Cell Biology

A 6-month-old boy is brought to the pediatrician by his parents, who are first cousins. This is their first child. Physical examination reveals a small, thin, lethargic infant with slightly misshapen long bones. His features are somewhat coarse. Joint movements are restricted, his corneas are clouded, and his gums are underdeveloped. His liver is not enlarged. Serum levels of acid hydrolases are found to be elevated. The child most likely has a defect in which of the following metabolic activities? A. Degradation of dermatan sulfate and heparan sulfate B. Degradation of gangliosides C. Degradation of glycogen D. Degradation of sphingomyelin E. Phosphorylation of mannose moieties F. Phosphorylation of tyrosine moieties Explanation: The correct answer is E. The patient has I-cell disease, also known as mucolipidosis II, which is due to a defective UDP-N-acetylglucosamine-1-phosphotransferase, the enzyme that phosphorylates mannose on enzymes destined for lysosomes. Proteins coded by nuclear DNA are synthesized on cytoplasmic ribosomes, which may be either "free" or associated with the endoplasmic reticulum to form the rough endoplastic reticulum (RER). Proteins synthesized on the RER are transferred into the Golgi apparatus, where they undergo further modifications that determine whether they remain part of the Golgi apparatus, become part of the plasma membrane, or are shipped to lysosomes or mitochondria. Proteins not marked for transport to a specific intracellular site follow the default pathway and are exported into the extracellular compartment. The signal for transport of the acid hydrolases (and probably other enzymes) to the lysosomes is phosphorylation of a terminal mannose moiety on an N-linked oligosaccharide to form mannose 6-phosphate. In I-cell disease, this terminal mannose moiety is not phosphorylated, and the acid hydrolases follow the default pathway and are secreted. Deficiency of alpha-L-iduronidase results in lysosomal accumulation of dermatan sulfate and heparan sulfate (choice A) in several conditions such as mucopolysaccharidosis I, Hurler's disease, or Hurler's/Scheie disease. Hexosaminidase A deficiency (Tay-Sachs disease) is one example of a condition in which ganglioside accumulation occurs (choice B). There are a number of diseases in which glycogen degradation (choice C) is defective. These are collectively termed glycogen storage diseases since they result in abnormal cellular accumulation of glycogen. In Pompe's disease, or type II glycogen storage disease, a lysosomal glucosidase is deficient, resulting in lysosomal glycogen accumulation. Deficiency of sphingomyelinase (choice D), an enzyme involved in degradation of sphingomyelin, results in Niemann-Pick disease. Phosphorylation of tyrosine moieties (choice F) is unrelated to lysosomes or lysosomal enzymes; however, decreased ability to phosphorylate tyrosine moieties might be associated with diabetes or dwarfism. 2

A cell that produces glycoproteins that contain 8-9 mannose residues per sugar chain has a glycosylation enzyme defect in which of the following organelles? A. Endoplasmic reticulum B. Golgi apparatus C. Lysosomes D. Mitochondria E. Plasma membrane Explanation: The correct answer is B. An oligosaccharide chain containing 9 mannoses is transferred from a dolichol carrier to the asparagine in an asn-X-ser/thr sequence during N-linked sugar assembly. This step occurs in the rough endoplasmic reticulum (choice A). The trimming of mannoses down to 5 residues occurs in the Golgi apparatus prior to the addition of complex sugars. A defect in one of the Golgi mannosidases would produce high-mannose sugar chains. No glycosylation occurs in the lysosomes (choice C), mitochondria (choice D), or plasma membrane (choice E). 3 Molecular genetic studies are performed on a family with known familial hypercholesterolemia. In this particular family, the defect in the LDL receptor gene involves a messenger mutation near the 11th exon, in the region of homology with epidermal growth factor receptor precursor. A defect at this site would be most likely to produce which of the following effects? A. Decreased transcription of LDL receptor gene B. Poor internalization of LDL bound to LDL receptor C. Poor retention of the LDL receptor in the membrane D. Reduced binding of LDL E. Trapping of the LDL receptor in the endoplasmic reticulum Explanation: The correct answer is E. Familial hypercholesterolemia, which is due to defective function of the LDL receptor, is an area of intense research. The molecular basis of LDL receptor abnormalities is becoming better understood, and more than 200 mutations in the gene for the LDL receptor have been identified. The gene has 5 general domains and 18 exons. Defects near exons 7 to 14 (including this case) are in the region of homology with epidermal growth factor receptor precursor. This region of the molecule is needed for dissociation of LDL from the receptor in the endosome. Receptors with a defect in this area (sometimes called class II mutations) also have trouble being initially transported to the Golgi complex (transport-deficiency alleles) and become trapped in endoplasmic reticulum.

