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ESSENTIAL third edition CELL BIOLOGY Alberts Bray Hopkin Johnson Lewis Raff .. After registering for This PDF overviews the contents of the DVD and. Request PDF on ResearchGate | On Jan 1, , Akif Uzman and others published Essential Cell Biology, Third Edition. Essential Cell Biology, Third Edition . The “Panel” sections are available on the DVD as pdf files; I hope that the authors might consider.
Make the bacteria overproduce chaperone proteins in addition to protein A. So the recipient molecule effectively acquires two hydrogen atoms. Could any of the following changes in mutant protein X explain your results? I should note that it is still one of the richest books for teaching students analytical thinking in cell and molecular biology. Reprinted material is quoted with permission, and sources are indicated. Reading frame two is the only reading frame that does not contain a stop codon.
A spectrophotometer was used to measure the initial rate of production of Y and Z by the reactions shown in Figure QA. The initial rates were measured for several independent reactions.
Given the data shown. Using Kinetics to Model and Manipulate Metabolic Pathways The product Y of an enzymatic reaction absorbs light at the wavelength nm and the product Z of another reaction absorbs at nm. Provided these two people are in close proximity and can communicate. In this case.
Does your answer to Part A make sense in light of this rate information? Activated Carrier Molecules and Biosynthesis Which of the following is NOT a crucial benefit of using enzymes to catalyze biological reactions? For instance. Using barter and money as analogies. In a simple economy. But in a more complex economy. It forms a high-energy intermediate of the same energy. What is the effect of substituting arsenate for phosphate in this reaction?
Figure Q a b c d e It forms a high-energy intermediate of lower energy. It increases the stability of the high-energy intermediate. Arsenate mimics phosphate and can also be incorporated into a similar high-energy intermediate Figure QB. It has no effect on the stability of the high-energy intermediate. The reaction profiles for the hydrolysis of these two high-energy intermediates are given in Figure QC.
It decreases the stability of the high-energy intermediate. Nucleotide 7. Acetyl CoA C. Carboxylated biotin E. To your dismay. Write the appropriate number beside each item in List 1. Glucose 4. S-adenosylmethionine List 2 1. ATP B. Amino acid You are about to give up when the following table from a biochemistry textbook catches your eye. Explain how this is achieved. As part of the process. Choice e is incorrect as living organisms are not closed systems.
They use special pathways for all their reactions that allow them to be energetically favorable. By consuming carbon dioxide and producing oxygen photosynthesis lessens global warming cause by the greenhouse effect choice e is false.
Burning is an uncontrolled oxidation in which the energy is all dissipated as heat. Choice d is incorrect: Catabolic reactions are the reactions in which a cell breaks down food molecules.
Choices a. Much of the mass of food is released as CO2 that is breathed out into the atmosphere or is released into the environment as waste products. Most of the energy contained in the chemical bonds of the food molecules is converted to energy to maintain order among molecules in the body.
Choice b is incorrect as living organisms do not use heat to power biochemical reactions. Photosynthesis harvests light energy from the sun and converts it into chemical bond energy. Heat is produced in the course of biochemical reactions. Living things. Choice a is false because food molecules and oxygen produced by photosynthesis are the sole source of the energy that powers nearly all living nonphotosynthetic organisms.
Choice a is incorrect as no system. By releasing heat to their environment. Answers No. Hydrogenation increases the number of C-H bonds in a molecule. The more reduced an atom becomes. Hydrogenation is a special kind of reduction reaction. Chemical reactions occur only when there is a loss of free energy. Enzymes act more selectively than other catalysts. The diameter of an atom is influenced by the amount of negative charge.
A redox reaction involves the complete or partial transfer of electrons from one molecule or atom to another. The donor is oxidized and the recipient is reduced in the reaction. A catalyst reduces the activation energy of a reaction. Change in free energy for the reaction is b minus c. Activation energy is a minus b. Choice c is an incorrect answer. Choice e is not a definition of equilibrium. Graph 4 is the same as the graph for the original reaction in terms of the relative energetic differences between substrates.
