1. The sequence of amino acids of the enzyme lysozyme is known. Below is a list of amino acids and the number of each in the lysozyme molecule. Type Molecular

1. The sequence of amino acids of the enzyme lysozyme is known. Below is a list of amino acids and the number of each in the lysozyme molecule. Type Molecular Weight Number in Lysozyme Alanine 89 12 Arginine 174 11 Asparagine 132 13 Aspartic acid 133 8 Cysteine 121 8 Glutamic acid 147 2 Glutamine 146 3 Glycine 75 12 stidine 155 1 Isoleucine 131 6 Leucine 131 8 Lysine 146 6 Methionine 149 2 Phenylalanine 165 3 Proline 115 2 Serine 105 10 Threonine 119 7 Tryptophan 204 6 Tyrosine 181 3 Valine 117 6 Questions: a. is the average molecular weight of the 20 amino acids? b. is the molecular weight of lysozyme? c. 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 sulphurs that 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 are neutral, which are positively charged, and which are negatively charged; then add up the individual 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. These groups (side chains) determine the charge on the protein and buffer the cytoplasm of the cell that 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 chain chemistry and the pH of the cellular compartment; other usually less important factors are temperature, ionic strength, and bulk microenvironments. The ratio of the charged to uncharged side chains is usually symbolized by K: K=c/u where 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 equal numbers of charged and uncharged molecules. We call the pH at which K is 1 (equal number of charged 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: Acidic pK Basic pK Aspartic acid 3.87 Lysine 10.53 Cysteine 8.33 Glutamic acid 4.25 Arginine 12.48 Tyrosine 10.07 stidine Polar pK 6.00 Cell compartments can vary in pH from about 4.0 to 8.5, but the pK values of the various amino acids 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 the pK 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 the information above, complete the following table: Amino acid in protein pK of the side chain Percent Charged at pH Percent Charged at pH = 7.2 = 7.4 Aspartic acid 3.87 ? ? stidine 6.0 ? ? Cysteine 8.33 ? ? Arginine 12.48 _________________________________________________ 3. Lysosomes are little sacs of acid in a cell. Their pH is about 5, and an electron micrograph suggests they have a diameter of 0.5 µm. The increased hydrogen ion concentration inside lysosomes is due to the pumping of hydrogen ions across the lysosomal 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 no buffering capacity inside the lysosome, how many hydrogen ions were moved to 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+] in moles/L in a lysosome at each pH (5 and 7.2), then determine the number of moles of hydrogen ions at each pH, and finally determine and compare the number of hydrogen ions at each pH). 4. Liposomes are laboratory-prepared artificial membranes. Liposomes can be made in a variety of sizes and can be made so that they have transmembrane proteins, which form membrane. Contents of the liposomes can also be known. For example, let’s say that one lab makes liposomes that are spheres with the diameter of 4 µm and that each liposome has an average of ten protein pores. Each liposome has an internal potassium ion concentration of 100 mM. Each protein pore transports 3x 10^6 potassium ions per second. The pores stay open an average of 0.3 second and stay closed an 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 potassium ions 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 µm^3 and the [K+] in mol/µm^3) b. How much time is required for the potassium ions in the liposome to reach equilibrium with their environment? Assume that this environment is relatively 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 in order 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). The protein component of glycophorin is 131 amino acids long and binds carbohydrates on the outside (noncytoplasmic side) of glycophorin. Then, approximately 100 modified sugar residues are attached near the end of each glycophorin; these account for about 60% of this macromolecule’s mass. The average molecular weight of an amino acid is 130 daltons. a. is the average molecular weight (in daltons) of each modified sugar residue on the glycophorin? b. An RBC contains an average of 6 x 10 ^5 glycophorin molecules. How many modified sugar residues are found attached to glycophorins in one RBC? c. How many grams does the protein component of glycophorin weigh in one RBC? ______________________ 6. Hydrogen peroxide is usually stored in a brown bottle away from sunlight because it spontaneously (but slowly) decomposes into O2 and water. In the brown bottle, the free energy of activation ∆G+ = 18 kcal/mol. In the presence of a catalyst, the decomposition is much faster. For each decrease of 1.36 kcal/mol in ∆G+, the rate of reaction is ten times faster. In the presence of catalase, an enzyme found in the blood, ∆G+ = 7 kcal/mol. In the presence 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 catalase than degradation under the influence of platinum? 