Decreased transcription of the LDL receptor gene (choice A) is considered a class I mutation and involves the signal sequence domain near exon 1. Poor internalization of LDL bound to LDL receptor (choice B) is considered a class IV mutation. Such mutations are associated with the membrane-spanning/cytoplasmic domain, specifically near exon 18. Poor retention of the LDL receptor in the membrane (choice C) is considered a class IV mutation and is associated with the membrane-spanning/cytoplasmic domain, specifically near exons 2-6. Reduced binding of LDL (choice D) is considered a class III mutation and involves the LDL binding domain near exons 2-6. 4 In order to study calcium fluxes in living cells, a researcher employs fluorescent calcium indicators and laser scanning confocal microscopy. In her experiments, she stimulates smooth muscle cells using a variety of pharmacological agents, then videotapes the calcium fluxes in the cell in real time using the confocal microscope. Which of the following drugs would most likely produce the strongest calcium signal in the smooth muscle cells? A. Clonidine B. Isoproterenol C. Phentolamine D. Phenylephrine E. Timolol Explanation: The correct answer is D. Two pieces of information are necessary to answer this question. First, you must know the drug classes represented by the answer choices. Second, you must know what second messengers are associated with these receptors. Phenylephrine is an α1 agonist; α1 receptors are coupled to Gq, which stimulates phospholipase C (PLC), leading to hydrolysis of phosphatidylinositol-4,5-bisphosphate (PIP2), a phospholipid in the plasma membrane. PIP2 is split into two second messengers: inositol-1,4,5-triphosphate (IP3) and diacylglycerol. IP3 diffuses through the cytoplasm to internal storage sites, where it triggers the release of calcium, thus providing the signal for this experiment. Diacylglycerol remains in the membrane and activates protein kinase C. Clonidine (choice A) is an α2 agonist. α2 receptors are coupled to Gi, which inhibits adenylate cyclase, leading to decreases in intracellular cAMP levels. Isoproterenol (choice B) is a nonselective β agonist. All β receptors are coupled to Gs, which stimulates adenylate cyclase, leading to elevated intracellular cAMP levels. Phentolamine (choice C) is a nonselective α antagonist. This drug would prevent the increase in calcium caused by stimulation of α1 receptors. Timolol (choice E) is a nonselective β antagonist. This drug would prevent β-induced increases in cAMP. 5 A rapid way to purify proteins that are targeted to the lysosomes would be to use affinity

