Genetics and Molecular Biology Lecture Notes

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Genetics and Molecular Biology S E C O N D E D I T I O N Robert Schleif Department of Biology The Johns Hopkins University Baltimore, Maryland The Johns Hopkins University Press Baltimore and LondonAn Overview of Cell Structure and Function 1 In this book we will be concerned with the basics of the macromolecular interactions that affect cellular processes. The basic tools for such studies are genetics, chemistry, and physics. For the most part, we will be concerned with understanding processes that occur within cells, such as DNA synthesis, protein synthesis, and regulation of gene activity. The initial studies of these processes utilize whole cells. These normally are followed by deeper biochemical and biophysical studies of individual components. Before beginning the main topics we should take time for an overview of cell structure and function. At the same time we should develop our intuitions about the time and distance scales relevant to the molecules and cells we will study. Many of the experiments discussed in this book were done with the bacterium Escherichia coli, the yeast Saccharomyces cerevisiae, and the fruit fly Drosophila melanogaster. Each of these organisms possesses unique characteristics making it particularly suitable for study. In fact, most of the research in molecular biology has been confined to these three organisms. The earliest and most extensive work has been done with Escherichia coli. The growth of this oranism is rapid and inexpen- sive, and many of the most fundamental problems in biology are displayed by systems utilized by this bacterium. These problems are therefore most efficiently studied there. The eukaryotic organisms are necessary for study of phenomena not observed in bacteria, but parallel studies on other bacteria and higher cells have revealed that the basic principles of cell operation are the same for all cell types. 12 An Overview of Cell Structure and Function Cell’s Need for Immense Amounts of Information Cells face enormous problems in growing. We can develop some idea of the situation by considering a totally self-sufficient toolmaking shop. If we provide the shop with coal for energy and crude ores, analogous to a cell’s nutrient medium, then a very large collection of machines and tools is necessary merely to manufacture each of the parts present in the shop. Still greater complexity would be added if we required that the shop be totally self-regulating and that each machine be self-assem- bling. Cells face and solve these types of problems. In addition, each of the chemical reactions necessary for growth of cells is carried out in an aqueous environment at near neutral pH. These are conditions that would cripple ordinary chemists. By the tool shop analogy, we expect cells to utilize large numbers of “parts,” and, also by analogy to factories, we expect each of these parts to be generated by a specialized machine devoted to production of just one type of part. Indeed, biochemists’ studies of metabolic pathways have revealed that an E. coli cell contains about 1,000 types of parts, or small molecules, and that each is generated by a specialized machine, an enzyme. The information required to specify the structure of even one machine is immense, a fact made apparent by trying to describe an object without pictures and drawings. Thus, it is reasonable, and indeed it has been found that cells function with truly immense amounts of information. DNA is the cell’s library in which information is stored in its sequence of nucleotides. Evolution has built into this library the information necessary for cells’ growth and division. Because of the great value of the DNA library, it is natural that it be carefully protected and preserved. Except for some of the simplest viruses, cells keep duplicates of the information by using a pair of self-complementary DNA strands. Each strand contains a complete copy of the information, and chemical or physical damage to one strand is recognized by special enzymes and is repaired by making use of information contained on the opposite strand. More complex cells further preserve their information by pos- sessing duplicate DNA duplexes. Much of the recent activity in molecular biology can be understood in terms of the cell’s library. This library contains the information necessary to construct the different cellular machines. Clearly, such a library contains far too much information for the cell to use at any one time. Therefore mechanisms have developed to recognize the need for particular portions, “books,” of the information and read this out of the library in the form of usable copies. In cellular terms, this is the regulation of gene activity. Rudiments of Prokaryotic Cell Structure A typical prokaryote, E. coli, is a rod capped with hemispheres (Fig. 1.1). -4 4 It is 1–3 µ (10 cm = 1 µ = 10 Å) long and 0.75 µ in diameter. Such aRudiments of Prokaryotic Cell Structure 3 Nuclear Cell Ribosomes region envelope 0.75µ Figure 