MBB322 tutorials

Question 1

Tutorial 1 1. The melting points of a series of 18-carbon fatty acids are: steric acid (18:0) 69.6°C oleic acid (18:1) 13.4°C linoleic acid (18:2) -5°C linolenic acid (18:3) -11°C What structural aspect of these 18 carbon fatty acids can be correlated with the melting point? Provide an explanation for this trend.

Answer 1

Tutorial 1 1. The number of cis double bonds. Each cis double bond causes a bend in the hydrocarbon chain and bent chain are less well packed than straight chains. The lower the extent of packing the lower the melting temperature.

Question 2

Tutorial 1 2. Branched chain fatty acids are found in some bacterial membrane lipids. Would their presence increase of decrease the fluidity of the membrane? Why?

Answer 2

Tutorial 1 2. Branched chain fatty acids will increase the fluidity of membranes (i.e.lower their melting point) because they decrease the extent of packing within the membrane. The effect of branches is similar to that of bends caused by double bonds.

Question 3

Tutorial 1 Q3. Biomembranes contain many types of lipid molecules. What are the three main type of lipid molecules found in biomembranes? How are the three types similar, and how are they different?

Answer 3

Tutorial 1 Q3. The three main types of lipid molecules in biomembranes are phosphoglycerides, sphingolipids, and steroids. All are amphipathic molecules having a polar head group and a hydrophobic tail, but the three types differ in chemical structure, abundance and function.

Question 4

Tutorial 2 Q1. Margarine is made from vegetable oil by a chemical process. Do you suppose this process converts saturated fatty acids to unsaturated fatty acids or vice versa? Explain your answer.

Answer 4

Tutorial 2 Q1. Vegetable oil is converted to margarine by reduction in double bonds (by hydrogenation) which converts unsaturated fatty acids to saturated ones. This changes allows the fatty acid chains in the lipid molecules to pack more tightly against each other, decreasing the fluidity.

Question 5

Tutorial 2 Q2. What are the common fatty acid chains in glycerophospholipids, and why to these fatty acid chains differ in their number of carbon atoms by multiples of 2?

Answer 5

Tutorial 2 Q2. The common fatty-acid chains in glycerophospholipids include myristate (14), palmitate (16) stearate (18) oleate (20) linoleate (22) arachidonate (24) These fatty acids differ in carbon atom number by multiples of 2 because they are elongated by the addition of 2 carbon units. For example, the acetyl group of acetyl CoA is a 2-carbon moiety. mpsola

Question 6

Tutorial 2 3. Although both faces of a biomembrane are composed of the same general types of macromolecules, principally lipids and proteins, the two faces of the bilayer are not identical. What accounts for the asymmetry between the two faces?

Answer 6

Tutorial 2 Q3. Since biomembranes form closed compartments, one face of bilayer is exposed to interior, and the other is exposed to the exterior. Each face interacts with different environments and carry out different functions. The different functions are directly dependent on the specific molecular composition of each face. For example, different types of phospholipids and lipid-anchored membrane proteins are typically present on the two faces. GPI anchored proteins – exo face Acylation – adding a fatty acyl group or a prenyl (isoprene group) = cyto face In addition, different domains of integral proteins are exposed on each face of the bilayer. Lipids and proteins of the exoplasmic face are often modified with carbohydrates

Question 7

Tutorial 31. The biosynthesis of cholesterol is a highly regulated process. What is the key regulated enzyme in cholesterol biosynthesis? This enzyme is subject to feedback inhibition. What is feedback inhibition? How does this enzyme sense cholesterol levels in the cell?

Answer 7

Tutorial 31. The key regulated enzyme in cholesterol biosynthesis is HMG (β- hydroxy-βmethylglutaryl)- CoA reductase. This enzyme catalyzes the ratecontrolling step in cholesterol biosynthesis. The enzyme is subject to negative feedback regulation by cholesterol. In fact, the cholesterol biosynthetic pathway as the first biosynthetic pathway shown to exhibit this type of end-product regulation. As the cellular cholesterol level rises, the need to synthesize additional cholesterol goes down. The expression and -enzymatic activity of HMGCoA reductase is suppressed. HMG-CoA reductase has eight transmembrane segments and, of these, five compose the sterol-sensing domain. Sterol sensing by this domain triggers the rapid, ubiquitin-dependent proteasomal degradation of HMG-CoA reductase. Homologous domains are found in other proteins such as SCAP (SREBP cleavageactivating protein) and Niemann-Pick C1 (NPC1) protein, which take part in cholesterol transport and regulation. Transcription is also controlled through the activity of SREBP. When cholesterol levels are low, SREBP and SCAP bud from the ER membrane and are transported to the Golgi. In the Golgi there are two proteases that cleave SREBP, liberating the bHLH DNA binding domain which can then translocate to the nucleus and transcriptionally regulate genes with an SRE element in their promoter.

