Unit 1 HW
Lipid bilayers are considered to be two-dimensional fluids. What does this mean? What drives the movement of lipid molecules and proteins within the bilayer? How can such movement be measured? What factors affect the degree of membrane fluidity?
Lipid bilayers are considered to be two-dimensional fluids because lipid molecules (and proteins if present) are able to rotate along their long axes and move laterally within each leaflet. Such movements are driven by thermal energy and maybe quantified by measuring fluorescence recovery after photobleaching, the FRAP technique. In this technique, specific membrane lipids or proteins are labeled with a fluorescent reagent, and then a laser is used to irreversibly bleach a small area of the membrane surface. The extent and rate at which fluorescence recovers in the bleached area, as fluorescent molecules diffuse back into the bleach zone and bleached molecules diffuse outward, can be measured. The extent of recovery is proportional to the fraction of labeled molecules that are mobile, and the rate of recovery is used to calculate a diffusion coefficient, which is a measure of the molecule's rate of diffusion within the bilayer. The degree of fluidity depends on factors such as temperature, the length and saturation of the fatty acid chain portion of phospholipids, and the presence/absence of specific lipids such as cholesterol.
GLUT1, found in the plasma membrane of erythrocytes, is a classic example of a uniporter. D)How do liver and muscle cells maximize glucose uptake without changing Vmax?
Liver cells convert glucose to glycogen, which maximizes the glucose gradient across the plasma membrane.
Name the three groups into which membrane-associated proteins may be classified. Explain the mechanism by which each group associates with a biomembrane.
Membrane-associated proteins may be classified as integral membrane proteins, lipid-anchored membrane proteins, or peripheral membrane proteins. Integral membrane proteins pass through the lipid bilayer and are therefore composed, of three domains: a cytosolic domain exposed on the cytosolic face of the bilayer; an exoplasmic domain exposed on the exoplasmic face of the bilayer; and a membrane-spanning domain, which passes through the bilayer and connects the cytosolic and exoplasmic domains. Lipid-anchored membrane proteins have one or more covalently attached lipid molecules, which embeds in one leaflet of the membrane and thereby anchors the protein to one face of the bilayer. Peripheral proteins associate with the lipid bilayer through interactions with either integral membrane proteins or with phospholipid heads on one face of the bilayer.
Name the four classes of ATP-powered pumps that produce active transport of ions and molecules. Indicate which of these classes transport ions only and which transport primarily small organic molecules. The initial discovery of one class of these ATP-powered pumps came from studying the transport of not a natural substrate but rather artificial substrates used as cancer chemotherapy drugs. What do investigators now think are common examples of the natural substrates of this particular class of ATP-powered pumps?
The four classes: P-class, V-class, F-class, and ABC superfamily Only the ABC superfamily members transport small organic molecules. All other classes pump cations or protons. The initial discovery of ABC superfamily pumps came from the discovery of multidrug resistance to chemotherapy and the realization that ultimately this was due to transport proteins (i.e., ABC superfamily pumps). Today, the natural substrates of ABC superfamily pumps are thought to be small phospholipids, cholesterol, and other small molecules.
An H+ ion is smaller than an H2O molecule, and a glycerol molecule, a three-carbon alcohol, is much larger. Both readily dissolve in H2O. Why do aquaporins fail to transport H+ whereas some can transport glycerol?
To be transported, a molecule must fit into the aquaporin channel and form hydrogen bonds with N-H groups of amino acids lining the channel. Although H+ is smaller than H2O, it cannot form the required hydrogen bonds. Glycerol is much larger than H2O, but the three-carbon chain is flexible and the three OH groups can form the required hydrogen bonds.
Much of what we know about cellular function depends on experiments utilizing specific cells and specific parts (e.g., organelles) of cells. What techniques do scientists commonly use to isolate cells and organelles from complex mixtures, and how do these techniques work?
Centrifugation: differential centrifugation- a cell homogenate is centrifuged at different speeds to separate the different components of the cell which settle at different times of the procedure according to their molecular weights. Equilibrium density-gradient centrifugation: density is the determining factor and the organelles migrate to a place in the density tube where the density of the molecule equals the density of the medium at that point. Using organelle-specific antibodies: monoclonal antibodies are used to separate organelles of similar density and molecular weights. Fluorescence-Activated Cell Sorter (FACS): the cells which possess cell surface proteins are separated by this technique.
