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BIOL 211G A20 - Cellular and Organismal Biology Sol Paper

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Question;Problem Set 1 - Problems on Biomolecules (2 problems):1. The sequence of amino acids of the enzyme lysozyme is known. Below is a list of amino acidsand the number of each in the lysozyme molecule.TypeMolecular WeightNumber in LysozymeAlanine8912Arginine17411Asparagine13213Aspartic acid1338Cysteine1218Glutamic acid1472Glutamine1463Glycine7512Histidine1551Isoleucine1316Leucine1318Lysine1466Methionine1492Phenylalanine1653Proline1152Serine10510Threonine1197Tryptophan2046Tyrosine1813Valine1176Questions:a. What is the average molecular weight of the 20 amino acids?b. What is the molecular weight of lysozyme?c. What is the average molecular weight of the amino acids in lysozome?d. How many S-S bonds are possible in lysozyme?(hint: look at the structures of each amino acid, and figure out which of these have sulphursthat are FREE to form bonds).e. Is the net charge on lysozyme positive or negative?(hint: again, look at the structures of each amino acid, and figure out which of these areneutral, which are positively charged, and which are negatively charged, then add up theindividual charges for each amino acid to get the net charge of this protein)2. Proteins (both enzymatic and structural) play a significant role in maintaining the pH in a cell.Seven of the 20 amino acids commonly found in proteins have an ionizable group. Thesegroups (side chains) determine the charge on the protein and buffer the cytoplasm of the cellthat contains it.Populations of proteins that contain ionizable side chains have a probability of being ?charged?or not. The probability of a side chain being charged is related primarily to the side chainchemistry and the pH of the cellular compartment, other usually less important factors aretemperature, ionic strength, and bulk microenvironments.The ratio of the charged to uncharged side chains is usually symbolized by K:K=c/uwhere c is the proportion of charged side chains and u is the proportion of uncharged side chains(so that c + u = 1).In the laboratory, the pH of a protein's environment can be altered so that we have equalnumbers of charged and uncharged molecules. We call the pH at which K is 1 (equal number ofcharged and uncharged side chains) the pK of the side chain.Below is a list of the seven amino acids whose charge is altered by the cellular pH:AcidicpKBasicpKAspartic acid3.87Lysine10.53 Cysteine8.33Glutamic acid4.25Arginine12.48 Tyrosine10.07HistidinePolarpK6.00Cell compartments can vary in pH from about 4.0 to 8.5, but the pK values of the various aminoacids remain fixed. When the pH of the environment is not equal to the pK of the side chain,the proportion of charged and uncharged side chains changes to as to satisfy the equation:pH = pK + log (c / u)As an example, for aspartic acid in a protein with a pH environment of 3.87, pH = pK,so log (c/u) = 0. Consequently, (c/u) = 10 ^ 0 = 1, and c = u. However, when the pH and thepK are different, the term log (c /u) will not be zero and the ratio (c/u) will be different from 1.Questions:a. The pH of healthy cell cytoplasm varies from 7.2 to 7.4. Using this information, and theinformation above, complete the following table:Amino acid in proteinpK of the side chainPercent Charged at pH Percent Charged at pH= 7.2= 7.4Aspartic acid3.87??Histidine6.0??Cysteine8.33??Arginine12.48_________________________________________________Problem Set 2 - Problems on Cell Structures and Membranes (3 problems):3.Lysosomes are little sacs of acid in a cell. Their pH is about 5, and an electronmicrograph suggests they have a diameter of 0.5 ?m. The increased hydrogen ionconcentration inside lysosomes is due to the pumping of hydrogen ions across thelysosomal membrane from the surrounding cytosol, which has a pH of 7.2.a. Assuming that a lysosome has the shape of a sphere and that there is nobuffering capacity inside the lysosome, how many hydrogen ions were movedto the inside of the lysosome to lead to an internal pH of 5?