Thus choices b and d are incorrect. An enzyme will make the value of a smaller and leave the values for b and c unchanged. The cell may directly couple the unfavorable reaction to a second. The binding energy is the standard free energy of the binding reaction. Choices a and b are false. The shape of the binding site affects the ability of the protein side chains to interact with portions of the substrate molecule.
Choice e is false. The strength of the protein-ligand interaction increases as the number of noncovalent bonds between the two increases. The standard free energy change. As the binding energy increases. Choice d is false. Under standard conditions equal concentrations of X and Y.
Both temperature and pH can disrupt noncovalent bonds that not only affect the binding. An increase in temperature will thus increase the reaction rate initially.
Favorable A. Figure A B. Graph 1 By increasing thermal motion. Y can accumulate to a very high level without causing significant amounts of X to build up. See Figure A for correct labeling of figure. Y will accumulate.. The higher the concentration and diffusion rate of substrate.
The diffusion coefficient of a molecule decreases with increasing mass shape is also a factor. The diffusion rate influences how often an enzyme will encounter and thus bind its substrate. Money is analogous to the storage of energy from a favorable reaction in the form of high-energy bonds in an activated carrier molecule.
Choice b is incorrect because a more positive free energy of binding indicates it is less energetically favorable and thus occurs less often. Choice e is likely to be false because many enzyme-substrate binding interactions rely on ionic bonds that are weakened by high salt concentrations.
Such activated carrier molecules can drive a huge variety of other unfavorable reactions in the cell. If half of the enzyme molecules are bound to the substrate. Although some binding reactions are diffusion-limited. The less tightly an enzyme binds its product. The diffusion rate is almost as fast in the cytoplasm and in water.
When [S] is substituted for KM in the equation.
Protons are always present in solution. In fact. Pyrophosphate does not look like ATP and is therefore unlikely to be used by the enzymes as an alternative energy source. The activation energy of the arsenate compound is extremely low. Choice a is false because oxidation of food molecules produces NADH.
Choices a and b are false as more energy is released by the hydrolysis of the arsenoanhydride bond as inferred by the greater difference in energy level between reactants and products in Figure Q so. B—7 acetyl group. Thus choices d and e are false. The other processes are thermodynamically spontaneous.
ADP will build up and inhibit the enzymes again. So the recipient molecule effectively acquires two hydrogen atoms. E—5 methyl group. D—1 carboxyl group. Whether e is true or not for any particular reaction is irrelevant. The reverse reactions are: A protein chain ends in a free amino group at the C-terminus. A protein is similar to a charm bracelet that has a linear linked chain or backbone with charms hanging off at regular intervals.
Similar to the amino acid sequence of a protein. The sequence of the atoms in the polypeptide backbone varies between different proteins. Nonpolar amino acids tend to be found in the interior of proteins.
The polypeptide backbone is free to rotate about each peptide bond. The genetically engineered bacteria make large amounts of protein A. For a protein with a stability of 7. Explain your reasoning. See Figure Q Make the bacteria overproduce chaperone proteins in addition to protein A. Make the bacteria synthesize protein A at a slower rate and in smaller amounts. Stability is a measure of the equilibrium between the folded F and unfolded U forms of the protein.
Dissolve the protein aggregate in urea. Which of the following procedures might help you to obtain soluble. On heating. Treat the insoluble aggregate with a protease. On removal of urea. Heat the protein aggregate to denature all proteins. Figure Q B. In the schematic diagram of a protein given in Figure Q The nonpolar amino acids are italicized..
Draw the hydrogen bonds as dashed lines. Indicate whether each structure is parallel or antiparallel. You find. Figure Q Calculate how many different amino acid sequences there are for a polypeptide chain 10 amino acids long. A protein such as hemoglobin. List the segments A—E of the protein that are most likely to be folded into compact.
You know the amino acid sequence of the protein and so can draw a map of where factor Xa and thrombin should cut it Figure Q From the arrangement of complementary binding surfaces. Although you still have no idea of what either a MAP kinase or Fus3p is. In response to your blank stare.
