytosis. B. active transport. C. simple diffusion. D. facilitated diffusion.

D Facilitated diffusion involves movement of molecules down a gradient with the involvement of a protein. The passage states that both bicarbonate and chloride ions are moving down a gradient, making this a case of facilitated transport (choice D is correct). Exocytosis does not generally involve ion transport and does not utilize membrane channels (eliminate choice A). Active transport involves moving ions or other molecules against a gradient (eliminate choice B). Simple diffusion is the movement of molecules down a gradient without a protein involved (eliminate choice C).

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.

A proton gradient is most directly related to the functioning of: A. the Na+/K+ ATPase. B. the collecting ducts in the nephron. C. voltage-gated calcium channels. D. ATP synthase.

D Proton gradients are established by the electron transport chain during aerobic respiration. The gradient is then used to power an ATP synthase (D is correct). The Na+/K+ ATPase hydrolyzes a molecule of ATP to move sodium and potassium against their concentration gradients (A is wrong), the collecting ducts in the nephron regulate osmotic balance of the body through interactions with ADH (B is wrong), and voltage-gated Ca2+ channels are regulated by voltage, not proton gradients (C is wrong).

Scientists are currently attempting to create a nose spray based on the inhibitory factor in Medium 3 to inhibit picorna virus and stop the common cold. In order for this drug to be effective, the inhibitory factor must be able to: I.bind to the target cell's ribosomes and attach to the host's mRNA. II.traverse the target cell membrane and enter its cytoplasm. III.stop the virus from replicating in the target cell. A. I only B. II only C. I and III only D. II and III only

D Since the drug is to inhibit viral replication (Statement III is true), it must be able to get across the host cell's membrane where the virus is being replicated (Statement II is true) and bind to the viral protein factor or enzyme, not the host-cell mRNA (Statement I is false).

The removal of the adrenal glands will result in a reduction of all of the following EXCEPT: A. K+ secretion in the nephron. B. Na+ reabsorption in the nephron. C. water reabsorption in the nephron. D. renin secretion by juxtaglomerular cells.

D The hormones secreted by the adrenal glands include epinephrine from the medulla, and cortisol, aldosterone, and low levels of sex steroids from the adrenal cortex. As stated in the passage, aldosterone increases potassium secretion and also increases sodium reabsorption (choices A and B are true and thus are eliminated). The increased Na+ reabsorption leads to increased water reabsorption, thereby increasing the blood volume (choice C is true and thus is eliminated). The loss of aldosterone would cause water and sodium loss, decreased blood volume and decreased blood pressure. In response to decreased blood pressure, renin secretion would increase, not decrease (choice D is incorrect and thus the correct answer choice).

Which of the following is an example of the type of defect that XP patients cannot repair? A. Guanine-guanine B. Cytosine-guanine C. Adenine-thymine D. Thymine-thymine

D The passage states that the problem with XP is the inability to repair pyrimidine dimers created by ultraviolet radiation. Cytosine, uracil, and thymine are the pyrimidines. Therefore, thymine-thymine is the only pyrimidine dimer listed in the answer choices.

A researcher has discovered a novel frameshift mutation in pyruvate carboxylase and hypothesizes that this could affect tumor metabolism. Which of the following would be the best experiment to perform to either confirm or disprove her hypothesis? A. Isolate genomic DNA from a cancer cell line and sequence the gene for pyruvate carboxylase, to confirm which amino acid is altered in the mutant form. B. Generate an animal model that expresses the mutant form of pyruvate carboxylase, and determine how the proton gradient is used by ATP synthase isolated from these animals. C. Perform western blot analysis on lysates from a brain tumor to confirm the mutation causes increased enzymatic activity. D. Generate a cell line that expresses the mutant form of pyruvate carboxylase and measure lactate secretion, and glucose and glutamine uptake compared to control cells that express the normal form of pyruvate carboxylase.

D The question stem states that a frameshift mutation in pyruvate carboxylase has already been found. There would therefore be little point in sequencing the gene again. In addition, "confirm[ing] which amino acid is altered in the mutant form" is a better match to confirming a point mutation rather than a frameshift mutation (choice A is wrong). Pyruvate carboxylase catalyzes the conversion of pyruvate to oxaloacetate in the first step of gluconeogenesis. Determining how the proton gradient is used by ATP synthase isn't relevant to this enzyme's function (choice B is wrong). Option C is a very tempting answer choice because it would be ideal to determine if the mutation is affecting enzymatic activity. Unfortunately, a western blot will not give information on this, since it can only measure protein levels (choice C is wrong). By process of elimination, the correct answer is choice D. Measuring lactate secretion, and glucose and glutamine uptake will give an indirect readout of whether basic metabolic processes are different in cells with the pyruvate carboxylase mutation versus in normal cells (choice D is correct).

If a patient with cystic fibrosis receives a double-lung transplant from a non-cystic fibrosis donor, would the new lungs be expected to develop cystic fibrosis? A. Yes, once you have cystic fibrosis, it develops in every organ of the body. B. Yes, since the primary defect is with respiratory secretions. C. No, because the infectious causes of the disease will be removed when the old lungs are taken out. D. No, since cystic fibrosis is due to a gene defect, the cells of the new lungs will have the normal CFTR gene.

D. Cystic fibrosis is a genetic disease based on the abnormal protein CFTR. In a set of lungs from a person without cystic fibrosis, the lung cells presumably have normal-functioning CFTRs. Therefore, the new lungs should not be subject to the development of cystic fibrosis (eliminate choices A and B). Cystic fibrosis is a multi-organ disease since the CFTR is used in secretions from several glands, but this does not cause normal CFTRs to become abnormal. The primary defect is the protein CFTR, not the pulmonary secretions or an infectious cause (choice C is incorrect and choice D is correct).

