Part III or IVBiological and Biochemical Foundations of Living Systems59 Questions95 Minutes

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Passage 1

Fermentation is an important mechanism for generating ATP in anaerobic conditions, when the cell cannot undergo cellular respiration. In yeast, fermentation is a key step in the beer making process. During fermentation, yeast converts sugars present in the pre-beer mixture into alcohol. This occurs in the absence of oxygen, and produces carbon dioxide as a by-product.

Glycolysis is the first step in this process, where the sugar molecules are broken down into pyruvate. Pyruvate then gets decarboxylated by the enzyme pyruvate decarboxylase. Pyruvate decarboxylase requires two co-factors during catalysis, magnesium and thiamine pyrophosphate (TPP), a vitamin B1 derivative. TPP has a thiazole ring, which is capable of forming a carbanion that nucleophilically attacks the substrate during pyruvate decarboxylase catalysis. The products of the decarboxylation reaction are acetaldehyde and carbon dioxide, which is released as a by-product of the reaction. Next, alcohol dehydrogenase converts acetaldehyde into ethanol. This step requires the co-factor nicotinamide adenine dinucleotide (NADH), which gets oxidized to NAD+ in the process.

Humans undergo fermentation in anaerobic conditions, for example, when muscles run out of oxygen during exercise. However, unlike in yeast cells, pyruvate gets catalyzed into lactic acid by the enzyme lactate dehydrogenase. This reaction also oxidizes NADH to NAD+. Lactic acid buildup is responsible for muscle fatigue, and, as muscles do not have a way to metabolize lactic acid, lactic acid is mainly removed by the bloodstream and subsequently broken down in the liver.

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Figure 1. UV absorption spectra of NAD+ and NADH

Attribution: "NADNADH" by Cronholm144, Wikimedia Commons.

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Figure 2. Reaction coordinate diagram of the lactate dehydrogenase reaction.

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1. If alcohol dehydrogenase, acetaldehyde, and NADH are mixed together in a reaction, which of the following measurements would be the best way to determine the activity of the enzyme?Please choose from one of the following options.

1 Measure the pH over time.2 Measure the amount of carbon dioxide produced over time.3 Measure the decrease in absorbance at 340nm over time.4 Measure the increase in ethanol concentration over time.

2. Figure 2 shows the reaction coordinate diagram of the lactate dehydrogenase reaction. Which of the following indicates the activation energy needed for the substrate to be converted to product in the presence of the enzyme?Please choose from one of the following options.

1 A2 B3 C4 D

3. In Figure 2, which of the following is equivalent to the energy at the transition state of the reaction without the enzyme present?Please choose from one of the following options.

1 A2 B3 C4 D

4. In humans, alcohol dehydrogenase metabolizes alcohol in the liver and lining of the stomach. Which of the following statements is NOT true?Please choose from one of the following options.

1 The quantity of alcohol dehydrogenase present in the liver and stomach affects how quickly a human can metabolize alcohol.

2 The acetaldehyde produced from this reaction is further metabolized by another enzyme.

3 The final products from this reaction are pyruvate and NAD+.4 A UV absorbance spectrum of this reaction over time would show an

increase of absorption at 340nm.

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5. Why might pyruvate fail to be converted into acetaldehyde at excessively low pH?Please choose from one of the following options.

1 Pyruvate is not present at low pH and therefore cannot be converted into acetaldehyde.

2 NADH cannot be oxidized to NAD+.3 Magnesium is not present at low pH and therefore cannot act as a

cofactor in the reaction.4 If TPP is protonated, it cannot form a carbanion to nucleophilically

attack the substrate.

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Passage 2

The exocrine component of the pancreas is an integral part of the digestive system. Though not a segment of the gastrointestinal (GI) tract itself, the pancreas is a major producer of enzymes necessary for the chemical digestion of food. These enzymes are synthesized in their zymogen forms and then sent to the small intestine where they are activated.

When pancreatic enzymes are activated before reaching the small intestine, they start to attack the pancreas itself, leading to inflammation, known as pancreatitis. Acute pancreatitis is a temporary condition, with about 80% of patients recovering completely. However, prolonged exposure to these enzymes will lead to irreversible cell death and chronic inflammation. This is called chronic pancreatitis, and it has been linked to pancreatic cancer and organ failure.

The leading cause of pancreatitis is alcohol abuse – the excessive consumption of alcohol over a prolonged period of time. Scientists are still trying to understand the development of the disease, and recently research focus has shifted to the acinar cells of the pancreas – the cells that produce the aforementioned pancreatic enzymes. The prevalent theory, illustrated in

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Figure 1, holds that byproducts of ethanol digestion weaken lysosomes (L) and zymogen granules (ZG) inside the acinar cell. When the contents of these organelles meet, pancreatic enzymes are activated. The presence of ethanol also leads to increased production of digestive enzymes.Figure 1. Hypothetical development of alcoholic pancreatitis in acinar cells. Ethanol metabolites destabilize lysosomes (L) and zymogen granules (ZG), and trigger increased transcription of pancreatic enzymes.

