This includes questions from Class 12 Biology chapters.
Q 1: Vegetative cell of pollen grain is
(1) Smaller than Generative cell
(2) Having abundant food reserve
(3) Spindle shaped with dense cytoplasm
(4) With a small spherical nucleus
✅ (2) Having abundant food reserve
▶️ A matured pollen grain contains 2 cells:
1. Vegetative cell: It is bigger, has abundant food reserve and a large irregularly shaped nucleus.
2. Generative cell: It is small and floats in the cytoplasm of the vegetative cell. It is spindle shaped with dense cytoplasm and a nucleus.
Q 2: Viability of pollen grains of some cereals such as rice, wheat etc. is
(1) 30 minutes
(2) 24-48 hrs
(3) Several months
(4) 1-2 years
✅ (1) 30 minutes
▶️ The viability period of pollen grains is variable. It depends on temperature and humidity.
▶️ Viability of pollen grains of some cereals (rice, wheat etc.) is 30 minutes.
▶️ Some members of Leguminosae, Rosaceae & Solanaceae have viability for months.
Q 3: Which is true regarding the cytoplasm of sperm and ovum?
(1) Ovum = sperm
(2) Ovum > sperm
(3) Sperm > ovum
(4) Both have no cytoplasm
✅ (2) Ovum > sperm
▶️ The sperm is small sized cell and has a small amount of cytoplasm.
▶️ The ovum is large sized cell with a large amount of cytoplasm.
Q 4: In oogenesis, second meiotic division occurs
(1) At embryonic stage
(2) Soon after birth
(3) Prior to ovulation
(4) During fertilization
✅ (4) During fertilization
▶️ At embryonic stage: Egg mother cells (oogonia) multiply to form primary oocytes. They enter prophase-I of the meiosis and get temporarily arrested at that stage.
▶️ Prior to ovulation: The primary oocyte undergoes first meiotic division to form a large secondary oocyte (n) & a tiny first polar body (n).
▶️ During fertilization: Secondary oocyte undergoes second meiotic division to form an ovum (ootid) and a second polar body.
Q 5: Which of the following sets of STIs are not completely curable?
(1) Hepatitis-B, Syphilis & HIV
(2) Syphilis, Gonorrhoea & HIV
(3) Gonorrhoea, chlamydiasis & HIV
(4) Hepatitis -B, genital herpes & HIV
✅ (4) Hepatitis -B, genital herpes & HIV
▶️ Diseases or infections transmitted through sexual intercourse are called Sexually transmitted diseases/ infections (STDs or STIs)/Venereal diseases (VD) or Reproductive tract infections (RTI).
▶️ Except hepatitis-B, genital herpes & HIV, other diseases are completely curable if detected early and treated properly.
Q 6: Four children belonging to same parents have the blood groups A, B, AB & O. The genotypes of the parents are
(1) Both parents are heterozygous for A group
(2) One parent is heterozygous for A and another parent is homozygous for B
(3) One parent is heterozygous for A and another parent is heterozygous for B
(4) Both parents are heterozygous for B group
✅ (3) One parent is heterozygous for A and another parent is heterozygous for B
▶️ If the children have A, B, AB, and O blood groups, then one parent must have the genotype IAi (heterozygous A) and the other parent must have the genotype IBi (heterozygous B).Parents: IAi X IBi
Q 7: What is correct about linked genes?
(1) Segregation occurs but no dominance
(2) No segregation but independent assortment
(3) No segregation, no dominance
(4) Segregation but no independent assortment
✅ (4) Segregation but no independent assortment
▶️ Segregation refers to the separation of two alleles of a gene during the formation of gametes.
▶️ Independent assortment refers to the independent segregation of genes located on different chromosomes.
▶️ Linked genes are genes that are physically close to one another on the same chromosome. This phenomenon is known as genetic linkage.
▶️ Linked genes segregate, but they do not assort independently because their close proximity on the chromosome causes them to be inherited together.
