Question

12. You are purifying a full-length IgG antibody using recombinant expression in E. coli cells. You are using two expression plasmids, one encoding the heavy chain (53 kDa) and the other encoding the light chain (23 kDa). You have added two different affinity tags to the C-termini of each gene (+5 kDa for the heavy chain and +8 kDa for the light chain). (a) What is the molecular weight of your native antibody? (4 pts) (b) Following affinity purification, you run a gel of your eluate in the presence or absence of β-ME (see below). Label each band in the gel (6 pts)-what do you think is going on here? (2 pts) STD +BME - BME 250 150 100 75 50 37 25 15 10

Answer 1

a. Molecular Weight is 178

Heavy Chain (53+5=58) and Light Chain (23+8=31)=89

Heavy Chain (53+5=58) and Light Chain (23+8=31)=89

                                                                Total                      = 178

b.

Beta mercaptoethanol involved in the reduction of disulphide bonds present in the immunoglobulin leads to the formation of heavy chains and light chains.

(b) Following affinity purification, you run a gel of your eluate in the presence or absence of B-ME (see below). Label each band in the gel (6 pts)-what do you think is going on here? (2 pts) STD + BME - BME 2Heavy chains +2Lgith chains 178 kD 250 150 100 75 1 Heavy chains 58 kD 50 1 light chains 31 kD 37 Tag protein 5+8+5+8-24 25 15 Tag protein 5+8+5+8-24 10

C. It can bind the anion exchange column

(d) You run another gel. Assuming your purification was successful, draw the result you expect in the presence and absence of B-ME. (6 pts) STD BME BME 2Heavy chains +2Lgith chains 250 150 100 75 152 kD 1Heavy chains 53 kD 50 37 25 1 light chains 23 kD 15 10

e

53 KD look D 2.3 kD AD haiv