As enzyme content increases and the barriers separating them weaken, it is more and more likely that the digestive juices meant for the small intestine activate in the acinar cells instead. This early enzyme activation begins a chain of events that eventually activates pancreatic stellate cells (PSCs). These PSCs are responsible for producing extracellular matrix – when activated in excess they can lead to the pancreatic fibrosis and permanent pancreatic damage characteristic of chronic pancreatitis.

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A major piece of support for this theory is the effect that alcohol and its metabolites have on fibrogenesis in PSCs. In a study conducted in 2000, rat PSCs were isolated and treated with ethanol and acetaldehyde, a product of ethanol oxidation. Each compound led to a significant increase in the production of collagen by the pancreatic cells.

Figure 2: Effects of ethanol and acetaldehyde on collagen synthesis in pancreatic stellate cells. Ethanol was added at concentrations of 10mmol/L (E10) and 50mmol/L (E50). Acetaldehyde was added at concentrations of 150μmol/L (A150) and 200μmol/L (A200) to cell preparations.

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6. What is the main mechanism that normally prevents pancreatic enzymes from being activated before they reach the small intestine?Please choose from one of the following options.

1 The higher pH of the pancreas destabilizes enzyme structure, rendering them inactive

2 Chaperone proteins in the pancreas bind the zymogens and release upon secretion into the pancreatic duct

3 Pancreatic enzymes do not activate until they come in contact with their target molecules

4 Enzymes in the small intestine are responsible for activating the pancreatic enzymes

7. Which biomolecule would be the least digested if the pancreatic juices no longer entered the small intestine?Please choose from one of the following options.

1 Fats2 Proteins3 Nucleic Acids4 Carbohydrates

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8. If scientists added 4MP, an alcohol dehydrogenase inhibitor, to the conditions found in Figure 2 , what are the expected results? (All results are shown in the same units as Figure 2)Please choose from one of the following options.

1

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9. In a follow-up experiment to the one seen in Figure 2, scientists added 4MP, an alcohol dehydrogenase inhibitor, to the ethanol preparation. What does this test?Please choose from one of the following options.

1 How alcohol metabolites affect collagen production2 How quickly the presence of ethanol activates PSCs3 The role of alcohol digestion in PSC activation4 Whether ADH is the enzyme at work in the PSC preparations

10. Which of the following treatment approaches would be LEAST effective in managing a patient with pancreatitis?Please choose from one of the following options.

1 Have the patient take lipase and amylase capsules2 Have the patient use proton pump inhibitors which reduce gastric acid

production3 Have the patient avoid antibiotics to promote normal flora in the

bowels4 Have the patient fast or avoid fatty meals

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Passage 3

Glioblastoma multiforme (GBM), the most common brain malignancy in adults, arises from genetic alterations in the glial cells of the central nervous system and carries a prognosis of certain death based on currently available treatments (surgical resection, chemotherapy, radiation). Recent advances in high-throughput genetic sequencing and bioinformatics have lead to the identification of a mutation in the gene for the enzyme NADP+-specific isocitrate dehydrogenase (referred to here as IDH) that is associated with increased survival for patients harboring the mutation (median survival of 15 months vs.31 months, Figure 1). Genetic analysis of tissue samples indicate that IDH mutation in GBM tumors is relatively rare, suggesting a prevalence of roughly 10%. However, the same studies have found that the IDH mutation is found in roughly 80% of GBM tumors that began as lower grade gliomas and progressed to GBM status over time (so-called secondary GBM).

Figure 1. Kaplan-Meier survival curve depicting probability of survival for adult GBM patients harboring the wild type (IDH wild-type, blue, 15 months) or mutant gene (IDH mutated, red, 31months) for isocitrate dehydrogenase (IDH).

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In order to determine the biological consequences of IDH mutations in GBM, researchers transfected a human oligodendroglioma cell line with a vector containing either the wild type (wt) IDH gene or the mutant IDH gene. The resulting cells were cultured for 48 hours in growth media, lysed, and centrifuged (the same process was also applied to cells transfected with a control vector that did not contain the IDH gene or its mutant variant). The supernatant from each of the three cell lines were separately combined with NADP+, buffer, and isocitrate. Resulting production of NADPH was assayed via spectrophotometry (Figure 2).

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Figure 2. Enzymatic activity of mutant and wild type IDH proteins as measured by spectrophotometry.

In addition to providing useful prognostic information to clinicians and patients, the discovery of IDH mutations in GBM tumors provides a potential biological explanation for the existence of secondary GBM.

Passage and figures adapted from: Yan, H., Parsons, D. W., Jin, G., McLendon, R., Rasheed, B. A., Yuan, W., . . . Bigner, D. D. (2009). IDH1 and IDH2 mutations in gliomas. N Engl J Med, 360(8), 765-773.

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11. Based on information presented in the passage, what is the best explanation for why the researchers chose to measure NADPH concentration in cells transfected with just the vector?Please choose from one of the following options.

1 Measurement of NADPH concentration in cells transfected with just the vector quantifies basal NADPH production in wild type eukaryotic cells, which lack the endogenous IDH gene.

