CBE 557 ECB Exam

20-3)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?

"In general, GAGs are strongly hydrophilic and tend to adopt highly extended conformations, which occupy a huge volume relative to their mass; thus GAGs act as effective "space fillers" in the extracellular matrix of connective tissues." If GAGs, and thus proteoglycans, contained less negative charge overall, the attraction of polar water and other positively charged ions would decrease, which would drastically decrease the overall volume and mass of the proteoglycans.

17-13)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.

1) Animal cells are much larger and more diversely shaped (2) Animal cells, and all other eukaryotic cells, have a nucleus that is shaped and held in place in the cell by intermediate filaments; the nuclear lamins attached to the inner nuclear membrane support and shape the nuclear membrane, and a meshwork of intermediate filaments surrounds the nucleus and spans the cytosol. (3) Animal cells can move by a process that requires a change in cell shape. Actin filaments and myosin motor proteins are required for these activities. (4) Animal cells have a much larger genome than bacteria; this genome is fragmented into many chromosomes. For cell division, chromosomes need to be accurately distributed to the daughter cells, requiring the function of the microtubules that form the mitotic spindle. (5) Animal cells have internal organelles.

1-10)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.

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.

2-13)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.

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 an H2S molecule is not strongly attracted to the hydrogen atoms in an adjacent H2S molecule, and the hydrogen bonds that are so predominant in water do not form.

2-20)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.

An amphipathic molecule has both hydrophilic and hydrophobic components, which means that in water it forms a system in which the hydrophobic portions avoid interaction with water and the hydrophilic portions are exposed to water. The image will pretty much be a circle of molecules with their hydrophobic ends touching each other in the center and the hydrophilic ends touching the water

11-18)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°C.

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

17-5)The formation of actin filaments in the cytosol is controlled by actinbinding 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.

Any actin-binding protein that stabilizes complexes of two or more actin monomers without blocking the ends required for filament growth will facilitate the initiation of a new filament (nucleation).

20-17)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.

while a carcinogen will immediately mutate a cell, a cell mutation does not immediately or directly cause cancer. Introduction to a carcinogen will create a genetic predisposition to cell mutation and over time mutated cells will replicate, creating an environment in which cancerous mutations can occur much more readily, and eventually do.

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3-3)Consider the analogy of the jiggling box containing coins that was described on page 83. 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,

A) In the reaction described by 100 coin flips, half being heads and half being tails, both the Gibbs free energy of reaction and the standard energy change would be 0. Since both the forward and backward reaction have the same free energy, the resulting free energy of a large enough sample size would be 0 as would the free energy of reaction, as the concentrations of "heads and tails" would be equal B) In the system with a box of coins described in the problem, the temperature of reaction could be compared to the height from which the coins are dropped or the amount of mixing the coins are put through before being dropped, both of which could affect the number of rotations possible and the end orientation. As far as the imaginary enzyme "jigglase", some enzyme of this type for the example could make the coins more likely to "react", or flip. For instance, "jigglase" could put the coins in a perfect position to flip, perhaps by positioning them on the edge of a table or on a spring surface.

5-12)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)Between nuclear divisions—that is, in interphase—the chromatin of the interphase chromosomes is in a relatively extended form in the nucleus, although some regions of it, the heterochromatin, remain densely packed and are transcriptionally inactive. B)During nuclear division—that is, in mitosis—replicated chromosomes become condensed into mitotic chromosomes, which are transcriptionally inactive and are designed to be readily distributed between the two daughter cells. 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 small 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 an octameric 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.

1-8)By now you should be familiar with the following cell 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. Golgi apparatus I. peroxisomes J. plasma membrane K. endoplasmic reticulum L. cytoskeleton M. ribosome

A)Contents of the main compartment of the cytoplasm, excluding membrane-enclosed organelles such as endoplasmic reticulum and mitochondria B)Contents of a cell that are contained within its plasma membrane but, in the case of eukaryotic cells, outside the nucleus C)Organelle in most cells in which chemical reactions associated with respiration and energy production occur. D)internal envelope that contains all genetic material of the cell, producing RNA, DNA, etc. E)plastid in plant cells containing chlorophyll where photosynthesis occurs, producing energy. F)an organelle in animal cells containing enzymes used to degrade other items enclosed in a membrane G)folded genetic stranding of nucleic acids and protein found in the nucleus of living cells H)a complex of vesicles and folded membranes within the cytoplasm; packages organelles and RNA I)Small membrane-enclosed organelle that contains enzymes that degrade lipids and destroy toxins. J)outer, flexible membrane of animal cells often with transport systems of various types K)system of membranous tubules around the nucleus, usually with ribosomes and responsible for protein and lipid synthesis L)microscopic system of protein filaments and tubules in the cytoplasm, giving structure M)Large macromolecular complex, composed of RNAs and proteins, that translates a messenger RNA into a polypeptide chain

20-9)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. Epidermal cells in the skin are continually shed and are renewed every few weeks; for a tattoo to be long-lasting, it is therefore necessary to deposit pigment below the epidermis. F. Although stem cells are not differentiated, they are specialized in the sense that they give rise only to specific cell types.

