Chapter 4: Cell Biology, Signal Transduction, and the Molecular Biology of Cancer

*The answer is F.* Niemann-Pick disease Type A is an autosomal recessive disorder most common in individuals of Ashkenazi Jewish descent. Affected infants present in the first year of life with hepatosplenomegaly and progressive hypotonia and mental retardation following a period of normal early development. The cause is deficiency of the sphingomyelinase enzyme, which in normal individuals is responsible for cleaving sphingomyelin into phosphorylcholine and ceramide. In infants with Niemann-Pick disease, sphingomyelin accumulates within phagocytes, producing characteristic "foamy histiocytes." These foamy appearing. Sphingomyelin-laden histiocytes accumulate in the liver and spleen causing massive hepatosplenomegaly. Progressive sphingomyelin accumulation in the central nervous system is responsible for the neurologic degeneration that occurs. Sphingomyelin deposition in the retina causes blindness as well. A cherry-red macular spot, similar to that seen in Tay-Sachs disease, is also often found. Death usually occurs before age three.

A 2-year old Caucasian male is being evaluated for progressive neurological deterioration. Laboratory evaluation, including leukocyte enzyme activity analysis suggest Niemann-Pick disease. Teh patient most likely has a deficiency of which of the following enzymes? (A) Arylsulfatase A (B) β-hexosaminidase A (C) α-galactosidase (D) β-glucosidase (E) Neuroaminidase (F) Sphinomyelinase (G) Ceramidase

*The answer is E.* This patient has Fabry disease (angiokeratoma corporis diffusum), which results from an inherited deficiency of alpha-galactosldase. In patients with Fabry disease, the globoside ceramic trihexoside accumulates in tissues. The earliest disease manifestations are hypohidrosis, acroparesthesia, and angiokeratomas. Acroparesthesia is episodic, often debilitating, burning neuropathic pain in the extremities. Angiokeratomas are punctuate, dark red, non-blanching macules and papules that classically occur between the umbilicus and the knees. Without enzyme replacement therapy, progressive renal insufficiency leading to renal failure and death may occur.

A 22-year-old Caucasian male is being evaluated for skin lesions on his lower abdomen. A sample of his fibroblasts is obtained and cultured as part of the investigative work-up. The cultured fibroblasts fail to metabolize ceramide trihexoside. This patient is at greatest Risk for developing which of the following? (A) Pulmonary emphysema (B) Hepatic cirrhosis (C) Spastic tetraparesis (E) Renal failure

*The answer is A.* This patient has follicular lymphoma, a non-Hodgkin lymphoma of the cleaved and noncleaved B-lymphocytes of the follicular center. The classic cytogenetic abnormality in follicular lymphoma is the t(14;18) translocation, which moves the Bc1-2 (B-cell lymphoma-2) protooncogene from chromosome 18 to chromosome 14, near the site of the immunoglobulin heavy chain enhancer element. Bcl-2 is considered a protooncogene because it has anti-apoptotic effects (prevents the release of pro-apoptotic factors within affected cells). When this proto-oncogene is positioned near the immunoglobulin enhancer element, the resultant Bcl-2 overexpression allows for cell immortality .

A 34-year-old male notes a lump in his neck that has grown slowly over the past several months. His social history is significant for smoking one pack of cigarettes per day and occasional alcohol use. On review of systems, he denies dysphagia, chest pain, weight loss or fever. Biopsy of the mass reveals abnormal cells with the t(14,18) chromosomal translocation. This chromosomal change in most likely to cause which of the following abnormalities in gene expression? A. BcI2 overexpression B. Bcr-abl hybrid formation C. C-myc overexpression D. Erb-B2 overexpression E. p53 inactivation

*The answer is B.* The patient is presenting with signs and symptoms suggestive of von Hippel Lindau (VHL) syndrome. The disease is autosomal dominant and is the result of deletion of the VHL gene on chromosome 3. VHL syndrome is characterized by excessive angiogenesis manifesting as cavernous hemangiomas in the skin (the patient's non-blanching, spongy skin lesion) and hemangioblastomas in the central nervous system, commonly in the cerebellum (causing ataxia), retina (causing vision problems), brainstem, and spine. Angiogenesis is the formation and expansion of newly created vasculature with an origin from normal, pre-existing blood vessels. This process occurs physiologically during in utero development and following lacerations to soft tissues or fractures of bone. However, angiogenesis also occurs pathologically in tumor development and in conditions like VHL syndrome. The VHL gene encodes the protein hypoxia-inducible factor-I alpha (HIF1A), a protein responsible for the production of erythropoietin (EPO) under conditions of hypoxia. In VHL Syndrome, HIF1A is pathologically activated, signaling a state of hypoxia when tissue is adequately oxygenated. Additional cytokines, including vascular endothelial growth factor receptor (VEGF) and platelet derived growth factor (PDGF), become activated and are essential in the process of angiogenesis. This combination of events leads to the formation of the hemangiomas that are characteristic of VHL syndrome. Additionally, VHL syndrome is associated with pheochromocytoma (as evidenced by the patients elevated blood pressure) and renal cell carcinoma.

A 34-year-old woman presents to the emergency department with the chief complaint of new-onset ataxia and inability to read distant traffic signs while driving. Physical exam reveals multiple non-blanching, spongy skin lesions and a blood pressure of 175/95 mm Hg_ Further discussion reveals that her father, who passed away when the patient was young, suffered from ataxia, seizures, and renal cancer. The mutation in this patient results in which of the following abnormalities? A. Loss of cell to basement membrane anchorage B. Excessive angiogenesis C. Hyperphosphorylation of cyclin D. Decreased DNA repair E. Inhibition of apoptosis

*The answer is D.* This is a family with Li-Fraumeni cancer syndrome, in which one copy of the p53 tumor suppressor gene carries a mutation. This is an autosomal dominant disorder in which family members have a significantly increased risk of malignancy as children or young adults. Typical cancers include osteosarcomas, soft tissue sarcomas, early onset breast cancers, adrenocortical tumors, and leukemia s.

A 40-year-old woman presents to her gynecologist after finding a lump in her right breast during a monthly self-examination. Her older sister, who just turned 45, was recently diagnosed with breast cancer. Mammography confirms the presence of a 3.2-cm density with irregular margins in the right breast. The gynecologist orders a biopsy, and the mass is determined to be malignant. The patient has a personal history of an adrenocortical cancer~ treated 18 months earlier. Additionally, she has a 29-year-old brother who was diagnosed with osteosarcoma as a child, a 37-year-old sister who was diagnosed with ovarian cancer 2 years ago and leukemia 4 years ago, and a father diagnosed with leukemia. The increased susceptibility to cancer in this family is caused by a mutation in which of the following genes? A. APC B. BRCA1 C. BRCA2 D. p53 E. Ret

*The answer is A.* The woman has myasthenia gravis, which is due to an autoimmune disorder in which antibodies directed against the acetylcholine receptor block the ability of acetylcholine to stimulate the muscle cells at the neuromuscular junction. Immunosuppressants can be taken to reduce the autoantibody production. Such drugs work, in part, through the activation of the tumor necrosis factor receptor, which activates apoptosis in the cells, leading to their destruction. Inhibiting apoptosis would exacerbate the problem, as the antibody-producing cells would survive longer and continue to produce the antibodies directed against the acetylcholine receptor. Drugs affecting the muscle would not help with this disorder, as it is a problem unique to the acetylcholine receptor expressed on the muscle surface. Stimulation or inhibition of cell growth does not stop the antibody-producing cells from continuing to make antibodies, and would not be an effective drug target for this disease.

A 42-year-old woman has slowly developed an inability to keep her eyes open at the end of the day. The eyelids droop, despite her best efforts to keep them open. This does not occur first thing in the morning. Further examination shows a generalized muscle weakness as the day progresses. A drug that may help to stabilize this condition would do which one of the following? (A) Stimulate apoptosis (B) Inhibit apoptosis (C) Stimulate cell growth (D) Inhibit cell growth (E) Induce muscle growth (F) Inhibit muscle growth

*The answer is A.* The condition described is chronic myelogenous leukemia (CML). It typically occurs in older patients and follows a chronic progressive course that can accelerate into a blast crisis. The abnormal gene product of the Philadelphia chromosome, t(9;22), produces a Bcr-Abl fusion protein that functions as a constitutively active tyrosine kinase receptor to promote leukemia growth. Presence of this chromosomal abnormality is believed to be the key pathophysiologic event in CML growth. Treatment for CML is imatinib (STI-571, or Gleevec), a tyrosine kinase inhibitor that is targeted specifically for the oncogenic protein made by the Philadelphia chromosome.

A 53-year-old man presents with complaints of fatigue, anorexia, and diaphoresis. His peripheral blood smear shows increased neutrophils and metamyelocytes. Cytogenetic studies show a t(9;22) t ranslocation . Which of the following describes the abnormal protein produced in this condition? A. A constitutively active tyrosine kinase B. A nuclear receptor C. A proangiogenic factor D. An acid phosphatase E. An antiapoptotic factor F. An immunoglobulin G. An overactive transcription factor

*The answer is D.* Nitrates may be metabolized to nitric oxide (NO) that activates a soluble guanyl cyclase in vascular smooth muscle. The increase in cGMP acti- vates protein kinase G and subsequently leads to vasodilation. Sildenafil inhibits cGMP= phosphodiesterase (PDE), potentiating vasodilation that can lead to shock and sudden death. Although sildenafil has much higher potency for the cGMP PDE isozyme in the corpora cavernosa, it can also inhibit the cGMP PDE in vascular smooth muscle. Nitric oxide synthase (choices A and B) is the physiologic source of nitric oxide in response to vasodilators such as acetylcholine, bradykinin, histamine, and serotonin.

A 58-year-old man with a history of angina for which he occasionally takes isosorbide dinitrate i having erectile dysfunction. He confides in a colleague, who suggests that sildenafil might help and gives him 3 tablets from his own prescription. The potentially lethal combination of these drugs relates to Left side: Isosorbide Dinitrate Right Side: Sildenafil

*The answer is B.* This child has I-cell disease. Failure to phosphorylate mannose residues to mannose-6-phosphate results in lysosomal enzymes failing to be delivered to the lysosomes. Ultimately, they are exocytosed, leading to a deficiency of lysosomal enzymes within the cell. Without these lysosomal enzymes, undegraded proteins accumulate, as evidenced by intracellular inclusion bodies.

A 6-month-old child develops failure to t hrive, developmental delay, and recurrent respiratory tract infections. The child is diagnosed with an autosomal recessive disease caused by a deficiency in N-acetylglucosamine-1-phosphotransferase, which phosphorylates mannose residues. As a result of this defect, what happens to the proteins that are normally the targets of this phosphotransferase? A. Destruction by the proteasome B. Exocytosis C. Expression at t he cell surface D. Retrograde transport to the endoplasmic reticulum E. Targeting to the lysosome

*The answer is E.* Duplication of cellular DNA occurs during the S phase. If there were a break in the DNA, the cell would not be able to replicate until the break is repaired. If the break is irreparable, the cell would undergo programmed cell death. Cells with high mitotic rates (cancer cells) will be affected by the drug more than normal cells. However, rapidly replicating cells such as hair cells and gastrointestinal endothelium will also be affected by the experimental drug, leading to adverse react ions such as alopecia and gastrointest inal upset.

A 60-year-old man with an 80-pack-year history presents to his physician with anorexia, weight loss, and hemoptysis. X-ray and CT scan reveal a mass in the right middle lobe of the lung, and a subsequent biopsy of the mass reveals locally advanced small-cell lung cancer. Previous X- rays did not show any masses. The man decides to participate in a trial of an experimental chemotherapeutic drug that preferentially inhibits the duplication of double-stranded DNA in tumor cells. What phase of the cell cycle is this drug most likely affecting? A. G0 B. G1 C. G2 D. M phase E. S phase

*The answer is B.* Lisinopril is an inhibitor of angiotensin-converting enzyme (ACE), which cleaves angiotensin I to form angiotensin II. Angiotensin II acts on vascular smooth muscle cells to cause them to contract in the efferent renal arteriole cells, this contraction increases the glomerular filtration rate. Angiotensin II works, in this case, by binding to angiotensin receptor type I, which is a seven-transmembrane domain receptor coupled with a G protein (Gq) that activates phospholipase C. Phospholipase C cleaves phosphatidylinositol (4,5)-bisphosphonate (PIP2) to form inositol phosphate 3 (IP3 ) and diacylglycerol (DAG). IP3 opens calcium channels, causing an intracellular increase in calcium, and DAG activates protein kinase C. Thus an ACE inhibitor will cause decreased activation of Gq and decreased activity of phospholipase C in vascular smooth muscle cells.

A 65-year-old man with hypertension, type 2 diabetes mellitus, and worsening renal function presents t o his primary care physician for a routine check-up. He reports that he has had worsening shortness of breath going up a flight of stairs, and physical examination reveals crackles at the bases of the lungs. Laboratory tests show the creatinine level has increased to 2.0 mg/dl from a baseline level of 1. 7 mg/ dl, and the B-type natriuretic peptide level has increased to 250 pg/ml from a baseline level of 75 pg/mL The physician decides to increase the patient's dosage of lisinopriL What will be the effect of the increased dosage of lisinopril on this patient's efferent renal arteriole smooth muscle cells? A. Decreased activity of adenylate cyclase B. Decreased activity of phospholipase C C. Increased activity of adenylate cyclase D. Increased activity of phospholipase C E. Increased activity of protein kinase C

*The answer is E.* The chronic myaloproliferative disorders are a group of bone marrow diseases characterized by the overproduction of myeloid cells. Primary myelofibrosis is caused by atypical megakaryocytic hyperplasia. The clonally expanded megakaryocytes activate fibroblast proliferation, resulting in progressive replacement of the marrow space by extensive collagen deposition. In the early stages, there is marrow hypercellularity with minimal fibrosis. As the disease progress, pancytopenia can result. Hepatomegaly and massive splenomegaly occur in myelofibrosis because the loss of bone marrow hematopoiesis is compensated for by extramedullary hematopoiesa. The peripheral smear characteristically shows teardrop-shaped red blood cells (dacrocytes) and nucleated red blood cells. With the exception of chronic myelogenous leukemia, the chronic myeloproliferative disorders (especially polycylhemia Vera) frequently harbor a mutation in the cytoplasmic tyrosine kinase, Janus kinase 2 (JAK2). This mutation (V617F) substitutes a bulky phenylalanine for a conserved valine at position 617, resulting in constitutive tyrosine phosphorylation activity, and consequently, cytokineindependent activation of the JAK-STAT pathway, A JAK2 inhibitor (ruxolitinib) has been approved for the treatment of primary myelofibrosis.

A 67-year-old man comes to the physician complaining of pallor, early satiety, and severe fatigue. Ha has also lost 20 pounds (9.07 kg) over the past 6 months. Physical examination reveals hepatomegaly and massive splenomegaly. A cytosolic protein recovered from his white bleed cells is found to have constitutive tyrosine phosphorylation activity. Consequently, there is persistent activation of STAT (signal transducers and activator of transcription) proteins. The patient is most likely suffering from which of the following disorders? A. Acute lymphocytic Leukemia B. Acute promyelocytic leukemia C. Chronic lymphocytic leukemia D. High-grade non-Hodgkin's lymphoma E. Myelofibrosis

*The answer is F.* Digoxin inhibits the Na+ /K+ pump, which typically extrudes 3 Na+ for every 2 K+ brought into the cell during depolarization. Because the pump is inhibited, the intracellular concentration of Na+ increases. The increase in intracellular Na+ inhibits the activity of the Na+;Ca2+ pump, which in turn increases the intracellular Ca2+ concentration. The increase in Ca2+ causes the increase in contractility.

A 78-year-old male, recently diagnosed with heart failure, has been prescribed a new medication, digoxin, and was informed concerning its potential toxicity. Some previously prescribed drugs, that may interact with digoxin, were replaced with other more appropriate medications. Which of the following intracellular ion concentrations increases as a direct result of digoxin? A. Bicarbonate B. Calcium C. Chloride D. Hydrogen E. Potassium F. Sodium

*The answer is E.* Growth factors can stimulate cell proliferation by altering the expression of certain genes in the nucleus. After a growth factor binds to its cell membrane receptor, signal transduction systems transfer the signal to the nucleus. Examples of signal transduction systems include; 1. MAP-kinase pathway 2. PI3K/AktJ/mTOR pathway 3. Inositol phospholipid pathway 4. cAMP pathway 5. JAK/STAT pathway The PI3K/Akt/mToR pathway is an intracellular signaling pathway that is important for cellular proliferation. This pathway is typically activated when a growth factor binds to its receptor tyrosine kinase, causing auto phosphorylation of specific tyrosine residues within the receptor. These phosphotyrosine residues activate phosphoinositide 3-kinase (PI3K), which then phosphorylates PlP2, found in the plasma membrane to PIP3. This leads to activation of a protein called Akt (or protein kinase B), a serine/threonine-specific protein kinase. Subsequently, Akt activates mTOR (mammalian target of rapamycin), which translocates to the nucleus to induce genes involved in cell survival, anti-apoptosis, and angiogenesis. mTOR activation is inhibited by PTEN (phosphatase and tensin homolog). a tumor suppressor protein that removes the phosphate group from PIP3. The PI3K/Akt/mTOR pathway is highly active in many cancer cells as a result of mutations causing increased activity of PI3K or Akt or loss of function of PTEN. Mutations involving certain growth factor receptors (eg, epidermal growth factor) can also enhance activity. Several drugs targeting this pathway (eg, mTOR inhibitors including rapamycin [sirolimus]} have shown benefit in treating certain cancers. *Educational Objective:* The Pl3K/Akt/mTOR pathway is an intracellular signaling pathway important for antiapoptosis, cellular proliferation, and angiogenesis. Mutations in growth factor receptors, Akt, mTOR, or PTEN that enhance the activity of this pathway contribute to cancer pathogenesis.

A group of investigators is studying the mechanisms involved in cancer pathogenesis. Their research focuses on the intracellular signaling cascades that begin after receptor tyrosine kinases are activated by their respective ligands. They find that the interaction of a certain growth factor with its receptor leads to the following sequence of events: Binding of growth factor Autophosphorylation of tyrosine residues Activation of phosphoinositide 3-kinase Activation of protein kinase B (Akt) Activation of X Which of the following is the most likely direct effect of X upon activation? A. Ca2+ efflux from endoplasmic reticulum B. cAMP accumulation C. Dimerization of STAT proteins D. Rapid decrease in cGMP levels E. Translocation to the nucleus and gene transcription

*The answer is E.* The cells described above are in metaphase of the mitosis (M) phase of the cell cycle, which is characterized by chromosomes migrating and lining up in the middle of the cell. In metaphase, the nuclear envelope has disintegrated and the mitotic spindle moves into the nuclear area. The chromosomes then become arranged in the plane of the spindle equator. Vincristine acts by binding to tubulin and blocking formation of microtubules, which are required to form the mitotic spindle.

A group of scientists at a pharmaceutical company are conducting in vitro experiments to investigate the effects of an antineoplastic drug. Under the microscope, it appears that with treatment, the majority of the cells are arrested at a stage in which their chromosomes are aligned in the vertical axis of the cells. Which antineoplastic agent has a mechanism of action similar to the one described? (A) 5-Fluorouracil (B) Cyclophosphamide (C) Etoposide (D) Methotrexate (E) Vincristine

*The answer is D.* The drug's action is during metaphase of the cell cycle and thus most likely serves to disrupt the mitotic spindle. Paclitaxel is a chemotherapeutic agent that prevents microtubule depolymerization. This stabilizes the mitotic spindle and disallows the migration of chromatids to their respective ends of the cell. Mitosis remains incomplete, leading to cell death. Paclitaxel is primarily used to treat advanced ovarian cancer and met astatic breast cancer.

A group of scientists is investigating the antineoplastic properties of a new drug. In an in vitro experiment, cancer cells are exposed to the drug and observed over time. Cells exposed to the drug are found to be uniformly arrested in metaphase with intact mitotic spindles. The mechanism of action of this new drug is similar to the action of which of the following drugs? A. 5-Fluorouracil B. Bleomycin C. Cyclophosphamide D. Paclitaxel E. Vincristine

*The answer is D.* Because the y-axis is 1/V, a smaller value for the 1/V means an increase in Vmax. An increase in Vmax (with no change in Km) means an increase in the number of enzymes (a kinase in this problem). Gene amplification (insertion of additional copies of the gene in the chromosome) is a well- known mechanism by which oncogenes are overexpressed and by which resistance to certain drugs is developed. For instance, amplification of the dihydrofolate reductase gene can confer resistance to methotrexate.