Q 8: The portion of a eukaryotic gene which is transcribed but not translated is
(1) Exon
(2) Intron
(3) Cistron
(4) Codon
✅ (2) Intron
▶️ Introns are non-coding regions of a gene that are initially transcribed into the pre-mRNA molecule but are later removed during RNA processing. The remaining coding regions (exons) are then spliced together to form the mature mRNA that is translated into a protein.
▶️ A cistron is a segment of DNA coding for a polypeptide.
▶️ A codon is a DNA or RNA sequence of three nucleotides that forms a unit of genomic information encoding a particular amino acid or signaling the termination of protein synthesis.
Q 9: Termination of polypeptide chain is brought about by
(1) UUG, UAG & UGG
(2) UAA, UAG & UGA
(3) UUA, UGA & UCA
(4) UAU, UAG & UGU
✅ (2) UAA, UAG & UGA
▶️ UAA, UAG, and UGA are known as Stop codons (termination codons or nonsense codons). These are nucleotide triplets in mRNA that signal the termination of protein synthesis during translation.
▶️ When a ribosome encounters a stop codon in the mRNA sequence, it signals the end of translation, and the polypeptide chain is released from the ribosome.
▶️ Stop codons do not code for any amino acids.
Q 10: Embryological support for evolution was disapproved by
(1) Karl Ernst von Baer
(2) Alfred Wallace
(3) Ernst Haeckel
(4) Louis Pasteur
✅ (1) Karl Ernst von Baer
▶️ Karl Ernst von Baer disagreed with Ernst Haeckel’s theory, which proposed that an organism’s embryological development repeats its evolutionary history.
▶️ Alfred Wallace came to the same conclusion of Natural selection theory proposed by Charles Darwin.
▶️ Louis Pasteur disproved Theory of spontaneous generation (Abiogenesis).
Q 11: Lamarckism was rejected because
(1) It is proved that there is no evidence for acquired characters
(2) It is proved that the characters are inherited only through genes
(3) It is proved that there was no short-necked giraffe on ancient earth
(4) All of these
✅ (2) It is proved that the characters are inherited only through genes
▶️ Lamarckism states that evolution of life forms occurred by the inheritance of acquired characters developed by use & disuse of organs.
▶️ E.g. Long neck of giraffe is due to continuous elongation to forage leaves on trees. This acquired character inherited to succeeding generations.
▶️ This theory was eliminated because it is proved that the characters are inherited only through genes.
Q 12: The infectious stage of Plasmodium that enters the human body is
(1) Trophozoites
(2) Sporozoites
(3) Female gametocytes
(4) Male gametocytes
✅ (2) Sporozoites
▶️ Plasmodium species cause Malaria in human.
▶️ The infectious stage of Plasmodium that enters the human body is Sporozoites.
▶️ These are transmitted to humans by the bite of female Anopheles mosquito.
▶️ Once in the human body, the sporozoites travel to the liver, where they invade liver cells and begin the process of replication.
Q 13: Which of the following sets of diseases is caused by bacteria?
(1) Pneumonia and Typhoid
(2) Typhoid and Malaria
(3) Tetanus and Filariasis
(4) Pneumonia and Common cold
✅ (1) Pneumonia and Typhoid
Diseases and pathogens:
▶️ Pneumonia – Streptococcus pneumoniae and Hemophilus influenzae (Bacteria).
▶️ Typhoid – Salmonella typhi (a bacterium).
▶️ Tetanus – Clostridium tetani (a bacterium).
▶️ Malaria – Plasmodium species (Protozoa).
▶️ Filariasis – Wuchereria sp. (Helminth).
▶️ Common cold – Rhinovirus (Virus)
Q 14: Choose the wrong combination.
(1) Azospirillum - Nitrogen fixation
(2) Cyanobacteria - Nostoc
(3) Trichoderma - Biofertilisers
(4) Methanobacterium - Rumen
✅ (3) Trichoderma - Biofertilisers
▶️ Trichoderma is used in biocontrol agents to protect plants from pathogens. Biocontrol is the use of biological methods for controlling plant diseases and pests.
▶️ Biofertilisers are organisms that enrich the nutrient quality of the soil. E.g., Rhizobium, Azospirillum and Azotobacter. They fix atmospheric N2.
