Princeton Review Test: Biology

A cancer-inducing virus was found that incorporates the sequence of a proto-oncogene into its DNA. The gene in the viral genome was transcribed in infected cells at a much higher rate than the same gene in the genome of the host cell. The induction of cancer is most likely due to: A. uncontrolled cell growth resulting from the overexpression of the gene product. Correct Answer B. mutation of the host-cell genome. C. tissue exhibiting contact inhibition. D. removal of oncogenes within the affected tissues.

A. A proto-oncogene being transcribed at a higher rate than normal could result in uncontrolled cell growth and cancer (choice A is correct). While mutation of the host-cell genome could cause cancer, this is not occurring in this case. The viral genome is merely being transcribed at a high rate (choice B is wrong). Contact inhibition is a means by which normal tissues can control their growth and division. A tissue exhibiting contact inhibition is not a cancerous tissue (choice C is wrong). Removal of oncogenes within the affected tissues would reduce the rate of cancer, not cause cancer (choice D is wrong).

Guevedoche children will develop a penis, scrotum with testicles, and other secondary male characteristics when they reach puberty. Initially perceived as females at birth, some are found to be males before puberty when an ultrasound of masses in their abdomen reveals undescended testicles and the absence of a uterus or fallopian tubes. Based on this information, it is most likely true that guevedoche children: A. produce MIF at normal levels. Correct Answer B. produce MIF at greatly elevated levels. C. produce MIF at greatly reduced levels. D. do not produce any MIF.

A. As stated in the passage, the uterus and fallopian tubes are derived from the Müllerian ducts. The absence of these organs indicates that MIF must be present (choice D is wrong), and at least at normal levels (choice C is wrong). There is no reason to assume that MIF levels would be greatly elevated, and in fact the passage states that while the testosterone-to-DHT ratio is elevated, the other hormones and enzymes are found at normal levels (choice A is correct and B is wrong).

Protostomes undergo spiral and determinant morula cleavage, where the two first divisions of the zygote are equal and the others are unequal. This results in some small and some large cells early on in morula formation, and means cell fate is decided early in the process. Deuterostomes develop differently. They perform radial cleavage, which results in totipotent cells that are all the same size. Which of the following is supported by this information? A. Humans are deuterostomes. Correct Answer B. The deuterostome morula is more differentiated than the protostome morula. C. The protostome morula cells are more similar to stem cells than the deuterostome morula cells D. The protostome morula has undergone less cell specialization than the deuterostome morula.

A. Based on information in the question, the protostome morula is more differentiated (or specialized) than the deuterostome morula (choices B and D can be eliminated). This means the deuterostome morula cells are more stem cell like (choice C can be eliminated). Cells in the human morula are all the same and are totipotent (choice A is correct).

Based on the passage, decreased Fru-2,6-P2 in the hepatocytes will stimulate which of the following? A. Gluconeogenesis Correct Answer B. Glycolysis Your Answer C. Krebs cycle D. Electron transport

A. Based on the passage Fru-2,6-P2 stimulates PFK-1 and inhibits FBPase-1. Therefore, lower levels of this metabolite will result in decreased glycolysis and increased gluconeogenesis. This makes answer choice A correct and B incorrect. Nothing in the passage suggests that Krebs cycle or electron transport will be affected by the levels of this metabolite.

All of the following statements regarding the digestive system of cystic fibrosis patients are true EXCEPT: A. flow of bile produced in the gallbladder may be prevented due to obstruction of small bile ducts. Correct Answer B. damage to exocrine tissue within the pancreas may hinder the secretion of lipases. C. secretions from both the salivary gland and pancreas are important for starch digestion. D. CF patients may have a normal digestive system.Your Answer

A. Bile is produced by the liver and is only stored in the gallbladder (choice A is not true and is the correct choice). Part of the exocrine function of the pancreas is to release digestive enzymes, such as lipases, into the small intestine. Damage to this tissue could prevent or hinder the release of these enzymes, and this may contribute to the difficulty seen in fat absorption (choice B is a true statement and can be eliminated). Starch digestion begins in the mouth, where salivary amylase breaks starch into polysaccharides. These polysaccharides are then broken down into disaccharides in the duodenum by pancreatic amylase. Note that this is true for anyone, CF patient or not (choice C is true and can be eliminated). Some CF patients may be completely normal with respect to the digestive system; the passage states that the CFTR gene is large and the gene can be mutated many different ways, leading to different CF phenotypes (choice D is true and can be eliminated).

Which of the following environmental factors will negatively impact the growth of Streptococcus faecalis? I. A decrease in the nutritional content of the milk II. The process of pasteurization III. A decrease in ambient oxygen concentration A. I only Correct Answer B. I and II only C. II and III only D. I, II, and III

A. Item I is correct: The relevant information is provided in paragraph 1. Streptococcus faecalis uses the sugar found in milk (lactose) for fermentation to generate ATP for the cells. Decreasing the lactose content of the milk would negatively impact bacterial growth. Item II is incorrect: Streptococcus faecalis is a spore-forming bacterium, and pasteurization kills only non-spore-forming bacteria. Item III is incorrect: Streptococcus faecalis is a facultative anaerobe, meaning that it does not require oxygen for growth.

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 Correct Answer 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).

Which of the following molecules contains an sp hybridized atom? A. Carbon dioxide Correct Answer B. Methanol C. cis-2-pentene D. Ammonia

A. Since the carbon atom in O=C=O has only two groups bonded to it and has no lone-pair electrons, CO2 must be a linear molecule, and the carbon atom must be sp hybridized.

Based on information provided in the passage, the uptake of Protein X by cultured Hurler's cells is most likely done by way of: A. receptor-mediated endocytosis. Correct Answer B. simple diffusion. C. active transport. D. osmosis.

A. Since the uptake of Protein X is selective, with about half the original concentration being taken up compared to other substances in the medium, a receptor must be involved (A is correct). Proteins cannot cross the membrane through simple diffusion (B is wrong), active transport requires the use of ATP which the passage does not mention is necessary (C is wrong), and osmosis is the movement of water down its concentration gradient (D is wrong).

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 Correct Answer 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 Correct Answer 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).

The degradation of connective tissue components in emphysema is most likely caused by which of the following? A. Proteases Correct Answer B. Lipases C. Monoanime oxidases D. Glycoside hydrolases

A. The connective tissue components damaged in emphysema include collagen and elastin, which are extracellular proteins. Choice A is correct. Therefore, proteases released from the alveolar macrophages would be the culprits. Lipases degrade lipids (choice B can be eliminated), monoanime oxidases degrade certain neurotransmitters such as norepinephrine and epinephrine (choice C can be eliminated), and glycoside hydrolases degrade carbohydrates (choice D can be eliminated).

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. Correct Answer B. elevated plasma osmolarity and blood pressure. C. an increased heart rate. D. elevated blood glucose level

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. Correct Answer 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. Correct Answer 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.

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. Correct Answer 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).

Using the data provided in the passage, what is the best explanation for the high viability rate of the DU145 prostate cancer cells when treated with Orlistat, compared to the other two prostate cancer cell lines? A. The DU145 cell line will not take up [1-C14] acetate. B. The DU145 cell line expresses less FAS than other cell lines. Correct Answer C. The DU145 cell line is more malignant than the other two prostate cancer cell lines. D. The LNCaP and PC3 cell lines contain a form of FAS more susceptible to Orlistat.

B. According to Figure 1 there is a relatively low expression of FAS in DU145, which likely accounts for the decreased impact of Orlistat treatment observed in Figure 2. This indicates that in these cancerous cells, FAS plays a less significant role in cancerous physiology (choice B is correct). Figure 1 demonstrates that [1-C14] acetate uptake does occur in DU145 cells, even if the uptake is lower than in the other two cell lines (eliminate choice A). Nowhere in the passage is it stated that DU145 is more malignant than the other cell lines (eliminate choice C). While it is possible that there is a FAS isozyme that is not inhibited by Orlistat, the data provided in the passage do not support this statement (eliminate choice D).

Which of the following conclusions is best supported by the data provided in the passage? A. Cells with nonsense mutations have an increased sensitivity to cisplatin. B. When compared to mice with missense mutations, mice with nonsense mutations in MLH1 have an increased incidence of colon tumors but do not exhibit decreased survival. Correct Answer C. T lymphocytes with nonsense mutations in MLH1 survive longer than wild type T lymphocytes. Your Answer D. When compared to mice with missense mutations, mice with nonsense mutations in MLH1 exhibit both an increased incidence of colon tumors and decreased survival.

B. According to Figure 2, mice with nonsense mutations (MLH1--/--) did have an increased incidence of colon tumors, when compared to mice with missense mutations (MLH1GY/GY). However, MLH1--/-- and MLH1GY/GY exhibit the same survival rate (and length). Thus, choice B is the correct answer, and choice D can be eliminated. Choice A is incorrect: according to Figure 3, cells with nonsense mutations (MLH1--/--) have the same rate of apotosis after treatment with cisplatin, and thus have a decreased sensitivity to cisplatin, not increased. Choice C can also be eliminated: T lymphocytes from mice with nonsense mutations (MLH1--/--) do not survive as long as wild type mice, as is shown in Figure 1.

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. Correct Answer C. competitive inhibitor by inhibiting the activity of PFK. D. noncompetitive inhibitor by stimulating the activity of PFK.

B. Based on Figure 1 of the passage, a high level of ATP inhibits the activity of PFK-1. This eliminates answer choices A and D. A competitive inhibitor would have the same the Vmax, which is not the case given the results of Experiment 1. This eliminates answer choice C, leaving choice B as the correct answer.

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+] Correct Answer 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.

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. Correct Answer 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).

Which of the following is the best technique for separating 2-butanol from propanoic acid? A. 1H NMR spectroscopy B. Fractional distillation Correct Answer 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 of the following describes a difference between the qPCR and microarray experiments that Dr. Rhev performed? A. While both experiments started with the isolation of RNA from cells, only the microarray experiment used reverse transcriptase. B. The qPCR data represents an arbitrary quantification of transcript expression, while the microarray data shows transcript levels (or fold changes) relative to a control.Correct Answer C. Microarray experiments are based on nucleic acid complementarity, since two samples of isolated RNA must compete for probe binding; qPCR instead uses a fluorescent dye that binds DNA. D. Data generated by qPCR is numerically larger than that of microarray data, since more RNA molecules are being tested by qPCR; microarrays instead generate relative fold change data, which is numerically smaller.

B. Both qPCR and microarray experiments start with the isolation of RNA from a cell. However, based on the passage, qPCR is done on cDNA which was generated from RNA and this is done via reverse transcriptase (eliminate choice A). qPCR data gives an arbitrary quantification of transcript expression. Microarray experiments are based on comparing transcript levels between two samples (in this case, a control cell line and an experimental cell line). Because of this, microarray data are presented as fold changes, or experimental with respect to control (choice B is correct). The first part of choice C is correct, but both microarrays and qPCR use fluorescent dyes that bind DNA (eliminate choice C). The qPCR data in Table 1 is numerically larger than the microarray data, but this is because it is based on an arbitrary scale, not because this experiment tests more RNA molecules (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. Correct Answer 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).

During DNA replication, which of the following enzymes is necessary for completion of the lagging strand? A. Helicase B. DNA ligase Correct Answer 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.

Which one of the following mutations would be most likely to convert a proto-oncogene into an oncogene? A. Silent mutation B. Missense mutation Correct Answer 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. Your Answer B. both proteins evolved from a single common ancestor.Correct Answer 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).

Based upon the results presented here, what is the Km (in mM) of PFK-1 for the uninhibited reaction? A. 0.05 B. 0.2 Correct Answer C. 0.5 D. 0.6 Your Answer

B. Km is defined as the substrate concentration required to reach one half of the maximal reaction rate. From the solid line (uninhibited reaction, low levels of ATP) in Figure 1, the Vmax = 50, so ½ Vmax = 25, and the substrate concentration required to get to 25 is approximately 0.2 mM.

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 activityCorrect Answer C. A person with increased osteocytic activity D. A person with higher than normal urinary calcium concentration Your Answer

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 proliferation of Lactobacillus in milk samples indicates: A. high lactose concentration. B. a drop in the pH. Correct Answer C. predominance of spore-forming bacteria. D. the absence of Streptococcus faecalis.

B. Paragraph 1 describes the environmental conditions that promote growth of Lactobacilli. A reduced pH via production of lactic acid by Streptococcus lactis, for example, provides a receptive environment for Lactobacilli (choice B is correct). Choice A should be eliminated because it contradicts paragraph 1, which tells you that lactose must be converted to lactic acid to spur growth of Lactobacilli. Choices C and D are wrong because in relation to the growth of Lactobacilli, the passage does not mention spore-forming bacteria or Streptococcus faecalis.

Which of the following is an accurate interpretation of the data from the experiment? A. There is significant improvement in survival of patients who received the gene replacement. B. There is no significant difference between the two groups in the incidence of new lesions. Correct Answer C. The experimental group had fewer new skin lesions at each measured interval. D. There are more patients who are lesion-free in the experimental group than in the control group.