(hint: first determine the volume of a lysosome in liters, then determine [H+] inmoles/L in a lysosome at each pH (5 and 7.2), then determine the number ofmoles of hydrogen ions at each pH, and finally determine and compare thenumber of hydrogen ions at each pH).4.Liposomes are laboratory-prepared artificial membranes. Liposomes can be made in avariety of sizes and can be made so that they have transmembrane proteins, which formmembrane. Contents of the liposomes can also be known.For example, let?s say that one lab makes liposomes that are spheres with the diameter of4 ?m and that each liposome has an average of ten protein pores. Each liposome has aninternal potassium ion concentration of 100 mM. Each protein pore transports 3x 10^6potassium ions per second. The pores stay open an average of 0.3 second and stay closedan average of 2 seconds, so, each pore opening and closing cycle takes about 2.3 seconds.a. Assuming that a liposome has the shape of a sphere, how many potassiumions are in a liposome initially?(hint: the method here is similar to what you used to solve problem 3 above,except find the volume of a liposome in ?m^3 and the [K+] in mol/?m^3)b. How much time is required for the potassium ions in the liposome to reachequilibrium with their environment? Assume that this environment isrelatively large and potassium-free.(hint: before calculating the total time it would take to reach this equilibrium,think about how many potassium ions would need to leak out of the liposome inorder to reach this equilibrium ? all of them, half of them, none of them, why?)5. Glycophorin is a single-pass transmembrane protein in red blood cells (RBCs). Theprotein component of glycophorin is 131 amino acids long and binds carbohydrates onthe outside (noncytoplasmic side) of glycophorin. Then, approximately 100 modifiedsugar residues are attached near the end of each glycophorin, these account for about60% of this macromolecule?s mass. The average molecular weight of an amino acid is130 daltons.a. What is the average molecular weight (in daltons) of each modified sugarresidue on the glycophorin?b. An RBC contains an average of 6 x 10 ^5 glycophorin molecules. How manymodified sugar residues are found attached to glycophorins in one RBC?c. How many grams does the protein component of glycophorin weigh in oneRBC?______________________Problem Set 3 - Problems on Enzymes and Cellular Respiration (3 problems):6. Hydrogen peroxide is usually stored in a brown bottle away from sunlight because itspontaneously (but slowly) decomposes into O2 and water. In the brown bottle, the freeenergy of activation?G+ = 18 kcal/mol. In the presence of a catalyst, the decompositionis much faster. For each decrease of 1.36 kcal/mol in?G+, the rate of reaction is tentimes faster.In the presence of catalase, an enzyme found in the blood,?G+ = 7 kcal/mol. In thepresence of the inorganic catalyst platinum,?G+ = 13 kcal/mol.a.How much faster is peroxide decomposition in the presence of a catalase?b. How much faster is peroxide decomposition in the presence of platinum?c. How many times faster is peroxide degradation under the influence of a catalasethan degradation under the influence of platinum?7. You eat a candy bar that has 180 calories. This energy is converted during respiration toATP. The reaction ADP + Pi? ATP requires 7.3 kcal/mol. One ?dietary? calorie isequal to 1000 ?chemical? calories. Respiration is maximally about 39% efficient inconverting substrate calories to ATP calories.a.Assuming that all of the energy in the candy actually gets to the respiratory sitein the cell, how many ATPs could your body make from this candy bar?8. The first two stages in respiration are glycolysis and Kreb?s cycle. For each molecule ofglucose input,10 NAD+ molecules are reduced2 FADs are reduced4 ADPs are phosphorylated.The free energy?G0? of the relevant ?respiration reactions? is:ReactionNADH + H+ + ? O2? NAD+ + H2OFADH2 + ? O2?FAD + H2OATP? ADP + PiGlucose + 6 O2? 6 CO2 + 6H2Oa.?G0? (kcal/mol)-51.7-43.4-7.3-686How much of the energy from one mole of glucose is converted into ATP duringthe first two stages?b. How much of the glucose energy is conserved in the reduced coenzymes NAD+and FAD?The third stage of respiration is oxidation phosphorylation.c.A total of 32 ATPs per glucose molecule are made during oxidativephosphorylation from the reduce coenzymes. How much of the glucose energy isconserved in ATP at the end of ALL three stages?d. What percentage of the total ATP energy is converted by the oxidativephosphorylation of the reduced coenzymes?e. What percentage of the glucose energy was lost was heat?