2 Measurement of NADPH concentration in cells transfected with just the vector quantifies basal NADPH production in wild type eukaryotic cells, which have the endogenous IDH gene.

3 Measurement of NADPH concentration in cells transfected with just the vector quantifies basal NADPH production in wild type prokaryotic cells, which have the endogenous IDH gene.

4 Measurement of NADPH concentration in cells transfected with just the vector quantifies basal NADPH production in wild type prokaryotic cells, which lack the endogenous IDH gene.

12. Like all types of cancer, GBM is best characterized by which of the following cellular pathologies?Please choose from one of the following options.

1 Changes in basal cellular protein levels leading to loss of cell cycle control

2 Gain of function mutations in tumor suppressor genes3 Gain of function mutations in genes controlling apoptosis4 Loss of function mutations in oncogenes

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13. Suppose that researchers chose to assay IDH enzyme activity using the exact same methods as stated in the passage, except that a cell line derived from a GBM patient harboring the IDH mutation was used in place of the cell line used in the original experiment. What changes would one most likely find with respect to the results depicted in Figure 2?Please choose from one of the following options.

1 Reduced NADPH concentration in the “vector + wt IDH” group, the “vector” group, and the “vector + mutant IDH” group.

2 No change in NADPH concentration in the “vector + wt IDH” group; no measurable NADPH concentration in the “vector” and “vector + mutant IDH” groups.

3 Reduced NADPH concentration in the “vector + wt IDH” group; no measurable NADPH concentration in the “vector” and “vector + mutant IDH” groups.

4 No change in the NADPH concentration in the “vector + wt IDH” group, the “vector” group, and the “vector + mutant IDH” group.

14. Assuming that the IDH mutation discussed in the passage confers a loss of function on the protein associated with the gene, which of the following would be the most likely immediate effect in cancer cells harboring the IDH mutation?Please choose from one of the following options.

1 Decreased concentration of electron-donating molecules.2 Decreased concentration of electron-accepting molecules.3 Decreased concentration of proton-donating molecules.4 Decreased concentration of proton-accepting molecules.

15. Based on information presented in the passage, a genetic mutation in which of the following cell types of the central nervous system would be least likely to lead to GBM?Please choose from one of the following options.

1 Neurons2 Oligodendrocytes3 Ependymal cells4 Astrocytes

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Questions 16. 17. 18. 19. 20 are NOT based on passages.16. Which of the following is the correct definition of a cation?Please choose from one of the following options.

1 A cation is a negatively charged ion.2 A cation is an ion that can have a positive or negative charge.3 A cation is an ion that cannot have a charge.4 A cation is a positively charged ion.

17. Which term describes the space between a neuron and its target cell?Please choose from one of the following options.

1 Axon terminal2 Post synaptic membrane3 Synaptic cleft4 Dendritic spine

18. Which of the following best describes the role of the membrane receptor?Please choose from one of the following options.

1 To aid communication between a cell and its environment2 To respond to the intracellular environment via second messengers3 To regulate cellular metabolism at the nuclear level4 To directly influence gene transcription via primary messengers

19. After consuming a banana split, which hormones would be expected to increase?Please choose from one of the following options.

1 Glucagon2 Parathyroid Hormone3 Insulin4 Prolactin

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20. Which heart valves are NOT actively closed by the contraction of muscular structures?Please choose from one of the following options.

1 Semilunar valves 2 Mitral valves3 Tricuspid valves4 Atrioventricular valves

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Passage 4

Traditional cancer therapy targets rapidly dividing cells to prevent malignant mitosis. Some of these drugs used in traditional cancer therapy, such as cisplatin, show limited specificity for cancerous cells, and instead affect diverse cell populations in the body, such as hematopoietic stem cells (HSC’s) in the bone marrow, skin cells, and gastrointestinal lining. Cisplatin is used for its ability to crosslink DNA by forming chemical bonds between nucleotides on complementary strands. Crosslinks in the DNA disrupt DNA replication and cell division. This gives cisplatin its limited specificity for cancer cells, since cancer cells tend to be replicating DNA and dividing more rapidly than most other cells in the body.

Recently, cancer therapy has targeted a mutant kinase protein known as Bcr-Abl. This fusion protein is a product of a mutation known as the Philadelphia chromosome, which occurs when chromosomes 9 and 22 undergo a reciprocal translocation. Unlike the “normal” product, Abl, Bcr-Abl is constitutively active. Bcr-Abl is involved in signal transduction pathways that lead to the activation of various downstream effector proteins within the cell. It influences both cell division and cell migration. Since cancer cells rapidly divide and undergo increased migration as they metastasize, Bcr-Abl is an excellent target for anti-cancer drugs.

A new chemotherapeutic agent, called imatinib, is a small molecule that functions as a highly specific inhibitor of Bcr-Abl. Imatinib is part of a larger family of tyrosine kinase inhibitors (TKI’s), and it competes with ATP for a binding site on Bcr-Abl. Imatinib is a very effective treatment option for patients with chronic myelogenous leukemia (CML), who most often harbor a Philadelphia chromosome. CML involves the proliferation of myeloid progenitor cells (which eventually give rise to a subset of white blood cells) deriving from the bone marrow.