B. Per the table, the numbers of new lesions in the two groups are extraordinarily similar. This indicates that there is no real difference between the two groups. There are several reasons as to why this may have occurred: (1) not all cells took up the adenovirus with the DNA, (2) those cells that did take up the adenovirus may not have expressed the DNA, and (3) more than one enzyme may play a role in excision repair. The survival curve shown illustrates that the two groups approach the same lesion-free percentage, so there is no survival benefit.

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

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).

In people suffering from color blindness, which of the following types of cells fails to function normally? A. Rod cells B. Cone cells Correct Answer 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).

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.Correct Answer 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).

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. Your Answer B. Streptomycin inhibits the small subunit (30S) of the ribosome. Correct Answer 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).

Based on Table 1, Protein E may be which of the following? I. γ-Globulins II. Hemoglobin III. Myoglobin A. I only B. II only Correct Answer C. II and III only Your Answer D. I, II, and III

B. Table 1 shows the function of Protein E to be gas transport. While both hemoglobin in blood and myoglobin in muscle can bind oxygen, only hemoglobin actually transports O2. Myoglobin simply stores O2to be used by the muscle cell (Statement II is true and III is false). γ-Globulins are just antibodies and are involved in immune function (Statement I is false).

Which of the following is NOT an appropriate resonance structure for the deprotonated form of acetic anhydride?

B. The deprotonated form of the molecule should have an overall net negative charge. Choice B is not a representation for the deprotonated form of acetic anhydride. It is simply a resonance form for acetic anhydride that has not been deprotonated.

Based on information in the passage, each of the following could be true EXCEPT: A. Lunatic, Manic and Radical fringe enzymes function as glycosyltransferases in the Golgi. B. Hey family proteins inhibit the transcription of Ifi22, c-myc and Bcl2. Correct Answer C. Hes proteins are helix-loop-helix transcription factors that function as transcriptional repressors. D. RBP-J is a DNA-binding protein that also contains at least one protein-protein interaction domain.

B. The passage says that Fringe enzymes modify Notch ligands and receptors and that this increases the molecular weight of the proteins. It is possible this involves glycosylation, which occurs primarily in the Golgi apparatus (choice A is possible and is thus not the answer). If Hey family proteins inhibit transcription of Ifi22, c-myc and Bcl2, the table should contain columns of data where Hey proteins have high expression, but Ifi22, c-myc and Bcl2 have low expression. While this is true for Ifi22, there are many columns in the chart that have high Hey expression and high c-myc/Bcl2 expression (choice B is not possible and is thus the correct answer). Based on information in the passage, Hes and Hey family proteins are Notch targets because they are highly expressed in the normal cell lines, expression increases when Notch signaling increases (i.e. when SHARP is mutated), and decreases when Notch signaling decreases (i.e. when Notch1, MAML1 or RBP-J are mutated). However, the passage contains no information on the cellular function of Hes and Hey proteins. It is possible they are helix-loop-helix transcription factors that function as transcriptional repressors for genes that are not included in Table 1 (choice C is possible and is thus not the answer). Based on information in the passage, RBP-J binds DNA and also either SHARP or the intracellular domain of Notch (choice D is possible and is thus not the answer). You may find this question challenging because the answer choices contain content for which you have no information; focus on finding an answer choice that contradicts the passage. Based on how this question is worded, three answer choices will be possible and one will be false. Remember that "possible" and "supported by the passage" are not the same thing.

Based on the passage, the influenza and X viruses probably differ from the picorna virus in that: A. the influenza and the X viruses contain an RNA genome whereas the picorna virus has a DNA genome. B. the ribosomes for the X and the influenza viruses dock at the end of the messenger RNA. Correct Answer C. the X and the influenza viruses contain viral enzymes whereas the picorna virus does not. D. the X and the influenza viruses consist of a capsid and nucleic acid whereas the picorna virus has a capsid, nucleic acid, and a tail.

B. The passage states that all the viruses were RNA viruses (A is wrong). It also states that the picorna virus' mRNA has an internal ribosome binding site and because of this requires an additional protein to act as a ribosome "receptor." This is the site of action of the drug. Since influenza and virus X are not affected by the drug, we can assume that they do not need this additional protein, and therefore bind ribosomes at the end of the mRNA (B is correct). There is nothing in the passage to suggest that the influenza and X viruses contain enzymes (C is wrong) or that their structure (which is completely irrelevant anyway) is different in any way (D is wrong).

Based on the passage, if the researcher wished to determine the amount of β-pleated sheet in Protein A, she would need to examine the protein's: A. primary structure. B. secondary structure. Correct Answer C. tertiary structure. D. quaternary structure.

B. The passage states that secondary protein structure includes α-helices and β-sheets.

A person who suffers partial destruction of the lateral lemniscus on the left side would be expected to suffer which of the following? A. Ipsilateral hearing loss Your Answer B. Bilateral hearing loss Correct Answer C. Total sensorineural deafness D. Defects in auditory processing and association

B. The passage states that the first decussation of the auditory pathway occurs at the connection between the cochlear nucleus and the superior olivary nucleus. Thus, any lesion that occurs past that point will result in binaural, rather than monaural, hearing loss (choice A is incorrect). By the same token, total sensorineural deafness would require bilateral destruction of the hearing pathway (choice C is incorrect). Auditory processing and association occur in the higher brain, after processing by the primary auditory cortex (choice D is incorrect). Bilateral hearing loss (choice B) is correct; a lesion on one side of the brain will result in partial but not complete hearing loss, since auditory information ascends bilaterally.

In a population of 10,000 people, how many carriers of XP are there? A. 99 B. 198 Correct Answer C. 900 D. 1,800

B. The passage states that the frequency of the recessive XP allele, q, is 1% (0.01). Therefore, the frequency of the dominant non-disease expressing allele, p, is 0.99. The number of carriers is determined by the expression 2pq. Therefore, in a population of 10,000 people, the number of carriers would be 2(0.01)(0.99)(10000), or 198.

The 11-cis-retinal chromophore associated with rhodopsin is derived from vitamin A. If a child is vitamin-A deficient, could this possibly lead to problems in night vision? A. Yes, because vitamins act as cofactors with proteins. B. Yes, because vitamin A is a precursor for 11-cis-retinal. Correct Answer C. No, because vitamins have no effect on visual excitation. D. No, because cones are responsible for night vision.

B. The question explains that 11-cis-retinal is derived from vitamin A, so vitamin A must be a precursor of 11-cis-retinal. A person deficient in vitamin A would have vision problems, particularly at night when the rod cells (which are responsible for dim, black and white vision) are more active (choice B is correct). Vitamins can act as cofactors but not in this case (eliminate choice A). Vitamins clearly have an effect on visual excitation in this case, as the excited pigment is derived from a vitamin (eliminate choice C), and cones are responsible for bright, color vision (eliminate choice D).

Synthesis of creatine requires: A. Glycine, arginine and cysteine Your Answer B. Glycine, arginine and methionine Correct Answer C. Alanine, lysine and cysteine D. Alanine, lysine and methionine

B. The two amino acids pictured on the left side of Figure 1 are glycine and arginine (choices C and D are wrong). The passage says that the third amino acid required in the synthesis of creatine has a thioether group. Only methionine has this functional group (choice A is wrong; choice B is correct).

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. Correct Answer 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.Correct Answer 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).

At an inter-particle distance of rm, what will be the value of the Lennard-Jones potential? A. Zero B. -ε Correct Answer C. 1/ε D. -ε6

B. When the inter-particle distance, r, is equal to rm then (choice B is correct; choices A, C, and D are 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.Correct Answer 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).

The most likely explanation for the inverse relationship between mammal size and pulse is that: A. most small mammals live under anaerobic conditions. B. large animals tend to consume proportionately more food than smaller animals. C. smaller mammals have a higher metabolic rate for a given mass of tissue than larger mammals. Correct Answer D. smaller mammals have reduced oxygen requirements for a given mass of tissue.

C. Choice A is wrong since all mammals use primarily aerobic respiration except for short periods of unusually strenuous activity. Choice B is wrong since it does not directly involve metabolic rate or pulse and would predict that the metabolic rate and pulse rate will remain the same despite size. Choice D is wrong since reduced oxygen requirements would predict a slower pulse. Choice C is the best choice: If tissues in smaller animals have a higher metabolic rate, they will require a greater supply of blood to provide oxygen and nutrients and carry away carbon dioxide. To meet these greater needs, the pulse rate could be increased.

Which of the following most appropriately describes the mechanism for the reaction shown? A. Decarboxylation B. Nucleophilic addition C. Addition-elimination Correct Answer D. Reduction

C. Decarboxylation requires two oxygens and a carbon to leave the compound generating carbon dioxide (eliminate choice A). This reaction is the formation of an imine; first the nucleophilic amine adds to the carbonyl group, breaking the bond. However, because the intermediate continues to react, eliminating water in the process to form the imine and form a new bond between the carbon and nitrogen, choice C is better than choice B. This is not a reduction because N, like O is more electronegative than C, so shifting from a carbonyl double bond to an imine double bond does not change the oxidation state of the carbon (eliminate choice D).

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.Correct Answer 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.

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.Correct Answer 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.

Guevedoche lack the ability to make 5-α-reductase in utero, and therefore do not develop a penis, scrotum, or prostate at that time. However, these male organs develop as they reach puberty. Which of the following is the most likely explanation for this phenomenon? A. The penis and scrotum that develop during adolescence are actually labial and clitoral swellings resulting from overexpression of the genes for the estrogen receptor. B. Because of a lack of 5-α-reductase, testosterone levels are greatly elevated, leading to masculinization of female external genitalia. Your Answer C. There are two genes on separate chromosomes for 5-α-reductase, and while one gene is normally expressed in utero, the other is not expressed until puberty. Correct Answer D. Because of a lack of 5-α-reductase, testosterone levels are elevated throughout childhood; then, at puberty, when testosterone and estrogen levels both increase even further, secondary male sex organs appear as a result of overdevelopment of the Wolffian ducts.

C. Guevedoche lack the ability to convert testosterone to DHT in utero, and therefore are unable at that time to develop the external male genitalia that depend on DHT for development (specifically the penis, prostate, and scrotum). There are actually two types of (5AR): type I 5AR is the product of a gene on Chromosome 5, and type II 5AR is the product of a gene on Chromosome 2. The mutation that leads to the guevedoche phenotype is found in type II 5AR; this gene is normally expressed in the penile, prostatic, and scrotal tissues in utero. However, other tissues (specifically scalp and skin) begin to express the gene for type I 5AR at puberty. The 5AR produced at puberty converts testosterone to DHT, and the DHT then binds to receptors on the tissues of the penis, scrotum, and prostate cells, causing them to develop (C is correct). Guevedoche have what appears to be labial swelling and an enlarged clitoris at birth, but these are actually just underdeveloped scrotal and penile tissues, which will develop further once levels of 5AR rise at puberty. In any case, expression levels of the estrogen receptor are not mentioned in the passage (choice A is wrong). The passage states that other hormones are found at normal levels, so testosterone levels can be expected to be normal (choices B and D are wrong). Only the ratio of testosterone to DHT is elevated, not the absolute level. Note also that the Wolffian ducts develop into the epididymis, vas deferens, and ejaculatory ducts, not the penis or scrotum (choice D is wrong).

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. Correct Answer 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.

Which of the following findings are likely to be seen on the ABR of a patient with bilateral acoustic neuromas? I. Decreased wave I amplitude II. Increased distance between waves I and III III. Increased distance between waves III and V A. I only B. II only C. I and II only Correct Answer D. II and III only

C. Item I is correct: In the legend for Figure 1, the text states that amplitude on ABR is correlated with number of neurons firing, while latency is correlated with speed of transmission. The passage states that acoustic neuromas decrease both number of functional neurons and speed of action potential propagation. Thus, bilateral acoustic neuromas will result in decreased wave I amplitude due to decreased numbers of functional neurons (eliminate choices B and D). Item II is correct: Since speed of action potential propagation within the vestibulocochlear nerve is slowed, measuring the time it takes signals to propagate through the vestibulocochlear nerve to another point in the brain will reveal a longer latency period (choice A is eliminated and choice C is correct). Item III is incorrect: Once a signal is within the brain, there is no reason to believe it will propagate any slower than it would in a normal individual.

The function of galactoside permease is to transport lactose across the bacterial cell membrane. Galactoside permease is necessary because: A. lactose is hydrophobic and cannot cross the lipid bilayer. B. lactose needs a protein carrier to move it against its concentration gradient. C. lactose is hydrophilic and cannot cross the lipid bilayer.Correct Answer D. lactose needs to be converted to allolactose to enter the cell.

C. Lactose, like other carbohydrates, is very hydrophilic, not hydrophobic (choice A is wrong). Hydrophilic molecules are unable to diffuse across a membrane into the cell (choice C is correct). There is no information to support B; in any case, it can be assumed that lactose is not being transported against a gradient (choice B is wrong). Allolactose only signals the presence of lactose, but it is lactose that is the energy source, and it is lactose that is imported into the cell (choice D is wrong).

