CBE 555

Apply the principle of exponential growth of a culture as described in Question 1-13 to the cells in a multicellular organism, such as yourself. There are about 10^13 cells in your body. Assume that one cell acquires a mutation that allows it to divide in an uncontrolled manner (i.e., it becomes a cancer cell). Some cancer cells can proliferate with a generation time of about 24 hours. If none of the cancer cells died, how long would it take before 1013 cells in your body would be cancer cells? (Use the equation N = N0 × 2t/G, with t, the time, and G, the length of each generation. Hint: 10^13 ≈ 2^43.)

10^13 = 2(t/1). Therefore, it would take only 43 days [t = 13/log(2)]. This explains why some cancers can progress extremely rapidly. Many cancer cells divide answers A:3 much more slowly, however, or die because of their internal abnormalities or because they do not have sufficient blood supply, and the actual progression of cancer is therefore usually slower.

"The nucleotide sequence of one DNA strand of a DNA double helix is 5'-GGATTTTTGTCCACAATCA-3'. What is the sequence of the complementary strand? "

5'-CCTAAAAACAGGTGTTAGT-3'.

A bacterium weighs about 10^-12 g and can divide every 20 minutes. If a single bacterial cell carried on dividing at this rate, how long would it take before the mass of bacteria would equal that of the Earth (6 × 10^24 kg)? Contrast your result with the fact that bacteria originated at least 3.5 billion years ago and have been dividing ever since. Explain the apparent paradox. (The number of cells N in a culture at time t is described by the equation N = N0 × 2^t/G, where N0 is the number of cells at zero time and G is the population doubling time.)

6 × 10^39 (= 6 × 10^27 g/10^-12 g) bacteria would have the same mass as the Earth. And 6 × 10^39 = 2^t/20, according to the equation describing exponential growth. Solving this equation for t results in t = 2642 minutes (or 44 hours). This represents only 132 generation times(!), whereas 5 × 10^14 bacterial generation times have passed during the last 3.5 billion years. Obviously, the total mass of bacteria on this planet is nowhere close to the mass of the Earth. This illustrates that exponential growth can occur only for very few generations, i.e., for minuscule periods of time compared with evolution. In any realistic scenario, food supplies very quickly become limiting. This simple calculation shows us that the ability to grow and divide quickly when food is ample is only one factor in the survival of a species. Food is generally scarce, and individuals of the same species have to compete with one another for the limited resources. Natural selection favors mutants that either win the competition or find ways to exploit food sources that their neighbors are unable to use.

In principle, there are many different, chemically diverse ways in which small molecules can be linked to form polymers. For example, the small molecule ethene (CH2=CH2) is used commercially to make the plastic polyethylene (...-CH2-CH2-CH2-CH2-CH2-...). The individual subunits of the three major classes of biological macromolecules, however, are all linked by similar reaction mechanisms, i.e., by condensation reactions that eliminate water. Can you think of any benefits that this chemistry offers and why it might have been selected in evolution?

A major advantage of condensation reactions is that they are readily reversible by hydrolysis (and water is readily available in the cell). This allows cells to break down their macromolecules (or macromolecules ofother organisms that were ingested as food) and to recover the subunits intact so that they can be "recycled," i.e., used to build new macromolecules.

"What, if anything, is wrong with the following statement: "DNA stability in both reproductive cells and somatic cells is essential for the survival of a species." Explain your answer."

A mutation in reproductive cells will be passed on to all cells in the body of the organism that develop from it, including the reproductive cells that are responsible for the production o fthe next generation of species. Hence the species won't be maintained. In somatic cells, DNA instability can gice rise to variant cells which can grow and divide in an uncontrolled way and could cause cancer.

The formation of actin filaments in the cytosol is controlled by actin- binding proteins. Some actin-binding proteins significantly increase the rate at which the formation of an actin filament is initiated. Suggest a mechanism by which they might do this.

A protein binding to actin that can stabilize a complex of at least two actin monomers without blocking the end will facilitate the formation of a new filament.

Simply judged by their potential for hydrogen-bonding, could any of these extraterrestrial bases replace terrestrial A, T, G, or C in terrestrial DNA? Explain your answer. ECB4 EQ5.08/Q5.07

A pyrimidine needs to pair with a purine to form the double helix. The eight possible combinations are V-A, V-G, W-A, W-G, X-C, X-T, Y-C, Y-T. But because of the positions of the H-bond donor and acceptor groups, none of those are stable enough. Thus no.

A. How many different molecules composed of (a) two, (b) three, and (c) four amino acids, linked together by peptide bonds, can be made from the set of 20 naturally occurring amino acids? B. Assume you were given a mixture consisting of one molecule each of all possible sequences of a smallish protein of molecular weight 4800 daltons. If the average molecular weight of an amino acid is, say, 120 daltons, how much would the sample weigh? How big a container would you need to hold it? C. What does this calculation tell you about the fraction of possible proteins that are currently in use by living organisms (the average molecular weight of proteins is about 30,000 daltons)?

A. (a) 400 (= 20^2); (b) 8000 (= 20^3); (c) 160,000 (= 20^4). B. A protein with a molecular weight of 4800 daltons is made of about 40 amino acids; thus there are 1.1 × 10^52 (= 20^40) different ways to make such a protein. Each individual protein molecule weighs 8 × 10-21 g (= 4800/6 × 1023); thus a mixture of one molecule of each weighs 9 × 1031 g (= 8 × 10-21 g × 1.1 × 1052), which is 15,000 times the total weight of the planet Earth, weighing 6 × 1024 kg. You would need a quite large container, indeed. C. Given that most cell proteins are even larger than the one used in this example, it is clear that only a minuscule fraction of the total possible amino acid sequences are used in living cells.

The average molecular weight of a protein in the cell is about 30,000 daltons. A few proteins, however, are much larger. The largest known polypeptide chain made by any cell is a protein called titin (made by mammalian muscle cells), and it has a molecular weight of 3,000,000 daltons. Estimate how long it will take a muscle cell to translate an mRNA coding for titin (assume the average molecular weight of an amino acid to be 120, and a translation rate of two amino acids per second for eukaryotic cells).

A. A titin molecule is made of 25,000 (3,000,000/120) amino acids. It therefore takes about 3.5 hours [(25,000/2 )* (1/60) * (1/60)] to synthesize a single molecule of titin in muscle cells.

The sketch in Figure Q15-22 is a schematic drawing of the electron micrograph shown in the third panel of Figure 15-19A. Name the structures that are labeled in the sketch.

A. Extracellular space B. Cytosol C. Plasma membrane D. Clathrin coat E. Membrane of deeply invaginated, clathrin-coated pit F. Captured cargo particles G. Lumen of deeply invaginated, clathrin-coated pit

Which of the following statements are correct? Explain your answers. A. An individual ribosome can make only one type of protein. B. All mRNAs fold into particular three-dimensional structures that are required for their translation. C. The large and small subunits of an individual ribosome always stay together and never exchange partners. D. RIbosomes are cytoplasmic organelles that are encapsulated by a single membrane. E. Because the two strands of DNA are complementary, the mRNA of a given gene can be synthesized using either strand as a template. F. An mRNA may contain the sequence ATTGACCCCGGTCAA. G. The amount of a protein present in a cell depends on its rate of synthesis, its catalytic activity, and its rate of degradation.

A. False. Ribosomes can make any protein that is specified by the particular mRNA that they are translating. After translation, ribosomes are released from the mRNA and can then start translating a different mRNA. It is true, however, that a ribosome can only make one type of protein at a time. B. False. mRNAs are translated as linear polymers; there is no requirement that they have any particular folded structure. In fact, such structures that are formed by mRNA can inhibit its translation, because the ribosome has to unfold the mRNA in order to read the message it contains. C. False. Ribosomal subunits exchange partners after each round of translation. After a ribosome is released from an mRNA, its two subunits dissociate and enter a pool of free small and large subunits from which new ribosomes assemble around a new mRNA. D. False. Ribosomes are cytoplasmic organelles, but they are not individually enclosed in a membrane. E. False. The position of the promoter determines the direction in which transcription proceeds and therefore which DNA strand is used as the template. Transcription in the opposite direction would produce an mRNA with a completely different (and probably meaningless) sequence. F. False. RNA contains uracil but not thymine. G. False. The level of a protein depends on its rate of synthesis and degradation but not on its catalytic activity.

Which of the following statements are correct? Explain your answers. A. T he hereditary information of a cell is passed on by its proteins. B. Bacterial DNA is found in the cytosol. C. P lants are composed of prokaryotic cells. D. A ll cells of the same organism have the same number of chromosomes (with the exception of egg and sperm cells). E. T he cytosol contains membrane-enclosed organelles, such as lysosomes. F. T he nucleus and mitochondria are surrounded by a double membrane. G. P rotozoans are complex organisms with a set of specialized cells that form tissues, such as flagella, mouthparts, stinging darts, and leglike appendages. H. L ysosomes and peroxisomes are the sites of degradation of unwanted materials.

A. False. The hereditary information is encoded in the cell's DNA, which in turn specifies its proteins (via RNA). B. True. Bacteria do not have a nucleus. C. False. Plants are composed of eukaryotic cells that contain chloroplasts as cytoplasmic organelles. The chloroplasts are thought to be evolutionarily derived from prokaryotic cells. D. True. The number of chromosomes varies from one organism to another, but is constant in all cells (except germ cells) of the same organism. E. False. The cytosol is the cytoplasm excluding all membrane-enclosed organelles. F. True. The nuclear envelope is a double membrane, and mitochondria are surrounded by both an inner and an outer membrane. G. False. Protozoans are single-celled organisms and therefore do not have different tissues or cell types. They have a complex structure, however, that has highly specialized parts. H. Somewhat true. Peroxisomes and lysosomes contain enzymes that catalyze the breakdown of substances produced in the cytosol or taken up by the cell. One can argue, however, that many of these substances are degraded to generate food molecules, and as such are certainly not "unwanted."

Which of the following statements are correct? Explain your answers. A. Proteins are so remarkably diverse because each is made from a unique mixture of amino acids that are linked in random order. B. Lipid bilayers are macromolecules that are made up mostly of phospholipid subunits. C. Nucleic acids contain sugar groups. D. Many amino acids have hydrophobic side chains. E. The hydrophobic tails of phospholipid molecules are repelled from water. F. DNA contains the four different bases A, G, U, and C.

A. False. The properties of a protein depend on both the amino acids it contains and the order in which they are linked together. The diversity of proteins is due to the almost unlimited number of ways in which 20 different amino acids can be combined in a linear sequence. B. False. Phospholipids assemble into bilayers in an aqueous environment by noncovalent forces. Lipid bilayers are therefore not macromolecules. C. True. The backbone of nucleic acids is made up of alternating ribose (or deoxyribose in DNA) and phosphate groups. Ribose and deoxyribose are sugars. D. True. About half of the 20 naturally occurring amino acids have hydrophobic side chains. In folded proteins, many of these side chains face toward the inside of a folded-up globular protein, because they are repelled from water E. True. Hydrophobic hydrocarbon tails contain only nonpolar bonds. Thus, they cannot participate in hydrogen-bonding and are repelled from water. We consider the underlying principles in more detail in Chapter 11. F. False. RNA contains the four listed bases, but DNA contains T instead of U. T and U are very much alike, however, and differ only by a single methyl group.

A. Describe the similarities and differences between van der Waals attractions and hydrogen bonds. B. Which of the two bonds would form (a) between two hydrogens bound to carbon atoms, (b) between a nitrogen atom and a hydrogen bound to a carbon atom, and (c) between a nitrogen atom and a hydrogen bound to an oxygen atom?

A. Hydrogen bonds form between two specific chemical groups; one is always a hydrogen atom linked in a polar covalent bond to an oxygen or a nitrogen atom, and the other is usually a nitrogen or an oxygen atom. Van der Waals attractions are weaker and occur between any two atoms that are in close enough proximity. Both hydrogen bonds and van der Waals attractions are short-range interactions that come into play only when two molecules are already in close proximity. Both types of bonds can therefore be thought of as means of "fine-tuning" an interaction, i.e., helping to position two molecules correctly with respect to each other once they have been brought together by diffusion. B. Van der Waals attractions would occur in all three examples. Hydrogen bonds would form only in (c).

The budding of clathrin-coated vesicles from eukaryotic plasma membrane fragments can be observed when adaptins, clathrin, and dynamin-GTP are added to the membrane preparation. What would you observe if you omitted (A) adaptins, (B) clathrin, or (C) dynamin? (D) What would you observe if the plasma membrane fragments were from a prokaryotic cell?

