Bihar Board Biology Sample Paper for Class 12

Jul 8 • Board Sample Papers • 23476 Views • 32 Comments on Bihar Board Biology Sample Paper for Class 12

Biology is one of the most interesting and easy subject for the exam preparation. With proper concentration one is able to earn good marks in this subject. Here we have provide Bihar Board Biology Sample Paper for Class 12 is prepared by setting an objective of providing an idea to get good marks in the board examination. Students gets confused to decide which type of questions are to prepare during exams. Thus try to solve as mush as sample papers to judge your performance level. So in this case Biology Sample Paper for Class 12, will help you select the type of questions asked in the exam.

BSEB Biology Sample Paper for Class 12

Solved Biology Sample Paper for Class 12

About Bihar Board: The Bihar school of examination board established for holding and conducting exam at the secondary level as well as elementary level. Usually  every year the Bihar School Examination Board organize the Annual Secondary School Examination in the month of February-March and Supplementary School Examination in the month of August-September on the basis of course as well as syllabus prescribed by the state Government.

Biology Sample Paper for Class 12

General Instruction:
a) This paper consists of two parts. PART-A and PART-B
b) Part-A contains of 24 objective type questions.
c) Part-B consists of 13 non-objective questions.
d) Instruction and marks are allotted to each questions.


Q. No.1 to 11 are of one mark questions. Select the correct answer to each questions.

Q1 GM Plants are useful in-
(A) increasing crop yield
(b) producing disease resistant plants
(c) ncreasing drought resistant plants
(D) all of these

Ans- b.  producing disease resistant plants

Q2. The theory of Origin of Species by natural selection was proposed by—
(A) Charles Darwin
(B) Lamarck
(C) Ernst Hacckel   
 (D) (Malthus


3. Name an enzyme used to cut DNA—
(A) Pectinase
(C) Restriction endonuclease
(D) Lysozyme

Ans-:  c Restriction endonuclease

4. What is the position of a tiger in a food cha in shown below ?
Grass → Deer → Tiger
(A) Producer
(B) Primary Consumer
(C) Secondary Consumer
(D) Decomposer

Ans-:c  Secondary Consumer

5. Replcation of DNA needs-:
(A) DNA Ligase
(B) DNA Ploymerase
(C) DNA Polymerase and DNA Ligase
(D) Translocase and RNA Polymerase

Ans-: c DNA Polymerase and DNA Ligase

6. Dudhwa National Park is situated in
(A) Assam
(B) U.P.
(C) Gujarat
(D) West Bengal

Ans-: b U.P.

7. Seeds of fruits are transformed
(A) Corolla
(B) style
(C) wall of ovary
(D) ovule

Ans-: d ovule

8. Which one of the following is oviparous?
(A) Flowering plants
(B) Crocodile
(C) Woman
(D) Monkey

Ans-: b Crocodile

9. Cancer of epithelial tissue is called
(A) Lymphona
(B) Liposa
(C) Leukaemia
(D) Carinoma

Ans-: d Carinoma

10. Which are of the following works as bio fertilizer in Paddy field ?
(A) blue-grane algae
(B)  yeast
(C) funge
(D) insect pest

Ans-:a blue-grane algae

In each question from Q. No. 11 to 15 two statements (Statement I and Statement II) are given choice the correct option for each question out of options (a), (b), (c) and (d) on the basis of given statements. 1×5
(a) If both the statements are true and statement Ii is the correct explanation of statement I.
(b) If both the statements are true and statement II is not the correct explanation of statement I.
(c) If statement I is true and statement Il is false.
(d) If statement I is false and statement II is true.

11. Statement I : Opium is a narcotic drug.
Statement  II : Opium is obtained from a plant Papaver sominiferum
Ans-: b

12. Statement I : Insulin produced by the application of biotechnology is known as humulin.
Statement II : Insulin is a type of hormone.

13. Statement I : Archacopteryx is a connecting link between-birds and mammals.
Statement II : Archaeopteryx is an extinct bird.

14. Statement I : Green plants are known as producers.
Statement II : They convert light energy into chemical energy.

15. Statement I : Male gametes are haploid cells
Statement II : Male gametes have n number of chromo somes.
Ans-: a

Q. No.16 to 18 is for right have more than one correct answer. Select the correct answer to each questions.

16. Surgical methods adopted for population control are
(A) Vasectomy
(B) tubeclomy
(D) Condom


17. Method of protein synthesis involves
(A) Replication
(B) Tanscription
(C) Translocation
(D) Duplication


18. Which one of the following is included in animal husbandry?
(A) Bee keeping
(B) Poultry farming
(C) Fish farming
(D) Organic farming