A)F, Gap junctions do not connect the cytoskeletons of two cells. They maintain gaps between the cells, allowing communication via small molecules only. B)True, often it means the cell has lost water content, like a tire might lose air C)False. Proteoglycans can withstand a large amount of compressive force but do not have a rigid structure. Their space-filling properties and ability to resist compression result from their tendency to absorb large amounts of water D)T E)T F)True. Stem cells stably express control genes that ensure that their daughter cells can only develop into certain differentiated cell types.

2-15)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. Lipids assemble into bilayers by noncovalent bonds. A membrane is therefore not a macromolecule. C)True. The backbone of nucleic acids is made up of alternating ribose (or deoxyribose in DNA) D)True. About half of the 20 naturally occurring amino acids have hydrophobic side chains E)T F)False, DNA contains bases A, G, T and C. U is present in RNA.

2-18) 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 require specific groups to interact; one is always a hydrogen atom linked by a polar covalent bond to an oxygen or a nitrogen, 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 close B) Van der Waals attractions would form in all three examples. Hydrogen bonds would form in (c) only.

11-10)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)If the phospholipids carried only one hydrocarbon tail rather than 2, the shape and size of the molecule would be affected and the lipid would become a detergent, affecting the overall function of the bilayer. B)The lipid bilayers formed would be much more fluid. 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 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)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.

17-3)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)The microtubule is shrinking because it has lost its GTP cap; that is the tubulin subunits at its end are all in their GDP-bound form. sufficient GTP-loaded subunits must be added quickly enough to cover up the GDP containing tubulin subunits at the microtubule end, a new GTP cap can form and regrowth is favored. B)The rate of addition of GTP-tubulin will be greater at higher tubulin concentrations. The frequency with which shrinking microtubules switch to the growing mode will therefore increase with increasing tubulin concentration. The consequence of this regulation is that the system is self-balancing: the more microtubules shrink (resulting in a higher concentration of free tubulin), the more frequently microtubules will start to grow again. Conversely, the more microtubules grow, the lower the concentration of free tubulin will become and the rate of GTP-tubulin addition will slow down; at some point, GTP hydrolysis will catch up with new GTP-tubulin addition, the GTP cap will be destroyed, and the microtubule will switch to the shrinking mode. C). If only GDP were present, microtubules would continue to shrink and eventually disappear, because tubulin dimers with GDP have very low affinity for each other and will not add stably to microtubules. D)If GTP is present but cannot be hydrolyzed, microtubules will continue to grow until all free tubulin subunits have been used up.

7-17)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

A)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)most harmful, the reading frame would be changed, and because this frameshift occurs near the beginning or in the middle of the coding sequence, much of the protein will contain a nonsensical and/or truncated sequence of amino acids. C)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—that is, they have no or only minor consequences for the organism. D)most harmful, the reading frame would be changed, and because this frameshift occurs near the beginning or in the middle of the coding sequence, much of the protein will contain a nonsensical and/or truncated sequence of amino acids. E) Substitution of one nucleotide for another is often completely harmless. In some cases, it will 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.

1-9)Which of the following statements are correct? Explain your answers. A. The hereditary information of a cell is passed on by its proteins. B. Bacterial DNA is found in the cytoplasm. C. Plants are composed of prokaryotic cells. D. With the exception of egg and sperm cells, all of the nucleated cells within a single multicellular organism have the same number of chromosomes. E. The cytosol includes membrane-enclosed organelles such as lysosomes. F. The nucleus and a mitochondrion are each surrounded by a double membrane. G. Protozoans are complex organisms with a set of specialized cells that form tissues such as flagella, mouthparts, stinging darts, and leglike appendages. H. Lysosomes and peroxisomes are the sites of degradation of unwanted materials.

A. False. Hereditary information of the cells is passed on via DNA B. True. Bacteria do not have a nucleus. C. False. Plants, like animals, are composed of eukaryotic cells. D. True. The number of chromosomes varies from one organism to another, but is constant in all nucleated cells (except germ cells) within the same multicellular 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. H. Somewhat true. Peroxisomes and lysosomes contain enzymes that catalyze the breakdown of substances produced in the cytosol or taken up by the cell.