A. The function of adaptins is to secure the clathrin coat to the vesicle membrane and help select cargo molecules for transport. In absence of adaptins, clathrin coats cannot assemble. At high clathrin concentrations and under the appropriate ionic conditions, clathrin cages assemble in solution, but they are empty shells, lacking other proteins, and they contain no membrane. This shows that the information to form clathrin baskets is contained in the clathrin molecules themselves, which are therefore able to selfassemble. B. Without clathrin, adaptins still bind to receptors in the membrane, but no clathrin coat can form and thus no clathrin-coated pits or vesicles are produced. C. Dynamin assembles as a ring around the neck of each deeply invaginated coated pit. Together with other proteins recruited to the neck of the vesicle, the dynamin causes the ring to constrict, thereby pinching off the vesicle from its parent membrane. In the absence of dynamin, deeply invaginated clathrin-coated pits form on the membrane, but they do not pinch off to form closed vesicles. D. Prokaryotic cells do not perform endocytosis. A prokaryotic cell therefore does not contain any receptors with appropriate cytosolic tails that could mediate adaptin binding. Therefore, no clathrin can bind and no clathrin coats can assemble.

A virus that grows in bacteria (bacterial viruses are called bacteriophages) can replicate in one of two ways. In the prophage state, the viral DNA is inserted into the bacterialchromosome and is copied along with the bacterial genome each time the cell divides. In the lytic state, the viral DNA is released from the bacterial chromosome and replicates many times in the cell. This viral DNA then produces viral coat proteins that together with the replicated viral DNA form many new virus particles that burst out of the bacterial cell. These two forms of growth are controlled by two transcription regulators, called c1 ("c one") and Cro, that are encoded by the virus. In the prophage state, cI is expressed; in the lytic state, Cro is expressed. In addition to regulating the expression of other genes, c1 represses the Cro gene, and Cro represses the c1 gene (Figure Q8-4). When bacteria containing a phage in the prophage state are briefly irradiated with UV light, c1 protein is degraded. A. What will happen next? B. Will the change in (A) be reversed when the UV light is switched off? C. Why might this response to UV light have evolved?

A. UV light throws the switch from the prophage to the lytic state: when cI protein is destroyed, Cro is made and turns off the further production of cI. The virus produces coat proteins, and new virus particles are made. B. When the UV light is switched off, the virus remains in the lytic state. Thus, cI and Cro form a gene regulatory switch that "memorizes" its previous setting. C. This switch makes sense in the viral life cycle: UV light tends to damage the bacterial DNA (see Figure 6-24), thereby rendering the bacterium an unreliable host for the virus. A prophage will therefore switch to the lytic state and leave the "sinking ship" in search for new host cells to infect.

The structure of a lipid bilayer is determined by the particular properties of its lipid molecules. What would happen if A. Phospholipids had only one hydrocarbon tail instead of two? B. The hydrocarbon tails were shorter than normal, say, about 10 carbon atoms long? C. All of the hydrocarbon tails were saturated? D. All of the hydrocarbon tails were unsaturated? E. The bilayer contained a mixture of two kinds of phospholipid molecules, one with two saturated hydrocarbon tails and the other with two unsaturated hydrocarbon tails? F. Each phospholipid molecule were covalently linked through the end carbon atom of one of its hydrocarbon tails to a phospholipid tail in the opposite monolayer?

A. You would have a detergent. The diameter of the lipid head would be much larger than that of the hydrocarbon tail, so that the shape of the molecule would be a cone rather than a cylinder, and the molecules would aggregate to form micelles rather than bilayers. B. Lipid bilayers formed would be much more fluid, as the tails would have less tendency to interact with one another. The bilayers would also be less stable, as the shorter hydrocarbon tails would be less hydrophobic, so the forces that drive the formation of the bilayer would be reduced. C. The lipid bilayers formed would be much less fluid. Whereas a normal lipid bilayer has the viscosity of olive oil, a bilayer made of the same lipids but with saturated hydrocarbon tails would have the consistency of bacon fat. D. The lipid bilayers formed would be much more fluid. Also, because the lipids would pack together less well, there would be more gaps and the bilayer would be more permeable to small, water-soluble molecules. E. If we assume that the lipid molecules are completely intermixed, the fluidity of the membrane would be unchanged. In such bilayers, however, the saturated lipid molecules would tend to aggregate with one another because they can pack so much more tightly and would therefore form patches of much-reduced fluidity. The bilayer would not, therefore, have uniform properties over its surface. F. The lipid bilayers formed would have virtually unchanged properties. Each lipid molecule would now span the entire membrane, with one of its two head groups exposed at each surface.

"A compact disc (CD) stores about 4.8 × 10^9 bits of information in a 96 cm2 area. This information is stored as a binary code—that is, every bit is either a 0 or a 1. A. How many bits would it take to specify each nucleotide pair in a DNA sequence? B. How many CDs would it take to store the information contained in the human genome?"

A. we need 2 bits (00, 01, 10, 11) B. Human genome: 3*10^9 nucleotide pairs 2CD are required (4.8*10^9*2/(3*10^9))

. Which of the following statements are correct? Explain your answers. A. Ribosomes are cytoplasmic structures that, during protein synthesis, become linked by an mRNA molecule to form polyribosomes. B. The amino acid sequence Leu-His-Arg-Leu-Asp-Ala-Gln-Ser-Lys-Leu-Ser-Ser is a signal sequence that directs proteins to the ER. C. All transport vesicles in the cell must have a v-SNARE protein in their membrane. D. T ransport vesicles deliver proteins and lipids to the cell surface. E. If the delivery of prospective lysosomal proteins from the trans Golgi network to the late endosomes were blocked, lysosomal proteins would be secreted by the constitutive secretion pathways shown in Figure 15-30. F. Lysosomes digest only substances that have been taken up by cells by endocytosis. G. N-linked sugar chains are found on glycoproteins that face the cell surface, as well as on glycoproteins that face the lumen of the ER, trans Golgi network, and mitochondria.

A.True. B. False. The signal sequences that direct proteins to the ER contain a core of eight or more hydrophobic amino acids. The sequence shown here contains many hydrophilic amino acid side chains, including the charged amino acids His, Arg, Asp, and Lys, and the uncharged hydrophilic amino acids Gln and Ser. C. True. Otherwise they could not dock at the correct target membrane or recruit a fusion complex to a docking site. D. True. E. True. Lysosomal proteins are selected in the trans Golgi network and packaged into transport vesicles that deliver them to the late endosome. If not selected, they would enter by default into transport vesicles that move constitutively to the cell surface. F. False. Lysosomes also digest internal organelles by autophagy. G. False. Mitochondria do not participate in vesicular transport, and therefore N-linked glycoproteins, which are exclusively assembled in the ER, cannot be transported to mitochondria.

Which of the following statements are correct? Explain your answers. A. Lipids in a lipid bilayer spin rapidly around their long axis. B. Lipids in a lipid bilayer rapidly exchange positions with one another in their own monolayer. C. Lipids in a lipid bilayer do not flip-flop readily from one lipid monolayer to the other. D. Hydrogen bonds that form between lipid head groups and water molecules are continually broken and re-formed. E. Glycolipids move between different membrane-enclosed compartments during their synthesis but remain restricted to one side of the lipid bilayer. F. margarine contains more saturated lipids than the vegetable oil from which it is made. G. some membrane proteins are enzymes. H. The sugar layer that surrounds all cells makes cells more slippery.

All of the statements are correct. - A, B, C, D. The lipid bilayer is fluid because its lipid molecules can undergo these motions. - E. Glycolipids are mostly restricted to the monolayer of membranes that faces away from the cytosol. - F. The reduction of double bonds (by hydrogenation) allows the resulting saturated lipid molecules to pack more tightly against one another and therefore increases viscosity - G. Many of the membrane enzymes involved in cell signaling are enzymes. - H. Polysaccharides are the main constituents of mucus and slime.

In cells, an enzyme catalyzes the reaction AB → A + B. It was isolated, however, as an enzyme that carries out the opposite reaction A + B → AB. explain the paradox.

All reactions are reversible, so if AB can go to A + B, then A + B can also go to AB. Which direction of the reaction dominates depends on the equilibrium constant of the reaction and concentrations of A, B, and AB. In the experiment, the enzyme was probably added to a solution containing much more A and B than AB, and helped generate more AB. However, in the cell, there could actually be way more AB than A and B, and the enzyme would work in the opposite direction. TL;DR An enzyme doesn't favor a direction of a reaction, but only speeds up reactions.

Explain why the hypothetical enzymes in figure 4-47 have a great advantage in opening the safe if they work together in a protein complex, as opposed to working individually in an unlinked, sequential manner.

All three proteins work together in a complex to contribute to the specificity. They help position each other correctly, provide mechanical to allow each other to perform tasks they could otherwise not do individually. Their functions are also coordinated in time.

Fatty acids are said to be "amphipathic." What is meant by this term, and how does an amphipathic molecule behave in water? Draw a diagram to illustrate your answer.

Amphipathic molecules have both a hydrophilic and a hydrophobic end. Their hydrophilic end can hydrogen-bond to water, but their hydrophobic end is repelled from water because it interferes with the water structure. Consequently, the hydrophobic ends of amphipathic molecules tend to be exposed to air at air-water interfaces, or, in the interior of an aqueous solution, they will always cluster together to minimize their contact with water molecules.

discuss the following statement: "if plant cells contained intermediate filaments to provide the cells with tensile strength, their cell walls would be dispensable."

An intermediate filament would not be able to prevent the plant cell membrance from bursting if there is a huge osmotic pressure from inside the cell hence the cell would be dispensible.

When bacteria are grown under adverse conditions, i.e., in the presence of a poison such as an antibiotic, most cells grow and proliferate slowly. But it is not uncommon that the growth rate of a bacterial culture kept in the presence of the poison is restored after a few days to that observed in its absence. Suggest why this may be the case.

Bacteria continually acquire mutations in their DNA. In the population of cells exposed to the poison, one or a few cells may harbor a mutation that makes them resistant to the action of the drug. Antibiotics that are poisonous to bacteria because they bind to certain bacterial proteins, for example, would not work if the proteins have a slightly changed surface so that binding occurs more weakly or not at all. These mutant bacteria would continue dividing rapidly while their cousins are slowed down. The antibioticresistant bacteria would soon become the predominant species in the culture.

Protein synthesis is very accurate: for every 10,000 amino acids joined together, only one mistake is made. What is the fraction of average-sized protein molecules and of titin molecules that are synthesized without any errors? (Hint: the probability P of obtaining an error-free protein is given by P = (1 - E)^n, where E is the error frequency and n the number of amino acids.)

Because of its large size, the probability of making a titiin molecule without any mistakes is only 0.08 [= (1 - 10-4)^25,000]; i.e., only 8 in 100 titin molecules synthesized are free of mistakes. In contrast, over 97% of newly synthesized proteins of average size are made correctly.

Some transcription regulators bind to DNA and cause the double helix to bend at a sharp angle. Such "bending proteins" can stimulate the initiation of transcription without contacting either the RNA polymerase, any of the general transcription factors, or any other transcription regulators. Can you devise a plausible explanation for how these proteins might work to modulate transcription? Draw a diagram that illustrates your explanation.

Bending proteins can help to bring distant DNA regions together that normally would contact each other only inefficiently(Figure A8-3). Such proteins are found in both prokaryotes and eukaryotes and are involved in many examples of transcriptional regulation.

Explain how phosphorylation and the binding of a nucleotide (such as ATP or GTP) can both be used to regulate protein activity. What do you suppose are the advantages of either form of regulation?

Both nucleotide binding and phosphorylation can induce allosteric changes in proteins. Some of the consequences are: -altered enzyme activity -drastic shape changes -changes in affinity for other small molecules and proteins Advantage of phosphorylation: -requires only a single amino acid on protein's surface (as opposed to specific binding site). Phosphates can therefore be added to many different side chains on the same protein thereby vastly increasing the complexity of regulation that can be achieved for a single protein. Advantage of nucleotide binding: -the fast rate with which a small nucleotide can diffuse to the protein (motor proteins, for example, require fast shape changes)

carefully consider the graph in Figure Q20-16, showing the number of cases of colon cancer diagnosed per 100,000 women per year as a function of age. Why is this graph so steep and curved, if mutations occur with a similar frequency throughout a person's life-span?

Cancer will result only if multiple mutations occur since a single mutation is not sufficient for cancer. If this was the case, the curve would have been straight. Hence the diagram shows that the cells with mutations in them increase with age and then it results into a cancer.

The drug Taxol, extracted from the bark of yew trees, has an opposite effect to the drug colchicine, an alkaloid from autumn crocus. Taxol binds tightly to microtubules and stabilizes them; when added to cells, it causes much of the free tubulin to assemble into microtubules. In contrast, colchicine prevents microtubule formation. Taxol is just as pernicious to dividing cells as colchicine, and both are used as anticancer drugs. Based on your knowledge of microtubule dynamics, suggest why both drugs are toxic to dividing cells despite their opposite actions.