Purified antibodies are useful for a variety of experimental purposes. Actin S—S bonds a are formed by the cross-linking of methionine residues. Collagen E. Hemoglobin F. Elastin D. Lysozyme C. Keratin B. Repeated condensation reactions with similar building blocks to form a growing chain 8. Addition of phosphate group 4.
Knowing this. What do you think might happen? Protease List 2 1. Hydrolysis of nucleotide triphosphate 9. Once the reaction is completed.
Hydrolysis of peptide bonds For each of the following sentences. Phosphatase C. Removal of phosphate group 5. Use each item in List 2 no more than once. Synthase H. Oxidation of substrate 2. Hydrolase E. ATPase G. Polymerase D. Dehydrogenase F. Cleavage of substrate using water 6. Kinase B.
Reduction of substrate 3. Condensation reaction in anabolic pathway 7. She generated an antibody against this transition state analog and mixed the antibody with chemical X. You are able to show that E inhibits enzyme V. How would you expect the mutant protozoan to behave? When this protein binds to another protein found at the base of the cilia. For these enzymes. ATP are regulated by phosphorylation. You have identified the threonine residue at which Speed is phosphorylated and changed it to an alanine residue.
This ciliar protein. Of the choices below. Mutations in the gene for Ras are found in many cancers. In its active form. Explain your answer. Why is this an important goal? Winnebago will continue to move from point A to point B. Choice b is untrue. Answers A protein is similar to a charm bracelet that has a linear linked chain or backbone with charms hanging off at regular intervals. Choice d is untrue. The part of a protein that is analogous to the bracelet chain is called the polypeptide backbone.
The final folded conformation adopted by a protein is that of lowest energy. The parts of a protein that are analogous to the charms are called the side chains. Choice a is untrue.
Choice e is untrue. Antiparallel Heating can lead to the partial denaturation and aggregation of proteins to form a solid gelatinous mass. Expressing the protein slowly and at lower levels might increase the amount of properly folded protein.
Since the protein you are expressing in bacteria is being made in large quantities. In the absence of chaperones. Some proteins require molecular chaperones in order to fold properly within the environment of the cell.
Treating the aggregate with a protease. Removing the urea slowly and gradually often allows the protein to refold. Overexpressing chaperone proteins might increase the amount of properly folded protein. Urea should solubilize the protein and completely unfold it. Nearly all of the amino acid side chains in this sequence are nonpolar or hydrophobic. A is parallel and B is antiparallel See Figure A A protein domain is the modular unit from which many larger single-chain proteins are constructed.
The three-dimensional conformation of a protein is its tertiary structure. Polar side chains P will exhibit the same pattern. Members of the same protein family have similar protein sequences.
Choice d is incorrect. Choices b and e are unlikely. So if the dog protein is a MAP protein kinase. Choice d is unlikely. Choice b is incorrect. If these sites are folded into the interior of a stable protein domain.
Defined multimer with four subunits. F—G Choice c is the answer. To cut the protein chain. From the sizes of the fragments produced by digestion of the protein with Factor Xa. Sheet Filament Dimers formed by a normal protein will run through the gelfiltration column faster than a mutant protein Y monomer. Choice e is incorrect for the reason stated in choice b. From the sizes of the fragments produced by thrombin. Choice a is incorrect since S—S bonds are formed between cysteines.
Factor Xa and thrombin must bind to their preferred cutting sites. Choice c is unlikely. Enzymes bind their substrates or inhibitors at the active site. In addition. Enzymes catalyze a chemical reaction by lowering the activation energy. An enzyme is generally not inhibited by its substrate. Such catalytic antibodies have been isolated and shown to catalyze a variety of reactions. Since BPG and carbon dioxide both bind to and stabilize the same form of hemoglobin.
It is more likely that C alone rather than B alone will inhibit T. H—9 Any substance that will bind to a protein is known as its ligand. The enzyme hexokinase is so specific that it reacts with only one of the two isomers of glucose.
Since BPG binding stimulates the dissociation of oxygen. The hypervariable structural element that forms the ligand binding site is comprised of several loops. Most proteins need to release their ligand at some point. When RacerX is not bound to Speed. A mutant missing RacerX choice c would not be able to swim fast at all and neither would one missing the protein kinase that phosphorylates Speed choice a or one lacking the Speed protein choice b.
