Assertion (A): ABO blood grouping in humans is an example of multiple allelism.
Reason (R): More than two genes in a population govern the same character in ABO blood grouping in humans.
Reason (R): More than two genes in a population govern the same character in ABO blood grouping in humans.
Show Hint
- Both (A) and (R) are true and (R) is the correct explanation of (A).
- Both (A) and (R) are true, but (R) is not the correct explanation of (A).
- (A) is true, but (R) is false.
- (A) is false, but (R) is true.
The Correct Option is C
Solution and Explanation
Step 2: The reason incorrectly states that more than two genes are involved. In reality, it is one gene with more than two alleles.
Step 3: Assertion is true, reason is false.
Top Questions on Principles of Inheritance and Variation
Thomas Hunt Morgan carried out several dihybrid crosses in Drosophila melanogaster to study genes that were sex linked.
(a) Give four major reasons for using the tiny fruit flies by Morgan for his experiments.
(b) How did Morgan and his group explain the physical association of the two genes on a chromosome to the frequency of recombination between gene pairs on the same chromosome?- CBSE CLASS XII - 2025
- Biology
- Principles of Inheritance and Variation
Answer either option-(A) or (B): (A)
- Name the two genes encoding for the Bt toxin protein used in biotechnology for the control of cotton bollworms.
- How does the expression of Bt toxin gene in Bt cotton plant help in providing resistance to cotton bollworms? Explain.
OR (B)
- Explain the given methods of introducing the alien DNA in the host cells:
- (I) Biolistics
- (II) Microinjection
- How is E. coli made 'competent' to take up a recombinant plasmid or DNA?
- CBSE CLASS XII - 2025
- Biology
- Principles of Inheritance and Variation
- A colourblind man marries a woman with normal sight who has no history of colourblindness in her family. What is the probability of their son being colourblind?
- CBSE CLASS XII - 2025
- Biology
- Principles of Inheritance and Variation
- Gregor Mendel conducted hybridisation experiments in garden pea for seven years and proposed the law of inheritance.
Why was he successful in his hybridisation experiments? Give two reasons.
State the law of independent assortment as proposed by Mendel after his dihybrid crosses.- CBSE CLASS XII - 2025
- Biology
- Principles of Inheritance and Variation
- In a pea plant (Pisum sativum) inflated pod shape is dominant over constricted pod shape. The expected ratio of phenotypes of the offspring in a cross between both the parents with heterozygous inflated pod shape will be:
- CBSE CLASS XII - 2025
- Biology
- Principles of Inheritance and Variation
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.