Cell division requires microtubules to be able to both polymerize and depolymerize. Both drugs prevents one or the other. In Taxol-treated-cells microtubule cannot depolymerize to reaarange themselves to build the spindle (no division then).

Gelatin is primarily composed of collagen, which is responsible for the remarkable tensile strength of connective tissue. it is the basic ingredient of jello; yet, as you probably experienced many times yourself while consuming the strawberry-flavored variety, jello has virtually no tensile strength. Why?

Collagen fibers are denaturesd when gelatin is boiled in water when making jello. When cooled, the disorderd fibers solidify in a mess. This resembles the collagen that is secered by the fibroblasts. Tensile strength is achieved when the fibers are aligned, bundled and crosslinked.

Discuss the relative advantages and disadvantages of light and electron microscopy. How could you best visualize (a) a living skin cell, (b) a yeast mitochondrion, (c) a bacterium, and (d) a microtubule?

Conventional light microscopy is much easier to use and requires much simpler instruments. Objects that are 1 μm in size can easily be resolved; the lower limit of resolution is 0.2 μm, which is a theoretical limit imposed by the wavelength of visible light. Visible light is nondestructive and passes readily through water, making it possible to observe living cells. Electron microscopy, on the other hand, is much more complicated, both in the preparation of the sample (which needs to be extremely thinly sliced, stained with electron-dense heavy metal, and completely dehydrated) and in the nature of the instrument. Living cells cannot be observed in an electron microscope. The resolution of electron microscopy is much higher, however, and biological objects as small as 1 nm can be resolved. To see any structural detail, microtubules, mitochondria, and bacteria would need to be analyzed by electron microscopy. It is possible, however, to stain them with specific dyes and then determine their location by light microscopy; if the dye is fluorescent, the stained objects can be seen with high resolution in a fluorescence microscope.

The plus ends of microtubules grow faster because they have a larger GTP cap.

False. The size of the GTP cap doesn't change the rate. However, as the + and - ends have different physical structure they do not have the same binding sites...

Draw a schematic diagram that shows a close-up view of two plasma membranes as they come together during cell fusion, as shown in Figure 11-30. Show membrane proteins in both cells that were labeled from the outside by the binding of differently colored fluorescent antibody molecules. Indicate in your drawing the fates of these color tags as the cells fuse. Will they remain on the outside of the hybrid cell after cell fusion and still be there after the mixing of membrane proteins that occurs during the incubation at 37°C? How would the experimental outcome be different if the incubation were done at 0°C?

For schematic, refer to Chapter 11 Question 16 Solution - Membrane fusion does not alter the orientation of the membrane proteins with their attached color tags - The portion of each transmembrane protein that is exposed to the cytosol always remains exposed to the cytosol and the portion exposed to the outside always remains exposed to the outside despite diffusional mixing. - At 0oC, the fluidity of the membrane is reduced, and the mixing of the membrane proteins is significantly slowed.

Protein A binds to protein B to form a complex, AB. At equilibrium in a cell the concentrations of A, B, and AB are all at 1 μM. a. referring to figure 3-19, calculate the equilibrium constant for the reaction A + B <=> AB. b. What would the equilibrium constant be if A, B, and AB were each present in equilibrium at the much lower concentrations of 1 nM each? c. How many extra hydrogen bonds would be needed to hold A and B together at this lower concentration so that a similar proportion of the molecules are found in the AB complex? (remember that each hydrogen bond contributes about 1 kcal/mole.)

From figure 3-19: equilibrium constant, K = [AB]/[A][B] a. K = [1 μM]/[1 μM][1 μM] = 10^6 liters/mole b. K = [1 nM]/[1 nM][1 nM] = 10^9 liters/mole c. 10^9 liters/mole / (x) = 10^6 liters/mol. x = 10^9/10^6 = 1000. According to table 3-1, this corresponds to -4.3 kcal of free energy. So, about 4-5 hydrogen bonds.

Your task in the laboratory of Professor Quasimodo is to determine how far an enhancer (a binding site for an activator protein) could be moved from the promoter of the straightspine gene and still activate transcription. You systematically vary the number of nucleotide pairs between these two sites and then determine the amount of transcription by measuring the production of Straightspine mRNA. At first glance, your data look confusing (Figure Q8-6). What would you have expected for the results of this experiment? Can you save your reputation and explain these results to Professor Quasimodo?

From our knowledge of enhancers, one would expect their function to be relatively independent of their distance from the promoter—possibly weakening as this distance increases. The surprising feature of the data (which have been adapted from an actual experiment) is the periodicity: the enhancer is maximally active at certain distances from the promoter (50, 60, or 70 nucleotides), but almost inactive at intermediate distances (55 or 65 nucleotides). The periodicity of 10 suggests that the mystery can be explained by considering the structure of double helical DNA, which has 10 base pairs per turn. Thus, placing an enhancer on the side of the DNA opposite to that of the promoter (Figure A8-6) would make It more difficult for the activator that binds to it to interact with the proteins bound at the promoter. At longer distances, there is more DNA to absorb the twist, and the effect is diminished.

What, if any, are the advantages in being multicellular?

In a multicellular organism, different cells take on specialized functions and cooperate with one another, so that any one cell type does not have to perform all activities for itself. Through such divisions of labor, multicellular organisms are able to exploit food sources that are inaccessible to single-celled organisms. A plant, for example, can reach the soil with its roots to take up water and nutrients, while at the same time, its leaves above ground can harvest light energy and CO2 from the air. By protecting its reproductive cells with other specialized cells, the multicellular organism can develop new ways to survive in harsh environments or to fight off predators. When food runs out, it may be able to preserve its reproductive cells by allowing them to draw upon resources stored by their companions—or even to cannibalize relatives (a common process, in fact).

Figure 8-17 shows a simple scheme by which three transcription regulators are used during development to create eight different cell types. How many cell types could you create, using the same rules, with four different transcription regulators? As described in the text, MyoD is a transcription regulator that by itself is sufficient to induce muscle-specific gene expression in fibroblasts. How does this observation fit the scheme in Figure 8-17?

In principle, you could create 16 different cell types with 4 different transcription regulators (all the 8 cell types shown in Figure 8-17, plus another 8 created by adding an additional transcription regulator). MyoD by itself is sufficient to induce muscle-specific gene expression only in certain cell types, such as some kinds of fibroblasts. The action of MyoD is therefore consistent with the model shown in Figure 8-17: if muscle cells were specified, for example, by the combination of transcription regulators 1, 3, and MyoD, then the addition of MyoD would convert only two of the cell types of Figure 8-17 (cells F and H) to muscle.

about 10^16 cell divisions take place in a human body during a lifetime, yet an adult human body consists of only about 10^13 cells. Why are these two numbers so different?

In the course of a life time, around 1000 cells are discarded by mechanisms like apoptosis for every 1 cell retained in the body. This explains the large number difference.

You have embarked on an ambitious research project: to create life in a test tube. You boil up a rich mixture of yeast extract and amino acids in a flask along with a sprinkling of the inorganic salts known to be essential for life. You seal the flask and allow it to cool. After several months, the liquid is as clear as ever, and there are no signs of life. A friend suggests that excluding the air was a mistake, since most life as we know it requires oxygen. You repeat the experiment, but this time you leave the flask open to the atmosphere. To your great delight, the liquid becomes cloudy after a few days and under the microscope you see beautiful small cells that are clearly growing and dividing. Does this experiment prove that you managed to generate a novel life-form? How might you redesign your experiment to allow air into the flask, yet eliminate the possibility that contamination is the explanation for the results? (For a ready-made answer, look up the classic experiments of Louis Pasteur.)

It is extremely unlikely that you created a new organism in this experiment. Far more probably, a spore from the air landed in your broth, germinated, and gave rise to the cells you observed. In the middle of the nineteenth century, Louis Pasteur invented a clever apparatus to disprove the then widely accepted belief that life could arise spontaneously. He showed that sealed flasks never grew anything if properly heat-sterilized first. He overcame the objections of those who pointed out the lack of oxygen or who suggested that his heat sterilization killed the life-generating principle, by using a special flask with a slender "swan's neck," which was designed to prevent spores carried in the air from contaminating the culture (Figure A1-3). The cultures in these flasks never showed any signs of life; however, they were capable of supporting life, as could be demonstrated by washing some of the "dust" from the neck into the culture.

One remarkable feature of the genetic code is that amino acids with similar chemical properties often have similar codons. Thus codons with U or C as the second nucleotide tend to specify hydrophobic amino acids. Can you suggest a possible explanation for this phenomenon in terms of the early evolution of the protein-synthesis machinery?

It is likely that in early cells the matching between codons and amino acids was less accurate than it is in present-day cells. The feature of the genetic code described in the question may have allowed early cells to tolerate this inaccuracy by allowing a blurred relationship between sets of roughly similar codons and roughly similar amino acids. One can easily imagine how the matching between codons and amino acids could have become more accurate, step by step, as the translation machinery evolved into that found in modern cells.

Natural selection is such a powerful force in evolution because cells with even a small proliferation advantage quickly outgrow their competitors. To illustrate this process, consider a cell culture that contains 1 million bacterial cells that double every 20 minutes. A single cell in this culture acquires a mutation that allows it to divide faster, with a generation time of only 15 minutes. Assuming that there is an unlimited food supply and no cell death, how long would it take before the progeny of the mutated cell became predominant in the culture? (Before you go through the calculation, make a guess: do you think it would take about a day, a week, a month, or a year?) How many cells of either type are present in the culture at this time? (The number of cells N in the culture at time t is described by the equation N = N0 × 2^t/G, where N0 is the number of cells at zero time and G is the generation time.)

It takes only 20 hours, i.e., less than a day, before mutant cells become more abundant in the culture. Using the equation provided in the question, we see that the number of the original ("wild-type") bacterial cells at time t minutes after the mutation occurred is 10^6 × 2^t/20. The number of mutant cells at time t is 1 × 2^t/15. To find out when the mutant cells "overtake" the wild-type cells, we simply have to make these two numbers equal to each other (i.e., 10^6 × 2^t/20 = 2^t/15). Taking the logarithm to base 10 of both sides of this equation and solving it for t results in t = 1200 minutes (or 20 hours). At this time, the culture contains 2 × 10^24 cells (10^6 × 2^60 + 1 × 2^80). Incidentally, 2 × 10^24 bacterial cells, each weighing 10^-12 g, would weigh 2 × 10^12 g (= 2 × 10^9 kg, or 2 million tons!). This can only have been a thought experiment.

Discuss the following statement: "The structure and function of a living cell are dictated by the laws of physics and chemistry."

Living cells evolved from nonliving matter, but grow and replicate. Like the material they originated from, they are governed by the laws of physics, thermodynamics, and chemistry. Thus, for example, they cannot create energy de novo or build ordered structures without the expenditure of free energy. We can understand virtually all cellular events, such as metabolism, catalysis, membrane assembly, and DNA replication, as complicated chemical reactions that can be experimentally reproduced, manipulated, and studied in test tubes. Despite this fundamental reducibility, a living cell is more than the sum of its parts. We cannot randomly mix proteins, nucleic acids, and other chemicals together in a test tube, for example, and make a cell. The cell functions by virtue of its organized structure, and this is a product of its evolutionary history. Cells always come from preexisting cells, and the division of a mother cell passes both chemical constituents and structures to its daughters. The plasma membrane, for example, never has to form de novo, but grows by expansion of a preexisting membrane; there will always be a ribosome, in part made up of proteins whose function it is to make more proteins, including those that build more ribosomes.

Why could covalent bonds not be used in place of noncovalent bonds to mediate most of the interactions of macromolecules?

Many of the functions that macromolecules perform rely on their ability to associate with and dissociate from other molecules readily. This allows cells, for example, to remodel their interior when they move or divide, and to transport components from one organelle to another. Covalent bonds would be too stable for such a purpose, requiring a specific enzyme to break each kind of bond.

Mutations are mistakes in the DNA that change the genetic plan from the previous generation. Imagine a shoe factory. Would you expect mistakes (i.e., unintentional changes) in copying the shoe design to lead to improvements in the shoes produced? Explain your answer.

Most random changes to the shoe design would result in objectionable defects: shoes with multiple heels, with no soles, or with awkward sizes would obviously not sell and would therefore be selected against by market forces. Other changes would be neutral, such as minor variations in color or in size. A minority of changes, however, might result in more desirable shoes: deep scratches in a previously flat sole, for example, might create shoes that would perform better in wet conditions; the loss of high heels might produce shoes that are more comfortable. The example illustrates that random changes can lead to significant improvements if the number of trials is large enough and selective pressures are imposed.

