Select the statements that are true for pollination mechanism in flowering plants from the given options:
Show Hint
- In Vallisneria, the female flowers are pollinated by pollen grains inside the water.
- In Zostera, pollen grains are released on the surface of water.
- In most of the water-pollinated species, pollen grains are covered by a mucilaginous coating.
Pollination by water is quite rare and limited to about 30 genera.
The Correct Option is D
Solution and Explanation
(ii) False: In Zostera, pollen grains are also released underwater, not on the surface.
(iii) True: Mucilaginous coating protects pollen in water-pollinated species.
(iv) True: Water pollination is rare and restricted to about 30 genera.
Top Questions on sexual reproduction in flowering plants
- Select the statements that are true for a typical megasporangium of flowering plants:
(i) It is attached to the placenta by hilum.
(ii) It has a chalaza end that represents the basal part of the ovule.
(iii) It has nucellus enclosed by integuments.
(iv) Its micropylar end is known as chalaza.
Choose the correct answer:- CBSE CLASS XII - 2025
- Biology
- sexual reproduction in flowering plants
Given below is a flower with its characteristic features specialised for the most common type of abiotic pollination.
Answer the following questions based on the above diagram:
(a) Name the mode of abiotic pollination that will be adopted by the given plant species in the above picture.
(b) State the need of exposed large feathery stigmas for the flower.
(c) What will be the two important adaptations in the pollen grains of the flowers pollinated by the above mode of pollination?
(d)What could be the probable reason for the petals being small and non-green?- CBSE CLASS XII - 2025
- Biology
- sexual reproduction in flowering plants
- Assertion (A): Perisperm is a diploid tissue.
Reason (R): Perisperm is the remains of nucellus which surrounds the embryo in certain seeds.- CBSE CLASS XII - 2025
- Biology
- sexual reproduction in flowering plants
(a) (i) Explain the structure of a typical monocotyledonous embryo of a flowering plant.
(ii) How are multiple embryos formed in a citrus fruit? What is the mechanism known as?OR
(b) (i) Name and explain the structural organization of the male sex accessory ducts in the human male reproductive system.
(ii) Describe the role of gonadotropin FSH (Follicle-Stimulating Hormone) in the regulation of spermatogenesis.- CBSE CLASS XII - 2025
- Biology
- sexual reproduction in flowering plants
- How many pollen grains and ovules are likely to be formed in the anther and ovary of an angiosperm bearing 50 microspore mother cells and 50 megaspore mother cells respectively?
- CBSE CLASS XII - 2025
- Biology
- sexual reproduction in flowering plants
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.