7-7)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 can 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 not individually enclosed in a membrane. E. False. The position of the promoter determines the direction in which transcription proceeds and therefore which of the two DNA strands is used as the template. Transcription of the other strand would produce an mRNA with a completely different (and in most cases 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.

5-1)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 commonly refers to the orientation of its sugar-phosphate backbone, one end of which contains a phosphate group and the other a hydroxyl group. B. True. G-C base pairs are held together by three hydrogen bonds, whereas A-T base pairs are held together by only two

5-7) A. 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? B. 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

A. The bases V, W, X, and Y can form a DNA-like doublehelical molecule with virtually identical properties to those of bona fide DNA. V would always pair with X, and W with Y. Therefore, the macromolecule could be derived from a living organism that uses the same principles to replicate its genome as those used by organisms on Earth. In principle, different bases, such as V, W, X, and Y, could have been selected during evolution on Earth as building blocks for DNA. (Similarly, there are many more conceivable amino acid side chains than the set of 20 selected in evolution that make up all proteins.) B. None of the bases V, W, X, or Y can replace A, T, G, or C. To preserve the distance between the two sugar- phosphate strands in a double helix, a pyrimidine always has to pair with a purine (see, for example, Figure 5-4). Thus, the eight possible combinations would be V-A, V-G, W-A, W-G, X-C, X-T, Y-C, and Y-T. Because of the positions of hydrogen-bond acceptors and hydrogenbond donor groups, however, no stable base pairs would form in any of these combinations, as shown for the pairing of V and A in Figure A5-7, where only a single hydrogen bond could form.

17-11)Which of the following statements are correct? Explain your answers. A. Kinesin moves endoplasmic reticulum (ER) membranes along microtubules so that the network of ER tubules becomes stretched throughout the cell. B. Without actin, cells can form a functional mitotic spindle and pull their chromosomes apart but cannot divide. C. Lamellipodia and filopodia are "feelers" that a cell extends to find anchor points on the substratum that it will then crawl over. D. GTP is hydrolyzed by tubulin to cause the bending of flagella. E. Cells having an intermediate-filament network that cannot be depolymerized would die. F. The plus ends of microtubules grow faster because they have a larger GTP cap. G. 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. H. 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.

A. True. A continual outward movement of ER is required; in the absence of microtubules, the ER collapses toward the center of the cell. B. True. Actin is needed to make the contractile ring that causes the physical cleavage between the two daughter cells, whereas the mitotic spindle that partitions the chromosomes is composed of microtubules. C. 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. D. False. To cause bending, ATP is hydrolyzed by the dynein motor proteins that are attached to the outer microtubules in the flagellum. E. False. Cells could not divide without rearranging their intermediate filaments, but many terminally differentiated and long-lived cells, such as nerve cells, have stable intermediate filaments that are not known to depolymerize. F. False. The rate of growth is independent of the size of the GTP cap. The plus and minus ends have different growth rates because they have physically distinct binding sites for the incoming tubulin subunits; the rate of addition of tubulin subunits differs at the two ends. G. True. Both are nice examples of how the same membrane can have regions that are highly specialized for a particular function. H. False. Myosin movement is activated by the phosphorylation of myosin, or by calcium binding to troponin.

6-3)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 D. Nuclease that removes RNA primers E. DNA helicase F. Primase

A. Without DNA polymerase, no replication can take place at all. B. In the absence of ligase, the newly replicated DNA strands will remain as fragments, but no nucleotides will be missing. C. Without the sliding clamp, the DNA polymerase will frequently fall off the DNA template that the cell will be unable to divide. D. In the absence of RNA-excision enzymes, the RNA fragments will remain covalently attached to the newly replicated DNA fragments. No ligation will take place, because the DNA ligase will not link DNA to RNA. E. Without DNA helicase, the DNA polymerase will stall because it cannot separate the strands of the template DNA ahead of it. Little or no new DNA will be synthesized. F. In the absence of primase, RNA primers cannot be made on either the leading or the lagging strand. DNA replication therefore cannot begin.

1-11)There are three major classes of protein filaments that make up the cytoskeleton of a typical animal cell. 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.

Actin filaments: 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 Microtubules: function as "railroad tracks" for many intracellular movements and are responsible for the separation of chromosomes during cell division

7-2) 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?

Actually, 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 not fully extended because they begin to fold up and interact with proteins as they are synthesized; this is why they are shorter than the corresponding DNA segments.