One that overproduced the protein phosphatase choice d would keep the Speed protein permanently dephosphorylated and thus would also be unable to swim fast. This is because the conformational change is coupled to ATP hydrolysis.
The lack of the protein phosphatase would mean that the Speed protein could remain phosphorylated all the time. It makes sense that cells would not want to have to phosphorylate their enzymes to turn them on when ATP levels are already low. Since the GTP-bound form is usually the active form.
The main purpose of glycolysis and the citric acid cycle is to generate ATP. G proteins are reset by nucleotide exchange. The conformational change driven by hydrolysis. All of the other changes will decrease the strength of the proliferative signal sent through the pathway by Ras. Because the altered Speed protein cannot be phosphorylated. The identification of the thousands of different protein folds—the structural units that form the basis of all proteins—may allow elucidation of the rules that determine the conformation adopted by each amino acid sequence.
If the enzymes are properly arranged spatially. A machine is only as good as each of its parts. Gathering the enzymes into a complex also makes it easier to regulate and move all of the enzymes together. Some terms may be used more than once. Cell-free extracts from S strain cells of S. Which are purines and which are pyrimidines?
Which bases pair with each other in double-stranded DNA? The bases in the figure are all drawn with the —NH— that attaches to the sugar at the bottom of the structure. How many base pairs per turn does a DNA helix have? Given the sequence of one strand of a DNA helix: Does the Tm increase or decrease as the length of DNA increases? Watson and Crick wrote. Is it better to cut the length of DNA in half so each person has a shorter length.
In principle what would be the minimum number of consecutive nucleotides necessary to correspond to a single amino acid to produce a workable genetic code? Assume that each amino acid is encoded by the same number of nucleotides. Do you think Tm is a constant or is dependent on other small molecules in the solution?
Do you think high salt concentrations increase. The temperature at which half of the duplex DNA molecules are intact and half have melted is defined as the Tm. Under standard conditions. You and a friend want to split a double-stranded DNA molecule so you each have half.
In the original publication describing the discovery of the structure of DNA. The human genome is comprised of 23 pairs of chromosomes found in nearly every cell in the body. Answer the quantitative questions below by choosing one of the numbers in the following list: Consider two different species of yeast that have similar genome size. What is the main function of the centromere? How many telomeres per cell?
What is their main function? How many replication origins per cell? Is it likely that they contain a similar number of genes? A similar number of chromosomes? The Structure of Eucaryotic Chromosomes For each of the following sentences.
The structure of the DNA-protein complex. In different parts of the same chromosome. Between different cells of the same organism. In different members of a pair of homologous chromosomes. What is the simplest possible explanation for this phenomenon? In different stages of the cell cycle. Do you think these two sets of histones will be the same? The lanes are as follows: Each vertical column.
Why is ATPase activity needed? Imagine that you have developed a method to isolate all of the histones bound to a single human chromosome. DNA spots near the top of the figure represent DNA molecules that are longer than those near the bottom.
After removing all the proteins. You then examine histones from the inactive X chromosome in an individual female and compare them to histones from her active X chromosome.
He grew half of his cells in the presence of glucose and the other half in the presence of galactose. He treated the DNA briefly with a low concentration of M-nuclease. A characteristic component of chromatin remodeling complexes is a powerful ATPase. Darker spots contain more DNA than fainter spots.
Your friend is working in a lab to study how cells adapt to growth on different carbon sources. Then he harvested the cells and isolated their DNA using a gentle procedure that leaves nucleosomes and some higher order chromatin structures intact. This was not observed in the other lanes.
Can you propose what it is and how it arose? What are the spots with longer lengths? Why is there a ladder of spots? Notice the faint spots and extensive smearing in lane 3. What about Salty? The lowest spot as observed in lanes 2.
What probably happened to the DNA to change the pattern between lanes 2 and 3? What kinds of enzymes might have been involved in changing the chromatin structure from lane 2 into lane 3? Do you think that gene expression of Sweetie is higher. Each fraction was then mixed with R strain cells of S. Answers Cell-free extracts from S strain cells of S. Only the fraction containing DNA was able to transform the R strain cells to pathogenic or S strain cells that could kill mice.