chromatography. An appropriate antibody to use on an affinity chromatography column would be one directed against A. acid hydrolases B. clathrin C. glucose-6-phosphate D. mannose-6-phosphate E. sialic acid Explanation: The correct answer is D. Proteins that are targeted for lysosomes have mannose-6-phosphate on their sugar chains. The mannose-6-phosphate is on an external site on the protein so that it can be recognized by the mannose-6-phosphate receptor on the lysosomal surface, and would also be easily accessible to an antibody directed against it. Acid hydrolases (choice A) are enzymes found inside the lysosomes. Coated pits (the cellular site for receptor-mediated endocytosis) contain clathrin (choice B). Production of glucose-6-phosphate (choice C) from glucose is the first step in the process of glycolysis. Sialic acid (choice E) is a terminal glycosylation product added to proteins (usually those destined for secretion) in the Golgi apparatus. 6 Liver cells in culture were kept at 0°C and treated with trypsin to digest the receptors on the cell surface. The temperature was then raised to 37°C and radioactive LDL was added to the culture media. Several hours later, the labeled LDL was found to be inside the cells. This specific process of LDL uptake is A. active transport B. facilitated diffusion C. phagocytosis D. pinocytosis E. receptor-mediated endocytosis Explanation: The correct answer is E. Even though the surface LDL receptors were digested by the trypsin, the recycling of unoccupied receptors to the cell surface provides a continual supply of new receptors to bind the labeled LDL. The LDL-receptor complex is internalized by receptormediated endocytosis. Active transport (choice A) is the energy-dependent movement of molecules across a membrane and against a concentration gradient.

Facilitated diffusion (choice B) is the transport of low-permeability molecules with the aid of a carrier protein. Phagocytosis (choice C) is the process by which cells such as macrophages and neutrophils engulf large particles. Pinocytosis (choice D) is the uptake of small molecules in solution, and is receptor independent. 7 Which of the following is an example of a glycosylated integral membrane protein with 7 transmembrane segments? A. Adenylate cyclase B. Beta-adrenergic receptor for epinephrine C. Cystic fibrosis transmembrane conductance regulator channel D. Glucose transporter E. Na+/K+ ATPase Explanation: The correct answer is B. This question requires that you recognize the family of receptors that interact with G proteins to initiate a signal transduction cascade. These receptors are all glycosylated integral membrane proteins that have 7 transmembrane segments. Beta-adrenergic receptors for epinephrine are an example. Adenylate cyclase (choice A) is an intracellular effector protein involved in signal transduction. It is not an integral membrane protein. The cystic fibrosis transmembrane conductance regulator channel (choice C) is a cyclic AMP activated chloride channel. Glucose transporters (choice D) are integral membrane proteins with 12 membrane-spanning domains. Na+/K+ ATPase (choice E) pumps sodium into and potassium out of the cell. It provides energy for active transport by hydrolyzing ATP. 8 A cell biologist wishes to examine the microstructure of an integral membrane protein. She solubilizes the protein by destabilizing its association with the membrane lipid bilayer. Which of the following techniques did she most likely employ? A. Alterations in pH B. Detachment of protein prenyl groups C. Dissociation of phospholipid polar head groups D. Increase in ionic strength

E. Interruption of hydrophobic interactions Explanation: The correct answer is E. This question is asking you what forces are responsible for retaining integral membrane proteins within the lipid bilayer, i.e., hydrophobic interactions. Amphipathic agents such as detergents are used to solubilize integral membrane proteins; they do so by disrupting hydrophobic interactions between integral membrane proteins and other membrane constituents, such as phospholipids. Choices A, B, C, and D would destabilize the association of peripheral membrane proteins with the membrane lipid bilayer. 9 A child with coarse facies, corneal clouding, joint stiffness and mental retardation is found to have a partial defect in N-acetylglucosaminotransferase. Damage to this enzyme directly affects which of the following biochemical functions? A. Formation of polysomes B. Microtubule attachment to centrioles C. Synthesis of ribosomal RNA D. Targeting of enzymes for lysosomes E. Transport of sugars into mitochondria Explanation: The correct answer is D. The disease is the type III form (pseudo-Hurler syndrome) of mucolipidosis, similar to I-cell disease. Specifically, the defect in I–cell disease and pseudo-Hurler syndrome is an inability of N-acetylglucosamine-1-phosphotransferase to add the recognition marker mannose phosphate to enzymes destined to enter lysosomes. Consequently, there is a generalized defect in lysosome function. For reference, Hurler's syndrome is the prototype mucopolysaccharidosis, characterized by skeletal deformities, mental retardation, deafness, blindness, and death in childhood. (Quasimodo, in the "Hunchback of Notre Dame" may have had a mucopolysaccharidosis). 10 A medical student is studying a liver biopsy taken from a regenerating liver following a partial hepatectomy. The student sees a dividing cell in which the chromosomes are aligned on a plate. Which of the following best describes the character of the chromosomes in the plate? A. 23 chromosomes with 2N DNA total and with each chromosome having one chromatid B. 23 chromosomes with 4N DNA total and with each chromosome having two chromatids C. 46 chromosomes with 2N DNA total and with each chromosome having one chromatid D. 46 chromosomes with 4N DNA total and each chromosome having two chromatids E. 92 chromosomes with 2N DNA total and with each chromosome having two chromatids Explanation:

