Category: Class 12 Biology Objective Questions

Biology Multiple Choice Questions on “Human Reproduction – Structure of a Sperm”.

1. The body of sperm is covered by _______
a) head
b) cell membrane
c) cell wall
d) cytoplasm
Answer: b
Clarification: As in the case of any other cell, sperms are enveloped by a phospholipid bilayer, which is also called as plasma membrane or cell membrane.

2. What part of sperm holds the haploid chromatin?
a) Acrosome
b) Head
c) Tail
d) Neck
Answer: b
Clarification: Sperm can be divided into 4 parts namely: head, neck, middle piece and tail. The haploid nucleus is present in the head. Acrosome is the cap-like structure of head which helps it to fertilize the ovum.

3. Egg is covered by a tough sheet of tissue that protects it from desiccation and infection by pathogens. But the same tissue also prevents sperm nuclei from encountering the egg nuclei. However, a part of sperm is known to release enzymes that digest this tough sheet. What part of sperm is it?
a) Tail end
b) Mitochondria
c) Acrosome
d) Sperm nuclei
Answer: c
Clarification: Acrosome is the cap-like structure present on the anterior-most end of the sperm head. It is filled with numerous enzymes like hyaluronic acid which are released upon encounter of egg in female oviduct. These enzymes function by digesting the membrane, thus allowing penetration of head and further release of male nuclei.

4. Rakesh and Reshma have difficulty conceiving a baby. They consulted a sex therapist. Sperm count of Rakesh was normal but the doctor observed that the motility of his sperm was less. What part of sperm do you think has the issue?
a) Tail
b) Nucleus
c) Mitochondria
d) Acrosome
Answer: c
Clarification: Mitochondria are present in the middle piece of the sperm. They are essential for producing energy required for movement of tail. This is essential for movement through the genital tract and oviduct before sperm encounters the egg. Rakesh might have less or poorly functional mitochondria, which prevent his normal count of sperms from moving along the genital tract.

5. Which of the following is not an essential feature of sperms that determine the fertility of a male?
a) Sperm count
b) Sperm motility
c) Sperm height
d) Sperm production rate
Answer: c
Clarification: Sperm count is essential as it determines the probability of a sperm encountering the egg. Sperm motility is essential to ensure the passage of sperms through the genital tract, uterus and oviduct. Sperm production rate determines the number of healthy coitus a male can have. Sperm height is not a determining factor of fertility.

6. Sperms are produced in _______
a) vas deferens
b) vasa efferentia
c) rete testis
d) seminiferous tubules
Answer: d
Clarification: Seminiferous tubules are lined by spermatogonia and Sertoli cells. Spermatogonia undergo meiosis to produce sperm that are nourished by Sertoli cells.

7. What is the consequence of low sperm count?
a) Death
b) Infertility
c) Abortion
d) Pregnancy
Answer: b
Clarification: Low sperm count leads to lesser amount of sperms entering the genital tract and passing successfully before encountering the egg in the oviduct. Thus less sperm count leads to lesser chances of fertilization and might result in infertility.

8. The major constituents of semen are _____ and _____
a) Sperms and RBCs
b) Sperms and Blood plasma
c) Sperms and seminal plasma
d) Sperms and WBCs
Answer: c
Clarification: Sperms are released out from seminiferous tubules facilitated by seminal plasma composed of secretions from epididymis, vas deferens, seminal vesicle, and prostate.

9. What doesn’t constitute to the seminal plasma?
a) Secretion of epididymis
b) Secretion of vas deferens
c) Secretion of vas efferens
d) Secretion of seminal vesicle
Answer: c
Clarification: Seminal plasma helps in the motility and maturation of sperm. Secretions produce it from vas deferens, epididymis, seminal vesicle and prostate.

10. What is not a function of the male sex hormone Testosterone?
a) Onset of spermatogenesis
b) Maintenance of accessory ducts
c) Release of semen
d) Maintenance of accessory glands
Answer: c
Clarification: Testosterone secreted by Leydig cells stimulates spermatogenesis. It also acts on accessory ducts and glands and maintains their proper function. Release of sperms and semen is not coordinated by testosterone.