A kinetic analysis of the tyrosine kinase activities in normal and transformed cells is shown below. Which of the following conclusions is best supported by these results? A. The tumor cell kinase has a higher-than normal affinity for ATP B. A kinase gene has been deleted from the tumor cell genome C. A noncompetitive inhibitor has been synthesized in the tumor cells D. A kinase gene has been amplified in the tumor cell genome E. The tumor cell kinase has a lower-than normal affinity for ATP

*The answer is E.* Radiation exposure results in cellular injury by damaging DNA. An affected cell responds by repairing the damage or by signaling apoptosis of the cell to prevent damaged nuclear material from being passed onto daughter cells. The mitochondrial (or intrinsic) pathway of apoptosis is the major mechanism of auto-destruction in mammalian cells. A family of proteins called cytochrome c are contained within the mitochondria of a healthy cell, as regulated by a balance of pro- and anti-apoptotic factors called the Bcl family. 'Survival signals," such as growth factors, stimulate the production of anti-apoptotic proteins (e.g., BcI-2) which control mitochondrial permeability and prevent the leakage of proteins that can trigger cellular suicide (apoptosis). However, in the setting of cell damage, "stress signals" stimulate the production of other Bcl-family proteins (e.g., Bim, Bid, Bad). These pro-apoptotic proteins activate two critical effectors called Bax and Bak, which insert into the mitochondrial membrane, creating channels through which mitochondrial proteins can leak out into the cytoplasm. When the protective function of anti-apoptotic proteins like Bcl.2 are overcome, cytochrome c is released into the cytoplasm. Cytochrome c binds to a cytosolic protein called apoptosis-activating factor-1 (Apaf-1), and this complex activates "initiator caspases" (e.g., caspase 9) which in turn activate "executioner caspases." Executioner caspases breakdown the cytoskeleton and activate endonucleases to digest nuclear material. Unlike with necrosis, the cell membrane remains intact with apoptosis, and there is an absence of accompanying inflammation.

A large retrospective study is being conducted to investigate whether the increased use of computed tomography (CT) as an imaging modality and the subsequent increased radiation exposure, is correlated with an increased incidence of malignancy. Which of the following is part of the normal response for removing damaged cells and preventing damaged DNA from being passed onto daughter cells? A. Decreased activity of Caspase 9 B. Decreased Fas receptor activation C. Increased activity of Bcl-2 D. Decreased lysosomal membrane permeability E. Increased activity of Apaf-1

*The answer is A.* Stimulation of α2-receptors antagonizes the actions of the sympathetic nervous system. It uses the Gi pathway, which results in inhibition of adenylyl cyclase, thus resulting in a decrease in cAMP. The other receptors use Gs as a secondary messenger, which activates adenylyl cyclase and results in an increase in cAMP.

A medication is created that blocks the action of adenylyl cyclase. Administration of the medication will have an intracellular effect similar to that caused by stimulating which of the following receptors? A. α2-Receptors B. β1 -Receptors C. β2-Receptors D. D1-Receptors E. H2-Receptors F. V2-Receptors

*The answer is A.* When activated, the Gq class of GPCRs results in activation of protein kinase C and an increase in int racellular ca lcium. Thus, like the TIY2 receptor, the a1-adrenergic receptor is a Gq-class GPCR. Other Gq-class GPCRs include muscarinic}, muscarinic3, and histamine1.

A new small-molecule drug for the treatment of HIV has just been discovered. This small molecule appears to bind to a newly discovered surface G-protein- coupled receptor (GPCR), TTY2, and sets off a cascading signal that leads to apoptosis of the cell. The TTY2 receptor is expressed in cells that have been infected by HIV but is not expressed in noninfected cells. Using in vitro experiments, scientists have been able to show that binding of the TTY2 receptor will act on phospholipase C, thereby leading to an increase in intracellular calcium. The TTY2 receptor is similar to which of the following receptors? A. α1 B. α2 c. β2 D. β3 E. Dopamine1

*The answer is A.* The description best fits the PIP2 system in which protein kinase C is activated.

A patient with manic depressive disorder is treated with lithium, which slows the turnover of inositol phosphates and the phosphatidyl inositol derivatives in cells. Which of the following protein kinases is most directly affected by this drug? A. Protein kinase C B. Receptor tyrosine kinase C. Protein kinase G D. Protein kinase A E. Protein kinase M

*The answer is B.* The drug described has the properties of a centrally acting o2 -receptor agonist, such as clonidine. α2 -Receptors are Gi receptors that act on adenylyl cyclase to decrease the levels of cyclic adenosine monophosphate and, thereby, protein kinase A activity. By acting on neurons in the medulla, this drug reduces sympathetic outflow, lowering blood pressure via depressed cardiac contractile force and reduced peripheral vascular tone.

A patient with refractory hypertension is changed to a new therapy by his primary care physician. Acting as a G-protein-coupled receptor agonist, this drug diminishes sympathetic outflow by decreasing levels of cAMP within neurons. What type of receptor is this drug most likely acting on? A. α1-Receptor B. α2-Receptor C. β1 -Receptor D. β2- Receptor E. Nicotinic acetylcholine receptor

*The answer is B.* The centromere is a region of condensed specialized chromatin that serves as a joining site for the centromere of a sister chromatid such that it assumes the classic "X" configuration shown in the diagram. Furthermore, it is at the centromere of the chromosome that the kinetochore attaches to the microtubules, allowing the chromatids to migrate to opposite sides of the cell toward the centrosomes (also known as microtubule organizing centers, MTOCs). The spindle apparatus is a cytoskeletal structure consisting of microtubules that function during cell division (mitosis or meiosis) to separate sister chromatids or homologous chromosomes. The assembly of the spindle apparatus begins during prophase, in which the centrosomes (MTOCs) migrate to opposite poles of the cell. Microtubule polymerization proceeds in a counterparallel fashion from each centrosome. Kinetochores that had preassembled on each chromosome's centromere capture some of the growing microtubules ends, leading to chromosomal alignment at the metaphase plate. Kinesins, guanosine triphosphate powered molecular motors, "walk" along the microtubules in a unidirectional manner to both separate the sister chromatids and further elongate the spindle apparatus as the microtubules themselves actively depolymerize and repolymerize. Therefore, a drug that inhibits the function of the microtubules, kinesins, kinetochores, or centromere would severely impair the assembly of the spindle apparatus and force the cells to arrest in prophase.

A pharmaceutical company is actively researching a novel compound that shows promise as a chemotherapeutic agent. After administering the drug to a culture of cancer cells, microscopic evaluation reveals arrest in the prophase of mitosis in all cells. Which of t he following intracellular targets was most likely inhibited by the drug? A. Actin B. Centromere C. Microfilament D. p53 E. Topoisomerase

*The answer is C.* Two pieces of information are critical here: first, β-tubulin is part of the microtubule structure, and second, microtubules form the mitotic spindle. Thus, mitosis, or the M phase, is most affected by the posited drug in the vignette.

A pharmacologist is developing novel chemotherapeutic agents. A prime mechanism of chemotherapeutic agents is the cessation of the cell cycle. A drug that antagonizes β-tubulin polymerization will affect which of the following phases of the cell cycle? A. G1 B. G2 C. M D. S

*The answer is C.* Non-polar, hydrophobic amino acids such as valine, alanine, isoleucine, methionine, and phenylalanine are generally located interiorly on globular proteins, where they are shielded from direct contact with water. The classic plasma membrane-spanning proteins are executors for glycoprotein hormones, such as TSH, LH, and FSH. These G-protein coupled membrane-bound receptors for glycoprotein hormones contain three major domains: extracellular (responsible for ligand binding), transmembrane (consisting of hydrophobic amino acids), and intracellular (coupled with G-proteins). The question stem describes repeating alpha-helical segments each composed of approximately 20 hydrophobic amino acids, this description is characteristic of the transmembrane region(s) of a protein (Choice C). (Choice A) The receptors for insulin, IGF-1, and several cytokines also have three domains: extracellular (ligand binding), transmembrane (composed of hydrophobic amino acids), and intracellular. The intracellular (cytosolic) domain of these membrane-associated receptor proteins contains a tyrosine kinase that is activated upon extracellular ligand binding. Once activated, tyrosine kinase will phosphorylate available tyrosine residues. (Choice B) Cyclic AMP formed during the activation of a G, protein can activate protein kinase A. Protein kinase A has the unique ability to phosphorylate and thereby activate proteins that are capable of translocating into the nucleus. Once within the nucleus, these special proteins bind to the promoter regions of DNA and modulate transcription. (The proteins that actually bind to DNA are not membrane proteins.) (Choice D) The binding of a ligand to the extracellular domain of a transmembrane protein causes indirect activation of the adenylate cyclase system, the opening of ion channels, direct activation of tyrosine kinase, activation of the calcium-calmodulin system, and activation of the inositol triphosphate system. The region of a transmembrane protein that interacts with the ligand is found in the extracellular space and therefore could not possibly be composed mainly of hydrophobic amino acids. (Choice E) This choice describes the intracellular iron-containing proteins (hemeproteins) such as hemoglobin, myoglobin, and cytochrome oxidase. Heme is a complex of protoporphyrin IX and iron. Hemoglobin A, the most common hemoglobin in adults, consists of two alpha and two beta globin chains held together by non-covalent interactions. Each subunit has stretches of alpha helices and a crevice lined by nonpolar amino acids, where heme binding occurs. *Educational Objective:* Integral membrane proteins contain transmembrane domains composed of alpha helices with hydrophobic amino acid residues such as valine, alanine, isoleucine, methionine, and phenylalanine.

A protein believed to play a role in signal transduction and the cellular response to thyroid stimulating hormone is studied. Special attention is paid to a region of this protein that contains several alpha-helical regions each composed of approximately 20 amino acid residues-consisting primarily of valine, alanine, and isoleucine. Which of the following functions does this particular region most likely perform? A. Phosphorylating tyrosine residues B. Binding to intranuclear DNA C. Spanning the cellular membrane D. Interacting with signaling substances E. Interacting with metal ions in transporting proteins

*The answer is E.* When an action potential arrives at a smooth muscle cell, the membrane depolarizes and voltage-gated calcium channels open. Calcium enters the cell and binds to calmodulin. This calcium-calmodulin complex activates myosin light-chain kinase, which serves to phosphorylate myosin light chains. Once they are phosphorylated, they facilitate cross-bridge formation and smooth muscle contraction.

A researcher studying peristaltic contractions in the small intestine notes that the mechanical force in the wall of the bowel is generated by smooth muscle contractions. The researcher notes that the intracellular mechanisms leading to the contraction are triggered by a cascade of molecular events. Initially, an action potential depolarizes the muscle membrane and facilitates calcium release from the sarcoplasmic reticulum. Calcium then binds to calmodulin, What is the function of the enzyme activated by the calcium-calmodulin complex in smooth muscles? A. Dephosphorylate actin thick filaments B. Dephosphorylate myosin light chains C. Dephosphorylate troponin C D. Phosphorylate act in thick filaments E. Phosphorylate myosin light chains F. Phosphorylate troponin C

*The answer is D.* Microtubules are involved in the transport of vesicles down the axons of neurons. Microtubules are made of polymerized tubulin. The microtubule is made up of two subunits, α and β, that polymerize to form a hollow tubule. The microtubule is also composed of transport tracks to which the molecular motor proteins known as dynein and kinesin at the tach. Dynein transports cellular cargo such as organelles towards the minus end of the microtubule, towards the cell center, whereas kinesin exhibits anterograde transport towards the plus end of the micotubule, away from its center.

A scientist is interested in studying how vesicles synthesized in the cell bodies of neurons are transported down t he axon to the terminal boutons. A deficiency in which of these proteins would impair this process? A. Clathrin B. Desmin C. Titin D. Tubulin E. Vimentin

*The answer is D.* The progress Of a cell through the cell cycle is largely regulated by two key classes Of molecules: cyclins and cyclin-dependent kinases (CDKs). Cyclins bind to CDKs and activate them. which in turn phosphorylate target proteins. Phosphorylation can either activate or inactivate target proteins to coordinate progress through the cell cycle, CDKs are always expressed within the cell, waiting for cyclin to be synthesized in response to various molecular signals. P21 is a cyclin-dependent kinase inhibitor, which inhibits various cyclin-CDK complexes. Thus, it normally inhibits progression through the cell cycle. preventing unregulated cell growth. Activity of p21 is mainly controlled by p53, a Strong tumor suppressor protein. P53 has several functions in moderating cell cycle activity and gene transcription. It has been described as "the guardian of the genome" because of its many anticancer functions, such as initiating repair of damaged DNA, inducing apoptosis, and regulating progression of the cell cycle.

A team of researchers conducted studies on biomarkers that could be used to detect lung cancer. They found that decreased levels of p21 is greater than 80% sensitive at detecting lung cancer in at risk patients. Under normal conditions, which of the following is most directly responsible for increasing p21 activity? A. Rb B. c-MYC C. pS3 D. cyclin D E. cyclin dependent kinases

*The answer is B.* This patient has I-cell disease, which is an inherited lysosomal storage due to a defect in N-acetylglucosaminyl-1-phosphotransferase. Clinical features of this condition include skeletal abnormalities, corneal clouding, restricted joint movement, coarse facial features, and severe psychomotor impairment. Death usually occurs in the first decade of life. Defective N-acetylglucosaminyl-1-phosphotransferase results in failure of the Golgi apparatus to phosphorylate mannose residues on glycoproteins. This dysfunction leads to extracellular protein excretion. When N-acetylglucosaminyl-1-phosphotransferase is functioning normally, proteins are delivered to lysosomes, not extra cellularly.

A term neonate is delivered by spontaneous vaginal delivery without complications. The mother did not receive prenatal care. On physical examination, the infant has bilateral hip dislocations restricted movement in shoulder and elbow joints, and coarse facial features. Laboratory results reveal elevated plasma levels of lysosomal enzymes. It is believed that the neonate has an inherited disorder. The enzyme that is dysfunctional in this disorder normally funct ions to phosphorylate proteins in what organelle? A. Endosomes B. Golgi apparatus C. Lysosomes D. Rough endoplasmic ret iculum (RER) E. Smooth endoplasmic reticulum (SER)

*The answer is D.* BRCA1 and BRCA2 are tumor-suppressor genes whose protein products function in DNA repair. Frameshift or nonsense mutations commonly occur in BRCA1 and BRCA2 and produce truncated protein products. Mutations in these genes result in a gene product that has less or no function, and can lead to DNA instability and subsequent gene rearrangements. Both alleles of these tumor-suppressor genes must be inactivated to cause loss of function of these genes. A woman with harmful BRCA1 or BRCA2 mutations has as much as a 60% risk of breast cancer during her lifetime. The BRCA gene mutations are the most common inherited causes of breast cancer. This is five times greater than the lifetime risk for the general population. Mutations to BRCA can also increase the risk of many cancers aside from breast cancer, including ovarian cancer. Women with the BRCA1 mutation have a greater lifetime risk of developing ovarian cancer than those with BRCA2 mutations.

A woman whose mother had high-grade invasive ductal carcinoma in both breasts develops breast cancer at age 26. While taking an extensive family history, it is revealed that the patient 's maternal great-aunt died of what was probably ovarian cancer at the age of 35. The patient's identical twin sister decides to undergo genetic testing to determine her chances of developing breast cancer. Which of the following mechanisms causes the genes most often responsible for familial breast cancer to become tumorigenic? A. Chromosomal rearrangement B. Dominant negative effect C. Gain of function D. Loss of function E. Viral insertion

*The answer is D.* The description of the mass in the child's right eye is typical of retinoblastoma, the most common primary intraocular tumor of childhood, with the majority of cases diagnosed in children less than two years of age. Leukocoria, an abnormal white reflection from the retina, is the most common presenting sign. Retinoblastoma is typically caused by mutation of the RBI gene. which codes for the Rb protein. Cases may be either familial or sporadic in nature; familial cases involve a germline RBI mutation, require only one somatic RBI mutation for carcinogenesis, and are typically multifocal and/or bilateral, while sporadic cases are due to two somatic RBI mutations and are unilateral. The Rb protein is a key regulator of the GI/S cell cycle checkpoint. It is a tumor suppressor that halts cell cycle progression by inactivating E2F.

An 18-month-old boy is brought to the physician for a routine check-up. He has been healthy since birth and is up to date on his vaccinations. There is no significant family history. During the physical exam, the physician notes a red reflex in the left eye of the child, but a white reflex in the right eye. Prompt ophthalmological evaluation reveals a solitary exophytic translucent mass adherent to the retina of the right eye. Further evaluation confirms the presence of a neoplastic pathology. Which of the following aberrant genetic processes is most likely responsible for this child's presentation? A. Decreased phosphorylation of cyclin-CDK complexes B. Decreased E2F transcription factor activity C. Increased phosphorylation of cyclin-CDK complexes D. Increased E2F transcription factor activity E. Increased p53 activity

*The answer is C.* The G-protein-coupled-receptors have a vary characteristic structure with seven transmembrane regions, an extracellular domain and an intracellular domain coupled with the trimeric G-protein. In their inactivated state, G-proteins exist as heterotrimers consisting of alpha, beta and gamma subunits with guanosine diphosphate (GDP) tightly bound to the alpha subunit G proteins are activated after binding of hormone to the extracellular domain, The first step in activation of a G~protein occurs when GDP is exchanged for GTP on the alpha subunit. Once bound to GTP, the alpha subunit dissociates from the beta and gamma subunits and exposes its catalytic domain for either adenylate cyclase or phospholipase C depending on the ligand. If the G-protein alpha subunit activates phospholipase C, then the degradation of phosphatidylinositol 4,5-bisphosphate to inositol 1,4,5 triphosphate (IP3) and diacylglycerd (DAG) occurs. Diacylglycerol stimulates protein kinase C, which is responsible for some intracellular effects. Inositol 1,4,5-triphosphate (IP3) produces most of the intracellular effects of this pathway by increasing intracellular calcium, and elevated intracellular calcium activates protein kinase C. If the action of IP3, were blocked as described in the question stem, then decreased activation of protein kinase C would occur upon hormone binding. (Choice C). (Choice A) The activity of phospholipase C would be unchanged if IP3 were blocked because phospholipase C exerts its effect before IP3, in the calcium phosphatidylinositol second messenger system. (Choice B) Lipoxygenase is an enzyme responsible for formation of leukotrienes from arachidonic acid It is not involved in intracellular signaling. (Choice D) Termination of the effects of hormones that act by cAMP or cGMP G-protein second messenger systems is carried out by the enzyme phosphodiesterese. Phosphodiesterase has no effect on the IP3, second messenger system. (Choice E) Activation of adenylate cyclase leads to the formation of cyclic AMP and the subsequent activation of protein kinase A. Protein kinase A activates the proteins that produce the intracellular effects of hormones. *Educational Objective*: After a hormone binds a G-protein coupled receptor that activates phospholipase C, the initial step of the IP3 second messenger system involves degradation of membrane lipids into diacylglycerol (DAG) and inositol triphosphate (lP3) by that enzyme. Protein kinase C is activated by DAG as well as calcium released from sarcoplasmic reticulum under the influence of lP3,

An agent that specifically blocks the interaction of inositol triphosphate with its intracellular receptor would most likely decrease the activity of: A. Phospholipase C B. Lipoxygenase C. Protein kinase C D. Phosphodiesterase E. Adenylate cyclase

*The answer is D.* Digoxin is a Na+-K+-ATPase inhibitor. In addition to its effects on myocardium, it has similar effects on vascular smooth muscle via the same mechanism. Inhibition of the pump in vascular smooth muscle will lead to an increase in intracellular [Na+]. This leads to an indirect inhibition of the Na+ /Ca2+ exchanger due to increased levels of Na+ within the cell. Usually the Na+ /Ca2+ exchanger transfers 1 Ca2 + out of the cell with every 3 Na+ entering but becomes inhibited with high levels of intracellular Na+. Though we do not use it to treat hypotension, the ensuing accumulation of Ca2+ inside the cell will then increase vascular resistance and systemic blood pressure.

An outbreak of infection in Ghana has nearly wiped out an entire village. A group of scientists travels to Ghana to determine the cause of this disease. As a result of their work, they discover that a new toxin-producing bacterium caused the outbreak. The toxin is noted to increase intracellular calcium and sodium. Which of the following drugs mimics the outcome of this toxin? A. Albuterol B. Botulinum toxin C. Cocaine D. Digoxin E. Procainamide

*The answer is B.* Cyclin dependent kinase (Cdk) levels remain relatively constant throughout the cell cycle, but their activity varies. They are inactive unless they are activated by binding the appropriate cyclin. Levels of cyclins change rapidly throughout the cell cycle. Appropriate cyclins are synthesized at the start of each phase of the cycle and destroyed before the cell moves to the next phase. This ensures that Cdks are only active at the appropriate time.