High levels of the female sex hormone estrogen increase the incidence of some forms of cancer. thus, some early types of contraceptive pills containing high concentrations of estrogen were eventually withdrawn from use because this was found to increase the risk of cancer of the lining of the uterus. male transsexuals who use estrogen preparations to give themselves a female appearance have an increased risk of breast cancer. High levels of androgens (male sex hormones) increase the risk of some other forms of cancer, such as cancer of the prostate. can one infer that estrogens and androgens are mutagenic?

Mutagenic cells are those that change the genetic information of the cell. The sex hormones increase the cell division rate and hence the mutation rate per cell. They also cause an increased number of cells in the breast, uterus and prostate regions which could increase the cells at rise. Hence in conclusion, they can favour the development of cancer but do not cause mutations directly.

. Which of the following types of mutations would be predicted to harm an organism? Explain your answers. A. Insertion of a single nucleotide near the end of the coding sequence. B. Removal of a single nucleotide near the beginning of the coding sequence. C. Deletion of three consecutive nucleotides in the middle of the coding sequence. D. Deletion of four consecutive nucleotides in the middle of the coding sequence. E. Substitution of one nucleotide for another in the middle of the coding sequence.

Mutations of the type described in (A) may or may not be harmful a reading-frame shift that occurs toward the end of the coding sequence, will result in a largely correct protein that may be functional. (B) harmful. the reading frame would be changed, and because this frameshift occurs near the beginning of the coding sequence, much of the protein will contain a nonsensical and/or truncated sequence of amino acids. (C) may or may not be harmful deletion of three consecutive nucleotides,leads to the deletion of an amino acid but does not alter the reading frame. The deleted amino acid may or may not be important for the folding or activity of the protein; in many cases such mutations are silent. (D) harmful the reading frame would be changed, and because this frameshift occurs in the middle of the coding sequence, much of the protein will contain a nonsensical and/or truncated sequence of amino acids. (E) harmless Substitution of one nucleotide for another will In some cases, not change the amino acid sequence of the protein; in other cases it will change a single amino acid; at worst, it may create a new stop codon, giving rise to a truncated protein.

Protein structure is determined solely by a protein's amino acid sequence. Should a genetically engineered protein in which the original order of all amino acids is reversed have the same structure as the original protein?

No, the structure wouldn't be the same or even similar, because the peptide bond has a polarity (hence the direction matters). Changing their order would put the side chains into different positions with respect to the peptide backbone and therefore change the way the polypeptide folds.

What are the forces that determine the folding of a macromolecule into a unique shape?

Noncovalent bonds form between the covalently linked subunits of a macromolecule such as a polypeptide or RNA chain causing the chain to fold into a unique shape. These noncovalent bonds include hydrogen bonds, ionic interactions, van der Waals attractions, and hydrophobic interactions. Because these interactions are weak, they can be broken with relative ease; thus, most macromolecules can be unfolded by heating, which increases thermal motion.

For the reactions shown in figure 3-21, sketch an energy diagram similar to that in figure 3-12 for the two reactions alone and for the combined reactions. indicate the standard free-energy changes for the reactions X → y, y → Z, and X → Z in the graph. indicate how enzymes that catalyze these reactions would change the energy diagram. FIGURE 3-21

Note that ΔG°X → Y is positive, whereas ΔG°Y → Z and ΔG°X → Z are negative. The graph also shows that ΔG°X → Z = ΔG°X → Y + ΔG°Y → Z. We do not know from the information given in Figure 3-12 how high the activation energy barriers are; they are therefore drawn to an arbitrary height (solid lines). The activation energies would be lowered by enzymes that catalyze these reactions, thereby speeding up the reaction rates (dotted lines), but the enzymes would not change the ΔG° values. FIGURE A3-4

To get a feeling for the size of cells (and to practice the use of the metric system), consider the following: the human brain weighs about 1 kg and contains about 10^12 cells. Calculate the average size of a brain cell (although we know that their sizes vary widely), assuming that each cell is entirely filled with water (1 cm^3 of water weighs 1 g). What would be the length of one side of this average-sized brain cell if it were a simple cube? If the cells were spread out as a thin layer that is only a single cell thick, how many pages of this book would this layer cover?

One average brain cell weighs 10^-9 g (= 1000 g/10^12). Because 1 g of water occupies 1 ml = 1 cm^3 (= 10^-6 m^3), the volume of one cell is 10^-15 m^3 (= 10^-9 g × 10^-6 m^3/g). Taking the cube root yields a side length of 10^-5 m, or 10 μm (10^6 μm = 1 m) for each cell. The page of the book has a surface of 0.057 m^2 (= 21 cm × 27.5 cm), and each cell has a footprint of 10^-10 m^2 (10^-5 m × 10^-5 m). Therefore, 57 × 10^7 (= 0.057 m^2/10^-10 m^2) cells fit on this page when spread out as a single layer. Thus, 10^12 cells would occupy 1750 pages (= 10^12/[57 × 10^7]).

"Which of the following changes takes place when a skeletal muscle contracts? A. Z discs move farther apart. B. Actin filaments contract. C. Myosin filaments contract. D. Sarcomeres become shorter."

Only (D) is correct. Upon contraction, the Z discs move closer together, and neither actin nor myosin filaments contract (see Figures 17-41 and 17-42).

Random mutations only very rarely result in changes in a protein that improve its usefulness for the cell, yet useful mutations are selected in evolution. Because these changes are so rare, for each useful mutation there are innumerable mutations that lead to either no improvement or inactive proteins. Why, then, do cells not contain millions of proteins that are of no use?

Only mutations that are beneficial to the organism (either for reproduction or survival) are passed along to future generations. The useless or harmful mutations are eliminated by natural selection.

Which of the following amino acids would you expect to find more often near the center of a folded globular protein? Which ones would you expect to find more often exposed to the outside? Explain your answers. Ser, Ser-p (a Ser residue that is phosphorylated), Leu, Lys, Gln, His, Phe, Val, Ile, Met, Cys-S-S-Cys (two cysteines that are disulfide- bonded), and Glu. Where would you expect to find the most n-terminal amino acid and the most c-terminal amino acid?

Polar amino acids are more likely to be fond on the outside, so: -Ser, Ser-P, Lys, Gln, His, and Glu Non-polar amino acids tend to be towards the center: -Leu, Phe, Val, Ile, Met, and Cys-S-S-Cys (because disulfide bond eleimates potential to form hydrogen bonds, thus making the molecule hydrophobic) The most N-terminal amino acid and the most C-terminal amino acid each contain a charged group (the amino and carboxyl groups, respectively, that mark the ends of the polypeptide chain) and hence are usually found on the protein's surface.

Similarly, predict the membrane orientation of a protein that is synthesized with an N-terminal cleaved signal sequence followed by a stop-transfer sequence, followed by a start-transfer sequence.

See Figure A15-4B. The N-terminal signal sequence initiates translocation of the N-terminal domain of the protein until translocation is stopped by the stop-transfer sequence. A cytosolic domain is synthesized until the start-transfer sequence initiates translocation again. The situation now resembles that described in (A), and the C-terminal domain of the protein is translocated into the lumen of the ER. The resulting protein therefore spans the membrane twice. Both its N-terminal and C-terminal domains are in the ER lumen, and a loop domain between the two transmembrane regions is exposed in the cytosol.

. What arrangement of signal sequences would enable the insertion of a multipass protein with an odd number of transmembrane segments?

See Figure A15-4C. It would need a cleaved signal sequence, followed by an internal stop-transfer sequence, followed by pairs of start- and stop-transfer sequences .

Which of the three 20-amino-acid sequences listed below in the single-letter amino acid code is the most likely candidate to form a transmembrane region (α helix) of a transmembrane protein? Explain your answer. A. I T L I Y F G N M S S V T Q T I L L I S B. L L L I F F G V M A L V I V V I L L I A C. L L K K F F R D M A A V H E T I L E E S

Sequence B is most likely to form a transmembrane helix. It is composed primarily of hydrophobic amino acids, and therefore can be stably integrated into a lipid bilayer.

To get an appreciation for the great speed of molecular diffusion, assume that a lipid head group is about the size of a ping-pong ball (4 cm in diameter) and that the floor of your living room (6 m × 6 m) is covered wall-to-wall with these balls. If two neighboring balls exchanged positions once every 10-7 second, what would their speed be in kilometers per hour? How long would it take for a ball to move from one side of the room to the opposite side?

Similarly, if a 4-cm ping-pong ball exchanged partners every 10-7 seconds and moved in a linear fashion it would reach the opposite wall in 1.5 × 10-5 sec (traveling at 1,440,000 km/hr. But a random walk would take longer. Using the equation above, we calculate the constant D in this case to be 8 × 107 cm2/sec and the time required to travel 6 m about 6002/(1.6 × 108)= 2 msec.

Hair is composed largely of fibers of the protein keratin. Individual keratin fibers are covalently cross- linked to one another by many disulfide (S-S) bonds. If curly hair is treated with mild reducing agents that break a few of the cross-links, pulled straight, and then oxidized again, it remains straight. Draw a diagram that illustrates the three different stages of this chemical and mechanical process at the level of the keratin filaments, focusing on the disulfide bonds. What do you think would happen if hair were treated with strong reducing agents that break all the disulfide bonds?

Strong reducing agents that break all the disulfide bonds would cause the keratin filaments to separate, so individual hairs would weaken and fragment. Mild reducing agents are used in treatments to either curl (if used with curlers) or straighten hair. FIGURE A4-4

through the exchange of small metabolites and ions, gap junctions provide metabolic and electrical coupling between cells. Why, then, do you suppose that neurons communicate primarily through synapses rather than through gap junctions?

Synapses allow signals to be modulated and be integrated with other signals received by the cell. They are complex relay devices allowing neurons to perform computations while gap junctions are not.

Why do you suppose epithelial cells lining the gut are renewed frequently, whereas most neurons last for the lifetime of the organism?

The cells in the gut are exposed to a hostile environment (e.g. they contain digestive enzymes) and hence frequent renewal protects it from harmful consequences sice the damadged cells are discarded. Neurons on the other hand are in a very protected environmnet and have very complex functions that are related to other neurons hence it is not easy to reconstruct the system if a neuron dies.

cells in the stem of a seedling that is grown in the dark orient their microtubules horizontally. How would you expect this to affect the growth of the plant?

The cells will grow in a vertical direction because the deposition of cellulose microfibroids will be horizontally oriented.

The electron micrographs shown in Figure Q17-20A were obtained from a population of microtubules that were growing rapidly. Figure Q17-20B was obtained from microtubules undergoing "catastrophic" shrinking. Comment on any differences between A and B, and suggest likely explanations for the differences that you observe.

The ends of the shrinking microtubule are visibly frayed, and the individual protofilaments appear to come apart and curl as the end depolymerizes. This micrograph therefore suggests that the GTP cap (which is lost from shrinking microtubules) holds the protofilaments properly aligned with each other, perhaps by strengthening the side-to-side interactions between αβ-tubulin subunits when they are in their GTP-bound form.

The molecular weight of all eukaryotic ribosomal proteins combined is about 2.5 *10^6 daltons. Would it be advantageous to synthesize them as a single protein?

The error rate limits the sizes of proteins that can be synthesized accurately. Similarly, if a eukaryotic ribosomal protein were synthesized as a single molecule, a large portion (87%) of this hypothetical giant ribosomal protein would be expected to contain at least one mistake. It is therefore more advantageous to make ribosomal proteins individually, because in this way only a small proportion of each type of protein will be defective, and these few bad molecules can be individually eliminated by proteolysis to ensure that there are no defects in the ribosome as a whole.

The phosphoanhydride bond that links two phosphate groups in ATP in a high-energy linkage has a ΔG° of -7.3 kcal/mole. Hydrolysis of this bond in a cell liberates from 11 to 13 kcal/mole of usable energy. how can this be? Why do you think a range of energies is given, rather than a precise number as for ΔG°?

The free energy of 11 to 13 kcal/mol (ΔG) liberated from this reaction depends on both ΔG° and the concentrations of the substrate and products. ΔG=ΔG°+RTln([product]/[substrate]). If the second term is negative (i.e. more substrate than product) then ΔG will be even more negative than ΔG° (i.e. releasing more energy). The reason a range is given is because the concentrations of ATP, ADP, and phosphate differ among cells, thus changing the value of the logarithm.

"Histone proteins are among the most highly conserved proteins in eukaryotes. Histone H4 proteins from a pea and a cow, for example, differ in only 2 of 102 amino acids. Comparison of the gene sequences shows many more differences, but only two change the amino acid sequence. These observations indicate that mutations that change amino acids must have been selected against during evolution. Why do you suppose that amino-acid-altering mutations in histone genes are deleterious?"

The functions of histone proteins must involve nearby all of their amino acids, such that a change in any position (mutation) is deletorious to the cell.