Biology Multiple Choice Questions on “Reproductive Health – Injectable Contraceptives and Implants”.

1. What contraceptives are placed under the skin?
a) Implants
b) Condoms
c) Pills
d) Intrauterine methods
Answer: a
Clarification: Implants are placed under the skin. They release hormonal doses to inhibit pregnancy. Although effective, they have many side-effects ranging from irregular menstruation, weight gain, abdominal pain, headaches, depression and so on.

2. Rape cannot result in pregnancy.
a) True
b) False
Answer: b
Clarification: Rape is an unethical but a sexual act. It is coitus, and if the female is in or near the ovulatory phase, rape can result in fertilization of egg and thus pregnancy.

3. What contraceptive cannot be used after unprotected intercourse?
a) Condoms
b) Implants
c) Pills
d) Injections
Answer: a
Clarification: Condoms are aimed at preventing the mixing of vaginal and seminal fluids. They will not prevent pregnancy after intercourse. Pills, implants, and injections can do the job.

4. What are the two hormones that can be used to prevent pregnancy in the form of injections or implants?
a) Progesterone and Estrogen
b) Progesterone and Testosterone
c) Estrogen and Testosterone
d) Estrogen and Oxytocin
Answer: a
Clarification: Implants and injections have more prolonged effects than oral contraceptives. They work similarly to pills. These are composed of mixtures of progesterone and estrogen.

5. Which of the following can serve as an emergency contraceptive?
a) Progestogen
b) Condoms
c) Spermicidal gels
d) Spermicidal lubes
Answer: a
Clarification: Progestogen and estrogen combinations are used as emergency contraceptives. They can be administered as pills, injections, or implants.

6. Progestogen and progestogen-estrogen combinations work only up to 72 hours after casual unprotected intercourse.
a) True
b) False
Answer: a
Clarification: Progestogen and progestogen-estrogen contraceptives work by preventing the release of egg or by inhibiting the entry of sperm into the fallopian tube or by preventing fertilization. Hence they should be administered within 72 hours of coitus. After 72 hours, the chances of egg being fertilized are very high.

Biology Multiple Choice Questions on “Inheritance of Two Genes-1”.

1. In order to study the inheritance of two genes, the parents should be true-breeding for _______
a) a single contrasting characters
b) a pair of contrasting characters
c) a pair of chromosomes
d) a single chromosome
Answer: b
Clarification: With the plants that are true-breeding for two contrasting characters, Mendel performed crosses. Based on these, he devised results for the inheritance of two genes.

2. If round seed (RR) was dominant over wrinkled seed (rr) in a monohybrid cross, and yellow pod (YY) was dominant over brown pod (yy), what should be the starting genotype of parents in a dihybrid cross?
a) RRrr and YYyy
b) RRYY and rryy
c) RRyy and YYrr
d) RrYy and RrYy
Answer: b
Clarification: The dihybrid cross is used to the inheritance pattern of two contrasting characters. Hence, they should be true-breeding. One for both dominant traits and the other for both recessive traits.

3. What is the expected phenotype of F1 progeny of a dihybrid cross for inflated green pod plants and constricted yellow pod plants?
a) Inflated green pods
b) Inflated yellow pods
c) Constricted green pods
d) Constricted yellow pods
Answer: a
Clarification: The F1 progeny will be heterozygotes. Hence the dominant characters will be expressed. Thus, all the F1 progeny will have inflated green pods.

4. Given the genotype of parents as AABB and aabb, what is the expected genotype of F1 progeny?
a) AABB
b) AAbb
c) aaBB
d) AaBb
Answer: d
Clarification: AABB parent will only be able to produce AB gametes, and aabb parent will only be able to produce ab gametes. Thus, the only possible genotype for the progeny is AaBb, a heterozygote.

5. What is the zygosity of the F1 progeny of a dihybrid cross?
a) Homozygosity
b) Heterozygosity
c) Hemizygostiy
d) Nullizygosity
Answer: b
Clarification: The F1 progeny of a dihybrid cross are formed by a cross between parents that are homozygous for two traits. Thus, they are all heterozygous.

