Cell Biology Final
mitochondria
Membrane-enclosed organelle that carries out oxidative phosphorylation and produces most ATP
lysosomes
Membrane-enclosed organelle that contains digestive enzymes active at a low pH to breaks down worn-out proteins, organelles, waste materials, and molecules taken up by endocytosis
How would the curve in the figure change if centrosomes were added at the outset?
centrosomes introduce nucleation sites that eliminate the lag phase A The rate of microtubule growth and the equilibrium level of free tubulin remain unchanged
that the citric acid cycle is the reverse of the normal carbon-fixation cycle? (T/F)
false, the chemical reactions are different even though the net effect is the same
How many electrons can be accommodated in the first, second, and third electron shells of an atom?
from the nucleus outward: 2, 8, 8
nucleus
prominent, rounded structure that contains the DNA of a eukaryotic cell; dense, positively charged center of an atom.
chloroplasts
site of photosynthesis in plant cells
peroxisomes
degrade lipids and destroy toxins.
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, and do not have a cell wall. Cytoskeletal elements provide mechanical strength and shape (2) Animal cells have a nucleus that is shaped and held in place by intermediate filaments; the nuclear lamins support and shape the nuclear membrane, and intermediate filaments surrounds the nucleus and spans the cytosol. (3) Animal cells move by a change in cell shape by Actin myosin (4) Animal cells have a much larger genome that is fragmented into many chromosomes. which require microtubules to assist in distribution during cell division (5) Animal cells have internal organelles whose location is dependent on motor proteins that move them along microtubules.
rarely, both sister chromatids of a replicated chromosome end up in one daughter cell. how might this happen? what could be the consequences of such a mitotic error?
(1) If the microtubules or their connections with a kinetochore were to break during anaphase (2) If microtubules from the same spindle pole attached to both kinetochores (3) If the cohesins that link sister chromatids were not degraded (4) If a duplicated chromosome never engaged microtubules and was left out of the spindle Some of these errors should activate a checkpoint mechanism to delay anaphase until fixed. if not fixed one daughter cell would contain only one copy of all the genes carried on that chromosome and the other daughter cell would contain three copies altering gene dosage causing nRNA and protein amount problems which is often detrimental to the cell. And if the single copy contains a defective gene is is not counteracted by a good gene
At many steps in the electron- transport chain, Fe ions are used as part of heme or FeS clusters to bind the electrons in transit. Why do these functional groups that carry out the chemistry of electron transfer need to be bound to proteins? Provide several different reasons why this is necessary.
(1) amino acid side chains provide the chemical environment for the redox potential of each Fe ion so electrons can be passed orderly giving up their energy in small steps and becoming more firmly bound as they proceed. (2) the proteins position the Fe ions so that the electrons can move efficiently between them. (3) proteins prevent electrons from skipping an intermediate step to channel the electron flow along a defined path. (4) proteins couple electron movement to the pumping of protons across the membrane, harnessing the energy and storing it in a proton gradient that is then used for ATP production.
what would happen in scenarios (1), (2), and (3) if the cells, in addition, produced normal tryptophan repressor protein from a second, normal gene?
(1) and (2), normal tryptophan repressor would completely restore the regulation of the tryptophan biosynthesis enzymes. (3) expression of the normal protein would have no effect because the tryptophan operator would remain occupied by the mutant protein
Are the formulas in Figure Q2-21 correct or incorrect? Explain your answer in each case.
(A) and (B) are both correct formulas of the amino acid phenylalanine. (B) is phenylalanine shown in the ionized form that exists in an aqueous solution, where the basic amino group is protonated and the acidic carboxylic group is deprotonated. (C) Incorrect. This structure of a peptide bond is missing a hydrogen atom bound to the nitrogen. (D) Incorrect. This formula of an adenine base features one double bond too many, creating a five-valent carbon atom and a four-valent nitrogen atom. (E) Incorrect. In this formula of a nucleoside triphosphate, there should be two additional oxygen atoms, one between each of the phosphorus atoms. (F) This is the correct formula of ethanol. (G) Incorrect. Water does not hydrogen-bond to hydrogens bonded to carbon. The lack of the capacity to hydrogen- bond makes hydrocarbon chains hydrophobic, i.e., water-hating. (H) Incorrect. Na and Cl form an ionic bond, but a covalent bond is drawn. (I) Incorrect. The oxygen atom attracts electrons more than the carbon atom; the polarity of the two bonds should therefore be reversed. (J) This structure of glucose is correct. (K) Almost correct. It is more accurate to show that only one hydrogen is lost from the -NH2 group, and the -OH group is lost from the -COOH group.
In which of the following reactions does the red atom undergo an oxidation? a. na → na+ (na atom → na+ ion) b. Cl → Cl- (Cl atom → Cl- ion) c. Ch3Ch2oh → Ch3Cho (ethanol → acetaldehyde) d. Ch3Cho → Ch3Coo- (acetaldehyde → acetic acid) e. Ch2=Ch2 → Ch3Ch3 (ethene → ethane)
(A) oxidation (B) reduction (C) largely unchanged but the neighboring carbon atom loses a hydrogen atom being oxidized (D) oxidized (E) reduced
complete the following sentence accurately, explaining your reason for accepting or rejecting each of the four phrases. the role of calcium in muscle contraction is: a. to detach myosin heads from actin. b. to spread the action potential from the plasma membrane to the contractile machinery. c. to bind to troponin, cause it to move tropomyosin, and thereby expose actin filaments to myosin heads. d. to maintain the structure of the myosin filament.
