Category: Class 12 Biology Objective Questions

Biology Multiple Choice Questions on “Flowering Plants Reproduction – Abiotic Agents of Pollination”.

1. Most common abiotic pollinator is ___
a) water
b) wind
c) soil
d) temperature
Answer: b
Clarification: Plants pollinated by wind produce a large (maybe billions) amount of pollen grains into the air so that they hit the target on other plants. Many crop plants like wheat are wind pollinated.

2. Pollination by ____ is rare.
a) wind
b) animals
c) soil
d) water
Answer: d
Clarification: In plants pollinated by the wind, flowers are small and inconspicuous. The pollen grains are made light weight and unwettable due to the presence of a mucilage covering. The water current acts as a pollen vector.

3. Pollination by wind is limited to about 30 genera of plants.
a) True
b) False
Answer: b
Clarification: Hydrophily occurs in about 30 genera of (majorly) monocots e.g., Vallisneria, Ceratophyllum, Zostera. In several aquatic plants with flowers protruding onto the surface of water, pollination may occur by wind or insects, e.g., Lotus, Water Hyacinth, Water Lily.

4. Pollination by wind is called ____
a) Autogamy
b) Xenogamy
c) Geitonogamy
d) Anemophily
Answer: d
Clarification: Anemophily is pollination by winds. It is a mode of cross pollination or transfer of pollen grains from a mature anther to the stigma of a pistil which is accomplished through the agency of wind, e.g., Coconut Palm, Date Palm, Maize, many grasses, Cannabis.

5. Pollination by water is called ______
a) Cleistogamous
b) Xenogamy
c) Hydrophily
d) Anemophily
Answer: c
Clarification: Hydrophily is pollination by water. It is the mode of pollination or transfer of pollen grains from the mature anther of a flower to the stigma of another flower which is accomplished through the agency of water.

6. Zoster is an example of ______
a) pollination by animals
b) pollination by wind
c) pollination by water
d) pollination by birds
Answer: c
Clarification: Zoster is an example of pollination by water. In Zostera, the marine angiosperm (Sea Grass), the pollen grains are long ribbon-like (upto 2500 μm) and without exine.

7. Which of the following is a dioecious, submerged, fresh water plant?
a) Vallisneria
b) Cannabis
c) Neelakurunji
d) Zoster
Answer: a
Clarification: Vallisneria is a dioecious, submerged, fresh water aquatic plant. The male flowers cut off and come to the surface of water. Mature female flowers rise to the surface by elongation of their stalks.

8. Pollen grains in plants that are transferred via wind are heavy and thorny.
a) False
b) True
Answer: a
Clarification: Wind pollinated plants are small, winged, light and dusty. These pollen grains can be blown off far and wide (a ball park figure of 1000 km). Pollen grains from Pinus are found several hundred kilometers from the parent plant.

9. In ______ pollination takes place on the surface of the water.
a) Hypohydrophily
b) Hydrophylly
c) Epihydrophily
d) Ceratophyllum
Answer: c
Clarification: Hydrophylly is of two types— hypohydrophily and epihydrophily. Hypohydrophily occurs below the surface of water, e.g., Zostera, Ceratophyllum. Epihydrophily takes place over the surface of water, e.g., Vallisneria.

10. ______ underwater plant has male and female parts in the same flower.
a) Zostera
b) Ceratophyllum
c) Vallisneria
d) Maize
Answer: b
Clarification: Ceratophyllum is a fresh water plant (submerged) which bears male and female, both, flowers on the same plant. A male flower has several (30-45) stamens. The mature anthers breaks, and rises upward and dehisce occurs on the surface. The pollen grains are round, with no exine. The pollens come in contact with long and sticky stigmas for pollination.

Biology Multiple Choice Questions on “Human Reproduction – Oogenesis”.

1. The onset of oogenesis occurs during _________
a) puberty
b) birth
c) adulthood
d) embryonic development
Answer: d
Clarification: Oogenesis starts right from the embryonic development stage. Oogonia divide to form primary oocytes that are arrested at Prophase I.

2. What is the process of the formation of a mature female gamete called?
a) Menstruation
b) Spermatogenesis
c) Oogenesis
d) Ovulation
Answer: c
Clarification: Spermatogenesis is the process of the production of mature sperms from spermatogonia, while oogenesis is the process of the formation of mature female gametes from oogonia.