6. All the progeny of F1 generation of a dihybrid cross are genotypically identical for the locus under study.
a) True
b) False
Answer: a
Clarification: The dihybrid cross initiates with true-breeding parents who are homozygous for two traits. Thus, each of them can only produce a single type of gamete. Hence the offsprings in the F1 generation are all genetically identical.

7. What is the expected phenotype of F1 generation obtained by the cross between true-breeding round yellow seed plants and true-breeding wrinkled green seed plants?
a) Round green seeds
b) Round yellow seeds
c) Wrinkled green seeds
d) Wrinkled yellow seeds
Answer: a
Clarification: Round seeds are dominant over wrinkled seeds. Also, yellow seeds are dominant over the green seeds. Thus, the progeny of these parents will be heterozygotic resulting expression of only the dominant characters.

8. Which of the following phenotypes is not obtained in the F2 generation?
a) Round yellow seeds
b) Wrinkled green seeds
c) Wrinkled round seeds
d) Round green seeds
Answer: c
Clarification: In the F2 generation of a dihybrid cross, all possible phenotypes are observed. Wrinkled and round are encoded by the same locus. Hence, they cannot occur together.

9. For the cross of round green and wrinkled yellow seeds, which of the following is incorrect with regards to the F2 generation?
a) Round seeds were three times more abundant than yellow seeds
b) Green seeds were three times less abundant than wrinkled seeds
c) Round seeds were as abundant as the green seeds
d) Yellow seeds were three times more abundant than green seeds
Answer: d
Clarification: The roundness and greenness of the seeds are encoded by different genes. According to the law of independent assortment, these two should segregate independently of each other. Thus in the F2 generation, green seeds: yellow seeds = 3:1, round seeds: wrinkled seeds = 3:1. From this, we can have, green seeds: wrinkled seeds = 3:1, green seeds: round seeds = 1:1, yellow seeds: round seeds = 1:3, yellow seeds: wrinkled seeds = 1:1.

10. A cross involving variation in two different traits is called _______
a) monohybrid cross
b) dihybrid cross
c) bihybrid cross
d) polyhybrid cross
Answer: a
Clarification: A dihybrid cross is the one where the partners involved are similar except for two traits under consideration. These were studied for the first time using true-breeding lines by Mendel.

11. The dihybrid cross was used by Mendel to derive the law of dominance.
a) True
b) False
Answer: b
Clarification: Mendel used the monohybrid cross to arrive at the laws of dominance and segregation. For the law of independent assortment, he did the dihybrid crosses.

12. For the cross of axial violet and terminal white flowers, which of the following is incorrect with regards to the F2 generation?
a) Axial violet are nine times more abundant than white flowers
b) Axial white flowers are as abundant as terminal violet flowers
c) Terminal white flowers are thrice abundant than the axial white flowers
d) Terminal violet flowers are three times more abundant than terminal white flowers
Answer: c
Clarification: This is a dihybrid cross. The genetics is Mendelian since the traits are present on different chromosomes and do not exhibit linkage. Thus, the expected phenotypic ratio is 9 axial violet: 3 axial white: 3 terminal violet: 1 terminal white. Hence axial violet flowers are nine times more abundant than white flowers; axial white flowers are as abundant as terminal violet flowers; terminal white flowers are 1/3rd as abundant as axial white flowers and terminal violet flowers are thrice abundant than terminal white flowers.

Biology Multiple Choice Questions on “Packaging of DNA Helix”.

1. What is the length of the DNA double helix, if the total number of bp (base pair) is 6.6 x 10⁹?
a) 2.2 m/bp
b) 2.5 m/bp
c) 2.2 m
d) 2.5 m
Answer: c
Clarification: The distance between 2 consecutive bp is 0.34 nm. The formula to calculate the length of the DNA double helix:
The length of the DNA double helix = The total number of bp x distance between two consecutive bp.
As a result, we get 2.2 m as the length of the DNA double helix.