11-8)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 are true A)The lipid bilayer is fluid because its lipid can undergo these motions B)The lipid bilayer is fluid because its lipid can undergo these motions C)Such exchanges require the action of a transporter. D)Hydrogen bonds are formed and broken by thermal motion E)Glycolipids are mostly restricted to the monolayer of membranes that faces away from the cytosol. Some special glycolipids, such as phosphatidylinositol (discussed in Chapter 16), are found specifically in the cytosolic monolayer. 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—that is, it turns oil into margarine. G)Examples include the many membrane enzymes involved in signaling (discussed in Chapter 16) H)Polysaccharides are the main constituents of mucus and slime; the carbohydrate coat of a cell, which is made up of polysaccharides and oligosaccharides, is a very important lubricant, particularly for cells that line blood vessels or circulate in the bloodstream

5-8)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. 5ʹ-GCGGGCCAGCCCGAGTGGGTAGCCCAGG-3ʹ 3ʹ-CGCCCGGTCGGGCTCACCCATCGGGTCC-5ʹ B. 5ʹ-ATTATAAAATATTTAGATACTATATTTACAA-3ʹ 3ʹ-TAATATTTTATAAATCTATGATATAAATGTT-5ʹ C. 5ʹ-AGAGCTAGATCGAT-3ʹ 3ʹ-TCTCGATCTAGCTA-5ʹ

As shown in Figure 5-4, an A-T pair contributes two hydrogen bonds, whereas a G-C pair contributes three hydrogen bonds. Therefore, helix C (containing a total of 34 hydrogen bonds) would melt at the lowest temperature, helix B (containing a total of 65 hydrogen bonds) would melt next, and helix A (containing a total of 78 hydrogen bonds) would melt last. Helix A is the most stable, largely owing to its high GC content

20-2)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). Would you expect collagen mutations to be detrimental if only one of the two copies of a collagen gene is defective?

As three collagen polypeptide chains have to come together to form the triple helix, a single defective polypeptide chain will impair assembly, even if normal chains are present at the same time. Collagen mutations are therefore dominant; that is, they have a deleterious effect even in the presence of a normal copy of the gene.

1-13)When bacteria are cultured under adverse conditions—for example, in the presence of a poison such as an antibiotic— most cells grow and divide slowly. But it is not uncommon to find that the rate of proliferation is restored to normal after a few days. Suggest why this may be the case.

Bacteria divide and replicate so often that mutations can be spread throughout a population rather quickly and quite effectively. In the presence of an antibiotic or poison, a bacterium with a genetic resistance to that poison will survive and replicate, passing along its genetic resistance while all other cells die. After some time, the surviving gene pool will grow at the same rate as it had previously been growing even in the presence of the poison, as the population is now resistant.

1-16)What, if any, are the advantages in being multicellular?

Being multicellular often allows for an organism or a system of cells to actively use various gene expressions that a single cell may not be able to. Furthermore, multicellular systems can often work in unison to allow for their common survival, as some cells may be responsible for advanced analysis of environment while other cells exist to process energy sources and supply all cells in the body to live prosperously. 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 division of labor, multicellular organisms are able to exploit food sources that are inaccessible to single-celled organisms.

17-9)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 filaments are composed of subunits in the form of protein dimers that are held together by coiled-coil interactions. Moreover, in both cases, the dimers polymerize through their coiled-coil domains into filaments. Whereas intermediate filament dimers assemble head-tohead, however, and thereby create a filament that has no polarity, all myosin molecules in the same half of the myosin filament are oriented with their heads pointing in the same direction. This polarity is necessary for them to be able to develop a contractile force in muscle.

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1-12)Natural selection is such a powerful force in evolution because organisms or cells with even a small reproductive advantage will eventually outnumber their competitors. To illustrate how quickly this process can occur, 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?

Calculation Recall: N=N0*2^(t/G)N15=N0,15x2^(t/15)=2^(t/15)N20=N0,20x2^(t/20)=1MM*2^(t/15)Set these two equations equal to each other to find the t at which they meet.t=1200 mins and at that time, N=2e24 cells.

17-17)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 depends on the ability of microtubules both to polymerize and to depolymerize. This is most obvious when one considers that the formation of the mitotic spindle requires the prior depolymerization of other microtubules to free up the tubulin required to build the spindle. This rearrangement is not possible in Taxoltreated cells, whereas in colchicine-treated cells, division is blocked because a spindle cannot be assembled. On a less obvious but no less important level, both drugs block the dynamic instability of microtubules and would therefore interfere with the workings of the mitotic spindle, even if one could be properly assembled.

17-7)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?