(B) and (C) The direct result of the action potential in the plasma membrane is the release of Ca2+ into the cytosol from the sarcoplasmic reticulum triggering muscle cell contraction Calcium ions bind to troponin, causing tropomyosin to move to expose myosin- binding sites on the actin filaments. (A) and (D) are wrong because Ca2+ has no effect on the detachment of the myosin head from actin or in maintaining the structure of the myosin filament.
the Ras protein functions as a molecular switch that is set to its "on" state by other proteins that cause it to expel its bound GdP and bind GtP. A GtPase-activating protein helps reset the switch to the "off" state by inducing Ras to hydrolyze its bound GtP to GdP much more rapidly than it would without this encouragement. thus, Ras works like a light switch that one person turns on and another turns off. you are given a mutant cell that lacks the GtPase-activating protein. What abnormalities would you expect to find in the way in which Ras activity responds to extracellular signals?
1. high background level of Ras activity, because Ras cannot be turned off efficiently. 2. Because some Ras molecules are already GTP-bound, Ras activity in response to an extracellular signal would be greater than normal but liable to saturate 3. response time is reduced, because the elevated background of GTP-bound Ras 4. increase in Ras activity would be prolonged compared to the response in normal cells.
if the cells of your body oxidize 1 mole of glucose, by how much would the temperature of your body (assume that your body consists of 75 kg of water) increase if the heat were not dissipated into the environment? [recall that a kilocalorie (kcal) is defined as that amount of energy that heats 1 kg of water by 1°C.]
322 kcal would be released as heat; heating the body by 4.3°C
What is the overall efficiency of ATP production from glucose?
53% (actual ATP/possible ATP)
how many ATP molecules could maximally be generated from one molecule of glucose, if the complete oxidation of 1 mole of glucose to Co2 and h2o yields 686 kcal of free energy and the useful chemical energy available in the high-energy phosphate bond of 1 mole of ATP is 12 kcal?
57 ATP molecules
The nucleotide sequence of one DNA strand of a DNA double helix is 5'-GGATTTTTGTCCACAATCA-3'. What is the sequence of the complementary strand?
5ʹ-TGATTGTGGACAAAAATCC-3ʹ.
What would the consequences be if the cells of your body could convert the energy in food substances with only 20% efficiency? Would your body as it is presently constructed—work just fine, overheat, or freeze?
80% of the available energy would be released as heat; The heat production would be more than 1.7-fold higher than normal and would overheat.
in the DNA of certain bacterial cells, 13% of the nucleotides are adenine. What are the percentages of the other nucleotides?
A - 13% T - 13% G - 37% C - 37%
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.
the budding of clathrin-coated vesicles from eukaryotic plasma membrane fragments can be observed when adaptins, clathrin, and dynamin-gtp are added to the membrane preparation. What would you observe if you omitted (a) adaptins, (b) clathrin, or (c) dynamin? (d) What would you observe if the plasma membrane fragments were from a prokaryotic cell?
A. Clathrin coats cannot assemble in the absence of adaptins that link the clathrin to the membrane. clathrin cages can assemble in solution, but are empty lacking proteins and a membrane B. Without clathrin, adaptins still bind to receptors in the membrane, but no clathrin coat can form and thus no clathrin-coated pits or vesicles are produced. C. Deeply invaginated clathrin-coated pits form on the membrane, but they do not pinch off to form closed vesicles D. Prokaryotic cells do not perform endocytosis and does not contain any receptors with cytosolic tails to mediate adaptin binding so no clatherin binding or coat
The curve shown in figure 3-24 is described by the Michaelis-Menten equation: rate (v) = Vmax [s]/([s] + KM) Can you convince yourself that the features qualitatively described in the text are accurately represented by this equation? in particular, how can the equation be simplified when the substrate concentration [s] is in one of the following ranges: (a) [s] is much smaller than the KM, (b) [s] equals the KM, and (C) [s] is much larger than the KM?
A. When [S] << KM, ([S] + KM) approaches KM so rate = Vmax[S]/ KM and the rate is proportional to [S]. B. When [S] = KM, [S]/([S] + KM) equals 1⁄2 so the reaction rate is half of the maximal rate Vmax. C. If [S] >> KM, ([S] + KM) approaches [S] so [S]/([S] + KM) equals 1 and the rate is at maximal rate Vmax.
The neurotransmitter acetylcholine is made in the cytosol and then transported into synaptic vesicles, where its concentration is more than 100-fold higher than in the cytosol. When synaptic vesicles are isolated from neurons, they can take up additional acetylcholine added to the solution in which they are suspended, but only when ATp is present. na+ ions are not required for the uptake, but, curiously, raising the ph of the solution in which the synaptic vesicles are suspended increases the rate of uptake. furthermore, transport is inhibited when drugs are added that make the membrane permeable to h+ ions. suggest a mechanism that is consistent with all of these observations.
Acetylcholine transport relies on an H+-acetylcholine antiport in the vesicle membrane. The H+ gradient is generated by an ATP-driven H+ pump in the vesicle membrane. Raising the pH decreases the H+ concentration increases the outward gradient causing an enhanced rate of acetylcholine uptake.
What are the similarities and differences between the reactions that lead to the activation of G proteins and the reactions that lead to the activation of Ras?
Activation in both cases depends on proteins that catalyze GDP-GTP exchange. G: activated GPCRs function directly Ras: enzyme-linked receptors assemble multiple signaling proteins into a signaling complex when the receptors are phosphorylated; one of these is an adaptor protein that recruits a guanine nucleotide exchange factor
what do you suppose happens in mutant cells that A. cannot degrade m-cyclin? b. always express high levels of p21? c. cannot phosphorylate rb?
All are unable to divide. A) enter mitosis but unable to exit mitosis. B) arrest permanently in G1 because the cyclin-Cdk complexes would be inactivated. C) cannot activate the transcription of genes required for cell division due to constant inhibition by unphosphorylated Rb.
Calculate the number of usable ATP molecules produced per pair of electrons transferred from NADH to oxygen, if (i) five protons are pumped across the inner mitochondrial membrane for each electron passed through the three respiratory enzyme complexes, (ii) three protons must pass through the ATP synthase for each ATP molecule that it produces from ADP and inorganic phosphate inside the mitochondrion, and (iii) one proton is used to produce the voltage gradient needed to transport each ATP molecule out of the mitochondrion to the cytosol where it is used.