_____________________Problem Set 4 - Problems on Photosynthesis (3 problems):9. The fixing of carbon in photosynthesis varies in regard to the amount of ATP needed for eachcarbon fixed.C3 plants use 3 ATPs per carbonC4 plants use 5 ATPs per carbonCAM plants use 5.5 ATPs per carbonIf ATP yields 7.3 kilocalories per mole, how many ATP calories are needed to create 1 moleof glucose (686 kcal/mol) by:a.b.c.d.a C3 plant? (hint: first, think about how many carbons are in 1 mole of glucose)a C4 plant?a CAM plant?Which type plant uses the most ATP energy in the making of glucose?10. When light is shined on a leaf, it causes hydrogen ions to be pumped into discs calledthylakoid lumens. The ions then diffuse out through a protein, and in the process an ATPmolecule is made for every three hydrogen ions. While illuminated, inside the disc, the pH canbe as low as 4. Outside the disc, the pH is about 7.2.A thylakoid lumen can be modeled as a short cylindrical rod that is 80 ? long and 5000 ? indiameter.a. How many hydrogen ions are found in one thylakoid lumen of this size at pH 4?b. How many are found at pH 7.2?c. How many more ATP molecules can be made from the disc described above,AFTER the light is turned off?11. Light is important in biology for photosynthesis. There are two different ways that light isdescribed in physics.In the first description, light travels in waves at a fixed speed c = 2.998 x 108 meters per second.The wavelength is the distance from peak to peak of a light wave, and corresponds to the color ofthe light. The wavelength is given by? (the Greek letter lambda). The wavelength varies from400 nm to 700 nm for light in the visible range, with blue light having?=450 nm and red lighthaving?=680 nm.The frequency is given by? (the Greek letter nu). The frequency is the number of peaks thatpass a point in a given time. Frequency is related to wavelength by the formula:? = c /?In the second description, light travels in particles called photons or quanta. Using thisdescription it makes sense to speak of a mole of light as 6.02 x 1023 photons.The energy of one photon of light is given byE = (hc) /? = h?where h is a conversion factor called Planck?s constant, h = 1.583 x 10-34 calorie seconds.In the laboratory, light with a very narrow wavelength range can be used for experiments. Onemole of an actinic light (activating light) that has a wavelength of 680 nm was used to excitechlorophyll, and caused fluorescence measured at a wavelength of 690 nm. The chlorophyll wasisolated, and therefore could do no photochemistry.a. What is the amount of energy (in kilocalories) in one mole of actinic red light?b. What is the amount of energy (in kcal) in the light that was fluoresced (assumingmaximal fluorescence)?c. What is the amount of energy (in kcal) that was lost as heat?d. What percentage of the red light energy was lost as heat?A photon of blue light will energize an electron from chlorophyll to a level comparable to aphoton of red light. Suppose blue light energy also caused fluorescence measured at awavelength of 690 nm.e. What percentage of the blue light energy was lost as heat (again assuming maximalfluorescence)?___________________________________________________________________________________________________Problem Set 5 - Problems on Plant Transport/Resource Acquisition, and Plant Response toStimuli (3 problems):12. Plasmodesmata are cytoplasmic connections across plant cell walls that connect adjacent cellcytoplasms. Some cells have few plasmodesmata connections while others have more, this isdue to genetics, age, and location within the plant.The density of plasmodesmata within the cell membrane ranges from a high of 25 per squaremicrometer to a low of 0.2 per square micrometer. The average plasmodesmata tube is 40 nm indiameter.Diagram of plasmodesmata between two plant cellsa. What percentage of the cell membrane surface area is composed of plasmodesmataat the high density? (Assume the area of one end of a plasmodesmal tube is a circle? see