Consider a transmembrane protein that forms a hydrophilic pore across the plasma membrane of a eukaryotic cell, allowing Na+ to enter the cell when it is activated upon binding a specific ligand on its extracellular side. It is made of five similar transmembrane subunits, each containing a membrane-spanning α helix with hydrophilic amino acid side chains on one surface of the helix and hydrophobic amino acid side chains on the opposite surface. Considering the function of the protein as a channel for Na+ ions to enter the cell, propose a possible arrangement of the five membrane-spanning α helices in the membrane.

The hydrophilic faces of the five membrane-spanning α helices, each contributed by a different subunit, are thought to come together to form a pore across the lipid bilayer that is lined with the hydrophilic amino acid side chains.

Imagine the two situations shown in Figure Q8-12. In cell I, a transient signal induces the synthesis of protein A, which is a transcriptional activator that turns on many genes including its own. In cell II, a transient signal induces the synthesis of protein R, which is a transcriptional repressor that turns off many genes including its own. In which, if either, of these situations will the descendants of the original cell "remember" that the progenitor cell had experienced the transient signal? Explain your reasoning.

The induction of a transcriptional activator protein that stimulates its own synthesis can create a positive feedback loop that can produce cell memory. The continued self-stimulated synthesis of activator A can in principle last for many cell generations, serving as a constant reminder of an event that took place in the past. By contrast, the induction of a transcriptional repressor that inhibits its own synthesis creates a negative feedback loop which ensures that the response to the transient stimulus will be similarly transient. Because repressor R shuts off its own synthesis, the cell will quickly return to the state that existed before the signal.

A motor protein moves along protein filaments in the cell. Why are the elements shown in the illustration not sufficient to mediate directed movement (Figure Q4-19)? With reference to figure 4-46, modify the illustration shown here to include other elements that are required to create a unidirectional motor, and justify each modification you make to the illustration. FIGURE Q4-19

The motor protein in the illustration can move just as easily to the left as to the right and so will not move steadily in one direction. However, if just one of the steps is coupled to ATP hydrolysis (for example, by making detachment of one foot dependent on binding of ATP and coupling the reattachment to hydrolysis of the bound ATP), then the protein will show unidirectional movement that requires the continued consumption of ATP. Note that, in principle, it does not matter which step is coupled to ATP hydrolysis. FIGURE A4-19

Proteoglycans are characterized by the abundance of negative charges on their sugar chains. How would the properties of these molecules differ if the negative charges were not as abundant?

The negative charges attract a cloud of cations like NA+ that are osmotically active and cause more water to be sucked into the matrix. Hence if absentm the swelling pressure would be low as well as the counter balanceing tension resulting into a matrix that isn't tough, resilient and resistant to compression.

A single nucleosome core particle is 11 nm in diameter and contains 147 bp of DNA (the DNA double helix measures 0.34 nm/bp). What packing ratio (ratio of DNA length to nucleosome diameter) has been achieved by wrapping DNA around the histone octamer? Assuming that there are an additional 54 bp of extended DNA in the linker between nucleosomes, how condensed is "beads-on-a-string" DNA relative to fully extended DNA? What fraction of the 10,000-fold condensation that occurs at mitosis does this first level of packing represent? ECB4 EQ5.16/Q5.15

The packing ratio within a nucleosome core is 4.5 [(147 bp × 0.34 nm/bp)/(11 nm) = 4.5]. If there is an additional 54 bp of linker DNA, then the packing ratio for "beads-on-a-string" DNA is 2.3 [(201 bp × 0.34 nm/ bp)/(11 nm + {54 bp × 0.34 nm/bp}) = 2.3]. This first level of packing represents only 0.023% (2.3/10,000) of the total condensation that occurs at mitosis.

Consider a protein that contains an ER signal sequence at its N-terminus and a nuclear localization sequence in its middle. What do you think the fate of this protein would be? Explain your answer.

The protein is translocated into the ER. Its ER signal sequence is recognized as soon as it emerges from the ribosome. The ribosome then becomes bound to the ER membrane, and the growing polypeptide chain is transferred through the ER translocation channel. The nuclear localization sequence is therefore never exposed to the cytosol. It will never encounter nuclear import receptors and the protein will not enter the nucleus.

Discuss whether the following statement is correct: "An ionic bond can, in principle, be thought of as a very polar covalent bond. Polar covalent bonds, then, fall somewhere between ionic bonds at one end of the spectrum and nonpolar covalent bonds at the other end."

The statement is correct. Both ionic and covalent bonds are based on the same principles: electrons can be shared equally between two interacting atoms, forming a nonpolar covalent bond; electrons can be shared unequally between two interacting atoms, forming a polar covalent bond; or electrons can be completely lost from one atom and gained by the other, forming an ionic bond. There are bonds of every conceivable intermediate state, and for borderline cases it becomes arbitrary whether a bond is described as a very polar covalent bond or an ionic bond.

There are three major classes of filaments that make up the cytoskeleton. What are they, and what are the differences in their functions? Which cytoskeletal filaments would be most plentiful in a muscle cell or in an epidermal cell making up the outer layer of the skin? Explain your answers.

The three major filaments are actin filaments, intermediate filaments, and microtubules. Actin filaments are involved in rapid cell movement, and are the most abundant filaments in a muscle cell; intermediate filaments provide mechanical stability and are the most abundant filaments in epidermal cells of the skin; and microtubules function as "railroad tracks" for intracellular movements, and are responsible for the separation of chromosomes during cell division. Other functions of all these filaments are discussed in Chapter 17.

Draw to scale the outline of two spherical cells, one a bacterium with a diameter of 1 μm, the other an animal cell with a diameter of 15 μm. Calculate the volume, surface area, and surface-to-volume ratio for each cell. How would the latter ratio change if you included the internal membranes of the cell in the calculation of surface area (assume internal membranes have 15 times the area of the plasma membrane)? (The volume of a sphere is given by 4πr^3/3 and its surface by 4πr^2, where r is its radius.) Discuss the following hypothesis: "Internal membranes allowed bigger cells to evolve."

The volume and the surface area are 5.24 × 10^-19 m^3 and 3.14 × 10^-12 m^2 for the bacterial cell, and 1.77 × 10^-15 m^3 and 7.07 × 10^-10 m^2 for the animal cell, respectively. From these numbers, the surface-to volume ratios are 6 × 10^6 m^-1 and 4 × 10^5 m^-1, respectively. In other words, although the animal cell has a 3375-fold larger volume, its membrane surface is increased only 225-fold. If internal membranes are included in the calculation, however, the surface-to-volume ratios of both cells are about equal. Thus, because of their internal membranes, eukaryotic cells can grow bigger and still maintain a sufficiently large membrane area, which—as we shall discuss in more detail in later chapters—is required for many essential functions.

As shown in figure 4-16, both α helices and the coiled-coil structures that can form from them are helical structures, but do they have the same handedness in the figure? Explain why? FIGURE 4-16

The α helix in the figure is right-handed, whereas the coiled-coil is left-handed. The reversal occurs because of the staggered positions of hydrophobic side chains in the α helix.

Look at the models of the protein in figure 4-12. is the red α helix right- or left-handed? are the three strands that form the large β sheet parallel or antiparallel? Starting at the n-terminus (the purple end), trace your finger along the peptide backbone. are there any knots? Why, or why not?

The α helix is right-handed. The three strands that form the large β sheet are antiparallel. There are no knots in the polypeptide chain, presumably because a knot would interfere with the folding of the protein into its three-dimensional conformation after protein synthesis.

In the membrane of a human red blood cell, the ratio of the mass of protein (average molecular weight 50,000) to phospholipid (molecular weight 800) to cholesterol (molecular weight 386) is about 2:1:1. How many lipid molecules are there for every protein molecule?

There are about (2/50,000)/(1/800 + 1/386)=100 lipid molecules for every protein molecule in the membrane.

Membrane lipid molecules exchange places with their lipid neighbors every 10^-7 second. A lipid molecule diffuses from one end of a 2-μm-long bacterial cell to the other in about 1 second. Are these two numbers in agreement (assume that the diameter of a lipid head group is about 0.5 nm)? If not, can you think of a reason for the difference?

There are about 4000 lipid molecules, each 0.5 nm wide, between one end of the bacterial cell and the other. So if a lipid molecule at one end moved directly in a straight line it would require only 4000 × 10^-7 = 4 × 10^-4 sec to reach the other end. In reality, however, the lipid molecule would move in a random path so that it would take considerably longer. We can calculate the approximate time required from the equation: t = x^2/2D where x is the average distance moved, t is the time taken, and D is a constant called the diffusion coefficient. Inserting step values x = 0.5 nm and t = 10-7 sec we obtain D = 1.25 × 10^-7 cm^2/sec. Using this value in the same equation but with distance x = 2 × 10^-4 cm (= 2 μm) gives the time taken t = 1.6 seconds.

What are the arguments that all living cells evolved from a common ancestor cell? Imagine the very early days of evolution of life on Earth. Would you assume that the primordial ancestor cell was the first and only cell to form?

There are many lines of evidence for a common ancestor. Analyses of modern-day living cells show an amazing degree of similarity in the basic components that make up the inner workings of otherwise vastly different cells. Many metabolic pathways, for example, are conserved from one cell to another, and the compounds that make up nucleic acids and proteins are the same in all living cells, even though it is easy to imagine that a different choice of compounds (e.g., amino acids with different side chains) would have worked just as well. Similarly, it is not uncommon to find that important proteins have closely similar detailed structures in prokaryotic and eukaryotic cells. Theoretically, there would be many different ways to build proteins that could perform the same functions. The evidence overwhelmingly shows that most important processes were "invented" only once and then became fine-tuned during evolution to suit the particular needs of specialized cells and specific organisms. It seems highly unlikely, however, that the first cell survived to become the primordial founder cell of today's living world. As evolution is not a directed process with a purposeful progression, it is more likely that there were a vast number of unsuccessful trial cells that replicated for a while and then became extinct because they could not adapt to changes in the environment or could not survive in competition with other types of cells. We can therefore speculate that the primordial ancestor cell was a "lucky" cell that ended up in a relatively stable environment in which it had a chance to replicate and evolve.

At the leading edge of a crawling cell, the plus ends of actin filaments are located close to the plasma membrane, and actin monomers are added at these ends, pushing the membrane outward to form lamellipodia or filopodia. What do you suppose holds the filaments at their other ends to prevent them from just being pushed into the cell's interior?

They are connected to the meshwork of the cell cortex, as cell contains actin-binding proteins that bundle and cross-link actin filament

. Some proteins shuttle back and forth between the nucleus and the cytosol. They need a nuclear export signal to get out of the nucleus. How do you suppose they get into the nucleus?

To get into the nucleus they must contain a nuclear localization signal as well. Proteins with nuclear export signals shuttle between the nucleus and the cytosol. Eg., is the A1 protein, which binds to mRNAs in the nucleus and guides them through the nuclear pores. Once in the cytosol, a nuclear localization signal ensures that the A1 protein is re-imported so that it can participate in the export of further mRNAs.

. Transcription occurs at a rate of about 30 nucleotides per second. Is it possible to calculate the time required to synthesize a titin mRNA from the information given here?

Transcription of the exons (25,000*3 = 75,000 nucleotides) requires about 42 minutes [(75,000/30) * (1/60)]. The time required to transcribe the entire gene ( (exons and introns)is likely to be considerably longer because introns can be quite large.

Without actin, cells can form a functional mitotic spindle and pull their chromosomes apart but cannot divide.

True. Actin is required to form the contractile ring that causes the physical cleavage between the two daughter cells

The transverse tubules in muscle cells are an extension of the plasma membrane, with which they are continuous; similarly, the sarcoplasmic reticulum is an extension of the endoplasmic reticulum.

True. Both are nice examples of how the same membrane can have regions that are highly specialized for a particular function.

Lamellipodia and filopodia are "feelers" that a cell extends to find anchor points on the substratum that it will then crawl over.

True. Both extensions are associated with transmembrane proteins that protrude from the plasma membrane and enable the cell to form new anchor points on the substratum.

Kinesin moves endoplasmic reticulum membranes along microtubules so that the network of ER tubules becomes stretched throughout the cell.

True. Otherwise, the ER would collapse towards the center of the cell

"the structure of an organism is determined by the genome that the egg contains." What is the evidence on which this statement is based? indeed, a friend challenges you and suggests that you replace the dna of a stork's egg with human dna to see if a human baby results. How would you answer him?

When the DNA is replaced, it is still not possible to get a human baby because appropriate conditions (like space and noursihment)for development do not exist. So DNA has the plans that specify the structure but they won't be executed.

Heavy smokers or industrial workers exposed for a limited time to a chemical carcinogen that induces mutations in dna do not usually begin to develop cancers characteristic of their habit or occupation until 10, 20, or even more years after the exposure. suggest an explanation for this long delay.

