Course Number 211g A20 Course Name Online Cellular And Organ
Course Number 211g A20course Name Online Cellular And Organismal
Write out your calculations and explanations for ALL problems. To solve the problems in this problem set, you will need to review (or look up, if you cannot remember) some geometry formulas. 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 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 mol/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 μm³ and the [K+] in mol/μm³) 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. What 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?
Paper For Above instruction
The biological processes governing cell membrane structure, function, and cellular components like lysosomes, liposomes, and membrane proteins are fundamental to understanding cellular biology. This paper systematically addresses each of the posed problems, offering detailed calculations and explanations based on core biochemical and biophysical principles.
Problem 3: Hydrogen Ion Movement in Lysosomes
Lysosomes are membrane-bound organelles with an acidic interior, essential for breakdown and recycling processes. The question asks how many hydrogen ions are transported to the lysosome's interior to achieve a pH of 5 from a cytosolic pH of 7.2. Assuming the lysosome is spherical with a diameter of 0.5 μm, and ignoring buffering capacity, calculations involve determining the lysosomal volume, hydrogen ion concentrations at different pH levels, and then the total moles of H⁺ ions involved.
First, calculate the volume of the lysosome. The volume \( V \) of a sphere is given by \( V = \frac{4}{3}\pi r^3 \). The radius \( r \) is half of the diameter: 0.25 μm. Therefore,
\[ V = \frac{4}{3} \pi (0.25)^3 \approx 0.06545 \, \text{μm}^3 \]
Next, convert this volume into liters. There are \( 10^{15} \) μm³ in 1 liter, so:
\[ V_{L} = \frac{0.06545 \, \text{μm}^3}{10^{15}} = 6.545 \times 10^{-17} \, \text{L} \]
Now, determine hydrogen ion concentrations at pH 5 and 7.2, based on the fact that pH = -log [H⁺]:
\[ [H^+]_{pH=5} = 10^{-5} \, \text{mol/L} \]
\[ [H^+]_{pH=7.2} = 10^{-7.2} \approx 6.31 \times 10^{-8} \, \text{mol/L} \]
Calculate the number of moles of H⁺ inside the lysosome at each pH:
\[ n_{pH=5} = (10^{-5}) \times 6.545 \times 10^{-17} \approx 6.545 \times 10^{-22} \, \text{mol} \]
\[ n_{pH=7.2} = 6.31 \times 10^{-8} \times 6.545 \times 10^{-17} \approx 4.13 \times 10^{-24} \, \text{mol} \]
The number of hydrogen ions moved into the lysosome is the difference between these two quantities, scaled to the number of molecules using Avogadro’s number (\(6.022 \times 10^{23}\)).
\[
\Delta n = n_{pH=5} - n_{pH=7.2} \approx 6.545 \times 10^{-22} - 4.13 \times 10^{-24} \approx 6.5 \times 10^{-22} \, \text{mol}
\]
Number of hydrogen ions transferred:
\[
N_{H^+} = \Delta n \times 6.022 \times 10^{23} \approx 6.5 \times 10^{-22} \times 6.022 \times 10^{23} \approx 39.1\, \text{ions}
\]
Hence, approximately 39 hydrogen ions are pumped into the lysosome to change its internal pH from 7.2 to 5.
Problem 4: Potassium in Liposomes and Equilibrium Time
Liposomes serve as model systems to study membrane transport. With a diameter of 4 μm, each liposome's volume can be computed, and from the internal potassium concentration, the initial number of potassium ions determined. Additionally, understanding the kinetics of potassium leakage to reach equilibrium involves considering the number of ions that need to exit and the rate at which they do so.