The correct answer is D. The cell division is mitotic rather than meiotic. The cell is in metaphase, as evidenced by alignment of chromosomes on the metaphase plate. Doubling of the DNA occurs much earlier, in prometaphase (late prophase), so the cell has "4N" DNA (double the normal 2N diploid DNA). The chromosomes at the metaphase plate consist of 2 chromatids each, but the total number of chromosomes is still the normal 46. 23 chromosomes with 2N DNA (choice A) is wrong because stages with 23 chromosomes only occur in meiosis. 23 chromosomes with 4N DNA (choice B) is wrong because stages with 23 chromosomes only occur in meiosis. Additionally, there is never a point in either meiosis or mitosis when 23 chromosomes contain 4N DNA. 46 chromosomes with 2N DNA, each chromosome having one chromatid (choice C), is wrong because the chromosomes have two chromatids and the DNA is therefore doubled (producing 4N DNA total) in metaphase. The cell normally has 46 chromosomes with 2N DNA only after mitosis is complete. 92 chromosomes (choice E) is wrong because in anaphase and telophase, at which points there are 92 chromosomes (in two cells which are not yet cleaved), there is a total of 4N DNA and each of the newly split chromosomes contains one chromatid. 11 A genetic mutation that results in abnormal stimulatory G protein (Gs) structure would adversely affect which of the following mechanisms? A. Active transport B. Facilitated diffusion C. Pinocytosis D. Receptor-mediated endocytosis E. Signal transduction Explanation: The correct answer is E. G proteins are involved in signal transduction, which allows macromolecules to affect a cell's biological functions without crossing the plasma membrane. The binding of an agonist, such as a hormone, to a receptor on the plasma membrane causes a conformational change in the receptor, which then interacts with a stimulatory G protein (Gs), accompanied by the binding of GTP to the alpha subunit of the G protein. The alpha subunit of the G protein (with attached GTP) dissociates from the beta and gamma subunits, and interacts with intracellular effector molecules such as adenylate cyclase, which it activates. Active transport (choice A) is the movement of molecules across a membrane from a region of low concentration to a region of high concentration; it requires both a carrier protein and the expenditure of energy (usually supplied by the hydrolysis of ATP by an ATPase). Facilitated diffusion (choice B) is the movement of a molecule down its concentration gradient by means of a protein carrier. This type of diffusion is suited for molecules with low permeability resulting from inappropriate size or polarity. Pinocytosis (choice C) is also known as fluid-phase endocytosis and refers to the uptake of molecules that are in solution. Receptors are not involved.