Appropriate cyclin dependent kinases are synthesized at the start of each phase of the cell cycle and destroyed before the cell proceeds to the next phase: true or false? (A) True (B) False

*The answer is A.* Calcium is maintained in high concentrations outside of the cell and in discrete compartments within the cell (eg, in mitochondria). Free intracellular calcium can activate several enzymes the cumulative effect of which is to induce significant cell injury. A few important enzyme classes include ATPases, which decrease the ATP supply; phospholipases, which decrease membrane stability; endonucleases, which induce DNA damage; and several proteases, responsible for protein breakdown.

As increased intracellular calcium is detrimental to the cell, calcium homeostasis is tightly regulated both across the cell membrane and within the cell via sequestration in the endoplasmic reticulum and mitochondria. In which of the following ways does increased intracellular calcium concentration cause the most cell damage? (A) Enzyme activation (B) Free radical generation (C) Increased membrane permeability (D) Inhibition of glycolysis (E) Inhibition of oxidative phosphorylation

*The answer is D.* Cyclins are destroyed at the end of each phase of the cell cycle by proteolysis in a proteasome. They are directed to proteasomes by the attachment of a small peptide signal molecule called ubiquitin, and ubiquitinylation of the cyclin is a necessary prelude to its destruction, but ubiquitin does not actually inactivate the cyclin. It simply marks it for destruction by the proteasome. Some cyclins (especially in M phase) can be inhibited by multiple phosphorylation, but this is reversible and the cyclins are only temporarily inactivated. The cyclins are re-activated within the same phase by removing the phosphates, so dephosphorylation actually activates cyclins in this instance.

At the end of each phase of the cell cycle cyclins activating Cdks in that phase are inactivated irreversibly by which of the following mechanisms? (A) Multiple phosphorylations. (B) Dephosphorylation. (C) Ubiquitinylation. (D) Destruction by proteolysis in a proteasome.

*The answer is A.* Tumour suppressor genes code for proteins which control progression through cell cycle checkpoints, or which inactivate growth signals. This inhibitory activity can normally take place provided that at least one of a pair of tumour suppressor genes produces an active protein. This is true of the Rb gene, where retinoblastoma, the characteristic cancer produced by inactivation of this gene, only occurs when both copies of the Rb gene are inactivated.

Both copies of a tumour suppressor gene must be inactivated by mutation in order to contribute to cancer development. True or false? (A) True (B) False

*The answer is C.* Niamenn-Pick disease Type A is an autosomal recessive disorder that typically presents in infants of Ashkenazi Jewish descent. The disease is characterized by a deficiency of sphinomyelinase that causes sphingomyelin to accumulate within phagocytes. The resultant "foamy histiocytes" accumulate in the liver, spleen, and skin. Additionally, there is gradual sphingomyelin deposition in the CNS, which causes neurologic degeneration. Affected infants classically present with loss of previously acquired motor capabilities following a period of normal development. The neurologic deterioration typically progresses to hypotonia and blindness by age 1. A cherry-red macular spot (as in Tay-Sachs disease) and hepatosplenomegaly are common findings on physical examination. Death usually before age 3.

Concerned parents bring their previously healthy 8-month old son for evaluation because he has been losing previously acquired motor skills. Hepatosplenomegaly is noted on physical examination. Liver biopsy reveals the presence of leukocytes with prominent intracellular sphingomyelin accumulations. The most likely diagnosis is: (A) Tay-Sachs disease (B) Hurler syndrome (C) Niemann-Pick disease (D) Gaucher disease (E) Von Gierke disease (F) Pompe disease (G) Fabry disease (H) Lesch-Nyhan syndrome

*The answer is E.* The patient described in the question stem is most likely suffering from a peroxisomal disease Peroxisomal diseases are rare inborn errors of metabolism where peroxisomes are either absent or nonfunctional. Very long chain and some branched chain fatty acids cannot undergo mitochondrial beta-oxidation These fatty acids are metabolized by a special form of beta oxidation (very long chain fatty acids) or by alpha Oxidation (branched chain fatty acids such as phytanic acid) within peroxisomes. When peroxisomes are absent or nonfunctional, these fatty acids accumulate within the tissues. One example of a peroxisomal disease is Zellweger syndrome. In this condition infants are unable to properly form myelin in the CNS. Symptoms of this disease include hypotonia and seizures as mentioned in the question stem as well as hepatomegaly. mental retardation, and early death within months of initial presentation. Refsum disease results from a defect in peroxisomal alpha oxidation and leads to neurologic disturbances in response to accumulation of phytanic acid within the body.

Cultured fibroblasts taken from an infant suffering from hypotonia and seizures show an impaired ability to oxidize very long chain fatty acids (VLCFA) and phytanic acid. The defect is most likely localized to: (A) Mitochondria (B) Rough endoplasmic reticulum (C) Proteasomes (D) Lysosomes (E) Peroxisomes (F) Golgi apparatus

*The answer is B.* Though cells with damaged DNA are killed by apoptosis if the damage cannot be repaired, most forms of damage, such as physical damage, and damage produced by oxygen deprivation produce necrosis. Here, membranes are ruptured and cell contents leak out causing inflammation in the surrounding tissue. Necrotic cell death is an accidental, not a regulated, process.

Damaged cells die by apoptosis. True or False? (A) True (B) False

*The answer is B.* Some Cdks are activated by different cyclins in different phases of the cell cycle. They can work at different checkpoints, depending which cyclin is binding to them. As well as activating the Cdk the cyclin can determine the substrate it phosphorylates, thus directing it to a different target. The cyclins, not the Cdks, are the main determinants of specificity.

Each cyclin dependent kinase (Cdk) is activated by a specific cyclin and works at one specific checkpoint in the cell cycle. True or false? (A) True (B) False

*The answer is D.* For step 1, you must know the second messenger pathways for e·ach of the G protein-coupled receptors. First, you need to know the three types of G proteins for the boards: Gs, Gi, and Gq. Then, associate each of the receptors with its respective pathway. One mnemonic that is helpful is "kiss and kick until you're sick of sex," spelled "qiss, qiq, siq, sqs," where α1, α2, β1, and β2 are "q-i-s-s," respectively; M1, M2 , and M3 are "q-i-q," respect ively; D1, D2, and H1 are "s-i-q," respectively; and s-q-s (super qinky sex) are H2, V1, and V2, respectively. The receptors stimulate the Gq pathway, which activates phospholipase C; phospholipase C cleaves membrane phospholipid PIP2 to IP3 and diacylglycerol (DAG). IP3 diffuses through the cytoplasm to bind its receptors in the endoplasmic reticulum; as a result, calcium is released. Calcium and DAG together activate protein kinase C.

Extracellular stimulation of M3 muscarinic receptors causes the cleavage of membrane-bound phospholipids. Through the action of second messengers, this stimulation results in which of the following? A. Activation of adenylyl cyclase B. Activation of protein kinase A C. Deactivation of protein kinase C D. Increase in intracellular calcium E. Increase in intracellular cAMP

*The answer is A.* The response of the hepatocytes to the stimulus described in the question stem is characteristic of the response of these cells to insulin. Insulin is an anabolic hormone that promotes the synthesis of glycogen, triglycerides, nucleic acids, and proteins. Insulin inhibits glycogenolysis and gluconeogenesis. Insulin acts via a tyrosine kinase mechanism. The insulin cell surface receptor is a transmembrane protein that also has cytosolic tyrosine kinase activity. The tyrosine kinase causes phosphorylation of a poorly characterized class of proteins known as insulin receptor substrates leading to activation of protein phosphatase. Protein phosphatase dephosphorylates glycogen synthase thereby activating that protein and promoting glycogen synthesis. Protein phosphatase also dephosphorylates fructose 1,6-bisphosphatase thereby inactivating that enzyme and inhibiting gluconeogenesis. This is also a good example of how phosphorylation and dephosphorylation of enzymes by second-messenger proteins can cause activation of some enzymes and inactivation of others. (Choice B) Protein kinase A is the primary intracellular effector enzyme in the G-protein adenylate cyclase second messenger system. Increased levels of cAMP stimulate protein kinase A to activate the necessary enzymes to carry out the intracellular actions of the hormone that bound the cell and activated adenylate cyclase in the first place. (Choice C) Phospholipase C is active in the G-protein / Inositol triphosphate (IP3) / Calcium second messenger system. Hormone binds its receptor and activates a G-protein that in turn activates phospholipase C to degrade phospholipids into inositol triphosphate and diacylglycerol. Both diacylglycerol and the increased intracellular calcium caused by IP3 will activate protein kinase C. (Choice D) Janus protein kinase (JAK) is a part of the second messenger system for peptide hormones such as some cytokines in a pathway referred to as JAK-STAT (signal transducers and activators of transcription). JAK has tyrosine kinase activity. (Choice E) Lipoxygenase is an enzyme involved in arachidonic acid metabolism and is responsible for the arm of that pathway that synthesizes leukotrienes. *Educational Objective:* Insulin is an anabolic hormone that acts via a tyrosine kinase second messenger system to stimulate the synthesis of glycogen, proteins, fatty acids and nucleic acids. Tyrosine kinase leads to the activation of protein phosphatase within cells, and protein phosphatase directly modulates the activity of enzymes in the metabolic pathways regulated by insulin.

Hepatocytes exposed to an external stimulus demonstrate a rapid increase in intracellular glycogen stores and a decrease in glucose release into the blood. Which of the following most likely promotes the effects described above? A. Protein phosphatase-1 B. Protein kinase A C. Phospholipase C D. Janus protein kinase IJAK) E. Lipoxygenase

*The answer is C.* Proteins with SH2 domains might bind to the insulin receptor substrate-1 (IRS-1) to transmit signals from the insulin receptor, a tyrosine kinase type of receptor. PI-3 kinase is an example of an SH2 domain protein. SH2 domains are not involved in DNA binding (choices A and D). Examples of protein= domains that bind DNA include zinc fingers (steroid receptors), leucine zippers (CREB protein), and helix-turn-helix proteins (homeodomain proteins).

In a DNA sequencing project, an open reading frame (ORF) has been identified. The nucleotide sequence includes a coding region for an SH2 domain in the protein product. This potential protein is most likely to A. bind to an enhancer region in DNA B. be a transmembrane hormone receptor C. transmit signals from a tyrosine kinase receptor D. bind to an upstream promoter element E. activate a soluble guanyl cyclase enzyme in vascular smooth muscle

*The answer is E.* The process described is called expression cloning. Expression cloning is a type of DNA cloning where the signals necessary for transcription and translation are included in the cloned DNA. This process allows bacteria to be used to produce large amounts of a protein of interest. First, mRNA is used as a template by reverse transcriptase to produce a cDNA strand containing promoter sequences (such as the Pribnow box, or -35 sequence in prokaryotes) and translation stimulatory sequences (Shine-Dalgamo). The gene is then incorporated into a plasmid and subsequently transcribed and translated into protein. In order to then be radiolabeled by a DNA probe, the protein must be able to bind DNA. Examples of proteins that are able to bind DNA include transcription factors, steroids, thyroid proteins, vitamin D receptors. retinoic acid receptors. DNA transcription and replication proteins, and others. Of the choices listed, N-myc is the only protein that can bind DNA. The MYC proteins are mammalian transcription factors. C-myc is the protein that is overexpressed in Burkitt lymphoma. *Educational Objective:* N-myc is a transcription factor that is, by definition, capable of binding DNA. Because of its DNA-binding ability, N-myc can be detected by DNA probes.

In a laboratory experiment, the enzyme reverse transcriptase uses a specific mRNA template to synthesize a strand of complementary DNA (DNA). The DNA is then integrated into a plasmid containing a bacteria! promoter, which allows the DNA to be expressed in bacterial cells. Large quantities of the target protein are obtained end subsequently identified using a radiolabeled DNA probe. The target protein is most likely which of the following? A. Insulin-like growth factor-1 B. Insulin receptor C. Protein kinase A D. Calmodulin E. N-myc protein F. k-RAS

*The answer is B.* The Ras gene codes for a GTPase switch protein that forms part of a growth factor signalling pathway. Stimulation of this protein via a ligand-bound growth factor receptor causes it to activate by exchanging a molecule of GDP for GTP. The activated Ras protein then activates downstream proteins which ultimately activate transcription factors. Normally the Ras signal is self-limiting because the protein has intrinsic GTPase activity, so it hydrolyses its bound GTP to GDP, switching the signal off. There are several known mutations of the Ras gene, but the in best studied example a point mutation of the gene produces a protein with a single amino acid change at the GTP binding site. This severely reduces the protein's GTPase activity, making it very slow to self-inactivate, once it has been activated. This prolongs the growth signal effectively allowing the cell to be stimulated without a growth factor signal.

In what way does the ras oncogene contribute to cancers? (A) Ras codes for an anti-apoptotic protein, which is produced in abnormally large amounts. (B) Ras codes for a GTPase switch protein, which in its mutated form cannot be switched off. (C) Ras codes for a transcription factor, which is produced in abnormally large amounts (D) Ras codes for a truncated form of a growth factor receptor, which is continually active.

*The answer is C.* Lysosomal membranes contain an enzyme which actively pumps protons into the organelle, thereby maintaining a low intraorganelle pH. This enzyme is the proton-translocating ATPase, as ATP hydrolysis provides the energy to pump protons against their concentration gradient. The removal of protons from the lysosome would raise pH, not lower it (thereby rendering answers A and B incorrect). Free diffusion of protons would not allow uptake of protons against a concentration gradient, as diffusion is the flow from a higher concentration to a lower concentration. Since the cytoplasmic pH is in the range of 7.2, if protons were freely diffusible across the lysosomal membrane, the protons would leave the lysosomes and enter the cytoplasm. The lysosomes do not synthesize large amounts of carboxylic acids (a weak acid) in order to lower the pH inside the organelle.

Lysosomal enzymes have a pH optimum between 4 and 6. The intralysosomal contents are kept at this pH by which of the following mechanisms? (A) The active pumping of protons out of the organelle (B) The free diffusion of protons out of the organelle (C) The active pumping of protons into the organelle (D) The free diffusion of protons into the organelle (E) The synthesis of carboxylic acids within the lysosome

*The answer is A.* A variety of hormone receptors are known to exert their intracellular effects via the phosphoinositol system. Examples include α-adrenergic, M1, and M3, cholinergic, V, (vasopressin), H, (histamine), oxytocin, angiotensin II, TRH, and GnasH receptors. This signal transduction pathway proceeds through the following steps: 1. Binding of a ligand to its cell surface receptor causes the exchange of GDP for GTP on the α-subunit of a Gq-protein associated with the receptor. The activated α-subunit undergoes a conformational change and exposes a phospholipase C (PLC) activating site. 2. After activation, PLC hydrolyzes phosphatidylinositol bisphosphate (PIP2) into diacylglycerol (DAG) and Inositol triphosphate (lP3). 3. DAG is able to directly stimulate protein kinase C (PKC), but the major activator of PKC is increased intracellular Ca2+ that occurs due to IP3, mediated-release of intracellular Car- stores from the endoplasmic reticulum. PKC is the major effector molecule in me pathway; it directly modulates the activity of other proteins via phosphorylation. *Educational Objective:* The phosphoinositol second messenger system begins with ligand-receptor binding and Gq-protein activation leading to activation of phospholipase C (PLC). PLC then hydrolyzes phosphatidylinositol bisphosphate and forms diacylglycerol and inositol triphosphate (lP3). Finally, IP3 activates protein kinase C via an increase in in intracellular Ca2+.

Molecular biologists studying signal transduction apply an agent to human cells that activates G-protein- dependent phospholipase C. Which of the following intracellular substances is most likely to increase immediately after exposure to this agent? A. Ca2+ B. cAMP C. CGMP D. Cl- E. mRNA F. NO

*The answer is B.* The Southern, Western, Northern, and Southwestern blot procedures are powerful techniques used to analyze and identify DNA fragments, proteins, mRNA, and DNA-bound proteins, respectively. The same basic technique underlies all of the blot procedures. First, the unknown sample is separated by gel electrophoresis. Separation occurs based on a molecule's size and charge. The separated molecules form bends on the gel that are then blotted onto a nitrocellulose membrane and incubated with a labeled probe to identify the specific DNA fragment, RNA molecule, or protein of interest. Southwestern blots are used to identify and isolate proteins that bind DNA. In this technique, the target protein binds to a labeled, double~stranded DNA probe that is homologous to the protein's regulatory sequence. of the molecules listed, c-Jun is the only DNA-binding protein. c-Jun and c-Fos are nuclear transcription factors that directly bind DNA via a leucine zipper motif. The genes that code for c-Jun and c- Fos are proto-oncogenes, genes that can become oncogenes following a mutation or with constitutive expression. *Educational Objective:* Southwestern blotting is used to detect DNA-binding proteins such as transcription factors, nucleases, and histones.

Molecular biologists undertake a series of experiments designed to classify proteins involved in various intracellular signaling pathways. During one of the experiments, a protein mixture obtained from a cell culture is separated by gel electrophoresis end subsequently transferred to a filter membrane. Labeled double-stranded DNA probes ere then used to detect a specific protein of interest in the sample. Which of the following proteins is most likely to be detected by this method? A. Ras B. c-Jun C. β1-adrenoreceptor D. S-100 E. Adenylate cyclase

*The answer is B.* Only phosphodiesterase participates as a signaling molecule in the visual cycle of photoreceptor cells

Retinitis pigmentosa (RP) is a genetically heterogeneous disease characterized by progressive photoreceptor degeneration and ultimately blindness. Mutations in more than 20 different genes have been identified in clinically affected patients. Recent studies have mapped an RP locus to the chromosomal location of a new candidate gene at 5q31. One might expect this gene to encode a polypeptide required for the activity of a(n) A. receptor tyrosine kinase B. cGMP phosphodiesterase C. phospholipase C D. adenyl cyclase E. protein kinase C

*The answer is C.* Receptors are subdivided into four major categories: steroid, ion channel, enzyme-linked, and G-protein-linked. Steroid receptors are located in the cell cytoplasm or nucleus, while the other three receptor types are found on the cell surface. Enzyme-linked receptors are proteins that span the call membrane, extruding an extracellular terminal that binds to the corresponding growth factor. Once bound, the receptor protein configuration is changed, which triggers a cascade of events. Some of the enzyme-linked receptors are enzymes themselves, and some activate enzymes present in the cytosol. The differences between these two types of enzyme-linked receptors are as follows: *With Intrinsic Enzyme Activity (Receptor Tyrosine Kinase):* - Extracellular domain (binds the growth factor) - Transmembrane domain - Cytosolic domain (enzyme) - MAP-kinase - Receptor autophosphorylates and triggers phosphorylation of Rad protein - Ex. Growth factor receptors EGF, PDGF, FGF, etc. *With Intrinisic Enzyme activity (Tyrosine-Kinase Associated Receptor):* - Extracellular domain - Transmembrane domain - Cytosolic domain (lacks enzymatic activity) - JAK/STAT - Receptor activates JAKs, which phosphorylate STATs (signal transducers and activators of transcription) - Receptors for cytokinase, growth hormone, prolactin, IL2

Some transmembrane receptors employ Janus Kinase (JAK) to stimulate enzymes in the cytoplasm. Which of the following substances is most likely to utilize this pathway? A. Insulin B. Platelet-derived growth factor (PDGF) C. Growth hormone D. Atrial natriuretic peptide E. Progesterone F. Gamma-aminobutyric acid (GABA)

*The answer is A, B, and D.* Replication protein A attaches to single stranded DNA, which is characteristic of DNA strand breaks or stalled replication forks. RPA attracts a complex of protein kinases which include ATM. These phosphorylate p53 preventing its breakdown and thus increasing its concentration. p53 inhibits DNA replication, preventing the start of S phase, until damage is repaired. If the DNA damage is not repaired in a reasonable time p53 will trigger apoptosis. Cyclins and cyclin dependent kinases are not involved in this control.

The cell is not allowed to pass the cell cycle restriction point if DNA damage is detected. Which of the following proteins are involved in detection of DNA damage and inhibition of the cycle at the restriction point? Please select all that apply. (A) Replication protein A (RPA). (B) ATM (ataxia telangiectasia mutated) protein. (C) Cyclin D. (D) p53.

*The answer is A.* The diagram indicates that the receptor activates a trimeric G-protein associated with the inner face of the membrane and that the G-protein subsequently signals an enzyme catalyzing a reaction producing a second messenger. Receptors that activate trimeric G-proteins have a characteristic seven-helix transmembrane domain. The other categories of receptors do not transmit signals through trimeric G-proteins.