Q 15: The sticky ends of DNA strands formed by the action of restriction enzymes facilitates action of
(1) DNA polymerase
(2) DNA ligase
(3) RNA primase
(4) DNA helicase
✅ (2) DNA ligase
▶️ DNA ligase is an enzyme that facilitates the joining of DNA strands together by forming a phosphodiester bond. In genetic engineering, DNA ligase is used to paste a DNA fragment of interest into a vector (ligation).
▶️ DNA Polymerase synthesizes new DNA by adding nucleotides to a DNA strand. It is a key player in DNA replication.
▶️ RNA Primase synthesizes a short RNA segment (primer) which serves as a starting point for DNA replication.
▶️ DNA Helicase unwinds the DNA helix during DNA replication, separating the two strands and allowing them to serve as templates for new DNA synthesis.
Q 16: The function of a selectable marker is to
(1) Eliminate transformants and non-transformants
(2) Eliminate non-transformants and permit transformants
(3) Eliminate transformants and permit non-transformants
(4) Identify and fix transformants
✅ (2) Eliminate non-transformants and permit transformants
▶️ Selectable marker (marker gene) is a gene that helps to select the transformants and eliminate the non-transformants.
▶️ If a piece of DNA is introduced in a host bacterium, it is called transformation. Such bacterium is transformant. If transformation does not occur, it is non-transformant.
▶️ Selectable markers of E. coli include the genes encoding resistance to antibiotics like ampicillin, tetracycline, chloramphenicol, kanamycin etc. Normal E. coli cells have no resistance against these antibiotics.
Q 17: The pro-insulin contains an extra stretch called
(1) A peptide
(2) B peptide
(3) C peptide
(4) D peptide
✅ (3) C peptide
▶️ Insulin consists of two short polypeptide chains (chain A & chain B) that are linked by disulphide bridges.
▶️ In mammals, insulin is synthesized as a pro-hormone (pro-insulin). It is processed to become mature and functional hormone.
▶️ The pro-insulin contains an extra stretch called C peptide. This is removed during maturation into insulin.
Q 18: If two species compete for the same resource, they could avoid competition by choosing different times for feeding or different foraging patterns. This is called
(1) Competitive release
(2) Interference competition
(3) Resource partitioning
(4) Struggle for existence
✅ (3) Resource partitioning
▶️ In Interference competition, the feeding efficiency of one species is reduced due to the interfering and inhibitory presence of other species, even if resources are abundant.
▶️ Competitive release: It is the expansion of distributional range of a species when the competing species is removed.
▶️ Struggle for existence: It is the competition among organisms for resources so that population size is limited.
Q 19: If 30 J of energy is trapped at producer level, then how much energy will be available to peacock as food in the following chain? Plant → Mice → Snake → Peacock
(1) 0.3 J
(2) 0.03 J
(3) 0.003 J
(4) 0.0003 J
✅ (2) 0.03 J
▶️ In a food chain, about 10% of the energy is transferred from one trophic level to the next. This is called 10% rule.
▶️ At the producer level (plants), there is 30 J of energy.
▶️ The energy available to primary consumer (mice) = 3 J. (10% of 30 J)
▶️ The energy available to secondary consumer (snake) = 0.3 J. (10% of 3 J)
▶️ The energy available to the tertiary consumer (peacock) = 0.03 J. (10% of 0.3 J)
Q 20: Which of the following is not broadly utilitarian argument?
(1) Pollination
(2) Provision of oxygen
(3) Aesthetic value of species
(4) Intrinsic value of species
✅ (4) Intrinsic value of species
Reasons for biodiversity conservation:
▶️ Narrowly utilitarian arguments: Human derive economic benefits from nature such as food, firewood, fibre, construction material, industrial products etc.
▶️ Broadly utilitarian arguments: Biodiversity has many ecosystem services such as Oxygen production, pollination, aesthetic pleasures etc.
▶️ Ethical arguments: Every species has an intrinsic value. We have a moral duty to care for their well-being.