Based on the information provided in Table 1, a reasonable hypothesis about proteins is that: A. globular proteins are always enzymes. B. fibrous proteins are composed only of β-sheets. C. globular proteins generally contain an active site.Correct Answer D. fibrous proteins generally have a regulatory function.

C. Protein C has a globular structure and is involved in immune function, not enzymatic catalysis (A is wrong). Proteins D and G are fibrous and are composed of α-helices (B is wrong), furthermore, most fibrous proteins are involved in structural roles (D is wrong). However, all the globular proteins described in the table (Proteins A, C, E, and F) contain an active site (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 Correct Answer 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).

Hurler's disease adheres to a pattern of autosomal recessive inheritance. A patient affected with Hurler's disease may have parents with which of the following genotypes? A. Both parents are carriers of Hurler's disease on one X chromosome each. B. One parent is a carrier of Hurler's disease on one X chromosome, and the other parent is normal. C. Both parents are carriers of Hurler's disease on a non-sex chromosome. Correct Answer D. One parent is affected with Hurler's disease, and the other parent is normal.

C. Since the disorder is autosomal, the defective gene will not be found on the X chromosome (A and B are wrong). Since the disorder is recessive, afflicted individuals must carry two defective genes, receiving one gene from each parent. Both parents must therefore be (at least) carriers (C is correct, and 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 Your Answer B. II only C. I and III only Correct Answer 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).

If one of the raw milk samples were slowly heated to 100°C and then held at that temperature for fifteen minutes, which of the following graphs would best represent resulting spore-forming and non-spore-forming bacterial populations during this time period?

C. The essential information is provided in paragraph 1, where the text states that pasteurization kills non-spore-forming bacteria. From this, infer that spore-forming bacteria are better able to resist the high temperatures of pasteurization (choice C is correct). Choice B should be eliminated because the graph it presents indicates high survival rates for the non-spore-forming bacteria. Paragraph 1 states that non-spore-forming bacteria do not survive pasteurization. Choices A and D are wrong because they indicate that pasteurization kills spore-forming bacteria.

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 Correct Answer 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).

Individuals with immune deficiencies are found to be more susceptible to cancer than are healthy individuals. Which of the following reasons is the most likely explanation? A. The lymph vessels contain tumor cells. B. These individuals produce an abundant amount of lymphocytes in their bone marrow. C. The immune system is compromised and cannot eliminate cancerous cells. Correct Answer D. The lymphocytes are no longer able to make antigens.

C. The job of the immune system is to remove any foreign substances (pathogens, toxins, etc.) from the body to prevent possible harm to the body. It also patrols the body for normal cells displaying abnormal functions (i.e., cancer cells) and eliminates them. If the immune system is deficient, it cannot do this very effectively (choice C is correct). Immune-deficient individuals would produce fewer lymphocytes than normal individuals, not more (choice B is wrong), and lymphocytes do not make antigens, they fight antigens (choice D is wrong). While choice A may be true, it is not the reason these individuals are more susceptible to cancer, but rather is the outcome of being more susceptible to cancer (choice A is wrong).

DAPT (N-[(3,5-difluorophenyl)acetyl]-L-alanyl-2-phenyl]glycine-1,1-dimethylethyl ester, C23H26F2N2O4) is a γ-secretase inhibitor. Treating TK103 cells with DAPT would most likely cause which of the following? A. CD25 transcript levels would increase due to elevated Notch signaling. B. Hes5 transcript levels would increase due to diminished Notch signaling. Your Answer C. Ifi22 transcript levels would increase due to diminished Notch signaling. Correct Answer D. Hey1 transcript levels would decrease due to elevated Notch signaling.

C. The passage says that γ-secretase functions in Notch receptor cleavage, and that this is an important step in generating the Notch intracellular domain. Treatment with DAPT will therefore decrease Notch signaling (eliminate choices A and D). This treatment will have a similar effect to the mutations in cell lines A04 and A18, since both of these have mutations that will lead to less Notch signaling. Hes5 transcript levels decrease in both these two cell lines (eliminate choice B), but Ifi22 transcript levels increase (choice C is correct).

According to the passage, which of the following may be associated with CF? I. Recurrent lung infection with bacteria that have a thick layer of peptidoglycan II. Deficiency in fat-soluble vitamins, such as vitamins A and E III. Inability to appropriately lower glucose levels after a meal A. I only B. II only C. II and III only Correct Answer D. I, II, and III

C. The passage states that CF patients have recurrent infections with Gram-negative bacteria (which have a thin layer of peptidoglycan and an outer membrane). Therefore, Item I is false, since it refers to Gram-positive bacteria. Item II is true: CF patients can have difficulty absorbing fats, and this could lead to problems absorbing the fat-soluble vitamins. Item III is also true: dysfunction (abnormal function) of the endocrine pancreas could lead to difficulties with insulin release (among the other hormones secreted by the pancreas), which would lead to an inability to reduce blood glucose levels after eating.

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

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 Correct Answer 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.

Deficiency of the 21-β-hydroxylase enzyme is an autosomal recessive disease that is the leading cause of congenital adrenal hyperplasia. Based on Figure 1, which of the following is least likely to be a symptom of 21-β-hydroxylase deficiency? A. Increased levels of 17-hydroxyprogesterone in the serum B. Affected females presenting with an enlarged clitoris, male pattern baldness, and early puberty C. Elevated levels of serum potassium with elevated levels of urine sodium D. Affected males presenting with severe hypertension in early childhood Correct Answer

D. 21-β-hydroxylase deficiency will lead to a lack of both aldosterone and cortisol due to the inability to synthesize these hormones, as well as increased levels of the substrates for this enzyme, namely progesterone and 17-OH-progesterone (choice A is likely to be a symptom and can be eliminated). The increased levels of 17-OH-progesterone will lead to the shunting of the pathway towards production of androgens, such as testosterone, which will lead to masculinization of female genital organs (choice B is likely to be a symptom and can be eliminated). The lack of cortisol interrupts the negative feedback pathway with ACTH and is the cause of the adrenal hyperplasia (enlargement of the adrenal glands in an effort to synthesize cortisol). The lack of aldosterone prevents reabsorption of sodium from the urine and secretion of potassium into the urine; consequently urine levels of sodium will increase, and serum levels of potassium will increase (choice C is likely to be a symptom and can be eliminated). However, the normal secretion of aldosterone and subsequent Na+ reabsorption is one of the primary means of regulating blood pressure. The reabsorption of sodium is followed by the reabsorption of water, as well as an increase in blood osmolarity and subsequent ADH secretion. The secretion of ADH causes even more water reabsorption, this time at the collecting duct of the nephron. The reabsorption of water increases blood volume and pressure. Thus, the lack of aldosterone due to the 21-β-hydroxylase deficiency would prevent the above from occurring, and would result in low blood pressure (choice D is not a possible symptom and is the correct answer choice).

If a single base pair mutation creates an oncogene from a proto-oncogene, this could possibly create all of the following outcomes EXCEPT: A. a protein that is more active. B. a protein that is more resistant to degradation than a normal protein. C. a tumor that metastasizes in the body. D. a tissue that undergoes the normal rounds of interphase.Correct Answer

D. A single base pair mutation in a gene could result in the replacement of an amino acid in the final protein produced from that gene with a new amino acid. This new protein could be more active or more resistant to degradation than the original protein (choices A and B are true and thus eliminated). The passage states that cancer (a metastasizing tumor) is the result of an activated oncogene (choice C is true and thus eliminated) and also that cancer cells lack the normal controls on cell growth and divide continuously (choice D is false and the correct answer).

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. Correct Answer

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 best compares the incidence of skin tumors in Group B and Group C mice? A. Group B mice have a higher incidence of skin tumors than Group C mice. B. Group C mice have a higher incidence of skin tumors than Group B mice. C. Group B mice have a higher incidence of skin tumors than Group C mice, but only when exposed to cisplatin. D. Group B and Group C mice exhibit the same incidence of skin tumors. Correct Answer

D. According to Figure 2, Group B (MLHGY/GY) mice have a lower rate of colon tumors than Group C (MLH1--/--) mice, but the same incidence of skin tumors (choice D is correct). There is no data in the passage that indicates skin tumor incidence as a function of cisplatin exposure (choice C can be eliminated). Choices A and B are both incorrect readings of Figure 2.

A couple comes to you for genetic counseling. They are considering having children, but the man is colorblind and a hemophiliac (both X-linked recessive traits). The woman does not have relatives with either disease. The couple wants to know how likely it is that they will have a child with either disease. What should be your response? A. There is a 50% chance that they will have a child with both diseases. B. There is a 25% chance that they will have a child with one of the two diseases. C. There is a 25% chance that, if they have a son, he will have both diseases. D. There is essentially no chance that they will have a child with either disease. Correct Answer

D. Both of these diseases are X-linked recessive; therefore, the father has an abnormal X chromosome. Since the mother has no relatives with either disease, it can be assumed that she has two normal X chromosomes. Therefore, children born to this couple could be: (1) normal boys, (2) normal girls, or (3) girls who are carriers for both diseases (carriers do not have the disease).

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 Correct Answer

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.

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. Correct Answer

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 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. Correct Answer

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.

Which of the following laboratory techniques could be used to detect prions? I. PCR II. ELISA III. Western blotting A. I only B. II only C. I and II only D. II and III only Correct Answer

D. Item I is false: PCR is used to amplify segments of DNA. Prions are proteins, and thus, PCR would not be an appropriate laboratory technique (choices A and C can be eliminated). Both remaining choices include Item II so it must be true: ELISA uses antibodies to test for proteins, and could be (and is) used to detect prion proteins. Item III is also true: Western blotting uses antibody probes to detect proteins (after separating them on a gel using gel electrophoresis and transferring them to a filter). Thus, western blotting can be (and is) used to detect prion proteins (choice B can be eliminated and choice D is correct).

Bilateral occlusion of the middle cerebral artery, which provides blood to the lateral temporal and parietal lobes, would be expected to cause an abnormality in which of the following components of a patient's ABR? I. Decreased wave III amplitude II. Decreased wave V amplitude III. Decreased wave VII amplitude A. I only B. I and II only C. II and III only D. III only Correct Answer

D. Item III is correct: The question describes a stroke which would obliterate cortical processing of auditory information in the auditory cortex and cause central deafness. Wave VII corresponds to the primary auditory cortex; since a stroke results in neuron death, it will show decreased amplitude (choices A and B can be eliminated). Items I and II are not correct: The rest of the pathway is not affected by the stroke, so amplitude of these areas should be normal (choice C is eliminated and choice D is correct).

Protein D might be any of the following EXCEPT: A. collagen. B. α-keratin. C. myosin. D. antibody. Correct Answer

D. Protein D plays a structural role. Collagen, α-keratin, and myosin can all play structural roles in cells (choices A, B, and C are wrong), but antibody proteins are secreted and play a role in immune function (choice D is correct).

All of the following may characterize Protein X EXCEPT that it is: A. an acid hydrolase. B. an enzyme. C. first secreted into the extracellular environment, where it is taken up by cells. D. produced by cells taken from Hurler's patients.Correct Answer

D. Protein X is clearly being secreted into the medium by normal cells, thus allowing it to be taken up by the Hurler's cells in order to correct the defect (C is true and eliminated). The lysosome is an organelle which functions in cellular digestion of endocytosed materials as well as old or worn-out cellular organelles. Digestion of this material requires enzymes, specifically acid hydrolases. Hurler's disease is characterized by excess GAGs building up in the lysosomes of affected patients; clearly, such individuals do not have the enzymes necessary to digest these macromolecules. Since Protein X is able to correct the problem, it must be missing in Hurler's individuals; furthermore, it must be an enzyme, and specifically an acid hydrolase (A and B are true and eliminated; D is false and the correct answer choice here).

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. Correct Answer

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).

What are the most likely patterns of inheritance for CF and achondroplasia? A. CF: X-linked recessiveAchondroplasia: autosomal dominant B. CF: autosomal dominantAchondroplasia: autosomal recessive C. CF: spontaneous mutation onlyAchondroplasia: X-linked recessive D. CF: autosomal recessiveAchondroplasia: autosomal dominant Correct Answer

D. 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. Based on this, we can conclude that CF is recessive and achondroplasia is dominant (choices B and C are wrong). Second, is the disease allele carried on an autosome or on one of the sex chromosomes? If significantly more men than women are affected, the disease is most likely sex-linked. Based on the pedigree, we can conclude that both CF and achondroplasia are autosomal (choice D is correct and A is wrong). Since both conditions are autosomal, the third question ("If sex-linked, is the disease carried on the X chromosome or the Y chromosome?") is irrelevant and unnecessary in this case.

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 Correct Answer

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).

What are the chances that the male indicated as #3 will have a child afflicted with PKU? A. 0% B. 25% C. 50% D. It cannot be predicted based on the information provided.Correct Answer

D. The male indicated as #3 is a heterozygote, thus the phenotype of his children will depend on the genotype of his eventual partner. If his partner is homozygous dominant, 0% of their children would be affected. If his partner is heterozygous, there is a 25% chance their children would be affected. If his partner is homozygous recessive, there is a 50% chance of affected children. Since we do not know the genotype of his partner, we are unable to predict the probability of affected children.