image above)b. What percentage of the cell membrane surface area is composed of plasmodesmataat the low density?13. Liquid water moves into and out of the cell by diffusion. Water vapor also moves from theinside of the plant leaf to the ambient air by diffusion, in a process known as transpiration. Thistranspiration causes dissolved minerals to be moved long distances in the plant.The time t in seconds for water to move such a long distance d in meters is given byt = (d2)/Dwhere D is the diffusion coefficient. A reasonable value for D is 2.4 x 10-5 m2/sec. (Note: watervapor diffuses in air much more rapidly than in liquid water).The path of the diffusion of water vapor from a leaf into the air varies considerably, but ameasured distance of 1 mm is reasonable (1 mm = 10-3 m).a. How long (in seconds) does it take a molecule of water vapor to be lost by this leaf?Hairiness of leaves is a genetic trait. Leaf hairs may double the distance water must diffuse.b. How long (in seconds) would it take to lose water from a hairy leaf?14. (This problem is split into two different problems ? 14 and 15 - to give you more credit forthe graphs.). One environmental factor that plants monitor is the duration of light. The length ofan uninterrupted dark period is often extremely important in the kind of growth (vegetative ascompared to flowering). Long-day (LD) plants require at least a certain day length before theybegin to flower, whereas short-day (SD) plants require at most a certain day length before theybegin to flower.In the corn belt in the United States, sunrise and sunset define the length of the day. Forexample, here are the times of sunrise and sunset of the first day of each month in northern Iowa:MonthJanFebMarAprMayJunJulAugSepOctNovDeca.Sunrise7:227:096:345:445:004:334:354:585:285:566:297:02Sunset4:455:195:526:246:547:237:337:146:325:434:584:35Plot the sunrise and sunset times on the same graph. (You can do this either on the computerwith Excel or by hand on graph paper. If you draw this by hand, make sure to scan your graphand then paste the image into the document that you will be submitting for grading. Pleaselabel your graph and axes properly: title below the graph (Figure 1. Give your own title here), xaxis = X title (in? ? hint: measured in what units?), y axis = Y title (in? units). Typically, inscience, graphs show the independent variable on the x-axis, and dependent variable on the yaxis. So, you will need to figure out which of these is the independent variable, and which is thedependent variable. The independent variable is the variable that is directly manipulated by ascientist during an experiment. The dependent variable is not directly manipulated by ascientist, and instead changes in response to a change in the independent variable.)b. Plot the duration of light per day and the duration of dark per day on the samesecond graph (see (a) for graphing instructions).15. This problem is a continuation of problem 14 above. Now, suppose you have inherited agreenhouse and you decided to become a farmer. You don?t want to pay for additional lighting,so you will use only sunlight. Also, you have a local market only for the following plants:PlantDillSpinachSoybeanCockleburSD or LDLDLDSDSDCritical Day Length111315.514a. When during the year would you expect each plant to be induced to flower?Explain how you determined this (hint: look at your graphs for problem 14)._________________________________________Problem Set 6 - Problems on Gene Expression/Repair and Protein Synthesis/Degradation (3problems):16. The DNA in a human non-gametic cell contains 6 billion base pairs. It is estimated thatabout 10,000 DNA changes occur in each cell in one day. These are quickly repaired so thatonly a few (1 to 5) mutations accumulate in one cell in a year.a. What percentage of the base pairs are altered each day?b. What percentage of the DNA changes that occur in one cell in one year escapethe proofreading and repair process ? calculate this in both cases: if 1 mutationaccumulates in one cell per year, and if 5 mutations accumulate in one cell peryear?17. In one mammalian cell it is estimated that 10,000 to 20,000 different types of mRNA can befound. Abundant mRNAs exist in many copies per cell (up to 12,000 copies/cell), but scarcemRNAs (5 to 15 copies/cell) can also be detected. At any instant, a snapshot of mRNA contentwould reveal a total of 360,000 mRNAs. The cell usually has about 10 times as many ribosomesas mRNAs. Assume 75% of the ribosomes are involved in protein synthesis at any instant intime.a. What is the maximum number of proteins that could be in the process ofsynthesis at any instant?b. How many scarce proteins (use 5 copies/cell) could be in the process of beingmade? (hint: use that 7.5 ribosomes are reading each of the 5 mRNAs, but explain orshow calculations for why you should use this).c. How many abundant proteins could be in the process of being made? (hint: usethat 7.5 ribosomes are reading each of the 12,000 mRNAs, but explain or showcalculations for why you should use this).d. What is the ratio of developing abundant proteins to developing scarce proteins?18. Protein molecules in the cytosol of a cell have different half-lives t ?. The half-life is thetime needed for 50% of the molecules to be lost or altered. Half-lives are determined by severalfactors, one of which is the ?marking? of the protein by ubiquitin, which signals that the proteinis intended for digestion by proteases. Ubiquitin binds differently to proteins due to differencesin their amino ends. The following table gives the half-lives t ? for some proteins with differentamino ends:Amino Acid End of ProteinMETSERTHRALAVALLEUPHEASPLYSARGt?> 20 hrs> 20 hrs> 20 hrs> 20 hrs> 20 hrs< 3 min< 3 min< 3 min< 3 min< 2 minHalf-life is related to the rate K of protein loss by the following equation:t ? = 0.693 / Kwhere 0.693 is the natural logarithm of 2, and K is measured in reciprocal time units.a.What is the rate of protein loss per minute when MET is the amino terminal?(Assume t ? = 24 hr).b. Compare this with the rate when ARG is the amino terminal end. (Assume t ? =2 min).c.What percentage of the protein with a MET amino end exists in the cell after 5half-lives?d. What percentage of the protein with an ARG amino end exists in the cell after 5half-lives?e. How much faster (as a percentage) is the rate of ARG-amino-end proteindegradation as compared to MET-amino-end degradation?f. Plot the loss of a population of protein molecules that have a MET amino end.Use time t on the horizontal axis and size of the population on the vertical axis,starting with 100% of the population at the starting time t = 0.g. On the same graph, plot the loss of a protein with ARG as the amino terminus.h. Hemoglobin exists in the cytoplasm of a red blood cell (RBC). Red blood cellslast about 120 days in the bloodstream. In your initial experiments you find that theamino terminus of one protein chain in RBCs is either a valine or a leucine (so adifference of only one methyl group). Which is more likely to be the correct aminoacid terminus of this protein? (Assume t ? = 24 hr for valine and 2 min for leucine)._______________________________________Problem Set 7 - Problems on Microorganisms/Bacteria (2 problems):19. Assume that you conducted the following experiment in order to examine how bacterialevolution occurs in jumps:An E. coli culture was maintained for 10,000 generations over 4 years. The liquid medium waschanged daily to maintain a constant environment. The average size of the cells at the start ofthe experiment was 0.35 x 10 -15 L. After 300 generations the size increased to 0.48 x 10 -15 L,and after another 300 generations the average increased to 0.49 x 10 -15 L. After 1200generations it increased to 0.58 x 10 -15 L and remained so until the end of the experiment.Volume (x 10 -15 L)E. coli Evolution0.70.60.50.40.30.20.100500100015002000Generations250030003500a. How many hours long is a generation of E. coli?b. What was the average size change over the course of the experiment in liters?c. What was the average size change in the first 300 generations in liters?d. How fast did the size increase over the course of the experiment in liters pergeneration?e. How fast did the size increase over the course of the experiment in liters per year?f. How fast did the size increase in the first 300 generations in liters per generation?g. How fast did the size incr

 

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