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Figure 1: The Philadelphia chromosome is caused by reciprocal translocation of a segment from chromosome 9 and a segment from chromosome 22. This situates the gene Abl, indicated by red circles, adjacent to the gene Bcr, indicated by yellow circles, resulting in the fusion protein Bcr-Abl.

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Figure 2: Activity of Bcr-Abl was measured based on its hydrolysis of ATP. The rate of reaction is shown as a function of ATP concentration in the presence and absence of imatinib.

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21. Which of the following scenarios is most likely to occur in the cells of a patient with the Philadelphia chromosome?Please choose from one of the following options.

1 The presence of an extra copy of chromosome 22, leading to more transcription of the Bcr-Abl gene and consequently higher protein activity.

2 A decrease in phosphorylated substrates downstream of Bcr-Abl, causing withdrawal from the cell cycle and entrance into the G0 phase.

3 An increase in the ratio of phosphorylated to dephosphorylated substrates downstream of Bcr-Abl.

4 Accumulation of activated protein substrates downstream of Bcr-Abl due to decreased phosphatase activity.

22. If imatinib competes with ATP for a binding site on Bcr-Abl, what effect would increasing ATP concentration have on the kinase activity of Bcr-Abl?Please choose from one of the following options.

1 In accordance with Le Chatlier’s Principle, increasing ATP concentration will lead to an increase in ADP concentration, resulting in the phosphorylation of fewer substrates by Bcr-Abl.

2 An increase in ATP will lead to phosphorylation of imatinib by the kinase Bcr-Abl, rendering it ineffective.

3 Because imatinib is a competitive inhibitor, increasing the concentration of ATP has no effect on Bcr-Abl activity.

4 By increasing the concentration of ATP, the relative concentration of bound imatinib can be effectively reduced to negligible levels.

23. Which of the following effects would most likely appear in a CML patient undergoing cisplatin treatment AND in a CML patient undergoing imatinib treatment?Please choose from one of the following options.

1 Reversal of the Philadelphia chromosome mutation2 Covalent bonds between strands of DNA3 An increased risk of bacterial infection4 Anemia due to lower levels of circulating red blood cells

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24. Suppose there is a related kinase to Abl called Abc, coded by the gene Abc. Which of the following genetic occurrences would be the LEAST likely to result in a constitutively active Abc kinase?Please choose from one of the following options.

1 A reciprocal translocation that separates two halves of an intron2 A deletion of the portion of the gene that codes for Abc’s regulatory

domain.3 A point mutation in the promoter region for the Abc gene4 A two nucleotide insertion at the beginning of the coding region.

25. How might the metabolism of Philadelphia chromosome positive cells (Ph+) differ from that of healthy cells in the same patient?Please choose from one of the following options.

1 Ph+ cells will have an increased metabolism to generate energy for processes like DNA replication and cell division.

2 Because Ph+ cells differ from healthy cells only by the presence of a single mutated kinase, there will be no differences in metabolism between these cell types.

3 Ph+ cells will conduct only glycolysis and anaerobic fermentation in order to generate the greatest number possible of ATP per molecule of glucose.

4 The Ph+ cells will have a lower metabolism because more of their proteins will be occupied for processes like DNA replication and cell division.

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Passage 5

Vesicles containing neurotransmitters are synthesized in the cell body, or “soma” of the neuron and carried down microtubule tracks by motor proteins. For exocytosis to occur, these vesicles must fuse with the membrane of the axon terminal, releasing their contents into the cleft.

NTKA is a newly discovered excitatory neurotransmitter. After its release into the synaptic cleft, NTKA is cleaved into two components, NT and KA, by an enzyme known as NTKAse. The KA component diffuses away into the extracellular space, while the NT component is brought back into the presynaptic neuron via specific membrane-bound transport proteins. Inside the axon terminal, NT is covalently bonded to another KA group, forming a new NTKA molecule that can be reused as a neurotransmitter.

A new disease, with an autosomal recessive mode of inheritance, is characterized by NTKA deficiency. Autosomal recessive NTKA deficiency is caused by a hyperactive NTKAse enzyme, “∆ NTKAse,” which binds NTKA with a much higher affinity than wild-type NTKAse does, quickening the depletion of NTKA from the synaptic cleft. Interestingly,∆ NTKAse and NTKAse do not appear to have different rates of catalysis at saturating substrate, “kcat”.

Three patients volunteer for a study involving NTKA deficiency. Via in vitro genetic manipulation, researchers are able to produce a large quantity of NTKAse protein from each of these patients and characterize them biochemically. Figure 1 shows a Lineweaver-Burke plot of NTKAse activity for three patients, each of whom has one or more relatives with NTKA deficiency. Patients 1 and 2 are of unknown genotype, while patient 3is homozygous recessive. The NTKAse activity of Patient 3 was measured in the presence of a drug known to limit the activity of ∆ NTKAse.