Receptor-mediated endocytosis (choice D) refers to the engulfment of a molecule that binds to a receptor on the surface of the cell, occurring commonly at the clathrin-coated pits on the plasma membrane. 12 Which of the following is a characteristic of steroid hormones? A. Activation of adenylate cyclase B. Activation of protein kinases C. Plasma membrane receptors D. Stimulation of cellular protein synthesis E. Termination of effects by phosphodiesterase Explanation: The correct answer is D. Peptide/protein hormones and steroid hormones act by very different mechanisms. Steroid hormones circulate in the bloodstream, then leave the bloodstream by dissolving in the lipid-rich plasma membrane. Steroids cross the plasma membrane and enter the cytoplasm, where they bind to a mobile protein receptor. The hormone/receptor complex then enters the nucleus, and triggers RNA synthesis leading to protein synthesis, thereby changing the cellular response. In contrast, protein and amine hormones bind to specific receptors on the outer surface of the plasma membrane (choice C), but do not enter the cytoplasm. Many activate adenylate cyclase (choice A) and thereby increase cAMP. The increased cAMP stimulates protein kinases (choice B), which change the target cell responses by phosphorylating intracellular proteins. Hydrolysis of cAMP by phosphodiesterases (choice E) terminates the effect of the hormone. 13 A 10-year-old child has a rare genetic disease that resembles Hurler syndrome, with “gargoyle facies,” multiple skeletal abnormalities, and cardiac, central nervous system, and liver involvement. The biochemical basis of the disease is an inability to add the recognition marker mannose phosphate to enzymes. In which of the following organelles does this step usually occur? A. Endoplasmic reticulum B. Golgi apparatus C. Lysosome D. Mitochondria E. Ribosome Explanation: The correct answer is B. The disease is I cell disease, which is a rare genetic disease that may receive disproportionate attention both because it is a mucolipidosis (a form of generalized lysosomal disorder) with accumulation of abnormal chemical material in lysosomes, and because the biochemical basis illustrates an interesting mechanism. Specifically, the Golgi apparatus in cells of these patients has an abnormal N-acetyl-glucosaminotransferase (N- acetylglucosamine-1phosphotransferase), and is not able to add the necessary recognition marker mannose phosphate

to enzymes usually destined to enter lysosomes. A complete deficiency of this enzyme (type I form of I cell disease) causes death early in life; partial deficiencies (Type III form) produce milder disease (pseudo-Hurler syndrome) with survival to adulthood. Endoplasmic reticulum (choice A) receives the growing peptide chain of the enzymes, but is not the site of addition of recognition markers. Lysosomes (choice C) are the normal destination for the enzymes with the mannose phosphate marker, but do not receive them if the marker is not present. Mitochondria (choice D) supply the ATP to drive these chemical reactions. Ribosomes (choice E) are the site of synthesis of the amino acid chains of proteins, but not the site of addition of the destination marker. 14 Which of the following amino acids is post-translationally hydroxylated in the cytoplasm of fibroblasts? A. Cysteine B. Glycine C. Proline D. Serine E. Tyrosine Explanation: The correct answer is C. Hydroxylation of proline in fibroblasts generates the modified amino acid hydroxyproline. This is an example of post-translational modification. Hydroxyproline is involved in stabilizing the three-dimensional triple helix structure of collagen. Cysteine (choice A) is unique in its ability to form a covalent disulfide bond with another cysteine residue elsewhere in the protein molecule, thereby forming a cystine residue. Such strong disulfide bonds stabilize the three-dimensional structure of the protein. Glycine (choice B) is abundant in fibroblasts since it constitutes every third amino acid in the primary sequence of collagen. However, glycine is not hydroxylated. Serine (choice D), tyrosine (choice E), and threonine can all be phosphorylated posttranslationally to form phosphoserine, phosphotyrosine, and phosphothreonine, respectively. These phosphorylated amino acids are believed to play a role in signal transduction. 15 A cell biologist wants to activate protein kinase C in cultured cells. Stimulation of which of the following receptor types would most likely produce the greatest activation? A. Alpha 1 adrenergic B. Beta 1 adrenergic C. Dopamine-2 (D2) D. Gamma-aminobutyric acid, type A (GABAA)

