Class 12 Biology Sample Paper With Solutions Set A

Sample Paper Class 12

Please refer to Class 12 Biology Sample Paper With Solutions Set A below. These Class 12 Biology Sample Papers will help you to get more understanding of the type of questions expected in the upcoming exams. All sample guess papers for Biology Class 12 have been designed as per the latest examination pattern issued by CBSE. Please practice all CBSE Sample Papers for Biology in Standard 12.

Sample Paper Class 12 Biology With Solutions Set A


1. Name the part of the flower which the tassels of the corn-cob represent.
Answer : The tassels of corn-cob represents stigma and style which wave in the wind to trap pollen grains.

2. Differentiate between parthenogenesis and parthenocarpy.
Answer : Differences between parthenogenesis and parthenocarpy are as follows: because seeds collected from hybrid plants, if sown subsequently, do not maintain hybrid characters due to segregation of traits.

3. Hybrid seeds have to be produced year after year. Give reason.
Answer : Hybrid seeds have to be produced year after year because seeds collected from hybrid plants, if sown subsequently, do not maintain hybrid characters due to segregation of traits.

4. Write the location and function of Sertoli cells in humans.
Answer : The cells that nourish the germ cells in the testes are called Sertoli cells. Sertoli cells are located in the germinal epithelium of the seminiferous tubules. 

5. Name the contrasting pod-related traits studied by Mendel in pea plant experiment.
Answer :  Traits related to pod studied by Mendel were pod shape and pod colour. Inflated (I) shaped and green coloured (G) pod were dominant traits while constricted (i) pod and yellow coloured (g) pod were recessive traits.

6. Write the level of biodiversity represented by a mangrove. Give another example falling in the same level.
Answer : Ecological diversity is found in mangrove. Rainforests also show same level of biodiversity.

7. Give an example of a human disorder that is caused due to a single gene mutation.
Answer : Sickle cell anaemia is due to inheritance of a defective allele coding for b-globin. It results in the transformation of HbA into HbS in which glutamic acid is replaced by valine at sixth position in each of two b-chains of haemoglobin.

8. Give two examples of potent bioweapon agents.
Answer : (i) Bacillus anthracis causing anthrax disease. (ii) Yersinia pestis causing plague.

9. Which enzymes are used during the breakdown of fungal cell and plant cell respectively in rDNA technology?
Answer : Fungal cells are treated with chitinase enzyme and plant cells are treated with cellulase enzyme. 10. The specific components that form deoxyguanosine are guanine (nitrogenous base) and deoxyribose (pentose sugar) linked together by glycosidic bond.

10. Name the specific components and the linkages between them that form deoxyguanosine.
Answer : The specific components that form deoxyguanosine are guanine (nitrogenous base) and deoxyribose (pentose sugar) linked together by glycosidic bond.

11. Assertion : Selectable marker is meant for distinguishing a recombinant from non-recombinant.
Reason : The recombinants can flourish in medium having both ampicillin and tetracycline, while the nonrecombinants
(a) Both assertion and reason are true, and reason is the correct explanation of assertion.
(b) Both assertion and reason are true, but reason is not the correct explanation of assertion.
(c) Assertion is true but reason is false.
(d) Both assertion and reason are false.
Answer : C


Assertion : Restriction enzymes cut the strand of DNA to produce sticky ends.
Reason : Stickiness of the ends facilitates the action of the enzyme DNA polymerase.
(a) Both assertion and reason are true, and reason is the correct explanation of assertion.
(b) Both assertion and reason are true, but reason is not the correct explanation of assertion.
(c) Assertion is true but reason is false.
(d) Both assertion and reason are false.
Answer : C

12. Assertion : The polynucleotide chains show polarity.
Reason : In a nucleotide, N-base is attached to deoxyribose sugar at C-1 and phosphoric acid is attached at
C-5 of sugar.
(a) Both assertion and reason are true, and reason is the correct explanation of assertion.
(b) Both assertion and reason are true, but reason is not the correct explanation of assertion.
(c) Assertion is true but reason is false.
(d) Both assertion and reason are false.
Answer : B