A Lineweaver-Burke plot can be used to characterize an enzyme’s activity in different concentrations of substrate. The reciprocal of the enzyme’s rate of reaction, 1/V , is plotted against the reciprocal of the substrate concentration,1/S. The enzyme’s maximum rate of reaction, Vmax, and the

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enzyme’s binding affinity for substrate, KM, can be extracted from the plot’s y- and x-intercepts, respectively. A low KM indicates high binding affinity, whereas a high KM indicates low binding affinity.

Figure 1: A Lineweaver-Burke plot was used to characterize theVmax and KM values of NTKAse in three patients. The vmax and Km values of each patient can be extracted by taking the inverses of the y- and x-intercepts respectively. The third patient’s assay includes an inhibitor of ∆NTKAse.

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26. How would the signaling activity of NTKA change at the post-synaptic neuron if a molecule that specifically bound the NTKAse active site were added to the synaptic cleft?Please choose from one of the following options.

1 NTKA would compete with this molecule for binding sites on the post-synaptic receptor, and fail to excite the post-synaptic neuron.

2 This molecule would block NT reuptake channels, leading to an increase in the synaptic concentration of NTKA.

3 NTKA would remain in the synapse for a longer period of time, leading to increased binding of post-synaptic receptors and more excitation.

4 This molecule would act as a competitive inhibitor of NTKAse, leading to more rapid degradation of NTKA in the synapse.

27. Based on the plot in Figure 1, which type of inhibition does the drug added to patient 3’s assay most likely correspond to?Please choose from one of the following options.

1 Non-competitive inhibition because the measured Km value of NTKAse increased while the measured Vmax value remained the same

2 Competitive inhibition because the measured Km value of NTKAse increased while the measured Vmax value remained the same

3 Non-competitive inhibition because the measured Vmax value of NTKAse decreased while the measured Km value remained the same

4 Competitive inhibition because the measured Vmax value of NTKAse decreased while the measured km value remained the same

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28. Given the enzymatic mechanism of NTKA deficiency, what are the most likely genotypes of patients 1 and 2, respectively?Please choose from one of the following options.

1 Heterozygous and homozygous dominant2 Homozygous dominant and heterozygous3 Homozygous recessive and homozygous dominant4 Heterozygous and homozygous recessive

29. NTKA functions in cells derived from which developmental structure?Please choose from one of the following options.

1 Archenteron2 Ectoderm3 Mesoderm4 Blastopore

30. What is the likelihood that a pair of female monozygotic twins will be affected by NTKA deficiency if their mother and father are both heterozygotes for the disease-causing allele?Please choose from one of the following options.

1 0%2 6.25%3 25%4 50%

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Passage 6

The heavy chain and light chain subunits of antibodies both contain three regions called complementarity determining regions (CDR1, CDR2, and CDR3). CDRs are responsible for the high specificity of the fragment antigen binding (Fab) of antibodies. CDRs are hypervariable among antibodies, allowing for an extensive range of antigens with which different antibodies can interact.

During B-Cell Maturation, the gene for CDR3 of the heavy chain is formed when a random VH segment, a randomDH segment, and a random JH segment are joined together (Figure1a). A similar process occurs with only VL and JL segments to form CDR3 of the light chain. To further randomize CDR3, DNA breaks are introduced at the junction sites, leading to nucleotide addition or subtraction before the breaks are sealed. Three different mechanisms cause random nucleotide alteration at the junction sites (Figure1b).

P-addition (palindromic-sequence addition) involves the creation of a DNA hairpin at the ends of gene segments, which causes unpairing of nucleotides to form a single strand. Repair enzymes then add complementary nucleotides, generating a new palindromic sequence. N-addition (“N” number of random nucleotides) entails the insertion of a random number of nucleotides (up to 15) in between two gene fragments. Lastly, in a phenomenon known as junctional flexibility, a random number of nucleotides may be lost on the adjoining end of either gene fragment. These random nucleotide alterations come with the risk of non-productive rearrangements, which occur when the altered nucleotide sequence disrupts the reading frame of the heavy chain or light chain gene, resulting in a non-functional antibody. Non-productive rearrangements cause the B-cell to initiate cell death.

Another mechanism for Fab variability is somatic hypermutation. Somatic hypermutation occurs after a B-cell encounters its specific antigen and involves a greatly increased chance of mutation in all CDRs. In a process known as affinity maturation, activated B-cells that develop a greater affinity

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for their antigen through somatic hypermutation are selected for based on their affinity for antigen presented on the surfaces of follicular dendritic cells. Those that have a weaker affinity for their antigen are signaled to apoptose.

Figure1: The VH, DH, and JH gene segments in the coding region for the Fab of the heavy chain are randomly selected and joined to create CDR3. P-addition and N-addition, followed by junctional flexibility further increase the hypervariability of the region.

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31. If there were 44 VH gene segments, 27DH gene segments, and 6JH gene segments, how many potential SEQUENCES (x) can the antibody heavy chain have?Please choose from one of the following options.