1.1 The dimensions of a typical E. coli cell. 2µ -13 -14 cell contains about 2 × 10 g of protein, 2 × 10 g of RNA that is mostly -15 ribosomal RNA, and 6 × 10 g of DNA. The cell envelope consists of three parts, an inner and outer mem- brane and an intervening peptidoglycan layer (Fig. 1.2). The outer surface of the outer membrane is largely lipopolysaccharides. These are attached to lipids in the outer half of the outer membrane. The polysac- charides protect the outer membrane from detergent-like molecules found in our digestive tract.outer membrane The outer membrane also consists of matrix proteins that form pores small enough to exclude the detergent-like bile salts, but large enough to permit passage of small molecules and phospholipids. Figure 1.2 Schematic drawing of the structure of the envelope of an E. coli cell. Matrix Lipopolysaccharide protein Lipoprotein Lipids Outer membrane Periplasmic Peptidoglycan space or cell wall Inner membrane Periplasmic protein Phospholipids Proteins Phospholipid4 An Overview of Cell Structure and Function GM M M M N-acetylglucosamine N-acetylmuramic acid G G G CH OH CH OH 2 2 O O H H H M M M H O OH O H H H OH H H NH H NH O GGG C O O H C CH C 3 M M M CH CH COOH 3 3 G G G (a) (b) Figure 1.3 Structure of the cell wall showing the alternating N-acetylglu- cosamine N-acetylmuramic acid units. Each N-acetylmuramic acid possesses a peptide, but only a few are crosslinked in E. coli. The major shape-determining factor of cells is the peptidoglycan layer or cell wall (Fig. 1.3). It lies beneath the outer membrane and is a single molecule containing many polysaccharide chains crosslinked by short peptides (Fig. 1.4). The outer membrane is attached to the pepti- 6 doglycan layer by about 10 lipoprotein molecules. The protein end of each of these is covalently attached to the diaminopimelic acid in the peptidoglycan. The lipid end is buried in the outer membrane. The innermost of the three cell envelope layers is the inner or cytoplasmic membrane. It consists of many proteins embedded in a phospholipid bilayer. The space between the inner membrane and the outer membrane that contains the peptidoglycan layer is known as the periplasmic space. The cell wall and membranes contain about 20% of the cellular protein. After cell disruption by sonicating or grinding, most of this protein is still contained in fragments of wall and membrane and can be easily pelleted by low-speed centrifugation. The cytoplasm within the inner membrane is a protein solution at about 200 mg/ml, about 20 times more concentrated than the usual cell-free extracts used in the laboratory. Some proteins in the cytoplasm may constitute as little as 0.0001% by weight of the total cellular protein whereas others may be found at levels as high as 5%. In terms of -8 -4 concentrations, this is from 10 M to 2 × 10 M, and in a bacterial cell this is from 10 to 200,000 molecules per cell. The concentrations of many of the proteins vary with growth conditions, and a current re- search area is the study of the cellular mechanisms responsible for the variations. The majority of the more than 2,000 different types of proteins found within a bacterial cell are located in the cytoplasm. One question yet to 30-60 units in E. coliRudiments of Prokaryotic Cell Structure 5 N-acetylmuramic acid OH O O C CH NHAc (CH ) OH 2 3 O O O H H O H C OH O H CH 3 OH O CH C C N C C N C CH CH C N C C N C C 3 2 2 H CH H H H 3 L-Ala D-Glu meso-DAP D-Ala N-acetylmuramic acid N-acetylmuramic acid L-Ala L-Ala D-Glu D-Glu meso-DAP meso-DAP D-Ala D-Ala Figure 1.4 Structure of the peptide crosslinking N-acetylmuramic acid units. DAP is diaminopimelic acid. be answered about these proteins is how they manage to exist in the cell without adhering to each other and forming aggregates since polypep- tides can easily bind to each other. Frequently when a bacterium is engineered for the over-synthesis of a foreign protein, amorphous precipitates called inclusion bodies form in the cytoplasm. Sometimes these result from delayed folding of the new protein, and occasionally they are the result of chance coprecipitation of a bacterial protein and the newly introduced protein. Similarly, one might also expect an occasional mutation to inactivate simultaneously two apparently unre- lated proteins by the coprecipitation of the mutated protein and some other protein into an inactive aggregate, and occasionally this does occur. The cell’s DNA and about 10,000 ribosomes also reside in the cyto- plasm. The ribosomes consist of about one-third protein and two-thirds RNA and are roughly spherical with a diameter of about 200 Å. The DNA in the cytoplasm is not surrounded by a nuclear membrane as it is in the cells of higher organisms, but nonetheless it is usually confined to a portion of the cellular interior. In electron micrographs of cells, the highly compacted DNA can be seen as a stringy mass occupying about one tenth of the interior volume, and the ribosomes appear as granules uniformly scattered through the cytoplasm.6 An Overview of Cell Structure and