Cells contain actin-binding proteins that bundle and cross-link actin filaments (see Figure 17−32). The filaments extending the lamellipodia and filopodia are firmly anchored in the filamentous meshwork of the cell cortex, thus providing the mechanical anchorage required for the growing rodlike filaments to deform the cell membrane.

17-1)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

Cells that migrate rapidly from one place to another, such as amoebae (A) and sperm cells (F), do not in general need intermediate filaments in their cytoplasm, since they do not develop or sustain large tensile forces. Plant cells (G) are pushed and pulled by the forces of wind and water, but they resist these forces by means of their rigid cell walls rather than by their cytoskeleton Epithelial cells (B), smooth muscle cells (C), and the long axons of nerve cells (E) are all rich in cytoplasmic intermediate filaments, which prevent them from rupturing as they are stretched and compressed by the movements of their surrounding tissues. All of the above eukaryotic cells possess intermediate filaments in their nuclear lamina. Bacteria, such as Escherichia coli (D), have none whatsoever

2-8)In principle, there are many different, chemically diverse ways in which small molecules can be joined together 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—that is, 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 over a linking chemistry such as that used to produce polyethylene?

Condensation reactions are quite usefully biologically due mainly to the abundance of water in the cell and they are easily reversed via hydrolysis reactions, as well, with the abundant water.

2-9)Why could covalent bonds not be used in place of noncovalent bonds to mediate most of the interactions of macromolecules?

Covalent bonds are inherently very stable, which makes them non-ideal for biological processes which often require frequent separations and have an inconsistent environment.

3-6)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

Enzymes operate with a variety of specific conditions, such as pH, temperature and surrounding polarity. Enzymes must be able to house both the individual reactants as well as the product, so under different conditions it is possible that the enzyme catalyzes the reverse reaction, as all reactions are reversible. It is also possible that the enzyme is able to catalyze the reverse reaction in a living cell, as concentration also plays a role in enzyme function.

20-7)Why do you suppose epithelial cells lining the gut are lost and replaced (renewed) frequently, whereas most neurons last for the lifetime of the organism?

Epithelial cells lining the gut are renewed frequently due to the severe and caustic environment in which they are located. These cells undergo severe damage and are replaced so as to assure that vital organs are protected and remain functioning over time. Neurons, on the other hand, are in a very protected and complex environment, requiring a very select homeostasis at all times to maintain the complex function of the brain.

20-8)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. How can you reconcile these apparently conflicting two numbers?

Every cell division generates one additional cell; so if the cells were never lost or discarded from the body, the number of cells in the body should equal the number of divisions plus one. The number of divisions is 1000-fold greater than the number of cells because, in the course of a lifetime, 1000 cells are discarded by mechanisms such as apoptosis for every cell that is retained in the body.

20-4)Analogs of hemidesmosomes are the focal contacts 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?

Focal contacts 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 the matrix or for crawling are generated by the actin cytoskeleton. In mature epithelia, focal contacts are presumably rare because the cells are largely fixed in place and have no need to crawl over the basal lamina or actively pull on it

2-19)What are the forces that determine the folding of a macromolecule into a unique shape?

Folding is caused by a variety of noncovalent bonds, which include hydrogen bonding, van der Waals attractions, ionic interactions and polar interactions.

20-12)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 chemical synapses (as shown in Figure 12−40) rather than through gap junctions?

Gap junctions are able to relay small molecules to allow for communications between cells, but this method of communication is simply nowhere nearly as effective as synapses in the brain. Neurons are responsible for instantaneous reactions to external stimuli, which requires a much more sophisticated and direct method for communication than the transfer of small molecules between cells

5-15)DNA forms a right-handed helix. Pick out the right-handed helix from those shown in Figure Q5-15.

Helix (A) is right-handed. Helix (C) is left-handed. Helix (B) has one right-handed strand and one left-handed strand. There are several ways to tell the handedness of a helix. For a vertically oriented helix, like the ones in Figure Q5-15, if the strands in front point up to the right, the helix is right-handed; if they point up to the left, the helix is left-handed

20-1)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?

Horizontal orientation of the microtubules would allow for the vertical growth of the plant, as it would allow for a vertical connection between the various horizontal microtubules. The growth of the cells will therefore be in a vertical direction, expanding the distance between the cellulose microfibrils without stretching them

20-19)Is cancer hereditary?

No, cancer is not hereditary. Generally speaking, cancer is caused by completely random mutations in genetic code. However, it is true that some mutations that can be passed down to the offspring making cancer-causing mutations more likely.

17-4)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 17−27). 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?