Why are the chemical stains required for visualizing cells and tissues with a basic light microscope? What advantage do fluorescent dyes and fluorescence microscopy provide in comparison to the chemical dyes used to stain specimens for light microscopy? What advantages do confocal scanning microscopy and deconvolution microscopy provide in comparison to conventional fluorescence microscopy?
Chemical stains are required for visualizing cells and tissues with the basic light microscope as cells and tissues are usually colorless and can't be seen properly without a satin. The advantage of fluorescence dyes and fluorescence microscopy instead of using chemical dyes in light microscopy is that we can observe proteins within a living cell. In some organisms naturally occurring fluorescent proteins are present. This can be tracked using a fluorescent microscope. Sometimes a protein is tagged with a fluorescent tag to see inside the cell. In light microscopy, we can't follow the movement of a protein inside a living cell. Only outer structures of the cell can be observed in live cells. In order to view organelles inside the cell, we need to stain the dead cell. The chemical stain used may affect the cell. The advantage of confocal scanning microscopy over fluorescence microscopy is that it overcomes the limitations seen in fluorescence microscopy. In fluorescence microscopy, the light emitted by molecules present above and below the focal plane overlap, and a blurred image is seen. In confocal scanning microscopy, this limitation is overcome by using a laser to scan the cell and a pinhole in front of the detector which blocks light not coming from the focal plane. In deconvolution microscopy, a series of images are taken at different focal planes of the same object. This is done by using a fluorescent microscope or a confocal scanning microscope. These images are then combined in three dimensions to give an image of the object.
Northern blotting, RT-PCR, and microarrays can be used to analyze gene expression. A lab studies yeast cells, comparing their growth in two different sugars, glucose and galactose. One student is comparing expression of the gene HMG2 under the different conditions. Which technique(s) could he use and why? Another student wants to compare expression of all the genes on chromosome 4, of which there are approximately 800. What technique(s) could she use and why?
Comparing expression of the gene HMG2 - for this study the student should use RT-PCR and Northern blotting as there is a comparison between only 2 expression levels. Since the student wants to compare the expression of a gene in two different samples (conditions), RT-PCR is the best for this, as it amplifies the gene of interest into multiple copies. If HMG2 is expressed under the test conditions of glucose or galactose, the products will be amplified. The number of copies of the gene product is directly proportional to the gene expression (comparative manner). The level of mRNA expressed in two different sugars must not be the same. The student can set up two separate experiments with two separate growth mediums, one having glucose and another one with galactose. Then isolate the mRNA for the HMG2 gene. Perform RT-PCR to amplify the isolated mRNA and proceed for Northern blotting using a probe for HMG2. Using this procedure he can compare the level of expression of the genes. Microarrays are best to compare the expression of all the genes on chromosome 4, of which there are approximately 800. This is because microarray is only used for studying and comparing the expression level of many genes at a time. Microarray or DNA chip analysis is used when the expression of thousands of genes has to be analyzed at the same time. It is not possible to use the Northern blot for these analyses. The data generated by DNA chip analyses is very huge and handled by a computer which is eventually impossible to interpret manually.
Proteins may be bound to the exoplasmic or cytosolic face of the plasma membrane by way of covalently attached lipids. What are the three types of lipid anchors responsible for tethering proteins to the plasma membrane bilayer, and which type is used by cell-surface proteins that face the external medium and by glycosylated proteoglycans?
Cytosolic proteins are anchored to the plasma membrane by acylation or prenylation. In the case of acylation, an N-terminal glycine residue of a protein is covalently attached to the 14-carbon fatty acid myristate (myristoylation) or a cysteine residue in a protein is attached to the 16-carbon fatty acid palmitate (palmitoylation). Prenylation occurs when the -SH group on a cysteine residue at or near the C-terminus of the protein is bound through a thioether bond to either a farnesyl or a geranylgeranyl (prenyl) group. Cell-surface proteins and heavily glycosylated proteoglycans are present on the exoplasmic face of the membrane and are linked there by a glycophosphatidylinositol (GPI) anchor.
GLUT1, found in the plasma membrane of erythrocytes, is a classic example of a uniporter. B)Glucose is a 6-carbon sugar while ribose is a 5-carbon sugar. Despite this smaller size, ribose is not efficiently transported by GLUT1. How can this be explained?