After an adult with cystic fibrosis ingests a carbohydrate-rich meal, which of the following would you expect to occur? A. High intracellular concentration of glucose B. Increased secretion of insulin C. Increased secretion of glucagon D. High extracellular concentration of glucose

D. The passage states that people with cystic fibrosis tend to have autodestruction of the pancreas. Autodestruction would eliminate the islets of Langerhans which secrete both glucagon and insulin (choices B and C are incorrect). After a carbohydrate-rich meal, the serum glucose concentration should rise (choice D is correct). In the absence of insulin, the serum glucose concentration will remain high until it can be cleared from the blood by the kidney (choice A is incorrect).

In which of the following structures is E. coli most likely to cause infection? A. Trachea B. Bladder C. Small intestine D. Large intestine

B The passage states that E. coli only causes disease outside of the intestinal tract (eliminate choices C and D), in the urinary tract, the biliary tract, and the nervous system in particular. The bladder (choice B is correct and eliminate choice A) is the best answer since this is part of the urinary tract, mentioned as a potential site of infection.

During spermatogenesis, spermatids: A. are frozen in meiosis II until after fertilization. B. have already undergone meiotic recombination. C. have four copies of the genome per cell. D. have no nucleus.

B. Spermatids are the sperm precursors that have completed meiosis but have not yet fully matured. A mature sperm has fully completed meiosis, and this is true of spermatids as well; it is the ova that are frozen in meiosis II until after fertilization (eliminate choice A). Recombination occurs in both oogenesis and spermatogenesis during meiotic prophase I, which the spermatid already completed (choice B is correct). The spermatid has passed through two reductive divisions in meiosis to end up with only one copy of the genome (eliminate choice C). The cell does have a nucleus. In fact, the sperm has virtually no cytoplasm, but the genome is still packaged into a nucleus (eliminate choice D).

Which of the following is expected to be true of children with Down syndrome? A. Aortic arterial blood carbon dioxide concentration is normal. B. Aortic arterial blood oxygen saturation is lower than normal. C. Pulmonary arterial blood pressure is lower than normal. D. Pulmonary arterial blood oxygen saturation is lower than normal.

B. In children with Down syndrome, one of the primary cardiac defects is a truncus arteriosus, which presents as a common arterial trunk coming off both the left and right ventricles. This allows for mixing of blood from the left and right circulations. Since the right circulation is relatively high in carbon dioxide and the left circulation is relatively low in carbon dioxide, the resultant mix will be somewhere between (i.e., not normal, choice A is wrong). By similar reasoning, the right circulation that is typically low in oxygen will mix with the left circulation that is typically higher in oxygen, so the overall aortic arterial oxygen saturation will be less than normal (choice B is correct) and the pulmonary arterial oxygen concentration will be higher than it is normally (choice D is wrong). Since a common arterial trunk is receiving blood from both the right and left ventricles, the pulmonary artery will be receiving more blood than usual and therefore will have a higher than normal blood pressure (choice C is wrong).

Fluoroquinolones are a new class of antibiotics that are extremely effective against a wide range of bacteria. It has been determined that fluoroquinolones enter cells along with water via a passive transport mechanism. After exposing both bacterial and human cells to high concentrations of fluoroquinolones, a researcher discovered that, in the bacterial cells only, the DNA was nicked and supercoiling had been disrupted. Of the following, the most likely explanation for this is that fluoroquinolones: A. are unable to diffuse into human cells. B. interfere with DNA polymerase, which is not present in human cells. C. interfere with DNA gyrase, which is not present in human cells. D. interfere with DNA helicase, which is not present in human cells.

C. Since fluoroquinolones diffuse into cells with water, they are able to enter all cells, eukaryote and prokaryote alike; they diffuse through special water channel proteins called porins (eliminate choice A). Both eukaryotes and prokaryotes utilize DNA polymerase and helicase (eliminate choices B and D). However, only prokaryotes use DNA gyrase to supercoil their DNA; eukaryotes wind DNA around histones through the action of other topoisomerases (choice C is correct).

Each of the following could increase the rate of a reaction EXCEPT: A. increasing the temperature. B. adding a catalyst. C. increasing the concentration of reactants. D. increasing the concentration of products.

D The rate of a reaction can be given by the product of the rate constant, k, and the concentration of the reactants raised to the correct exponents. The reaction rate typically depends on the concentration of some or all of the reactants, so choice C is true and can be eliminated. The rate constant is given by the Arrhenius equation, k = Ae-Ea/RT. Increasing the temperature would increase k, so choice A is also true, and can be eliminated. Adding a catalyst would decrease the activation energy of the reaction and also increase k, so choice B is true and can be eliminated. Increasing the concentration of products will not increase the reaction rate, making choice D the correct answer.

In a given population, what is the frequency of carriers of Patau syndrome? A. 1/100 B. 198/10,000 C. 9,801/10,000 D. None of the above

D. The Hardy-Weinberg equation can only be used to describe the frequencies of autosomal recessive or dominant traits or conditions, not changes in chromosome number. Since trisomies are not autosomal recessive conditions, the Hardy-Weinberg equation cannot be used, and one cannot predict the occurrence of the carriers in a population (choice D is correct and choices A, B, and C are eliminated). Also, because trisomies are not based on dominant or recessive expressions, there is no "carrier" state.

What to remember when attacking pedigree problems?

Remember to answer the three basic questions before tackling any pedigree problem, and remember that in pedigrees showing two different conditions, the three questions must be answered separately for each condition. First, is the condition/disease caused by a dominant allele or a recessive allele? If the disease skips generations it is most likely recessive.