The diagram represents a signal transduction pathway associated with hormone X. The receptor for hormone X is most likely to be characterized as a(n) A. seven-helix transmembrane domain receptor B. intracellular receptor with a zinc-finger domain C. helix-turn-helix transmembrane domain receptor D. transmembrane receptor with a guanyl cyclase domain E. tyrosine kinase domain receptor

*The answer is D.* Growth factors can stimulate call proliferation by altering the expression of certain genes. This requires the use of signal transduction systems that can transfer the signal to the nucleus. Examples of such systems include: 1. Ras-MAP kinase pathway 2. PI3K/Akt/mTAR pathway 3. Inositol phospholipid pathway 4. cAMP pathway 5. JAK/STAT pathway The Ras-MAP kinase pathway starts with a growth factor ligand binding to the receptor tyrosine kinase, causing auto-phosphorylation of the receptor. Phosphotyrosine produced in this reaction interacts with a number of proteins (such as SH2-domain proteins and SOS protein), leading to Ras activation. Ras is a G-protein that exists in active and inactive forms. Inactive Ras contains GDP, while the active form is bound to GTP. Activated Ras begins a phosphorylation cascade starting with activation of Raf kinase. This cascade results in the activation of MAP (mitogen-activated protein) kinase, which enters the nucleus to influence gene transcription. The Ras protein exists in a balance between its active and inactive forms. Inactive (GDP-containing) Ras is activated by a signal originating from the receptor tyrosine kinase. Active (GTP-containing) Ras is inactivated by GAP (GTPase-activating protein), which induces the hydrolysis of GTP into GDP. Mutation of Ras can read to an inability to split GTP, the resultant permanently activated Ras stimulates cell proliferation and can lead to cancer.

The interaction of certain growth factors with their receptors leads to the following sequence of events: Binding of growth factor ligand Autophosphorylation of tyrosine residues Interaction with SOS protein Activation of "X" protein Activation of Raf kinase Activation of MAP kinase Gene transcription. Protein "X" becomes activated when It binds which of the following: (A) ATP (B) cAMP (C) IP3 (D) GTP (E) Ca2+

*The answer is B.* Cyclin-dependent kinases (Cdks) are the key enzymes that carry out the phosphorylations that regulate progression through the cell cycle. The activity of Cdks is closely regulated and varies continuously throughout the cycle, but the actual amounts of Cdks present do not vary very much. Instead, their activity is controlled by cyclins and by phosphorylation. Binding of cyclins and phosphorylation by activating kinases activates Cdks, and they may also be inactivated by Cdk inhibitor proteins.

The passage of a cell through the stages of the cell cycle is controlled by protein kinases that phosphorylate many different proteins at appropriate times. What are these protein kinases called? (A) Cdk activating kinases (B) Cyclin-dependent kinases (C) Cyclins (D) Tyrosine kinases

*The answer is E.* If the GTPAse activity of the Ras protein would be less active, Ras would remain active for longer periods and there would be more stimulation of the MAP kinase pathway.

The protein product of the Ras oncogene is a mutated Ras protein. All of the following would be true EXCEPT (A) The Ras protein is a G-protein and functions as an internal clock (B) G-proteins have evolved to stay active for a certain length of time (C) Ras protein is active in cell growth and division (D) Ras can mutate so that it is less active as a GTPase (E) A less active GTPase would mean less stimulation of the MAP kinase pathway

*The answer is B.* NF-kB is activated by the phosphorylation of IkB which becomes ubiquitinated and degraded in the proteasome. This unmasks the nuclear localization sequence (NLS) on NF-kB, allowing the import to the nucleus.

The transcription factor NF-kB is activated by a protein kinase that (A) phosphorylates NF-kB directly to activate it (B) phosphorylates the inhibitory factor IkB, causing it to be degraded and to release NF-kB (C) phosphorylates a MAP kinase to initiate its cascade (D) activates a phosphatase that removes an inhibiting phosphate from the NF-kB (E) phosphorylates a protease that then degrades IkB, releasing NF-kB

*The answer is C. Bax is a protein of the Bcl-2 family. This family of proteins all have a role in the apoptosis pathway. Some are pro-apoptotic and some anti-apoptotic and the balance between them determines whether or not apoptosis is activated via the intrinsic pathway. Bax and Bad are pro-apoptotic proteins. The protein which names the family, Bcl-2, is anti-apoptotic. The key event which triggers the start of the intrinsic pathway of apoptosis is the release of cytochrome c from the mitochondrion. Cytochrome c is therefore pro-apoptotic.

The triggering of the intrinsic pathway of apoptosis involves a balance between pro-apoptotic and anti-apoptotic proteins. Which of the following is anti-apoptotic? (A) Bax. (B) Bad. (C) Bcl-2. (D) Cytochrome c.

*The answer is E.* Although any of the listed options might be encoded by an oncogene, the "tyrosine kinase" description suggests it is likely to be a growth factor receptor.

Tumor cells from a person with leukemia have been analyzed to determine which oncogene is involved in the transformation. After partial sequencing of the gene, the predicted gene product is identified as a tyrosine kinase. Which of the following proteins would most likely be encoded by an oncogene and exhibit tyrosine kinase activity? A. Nuclear transcriptional activator B. Epidermal growth factor C. Membrane-associated G protein D. Platelet-derived growth factor E. Growth factor receptor

*The answer is B.* The intrinsic apoptotic pathway is tightly regulated in mammalian cells by two groups of proteins, one favouring apoptosis and the other inhibiting it. They all belong to a single protein family, the Bcl-2 family. Bax and Bad are pro-apoptotic members of the family and are believed to be involved in the initial formation of pores in the mitochondrial membrane. Bcl-2 (the first discovered member) and Bcl-xLare important anti-apoptotic proteins that prevent pore formation and cytochrome c release. Whether a stimulus that causes cell stress produces cell destruction will therefore depend on the balance between the pro-apoptotic and anti-apoptotic proteins.

What roles in regulating the intrinsic pathway of apoptosis are played by the Bcl-2 protein family members Bax and Bcl-2? (A) Bax inhibits apoptosis while Bcl-2 stimulates apoptosis. (B) Bax stimulates apoptosis while Bcl-2 inhibits apoptosis. (C) Both Bax and Bcl-2 inhibit apoptosis. (D) Both Bax and Bcl-2 stimulate apoptosis.

*The answer is C.* Ras is the key controlling protein of one of the message transduction pathways operating in response to growth factor /cytokine signals. It is a small monomeric GTPase time switch protein. It is activated when the active receptor binds a GDP-GTP exchange factor, bringing it close to the membrane where Ras is anchored. This causes the GDP bound to Ras to be exchanged for GTP. The change activates the Ras molecule, which can then activate a series of downstream kinases finally culminating in activation of a nuclear transcription factor. The activation of Ras is self-limiting because the protein is a GTPase. It can hydrolyse its bound GTP to GDP, which returns it to the inactive state. However, it works only slowly so that it stays in the active form for a short time period - in fact it works as a time switch. G-proteins are trimeric proteins containing a subunit with GTPase activity. They carry out a role similar to Ras in a different message transduction system.

What type of protein is Ras? (A) A tyrosine kinase (B) A serine-threonine kinase (C) A small monomeric GTPase switch protein (D) A G protein switch

*The answer is A.* The Ras protein, activated by growth factor receptor, activates a series of protein kinases which finally phosphorylate and activate a transcription factor (or factors) in the nucleus. Ras itself is anchored to the inside of the plasma membrane by means of a long chain lipid (hydrophobic) group covalently bound to it. Ras is a GTPase switch protein which is active when it has GTP bound to it, but in an inactive form when GDP is bound. Because it has GTPase activity Ras will normally be in inactive (GDP bound) form. An activated growth factor receptor activates Ras by first binding an adaptor protein, GRB. GRB then itself binds the GDP-GTP exchange protein, SOS, locating it close to the membrane where it can come into contact with Ras and catalyse the removal of GDP from Ras and its replacement with GTP. This activates Ras for a time, until the intrinsic GTPase activity of Ras hydrolyses the GTP back to GDP, turning the process off.

When it activates the Ras signal transduction pathway, the function of the activated receptor is to locate a GDP-GTP exchange protein close to the membrane where the Ras protein is located. (A) True (B) False

*The answer is D.* An oncogene is a mutated form of a normal cellular gene (the proto-oncogene) which codes for a protein which is either controlled abnormally so that it is expressed in abnormally large amounts or has gained activity so that it is more active than the normal protein. In either case only one gene of a pair needs to become oncogenic for the activity to be expressed, so the mutation is dominant. Some oncogenes code for a protein which forms part of a signal transduction pathway. A 'gain of function' mutation in such a gene can make the pathway more active. Other oncogenes code for a protein which protects the cell against apoptosis. Increased production of this protein can prevent the cell entering apoptosis. However either of these examples does not give a full definition of an oncogene. Cell cycle control proteins prevent cell division and are anti-oncogenic.

Which of the following best defines an oncogene? (A) An oncogene codes for a cell cycle control protein. (B) An oncogene codes for a mutated form of a protein that forms part of a signal transduction pathway. (C) An oncogene codes for a protein that prevents the cell from undergoing apoptosis. (D) An oncogene is a dominantly expressed mutated gene that gives a cell a growth or survival advantage.

*The answer is A, B, and C.* Carcinogenic (potentially cancer causing) chemicals can be present in some foods. These chemicals may be mutagenic in their native form or be converted to mutagens in the body. Ionising and uv radiation can both cause DNA damage. Reactive oxygen species which are generated in cells from molecular oxygen can also modify DNA, causing mutations. Retroviruses (HIV is a retrovirus with a single stranded RNA genome) are known to introduce mutated genes into cells, contributing to some cancers, and study of this has been important in developing our understanding of cancer and HIV is associated with development of cancers. However, there is no evidence that HIV virus introduces mutations into cells. Its contribution to development of cancers is probably that it suppresses the immune response, making immune surveillance less effective.

Which of the following can cause mutations which contribute to development of cancers? Please select all that apply. (A) Chemicals in food. (B) UV and ionising radiation. (C) Reactive oxygen species. (D) HIV virus.

*The answer is A.* Cyclic AMP binds to a tetrameric enzyme, protein kinase A, which is inactive because active sites on the two catalytic subunits are masked by regulatory subunits. Cyclic AMP binding causes the molecule to dissociate, releasing monomeric subunits in which the active site is unmasked. Protein kinase B is activated by phosphatidyl inositol 3,4,5 triphosphate. Protein kinase C is activated by 1,2 diacyl glycerol, aided by increased cytoplasmic Ca++ concentration caused by inositol triphosphate. Inositol triphosphate and diacyl glycerol are the two messengers produced when phosphatidyl inositol 4,5 bis phosphate is split by phospholipase C, so both messengers are released at the same time, and, in this instance, work synergistically. G-protein receptor kinase is a control protein which phosphorylates the receptor protein which causes activation of adenylate cyclase. This helps desensitise the receptor since, when it has been phosphorylated, an inhibitory protein β-arrestin can bind to it.

Which of the following enzymes is activated by cyclic AMP, passing on the hormonal signal? (A) Protein kinase A. (B) Protein kinase B. (C) Protein kinase C. (D) G protein receptor kinase.

*The answer is A, B, and C.* In order to grow and divide most somatic cells must be stimulated by signalling molecules called growth factors. Cancer cells acquire mutations which stimulate them to grow without this stimulation. Cells with this type of mutation are said to be transformed. Normal cells can only divide a limited number of times and then senesce and die by apoptosis. Cancer cells avoid this limitation, and can replicate an unlimited number of times. When the DNA of normal cells becomes damaged, cell division is halted until the DNA is repaired. If a repair cannot be carried out the cell is pushed into apoptosis. All cancer cells have damaged DNA. This can be extensive and may include loss of chromosomes, but this does not halt their cell cycle and does not cause apoptosis, and they continue to grow and divide. However, cancer is not infectious, and cancer cells do not cause other cells to become cancerous.

Which of the following is a characteristic of a cancer cell? Please select all that apply. (A) Replicates an unlimited number of times. (B) Grows and divides without stimulation by a growth factor. (C) DNA damage does not halt cell division or stimulate apoptosis. (D) Releases factors which cause nearby cells to become cancerous.

*The answer is A.* Tumours which remain at a single site can usually be dealt with clinically and are therefore regarded as benign. However, even the cells of benign tumours can grow without receiving a growth signal, and can divide indefinitely. A cell mass cannot exceed a diameter of about 2mm unless it develops blood vessels to supply food and oxygen, so benign tumours will also develop blood vessels (angiogenesis). The key point at which a tumour becomes malignant is when cells break free from the initial tumour and spread, via the blood and lymphatic system, to establish themselves in other sites, forming secondary tumours. This is the process of metastasis.

Which of the following is characteristic of a malignant rather than a benign tumour? (A) Undergoes metastasis. (B) Develops a blood supply. (C) Cells divide an unlimited number of times. (D) Grows without needing a growth signal.

*The answer is B.* The protein encoded by the retinoblastoma gene (Rb), like p53 also blocks the cycle at the G1 checkpoint. Rb is not itself a transcription factor. It combines with a transcription factor called E2F and prevents E2F from activating genes required for progression into S phase. In normal cells, when a mitogenic stimulus is received, the genes for G1 cyclin synthesis are activated so that the Cdk activity increases. The Cdk phosphorylates and inactivates the Rb protein, preventing it from combining with E2F so that the cycle can advance to S phase. However if the cycle is proceeding in an aberrant way due to an oncogene bypassing the need for a mitogenic signal, since the normally required Cdks are not active, the Rb is not phosphorylated and remains bound to E2F, arresting the cycle. Rb if present and active can therefore restrain the cell cycle from proceeding abnormally. Retinoblastoma gene defects are known to be associated with certain types of cancer.

Which of the following statements about the Rb tumour suppressor protein is correct? (A) Rb is activated when phosphorylated by Cdk. (B) Rb binds the transcription factor E2F and thus prevents the cell from entering S phase until a mitogenic signal is received. (C) Rb is a transcription factor. (D) When a mitogenic signal is received, Rb binds the transcription factor E2F and thus stimulates the cell to enter S phase

*The answer is A.* An oncogene is derived from a normal cellular gene (the proto-oncogene) which becomes mutated in such a way that the protein product of the gene is either continuously active or more active than the proto-oncogene product. This is not possible if the product of the mutated gene is an inactive protein.

Which of the following types of mutation cannot give rise to an oncogene? (A) Addition or deletion of a base producing a nonsense message and an inactive protein product. (B) A point mutation changing just one amino acid in the protein product. (C) A translocation which puts the gene under the control of a strong promoter, producing over-expression. (D) A point mutation which produces a stop codon, prematurely terminating the message and producing a truncated protein.

*The answer is D.* Tumour-suppressor genes code for proteins that normally control progression through a cell cycle checkpoint. These genes only contribute to a cancer when both copies are mutated in a way that produces an inactive protein product. A single mutation inactivating one gene only has no effect. Tumour-suppressor genes contribute to cancer because the absence of the protein product removes controls on cell division. When genes coding for proteins forming part of a growth factor signalling pathway, or coding for proteins which help prevent apoptosis become mutated and contribute to cancers, they become oncogenes. Oncogenes produce proteins that have gained a function. They contribute to cancers by stimulating cell growth, or by preventing a damaged cell from killing itself. Mutations in genes coding for DNA repair enzymes contribute to cancers in the same way that tumour-suppressor genes do, because in their absence damaged DNA cannot be repaired, but they are not conventionally regarded as tumour suppressors.

Which of the following types of protein could be coded by a tumour-suppressor gene? (A) A protein that forms part of a growth factor signaling pathway. (B) A protein that codes for a DNA repair enzyme. (C) A protein that helps prevent apoptosis. (D) A protein that controls progression through the cell cycle.

*The answer is B.* The p53 gene is a tumour suppressor gene which is inactivated in at least 50% of all cancers. Its role is to prevent the replication of cells with damaged DNA. p53 protein is normally present at very low levels in the cell, since it is broken down very quickly after its synthesis. However, proteins that recognize lengths of single stranded DNA formed by strand breaks or stalled transcription, phosphorylate p53 and stabilize it, increasing its concentration. p53 protein is a transcription factor, but it causes the production of proteins which halt the cell cycle. It also stimulates production of DNA repair enzymes, but not the ones that replace telomere sequences lost during cell division. The DNA repair enzymes produced give the cell the opportunity to repair its damaged DNA, but if it fails to do so in a reasonable time, p53 protein stimulates production of proteins which push the cell into apoptosis.

Which property of p53 enables it to prevent the development of cancer? (A) p53 is a transcription factor that causes production of proteins that stimulate the cell cycle. (B) p53 prevents the replication of cells with damaged DNA. (C) p53 prevents cells from triggering apoptosis. (D) p53 stimulates synthesis of DNA repair enzymes that replace telomere sequence lost during cell division.

*The answer is D.* Most intracellular second messengers work via protein kinases. Cyclic AMP, cyclic GMP and 1,2 diacyl glycerol all activate specific protein kinase enzymes (protein kinases A, G and C respectively for these three second messengers). Inositol triphosphate is different. It is a water soluble molecule and is released into the cytosol, where it binds to a Ca++ channel protein in the endoplasmic reticulum, opening the channel. Ca++ is maintained at a very low concentration in the cytosol (about 10-7M) but in the endoplasmic reticulum its concentration is around 10-3M. When a Ca++channel opens through the endoplasmic reticulum membrane, Ca++ is released into the cytosol, raising the concentration to around 10-5M. This increase is detected by Ca++ binding proteins, the most common of which is calmodulin. It undergoes a large conformational shape change on binding Ca++, allowing it to activate other proteins with which it is closely associated.

Which second messenger signals the release of Ca++ from the endoplasmic reticulum? (A) Cyclic AMP (B) Cyclic GMP (C) 1,2 diacyl glycerol (D) Inositol triphosphate

*The answer is C. The patient exhibits the Philadelphia chromosome (a translocation between chromosomes 9 and 22), which creates the bcr-abl protein, a fusion protein of two normal cellular proteins. The abl protein is a tyrosine kinase; when the gene is moved from chromosome 9 to 22 and put under the control of the bcr gene, it is misexpressed and constitutively active and leads to cellular growth in the blood cells in which the protein is expressed. The patient has chronic myelogenous leukemia (CML). Since this is a dominant activity, there is no loss of a tumor suppressor gene. The translocation does not interfere with DNA repair, so there is no direct increase in mutation rate due to this translocation. There is no loss of tyrosine kinase activity (there is actually a gain of activity), nor is there gain of a ser/thr kinase (as abl is a tyrosine kinase).

You have a patient with an elevated white blood cell count and a feeling of malaise. Molecular analysis of the white cells detects the presence of the following karyotype. The molecular basis of this disease is which of the following? (A) Loss of an essential tumor suppressor activity (B) Increased rate of DNA mutation due to loss of DNA repair enzymes (C) Creation of a fusion protein not normally found in cells (D) Loss of a critical tyrosine kinase activity (E) Gain of a critical ser/thr kinase activity

*The answer is C.* Myasthenia gravis is an autoimmune disease where antibodies attack acetylcholine receptors. Although enough acetylcholine is released, the neurotransmitter cannot bind to the receptor. As a result, there is trouble depolarizing the postsynaptic membrane. This disease can be treated with drugs that inhibit

Your patient has myasthenia gravis. You might tell her all of the following EXCEPT (A) She has an autoimmune disease (B) She does not have enough acetylcholine receptors (C) She has too much acetylcholine released (D) She has trouble depolarizing the postsynaptic membrane (E) Drugs that inhibit acetylcholine esterase can help her

*The answer is E.* The most common mutation in cystic fibrosis is the 3-base pair deletion in the CFTR gene that removes phenylalanine at amino acid position 508. This removes phenylalanine at the 508th position of the CFTR protein and causes defective post-translational folding and glycosylation. Cystic fibrosis is the most common lethal genetic disease of the white population. It is caused by an autosomal recessive defect in the CFTR gene on chromosome 7. The mutation causes defective post-translational folding and glycosylation, resulting in degradation of the CFTR channel before reaching the cell surface. The CFTR gene codes for the CFTR channel which actively secretes Cl- in the lung and GI tract and actively reabsorbs Cl- from sweat. This defective Cl- channel causes secretion of abnormally thick mucus that plugs the lungs, pancreas, and liver and leads to the variety of symptoms seen in CF including recurrent pulmonary infection, pancreatic insufficiency, and meconium ileus in newborns.

A 1-year-old child who was born outside of the United States is brought to a pediatrician for the first time because she is not gaining weight. Upon questioning, the pediatrician learns that the child has had frequent pulmonary infections since birth, and on exam the pediatrician appreciates several nasal polyps. Genetic testing is subsequently ordered to confirm the suspected diagnosis. Testing is most likely to show absence of which of the following amino acids from the protein involved in this child's condition? (A) Histidine (B) Leucine (C) Lysine (D) Valine (E) Phenylalanine

*The answer is E.* The boy has Type 1 diabetes, and is producing no insulin. One of insulin's effects is to stimulate the translocation of GLUT4 transporters from internal vesicles to the plasma membrane of muscle and fat cells. The increase in glucose transport molecules on the cell surface is important for rapidly reducing blood glucose levels. The GLUT4 transporter is for facilitative diffusion, and is not dependent on an ion gradient across the membrane for effective transport, as are the glucose transporters in the small intestine.

