(a) (i) Explain how does double fertilisation take place in a flowering plant.
(ii) Write the fate of the products of double fertilization in these plants.
OR
(b) (i) Explain the structure of testicular lobules in the human male reproductive system. Name the two types of cells present in the seminiferous tubules and state their role.
(ii) Describe the role of hypothalamic hormone GnRH in spermatogenesis.
(a) (i) Explain how does double fertilisation take place in a flowering plant.
(ii) Write the fate of the products of double fertilization in these plants.
OR
(b) (i) Explain the structure of testicular lobules in the human male reproductive system. Name the two types of cells present in the seminiferous tubules and state their role.
(ii) Describe the role of hypothalamic hormone GnRH in spermatogenesis.
Show Hint
Solution and Explanation
(a) (A) Double fertilisation in flowering plants: It is a unique event in angiosperms where two male gametes are involved. One male gamete fuses with the egg cell to form a diploid zygote (syngamy), and the other fuses with the two polar nuclei to form a triploid primary endosperm nucleus (PEN). This process is called triple fusion. Together, syngamy and triple fusion constitute double fertilization.
(B) Fate of products:
OR
(b) (A) Structure of testicular lobules: Each testis has about 250 compartments called testicular lobules. Each lobule contains 1–3 highly coiled seminiferous tubules. These tubules are the sites of spermatogenesis. Two main types of cells are present:(D) Role of GnRH in spermatogenesis: The hypothalamus releases GnRH (Gonadotropin Releasing Hormone), which stimulates the anterior pituitary to secrete:
Top Questions on Human Reproduction
- Given below are few statements with respect to the process of oogenesis in a human ovary:
(i) Each granulosa cell gets surrounded by a primary oocyte to form the primary follicle.
(ii) A couple of million of oogonia are formed within each fetal ovary during the embryonic development.
(iii) Tertiary follicle is characterised by antrum and theca layers.
(iv) Secondary follicles form a zona pellucida layer around it.
(v) Mature graafian follicle ruptures to release the secondary oocyte from the ovary.
Choose the option with all true statements from the given options:- CBSE CLASS XII - 2025
- Biology
- Human Reproduction
Student to attempt either option (A) or (B):
(A) Write two features of an ideal contraceptive. Explain any one natural contraceptive method that makes the chances of conception almost nil.
OR
(B) Explain GIFT and ICSI.- CBSE CLASS XII - 2025
- Biology
- Human Reproduction
- Explain the process of formation of placenta in a human female after the implantation of the blastocyst in the endometrium of the uterus.
- CBSE CLASS XII - 2025
- Biology
- Human Reproduction
- Assertion (A): ‘Saheli’, an oral contraceptive inhibits ovulation and increases phagocytosis of sperms.
Reason (R): It is a non-steroidal preparation.- CBSE CLASS XII - 2025
- Biology
- Human Reproduction
- Given below are a few statements with reference to the accessory ducts of the human male reproductive system:
- [(i)] The seminiferous tubules of the testes open into rete testis then into the vas deferens.
- [(ii)] The vasa efferentia leave the testes and open into the epididymis.
- [(iii)] The epididymis leads to vas deferens that ascends into the abdomen.
- [(iv)] The vas deferens receives a duct from the prostrate gland and opens into the urethra as ejaculatory duct.
- [(v)] The urethra originates from the urinary bladder and extends through the penis to its external opening, urethral meatus.
- CBSE CLASS XII - 2025
- Biology
- Human Reproduction
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.