Despite its smaller size, ribose cannot bind to GLUT1 as glucose does because it cannot form the same noncovalent bonds—and thus cannot be transported.
What are detergents? How do ionic and nonionic detergents differ in their ability to disrupt cell membrane structure?
Detergents are amphipathic molecules. The hydrophobic part of a detergent molecule readily interacts with the hydrophobic tails of the phospholipids disrupting their interaction with each other; the hydrophilic part readily associates with water. This breaks up the organization of the lipid bilayer, ultimately leading to the formation of micelle droplets, composed of a single phospholipid layer with the polar heads in contact with water and a hydrophobic core excluding water. Ionic detergents, like all detergents, bind to both the hydrophilic and hydrophobic regions of membrane proteins that have been exposed after lipid bilayer disruption. Because of their charge, they can also disrupt the ionic and hydrogen bonds holding together the secondary and tertiary structure of a protein and are thus useful for completely denaturing a protein. Non-ionic detergents do not denature proteins and are therefore useful for extracting membrane proteins while maintaining their native conformation. At concentrations below the critical micelle concentration, they also prevent the hydrophobic regions of proteins that have been extracted from the cell membrane from interacting with each other and forming insoluble aggregates.
GLUT1, found in the plasma membrane of erythrocytes, is a classic example of a uniporter. A)Design a set of experiments to prove that GLUT1 is indeed a glucose-specific uniporter rather than a galactose- or mannose-specific uniporter.
Determination of the rate of erythrocyte (GLUT1) transport of substrate versus concentration allows the determination of Km for glucose versus galactose versus mannose. The Km for glucose will be lowest, indicating that GLUT1 is glucose-specific, not galactose- or mannose-specific.
Explain why the coupled reaction ATP ADP + Pi in the P-class ion pump mechanism does not involve direct hydrolysis of the phosphoanhydride bond.
Direct hydrolysis of the phosphoanhydride bond would result in the release of the bond energy as heat, which would thus be lost. By first transferring the phosphate bond to an aspartate (D) residue, the P-class ATPase uses the released bond energy to drive a conformational change in the protein from the E1 to the E2 state.
Two methods for functionally inactivating a gene without altering the gene sequence are by dominant-negative mutations and RNA interference (RNAi). Describe how each method can inhibit the expression of a gene.
Dominant-Negative mutations inactivate a gene by removing specific cellular domains. An example of this would be the removal of an activation domain while still containing a binding domain. This will lead to reduced levels of gene activation. The removal of the functional domain would create a dominant-negative phenotype. RNAi inhibits the expression of a gene by targeting certain mRNA molecules for neutralization. Double-stranded RNA (dsRNA) is cut up by a dicer protein and the small portions separate and bind to certain proteins from special families. By binding to both sides, the proteins can either cleave the mRNA all together or they can regulate the target sequence with other proteins.
Phospholipid biosynthesis at the interface between the endoplasmic reticulum (ER) and the cytosol presents a number of challenges that must be solved by the cell. Explain how each of the following is handled. C) Many membrane systems in the cell, for example, the plasma membrane, are unable to synthesize their own phospholipids, yet these membranes must also expand if the cell is to grow and divide.
Even though membrane systems like the plasma membrane are unable to synthesize their own phospholipids, they can still expand by incorporating phospholipids that have been synthesized elsewhere. More specifically, vesicles can transport phospholipids to the membrane system, some phospholipids can be transferred from other membranes, and lipid transfer proteins can also help disperse phospholipids.
Patch clamping can be used to measure the conductance properties of individual ion channels. Describe how patch clamping can be used to determine whether or not the gene coding for a putative K+ channel actually codes for a K+ or Na+ channel.
Expression of a channel protein in a normally nonexpressing cell permits the patch-clamp assessment of channel properties. Typically, the cell used is a frog oocyte. Frog oocytes do not normally express plasma membrane channel proteins. Channel protein expression may be induced by microinjection of in vitro-transcribed mRNA encoding the protein. Frog oocytes are large and hence technically easier to inject and to patch-clamp than other cells. One can then vary the ionic composition of the medium and determine whether the presence of Na+ or of K+ supports ionic movement through the channel.
