Question

Please answer all questions below.

3. Joe and Jill are graduate students in a lab studying the eukaryotic ribosome. Each has the task of using sedimentation to isolate one of the large subunits that make up the ribosome. A) If Jill is in charge of the 40S subunit (has an S of 40S), and Joe, the 60S subunit (has an S of 60S), assuming they both use the same centrifuge tubes and rotor speed, whose sedimentation run will take longer and why? B) How many times longer will the slower run take than the faster run? Use equations to justify. C) Joe realizes he has to get home early, and needs his run to take half the time it usually takes. In which direction and how many fold should Joe change his rotor speed? Use equations to justify. D) After Joe and Jill have purified their subunits, they attempt to reconstitute the full eukaryotic ribosome. To characterize their assembly product, they submit it to velocity sedimentation and find that their product is 80S, not the 100S they were expecting. Assuming a 1:1 stoichiometry of large and small subunits, did something go wrong? Justify your answer. E) Can the sedimentation coefficient for the association of two species be the sum of the sedimentation coefficient for two individual species? Derive what the friction coefficient would have to be in terms of the friction coefficients and masses of each of the individual species, assume the individual components have the same partial specific volume.

Answer 1

a.) The sedimentation rate of the 60S subunit will be longer as compared to 40S subunit. This is because the S is actually Svedberg's unit of sedimentation coefficien t(1S) which is equal 10^-13 seconds. While deriving the equation the sedimentation coefficient is directly proportional to molecular weight so the 60S subunit will take longer to sediment.

sedimentation coefficient is written in equation as:

s=vt/ a

where vt is the terminal velocity. Terminal velocity of a particle in a given medium is constant because the force of gravitation (or centrifugation) is cancelled by the viscous force of the medium. In such case, terminal velocity is calculated as ratio of centrifugal force to the viscous resistance (from Stoke's law) and its equation becomes:

vt=mrω²/ 6πηr0

b.) the slower run will take double the time of the faster run i.e.; 60S subunit (MW is higher) will take double the time of the 40S subunit

BOS MW 2.500.000 MW 4,200,000 505 305 60S MW 1,600,000 MW 900,000 MW 2,800,000 MW 1,400,000 23S rRNA 16S rRNA 5S rRNA C 5S rRNA C 18S rRNA 120 nucleotides 2900 nucleotides 1540 nucleotides 28S rRNA 5.8S rRNA G 160 - nucleotides 4700 nucleotides 120 nucleotides 1900 nucleotides --33 proteins 34 proteins 21 proteins PROCARYOTIC RIBOSOME 49 proteins EUCARYOTIC RIBOSOME