An electron pair causes 10 H+ to be pumped across the membrane when passing from NADH to O2. 4 (3 for synthesis and 1 for export) H+ are needed to make each ATP. Therefore, 2.5 ATP molecules are synthesized from each NADH
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 and require many unsatured phospholipids in order to keep their membranes fluid
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. 5'-GCGGGCCAGCCCGAGTGGGTAGCCCAGG-3' 3'CGCCCGGTCGGGCTCACCCATCGGGTCC-5' 5'-ATTATAAAATATTTAGATACTATATTTACAA-3' 3'TAATATTTTATAAATCTATGATATAAATGTT-5' 5'-AGAGCTAGATCGAT-3' 3'-TCTCGATCTAGCTA-5'
As the two strands are held together by hydrogen bonds between the bases, the stability of a DNA double helix is largely dependent on the number of hydrogen bonds that can be formed. helix C (34 hydrogen bonds) would melt at the lowest temperature helix B (65 hydrogen bonds) would melt next helix A (78 hydrogen bonds) would melt last
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
B; It is composed primarily of hydrophobic amino acids, and therefore can be stably integrated into a lipid bilayer (A contains many polar amino acids and C contains many charged amino acids)
When bacteria are grown under adverse conditions, i.e., in the presence of a poison such as an antibiotic, most cells grow and proliferate slowly. But it is not uncommon that the growth rate of a bacterial culture kept in the presence of the poison is restored after a few days to that observed in its absence. suggest why this may be the case.
Bacteria continually acquire mutations in their DNA. In the population of cells exposed to the poison, one or a few cells may harbor a mutation that makes them resistant to the action of the drug. Antibiotics that are poisonous to bacteria because they bind to certain bacterial proteins, for example, would not work if the proteins have a slightly changed surface so that binding occurs more weakly or not at all. These mutant bacteria would continue dividing rapidly while their cousins are slowed down. The antibiotic- resistant bacteria would soon become the predominant species in the culture.
in his highly classified research laboratory, dr. lawrence m. is charged with the task of developing a strain of dog-sized rats to be deployed behind enemy lines. in your opinion, which of the following strategies should dr. m. pursue to increase the size of rats? A. block all apoptosis. b. block p53 function. c. overproduce growth factors, mitogens, or survival factors. d. obtain a taxi driver's license and switch careers. explain the likely consequences of each option.
C, which should result in an increase in cell numbers. but cell numbers must increase similarly to maintain balanced proportions yet different cells respond to different growth factors. The growth will occur but not to the proportion of a dog-sized rate
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. the formation of the mitotic spindle requires depolymerization (inhibited by Taxol) and polarization (blocked by colchicine) the dynamic instability of microtubules is inhibited so if they could form they would not work
if compound X resembles S and binds to the active site of the enzyme but cannot undergo the reaction catalyzed by it, what effects would you expect the addition of X to the reaction to have? compare the effects of X and of the accumulation of p.
Compound X would act as an inhibitor of the reaction and work similarly by forming an EX complex. since P has to be made before it can inhibit the reaction, it takes longer to act than X, which is present from the beginning of the reaction.
cytoplasm
Contents of a cell that are contained within its plasma membrane but contained outside the nucleus.
cytosol
Contents of the cytoplasm, excluding membrane-enclosed organelles
Which of the following reactions will occur only if coupled to a second, energetically favorable reaction? a. glucose + o2 → Co2 + h2o b. Co2 + h2o → glucose + o2 C. nucleoside triphosphates → dna d. nucleotide bases → nucleoside triphosphates e. adP + Pi → aTP
Coupled: B, D, and E In each case, higher-order structures are formed that are more complicated and have higher-energy bonds than the starting materials. Reaction A is a catabolic reaction that leads to compounds in a lower energy state and will occur spontaneously. Reaction C contain enough energy to drive DNA synthesis
Dinitrophenol (DNP) is a small molecule that renders membranes permeable to protons. In the 1940s, small amounts of this highly toxic compound were given to patients to induce weight loss. DNP was effective in melting away the pounds, especially promoting the loss of fat reserves. Can you explain how it might cause such loss? As an unpleasant side reaction, however, patients had an elevated temperature and sweated profusely during the treatment. Provide an explanation for these symptoms
DNP diminishes the proton gradient across the inner mitochondrial membrane. Cells continue to oxidize food molecules but H+ ions pumped across the membrane flow back into the mitochondria in a futile cycle. the energy of the electrons cannot drive ATP synthesis, and instead is released as heat. fat reserves are used to feed the electron- transport chain, and the whole process simply "wastes" energy as heat.
Why is the first step shown with bidirectional arrows and the second step as a unidirectional arrow?
Enzyme and substrate are in equilibrium between their free and bound states; once bound to the enzyme, a substrate molecule may either dissociate again or be converted to product. As the substrate is converted to product a reaction usually proceeds strongly in the forward direction, as indicated by the unidirectional arrow.
some membrane proteins are enzymes.
Examples include the many membrane enzymes involved in cell signaling
n-linked sugar chains are found on glycoproteins that face the cell surface, as well as on glycoproteins that face the lumen of the er, trans golgi network, and mitochondria. (T/F)
False, Mitochondria do not participate in vesicular transport, and therefore N-linked glycoproteins cannot be transported to mitochondria.
the amino acid sequence leu-His-arg-leu-asp-ala-gln- ser-lys-leu-ser-ser is a signal sequence that directs proteins to the er. (T/F)
False. ER directing signal sequences contain a core of eight or more hydrophobic amino acids
a bacterial replication fork is asymmetrical because it contains two dna polymerase molecules that are structurally distinct. (T/F)
False. Identical DNA polymerase molecules catalyze DNA synthesis; The replication fork is asymmetrical because the lagging strand is synthesized in pieces that are then stitched together.