Question 8

Tutorial 32. An early step in fatty acid synthesis is that acetyl-CoA is carboxylated, to be immediately followed by decarboxylation of the product. What is the carboxylated product called? What do you think is the point of this?

Answer 8

Tutorial 32. Fatty acids are synthesized two carbon atoms at a time, but the donor ofthese is a three carbon unit, malonyl-CoA. Acetyl-CoA is converted tomalonyl-CoA by an ATP dependent carboxylation. The subsequentdecarboxylation results in a large negative ΔG value. Thus, the point ofcarboxylation and decarboxylation is to make the process of additing twocarbon units to the growing fatty acid chain irreversible.

Question 9

Tutorial 33. Although clinical trials have not yet been carried out, some physicians have suggested that patients being treated with statins also should take a supplement of coenzyme Q (ubiquinone). Suggest a rationale for this recommendation.

Answer 9

Tutorial 33. Statins inhibit HMG-CoA reductase. This enzyme is in the pathway thatleads to the production of farnesyl pyrophosphate. Farnesyl pyrophosphateis a precursor of ubiquinone. Thus, statins might reduce the amount ofubiquinone available for mitochondrial respiration. Ubiquinone can beobtained in the diet as well as directly synthesized. However, it is not clearhow dietary sources can substitute for reduced synthesis.

Question 10

Tutorial 34. True/False. Some membrane proteins are attached to the cytoplasmic surface of the plasma membrane through a C-terminal linkage to a GPI anchor. Explain your answer.

Answer 10

Tutorial 34. False. Proteins that are linked to GPI anchors are attached to the externalsurface of the plasma membrane. The attachment occurs in the lumen of theER which is topologically equivalent to the outside of the cell.

Question 11

Tutorial 35. You have sequenced an oligopeptide containing an RGD sequence surrounded by other amino acids. What is the effect of this peptide when added to a fibroblast cell culture grown on a layer of fibronectin adsorbed to the tissue culture dish? Why does this happen?

Answer 11

Tutorial 35. The RGD sequence on fibronectin mediates binding to integrin proteins. IfRGD-containing peptides were added to a layer of fibroblasts grown on afibronectin substrate in tissue culture, the RGD peptides would compete withfibronectin for binding to the integrins present in the fibroblast extracellularmatrix. As a result, the fibroblasts would likely lose adherence to thefibronectin substrate.

Question 12

Tutorial 42. One type of vertebrate cell that is thought to lack integrins is theerythrocyte (red blood cell). Does this surprise you? Why or why not?

Answer 12

Tutorial 42. This is not surprising. Erythocytes remain as free circulating cells. Theyare characterized by a lack of interaction with other cells or an extracellularmatrix.

Question 13

Tutorial 43. The extracellular matrix is a relatively inert scaffolding that stabilizes thestructure of tissues. True or False and explain why.

Answer 13

Tutorial 43. False. The extracellular matrix plays an active role influencing thedevelopment, migration, proliferation, shape, and metabolism of cells thatcontact it.

Question 14

Tutorial 44. The glycosaminoglycan polysaccharide chains that are linked to specificcore proteins to form the proteoglycan components of the basal lamina arehighly negatively charged. How do you suppose these negatively chargedpolysaccharide chains help to establish a hydrated gel-like environmentaround the cell? How would the properties of these molecules differ if thepolysaccharide chains were uncharged?

Answer 14

Tutorial 44. The high density of negative charges on the polysaccharide componentsof proteoglycans causes the sugar chains to be very extended, occupying alarge volume and providing a hydrated gel-like environment around the cell.The negative charges attract positively charged ions, mainly sodium ions,which in turn draw in water molecules by osmosis. In the absence of thenegative charges, the sugar chains would collapse into fibers or granules,dramatically altering the properties of the basal lamina.

