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E SSE N TIA L C E LL B IO LO G Y third ed itio n ESSENTIAL CELL BIOLOGY third edition ECB3 interactive DVD-ROM inside Alberts Bray Hopkin Johnson. ESSENTIAL third edition CELL BIOLOGY Alberts Bray Hopkin Johnson Lewis Raff .. After registering for This PDF overviews the contents of the DVD and. EDITION FOURTH FOURTH EDITION ESSENTIAL CELL BIOLOGY CELL BIOLOGY cell biology / Bruce Alberts [and seven others]. an informa business, Third . The “References” PDF document is available on both the instructor and.
A string of nucleosomes coils up with the help of histone H1 to form the more compact structure of the 30 nm fiber. On heating. How would you expect the mutant protozoan to behave? An Exo— polymerase will be unable to proofread and thus will hydrolyze fewer nucleotides than one that can proofread. Under standard conditions equal concentrations of X and Y. The reasoning is as follows.
Mitochondria are found in aerobic eucaryotes. Mitochondria take in oxygen and release carbon dioxide. Mitochondria are thought to be derived from aerobic bacteria. Mitochondria are enclosed by two membranes. Despite the diversity. Transparency is increased by slicing them into thin sections. Answers b Cells can be very diverse: The wavelength of visible light.
Although viruses contain the same type of molecules as cells. Most tissues are not transparent enough to be examined directly in a light microscope. For example. The reasoning is as follows. On the other hand. In all the cells of the offspring. But all of the chloroplasts in the offspring and thus all of the chloroplast DNA must derive from those in the female egg cell. Genetic information is only carried in DNA. According to the theory of mitochondrial origin outlined in this chapter.
By definition. Another organelle found in essentially all eucaryotic cells is the mitochondrion. If we were to strip away the plasma membrane from a eucaryotic cell and remove all of its membrane-enclosed organelles. Golgi apparatus—2 Endoplasmic reticulum—4 Bacteria have cell walls. Bacteria have no cytoskeleton.
Mice reproduce relatively rapidly and the extensive scientific community that works on mice have developed techniques to facilitate genetic manipulations. Chloroplasts b Mice are likely to provide the best model system. Mice have teeth and have long been used as a model organism. Like plant cells. Humans are not a model system. The ancestral cell therefore possessed sets of proteins to carry out these essential functions.
Genes in different species that trace their descent back to a common ancestral gene in this way that is. The genomic information is encoded in nucleic acids. This means that their individual genes are also descended from common ancestral genes. Many of the essential challenges facing modern-day cells are the same as those facing the ancestral cell.
These compounds are not easily created in the absence of life. Almost any mutation altering the amino acid sequence of such a protein will be harmful and will be eliminated by natural selection. While these proteins often show some speciesspecific diversification. Because the basic biochemical characteristics are retained.
As a result. A particular set of twenty amino acids is used to make protein. For example: Solutions to many of the essential challenges that face a cell such as the synthesis of proteins. Phospholipids are used to create cell membranes. Nucleic acids. If one were to discover a new life form that did not contain these compounds.
Temperaturesensitive mutations permit the organisms to be propagated at the permissive temperature where the proteins function normally and allow the scientist to study the consequences of a lack of the essential gene function at the restrictive temperature where the protein is defective. A temperature-sensitive mutant usually has a mutation in a gene that results in the production of a protein that does not function properly at a certain temperature the restrictive temperature.
In order to study an organism. If a scientist is interested in an essential process. At the permissive temperature. How does it differ in structure from nonradioactive carbon? How does this difference affect its chemical behavior?
What calculation can you show your friend to convince him you are right? What is its valence? What is the concentration. What are its atomic number and atomic weight? How many electrons does it have? Carbon with an atomic weight of 14 is radioactive. How many molecules are there in 1 mole of glucose? Once you learn that each human contains more bacterial cells in the digestive system than human cells.
Use Figure Q if necessary. Figure Q a b c d e Hydrogen and hydrogen Magnesium and chlorine Carbon and oxygen Sulfur and hydrogen Carbon and chlorine If one atom participating in the bond has a stronger affinity for the electron. In what scientific units is the strength of a chemical bond usually expressed? Not all words or phrases will be used.
On a sketch of a single water molecule. What is the pH of pure water? What concentration of hydronium ions does a solution of pH 8 contain? Complete the following reaction: Sketch three different ways three water molecules could be held together by hydrogen bonding. How many hydrogen bonds can a hydrogen atom in a water molecule form?