Its ability to change these into cells with pathogenic properties resembling the S strain cells was tested by injecting the mixture into mice. Figure A Cytosine pairs with guanine and adenine with thymine.
Choice c is incorrect. Since the sequence of nucleotides in the DNAs from different species varies considerably. There are approximately 10 base pairs per turn when the DNA has the standard conformation. This double-stranded DNA molecule has the same sequence whether read forward or backward.
Choice e is incorrect because all genomic DNA is double stranded except in some viruses. Yet since A must always pair with T and G with C. As there are 20 amino acids used in proteins. The thermal energy required for melting depends on how many hydrogen bonds between the strands must be broken. In eucaryotic chromosomes. Each G-C base pair contributes three hydrogen bonds.
A code of three consecutive nucleotides has 64 43 different members and thus can easily accommodate the 20 amino acids plus a signal to stop protein synthesis. DNA is complexed with proteins to form chromatin. The paternal and maternal copies of human Chromosome 1 are homologous. Watson and Crick meant that the complementary base pairing of the strands allows a single strand to contain all of the information necessary to direct the synthesis of a new complementary strand.
Fluorescent molecules can be used to paint a chromosome by virtue of DNA hybridization. It is better to separate the stands and each take a single strand. High salt concentrations are more effective at shielding the two negatively charged phosphate-sugar backbones in the double helix from each other.
Nucleosomes are present in eucaryotic chromosomes. There are 92 telomeres per cell. A similar genome size indicates relatively little about the number of genes and virtually nothing about the number of chromosomes. This is generally true in most organisms. The centromeres play a key role in the distribution of chromosomes to daughter cells during mitosis. There are far more than replication origins in a human cell. Telomeres serve to protect the ends of chromosomes and to enable complete replication of the DNA of each chromosome all the way to its tips.
A nucleosome core particle contains a core of histone with DNA wrapped around it approximately twice. Nucleosome formation compacts DNA into approximately one-third of its original length. Nucleosomes are aided in their formation by the high proportion of basic amino acids in histone proteins. This is accomplished by binding to proteins that help package the DNA in an orderly manner so it can fit in the small space delimited by the nuclear envelope.
A nucleosome contains two molecules each of histones H2A and H2B. There are 46 centromeres per cell. Yet Sc has genes packaged into 16 chromosomes and Sp has genes in 3 chromosomes. A gene is a segment of DNA that stores the information required to specify the particular sequence found in a protein or. Genes that are being transcribed are thought to be packaged in a less condensed type of euchromatin. The 30 nm chromatin fiber is further compacted by the formation of loops that emanate from a central axis.
But even within the same interphase chromosome. If the sectored coloring is due to X-chromosome inactivation. Because different cell types in the same organism are expressing different genes. Once inactivated. It may take extra time to remodel the chromatin to make it more accessible to the proteins required to initiate and carry out DNA replication.
The fact that sectoring results from X-chromosome inactivation cannot by itself distinguish between the possibilities that coat color is determined by a single gene or by multiple genes choice e is false.
Female calico cats receive an X chromosome from the father and another from the mother. The zigzag model describes the structure of the 30 nm fiber. Since the pattern of X-chromosome inactivation is established randomly over several days of embryonic development. The other statements are false. In the cells of female mammals. Since males have only one X chromosome that they receive from their mother and that never becomes inactivated. Chromatin differs most dramatically during the different stages of the cell cycle.
A string of nucleosomes coils up with the help of histone H1 to form the more compact structure of the 30 nm fiber. Nucleosome core particles are separated from each other by stretches of linker DNA.
In support of this proposal. For DNA to be accessible—for transcription or some other process—noncovalent bonds must be broken in order to change interactions between DNA and its packaging proteins. Perhaps the Sweetie gene contains instructions for a protein that is required for cells to metabolize galactose but not glucose.