A 12-year-old boy is admitted to the hospital in ketoacidosis with a blood glucose level of 700 mg/dL (normal fasting levels are between 80 and 100 mg/dL). The boy is shown to have no detectable C-peptide upon further testing. A potential reason for the elevated blood glucose is which one of the following? (A) A lack of a sodium gradient across cellular membranes (B) A lack of a calcium gradient across cellular membranes (C) A lack of a chloride gradient across cellular membranes (D) A reduced number of glucose transport molecules in the brain membrane (E) A reduced number of glucose transport molecules in the muscle membrane (F) A reduced number of glucose transport molecules in the liver membrane

*The answer is C.* The mitochondria are organelles of fuel oxidation and ATP generation. Lysosomes contain hydrolytic enzymes that degrade proteins and other large molecules. The Golgi form vesicles for transport of molecules to the plasma membrane and for secretion. The nucleus carries out gene replication and transcription of DNA.

A 12-year-old boy is admitted to the hospital in ketoacidosis with a blood glucose level of 700 mg/dL (normal fasting levels are between 80 and 100 mg/dL). The boy is shown to have no detectable C-peptide upon further testing. The child is treated appropriately such that the glucose levels have been reduced, and he does not become dehydrated. Once glucose is transported into his cells, which organelle is responsible for generating energy from the oxidation of glucose to carbon dioxide and water? (A) Lysosome (B) Golgi complex (C) Mitochondria (D) Nucleus (E) Peroxisome

*The answer is A.* The boy has hereditary spherocytosis, which is due to a mutation in a red blood cell cytoskeletal protein. The most common mutation is in spectrin, although mutations in ankyrin, band 3, and protein 4.2 can also lead to this phenotype. Owing to the mutation in the cytoskeletal protein, the membrane shape becomes spherical instead of concave. This leads to the removal of the spherical cells by the spleen, leading to both anemia and splenomegaly. Mutations in the proteins in the plasma membrane or the endoplasmic reticulum will not lead to this disorder. Red blood cells do not have a nucleus or mitochondria.

A 12-year-old boy is displaying tiredness and lethargy, and is found to have a hypochromic, microcytic anemia. Microscopic examination of the boy's red blood cells demonstrated a spherical shape, rather than concave. The mutation in this child is most often found in a protein located in which part of the red blood cell? (A) The cytoskeleton (B) The nucleus (C) The mitochondria (D) The endoplasmic reticulum (E) The plasma membrane

*The answer is D.* Cytokines are the messengers of the immune system. Neuropeptides and biogenic amines (small-molecule neurotransmitters) are messengers of the nervous system. Steroid hormones are messengers of the endocrine system. Amino acids (such as glycine and glutamate) can act as mediators within the nervous system. Once the shot is given, immune cells secrete cytokines to induce the synthesis of antibodies against the antigens injected into the girl.

A 15-month-old girl has been given an MMR immunization. Which of the following chemical messengers is responsible for the body's ability to mount an immune response to this vaccination? (A) Neuropeptides (B) Biogenic amines (C) Steroid hormones (D) Cytokines (E) Amino acids

*The answer is D.* The patient is exhibiting complete androgen insensitivity syndrome (CAIS), an inability to respond to androgens due to a lack of the androgen receptor. Thus, while genotypically a male, and producing normal levels of male hormones, the body cannot respond to the hormones, and female sexual characteristics develop. Androgen synthesis remains normal, however, which leads to high circulating testosterone levels in the patient.

A 17-year-old girl was seen by her family physician due to a lack of menses. On physical exam the girl appeared to be well-developed, but with a striking lack of pubic hair. Blood work indicated very high levels of testosterone, with elevated estradiol levels as well. A karyotype of the patient is shown below. The underlying biochemical defect in this patient is which of the following? (A) Inability to synthesize estrogen (B) Lack of estrogen receptor (C) Lack of cortisol receptor (D) Lack of androgen receptor (E) Inability to synthesize aldosterone

*The answer is D.* The patient's ambiguous secondary sex characteristics and lack of menstrual activity suggest the possibility of an androgen resistance syndrome. The male karyotype and blood testosterone levels confirm this. This clinical condition might have arisen as a result of steroid 5α-reductase deficiency or inherited defects in the androgen receptor (testicular feminization).

A 19-year-old woman has been referred to an endocrinologist by her gynecologist because of delay in the initiation of her menstrual periods. Physical examination reveals underdeveloped breasts, an enlarged clitoris (rudimentary penis), and the presence of small masses within the labia majora. Blood testosterone is in the normal range for males and a chromosome spread indicates a karyotype of 46,XY. This patient most likely has a defect in signaling through a pathway involving which of the following? A. Cyclic AMP-dependent protein kinase (PKA) B. Protein kinase C (PKC) C. A cell-surface tyrosine kinase receptor D. A nuclear receptor E. A heterotrimeric G protein

*The answer is C.* Many cytokine receptors contain multiple subunits, and some subunits are shared by multiple receptors. The boy has X-linked severe combined immunodeficiency syndrome (SCID), which is a defect in a cytokine receptor component, the γ-chain, common to a large number of cytokine receptors. Lack of this component renders most cytokine receptors inoperable, leading to lack of response to cytokines and inability to mount defenses against an immune attack. The defect is not in the TGF-β/SMAD pathway, nor in the JAK/STAT family of proteins. It is also not related to a ubiquitin ligase, which targets proteins for proteolysis by the proteasome.

A 2-year-old boy had frequent infections and a number of pneumonia episodes. Blood work showed greatly reduced levels of both T and B lymphocytes. X-rays showed normal thymus development. The primary mutation in this patient is likely to be which of the following? (A) Mutated JAK proteins (B) Mutated STAT proteins (C) Defective cytokine receptor component (D) Mutated SMAD proteins (E) Mutated ubiquitin ligase

*The answer is C.* The patient has presented with classic symptoms of bipolar disorder. Lithium is a first-line treatment of bipolar disorder whose mechanism of action is to interrupt the PI cycle by blocking the action of inositol monophosphatases, the enzyme that converts inositiol phosphate to free inositol, such that phosphatidylinositol can be resynthesized from CDP-diacylglycerol and inositol. Through an interruption in the cycle, the key PI-cycle second messengers cannot be continually generated, leading to a reduction in signaling capabilities.

A 21-year-old patient is being evaluated for a major depressive disorder. During the interview, he admits to having several episodes in the past of feeling "on top of the world," able to function very well with only 4 hours of sleep per night, maxing out his credit cards (a very unusual characteristic for him), and "indiscriminate" sexual encounters with multiple partners. The elemental medication most commonly used to treat this patient's disorder will lead to the accumulation of which one of the following compounds? (A) Inositol (B) Phosphatidylinositol (C) Inositol phosphate (D) Inositol bisphosphate (E) Inositol trisphosphate (F) Diacylglycerol

*The answer is B.* The patient is experiencing an anxiety disorder and panic attacks. These symptoms are often treated with psychotherapy and benzodiazepams. Patients with anxiety disorders have low gamma aminobutyric acid (GABA). Benzodiazepams, such as diazepam, increase the efficiency of the synaptic transmission of GABA, helping to make any existing GABA more efficacious. This is through the drug binding to GABA receptors such that when GABA binds to the receptor, the response to GABA is greater than in the absence of the drug (one effect is to leave chloride channels open for greater periods of time in response to GABA, thereby depolarizing the membrane and sending an inhibitory signal). GABA is the chief inhibitory neurotransmitter. GABA is a biogenic amine or "small-molecule" neurotransmitter, and is derived from the decarboxylation of glutamate. Neuropeptides are the other type of chemical messenger secreted by the nervous system, and are usually small peptides. Cytokines are small protein messengers of the immune system. G-proteins aid in transmitting the signals induced by proteins that bind to heptahelical receptors (such as the epinephrine or glucagon receptors). GABA does not transmit its signal through a G-protein.

A 25-year-old female presents with intense fear whenever she has to drive her car through a tunnel. She feels faint, sweats profusely, has palpitations, and hyperventilates. She is prescribed diazepam to reduce her symptoms. The type of chemical messenger enhanced by this treatment is best described as which one of the following? (A) Neuropeptide (B) Biogenic amine (C) Large-molecule neurotransmitter (D) Cytokine (E) G-protein

*The answer is D.* The pathologist has obtained a prion disease, which results from the prion protein adopting an alternative conformation and precipitating in neural tissue. The normal prion protein is a GPI-anchored protein.

A 57-year-old pathologist, who had often cut himself while performing autopsies, develops blurred vision, dementia, personality changes, and muscle twitching in a very short period of time. The protein that is leading to these behavioral changes is best described as which one of the following? (A) A soluble, cytoplasmic protein (B) A peripheral membrane protein (C) An embedded membrane protein (D) A GPI-anchored membrane protein (E) A secreted protein

*The answer is A.* Infection by the Epstein-Barr virus will lead to the synthesis of a Bcl-2-like factor that antagonizes apoptosis, and allows the virus-infected cells to survive and continue producing more viruses. This factor is not present in the other viruses listed as potential answers.

A 29-year-old female presents with chronic fatigue for the past 9 months. She did have mononucleosis in the past, and blood work reveals a chronic viral infection. An analysis of a liver biopsy indicated that when placed under conditions in which apoptosis should be initiated, the cells continued to grow. The viral infection was most likely caused by which one of the following? (A) Epstein-Barr virus (B) Influenza virus (C) Simian sarcoma virus (D) Polio virus (E) Herpes simplex virus

*The answer is C.* The patient has achondroplasia, the most common form of short limb dwarfism in humans. This disorder is due to constitutive activation of FGFR3 (a receptor for FGF), which particularly plays havoc with the cartilaginous growth plate in limbs. Full details of how constitutive activation of this pathway leads to all of the phenotypic effects of this form of dwarfism are not yet fully appreciated. This disorder is not related to alterations in the EGF, TGF-β, or JAK/STAT pathways; it is unique to constitutive activation of the FGF pathway via FGF receptor 3.

A 3-month-old child of normal parents presents with shortened arms and legs, a large head with a prominent forehead, a curved lower spine, and bowed lower legs. The child is in the fifth percentile for height. The child's condition is most likely the result of which of the following? (A) Constitutive activation of the EGF pathway in bone (B) Inactivation of the TGF-β pathway in bone (C) Constitutive activation of the FGF pathway in bone (D) Inactivation of the FGF pathway in bone (E) Constitutive activation of the JAK/STAT pathway in bone

*The answer is C.* The infant has Pompe disease, a loss of liver α-glucosidase activity. This is glycogen storage disease II. The finding of normal glycogen structure eliminates liver debranching and branching activities as being deficient. The missing enzyme is a lysosomal enzyme, and nondegraded glycogen accumulates in the lysosome, interfering with lysosomal function (hence, a lysosomal storage disease). The malfunctioning of the lysosomes is what leads to the muscle and liver problems. A defect in glycogen phosphorylase (liver) would lead to fasting hypoglycemia, and an enlarged liver, but not the muscle problems exhibited by the child. A defect in glycogen synthase would also lead to fasting hypoglycemia, but would not lead to severe muscle and liver disease. Additionally, in an individual with a defect in glycogen synthase, glycogen would not be found in the liver biopsy since it could not be formed.

A 3-month-old infant was brought to the pediatrician due to muscle weakness (myopathy) and poor muscle tone (hypotonia). Physical exam revealed an enlarged liver and heart, and heart failure. The infant had always fed poorly, had failure to thrive, and had breathing problems. He also had trouble holding up his head. Blood work indicated early liver failure. A liver biopsy indicated that glycogen was present and of normal structure. A potential defect in this child is which of the following? (A) Liver glycogen phosphorylase (B) Liver glycogen synthase (C) Liver α-glucosidase (D) Liver debranching enzyme (E) Liver branching enzyme

*The answer is C.* The patient has Fifth disease, a viral illness caused by parvovirus B19. The "slapped cheek" appearance of this rash is very distinctive. The eicosanoids control cellular function in response to injury (in this case, a viral infection). In response to the viral infection, vascular endothelial cells will secrete prostaglandins that act on smooth muscle cells to cause vasodilation, which leads to the reddish appearance of the infected individual. The release of eicosanoids may also be responsible for the fever that sometimes accompanies Fifth disease. Neuropeptides, cytokines, steroid hormones, or amino acids are not responsible for the vasodilation that occurs in this disease.

A 3-year-old boy presents with 3 days of a low-grade fever, joint pain, and a "lacy-" appearing rash on his arms and legs. His rash began on his face and he appeared to have "slapped cheeks." The chemical messenger that caused the symptoms (vasodilatation presenting clinically as a "rash") can be classified as which one of the following? (A) Cytokine (B) Neuropeptide (C) Eicosanoid (D) Steroid hormone (E) Amino acid

*The answer is C.* The patient has hyperthyroidism, or Grave disease, an overproduction of thyroid hormone. Thyroid hormone is derived, in part, from tyrosine, which is iodinated to produce the active forms of thyroid hormone, T3 and T4. Cholesterol is a precursor to steroid hormones. Arachidonic acid is a precursor to eicosanoids (prostaglandins). Tryptophan is a necessity in the production of serotonin, and glutamate is needed to produce GABA. The symptoms described do not occur if there is an overproduction of steroid hormones, eicosanoids, serontonin, or GABA.

A 30-year-old female presents with a 15-pound weight loss over 1 month, heat intolerance, tachycardia, tremor, bilateral exophthalmos, and a mass in the anterior neck. The hormone overproduced in this condition requires which one of the following? (A) Arachidonic acid (B) Cholesterol (C) Tyrosine (D) Tryptophan (E) Glutamate

*The answer is B.* Hereditary breast cancer is due to inherited mutations in either of the tumor suppressor genes BRCA1 or BRCA2. These genes encode proteins that play important roles in DNA repair (primarily single- and double-strand breaks), and it is the loss of this function that predisposes the patient to breast and ovarian cancers. The inability to repair the breaks in the backbone leads to errors during replication, and mutations will develop that eventually lead to a loss of growth control. This is a loss-of-function disorder, which characterizes the genes involved as tumor suppressors. Inheriting one mutated copy of BRCA1 means that the other, normal copy of BRCA1 must be lost in a particular cell in order to initiate the disease (loss of heterozygosity). For breast cancer, this occurs 85% of the time (penetrance upon inheriting a BRCA1 or BRCA2 mutation). An oncogene is a dominant gene, so only one mutated copy can bring about the disease. BRCA1 or BRCA2 mutations do not directly lead to a loss of apoptosis, or to a constitutively active MAP kinase pathway.

A 32-year-old female has developed breast cancer. Her mother and one maternal aunt had breast cancer and her maternal grandmother had ovarian cancer. Which of the following best describes the mechanism behind this inherited problem? (A) A tumor suppressor leading to loss of apoptosis (B) A tumor suppressor leading to an inability to repair DNA (C) A tumor suppressor leading to a constitutively active MAP kinase pathway (D) An oncogene leading to loss of apoptosis (E) An oncogene leading to an inability to repair DNA (F) An oncogene leading to a constitutively active MAP kinase pathway

*The answer is B.* The patient has myasthenia gravis, in which she generates antibodies against the acetylcholine receptor. Treatment with edrophonium chloride leads to a transient increase in acetylcholine levels (through the temporary inactivation of acetylcholinesterase) such that acetylcholine can bind to receptors (via competition with the antibodies). Normal levels of acetylcholine are too low for such competition to be successful. This disorder does not generate antibodies which lead to acetylcholine destruction, inhibition of acetylcholinesterase, inhibition of acetylcholine synthesis, or release of acetylcholine at the synapse.

A 37-year-old female has trouble keeping her eyes open and swallowing and is beginning to slur her speech. The patient has also noticed a weakness in her arms and legs. Treatment with edrophonium chloride results in a temporary relief of symptoms. The underlying etiology of this disorder involves auto-antibodies that do which of the following? (A) Destroy acetylcholine (B) Block acetylcholine receptors (C) Inhibit acetylcholinesterase (D) Inhibit acetylcholine synthesis (E) Stimulate acetylcholine release into the synapse

*The answer is A.* Many classes of drugs, including some antihistamines (e.g., Benadryl), some antipsychotics (e.g., olanzapine), tricyclic antidepressants (e.g., amitriptyline), and atropinelike drugs (e.g., atropine, scopolamine) have anticholinergic effects or side effects, and function as antagonists to the acetylcholine receptor. Muscarinic acetylcholine receptors act through G-protein activation, whereas nicotinic acetylcholine receptors act as an ion channel, allowing sodium to flow through the receptor once it has been activated. The drug overdose in this case is inhibiting the muscarinic receptors, which occur in the autonomic and central nervous symptoms. This is a very typical case of anticholinergic overdose, with the "classic" symptoms classified as "blind as a bat," "dry as a bone," "red as a beet," "mad as a hatter," and "hot as a hare." The overdose is not affecting the nicotinic acetylcholine receptors, or receptors for GABA, serotonin, or catecholamines.

A 39-year-old man is brought to the ER for a suspected suicide attempt. He has blurred vision; very dry, hot, red skin; dry mouth; urinary retention; confusion; hallucinations; loss of balance; and tachycardia. Emergency medical technicians (EMTs) found an empty bottle of amitriptyline in his apartment. The date on the bottle was just 1 week ago, yet all the pills were missing. The effects the man is experiencing is due to an inhibition of which of the following processes? (A) Muscarinic acetylcholine receptor signaling (B) Nicotinic acetylcholine receptor signaling (C) GABA signaling (D) Serotonin signaling (E) Catecholamine signaling

*The answer is D.* The boy is exhibiting the symptoms of cystic fibrosis, which is due to a mutation in the CFTR. The CFTR is required for chloride transport across the membrane, is activated by phosphorylation by the cAMP-activated protein kinase, and when activated allows chloride to flow down its electrochemical gradient. A defective CFTR also alters the ion composition of mucus, reducing its ability to absorb water through osmosis, leading to the drying of mucus in various ducts and tissues, including the lung cells. The lung cells normally secrete a thin, watery mucus designed to trap small particles, which are moved through the lung so they can be swallowed or removed by coughing. When water cannot leave the lung cells, the mucus dries out, leading to pulmonary dysfunction due to clogged bronchi.

A 4-year-old boy has had multiple episodes of pneumonia, steatorrhea, and has fallen off his normal growth curve. A sweat test was positive for chloride ions. The reason this boy is at risk for repeat episodes of pneumonia is which one of the following? (A) Elastase destruction of lung cells (B) Defective α1-antitrypsin activity (C) Excessive water in the lungs (D) Dried mucus accumulation in the lungs (E) Loss of lung cells due to a defect in DNA repair

*The answer is A.* Cholera is caused by Vibrio cholerae, found in fecally contaminated food or water and in shellfish. Cholera toxin, which is composed of multiple subunits, utilizes some subunits to allow one particular subunit with enzymatic activity to enter the intestinal epithelial cell. This toxin catalyzes the ADP-ribosylation of Gαs, inhibiting the GTPase activity of the α subunit of the G-protein, leading to constitutive activation of adenylate cyclase, and high cAMP levels. This leads to the activation of ion channels, having potassium, sodium, and chloride ions leave the intestinal epithelial cells into the lumen, along with water, leading to the watery diarrhea. The treatment consists of rehydration with electrolytes. Owing to the volume of water lost, the disease is usually self-limiting, as the bacteria causing the disorder are washed out of the intestine.

A 4-year-old boy went to the beach with his parents, and they found some clams, which they later ate for dinner. A few hours later, the boy developed a fever, started vomiting, and had profuse watery diarrhea. After being taken to the emergency room (ER), the boy was treated for potential dehydration, and recovered uneventfully. The root molecular cause of his symptoms was which one of the following? (A) ADP-ribosylation of a Gαs protein (B) Phosphorylation of a Gαs protein (C) Acetylation of a Gαs protein (D) ADP-ribosylation of a Gαi protein (E) Phosphorylation of a Gαi protein (F) Acetylation of a Gαi protein

*The answer is C.* The boy has contracted cholera. The bacterium which causes cholera (Vibrio cholerae) produces a toxin that ADP-ribosylates a stimulatory G protein, leading to permanent activation of the G protein (via inhibition of the intrinsic GTPase activity). This leads to constant cAMP production, which in turn leads to the secretion of sodium and water into the lumen of the intestine, producing watery diarrhea. If not treated, dehydration rapidly results, and death can occur.