Function Rudiments of Eukaryotic Cell Structure A typical eukaryotic cell is 10 µ in diameter, making its volume about 1,000 times that of a bacterial cell. Like bacteria, eukaryotic cells contain cell membranes, cytoplasmic proteins, DNA, and ribosomes, albeit of somewhat different structure from the corresponding prokaryotic ele- ments (Fig. 1.5). Eukaryotic cells, however, possess many structural features that even more clearly distinguish them from prokaryotic cells. Within the eukaryotic cytoplasm are a number of structural proteins that form networks. Microtubules, actin, intermediate filaments, and thin filaments form four main categories of fibers found within eu- karyotic cells. Fibers within the cell provide a rigid structural skeleton, participate in vesicle and chromosome movement, and participate in changing the cell shape so that it can move. They also bind the majority of the ribosomes. The DNA of eukaryotic cells does not freely mix with the cytoplasm, but is confined within a nuclear membrane. Normally only small pro- teins of molecular weight less than 20 to 40,000 can freely enter the nucleus through the nuclear membrane. Larger proteins and nuclear RNAs enter the nucleus through special nuclear pores. These are large structures that actively transport proteins or RNAs into or out of the nucleus. In each cell cycle, the nuclear membrane dissociates, and then later reaggregates. The DNA itself is tightly complexed with a class of proteins called histones, whose main function appears to be to help DNA retain a condensed state. When the cell divides, a special apparatus called the spindle, and consisting in part of microtubules, is necessary to pull the chromosomes into the daughter cells. Eukaryotic cells also contain specialized organelles such as mito- chondria, which perform oxidative phosphorylation to generate the cell’s needed chemical energy. In many respects mitochondria resemble bacteria and, in fact, appear to have evolved from bacteria. They contain DNA, usually in the form of a circular chromosome like that of E. coli Figure 1.5 Schematic drawing of a eukaryotic cell. Plasma membrane Mitochondrion Fibers Nuclear 10µ membrane Nucleus Endoplasmic reticulum Golgi apparatusn Packing DNA into Cells 7 and ribosomes that often more closely resemble those found in bacteria than the ribosomes located in the cytoplasm of the eukaryotic cell. Chloroplasts carry out photosynthesis in plant cells, and are another type of specialized organelle found within some eukaryotic cells. Like mitochondria, chloroplasts also contain DNA and ribosomes different from the analogous structures located elsewhere in the cell. Most eukaryotic cells also contain internal membranes. The nucleus is surrounded by two membranes. The endoplasmic reticulum is an- other membrane found in eukaryotic cells. It is contiguous with the outer nuclear membrane but extends throughout the cytoplasm in many types of cells and is involved with the synthesis and transport of membrane proteins. The Golgi apparatus is another structure contain- ing membranes. It is involved with modifying proteins for their trans- port to other cellular organelles or for export out of the cell. Packing DNA into Cells The DNA of the E. coli chromosome has a molecular weight of about 2 9 6 × 10 and thus is about 3 × 10 base pairs long. Since the distance between base pairs in DNA is about 3.4 Å, the length of the chromosome 7 4 is 10 Å or 0.1 cm. This is very long compared to the 10 Å length of a bacterial cell, and the DNA must therefore wind back and forth many times within the cell. Observation by light microscopy of living bacterial cells and by electron microscopy of fixed and sectioned cells show, that often the DNA is confined to a portion of the interior of the cell with dimensions less than 0.25 µ . To gain some idea of the relevant dimensions, let us estimate the number of times that the DNA of a bacterium winds back and forth within a volume we shall approximate as a cube 0.25 µ on a side. This will provide an idea of the average distance separating the DNA duplexes and will also give some idea of the proportion of the DNA that lies on Figure 1.6 Calculation of the num- ber of times the E. coli chromosome winds back and forth if it is confined within a cube of edge 0.25 µ . Each of n the n layers of DNA possesses n seg- ments of length 0.25 µ . .25µ 3 2 n 0.25 = 10 µ n 60 .25µ + .25µ8 An Overview of Cell Structure and Function the surface of the chromosomal mass. The number of times, N, that the DNA must wind back and forth will then be related to the length of the DNA and the volume in which it is contained. If we approximate the path of the DNA as consisting of n layers, each layer consisting of n 2 segments of length 0.25 µ (Fig. 1.6), the total number of segments is n . 