What type of cell is the origin of an acoustic neuroma? A. Microglia B. Astrocyte C. Oligodendrocyte Your Answer D. Schwann cell Correct Answer

D. The passage states that an acoustic neuroma is a cancer of the myelin-producing cells of the vestibulocochlear nerve, a cranial nerve. Cranial nerves are part of the peripheral nervous system, so they are myelinated by Schwann cells (choice D is correct) rather than oligodendrocytes (eliminate choice C). Microglia have an immune role in the CNS (eliminate choice A), while astrocytes carry out various maintenance and regulatory activities in the CNS (eliminate choice B).

Transformation of a normal cell to a cancerous cell may begin with any of the following EXCEPT: A. a series of mutations. B. conversion of a proto-oncogene to an oncogene. C. infection of a cell with a virus that changes the nuclear genome. D. a delay in prophase of mitotic division. Correct Answer

D. The passage states that mutation of genes and conversion of proto-oncogenes into oncogenes are two of the main causes of cancer (choices A and B are true and thus eliminated). Viruses that change (mutate) the nuclear genome could therefore also cause cancer (choice C is true and thus eliminated). A delay in mitotic division would not change the DNA in any way and would not induce cancer (choice D is false and the correct answer).

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 Correct Answer

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.

If the frequency of the CF allele in a randomly mating population is 0.02, what is the frequency of individuals who do NOT manifest symptoms of CF? A. 0.0004 B. 0.9604 Your Answer C. 0.98 D. 0.9996 Correct Answer

D. This is a Hardy-Weinberg question and requires use of the phenotypic frequency equation p2 + 2pq + q2 = 1. Here, p represents the frequency of the dominant allele, and q represents the frequency of the recessive allele. CF is an autosomal recessive disorder, so for individuals to express CF, they must be homozygous recessive, or qq(= q2). The question states that the frequency of the CF allele is 0.02, so q = 0.02 and thus q2 = 0.0004. But remember, these are the affected individuals, so the frequency of unaffected individuals is 1 - 0.0004 = 0.9996.

Previous work has shown that the Notch signal is terminated via covalent modification. The Notch intracellular domain is phosphorylated and then ubiquitinated, targeting the protein for degradation. Based on this information, which of the following is most likely true? A. Phosphorylation of the Notch intracellular domain by Cdk8 is a more crucial step in negative regulation than addition of ubiquitin tags by Fbxw7. B. Fbxw7, Cdk8 and SHARP proteins function in this negative regulation of Notch1, since ablation of these proteins leads to elevated transcription of all Notch target genes. Your Answer C. Fbxw7 has been shown to add ubiquitin tags to several signaling proteins in the cell and targets them for degradation; Cdk8 is a well-studied kinase that transfers phosphate groups from cytoplasmic pools onto biological macromolecules. D. Notch is negatively regulated by the kinase Cdk8 and Fbxw7, an E3 ubiquitin ligase; both of these proteins are localized to the nucleus and thus likely contain a nuclear localization signal. Correct Answer

D. When Cdk8 is mutated (cell line A49), Notch signaling increases. However, this effect is more profound when Fbxw7 is mutated in cell line A22 (eliminate choice A). There are two reasons to eliminate choice B. First, the passage describes the cellular function of SHARP as a scaffold protein that represses transcription. It doesn't say anything about SHARP negatively regulating Notch signaling. In addition, decreasing expression of Fbxw7 (cell line A22), Cdk8 (cell line A49), and SHARP (cell line A56) causes increased transcription of many Notch target genes in Table 1. However, not all genes are affected in each case. You can also therefore eliminate choice B because of its extreme tone. Choice C is promising until it describes how a kinase transfers phosphate groups from cytoplasmic pools onto biological macromolecules. Kinases use phosphate groups from ATP, not cytoplasmic pools (eliminate choice C). Since the Notch intracellular domain is controlling transcription, it must be located in the nucleus. It makes sense that Fbxw7 and Cdk8 are both also localized to the nucleus and to do this, they likely contain a nuclear localization signal (choice D is correct).

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. Correct Answer 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.

Question Explanation 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.

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. Correct Answer

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.

Isolated damage to the left side of the visual cortex would: A. impair vision of the right side of the visual field.Correct Answer B. impair vision of the left side of the visual field. C. impair perception of complex patterns only. D. impair color vision only.

A. According to the passage, the left side of the visual cortex controls the right side of the visual field, and vice versa (choice A is correct and choice B can be eliminated). Perception of complex patterns is a function of V2, which is located in the both hemispheres, and not the left side only (choice C can be eliminated). Choice D can also be eliminated: color vision is related to a variety of processes from the eye to the brain, and is not limited to the left side of the brain only.

If a drug is applied to a rod cell which blocks sodium channels in the absence of light, this will result in: A. the hyperpolarization of the retinal rod plasma membrane. Correct Answer B. the repolarization of the retinal rod plasma membrane. C. an increase in the sodium concentration in the cell. D. the conversion of all trans-retinal to 11-cis-retinal.

A. Blocking the sodium channels in the absence of light would have the same effect on the cell as a burst of light. The membrane would hyperpolarize due to the fact that sodium can no longer enter the cell to depolarize it (choice A is correct; eliminate choice B). Sodium concentration would decrease, not increase, because the channel is blocked (eliminate choice C), and, finally, the conversion of all trans-retinal to 11-cis-retinal would occur only in the presence of light (eliminate choice D).

Creatine supplementation would cause which of the following effects? A. Increased activity of satellite cells and lower serum myostatin levels Correct Answer B. Decreased activity of satellite cells and lower serum myostatin levels C. Increased activity of satellite cells and higher serum myostatin levels D. Decreased activity of satellite cells and higher serum myostatin levels

A. Creatine supplementation is taken by athletes to increase performance. The passage says that satellite cells proliferate and donate additional myonuclei to their parent muscle fiber, which would be a good thing when training (choices B and D are wrong). The passage also says that myostatin inhibits myogenesis, which would hinder muscle building. Creatine therefore likely inhibits myostatin release (choice C is wrong; choice A is correct).

If the researchers had used a luminescent probe to visualize the location of FAS itself in Experiment 3, where would it predominantly be found? A. Cytosol Correct Answer B. Nucleus C. Mitochondria D. Smooth ER

A. Fatty acid synthesis takes place in the cytosol and requires the presence of FAS (choice A is correct). No part of the fat metabolism takes place in the nucleus of the cell (eliminate choice B). The mitochondrial matrix is the location for beta-oxidation (fatty acid break down; eliminate choice C). Elongation and desaturation of fatty acids takes place inside the smooth ER using separate enzymes. FAS is not used during extensions of fatty acids beyond carbon-16 nor is it involved in the introduction of any double bonds (eliminate choice D).

If the plasma glucose falls below normal, which of the following conditions is true regarding pancreatic activity? A. High levels of glucagon will be produced from alpha cells.Correct Answer B. High levels of insulin will be produced from beta cells. C. Pancreatic acinar cells will secrete high levels of somatostatin. D. The pancreas will produce high levels of both insulin and glucagon.

A. Glucagon is secreted by alpha cells in the pancreas and acts to elevate plasma glucose if it falls too low (choice A is correct). Insulin acts to lower blood glucose and will not be secreted in response to low blood glucose (choices B and D are wrong).Somatostatin is secreted in response to raised glucose or other nutrients and acts to repress digestive functions, so it will not be elevated in response to decreased glucose (choice C is wrong).

Like α(1→4)-glycosidase, which of the following enzymes acts to cleave α(1→4) linkages between glucose molecules? A. Glycogen phosphorylase Correct Answer 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.

In addition to the missense and nonsense mutations described in the passage, three other types of mutations were characterized. Which of the following mutation types would be most likely to result in the same phenotype as Group C mice? Deletion of the MLH1 gene Nonsense mutation in the last codon Insertion of three nucleotides at the end of the gene A. I only Correct Answer B. II only C. III only D. I, II, and III

A. Group C mice have nonsense mutations that caused a severe phenotype (increased tumor incidence and inability to respond to cisplatin). The increase in the severity of the phenotype, when compared to Group B mice (who had missense mutations) is likely due to the fact that this nonsense mutation early on in the gene caused the loss of the entire protein. This question is looking for mutations that would result in a similar result (effective loss of the protein product). Item I is correct: deletion of the MLH1 gene would certainly cause a loss of MLH1. This eliminates choices B and C, which do not include Item I. Item II is incorrect: a nonsense mutation in the last codon would only result in the loss of one amino acid, and would be unlikely to cause the severe phenotype noted in Group C mice. This eliminates choice D, leaving only choice A. To confirm the choice, Item III is also incorrect: insertion of three nucleotides at the end of the gene would not cause loss of the protein, and is likely to have no effect.

B lymphocytes are involved in a process called: A. humoral immunity. Correct Answer B. cell-mediated immunity. C. passive immunity. D. the complement system.

A. Humoral immunity refers to immunity through fluid; in this case, the blood. This is where B cells have their effect: They release antibodies into the bloodstream (choice A is true). Cell-mediated immunity is the process that cytotoxic T cells are involved in (choice B is wrong). Passive immunity refers to an injection of antibodies which can act as a temporary supply and help fight off a disease until a person's own immune system can start producing antibodies (choice C is wrong). The complement system is a system of proteins and enzymes which, when activated, can cause cell lysis (choice D is wrong).

All of the following are characteristics of both Strep. pneumoniae and Strep. viridans EXCEPT: A. they are sensitive to optochin. Correct Answer B. they are α-hemolytic. C. they are Gram-positive. D. they are catalase-production negative.

A. It can be observed from Figure 1 that both Strep. viridans and Strep. pneumoniae are α-hemolytic (choice B is wrong), Gram-positive (as are all bacteria in the figure; choice C is wrong), and catalase-production negative (from the first test on the left side; choice D is wrong). The test that distinguishes these two bacteria is sensitivity to optochin (choice A is correct).

Which of the following developmental processes occur among annelids? I. The blastopore develops into the mouth II. Spiral cleavage III. Indeterminate cleavage A. I and II only Correct Answer B. I and III only C. II and III only D. I, II, and III

A. Item I is true: annelids are protostomes, in which the cells near the blastopore develop into the mouth (eliminate choice C). Item II is true: they undergo spiral cleavage (eliminate choice B). Item III is false: cleavage is determinate (choice A is correct; eliminate choice D).

Which of the following conclusion(s) is/are supported by the results of the experiment shown in Figure 1? Energetic frustrations result in movement of the energy landscape to favor native interactions. I. The Go model in the absence of energetic frustrations has a higher II. Ea to move from an unfolded to a folded state for the H5A protein. III. Introduction of energetic frustrations results in an increase in the free energy at all stages of protein folding. A. I only. Correct Answer B. II only. C. I and III only. D. II and III only.

A. Item I is true: the results in Figure 1 show that the introduction of energetic frustrations results in a shift of the energy landscape to the right, toward increasing values of Q, which represents the fraction of native conformations present in the protein. Thus, a shift to the right represents an increase in native conformations (choices B and D are wrong). Item II is false: the Go model without frustrations for protein H5A shows an overall lower, not higher Ea, in moving from an unfolded to folded state (moving toward more native interactions). This is seen as a smaller "hump" at a Q value of 0.5. Item III is false: the passage states that the -ln(P(Q)) is proportional to free energy, and the Go model without frustrations is higher in free energy than the model with frustrations for certain values of Q and lower in free energy for other values (choice C is wrong and choice A is correct).

Contraction of cardiac muscle begins just after depolarization and lasts about as long as the action potential. During a contraction: A. Ca2+ enters the cell through voltage-gated ion channels.Correct Answer B. sarcomere length increases. C. polymerized actin becomes depolymerized. D. myosin binds irreversibly to actin.

A. Part of the cardiac action potential is the opening of slow voltage-gated calcium channels to create the plateau in depolarization (Phase 2) so choice A is correct. When the voltage-gated calcium channels open, calcium enters the cell and can play a role in the contraction of the cardiac muscle. Sarcomere length decreases during each contraction, and does not increase (choice B is wrong). Actin filaments in contractile tissue are not dynamic as they are in other processes and do not spontaneously depolymerize and repolymerize (choice C is wrong). Myosin binds actin reversibly, not irreversibly (choice D is wrong).

Cells from the pigmented region of a frog gastrula are transplanted to a light gastrula region. The light region surrounding the graft is induced to form a second incomplete tadpole. Cells transplanted from a light region to the same site do not induce formation of a second tadpole. Based on the results of this experiment, the developmental fate of the region surrounding the graft most likely depends on: A. the region from which the graft cells were taken.Correct Answer B. the region to which the graft cells are transplanted. C. the expression of pigment genes. D. the migration of the graft cells throughout the transplant region.

A. Pigmented cells induce development of an incomplete tadpole, while cells from the light region do not when transplanted to the same area. The region from which the cells are taken appears more important for developmental fate than the area they are transplanted to (choice A is correct, and choice B is wrong). Pigmented cells appear to be the key factor, but pigmentation itself is likely to be coincidental and not the cause of the developmental fate of the region (choice A is a better answer than choice C). There is no evidence presented that migration of cells is important (choice D is wrong).

