Online Cellular And Organismal Biology

Online Cellular And Organismal B

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 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. 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? Course Number: 211G A20 Course Name: Online – Natural History of Life Instructor: Catherine Hartkorn Individual/Partner Project - Biology Math Problems Assignment Remember to write out your calculations and explanations for ALL problems. As usual, if you need help with these problems, you may ask me for help either privately through Blackboard email, chat or by calling me, or you may also post your questions directly under the discussion for this project.

We have already gone over how to graph in Excel in the online 211 laboratory course; so if you are in my 211 laboratory course, I would prefer that you create your graphs for the problems below in Excel. If you are not in my 211 laboratory course, you may either use Excel or you may create your graphs by hand on graph paper and then scan in and attach images of your graphs to your answers. If you don’t know how to use Excel for graphing but would like to learn: please let me know, and I will go over it with you. Problem Set 5 - Problems on Plant Transport/Resource Acquisition, and Plant Response to Stimuli (3 problems): 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. What 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.

What 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 = (d 2 )/D where D is the diffusion coefficient. A reasonable value for D is 2.4 x 10 -5 m 2 /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 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 SD or LD Critical Day Length Dill LD 11 Spinach LD 13 Soybean SD 15.5 Cocklebur SD 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). Course Number: 211G A20 Course Name: Online – Natural History of Life Instructor: Catherine Hartkorn Individual/Partner Project - Biology Math Problems Assignment Remember to write out your calculations and explanations for ALL problems. 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 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. a C3 plant? (hint: first, think about how many carbons are in 1 mole of glucose) b. a C4 plant? c. a CAM plant? d. 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 à… long and 5000 à… 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 10 8 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 10 23 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 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 (assuming maximal 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 a photon of red light. Suppose blue light energy also caused fluorescence measured at a wavelength of 690 nm. e. What percentage of the blue light energy was lost as heat (again assuming maximal fluorescence)?

Paper For Above instruction

Calculating the number of hydrogen ions moved into lysosomes to alter internal pH involves several steps. First, determine the volume of a spherical lysosome with a diameter of 0.5 µm. The volume (V) of a sphere is given by V = (4/3)πr³. Here, the radius r = 0.25 µm, so:

V = (4/3) × 3.1416 × (0.25)³ ≈ (4/3) × 3.1416 × 0.015625 ≈ 0.0654 µm³.

Convert this volume to liters: since 1 µm³ = 10^-15 liters,

V = 0.0654 × 10^-15 liters = 6.54 × 10^-17 liters.

Next, determine the concentration of H+ ions at pH 5 and pH 7.2:

• [H+] at pH 5: 10^-5 mol/L
• [H+] at pH 7.2: 10^-7.2 mol/L ≈ 6.31 × 10^-8

Calculate the moles of H+ at each pH by multiplying concentration by volume:

At pH 5:

n = (10^-5) × (6.54 × 10^-17) = 6.54 × 10^-22 mol

At pH 7.2:

n = (6.31 × 10^-8) × (6.54 × 10^-17) ≈ 4.13 × 10^-24 mol

Since the lysosome initially had the hydrogen ions, the number of ions moved in is approximately:

Difference = 6.54 × 10^-22 - 4.13 × 10^-24 ≈ 6.50 × 10^-22 mol.

Number of hydrogen ions moved:

Number of ions = (6.50 × 10^-22) mol × (6.022 × 10²³) ions/mol ≈ 39.2 ions

Practically, this is about 39 hydrogen ions moved to produce the pH change inside the lysosome.

For liposomes with a diameter of 4 µm, the process is similar, but volume calculation differs. The radius r = 2 µm.

Volume V = (4/3)πr³ ≈ (4/3) × 3.1416 × 8 ≈ 33.51 µm³.

Convert to molar concentration: 100 mM = 0.1 mol/L.

Convert volume to liters: 33.51 × 10^-15 liters.

Calculate initial moles of K+ ions:

n = (0.1 mol/L) × (33.51 × 10^-15) ≈ 3.351 × 10^-15 mol.

The time for ions to reach equilibrium depends on the ion leak rate, but generally, the total number of ions to be lost equals the initial amount, assuming the external potassium concentration remains negligible. Since the pores operate in cycles (0.3 sec open, 2 sec closed), the cycle duration is 2.3 seconds.

For glycophorin's molecular weight of amino acids:

Number of amino acids = 131.

Mass of amino acid component:

Mass = 131 × 130 Da = 17,030 Da.

Modified sugars per molecule: approximately 100 residues, each about the same mass as amino acids, thus:

Mass of sugars (60% of total mass) per molecule = 17,030 Da / (1 - 0.6) = 17,030 Da / 0.4 ≈ 42,575 Da.

Mass per sugar residue:

Average weight = 42,575 Da / 100 residues ≈ 425.75 Da.

Number of sugar residues in one RBC:

6 × 10⁵ glycophorin molecules × 100 residues ≈ 6 × 10⁷ residues.

Mass of protein component in one RBC:

Total amino acids per molecule: 131 × 130 Da ≈ 17,030 Da, total per RBC:

Mass = 6 × 10⁵ molecules × 17,030 Da ≈ 1.022 × 10¹⁰ Da.

Convert to grams: 1 Da = 1.6605 × 10^-24 g, so:

Mass (g) = 1.022 × 10¹⁰ Da × 1.6605 × 10^-24 g/Da ≈ 1.696 × 10^-14 g.

Plasmodesmata surface area calculations involve the total number at high and low density, as well as the clone area of one plasmodesma end (circle):

Area of one plasmodesma end: A = πr², with r = 20 nm = 20 × 10^-3 µm.

A = 3.1416 × (20 × 10^-3)² ≈ 3.1416 × 400 × 10^-6 = 0.00125664 µm².

High density (25 per µm²):

Percentage of membrane occupied: (Number of plasmodesmata × area per end) / total membrane area.

Assuming total membrane area per µm²: At high density, total area contributed by plasmodesmata per µm² is:

25 × 0.00125664 µm² ≈ 0.0314 or 3.14%. Similar calculations follow for the low density of 0.2 per µm², resulting in approximately 0.0025 or 0.25%.

Diffusion time calculations for water vapor movement: For d = 1 mm = 0.001 m, the time t = d² / D = (0.001)² / 2.4 × 10^-5 ≈ 0.0417 seconds. For hairy leaves, distance doubles to 2 mm, so time doubles to approximately 0.0833 seconds.