all of the energy produced is in the form of heat. (T/F)
False. If this were the case, then the reaction would be useless for the cell. No chemical energy would be harvested in a useful form to be used for metabolic processes
lysosomes digest only substances that have been taken up by cells by endocytosis. (T/F)
False. Lysosomes also digest internal organelles by autophagy.
none of the produced energy is in the form of heat. (T/F)
False. No energy-conversion process can be 100% efficient. Because entropy must increase
many steps in the oxidation of sugar molecules involve reaction with oxygen gas. (T/F)
False. O2 is used only in the last step
okazaki fragments are removed by a nuclease that degrades rna. (T/F)
False. Only the RNA primers are removed by an RNA nuclease; Okazaki fragments are pieces of newly synthesized DNA on the lagging strand that are eventually joined together by DNA ligase.
plants are composed of prokaryotic cells. (T/F)
False. Plants are composed of eukaryotic cells that contain chloroplasts as cytoplasmic organelles. The chloroplasts are thought to be evolutionarily derived from prokaryotic cells.
protozoans are complex organisms with a set of specialized cells that form tissues, such as flagella, mouthparts, stinging darts, and leglike appendages. (T/F)
False. Protozoans are single-celled organisms and therefore do not have different tissues or cell types. They have a complex structure, however, that has highly specialized parts.
the cytosol contains membrane-enclosed organelles, such as lysosomes. (T/F)
False. The cytosol is the cytoplasm excluding all membrane-enclosed organelles.
the hereditary information of a cell is passed on by its proteins (T/F)
False. The hereditary information is encoded in the cell's DNA, which in turn specifies its proteins (via RNA).
A DNA strand has a polarity because its two ends contain different bases. (T/F)
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.
The reaction supplies the cell with essential water. (T/F)
False. produce some water but a minimal amount in comparison
Glycolipids move between different membrane-enclosed compartments during their synthesis but remain restricted to one side of the lipid bilayer.
Glycolipids are mostly restricted to the monolayer of membranes that faces away from the cytosol. Some special glycolipids, such as phosphatidylinositol are found specifically in the cytosolic monolayer.
What do the answers tell you about the reactivity of helium and the bonds that can form between sodium and chlorine?
Helium with its fully occupied electron shell is chemically unreactive. Sodium and chlorine, on the other hand, are extremely reactive and readily form Na+ and Cl- ions, which can form ionic bonds to produce NaCl (table salt).
Assuming that the histone octamer (shown in Figure 5-21) forms a cylinder 9 nm in diameter and 5 nm in height and that the human genome forms 32 million nucleosomes, what volume of the nucleus (6 µm in diameter) is occupied by histone octamers? (Volume of a cylinder is πr2h; volume of a sphere is 4/3 πr3.) What fraction of the total volume of the nucleus do the histone octamers occupy? how does this compare with the volume of the nucleus occupied by human DNA?
Histone octamers has a volume of 1.02x10^10 nm and occupy 9% of the volume of the nucleus Because the DNA also occupies about 9% of the nuclear volume, together they occupy about 18% of the volume of the nucleus.
Describe the similarities and differences between van der Waals attractions and hydrogen bonds.
Hydrogen bonds form between two specific chemical groups; one is always a hydrogen atom linked in a polar covalent bond to an oxygen or a nitrogen atom, and the other is usually a nitrogen or an oxygen atom. Van der Waals attractions are weaker and occur between any two atoms that are in close enough proximity. Both hydrogen bonds and van der Waals attractions are short-range interactions that come into play only when two molecules are already in close proximity. Both types of bonds can therefore be thought of as means of "fine-tuning" an interaction, i.e., helping to position two molecules correctly with respect to each other once they have been brought together by diffusion.
What would happen if the solution contained an analog of GtP that cannot be hydrolyzed?
If GTP is present but cannot be hydrolyzed, microtubules will continue to grow until all free tubulin subunits have been used up.
What would happen if only GdP, but no GtP, were present in the solution?
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.
A rise in the intracellular ca2+ concentration causes muscle cells to contract. in addition to an ATp- driven ca2+ pump, muscle cells that contract quickly and regularly, such as those of the heart, have an additional type of ca2+ pump - an antiport that exchanges ca2+ for extracellular na+ across the plasma membrane. The majority of the ca2+ ions that have entered the cell during contraction are rapidly pumped back out of the cell by this antiport, thus allowing the cell to relax. ouabain and digitalis are used for treating patients with heart disease because they make heart muscle cells contract more strongly. both drugs function by partially inhibiting the na+ pump in the plasma membrane of these cells. can you propose an explanation for the effects of the drugs in the patients? What will happen if too much of either drug is taken?
If the Na+ pump is partially inhibited by ouabain or digitalis, the electrochemical gradient of Na+ is less steep making the Ca2+-Na+ antiport works less efficiently reducing the Ca2+ that is removed from the cell leaving an elevated level of Ca2+ in the cytosol leading to higher Ca2+ causing a stronger and longer-lasting muscle contraction. The drugs are deadly poisons if too much is taken.
Carefully consider the result shown in figure Q5-13. each of the two colonies shown on the left is a clump of approximately 100,000 yeast cells that has grown up from a single cell, which is now somewhere in the middle of the colony. The two yeast colonies are genetically different, as shown by the chromosomal maps on the right. The yeast Ade2 gene encodes one of the enzymes required for adenine biosynthesis, and the absence of the Ade2 gene product leads to the accumulation of a red pigment. At its normal chromosome location, Ade2 is expressed in all cells. When it is positioned near the telomere, which is highly condensed, Ade2 is no longer expressed. how do you think the white sectors arise? What can you conclude about the propagation of the transcriptional state of the Ade2 gene from mother to daughter cells?