Which of the following salts, when dissolved in water, would produce the solution with the highest pH? A. Na2CO3 Correct Answer 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).

Acetylcholine binding to AChR directly leads to which of the following effects? A. Influx of Na+ Correct Answer B. Release of Ca2+ from the sarcoplasmic reticulum C. Influx of Cl- D. Efflux of K+

A. The nicotinic AChR in NMJ is a ligand-gated cation channel on the sarcolemma (choice B can be eliminated). Therefore, when acetylcholines binds to it, there is an influx of cations, especially Na+(choice A is correct), which causes depolarization of the muscle fiber (choices C and D can be eliminated).

A fission yeast mutant is isolated in the lab and examined under the microscope. The cells are much smaller than wild type control cells. This mutant most likely has a: A. mutation in the promoter of Wee1 that causes decreased transcription. Correct Answer B. silent mutation in Cdc13. C. mutation in the promoter of Sal3 that causes transcription silencing. D. nonsense mutation in Cdc25. Your Answe

A. The passage states that Wee1 inhibits the activity of Cdc2. This would keep the cell in G2 and prevent mitosis. Decreased expression of Wee1 would cause the cells to pass through G2 and enter mitosis early; they would therefore be smaller than wild type cells (choice A is correct). Sal3 and Cdc25 promote Cdc2 activation and therefore help push the cell through G2 and into M-phase. Decreased Sal3 levels would cause the cell to stay in G2 phase longer, resulting in larger or elongated cells (choice C is incorrect), and a nonsense mutation in Cdc25 would cause the same phenotype (choice D is incorrect). A silent mutation by its very definition has no effect (choice B is incorrect).

Based on the passage, which of the following graphs would represent the flow rate in COPD patients compared to normal patients?

A. The passage states that in COPD the flow rate is decreased (eliminate choice D) and the overall lung volume is increased (choice B and C can be eliminated). The correct answer is choice A.

Under physiological conditions, which of the following amino acids is most likely to be found on the exterior of a body protein after folding is complete? A. E Correct Answer B. I C. M D. Y Your Answer

A. The passage states that one of the major driving factors for protein folding is the hydrophobic effect, whereby hydrophobic (nonpolar) residues fold in such a way as to minimize their interaction with water. As such, the majority of hydrophobic residues prefer to face toward the protein interior, away from water present at the protein surface. In contrast, hydrophilic residues prefer to face toward the protein exterior and are therefore likely to be found on the protein surface. At physiological pH, glutamate (E) will be in its deprotonated form with a full negative charge, making the side chain extremely polar and most likely to be found in the exterior of the protein (choice A is correct). Both isoleucine (I) and methionine (M) are nonpolar, hydrophobic amino acid residues that are expected to be found on the protein interior (eliminate choices B and C). While the side chain of tyrosine (Y) contains a hydroxyl group (OH) capable of forming hydrogen bonds, it also contains a large hydrophobic ring that makes it considerably less polar and thus less likely to be found in the protein exterior than glutamate (eliminate choice D; choice A is better).

In which cellular organelle are you most likely to find an abnormal accumulation of deposited glycogen in patients affected by Pompe disease? A. Lysosomes Correct Answer B. Mitochondria C. Golgi Body D. Nuclei

A. The passage states that the organelle lacking sufficient functional acid maltase and in which abnormal glycogen accumulates contains an acidic environment. Of the four choices, lysosomes have the most acidic environment making choice A true and the correct answer (eliminate choices B, C and D).

The graph below shows the membrane potential of a rod-cell plasma membrane: Which one of the following is represented at time t = 1 sec? A. Hyperpolarization Correct Answer B. Depolarization C. Repolarization D. Nonpolarization

A. The resting potential of rod cells is much higher than that of normal cells due to the constant influx of sodium. Hyperpolarization is defined as the membrane potential moving away from rest potential in the negative direction, and this is what is occurring at time t = 1 sec on the graph (choice A is correct). Depolarization is movement of the membrane potential away from rest in the positive direction (eliminate choice B), and repolarization is a return to rest potential (eliminate choice C). There is no such as thing as nonpolarization (eliminate choice D).

Excess calcitonin in the blood would most likely result in which of the following abnormalities? A. Abnormally weak bones B. Abnormally dense bones Correct Answer C. A greater than normal number of osteoclasts D. A decrease in the collagen content in bones

B. Calcitonin reduces blood calcium levels through three processes: decreasing calcium reabsorption by the kidney, decreasing calcium absorption by the small intestine, and activating osteoblasts, which store calcium as bone. This last function increases bone density (choice B is correct; eliminate choice A). Osteoclasts degrade bone and are stimulated by parathyroid hormone, not calcitonin (eliminate choice C). Decreasing the collagen content in bones would also result in weaker bones (eliminate choice D), and in any case, the amount of collagen is not dependent on calcium or its regulatory hormones.

Which of the following best describes the change in vision experienced with repetitive visual stimulation in Figure 3? A. Recognition of colors improved more than recognition of shapes and patterns. B. Recognition of colors and shapes improved, but recognition of patterns did not. Correct Answer C. Recognition of patterns improved more than recognition of shapes and colors. D. Recognition of colors, shapes, and patterns all improved.

B. According to Figure 3, recognition of colors and shapes improved with time, but recognition of patterns did not (choice B is correct). Recognition of shapes improved the most (choice A can be eliminated). Since recognition of patterns did not improve with time, choices C and D can be eliminated.

Upon exposure to cisplatin, wild-type cells are expected to: A. replicate uncontrollably, causing the formation of skin or colon tumors. B. arrest in G2 and induce apoptosis. Correct Answer C. decreased expression of tumor suppression genes. D. enter the M phase of the cell cycle.

B. According to the passage, cisplatin is an anti-cancer drug, and administration of cisplatin causes cells that do not have mismatch repair (MMR) mutations to arrest in G2 and induce apoptosis (choice B is correct). None of the other answer choices are consistent with the fact that cisplatin is an anti-cancer drug, nor are they consistent with the mechanism described in the passage. Choice A can be eliminated, as cisplatin would cause cancer if its administration caused the formation of skin or colon tumors. The same logic can be used to eliminate choices C and D. If cisplatin caused decreased tumor suppression expression or induction of mitosis, it would not work as an anti-cancer drug.

Diabetes insipidus is a condition in which an individual fails to produce antidiuretic hormone. If a person suffering from diabetes insipidus were to consume large amounts of a sugar-containing beverage, which of the following would most likely be observed after one hour? A. Production of concentrated urine B. Production of dilute urine Correct Answer C. Elevated plasma glucose levels D. Elevated urine glucose levels

B. An individual who cannot make ADH will be unable to reabsorb water from the collecting duct of the nephron, resulting in urine that is very dilute (choice B is correct, and choice A is wrong). Since diabetes insipidus is strictly a disorder involving ADH production, no effect would be seen on plasma glucose levels (very low after one hour due to the effects of insulin) or urine glucose levels, which should always be zero in a normal individual. Do not confuse this disorder with diabetes mellitus, a disorder involving insulin in which blood and urine glucose levels are abnormally high (choices C and D are wrong).

A strain of E. coli develops a mutation that results in the production of a repressor protein which is no longer able to bind allolactose. This mutation would most likely be: A. harmful, because the repressor would not be able to bind to the operator. B. harmful, because the repressor would not be able to be inactivated. Correct Answer C. beneficial, because allolactose could be converted to lactose. D. beneficial, because lactose could more easily be transported into the cell.

B. Binding to allolactose is required for inactivation of the repressor protein. If the repressor protein were mutated such that it could no longer bind to allolactose, it would never be inactivated (that is, removed from the operator; choice A is wrong), and the genes in the lac operon would never be transcribed. This would be harmful to the bacteria, because they would be unable to utilize lactose as an energy source (choice B is correct). The repressor has nothing to do with converting allolactose (choice C is wrong) nor with transporting lactose into the cell (choice D is wrong).

Diffusion capacity of carbon monoxide (DLCO) is another parameter used for diagnosis of pulmonary diseases because carbon monoxide is diffusion limited. According to the passage, which of the following would be true in emphesyma compared to a normal patient? A. The DLCO would higher. B. The DLCO would be lower. Correct Answer C. The DLCO would remain unchanged. D. The DLCO would depend on the amount of RBCs at the alveolar interface.

B. Carbon monoxide is diffusion limited; therefore, the rate of diffusion of CO is only limited by factors affecting diffusion across the membrane (choice D can be eliminated). The factors would include thickness of the diffusion barrier and the effective surface area. In emphesyma, the effective surface area is reduced due to loss of alveolar sacs, thus DLCO would be decreased (choice B is correct; choice A and C can be eliminated).

In a follow-up experiment, normal cells were put under four different growth conditions. Which of the following best matches an experiment to its hypothesized outcome? A. Cells grown in low oxygen conditions down-regulate the function of all IDH isoforms. B. Cells stimulated with high ADP levels in their growth media have high IDH3 activity. Correct Answer C. Cells stimulated with high levels of ATP in their growth media have high levels of IHD3. D. Cell stimulated with high levels of NADH in their growth media have low IDH1 activity.

B. Cells grown in low oxygen conditions will resort to fermentation to survive, meaning IDH3 function would likely be low. However, IDH1 and IDH2 are not involved in the Krebs cycle, so there is no reason for the cell to also down-regulate these enzymes (eliminate choice A). ADP accumulation stimulates cellular respiration, including IDH3 activity (choice B is correct). High ATP levels inhibit cellular respiration, so IDH3 activity would be negatively regulated (eliminate choice C). High NADH negatively regulates IDH3, but IDH1 and IDH2 use NADPH as a cofactor. NADH would therefore have little effect on IDH1 or IDH2 activity (eliminate choice D).

Novobiocin is an antibiotic that inhibits bacterial DNA gyrase. Its antibiotic properties are based on the fact that it: A. prevents DNA helix formation. Your Answer B. inhibits DNA replication. Correct Answer C. prevents RNA transcription. D. inhibits protein translation.

B. DNA gyrase supercoils bacterial DNA (choice B is correct). It does not affect helix formation; the DNA is already in a double-helix before it is supercoiled (choice A is wrong). It does not affect protein synthesis, since that has to do with RNA and ribosomes, not DNA (choice D is wrong). The job of DNA gyrase is to continually introduce negative supercoils into the circular bacterial genome. Then, when the helix is opened for replication and positive supercoils are introduced (the DNA gets more tightly wound at the ends of the replication fork), the already-present negative supercoils cancel out the newly introduced positive supercoils, and DNA tension is kept to a minimum. If this does not occur, the DNA strand would become too tightly wound for replication to continue, and in the absence of replication, the bacteria cannot reproduce. Note that the same could be argued for transcription, but since that involves such a small region of the genome, the positive supercoiling is not a significant effect (choice C is wrong).

If a nucleotide is deleted during the transcription process this event would most likely lead to: A. no change in the protein. B. the destruction of the correct reading frame. Correct Answer C. an mRNA without a cap. D. a DNA gene with the incorrect base pairs.

B. Deleting a nucleotide during transcription would result in a serious mutation called a frameshift mutation, which would completely destroy the protein's amino acid sequence and structure (choice B is correct and choice A is wrong). It is unlikely that it would cause the mRNA to not be capped, so choice C is eliminated. Finally, removing a base from RNA would have no effect on the DNA, as DNA replication does not depend on RNA at all (eliminate choice D).

Which of the following describes how the brain uses parallel processing to process a visual stimulus? A. A different area of the brain is activated by each type of image (images, letters, etc.). B. Visual areas V1-V5 are utilized to analyze different aspects of an image simultaneously. Correct Answer C. Images are simultaneously processed using both top-down and bottom-up processing. D. Visual areas V1-V5 are utilized to analyze each aspect of an image in succession.

B. Parallel processing describes the simultaneously processing of different aspects of a stimulus (choice B is correct), instead of analyzing the pieces of a stimulus step-by-step (choice D can be eliminated). Choice A is the definition of feature detection theory, and can be eliminated. Top-down and bottom-up processing are elements of gestalt psychology; the simultaneous use of both top-down and bottom-up processing is characteristic of most sensory processing, but is not the definition of parallel processing (choice C can be eliminated).

Which of the following is true based on the chart below? A. Lipid and protein macromolecules can both contain serine, while lipids and nucleic acids have no biological building blocks in common. B. Components of the plasma membrane can contain nitrogen atoms and amino acids. Correct Answer C. Phospholipids are amphoteric, in that they have hydrophilic and hydrophobic properties or regions. D. At least some phospholipids contain three fatty acids connected to glyercol via ester linkages.

B. Proteins and phosphatidylserine both contain serine, and phospholipids and nucleic acids both contain phosphate groups (eliminate choice A). Based on the chart in the question stem, choice B is correct. Phospholipids can contain nitrogen atoms (as in ethanolamine and serine) and also amino acids (as in serine). Lipids are amphipathic because they contain both hydrophilic and hydrophobic regions. Amphoteric means that a molecule can act as either an acid or a base (eliminate choice C). Phospholipids contain a backbone (usually glycerol) with two fatty acids and a phosphate group. The phosphate group can be linked to other polar groups. Lipids that contain a glycerol backbone and three fatty acids are called triacylglycerides (eliminate choice D).