When they are exposed to carcinogen, mutations are induced but they are not enoughto convert the cell into a cancer cell. These cells that have mutations in them accumulate with time and eventually one turns into a cancer cell.

The elements oxygen and sulfur have similar chemical properties because they both have six electrons in their outermost electron shells. Indeed, both elements form molecules with two hydrogen atoms, water (H2O) and hydrogen sulfide (H2S). Surprisingly, at room temperature, water is a liquid, yet H2S is a gas, despite sulfur being much larger and heavier than oxygen. Explain why this might be the case.

Whether a substance is a liquid or gas at a given temperature depends on the attractive forces between its molecules. H2S is a gas at room temperature and H2O is a liquid because the hydrogen bonds that hold H2O molecules together do not form between H2S molecules. A sulfur atom is much larger than an oxygen atom, and because of its larger size, the outermost electrons are not as strongly attracted to the nucleus of the sulfur atom as they are in an oxygen atom. Consequently, the hydrogen-sulfur bond is much less polar than the hydrogen-oxygen bond. Because of the reduced polarity, the sulfur in a H2S molecule is not strongly attracted to the hydrogen atoms in an adjacent H2S molecule, and hydrogen bonds, which are so predominant in water, do not form.

mutations in the genes encoding collagens often have detrimental consequences, resulting in severely crippling diseases. Particularly devastating are mutations that change glycines, which are required at every third position in the collagen polypeptide chain so that it can assemble into the characteristic triple-helical rod (see Figure 20-9). a. Would you expect collagen mutations to be detrimental if only one of the two copies of a collagen gene is defective? B. a puzzling observation is that the change of a glycine residue into another amino acid is most detrimental if it occurs toward the amino terminus of the rod-forming domain. suggest an explanation for this.

a) Yes, the assembly will be impaired and this can have a detrimental effect even if the normal copy of the gene is present. B) The assembly into the triple-helical rod starts from the amino-terminal ends hence if a mutation is already present here, only short rods would be formed unlike a situation in which the mutation was further away to produce longer normal rods.

a gene encoding one of the proteins involved in dna replication has been inactivated by a mutation in a cell. in the absence of this protein, the cell attempts to replicate its dna. What would happen during the dna replication process if each of the following proteins were missing? a. dna polymerase B. dna ligase c. Sliding clamp for dna polymerase d. nuclease that removes rna primers e. dna helicase F. primase

a) replication will not talke place becare the RNA primers will be laid down at the origin of replication. B) 5' phosphate end of one DNA fragment wouldn't be joind to the adajacent 3' hydroxyl end of the next hence new DNA strands will remain as fragments. c) the sliding clap keeps the DNA polymerase firmly attached to the template while it is synthesizing new DNA strands. if these didnt exist, most DNA polymerase molecules will synthesize only a short string of nucleotides before falling off the DNA tempelate strand. d) there won't be ligation since the DNA ligase will not link DNA to RNA and the RNA fragments will remail covalently attached to the new DNA fragments. e) They separate the double helix strands hence if absent, the strands wouldn't separate and little or no DNA will be synthesized. f) The RNA primer wouldn't be synthesized hence DNA replication can't begin.

"Look carefully at the micrograph and drawing 2 in Figure 6-9. A. Using the scale bar, estimate the lengths of the DNA strands between the replication forks. Numbering the replication forks sequentially from the left, how long will it take until forks 4 and 5, and forks 7 and 8, respectively, collide with each other? (Recall that the distance between the bases in DNA is 0.34 nm, and eukaryotic replication forks move at about 100 nucleotides per second.) For this question, disregard the nucleosomes seen in the micrograph and assume that the DNA is fully extended. B. The fly genome is about 1.8 × 108 nucleotide pairs in size. What fraction of the genome is shown in the micrograph?"

a)Between 2 and 3, it is around 200nm, 4 and 5, it is around 300nm and between 6 and 7 it is around 150nm. 0.34nm corresponds to 1 nucleotide hence 300nm would correspond to 882 nucleotides. It takes 1 second to cross 100 nucleotides so for 882, it would take aroind 8.8 seconds for forks 4 and 5. forks 7 and 8 would never meet because they are going in opposide directions. b) The total length in the micrograph is 1500nm corresponding to 4412 nucleotides. this is 0.0024%

Which of the following statements are correct? explain your answers. a. Gap junctions connect the cytoskeleton of one cell to that of a neighboring cell or to the extracellular matrix. B. a wilted plant leaf can be likened to a deflated bicycle tire. c. Because of their rigid structure, proteoglycans can withstand a large amount of compressive force. d. the basal lamina is a specialized layer of extracellular matrix to which sheets of epithelial cells are attached. e. skin cells are continually shed and are renewed every few weeks; for a permanent tattoo, it is therefore necessary to deposit pigment below the epidermis. F. although stem cells are not differentiated, they are specialized and therefore give rise only to specific cell types.

a)false: gap junctions are not connected to the cytoskeleton. They help provide cell to cell communication by allowing small molecules to pass through the junctions. B)true: the turgor pressure is reduced and cell walls are not rigid because of low compressive strength. c)false: proteoglycans can withstand a large abount of compressive force but are not rigid. They have a tendancy to absorb large amounts of water. d)true. e)true f) true: stem cells express control genes that ensure that their daughter cells will be of differentiated cell types

In a simple reaction A <=> A', a molecule is interconvertible between two forms that differ in standard free energy G° by 4.3 kcal/mole, with A' having the higher G°. a. Use Table 3-1 (p. 98) to find how many more molecules will be in state A' compared with state A at equilibrium. b. If an enzyme lowered the activation energy of the reaction by 2.8 kcal/mole, how would the ratio of A to A' change?

a. According to table 3-1, a change in free energy of 4.3 kcal/mol corresponds to a change in K of 10^(-3). So, [A']/[A] = 10^(-3), meaning at equilibrium, there will be 1000 times as many A molecules as there are A' molecules. b. The ratio would not change. Enzymes only affect the rate of the reaction, not the equilibrium state.

Simple enzyme reactions often conform to the equation E + S <=> ES → EP <=> E + P where E, S, and P are enzyme, substrate, and product, respectively. a. What does ES represent in this equation? b. Why is the first step shown with bidirectional arrows and the second step as a unidirectional arrow? c. Why does E appear at both ends of the equation? d. One often finds that high concentrations of p inhibit the enzyme. Suggest why this might occur. e. If compound X resembles S and binds to the active site of the enzyme but cannot undergo the reaction catalyzed by it, what effects would you expect the addition of X to the reaction to have? Compare the effects of X and of the accumulation of P.

a. ES represents the enzyme-substrate complex. b. An enzyme binding to a substrate can either dissociate or be converted to product. The substrate to product reaction, however, proceeds strongly in the forward direction. c. Enzymes don't get affected--they are catalysts and are liberated after the reaction. d. Often, the product of a reaction resembles the substrate sufficiently that it can also bind to the enzyme. Any enzyme molecules that are bound to the product (i.e., are part of the EP complex) are unavailable for catalysis; excess P therefore inhibits the reaction by lowering the concentration of free E. e. Compound X would act as an inhibitor of the reaction and work similarly by forming an EX complex. However, since P has to be made before it can inhibit the reaction, it takes longer to act than X, which is present from the beginning of the reaction.

Which of the following statements are correct? explain your answers. a. The active site of an enzyme usually occupies only a small fraction of the enzyme surface. b. Catalysis by some enzymes involves the formation of a covalent bond between an amino acid side chain and a substrate molecule. c. a β sheet can contain up to five strands, but no more. d. The specificity of an antibody molecule is contained exclusively in loops on the surface of the folded light-chain domain. e. The possible linear arrangements of amino acids are so vast that new proteins almost never evolve by alteration of old ones. f. Allosteric enzymes have two or more binding sites. g. Noncovalent bonds are too weak to influence the three- dimensional structure of macromolecules. h. Affinity chromatography separates molecules according to their intrinsic charge. i. Upon centrifugation of a cell homogenate, smaller organelles experience less friction and thereby sediment faster than larger ones.

a. True. Only a few amino acid side chains contribute to the active site. b. True. Some enzymes form covalent intermediates with their substrates. However, in all cases the enzyme is restored to its original structure after the reaction. c. False. β sheets can, in principle, contain any number of strands because the two strands that form the rims of the sheet are available for hydrogen-bonding to other strands. (β sheets in known proteins contain from 2 to 16 strands.) d. False. It is true that the specificity of an antibody molecule is exclusively contained in polypeptide loops on its surface; however, these loops are contributed by both the folded light and heavy chains. e. False. The possible linear arrangements of amino acids that lead to a stably folded protein domain are so few that most new proteins evolve by alteration of old ones. f. True. Allosteric enzymes generally bind one or more molecules that function as regulators at sites that are distinct from the active site. g. False. Although single noncovalent bonds are weak, many such bonds acting together are a major contributor to the three-dimensional structure of macromolecules. h. False. Affinity chromatography separates specific macromolecules because of their interactions with specific ligands, not because of their charge. i. False. The larger an organelle is, the more centrifugal force it experiences and the faster it sediments, despite an increased frictional resistance from the fluid through which it moves.

The curve shown in figure 3-24 is described by the Michaelis-Menten equation: rate (v) = Vmax [s]/([s] + KM) Can you convince yourself that the features qualitatively described in the text are accurately represented by this equation? In particular, how can the equation be simplified when the substrate concentration [s] is in one of the following ranges: (a) [s] is much smaller than the KM (b) [s] equals the KM (C) [s] is much larger than the KM?

a. if s << KM, then v -> Vmax ([s]/KM) b. if [s] = KM, then v -> Vmax/2

Consider the analogy of the jiggling box containing coins that was described on page 85. The reaction, the flipping of coins that either face heads up (h) or tails up (T), is described by the equation h ↔ T, where the rate of the forward reaction equals the rate of the reverse reaction. a. What are ΔG and ΔG° in this analogy? b. What corresponds to the temperature at which the reaction proceeds? What corresponds to the activation energy of the reaction? assume you have an "enzyme," called jigglase, which catalyzes this reaction. What would the effect of jigglase be and what, mechanically, might jigglase do in this analogy?

a. ΔG° = 0 because H and T have the same probability so there is no driving force (i.e. no energy difference) that favors going from one state to the other. The concentration difference between H and T initially is what causes a ΔG, and will drive the reaction towards equilibrium, where H = T and ΔG = 0. b. The amount of shaking the box corresponds to temperature. The activation energy is the amount of energy required to flip a coin 90 degrees (i.e. from flat to on its edge--because after that it will spontaneously fall on one side or the other). "Jigglase" would be something that speeds up the reaction, so in the coin case, something that makes more flipping happen for some fixed amount of jiggling. One example could be a magnet above the box which lowers the activation energy by reducing the effective weight of the coins.

By now you should be familiar with the following cellular components. Briefly define what they are and what function they provide for cells. A. cytosol B. cytoplasm C. mitochondria D. nucleus E. chloroplasts F. lysosomes G. chromosomes H. G olgi apparatus I. peroxisomes J. plasma membrane K. endoplasmic reticulum L. cytoskeleton

cytosol Contents of the main compartment of the cytoplasm, excluding membrane-enclosed organelles such as endoplasmic reticulum and mitochondria. The cell fraction remaining after membranes, cytoskeletal components, and other organelles have been removed. cytoplasm Contents of a cell that are contained within its plasma membrane but, in the case of eukaryotic cells, contained outside the nucleus. mitochondrion (plural mitochondria) Membrane-enclosed organelle, about the size of a G:16 GLOSSARY bacterium, that carries out oxidative phosphorylation and produces most of the ATP in eukaryotic cells. nucleus In biology, refers to the prominent, rounded structure that contains the DNA of a eukaryotic cell. In chemistry, refers to the dense, positively charged center of an atom. chloroplast Specialized organelle in algae and plants that contains chlorophyll and serves as the site in which photosynthesis takes place. chromosome Long, threadlike structure composed of DNA and proteins that carries the genetic information of an organism; becomes visible as a distinct entity when a plant or animal cell prepares to divide lysosome Membrane-enclosed organelle that breaks down worn-out proteins and organelles and other waste materials, as well as molecules taken up by endocytosis; contains digestive enzymes that are typically most active at the acid pH found inside these organelles. Golgi apparatus Membrane-enclosed organelle in eukaryotic cells that modifies the proteins and lipids made in the endoplasmic reticulum and sorts them for transport to other sites. peroxisome Small membrane-enclosed organelle that contains enzymes that degrade lipids and destroy toxins. plasma membrane The protein-containing lipid bilayer that surrounds a living cell. endoplasmic reticulum (ER) Labyrinthine membrane enclosed compartment in the cytoplasm of eukaryotic cells where lipids and proteins are made.