In the lower colony of Figure Q5-13, the Ade2 gene is inactivated when placed near a telomere, but apparently it can become spontaneously activated in a few cells, which then turn white. Once activated in a cell, the Ade2 gene continues to be active in the descendants of that cell, resulting in clumps of white cells in the colony. This result shows both that the inactivation of a gene positioned close to a telomere can be reversed and that this change is passed on to further generations. This change in Ade2 expression probably results from a spontaneous decondensation of the chromatin structure around the gene.
chromosomes
Long, threadlike structure composed of DNA and proteins that carries the genetic information of an organism; visible during division.
one of the functions of m-cdk is to cause a precipitous drop in m-cyclin concentration halfway through m phase. describe the consequences of this sudden decrease and suggest possible mechanisms by which it might occur.
Loss of M cyclin leads to inactivation of M-Cdk. so the cells exit from mitosis. The M cyclin is degraded by ubiquitin-dependent destruction in proteasomes, and the activation of M-Cdk leads to the activation of the APC, which ubiquitylates the cyclin, but with a substantial delay.
golgi apparatus
Membrane-enclosed organelle in eukaryotic cells that modifies the proteins and lipids made in the endoplasmic reticulum and sorts them for transport to other sites.
Unlike mitochondria, chloroplasts do not have a transporter that allows them to export ATP to the cytosol. How, then, do plant cells obtain the ATP that they need to carry out energy- requiring metabolic reactions in the cytosol?
Mitochondria supply the cell with ATP derived by oxidative phosphorylation. Glyceraldehyde 3-phosphate from chloroplasts is eventually used as a source of energy for ATP production
an animal cell, roughly cubical in shape with side length of 10 µm, uses 109 aTP molecules every minute. assume that the cell replaces this aTP by the oxidation of glucose according to the overall reaction 6o2 + C6h12o6 →6Co2 + 6h2o and that complete oxidation of each glucose molecule produces 30 aTP molecules. how much oxygen does the cell consume every minute? how long will it take before the cell has used up an amount of oxygen gas equal to its own volume?
One minute the cell will consume 3.3 × 10-16 moles of oxygen consuming 0.7% of its O2 volume and its own volume of O2 gas in 2.25 hours
How does PI 3-kinase activate the Akt kinase after activation of RtK?
PI 3-kinase is activated by activated RTK, it phosphorylates a specific inositol phospholipid in the plasma membrane. The resulting phosphorylated inositol phospholipid then recruits to the plasma membrane both Akt and another protein kinase to phosphorylate and activate Akt. A third kinase that is permanently associated with the membrane also helps activate Akt
explain the different parts of the curve (labeled a, b, and c). draw a diagram that shows the behavior of tubulin molecules in each of the three phases.
Phase A: lag phase - tubulin molecules assemble to form nucleation centers Phase B: rapid rise to a plateau as tubulin dimers add to the ends of the elongating microtubules Phase C: equilibrium is reached
The sugar layer that surrounds all cells makes cells more slippery.
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
Explain why cyclic AmP must be broken down rapidly in a cell to allow rapid signaling.
Rapid breakdown keeps the intracellular cyclic AMP concentrations low. The lower the cAMP levels are, the larger and faster the increase achieved upon activation of adenylyl cyclase, which makes new cyclic AMP.
in a clever experiment performed in 1962, a cysteine already attached to its tRNA was chemically converted to an alanine. these "hybrid" trNA molecules were then added to a cell-free translation system from which the normal cysteine-trNAs had been removed. When the resulting protein was analyzed, it was found that alanine had been inserted at every point in the polypeptide chain where cysteine was supposed to be. Discuss what this experiment tells you about the role of aminoacyl- trNA synthetases during the normal translation of the genetic code.
Ribosome does not check the amino acid that is attached to a tRNA and will "blindly" incorporate that amino acid into the position according to the match between the codon and anticodon. Synthetase enzymes that correctly match tRNAs and amino acids are responsible for correct genetic reading
lysosomes and peroxisomes are the sites of degradation of unwanted materials. (T/F)
Somewhat true. Peroxisomes and lysosomes contain enzymes that catalyze the breakdown of substances produced in the cytosol or taken up by the cell. One can argue, however, that many of these substances are degraded to generate food molecules, and as such are certainly not "unwanted."
If ATP synthase making ATP can be likened to a water-driven turbine producing electricity, what would be an appropriate analogy when it works in the opposite direction?
Such a turbine running in reverse is an electrically driven water pump to pump water against its natural flow
Which of the following amino acids would you expect to find more often near the center of a folded globular protein? Which ones would you expect to find more often exposed to the outside? explain your answers. Ser, Ser-p (a Ser residue that is phosphorylated), Leu, Lys, Gln, his, phe, Val, ile, Met, cys-S-S-cys, and Glu. Where would you expect to find the most n-terminal amino acid and the most c-terminal amino acid?
Surface: polar amino acids Ser, Ser-P, Lys, Gln, His, and Glu Interior: hydrophobic amino acids Leu, Phe, Val, Ile, and Met, oxidation of two cysteine residues to form a disulfide bond the most N-terminal amino acid and the most C-terminal amino acid each contain a charged group and hence are usually found on the protein's surface.
during movement, muscle cells require large amounts of aTP to fuel their contractile apparatus. These cells contain high levels of creatine phosphate which has a standard free-energy change (ΔG°) for hydrolysis of its phosphate bond of -10.3 kcal/mole. why is this a useful compound to store energy? Justify your answer with the information shown in figure 13- 8.
The free energy stored in the phosphate bond in creatine phosphate is larger than that of the anhydride bonds in ATP. Hydrolysis of creatine phosphate can therefore be directly coupled to the production of ATP. creatine phosphate + ADP → creatine + ATP (ΔGo = -3 kcal/mole)
a resting human hydrolyzes about 40 kg of ATP every 24 hours. The oxidation of how much glucose would produce this amount of energy?
The chemical formula of ATP is C10H12O13N5P3, and its molecular weight is 503 g/mole. A resting body hydrolyzes about 80 moles of ATP in 24 hours. This amount of energy can be produced by oxidation of 480 g glucose
Why does e appear at both ends of the equation?