An error is made during production of an RNA transcript which results in a change in a single amino acid in the final protein. This error is most likely a result of: A. incorrect translation of the mRNA into protein by the ribosomes. B. the poor editing ability of RNA polymerase. Correct Answer C. a frameshift mutation occurring during transcription. D. incorrect translation of the DNA into RNA by RNA polymerase.

B. RNA polymerases have no editing function since RNA is a transient molecule and its information is not passed on to offspring (choice B is correct). The question states that the error is made during production of the RNA, so the change in amino acid sequence reflects the new nucleotide sequence and is not due to incorrect translation of the mRNA (choice A is wrong). Frameshift mutations are very severe and would result in all of the amino acids from the error on being changed (choice C is wrong), and DNA is not translated by RNA polymerase, it is transcribed (choice D 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. Correct Answer 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 FADH2 will 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.

Why is there a latent period in the production of antibodies during the primary response? A. The pathogen must first proliferate before the immune system can mount a response. B. Time is required for B lymphocytes to proliferate. Correct Answer C. Antibodies must be produced by T lymphocytes. D. The antigen must infect B lymphocytes prior to the production of antibodies.

B. The passage describes the latent period as the time required for the lymphocytes to begin to divide and differentiate into plasma and memory cells (choice B is correct). The pathogen need not proliferate for an immune response to be mounted; it only needs to be present (choice A is wrong). Antibodies are produced by B lymphocytes, not T lymphocytes (choice C is wrong), and an antigen does not infect the B cells, it must only bind to them (choice D is wrong).

Each of the following experiments preceded the work described in the passage EXCEPT: A. mRNA was harvested from bovine tissue and treated with reverse transcriptase. B. restriction fragment length polymorphism analysis was done, to determine which introns and exons are retained in each cDNA. Correct Answer C. affinity chromatography using an IDH3 antibody was used to isolate the two proteins that cooperate with IDH3. D. mass spectrometry was used to identify the proteins that function with IDH3 in a trimer. Your Answer

B. The passage says that bovine cDNA was studied to determine how IDH1 was spliced. cDNA (or complementary DNA) is made by reverse transcribing mRNA (choice A is true and eliminated). RFLP uses restriction endonucleases to cut 10-100 base-pair stretches of polymorphic DNA, called minisatellites, into small fragments. It can be used for DNA fingerprinting, but would not be useful here (choice B is false and is the correct answer). Affinity chromatography can be used to specifically isolate a biomolecule and mass spectroscopy can be used to identify peptide chains (choices C and D are true and eliminated).

According to the information in the passage, all post-transcriptional processing occurs during which of the following steps? A. DNA RNA transcript B. RNA transcript processed mRNA Correct Answer C. mRNA processed mRNA Your Answer D. processed mRNA protein

B. The passage states that post-transcriptional modification alsoinvolves the addition of a 5' cap and a 3' poly-A tail, implying that intron excision is included in the definition of post-transcriptional modification (choice B is correct). Choice A is wrong since this only creates a transcript; it does not modify it. Choice C is wrong since it does not include exon splicing, and choice D is wrong since it occurs after modification.

According to the passage, when light strikes rhodopsin, the concentration of sodium in the cell: A. increases, because of the sodium concentration gradient. B. decreases, because less sodium will cross the plasma membrane. Correct Answer C. remains the same, because sodium continues to enter the plasma cell at the same rate. D. decreases, because all trans-retinal has been isomerized.

B. The passage states that the rod-cell membrane is highly permeable to sodium. Therefore, due to a sodium concentration gradient, sodium is constantly flowing into the cell. When light strikes the pigment, the sodium channels close, and no more sodium can enter the cell (choice B is correct; eliminate choices A and C). Also, when light strikes the pigment, it is 11-cis-retinal that has been isomerized, not all trans-retinal (eliminate choice D).

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 Your Answer B. 3 Glycerol and 1 potassium linoleate C. 1 Glycerol and 3 potassium linoleate Correct Answer 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).

All of the following observations would support the energy landscape theory of protein folding EXCEPT: A. Lys and Glu residues that interact in the protein's native structure are the first amino acids to interact during protein folding. B. Two non-native amino acids, Phe and His, temporarily interact near the end of protein folding. C. Val and Ser, two non-native residues, interact early in the folding process, and drive further folding of native and non-native residues Correct Answer D. The significant majority of Ile residues are found on the protein interior once protein folding is complete.

C. According to the passage, the energy landscape theory of protein folding suggests that certain native interactions in the beginning of the protein folding process lead to further energetically favorable conformational shifts and that energetically unfavorable interactions are kept to a minimum. If it were found that two non-native residues interact early in the folding process and drive further protein folding, this would directly oppose the supposition of energy landscape theory that early native interactions drive protein folding (choice C would not support the theory and is the correct answer choice). If two native residues interact in the beginning of the protein process, it would confirm, not oppose the energy landscape theory (choice A supports the theory and can be eliminated). Non-native interactions are kept to a minimum, but are not absent altogether, and so non-native interactions near the end of the protein folding process would confirm, not deny the energy landscape theory (choice B supports the theory and can be eliminated). In general, proteins fold to place hydrophobic residues, like isoleucine, facing their interior, and so this expected finding supports the energy landscape theory (choice D supports the theory and can be eliminated).

Of the following explanations for the differing age of onset and severity of Pompe disease, which of the following is most reasonable? A. Trinucleotide repeats present within a protein-coding region lead to production of a mutant protein with a gain of function. B. Increased circulating levels of the enzyme β(1→4)-glucosidase compensate for the lack of functional α(1→4)-glucosidase in individuals demonstrating later onset. C. Greater concentrations of functional α(1→4)-glucosidase are present in individuals demonstrating later onset of the disease. Correct Answer D. Stored glycogen is utilized locally by skeletal muscle, but not by cardiac muscle

C. According to the passage, the symptoms of Pompe disease are due to an abnormal accumulation of glycogen. The passage also states that acid maltase acts to degrade glycogen in the affected organelle. This suggests that greater concentrations of functional acid maltase (α(1→4)-glucosidase) might delay the accumulation and thus deposition of glycogen and also delay the onset of disease symptoms, making choice C true and the correct answer. The passage indicates that accumulation of glycogen in organelles is due to the lack of functional acid maltase; protein products that have lost functionality is inconsistent with a gain of function mutation, making choice A false and an incorrect answer. β(1→4)-glucosidase activity is specific to cleavage of beta-glycosidic linkages and would be unable to cleave the alpha glycosidic linkages present in glycogen, making choice B false and an incorrect answer. While choice D is generally correct (cardiac muscle stores relatively little glycogen), the statement does not address the question posed in the stem, making choice D incorrect.

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. Correct Answer 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).

Hydrolysis of the phosphocreatine bond releases 43.1 kJ/mol of energy. What is the standard free energy of hydrolysis for ATP? A. -87.4 kJ/mol B. -43.1 kJ/mol Your Answer C. -30.5 kJ/mol Correct Answer D. +6.7 kJ/mol

C. The passage says that creatine kinase can transfer a phosphate group from phosphocreatine to ADP via a spontaneous reaction. In other words, phosphocreatine dephosphorylation can be coupled to ADP phosphorylation and the net overall reaction is spontaneous, or has a -ΔG. If phosphocreatine dephosphorylation releases 43.1 kJ/mol, ADP phosphorylation must use less energy than this (choice C is correct). If ATP hydrolysis released 87.4 kJ/mol or 43.1 kJ/mol, the transfer of a phosphate group from phosphocreatine to ADP would not be a spontaneous reaction (choices A and B are wrong). ATP hydrolysis releases energy to the cell, so must have a negative standard free energy (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. Correct Answer 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 antibody-mediated actions is NOT supported by the passage? A. Antibody binding mediates destruction of AChRs and NMJ by immune cells. Your Answer B. Antibody binding to the AChRs interferes with acetylcholine binding. C. Antibody binds to an inhibitory element on the DNA and down-regulates the expression of AChR gene, thus decreasing the synthesis of AChR. Correct Answer D. Antibody binding causes internalization and degradation of the AChRs.

C. Antibodies are proteins, therefore they cannot cross the membrane and act as transcription factors. They bind to the receptor and thereby hinder the binding of the acetylcholine; this conclusion is supported by the results of Experiment I, demonstrated in the relatively dip in muscle action potential shown in Figure 1 (choice B can be eliminated). Supported directly by the result of Experiment III, the passage mentions that the number of AChRs on the membrane is decreased in EAMG, implying that the antibodies are acting as opsins and targeting the receptors for both intracellular and extracellular degradation (choice A and D can be eliminated; choice C is correct).

Which of the following conditions leads to a reduction in water reabsorption by the kidney? A. High plasma levels of vasopressin B. Increased osmolarity of interstitial fluid in the kidney medulla C. Decreased permeability of the collecting duct to waterCorrect Answer D. An ACTH-secreting tumor

C. Factors that reduce water resorption would tend to increase urinary output, decrease blood volume, and decrease blood pressure. High levels of vasopressin (ADH) would increase water resorption and plasma volume (choice A is wrong). The greater the osmolarity of the interstitial fluid in the medulla of the kidney, the greater the water resorption as filtrate passes into the medulla in the descending loop of Henle (choice B is wrong). As filtrate passes through the collecting duct, water can be reabsorbed and the urine concentrated. If the collecting duct is impermeable to water, however, then less water will be reabsorbed, and the filtrate will remain dilute (choice C is correct). ACTH is the hormone that controls secretion of aldosterone (as well as cortisol) by the adrenal cortex. An ACTH-secreting tumor will cause elevated secretion of aldosterone, increased Na+ reabsorption, and a subsequent increase in water reabsorption (choice D is wrong).

Fatty acid synthase is most likely suppressed by all of the following hormones EXCEPT: A. glucagon. B. thyroid hormone. C. insulin. Correct Answer 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 would be increased in a patient with COPD? I. FEV1/FVC II. Total lung capacity (TLC) III. PCO2 in blood A. I only B. II only C. II and III only Correct Answer D. I, II, and III

C. Item I is incorrect: FEV1/FVC would be decreased as FEV1 and FVC are significantly decreased in COPD (Figure 1). Eliminate choices A and D. Item II is correct: TLC would be increased because of high RV and hyperinflation. Eliminate choice A. Item III is correct. The decrease in the ability to breathe air out would result in higher PCO2 in the blood. Eliminate choice B and choice C is correct.

Which of the following can be inferred based on information in the passage? I. Creatine more easily crosses the mitochondrial membrane than the nucleotides of adenine. II. Skeletal muscle cells contain distinct cytosolic and mitochondrial isoforms of creatine kinase. III. Creatine supplementation would likely be an effective aid when training for an endurance event, such as a marathon. A. I only B. II only C. I and II only Correct Answer D. I, II, and III

C. Item I is true. The passage describes a phosphocreatine shuttle system that allows transfer of high-energy phosphate bonds from the mitochondria to the sarcoplasm of a muscle cell: creatine kinase transfers a phosphate group from ATP to creatine in the mitochondria. Phosphocreatine is then translocated to the sarcoplasm, and the phosphate group is transferred back to a molecule of ADP via a spontaneous reaction mediated by another creatine kinase. If adenine nucleotides could easily cross the mitochondrial membrane, there would be little need for this system (choice B is wrong). Item II is true. The last paragraph of the passages discusses two different creatine kinase enzymes, with different locations (mitochondria and sarcoplasm) and functions (choice A is wrong). Item III is false. The passage says that the phosphocreatine shuttle system allows for a rapid and high level of ATP generation in the sarcoplasm for a short period of time, and that this is useful during high-intensity physical exercise. Creatine supplementation is therefore most effective when performing short-term and high intensity tasks, such as sprinting (choice D is wrong; choice C is correct).

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

Which of the following best characterizes an energetic frustration in protein folding? A. An unfavorable interaction between native residues resulting in a local energy minimum. B. A favorable interaction between non-native residues resulting in a local energy minimum. C. An unfavorable interaction between non-native residues resulting in a local energy maximum. Correct Answer D. A favorable interaction between native residues in resulting in a local energy maximum.

C. Note that this is a 2x2 question. The passage states that according to energy landscape theory, frustrations, which manifest as local energy maxima (choices A and B can be eliminated), are limited. The passage also states that unfavorable interactions are the ones that result in energy maxima (choice D is wrong and choice C is correct).

Which one of the following graphs best depicts the titration curve for the addition of a strong base to a strong acid?

C. Since a strong acid is being titrated, the pH would be low at the outset and increase as base is added; this eliminates choices A and B. Titration curves are characterized by a sharp change in pH at the equivalence point, and for a strong acid- strong base titration, this change is not only sharp but also large. Graph C is best.