Fatty acids must associate with lipid chaperones in order to move within the cell. Why are these chaperones needed, and what is the name given to a group of proteins that are responsible for this intracellular trafficking of fatty acids? What is the key distinguishing feature of these proteins that allows fatty acids to move within the cell?
Fatty acids have very low solubility inside an aqueous-rich intracellular environment. Therefore, they associate with fatty-acid binding proteins (FABPs), which are cytosolic proteins that contain a hydrophobic pocket or barrel, lined by beta-sheets. This pocket provides a haven for the long-chain fatty acid, where it interacts in a noncovalent fashion with the FABP.
Chromatography is an analytical method used to separate proteins. Describe the principles for separating proteins by gel filtration, ion-exchange, and affinity chromatography
Gel filtration: The gel filtration is a type of chromatography that separates proteins based on size. A porous bed of beads is laid down and molecules are allowed to pass through them. Smaller molecules diffuse faster into pores of beads and therefore move ahead. The larger molecules tend to slow down and lag behind. The proteins of different sizes can be collected as they pass from the column. Ion exchange: It separated proteins based on their ionic charge. A column is prepared with a specific charge say cation. The proteins when passing through the column are subjected to ionic attractions. Cations in the column will bind to the anions in the protein samples. Thus from a mixture of protein, one particular charged molecules can be separated out. Sample molecules are exchanged with the buffer ions as they compete with the binding site in the column. The length of retention depends on the strength of its charge. Most weary charged are elated first followed by successively stronger ones. Ex. Nickel affinity column. Affinity chromatography: Here, the separation is based on the molecular conformation of the proteins in the sample. The column is coated with a ligand that has a specific binding affinity to proteins in the sample like a lock and key fit. For example, the glycoproteins can be purified using lectin in ten columns. Lectin has a higher affinity for glycoproteins. The proteins which do not bind are washed away. Th bound molecules can be eluted later using buffers.
Describe the symport process by which cells lining the small intestine import glucose. What ion is responsible for the transport, and what two particular features facilitate the energetically favored movement of this ion across the plasma membrane?
Glucose uptake from the intestinal lumen into the epithelial cells is driven by symport with two Na+ ions by a two Na+/glucose symporter. Binding of two Na+ ions and one glucose molecule to high-affinity, outward-facing sites in the protein causes a series of conformational changes in the symporter that eventually allows Na+ and glucose to be released from low-affinity sites facing the cytosol. Transport by this symporter is energetically favorable because the movement of Na+ ions into the cell is driven by both its concentration gradient and the transmembrane voltage gradient. Transport of two Na+ ions into the cell provides ~6 kcal of energy—enough to generate an intracellular glucose concentration that is 30,000 times higher than in the intestinal lumen.
A number of foreign proteins have been expressed in bacterial and mammalian cells. Describe the essential features of a recombinant plasmid that are required for expression of a foreign gene. How can you modify the foreign protein to facilitate its purification? What is the advantage of expressing a protein in mammalian cells versus bacteria?
In order for foreign genes to be expressed, an expression vector must possess the correct translation initiation sequence, promoter, and start/termination codons. In order to facilitate purification, cDNA can be modified to separate the desired proteins from the cellular proteins. Mammalian cells are more difficult to work with than bacterial cells but they are better for studying mammalian proteins which require organelles that bacteria do not possess. Bacterial cells are less fragile, however, and are typically able to produce more proteins.
Nitric oxide (NO) is a gaseous molecule with lipid solubility similar to that of O2 and CO2. Endothelial cells lining arteries use NO to signal surrounding smooth muscle cells to relax, thereby increasing blood flow. What mechanism or mechanisms would transport NO from where it is produced in the cytoplasm of an endothelial cell into the cytoplasm of a smooth muscle cell where it acts?
Like O2 and CO2, NO passively diffuses through membranes. As it is produced by an enzyme and accumulates in the endothelial cell cytosol, NO passively diffuses down its concentration gradient through the endothelial cell plasma membrane out of the cell and then passively diffuses through the plasma membrane into the cytoplasm of the smooth muscle cell, where it acts to decrease contraction.
Acetic acid (a weak acid with a pKa of 4.75) and ethanol (an alcohol) are each composed of two carbons, hydrogen, and oxygen, and both enter cells by passive diffusion. At pH 7, one is much more membrane permeable than the other. Which is more permeable and why? Predict how the permeability of each is altered when the pH is reduced to 1.0, a value typical of the stomach.