Question 15

Tutorial 4 5. What is the normal function of tight junctions? What can happen to tissues when tight junctions to not function properly?

Answer 15

Tutorial 4 5. Tight junctions help to hold cells together in tissues and control the flow of solutes between cells in an epithelial sheet.

Question 16

Tutorial 51. Propose specific types of mutation in the gene for the regulatory subunitof cyclic AMP-dependent protein kinase (PKA) that could lead to either apermanently active PKA or a permanently inactive PKA.

Answer 16

Tutorial 51. Any mutation that generated a regulatory subunit incapable of binding to the catalyticsubunit would produce a permanently active PKA. When the catalytic subunit is notbound to the regulatory subunit, it is active.Two general types of mutation in the regulatory subunit could produce a permanentlyinactive PKA. A regulatory subunit that was altered so that it could bind the catalyticsubunit, but not bind cyclic AMP, would not release the catalytic subunit, rendering PKApermanently inactive. Similarly, a mutant regulatory subunit that could bind cyclic AMP,but not undergo the conformational change needed to release the catalytic subunit, wouldpermanently inactivate PKA.

Question 17

Tutorial 52. Why do you suppose cells use Ca2+ (intracellular concentration ~10-7 M)for signaling rather than the more abundant Na+ (intracellular concentration~ 10-3M)?

Answer 17

Tutorial 52. Because the intracellular concentration of Ca2+ is so low, an influx of relatively fewCa2+ ions leads to large changes in its cytosolic concentration. Thus, a 10-fold increasein Ca2+ can be achieved by raising the concentration of Ca2+ into the micromolar range,which would require entry of relatively few ions into the cytosol. By contrast, many moreNa+ ions (104 more) would need to enter the cytosol to change its concentration by thesame amount. In muscle, more than a 10-fold change in Ca2+ can be achieved inmicroseconds by releasing Ca2+ from the intracellular stores of the sarcoplasmicreticulum, a task that would be difficult to accomplish if changes in the millimolarrange were required.

Question 18

Tutorial 53. Both rhodopsin in vision and the muscarinic acetylcholine receptor systemin cardiac muscle are coupled to ion channels via G proteins. Describe thesimilarities and differences between these two systems.

Answer 18

Tutorial 53. Activation of muscarinic acetylcholine receptors in cardiac muscle slows the rate ofheart muscle contraction. These receptors are coupled to an inhibitory G protein.Activation of this system causes a decrease in cAMP in the cell that leads to opening ofK+ channels on the cell membrane. The muscle cell becomes hyperpolarized, whichreduces the frequency of muscle contraction. Rhodopsin is a G protein–coupled receptorthat is activated by light. Rhodopsin contains a light absorbing pigment, 11-cis-retinal,that is covalently linked to opsin. In the presence of light, 11-cis-retinal is converted toall-transretinal. This activated opsin then interacts and activates transducin, an associatedG protein. The activated Gα-GTP complex binds to the inhibitory subunit of aphosphodiesterase. The released catalytic subunits of the phosphodiesterase hydrolyzescGMP to 5′-GMP. As a result, the cGMP level declines, leading to the closing of anucleotide-gated ion channel. As with the cardiac muscle system, signal activationultimately results in hyperpolarization of the photoreceptor cells.

Question 19

Tutorial 54. Excitatory neurotransmitters open Na+ channels, while inhibitoryneurotransmitters open either Cl- or K+ channels. Rationalize thisobservation in terms of the effects of these ions on the firing of an actionpotential.

Answer 19

Tutorial 54. Opening Na+ channels allows an influx of Na+ ions that depolarizes the membranetoward the threshold potential for firing an action potential. By contrast, opening eitherCl– or K+ channels opposes membrane depolarization.Both Cl– and K+ ions are near their equilibrium distribution across the membrane: theresting membrane potential (negative inside) balances their concentration differencesacross the membrane (Cl– high outside and K+ high inside). As the membrane begins todepolarize (that is, as the membrane potential becomes more positive), both ions will tendto move down their concentration gradients (Cl– ions into the cell, K+ ions out of thecell). If a channel for either ion is opened, its movement across the membrane will makethe inside of the cell more negative, tending to reestablish the original membranepotential and suppressing the firing of an action potential.

