Why would predicted molecular weights of heavy and light chains not add up evenly to match the molecular weight of undenatured IgG antibody
The strucuture of IgG antibody shows to have two major parts, a pair of heavy and light chains. The heavy chains extend upto the length of the antibody whereas the light chains are found exclusively on the antign-binding site of the antibody. Thus, it is expected that the total summation of molecular weights of the pairs of heavy and light chains must add up to give the total molecular weight of the antibody. However, structural composition of the antibody revelas that summation of the individual chains of the antibody is different from the actual weight of the antibody. This is because the heavy and light chains are bound to each other by sulphide linkages and appear to have excess weight. On the contrary, degradation/denaturation of an antibody releases the heavy and light chains also causes degradation of some of the structural proteins as well. Thus, the final weight of the net antibody counts less than the expected summation weight.
Why would predicted molecular weights of heavy and light chains not add up evenly to match...
Antibodies are composed of two types of peptides, heavy chains and light chains. One of the major classes of antibody molecules is immunoglobulin G (IgG). Two heavy chains and two light chains form one IgG molecule. Classify each statement as describing the heavy chains, light chains, or both chains of IgG. Heavy chains Light chains Both chains
Predict the patterns (number of bands and apparent molecular weights) of the following proteins on SDS gels: a. A monomeric protein with a molecular weight of 35,000 Da b. A trimetric protein containing three chains, each with a molecular weight of 60,000 Da c. Immunoglobulin G (Nelson and Cox, page 178) in a non-reducing gel (no beta -mercaptoethanol added in the sample solution) - the light chains have a molecular weight of 25,000 Da and the heavy chains 50,000 Da...
i dont understand why the molecular weight i calculated (42.012 g/mol) for my unknown liquid (acetone) is less than the actual molecular weight (58.08 g/mol). what errors could i have made in my lab that would account for the difference. Molecular Weight of a Volatile Liquid In this experiment, an amount of liquid more than sufficient to fill the flask when vaporized is placed in a flask of measured volume and mass. The flask is then heated in a boiling...
Thumbs up for good answer. Why does the liquid in the round bottom flask turn light pink?, please explain (This happened during/after step 2). This is the procedure and the experiment is called Synthesis of Artificial Flavorings by Fischer Esterification. Thank you! SAMPLE PROCEDURES: 1. Isoanvl acetate (banana) + H2O OH HO 1. Mix 6 mL of isoamyl alcohol (0.809 g/mL, 88.148 g/mol) and 10 ml ofglacial acetic acid (1.05 g/ml 60.05 g/mol) in a 100 mL round bottom flask....
help with questions 5 to 10 please PCB 3023L Lab #4 Protocol & Worksheet (30pt) You may work in your lab groups durine class. but all written answers must be completed individually in your own words. 1) Using the plasmid map for orientation 1 and the cDNA map as a guide, complete the plasmid map for orientation #2. (4pt) 612 1318 1 - EcoRi EcoRI Xbal ECORV -Xbal- 1662 +Bell EcoRI EcoRV Not FP -- Xhol X 2015 PRSP +...