2. The DNA of the E. coli is scattered throughout the cell.
a) True
b) False
Answer: b
Clarification: E. coli is a prokaryotic cell. Prokaryotic cells do not have a well-defined nucleus. In eukaryotes, the genetic material which is the DNA is present in the nucleus. Even though the prokaryotes lack a nuclear structure, their DNA is not scattered throughout the cell. The nuclear material in a prokaryote is called as a nucleoid. The nucleoid has the DNA inside of it. It will be circular in shape. And so, the DNA is not scattered throughout the cell.

3. What are the set of positively charged basic proteins called as?
a) Histidine
b) DNA
c) RNA
d) Histones
Answer: d
Clarification: Histones are rich in basic amino acid (AA) residues such as lysine and arginine. These two amino acids carry a positive charge at their side chains. Due to this reason, histones are called as the set of positively charged basic proteins.

4. What are the thread-like stained structures present in the nucleus known as?
a) Chromosome
b) Chromatid
c) Chromatin
d) Chloroplast
Answer: c
Clarification: A chromatin is a long thread-like structure which contains the nucleosome. Chromosome is a form which is achieved by the chromatin when it is undergoing cell division. In this form, both of the sister chromatids will be attached together at the centromere. Chromatids or sister chromatids are the thread like structures which are achieved when the chromosome is cut longitudinally during cell division. Chloroplast is a plastid that contains the green pigment chlorophyll.

5. How many bp are present in a typical nucleosome?
a) 200 bp
b) 100 bp
c) 300 bp
d) 90 bp
Answer: a
Clarification: Nucleosomes are the bead-like structures which are present on a chromatin. These nucleosomes contain the DNA, which is the genetic material. Such a typical nucleosome contains nearly 200 bp.

6. When the negatively charged DNA combines with the positively charged histone octamer, which of the following is formed?
a) Nucleus
b) Nucleoid
c) Nucleosome
d) Nucleosome
Answer: c
Clarification: Nucleosomes are the bead-like structures which are present on the strands of chromatin. These nucleosomes are formed when the DNA combines with the histone octamer. Less than 2 DNA turns can be seen which are wound around the octameric protein (Histones).

7. How many nucleosomes are present in a mammalian cell?
a) 20 million
b) 30 million
c) 40 million
d) 10 million
Answer: b
Clarification: The diploid cell of a eukaryotic mammal contains 6.4 x 10⁹ nucleotides. A single cell of such type will 30 million nucleosomes in it. Nucleosomes containing chromatin when viewed under an electron microscope appear to be like “Beads-on-string”.

8. Which of the following chromatins are said to be transcriptionally active and inactive respectively?
a) Euchromatin, Heterochromatin
b) Euchromatin, Prochromatin
c) Prochromatin, Euchromatin
d) Heterochromatin, Euchromatin
Answer: a
Clarification: Inside the typical structure of a nucleus, the part of the chromatin which is considerably loose and stains light is said to be the Euchromatins. Likewise, the part of the chromatin which is considerably dense and stains dark is said to be the Heterochromatins. The part which is loose is said to be transcriptionally active, and the dense part is said to be transcriptionally inactive.

9. What are the additional set of proteins which are required for the packaging of chromatin at the higher levels known as?
a) Histone proteins
b) Non-Histone proteins
c) Histone chromosomal proteins
d) Non-Histone chromosomal proteins
Answer: d
Clarification: NHC or Non-Histone Chromosomal proteins are the collection of proteins that act in a eukaryotic nucleus. They are responsible for the processes of DNA replication, RNA synthesis, RNA processes and also nuclear processes.

10. What is the typical distance between two base pairs in nm?
a) 0.34 nm
b) 0.32 nm
c) 0.33 nm
d) 0.35 nm
Answer: a
Clarification: The double stranded helical structure of the DNA, takes a turn after every 10 base pairs which is noted to be as 3.4 nm. This is called as a pitch (distance). 0.34 nm is considered to be the distance between 2 base pairs.

11. What is the structure present inside the nucleus known as?

a) Chromosome
b) Chromatid
c) Ribosome
d) Lysosome
Answer: d
Clarification: The structure present inside the nucleus is known as the chromatid. The small circular structures which are present on it are called as the nucleosomes. Chromosomes are the condensed form of the chromatids. Ribosome are the components which help in the synthesis of proteins. Lysosome are the cell components which are also known are the suicidal bags of the cells.