This energetically unfavorable feat can be accomplished by harnessing the energy released by hydrolysis of ATP. Perhaps the wrapping of DNA within the nucleosomes has been loosened considerably as in Figure A nucleosome is held together by a large number of weak. Based on the information presented in Chapter 5. As the chromatin appears to have been loosened near Sweetie. The ladder of bands with longer lengths probably corresponds to stretches of DNA associated with increasing numbers of nucleosomes 1.
This change in the nucleosomes must be specific to the Sweetie gene as it is not seen at the Salty gene or throughout the genome. This interpretation must mean that the M-nuclease digestion did not go to completion. Based on the ability of M-nuclease to cut anywhere near Sweetie after growth in galactose. The main candidates for enzymes that catalyzed the nucleosome alterations near Sweetie are chromatin remodeling complexes and enzymes that covalently modify histone tails with methyl.
On a DNA strand that is being used as a template. On a DNA strand that is being synthesized. Indicate where the most recent DNA synthesis has occurred use S. Number the Okazaki fragments on each strand 1. Indicate the direction of movement of the replication forks with arrows. Indicate where the origin of replication was located use O. Label the leading-strand template and the lagging-strand template of the righthand fork [R] as X and Y.
Explain your answer with a diagram illustrating sequential snapshots of the meeting of two adjacent replication forks. Indicate by arrows the direction in which the newly made DNA strands indicated by dark lines were synthesized.
When bidirectional replication forks from adjacent origins meet. You are studying a strain of bacteria that carries a temperature-sensitive mutation in one of the genes required for DNA replication. The bacteria grow normally at the lower temperature. When the strands of this DNA are separated by heating.
When you analyze the remains of the bacterial cells grown at the higher temperature you find evidence of partly replicated DNA. Which of the following properties do you expect the mutant polymerase to have? Which of the proteins listed below are most likely to be impaired in these mutant bacteria?
Next to the proteins listed below. Explain your answers. Many of the mice in her litter are deformed. DNA Repair A pregnant mouse is exposed to high levels of a chemical. Where do you think the block to replication arises? Choose the protein or protein complex below that is most likely responsible for the failure to replicate bacterial DNA. An important step in the conversion of a normal cell into a cancer cell.
Give an explanation for your answer. Which TWO of the following statements could explain these results? Explain why telomerase might be necessary for the ability of cancer cells to divide over and over again. What proportion of its progeny would you expect to contain the mutation after three more rounds of DNA replication and cell division? Which one of the following is the most likely explanation for this phenomenon? After this original bacterium divides once. Imagine this error was not corrected and has no effect on the ability of the progeny to grow and reproduce.
The results are summarized in Figure Q You notice that the coding sequence that actually directs the sequence of amino acids in the enzyme is very similar in the two organisms but that the surrounding sequences vary quite a bit. You test each mutant for its hypersensitivity to three DNA-damaging agents: What is the most likely explanation for this? What aspect of repair is most likely to be affected in the other mutants? You are examining the DNA sequences that code for the enzyme phosphofructokinase in skinks and Komodo dragons.
Viruses probably evolved from intracellular mobile genetic elements. Site-specific recombination can repair sites of damaged DNA. Is it identical to the chromosome 3 that she received from her mother her maternal chromosome or identical to the chromosome 3 she received from her father her paternal chromosome or neither? What does this indicate about your resemblance to your grandfather and grandmother?
Genes that contain instructions for making motor proteins are called mobile genetic elements. Starting with the representation in Figure Q of the double stranded maternal and paternal chromosomes found in your mother.
During meiosis. Homologous recombination results in the accumulation of mobile genetic elements. If there is not at least one occurrence of crossing-over within each pair of homologous chromosomes during meiosis. Mobile genetic elements comprise nearly half of the human genome.
Which of the following statements about them are TRUE? Reverse transcriptase is used only by viruses. Reverse transcriptase uses ribonucleotides. Why might these mobile genetic elements have evolved this strategy? You may choose more than one option.
Choice a is incorrect because the Drosophila genome is bigger than the E. If the strands were parallel.
Answers Choice c is the answer. Although choices d and e are correct statements. Bacteria have one origin of replication and Drosophila has many. Choices c and e are technically correct statements. Consider a single replication origin: The fork moving in one direction synthesizes a daughter strand continuously as part of leading-strand synthesis.