What is the FRC for an individual with the following lung volumes: total lung capacity (TLC) = 5700 mL, vital capacity (VC) = 4700 mL, inspiratory reserve volume (IRV) = 3000mL and tidal volume (VT) = 500 mL? A. 3500 mL B. 1000 mL Your Answer C. 2200 mL Correct Answer D. 1200 mL

C. The passage states that FRC = ERV + RV. TLC = VC + RV RV= TLC - VC = 5700 mL - 4700 mL = 1000 mL VC = VT + IRV + ERV ERV = VC - VT - IRV = 4700 mL - 500 mL - 3000 mL= 1200 mL FRC = ERV + RV= 1200 mL + 1000 ml = 2200 mL Thus choice C is correct and choices A, B and D are eliminated.

A series of cell lines were generated, each missing expression of one or more IDH isoform. When grown in the presence of 14C-D-glucose: A. an IDH2/3-null line generates large amounts of 14CO2, which is being made in the mitochondrial matrix. B. IDH1-null lines release large amounts of 14CO2 into their growth media, because CO2 can passively diffuse across the plasma membrane. C. large amounts of 14C-lactate is detectable in the growth media of IDH3-null lines but not IDH1 or IDH2-null lines.Correct Answer D. glucose-6-phosphatase and pyruvate carboxylase expression are up-regulated in IDH3-null cells. Your Answer

C. The passage states that IDH (all isoforms) is responsible for oxidative decarboxylation, a reaction that produces CO2. IDH3 is located in the mitochondria (where the TCA cycle occurs), and based on the data in Figure 1, IDH2 is also in the mitochondria. Cells missing IDH2 or IDH3 would not be able to generate CO2 (choice A is wrong). IDH1 is said to perform a similar function outside of the context of the TCA cycle; cell lines that lack IDH1 would also not be able to make CO2 (choice B is wrong). Since IDH3 is important in the TCA cycle, cells that lack IDH3 will resort to anaerobic metabolism to survive. This will generate large amounts of lactate (choice C is correct). Glucose-6-phosphatase and pyruvate carboxylase are gluconeogenesis enzymes. If cells are running high levels of glycolysis, they will not also run gluconeogenesis (choice D is wrong).

A lab technician observes (on blood agar) a round culture of Gram-positive cocci surrounded by a clear zone. To determine the species of bacteria, the technician should: A. perform a catalase test. Your Answer B. test sensitivity to optochin. C. test sensitivity to bacitracin. Correct Answer D. test growth on novobiocin.

C. The passage states that clear zones of lysis are produced by β-hemolytic colonies. β-Hemolytic bacteria could be either Strep. pyogenes or Group B streptococci, which can be distinguished by sensitivity to bacitracin (choice C is correct; eliminate choices A, B and D).

If an individual is reinfected with the same virus the immune system will: A. elicit a slow immune response. B. require a greater amount of the virus than the first response. C. destroy the pathogen before it has time to cause disease. Correct Answer D. reject the pathogen before it has a chance to proliferate in the body.

C. The passage states that the response to a second infection with the same antigen is much more rapid than the primary response and also requires much less antigen (choices A and B are wrong). Furthermore, the pathogen is not rejected by the body; it is destroyed by the body (choice D is wrong, and choice C is correct).

A graduate student in a biochemistry lab studies D-amino acid oxidase, an FAD-containing peroxisomal enzyme. She commonly uses a series of D-amino acid oxidase inhibitors in her experiments, but a new volunteer in the lab mixed up the tubes of inhibitors and she is not clear which inhibitor is in which tube. She runs a series of experiments to sort out the inhibitors. The data look like this: Which of the following is true? A. The solid line on the data graph is uninhibited D-amino acid oxidase, the dotted line is a competitive inhibitor, and the dashed line is a mixed inhibitor. B. The dotted line on the data graph is uninhibited D-amino acid oxidase, the solid line is a noncompetitive inhibitor, and the dashed line is an uncompetitive inhibitor. C. The dotted line on the data graph is uninhibited D-amino acid oxidase, the solid line is a competitive inhibitor, and the dashed line is an uncompetitive inhibitor. Correct Answer D. The dotted line on the data graph is uninhibited D-amino acid oxidase, the solid line is a competitive inhibitor, and the dashed line is a noncompetitive inhibitor. Your Answer

C. The solid line and the dotted line have the same Vmax. Since there are no enzyme inhibitors that increase Vmax, one of these must be the control line and the other must be due to a competitive inhibitor. Since competitive inhibitors increase the Km of the reaction, the dotted line must be uninhibited D-amino acid oxidase (choice A is wrong) and the solid line must be due to a competitive inhibitor (choice B is wrong). The uninhibited dotted line has a Vmax of about 90 nmol/min, so a ½Vmax of about 45 nmol/min. This corresponds to a Km of about 10 nmol. The dashed line has a lower Vmax (about 65 nmol/min), so a lower ½Vmax of (32 nmol/min). This corresponds to a Km closer to 5 nmol. A decreased Vmax and decreased Km means this line is due to an uncompetitive inhibitor, not a noncompetitive inhibitor (choice C is correct and choice D is wrong).

Which of the following best describes the cascade that leads to cellular apoptosis after exposure to intra- or extracellular death signals? A. Effector kinases cluster together and activate each other. The activation of effector kinases leads to the activation of initiator caspases, which cleave proteins at aspartic acid sites, triggering apoptosis. B. Effector caspases cluster together and activate each other. The activation of effector caspases leads to the activation of initiator caspases, which cleave proteins at aspartic acid sites, triggering apoptosis. C. Initiator caspases cluster together and activate each other. The activation of initiator caspases leads to the activation of effector caspases, which cleave proteins at aspartic acid sites, triggering apoptosis. Correct Answer D. Effector kinases cluster together with initiator capsases, activating the initiator capsases. The activation of effector kinases leads to the deactivation of initiator caspases, which cleave proteins at aspartic acid sites, triggering apoptosis.

C. This is a recall question pertinent to concepts discussed in the passage (apoptosis). Apoptosis is carried out by proteases called caspases, which cleave targets at aspartic acid residues; apoptosis is not started by effector kinases (eliminate choices A and D). Initiator caspases cluster together and initiate apoptosis (choice C is correct). These initiator caspases can then activate effector caspases, which cleave proteins at aspartic acid residues; this triggers apoptosis. Choice B is incorrect, and confuses the role of initiator and effector caspases.

During the blastula stage, neighboring cells are in close contact with each other. The formation of the blastula is contemporaneous with the formation of tight junctions. The function of tight junctions is to: A. provide a means through which induction occurs. B. act as pores connecting the cytoplasm of adjacent cells. C. seal off the blastocoel and unite connected cells into a tissue. Correct Answer D. transmit action potentials between cells.

C. Tight junctions hold cells together and can form a seal so that fluid does not leak between cells (choice C is correct). Induction involves communication between cells but can be performed by secreted factors and does not require tight junctions for the communication to happen (choice A is wrong). Tight junctions do not, however, form a junction between the cytoplasm of cells (choice B is wrong), nor do they allow the transmission of action potentials (choice D is wrong).

Rad21 functions in double-strand break repair and is also a subunit of the chromatid cohesin complex, necessary to keep sister chromatids connected. Cells with non-functional Rad21 due to mutation will display all the following phenotypes EXCEPT: A. increased sensitivity to UV and reactive oxygen species. B. decreased ability to maintain chromosomes in the 2x state. C. faster than usual mitotic phase due to increased chromosome motility. Correct Answer D. increased probability of cell cycle arrest at the G2/M transition. Your Answer

C. UV light and reactive oxygen species can both induce double-strand breaks in DNA. If cells have defective Rad21, they will have a diminished ability to repair these breaks (choice A is true and can be eliminated). Since Rad21 is important in chromatid cohesion, without functional Rad21, the sister chromatids may not be able to remain joined after DNA replication. This means cells would revert to a 4n(tetraploid) 1x (one chromatid per chromosome) state instead of the 2n 2x state that is normal in G2 and the first part of mitosis (choice B is true and can be eliminated). If double-strand break repair cannot occur to full capacity, the cell would likely arrest at the G2/M transition (choice D is true and is can be eliminated). However, there is no reason why decreased double-strand break repair or poor chromatid cohesion would cause a faster mitotic phase or increased chromosome motility. If anything, mitosis would be arrested as the cell tried to deal with disorganized chromosomes (choice C is a false statement and is therefore the correct answer).

The conformational change of a regulatory protein after the binding of a repressor most likely represents an alteration of the protein's: A. amino acid composition. B. primary structure. C. secondary structure. D. tertiary structure. Correct Answer

D. A conformational change in a protein is a change in the larger scale folding of a protein, with changes in the relative positions in space of amino acids located far from each other in the linear polypeptide chain. The tertiary structure of a protein involves large-scale structure within a polypeptide chain that is stabilized by interactions between amino acids that can be distant from each other in the linear sequence; thus, this is the most likely level of structure altered during a conformational change (choice D is correct). The amino acid composition and primary structure (the linear sequence of amino acid residues in the polypeptide chain) cannot be changed without breaking covalent peptide bonds, which does not occur in a conformational change (choices A and B are wrong). The secondary structure of a protein involves folding that is stabilized by the nearest neighbors in the polypeptide chain, including structures such as α helix and β sheet. Conformational changes can alter secondary structures somewhat, but mostly alter tertiary structure (choice D is a better answer than choice C).

In a male individual with Down's Syndrome (trisomy 21), how many chromosomes would be visible at metaphase I of spermatogenesis? A. 23 B. 24 C. 46 D. 47 Correct Answer

D. An individual with trisomy 21 has an extra copy of chromosome 21 (three total copies). During metaphase I, the developing gametes are still diploid (separation of homologues has not yet occurred), so this individual would have the normal 46 chromosomes plus the extra copy of chromosome 21, for a total of 47 chromosomes (choice D is correct; eliminate choices A, B and C).

The enzyme lysozyme degrades peptidoglycan. If lysozyme is mixed in blood agar prior to addition of Strep. viridans, which of the following would be observed? A. Colonies surrounded by a clear zone B. Colonies surrounded by a green zone C. Colonies without a zone of hemolysis around them D. No colonies Correct Answer

D. Bacterial cell walls are made of peptidoglycan and are susceptible to degradation by lysozyme. Gram-positive strains (such as Strep. viridans) are particularly vulnerable since their cell wall is unprotected by membrane like the cell walls of Gram-negative strains. Thus the cell wall of the Strep. viridans will be damaged by the lysozyme, so that the bacteria will not survive and no colonies will form (choice D is correct). Note that the hemolysis observed around some colonies is caused by the ability of bacteria to lyse red blood cells and is not related to lysozyme activity (choices A, B, and C are wrong).

All of the following are true concerning the side chain of cysteine EXCEPT: A. the sulfur is sp3 hybridized. B. it can help stabilize tertiary protein structure. C. the sulfur has two lone pairs of electrons. D. it undergoes hydrogen bonding. Correct Answer

D. Because the S has two lone electron pairs, it is sp3 hybridized (eliminate choices A and C). When two cysteine residues are close in space in a protein, the formation of a disulfide bond between them helps to stabilize the globular structure of the protein (eliminate choice B). The thiol group of cysteine does not undergo hydrogen bonding because it contains no O—H, N—H, or F—H bonds.

Catalase is an enzyme that catalyzes the following reaction: 2 H2O2 2 H2O + O2 H2O2 is toxic to bacteria and is found in the phagosomes of macrophages. Which of the following bacteria is most likely to be resistant to macrophage attack? A. Group B Streptococci B. Strep. viridans C. Strep. pneumoniae D. Staph. aureus Correct Answer

D. Catalase inactivates peroxide and macrophages produce peroxides to kill bacteria. If a bacterium makes catalase, it may be more resistant to macrophages. The bacteria in the lower half of the figure are positive in the catalase test, which includes Staph. aureusand none of the other species (choice D is correct; eliminate choices A, B and C).

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. Correct Answer

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.

A gene has 900 base pairs, and the protein translated from this gene has 210 amino acids. This is mostly likely due to the fact that: A. exons have been removed from the RNA transcript. B. the stop codon in mRNA is found at the 210th base pair. C. the DNA molecule was not synthesized with the appropriate enzymes. D. introns have been removed in the formation of the mRNA. Correct Answer

D. Exons are not removed from the transcript; they are ligated together to form the mature mRNA (choice A is wrong). A stop codon at the 210th base pair would result in a protein only 70 amino acids long (3 nucleotides = 1 amino acid), so choice B is wrong. There is only one enzyme which can synthesize DNA (DNA polymerase) and in any case, if the DNA molecule was synthesized incorrectly, it would be highly unlikely that any protein could be made from that gene at all (choice C is wrong). Choice D is correct: In the formation of mature mRNA, introns are removed and exons are ligated together. To make a protein of 210 amino acids, exons amounting to 210 × 3 = 630 base pairs must have been ligated together, along with 3 base pairs for a stop codon (633bp total); introns amounting to 900 - 633 = 267 base pairs were spliced out.

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 Correct Answer

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.