E. Nicotinic cholinergic Explanation: The correct answer is A. It is a good idea to know the mechanism of action of various receptors. Alpha-1 adrenergic receptors are coupled to the G protein, Gq. This G protein causes breakdown of the membrane phospholipid phosphatidylinositol bisphosphate (PIP2), forming the products diacylglycerol, which stimulates protein kinase C, and inositol triphosphate (IP3), which releases calcium from the endoplasmic reticulum. Beta-1 adrenergic receptors (choice B), like all beta receptors, are coupled to the Gs G protein, leading to the activation of adenylate cyclase. This increases intracellular concentrations of cAMP by converting ATP to cAMP. cAMP, in turn, stimulates protein kinase A, also known as cAMP-dependent protein kinase. Dopamine-2 receptors (choice C) are coupled to Gi. Stimulation of these receptors inhibits adenylate cyclase, thus lowering cAMP levels and protein kinase A activity. Gamma-aminobutyric acid, type A (GABAA) receptors (choice D) are not G-protein-coupled receptors, but are instead ligand-gated ion channel receptors. Stimulation of these inhibitory receptors causes a chloride flux, thus stabilizing the membrane potential of the cell. Nicotinic cholinergic receptors (choice E) are not G-protein-coupled receptors, but are instead ligand-gated ion channel receptors. Stimulation of these receptors causes sodium influx, leading to excitation of the cells. 16 A child with retinoblastoma is found to have a 13q14 deletion. The Rb gene, which resides at this locus, produces which kind of tumor-associated protein? A. Cell cycle regulator B. Growth factor C. Growth factor-binding protein D. Growth factor receptor E. Transcription activator Explanation: The correct answer is A. The Rb gene is an example of a tumor suppressor gene. Tumor suppressor genes encode proteins that downregulate cell growth; consequently, their deletion leads to the development of cells with a growth advantage over normal cells. Even if you know nothing about the Rb protein, choice A is still the only logical answer because it is the only example of a protein that, if absent, would favor cell growth. The Rb protein binds to transcription factors in the nucleus, preventing cells from progressing from the S1 to M stages of the cell cycle. Children born with a 13q14 deletion have only one chromosome encoding Rb; therefore only a single "hit " is required to completely knock out Rb production and lead to the development of retinoblastoma. All of the incorrect choices are proteins encoded by oncogenes, rather than tumor suppressor genes. Oncogenes favor tumorigenesis through overexpression, not deletion. Growth factors (choice B) are oncoproteins that are produced by tumors and have a positive feedback effect. Examples of growth factors are PDGF and fibroblast growth factor; the oncogenes encoding them are sis and hst-1, respectively.

The prototypical growth factor-binding protein (choice C) is ras, which is mutated in a large variety of cancers. Ras normally functions as an activator of protein kinases that regulate cell growth. Overactivity of the ras protein is highly mitogenic. Growth factor receptors (choice D) are either expressed as mutant forms or overexpressed in tumors, leading to upregulation of growth. An example of a growth factor receptor oncogene is erb-B2, present in some breast cancers. Transcription activators (choice E) are DNA-binding proteins that promote DNA transcription. Amplification of these oncogenes causes cancer by promoting the transcription of growth-related genes. 17 At which of the following sites is the characteristic triple helical structure of the collagen initially formed? A. Extracellular space B. Golgi body C. Nucleus D. Rough endoplasmic reticulum E. Smooth endoplasmic reticulum Explanation: The correct answer is B. Collagen formation begins with transcription of mRNA from appropriate DNA genes in the nucleus. While still within the nucleus, the mRNA is spliced. It is then transported through the cytoplasm to the ribosomes on the rough endoplasmic reticulum. Individual chains are translated on the ribosomes, with the ends feeding into the endoplasmic reticulum lumen. Within the lumen, glycosylation of the individual chains occurs. The material then moves toward the Golgi bodies (whose lumens are connected to the endoplasmic reticulum) where the triple helices of procollagen form. The procollagen is then secreted into the extracellular space, where cleavage of pro-peptides and cross- linking of different triple helices occurs, maturing the collagen. The extracellular space (choice A) is the site of procollagen cleavage and cross- linking. The nucleus (choice C) is the site of mRNA transcription and splicing. The rough endoplasmic reticulum (choice D) is the site of chain translation and glycosylation. The smooth endoplasmic reticulum (choice E) does not participate in collagen synthesis.