PR Practice Passages - Biology

The results shown in Figure 1 indicate that: A. p16 dimerization was prevented by DTT because it prevents the formation of, or breaks, disulfide bonds. B. the additional band at 240 minutes in the absence of DTT must be due to dimerization with a different protein. C. the absence of reducing conditions prevented the dimerization of p16. D. because the band indicating dimerization is less than 30 kDa, no conclusion can be made about the effect of DTT on dimerization.

A. DTT (dithiothreitol) is a reducing agent, thus it would break, or prevent the formation of, disulfide bonds. Figure 1 supports this, showing that in the absence of DTT, dimerization of p16 occurs beginning at 120 minutes (the presence of a second, larger band). This band is absent in the presence of DTT (choice A is correct and choice C is wrong). Since only p16 was immunoprecipitated, no other proteins should be present, plus the band at 14 kDa in the absence of DTT gets smaller at 120 and 240 minutes, indicating that the amount of that protein decreases. The band at about 28 kDA gets bigger, indicating that the amount of that protein increases; this must be due to loss of the individual p16 polypeptides as they dimerize (choice B is wrong). Just because the band indicating dimerization is less than 30 kDa doesn't mean we can't form conclusions. p16 was described as being 14-16 kDa, thus a dimer would have a size in the range of 28-32 kDa. The band in no DTT at 120 and 240 minutes is about that size (choice D is wrong).

A researcher identified protein disulfide isomerase (PDI) as the enzyme responsible for catalyzing the formation of disulfide bonds in the ER. She combines viral p16 monomers with PDI in vitro and detects the formation of p16 dimers. Which of the following is true once the solution has reached equilibrium? A. The rate of p16 dimer formation is the same the rate of p16 dimer dissociation. B. Removal of PDI from solution will decrease the amount of p16 dimers in solution. C. Addition of more PDI will increase the amount of p16 dimers in solution. D. p16 monomers have ceased to associate together to form p16 dimers.

A. At equilibrium, the rate of the forward reaction is equal to that of the reverse reaction (choice A is correct). It's not that p16 dimers are not forming (choice D is wrong) but rather that p16 dimers are forming just as fast as p16 dimers are breaking down. As a result, the net amount of p16 dimers is not changing. Enzymes decrease the amount of time it takes for a reaction to reach equilibrium. However, once a reaction has reached equilibrium, the addition or removal of the enzyme will have no affect (choices B and C are wrong).

Which of the following conclusions about the insulinogenic index between 30 and 60 minutes after lunch is the most likely? A. Aspartame has a higher insulinogenic index than sucrose. B. Stevia has a lower insulinogenic index than sucrose. C. Aspartame has a lower insulinogenic index than sucrose. D. Stevia has a higher insulinogenic index than sucrose.

A. Equation 1 shows that the insulinogenic index is a ratio of the change in insulin to the change in glucose over a specified time interval. Looking at Figure 2, the change in glucose levels is relatively similar for all three sweeteners 30 to 60 minutes after lunch; however, looking at Figure 3, the greatest change in insulin levels during that same time period is for those consuming aspartame. Thus, the insulinogenic index for aspartame should be higher than for the other two sweeteners (choice A is correct). With little change in glucose levels between 30 and 60 minutes, the similar changes in insulin level for both stevia and sucrose consumption as shown in Figure 3 would yield similar insulinogenic indices for these two conditions (choices B and D can be eliminated). The larger change in insulin level with aspartame consumption in the presence of similar glucose levels indicates that aspartame would have a higher, not lower, insulinogenic index as described above (choice C can be eliminated).

In the experiment performed in the passage, suppose that instead of cDNA constructs being mutated, wild-type DNA constructs were mutated at the same physical site as that of the cDNA constructs. How would the ASM activity of both constructs be affected? A. No change in ASM activity of both B. Increase in ASM activity of both C. Decrease in ASM activity of both D. Increase in ASM activity in one of the constructs only

A. In wild-type DNA, after transcription of the SMPD1 gene, the hnRNA is modified and introns are spliced out. The resulting mRNA is then translated into acid sphingomyelinase (ASM). The cDNA constructs in the experiment are created by reverse transcription of the mRNA post-splicing, followed by site-directed mutation. However, when these constructs are transfected into cell lines and then transcribed into mRNA, no further splicing occurs; the introns were already spliced out to make the cDNA in the first place. Thus, whether using wild-type DNA or using constructed cDNA, the ASM that is produced is a result of translation of already-spliced mRNA in both cases. Because of this, mutating wild-type DNA in the same location as the cDNA constructs will have no effect on the amount of ASM activity (choice C is correct, choices A, B, and D can be eliminated).

Phosphomimetics are amino acid substitutions that mimic a phosphorylated protein in structure and charge and, often, biological activity. Which of the following is most likely a true statement? A. A point mutation at residue 473 of PKB (S to D) would be phosphomimetic. B. A nonsense mutation at residue T19 of MLC or S473 of PKB will closely match the outcome of a phosphomimetic at the same site. C. Changing residue 19 of MLC to amino acid A instead of T will have the same effect as a phosphomimetic. D. None of the above statements is likely true.

A. Phosphate groups are deprotonated at physiological pH and thus have negative charge. The same is true of R-groups on the two acidic amino acids (aspartic acid or D, and glutamic acid or E). These two amino acids can be used to mimic phosphorylated amino acids (choice A is correct and choice D is wrong). A nonsense mutation will result in an early stop codon and a truncated protein. This will not have the same effect as a phosphomimetic (choice B is wrong). Alanine (A) is a small and hydrophobic amino acid that cannot be phosphorylated. If T19 of MLC or S473 of PKB were mutated to alanine, they could not be phosphorylated and this would not have the effect of a phosphomimetic (choice C is wrong).

Which of the following would be most likely exacerbate early decelerations observed during RHR monitoring? A. Administration of oxytocin to mother B. Decreasing pressure on the cervix C. Decreasing amniotic fluid volume D. Injection of epinephrine (a sympathetic agonist) in the umbilical vein

A. According to the passage, early decelerations are usually due to head compression from uterine contraction. Administration of oxytocin, a hormone which increases the intensity of uterine contraction during labor would exacerbate this effect (choice A is correct). Decreasing pressure on the cervix and decreasing amniotic fluid volume would both result in a decrease in uterine contraction and would likely decrease early decelerations (choices B and C are wrong). Epinephrine, a sympathetic agonist, would serve to increase heart rate resulting in tachycardia (choice D is wrong).

Is it possible for both men and women to develop Prader-Willi syndrome? A. Yes, both sexes have an equal probability of disease. B. Yes, both sexes can have the disease, but it is found predominantly in women because of maternal imprinting. C. No, because it is only maternally imprinted. D. No, because imprinting is reset with each successive generation.

A. As described in the passage, Prader-Willi is an autosomal syndrome resulting from the inheritance of a maternally-imprinted gene in combination with a deletion/mutation of the homologous paternal gene. Therefore, both genders can theoretically be afflicted with the disease equally (choices C and D are wrong). Maternal imprinting does not confer an increased risk of women being affected, particularly given that the disease is autosomal; all children have the same probability of inheriting an imprinted gene regardless of gender (choice B is wrong and choice A is correct).

Which of the following steroids would be best used for a cancer patient whose weight has significantly decreased? A. Oxyandrolone B. Oxymetholone C. Testosterone D. Fluoxymesterone

A. Because the medicinal use would be to increase body weight, a higher anabolic effect versus androgenic effect would be desired. The steroid with the smallest androgenic-anabolic ratio is oxyandrolone (choice A is correct). Oxymetholone and fluoxymesterone have intermediate ratios and should be eliminated (choices B and D are wrong). Testosterone has the highest androgenic-anabolic ratio, which is opposite of the desired effect in this case (choice C is wrong).

Aberrant DNA methylation patterns (hypermethylation and hypomethylation) have been shown to be associated with human cancers. Hypermethylation typically occurs at CG islands in the promoter region and is associated with gene inactivation. Hypomethylation has also been implicated in cancer by different mechanisms and may lead to the overexpression of genes. Which of the following is the most likely pattern of methylation that would lead to cancer? A. Hypermethylation of tumor suppressor genes and hypomethylation of proto-oncogenes B. Hypermethylation of proto-oncogenes and hypomethylation of tumor suppressor genes C. Hypermethylation of both tumor suppressor genes and proto-oncogenes D. Hypomethylation of both tumor suppressor genes and proto-oncogenes

A. Cancers are typically associated with inactivation of tumor suppressor genes, and the question stem states that hypermethylation can lead to gene inactivation; thus, tumor suppressor genes must be hypermethylated (choices B and D can be eliminated). It is suggested that hypomethylation may lead to the overexpression of genes, thus the hypomethylation of a protooncogene could lead to its upregulation and conversion into an oncogene (choice C can be eliminated and choice A is correct).

In the development of an STI for the treatment of FOP, a researcher discovers an increase in muscular intracellular calcium in an animal model, which results in increased muscular tone. Which of the following is the LEAST likely cause? A. The STI acts as a receptor antagonist on the motor endplate. B. The STI inhibits acetylcholinesterase. C. The STI decreases the sarcoplasmic reticulum membrane integrity. D. The STI inhibits the Ca2+-ATPase of the sarcoplasmic reticulum.