How many hydrogen bonds can the oxygen atom in a water molecule form? Will the reaction in C occur more readily be driven to the right if the pH of the solution is high? Aldehyde 3. Carboxyl 5. To answer this question you measure enzyme activity over a range of pH. Which form of histidine is necessary for the active enzyme? Figure Q Molecules in Cells Match the chemical groups shown in the first list with their names selected from the second list.
Depending on the pH of its environment. It is not clear whether this histidine is required in its protonated or unprotonated state. Hydroxyl Phosphate 4. List 1 A. Amino 2. Methyl 7.
Carbonyl ketone 6. Ester 9. Consider an enzyme with a histidine side chain that is known to play an important role in the function of the enzyme. Amido 8. How many carbon atoms does the molecule represented in Figure Q have? How many hydrogen atoms? What type of molecule is it? Figure Q Steroids Which of the following are examples of isomers? DNA C. Fatty acids 7. Amino acids 2. RNA D. Pyrophosphates 5.
Ribonucleotides 6. Deoxyribonucleotides 3. Lipids List 2 1. Sugars 8. Proteins E. Polysaccharides B. Aldehydes 4. Write out the sequence of amino acids in the following peptide using the full names of the amino acids.
Phosphate Nonpolar head group Glycerol Polar head group Saturated fatty acid Acetic acid Sugar Hydrophobic region Hydrophilic region Nonsaturated fatty acid Figure Q Phospholipids can form bilayer membranes because they are a hydrophobic. According to the conventional way of writing the sequence of a peptide or a protein.
Write the same sequence using the single letter code for amino acids. The bond shown in the box in the acetylated lysine side chain is most like a n Figure Q a b c d f ester. After having read this chapter. Forming hydrogen bonds to aid solubility in water C. One of your classmates says the two proteins must be identical. From the list provided. A protein Z binds to DNA through noncovalent ionic interactions involving lysines.
Packing tightly the hydrophobic interior core of a globular protein acidic basic nonpolar uncharged polar DNA is negatively charged at physiological pH. What will be the effect of acetylation of the lysine side chains see Figure Q in protein Z on the strength of this binding? Binding to another water-soluble protein D. Such noncovalent bonds are also critical for interactions with other proteins and cellular molecules. Forming ionic bonds with negatively charged DNA B.
Could any of the following changes in mutant protein X explain your results? If so. You have isolated a strain of the microorganism that produces a mutant form of protein X. Carbon14 has two additional neutrons in its nucleus. The atomic weight. Magnesium has a valence of two and chlorine has a valence of one. The number of electrons.
All ions in this salt will have full outermost electron shells. The first shell can accommodate two electrons and the second shell. Carbon therefore has a valence of four because it needs to gain four additional electrons or would have to give up four electrons to obtain a full outermost shell. Carbon is most stable when it shares four additional electrons with other atoms including other carbon atoms by forming four covalent bonds.
A mole of a substance has a mass equivalent to its molecular weight expressed in grams. Answers 16 proteins and 19 neutrons A. As its electrons determine the chemical properties of an atom.
The valence is the minimum number of electrons that must be lost or gained to fill the outer shell of electrons. If you multiply the number of people on Earth by the number of cells in the human body. These polar covalent bonds should not be confused with the weaker noncovalent bonds that are critical for the three-dimensional structure of biological molecules and for interactions between these molecules.
Choice a is false because hydrogen bonds are not covalent. Hydrogen bonds are critical for maintaining the conformation. See Figure AB. Choice d is false because many molecules besides water can form hydrogen bonds and do so regardless of whether or not water is present.
Choice e is the answer. These covalent bonds have a characteristic bond length and become stronger and more rigid when two electrons are shared in a double bond. Choice c is essentially another way of stating choice b and thus is false. See Figure AA. Choice b is false because the nonpolar-CH groups on hydrocarbons cannot form good hydrogen bonds.
Equal sharing of electrons yields a nonpolar covalent bond. Choices b. In methanol one of the hydrogens of a water molecule has been replaced by a nonpolar methyl group. Salts are therefore less soluble in methanol. Adenine and guanine are bases containing different numbers of nitrogen and oxygen atoms. E—1 A—7. Therefore the rightward reaction. The enzyme is active only at low. If the pH is high. As methanol makes fewer hydrogen bonds. Glucose and galactose are both six-carbon sugars and thus both have the formula C6H12O6.