2 7 Therefore, 2,500n Å = 10 Å and n = 60. The spacing between adjacent segments of the DNA is 2,500 Å/60 = 40 Å. The close spacing between DNA duplexes raises the interesting prob- lem of accessibility of the DNA. RNA polymerase has a diameter of about 100 Å and it may not fit between the duplexes. Therefore, quite possibly only DNA on the surface of the nuclear mass is accessible for transcrip- tion. On the other hand, transcription of the lactose and arabinose operons can be induced within as short a time as two seconds after adding inducers. Consequently either the nuclear mass is in such rapid motion that any portion of the DNA finds its way to the surface at least once every several seconds, or the RNA polymerase molecules do penetrate to the interior of the nuclear mass and are able to begin transcription of any gene at any time. Possibly, start points of the arabinose and lactose operons always reside on the surface of the DNA. Compaction of the DNA generates even greater problems in eu- karyotic cells. Not only do they contain up to 1,000 times the amount of the DNA found in bacteria, but the presence of the histones on the DNA appears to hinder access of RNA polymerase and other enzymes to the DNA. In part, this problem is solved by regulatory proteins binding to regulatory regions before nucleosomes can form in these positions. Apparently, upon activation of a gene additional regulatory proteins bind, displacing more histones, and transcription begins. The DNA of many eukaryotic cells is specially contracted before cell division, and at this time it actually does become inaccessible to RNA polymerase. At all times, however, accessibility of the DNA to RNA polymerase must be hindered. Moving Molecules into or out of Cells Small-molecule metabolic intermediates must not leak out of cells into the medium. Therefore, an impermeable membrane surrounds the cytoplasm. To solve the problem of moving essential small molecules like sugars and ions into the cell, special transporter protein molecules are inserted into the membranes. These and auxiliary proteins in the cytoplasm must possess selectivity for the small-molecules being trans- ported. If the small-molecules are being concentrated in the cell and not just passively crossing the membrane, then the proteins must also couple the consumption of metabolic energy from the cell to the active transport. The amount of work consumed in transporting a molecule into a volume against a concentration gradient may be obtained by consider- ing the simple reaction where A is the concentration of the molecule o outside the cell and A is the concentration inside the cell: iMoving Molecules into or out of Cells 9 A A oi → A A o i ← This reaction can be described by an equilibrium constant A i K = eq A o The equilibrium constant K , is related to the free energy of the reaction eq by the relation ∆ G = RTlnK eq . where R is about 2 cal/degmole and T is 300° K (about 25° C), the temperature of many biological reactions. Suppose the energy of hy- drolysis of ATP to ADP is coupled to this reaction with a 50% efficiency. Then about 3,500 of the total of 7,000 calories available per mole of ATP hydrolyzed under physiological conditions will be available to the transport system. Consequently, the equilibrium constant will be ∆ G − K = e eq RT 3,500 =e 600 = 340. One interesting result of this consideration is that the work required to transport a molecule is independent of the absolute concentrations; it depends only on the ratio of the inside and outside concentrations. The transport systems of cells must recognize the type of molecule to be transported, since not all types are transported, and convey the molecule either to the inside or to the outside of the cell. Further, if the molecule is being concentrated within the cell, the system must tap an energy source for the process. Owing to the complexities of this process, it is not surprising that the details of active transport systems are far from being fully understood. Four basic types of small-molecule transport systems have been discovered. The first of these is facilitated diffusion. Here the molecule10 An Overview of Cell Structure and Function Cytoplasm Cytoplasmic membrane PEP Enzyme I Phosphoenzyme I Pyruvate HPr Enzyme I Phospho- Enzyme HPr III-X Phosphoenzyme HPr Enzyme III-X II-X Sugar-X Phosphoenzyme Enzyme II-X III-X Enzyme II-X Phosphosugar-X Figure 1.7 The cascade of reactions associated with the phosphotransferase sugar uptake system of E. coli. must get into or out of the cell on its own, but special doors are opened for it. That is, specific carriers exist that bind to the molecule and shuttle it through the membrane. Glycerol enters most types of bacteria by this mechanism. Once within the cell the glycerol is phosphorylated and cannot diffuse back out through the membrane, nor can it exit by using the glycerol carrier protein that carried the glycerol into the cell. A second method of concentrating molecules within cells is similar to the facilitated diffusion and phosphorylation of glycerol. The phos- photransferase system actively rather than passively carries a number of types of sugars across the cell membrane and, in the process, phos- phorylates them (Fig. 1.7). The actual energy for the transport comes from phosphoenolpyruvate. The phosphate