A. If the STI acted as an antagonist (inhibitor) at the motor endplate, fewer ligand-gated ion channels would open and the cell may fail to depolarize enough to stimulate the opening of voltage-gated calcium channels. This would result in decreased cytosolic calcium (choice A is least likely to explain an increase in intracellular calcium and increased muscular tone and is thus the correct answer). Acetylcholinesterase (AChE) is responsible for the destruction of acetylcholine at the neuromuscular junction. Inhibiting AChE would result in increased levels of ACh, increased muscular stimulation, and increased intracellular calcium levels (choice B is a possible cause and can be eliminated). Because calcium is stored in the sarcoplasmic reticulum (SR), disruption of the SR membrane integrity could result in an increase in intracellular calcium (choice C is a possible cause and can be eliminated). The Ca2+-ATPase is responsible for removing calcium from the cytosol and returning it to the sarcoplasmic reticulum. Its inhibition would result in increased intracellular calcium (choice D is a possible cause and can be eliminated).

A culture from a sputum sample develops into colonies of round bacterial cells. Which of the following could be the pathogen causing infection in this patient? A. Streptococcus sp. (streptococcal pharyngitis; "strep throat") Correct Answer B. Mycobacterium tuberculosis (tuberculosis, TB, tubercle bacillus) C. Bacillus anthracis (anthrax) D. Spirillum minus (rat-bite fever)

A. Round bacterial cells are known as cocci. Streptococcus bacteria causing streptococcal pharyngitis, commonly called strep throat, would develop into a culture of round bacterial cells (choice A is correct). Mycobacterium tuberculosis is rod-shaped, as indicated by the "bacillus" in tubercle bacillus (choice B is wrong). The passage explicitly states that Bacillus anthracis is rod-shaped, and this can also be inferred from the genus Bacillus (choice C is wrong). Spirillum minus is a spiral-shaped bacterium (choice D is wrong).

What would be the immune system profile of the offspring of a pair of hu-SCID chimera mice? A. Offspring would have the genetic profile for SCID and lack any human immune system cells. Correct Answer B. Offspring would have the genetic profile for SCID and carry some human immune system cells. C. Offspring might have the genetic profile for SCID and would lack any human immune system cells. D. Offspring might have the genetic profile for SCID and would carry some human immune system cells.

A. Since the mutation that leads to the SCID condition is caused by a recessive allele, SCID mice must be homozygous recessive for the mutation. Thus, if both parents have the SCID phenotype, all offspring will also be SCID (choices C and D can be eliminated). The human cells that are placed in the chimeras do not become part of the germ cell line and therefore will not be passed on (choice B can be eliminated and choice A is the best answer).

If future experiments confirm that both MBG-1 and MBG-2 promote mitochondrial biogenesis, it will be most likely true that: A. Some tissues (such as heart, muscle, pancreas and kidney) have high levels of MBG-1, MBG-2, and mitochondrial biogenesis. Correct Answer B. Cell stress, proliferation pathways and pro-apoptotic pathways can promote mitochondrial biogenesis. C. Mitochondrial biogenesis occurs primarily in G1 of the cell cycle, with negligible levels in G2. D. MBG-1 and MBG-2 are both synthesized by 80S ribosomes docked on the rough endoplasmic reticulum membrane. Your Answer

A. Some animal tissues have high number of mitochondria: heart and skeletal muscle require large amounts of energy for mechanical work, the pancreas is responsible for huge amounts of biosynthesis, and the kidney processes a huge volume of fluid every day. Since mitochondria generate energy (they are the "powerhouse" of the cell), these tissues will likely have many of these organelles. Mitochondria are made via biogenesis, so it makes sense that these tissues will have high levels of MBG-1, MBG-2, and mitochondrial biogenesis (choice A is correct). Pro-apoptotic pathways will promote cell death. If a cell is dying, it will not likely promote mitochondrial biogenesis (choice B is wrong). Organelle replication and general housekeeping activities of the cell are predominantly performed in the gap phases of the cell cycle (both G1 and G2, choice C is wrong). The passage says that both MBG-1 and MBG-2 have mitochondrial localization signals. These proteins will not go through the secretory protein pathway (choice D is wrong).

How could a cell-mediated PID contribute to impaired B cell function? A. B cells would not receive stimulatory signals from helper T cells and would not be triggered to proliferate. B. Macrophages would be unable to present antigen to B cells in order to induce appropriate antibody production. C. Abnormally growing cells would fail to display intracellular peptides and would go unrecognized by B cells. D. The complement cascade would not be induced so B cells would lack a necessary signal for proliferation.

A. The answer needs to link the function of T cells (the cell -mediated arm of the immune system) to that of B cells. B cells receive co-stimulation from helper T cells and the absence of that signal would impair B cell proliferation and function (choice A is the best answer). While macrophages do present antigen, they present to helper T cells, not B cells (choice B is wrong). Abnormally growing cells are targeted by killer T cells, but this does not involve B cells (choice C is wrong). The complement cascade is not required for B cell proliferation (choice D is wrong).

Activation of the sympathetic autonomic nervous system would have what effect on esophageal peristalsis? A. No effect Correct Answer B. Increased contractile rate C. Decreased contractile rate Your Answer D. Decreased contractile force

A. While sympathetic stimulation generally results in a decrease in GI activity, esophageal peristalsis does not change with sympathetic activity; basic maintenance activities, like swallowing saliva, need to be able to continue, even if eating and drinking are not occurring as part of a fright-or-flight response (choice A is correct).

Which of the following is true regarding the housekeeper protein used in Figure 1? A. β-actin was not a good control to use, since it is found predominantly in intermediate filaments and expression of these cytoskeleton components varies with the cell cycle. B. β-actin was a good housekeeper protein to use because it is expressed in most cells and is a major component of microfilaments. C. β-actin was a good housekeeper protein to use because it is commonly expressed in many cell types; it was used to make sure lysate separation occurred during gel electrophoresis. D. Since there was no housekeeper protein used in this experiment, the researchers should have quantified Western blot signals and compared each phospho-protein to the total amount of that protein.

B. Since most cells express β-actin, it is a good housekeeper protein to use in Western blot analysis (choice D is wrong). The purpose of checking housekeeper protein (β-actin) levels across the different samples is to ensure each lane of the Western blot contains the same amount of total lysate. This is important if lanes are going to be compared to each other. As the passage explains, if β-actin levels were not the same across the different lanes, Western blot signals would be quantified and normalized with respect to β-actin (or total lysate) amounts. Since the β-actin levels were the same, it was not necessary to do this; β-actin was a good housekeeper protein. Actin is the major components of microfilaments, not intermediate filaments (choice A is wrong). Intermediate filaments are made of proteins such as keratin, vimentin, lamin and desmin. The purpose of a housekeeper protein is not to make sure separation occurred during gel electrophoresis. This could be checked by looking at how the molecular weight ladder separated on the gel or by staining the gel to look at how the lysate samples have separated after electrophoresis (choice C is wrong).

The authors were able to confirm that there was no difference in post-prandial satiety and satisfaction across the three conditions. Assuming that the authors' other results can be externally validated in the context of participants' dietary routines at home, which of the following outcomes would be most likely over the long term? A. Supplementation with aspartame would decrease incidence of type 2 diabetes mellitus. B. Supplementation with stevia would decrease incidence of type 2 diabetes mellitus. C. Supplementation with stevia would increase incidence of obesity. D. Supplementation with aspartame would increase incidence of obesity.

B. Type 2 diabetes mellitus (adult-onset) is thought to be caused by persistently elevated levels of post-prandial glucose which lead to persistently elevated insulin levels and eventual insulin resistance; because the stevia was found to have consistently lower post-prandial blood glucose and insulin levels, it can be predicted that supplementation of one's diet with stevia-sweetened food would produce similar levels of satisfaction with a decreased risk of type 2 diabetes over the long term (choice B is correct). The aspartame condition was found to significantly decrease levels of blood glucose and insulin 20 minutes after the pre-load, but there was no difference after the meal; these results are not as strong as those for stevia which had decreased levels both post pre-load and post-meal (choice A can be eliminated). Figure 1 in combination with the information in the question stem show that the consumption of artificial sweeteners resulted in lower caloric intake for the same level of satisfaction and satiety; thus, it would be expected that decreased caloric intake would lead to more normal weight over the long term (choices C and D can be eliminated).

Which of the following would be expected in a BMP7-knockout mouse embryo compared to a normal mouse embryo? A. An increased amount of PRDM16 mRNA B. A marked paucity of brown fat and almost a complete absence of UCP1 protein C. An increase in PGC-1α activity D. decrease in the expression of the genes resistin and serpin3ak

B. As the passage states, BMP7 activates the formation of brown fat, including induction of PRDM16 and PGC-1α. Therefore, a mouse that lacked BMP7 would be expected to not express the PRDM16 gene, which would lead to decreased levels of PRDM16 mRNA (choice A is wrong). Since PRDM16 is the "master switch" for brown adipose tissue differentiation (as stated in the passage), the mouse would lack brown adipose tissue and would not express UCP1 (choice B is correct). Since BMP7 also leads to the expression of PGC-1α, it follows that lack of BMP7 would lead to decreased PGC-1α activity (choice C is wrong). Also, the passage states that PRDM16 inhibits the expression of the genes resistin and serpin3ak. Thus a lack of BMP7, which leads to a lack of PRDM16, would result in an increased expression of resistin and serpin3ak (choice D is wrong).

Based on Figure 1, thyroid hormone (a peptide hormone) exerts its effect on brown fat cells via amechanism of action that is most analogous to the mechanism of action of which of the following? A. Adrenocorticotropic hormone (ACTH) B. Cortisol C. Antidiuretic hormone (ADH) D. Insulin

B. Despite the fact that thyroid hormone is a peptide hormone, Figure 1 supports the fact that this hormone works intracellularly to exerts its effect on the cell at the level of gene expression. This is the mechanism of action primarily used by steroid hormones, such as cortisol (choice B is correct), aldosterone, testosterone, estrogen and progesterone. Peptide hormones, including ACTH (choice A), ADH (choice C) and insulin (choice D), and several others, exert their effect by binding to cell surface receptors and modifying intracellular enzyme activity.