3. What is the stage of the cell cycle at which primary oocytes are arrested?
a) Prophase I
b) Metaphase I
c) Prophase II
d) Metaphase II
Answer: a
Clarification: Oogonia are formed during fetal development. Soon they start undergoing meiosis but are arrested at Prophase I awaiting further cues of division. Oogonia at this stage are called primary oocytes.

4. A person with tetraploidy will have _______ set of chromosomes in their first polar body.
a) haploid
b) triploid
c) diploid
d) polyploid
Answer: c
Clarification: Oogenesis is the production of a haploid secondary oocyte from a diploid oogonium and in the process resulting in the production of two polar bodies, each of which is haploid. Naturally, a tetraploid person having tetraploid oogonia will produce diploid polar bodies and a diploid oocyte.

5. If a female is born with 6 million primary oocytes in each ovary, the maximum number of eggs that she can produce in her lifetime is 6 million.
a) True
b) False
Answer: a
Clarification: Oogenesis begins with a single diploid oogonium arresting at Prophase I resulting in a primary oocyte. Each primary oocyte finishes the first round of meiosis to form one secondary oocyte and the first polar body. The secondary oocyte then undergoes a second round of meiosis to create one egg and a second polar body. And no more primary oocytes are added during the lifetime of the female. Instead, most primary oocytes degenerate as the female attains puberty. This limits the number of eggs that can be produced to less than 6 million.

6. The secondary oocyte has 46 chromatids, while the ovum has only 23 chromatids.
a) True
b) False
Answer: a
Clarification: Secondary oocyte is formed after the first round of meiosis and thus has a haploid set of chromosomes i.e., 23 chromosomes. But each chromosome has two daughter chromatids, thus making a total of 46 chromatids. Ova are formed after the second round of meiosis, and each ovum gets a single chromatid, thus having a total of 23 chromatids.

7. The division of primary oocyte into the secondary oocyte and first polar body is an example of _______
a) symmetric division
b) asymmetric division
c) Cell death
d) asexual reproduction
Answer: b
Clarification: Asymmetric division refers to the unequal distribution of cytoplasm between the daughter cells. Primary oocyte undergoes the first round of meiosis, producing a large haploid secondary oocyte and a tiny first polar body. Thus this is an instance of asymmetric division.

8. What are the cells that primary oocyte divides into called?
a) Secondary oocyte and first polar body
b) Secondary oocyte and second polar body
c) First polar body and second polar body
d) Ovum and second polar body
Answer: a
Clarification: The primary oocyte is arrested at Prophase I of meiosis, and the division is completed only in the tertiary follicle. This division is asymmetric and produces a large haploid secondary oocyte and a small first polar body.

9. What are the cells that secondary oocyte divides into called?
a) Ovum and first polar body
b) Ovum and second polar body
c) First polar body and second polar body
d) Primary oocyte and second polar body
Answer: b
Clarification: Secondary oocyte is formed as a result of the asymmetric division of primary oocyte, which also produces a tiny first polar body. Secondary oocyte further undergoes another asymmetric division producing a larger ovum and a smaller second polar body.

10. Which of the following is the correct set of ploidy and cell type?
a) Primary oocyte: Diploid; Secondary oocyte: Haploid; Ovum: Haploid
b) Primary oocyte: Haploid; Secondary oocyte: Haploid; Ovum: Haploid
c) Oogonium: Diploid; Primary oocyte: Diploid; Secondary oocyte: Diploid
d) Oogonium: Diploid; Primary oocyte: Haploid; Secondary oocyte: Haploid
Answer: a
Clarification: The diploid oogonium goes through the first round of meiosis to form primary oocyte, which is arrested at Prophase I. Thus, the primary oocyte is also diploid. The division is completed after the formation of the secondary oocyte, which is haploid. The secondary oocyte undergoes another round of meiosis to form haploid ovum.

11. What stage is the oocyte released from the ovary?
a) Primary oocyte
b) Secondary oocyte
c) Tertiary oocyte
d) Ovum
Answer: b
Clarification: Primary oocyte is formed during fetal development. The secondary oocyte is formed during puberty and is present in the ovary until it is released after the rupture of the Graafian follicle that holds it. Ovum is formed in the fallopian tube.