Most synthetic reactions in cells are catalyzed by enzymes thus choice e is correct. They are thus isomers of each other.
The ability of water to dissolve salts is a direct consequence of its ability to make ionic interactions. E—6 e Choice b is the answer. Methanol will form fewer hydrogen bonds thus choice a is false and make fewer ionic interactions than water does. Glycogen and cellulose are different polymers of glucose. A condensation reaction releases a water molecule when forming polymers like polysaccharide energy reserves from monomeric units like simple sugars.
Alanine and glycine are amino acids with quite different side chains. It is through these two groups that they form peptide bonds. All the other statements about ATP are true. Figure A 1—D.
Choice d is false because only L-amino acids are found in proteins. But synthesis of DNA requires the deoxyribose form of the nucleotide. They also differ in one of the pyrimidine bases used. N-terminal is proline. There are 20 common amino acids choice a is false. Each amino acid forms only two covalent bonds with other amino acids.
Like a peptide bond. All the other features are the same. RNA contains the pyrimidine uracil. As their name implies. Noncovalent ionic interactions such as those that hold two proteins together are most likely to be disrupted by salt.
Choices b and d are examples of covalent bonds. The male will. If protein X is completely unfolded at elevated temperatures it will not be able to bind to protein Y. Choice e is incorrect. Because acetylation neutralizes the positive charge thus choice d is incorrect. Choice f would result in a strain that could not turn pink at any temperature. Choice a would produce a protein X that would bind to protein Y more tightly than the normal protein.
In contrast. Unmodified lysine side chains are positively charged and hence attractive to the negatively charged DNA thus choice c is incorrect. Thermal motion is one of the forces that can disrupt the weak noncovalent bonds responsible for holding X and Y together. The activities of these two proteins will definitely be different.
The reversed protein chains cannot make the same noncovalent interactions during folding and thus will not adopt the same 3-D structure as the original protein.
If a covalent bond was made choice c.
Figure A A. Weak noncovalent bonds are also responsible for folding X into the proper 3-D structure. If protein X makes fewer hydrogen bonds to protein Y. It is unlikely that the reverse chain will fold into any well-defined. See Figure A Why or why not? What happens to the atoms contained in the food and the energy stored in the chemical bonds of food molecules?
The oxidation state of an atom influences its diameter. How does it differ? For each of the pairs A—D in Figure Q If oxidation occurs in a reaction. The hydrogenation of an unsaturated fatty acid to a saturated fatty acid.
Complete the equation for respiration: Draw a new curve on the graph to indicate how an enzyme that converts S to P will change the energetics of the reaction. Figure Q Energetically favorable reactions are those that a decrease the entropy of a system.
Indicate the following on the graph. The activation energy for the reaction B. The free energy change for the reaction C. The top diagram shows the original energy profile for the reaction. For each description of an altered reaction below. Figure Q A. Describe two ways that this may be accomplished. Is the standard free energy change of this reaction positive or negative?
At equilibrium. What is the value of the standard free energy? Refer to Table in the textbook or use a calculator. What is the value of Keq for this reaction? Is conversion of X to Y favorable? Will it happen quickly? Using the equations below and your new knowledge.
S and I. The binding reactions are described by the following equations and values: Draw lines near the composite reaction curve to indicate the free energy change for each of the three reactions: Is the composite reaction favorable or unfavorable?
Which of the following molecules would be expected to accumulate in large amounts? The rate of diffusion determines the association rate of all binding reactions. If you plot reaction rate against temperature. The diffusion rate for small molecules is much slower in the dense gel of the cytoplasm than in pure water.
You are measuring the effect of temperature on the rate of an enzyme-catalyzed reaction. The diffusion rate is faster for large macromolecules than for small molecules. Explain why temperature has this effect. The number of reactions per second that an enzyme catalyzes is independent of diffusion.
Diffusion causes a molecule to move along a fairly straight path. A graph was made of the initial reaction rate v plotted against the concentration of X [X] Figure Q The turnover number of enzyme A is greater than the turnover number of enzyme B. 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?
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Molecular biology. Alberts, Bruce. E78 Peter Walter Narrated by: Julie Theriot Producer: Michael Morales Interface Design: He is the editor-in-chief of Science magazine. For 12 years he served as President of the U. National Academy of Sciences Dennis Bray received his Ph. In he was awarded the Microsoft European Science Award. Karen Hopkin received her Ph.
Alexander Johnson received his Ph. Julian Lewis received his D. Martin Raff received his M.