List I | List II | ||
---|---|---|---|
A | Clostridium butylicum | I | Ethanol |
B | Saccharomyces cerevisiae | II | Streptokinase |
C | Trichoderma polysporum | III | Butyric acid |
D | Streptococcus sp. | IV | Cyclosporin-A |
(A) Clostridium butylicum: This microorganism is known for producing butyric acid, a short-chain fatty acid.
(B) Saccharomyces cerevisiae: This yeast is widely used in fermentation processes to produce ethanol.
(C) Trichoderma polysporum: This fungus is used in the production of cyclosporin-A, an immunosuppressant drug.
(D) Streptococcus sp.: This bacterium produces streptokinase, an enzyme used in medical treatments to dissolve blood clots.
The correct matching is shown below:
\[ \begin{array}{|c|c|} \hline \textbf{Microorganism} & \textbf{Product} \\ \hline \textbf{A.} \ Clostridium\ butylicum & \text{Butyric Acid (III)} \\ \textbf{B.} \ Saccharomyces\ cerevisiae & \text{Ethanol (I)} \\ \textbf{C.} \ Trichoderma\ polysporum & \text{Cyclosporin-A (IV)} \\ \textbf{D.} \ Streptococcus\ sp. & \text{Streptokinase (II)} \\ \hline \end{array} \]
Thus, the correct match is:
\[ A \to III, \quad B \to I, \quad C \to IV, \quad D \to II \]
Explanation:
This concludes our solution.
The output (Y) of the given logic gate is similar to the output of an/a :
A | B | Y |
0 | 0 | 1 |
0 | 1 | 0 |
1 | 0 | 1 |
1 | 1 | 0 |
List I (Spectral Lines of Hydrogen for transitions from) | List II (Wavelength (nm)) | ||
A. | n2 = 3 to n1 = 2 | I. | 410.2 |
B. | n2 = 4 to n1 = 2 | II. | 434.1 |
C. | n2 = 5 to n1 = 2 | III. | 656.3 |
D. | n2 = 6 to n1 = 2 | IV. | 486.1 |