12. The membrane surrounding secondary oocyte is _______
a) Theca interna
b) Granulosa
c) Zona pellucida
d) Theca externa
Answer: c
Clarification: Zona pellucida is a transparent extracellular matrix formed outside the secondary oocyte. It enhances interactions of egg with the follicle cells and also protects eggs during the development.

Biology Multiple Choice Questions on “Reproductive Health – Surgical Methods of Contraception”.

1. What are the invasive methods used to prevent pregnancy called?
a) Natural methods
b) Barrier methods
c) Oral methods
d) Surgical methods
Answer: d
Clarification: Surgical methods are invasive. They work by using a surgical intervention to render one or both partners incompetent.

2. Surgical methods are permanent.
a) True
b) False
Answer: a
Clarification: Surgical methods work by operating male or female partners. This renders them incompetent to fertilize. These are permanent methods.

3. Surgical methods are also called as _________
a) fertilization
b) sterilization
c) ejaculation
d) emission
Answer: b
Clarification: Surgical methods are non-reversible. Hence these render the person sterile. These methods are therefore called sterilization.

4. Which of the following is prevented by surgical intervention?
a) Fusion of male and female gametes
b) Synthesis of gametes
c) Motility of sperms
d) Transport of gametes
Answer: d
Clarification: Surgical interventions involve the removal/breakage of connection between the site of synthesis and delivery of gamete. Thus, transport of gamete is inhibited.

5. What is the procedure for sterilization in males?
a) Vasectomy
b) Tubectomy
c) Decapitation
d) Circumcision
Answer: a
Clarification: Vasectomy is the procedure used to sterilize males. It involves the removal of a part of vas deferens or it being tied up through an incision on the scrotum.

6. What is the procedure for sterilization in females?
a) Vasectomy
b) Tubectomy
c) Decapitation
d) Circumcision
Answer: b
Clarification: Tubectomy is the procedure that sterilizes females. It involves removal of a part of fallopian tube. It can also involve tying the tube via an incision in the abdomen.

7. Surgical methods have meager rates of success as contraceptive methods.
a) True
b) False
Answer: b
Clarification: Surgical methods render the partner sterile. Hence these are probably the most effective of all contraceptive methods. However, these are permanent methods and hence can be used only if the couple desires no more pregnancy.

8. In vasectomy, the incision is made through which body part?
a) Abdomen
b) Penis
c) Scrotum
d) Breasts
Answer: c
Clarification: Vasectomy is a sterilization procedure for males. It involves tying off the vas deferens through an incision made in the scrotum.

9. In tubectomy, the incision is made through which body part?
a) Abdomen
b) Penis
c) Scrotum
d) Breasts
Answer: a
Clarification: Tubectomy is a sterilization procedure for females. In involves tying off the fallopian tube through an incision made in the abdomen or through the vagina.

Biology Question Bank on “Inheritance of Two Genes-2”.

1. What is the phenotypic ratio of F2 generation in a dihybrid cross?
a) 3:1
b) 1:2:1
c) 9:3:1
d) 9:3:3:1
Answer: d
Clarification: The F2 generation in a dihybrid cross produces all possible phenotypes. The ratio obtained for it is 9:3:3:1.

2. In Mendel’s experiments on garden pea plants, he performed a dihybrid cross of round yellow and green wrinkled seed plants. In the F2 generation, he sampled 1600 plants. Which of the following represents the correct number of plants of each phenotype?
a) 900 round yellow, 300 round green, 300 wrinkled yellow and 100 wrinkled green
b) 900 round green, 300 round yellow, 300 wrinkled yellow and 100 wrinkled green
c) 900 wrinkled green, 300 round yellow, 300 wrinkled yellow and 100 round green
d) 900 round green, 300 wrinkled green, 300 wrinkled yellow and 100 round yellow
Answer: a
Clarification: The F2 progeny of a dihybrid Mendelian cross has a phenotypic ratio of 9:3:3:1. Thus of the 1600 plants sampled, 900 will be round yellow, 300 round green, 300 wrinkled yellow and 100 wrinkled green

3. Which law was proposed by Mendel based on his dihybrid cross studies?
a) Law of Dominance
b) Law of Recessiveness
c) Law of Segregation
d) Law of Independent Assortment
Answer: d
Clarification: The first two laws of Mendel were based on his studies on monohybrid crosses. These were the law of dominance and the law of segregation. The dihybrid cross studies supported them, but also led to the formulation of a third law: the law of independent assortment.