Analogs of hemidesmosomes are the focal contact sites described in Chapter 17, which are also sites where the cell attaches to the extracellular matrix. These junctions are prevalent in fibroblasts but largely absent in epithelial cells. On the other hand, hemidesmosomes are prevalent in epithelial cells but absent in fibroblasts. In focal contact sites,intracellular connections are made to actin filaments, whereas in hemidesmosomes connections are made to intermediate filaments. Why do you suppose these two different cell types attach differently to the extracellular matrix?

hemidesmosomes bind epithelial cells to the basal lamina by the intermediate filaments. This is because In mature epithelium, focal contact sites are presumably less prominent because the cells are largely fixed in place and have no need to crawl over the basal lamina or actively pull on it. Focal contact sites on the other hand are common in connective tissue, where fibroblasts exert traction forces on the extracellular matrix, and in cell culture, where cell crawling is observed. The forces for pulling on matrix or for driving crawling movement are generated by the actin cytoskeleton.

As shown in the drawings in Figure 15-3, the lipid bilayer of the inner and outer nuclear membranes forms a continuous sheet, joined around the nuclear pores. As membranes are two-dimensional fluids, this would imply that membrane proteins can diffuse freely between the two nuclear membranes. Yet each of these two nuclear membranes has a different protein composition, reflecting different functions. How could you reconcile this apparent contradiction?

• Although the nuclear envelope forms one continuous membrane, it has specialized regions that contain special proteins and have a characteristic appearance. One such specialized region is the inner nuclear membrane. • Membrane proteins can indeed diffuse between the inner and outer nuclear membranes, at the connections formed around the nuclear pores. • Those proteins with particular functions in the inner membrane, however, are usually anchored there by their interaction with other components such as chromosomes and the nuclear lamina.

Explain how an mRNA molecule can remain attached to the ER membrane while individual ribosomes translating it are released and rejoin the cytosolic pool of ribosomes after each round of translation.

• An mRNA molecule is attached to the ER membrane by the ribosomes translating it. This ribosome population, however, is not static; the mRNA is continuously moved through the ribosome. • Those ribosomes that have finished translation dissociate from the 3ʹ end of the mRNA and from the ER membrane, but the mRNA itself remains bound by other ribosomes, newly recruited from the cytosolic pool, that have attached to the 5ʹ end of the mRNA and are still translating the mRNA. • Depending on its length, there are about 10-20 ribosomes attached to each membrane-bound mRNA molecule.

Why do eukaryotic cells require a nucleus as a separate compartment when prokaryotic cells can manage perfectly well without?

• Eukaryotic gene expression is more complicated than prokaryotic gene expression. • Prokaryotic cells do not have introns that interrupt the coding sequences of their genes, so that an mRNA can be translated immediately after it is transcribed, without a need for further processing. In fact, in prokaryotic cells, ribosomes start translating most mRNAs before transcription is finished. • In eukaryotic cells, on the other hand, RNA transcripts have to be spliced before they can be translated. The nuclear envelope separates the transcription and translation processes in space and time: a primary RNA transcript is held in the nucleus until it is properly processed to form an mRNA, and only then is it allowed to leave the nucleus so that ribosomes can translate it.

Influenza viruses are surrounded by a membrane that contains a fusion protein, which is activated by acidic pH. Upon activation, the protein causes the viral membrane to fuse with cell membranes. An old folk remedy against flu recommends that one should spend a night in a horse's stable. Odd as it may sound, there is a rational explanation for this advice. Air in stables contains ammonia (NH3) generated by bacteria in the horse's urine. Sketch a diagram showing the pathway (in detail) by which flu virus enters cells, and speculate how NH3 may protect cells from virus infection. (Hint: NH3 can neutralize acidic solutions by the reaction NH3 + H+ → NH4 +.)

• Influenza virus enters cells by endocytosis and is delivered to endosomes, where it encounters an acidic pH that activates its fusion protein. • The viral membrane then fuses with the membrane of the endosome, releasing the viral genome into the cytosol (Figure A15-11). • NH3 is a small molecule that readily penetrates membranes. Thus, it can enter all intracellular compartments, including endosomes, by diffusion. • Once in a compartment that has an acidic pH, NH3 binds H+ to form NH4 +, which is a charged ion and therefore cannot cross the membrane by diffusion. NH4 + ions therefore accumulate in acidic compartments, raising their pH. When the pH of the endosome is raised, viruses are still endocytosed, but because the viral fusion protein cannot be activated, the virus cannot enter the cytosol.

In the electron micrograph in Figure 7-8, are the RNA polymerase molecules moving from right to left or from left to right? Why are the RNA transcripts so much shorter than the DNA segments (genes) that encode them?

• The RNA polymerases are not moving at all in the micrograph, because they have been fixed and coated with metal to prepare the sample for viewing in the electron microscope. • However, before they were fixed, they were moving from left to right, as indicated by the gradual lengthening of the RNA transcripts. • The RNA transcripts are shorter because they begin to fold up (i.e., to acquire a three-dimensional structure) as they are synthesized (see Figure 7-5), whereas the DNA is an extended double helix.

A sequence of nucleotides in a DNA strand—5′-TT AACGGCTTTTTTC-3′— was used as a template to synthesize an mRNA that was then translated into protein. Predict the C-terminal amino acid and the N-terminal amino acid of the resulting polypeptide. Assume that the mRNA is translated without the need for a start codon.

• The mRNA will have a 5ʹ-to-3ʹ polarity, opposite to that of the DNA strand that serves as the template. Thus the mRNA sequence will read 5ʹ-GAAAAAAGCCGUUAA-3ʹ. (For RNA the complementary base pairs are C-G and A-U ( as opposed to A-T in DNA)) • The N-terminal amino acid coded for by GAA is glutamic acid. • UAA specifies a stop codon, so the C-terminal amino acid is coded for by CGU and is an arginine.

. Iron (Fe) is an essential trace metal that is needed by all cells. It is required, for example, for synthesis of the heme groups and iron-sulfur centers that are part of the active site of many proteins involved in electron-transfer reactions; it is also required in hemoglobin, the main protein in red blood cells. Iron is taken up by cells by receptor-mediated endocytosis. The iron-uptake system has two components: a soluble protein called transferrin, which circulates in the bloodstream; and a transferrin receptor—a transmembrane protein that, like the LDL receptor in Figure 15-33, is continually endocytosed and recycled to the plasma membrane. Fe ions bind to transferrin at neutral pH but not at acidic pH. Transferrin binds to the transferrin receptor at neutral pH only when it has an Fe ion bound, but it binds to the receptor at acidic pH even in the absence of bound iron. From these properties, describe how iron is taken up, and discuss the advantages of this elaborate scheme.

• Transferrin without Fe bound does not interact with its receptor and circulates in the bloodstream until it catches an Fe ion. • Once iron is bound, the iron- transferrin complex can bind to the transferrin receptor on the surface of a cell and be endocytosed. • Under the acidic conditions of the endosome, the transferrin releases its iron, but the transferrin remains bound to the transferrin receptor, which is recycled back to the cell surface, where it encounters the neutral pH environment of the blood. • The neutral pH causes the receptor to release the transferrin into the circulation, where it can pick up another Fe ion to repeat the cycle. The iron released in the endosome, moves on to lysosomes, from where it is transported into the cytosol. • Advantage: The system allows cells to take up iron efficiently even though the concentration of iron in the blood is extremely low. The iron bound to transferrin is concentrated at the cell surface by binding to transferrin receptors; it becomes further concentrated in clathrin-coated pits, which collect the transferrin receptors. In this way, transferrin cycles between the blood and endosomes, delivering the iron that cells need to grow.

Why do eukaryotic cells, and especially animal cells, have such large and complex cytoskeletons? List the differences between animal cells and bacteria that depend on the eukaryotic cytoskeleton.

"* Animal cells are larger and more diversely shaped, no cell wall * Animal cells have a nucleus, that is shaped and held in place by intermediate filaments * Animal cells can move by changing their shape (thanks to Actin-Myosin) * Animal cells have much larger genome stored as chromosoms and that is passed on to daughter cells thanks to microtubules * Animal cells have internal organelles, whose positions depends on molecular motors"

"Which of the following statements are correct? Explain your answers. A. A DNA strand has a polarity because its two ends contain different bases. B. G-C base pairs are more stable than A-T base pairs."

"A. False. The polarity of a DNA strand usually refers to the orientation of its sugar-phosphate backbone: one end contains a phosphate group, the other end contains a hydroxyl group. B. True. G-C: 3 H-bonds vs A-T: only 2 H-bonds "

"A typical time course of polymerization of purified tubulin to form microtubules is shown in Figure Q17-19. A. Explain the different parts of the curve (labeled A, B, and C). Draw a diagram that shows the behavior of tubulin molecules in each of the three phases. B. How would the curve in the figure change if centrosomes were added at the outset?"

"A. Phase A: lag phase - tubulin molecules are assembling to form nucleation centers. Then we have the elongation phase (filaments are growing). Then equilibrium is reached (some filaments are growing as others are shrinking) Diagram: see textbook B. No lag phase, rest is unchanged"

"Dynamic instability causes microtubules either to grow or to shrink rapidly. Consider an individual microtubule that is in its shrinking phase. A. What must happen at the end of the microtubule in order for it to stop shrinking and to start growing again? B. How would a change in the tubulin concentration affect this switch? C. What would happen if only GDP, but no GTP, were present in the solution? D. What would happen if the solution contained an analog of GTP that cannot be hydrolyzed?"

"A. Shrinking because it has lost its GTP cap. We need a lot of GTP subunits added quickly enough to cover the GDP at the microtubule's end, to form a new GTP cap B. higher tubulin concentration -> higher GTP-tubulin addition rate. The concentration will decrease and reach a level where the GTP addition rate is slow enough to switch back to shrinking and so on. It is self-balancing C. Microtubules will continue to shrink and will eventually disappear D. The microtubules will continue to grow until all GTP subunits have been used up"

"Define the following terms and their relationships to one another: A. Interphase chromosome B. Mitotic chromosome C. Chromatin D. Heterochromatin E. Histones F. Nucleosome"

"A. State in which a eukaryotic chromosome exists when the cell is between divisions; more extended and transcriptionally active than mitotic chromosomes. B. Highly condensed duplicated chromosome in which the two new chromosomes (also called sister chromatids) are still held together at the centromere. The structure chromosomes adopt during mitosis. C. Complex of DNA and proteins that makes up the chromosomes in a eukaryotic cell. D. Highly condensed region of an interphase chromosome; generally gene-poor and transcriptionally inactive. E. One of a group of abundant highly conserved proteins around which DNA wraps to form nucleosomes, structures that represent the most fundamental level of chromatin packing. F. Beadlike structural unit of a eukaryotic chromosome composed of a short length of DNA wrapped around a core of histone proteins; includes a nucleosomal core particle (DNA plus histone protein) along with a segment of linker DNA that ties the core particles together. E is part of F D-A C-B"

"Which of the following types of cells would you expect to contain a high density of intermediate filaments in their cytoplasm? Explain your answers. A. Amoeba proteus (a free-living amoeba) B. Skin epithelial cell C. Smooth muscle cell in the digestive tract D. Escherichia coli E. Nerve cell in the spinal cord F. Sperm cell G. Plant cell"

"B-C-E: preventing them from rupturing as they are stretched and compressed by the movements of surrounding tissue A-F: no need for those filaments as they migrate rapidly and do not sustain large tensile forces G: they resists the forces thanks to their rigid cell walls D: bacteria, none whatsoever"

Compare the structure of intermediate filaments with that of the myosin-II filaments in skeletal muscle cells. What are the major similarities? What are the major differences? How do the differences in structure relate to their function?

"Both are composed of subunits, which are protein dimers held together by coil-coiled interaction. Intermediate filaments dimers assemble head to head (no polarity) Myosin dimers assemble with their heads pointing in the same direction (polarity as required to give a contractile force in muscle)"

Dynein arms in a cilium are arranged so that, when activated, the heads push their neighboring outer doublet outward toward the tip of the cilium. Consider a cross section of a cilium (see Figure 17-26). Why would no bending motion of the cilium result if all dynein molecules were active at the same time? What pattern of dynein activity can account for the bending of a cilium in one direction?

"It they were both active at the same time, we would observe no relative motion of one microtubule to the other as required for bending. Only a few dynein molecules must be activated on one side of the cilium, which will bend away from this ""active"" side."

A useful technique for studying microtubule motors is to attach them by their tails to a glass coverslip (which can be accomplished quite easily because the tails stick avidly to a clean glass surface) and then allow them to settle. The microtubules may then be viewed in a light microscope as they are propelled over the surface of the coverslip by the heads of the motor proteins. Because the motor proteins attach at random orientations to the coverslip, however, how can they generate coordinated movement of individual microtubules rather than engaging in a tug-of-war? In which direction will microtubules crawl on a 'bed' of kinesin molecules (i.e., will they move plus-end first, or minus-end first)?