12. The DNA helical structure is linked to which type of histone protein in the following diagram?

a) H2A histones
b) H1 histones
c) H2B histones
d) H3 histones
Answer: b
Clarification: H1 histones are commonly known as the linker histones. These histones can be located at the passage ways of the DNA, namely the entry and the exit sites. It is mostly found in protists and bacteria. It is also commonly known as nucleoproteins.

Biology Multiple Choice Questions on “A Brief Account of Evolution of Animals”.

1. Invertebrates were formed and active about _______ million years ago.
a) 350
b) 500
c) 320
d) 200
Answer: b
Clarification: It was 500 million years back when the invertebrates were formed and active. Jawless fish and other fishes with strong stouts evolved around 350 million years ago. Seaweeds and few plants evolved about 320 million years ago.

2. The animals which evolved into the first amphibian that lived on both land and water, were _____
a) Jawless fish
b) Lobefin
c) Ichthyosaurs
d) Shrew
Answer: b
Clarification: Lobefin were the first animals to be declared as amphibians. The first mammals resembled shrews whereas jawless fish evolved around 350 million years ago. Ichthyosaurs were fish-like reptiles that existed 200 million years ago.

3. Which of these are correct regarding the evolution of animals?
a) Lobefins were ancestors of modern-day frogs and salamanders
b) Reptiles evolved into amphibians
c) Reptiles lay thin-shelled eggs
d) Reptiles of same shapes and sizes evolved
Answer: a
Clarification: It is said to believe that modern-day frogs and salamanders evolved from lobefins. Amphibians evolved into reptiles also reptiles lay thick-shelled eggs. Reptiles of different shapes and sizes evolved during animal evolution.

4. The appearance of first amphibians was during the period of ______
a) Ordovician
b) Louis Pasteur
c) Cambrian
d) Devonian
Answer: d
Clarification: Appearance of first amphibians were during the period of Devonian. Bony fishes and lungfishes also evolved during this period. It occurred about 5 crore years ago.

5. Origin and evolution of sharks were during the Carboniferous period.
a) True
b) False
Answer: b
Clarification: It is a wrong statement. Origin and evolution of sharks were during the Devonian period. It is known as the “Age of fishes”. They belong to the Palaeozoic era.

6. Which of the following did not originate during the Triassic period?
a) Oviparous prototheria
b) Mammals
c) Dinosaurs
d) Apes
Answer: d
Clarification: Origin of first oviparous prototheria, mammals, dinosaurs, flying reptiles. It occurred at about 3.5 crore years ago. Apes did not evolve during this period.

7. Evolution of toothed birds occurred about ______ crore years ago.
a) 5.4 – 5.9
b) 5 – 6
c) 3.5
d) 7.1
Answer: a
Clarification: Evolution of toothed birds occurred about 5.4 – 5.9 years ago. Birds like archaeopteryx evolved during this period. Also, marsupials evolved from prototherians. This all happened during the Jurassic period.

8. What occurred during the Cretaceous period of animal evolution?
a) Dinosaurs evolved
b) Extinction of dinosaurs
c) Evolution of primitive birds
d) Modern birds disappeared
Answer: b
Clarification: Extinction of dinosaurs and primitive birds occurred. First modern birds appeared during the Cretaceous period. Also, primitive placental mammals like insectivorous shrews also originated. This all occurred 7.1 crore years ago.

9. What evolved during Oligocene epoch of animal evolution?
a) Anthropoid apes
b) Humans
c) Birds
d) Mammals
Answer: a
Clarification: Anthropoid apes evolved during the Oligocene epoch of animal evolution. These apes are said to evolve from monkeys and crocodiles. It occurred at about 1.9 crore years ago during the Tertiary period.

10. Humans appeared during which epoch?
a) Holocene
b) Pleistocene
c) Pliocene
d) Miocene
Answer: b
Clarification: Human appeared during the Pleistocene epoch. The appearance of the social life of man was also observed. Miocene is the golden age of mammals. The apelike ancestor of human appeared during Pliocene epoch. Holocene is known as the age of man.