A 4-year-old boy went to the shore with his family and ate clams for dinner, but shortly thereafter developed severe watery diarrhea, vomiting, and leg cramps. The molecular basis of his disorder is which of the following? (A) Inhibition of protein synthesis (B) Inhibition of a stimulatory G protein (C) Activation of a stimulatory G protein (D) Inhibition of an inhibitory G protein (E) Activation of an inhibitory G protein

*The answer is B.* In cystic fibrosis, at the 508th position of the CFTR protein and causes defective post-translational folding and glycosylation.

A newborn is found to have cystic fibrosis during routine newborn screening. The parents, both biochemists, are curious about the biochemical basis of their newborn's condition. The pediatrician explains that the mutation causing cystic fibrosis affects the CFTR gene which codes for the CFTR channel. Which of the following correctly describes the pathogenesis of the most common CFTR mutation? (A) Insufficient CFTR channel production (B) Defective post-translational glycosylation of the CFTR channel (C) Excess CFTR channel production (D) Defective post-translational hydroxylation of the CFTR channel (E) Defective post-translational phosphorylation of the CFTR channel

*The answer is D.* The woman has myasthenia gravis, which is due to autoantibodies directed against the acetylcholine receptor. As such, acetylcholine stimulation of muscle cells is decreased, owing to a reduced number of functional acetylcholine receptors at the neuromuscular junction. One way to treat this condition is to inhibit acetylcholinesterase, the enzyme that degrades acetylcholine at the neuromuscular junction. By keeping the levels of acetylcholine high at the junction, there is a greater probability that the receptors that are active are occupied, and the signal from the neuron can be transmitted. Inhibiting the production of acetylcholine would exacerbate the problem, as would stimulating the production of immune cells (more autoantibodies would potentially be generated). Epinephrine is not involved at the neuromuscular junction, and stimulation of catechol-O transferase is a mechanism to inhibit the action of catecholamines in nonneuronal tissues, and does not contribute to the progression of myasthenia gravis.

A 42-year-old woman has slowly developed an inability to keep her eyes open at the end of the day. The eyelids droop, despite her best efforts to keep them open. This does not occur first thing in the morning. Further examination shows a generalized muscle weakness as the day progresses. A drug that may help to stabilize this condition would do which one of the following? (A) Stimulate the production of immune cells (B) Stimulate the production of epinephrine (C) Inhibit the production of acetylcholine (D) Inhibit acetylcholinesterase (E) Stimulate catechol-O-transferase

*The answer is B.* The woman has myasthenia gravis, which is due to the presence of autoimmune antibodies directed against the acetylcholine receptor in the neuromuscular junction. To confirm the diagnosis by Western blot, a sample of acetylcholine receptor would be run through a polyacrylamide gel, the protein transferred to filter paper, and the filter paper incubated with the patient's sera. If the sera contain antibodies that bind to the acetylcholine receptor, the antibodies will be bound to the filter paper, and then visualized using a secondary antibody linked to a reporter enzyme. Controls would be done to indicate that sera from an individual who did not have myasthenia gravis did not allow for the formation of a band on the Western blot. Running acetylcholinesterase, the epinephrine receptor, catechol-O-methyltransferase, the glucocorticoid receptor, or HMG-CoA reductase on the gel would not allow detection of antibodies against the acetylcholine receptor in the patient's blood sample.

A 42-year-old woman has slowly developed an inability to keep her eyes open at the end of the day. The eyelids droop, despite her best efforts to keep them open. This does not occur first thing in the morning. Further examination shows a generalized muscle weakness as the day progresses. In order to diagnose the disease the patient is experiencing, a Western blot was run using the patient's sera as the source of antibodies. The protein run on the gel would need to be which one of the following? (A) Acetylcholinesterase (B) Acetylcholine receptor (C) Epinephrine receptor (D) Catechol-O-methyltransferase (E) Glucocorticoid receptor (F) HMG-CoA reductase

*The answer is B.* B. The patient has hereditary colon cancer, specifically adenomatous polyposis coli, which presents in the fourth or fifth decade of life, with multiple polyps lining the lumen of the colon. The defective protein is APC, which regulates β-catenin activity. The loss of APC function leads to inappropriately activated β-catenin, which is a transcription factor and can stimulate the expression of myc and cyclin D1, to promote cell growth. The inappropriate expression of myc and cyclin D1 due to the loss of APC is an initiating event in tumorigenesis. The APC mutation does not affect BCl-2 activity, which is an antiapoptotic activity. While the loss of APC activity will lead to a gain-of-cyclin expression, that gain is due to the activation of β-catenin, which would be the initiating event for tumor formation.

A 45-year-old man presents with blood in his stool. Workup reveals a stage 3 (Dukes 3) colon carcinoma, with multiple polyps within the colon. A family history reveals that his father and grandfather both had colon cancer in their fifth decade of life. A potential initiating activating event in the development of this tumor is which one of the following? (A) Loss of β-catenin activity (B) Activation of β-catenin activity (C) Loss of transcription factor myc activity (D) Activation of Bcl-2 activity (E) Loss of cyclin expression (F) Gain-of-cyclin expression

*The answer is D.* The child has the symptoms of X-linked adrenoleukodystrophy, which is an X-linked disorder with a mutation in the ABCD1 gene. The ABCD1 gene is required for the transport of very long-chain fatty acids into the peroxisome for catabolism. In the absence of this activity, the very long-chain fatty acids accumulate, become incorporated into phospholipids, and alter the structure of myelin, leading to the neurological disorders observed. The lysosomes, nucleus, and Golgi apparatus are not involved in very long-chain fatty acid oxidation. The nucleolus is found in the nucleus and is the site of ribosome formation. Mitochondria oxidize fatty acids, but not when they are very long-chain fatty acids (greater than 20 carbons). In those cases, the initial steps of oxidation occur in the peroxisome, and when the chain length has been reduced, the partially oxidized fatty acid is transferred to the mitochondria to finish the oxidation of the compound.

A 5-year-old boy begins to regress in terms of developmental milestones, particularly neurologically. Shortly thereafter, the child enters a coma, and dies 2 years into the coma. Upon autopsy, the myelin sheath in the brain was found to be abnormal, as it contained a large quantity of very long-chain fatty acids in its phospholipids. The adrenal glands were also abnormal in appearance. The child, at the molecular level, had inherited a mutation that led to an inability to catalyze reactions that occur in which one of the following intracellular organelles? (A) Lysosomes (B) Nucleus (C) Mitochondria (D) Peroxisomes (E) Golgi apparatus (F) Nucleolus

*The answer is D.* During the development of an atherosclerotic plaque, PDGF is released by platelets at the site of injury, which aids in the stimulation of smooth muscle cell proliferation as the damage is being repaired. PDGF is also synthesized by smooth muscle cells, in an autocrine fashion, to further stimulate their own proliferation. FGF and EGF do not appear to play a role in the development of such plaques. HDL is a cholesterol carrier that removes cholesterol from tissues, and VLDL is the lipid carrier released by the liver for delivery of triglyceride to the tissues.

A 52-year-old man suddenly collapses and is rushed to the hospital, where it is found that his blood troponin-C levels are elevated. An angiogram indicates an 87% blockage of three major arteries to the heart. If these arteries were examined at a molecular level, excessive smooth muscle cell proliferation would be noticed. This proliferation has occurred over the past 25 years due to the secretion and action of which of the following factors? (A) VLDL (B) HDL (C) FGF (D) PDGF (E) EGF

*The answer is C.* The child has a form of Gaucher disease, which is a defect in a glucosidase which removes glucose from glucosylceramide. The accumulation of glucosylcerebroside in the lysosomes leads to the observed symptoms. Defects in degrading galactosylceramide lead to Krabbe disease, which does not result in hepatomegaly and splenomegaly. A defect in degrading sulfatide leads to metachromatic leukodystrophy, which has different symptoms than what the child is experiencing. A defect in the degradation of sphingomyelin leads to Niemann-Pick disease, with a different set of symptoms than that seen with Gaucher disease. A defect in degrading ceramide leads to Farber disease, a defect in ceramidase, with more severe symptoms than those observed in Gaucher disease.

A 6-month-old boy is brought to the pediatrician due to a large stomach. The doctor noticed splenomegaly, with no pain. The boy was always tired and had anemia. The boy also has thrombocytopenia and bruises easily. X-rays show a deformity of the distal femur, as shown below. This disorder is caused by an accumulation of which of the following in macrophage lysosomes? A) Galactosylceramide (B) Sulfatide (C) Glucosylceramide (D) Sphingomyelin (E) Ceramide

*The answer is A.* This man originally displayed an actinic keratosis that has, over the intervening 3-year period, become a squamous cell carcinoma. Actinic keratosis develops in areas of the skin that are frequently exposed to sunlight, such as the top of the ear. The frequent exposure to UV light led to the creation of pyrimidine dimers in the DNA. If the cells cannot repair the DNA damage rapidly enough, cancerous changes do occur over time. The presentation of actinic keratosis may represent a precancerous state toward squamous cell carcinoma. Removal of the actinic keratosis would have prevented the development of the tumor. Hydroxyl radicals are created by ionizing radiation such as X-rays, not by sunlight. Oncogenic RNA viruses such as HTLV-1 could cause lymphomas or leukemias, but have not been implicated in squamous cell carcinoma. TNF receptor mutations can occur in immune system cells, leading to apoptosis, which would lead to an immune defect, but not the cancer observed. UV light does not lead to the creation of double-strand breaks in DNA.

A 62-year-old male has a reddish, rough patch with white scales on the top of his ear. He does not get this treated, and 3 years later it has become an enlarged, raised lesion with a central ulcerated area that will not heal. Of the following, which is the most likely causative factor for this malignancy? (A) Creation of pyrimidine dimers (B) Creation of hydroxyl radicals (C) Oncogenic RNA virus (D) TNF receptor mutation (E) Double-strand breaks in the DNA

*The answer is B.* The patient has polycythemia vera, a myeloproliferative disorder, which has recently been shown to result from constitutive JAK2 activity. Under normal conditions, JAK2 is only activated when an appropriate cytokine binds to its receptor. Once activated, JAK2 phosphorylates STAT proteins, which translocate to the nucleus and initiate gene transcription, leading to a proliferative response. In this disorder, JAK2 is always active, and the cells proliferate even in the absence of a signal. SMAD proteins transmit signals from TGF-β signaling and are not involved in this disorder. Her2 is the EGF receptor, which also does not play a role in this disorder.

A 62-year-old man visits his physician for feeling fatigued. It is noticed that the spleen is enlarged and slightly tender when palpated. He has abdominal pain and echymoses on his skin. His blood pressure is high. Blood work indicates an abnormally high red blood cell, white blood cell, and platelet counts. A bone marrow sample from the patient is shown below. The suggested treatment is phlebotomy to reduce the thickness of the blood. This disorder can result from which of the following? (A) A loss of function of JAK2 (B) A gain of function of JAK2 (C) A loss of function of SMAD4 (D) A gain of function of SMAD4 (E) Constitutive Her2 activity

*The answer is A.* No energy is needed for simple diffusion, which is the case if this is a steroid hormone. The other answer choices require ATP, which would be necessary for an active transport process, whether it be activation of a channel by phosphorylation, or generation of a gradient across the membrane for cotransport. Simple and facilitated diffusion do not require any energy sources for transport.

A bodybuilder has gained 50 pounds of muscle over the last 6 months, facilitated by both increased weight lifting and black-market pharmaceutical injection of one substance. He never experienced hypoglycemia during this time frame. How much energy is needed for the transport of the majority of the illegal substance that the bodybuilder is using for the drug to enter cells? (A) No energy is required. (B) One ATP molecule is used for each molecule of the substance transported. (C) Only a few ATP molecules are being used to open and close the channel through which many substance molecules diffuse. (D) This is an example of cotransport, in which the energy generates a sodium gradient across the membrane, and it is difficult to calculate an exact energy amount. (E) The transporter has to be phosphorylated once to allow transport to occur for many solutes.

*The answer is A.* The bodybuilder is injecting (most probably) testosterone, a steroid hormone, which aids in building muscle mass. Steroid hormones are lipid-soluble substances, and cross membranes by simple diffusion. The receptor for steroid hormones is present inside the cell (either the cytoplasm or nucleus), and once the steroid hormone enters the cell, it will bind to the receptor in a saturable manner. Once the concentration of the hormone inside and outside the cells is equal, transport will stop. Active transport refers to using energy to concentrate a solute against its concentration gradient, which is not the case for steroid hormone transport across the membrane. Facilitative diffusion requires a membrane-bound carrier (no energy), but as indicated previously, the carrier (receptor) for these hormones is intracellular. Steroid hormones do not enter cells through either endocytosis or pinocytosis. The fact that the bodybuilder never became hypoglycemic after taking the drug suggests that it was not insulin being injected.

A bodybuilder has gained 50 pounds of muscle over the last 6 months, facilitated by both increased weight lifting and black-market pharmaceutical injection of one substance. He never experienced hypoglycemia during this time frame. The mechanism whereby the illegal substance is entering the muscle cells of the bodybuilder is most likely which one of the following? (A) Simple diffusion (B) Active transport (C) Endocytosis (D) Facilitative diffusion (E) Pinocytosis

*The answer is D.* Protein kinase B is recruited to the membrane, along with the phosphoinositide- dependant kinase, through binding to PIP3 in the membrane. The PIP3 is generated from PIP2 through the actions of PI-3′-kinase. If PI-3′-kinase had lost its SH2 domains, this protein would no longer be able to bind to phosphorylated IRS-1, and in turn be activated by the insulin receptor tyrosine kinase activity. Proteins containing SH2 domains are capable of binding to other proteins which contain phosphotyrosine residues. In the absence of PI-3′-kinase activity, protein kinase B could not be activated, as both PKB and PDK1 (phosphoinositide-dependant kinase 1) require phosphatidylinositol trisphosphate in the membrane as a docking station, through the proteins pleckstrin homology domain. None of the other choices listed (ras, raf, GAP, GRB2) require PI-3′-kinase activity in order to be activated.

A cell line has been generated in which a protein important in signal transduction was mutated such that its SH2 domain was no longer functional. After adding insulin to this cell line, protein kinase B could no longer be activated. The protein which was mutated is which of the following? (A) Ras (B) Raf (C) GAP (D) PI-3′-kinase (E) GRB2

*The answer is B.* If Ras has a decreased ability to hydrolyze GTP, it will remain active. As a result, the MAP Kinase pathway will remain active and the cell growth and division will be stimulated.

A cell that contains a mutation in Ras that decreases its ability to hydrolyze GTP (without affecting its ability to bind GTP) will: (A) be less likely to divide (B) have increased activation of the map kinase pathway (C) have decreased expression of cyclin D (D) have increased Glut4 on the plasma membrane (E) none of the above

*The answer is E.* Tay-Sachs is a defect in hexosaminidase A, which removes the terminal N-acetylgalactosamine residue from ganglioside GM2, producing the free sugar and GM3. Hexosaminidase A does not cleave glucose, ceramide, sphingosine, or the fatty acyl component of ceramide from GM2; it is specifi c for N-acetylgalactosamine.

A child has been diagnosed with Tay-Sachs disease, in which a particular lipid accumulates within the lysosomes. The component of this lipid which cannot be removed in the lysosome is which of the following? (A) Ceramide (B) Sphingosine (C) Fatty acid (D) Glucose (E) N-acetylgalactosamine

*The answer is E.* Lithium primarily inhibits the phosphatase which converts inositol phosphate to free inositol, thereby disrupting the phosphatidylinositol cycle, leading to increased levels of the intermediates of the cycle, which are often signaling molecules. Lithium does not affect the generation of diacylglycerol, inositol trisphosphate (IP3), inositol bisphosphate (IP2), or inositol phosphate (IP); it affects just the conversion of IP to free inositol and a phosphate.

A depressed patient is prescribed lithium by his psychiatrist. The effect of lithium is to block the generation of which of the following? (A) Diacylglycerol (B) Inositol trisphosphate (C) Inositol bisphosphate (D) Inositol phosphate (E) Inositol

*The answer is B.* Steroid hormones must enter the cell by passive diffusion, and if the membrane is less fluid, it will be more difficult for the hormone to enter the cell to bind to its receptor. Cytokines, transforming growth factors (TGFs), insulin, and glucagon all bind to transmembrane receptors, which transmit their signal to the cytoplasmic portion of the receptor. It is less likely that those conformational signals will be affected by the fluidity of the membrane than the passage of the steroid hormone through the membrane. Decreased membrane fluidity may impair dimerization of the receptors, but the initial binding events should still occur normally.

A message transmitted by which example of a chemical messenger would most likely be negatively affected by a mutation that greatly reduced the fluidity of the plasma membrane? (A) Cytokine (B) Steroid hormone (C) A transforming growth factor (D) Insulin (E) Glucagon

*The answer is C.* The silencing RNA will ablate MEK expression in this cell line. Effects which are upstream to MEK activation will still occur; effects downstream of MEK activation will be blocked. MEK activates ERK, which leads to alterations in gene transcription, such as myc and fos. Tyrosine kinase receptors do not activate STAT proteins (that occurs with cytokine receptors). PLC-γ activation occurs upon receptor tyrosine kinase activation, and binding of the phospholipase to the receptor via its SH2 domains. Activation of PLC-γ does not require MEK activation. The ras-raf pathway leading to the activation of myc and fos is shown below.

A mouse fibroblast cell line is treated with an siRNA targeted toward MEK. When such cells are treated with a growth factor which works through a tyrosine kinase receptor, which one of the following effects would occur? (A) Activation of ERK (B) Activation of myc (C) Activation of phospholipase (PLC-γ) (D) Activation of fos (E) Activation of STAT

*The answer is B.* Cyclin-dependent kinase inhibitors (CKI) act to block the action of kinases that are activated by cyclins. When such an activity is lost (meaning that the gene products from both chromosomes are inactive), uncontrolled cell proliferation can result. Since the activity must be lost, such genes are classified as tumor suppressors, as opposed to the dominant oncogenes, in which an activity is gained via mutation or inappropriate gene regulation. The CKIs are not involved in apoptosis, nor do they act as growth factors.

A patient has been diagnosed with a melanoma, and molecular analysis has indicated that the tumor has sustained a loss of p16(INK4) activity (inhibitor of cyclin-dependent kinase 4). Such a gene would be best classified as which of the following? (A) A dominant oncogene (B) A tumor suppressor (C) A proapoptotic factor (D) An antiapoptotic factor (E) A growth factor

*The answer is D.* The phagosomes fuse with lysosomes, where the acidity and digestive enzymes within the lysosomes destroy the contents of the phagosome (in this case, the bacteria within the phagosome). The digestive enzymes have a pH optimum of 5.5, which is maintained within the lysosome through the actions of a proton pump (a proton translocating ATPase activity). The nucleus and mitochondria are not involved in the lysosomal digestion of phagosome contents. Blocking either the sodium ATPase, potassium ATPase, or a calcium-activated ATPase will greatly affect other organs, but will not affect the ability of the lysosome to degrade its contents.

A person is diagnosed with group A streptococcal bacteremia. One of the body's major defenses in this type of disease is for eosinophils to phagocytize the bacteria. Once internalized, the bacteria are destroyed by fusing the phagosome with a particular intracellular organelle. Which one of the following would destroy the activity of that organelle such that the bacteria would not be incapacitated? (A) Inhibiting sodium-potassium ATPase activity (B) Interfering with mitochondrial protein synthesis (C) Blocking transport through nuclear pores (D) Inhibiting a proton-translocating ATPase (E) Inhibiting a calcium-activated ATPase

*The answer is D. The patient has Burkitt lymphoma, which in 90% of the cases is due to altered regulation of the myc gene (constitutive activation of transcription), due to a translocation of the myc gene such that it is controlled by an immunoglobulin promoter (which is why this disorder results in abnormal blood cell proliferation, as these are the cells that produce the immunoglobulins). While Epstein-Barr virus is thought to render individuals susceptible to Burkitt lymphoma, the oncogenic event is the misexpression of the myc gene. Bcr-abl is associated with chronic myelogenous leukemia. Bcl-2 overexpression leads to a loss of apoptotic potential and is not associated with Burkitt lymphoma. EGFreceptor activation (similar to the erbB oncogene) also does not lead to these symptoms.

A physician in a rural African clinic sees a child with swelling of the jaw, loosening of the teeth, and swollen lymph nodes (see the picture below). Karyotype analysis of blood cells shows a translocation between chromosomes 8 and 14. This rapidly growing tumor is most likely due to which of the following? (A) EBV activation (B) Bcr-abl activation (C) BCl-2 activation (D) Constitutive myc expression (E) EGF-receptor activation

*The answer is D.* The chimeric receptor was created such that when estrogen bound to the receptor, the receptor would then bind to testosterone response elements in promoters in the genome (the DNA-binding domain was specific for the testosterone receptor). The steroid hormone receptors do not have tyrosine kinase activity, nor do they activate PKA. Since the cells being used for this experiment lack the estrogen receptor, there would not be any effect of adding estrogen on estrogen-specific gene expression.