Of the two at neutral pH, ethanol is the much more membrane permeant. It has no acidic or basic group and is uncharged at a pH of 7. The carboxyl group of acetic acid is predominantly dissociated at this pH and hence acetic acid exists predominantly as the negatively charged acetate anion. It is nonpermanent. At a pH of 1, ethanol remains uncharged and membrane-permeant. The carboxyl group of acetic acid is now predominantly nondissociated and uncharged. Hence, acetic acid is now membrane permeant. Any difference in permeability is very small.
Various methods have been developed for detecting proteins. Describe how radioisotopes and autoradiography can be used for labeling and detecting proteins. How does Western blotting detect proteins?
Proteins can be made radioactive by the incorporation of radioactively labeled amino acids during protein synthesis. Methionine or cysteine labeled with sulfur-35 is two commonly used radioactive amino acids, although many others have also been used. The radioactively labeled proteins can be detected by autoradiography. In one example of this technique, cells are labeled with a radioactive compound and then overlaid with a photographic emulsion sensitive to radiation. The presence of radioactive proteins will be revealed as deposits of silver grains after the emulsion is developed. A Western blot is a method for detecting proteins that combines the resolving power of gel electrophoresis, the specificity of antibodies, and the sensitivity of enzyme assays. In this method, proteins are first separated by size using gel electrophoresis. The proteins are then transferred onto a nylon filter. A specific protein is then detected by the use of an antibody specific for the protein of interest (primary antibody) and an enzyme-antibody conjugate (secondary antibody) that recognizes the primary antibody. The presence of this protein-primary antibody-enzyme-conjugated secondary antibody complex is detected using an assay specific for the conjugated enzyme.
A number of techniques can separate proteins on the basis of their differences in mass. Describe the use of two of these techniques, centrifugation and gel electrophoresis. The blood proteins transferrin (MW 76 kDa) and lysozyme (MW 15 kDa) can be separated by rate-zonal centrifugation or SDS-polyacrylamide gel electrophoresis. Which of the two proteins will sediment faster during centrifugation? Which will migrate faster during electrophoresis?
Proteins can be separated by mass by centrifuging them through a solution of increasing density, called a density gradient. In this separation technique, known as rate-zonal centrifugation, proteins of a larger mass generally migrate faster than proteins of smaller mass. However, this is not always true because the shape of the protein also influences the migration rate. Gel electrophoresis can also separate proteins based on their mass. In this technique, proteins are separated through a polyacrylamide gel matrix in response to an electric field. Because the migration of proteins through a polyacrylamide gel is also influenced by the shape of proteins, the ionic detergent sodium dodecyl sulfate is added to denature proteins and force proteins into similar conformations. During rate-zonal centrifugation, a protein of larger mass (transferrin) will sediment faster during centrifugation, whereas a protein of smaller mass (lysozyme) will migrate faster during electrophoresis.
Restriction enzymes and DNA ligase play essential roles in DNA cloning. How is it that a bacterium that produces a restriction enzyme does not cut its own DNA? Describe some general features of restriction-enzyme sites. What are the three types of DNA ends that can be generated after cutting DNA with restriction enzymes? What reaction is catalyzed by DNA ligase?
Restriction enzyme sites function as a defense against bacteriophages. In order to keep restriction enzymes from cutting their own DNA, the bacteria coat the restriction sites with methyl groups. After DNA is cut with a restriction enzyme it can produce a "sticky" end which has an overhang, or a blunt end which is flush. DNA ligase catalyzes the reaction which links the 5' phosphate group of one DNA strand to the 3' end of another. This will produce a sugar-phosphate backbone.
shRNAs and siRNAs can be used to successfully knock down the expression of any specific protein in a given cell line or organism. The utility of one over the other is debatable, but there are merits to using one for therapeutic applications in a living organism. Which of the two methods is likely to be more advantageous in the long-term and what is one of its limitations?
ShRNA is better long term. The limitation is at high concentration, shRNA can saturate Dicer and RISC leading to toxicity.
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?
