Question
Percentage of $$Se$$ in peroxidase anhydrase enzyme is $$0.5\% $$ by weight $$\left( {at.{\text{ weight = 78}}{\text{.4}}} \right),$$ then minimum molecular weight of peroxidase anhydrase enzyme is
A.
$$1.568 \times {10^3}$$
B.
$$15.68$$
C.
$$2.168 \times {10^4}$$
D.
$$1.568 \times {10^4}$$
Answer :
$$1.568 \times {10^4}$$
Solution :
Suppose the molecular weight of enzyme $$ = x$$
$$0.5\% $$ by weight means in $$100g$$ of enzyme weight of $$Se = 0.5g$$
$$\therefore $$ In $$xg$$ of enzyme weight of $${\text{ }}Se = \frac{{0.5}}{{100}} \times x$$
$$\eqalign{
& {\text{Hence,}}\,{\text{78}}{\text{.4 = }}\frac{{0.5 \times x}}{{100}} \cr
& \therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 15680 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1.568 \times {10^4} \cr} $$