group and part of the chemi- cal energy contained in the phosphoenolpyruvate is transferred down a series of proteins, two of which are used by all the sugars transported by this system and two of which are specific for the particular sugar being transported. The final protein is located in the membrane and is directly responsible for the transport and phosphorylation of the trans- ported sugar. Protons are expelled from E. coli during the flow of reducing power + from NADH to oxygen. The resulting concentration difference in H ions between the interior and exterior of the cell generates a proton motive force or membrane potential that can then be coupled to ATP synthesis or to the transport of molecules across the membrane. Active transport systems using this energy source are called chemiosmotic systems. In the process of permitting a proton to flow back into the cell, another small molecule can be carried into the cell, which is called symport, or carried out of the cell, which is called antiport (Fig. 1.8).Moving Molecules into or out of Cells 11 Outside + + 2H 2H 2e- 2e- Cytoplasmic membrane + + + + NADH H NAD 2H 1/2 O + 2H H O 2 2 Inside In many eukaryotic cells, a membrane potential is generated by the sodium-potassium pump. From the energy of hydrolysis of one ATP + + molecule, 3 Na ions are transported outside the cell and 2 K ions are transported inside. The resulting gradient in sodium ions can then be coupled to the transport of other molecules or used to transmit signals along a membrane. Study of all transport systems has been difficult because of the necessity of working with membranes, but the chemiosmotic system has been particularly hard due to the difficulty of manipulating membrane potentials. Fortunately the existence of bacterial mutants blocked at + Figure 1.8 Coupling the excess of H ions outside a cell to the transport of a specific molecule into the cell, symport, or out of the cell, antiport, by specific proteins that couple the transport of a proton into the cell with the transport of another molecule. The ATPase generates ATP from ADP with the energy derived from permitting protons to flow back into the cell. + + + + H + H H H H H + + + H + + H + H H H H Antiport Symport H Membrane H + H H H comes in while H and come ADP + Pi goes out. in together. ATP12 An Overview of Cell Structure and Function various steps of the transport process has permitted partial dissection of the system. We are, however, very far from completely understanding the actual mechanisms involved in chemiosmotic systems. The binding protein systems represent another type of transport through membranes. These systems utilize proteins located in the periplasmic space that specifically bind sugars, amino acids, and ions. Apparently, these periplasmic binding proteins transfer their substrates to specific carrier molecules located in the cell membrane. The energy source for these systems is ATP or a closely related metabolite. Transporting large molecules through the cell wall and membranes poses additional problems. Eukaryotic cells can move larger molecules through the membrane by exocytosis and endocytosis processes in which the membrane encompasses the molecule or molecules. In the case of endocytosis, the molecule can enter the cell, but it is still separated from the cytoplasm by the membrane. This membrane must be removed in order for the membrane-enclosed packet of material to be released into the cytoplasm. By an analogous process, exocytosis releases membrane-enclosed packets to the cell exterior. Releasing phage from bacteria also poses difficult problems. Some types of filamentous phage slip through the membrane like a snake. They are encapsidated as they exit the membrane by phage proteins located in the membrane. Other types of phage must digest the cell wall to make holes large enough to exit. These phage lyse their hosts in the process of being released. An illuminating example of endocytosis is the uptake of low density lipoprotein, a 200 Å diameter protein complex that carries about 1,500 molecules of cholesterol into cells. Pits coated with a receptor of the low density lipoprotein form in the membrane. The shape of these pits is guided by triskelions, an interesting structural protein consisting of three molecules of clathrin. After receptors have been in a pit for about Figure 1.9 Endocytosis of receptor-coated pits to form coated vesicles and the recycling of receptor that inserts at random into the plasma membrane and then clusters in pits. Receptors Coated vesicle Coated pit Clathrin cage Plasma membraneDiffusion within the Small Volume of a Cell 13 ten minutes, the pit pinches off and diffuses through the cytoplasm (Fig. 1.9). Upon reaching the lysosome, the clatherin cage of triskelions is disassembled, cholesterol is released, and the receptors recycle. Diffusion within the Small Volume of a Cell Within several minutes of adding a specific inducer to bacteria or eukaryotic cells, newly synthesized active enzymes can be detected. These are the result of the synthesis of the