Since biomembranes form closed compartments, one face of the bilayer is automatically exposed to the interior of the compartment while the other is exposed to the exterior of the compartment. Each face, therefore, interacts with different environments and performs different functions. The different functions are in turn 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. In addition, different domains of integral proteins are exposed on each face of the bilayer. Finally, in the case of the plasma membrane, the lipids and proteins of the exoplasmic face are often modified with carbohydrates.
Explain the following statement: The structure of all biomembranes depends on the chemical properties of phospholipids, whereas the function of each specific biomembrane depends on the specific proteins associated with that membrane.
The amphipathic nature of phospholipid molecules (a hydrophilic head and hydrophobic tail) allows these molecules to self-assemble into closed bilayer structures when in an aqueous environment. The phospholipid bilayer provides a barrier with selective permeability that restricts the movement of hydrophilic molecules and macromolecules across the bilayer. The different types of proteins present on the two faces of the bilayer contribute to the distinctive functions of each membrane and control the movement of selected hydrophilic molecules and macromolecules across it.
Mass spectrometry is a powerful tool in proteomics. What are the four key features of a mass spectrometer? Describe briefly how MALDI and two-dimensional polyacrylamide gel electrophoresis (2D-PAGE) could be used to identify a protein expressed in cancer cells but not in normal healthy cells.
The four features of a mass spectrometer are 1) the ion source, 2) the mass analyzer, 3) the detector, and 4) a computerized data system. Basically, the investigator would collect protein samples from the cancerous cells and from the normal healthy cells, the latter serving as a control. Samples would be prepared for 2D PAGE and after electrophoresis, the gels would be dyed and the profiles compared. If a protein "spot" were present in the sample from the cancer cell and not the control, it would be isolated out of the gel, protease-digested using trypsin to generate peptides that are mixed with a matrix, and applied to a metal target. A laser is used to ionize the peptides, which are vaporized into singly charged ions. In the case of a time of flight (TOF) mass analyzer, the time it takes the ions to pass through the analyzer before reaching the detector is inversely proportional to its mass and directly proportional to the charge they carry, generating a spectrum in which each molecule has a distinct signal, allowing the investigator to calculate each ion's mass. The fourth essential component is a computerized data system that acquires and stores the data, which are then compared to information in databases. The mass and charge signature, or fingerprint, of the unknown, is compared to that of peptides in a database and the best match protein is identified.
The ability to selectively modify the genome in the mouse has revolutionized mouse genetics. Outline the procedure for generating a knockout mouse at a specific genetic locus. How can the loxP-Cre system be used to conditionally knock out a gene? What is an important medical application of knockout mice?
The gene to be knocked out is isolated from a mouse gene library. Then a new DNA sequence is engineered which is very similar to the original gene and its immediate neighbor sequence, except that it is changed sufficiently to make the gene inoperable. Usually, the new sequence is also given a marker gene, a gene that normal mice don't have, and that confers resistance to a certain toxic agent (e.g., neomycin) or that produces an observable change (e.g. color or fluorescence). In addition, a second gene, such as herpes tk+, is also included in the construct in order to accomplish a complete selection. Some of the newborn chimera mice will have gonads derived from knocked-out stem cells, and will therefore produce eggs or sperm containing the knocked-out gene. When these chimera mice are crossbred with others of the wild type, some of their offspring will have one copy of the knocked-out gene in all their cells. These mice will be entirely white and are not chimeras, however, they are still heterozygous. Some of the newborn chimera mice will have gonads derived from knocked-out stem cells, and will therefore produce eggs or sperm containing the knocked-out gene. When these chimera mice are crossbred with others of the wild type, some of their offspring will have one copy of the knocked-out gene in all their cells. These mice will be entirely white and are not chimeras, however, they are still heterozygous.
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 a cell?
The key regulated enzyme in cholesterol biosynthesis is HMG (b-hydroxy-bmethylglutaryl)-CoA reductase. This enzyme catalyzes the rate-controlling step in cholesterol biosynthesis. The enzyme is subject to negative feedback regulation by cholesterol. In fact, the cholesterol biosynthetic pathway was 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 HMG-CoA reductase are 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 cleavage activating protein) and Niemann-Pick C1 (NPC1) protein, which takes part in cholesterol transport and regulation.
Describe how complementation analysis can be used to reveal whether two mutations are in the same or in different genes. Explain why complementation analysis will not work with dominant mutations.