Of the following sources, which can damage DNA in a similar manner to that of zinc finger nucleases?I. UV radiationII. X-ray radiationIII. Benzene A. I only B. II only C. I and II only D. III only

B. Item I is a good place to start because it appears twice in the answer choices; determining whether to keep or eliminate it turns the question into a 2X2 elimination. Item I is false: as stated in the paragraph 2, zinc finger nucleases cause double stranded breaks within DNA. UV radiation damages DNA by causing pyrimidine dimers (choices A and C can be eliminated). Both remaining choices include Item II so it must be true and we can consider Item III. Item III is false: benzene, an aromatic compound, results in intercalation of DNA bases rather than causing double-stranded breaks (choice D can be eliminated and choice B is correct). Note that Item II is true: X-ray radiation, like zinc finger nucleases, can cause double stranded breaks.

Glycerin trinitrate likely forms which of the following in the body to treat diffuse esophageal spasm? A. Glyceryl trinitrate B. NO Correct Answer C. NO2 Your Answer D. NO3-

B. According to the passage, diffuse esophageal spasm is linked to malfunction of endogenous nitric oxide (NO) production. Since glycerin trinitrate successfully improves peristalsis and reduces chest pain, it is likely that it forms NO in the body (choice B is correct). In fact, it is a potent vasodilator and serves as a treatment for both DES and heart disease. Glyceryl trinitrate is another name for glycerin trinitrate (choice A is wrong), NO2 is nitrous oxide ("laughing gas," choice C is wrong), and NO3- is nitrate (choice D is wrong).

How will adding more zinc finger domains to a zinc finger nuclease complex likely impact the quantity of resulting fragments when mixed with DNA? A. Increase number of fragments B. Decrease number of fragments C. No effect on number of fragments D. Increase number of fragments only if DNA is palindromic

B. As described in Paragraph 2, one or more zinc finger domains can be tethered to an endonuclease to make the zinc finger nuclease complex. The effect of adding additional zinc finger domains will be a higher specificity of the zinc finger nuclease for the s trands of DNA. This is because the DNA strand complementary to the combined zinc finger domains will need to be longer in order to bind. Thus, there will be fewer locations on the DNA where the zinc finger nuclease can bind. If there are fewer binding locations, there will be fewer cuts along the DNA, and thus the number of resulting fragments will decrease (choice B is correct; choices A, C, and D are wrong).

A physician collects two blood samples and sends them to the lab for culture and an ELISA. The culture comes back negative but the ELISA is positive for human anti-PA IgG. What is a possible reason for this result? A. The patient is infected with a virus producing similar symptoms. B. The patient does not have an active infection. Correct Answer C. The ELISA results are not conclusive because the test is nonspecific. D. B. anthracis do not grow in culture.

B. Bacteria can be cultured from patient blood only when an active infection is present in the blood. This patient could have already cleared the infection and mounted an immune response, resulting in a positive ELISA but negative culture result (choice B is correct). The ELISA test is very specific (choice C is wrong), and there is no reason to believe that a human would develop the same or similar antibodies against a B. anthracis infection and a viral infection. Even if the patient was infected with a virus that cause symptoms similar to a B. anthracis infection, he or she would not have a positive ELISA given the specificity of antibodies (choice A is wrong). According to the passage, B. anthracis can be grown in culture (choice D is wrong).

The experiments in the passage would require each of the following procedures EXCEPT: A. Quantitative PCR B. Transformation Correct Answer C. Cell lysis Your Answer D. Western blot

B. In the experiment, siRNA pools would be introduced into the growing cells by transfection (the transfer of nucleic acids into eukaryotic cells). Transformation is the transfer of nucleic acids into a bacterial cell, not a eukaryotic cell (choice B would not be required by the experiments and is the correct answer choice). The data in Figure 2 would require cell lysis (choice C is required and can be eliminated), extraction of DNA from different subcellular fractions, and quantitative PCR to quantify the genomes (choice A is required and can be eliminated). The cells would be lysed to generate protein lysates, and protein levels are determined via western blots (choice D is required and can be eliminated).

Which of the following is correct about anabolic-androgenic steroids? A. They are difficult to detect because they are structurally similar to naturally produced hormones. B. Marathon runners and anemic patients benefit from the same effect of anabolic-androgenic steroids. C. Medicinal uses only take advantage of the anabolic effects. D. Androgenic side effects only occur in men.

B. Marathon runners would want to use these steroids to enhance endurance via an increased red blood cell count, which is also why anemic patients would take these steroids (choice B is correct). The passage states that these steroids are easily detected (choice A is wrong), and that the type of steroid used medicinally depends on whether the anabolic or androgenic effects are required (choice C is wrong). It is not indicated that the androgenic side effects only occur in men; anabolic-androgenic steroid use can cause typically male characteristics, such as growth of body hair, to be evident in females (choice D is wrong).

All of the following statements about the SCID or hu-SCID mice are true EXCEPT: A. by utilizing human bone marrow from a wide variety of sources when creating the hu-SCID chimera, different human autoimmune diseases could be introduced into the SCID mouse. B. helper T cells of the hu-SCID chimera are not able to be infected by HIV. Correct Answer C. SCID mice have an extremely low rate of organ transplant rejection. Your Answer D. SCID mice have impaired complement immunity.

B. Since the helper T cells in the hu-SCID chimera come from humans in the form of a bone marrow transplant and since human helper T cells are capable of being infected with HIV, the helper T cells of the hu-SCID chimera should be able to be infected by HIV (choice B is not true and is the correct answer). One of the nice things about the SCID mouse is that it can be "custom designed" by using different human bone marrow donors. If the bone marrow from a normal (non-autoimmune) individual is used, then the hu-SCID chimera would have a normal human immune system. But, if the bone marrow from an individual with an autoimmune disease is used, then the hu-SCID chimera could also display that disease (choice A is true and can be eliminated). Since SCID mice lack a functional immune system they are unable to reject transplanted organs (choice C is true and can be eliminated). Finally, the complement system can be activated by antibodies (among other ways of activation), and, since the SCID mouse is unable to produce antibodies, it is likely that the SCID mouse would have some degree of compromised complement activation (choice D is true and can be eliminated).

Which individual is most likely to present with an immune system profile like that of a person with SCID? A. A 54-year old man who underwent a thymectomy as an infant B. A 3-year old girl undergoing extensive radiation therapy as part of cancer treatment C. A 27-year old woman positive for the human immunodeficiency virus (HIV) D. A 41-year old man with Type I diabetes

B. Since the question is looking for a condition most analogous to SCID, the correct answer needs to represent the individual with deficits in both cell-mediated and humoral immunity. Someone who had their thymus removed as a child would be expected to have deficits in cell-mediated immunity to the lack of T cell education and the potential for more auto-reactive T cells but would not have the full scale failure of SCID (choice A can be eliminated). An HIV patient will experience deficits due to infection of T helper cells. However, there is no additional information to describe the state of the patient's humoral immunity (choice C can be eliminated). An adult with Type I diabetes has likely been living with the disorder for much of his or her life. Diabetes leads to high blood sugar (and a number of other effects such as kidney failure and poor wound healing), but this is not the same as having a non-functional immune system (choice D can be eliminated). Choice B represents the most profound lack of immune system cells in the most vulnerable patient; radiation will kill the progenitor cells that would normally be creating the immune system of a young child.

Ventricular septal defects (VSDs) are the most common congenital heart malformations. VSD is due to a failure of the ventricular septum to fully close during development. Which of the following primary germ layers is responsible for forming the septum of the ventricles of the heart? A. Endoderm Your Answer B. Mesoderm Correct Answer C. Ectoderm D. Cardioderma

B. The mesoderm is responsible for all muscle, bone, connective tissue, urogenital organs and the entire cardiovascular/lymphatic system, including blood and the heart (choice B is correct). The ectoderm is responsible for the nervous system as well as the epidermis of skin and its derivatives (hair, nails, sweat glands, sensory receptors) and nasal, oral, and anal epithelium (choice A is incorrect). The endoderm forms the GI tract epithelium (except the mouth and anus), the GI glands (liver, pancreas, etc.), urinary bladder and the epithelial lining of the urogenital organs and ducts (choice C is incorrect). Cardioderma is not a primary germ cell layer at all (choice D is incorrect). Cardioderma is actually the genus name of the "heart-nosed bat" (Cardioderma cor).

Is it possible for a man with Prader-Willi syndrome who marries an unaffected woman to have children with Prader-Willi syndrome? A. Yes, the children will either inherit a silenced (imprinted) chromosome or a mutated chromosome from him. B. Yes, the children have a 50% chance of inheriting the father's mutated chromosome and will all inherit a silenced (imprinted) chromosome from their mother. C. No, the mutated chromosome can only be inherited from the mother. D. No, they could inherit the mutated chromosome from him, but could inherit a normal chromosome from their mother.

B. The passage states that Prader-Willi syndrome is caused by the combination of maternal imprinting (gene silencing) and the inheritance of a mutated chromosome from one's father. The man in this question must have one mutated chromosome and one imprinted chromosome. However, since the gene must be maternally imprinted, he will not be able to pass the imprinting to his children; they will inherited either a mutated chromosome or a normal (unimprinted) chromosome from him (choice A is wrong). Since the gene is maternally imprinted, all children will inherit an imprinted chromosome from their mother; if they are unlucky enough to get the mutated chromosome from their father, they will also have Prader-Willi syndrome (choice B is correct). The mother does not have any mutated chromosomes (she is unaffected, choice C is wrong), and all chromosomes inherited from the mother will be imprinted and thus silenced (choice D is wrong).