Choice a is unlikely because it postulates an entirely new function for the defective polymerase. Choices c and e are true but are not the reason DNA polymerase requires a primer. Figure A d Each newly synthesized strand in a daughter duplex was synthesized by a mixture of continuous and discontinuous DNA synthesis from multiple origins.
Choices a and d are false statements. See Figure A for an illustration of the meeting of two adjacent replication forks. Choice b is incorrect because if the rate of polymerization changes.
An Exo— polymerase will be unable to proofread and thus will hydrolyze fewer nucleotides than one that can proofread. Like other RNA polymerases. The length of the primer does not correlate with the length of the DNA strand that is synthesized. The nucleotide fragments that accumulate in the mutant are likely to be Okazaki fragments. Only one or two molecules of each of the other proteins or protein complexes are required at each replication fork.
The leading strand requires primase to initiate DNA synthesis. Then enzymes involved in repair synthesis and ligation join the newly synthesized stretches of DNA to make a continuous strand containing only DNA. These sequences differ between bacteria and yeast. DNA from all organisms is chemically identical except for the sequence of nucleotides. The proteins listed in choices a through d can act on any DNA regardless of its sequence.
These proteins are required to remove the RNA primer. With each round of DNA replication. Choice b is the answer. The proportion of progeny containing the mutation will. Neither can choice c. Choice c is incorrect because mismatch repair repairs the newly replicated leading strand or the lagging strand to match its template strand.
So at this stage. Choice a is incorrect. So one daughter cell of that cell division will carry a completely normal DNA molecule. Without telomeres capping the chromosome ends.
At the next round of DNA replication and cell division. Cells whose DNA lacks telomeres will stop dividing or die. If the original mouse were defective in DNA repair choice e. Subsequent cell divisions of these mutant bacteria will give rise only to mutant bacteria.
When the two strands are crosslinked and thus both strands are damaged. Choice e describes the way that most cancers arise in people late in life. A defect in DNA synthesis or nucleotide biosynthesis would likely be lethal choices b and d are incorrect. Dracula and Mole are likely to be defective in the recognition or excision of thymidine dimers.
These nicks are sealed soon after replication. The other mutants are specific for a particular type of damage. The repair pathways for all three kinds of damage are similar in the later steps.
In most eucaryotic cells. When mismatch repair occurs at other times. Faust is likely to be defective in the recognition or excision of U-G mismatched base pairs. Self-Destruct is more likely than the other mutants to be defective in the DNA repair polymerase because Mr. Thus the mutations are likely to be in genes required for the first stage of repair. Self-Destruct is defective in repair of all three kinds of DNA damage.
Mutations arising in somatic cells are not inherited choice a is incorrect. Mismatches occur most often as a result of replication errors.
Site-specific recombination. Homologous recombination does sometimes aid in the repair of DNA damage caused by the excision of a mobile genetic element from the chromosome. If there were no crossing over. The copy of chromosome 3 you received from your mother is a hybrid of the ones she received from her mother and her father.
Forty-five percent of the human genome is comprised of mobile genetic elements. Due to extensive crossing over. Site-specific recombination is used mostly in the mobilization of mobile genetic elements. Since skinks and Komodo dragons share a common lizard ancestor. Choice e is true. Mutations—whether they arise by mistakes in replication or by damage to the DNA that remains unrepaired—tend to hit the DNA fairly randomly choices a and b are false.
Figure A C. Cellular motor proteins are completely unrelated to mobile genetic elements. The right answers include any chromosome in which a portion matches the information from the paternal chromosome and the remainder matches the information from the maternal chromosome.
Mutations in noncoding sequences are more likely to have no effect on the functioning of the organism and thus frequently get passed along to progeny.
A good guess for how viruses evolved is that some mobile genetic elements acquired genes encoding coat proteins and other proteins required for packaging and cellular escape of the nucleic acids of mobile genetic elements. Homologous recombination. Mobile genetic elements often have no replication origin. Please check your email for instructions on resetting your password. If you do not receive an email within 10 minutes, your email address may not be registered, and you may need to create a new Wiley Online Library account.
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