The process of genetic complementation test is applicable for recessive traits only. Genetic complementation produces a wild type phenotype when two strains of an organism with different homozygous recessive mutations are crossed or mated. Complementation will take place and wild type phenotype will be formed only if the recessive mutations are present in different genes because each strain's genome will give its the wild type allele to other strain's genome so that its mutated recessive allele is complemented with the wild type. As the mutations are recessive, the offspring will possess the wild-type phenotype. Complementation will never occur if the mutations are in the same gene. So we can conclude that the production of wild phenotype means complementation has occurred and two different recessive mutations are in different genes and when there is no complementation, it means recessive mutations are present in the same gene.
In the case of the bacterial sodium/leucine transporter, what is the key distinguishing feature about the bound sodium ions that ensures that other ions, particularly K+, do not bind?
The six oxygens in the main-chain carbonyl or side-chain carboxyl groups that bind each of the two Na+ ions in the transporter are exquisitely positioned with a geometry similar to that of the water molecules with which Na+ associates in solution. At one site, the carboxyl group of the bound leucine provides one of the coordinating oxygens. When Na+ ions bind to the oxygens, they lose their water of hydration. The increase in entropy that occurs when hydration water molecules are freed promotes Na+ ions binding at both sites. K+ ions (and water molecules themselves) are too big to bind the six oxygens in the proper geometry and so do not compete with Na+.
Phospholipid biosynthesis at the interface between the endoplasmic reticulum (ER) and the cytosol presents a number of challenges that must be solved by the cell. Explain how each of the following is handled. A) The substrates for phospholipid biosynthesis are all water-soluble, yet the end products are not.
The substrates for phospholipid biosynthesis are all water-soluble because they are synthesized at the interface between the ER and the cytosol, an area where the production of hydrophobic molecules (like the end products) would not be favorable because of the hydrophilic environment at the interface. Thus, water-soluble, hydrophilic substrates for phospholipid biosynthesis are synthesized in the cytosol. From there, the enzymes from the ER connect these substrates to end up creating the hydrophobic, phospholipid end products.
Name the three classes of transporters. Explain which one or more of these classes is able to move glucose and which move bicarbonate (HCO3-) against an electrochemical gradient. In the case of bicarbonate, but not glucose, the ΔG of the transport process has two terms. What are these two terms, and why does the second not apply to glucose? Why are cotransporters often referred to as examples of secondary active transport?
The three classes of transporters are uniporters, symporters, and antiporters. Both symporters and antiporters are capable of moving organic molecules against an electrochemical gradient by coupling an energetically unfavorable movement to the energetically favorable movement of a small inorganic ion. The ΔG for bicarbonate has two terms, a concentration term, and an electrical term because bicarbonate is an anion. Glucose is neutrally charged and hence its ΔG for transport has only a concentration term. Unlike pumps, neither symporters nor antiporters hydrolyze ATP or any other molecule during transport. Hence, these cotransporters are better referred to as examples of secondary active transporters rather than as actual active transporters. The term active transporter is restricted to the ATP pumps where ATP is hydrolyzed in the transport process.
Certain proton pump inhibitors inhibit secretion of stomach acid and are among the most widely sold drugs in the world today. What pump does this type of drug inhibit, and where is this pump located?
These drugs irreversibly inhibit the H+/K+ATPase in the apical membrane of stomach parietal cells. Although the inhibition of a given H+/K+ATPase is irreversible, the cells eventually make more of the pump.
Phospholipid biosynthesis at the interface between the endoplasmic reticulum (ER) and the cytosol presents a number of challenges that must be solved by the cell. Explain how each of the following is handled. B)The immediate site of incorporation of all newly synthesized phospholipids is the cytosolic leaflet of the ER membrane, yet phospholipids must be incorporated into both leaflets.
This challenge is handled by flippases. That is, newly synthesized phospholipids in the cytosolic leaflet of the ER membrane can be flipped and then incorporated into the exoplasmic leaflet with the help of flippases.
Biomembranes contain many different types of lipid molecules. What are the three main types of lipid molecules found in biomembranes? How are the three types similar, and how are they different?