7. You eat a candy bar that has 180 calories. This energy is converted during respiration to ATP. The reaction ADP + Pi  ATP requires 7.3 kcal/mol. One “dietary” calorie is equal to 1000 “chemical” calories. Respiration is maximally about 39% efficient in converting substrate calories to ATP calories. a. Assuming that all of the energy in the candy actually gets to the respiratory site in 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 of glucose input, 10 NAD+ molecules are reduced 2 FADs are reduced 4 ADPs are phosphorylated. The free energy ∆G0’ of the relevant “respiration reactions” is: Reaction NADH + H+ + ½ O2  NAD+ + H2O FADH2 + ½ O2 FAD + H2O ATP  ADP + Pi Glucose + 6 O2  6 CO2 + 6H2O a. ∆G0’ (kcal/mol) -51.7 -43.4 -7.3 -686 How much of the energy from one mole of glucose is converted into ATP during the 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 oxidative phosphorylation from the reduce coenzymes. How much of the glucose energy is conserved in ATP at the end of ALL three stages? d. percentage of the total ATP energy is converted by the oxidative phosphorylation of the reduced coenzymes? e. percentage of the glucose energy was lost was heat? _____________________ 9. The fixing of carbon in photosynthesis varies in regard to the amount of ATP needed for each carbon fixed. C3 plants use 3 ATPs per carbon C4 plants use 5 ATPs per carbon CAM plants use 5.5 ATPs per carbon If ATP yields 7.3 kilocalories per mole, how many ATP calories are needed to create 1 mole of 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 called thylakoid lumens. The ions then diffuse out through a protein, and in the process an ATP molecule is made for every three hydrogen ions. While illuminated, inside the disc, the pH can be 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 Å long and 5000 Å in diameter. 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 is described 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 of the light. The wavelength is given by λ (the Greek letter lambda). The wavelength varies from 400 nm to 700 nm for light in the visible range, with blue light having λ=450 nm and red light having λ=680 nm. The frequency is given by ν (the Greek letter nu). The frequency is the number of peaks that pass a point in a given time. Frequency is related to wavelength by the formula: ν = c / λ In the second description, light travels in particles called photons or quanta. Using this description 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 by E = (hc) / λ = hν 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. One mole of an actinic light (activating light) that has a wavelength of 680 nm was used to excite chlorophyll, and caused fluorescence measured at a wavelength of 690 nm. The chlorophyll was isolated, and therefore could do no photochemistry. a. is the amount of energy (in kilocalories) in one mole of actinic red light? b. is the amount of energy (in kcal) in the light that was fluoresced (assuming maximal fluorescence)? c. is the amount of energy (in kcal) that was lost as heat? d. 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 a photon of red light. Suppose blue light energy also caused fluorescence measured at a wavelength of 690 nm. e. percentage of the blue light energy was lost as heat (again assuming maximal fluorescence)? ___________________________________________________________________________________________________ 12. Plasmodesmata are cytoplasmic connections across plant cell walls that connect adjacent cell cytoplasms. Some cells have few plasmodesmata connections while others have more; this is due to genetics, age, and location within the plant. The density of plasmodesmata within the cell membrane ranges from a high of 25 per square micrometer to a low of 0.2 per square micrometer. The average plasmodesmata tube is 40 nm in diameter. Diagram of plasmodesmata between two plant cells a. percentage of the cell membrane surface area is composed of plasmodesmata at the high density? (Assume the area of one end of a plasmodesmal tube is a circle – see image above) b. percentage of the cell membrane surface area is composed of plasmodesmata at the low density? 13. Liquid water moves into and out of the cell by diffusion. Water vapor also moves from the inside of the plant leaf to the ambient air by diffusion, in a process known as transpiration. This transpiration 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 by t = (d2)/D where D is the diffusion coefficient. A reasonable value for D is 2.4 x 10-5 m2/sec. (Note: water vapor 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 a measured 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 for the graphs.). One environmental factor that plants monitor is the duration of light. The length of an uninterrupted dark period is often extremely important in the kind of growth (vegetative as compared to flowering). Long-day (LD) plants require at least a certain day length before they begin to flower, whereas short-day (SD) plants require at most a certain day length before they begin to flower. In the corn belt in the United States, sunrise and sunset define the length of the day. For example, here are the times of sunrise and sunset of the first day of each month in northern Iowa: Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec a. Sunrise 7:22 7:09 6:34 5:44 5:00 4:33 4:35 4:58 5:28 5:56 6:29 7:02 Sunset 4:45 5:19 5:52 6:24 6:54 7:23 7:33 7:14 6:32 