13. Assertion : Dodo, Passenger pigeon, Steller’s sea cow have become extinct.
Reason : Excessive exploitation of a species, whether animal or plants reduces size of its population so that
it becomes vulnerable to extinction.
(a) Both assertion and reason are true, and reason is the correct explanation of assertion.
(b) Both assertion and reason are true, but reason is not the correct explanation of assertion.
(c) Assertion is true but reason is false.
(d) Both assertion and reason are false.
Answer : A

14. Assertion : Very small animals are rarely found in polar areas.
Reason : Small animals have a larger surface area relative to their volume, so they tend to lose body heat
very fast in cold areas.
(a) Both assertion and reason are true, and reason is the correct explanation of assertion.
(b) Both assertion and reason are true, but reason is not the correct explanation of assertion.
(c) Assertion is true but reason is false.
(d) Both assertion and reason are false.
Answer : A

15. Read the following and answer any four questions from 15(i) to 15(v) given below:
A 5 year old son lives with his mother. Suddenly 2 persons claimed to be his father and filed a case. Police had ordered an investigation to solve the issue and to provide custody to right person. Forensic lab performed DNA fingerprinting and following pattern of bands is obtained.

Class 12 Biology Sample Paper With Solutions Set A

Answer : 

(i) How are DNA fragments separated?
(a) By gel electrophoresis according to size
(b) By gel electrophoresis according to charge
(c) By PCR according to size
(d) By Southern blotting according to speed
Answer : A

(ii) The method of hybridisation of DNA with radioactive probe is
(a) Northern blotting
(b) Western blotting
(c) Southern blotting
(d) none of these.
Answer : C

(iii) Identify the correct sequence of various steps of DNA fingerprinting.
A. Restricted DNA fragments are separated by electrophoresis
B. DNA band pattern is transferred to nylon membrane
C. DNA probe binds to radioactive probe
D. The X-ray film is developed to visualise the bands
(a) D → C → B → A
(b) A → D → C → B
(c) D → A → C → B
(d) A → D → B → C
Answer : D

(iv) Assertion : Claimant father 2 is the biological father of the child.
Reason : DNA bands of claimant father 2 are mere similar to DNA bands of mother.
(a) Both assertion and reason are true, and reason is the correct explanation of assertion.
(b) Both assertion and reason are true, but reason is not the correct explanation of assertion.
(c) Assertion is true but reason is false.
(d) Both assertion and reason are false.
Answer : C

(v) DNA fingerprinting is useful in
(a) solving paternity/maternity disputes
(b) identifying genes connected with heredity disorders
(c) to detect crime and legal pursuits
(d) all of these
Answer : D

16. Read the following and answer any four questions from 16(i) to 16(v) given below:
Two insect species were used in laboratory experiment. For one treatment, both species were grown by themselves (in separate chambers) on a suitable food source. For the second treatment the two species were grown together (in the same chamber) on the same type of amount of food as in the first treatment. 
The given graph shown the results (the number of individuals of each species in the two treatments) at the end of the experiment. 

Class 12 Biology Sample Paper With Solutions Set A

(i) The two species A and B should be classified as
(a) competitors
(b) mutualist
(c) predators of pathogens
(d) commensals.
Answer : A

(ii) Which of the following is correct regarding the experiment result?
(a) Species A grown more if it is grown alone.
(b) Species B grown more if it is grown alone.
(c) Species B grown more than the species A.
(d) Both species A and B grown more when they are grown together.
Answer : D

(iii) Which of the following could be another example of the given interaction?
(a) Interaction between alga and fungus
(b) Interaction between sea anemone and clown fish
(c) Interaction between tick bird with rhinoceros
(d) Interaction between crocodile with bird
Answer : A