A researcher was investigating the mechanism of action of steroid hormone receptors. He created a chimeric receptor that contained the estrogen ligand- binding domain, and also the testosterone DNA-binding domain. The transactivation domain was also from the testosterone receptor. When this chimeric receptor was expressed in eukaryotic cells lacking estrogen receptors, which of the following would you expect to occur when estrogen is added to the cells? (A) Increase in tyrosine kinase activity (B) Activation of PKA (C) Induction of estrogen-specific genes (D) Induction of testosterone-specific genes (E) Inhibition of estrogen-specific genes

*The answer is A.* Overexpression of ras leads to transformation because there is insufficient GAP (GTPase-activating protein) to inactivate the ras in the cell. This can be overcome by adding GAP to the cells, such that the ras to GAP ratio is closer to one. GAP can thus inhibit ras activity and block transformation. This will not work with oncogenic ras, as oncogenic ras has obtained a mutation which has inactivated its built-in GTPase activity. Thus, GAP cannot activate oncogenic ras and cannot block ras function. SOS is a guanine nucleotide exchange protein that leads to ras activation; it cannot turn off activated ras. Raf, fos, and PKA do not play a role in regulating ras activity.

A researcher was studying a tumor in mice that was induced by transfection of a vector, which led to overproduction of c-ras within the tumor. The researcher wanted to now add a second expression vector that would counteract the effects of the overexpressed c-ras. Which one of the following genes should be investigated? (A) GAP (B) SOS (C) Raf (D) Fos (E) PKA

*The answer is A.* The conversion of an alanine residue to a glutamate residue results in the insertion of a negative charge into this region of the protein (the same overall effect that phosphorylation has). The introduction of the negative charge would lead to similar conformational changes that occur when a phosphate is introduced in this part of the protein. Thus, the glutamate side chain will mimic the effects of phosphorylation. Converting a serine to an alanine does not insert a negative charge into this area of the protein (the side chain of alanine is a methyl group); thus, the required conformation changes cannot occur. The same rationale applies for conversion of a threonine to an alanine; the alanine side chain cannot be modified to contain negative charges. Conversion of the alanine to a threonine introduces a hydroxyl group to the region, but the hydroxyl group has a high pKa and will not deprotonate at physiological pH. This is similar to the serine to tyrosine conversion; tyrosine also has a hydroxyl group, but the pKa is too high to lead to significant deprotonation and introduction of a negative charge into this area.

A signaling protein is activated by phosphorylation. The site of phosphorylation within the protein falls within the sequence-S-A-T-; either the serine or threonine residue can be phosphorylated. Phosphorylation of either residue results in an active protein. In certain tumors, however, an activating mutation is found in this region that renders the protein active in the absence of an phosphorylation event. Which one of the following amino acid mutations is most likely an activating mutation? (A) A to E (B) S to A (C) T to A (D) A to T (E) S to Y

*The answer is C.* The TGF-β receptors work together to initiate phosphorylation of SMAD transcription factors. Once phosphorylated, the specific SMAD factor dimerizes with a common SMAD, SMAD4, for transport to the nucleus and binding to the appropriate response elements in DNA. A lack of SMAD4 would account for the total lack of response to TGF-β, as all signaling through TGF-β requires this component. JAK and TYK work with cytokine receptors, not with TGF-β receptors, and SMAD1 and -2 are specific receptor SMADS; loss of either of them would diminish certain types of responses to TGF-β, but not all responses to TGF-β. The pathway of TGF-β signaling is shown below.

A variant cell line was discovered that would not respond to any form of TGF-β, yet both type I and type II receptors were present and functional on the cell surface, as determined by in vitro experiments. A potential mutation in which one of the following proteins can lead to this phenotype? (A) SMAD2 (B) SMAD3 (C) SMAD4 (D) JAK1 (E) TYK1

*The answer is B.* The sprinter has released epinephrine in anticipation of the race. Epinephrine binds to a heptahelical receptor, which is linked to a stimulatory G protein. Upon epinephrine binding the G protein is activated, and the Gαs subunit binds GTP and travels to, and activates, adenylate cyclase, which raises intracellular cAMP levels. Epinephrine binding to such muscle receptors does not result in an increase of tyrosine kinase, inhibitory G protein, phospholipid-specific G protein (Gq), and cAMP phosphodiesterase activities.

A world-class sprinter, while in the starting blocks waiting for the race to start, stimulates which of the following muscle proteins in response to hormone release? (A) A tyrosine kinase (B) A Gαs protein (C) A Gαi protein (D) A Gq protein (E) cAMP phosphodiesterase

*The answer is B. The child has I-cell disease (mucolipidosis type II), which is a deficiency in protein sorting, particularly of sending lysosomal enzymes to the lysosome (a lysosomal storage disease). The I of I-cell disease stands for inclusion bodies. If the child develops these clinical and radiologic symptoms later in life, one would consider the diagnosis of Hurler syndrome (mucopolysaccharidosis). Lysosomal enzymes are tagged with mannose-6-phosphate (M6P) during posttranslational modification. Enzymes containing M6P then bind to a M6P receptor, which transports the enzymes to the lysosomes. Lacking such a signal, patients with I-cell disease secrete their lysosomal contents into the plasma and interstitial fluids. This leads to lysosomal dysfunction and cellular and tissue destruction. This disease is not peroxisomal, nor does it enhance protein secretion.

A young child exhibits the following symptoms: Coarse facial features, congenital hip dislocation, inguinal hernias, and severe developmental delay. These symptoms are fully evident at the child's age of 1. Cellular analysis demonstrated the presence of inclusion bodies within the cytoplasm of liver cells. The inclusion bodies are the result of which of the following? (A) Enhanced lysosomal enzyme activity (B) Reduced lysosomal enzyme activity (C) Enhanced peroxisomal enzyme activity (D) Reduced peroxisomal enzyme activity (E) Enhanced protein secretion

*The answer is E.* When active, Gαs activates adenylate cyclase, which increases the production of cAMP. When active, Gαi inhibits adenylate cyclase. When active, Gaq activates phospholipase C, which increases the production of inositol triphosphate.

All G-proteins are activated by binding to GTP and lose their activity when they hydrolyze GTP to GDP. However, their activity is quite different. Which of the following is not correct? (A) When active, Gαs activates adenylate cyclase (B) When active, Gαs results in increased production of cAMP (C) When active, Gαi results in the inhibition of adenylate cyclase (D) When active, Gαq results in the activation of phospholipase C (E) When active, Gαq results in the increased production of phosphatidylinositol trisphosphate

*The answer is E.* Phosphatidylinositol triphosphate serve as a docking site for proteins containing a Src homology 2 domain. This is a function of the phosphorylated tyrosine sequences on the receptor.

All of the following are functions served by the phosphatidylinositol phosphates EXCEPT (A) Phosphatidylinositol bisphosphate can be hydrolyzed to produce diacylglycerol, a second messenger (B) Phosphatidylinositol bisphosphate can be hydrolyzed to produce inositol trisphosphate, a second messenger (C) Phosphatidylinositol bisphosphate can be phosphorylated to produce phosphatidylinositol trisphosphate (D) Phosphatidylinositol trisphosphate can serve as a plasma membrane docking site for signal transduction proteins (E) Phosphatidylinositol triphosphate will serve as a docking site for proteins containing a Src homology 2 domain

*The answer is B.* Phosphatidylinositol 3' kinase breaks down phosphatidylinositol bisphosphate (PIP2) to inositol triposphate (IP3) and DAG. ATP is hydrolyzed into ADP.

All of the following are substrates and products of the reaction catalyzed by phosphatidylinositol 3' kinase EXCEPT (A) Phosphatidylinositol bisphosphate is a substrate (B) Inositol trisphosphate is a substrate (C) ATP is a substrate (D) Phosphatidylinositol trisphosphate is a product (E) ADP is a product

*The answer is A.* When a growth factor binds to its receptor in the RAS and MAP pathway, the tyrosine kinase receptor dimerizes and autophosphorylates tyrosine resides.

All of the following are true in the Ras and MAP pathway EXCEPT. When a growth factor binds to its receptor in the RAS and MAP pathway, (A) The occupied growth factor receptor autophosphorylates seryl residues (B) Grb2 uses its Src homology domain to bind to the phosphorylated receptor (C) Grb2 changes conformation and binds to the next protein (SOS) in the signal transduction pathway (D) Ras exchanges GDP for GTP and activates the Raf, the first kinase in the MAP kinase pathway (E) The MAP kinase pathway regulates transcription factors that regulate mRNA synthesis

*The answer is D.* Autocrine stimulation is defined as one cell both secreting and responding to a growth factor. During the development of an atherosclerotic plaque, the smooth muscle cells both secrete and respond to PDGF. When tumors secrete FGF, they stimulate endothelial cell proliferation, which aids in angiogenesis (the tumor cells are not responding to the FGF--the endothelial cells are). Glucagon release from the pancreas is an endocrine process (the target for glucagon are cells far from the pancreas), as is insulin stimulation of glucose transport in muscle. Platelets secreting PDGF is not autocrine, as the platelets do not respond to PDGF.

An example of an autocrine stimulatory pathway is which of the following? (A) Blood vessel growth within a tumor (B) Glucagon release from the pancreas (C) Insulin stimulation of glucose transport into adipocytes (D) Injured smooth muscle cells producing PDGF (E) Platelets secreting PDGF at internal blood vessel injuries

*The answer is B.* Lysosomes contain a single membrane (so there is no outer membrane, such as in mitochondria) that contains a proton translocating ATPase. The ATPase will concentrate protons inside of the lysosome, at the expense of ATP hydrolysis, to acidify the intraorganelle pH such that the lysosomal enzymes will be active. If the intravesicular pH cannot be lowered, the digestive enzymes will be inactive, and no digestion will take place. Targeting a chloride pump in the lysosomal membrane will not affect the activity of the lysosomal enzymes. If the lysosomal enzymes were not marked with a mannose-6-phosphate residue in the Golgi apparatus, they would not be able to bind to the mannose-6-phosphate receptor to be targeted to the lysosomes. If the drug altered either of those proteins (the enzyme responsible for adding the mannose-6-phosphate or the mannose-6-phosphate receptor), then the lysosomal enzymes would not be in the lysosomes, which is not the case. Such a drug would bring about the symptoms of I-cell disease.

An experimental drug has been added to a eukaryotic cell, and while the drug was designed to interfere with a membrane transport process, the investigators found that in cells treated with the drug the lysosomes quickly turn into inclusion bodies. None of the material directed to the lysosomes for removal was being digested in the lysosome, and remained intact inside the organelle. An analysis of lysosomal contents in drug-treated cells indicated that the full complement of lysosomal enzymes were present in the organelle. Assuming that the drug is targeting just one protein, which one of the following proteins is most likely the target? (A) An outer membrane protein that allows the lysosomal membrane to become permeable to small molecules (B) A proton-translocating ATPase in the lysosomal membrane (C) A chloride pump in the lysosomal membrane (D) The enzyme that adds mannose- 6-phosphate to lysosomal enzymes (E) The mannose-6-phosphate receptor

*The answer is A.* Cyclin kinase inhibitors act as brakes on the cell cycle. If the cyclin kinase inhibitor can be removed from the cell, then the cell cycle could proceed in an uncontrolled fashion. MicroRNAs reduce the amount of protein product formed from the target mRNA. In order to eliminate the production of the cyclin kinase inhibitor, the microRNA would need to be overexpressed, such that all target mRNAs are bound, and translation of the gene product is halted. Reducing the expression of the microRNA would lead to overexpression of the cyclin kinase inhibitor, and more control of the cell cycle. This is also the case if the microRNA lost all activity (overproduction of its target), or lost its specificity (again, overproduction of the target).

Assume there is a microRNA that participates in regulating the expression of a particular cyclin kinase inhibitor. How might an alteration in this microRNA lead to uncontrolled cell proliferation? (A) Overexpression of the microRNA, so it acts as an oncogene. (B) Reduced expression of the microRNA, so it acts as a tumor suppressor. (C) A total loss of activity of the microRNA, so it cannot bind to its target mRNA. (D) A loss of specificity of the microRNA for its target, so different mRNAs are not targeted. (E) No change in microRNA activity can lead to uncontrolled cell proliferation.

*The answer is A.* The G1 phase is the most variable in length in the cell cycle, and usually the longest. The G1 - S checkpoint is called the restriction point. Once the cell has passed this point it is committed to go through to mitosis, and can only avoid this by undergoing apoptosis, which will be triggered if the cell cannot meet any of the requirements for passing later checkpoints in the cell cycle. If the cell's DNA is damaged this will normally be detected by proteins which recognize and bind to sections of single stranded DNA which is characteristic of strand breaks, or stalled replication forks. These trigger a mechanism which holds the cell at the restriction point, giving time for repair enzymes to repair the damaged DNA. If the damage is not repaired within a reasonable time the same mechanism will force the cell into apoptosis.

At which cell cycle checkpoint is the cell cycle halted if the cell's DNA is damaged? (A) G1 - S (B) S - G2 (C) G2 - M (D) G0 - G1

*The answer is C.* Membrane fluidity needs to be constant at the two different temperatures for the bacteria to grow. At the lower temperature, a higher percentage of unsaturated fatty acids and short-chain saturated fatty acids would be present as those fatty acids have a lower melting point than long-chain saturated fatty acids. Cholesterol is not found in bacterial membranes.

Bacteria grown at 15°C contain a different fatty acid composition in their membranes as compared to bacteria grown at 37°C. Which one of the following would best represent the composition of the fatty acids at these two different temperatures? (A) Bacteria at the lower temperature would contain a higher percentage of saturated fatty acids than bacteria grown at the higher temperature. (B) Bacteria grown at the lower temperature would have a higher percentage of long-chain fatty acids than bacteria grown at the higher temperature. (C) Bacteria grown at the lower temperature would have a higher percentage of unsaturated fatty acids than bacteria grown at the higher temperature. (D) Bacteria grown at the lower temperature would have an increased level of cholesterol as compared to bacteria grown at the higher temperature. (E) Bacteria grown at the lower temperature would have a decreased level of cholesterol as compared to bacteria grown at the higher temperature.

*The answer is C.* In I-cell disease, the lysosomal hydrolases are mistargeted and are excreted from cells into the circulation. As the pH of the blood is above 7 and the pH optimum of these enzymes is around 5, there is no activity of the hydrolases in blood. In Hurler syndrome, a defect in the degradation of mucopolysaccharides, there is an accumulation of dermatan and heparan sulfate in the urine, but not in the blood. Short-chain dicarboxylic acids are produced with a defect in medium chain acyl-CoA dehydrogenase, and cytochrome c release into the cytoplasm of cells from mitochondria is the signal to initiate apoptosis.

Children with either I-cell disease or Hurler syndrome show very similar clinical features. One method to distinguish between the two is to find which of the following elevated in the blood? (A) Heparan sulfate (B) Short-chain dicarboxylic acids (C) Lysosomal hydrolases (D) Dermatan sulfate (E) Cytochrome c

*The answer is B.* In cholera, the G-protein loses its GTPase activity. As a result, the G-protein will remain bound to GTP, which means that it will always be active. Thus, the messengers and signals after the G-protein are also active. In the cAMP-dependent pathway, G-proteins activate adenylyl cyclase, which then converts AMP to cAMP. cAMP then activates protein kinase A, which then phosphorylates phosphorylase kinase (now active) and glycogen synthetase (now inactive). Phosphorylase kinase than activates glycogen phosphorylase.

Cholera toxin inhibits the ability of the alpha subunit of Gs to hydrolyze GTP. The resulting effect is the ______ of adenylyl cyclase. (A) stimulation (B) inhibition (C) molecular degradation (D) increased synthesis

*The answer is D.* For most translocations that lead to uncontrolled cell growth, a gene is inappropriately expressed because it has been moved adjacent to a constitutive promoter (such as the myc gene next to the immunoglobulin promoter in Burkitt lymphoma). The dysregulation of cell proliferation does not occur owing to problems with mitosis or DNA replication, nor with crossing over. In most cases, the problem is an increased or inappropriate expression of the translocated gene, and not a loss of gene expression.

Chromosomal translocations can lead to uncontrolled cell growth due to which one of the following? (A) Interference with mitosis (B) Interference with DNA synthesis (C) Unequal crossing over during mitosis (D) Inappropriate expression of translocated genes (E) Loss of gene expression

*The answer is D.* Philadelphia chromosome (a translocation of chromosomes 9 and 22) produces a new gene product, a fusion protein of bcr and abl (bcr from chromosome 22 and abl from chromosome 9, with the fusion protein being produced from the shorter chromosome 22). Abl is a tyrosine kinase, and when fused with bcr, it is constitutive and no longer properly regulated. The presence of this unregulated kinase leads to a loss of cellular growth control. The bcr-abl protein is not a transcription factor, growth factor receptor, growth factor, or ser/thr kinase.

Chronic myelogenous leukemia, due to the presence of Philadelphia chromosome, leads to transformation due to the creation of a factor which is nonregulated and aberrant. This factor is best described as which of the following? (A) Transcription factor (B) Growth factor receptor (C) Growth factor (D) Tyrosine kinase (E) Ser/thr kinase

*The answer is C.* Cyclin dependent kinases are inactive alone because their active site is inaccessible to substrate. Binding of cyclins partially activates Cdks, but for full activity they also need to have a threonine residue phosphorylated by a Cdk activating protein kinase. A tyrosine kinase is part of the signalling mechanism by which mitogenic growth factors transmit their signal into the cell, and they will be active when synthesis of cyclins is signalled, but they play no part in the direct control of the cell cycle.

Cyclin dependent kinases which control progression through cell cycle checkpoints are fully activated by which of the following: (A) binding to cyclins. (B) phosphorylation by Cdk activating protein kinase. (C) binding to cyclin, plus phosphorylation by a Cdk activating protein kinase. (D) phosphorylation by a tyrosine kinase.

*The answer is B.* In the phospholipase C-IP₃ pathway, a signal molecule, such as epinephrine binds to the G protein-coupled receptor. The Gαq protein is then activated when GTPase activity is used to exchange GDP to GDP. This then activates Phospholipase C, which breaks down PIP₂ to IP3 and IP₃. DAG remains attached to the membrane and IP₃ is released where it binds to the sarcoplasmic reticulum, which triggers the release of calcium. Calcium and DAG activate protein kinase C, and Calcium alone activates Calcium-Calmodulin.

Diacylglycerol and calcium activate (A) protein kinase A (B) protein kinase C (C) PI 3-kinase (D) Ca2+/calmodulin-dependent kinase (E) a tyrosine kinase

*The answer is B.* After phospholipase C breaks down PIP₂ to DAG and IP₃, IP₃ is released it binds to the sarcoplasmic reticulum, which triggers the release of calcium.

The IP₃ produced by phospholipase C acts to (A) open calcium channels in the plasma membrane (B) open calcium channels in the endoplasmic reticulum (C) activate IP₃ dependent kinase (D) open channels in mitochondria

*The answer is E.* Executioner caspases are proteases which cleave thousands of specific targets (e.g. nuclear lamins, components of the cytoskeleton) resulting in controlled cell death aka apoptosis. This may be activated by the loss of Akt signaling, initiator caspases, the leakage of cytochrome c from mitochondria, and movement of a BH3-only protein to the mitochondria.

Executioner caspases may be activated by: (A) loss of Akt signaling (B) initiator caspases (C) the leakage of cytochrome c from mitochondria (D) movement of a BH3-only protein to the mitochondria (E) all of the above

*The answer is B.* I-cell disease is a lysosomal storage disease that is a result of a a deficiency in targeting lysosomal enzymes to the organelle. Targeted proteins, such as lysosome hydrolases, lack mannose-6-phosphate (M6P) signal and cannot transfer from the lysosome to the Golgi apparatus. As a result, the proteins and enzymes that are meant to be delivered to the lysosome are secreted out of the cell. And the lysosome will not be able to break down particles.

I cell disease is a severe type of lysosomal storage disease due to defective N-acetylglucosamine phosphotransferase. Which of the following are not a likely outcome of the disease? (A) Trafficking of some acid hydrolases to the lysosome will be affected (B) Recycling of Mannose 6 phosphate receptors will be affected (C) Breakdown of endocytosed material will be affected (D) Some acid hydrolases will get secreted (E) The formation of Mannose 6 phosphate will be affected

*The answer is D.* Cholera toxin inhibits the GTPase activity of the G protein alpha subunit. In that way the adenylate cyclase remains active and keeps producing cyclic AMP which causes uncontrolled secretion of sodium ions and water into the intestine leading to diarrhea and vomiting and dehydration. It does not inhibit the binding of VIP to the receptor but it perpetuates the signal even after VIP has become detached from the receptor. There is no involvement of a Gi protein.