5:43 4:58 4:35 Plot the sunrise and sunset times on the same graph. (You can do this either on the computer with Excel or by hand on graph paper. If you draw this by hand, make sure to scan your graph and then paste the image into the document that you will be submitting for grading. label your graph and axes properly: title below the graph (Figure 1. Give your own title here), x axis = X title (in ? – hint: measured in what units?); y axis = Y title (in? units). Typically, in science, 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 the dependent variable. The independent variable is the variable that is directly manipulated by a scientist during an experiment. The dependent variable is not directly manipulated by a scientist, 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 same second graph (see (a) for graphing instructions). 15. This problem is a continuation of problem 14 above. Now, suppose you have inherited a greenhouse 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: Plant Dill Spinach Soybean Cocklebur SD or LD LD LD SD SD Critical Day Length 11 13 15.5 14 a. 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). _________________________________________ 16. The DNA in a human non-gametic cell contains 6 billion base pairs. It is estimated that about 10,000 DNA changes occur in each cell in one day. These are quickly repaired so that only a few (1 to 5) mutations accumulate in one cell in a year. a. percentage of the base pairs are altered each day? b. percentage of the DNA changes that occur in one cell in one year escape the proofreading and repair process – calculate this in both cases: if 1 mutation accumulates in one cell per year, and if 5 mutations accumulate in one cell per year? 17. In one mammalian cell it is estimated that 10,000 to 20,000 different types of mRNA can be found. Abundant mRNAs exist in many copies per cell (up to 12,000 copies/cell); but scarce mRNAs (5 to 15 copies/cell) can also be detected. At any instant, a snapshot of mRNA content would reveal a total of 360,000 mRNAs. The cell usually has about 10 times as many ribosomes as mRNAs. Assume 75% of the ribosomes are involved in protein synthesis at any instant in time. a. is the maximum number of proteins that could be in the process of synthesis at any instant? b. How many scarce proteins (use 5 copies/cell) could be in the process of being made? (hint: use that 7.5 ribosomes are reading each of the 5 mRNAs; but explain or show calculations for why you should use this). c. How many abundant proteins could be in the process of being made? (hint: use that 7.5 ribosomes are reading each of the 12,000 mRNAs; but explain or show calculations for why you should use this). d. 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 ½. The half-life is the time needed for 50% of the molecules to be lost or altered. Half-lives are determined by several factors, one of which is the “marking” of the protein by ubiquitin, which signals that the protein is intended for digestion by proteases. Ubiquitin binds differently to proteins due to differences in their amino ends. The following table gives the half-lives t ½ for some proteins with different amino ends: Amino Acid End of Protein MET SER THR ALA VAL LEU PHE ASP LYS ARG t½ > 20 hrs > 20 hrs > 20 hrs > 20 hrs > 20 hrs < 3 min < 3 min < 3 min < 3 min < 2 min Half-life is related to the rate K of protein loss by the following equation: t ½ = 0.693 / K where 0.693 is the natural logarithm of 2, and K is measured in reciprocal time units. a. is the rate of protein loss per minute when MET is the amino terminal? (Assume t ½ = 24 hr). b. Compare this with the rate when ARG is the amino terminal end. (Assume t ½ = 2 min). c. percentage of the protein with a MET amino end exists in the cell after 5 half-lives? d. percentage of the protein with an ARG amino end exists in the cell after 5 half-lives? e. How much faster (as a percentage) is the rate of ARG-amino-end protein degradation 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 cells last about 120 days in the bloodstream. In your initial experiments you find that the amino terminus of one protein chain in RBCs is either a valine or a leucine (so a difference of only one methyl group). Which is more likely to be the correct amino acid terminus of this protein? (Assume t ½ = 24 hr for valine and 2 min for leucine). _______________________________________ 19. Assume that you conducted the following experiment in order to examine how bacterial evolution occurs in jumps: An E. coli culture was maintained for 10,000 generations over 4 years. The liquid medium was changed daily to maintain a constant environment. The average size of the cells at the start of the 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 1200 generations 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 Evolution 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0 500 1000 1500 2000 Generations 2500 3000 3500 a. How many hours long is a generation of E. coli? b. was the average size change over the course of the experiment in liters? c. 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 per generation? 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 increase in the first 300 generations in liters per year?

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