(iv) The association between two species A and B is a type of association where
(a) both species are harmed
(b) both species are benefitted
(c) one species is harmed and other is benefitted
(d) one species is harmed and other is unaffected.
Answer : B

(v) Assertion : Interaction between two species A and B represent commensalism.
Reason : Interaction between species A and B results negative effects on growth and survival of both the population.
(a) Both assertion and reason are true, and reason is the correct explanation of assertion.
(b) Both assertion and reason are true, but reason is not the correct explanation of assertion.
(c) Assertion is true but reason is false.
(d) Both assertion and reason are false.
Answer : D


17. Draw and label the parts of the head region only of a human sperm.
Answer : The head region of human sperm is as follows:   

Class 12 Biology Sample Paper With Solutions Set A

18. Mention how the various factors affect linkage.
Answer : Linkage is affected by the following factors:
(i) The strength of linkage increases with the increase in age.
(ii) A rise in temperature decreases the strength of linkage.
(iii) Radiations like UV rays and X rays decrease the strength of linkage.
(iv) Distance between two genes if increased, then the linkage strength decreases.

19. Write the names of the microbes from which the following enzymes are obtained.
(i) Protease (ii) Amylase (iii) Rennet
(iv) Lactase (v) Streptokinase (vi) Lipase
Answer : (i) Aspergillus (ii) Rhizopus
(iii) Mucor (iv) Saccharomyces fragilis
(v) Streptococcus (vi) Candida lipolytica


(a) Name the two microbes used as biocontrol agents.
(b) Explain how the two microbes act as biocontrol agents. Describe why biopesticides are preferred over chemical pesticides.
Answer : (a) (i) Nucleopolyhedrovirus (ii) Bacillus thuringiensis
(b) Nucleopolyhedrovirus a genus of baculoviruses are useful in controlling many insects and other arthropods. They are species specific narrow spectrum bioinsecticides with no side effects on plants, mammals, birds, fish and non-target insects. Bacillus thuringiensis is a soil bacterium. The spores of this bacterium possess insecticidal cry protein, hence, kill larvae of certain insects. This bacterium was first biopesticide to be used on commercial scale. The commercial preparations of B. thuringiensis contain a mixture of spores, cry protein and inert carrier. Chemical pesticides used in agricultural fields are toxic and biocides. They kill even useful organisms along with harmful ones, harm human beings and animals, pollute soil and water and crop plants. It is estimated that despite the use of chemical pesticides 30% of the agricultural produce is lost to pathogens and pests because these continue to develop resistance against various pesticides. On the other hand biopesticides are the biological agents that control the growth of weeds, insects and pathogens in an agricultural field. They have targeted actions and are harmless to the crop plants, other beneficial field animals and humans. Therefore, nowadays biopesticides are preferred over chemical pesticides.

20. What are the basic requirements of a polymerase chain reaction?
Answer : The basic requirements of a polymerase chain reaction are: (i) DNA template – The desired segment of target DNA molecule that is to be amplified. (ii) Primers – Two nucleotide primers, usually 10-18 nucleotides long and complementary to the sequences present at the 3′ end of the target DNA segment. (iii) Enzyme – Taq DNA polymerase which is stable at high temperature is required to carry out synthesis of new DNA.


How Agrobacterium tumefaciens has been made a useful cloning vector to transfer DNA to plant cells?
Answer : Agrobacterium tumefaciens is a soil-inhabiting bacterium that may invade growing plants at the junction of root and stem, where it can cause a cancerous growth known as a crown gall. The bacterium, which infects dicotyledonous plants only, contains plasmid (known as a Ti plasmid) that carries the genes for tumour formation. When the bacterium invades the host cells, the Ti plasmid enters the host nuclei. Part of the Ti plasmid become inserted into the chromosome, where it produces copies of itself. For using Ti plasmid as a vector, researchers have eliminated its tumour causing properties while keeping its ability to transfer DNA into plant cells. This bacterium is called natural genetic engineer because genes carried by Ti plasmid produce effect in several parts of the plant.