4. What law talks about the segregation of two or more traits independent of one another?
a) Law of Dominance
b) Law of Recessiveness
c) Law of Segregation
d) Law of Independent Assortment
Answer: d
Clarification: The Law of Independent Assortment states the independent segregation of traits that are unrelated. Mendel drew this conclusion based on his studies on garden pea plants.

5. Which of the following is a correct interpretation of the law of independent assortment?
a) Factors exist in pairs
b) Factors segregate such that each gamete gets a single copy
c) For multiple traits under consideration, each segregate independently of one another
d) For multiple traits under consideration, they segregate in a specific pattern
Answer: c
Clarification: The law of independent assortment was the third law of Mendel. It states that when multiple traits are under consideration, each trait segregates independently of the other.

6. If x is the phenotypic ratio of monohybrid cross for trait A and y is the phenotypic ratio of monohybrid cross for trait B, what would be the phenotypic ratio of a dihybrid cross involving traits A and B?
a) xy
b) x + y
c) x² + y²
d) (xy)²
Answer: a
Clarification: For a dihybrid cross involving two traits, the independent law of assortment states that the traits would segregate independently of each other. Thus, the phenotypic ratio would be xy.

7. The law of independent assortment states that a dihybrid heterozygote would produce four types of gametes.
a) True
b) False
Answer: a
Clarification: The law of independent assortment states that the segregation of traits is independent of one another in a dihybrid cross. Thus, a heterozygote AaBb will produce four types of gametes: AB, Ab, aB, and ab.

8. Which of the following are the correct gametes produced by TtYy?
a) Tt, Yy, TY, ty
b) TY, ty, Tt, Yt
c) Yt, yt, YT, yT
d) YT, yt, yY, tT
Answer: c
Clarification: TtYy is a dihybrid heterozygote. The independent law of assortment applies to it. Thus, it will produce four types of gametes. These will be TY, Ty, tY, and ty.

9. For a dihybrid cross, the number of squares is 16 in a Punnett square of F2 generation.
a) True
b) False
Answer: a
Clarification: For a dihybrid cross, four types of gametes are produced by the heterozygotic generation. This gives rise to 4*4=16 combinations. Hence 16 squares are required in the Punnett square.

10. For a poly hybrid cross concerning N trait, what is the number of squares required in the Punnett square to check for the F2 generation?
a) 2^N
b) N²
c) 2^(2N)
d) (2N)²
Answer: c
Clarification: For a poly hybrid cross involving N traits, according to the law of independent assortment, 2^N types of gametes will be produced by each parent. Therefore, the cross of these progeny will give (2^N)*(2^N) = 2^(2N).

11. For a poly hybrid cross concerning N traits, what is the number of types of gametes produced by the F1 generation?
a) 2^N
b) N²
c) 2^(2N)
d) (2N)²
Answer: a
Clarification: In a poly hybrid cross, each F1 is a heterozygote. According to the independent law of assortment, each gamete will get one copy of all N factors. Thus, there are 2^N types of gametes possible.

To practice Biology Question Bank,

Biology Assessment Questions for MBBS Entrance Exams on “The Search for Genetic Material”.

1. Which of the following bacterium is responsible for causing pneumonia?
a) Staphylococcus saprophyticus
b) Streptococcus pyogenes
c) Staphylococcus aureus
d) Streptococcus pneumoniae
Answer : d
Clarification: Streptococcus pyogenes causes tonsillitis and rheumatic fever. Staphylococcus aureus causes abscesses. Streptococcus pneumoniae causes pneumonia. Staphylococcus saprophyticus causes urinary tract infections.

2. When Streptococcus pneumoniae were cultured in a culture plate by Frederick Griffith, which among the following were produced?
a) Smooth colonies(S) and Rough colonies (R)
b) Shiny colonies (S) and Rough colonies (R)
c) Smooth colonies (S) and Rigorous colonies (R)
d) Silky colonies (S) and Rigorous colonies (R)
Answer: b
Clarification: In 1928, Frederick Griffith cultured the pneumococcus bacterium. He observed that those bacteria were producing different colonies. These colonies were named as the S and R colonies. The S colonies are the Shiny colonies because they had the mucous(polysaccharide) coat. The R colonies are the Rough colonies because they did not possess the mucous coat.