The enzyme is a catalyst and is therefore liberated in an unchanged form after the reaction
using the nernst equation and the ion concentrations given in Table 12-1 (p. 385), calculate the equilibrium membrane potential of K+ and na+ that is, the membrane potential where there would be no net movement of the ion across the plasma membrane (assume that the concentration of intracellular na+ is 10 mm). What membrane potential would you predict in a resting animal cell? explain your answer. What would happen if a large number of na+ channels suddenly opened, making the membrane much more permeable to na+ than to K+? (note that because few ions need to move across the membrane to change the charge distribution across the membrane drastically, you can safely assume that the ion concentrations on either side of the membrane do not change significantly.) What would you predict would happen next if the na+ channels closed again?
The equilibrium potential of K+ is -90 mV and Na+ is +72 mV. The K+ leak channels are the main ion channels open so the membrane potential is close to -90 mV. When Na+ channels open, Na+ rushes in and change sthe potential to +72 mV Upon closure of the Na+ channels, the K+ leak channels change the potential back to -90 mV.
It seems paradoxical that a lipid bilayer can be fluid yet asymmetrical. Explain.
The fluidity of the bilayer is strictly confined to one plane: lipid molecules can diffuse laterally in their own monolayer but do not readily flip from one monolayer to the other, lipid molecules in one monolayer remain in it unless they are actively transferred by flippase.
GPCRs activate G proteins by reducing the strength of GdP binding to the G protein. this results in rapid dissociation of bound GdP, which is then replaced by GtP, because GtP is present in the cytosol in much higher concentrations than GdP. What consequences would result from a mutation in the α subunit of a G protein that caused its affinity for GdP to be reduced without significantly changing its affinity for GtP? Compare the effects of this mutation with the effects of cholera toxin.
The mutant G protein would be almost continuously activated, because GDP would dissociate spontaneously, allowing GTP to bind even in the absence of an activated GPCR. similar to cholera toxin, which cannot hydrolyze GTP to shut itself off. In contrast to the cholera toxin case, the mutant G protein would not stay permanently activated but would instantly become activated after inactivation
plasma membrane
The protein-containing lipid bilayer that surrounds a living cell. (1) endoplasmic reticulum — Labyrinthine membrane- lipids and proteins production. (2) cytoskeleton — gives the cell shape and the capacity for directed movement (actin, microtubules, and intermediate filaments)
How would a change in the tubulin concentration affect this switch?
The rate of addition of GTP-tubulin will be greater at higher tubulin concentrations. so microtubules will switch to the growing mode. the system is self-balancing: the more microtubules shrink, the more frequently microtubules will start to grow again. and visa versa.
Write the chemical formula for a condensation reaction of two amino acids to form a peptide bond. Write the formula for its hydrolysis.
The reactions are diagrammed in Figure A2-14, where R1 and R2 are amino acid side chains. hydrolysis adds water and separates C and N ends; condensation removes water and binds C and N ends
margarine contains more saturated lipids than the vegetable oil from which it is made.
The reduction of double bonds allows the resulting saturated lipid molecules to pack more tightly against one another and therefore increases viscosity— that is, it turns oil into margarine.
The rate of a simple enzyme reaction is given by the standard Michaelis-Menten equation: rate = Vmax [s]/([s] + KM) if the Vmax of an enzyme is 100 µmole/sec and the KM is 1 mM, at what substrate concentration is the rate 50 µmole/sec? Plot a graph of rate versus substrate (s) concentration for [s] = 0 to 10 mM. Convert this to a plot of 1/rate versus 1/[s]. Why is the latter plot a straight line?
The substrate concentration is 1 mM. A plot of 1/rate versus 1/[S] is a straight line because rearranging the standard equation yields the equation listed in Question 3-23B.
The resting membrane potential of a typical animal cell is about -70 mv, and the thickness of a lipid bilayer is about 4.5 nm. What is the strength of the electric field across the membrane in v/ cm? What do you suppose would happen if you applied this field strength to two metal electrodes separated by a 1-cm air gap?
The voltage gradient across the membrane is about 150,000 V/cm (70 × 10-3 V/4.5 × 10-7 cm) which is an extremely powerful electric field; A voltage of 150,000 V would instantly discharge in an arc across a 1-cm-wide gap
what would happen to the regulation of the tryptophan operon in cells that express a mutant form of the tryptophan repressor that (1) cannot bind to dnA, (2) cannot bind tryptophan, or (3) binds to dnA even in the absence of tryptophan?
Transcription of the tryptophan operon would not be regulated by tryptophan; permanently turned on in scenarios (1) and (2) and permanently shut off in scenario (3)
iron (Fe) is an essential trace metal that is needed by all cells. it is required, for example, for synthesis of the heme groups and iron-sulfur centers that are part of the active site of many proteins involved in electron-transfer reactions; it is also required in hemoglobin, the main protein in red blood cells. iron is taken up by cells by receptor-mediated endocytosis. the iron uptake system has two components: a soluble protein called transferrin, which circulates in the bloodstream; and a transferrin receptor a transmembrane protein that, like the ldl receptor in Figure 15-33, is continually endocytosed and recycled to the plasma membrane. Fe ions bind to transferrin at neutral pH but not at acidic pH. transferrin binds to the transferrin receptor at neutral pH only when it has an Fe ion bound, but it binds to the receptor at acidic pH even in the absence of bound iron. From these properties, describe how iron is taken up, and discuss the advantages of this elaborate scheme.
Transferrin without Fe bound does not interact with its receptor and circulates in the bloodstream until it catches an Fe ion. Once iron is bound, the iron- transferrin complex can bind to the transferrin receptor on the surface of a cell and be endocytosed. Under the acidic conditions of the endosome, the transferrin releases its iron, but the transferrin remains bound to the transferrin receptor, which is recycled back to the cell surface, where it encounters the neutral pH environment of the blood. The neutral pH causes the receptor to release the transferrin into the circulation, where it can pick up another Fe ion to repeat the cycle. The iron released in the endosome moves on to lysosomes, from where it is transported into the cytosol. transferrin cycles between the blood and endosomes, delivering the iron that cells need to grow.