Vinca alkaloids are a class of anti-cancer drugs derived from the periwinkle plant. Once absorbed into a cell, they interfere with the polymerization of microtubules. These drugs can prevent cancer from spreading by disrupting: A. pseudopod formation, thereby preventing cellular locomotion. B. prophase, thereby halting mitosis. C. transcription, thereby halting production of crucial cell proteins. D. metaphase, thereby halting tumor growth. Correct Answer

D. Formation of microtubules is crucial for formation of the metaphase plate. Microtubules polymerize from the centrioles outward. They contact centromeres (to become kinetochore fibers) and "push" the chromosomes towards the center of the cell to form the metaphase plate. Without proper microtubule polymerization, metaphase (and the rest of mitosis) cannot occur. If mitosis cannot occur, then neither can tumor growth. Note that microtubules are also required during prophase in forming the mitotic spindle; however, some parts of prophase can still occur in the absence of microtubule formation (DNA condensation, loss of the nuclear membrane). This makes choice D a better choice than choice B, since virtually all of metaphase depends on proper microtubule polymerization. Pseudopod formation requires the growth of microfilaments (actin fibers), not microtubules (choice A is wrong), and transcription (RNA polymerization) does not require microtubules at all (choice C is wrong).

A physician treating a patient with Von Gierke's disease notes elevated lipid, lactic acid and uric acid levels in the patient's blood. Which of the following additional findings regarding the patient is the physician most likely to also note? A. An inability to metabolize orally-administered glucose via normal metabolic means. B. Neutropenia, a disorder characterized by an unusually low number of neutrophils, a leukocyte. C. Increased branching of liver glycogen stores. Your Answer D. Decreased blood glucose concentrations Correct Answer

D. Glucose-6-phosphatase, the enzyme that the passage states is deficient in Von Gierke's disease patients, hydrolyzes glucose-6-phosphate, resulting in the release of a phosphate group and free glucose for export from the cell. When this enzyme is absent, glucose-6-phosphate, a product of the breakdown of glycogen in the liver, will remain trapped within the liver and will not pass into circulation. As a result, Von Gierke's patients are at an increased risk of hypoglycemia and require regular administration of exogenous glucose to meet metabolic demands. This is consistent with choice D which is the correct answer. Von Gierke's patients do not demonstrate an enzyme deficiency that would interfere with the normal metabolism of glucose, making choice A false and the incorrect answer. Choice C is consistent with information provided in the passage regarding the form of the disease related to Von Gierke's disease, in which affected patients suffer recurrent bouts of infection, but is inconsistent with passage information regarding Von Gierke's disease, thus choice B is false and an incorrect answer.

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 FADH2 and NADH by the electron transport chain proteins. D. It causes a decrease in the electromotive potential built up by the electron transport chain. Correct Answer

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). FADH2 and 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).

Which of the following events takes place during Phase 0? A. A decrease in intracellular Ca2+ B. An efflux of intracellular K+ C. Closure of voltage-gated channels D. An influx of extracellular Na+ Correct Answer

D. In both cardiac and skeletal muscle, as in neurons, the initial depolarization (Phase 0) is caused by the opening of voltage-gated sodium channels. When the channels open, sodium flows into the cell down the gradient maintained as part of the resting membrane potential (choice D is correct). Calcium and potassium flux do not occur in this period, and voltage-gated channels are opening, not closing (choices A, B, and C are wrong).

Which of the following is/are the disadvantages of passive induction of EAMG? Requires regular infusion of antibody Breaks host tolerance for AChR by stimulating T cells Needs large doses of IgG A. I only Your Answer B. II only C. I and II only D. I and III only Correct Answer

D. Item I is correct: The antibodies have a certain half-life, so in order to maintain EAMG in the rats through the passive method, they need to be injected with the antibodies on a regular basis (choice B can be eliminated). Item II is incorrect: Administration of AChR would break the tolerance but anti-AChR would not produce reactive T cells against AChR (choice C can be eliminated). Item III is correct: Higher dosage of the antibodies is required to have a significant effect, as inhibition of few AChR would not produce pathology (choice A can be eliminated and choice D is correct).

Thirteen amino acids, including methionine, valine and proline, areglucogenic in humans. This means their α-keto acid carbon skeleton is converted to pyruvate during amino acid catabolism. After deamination, valine can therefore: I. Be converted into CO2 and H2O to generate ATP. II. Generate at least three NADH and two FADH2. III. Enter gluconeogenesis to generate glucose. A. I only B. III only C. I and II only D. I and III only Correct Answer

D. Item I is true: Amino acids are catabolized via deamination into α-keto acids and ammonia. Based on the information in the question stem, the α-keto acid formed from valine will be converted to pyruvate. Pyruvate can keep going through cellular respiration to generate CO2, H2O and ATP. Eliminate choice B. Item II is false: Pyruvate is converted into one acetyl-CoA (during which 1 NADH is made), and the acetyl-CoA would then generate three NADH and only one FADH2 as it cycles through the Krebs cycle. Eliminate choice C. Item III is true: Pyruvate can also enter gluconeogenesis to generate glucose. Eliminate choice A and choice D is correct.

If the Control 4 western blot tested for α-tubulin expression, which of the following is/are true? I. COX4, or cytochrome c oxidase subunit IV would be a good protein for Control 3. II. IDH1 is expressed in the cytosol and peroxisomes. III. Control 1 could probe for histone H3, and control 2 could probe for catalase expression. A. I and II Your Answer B. I and III C. II and III D. I, II and III Correct Answer

D. Item I is true: The passage says that IDH3 catalyzes the fourth step of the tricarboxylic acid cycle, which occurs in the mitochondria, so fraction/control 3 must be the mitochondrial compartment. Cytochrome c oxidase subunit IV isn't a protein you've necessarily heard of before, but cytochrome c is part of the electron transport chain and is thus localized to the mitochondria. It makes sense this enzyme is localized there too (eliminate choice C). Item II is true: The passage says that fraction/control 2 are peroxisomal and α-tubulin is a component of microtubules in the cytoskeleton. IDH1 is expressed in these two compartments (eliminate choice B). Item III is true: The passage says that IDH is not expressed in the nucleus and none of the isoforms have expression in fraction 1. This fraction is thus nuclear, and histone H3 would be a good marker. Histone proteins function in eukaryotic genome packaging. Catalase is a peroxisomal protein (eliminate choice A and choice D is correct).

NADPH is produced in which of the following pathways? Glycolysis Krebs cycle Pentose phosphate pathway (PPP) A. I only B. II only C. I and II D. III only Correct Answer

D. Items I and II are incorrect: glycolysis and the Krebs cycle produce NADH and not NADPH (eliminate choices A, B, and C). Item III is correct: the pentose phosphate pathway produces NADPH that is used in many anabolic pathways (choice D is correct).

Which of the following would most likely become a major concern of mortality in MG patients? A. Cardiac failure B. Muscle weakness in the upper limbs C. Paralysis of the facial muscles D. Respiratory distress Correct Answer

D. MG is a disease affecting skeletal muscle; therefore, all skeletal muscles in the body are susceptible (eliminate choice A). A very important skeletal muscle in humans is the diaphragm; in severe MG patients, death can occur due to respiratory distress or arrest. Paralysis of facial muscles and weakness of upper limbs can occur; however, they are generally not the primary cause of mortality (eliminate choices B and C; choice D is correct).

Complete reduction of acetic acid will produce which of the following? A. O2(g) B. CO2(g) C. CH3CHO D. CH3CH2OH Correct Answer

D. Reduction of acetic acid will initially produce the aldehyde given as choice C, where the carbonyl carbon has been reduced from an oxidation state of +3 to +2. Choice D is the result of further reduction to an oxidation state of +1, while choices A and B do not correspond to the reduction of acetic acid.

A chemist prepares a 1 M solution of sulfuric acid. Which of the following gives the relative concentrations of the species in solution? A. [H2SO4] > [HSO4-] > [SO42-] B. [SO42-] > [HSO4-] > [H2SO4] Your Answer C. [HSO4-] > [H2SO4] > [SO42-] D. [HSO4-] > [SO42-] > [H2SO4] Correct Answer

D. Since sulfuric acid is a strong acid and dissociation of the first proton is complete, undissociated H2SO4 will have the lowest relative concentration, eliminating choices A and C. Since the dissociation constant for the loss of the second proton is well below 1 (~ 10^-2), HSO4- will exist in greater concentration than SO42-, making choice D the correct answer.

Which of the following visual areas are included in the ventral stream? I. V1 II. V3 III. V4 A. I only B. I and II only C. III only Your Answer D. I, II, and III Correct Answer

D. Start by identifying the item that is found in exactly two of the answer choices, and start analyzing that one. Both Items II and III are found in exactly two answer choices, so start with Item II. Item II is true: V3 is located in both the dorsal and ventral streams (choices A and C can be eliminated). Notice that both remaining choices including Item I, so it must be true, and we can look at Item III. Item III is true: V4 is also found in the ventral stream (choice B can be eliminated and choice D is correct). Note that Item I is true: V1 is located in both the dorsal and ventral streams.

The regulation of CDKs includes: A. post-translational modification and allosteric regulation. B. allosteric regulation and proteolytic cleavage. C. proteolytic cleavage and protein associations. D. protein associations and post-translational modification.Correct Answer

D. The passage states that the activity of Cdc2, a CDK, is regulated by cyclin binding (cyclins are proteins, choices A and B can be eliminated) and phosphorylation, a type of post-translational modification (choice D is correct). There is no information in the passage on allosteric regulation of Cdc2 or proteolytic cleavage (choice C is eliminated).

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. Your Answer B. Equilibrium is reached faster in flask Y because ΔGY is 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. Correct Answer

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.

All of the following tissues are part of the immune response EXCEPT the: A. thymus. B. spleen. C. lymph nodes. D. pancreas. Correct Answer

D. The thymus is the site of T-cell maturation (choice A can be eliminated), the spleen is the site of B-cell maturation and also filters antigens from blood (choice B is wrong), and lymph nodes are sites of both B- and T-cell proliferation (choice C can be eliminated). The pancreas produces digestive enzymes, bicarbonate, and hormones but is not involved in the immune response (choice D is not true and is thus the correct answer choice).

A researcher discovered a strain of yeast with a mutation in the Sal3 promoter that causes its constitutive activation. Cells with this mutation would be: A. larger than wild type cells and spend more time in G2. B. smaller than wild type cells and spend more time in G2. C. larger than wild type cells and spend less time in G2.Your Answer D. smaller than wild type cells and spend less time in G2.Correct Answer

D. This is a 2X2 question because you need to decide if the cell will be smaller or larger, and if it spends more or less time in G2. Two of the answer options are relatively easy to eliminate: choices B and C are not possible as a cell cannot spend more time in a growth phase and then be smaller than wild type cells, and vice versa (choices B and C are wrong). If a fission yeast cell over-expresses Sal3, it will be pushed through G2 and will enter M-phase early. Less time in G2means the cells will be smaller (choice D is correct and choice A is wrong).

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

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.

All of the following are true of eukaryotic mRNA EXCEPT that: A. transcription occurs in the 5' 3' direction. B. the 5' end is usually capped. C. the mRNA is smaller than the initial transcript. D. it is polycistronic. Correct Answer

D. Transcription always occurs in the 5' to 3' direction, and eukaryotic mRNA is usually capped before translation (choices A and B are true and therefore eliminated). Eukaryotic mRNA is smaller than the original RNA transcript due to the fact that the introns must be removed and the exons spliced together (choice C is true and eliminated). However, eukaryotic mRNA is monocistronic, meaning that only a single type of protein can be translated from a given mRNA (choice D is incorrect and therefore the correct answer). Polycistronic mRNA is a feature of prokaryotes, which must be more efficient with the limited amount of DNA they have to work with and can translate several different proteins from the same mRNA.

A drug is developed that mimics the effect tryptophan has on the bacterial repressor protein. After administration of this drug, bacterial tryptophan synthesis will: A. increase, because the promoter for the trp operon binds RNA polymerase. B. increase, because the regulatory protein is inactivated. C. decrease, because the promoter binds RNA polymerase. D. decrease, because the regulatory protein is activated.Correct Answer

D. Tryptophan's effect on the bacterial repressor protein is to bind to and activate it. When the repressor is activated, it binds to the operator region of the trp operon and turns off transcription of the genes in that operon. Since these are genes for enzymes involved in the synthesis of tryptophan, binding of the repressor would result in a decrease in tryptophan synthesis (choice D is correct, and choice B is wrong). Note that the binding of RNA polymerase to the promoter region alone is not enough to initiate or regulate transcription (and subsequent synthesis of tryptophan); the repressor protein must also be involved (choices A and C are wrong).

A biochemist determines reaction rate using varying substrate concentrations and two units of dihydrolipoyl transacetylase, the E2 enzymatic component of the pyruvate dehydrogenase complex, at standard state. The following data are obtained. What is the Km and Vmax of this enzyme under these conditions? A. Km = 17.5 µmol, Vmax = 35 µmol/min B. Km = 9.2 µmol, Vmax = 35 µmol/min C. Km = 17.5 µmol, Vmax = 44 µmol/min D. Km = 9.2 µmol, Vmax = 44 µmol/min Correct Answer

D. Vmax is the maximum reaction rate, or 44 µmol/min (eliminate choices A and B). Km is the amount of substrate required to get to ½Vmax (22 µmol/min). Based on data in the chart, Km will be a little below 10 µmol (eliminate choice C, choice D is correct).