Three main types: steroids (cholesterol), sphingolipids, and phosphoglycerides. While all three are similar in that they have both hydrophobic and hydrophilic regions (amphipathic, polar head, and hydrophobic tail), they differ from each other in "chemical structure, abundance, and function." (Lodish et al). Phosphoglycerides are unique in that they are the most abundant, have a hydrophobic tail with two fatty acid chains, a hydrophilic head connected to a phosphate group, and function as signaling molecules. Sphingolipids are unique in that they are not as abundant as phosphoglycerides, have a hydrophobic tail with one fatty acid chain which is connected to hydrophilic head (which may or may not contain a phosphate group) via an amide link, and have a protective structural function. Cholesterol (steroids) is unique in that it is most abundant in plasma membranes, has a four-ring structure, and functions as a manager of membrane fluidity.
GLUT1, found in the plasma membrane of erythrocytes, is a classic example of a uniporter. E)Tumor cells expressing GLUT1 often have a higher Vmax for glucose transport than do normal cells of the same type. How could these cells increase the Vmax?
Tumor cells often express a higher number of glucose transporters than normal cells.
Uniporters and ion channels support facilitated diffusion across biomembranes. Although both are examples of facilitated diffusion, the rates of ion movement via an ion channel are roughly 104- to 105-fold faster than that of molecule movement via a uniporter. What key mechanistic difference results in this large difference in transport rate? What contribution to free energy (ΔG) determines the direction of transport?
Uniporters are slower than channels because they mediate a more complicated process. The transported substrate both binds to the uniporter and elicits a conformational change in the transporter. A uniporter transports one substrate molecule at a time. In contrast, channel proteins form a protein-lined passageway through which multiple water molecules or ions move simultaneously, single file, at a rapid rate. The major contributor to the free-energy-driving transport through a uniporter is the entropy increase as a molecule moves from a high concentration to a low concentration.
GLUT1, found in the plasma membrane of erythrocytes, is a classic example of a uniporter. F)Fat and muscle cells modulate the Vmax for glucose uptake in response to insulin signaling. How?
When insulin is low, GLUT4 is stored in intracellular vesicles. Insulin induces a rapid increase in the Vmax for glucose uptake by stimulating fusion of these vesicles with the plasma membrane, thereby increasing the number of plasma membrane glucose transporters.
Physical methods are often used to determine protein conformation. Describe how x-ray crystallography, cryoelectron microscopy, and NMR spectroscopy can be used to determine the shapes of proteins. What are the advantages and disadvantages of each method? Which is better for small proteins? Large proteins? Huge macromolecular assemblies?
X-ray crystallography can be used to determine the three-dimensional structure of proteins. In this technique, x-rays are passed through a protein crystal. The diffraction pattern generated when atoms in the protein scatter the x-rays is a characteristic pattern that can be interpreted into defined structures. Cryoelectron microscopy involves the rapid freezing of a protein sample and examination with a cryoelectron microscope. A low dose of electrons is used to generate a scatter pattern that can be used to reconstruct the protein's structure. In nuclear magnetic resonance (NMR) spectroscopy, a protein solution is placed in a magnetic field and the effects of different radio frequencies on the spin of different atoms are measured. From the magnitude of the effect of one atom on an adjacent atom, the distances between residues can be calculated to generate a three-dimensional structure. X-ray crystallography can provide extremely high-resolution structural information on molecules and molecular complexes of any size. The principal disadvantage of x-ray crystallography is the challenge of producing samples in the form of single crystals suitable for diffraction experiments. NMR spectroscopy gives high-resolution information on protein structures in solution. It also is ideal for monitoring protein dynamics. However, NMR spectroscopy is limited in its ability to conclusively determine the structures of very large proteins and symmetrical macromolecular assemblies. The principal advantage of electron microscopy is the relative ease of sample preparation. However, the structural resolution is generally not so high as with the other methods, especially for asymmetric assemblies. NMR is better for small proteins. Electron microscopy and x-ray crystallography, so long as a suitable crystal can be obtained, are ideal for large proteins and macromolecular assemblies.
GLUT1, found in the plasma membrane of erythrocytes, is a classic example of a uniporter. C)A drop in blood sugar from 5 mM to 2.8 mM or below can cause confusion and fainting. Calculate the effect of this drop on glucose transport into cells expressing GLUT1.
at 5 mM, GLUT1-expressing cells transport glucose at a 77% maximal rate, whereas at 2.8 mM, the cells transport at a 65% maximal rate.
