Calculate the time of sedentary life and the frequency of hoppings from one membrane layer to another one of the lipids of sarcoplasmic reticulum membrane if the coefficient of lateral diffusion is D=45 µm2/sec and the area of one phospholipid molecule is А=1,9 nm2 .
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Calculate the time of sedentary life and the frequency of hoppings from one membrane layer to another one of the lipids of sarcoplasmic reticulum membrane if the coefficient of lateral diffusion is D=45 µm2/sec and the area of one phospholipid molecule is А=1,9 nm2 .
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- Phospholipid lateral motion in membranes is characterized by a diffusion coefficient of about 1 x 10-8 cm2/sec. The distance traveled in the membrane in a given time is r = √4Dt, where r is the distance traveled in centimeters is the diffusion coefficient, and t is the time during which diffusion occurs. Calculate the distance (in nanometers) traveled by a phospholipid in a bilayer in 25 msec (milliseconds).The sarcoplasmic reticulum Ca2+-ATPase, pumps 2 mol Ca2+ out of sarcomeres per mol ATP hydrolyzed. Given the following steady-state concentrations and a membrane potential of 67 mV (inside negative), calculate ΔG for the following active transport process at 37 ∘C and pH=7.4:2Ca2+(in)+ATP+H2O→2Ca2+(out)+ADP+Pi+H+Uniporters and ion channels support facilitated transport across cellular membranes. Although both are examples of facilitated transport, the rates of ion movement via an ion channel are roughly 104 - to 105 -fold faster than the rates of molecule movement via a uniporter. What key mechanisticdifference results in this large difference in transport rate?What contribution to free energy (ΔG) determines the direction of transport?
- Uniporters and ion channels support facilitated transport across cellular membranes. Although both are examples of facilitated transport, the rates of ion movement via an ion channel are roughly 104- to 105-fold faster than the rates of molecule movement via a uniporter. What key mechanistic difference results in this large difference in transport rate? What contribution to free energy (ΔG) determines the direction of transport?The sarcoplasmic reticulum Ca2+-ATPase, pumps 2 mol Ca2+ out of sarcomeres per mol ATP hydrolyzed. Part A: Given the following steady-state concentrations and a membrane potential of 64 mV (inside negative), calculate ΔG for the following active transport process at 37 ∘C and pH=7.4: 2Ca2+(in)+ATP+H2O→2Ca2+(out)+ADP+Pi+H+ ATP=2.8mM,ADP=206μM,Pi=5.4mM,Ca2+(in)=34μM,Ca2+(out)=2.2mM The answer to part A was -7.4 kJ/mol I need help with Part B: Part B: The activity of the Ca2+-ATPase is regulated reversibly under normal conditions to maintain homeostatic concentrations of Ca2+ inside the sarcomere. However, in a rare genetic disorder, irreversible activation of the Ca2+-ATPase can occur. Assuming 37 ∘C, pH=7.4, and the steady-state concentrations for ATP, ADP Pi, and Ca2+(out) given in part (a) calculate the minimum [Ca2+] inside a sarcomere that has irreversibly activated Ca2+-ATPase (i.e., the Ca2+-ATPase activity is always “on”).Diffusion and osmosis classification Classify the following characteristics based on whether they are describing diffusion, osmosis, or both. Diffusion Results in an Can occur with equal distribution of solute molecules or without a membrane Always involves the movement of water Requires a semi-permeable membrane Osmosis Involves the movement of gases, ions, and small water soluble molecules Passive form of movement that requires no energy Moves from areas of high concentration to low Both Diffusion and Osmosis Responsible for gas exchange in the lungs concentration A 3 of 15 Next > Cation to open the document "Epicinstaller-13.0.0-fortnite-a8e4f12cada646caa706d8be407be69f (3).msi". tv 22
- Estimate the flux (mg/cm2/s) by diffusion of estrogen (a steroid) through a lipid bilayer cell membrane when assuming the diffusion coefficient for estrogen across the lipid bilayer is 10^–6 cm2/s, and that the initial concentration of estrogen in the extracellular fluid is 1 ng/mL and 0 in the cytoplasm.Describe what gap junction are, emphasizing on how they are different two other cellular points of contact, their protein components and cellular function. Describe the three main sources of membrane fluidity. Describe in detail what proteoglycans are, including their hygroscopicity and biological roles.Analogs of hemidesmosomes are the focal contact sites, which are also sites where the cell attaches to the extracellular matrix. These junctions are prevalent in fibroblasts but largely absent in epithelial cells. on the other hand, hemidesmosomes are prevalent in epithelial cells but absent in fibroblasts. In focal contact sites, intracellular connections are made to actin filaments, whereas in hemidesmosomes connections are made to intermediate filaments. Why do you suppose these two different cell types attach differently to the extracellular matrix?
- For each of the following scenarios described where a molecule or ion is moving from one side of a membrane to the other, select the method by which the molecule or ion is moving. Each answer can be used more than once, or not at all. - Simple Diffusion - Facilitated diffusion by a channel protein - Facilitated diffusion by a carrier/transport protein - Active transport by a pump - Could be facilitated diffusion by a channel or a carrier; not enough information is given A- While water can freely diffuse across the membrane, it does not do so fast enough for living organisms to function properly. Therefore, membrane proteins known as aquaporins can increase the rate at which water moves across the membrane. The movement of water across the membrane via aquaporins (which do not change shape) is an example of which type of transport? B-Many snake venoms induce paralysis by acting on acetylcholine receptors. Nicotinic acetylcholine receptors are transmembrane proteins that allow Na+, K+…Usually , rates of diffusion vary inversely with molecular weights; so smaller molecules diffuse faster than do larger ones. In cells, however, calcium ion diffuses more slowly than does cAMP. Propose a possible explanation.The average time it takes for a molecule to diffuse adistance of x cm is given byt = x2/2D where t is the time in seconds and D is the diffusioncoefficient. Given that the diffusion coefficient ofglucose is 5.7 × 10−7cm2/s, calculate the time it wouldtake for a glucose molecule to diffuse 10 μm, which isroughly the size of a cell.