Which of the following treatments would help decrease the false positive rate of diagnostic 18F-FDGPET scans for cancers?I. Warming the room occupied by the study subjectII. Administering beta-blockers (medications that block the β-adrenergic receptors)III. Administering a synthetic version of T4 A. I only B. I and II C. III only D. I, II, and III

B. The passage states that the increased metabolic activity and uptake of substrates, including glucose, by activated brown fat cells leads to "false positives" on 18F-FDG PET scans for cancers. Therefore, in order to decrease the uptake of radiolabeled glucose at the brown fat cells, the intervention should decrease brown fat cell metabolic activity. Item I is true: The passage states that exposure to cold temperatures rapidly activates brown fat cells; thus, warming the room would lead to decreased brown fat activity (choice C can be eliminated). Item II is true: Since activity of the β3 receptor leads to increased brown fat cell metabolic activity, the administration of beta-blockers should lead to decreased activity (choice A can be eliminated). Item III is false: The passage and figure support that increased thyroid hormone (T4) levels lead to increased brown fat cell metabolic activity. Thus, the administration of a synthetic T4 would lead to increased metabolic activity and be counterproductive (choice D can be eliminated and choice B is correct).

Which of the following is NOT a trigger for labor? A. Increased uterine stretch B. Decreased levels of hCG C. Increased stretch of the cervix D. Decreased levels of estrogen

B. There are several things that can lead to an excitable uterus and possible labor. Increased stretch on the uterus can trigger its contraction (choice A could be a trigger for labor and can be eliminated). Increased stretch on the cervix, typically caused by the baby's head, can trigger the release of oxytocin which in turn stimulates uterine contractions (choice C could be a trigger for labor and can be eliminated). Due to the deterioration of the placenta, decreased levels of estrogen can stimulate contractions (choice D could be a trigger for labor and can be eliminated). However, hCG is virtually absent from the system from about five months of gestation onward; its levels are high during early pregnancy to maintain the corpus luteum (and thus the levels of estrogen and progesterone), but hCG level begins to fall once the placenta is formed (about 3 months gestation) and are virtually absent by full term labor (choice B is not a trigger for labor and is the correct answer choice).

Which of the following is/are medicinal use(s) of anabolic-androgenic steroids?I. Induction of pubertyII. Preventing lean muscle mass lossIII. Acne treatment A. I only B. II only C. I and II only D. I, II, and III

C. Items I and II are true: both induction of puberty and increased lean muscle mass are effects of anabolic-androgenic steroid use, and can be used as medical treatment (choices A and B can be eliminated). Anabolic-androgenic steroids can cause acne, making Item III false (choice D can be eliminated; choice C is the correct answer).

Which of the following cofactors is NOT produced during glycolysis? A. ADP B. ATP C. NAD+ D. NADH

C. In the conversion of glyceraldehyde-3-phosphate to 1,3-bisphosphoglycerate, both NAD+ and inorganic phosphate (Pi) are consumed and NADH is produced; thus, NAD+ is not generated in glycolysis (choice D can be eliminated and choice C is correct). ADP is produced twice (choice A can be eliminated), first in the step in which glucose is converted to glucose-6-phosphate and second when fructose-6-phosphate is converted to fructose-1,6-bisphosphate. In both of these steps, ATP provides the Pi to the carbohydrate undergoing metabolism. In addition to generating pyruvate for the Krebs cycle, the primary purpose of glycolysis is to generate ATP (choice B can be eliminated).

Which of the following statements best describes the mechanism by which insulin decreases blood glucose levels? A. Insulin binds directly to individual molecules of glucose in the bloodstream, facilitating uptake. B. Insulin passes easily through the lipophilic cell membrane and travels to the nucleus where it upregulates the synthesis of glucose transporter proteins. C. Insulin binds to the extracellular portion of the insulin receptor, triggering an intracellular cascade resulting in increased expression of glucose transporters on the cell surface. D. Insulin binds to a receptor in the brain, leading to increase autonomic nervous system activity and, thus, cellular respiration.

C. Insulin is a peptide hormone which will bind to a receptor on the extracellular surface of the cell membrane to induce an intracellular biochemical cascade for the desired effect, in this case, increasing expression of glucose (GLUT4) channels on the cell membrane surface. These channels then allow for increased influx of glucose from the blood into the intracellular space (choice C is correct). Insulin does not bind directly to glucose in the blood (choice A is wrong). Insulin is not a steroid hormone and does not pass easily through lipophilic membranes (choice B is wrong). While insulin actually does bind to receptors in the brain to induce increased autonomic system activity, this is not known to be a major contributor to glucose metabolism (choice D is wron

Which of the following is most likely true if the formation of the p21 dimer in the ER is an exothermic process? A. The formation of a p21 dimer is spontaneous at all temperatures. B. The formation of a p21 dimer is nonspontaneous at all temperatures. C. The formation of a p21 dimer is spontaneous only at low temperatures. D. The formation of a p21 dimer is spontaneous only at high temperatures.

C. The formation of a dimer is associated with a decrease in entropy (ΔS < 0). The question stem states the formation of a protein dimer is an exothermic process (ΔH < 0). A reaction is spontaneous if ΔG < 0. Based on ΔS and ΔH values and the equation ΔG = ΔH - TΔS, ΔG for the formation of the p21 dimer will be less than 0 only at low temperatures.

Given that most disulfide bond formation occurs in the lumen of the ER and not in the cytoplasm, which of the following statements is true? A. Both the ER lumen and the cytoplasm are reducing environments. B. Both the ER lumen and the cytoplasm are oxidizing environments. C. The ER lumen is an oxidizing environment while the cytoplasm is a reducing environment. D. The ER lumen is a reducing environment while the cytoplasm is an oxidizing environment.

C. The formation of disulfide bonds requires an oxidation, thus the ER lumen must be an oxidizing environment (choices A and D can be eliminated). Since disulfide bond formation generally does not occur in the cytoplasm, it must be a reducing environment (choice B can be eliminated and choice C is correct).

A patient with a pheochromocytoma, a tumor of the adrenal gland that leads to hypersecretion ofepinephrine, would be expected to have which of the following? A. Decreased brown adipose tissue intracellular levels of T3 B. Decreased brown adipose tissue Krebs cycle activity C. Increased brown adipose tissue levels of D2 mRNA D. Increased fat cell activation, which would lead to increased weight gain

C. As described in the first paragraph and the figure, increased epinephrine would lead to activation of the β3 receptor, a GPCR which triggers the production of D2 (choice C is correct). The increased D2 would convert more T4 to T3 (choice A is wrong). This would lead to increased metabolic activity, including the Krebs cycle (choice B is wrong). Ultimately be in the form of uncoupled oxidative respiration, this increased metabolic activity would be expected to lead to weight loss rather than weight gain (choice D is wrong).

Nondisjunction during anaphase of meiosis may lead to a trisomy known as Klinefelter's syndrome (XXY). The resulting individual has a male phenotype but has testicular atrophy, a slender body shape, gynecomastia (breast development in males), and female hair distribution. This syndrome usually results in dysgenesis of the seminiferous tubules, as well as abnormal Leydig cell function. Which of the following hormones would be expected to be increased compared to a normal male in a patient with Klinefelter's syndrome? A. Inhibin B. Testosterone C. FSH D. Prolactin

C. A patient with Klinefelter's syndrome has testicular atrophy, thus any hormone normally produced by the testes (testosterone and inhibin) would be decreased and can be eliminated (choices A and B are wrong). The decrease in testosterone and inhibin would lead to decreased negative feedback on the anterior pituitary, causing increased levels of LH and FSH (choice C is correct). Prolactin should be unaffected in patients with Klinefelter's syndrome; the gynecomastia in this case is due to the increased levels of estrogen (choice D can be eliminated).

In the creation of the hu-SCID chimera, why is it necessary to place human thymus within the mouse? A. The human thymus ablates any remaining immune system cells within the SCID animal. B. The human thymus provides a source of lysozyme and thus bolsters innate immune protection. C. The human thymus acts as the source of self-antigens, eliminating auto-reactive cells from those produced by the bone marrow progenitors. D. The human thymus prepares the mouse's body to accept the introduction of other tissues and prevents graft rejection.

C. A primary role for the thymus in the immune system is to present self-antigens to developing T cells and remove those that react to self-antigens. This helps prevent autoimmune responses (choice C is the best answer). The original SCID mouse is better able to avoid graft rejection due to its non-functional immune system (choice D is eliminated) and there is no evidence to support the idea that the thymus will kill off native immune system cells within the mouse (choice A is eliminated) or provide the enzyme lysozyme (choice B is eliminated).

According to the passage, which of the following is the most effective treatment for FOP? A. Surgery removing the overgrown bone B. Sonification of overgrown bone to aid in bone particle removal C. Systemic glucocorticoids to reduce inflammation D. Oral calcitonin

C. According to the passage, current treatment for FOP focuses on supportive therapy and management of inflammation. Glucocorticoids reduce inflammation and could help decrease the degree of heterotropic ossification in the patient (choice C is a viable treatment). Both surgical removal of bone and sonification would likely result in trauma which may worsen the condition (choices A and B are not viable treatments). Oral calcitonin, a polypeptide, would be digested in the stomach and would not have significant therapeutic benefit; more importantly, calcitonin helps stimulate osteoblasts and bone formation which is clearly not the best option for treating FOP (choice D is wrong).