3. The mice infected with the R strain of the bacterium dies, but when infected with the S strain, they do not die.
a) False
b) True
Answer: a
Clarification: The S strain produced by the bacterium is said to be the virulent strain. The mice infected with the S strain will die due to its virulent nature. Whereas, the one infected with the R strain doesn’t die due to its avirulent.
Mice -> S strain -> Death
Mice -> R strain -> Alive

4. Which of the following is the wrong sequential order, when the S or the R strain of the bacterium is injected into the mice?
a) Mice -> S strain -> Dead
b) Mice -> R strain -> Alive
c) Mice -> Heat killed S strain -> Alive
d) Mice -> Heat killed S strain + live R strain -> Alive
Answer: d
Clarification: When the heat killed S-strain of the bacterium is given along with the R strain to the mice, a process of transformation occurs. The heat killed S strain (avirulent) was able to transform the R strain (avirulent) to become a virulent strain. During this process of transformation, the heat killed S strained bacterium, transforms the R strain to produce the mucous(polysaccharide) and become virulent.

5. Which of the following is responsible for transforming the R strain into the S strain?
a) Purified bio-chemicals from S-strain
b) Purified bio-chemicals from R-strain
c) Purified bio-chemicals from heat killed S-strain
d) Purified bio-chemicals from heat killed R-strain
Answer: c
Clarification: S-strains are the virulent ones which have the ability to kill a mouse when injected into it. But, when the heat killed S-strain is injected into the mouse along with the R-strain, the mouse happens to die due to transformation. The purified bio-chemicals (protein, DNA and RNA) from the heat killed S-strain, transforms the R-strain into the S-strain (virulent).

6. Which of the following is responsible for the inhibition of transformation in organisms?
a) DNase
b) RNase
c) Protease
d) Nuclease
Answer: a
Clarification: The presence of protease (protein digesting enzymes) and RNase (RNA digesting enzymes) in the presence of the virulent or avirulent strains will not affect the process of transformation. But on the contrary, when the DNase is present, the process of transformation will not occur. This proves that the transformation process cannot occur as long as the DNase enzyme is present. The role of nuclease is to cleave the phosphodiester bonds between the nucleotides of nucleic acids.

7. What happens when S-strain and the biomolecule (Lipid) are together, when testing the transformation with regards to the R-strain?

a) R-strain
b) S-strain
c) R-strain and S-strain
d) Dead R-strain
Answer: a
Clarification: Only DNase has the ability to inhibit transformation. The other enzymes like RNase and lipase do not possess that ability. The process of transformation is not seen when the biochemical DNase is with S-strain. Lipase is an enzyme that catalyzes the hydrolysis (H₂O addition) of lipids and fats. The process of transformation will occur in lipids. As a result, R-strains will be produced in lipids.

8. Which of the following is considered to be as an incorrect difference between DNA and DNase?

a) Only I
b) Both II and III
c) Both III and IV
d) Only IV
Answer: d
Clarification: DNA is the genetic material of an organism. It contains all the required information for heredity. DNase will not promote the DNA. It is used to cleave the phosphodiester bonds by hydrolysis. As a result of this, the DNA biomolecule is degraded.

9. What will be the state of the mouse that has been injected with the heat killed S-strain from the Staphylococcus pneumoniae?
a) Alive
b) Dead
c) On the verge of dying
d) Lives with pneumonia
Answer: a
Clarification: The mouse fed with the S-strain from the Staphylococcus pneumoniae, happens to naturally die. This is on an account that the S-strains are virulent. Under such conditions, the mouse will not be able to survive. But, when the heat killed S-strains from Staphylococcus pneumoniae are injected into the mouse, it will live.

11. What will be the outcome when R-strain is injected into the mice?

a) Mice dies as soon as it is injected
b) Mice lives with the bacteria and dies suddenly
c) Mice dies after ejecting out the bacteria
d) Mice lives even after injecting the R-strain of the bacterium
Answer: d
Clarification: R-strains are the non-virulent strains. So, when they are injected into the mice, they won’t cause any damage to it. As a result, the mice will be able to live without complications. The bacteria contain both the S and R strains, but, only the R-strain is being injected in this scenario.

