A 10-fold diffusion of the rabbit anti-human Netprin antibody, at a starting con

Question

A 10-fold diffusion of the rabbit anti-human Netprin antibody, at a starting concentration of 1.0mg/ml would yield which of the following concentrations? 0.1 mg/ml 0.01 mg/ml 0.001 mg/ml 1 mg/ml Use the standard graph below to estimate the molecular weight of paramyosin which migrates 7 in an SOS-agarose gel 105,000 6, 500 32,000 55,000 Ricin is a highly toxic, naturally occurring lectin that is produced in the seeds of the castor oil plant. Ricin binds complex carbohydrates on the surface of eukaryotic cells. To prevent ricin toxicity you must identify the carbohydrate that Ricin binds. To achieve this you perform an experiment on your cheek epithelial cells like you did in cell bio lab using peroxidase linked to ricin. The four carbohydrates you test are fucose, glucose, mannose and galactose. You find that a purple color develops when you add fucose, glucose and mannos in excess but there is no color when you add galactose. Which sugar residue does Ricin normally bind to? fucose glucose galactose mannose In your tissue printing lab some vegetables stained poorly for peroxidase (purpie) but showed intense staining with the protein stain (red). Which statement below is the best conclusion of this observation? There are low quantities of peroxidase compared to the amount of total protein present There are high quantities of peroxidase compared to the amount of total protein present There are equal quantities of peroxidase and other proteins present You cannot make a conclusion about peroxidase and protein quantities based on this observation Which structure in the trachea contains the highest concentration of tubulin? cilia of the epithelium cartilage layer muscle layer nuclei of the epithelium

Explanation / Answer

1.c(1/10x1/100=0.001)

2.Rf = migration distance of the protein
Migration distance of the dye front

the unknown protein has an Rf of 0.7084. Using the equation for the linear plot we can calculate the Log (MW). (-2.0742 x 0.7084) +2.8 = 1.3305. So the inverse log is 10^1.3305= 21.4kDa for the molecular weight of the unknown protein.

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