"dna stability in both reproductive cells and somatic cells is essential for the survival of a species." (T/F)
True If the DNA in somatic cells is not sufficiently stable, the organism dies, 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, mutations will accumulate and be passed on, destroying the species
ribosomes are cytoplasmic structures that, during protein synthesis, become linked by an mrna molecule to form polyribosomes. (T/F)
True.
transport vesicles deliver proteins and lipids to the cell surface. (T/F)
True.
Bacterial dna is found in the cytosol. (T/F)
True. Bacteria do not have a nucleus.
G-C base pairs are more stable than A-T base pairs. (T/F)
True. G-C base pairs are held together by three hydrogen bonds, whereas A-T base pairs are held together by only two.
none of the aberrant bases formed by deamination occur naturally in dna. (T/F)
True. If a damaged nucleotide also occurred naturally in DNA, the repair enzyme would have no way of identifying the damage. It would therefore have only a 50% chance of fixing the right strand.
in cells, the reaction takes place in more than one step. (T/F)
True. If only one step, then all the energy would be released at once making it impossible to harness
of the major control points in gene expression transcription initiation is one of the most common. (T/F)
True. It is advantageous to exert control at the earliest possible point in a pathway to conserves metabolic energy
if the delivery of prospective lysosomal proteins from the trans Golgi network to the late endosomes were blocked, lysosomal proteins would be secreted by the constitutive secretion pathways shown in Figure 15-30. (T/F)
True. Lysosomal proteins are selected in the trans Golgi network to be delivered to the late endosome. If not selected, they are moved to the cell surface.
in the absence of dna repair, genes are unstable. (T/F)
True. Mutations would accumulate rapidly, inactivating many genes.
some organisms carry out the reverse reaction. (T/F)
True. Plants convert CO2 into sugars by photosynthesis producing O2
in bacteria, but not in eukaryotes, many mrnAs contain the coding region for more than one gene. (T/F)
True. Prokaryotic mRNAs are often transcripts of entire operons. Ribosomes can initiate translation at internal AUG start sites of these "polycistronic" mRNAs
The energy is produced by a process that involves the oxidation of carbon atoms. (T/F)
True. The carbon atoms in glucose are in a reduced state compared to fully oxidized CO2
the nucleus and mitochondria are surrounded by a double membrane. (T/F)
True. The nuclear envelope is a double membrane, and mitochondria are surrounded by both an inner and an outer membrane.
all cells of the same organism have the same number of chromosomes (with the exception of egg and sperm cells). (T/F)
True. The number of chromosomes varies from one organism to another, but is constant in all cells (except germ cells) of the same organism.
cancer can result from the accumulation of mutations in somatic cells. (T/F)
True. Usually, multiple mutations of specific types need to accumulate in a somatic cell lineage to produce a cancer. A mutation in a gene that codes for a DNA repair enzyme can make a cell more liable to accumulate further mutations, thereby accelerating the onset of cancer.
the error rate of dna replication is reduced both by proofreading by dna polymerase and by dna mismatch repair. (T/F)
True. With proofreading and DNA mismatch repair makes 99% of its errors corrected (error rate to one in 109)
most dnA-binding proteins bind to the major groove of the dnA double helix. (T/F)
True. major groove is wide to allow a protein surface access to the base pairs. The sequence of H-bond donors and acceptors can be "read" by the protein to determine the sequence and orientation of the DNA.
all transport vesicles in the cell must have a v-snare protein in their membrane. (T/F)
True. required to dock at the correct target membrane and recruit a fusion complex
some cells that grow in the absence of o2 produce Co2. (T/F)
True. yeast cells convert pyruvate to ethanol and CO2
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?
Van der Waals attractions would occur in all three examples. Hydrogen bonds would form only in (c).
What determines the direction in which the ATP synthase operates?
When ATP is used ADP increases and ATP synthesis is initiated utilizing the proton gradient to restore the ATP/ADP ratio. When the electrochemical proton gradient drops below the equilibrium point, ATP synthase uses ATP in the matrix to restore this gradient.
the genes that encode the enzymes for arginine biosynthesis are located at several positions around the genome of E. coli, and they are regulated coordinately by a transcription regulator encoded by the ArgR gene. the activity of the Argr protein is modulated by arginine. upon binding arginine, Argr alters its conformation, dramatically changing its affinity for the dnA sequences in the promoters of the genes for the arginine biosynthetic enzymes. given that Argr is a repressor protein, would you expect that Argr would bind more tightly or less tightly to the dnA sequences when arginine is abundant? if Argr functioned instead as an activator protein, would you expect the binding of arginine to increase or to decrease its affinity for its regulatory dnA sequences? explain your answers.
When arginine is abundant, expression of the biosynthetic genes should be turned off. If ArgR acts as a gene repressor then binding of arginine should increase its affinity for its regulatory sites, allowing it to bind and shut off gene expression. If ArgR acted as a gene activator the binding of arginine would reduce its affinity for its regulatory DNA, preventing its binding and thereby shutting off expression of the Arg genes.
The elements oxygen and sulfur have similar chemical properties because they both have six electrons in their outermost electron shells. Indeed, both elements form molecules with two hydrogen atoms, water (H2O) and hydrogen sulfide (H2S). Surprisingly, at room temperature, water is a liquid, yet H2S is a gas, despite sulfur being much larger and heavier than oxygen. Explain why this might be the case.
Whether a substance is a liquid or gas at a given temperature depends on the attractive forces between its molecules. H2S is a gas at room temperature and H2O is a liquid because the hydrogen bonds that hold H2O molecules together do not form between H2S molecules. A sulfur atom is much larger than an oxygen atom, and because of its larger size, the outermost electrons are not as strongly attracted to the nucleus of the sulfur atom as they are in an oxygen atom. Consequently, the hydrogen-sulfur bond is much less polar than the hydrogen-oxygen bond. Because of the reduced polarity, the sulfur in a H2S molecule is not strongly attracted to the hydrogen atoms in an adjacent H2S molecule, and hydrogen bonds, which are so predominant in water, do not form.