appropriate messenger RNA, its translation into protein, and the folding of the protein to an active conformation. Quite obviously, processes are happening very rapidly within a cell for this entire sequence to be completed in several minutes. We will see that our image of synthetic processes in the cellular interior should be that of an assembly line running hundreds of times faster than normal, and our image for the random motion of molecules from one point to another can be that of a washing machine similarly running very rapidly. The random motion of molecules within cells can be estimated from basic physical chemical principles. We will develop such an analysis since similar reasoning often arises in the design or analysis of experi- _ _ 2 ments in molecular biology. The mean squared distance R that a _ _ 2 molecule with diffusion constant D will diffuse in time t is R = 6Dt (Fig. 1.10). The diffusion constants of many molecules have been measured and are available in tables. For our purposes, we can estimate a value KT for a diffusion constant. The diffusion constant is D = ⁄f , where K is -16 the Boltzmann constant, 1.38 × 10 ergs/degree, T is temperature in degrees Kelvin, and f is the frictional force. For spherical bodies, f = 6πηr , where r is the radius in centimeters and η is the viscosity of the medium in units of poise. -2 The viscosity of water is 10 poise. Although the macroviscosity of the cell’s interior could be much greater, as suggested by the extremely high viscosity of gently lysed cells, the viscosity of the cell’s interior with Figure 1.10 Random motion of z a particle in three dimensions be- ginning at the origin and the definition of the mean squared _ _ 2 distance R . R y x14 An Overview of Cell Structure and Function respect to motion of molecules the size of proteins or smaller is more likely to be similar to that of water. This is reasonable, for small molecules can go around obstacles such as long strands of DNA, but large molecules would have to displace a huge tangle of DNA strands. A demonstration of this effect is the finding that small molecules such as amino acids readily diffuse through the agar used for growing bacterial colonies, but objects as large as viruses are immobile in the agar, yet diffuse normally in solution. -7 KT Since D = ⁄ πηr, then D = 4.4 × 10 for a large spherical protein of 6 radius 50 Å diffusing in water, and the diffusion constant for such a protein within a cell is not greatly different. Therefore _ _ 2 −7 R = 6 × 4.4 × 10 t, and the average time required for such protein molecules to diffuse the length of a 1 µ bacterial cell is 1/250 second and to diffuse the length of a 20 µ eukaryotic cell is about 2 seconds. Analogous reasoning with respect to rotation shows that a protein rotates about 1/8 radian (about 7°) in the time it diffuses a distance equal to its radius. Exponentially Growing Populations Reproducibility from one day to the next and between different labora- tories is necessary before meaningful measurements can be made on growing cells. Populations of cells that are not overcrowded or limited by oxygen, nutrients, or ions grow freely and can be easily reproduced. Such freely growing populations are almost universally used in molecu- lar biology, and several of their properties are important. The rate of increase in the number of cells in a freely growing population is proportional to the number of cells present, that is, dN µ t = µ N, or N(t) = N(0)e . dt In these expressions µ is termed the exponential growth rate of the cells. The following properties of the exponential function are frequently useful when manipulating data or expressions involving growth of cells. alnx a e = x d ax ax e = ae dx ∞ n x x e = ∑ n n = 0 µ t Quantities growing with the population increase as e Throughout . this book we will use µ as the exponential growth rate. The time required for cells to double in number, T , is easier to measure experimentally as d well as to think about than the exponential growth rate. Therefore we often need to interconvert the two rates T and µ . Note that the number d of cells or some quantity related to the number of cells in freely growingComposition Change in Growing Cells 15 t/Td ln2 populations can be written as Q(t) = 2 , and since 2 = e , Q(t) can (ln2/ also be written as Q(t) = e Td)t, thereby showing that the relation between T and µ is µ = ln2/T . d d Composition Change in Growing Cells In many experiments it is necessary to consider the time course of the induction of an enzyme or other cellular component in a population of growing cells. To visualize this, suppose that synthesis of an enzyme is initiated at some time in all the cells in the population and thereafter the synthesis rate per cell remains constant. What will the enzyme level per cell be at later times? The Relationship between Cell Doublings, Enzyme Doublings, and Induction Kinetics Time t=0 t=T t=2T t=3T d d d Cell Mass 1 2 4 8 Enzyme present if synthesis began long ago A 2A 4A 8A Enzyme synthesized during one doubling time A 2A 4A Enzyme