1 x = 44 x 27 x 62 x = 13 x < 44 x 27 x 64 x > 44 x 27 x 6

32. Which B-cell process(es) would one expect to occur in a germinal center of a lymph node?I. affinity maturationII.B-cell maturationIII. somatic hypermutationIV. productive rearrangement of VH, DH, JH gene fragmentsPlease choose from one of the following options.

1 I only2 I, II, IV only3 I and III only4 II and IV only

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33. Suppose the last six nucleotides of a DH, gene segment are5’-G-T-C-G-A-T-3’. A P-addition was made, forming a palindromic sequence on the 3’ end. If the original double-stranded DNA was cleaved three nucleotides from the 5’ end, what sequence of complementary nucleotides would repair enzymes add to the resulting open strand?Please choose from one of the following options.

1 5’-C-T-A-A-T-C-3’

2 5’-C-T-A-T-A-G-3’

3 5’-G-A-T-T-A-G-3’

4 5’-G-A-T-A-T-C-3’

34. Antibodies against a bacterial strain were isolated from a patient one week after infection and again four weeks after infection. How will the dissociation constant, Kd , of the antibodies isolated after four weeks compare to the Kd of the antibodies isolated after one week?Please choose from one of the following options.

1 It will be much larger.2 It will be smaller.3 It will be slightly larger.4 It will remain the same.

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35. A high diversity of antibodies is an essential part of a healthy adaptive immune system. Which of the following contribute to antibody diversity?I. somatic hypermutationII.gene fragment rearrangementIII. junctional flexibilityIV. affinity maturationPlease choose from one of the following options.

1 I, II, III and IV only2 II and III only3 I and IV only4 I, II and III only

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Questions 36. 37. 38. 39. 40. are NOT based on passages.

36. A couple is told by their doctor that the reason why they cannot have children is that the sperm of the male lack motility because it does not have the structures responsible for propulsion. Which cellular structures are likely to be the underlying cause of this lack of motility?Please choose from one of the following options.

1 Microtubules2 Vacuoles3 Endoplasmatic reticulum4 Golgi apparatus

37. Methicillin-resistant Staphylococcus aureus (MRSA) is an antibiotic-resistant “superbug” that can cause deadly infections in humans. What would these Gram-positive bacteria look like under a microscope?Please choose from one of the following options.

1 Clear rods2 Pink rods3 Purple spheres4 Purple spirals

38. Which factor may help determine the antigenicity of a virus?Please choose from one of the following options.

1 The internal proteins2 The whole capsid3 The size of the virus4 Some of the capsomeres

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39. What would be a direct consequence of a single point mutation that modifies the proteins inside the acrosome?Please choose from one of the following options.

1 Inability to dissolve zona pellucida's glycocalyx 2 Increased polyspermy3 Failure to complete spermatogenesis4 Impairment of sperm motility

40. Norepinephrine is a neurotransmitter of the sympathetic nervous system. What type of neuron releases it?Please choose from one of the following options.

1 Interneuron innervating the stomach2 Autonomic neuron innervating the heart3 Motor neuron innervating the biceps4 Afferent neuron innervating the pupils

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Passage 7

When studying DNA, researchers often amplify regions of interest using the polymerase chain reaction (PCR). PCR replicates specific fragments between two primers, which are complementary to short segments of DNA.

Researchers can study gene transcription using reverse transcriptase PCR (RT-PCR), which involves exploiting the ability of reverse transcriptase to synthesize complementary DNA (cDNA) from mRNA transcripts, and then using PCR to amplify regions of interest (Figure 1). An important step in RT-PCR is the purification of mRNA from the cell extracts. Without this purification step, genomic DNA can contaminate the PCR reaction and become amplified along with mRNA. Like other DNA polymerases, reverse transcriptase requires a primer. Researchers must take care not to contaminate their RNA samples with widely prevalent RNAse enzymes, which degrade RNA and can ruin experiments. Inhibitors of RNAse can be used to prevent such degradation in the case of RNAse contamination.

A graduate student interested in skin cancer and melanin synthesis used RT-PCR to measure how UV light exposure stimulated production of melanin. Specifically, she wanted to determine the effect of UV light on the transcription of the gene for pro-opiomelanocortin (POMC) in keratinocytes. As a post-translational modification, POMC is cleaved to yield multiple active peptides, including melanocyte-stimulating hormone (α-MSH), which can then trigger synthesis of melanin in nearby melanocytes.

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Figure 1: Schematic of RT-PCR experiment. The position of the reverse transcriptase primer (Primer 1) and the positions of the PCR primers (Primers A and B) are shown.

The student exposed keratinocytes to ultraviolet light of a constant intensity for various lengths of time. After this UV exposure, she harvested cells and conducted RT-PCR, using primers specific for a fragment of POMC spanning an intron (Figure 1). She then loaded her cDNA samples into an agarose gel and applied voltage to separate the fragments by size. The brightness of the bands (Figure 2) represents the quantity of amplified cDNA present in the gel for that sample, and therefore the quantity of POMC mRNA present in the original sample.

To determine size of her fragments, the student added a DNA ladder to the leftmost lane, which included 8 DNA fragments at 100 base pair intervals

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ranging from 100 to 800 base pairs. She also included positive and negative controls.