Part (a): Initial number of potassium ions
Again, compute the volume of a liposome:
\[ r = 2\, \text{μm} \]
\[ V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (2)^3 = \frac{4}{3} \pi \times 8 \approx 33.51\, \text{μm}^3 \]
Convert volume to molar amount: knowing that 1 μm³ corresponds to \(10^{-15}\) liters, the volume in liters is:
\[ V_{L} = 33.51 \times 10^{-15} = 3.351 \times 10^{-14} \, \text{L} \]
With an internal potassium concentration of 100 mM (0.1 mol/L), the initial moles of K⁺ in the liposome are:
\[ n_{K} = 0.1 \times 3.351 \times 10^{-14} = 3.351 \times 10^{-15} \, \text{mol} \]
Number of potassium ions initially present:
\[
N_{K} = 3.351 \times 10^{-15} \times 6.022 \times 10^{23} \approx 2.017 \times 10^{9} \, \text{ions}
\]
Part (b): Time to reach equilibrium
Assuming the environment is potassium-free and the liposome starts with approximately 2 billion potassium ions, the ions needed to leak out to reach equilibrium is essentially all of them, as the external potassium is negligible. The leakage rate is determined by the total number of pores, their transport rate, and the opening cycle.
Total potassium ions transported per cycle per pore:
\[
3 \times 10^{6} \, \text{ions/sec} \times 0.3\, \text{sec} = 9 \times 10^{5} \, \text{ions}
\]
per pore per cycle.
For 10 pores, total ions per cycle:
\[
9 \times 10^{5} \times 10 = 9 \times 10^{6} \, \text{ions}
\]
Number of cycles needed:
\[
\frac{N_{K}}{9 \times 10^{6}} \approx \frac{2.017 \times 10^{9}}{9 \times 10^{6}} \approx 224
\]
cycles.
Since each cycle takes 2.3 seconds, total time:
\[
224 \times 2.3 \approx 515\, \text{seconds} \approx 8.58\, \text{minutes}
\]
Thus, approximately 8.6 minutes are required for potassium ions to leak out and reach equilibrium. This simplified calculation assumes continuous operation and no additional retaining mechanisms.
Problem 5: Glycophorin Structure and Sugar Residues
Glycophorin is a single-pass transmembrane protein in red blood cells, with a length of 131 amino acids. It has carbohydrate attached residues, amounting to about 100 per molecule, which constitute about 60% of its mass. The average molecular weight of an amino acid is 130 Daltons.
Part (a): Molecular weight of modified sugar residues
The protein component length in amino acids: 131 amino acids.
Total protein mass:
\[
131 \times 130 \, \text{Da} = 17,030\, \text{Da}
\]
Assuming the total mass per glycophorin molecule is:
\[
\text{Mass of sugars} = 0.60 \times \text{Total mass}
\]
Total mass:
\[
\text{Total} = \text{protein mass} + \text{sugar mass}
\]
Let the mass of the sugars be \( M_s \):
\[
M_s = 0.60 \times M_{total}
\]
and
\[
M_{total} = M_{protein} + M_s = 17,030\, \text{Da} + M_s
\]
which yields:
\[
M_s = 0.60 \times (17,030 + M_s) \Rightarrow M_s = 0.60 \times 17,030 + 0.60 \times M_s
\]
\[
M_s - 0.60 \times M_s = 0.60 \times 17,030
\]
\[
0.40 \times M_s = 10,218
\]
\[
M_s = \frac{10,218}{0.40} = 25,545\, \text{Da}
\]
Number of sugar residues:
\[
\text{Number} = 100
\]
Total sugar mass:
\[
\boxed{
\text{Mass per sugar residue} = \frac{25,545\, \text{Da}}{100} = 255.45\, \text{Da}
}
\]
Answer: The average molecular weight of each modified sugar residue is approximately 255.45 Daltons.
Part (b): Number of modified sugar residues per RBC
Number of glycophorin molecules per RBC:
\[
6 \times 10^5
\]
Total sugar residues:
\[
6 \times 10^5 \times 100 = 6 \times 10^7
\]
Part (c): Mass of the glycophorin protein component in one RBC
Mass per protein component:
\[
17,030 \, \text{Da} = 17,030\, \text{amu}
\]
Expressed in grams:
\[
17,030\, \text{Da} \times 1.66054 \times 10^{-24}\, \text{g/Da} \approx 2.83 \times 10^{-20}\, \text{g}
\]
Total mass in one RBC:
\[
6 \times 10^5 \times 2.83 \times 10^{-20} \approx 1.70 \times 10^{-14}\, \text{g}
\]
Hence, the protein component of glycophorin in one RBC weighs approximately \(1.70 \times 10^{-14}\) grams.
References
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