Which of the following is the most likely cause for the difference in fetal and adult heart rate? A. Increased parasympathetic activity at the adult AV node Your Answer B. Increased norepinephrine release at the adult AV node C. Decreased parasympathetic activity at the fetal SA node Correct Answer D. Decreased norepinephrine release at the fetal SA node

C. According to the passage, fetal heart rate is normally between 110-150 bpm which is significantly elevated compared to adult heart rate. The SA node serves as the pacemaker in the healthy heart (choices A and B are wrong) and a decrease in parasympathetic activity at the fetal SA node relative to the adult SA node would result in an elevated fetal heart rate (choice C is correct). Norepinephrine serves as the most common neurotransmitter at the final synapse in the sympathetic autonomic nervous system, so a decrease in sympathetic activity would serve to slow the fetal heart rate which does not account for the observation that it is faster than in the adult (choice D is wrong).

Researchers have characterized several compounds affecting the activity of esophageal smooth muscle activity. According to the figure, which of the following would lead to smooth muscle contraction in the esophagus? A. Increased acetylcholinesterase release Your Answer B. Destruction of the rostral potion of the dorsal motor nucleus (DMN) C. Addition of a Substance P (SP) agonist Correct Answer D. Increased vasoactive intestinal peptide (VIP)

C. An agonist produces or increases the physiological effect of interest. From the figure, SP activates smooth muscle contraction, thus its agonist would also do so (choice C is correct). Acetylcholinesterase removes ACh from the synaptic cleft, so increasing acetylcholinesterase would decrease stimulation to the smooth muscle cell (choice A is wrong). Destruction of the rostral portion of the dorsal motor nucleus would damage the vagus nerve and would prevent smooth muscle contraction (among a list of other, far more severe problems; choice B is wrong). According to the figure, VIP inhibits smooth muscle contraction (choice D is wrong).

A researcher creates an inducible knockout mouse designed for limiting levels of calmodulin expression in the digestive tract. Which of the following would be a likely effect in the digestive tract due to calmodulin knockout? A. Diarrhea B. Increased myosin light-chain phosphorylation C. Dysphagia (painful/difficulty swallowing) D. Increased cross-bridging

C. As described in the passage, calmodulin binds calcium and results in the activation of myosin light- chain kinase and subsequent phosphorylation of the myosin light-chain. Phosphorylation allows for cross bridging and smooth muscle contraction. By limiting the amount of calmodulin present, smooth muscle contraction in the digestive tract cannot take place as readily, and peristalsis cannot occur regularly. This leads to discomfort when trying to swallow food (choice C is correct). Diarrhea would be expected in a situation of increased digestive tract motility (choice A is wrong). With a deficiency of calmodulin, decreased myosin light-chain phosphorylation would be expected (choice B is wrong), and thus decreased subsequent cross-bridge formation (choice D is wrong).

Since the drug azodicarbonamide is electrophilic, it will most likely be reactive at the molecular level with which of the following components of the zinc finger domain? A. Zinc cation B. Valine side-chains within the zinc finger C. Cysteine residues coordinating the zinc cation D. Nitrogen lone-pairs within the peptide bonds of the peptide backbone

C. Azodicarbonamide is an electrophilic molecule that reacts with the zinc finger domain to inhibit the nucleocapsid formation within HIV. The component most reactive with the electrophilic drug will be one that is nucleophilic. Nucleophiles are electron-rich and have a lone-pair to donate. Cysteine residues are nucleophilic, and if cysteine attacks the azodicarbonamide drug, its coordination with the zinc cation of the HIV zinc finger will be disrupted, making it a reasonable mechanism of drug action (choice C is correct). The zinc cation is a positively charged divalent ion and as a result is not electron-rich (choice A is wrong). Valine side chains cannot act as nucleophiles because they do not have free lone pairs to attack electrophiles (choice B is wrong). While the nitrogen atoms in the peptide bonds do have free lone pairs, they are not as nucleophilic, as the lone pairs are delocalized through resonance with the carbonyl oxygen in the peptide backbone (choice D is wrong).

The data presented in the passage best supports which of the following conclusions? A. Both siRNA pools were effective at binding their target transcript and promoting its degradation. B. siRNA pool 1 directly caused the degradation of MBG-1 protein, but siRNA pool 2 was not as effective. Your Answer C. siRNA pool 1 caused degradation of the MBG-1 transcript; siRNA pool 2 is either not effective or MBG-2 proteins have a long half life. Correct Answer D. Neither siRNA pool was effective at knocking down mRNA levels, but pool 1 was more effective at inhibiting translation.

C. Figure 1 shows that siRNA pool 1 effectively decreases the amount of MBG-1 protein in the cells, while siRNA pool 2 was not effective at decreasing the amount of MBG-2 protein. siRNAs bind complementary sequences on mRNAs, and this double stranded RNA is then degraded. The amount of transcript in the cell decreases and gene expression is thus negatively regulated. In the experiment in the passage, MBG-1 transcript will be made and degraded when siRNA pool 1 was added. Protein levels are low because the mRNA is degraded, not because the protein is degraded (choice B is wrong) or because translation is inhibited (choice D is wrong). It is possible that MBG-2 proteins have a long half-life in the cell. In this case, even if MBG-2 siRNAs bind and degrade the transcript, there will be minimal effect of the siRNA pool on protein levels (choice C is the correct answer). Choice A is possible, but without quantifying the amount of mRNA in the cell, you cannot be sure this is happening (choice C is better than choice A).

Researchers have shown that peristalsis in the esophagus is preceded by a wave of smooth muscle relaxation (or inhibition of contraction) known as deglutitive inhibition. What is a possible reason for this wave of inhibition? A. To eat large quantities of food in one sitting B. To prevent potential esophageal damage associated with overuse Your Answer C. To allow for the normal drinking of a glass of water Correct Answer D. To allow for increased food storage capacity in the digestive tract

C. From the passage, we know that esophageal peristalsis contractions last for 8-10 seconds, but drinking water occurs more rapidly than that and thus this would obstruct the next bolus of water from passing down the esophagus. By causing inhibition along the esophagus during repeated swallowing, it allows fluids to flow down the esophagus without interference from peristalsis (choice C is correct). Deglutitive inhibition will not increase the total quantity of the food that can be eaten in one sitting (although it may affect rate), nor will it increase storage capacity of the digestive tract as food is not stored in the esophagus (choices A and D are wrong). Deglutitive inhibition may prevent damage associated with food being forced down a contracted esophagus, but this would not be related to overuse (choice B is wrong).

One of the potential complications of inhalation or pulmonary anthrax is the development of hypotension. Which of the following would not cause hypotension in a healthy patient? A. SA node depression B. Destruction of the posterior pituitary C. ACTH secreting lung tumor Correct Answer D. Hemorrhage

C. Hypotension is a reduction in blood pressure. ACTH stimulates the release of cortisol and aldosterone from the adrenal cortex, and aldosterone causes an increase in the reabsorption of Na+ from the distal tubules of the kidneys. This would lead to an increase in blood pressure, or hypertension (choice C does not cause hypotension and is the correct answer choice). SA node depression would result in a decrease in heart rate and a subsequent decrease in blood pressure (choice A is a possible cause of hypotension and can be eliminated). Destruction of the posterior pituitary would lead to a drop in ADH and a subsequent inability to reabsorb water from the collecting ducts of the kidneys. This would result in a decrease in blood volume and thus a drop in blood pressure (choice B is a possible cause of hypotension and can be eliminated). Hemorrhage would also lead to a decrease in blood volume and result in hypotension (choice D can be eliminated).

Mr. A has hemophilia; he and his wife (Ms. B, who is neither affected by hemophilia nor a carrier) produce one unaffected son (Mr. C) and one unaffected daughter (Ms. D). The son of Ms. D marries an unaffected, non-carrier woman. What is the probability they will produce a child who is a carrier of hemophilia? A. 0 B. 1/8 Your Answer C. 1/4 Correct Answer D. 1/2

C. If Mr. A has hemophilia (XhY), then his unaffected daughter, Ms. D, is a carrier of hemophilia (XHXh). Her son may or may not be affected depending on which X chromosome he inherits; he as a 1/2 probability of inheriting Xh. If he is affected and marries an unaffected, non-carrier woman, then all of his daughters will be carriers of hemophilia, and none of his sons would be affected. He has a 1/2 probability of being affected by hemophilia, and a 1/2 probability of producing a daughter, so the total probability of producing a carrier child is 1/4 (1/2 x 1/2 = 1/4).

Many of the side effects of anabolic-androgenic steroids result from feedback on the hypothalamic pituitary axis. Which of the following is true regarding the nature of this interaction? A. It is a positive feedback cycle that results in an increase in hormone release from the anterior pituitary. B. It is a negative feedback cycle that results in an increase in hormone release from the anterior pituitary. C. It is a negative feedback cycle that results in a decrease in hormone release from the anterior pituitary. D. It is a negative feedback cycle that results in a decrease in hormone release from the posterior pituitary.

C. Naturally occurring anabolic-androgenic steroids, including testosterone, are products of the hypothalamic-pituitary axis. The secretion of gonadotropin releasing hormone from the hypothalamus stimulates the release of LH and FSH from the anterior pituitary, which in turn stimulate testosterone production. A build-up of hormones with similar chemical structures to testosterone will induce negative feedback on the hypothalamic-pituitary axis (choice A is wrong). Negative feedback tends to inhibit the release of hormones at the target organs (choice B is wrong). In the case of anabolic-androgenic steroids, the target organ is the anterior pituitary gland, not the posterior pituitary (choice D is wrong and choice C is correct).