If all the dynein arms were equally active, there could be no significant relative motion of one microtubule to the other as required for bending. (Think of a circle of nine weightlifters, each trying to lift his neighbor off the ground: if they all succeeded, the group would levitate!). Thus, a few ciliary dynein molecules must be activated selectively on one side of the cilium. As they move their neighboring microtubules toward the tip of the cilium, the cilium bends away from the side containing the activated dyneins.

11-9)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.

5-3)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 aminoacid-altering mutations in histone genes are deleterious?

In contrast to most proteins, which accumulate amino acid changes over evolutionary time, the functions of histone proteins must involve nearly all of their amino acids, so that a change in any position would be deleterious to the cell.

20-6)Why does ionizing radiation stop cell division?

Ionizing radiation stops cell division due to the massive disturbance this radiation causes with electrons and, thus, chemical bonding. The structure of DNA is extremely dependent on chemical and ionic bonding, so electron disturbance is fatal for DNA structure.

1-3)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 lifeform? How might you redesign your experiment to allow air into the flask, yet eliminate the possibility that contamination by airborne microorganisms is the explanation for the results? (For a readymade 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. The experiment could use an intricate, sealed system with O2 and CO2 cycling to prevent the introduction of ambient life forms that would be able to reproduce in the favorable environment. Or 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

7-13)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.

1-6)Discuss the relative advantages and disadvantages of light and electron microscopy. How could you best visualize a living skin cell, a yeast mitochondrion, a bacterium, and a microtubule?

Light : -(advantages)simpler, nondestructive so can see living cells -disadvantage: low resolution Electron: -Advantage- high resolution -disadvantage: complex, cannot look at living cells To resolve both a living skin cell and bacterium, it is probably best to use a light microscope to observe the living features of both cells. An electron microscope could be used to resolve smaller features of the cells, namely a yeast mitochondrion and microtubule.

1-14)Apply the principle of exponential growth of a population of cells in a culture (as described in Question 1-12) to the cells in a multicellular organism, such as yourself. There are about 1013 cells in your body. Assume that one cell has acquired mutations that allow it to divide in an uncontrolled manner to become 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?

Look at calculation

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 the fluorescent labels 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?

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 (Figure A11-16). At 0°C, the fluidity of the membrane is reduced, and the mixing of the membrane proteins is significantly slowed. Drawing in doc

1-2)Mutations are mistakes in the DNA that change the genetic plan from that of 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. 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: the loss of high heels might produce shoes that are more comfortable (and less dangerous). The example illustrates that random changes can lead to significant improvements if the number of trials is large enough and selective pressures are imposed.

17-18)A useful technique for studying microtubule motors is to attach them by their tails to a glass cover slip (which can be accomplished quite easily because the tails stick avidly to a clean glass surface) and then allow them to settle. Microtubules may then be viewed in a light microscope as they are propelled over the surface of the cover slip by the heads of the motor proteins. Because the motor proteins attach at random orientations to the cover slip, 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 are unidirectional in their action; kinesin always moves toward the plus end of a microtubule and dynein toward the minus end. Thus if kinesin molecules are attached to glass, only those individual motors that have the correct orientation in relation to the microtubule that settles on them can attach to the microtubule and exert force on it to propel it forward. Since kinesin moves toward the plus end of the microtubule,the microtubule will always crawl minus-end first over the cover slip

17-23)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).

17-6)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.

Only fluorescent actin molecules assembled into filaments are visible, because unpolymerized actin molecules diffuse so rapidly that they produce a dim, uniform background. Since, in your experiment, so few actin molecules are labeled (1:10,000), there should be at most one labeled actin monomer per filament (see Figure 17−30). The lamellipodium as a whole has many actin filaments, some of which overlap, and it therefore shows a random, speckled pattern of actin molecules, each marking a different filament. This technique (called "speckle fluorescence") can be used to follow the movement of polymerized actin in a migrating cell. If you watch this pattern with time, you will see that individual fluorescent spots move steadily back from the leading edge toward the interior of the cell, a movement that occurs whether or not the cell is actually migrating. Rearward movement takes place because actin monomers are added to filaments at the plus end and are lost from the minus end (where they are depolymerized) (see Figure 17−35B). In effect, actin monomers "move through" the actin filaments, a phenomenon termed "treadmilling." Treadmilling has been demonstrated to occur in isolated actin filaments in solution and also in dynamic microtubules, such as those within a mitotic spindle.

11-11)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?

Phospholipids are more cylindrical in shape than detergents, which are best described as cones. The main difference between the two is that phospholipids have two hydrophobic tails, while detergents have only 1. Therefore, to make a phospholipid into a detergent, it is possible to simply remove one of the tails.