Question 20

Tutorial 5 5. Normally an action potential is propagated unidirectionally along the length of an axon until the axon terminus. What would you predict would be the effect of artificially depolarizing an axon in the middle of its length? describe the action potential

Answer 20

Tutorial 5 5. During an action potential, the voltage dependent Na+ channels open in advance of the voltage-dependent K+ channels, and they close in response to the depolarization of the membrane potential that they produce. This closing is called the refractory period during which the channels cannot reopen. Because most action potentials proceed from the cell body to the axon terminus, this refractory period insures that the depolarizing signal moves in only one direction. When an axon is artificially depolarized in its middle, the channels on either side of the stimulation site are not in a refractory state. Therefore, a signal can be generated that is propagated in both directions from the point of stimulation.

Question 21

Tutorial 56. The outer segments of rod photoreceptor cells can be broken off, isolated,and used to study the effects of small molecules on visual transductionbecause the broken end of each segment remains unsealed. How would youexpect the signaling pathway to be affected following the addition of anonhydrolyzable analog of GTP?

Answer 21

Tutorial 56. A nonhydrolyzable analog of GTP would lead to prolonged activation oftransducin (the G-protein) in response to activated rhodopsin. Continued activationof transducin would keep cyclic GMP phosphodiesterase high, which wouldlead in turn to a protracted decrease in cyclic GMP, a prolonged hyperpolarizationof the membrane.

Question 22

Tutorial 61. A mutation in the Ras protein renders Ras constitutively active (RasD).What is constitutive activation? How is constitutively activate Ras cancerpromoting? What type of mutation might render MAPK constitutivelyactive?

Answer 22

Tutorial 61. Constitutive activation is the alteration of a protein or signaling pathway such that it isfunctional or engaged even in the absence of an upstream activating event. For example,RasD is constitutively active because it cannot bind GAP and therefore remains in theGTP-bound, active state even when cells are not stimulated by growth factor to activate areceptor tyrosine kinase.Constitutively active Ras is cancer promoting because cells will proliferate in the absenceof growth factors, and thus normal regulatory mechanisms for cell proliferation arebypassed.A mutation that made MAPK active as a kinase and able to enter the nucleus withoutbeing phosphorylated by MEK would render MAPK constitutively active.

Question 23

Tutorial 62. 90% of cancer deaths are caused by metastasis. Define metastasis.Explain the rationale for the following new cancer treatment:A. batimastat, an inhibitor of matrix metalloproteinases and of theplasminogen activator receptor.

Answer 23

Tutorial 62. Metastasis is the process by which cancer cells escape their tissue of origin, travelthrough the circulation, and invade and proliferate within another tissue or organ.A. Batimastat inhibits enzymes that degrade the extracellular matrix, and thus cancercells will be unable to digest the basement membrane and escape the tissue of origin.

Question 24

Tutorial 63. Hereditary retinoblastoma generally affects children in both eyes, whilespontaneous retinoblastoma usually occurs during adulthood only in oneeye. Explain the genetic basis for the epidemiological distinct between thesetwo forms of retinoblastoma Explain the apparent paradox: loss-of-functionmutations in tumor suppressor genes at recessively, yet hereditaryretinoblastoma is injerited as an autosomal dominant.

Answer 24

Tutorial 63. In hereditary retinoblastoma, individuals have inherited one mutated copy of the RBgene, and therefore require only a spontaneous mutation in the other copy to lackfunctional Rb protein. The relative frequency of a single spontaneous mutation is highenough that these individuals develop retinoblastoma early in life in both of their eyes.However, in spontaneous retinoblastoma, individuals have inherited two normal copies ofthe RB gene. Therefore, spontaneous mutations in each copy of RB must occur within asingle cell for it to lack functional Rb. The likelihood of a cell’s possessing bothmutations is extremely low, and thus these mutations rarely occur until adulthood andthen usually in a single eye. Because the chance of an individual with hereditaryretinoblastoma receiving an inactivating mutation in the other copy of the RB gene in anyone of the susceptible cells is quite high, the disease is inherited in a dominant manner.

Question 25

Tutorial 64. Normal cells greatly slow their rates of cell division after filling a culturedish with a layer one cell deep (a monolayer). If a circular group of cells isremoved from the middle of the filled culture dish, the cells on the edges ofthe “wound” will begin to divide, fill the space and then stop dividing again.How does the behavior of transformed cells differ?