Figure 2: Gel electrophoresis. Samples added at “top” of gel in the photo below. Lanes 1-5 contain RT-PCR products from keratinocytes exposed to different durations of UV exposure. Lane 6 = positive control; Lane 7 = negative control; Ladder = 8 DNA fragments ranging from 100 to 800 base pairs in length.

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41. Assuming all are approximately 8 nucleotides in length, which of the following would be effective as primers during the reverse transcription step of the experiment described above?I. A multitude of random, scrambled primersII.Primers specific to the first intron of POMCIII. A string of thymine nucleotidesIV. the same set of primers as used in the PCR amplification stepPlease choose from one of the following options.

1 I, II, and III2 I, III and IV3 II and IV4 IV only

42. What would be the most effective negative control for the experiment described above?Please choose from one of the following options.

1 Extract from keratinocytes that had been killed with UV light2 Extract from melanocytes post-UV exposure3 Extract from keratinocytes that had NOT been treated with UV light4 A known sample of POMC mRNA

43. One interpretation for the negative result in lane 1 is that a 15-minute exposure time to UV light is insufficient for induction of POMC transcription. Which of the following is a necessary assumption of this interpretation?Please choose from one of the following options.

1 The RNAse inhibitors used in the experiment protected the mRNA from degradation.

2 Light of a sufficiently short wavelength was used to induce POMC transcription.

3 POMC transcription rate is inversely related to the wavelength of the applied light.

4 Treatment with UV light did not induce apoptosis in any of the keratinocytes.

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44. Why might lane 3 in Figure 2 have two bands, while other lanes have one or none at all?Please choose from one of the following options.

1 The sample in lane 3 was contaminated with RNAses, which degraded the mRNA product to give two fragments.

2 The primers hybridized to each other and were amplified, generating many “primer dimers,” along with the desired fragment.

3 The sample in lane 3 was contaminated with genomic DNA.4 Alternative splicing gave an mRNA that lacked intron 2, but contained

intron 1.

45. Suppose that α-MSH is found to activate POMC transcription. With this information in mind, which is the most accurate graph of the data from Figure 2?Please choose from one of the following 3 options.

1

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2

3

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Passage 8

Sandy and Danny have come to a clinic because they are considering having another child but are worried by the history of Marfan syndrome in Sandy’s family. Marfan syndrome is a genetic disorder of the connective tissue – mutations on the FBN1 gene cause the protein it codes for to misfold. This protein is fibrillin-1, a large glycoprotein essential in creating the extracellular matrix (ECM). Fibrillin-1 helps provide structural support in elastin fibers, which are involved in helping tissues regain their shape after stretching. It also sequesters transforming growth factor beta (TGF-β), a growth factor responsible for cell proliferation and differentiation. Low fibrillin-1 levels can lead to elevated TGF-β, which actually leads to the degradation of the ECM.

In particular, Marfan syndrome affects the ocular, skeletal, and cardiovascular systems. Elastin fibers are key components of the ligaments responsible for suspending the lens of the eye, as well as aortic root. Patients with Marfan syndrome can have serious heart valve and aortic defects. They are often very tall and thin, and can suffer from arachnodactyly (long, thin fingers). Many patients have some form of myopia, and have an increased risk of retinal detachment.After questioning Sandy further, the geneticist learns that both of her brothers are very tall and each has had lens dislocations. Her niece has mitral valve prolapse, a heart murmur due to a thicker heart valve. Sandy’s mother was tall and had long fingers, though she had no cardiovascular or ocular issues. Sandy’s maternal grandfather passed away due to an aortic aneurysm – an enlargement of the aorta to more than 150% its normal size. However, Sandy’s sister has never shown any sign of the disease and has a perfectly healthy son.

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Figure 1. Pedigree of Sandy’s family.

46. In Figure 1, who is Sandy?Please choose from one of the following options.

1 42 8

3 7

4 6

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47. Which of the following could explain the variable expressivity shown in Sandy’s family?Please choose from one of the following options.

1 The FBN1 allele exhibits incomplete penetrance2 Each individual has different genetic and environmental factors that

contribute to the disease's phenotype3 The FBN1 allele exhibits incomplete dominance4 Each individual inherits different mutations of FBN1 gene

48. What type of disease is Marfan syndrome?Please choose from one of the following options.

1 Autosomal recessive2 X-linked recessive3 Autosomal dominant4 Mitochondrial

49. It turns out that Marfan syndrome has a penetrance of 80% and Sandy’s sister actually has the FBN1 mutation. What are the chances of the sister’s next child having Marfan syndrome?Please choose from one of the following options.

1 80%2 50%3 40%4 0%

50. Which of the following symptoms would you not associate with Marfan syndrome?Please choose from one of the following options.

1 Tight joints2 Mitral valve leakage3 Finger hyperextension4 Hip dislocation

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Passage 9

Glucocorticoids (GCs) refer to receptor-specific steroid hormones acting on glucose metabolism, binding to the GR (glucocorticoid receptor). The process of GC signaling involves a molecule composed of glucose, a cortex, and a steroid. GCs are produced in various responses mediated by neuroendocrine and endocrine signals along the HPA (hypothalamic-pituitary-adrenal axis).