20-11)Discuss the following statement: "If plant cells contained intermediate filaments to provide the cells with tensile strength, their cell walls would be dispensable."

Plants are exposed to extreme changes in the environment, which often are accompanied by huge fluctuations in the osmotic properties of their surroundings. An intermediate-filament network as we know it from animal cells would not be able to provide full osmotic support for cells: the sparse, rivetlike attachment points would not be able to prevent the membrane from bursting in response to a huge osmotic pressure applied from the inside of the cell

11-4)Explain why the polypeptide chain of most transmembrane proteins crosses the lipid bilayer as an α helix or a β barrel.

Polypeptide chains contain hydrophobic side chains and hydrophilic peptide chains, which requires the formation of an α-helix in order to shield the hydrophilic chains from the hydrophobic tails of the phospholipids bilayer. Internal hydrogen bonds between the peptide bonds stabilize the α helix and β barrel

11-7)Describe the different methods that cells use to restrict proteins to specific regions of the plasma membrane. Can a membrane with many of its proteins restricted still be fluid?

Proteins can be tethered (A) to the cell cortex inside the cell, (B) to extracellular matrix molecules outside the cell, or (C) to proteins on the surface of another cell.The mobility of the membrane proteins is drastically reduced if they are bound to other proteins such as those of the cell cortex or the extracellular matrix. Some membrane proteins are confined to membrane domains by barriers, such as tight junctions. The fluidity of the lipid bilayer is not significantly affected by the anchoring of membrane proteins; the sea of lipid molecules flows around anchored membrane proteins like water around the posts of a pier.

7-11)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 copying a DNA sequence. 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

11-19)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.

1-18)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?

The common argument that living cells evolved from a common ancestor is that many different cells types share many traits between them. For instance, it appears almost certain that the mitochondria of eukaryotes evolved from prokaryotic cells, implying that eukaryotes most likely evolved from the single celled prokaryote. However, it is unlikely that the primordial ancestor was the first and only cell to form. Based on general evolution principles, it is likely that many different cells types originally formed and the strongest cell was able to survive. The sheer odds of a single cell survival would be too small to assume the ancestor was the first and only.

17-20)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.

1-15)"The structure and function of a living cell are dictated by the laws of chemistry, physics, and thermodynamics." Provide examples that support (or refute) this claim.

The laws of physics and chemistry govern how the world around living things behaves and unfolds. It is only fair to assume that all living things obey the same laws that everything around them obey. All activity in living cells can be explained by or is bound by the laws of physics and chemistry, whether it is a chemical reaction in the mitochondria or a lysosome or the physical characteristics of microtubules during cell separation.

11-17)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 (described in Chapter 4, pp. 121-122 and pp. 127-128).

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. First, they can form transmembrane segments that span a lipid bilayer. This requires about 20 of them to be located sequentially in a polypeptide chain. 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. In either case, the hydrophobic forces in the lipid bilayer or in the interior of a protein are based on the same principles.

3-8)The phosphoanhydride bond that links two phosphate groups in ATP in a high-energy linkage has a ΔG° of -30.5 kJ/mole. Hydrolysis of this bond in a cell liberates from 46 to 54 kJ/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 ΔG derived from ATP hydrolysis depends on both the ΔG° and the concentrations of the substrate and products. For example, for a particular set of concentrations, one might have ΔG = -50 kJ/mole = -30.5 kJ/mole + 2.58 ln [ADP] × [Pi] [ATP] ΔG is smaller than ΔG°, largely because the ATP concentration in cells is high (in the millimolar range) and the ADP concentration is low (in the 10 μM range). The concentration term of this equation is therefore smaller than 1 and its logarithm is a negative number. ΔG° is a constant for the reaction and will not vary with reaction conditions. ΔG, in contrast, depends on the concentrations of ATP, ADP, and phosphate, which can be somewhat different between cells

11-14)Consider a transmembrane protein that forms a hydrophilic pore across the plasma membrane of a eukaryotic cell. When this protein is activated by binding a specific ligand on its extracellular side it allows Na+ to enter the cell. The protein 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 membranespanning 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 (Figure A11-14). Ions can pass through this hydrophilic pore without coming into contact with the lipid tails of the bilayer. The hydrophobic side chains interact with the hydrophobic lipid tails.

11-3)It seems paradoxical that a lipid bilayer can be fluid yet asymmetrical. Explain.

The lipid bilayer is asymmetrical for a variety of reasons, including the presence of various transport proteins in the bilayer as well as the non-uniform dispersion of phospholipids. Regardless of the bilayer's fluidity, the membrane remains asymmetric about a given axis due to the large variations in the membrane composition.

