A 25,000 Da protein contains a single binding site for a molecule (ligand), whose molecular weight is 2,500 Da. Assuming high affinity and physiologically irreversible binding, the amount of the ligand required to occupy all the binding sites in 10 mg protein will be

Question

A 25,000 Da protein contains a single binding site for a molecule (ligand), whose molecular weight is 2,500 Da. Assuming high affinity and physiologically irreversible binding, the amount of the ligand required to occupy all the binding sites in 10 mg protein will be (A) 0.1 mg (B) 1 mg (C) 10 mg (D) 100 mg

Tardigrade Admin · Accepted Answer

Assuming x as the amount of ligand to occupy all the binding sites in 10 mg protein. (x mg /Protein in grams) = (Ligand molecular weight/Protein molecular weight) x=(2500/25000)× 10=1mg

Q. A 25,000 Da protein contains a single binding site for a molecule (ligand), whose molecular weight is 2,500 Da. Assuming high affinity and physiologically irreversible binding, the amount of the ligand required to occupy all the binding sites in 10 mg protein will be

0.1 mg

1 mg

10 mg

100 mg

Assuming $x$ as the amount of ligand to occupy all the binding sites in 10 mg protein.
$\frac{x m g}{P ro t e in in g r am s}$ $= \frac{L i g an d m o l ec u l a r w e i g h t}{P ro t e inm o l ec u l a r w e i g h t}$
$x = \frac{2500}{25000} \times 10 = 1 m g$

Q. A 25,000 Da protein contains a single binding site for a molecule (ligand), whose molecular weight is 2,500 Da. Assuming high affinity and physiologically irreversible binding, the amount of the ligand required to occupy all the binding sites in 10 mg protein will be

0.1 mg

1 mg

10 mg

100 mg

Assuming x as the amount of ligand to occupy all the binding sites in 10 mg protein. Proteiningramsxmg​ =ProteinmolecularweightLigandmolecularweight​ x=250002500​×10=1mg

Assuming $x$ as the amount of ligand to occupy all the binding sites in 10 mg protein.
$\frac{x m g}{P ro t e in in g r am s}$ $= \frac{L i g an d m o l ec u l a r w e i g h t}{P ro t e inm o l ec u l a r w e i g h t}$
$x = \frac{2500}{25000} \times 10 = 1 m g$