Figure 1. The HPA (hypothalamic-pituitary-adrenal axis)

Under normal conditions, the hypothalamus sends corticotropin releasing hormone (CRH) to the anterior pituitary gland. The CRH then stimulates the anterior pituitary to release adrenocorticotropic hormone (ACTH). The

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ACTH then stimulates the adrenal cortex, located above the kidneys, to secrete cortisol. Cortisol in the blood will normally regulate the hypothalamus and anterior pituitary production of CRH and ACTH respectively through negative feedback inhibition.

Cushing’s syndrome results from excessive levels of cortisol circulating in the body for extended periods of time. This can be caused by taking medicinal glucocorticoids, such as prednisone, for the treatment of asthma, lupus and other autoimmune diseases. Cushing’s syndrome is similar to Cushing’s disease, however, in Cushing’s Disease, hypercortisolism is resulting from the overproduction of ACTH from the pituitary. This could be caused by either a pituitary adenoma or a non-pituitary tumor.

In humans, the symptoms of Cushing’s syndrome can vary between individuals. The classic signs of Cushing’s syndrome in humans include: rounded face, upper body obesity, fat deposits on neck, and thin arms and legs.

To study the effects of the hypothalamo-pituitary-adrenal axis, groups of mice were injected with different hormones. Group A mice were injected with cortisol to mimic effects of Cushing’s syndrome. Group B mice were injected with hormone X. Group C mice were injected with a saline solution. Blood samples were later taken from the various groups and average hormone levels were measured and recorded in Table 1.

Table 1. Levels of hormones (in nmol/L) found in blood sample taken from experimental mice groups.

CRH ACTH Cortisol

Group A 20 150 900

Group B 45 430 760

Group C 30 230 400Data adapted from: Cushing's Disease. UCLA Pituitary Tumor Program, Pituitary Tumor Surgery

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51. According to the results of the experiment, which is the most likely identity of hormone X?Please choose from one of the following options.

1 CRH, because Group B’s concentration of ACTH and cortisol is higher than that of the control group.

2 ACTH, because Group C’s concentration of ACTH and cortisol is lower than that of the control group.

3 ACTH, because Group B’s concentration of ACTH and cortisol is higher than that of the control group.

4 CRH, because Group C’s concentration of ACTH and cortisol is lower than that of the control group.

52. Why does a pituitary adenoma cause a patient to have an excess level of cortisol?Please choose from one of the following options.

1 It increased the size of the hypothalamus.2 Its cells did not respond to CRH.3 Its cells did not respond normally to cortisol.4 It decreased the level of ACTH circulating in the body.

53. Which of the following can result in a chronic increase in a patient’s ACTH and CRH levels?Please choose from one of the following options.

1 Pituitary tumor.2 Destruction of the adrenal glands.3 Taking medicinal glucocorticoids, such as prednisone.4 Hypersecretion of cortisol from the hypothalamus.

54. Which of the following would exacerbate the symptoms of Cushing’s disease?Please choose from one of the following options.

1 Taking glucocorticoids to treat asthma.2 Taking a glucocorticoid receptor antagonist.3 Somatic cells not responding to cortisol.4 Radiation therapy to treat a pituitary adenoma.

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55. Which of the following conditions would lead a patient to have upper body obesity, fat deposits on his neck, and thin arms and legs?Please choose from one of the following options.

1 The patient has an autoimmune disease damaging the adrenal cortex, leading to decreased cortisol levels.

2 The patient is suffering from hyposecretion of ACTH, leading to decreased stimulation of the adrenal cortex and increased cortisol levels.

3 The patient has an infection damaging the anterior pituitary, leading to decreased stimulation of the adrenal cortex and decreased cortisol levels.

4 The patient had chronic exposure to medicinal glucocorticoid, leading to increased cortisol levels.

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Questions 56. 57. 58. 59 are NOT based on passages.

56. Which statement best describes how cholesterol affects cell membrane fluidity?Please choose from one of the following options.

1 Cholesterol decreases fluidity at high temperatures and decreases fluidity at low temperatures.

2 Cholesterol decreases fluidity at high temperatures and increases fluidity at low temperatures.

3 Cholesterol increases fluidity at high temperatures and increases fluidity at low temperatures.

4 Cholesterol increases fluidity at high temperatures and decreases fluidity at low temperatures

57. Extracellular fluids are LEAST likely to move through the space between cells that are joined by which type of intercellular junction?Please choose from one of the following options.

1 Tight Junction2 Desmosome3 Macula adherens4 Gap junction

58. The nephron reabsorbs glucose through a sodium / glucose transporter. What sort of transporter is it?Please choose from one of the following options.

1 Symporter2 Antiporter3 Sodium pump4 Protein channel

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59. Which of the following is NOT a premise of cell theory?I. All cells arise from other cells.II. All living cells require water for survival.III. All living things are only composed of cells.Please choose from one of the following options.

1 I only2 II only3 I and II4 II and III

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