Answer 25

Tutorial 64. Transformed cells do not exhibit such contact inhibition. When the dish fillsup, cell division does not stop; it continues. The transformed cells would pileup on top of one another on the dish forming multiple layers, instead of themonolayer formed by normal, nontransformed cells.

Question 26

Tutorial 65. You are studying a gene that is involved in the development of tumors.The gene is represented once in the affected cells. The allele of the samegene on the homologous chromosome is normal. When the gene is insertedinto normal cells, the cells are transformed. What type of gene is this?

Answer 26

Tutorial 65. Oncogene

Question 27

Tutorial 71. Imagine that you could microinject cytochrome c into the cytosol of wildtypecells and of cells that were doubly defective for Bax and Bak. Wouldyou expect one, both or neither to undergo apoptosis? Explain yourreasoning.

Answer 27

Tutorial 71. Upon microinjection of cytochrome c, both types of cell undergoapoptosis.The presence of cytochrome c in the cytosol is a signal for theassembly ofapoptosomes and the downstream events that lead to apoptosis.Cells that are defective for both Bax and Bak cannot release cytochrome cfrom mitochondria in response to upstream signals, but there is no defect inthe downstreamnpart of the pathway that is triggered by cytosoliccytochrome c. Thus, microinjection bypasses the defects in the doublydefective cells, triggering apoptosis.

Question 28

Tutorial 72. Why do you suppose that cells have evolved a special G0 state to exit thecell cycle rather than just stopping in G1 at the G1 checkpoint?

Answer 28

Tutorial 72. For multicellular organisms, the control of cell division is extremelyimportant. Individual cells must not proliferate unless it is to the benefit ofthe whole organism. The G0 state offers protection from aberrant activationof cell division, because the cell-cycle control system is partly or completelydismantled. If a cell just paused in G1, it would still contain all the cell-cyclemachinery and might still be induced to divide. It would also have to remakethe ‘decision’ not to divide almost continuously. To reenter the cell cyclefrom G0, a cell has to resynthesize the components that have disappeared,which is unlikely to occur by accident.

Question 29

Tutorial 73. Many cell –cycle genes from human cells function perfectly well whenexpressed in yeast cells. Why do you suppose that this is consideredremarkable? After all, many human genes encoding enzymes for metabolicreactions also function in yeast and no one thinks this is remarkable.

Answer 29

Tutorial 73. Enzymes for most metabolic reactions function in isolation; that is, theirenzymatic competence does not depend on critical interactions with otherproteins. So long as the enzyme folds properly and its small moleculesubstrate is present, the reaction will proceed. By contrast, cell-cycleproteins must interact with many other proteins to form complexes that arecritical for the coordinated progression through the cell cycle. The ability ofmany human cell-cycle proteins to interact with yeast components impliesthat the binding surfaces responsible for these interactions have beenpreserved through more than a billion years of evolution. This isremarkable!

Question 30

Tutorial 81. One important biological effect of a large dose of ionizing radiation is tohalt cell division.A. How does a large dose of ionizing radiation stop cell division?B. What happens if a cell has a mutation that prevents it from halting celldivision after being irradiated?C. An adult human that has reached maturity will die within a few days ofreceiving a radiation dose large enough to stop cell division. What does thistell you?

Answer 30

Tutorial 81.A. Radiation leads to DNA damage, which activates ATM and ATRkinases, which phosphorylate and activate Chk1 and Chk2 kinases, whichphosphorylate and stabilize p53, which induces expression of p21, whichbinds to and inactivates G1/S-Cdk and S-Cdk, which stops progressionthrough the cell cycle.B. In the absence of a functional DNA damage checkpoint, the cell willreplicate the damaged DNA, introducing mutations into the genomesinherited by the daughter cells.C. Cell division is an ongoing process that does not cease when we reachmaturity. Blood cells, epithelial cells in the skin or lining the gut, and thecells of the immune system, for example, are being constantly produced bycell division to meet the body’s needs. Our bodies produce about 1011 newblood cells each day.

