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Atul Ghose
Asked: 2022-11-11T21:37:27+05:30 In:

Percentage of Se in peroxidase anhydrase enzyme is `0.5%` by weight (at. Wt. `=78.4)`, then minimum molecular weight of peroxidase anhydrase enzyme is:
A. (a)`1.576xx10^(4)`
B. (b)`1.576xx10^(3)`
C. (c )`15.76`
D. (d)`2.136xx10^(4)`

Percentage of Se in peroxidase anhydrase enzyme is `0.5%` by weight (at. Wt. `=78.4)`, then minimum molecular weight of peroxidase anhydrase enzyme is:
A. (a)`1.576xx10^(4)`
B. (b)`1.576xx10^(3)`
C. (c )`15.76`
D. (d)`2.136xx10^(4)`
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  1. Correct Answer – c
    The enzyme must contain at least one atom of Se.
    `:. 0.5 g` of Se is present in `100 g` of enzyme
    `:. 78.4 g` of Se will be present in `(100xx78.4)/0.5`
    `=1.576xx10^(4) g` of enzymes

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Julie Gaurav More
Asked: 2022-10-31T19:48:23+05:30 In:

Percentage of Se in peroxidase anhydrase enzyme is `0.5%` by weight (at. Wt. `=78.4)`, then minimum molecular weight of peroxidase anhydrase enzyme is:
A. (a)`1.568xx10^(3)`
B. (b)`15.68`
C. (c )`1.568xx10^(4)`
D. (d)`2.136xx10^(4)`

Percentage of Se in peroxidase anhydrase enzyme is `0.5%` by weight (at. Wt. `=78.4)`, then minimum molecular weight of peroxidase anhydrase enzyme is:
A. (a)`1.568xx10^(3)`
B. (b)`15.68`
C. (c )`1.568xx10^(4)`
D. (d)`2.136xx10^(4)`
Parabola
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Leave an answer

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1 Answer

  1. Correct Answer – C
    Suppose the mol. Wt. of enzyme =x
    Given `100 g` of enzyme wt of `Se=0.5 g`
    `:.` In x g of enzyme wt. of Se` =0.5/100 xx x`
    Hence `78.4=(0.5xx x)/100`
    `:. X=15680=1.568xx10^(4)`

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