Which of the following takes place in the lymph nodes? A. Differentiation of B cells B. Release of circulating antibodies by natural killer cells C. Nonspecific phagocytosis by macrophages D. Removal of self-reacting B cells

C. The lymph nodes serve as a site for nonspecific filtration of the lymph by macrophages as well as a site for storage and activation of B and T cells (choice C is correct). B cell differentiation and screening for self-reacting cells takes place in the bone marrow (choices A and D are wrong). Natural killer cells are part of the innate immune system and do not release antibodies (choice B is wrong).

Anabolic steroids are used by some athletes to increase muscle mass. Which of the following is the best explanation for this effect? A. They only act to increase the number of muscle fibrils. B. They increase red blood cell count. C. They increase protein metabolism in muscles and also increase the size and number of muscle cells. D. They lead to an increase in testosterone production.

C. The passage states that the anabolic effects of steroids can lead to increased protein synthesis as well as an increase in the size and number of muscle cells (choice C is correct); this confirms that their effects go beyond only increasing the number of muscle fibrils (choice A is wrong). Although endurance athletes benefit from an increased red blood cell count, this does not increase muscle mass (choice B is a true statement but does not offer an explanation for increased muscle mass and can be eliminated). The passage states that steroid use decreases testosterone production (choice D is wrong).

Which of the following is a potential molecular mechanism by which a patient could develop fibrodysplasia ossificans progressiva? A. A mutation that generates a stop codon early in the coding region B. A mutation in the promoter region of the gene C. A mutation in the coding region that changes an amino acid D. An insertion mutation in an intron

C. The passage states that the gene responsible for FOP has a missense mutation in the protein's activation domain. A missense mutation is a point mutation that results in a single amino acid substitution (choice C is correct). In the case of FOP specifically, histidine is substituted for arginine at position 206 of the protein. A mutation that generates an early stop codon, which would truncate the protein, is called a nonsense mutation (choice A is wrong). A mutation in the promoter region of the gene would affect whether the gene is transcribed, but would not affect the conformation of the resulting protein (choice B is wrong). Do not be fooled by the wording in the passage, which describes a mutation in the activation domain of the protein; the promoter is a part of the DNA, not the protein. A mutation in an intron will be spliced out during RNA processing and should not affect the resulting protein; it is a silent mutation (choice D is wrong).

Which of the following could support the conclusions drawn by the collaborating neurobiology lab mentioned in the passage? I. Pharmacologically inhibiting ILK activity in cultured sympathetic neurons reduces dendrite formation. II. Expressing a dominant negative GSK3 allele causes robust dendrite initiation. III. Inhibition of GSK3 by genetic knockdown promotes dendrite initiation even when ILK is simultaneously inhibited. A. I only B. I and II C. II and III D. I, II and III

D. Since Item III appears in only two of the answer choices, start by analyzing Item III. This will turn the question in to a 50:50 elimination. Item III is true: the collaborating lab has shown that a decrease in GSK3 activity promotes dendrite initiation and growth. If the inhibition of GSK3 by genetic knockdown (i.e., there is no GSK3 and thus no GSK3 activity) produces the same effect, this supports their conclusion, regardless of the activation or inhibition of ILK (choices A and B can be eliminated). Both remaining choices include Item II so it must be true, and we can look at Item I. Item I is true: the collaborating neurobiology lab has data indicating that ILK phosphorylates GSK3 in sympathetic neurons, and that in these cells, this results in decreased GSK3 activity to promote dendrite initiation and growth. In other words, they believe that ILK negatively regulates GSK3, thereby positively regulating dendrite initiation and growth. If ILK were inhibited, it would no longer be able to negatively regulate GSK3, and there would be a reduction in dendrite formation. Item I supports their conclusion (choice C can be eliminated and choice D is correct). Note that Item II is in fact true: a "dominant negative" GSK3 allele means that either GSK3 will not be expressed, or, if it is expressed, it will be non-functional. This is consistent with the idea of inhibiting GSK3 and seeing dendrite growth.

A pathologist examines a biopsied tissue sample under high resolution and observes numerous red, mononucleated, striated cells. Is this an appropriate tissue sample for determining whether the patient is suffering from FOP? A. Yes, because heterotropic bone replaces skeletal muscle. B. Yes, because heterotropic bone replaces connective tissue. C. No, because it is not a bone sample. D. No, because FOP does not affect cardiac muscle.

D. This biopsy is of cardiac muscle (mononucleated, striated cells) and would not undergo heterotropic ossification (HO); therefore, we cannot conclude whether this patient is suffering from FOP (choice D is correct). Skeletal muscle is striated, but the cells are multinucleated (choice A is wrong). Connective tissue consists of cells within a matrix, and does not exhibit the striated organization of skeletal and cardiac muscle (choice B is wrong). A pathologist would be looking for ossification of tissues other than bone to diagnose FOP (choice C is wrong).

When sphingomyelin is broken down by sphingomyelinases, it is degraded into two components: phosphocholine and ceramide. One of these components diffuses through the cell membrane and participates in the apoptotic signaling pathway. The component that most likely participates in the apoptotic signaling pathway is: A. phosphocholine, because its size relative to ceramide allows it to diffuse through the cell membrane more easily. B. phosphocholine, because its polarity relative to ceramide allows it to diffuse through the cell membrane more easily. C. ceramide, because its size relative to phosphocholine allows it to diffuse through the cell membrane more easily. D. ceramide, because its polarity relative to phosphocholine allows it to diffuse through the cell membrane more easily.

D. This is a 2x2 type of question. As stated in the question stem, the products of sphingomyelin degradation are phosphocholine (Figure 1) and ceramide (everything in sphingomyelin except for phosphocholine). The cell membrane is comprised of a lipid bilayer with an interior hydrophobic core. Because of this hydrophobic core, charged, polar molecules such as phosphocholine will not be permeable (choices A and B are wrong). In general, smaller molecules diffuse through the membrane faster than larger molecules. However, ceramide is a larger molecule than phosphocholine, so size is not a reasonable explanation as to why ceramide can diffuse through the membrane (choice C is wrong; choice D is correct). Because of its long, hydrocarbon chains, the hydrophobic ceramide can diffuse through the also hydrophobic membrane interior and participate in the apoptotic signaling pathway.

The rate of which of the following biochemical processes would increase with increasing insulin secretion? A. Lipolysis B. Proteolysis C. Gluconeogenesis D. Glycogen synthesis

D. Insulin release and elevated blood insulin levels are typically associated with an anabolic state in which fats, proteins, and complex carbohydrates are synthesized by the body; thus, it is reasonable to expect that glycogen (a complex, branching carbohydrate) will be synthesized during these times (choice D is correct). Lipolysis and proteolysis will primarily take place during catabolic states, times when the body needs to break down stored fats and proteins for energy in a fasting state (choices A and B are wrong). Gluconeogenesis produces glucose which can be released into the blood in a fasting state (choice C is wrong).

In one experiment, an adenovirus vector was used to increase BMP7 expression in mice. Which of the following would be LEAST likely to occur? A. A reduction in weight gain B. An increase in brown fat mass C. An increase in energy expenditure D. An increase in fasting plasma glucose levels

D. Since BMP7 induces brown fat cell metabolism, an increase in BMP7 would likely lead to weight loss due to increased metabolism and uncoupled cellular respiration (choice A is likely and can be eliminated). This would result in increased energy expenditure (choice C is likely and can be eliminated). BMP7 also induces many genes, including PRDM16, which would lead to brown fat cell differentiation; thus, the total brown fat cell mass would increase (choice B is likely and can be eliminated). However, the passage states that brown fat cell activity is inversely related to fasting plasma glucose; therefore, an increase in brown fat cell activity would lead to a decrease in fasting plasma glucose levels (choice D is not likely and the correct answer choice).

According to the passage, which of the following samples is LEAST likely to result in an anthrax diagnosis? A. Blood B. Sputum C. Feces D. Cerebrospinal fluid Correct Answer

D. A positive result from a sample of cerebrospinal fluid would result in a diagnosis of meningitis, not anthrax (choice D is correct). The passage states that an infection with B. anthracis can manifest as cutaneous, pulmonary, or gastrointestinal anthrax, which indicates that a positive result from samples of blood, sputum, or feces could lead to an anthrax diagnosis (choices A, B, and C can be eliminated).

A patient with fibrodysplasia ossificans progressiva presents at the doctor's office. Which of the following physical exam findings would you expect to find in this patient? A. Tongue deviates to one side when the patient is instructed to stick it out B. Nystagmus (rapid involuntary oscillations of the eye) C. Severe dyspnea (labored respiration) D. Nuchal rigidity (neck stiffness)

D. According to the passage, FOP spares the diaphragm, tongue, and extraocular muscles, meaning we would expect normal physical exam findings in these areas (choices A, B, and C are not potential areas of ossification and are less likely to result in abnormal findings). Nuchal rigidity (neck stiffness) could occur due to ossification of the skeletal muscle in the neck (choice D is correct).

In the PCR detection of the bacterium, a laboratory technician obtains a positive result. How many bands should the technician detect on the gel? A. 2 B. 3 C. 4 D. 5 Correct Answer

D. According to the passage, the PCR assay three loci on the bacterial chromosome and one on each of the virulence plasmids (2 total from the plasmids). This means there should be five DNA fragments amplified by PCR, and the technician should have seen a total of five bands on the electrophoresis gel (choice D is correct).