present if synthesis begins at t=0 0 A 3A 7A One way to handle this problem is to consider a closely related problem we can readily solve. Suppose that synthesis of the enzyme had begun many generations earlier and thereafter the synthesis rate per cell had remained constant. Since the synthesis of the enzyme had been initiated many cell doublings earlier, by the time of our consideration, the cells are in a steady state and the relative enzyme level per cell remains constant. As the cell mass doubles from 1 to 2 to 4, and so on, the amount of the enzyme, A, also doubles, from A to 2A to 4A, and so on. The differences in the amount of the enzyme at the different times give the amounts that were synthesized in each doubling time. Now consider the situation if the same number of cells begins with no enzyme but instead begins synthesis at the same rate per cell as the population that had been induced at a much earlier time (see the last row in the table). At the beginning, no enzyme is present, but during the first doubling time, an amount A of the enzyme can be synthesized by the cells. In the next doubling time, the table shows that the cells can synthesize an amount 2A of the enzyme, so that after two doubling times the total amount of enzyme present is 3A. After another doubling time the amount of enzyme present is 7A. Thus at successive doublings after 1 3 7 induction the enzyme level is ⁄ , ⁄ , ⁄ ,... of the final asymptotic value. 2 4 8 Age Distribution in Populations of Growing Cells The cells in a population of freely growing cells are not all alike. A newly divided cell grows, doubles in volume, and divides into two daughter16 An Overview of Cell Structure and Function p(a) Figure 1.11 Age distribution in an exponentially growing population in which all cells divide when they reach age 1. Note that the popula- tion contains twice as many 0 1 Age zero-age cells as unit-age cells. cells. Consequently, freely growing populations contain twice as many cells that have just divided as cells about to divide. The distribution of cell ages present in growing populations is an important consideration in a number of molecular biology experiments, one of which is men- tioned in Chapter 3. Therefore we will derive the distribution of ages present in such populations. Consider an idealized case where cells grow until they reach the age of 1, at which time they divide. In reality most cells do not divide at exactly this age, but the ages at which cell division occurs cluster around a peak. To derive the age distribution, let N(a,t)da be the number of cells with age between a and a + da at time t. For convenience, we omit writing the da. Since the number of cells of age a at time t must be the same as the number of zero-age cells at time t-a, N(a,t) = N(0,t -a). Since the µ t, numbers of cells at any age are growing exponentially, N(0,t) = N(0,0)e -µ a and N(a,t) = N(0,t-a) = N(0,t)e . Therefore the probability that a cell is -µ a -a/Td of age a, p(a), is p(0)e = p(0)2 (Fig. 1.11). Problems 1.1. Propose an explanation for the following facts known about E. coli: appreciable volume exists between the inner membrane and the peptidoglycan layer; the inner membrane is too weak to withstand the osmotic pressure of the cytoplasm and must be supported by a strong, rigid structure; and no spacers have been discovered that could hold the inner membrane away from the peptidoglycan layer. + 1.2. If the E. coli interior were water at pH 7, how many H ions would exist within the cell at any instant? 1.3. If a population of cells growing exponentially with a doubling 7 time T were contaminated at one part in 10 with cells whose doubling d time is 0.95 T , how many doublings will be required until 50% of the d cells are contaminants? Problems 17 1.4. If an enzyme is induced and its synthesis per cell is constant, show that there is a final upper-bound less than 100% of cellular protein that this enzyme can constitute. When the enzyme has reached this level, what is the relation between the rate of synthesis of the enzyme and the rate of dilution of the enzyme caused by increase of cellular volume due to growth? 1.5. In a culture of cells in balanced exponential growth, an enzyme was induced at time t = 0. Before induction the enzyme was not present, and at times very long after induction it constituted 1% of cell protein. What is the fraction of cellular protein constituted by this protein at any time t 0 in terms of the cell doubling time? Ignore the 1 min or so lag following induction until the enzyme begins to appear. 1.6. In a culture of cells in balanced exponential growth, an enzyme was fully induced at some very early time, and the level of enzyme ultimately reached 1% of total protein. At time t = 0 the synthesis of enzyme was repressed. What fraction of cellular protein is constituted by the enzyme for t 0 (a) if the repressed rate of synthesis is 0 and (b) if the repressed rate of synthesis is 0.01 of the fully induced rate? -3 1.7. If the concentration of a typical amino acid in a bacterium is 10 M, estimate how long this quantity, without replenishment, could -13 support protein synthesis at the rate that yields 1 × 10 g of newly synthesized protein with a cell doubling time of 30 min. 1.8. If a typical protein can diffuse from one end to the other of a cell in 1/250 sec when it encounters viscosity the same as that of water, how long is required if the viscosity is 100 times greater? 