12. Which was considered to be as the genetic material prior to the works done by Oswald Avery, Colin MacLeod and Maclyn McCarty?
a) Nucleoids
b) Nucleons
c) Protein
d) Chromosome
Answer: c
Clarification: Prior to the works done by Oswald Avery, Colin MacLeod and Maclyn McCarty, the genetic material was thought to be as the protein. Nucleoids are the irregular shaped structures in the prokaryotic cell. They contain nearly all of the genetic material of the prokaryotes. Nucleons are present inside the nucleus of an atom, alongside with the protons. Chromosome carries the genetic information of the genes.

To practice Biology Assessment Questions for MBBS Entrance Exams,

Biology Multiple Choice Questions on “Origin and Evolution of Man – 1”.

1. Man belongs to the super family ______
a) Hominidae
b) Hominoidae
c) Primates
d) Mammalia
Answer: b
Clarification: Hominoidae is the super family of man. Man belong to the family Hominidae. The order of man is Primates and class is Mammalia.

2. Who called humans as Homo sapiens wiseman?
a) Carolus Linnaeus
b) Hugo de Vries
c) Joseph Walter
d) Charles Darwin
Answer: a
Clarification: The first scientific name of man was given by Carolus Linnaeus. He called them Homo sapiens wiseman. He added wiseman to the name as it was the only organism with high brain capacity.

3. Huxley explained origin of man in his book _______
a) the man’s place in world
b) the man’s place in earth
c) the man’s place in nature
d) the man’s place in universe
Answer: c
Clarification: Origin of man was explained in Huxley’s book called “The man’s place in nature”. It shows the connection of man with nature and how he adapted to survive in nature.

4. What did Darwin explain in his book “The descent of man”?
a) Ancestry of man
b) Evolution of organism
c) Origin of life
d) Production of man
Answer: a
Clarification: Charles Darwin wrote the book “The descent of man”. In his book, he mainly focusses on the ancestry of man and how man originated into its current form. He talks about the development of man and also the different races within man.

5. Human is a member of order mammalia and class primates.
a) True
b) False
Answer: b
Clarification: It is a wrong statement. The order of humans is primates and they belong to the class mammalia. Their genus is Homo and species Sapiens.

6. Primates originated during which era?
a) Mesozoic
b) Cenozoic
c) Paleozoic
d) Azoic
Answer: b
Clarification: They originated 80 – 100 million years ago. They belong to palaeocene epoch of coenozoic era. Primates originated from elephant shrews but they were not real primates.

7. The first real primates were _______
a) elephant shrews
b) tree shrews
c) monkey shrews
d) ape shrews
Answer: b
Clarification: Tree shrews were the first real primates originated in palaeocene epoch. They have two pairs of mammary glands, V-shaped jaw, shorter snout. Claws are absent and nails present on digits.

8. Humans, apes and monkeys belongs to ______
a) Anthropoids
b) Prosimians
c) Permian
d) Shrews
Answer: a
Clarification: Prosimians and Anthropoids are two sub orders of primates. Humans, apes and monkeys belong to Anthropoids whereas Lemur, loris and tarsiers belong to Prosimians.

9. Rhesus monkey belongs to _______
a) new world monkeys
b) old world monkeys
c) parallel world monkeys
d) future world monkeys
Answer: b
Clarification: Rhesus also known as Macaca belongs to old world monkeys. They belong to Africa and Asia. They are characterized by a narrow flat nose with downward direction of nostril.

10. New world monkey evolved from _______
a) South and Middle America
b) North and Middle East
c) South Africa
d) Asia
Answer: a
Clarification: Spider monkey and marmosets belongs to the new world monkeys. The new world monkey evolved from South and Middle America. They are characterized by a protruding nose with upward direction of nostril.

11. The following figure shows an example of a primate, commonly known as ______

a) Elephant shrew
b) Tree shrew
c) Lemur
d) Monkey
Answer: b
Clarification: Tupaia, a tree shrew is shown in the diagram. Claws are absent and nails are present on digits. By nature, they are insectivorous.