Q. No. 19 to 22 are of 2 column matching type. Match correctly. 4 × 1 = 4
19. Global warming                                           (A) Hybridoma technology
20. Monoclonal antibodies                                (B) Bacteria and Fungi
21. Clitoris                                                        (C) CO2
22. Antibiotics                                                   (D) Homologons to penis

Ans-:   19-c

Q. No. 23 to 25 are comprehensive type. Read carefully the passage and select one correct of given four questions. 3 × 2 = 6

23. Which one of the following is the carrier of hereditary characters ?
(A) Chromosome
(B) Mitochondria
(C) Nucleous
(D) Protein molecule

Ans-: Chromosome

24.The differences found in the similar aft springs reproduced by common parents are called
(A) Evolution
(B) Variation
(C)  Mutation
(D) homologous structures

Ans-: Variation

25. By which phenomenon each organism maintains its continuity ?
(A) Evolution
(B) Heredity
(C) Reproduction
(D) Variation

Ans-:  Reproduction


Q. No. 1 to 11 are short answer type questions. Each question carries 2 marks. 11 × 2 = 22

1. Describe three advantages and three disadvantages of cross-pollination in plants
Ans-: Advantages of cross pollination-:
It overcomes self-sterility.
It eliminates defective tracts
It introduces variations due to genetic recombination.

Disadvantages are-:
It is highly wasteful process.
There is always a chance factor for it.
Good characters can be diluted and undesirable characters will enter in the progeny.

2. Write a short notes on human ovary.
Ans-:   There consist of a pair of ovaries in females which is situated in the lower abdomen. Ovum is produced inside ovary by the process oogenesis Ovary in lined with a layer of connective tissue called tunica albuginea. There is a germinal epithelium layer beneath it. Stroma is made of connective tissue & fibers present in cavity of ovary, germinal epithelial cells form follicles by repeated division. One of its cell enlarge to form oocyte. It makes primary, secondary and  graafian follicle later on. The mature follicle is called gaafian follicl.

3. What is Mendel’s law of segregation? Explain
Ans-: Garden pea had been selected by Mendel when he performed his experiment on segregation he concluded that the dwarfness never get lost or change it remains in the hybrid unit as a dominant trait.

4. What are antibiotics ?
Ans-: These are chemical that are synthesized by microbes.

5. What do you mean by tissue culture?
Ans-: Plant tissue culture is the technique of invitro maintenance & growth of plant cells, tissues or organs on a suitable artificial culture medium contained in small containers under controlled conditions. An excised fragment of tissue plant part used for raising a culture in called explant. It can be a part of root, stem, leaf, seed, embryo, embryo sac, ovary, anthers etc. Explant in inoculated in culture tube with culture medium. It is done in incubation chamber. These cells divide repeatedly & form a group of tissue.

6. Describe the difference between species and population.
Ans-:  Species is a bigger aggregation of similar individuals whereas Population is an aggregation of individuals of the same species at the same time in a particular area or space.

7. What is the role of producer is an ecosystem?
Ans-: The organisms which synthesize food & produce energy in an ecosystem are called producers.

8. What do you know about Bhopal gas tragedy?
Ans-: It was 3rd Dec. 1984 midnight, when a poisonous gas leaked from union carbide company factory making methyl  iso-cyanate related pesticides. This incident is known as Bhopal gas tragedy.

9- Why is human placenta referred to as haemo chorial type ?
Ans-: Placenta is foeto maternal connective that develops during pregnancy and forms a temporary association between foetal & maternal tissues for supporting the foetus during development. Foetal part is made of chorion & allantois. Maternal part is called decidua basalis. In the region of contact, the epithelial connective tissue & endothelial lining in uterine mucosa get discarded so only foetal barriers persist. So it is called as haemochorial placenta. It provides for rapid exchange of materials between foetal & maternal blood with barriers acting as ultrafilter. Placenta secretes relaxin hormone at the time of paturitin which relaxes & dilates pubic symphysis, cervix & vaginal tube.

10- Write down the differences between predators & parasites
Ans-:  Predators are larger & stronger  animal which kill & consume prey where as parasite are Small or microscopic organisms depending on the host.

11- what are carcinogens?
Ans-:Carcinogens are agents that tend to favour cancer development or produces cancer. They can be physical irritants, chemical agents, radiations, or biological agents.
Two chemical carcinogens are-
(i) Cigarette smoke (N-nitrosodimenthylene) affect lungs.
(ii) Mustard gas also affects lungs.

Q. No. 12 to 13 are of long answer questions. Each question carries 5 marks. 2× 5 = 20

12. Describe sex-determination in human being.
Ans-: Lamarck was the French Soldier which turned Bank Employee turned Medical practitioner turned Naturalist and is known for the complete theory of the evolution. He discussed this in his important publication – Philosophic Zoologique (1809) Postulates of Lamarck :

There are 4 postulates
(a) Growth Principle :- Internal forces of life tend to increase size of the organism.
(b) Change in environment and formation of organ- To fulfil new demands created by change individuals develop new organs at their will.
(c) Use and disuse of organ :- Organs develop by use and atrophy by disuse.