Question 31

Tutorial 82. What is the functional definition of restriction point? Cancer cellstypically lose restriction-point controls. Explain how the followingmutations, which are found in some cancer cells, lead to a bypass ofrestriction point controls.A. overexpression of cyclin DB. injection by retroviruses encoding v-Fos and v-Jun

Answer 31

Tutorial 82. The restriction point is the place in the cell cycle beyond which cells arecommitted to completing DNA replication and mitosis even if growthfactors, or mitogens, are removed. To enter the cell cycle, quiescent cells inG0 require growth factors, which bind to cell-surface receptors andtrigger a signaling cascade that leads to the transcription of early-responsegenes and then delayed-response genes. Among the delayed responsegenes is the cyclin D gene, which partners with CDKs 4 and 6, and this mid-G1 cyclin-CDK complex phosphorylates the Rb protein. When Rb isphosphorylated by the mid- G1 cyclin-CDK, it can no longer bind thetranscription factor E2F. When E2F is released from Rb, then it inducestranscription of the genes that promote entry into S phase.A. High levels of cyclin D bypass the requirement for growth factors,which normally induce synthesis of cyclin D.B. Virally-encoded Fos and Jun bypass the requirement for growthfactors to induce expression of cellular fos and jun, which are early responsegenes.

Question 32

Tutorial 83. What is a cell cycle checkpoint? How do cell cycle checkpoints help topreserve the fidelity of the genome?

Answer 32

Tutorial 83. A cell-cycle checkpoint is a place in the cell cycle where a cell’s progressthrough the cycle is monitored, and, if the current process has not beencompleted properly, further progression through the cell cycle is inhibited;the cell cycle is arrested at this checkpoint until the process in question iscompleted successfully. Checkpoints exist at G1 and S phases to assessDNA damage, at G2 to assess DNA damage and to determine whether DNAreplication is complete, and at M phase to identify any problems withassembly of the mitotic spindle or chromosome segregation. Because thesecheckpoints identify problems with the genome (unreplicated, damaged, orimproperly segregated DNA) and arrest the cell cycle so that these problemscan be fixed, checkpoints can prevent the propagation of mutations into thenext cell generation and thereby preserve the fidelity of the genome.

Question 33

Tutorial 84. What are the main similarities and differences in the general propertiesand roles of macrophages and neutrophils?

Answer 33

Tutorial 84. Similarities:Both are phagocytic cells produced by the bone marrow. Both injectintracellular pathogens and destroy them intracellularly. They both carryreceptors on their cell surfaces that recognize pathogens and facilitatepathogen update.Differences:Macrophages are mainly resident in tissues while neutrophils circulate in theblood and enter tissues after an infection has become established.Macrophages are long lived and have important functions other than justphagocytosis. Once activated by the presence of pathogens they producecytokines that induce an inflammatory response. This helps attractneutrophils to the site of infection. They also help initiate an adaptiveimmune response by functioning as antigen presenting cells. In contrast,one neutrophils are in tissues that are short lived cells. Their only function isto take up and kill pathogens.

Question 34

Tutorial 85. What strategy ensures that passage through the cell cycle is unidirectionaland irreversible? What is the molecular machinery underlying this strategy?

Answer 34

Tutorial 85. The unidirectional and irreversible passage through the cell cycle isbrought about by the degradation of critical protein molecules at specificpoints in the cycle. Examples are the proteolysis of securin at the beginningof anaphase, proteolysis of cyclin B in late anaphase, and proteolysis of theS-phase CDK inhibitor at the start of S phase. The proteins are degraded bya proteasome, a multiprotein complex. Proteins are marked for proteolysisby the proteasome by the addition of multiple molecules of ubiquitin to oneor more lysine residues in the target protein. Securin and cyclin B are bothpolyubiquitinylated by the APC/C complex. The S-phase CDKinhibitor is polyubiquitinylated by SCF.

Question 35

Tutorial 86. What is opsonization? What is the role of antibodies in this process?

Answer 35

Tutorial 86. Decoration of particulate antigens with antibodies enhances phagocytosis.This process is called opsonization. In this process, antibodies attach to avirus or microbial surface by binding to their cognate antigen. Specializedphagocytic cells such as dendritic cells or macrophages can recognizethe constant regions of bound antibodies by means of Fc receptors. Fcreceptor-dependent events allow the dendritic cells and macrophages tomore readily inject and destroy antigenic particles.