The technique for the early detection of a disease based on the principle of antigen-antibody interaction is:
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- RNAi
- EST
- PCR
- ELISA
The Correct Option is D
Solution and Explanation
Top Questions on Biotechnology
Assertion (A): The meristems are grown ‘in vitro’ to obtain virus-free plants from an infected plant.
Reason (R): If the plant is infected with a virus, the roots and the stems are free of virus.- CBSE CLASS XII - 2025
- Biology
- Biotechnology
The basic scheme of the essential steps involved in the process of recombinant DNA technology is summarized below in the form of a flow diagram. Study the given flow diagram and answer the questions that follow.
(a) Name the enzyme used in Step-1 to join the cut plasmid and alien DNA.
(b) State the technical term used for Step-3.
(c) Justify the use of the same Restriction Enzyme EcoRI to cut both the vector DNA and the alien DNA.- CBSE CLASS XII - 2025
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(a) Name and explain the biotechnological strategy wherein the infection by the nematode Meloidogyne incognitia can be prevented using Agrobacterium vectors in the roots of tobacco plant by RNA interference.
- CBSE CLASS XII - 2025
- Biology
- Biotechnology
- Enlist one advantage and two disadvantages of Green Revolution.
- CBSE CLASS XII - 2025
- Biology
- Biotechnology
- Explain the amplification of gene of interest using the technique of Polymerase Chain Reaction (PCR).
- CBSE CLASS XII - 2025
- Biology
- Biotechnology
Questions Asked in CBSE CLASS XII exam
- An electric dipole of dipole moment \( \vec{p} = (0.8\,\hat{i} + 0.6\,\hat{j}) \times 10^{-29}\,\text{Cm} \) is placed in an electric field \( \vec{E} = 1.0 \times 10^7\,\hat{k}\,\text{V/m} \). Calculate the magnitude of the torque acting on it and the angle it makes with the x-axis, at this instant.
- CBSE CLASS XII - 2025
- Electric Dipole
- Calculate the binding energy per nucleon (in MeV) of a helium nucleus \(\left( ^4_2\text{He} \right)\). Given: \[ \begin{aligned} m\left(^4_2\text{He}\right) &= 4.002603\,\text{u} \\ m_n &= 1.008665\,\text{u} \\ m_H &= 1.007825\,\text{u} \\ 1\,\text{u} &= 931.5\,\text{MeV}/c^2 \end{aligned} \]
- CBSE CLASS XII - 2025
- Nuclear physics
- An electric field \( \vec{E} \) is given by:
\[ \vec{E} = \begin{cases} +100\, \hat{i} \dfrac{\text{N}}{\text{C}} & \text{for } x>0 \\ -100\, \hat{i} \dfrac{\text{N}}{\text{C}} & \text{for } x<0 \end{cases} \] A right circular cylinder of length \( 10\, \text{cm} \) and radius \( 2\, \text{cm} \), is placed such that its axis coincides with the x-axis and its two faces are at \( x = -5\, \text{cm} \) and \( x = 5\, \text{cm} \). Calculate: (a) the net outward flux through the cylinder, and (b) the net charge inside the cylinder.- CBSE CLASS XII - 2025
- Electric Flux
In a metallic conductor, an electron, moving due to thermal motion, suffers collisions with the heavy fixed ions but after collision, it will emerge with the same speed but in random directions. If we consider all the electrons, their average velocity will be zero. When an electric field is applied, electrons move with an average velocity known as drift velocity (vd). The average time between successive collisions is known as relaxation time (τ ). The magnitude of drift velocity per unit electric field is called mobility (μ). An expression for current through the conductor can be obtained in terms of drift velocity, number of electrons per unit volume (n), electronic charge (−e), and the cross-sectional area (A) of the conductor. This expression leads to an expression between current density (-j) and the electric field (E→ ). Hence, an expression for resistivity (ρ) of a metal is obtained. This expression helps us to understand increase in resistivity of a metal with increase in its temperature, in terms of change in the relaxation time (τ ) and change in the number density of electrons (n).
- CBSE CLASS XII - 2025
- Current electricity
When light travels from an optically denser medium to an optically rarer medium, at the interface it is partly reflected back into the same medium and partly refracted to the second medium. The angle of incidence corresponding to an angle of refraction 90° is called the critical angle (ic) for the given pair of media. This angle is related to the refractive index of medium 1 with respect to medium 2. Refraction of light through a prism involves refraction at two plane interfaces. A relation for the refractive index of the material of the prism can be obtained in terms of the refracting angle of the prism and the angle of minimum deviation. For a thin prism, this relation reduces to a simple equation. Laws of refraction are also valid for refraction of light at a spherical interface. When an object is placed in front of a spherical surface separating two media, its image is formed. A relation between object and image distance, in terms of refractive indices of two media and the radius of curvature of the spherical surface can be obtained. Using this relation for two surfaces of lens, ’lensemaker formula’ is obtained.