The gene responsible for Angelman syndrome is UBE3A. In a person with this syndrome, what is the difference between the maternal and paternal chromosome? A. There is no difference; both homologous genes are methylated. B. There is no difference; both homologous genes are unmethylated. C. The maternal gene is methylated and the paternal one deleted or mutated. D. The paternal gene is methylated and the maternal one deleted or mutated. Correct Answer

D. As stated in the passage, methylation is the mechanism underlying imprinted diseases whereby methylated genes are repressed. As such, the paternal and maternal chromosomes will differ in terms of methylation patterns (choices A and B are wrong). The passage states that Angelman syndrome is a paternally imprinted disease, thus a segment of the paternal chromosome is silenced (imprinted via methylation) and the maternal gene in this region is deleted or mutated (choice D is correct and choice C is wrong).

The pharmacodynamics of steroid hormones and peptide hormones differ due to the different chemical properties of these two classes of hormones. Which of the following is true regarding these differences? A. Steroid hormones need to be taken intravenously, whereas peptide hormones are readily taken orally. B. Steroid hormones affect posttranslational changes, whereas peptide hormones only affect transcription. C. Steroids begin their action with the production of second messengers, whereas peptide hormones begin their action when bound to their receptor. D. Steroids activate their target intracellularly, whereas peptide hormones activate their target extracellularly.

D. Because they are hydrophobic, steroid hormones can penetrate the cellular membrane to bind to intracellular receptors. In contrast, peptide hormones typically bind to receptors on the cell surface and produce second messengers (choice D is correct and choice C is wrong). Also, due to their hydrophobicity, steroids can be taken orally because they will readily cross the gastrointestinal membrane. Peptide hormones can be broken down by gastrointestinal peptidases and must be injected (choice A is wrong). Steroid hormones act as gene expression regulators (i.e., they affect transcription) when bound to their receptor, whereas peptides modify existing enzymes (choice B is wrong).

Which of the following is the most probable cause of the FHR pattern seen in Figure 2? A. Fetal movements due to increased maternal activity B. Head compression from normal contractions C. Maternal fever D. Premature rupture of amniotic membranes (PROM) resulting in oligohydramnios

D. Figure 2 shows a FHR pattern consistent with variable decelerations, since there is an erratic FHR with abrupt, deep decelerations. As the passage indicates, this is commonly caused by umbilical cord compression which may result from oligohydramnios (decreased amniotic fluid). Thus, a premature rupture of the amniotic membranes would lead to decreased amniotic fluid; this would be an explanation for the umbilical cord compression and resulting variable decelerations (choice D is correct). Fetal movements would be expected to cause normal accelerations in the pattern (choice A is wrong). Head compression from normal contractions would cause early decelerations and an expected increase in amniotic pressure which would coincide with the contractions themselves (choice B is wrong) and maternal fever would cause a baseline tachycardia (FHR > 160 bpm, choice C is wrong).

A medical student on her internal medicine clerkship quickly eats a roll from the cafeteria but, after taking a very large final bite, feels as though she cannot move it down her entire esophagus. She is not having any difficulty breathing but is experiencing sharp pains in the center of her chest every few seconds. The pain increases for several minutes and then suddenly ends. What is the likely source of pain and why is she having difficulty swallowing the roll? A. Primary peristaltic wave, increased pyloric sphincter tone B. Primary peristaltic wave, increased cardiac sphincter tone C. Secondary peristaltic wave, increased pyloric sphincter tone D. Secondary peristaltic wave, increased cardiac sphincter tone

D. Given the medical student's initial inability to completely swallow the roll, primary peristaltic waves must have failed to move the roll effectively down the esophagus (choices A and B can be eliminated). Thus, secondary peristaltic waves are needed to drive the food down the esophagus and (ideally) into the stomach. Increased cardiac sphincter tone (regulating entrance of food from the esophagus into the stomach) would initially prevent food from entering the stomach and lead to discomfort with each secondary peristaltic wave (choice D is correct). The pyloric sphincter is at the end of the stomach and regulates entrance of chyme into the duodenum, thus increased tone at this sphincter is unlikely to cause the discomfort in this instance (choice C is wrong).

A new blood sample arrives at a lab and grows a gram- positive bacterium later identified as B. anthracis; however the ELISA registers a negative result. Is this patient infected with B. anthracis and what most likely accounted for this result? A. No; the laboratory sample was contaminated. B. No; the cultured bacteria were part of the natural flora of the body. C. Yes; the tested blood sample would register a positive ELISA in approximately seven days. D. Yes; the infection was recently acquired. Correct Answer

D. If the patient has very recently been infected with B. anthracis for the first time, the body may not yet have mounted a detectable immune response (choice D is correct). A significant humoral response in the form of elevated serum antibodies will be detected approximately one week after infection. However, the already-tested blood sample would not register a positive result after seven days, since outside of the body blood lacks the machinery to produce antibodies; a fresh blood sample would be required (choice C is wrong). A positive blood culture could result from contamination, but this is less likely than an active infection (choice A is possible but less likely than choice D). B. anthracis is not part of the body's normal flora; it is a disease-causing agent (choice B is wrong).

Mitochondrial genes are located in: A. only in the nuclear genome, specifically those in heterochromatin regions. B. only in the nuclear genome, specifically those in euchromatin regions. C. only in the mitochondrial genome, which is circular and made of double stranded DNA. Your Answer D. in both the nuclear and mitochondrial genome, both of which are made of double stranded DNA. Correct Answer

D. Mitochondrial proteins can be coded by either the nuclear or the mitochondrial genome (choice D is correct; choices A, C, and C are wrong). If coded for by the nuclear genome, the protein will contain a mitochondrial import signal. The human mitochondrial genome is small (you don't need to know these numbers, but to give you a sense of scale, it has 16,569 base pairs of double stranded DNA and codes for 37 genes). Many biochemical and cellular processes occur in the mitochondria, so this organelle must contain many other proteins (again, you don't need to know this number but mitochondria typically contain about 3000 different proteins!).

In order to better interpret the data presented in the passage, which additional experiments should have been conducted?I. A positive control, where scrambled siRNA was transferred into cells, to determine only the potential cell toxicity effects of the experimental outline.II. Quantify the amount of MBG-1 and MBG-2 mRNA in the cells, via reverse transcriptase quantitative PCR.III. Transfer siRNA molecules individually instead of in pools, to rule out synergistic effects of multiple siRNA molecules against the same target. A. II only B. I and II only Your Answer C. I and III only D. II and III only Correct Answer

D. Since Item III is found in two of the answer choices, start by analyzing Item III to turn this into a 50:50 question. Item III is true: testing pools of siRNA can be a good first step, but it is possible that a few of the molecules in the mix are cooperating to cause the results presented. The experiment is being performed to determine the effects of these two proteins on mitochondria, and a specific way to target MBG expression in cells. Next then, the siRNA molecules should be applied individually (choices A and B can be eliminated). You can analyze either of the remaining items next to get the correct answer. Item I is false: the experiments in the passage are missing both positive and negative controls. Transfecting cells with scrambled siRNA to determine potential cell toxicity effects would be a good experiment to do, but this is a negative control (expected not to work), not a positive control (expected to work, choice C can be eliminated and choice D is correct). Note that Item II is in fact true: Figure 1 shows how MBG-1 and MBG-2 protein levels change after application of the siRNA pools, but protein levels don't always correlate with mRNA levels. For example, if a certain protein has a very long half life in the cell, siRNA may still effectively be targeting mRNA but this will not result in a change in protein levels. Knockdown experiments will ideally show the effects of siRNA on both RNA and protein. Reverse transcriptase quantitative PCR (RT-qPCR) is a good technique to quantify the amount of mRNA.

Which of the following is the most likely to repair method for DNA damage caused by zinc finger nucleases? A. Nucleotide excision repair B. Direct reversal C. Mismatch repair pathway D. Non-homologous end joining

D. Zinc finger nucleases cause double stranded breaks within DNA so this is the damage that needs to be addressed by the repair mechanism. Nucleotide excision repair and the mismatch repair pathway tends to occur when only one of the two strands has a defect as the other strand is used as a template to repair the damaged strand (choices A and C can be eliminated). Direct reversal is a repair mechanism that occurs when there is no breakage of the phosphodiester backbone, such as the formation of dimers (choice B can be eliminated). Non-homologous end joining is a repair mechanism that frequently occurs to repair a double stranded break (choice D is correct).

Based on the data in Figure 1, which of the following is true? A. ILK phosphorylates T19 of MLC in kidney proximal tubule epithelial cells. B. S473 is phosphorylated solely by ILK in kidney proximal tubule epithelial cells. C. Each lane of the Western blot contained the same amount of β-actin, indicating equal amounts of each lysate were used and eliminating the need for normalization. D. The variation in the total MLC Western blot could be due to protein degradation of some lysate samples but not others.

Figure 1 shows that phospho-MLC increased in both JV32 and NA13 cells when growth factor was added. This change was similar in the two cells types in spite of the fact that NA13 cells do not express ILK. Also notice that total MLC protein levels change in a similar way for both cells types. This suggests growth factor treatment is causing an increase in total MLC in kidney proximal tubule epithelial cells and that this effect is independent of ILK expression. While the passage says that ILK phosphorylates MLC on T19, this must occur in other cell types and not the cells used here (choice A can be eliminated). ILK is causing phosphorylation of PKB in JV32 cells, but, since phospho-PKB levels also increase slightly in NA13 cells, there must be other proteins contributing to PKB phosphorylation in these cells (choice B can be eliminated). β-actin is the housekeeper protein used in this experiment to determine how much lysate is present in each lane. Since each lane has a similar amount of β-actin, it also likely contains the same amount of overall lysate (choice C is correct). The passage says that protease inhibitors were added to the lysis buffer. It is, therefore, unlikely for protein degradation to occur in some samples but not others (choice D is wrong).

