Register

GATE BT 2022 | Question: 30

• edited by
0 votes
0 votes

Match the stationary phase (Column I) with its corresponding chromatography technique (Column II).

$$\begin{array} {lll} \textbf{Column I} & & \textbf{Column II} \\ \text{P. Protein A}  & & \text{1. Size exclusion chromatography} \\ \text{Q. Sephadex} & &  \text{2. IOn-exchnage chromatography} \\ \text{R. Phenylsepharose} & &  \text{3. Affinity chromatography} \\ \text{S. Diethylaminoethyl cellulose} & & \text{4. Hydrophobic interaction chromatography} \end{array}$$

  1. $\text{P-1, Q-4, R-2, S-3}$
  2. $\text{P-3, Q-1, R-4, S-2}$ 
  3. $\text{P-3, Q-4, R-2, S-1}$
  4. $\text{P-4, Q-1, R-3, S-2}$
• edited by
User Avatar

Please log in or register to answer this question.

Voting & Accepting an answer helps improve the community
Answer:

Related questions

0 votes
0 votes
0 answers
1
Lakshman Bhaiya asked Mar 20, 2022
GATE BT 2022 | Question: 1
What is the order of the differential equation given below? $\dfrac{d^{2}y}{dx^{2}} – 6x = 3x^{4} – 2x^{3} + 2$ $1$ $2$ $3$ $4$
What is the order of the differential equation given below?$\dfrac{d^{2}y}{dx^{2}} – 6x = 3x^{4} – 2x^{3} + 2$$1$$2$$3$$4$
0 votes
0 votes
0 answers
2
Lakshman Bhaiya asked Mar 20, 2022
GATE BT 2022 | Question: 2
If the eigenvalues of a $2 \times 2$ matrix $P$ are $4$ and $2,$ then the eigenvalues of the matrix $P^{-1}$ are $0, 0$ $0.0625, 0.25$ $0.25, 0.5$ $2, 4$
If the eigenvalues of a $2 \times 2$ matrix $P$ are $4$ and $2,$ then the eigenvalues of the matrix $P^{-1}$ are$0, 0$$0.0625, 0.25$$0.25, 0.5$$2, 4$
0 votes
0 votes
0 answers
4
Lakshman Bhaiya asked Mar 20, 2022
GATE BT 2022 | Question: 4
Match the media component (Column I) with its role (Column II). ... $\text{P-3, Q-2, R-4, S-1}$ $\text{P-3, Q-4, R-2, S-1}$
Match the media component (Column I) with its role (Column II).$$\begin{array} {ll} \textbf{Column I} & \textbf{Column I} \\ \text{P. Sucrose} & \text{1. Anti-foam agen...
0 votes
0 votes
1 answer
5
Lakshman Bhaiya asked Mar 20, 2022
GATE BT 2022 | Question: 5
The binding free energy of a ligand to its receptor protein is $-11.5 \; \text{kJ mol}^{-1}$ at $300 \; \text{K.}$ What is the value of the equilibrium binding constant? Use $\text{R = 8.314 J mol}^{-1} \text{K}^{-1}.$ $0.01$ $1.0$ $4.6$ $100.5$
The binding free energy of a ligand to its receptor protein is $-11.5 \; \text{kJ mol}^{-1}$ at $300 \; \text{K.}$ What is the value of the equilibrium binding constant?U...
Link Copied!