In: Chemistry
Proteins can move laterally in biomembranes and the lateral distance (r) traveled in a given time (t) is given by, The diffusion constant for (D) for the membrane protein fibronectin is approximately 0.7 x 10-12 cm2·sec-1, whereas that for rhodopsin is about 3.0 x 10-9 cm2·sec-1. a. calculate the distance (r) traveled by each of these proteins in 10 msec (answer in nm; be sure to circle your answers) b. what can you reasonably assume about the interactions of these two proteins with other membrane components? Your answer should be 1-4 sentences.
Average distance calculated by formula( refer equation below)
Where x is the distance travelled by protein, D is diffusion constant anf t is time in seconds
Putting values given in the question
X2 = 2 × D × t
For fibronectin ------- X2 = 2 × 0.7 x 10-12 cm2•sec-1 × 10 msec
X = 1.1832 × 10-7 cm = 1.1832 nm-----------------------------------------------------1
For rhodopsin ------ X2 = 2 × 3.0 x 10-9 cm2•sec-1 × 10 msec
X =0.0000077459 cm = 77.459 nm ----------------------------------------------------2
Biological membranes are not rigid structure in contrast lipids of biological membrane and proteins present in the membrane are in constant lateral motion called as lateral diffusion. From the above calculation it is clear that rhodopsin travels a longer distance than fibronectin in given time. Also diffusion constant for rhodopsin is higher than fibronectin. This may be due to amino acid composition and structure of rhodopsin. Rhodopsin is known to undergo rotational and lateral Brownian motion in the plane of photoreceptor membrane. This movement of protein is very essential in phototransduction cascade. Mobility of protein differs depending upon its function in the cell. Now fibronectin is having low mobility because it is anchored to actin filaments on the inside of the plasma membrane through integrin, a transmembrane protein that links the extracellular matrix to the cytoskeleton whereas the rapid movement of rhodopsin is essential for fast signalling.