This is supported by following examples :
(i) LOSS of limbs in snakes..
(ii) Both eyes single-sided in flat-fish.
(iii) Vestigeal organs in animals due to disuse.
(iv) Lengthening of neck in Giraffe to reach to the leaves of tall plants.

(i) Mutilation experiments by Weismann in which loss of tail in white rats is not transmitted,
(ii) Boring of ear not transmitted.
(iii) Wearing of Iron shoes by Chinese women to shorten their feet.
(iv) Artificial parthenogenesis in sea urchin eggs are not followed in later generation
(v) Inheritance of all characters mentioned in use & disuse can be explained scientifically on the basis of Darwinism. Neck of Giraffe was lengthened because populations with long-neck had selective advantage over those who had no long-necks Thus gradually long-necked Giraffe replaced short-necked Giraffe.

Support–Many experiments have been conducted but none is up to the mark.
(a) McDougall conducted experiment on maze-leaming in rats and found that learning is transmitted. It has been contradicted by Agar.
(b ) Sumner conducted experiment on thermal induction of long tails in rats. Temperature effects germ cells also hence it is not an example of the law.
(c) Kemmerer conducted experiment on pigmentation in Salamanders but he used china-ink to prove his points.
(d) Lysenko conducted experiment on wheat cultivation in Russia and damaged us economy.
(e) Guyer & Smyth punctured vessels around lens in male rats and bred it with normal female rats. Many of the progeny were blind. It is claimed that the lens antigen induced antibody formation which reacted with its respective DNA to bring about blindness-RNA to DNA is possible through reverse transcription. If protein to RNA becomes true this can happen .

13. Describe the structure of DNA.
Ans-:  Structure of DNA. DNA is long double chain on duplex ‘molecule’ formed of millions of deoxyribonucleotides. Length of DNA is characteristic of paganism and its chromosomes. Bacteriophage Φ × 174 has 5386 nucleotides. Bacteriophage lambda possesses DNA saving all 502 base pairs (bp). Chichester coil has DNA consisting of 4.6 × 10º bp. A single genome, consisting of 23 chromosomes, possesses 3.165 10º bp in case of human beings. Single stranded DNA molecules occur in some viruses, e.g., coliphage Φ × 174. DNa duples has a diameter of 20Å. The duplex is called plectonemically in a right handed manner just as a tope stair is twisted to form a spiral. This coiling produces alternate major (length 22 Å) and minor (length 12 Å) genoves.

A phosphodiester linkage (two ester formations by same phosphate radical) is …between sugar rosidues of adjacent nucleotides. In one chain of the DNA duplex the last deoxyribose of one end has its carbon 5′ free while in the other chain the last deoxyribose of this end has its carbon3′ free. The ..of the other and of the doplex is feverse. The opposite and complementary nitrogen bases are held together by hydrogen bonds, two between A and T (at positions 1→3, 6→4) and there between C and G (at positions 1→4, 2→6 and 6→2). Two types of forces stabilise the duplex and hold the two DNA chains together :
(i) Hydrogen bonds between the complementary nitrogen bases of the two chains.
(ii) Hydropholie internetions between nitrogen bases. The latter are also kept stacked inside the helix while  the polar groups are kept on the outside in contact with water.
Base Pairing. It is the pairing formed in DNA double helix between purine of one strand and pyrimidine of the second strand. Base pairing is specifie with adenine lying opposite thymine and cytosine occurring opposite guanine.

A proper base pairing is required for two purposes.
(i) The two DNA strands can remain exactly paralled only when the space between them remains uniform. Presens of two purines as base pair will make the DNA double helix wide due to bulging out while two pyrimidines as base pair will make the duples non rows.
(ii) Spase between the two DNA strands of the duplex is 20Å. The one available for the two nitrogen bases is about 11Å. It can accommodate neither two paurines, nor two pyrimidines. The space is sufficient for one pyriminline and one putine with a small area in between for forming hydrogen bonds.
(iii) Only adenine-thymine and cytosine-guanine base paired have proper apatial arrangements and configurations to develop hydrogen bonds, two between A and T and three between C and G.

For more Biology Sample papers in pdf you can click the following:
Bihar Board Biology Sample Paper Class XII
Class XII Biology Sample Paper Bihar Board

For more Bihar Board sample papers, follow the given links:
1) Sample Paper of Mathematics class XII of Bihar Board
2) Physics Sample Paper for class XI of Bihar Board
3) Sample Paper of Mathematics class XII of Bihar Board

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    12 th

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    Inter science

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    Board preparation of 12

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