7-5)A sequence of nucleotides in a DNA strand—5ʹ-TTAACGGCTTTTTTC-3ʹ— 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. 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. Note that the usual convention in describing the sequence of a gene is to give the sequence of the DNA strand that is not used as a template for RNA synthesis; this sequence is the same as that of the RNA transcript, with T written in place of U.

2-3)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 direct sharing of electrons between two atoms, which means that an extremely polar covalent bond is essentially the same as an ionic bond, in that it shares electrons between two very differently charged atoms.

6-12)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.

The statement is correct. If the DNA in somatic cells is not sufficiently stable (that is, if it accumulates mutations too rapidly), the organism dies (of cancer, for example), and because this may often happen before the organism can reproduce, the species will die out. If the DNA in reproductive cells is not sufficiently stable, many mutations will accumulate and be passed on to future generations, so that the species will not be maintained.

20-14)"The structure of an organism is determined by the genome that the fertilized 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?

The statement made in the question can be seen simply by comparing the variety of human beings and species that exist. Small differences in the genome in the egg can create large differences in the offspring. However, genomic code is not the only prerequisite for successful birth, the method by which the egg is fertilized and cared for is obviously crucial as well, for instance a human would be unable to develop in an ostrich egg, as the environment does not allow for the attachment of an umbilical cord or other necessary features.

11-5)For the two detergents shown in Figure 11-26, explain why the blue portions of the molecules are hydrophilic and the red portions hydrophobic. Draw a short stretch of a polypeptide chain made up of three amino acids with hydrophobic side chains (see Panel 2-6, pp. 76- 77) and apply a similar color scheme. Indicate which portions of your polypeptide would form hydrogen bonds with water

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 also form hydrogen bonds with water molecules and are therefore hydrophilic. In contrast, the red portions of the detergents are either hydrocarbon chains or aromatic rings, neither of which has polar groups that could form hydrogen bonds with water molecules; they are therefore hydrophobic. Drawing included in doc

1-4)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 × 1024 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 × 2t/G, where N0 is the number of cells at zero time, and G is the population doubling time.)

This paradox can be explained by the fact that food and energy sources are not plentiful enough to support this population of bacteria and that cell death also plays a role in the process. (calculation in notes)

11-13)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

17-2)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

Two tubulin dimers have a lower affinity for each other (because of a more limited number of interaction sites) than a tubulin dimer has for the end of a microtubule.Thus, to initiate a microtubule from scratch, enough tubulin dimers have to come together, and remain bound to one another for long enough, for other tubulin molecules to add to them. Only when a number of tubulin dimers have already assembled will the binding of the next subunit be favored. The formation of these initial "nucleating sites" is therefore rare and does not occur spontaneously at cellular concentrations of tubulin. Centrosomes contain preassembled rings of γ-tubulin (in which the γ-tubulin subunits are held together in much tighter side-to-side interactions than αβ-tubulin can form) to which αβ-tubulin dimers can bind. The binding conditions of αβ-tubulin dimers resemble those of adding to the end of an assembled microtubule. The γ-tubulin rings in the centrosome can therefore be thought of as permanently preassembled nucleation sites.

1-1)"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.

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 is, 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.

11-1)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 molecules arrange themselves around a hydrophobic molecule such that their interactions with the hydrophobe are minimized. When frozen, water molecules spread out, forming a wider structure of ice. By creating a "cage-like" structure, the water is unknowingly interacting only with itself, as there are no interactions to be had with the hydrophobe. Obviously this arrangement is the lowest energy state of H2O, as it mimics the solid state. Molecular movement is limited severely.

20-13)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?

When making Jello, the gelatin with the collagen is boiled, denaturing the collagen. When the collagen is allowed to cool, it does not retain its perfectly bundled structure and forms a less tensile gel.

20-18)High levels of the female sex hormone estrogen increase the risk 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?

While it is clear that sex hormones can be classified as carcinogenic, it is not necessarily clear that they directly cause mutations. Since sex hormones occur naturally in the body, it is unlikely that they have evolved in the human body while causing plentiful cancer-causing mutations. It is possible that the sex hormones encourage rapid cell division (as they would during, say, puberty) which increases the odds of mutation and are not necessarily a direct cause for mutation

20-16)Carefully consider the graph in Figure 20−43, which shows 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

While mutations occur with similar frequency throughout life, cancer-causing mutations do not. As more and more mutations occur over time, the odds that one of those mutations is cancer causing increases with time; eventually mutated cells will become cancerous.

3-4)see figure

see figure