The reaction catalyzed by diglyceride acyltransferase involves the formation of: A. an ester bond. B. a peptide bond. C. a disulfide bond. D. a glycosidic bond.

The passage states that diglyceride acyltransferase converts a diglyceride into a triglyceride. Diglycerides are made up of a glycerol molecule with two fatty acids attached; conversion into a triglyceride means that another fatty acid would be attached (hence the name acyltransferase). The connection of a fatty acid to a glycerol molecule involves the formation of an ester bond (choice A is correct). Peptide bonds are found between amino acids in a protein (choice B is wrong), disulfide bonds are covalent bonds between sulfur atoms in tertiary or quaternary protein structure (choice C is wrong), and glycosidic bonds join monosaccharides in di- and polysaccharide structures (choice D is wrong).

Subject A and Subject B each have siblings with Niemann-Pick disease but are themselves unaffected. The parents of both subjects are also unaffected. If Subject A and a separate individual have a child with Niemann-Pick disease, what is the chance that a child from Subject A and Subject B will also have Niemann-Pick disease? A. 0 B. 4/9 C. 1/6 D. 1/4

Niemann-Pick disease is inherited in an autosomal recessive fashion, so only homozygous recessive individuals will be affected. If two carriers (heterozygous individuals) mate, there will be a 1/4 chance of the child having the disease. Since Subject A had a child with the disease, she must be a carrier. Subject B could either be homozygous dominant or a carrier. Because we know Subject B does not have the disease (is not homozygous recessive) and both Subject B's parents are carriers (since Subject B has siblings with the disease), there is a 2/3 chance that Subject B is heterozygous. The likelihood that Subject B is a carrier and that Subject B and Subject A have a child with the disease is therefore 2/3 x 1/4 = 1/6 (choice C is correct, choices A, B, and D are wrong).

Which of the following will result from an increase in levels of acid sphingomyelinase on neuronal communication?I. Increased length of time for voltage-gated Na+ channel inactivationII. Increased leakage of Na+ into the extracellular fluid following an action potentialIII. Decrease in maximum membrane potential during depolarization A. I only B. II only C. III only D. II and III only

Since Items II and III both appear in the answer choices twice, it is best to start by analyzing one of these items first. Identification as either true or false will eliminate half the answer choices. Item II is true: acid sphingomyelinase (ASM) is involved in the breakdown of sphingomyelin. Sphingomyelin is a key component in neuronal myelin sheaths. Consequently, if ASM levels are increased, we would expect a decrease in the integrity of the myelin sheath. This will allow increased leakage of sodium ions out of the axon into the extracellular fluid and consequently decrease conduction velocity. This is because the myelin sheath serves as an insulator of sodium ions (choices A and C can be eliminated). Since Item I does not appear in either remaining choice it must be false, we can look at Item III. Item III is false: action potentials are an "all-or-none" response; a decrease in myelin will not change the maximum depolarization potential during an action potential. The magnitudes of depolarization for action potentials in neurons are equivalent as long as threshold potential is reached, regardless of the strength of the stimulus (choice D can be eliminated and choice B is correct). Note that Item I is false: decreasing the amount of myelin will not increase the length of time it takes for opened sodium channels to inactivate. Once voltage gated sodium channels reach threshold potential, the amount of time it takes for inactivation is a constant.

An experiment is performed in which four strains of a lysogenic (+) -ss RNA virus are reverse transcribed to ultimately form dsDNA. Each DNA sample is then heated to separate the strands. Which initial viral sequence results in a dsDNA molecule with the most negative change in enthalpy during separation? A. 5'-GAUAAUAAA-3' Correct Answer B. 5'-AAAGGAUUU-3' C. 5'-ACACACACA-3' D. 5'-CCAAUUUCC-3' Your Answer

The change in enthalpy necessary to separate the dsDNA represents the energy required to break the hydrogen bonds of the base pairs. The more hydrogen bonds there are, the more energy will be required to break them. However, the question asks for the DNA with the most negative change in enthalpy, so the correct answer will be the viral strand yielding the DNA strand with the fewest hydrogen bonds formed. GC base pairs are held together by three hydrogen bonds while AT base pairs are held together by two; thus, the fewer GC base pairs a strand has, the easier it will be to separate. Out of the four candidate sequences, the first will form only one GC bond and hence will have the fewest hydrogen bonds in the DNA (choice A is correct). Choice B would generate three GC bonds while choices C and D would generate four each.

Which of the following represents substrate residue structures in MLC and PKB respectively, after the ILK-mediated reaction?

The passage says that ILK phosphorylates T (threonine) 19 of MLC and S (serine) 473 of PKB. Choice A shows these two amino acids in their phosphorylated forms (choice A is correct). Choice B shows one type of phospho-histidine and phospho-threonine (eliminate choice B), Choice C shows phospho-serine and phospho-tyrosine (eliminate choice C), and choice D shows ribose 5-phosphate and a different type of phospho-histidine (eliminate choice D). Note that you did not have to know the absolutely structures of the amino acids; simply knowing that threonine and serine do not contain any ring structures allows the elimination of choices B, C, and D.

Is the hypothesis reached by the ILK research team supported by the data in Figure 1? A. Yes: in both JV32 and NA13, phospho-GSK3 levels increased with growth factor treatment, while DK5 cells had low phospho-GSK3. B. No: if ILK is the only protein phosphorylating GSK3, phospho-GSK3 levels should increase in JV32 cells when treated with growth factor but should not increase in NA13 cells. C. No: There is no way to determine the effect of ILK on GSK3 phosphorylation, since the researchers did not show if the total amount of GSK3 changed with growth factor treatment. D. No: The researchers cannot determine the effect of ILK on GSK3 without knowing how phospho- GSK3 changes in DK5 cells after growth factor treatment.

The passage states that growth factors can have a number of different effects on cells (including proliferation), mediated by a number of different proteins (including, but not limited to, various kinases). The results in the Western blot for phospho-GSK3 show an increase in phospho-GSK3 on treatment with a growth factor in both JV32 cells (that express ILK) and in NA13 cells (that do not express ILK); this suggests first that ILK is not the only protein involved in phosphorylating GSK3 (choices A and B are wrong). However the researchers do not show a Western blot that gives information on the total amount of GSK3, and if it is also changing on treatment with growth factors. Without this information, researchers cannot claim that phospho-GSK3 levels are changing only due to ILK effects; it may be, for example, that total cell counts (and thus total cell protein) is increased on treatment with growth factors, and that alone contributes to the increase in phospho-GSK3. Consider the results in Figure 1 for phospho-PKB (PKBpS473). These results match the idea that ILK phosphorylates PKB on amino acid S473; phospho-PKB increases in JV32 cells after growth factor treatment and the total amount of PKB does not change across the samples. This suggests that the increase in phospho-PKB is due to the activity of ILK and not due to increased cell numbers (and thus cell protein levels). Note also that this response is more robust than what happens in NA13 cells (which have no ILK) , further supporting the idea that ILK is not the only protein phosphorylating GSK3. DK5 cells express a dominant-negative isoform of ILK, meaning that ILK will be nonfunctional in these cells. Knowing how these cells respond to growth factors would strengthen the experiment but is not strictly necessary in order to conclude how ILK is affecting GSK3 (eliminate choice D).

Based on the passage, which of the following may be inferred? A. The regulation of brown fat cells appears to be a potential important therapeutic target for obesity. B. Activation of brown fat cells leads to what may be considered a "local hypothyroid state". C. A patient with diabetes that has frequent hyperglycemia would be likely to have a high state of brown adipose tissue metabolic activity. D. The PRDM16 gene is selectively present in brown fat cell genomes.

The regulation of brown fat cells as a potential therapeutic target for obesity is supported in the passage by various statements, including the fact that brown fat cell activity increases metabolism, dissipates energy, and also because it is inversely correlated with body-mass index (BMI). Therefore, choice A is correct. The increased levels of D2 in the brown fat cells leads to increased levels of active T3 hormone; therefore, it is considered to be a "local hyperthyroid state" (choice B is wrong). The passage directly states that the level of brown adipose tissue activity is inversely correlated with the serum glucose levels (choice C is wrong). While the PRDM16 gene is selectively expressed in brown fat cells, the gene is present in the genome of all cells (choice D is wrong). Remember that all of an individual's genes are present in their genome and that specific genes are expressed (i.e., transcribed and translated) in specific tissue types.

From the results of the experiment, what is the most reasonable conclusion that can be made from the data? A. Only the mutation represented by construct C1460T could be responsible for symptoms of Niemann-Pick disease. B. Only the mutation represented by construct C973G could be responsible for symptoms of Niemann-Pick disease. C. Both mutations represented by the two constructs could be responsible for symptoms of Niemann-Pick disease. D. Neither of the mutations represented by the two constructs could be responsible for symptoms of Niemann-Pick disease.

This is a data-interpretation type of question. From the data table, the mutation represented by cDNA construct C1460T resulted in no change in ASM activity from the non-mutated wild-type DNA; the construct produces ASM with 100% of wild-type ASM activity (choices A and C are wrong). In contrast, the mutation represented by cDNA construct C973G has a large decrease in ASM activity, having only 3% of wild-type activity. Niemann-pick disease results in a build-up of sphingomyelin due to decreased ASM activity, thus the mutation in this second construct is more likely to be responsible for symptoms of Niemann-Pick disease (choice B is correct and choice D is wrong).