"Motor proteins have one and only one direction. Kinesin moves from the minus end to the plus end, wherease dynein does the opposite. If kinesin is attached to glass, only those now individual motors with the right orientation in relation to te microtubule can attach to it and create a force to propel it forward. The microtubule would then always crawl minus end first over the coverslip"

"The average time taken for a molecule or an organelle to diffuse a distance of x cm is given by the formula t = x^2/2D where t is the time in seconds and D is a constant called the diffusion coefficient for the molecule or particle. Using the above formula, calculate the time it would take for a small molecule, a protein, and a membrane vesicle to diffuse from one side to another of a cell 10 μm across. Typical diffusion coefficients in units of cm2/sec are: small molecule, 5 × 10-6; protein molecule, 5 × 10-7; vesicle, 5 × 10-8. How long would a membrane vesicle take to reach the end of an axon 10 cm long by free diffusion? How long would it take if it was transported along microtubules at 1 μm/sec? "

"The average time taken for a small molecule (such as ATP) to diffuse a distance of 10 μm is given by the calculation 2 / (2 × 5 × 10-6) = 0.1 seconds Similarly, a protein takes 1 second and a vesicle 10 seconds on average to travel 10 μm. A vesicle would require on average 109 seconds, or more than 30 years, to diffuse to the end of a 10-cm axon. This calculation makes it clear why kinesin and other motor proteins evolved to carry molecules and organelles along microtubules. "

"The two strands of a DNA double helix can be separated by heating. If you raised the temperature of a solution containing the following three DNA molecules, in what order do you suppose they would "melt"? Explain your answer. A. 5'-GCGGGCCAGCCCGAGTGGGTAGCCCAGG-3' 3'-CGCCCGGTCGGGCTCACCCATCGGGTCC-5' B. 5'-ATTATAAAATATTTAGATACTATATTTACAA-3' 3'-TAATATTTTATAAATCTATGATATAAATGTT-5' C. 5'-AGAGCTAGATCGAT-3' 3'-TCTCGATCTAGCTA-5'"

"The melting temperature depends on both the size of the strand and the number of H-bonds (the more of them the higher the melting temperature) Hence: C-B-A (from lowest to highest)"

Suppose you had a method of cutting DNA at specific sequences of nucleotides. How many nucleotides long (on average) would such a sequence have to be in order to make just one cut in a bacterial genome of 3 × 10^6 nucleotide pairs? How would the answer differ for the genome of an animal cell that contains 3 × 10^9 nucleotide pairs?

"To specify a unique N nucleotide long sequence we need 4^N>3*10^6, Thus N~11 Animal cell: N~16"

Why do you suppose it is much easier to add tubulin to existing microtubules than to start a new microtubule from scratch? Explain how γ-tubulin in the centrosome helps to overcome this hurdle.

"Tubulin dimers have less affinity for each other than for the end of a microtubule and its several interaction sites + random movement (a lot of dimers need to collide and interavt with each other building a microtubule before they come appart...) γ-tubulin present in the centrosome act like an existing microtubule to which the dimers can bind easily"

Suppose that the actin molecules in a cultured skin cell have been randomly labeled in such a way that 1 in 10,000 molecules carries a fluorescent marker. What would you expect to see if you examined the lamellipodium (leading edge) of this cell through a fluorescence microscope? Assume that your microscope is sensitive enough to detect single fluorescent molecules.

"Unpolymerized actin monomers are randomly moving creating a dim uniform background. However, the filaments being relatively steady will be visible. The chance of having two marked actin monomer on the same filament being really small, every filament should have at leat one of those. We would then see a speckled pattern of actin molecules, each marking a different filament."

A macromolecule isolated from an extraterrestrial source superficially resembles DNA, but closer analysis reveals that the bases have quite different structures (Figure Q5-7). Bases V, W, X, and Y have replaced bases A, T, G, and C. Look at these structures closely. Could these DNA-like molecules have been derived from a living organism that uses principles of genetic inheritance similar to those used by organisms on Earth?

"V-X // W-Y Hence, yes it could be derived from a living organism using earthlike principles of genetic inheritance"

. Predict which one of the following organisms will have the highest percentage of unsaturated phospholipids in its membranes. Explain your answer. A. Antarctic fish B. Desert snake C. Human being D. Polar bear E. Thermophilic bacterium that lives in hot springs at 100 degrees

(A) Antarctic fish live at subzero temperatures and are cold-blooded. To keep their membranes fluid at these temperatures, they have an unusually high percentage of unsaturated phospholipids.

. It seems paradoxical that a lipid bilayer can be fluid yet asymmetrical. Explain.

- Fluidity is in one plane - Lipid molecules can diffuse laterally in their own monolayer but do not readily flip from one monolayer to the other. - Specific types of lipid molecules inserted into one monolayer therefore remain in it unless they are actively transferred by flippase.

What is meant by the term "two-dimensional fluid"?

- In a two-dimensional fluid, the molecules are free to move only in one plane - The molecules in a normal fluid, in contrast, can move in three dimensions.

Explain why the polypeptide chain of most transmembrane proteins crosses the lipid bilayer as an α helix or a β barrel.

- In both an α helix and a β barrel, the polar peptide bonds of the polypeptide backbone can be completely shielded from the hydrophobic environment of the lipid bilayer by the hydrophobic amino acid side chains. - Internal hydrogen bonds between the peptide bonds stabilize the α helix and β barrel

What are the differences between a phospholipid molecule and a detergent molecule? How would the structure of a phospholipid molecule need to change to make it a detergent?

- Phospholipid molecules are approximately cylindrical in shape. - Detergent molecules, by contrast, are conical or wedge-shaped. - A phospholipid molecule with only one hydrocarbon tail, for example, would be a detergent. - To make a phospholipid molecule into a detergent, you would have to make its hydrophilic head larger or remove one of its tails so that it could form a micelle.

Compare the hydrophobic forces that hold a membrane protein in the lipid bilayer with those that help proteins fold into a unique three-dimensional structure.

- The exposure of hydrophobic amino acid side chains to water is energetically unfavorable. - There are two ways that such side chains can be sequestered away from water to achieve an energetically more favorable state. o First, they can form transmembrane segments that span a lipid bilayer. This requires about 20 of them to be located sequentially in the polypeptide chain. o Second, the hydrophobic amino acid side chains can be sequestered in the interior of the folded polypeptide chain. This is one of the major forces that lock the polypeptide chain into a unique three-dimensional structure.

Describe the different methods that cells use to restrict proteins to specific regions of the plasma membrane. Is a membrane with many of its proteins restricted still fluid?

- The lateral mobility of plasma membrane proteins can be restricted in several ways: o Proteins can be tethered to the cell cortex inside the cell o to extracellular matrix molecules outside the cell o to proteins on the surface of another cell o Diffusion barriers can restrict proteins to a particular membrane domain - The fluidity of the lipid bilayer is not significantly affected by the anchoring of membrane proteins

For the two detergents shown in Figure 11-25, explain why the red portions of the molecules are hydrophilic and the blue portions hydrophobic. Draw a short stretch of a polypeptide chain made up of three amino acids with hydrophobic side chains (see Panel 2-5, pp. 74-75) and apply a similar color scheme.

- The sulfate group in SDS is charged and therefore hydrophilic. - The OH group and the C-O-C groups in Triton X-100 are polar; they can form hydrogen bonds with water molecules and are therefore hydrophilic. - In contrast, the blue portions of the detergents are either hydrocarbon chains or aromatic rings, which don't have polar groups that could form hydrogen bonds with water molecules, making them hydrophobic.

Why does a red blood cell plasma membrane need transmembrane proteins?

- Transmembrane proteins anchor the plasma membrane to the underlying cell cortex, strengthening the membrane so that it can withstand the forces on it when the red blood cell is pumped through small blood vessels. - Transmembrane proteins also transport nutrients and ions across the plasma membrane.

Water molecules are said "to reorganize into a cagelike structure" around hydrophobic compounds (e.g., see Figure 11-9). This seems paradoxical because water molecules do not interact with the hydrophobic compound. So how could they "know" about its presence and change their behavior to interact differently with one another? Discuss this argument and, in doing so, develop a clear concept of what is meant by a "cagelike" structure. How does it compare to ice? Why would this cagelike structure be energetically unfavorable?

- Water is a liquid, and thus hydrogen bonds between water molecules are not static; they are continually formed and broken again by thermal motion. - When a water molecule happens to be next to a hydrophobic molecule, it is more restricted in motion and has fewer neighbors with which it can interact because it cannot form any hydrogen bonds in the direction of the hydrophobic molecule. It will therefore form hydrogen bonds to the more limited number of water molecules in its proximity. - Bonding to fewer partners results in a more ordered water structure, which represents the cagelike structure.

A. Predict the membrane orientation of a protein that is synthesized with an uncleaved, internal signal sequence (shown as the red start-transfer sequence in Figure 15-17) but does not contain a stop-transfer sequence.

. See Figure A15-4A. The internal signal sequence functions as a membrane anchor, as shown in Figure 15-17. Because there is no stop-transfer sequence, however, the C-terminal end of the protein continues to be translocated into the ER lumen. The resulting protein therefore has its N-terminal domain in the cytosol, followed by a single transmembrane segment, and a C-terminal domain in the ER lumen.

"Life" is easy to recognize but difficult to define. According to one popular biology text, living things: 1. Are highly organized compared to natural inanimate objects. 2. Display homeostasis, maintaining a relatively constant internal environment. 3. Reproduce themselves. 4. Grow and develop from simple beginnings. 5. Take energy and matter from the environment and transform it. 6. Respond to stimuli. 7. Show adaptation to their environment. Score a person, a vacuum cleaner, and a potato with respect to these characteristics.

1 Trying to define life in terms of properties is an elusive business, as suggested by this scoring exercise (Table A1-1). Vacuum cleaners are highly organized objects, and take matter and energy from the environment and transform the energy into motion, responding to stimuli from the operator as they do so. On the other hand, they cannot reproduce themselves, or grow and develop—but then neither can old animals. Potatoes are not particularly responsive to stimuli, and so on. It is curious that standard definitions of life usually do not mention that living organisms on Earth are largely made of organic molecules, that life is carbon based. As we now know, the key types of "informational macromolecules"—DNA, RNA, and protein— are the same in every living species.

How many possible nucleotide sequences are there for a stretch of DNA that is N nucleotides long, if it is (a) single- stranded or (b) double-stranded?

"(a) 4^N (b) As someof the 4^N sequence will be complement to another one, less than 4^N If N is odd, every stand will complement another: 0.5*4^N If N is even, some strand are self-complementary: 0.5(4^N+4^(N/2))"

Why does ionizing radiation stop cell division?

It tears through matter and breaks chemical bonds. It damages the DNA and sometimes the damage could be severe that it can't be repaired and they may eventually undergo apoptosis (Programmed cell death)

. List the ordinary, dictionary definitions of the terms replication, transcription, and translation. By their side, list the special meaning each term has when applied to the living cell.

Replication. Dictionary definition: the creation of an exact copy; molecular biology definition: the act of duplicating DNA. Transcription. Dictionary definition: the act of writing out a copy, especially from one physical form to another; molecular biology definition: the act of copying the information stored in DNA into RNA. Translation. Dictionary definition: the act of putting words into a different language; molecular biology definition: the act of polymerizing amino acids into a defined linear sequence using the information provided by the linear sequence of nucleotides in mRNA.

is cancer hereditary?

no. Cancer does not arise from mutations that are inherithed but from those mutations that occur in our somatic cells.

GTP is hydrolyzed by tubulin to cause the bending of flagella.

False. ATP is hydrolyzed by the dynein motor protein

T/F? Activation of myosin movement on actin filaments is triggered by the phosphorylation of troponin in some situations and by Ca2+ binding to troponin in others.

False. Myosin movement is activated by the phosphorylation of myosin, or by calcium binding to troponin.

Cells having an intermediate-filament network that cannot be depolymerized would die.

False. Nerve cells are a good example. As they are long-living and not dividing

In the DNA of certain bacterial cells, 13% of the nucleotides are adenine. What are the percentages of the other nucleotides?

"%A=%T=13% %G=%C=(100-2*%A)/2=37%"

DNA forms a right-handed helix. Pick out the right-handed helix from those shown in Figure Q5-15.

"A. 2 right-handed B. 1 right-handed C. none"

Chapter 1 Question 11) Identify the different organelles indicated with letters in the electron micrograph of a plant cell shown below. Estimate the length of the scale bar in the figure.

11 In this plant cell, A is the nucleus, B is a vacuole, C is the cell wall, and D is a chloroplast. The scale bar is about 10 μm, the width of the nucleus.