In cholera, there is uncontrolled secretion of sodium ions and water into the intestinal lumen because of the action of cholera toxin on a G protein coupled receptor system. How does the toxin act? (A) Cholera toxin activates a Gi (inhibitory) protein. (B) Cholera toxin inhibits phosphodiesterase so that the signal is not switched off. (C) Cholera toxin inhibits the binding of vasoactive intestinal polypeptide to the receptor. (D) Cholera toxin inhibits the GTPase activity of the G protein alpha subunit.

*The answer is B.* Paracrine signaling involves diffusion of a substance locally from one cell to another via the interstitial space rather than through blood vessels. Endocrine signaling would require that VEGF travel through the blood to reach the endothelial target cells. Autocrine signaling requires that the same cell both send the signal and respond to it. Juxtacrine signaling requires that the VEGF be displayed from the surface of one cell and bound by a receptor on another. Synaptic signaling is reserved for neurons. None of these other signaling modes fit the description for the mechanism of action of VEGF.

In order for a solid tumor to grow beyond a certain size, it must develop a blood supply by elaborating factors such as vascular endothelial growth factor (VEGF). VEGF secreted by the tumor cells diffuses to nearby endothelial cells, which respond by dividing and migrating toward the tumor to eventually develop into blood vessels and vascularize the tumor. Which of the following modes of intercellular signaling is operative in the case of VEGF? A. Endocrine B. Paracrine C. Autocrine D. Juxtacrine E. Synaptic

*The answer is C.* The cell is stimulated to enter the cell cycle from outside by cytokines and growth factors, which are peptides secreted by nearby cells. In some cells, particularly in the immune system, the growth factor may be a membrane protein of the stimulating cell. These factors activate the cell by binding to receptors in the plasma membrane. As a result, signal transduction mechanisms initiate the cell cycle by stimulating synthesis of cyclins, which are able to activate constitutively inactive cyclin-dependent kinases.

In order to enter the cell cycle a cell must be stimulated from outside. What type of molecule provides this stimulation? (A) Cyclins (B) Cyclin-dependent kinases (C) Cytokines and growth factors (D) Tyrosine kinases

*The answer is A.* The insulin signal would be prematurely terminated prematurely. Neither the ras nor the PI 3-kinase pathways would be activated and GLUT 4 transporters would not be recruited to the cell membrane. IRS phosphorylation on serine or threonine residues is one of the mechanisms that switch off the insulin signal and they are a possible cause of insulin resistance.

Insulin receptor substrates IRS 1 /2 are phosphorylated by the receptor tyrosine kinase domain. What would happen if the IRS 1/2 were phosphorylated on a serine/threonine residue? (A) The insulin signal would be prematurely terminated leading to insulin resistance. (B) The ras pathway would be activated prematurely and the insulin signal amplified. (C) The PI 3-kinase pathway would be activated prematurely and insulin metabolic effects would be amplifies. (D) GLUT 4 transporters would remain fixed on the cell membrane and would not be able to be internalised.

*The answer is B.* X-linked SCID is due to the lack of a common cytokine receptor subunit (the γ subunit), which affects the ability of a variety of cytokines to transmit signals to hematopoietic cells. Adrenoleukodystrophy is due to the buildup of very long-chain fatty acids, for a variety of reasons. Influenza is due to a virus. Myasthenia gravis is due to the production of autoantibodies directed against the acetylcholine receptor. Type 1 diabetes results from an inability to produce insulin.

Interference in generalized cytokine signaling can lead to which one of the following disorders? (A) Adrenoleukodystrophy (B) SCID (C) Influenza (D) Myasthenia gravis (E) Type 1 diabetes

*The answer is D.* In response to binding of a growth factor to its cell surface receptor, the receptor forms a dimer that stimulates its intrinsic kinase activity to phosphorylate tyrosine residues on the cytoplasmic region. These phosphotyrosine sites allow docking of the adaptor complex GRB2-SOS, which binds and thereby activates Ras through GDP to GTP exchange. Constitutively activated Ras is unable to hydrolyze bound GTP and thus cannot respond to the binding of Ras-GAP. Raf-1, MEK, MAP kinase, and Elk-1 all are downstream elements of the signaling pathway that depend on the activity of Ras.

It is estimated that mutations of RAS occur in over 30% of human cancers. In most of these cases, the mutations interfere with the intrinsic GTPase activity of Ras so that the protein becomes constitutively or continuously active, irrespective of whether growth factors are present. Constitutively activated Ras has become insensitive to which of the following elements of the growth factor signaling pathway? A. Raf-1 B. MEK C. MAP kinase D. Ras-GAP E. Elk-1

*The answer is A.* Li-Fraumeni syndrome results from an inherited mutation in p53, the guardian of the genome. This protein monitors the DNA for damage, and if damage is found, acts as a transcription factor to arrest the cell cycle, allow the DNA damage to be repaired, and then to allow the cycle to proceed. If the DNA damage cannot be repaired, then apoptosis is induced so that the cell will not replicate the damaged DNA. P53 does not regulate the CDKs, tyrosine kinases, or G-proteins. Loss of p53 expression will alter gene transcription (repair enzymes will not be induced, nor will apoptosis be induced if the damage cannot be repaired), but does not hinder the normal regulation of gene transcription.

Li-Fraumeni syndrome results from which one of the following? Choose the one best answer. (A) Inability to recognize DNA damage (B) Inability to regulate CDKs (C) Inability to regulate a tyrosine kinase (D) Inability to regulate gene transcription (E) Inability to activate a heterotrimeric G-protein

*The answer is F.* Most lysosomal hydrolases have their highest activity near an acidic pH of 5.5 (pH optimum) and little activity in a neutral or basic environment. The intralysosomal pH is maintained near 5.5 by vesicular ATPases, which actively pump protons into the lysosome. The cytosol and other cellular components have a pH near 7.2, and are therefore protected from escaped hydrolases. The pH of the blood is maintained between 7.2 and 7.4, so the escaped lysosomal enzymes will have no activity at that pH, and will not affect the proteins and cells in the circulation. I-cell disease results from the inability to appropriately target lysosomal proteins, and it is a lysosomal storage disease.

Lysosomal hydrolases are targeted to the lysosome by the addition of a carbohydrate residue to the protein. An inability to add this carbohydrate leads to a disease in which the lysosomal hydrolases are treated as secreted proteins, and are exported from the cell, rather than taken to the lysosomes. The secreted proteins will have which one of the following effectson the cells and proteins in the circulation? (A) The blood cells will have their membrane proteins digested. (B) The blood cells will have their carbohydrates on the cell surface removed. (C) The blood cell membranes will become leaky, leading to the death of the blood cells. (D) Circulating proteins will be degraded, whereas the blood cells will be protected against the enzymes. (E) Circulating proteins will be targeted to the spleen for removal. (F) There will be no effect on the proteins and cells in the circulation.

*The answer is A.* Receptor-mediated endocytosis refers to the clustering of receptors over clathrin-coated pits in the inner membrane, and then invagination of the membrane to form an intracellular vesicle that contains the receptor-growth factor complex. Exocytosis is the opposite effect-an intracellular vesicle fuses with the plasma membrane to release its contents into the extracellular space. Pinocytosis refers to endocytosis without the receptors-small particles can be taken into the cell through vesicle formation on the cell surface. Potocytosis refers to receptor-mediated entry into a cell through caveolae, and not through clathrin-coated pits. Phagocytosis refers to the forming of a membrane around a particle, and then the endocytosis of that membrane containing the particle (or bacteria).

Many growth factors, upon binding to their receptor, exhibit downregulation, in which the receptor number on the cell surface decreases. This occurs due to which one of the following processes? (A) Endocytosis (B) Exocytosis (C) Pinocytosis (D) Potocytosis (E) Phagocytosis

*The answer is A.* Most of the drugs that target specific types of G protein-coupled receptors are either agonists that bind to the ligand binding site and stimulate receptor activity or are antagonists that bind to the receptor and prevent ligand binding. The G protein α and βγ subunits, adenylate cyclase, and phospholipase C are all elements shared among many types of receptors.

Many of the drugs used in the treatment of hypertension and cardiovascular disease are designed to interfere with the action of cell surface receptors that couple to heterotrimeric G proteins. In order for these drugs to operate in a specific manner so that cellular responses to only one type of receptor are affected, the drug would need to be targeted toward which element of the pathway? A. The ligand binding site of the receptor B. The βγ complex of the G protein C. The α subunit of the G protein D. Adenylate cyclase E. Phospholipase C

*The answer is B.* The child has hereditary retinoblastoma, which is due to an inherited mutation in the rb gene. As the rb gene is a tumor suppressor gene, once loss of heterozygosity occurs, the function of rb in the cell cycle is lost. Rb helps to regulate the E2F family of transcription factors. Once cyclin D is synthesized, and activates a pair of CDKs, rb protein is phosphorylated, which causes it to leave a complex with the E2F factors. The removal of rb from the protein complex activates the E2F proteins, which initiate new gene transcription to allow the cell to transition to the S phase of the cell cycle. In the absence of any functional rb gene product, the transition to S phase is unregulated, and occurs continuously, leading to tumor growth. The rb gene product is not required for any other checkpoints in the cell cycle.

On a routine newborn exam, it is noted that the red reflex is absent in one eye. An MRI shows a tumor blocking the retina. Regulation at which phase of the cell cycle would be affected by the mutation that leads to this tumor? (A) G0 to G1 (B) G1 to S (C) S to G2 (D) G2 to M (E) M to G1 (F) G1 to G0

*The answer is C.* The occurrence of a variety of cancers at fairly early ages in this family, particularly the finding of osteosarcoma in such a young girl, suggests the possibility of an inherited disorder of a tumor suppressor gene. Since the tumors are not associated with the eye, RB1 is unlikely as the cause. The spectrum of cancers in the family is consistent with the Li-Fraumeni syndrome, which involves inheritance of a loss-of-function mutant form of the tumor suppressor gene, TP53, encoding p53.

Osteosarcoma has recently been diagnosed in a 12-year-old girl. Family history indicates that her paternal aunt died of breast cancer at age 29 after having survived treatment for an adrenocortical carcinoma. An uncle died of a brain tumor at age 38 and the patient's father, age 35, has leukemia. 6. An analysis of this patient's DNA would most likely reveal a mutation in which of the following genes? A. RB1 B. RAS C. TP53 D. c-ABL E. PKC

*The answer is B.* p53 is a protein which scans the genome for damage, and when damage is spotted, it induces the synthesis of genes which will stop the cell from continuing through the cell cycle. p53 will also lead to the synthesis of genes involved in repairing the DNA damage. Once the damage has been repaired, the cell will resume its passage through the cell cycle. If the damage cannot be repaired, apoptosis will be initiated in the cell. If p53 is missing, or mutated such that its functions are lost, damaged DNA will be replicated, and at times, the replisome will make errors repairing the damage. This will increase the overall mutation rate of the cell such that eventually mutations will appear in genes which regulate cell proliferation, and a cancer will develop. p53 is not involved in Wnt signaling or activation of MAP kinases. Functional p53 can increase apoptotic events, but the lack of p53 will actually decrease the frequency of apoptosis in cells. This protein is also not involved in transcriptioncoupled DNA repair.

Over 50% of human tumors have developed an inactivating mutation in p53 activity. The lack of this activity contributes to tumor cell growth via which one of the following mechanisms? (A) Loss of Wnt signaling (B) Increase in DNA mutation rates (C) Activation of MAP kinases (D) Increase in apoptotic events (E) Increase in transcription-coupled DNA repair

*The answer is E.* At the cellular level, the RB1 gene is recessive because loss of function affecting both alleles must occur to produce disease. This patient has inherited a defective RB1 allele from her father and is thus heterozygous at the RB1 locus. Most of her retinal precursor cells have one functional RB1 allele and those cells proliferate under normal growth restraints. However, these cells are susceptible to mutations affecting pRb function or an error leading to loss of the remaining functional RB1 allele. These mutations occur by chance during cell division and lead to a tumor by clonal outgrowth. The process by which the sole functional allele is lost or mutated is referred to as loss of heterozygosity (LOH).

Patients with retinoblastoma suffer from a high incidence of tumors arising from clonal outgrowth of some retinal precursor cells due to mutation of the tumor suppressor gene RB1. Analysis of cells from these tumors indicates that both copies of the RB1 gene are mutated or lost, whereas the surrounding retinal cells have at least one functional RB1 allele. Which of the following terms best describes the genetic phenomenon that leads to tumor development in retinoblastoma patients? A. Loss of imprinting B. Deregulated expression C. Incomplete penetrance D. Gain of function E. Loss of heterozygosity

*The answer is D.* Lysosomes depend on a proton gradient to acidify their intracellular milieu, such that the lysosomal hydrolases will be at their pH optima (around 5.5). Mitochondria require a proton gradient across their inner membrane in order to synthesize ATP via oxidative phosphorylation. The nucleus does not concentrate protons; the intranuclear space has the same pH as the cytoplasm.

Proton gradients across membranes are essential for the functions of which of the following organelles? Choose the one best answer. (A) Lysosomes (B) Mitochondria (C) Nucleus (D) Lysosomes and mitochondria (E) Lysosomes and nucleus (F) Nucleus and mitochondria

*The answer is A.* The SH2 (Src Homology 2) domain is a structurally conserved protein domain contained within the Src oncoprotein and in many other intracellular signal-transducing proteins. SH2 domains allow proteins containing those domains to dock to phosphorylated tyrosine residues on other proteins. SH2 domains are commonly found in adaptor proteins that aid in the signal transduction of receptor tyrosine kinase pathways.

SH2 domains are (A) protein domains that bind phosphotyrosine-containing peptides (B) the domains on receptor protein-tyrosine kinases that contain the phosphorylated tyrosine (C) domains that mediate the dimerization of receptor protein-tyrosine kinases (D) the domains on receptor protein-tyrosine kinases that possesses the kinase activity

*The answer is C.* The heterotrimeric G-proteins bind GTP on the α subunit, which activates the subunit. The activation is self-controlled by a built-in GTPase that is present within the α subunit. This activity slowly hydrolyzes the bound GTP to GDP, thereby inactivating the subunit, and allowing the heterotrimer to re-form. Similar events occur in protein synthesis with both initiation factors and elongation factors. These factors are active when GTP is bound to them, and a built-in GTPase activity within these factors limits the length of time they are active. Such control systems are not observed in DNA or RNA synthesis, or in carrier mediated transport across a cellular membrane.

The regulation of heterotrimeric G-proteins is similar to the regulation of which one of the following processes? (A) DNA synthesis (B) RNA synthesis (C) Protein synthesis (D) Active transport (E) Facilitated transport

*The answer is C.* The combination of TGF-α and TGF-β leads to collagen production in fibroblasts. The collagen is secreted into the soft agar and provides an adherence point for the fibroblasts, which allows them to proliferate. The cells will not grow if this adherence point is not present. The effects of TGF-α and TGF-β do include activation of cell growth, but this can only occur in soft agar if collagen has been synthesized and secreted. The factors do not block ras expression or inhibit LETS (fibronectin) synthesis. Oxidative phosphorylation is also not directly affected by these growth factors.

Treatment of fibroblasts with TGF-β and TGF-α will allow such cells to grow in soft agar, a property of transformed cells. This occurs due to which of the following? (A) Activation of cell growth (B) Activation of oxidative phosphorylation (C) Activation of collagen synthesis (D) Inhibition of ras expression (E) Inhibition of LETS synthesis

*The answer is C.* The first thing that happens when the receptor binds its specific signal molecule is that a change in conformational structure of the extracellular domain causes receptors to dimerise. This brings two receptor cytoplasmic regions close together, allowing the tyrosine kinase domain of each partner to phosphorylate tyrosines in the C-terminal 'tail' of the other. The phosphorylated tyrosines are then recognised by proteins with SH2 domains, passing on the message. Events happen in this order whether the receptor is one with its own tyrosine kinase domain or one which associates with a cytoplasmic tyrosine kinase. The tyrosine kinase domain of the receptor does not need to be activated. Since the signal binding domain and the tyrosine kinase domain are separated by a membrane it is not possible for a conformational change in the signal binding domain to cause an activating conformational change in the tyrosine kinase domain.

What is the first event which happens to a growth factor/ cytokine receptor after binding its cognate signal molecule? (A) Binding of signal activates the receptor's cytosolic tyrosine kinase domain. (B) Binding of signal causes the receptor to pick up a tyrosine kinase from the cytosol. (C) Binding of signal causes receptors to dimerise. (D) Binding of signal causes tyrosines in the receptor's C-terminal tail to become phosphorylated.

*The answer is B.* When the blood concentration of insulin is increased, less glucagon is produced and there is an increase in the activity of cAMP phosphodiesterase, which coverts cAMP to AMP. Glucagon stimulates the activity of adenylate cyclase, production of cAMP, and protein kinase A activity.

When the blood concentration of insulin is increased, less glucagon is produced and there is a(n) (A) Increases in the activity of adenylate cyclase (B) Increase in the activity of cAMP phosphodiesterase (3', 5'-phosphodiesterase) (C) Increase in binding of cAMP to the inhibitory subunits of protein kinase A (D) Increase in cAMP (E) Increase in protein kinase A activity

*The answer is C.* The initiating procaspases of the intrinsic apoptosis pathway are attached to the outer membrane of the mitochondria. They are released when the intrinsic pathway is initiated, forming a multimolecular cluster called the apoptosome in which the clustered procaspases can self activate. Thus the mitochondria are the organelles involved in the initiation of internally stimulated apoptosis. In some circumstances incorrectly folded proteins that accumulate in the endoplasmic reticulum can be transported to the cytoplasm where they trigger apoptosis directly. However, this is an unusual mode of initiation which occurs rarely.

Which cellular organelles are involved in the initiation of the intrinsic pathway of apoptosis? (A) endoplasmic reticulum (B) lysosomes (C) mitochondria (D) peroxisomes

*The answer is A.* Almost all signal molecules are water soluble, either small molecules, such as adrenaline, or small to medium sized peptides like epidermal growth factor and interferon. The lipid bilayer of the outer cell membrane presents a barrier which these molecules cannot cross. These signals affect the cell without entering it, by binding to receptors located on the cell surface facing outwards. However there is one quite small group of hormones and signal molecules which work differently. These are mostly molecules synthesised from the sterol cholesterol and called steroid hormones, though the group includes two other molecules, the thyroid hormones and retinoic acid, which are of similar size and solubility properties but different structure. They are hydrophobic molecules and can diffuse through the outer membrane freely. They enter the cell and bind to a receptor which is normally located in the cytosol, but in a few cases is in the nucleus. Progesterone is a steroid hormone.

Which of the following signalling molecules binds to a receptor situated in the cytosol, rather than the outer membrane of the cell (A) Progesterone. (B) Adrenaline (Epinephrine). (C) Epidermal growth factor. (D) Interferon.

*The answer is B.* Sequencing of the human genome has shown more genes coding for serpentine receptors than for any other single type of protein, indicating the importance of this type of message transfer. These receptors can activate a range of different heterotrimeric G-proteins which work through different message transduction pathways. Glucagon activates a membrane enzyme, adenylate cyclase, via a G-protein producing the second messenger cyclic AMP from ATP. Light photons activate a receptor in rod and cone cells which cause a G-protein called transducin to activate a cyclic GMP phosphodiesterase enzyme, while vasopressin activates the membrane enzyme phospholipase C via a G-protein, generating the second messengers inositol triphosphate and diacyl glycerol from the membrane phospholipid phosphatidyl inositol 4,5 bis phosphate. Insulin does not work via this type of receptor however, but binds to a receptor which is similar to the growth factor/cytokine type.

Which of the following signals does not activate a serpentine receptor which in turn activates a heterotrimeric G-protein? (A) Glucagon. (B) Insulin. (C) Light in retinal rod and cones cells. (D) Vasopressin.

*The answer is B.* In the MAP Kinase pathway, a growth factor binds to the tyrosine receptor, which leads to the dimerization of the receptor. The tyrosine residues are then phosphorylated and then Growth Factor Receptor Bound Protein 2 (GRB2) binds to the tyrosine phosphorylated sequences via its SH2 domain. SOS (GEF) is then recruited and binds to GRB2 and inactive Ras/GDP binds to SOS. SOS then activates Ras/GDP to Ras/GTP by exchanging the GDP on to GTP. Ras then binds to effector proteins, such as Raf, which activates a series fo downstream kinases - MAP kinases.

Which of the following statements is true? (A) MAP kinase is important for phosphorylating MAP kinase kinase (B) SOS acts as a GEF to activate Ras (C) Ras becomes activated when an RTK phosphorylates its bound GDP to create GTP (D) Raf phosphorylates transcription factors located in the nucleus