What effects would you expect if ddcMp were added under the same conditions?
ddCMP lacks the 5ʹ-triphosphate group as well as the 3ʹ-hydroxyl group so it cannot provide the energy that drives the polymerization reaction of nucleotides into DNA and is not incorporated into the replicating DNA
What would you expect if ddctp were added to a dna replication reaction in large excess over the concentration of the available deoxycytosine triphosphate (dctp), the normal deoxycytosine triphosphate?
ddCTP is identical to dCTP, except it lacks the 3ʹ-hydroxyl group on the sugar ring. ddCTP becomes incorporated into DNA but creates a dead end to which no further nucleotides can be added. if ddCTP is added in large excess, new DNA strands will be synthesized until the first G is encountered on the template strand when ddCTP will then be incorporated instead of C, and the strand will be terminated.
What does ES represent?
enzyme-substrate complex
How many electrons would atoms of the elements listed below have to gain or lose to obtain a completely filled outer shell? helium oxygen carbon sodium chlorine
helium - already full oxygen - gain 2 carbon - gain or lose 4 sodium - lose 1 chlorine - gain 1
name the three ways in which an ion channel can be gated.
ligand, voltage, or mechanically (stress) gated
A sequence of nucleotides in a DNA strand 5-ttAACggCttttttC-3 was used as a template to synthesize an mrNA that was then translated into protein. predict the C-terminal amino acid and the N-terminal amino acid of the resulting polypeptide. Assume that the mrNA is translated without the need for a start codon.
mRNA sequence: 5ʹ-GAAAAAAGCCGUUAA-3ʹ. N-terminal amino acid: (GAA) glutamic acid C-terminal amino acid: (CGU) arginine
What must happen at the end of the microtubule in order for it to stop shrinking and to start growing again?
microtubule shrinks because it has lost its GTP cap, and the tubulin subunits at its end are all in their GDP-bound form causing all added subunits to be hydrolyzed or fall off. but If enough GTP-loaded subunits are 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.
How do cells in plant roots survive, since they contain no chloroplasts and are not exposed to light?
sucrose is transported from photosynthetic cells by sap to root cells which is then oxidized by glycolysis to produce ATP and be used for other metabolites.
A mutation in DNA generates a UGA stop codon in the middle of the mrNA coding for a particular protein. A second mutation in the cell's DNA leads to a single nucleotide change in a trNA that allows the correct translation of the protein; that is, the second mutation "suppresses" the defect caused by the first. the altered trNA translates the ugA as tryptophan. What nucleotide change has probably occurred in the mutant trNA molecule? What consequences would the presence of such a mutant trNA have for the translation of the normal genes in this cell?
tRNA anticodon is UCA; will recognize a UGA codon and incorporate a tryptophan residue Resulting in some proteins made with additional amino acids at their C-terminal end that were translated until a non- UGA stop codon in the mRNA in the reading frame was encountered
one often finds that high concentrations of p inhibit the enzyme. Suggest why this might occur.
the product of a reaction resembles the substrate sufficiently that it can also bind to the enzyme. Any enzyme molecules that are bound to the product are unavailable for catalysis; excess P therefore inhibits the reaction by lowering the concentration of free E.
When an inhibitory neurotransmitter such as gAbA opens cl- channels in the plasma membrane of a postsynaptic neuron, why does this make it harder for an excitatory neurotransmitter to excite the neuron?
the sole function of inhibitory neurotransmitters on transmittergated Cl- channels that open in the plasma membrane of a postsynaptic neuron is minute but excitatory neurotransmitter at the same time causes a depolarization by Na+ and Cl- movement into the cell inhibitory neurotransmitters suppress an action potential by making the membrane harder to depolarize.
What would happen if ddctp were added at 10% of the concentration of the available dctp?
there is a 1 in 10 chance of its being incorporated whenever a G is encountered on the template strand producing a population of DNA fragments. The lengths of the fragments can be used to deduce where the G residues are located on the template strand. This is used in DNA sequencing; 3ʹ-azido-3ʹ-deoxythymidine (AZT) is used in HIV-infected patients to treat AIDS. AZT is converted into the triphosphate form and is incorporated into the growing viral DNA. It lacks a 3ʹ-OH group, it blocks DNA synthesis and replication of the virus.
consider a protein that contains an er signal sequence at its n-terminus and a nuclear localization sequence in its middle. What do you think the fate of this protein would be? explain your answer.
translocated into the ER. ER signal sequence is recognized immediately binding the ribosome to the ER membrane through the ER translocation channel. The nuclear localization sequence is therefore never exposed to the cytosol and will never encounter nuclear import receptors, and the protein will not enter the nucleus.
Hydrogen bonds that form between lipid head groups and water molecules are continually broken and re-formed. (T/F)
true
Lipids in a lipid bilayer do not flip-flop readily from one lipid monolayer to the other. (T/F)
true
Lipids in a lipid bilayer rapidly exchange positions with one another in their own monolayer. (T/F)
true
Lipids in a lipid bilayer spin rapidly around their long axis. (T/F)
true
the activation of electrons by the photosystems enables chloroplasts to drive electron transfer from H2O to carbohydrate, which is the opposite direction of electron transfer in the mitochondrion? (T/F)
true
the products of chloroplasts are the substrates for mitochondria? (T/F)
true
Under what conditions would one expect the ATP synthase to stall, running neither forward nor backward?
when the energy from the proton gradient equals the ΔG required to make ATP
As shown in figure 4-16, both α helices and the coiled-coil structures that can form from them are helical structures, but do they have the same handedness in the figure? explain why?
α helix - right-handed coiled-coil - left-handed. The reversal occurs because of the staggered positions of hydrophobic side chains in the α helix.