1.9. A protein of molecular weight 30,000 daltons is in solution of 200 mg/ml. What is the average distance separating the centers of the molecules? If protein has a density of 1.3, what fraction of the volume of such a solution actually is water? + + 1.10. How can the existence of the Na -K pump in eukaryotic cells be demonstrated? 1.11. How can valinomycin be used to create a temporary membrane potential in cells or membrane vesicles? 1.12. Suppose the synthesis of some cellular component requires synthesis of a series of precursors P , P , P proceeding through a series 1 2 n of pools S . i P →P →P → →P 1 2 3 n S →S →S → →S 1 2 3 n Suppose the withdrawal of a precursor molecule P from pool S , and i i its maturation to S is random. Suppose that at t = 0 all subsequently i+1 synthesized precursors P are radioactively labeled at constant specific 1 activity. Show that at the beginning, the radioactive label increases n proportional to t in pool S . n 1.13. Consider cells growing in minimal medium. Suppose a radio- active amino acid is added and the kinetics of radioactivity incorpora-18 An Overview of Cell Structure and Function tion into protein are measured for the first minute. Assume that upon addition of the amino acid, the cell completely stops its own synthesis of the amino acid and that there is no leakage of the amino acid out of the cell. For about the first 15 sec, the incorporation of radioactive 2 amino acid into protein increases as t and thereafter as t. Show how this delayed entry of radioactive amino acids into protein results from the pool of free nonradioactive amino acid in the cells at the time the radioactive amino acid was added. Continue with the analysis and show how to calculate the concentration of this internal pool. Use data of Fig. 2 in J. Mol. Bio. 27, 41 (1967) to calculate the molarity of free proline in E. coli B/r. 1.14. Consider a more realistic case for cell division than was consid- ered in the text. Suppose that cells do not divide precisely when they reach age 1 but that they have a probability given by the function f(a) of dividing when they are of age a. What is the probability that a cell is of age a in this case? References Recommended Readings Role of an Electrical Potential in the Coupling of Metabolic Energy to Active Transport by Membrane Vesicles of Escherichia coli, H. Hirata, K. Altendorf, F. Harold, Proc. Nat. Acad. Sci. USA 70, 1804-1808 (1973). Coated Pits, Coated Vesicles, and Receptor-Mediated Endocytosis, J. Goldstein, R. Anderson, M. Brown, Nature 279, 679-685 (1979). Osmotic Regulation and the Biosynthesis of Membrane-Derived Oligosac- charides in Escherichia coli, E. Kennedy, Proc. Nat. Acad. Sci. USA 79, 1092-1095 (1982). Cell Structure Sugar Transport, I. Isolation of A Phosphotransferase System from E. coli, W. Kundig, S. Roseman, J. Biol. Chem. 246, 1393-1406 (1971). Localization of Transcribing Genes in the Bacterial Cell by Means of High Resolution Autoradiography, A. Ryter, A. Chang, J. Mol. Biol. 98, 797-810 (1975). The Relationship between the Electrochemical Proton Gradient and Ac- tive Transport in E. coli Membrane Vesicles, S. Ramos, H. Kaback. Biochem. 16, 854-859 (1977). Escherichia coli Intracellular pH, Membrane Potential, and Cell Growth, D. Zilberstein, V. Agmon, S. Schuldiner, E. Padan, J. Bact. 158, 246-252 (1984). Ion Selectivity of Gram-negative Bacterial Porins, R. Benz, A. Schmid, R. Hancock, J. Bact. 162, 722-727 (1985). Measurement of Proton Motive Force in Rhizobium meliloti with Es- cherichia coli lacY Gene Product, J. Gober, E. Kashket, J. Bact. 164, 929-931 (1985). Internalization-defective LDL Receptors Produced by Genes with Non- sense and Frameshift Mutations That Truncate the Cytoplasmic Do-References 19 main, M. Lerman, J. Goldstein, M. Brown, D. Russell, W. Schneider, Cell 41, 735-743 (1985). Escherichia coli and Salmonella typhimurium, Cellular and Molecular Biology, eds. F. Neidhardt, J. Ingraham, K. Low, B. Magasanik, M. Schaechter, H. Umbarger, Am. Society for Microbiology (1987). Introduction of Proteins into Living Bacterial Cells: Distribution of La- beled HU Protein in Escherichia coli, V. Shellman, D. Pettijohn, J. Bact. 173, 3047-3059 (1991). Characterization of the Cytoplasm of Escherichia coli as a Function of External Osmolarity, S. Cayley, B. Lewis, H. Guttman, M. Record, Jr., J. Mol. Biol. 222, 281-300 (1991). Estimation of Macromolecule Concentrations and Excluded Volume Ef- fects for the Cytoplasm of Escherichia coli, S. Zimmerman, S. Trach, J. Mol. Biol. 222, 599-620 (1991). Inside a Living Cell, D. Goodsell, Trends in Biological Sciences, 16, 203-206 (1991).

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