21. What are bioreactors? List five growth conditions that a bioreactor provides for obtaining the desired product.
Answer : A bioreactor is a vessel in which raw materials are biologically converted into specific products by microbes, plant and animal cells and their enzymes. These are used for food processing, fermentation, waste treatment, etc. Growth conditions that a bioreactor provides for obtaining the desired products are as follows:
(i) Controlled environment for optimum product yield
(ii) Aseptic fermentation for number of days and prevention of escape of viable cells
(iii) Adequate mixing and aeration for optimum growth and production, without damaging the microorganism
(iv) Easy and dependable temperature control
(v) Facility of sampling

22. Differentiate between type I, II and III restriction endonucleases.
Answer : Differences between type I, type II and type III restriction endonucleases are as follows: 

Class 12 Biology Sample Paper With Solutions Set A
Class 12 Biology Sample Paper With Solutions Set A

23. Name the two types of immune systems in a human body. Why are cell mediated and humoral immunities so called?
Answer : Two types of immune systems in human body are : Antibody Mediated Immune System (AMIS) and Cell Mediated Immune System (CMIS). Antibodymediated (or humoral) immunity is associated with the appearance of antibodies, secreted by B-lymphocytes, in the extracellular fluids such as plasma, lymph and external secretions. Cell-mediated immunity is mediated by T-lymphocytes that defend against pathogens including protists and fungi that enter the cells.

24. Name an allergen and write the response of the human body when exposed to it.
Answer : Pollen grain is an allergen. Exposure to pollen causes hay fever. It is the form of allergy due to pollen of grasses, trees and other plants. It is characterised by inflammation of the membrane lining the nose and sometimes of the conjunctiva. The symptoms are sneezing, running nose and watering eyes due to histamine release.

25. Name the type of interaction in each of the following.
(i) Clown fish living among the tentacles of sea anemone
(ii) Nepenthes capturing an insect
(iii) Mycorrhizae living on the roots of higher plants
(iv) Lice living on skin of humans
(v) Superior barnacle (Balanus) dominating the rocky sea coasts of Scotland and excluding the smaller barnacle Chathamalus
Answer :  (i) Commensalism (ii) Predation (iii) Mutualism (iv) Parasitism (v) Competition 2


26. A woman has certain queries as listed below, before starting with contraceptive pills. Answer them :
(i) What do contraceptive pills contain and how do they act as contraceptives?
(ii) What schedule should be followed for taking these pills?
Answer : (i) Contraceptive pills may contain either hormone progestogen or a combination of both progestogen and estrogen hormones. Oral (contraceptive) pills help to prevent pregnancy by inhibiting body’s natural cyclical hormones. They usually stop the body from ovulating, change cervical mucus to make it difficult for sperm to go through cervix and prevent implantation of the fertilised egg.
(ii) Contraceptive pills have to be taken daily for 21 days starting within the first five days of menstrual cycle and this has to be repeated after a gap of 7 days in the same pattern till the female desires to prevent conception. 

27. A cross between a normal couple resulted in a son who was haemophilic and a normal daughter. In course of time, when the daughter was married to a normal man, to their surprise, the grandson was also haemophilic.
(i) Represent this cross in the form of a pedigree chart. Give the genotypes of the daughter and her husband.
(ii) Write the conclusion you draw form the inheritance pattern of this disease.
Answer : (i) The pedigree chart of the cross is represented as:

Thus, from the above pedigree, the genotypes of daughter and her husband can be deduced as XXh and XY respectively. (ii) From this, it can be concluded that haemophilia is a sex-linked disease showing criss-cross pattern of inheritance where a parent passes the traits to the grandson. Here , the female is the carrier of the disease haemophilia. 

28. (a) Plot a logistic growth curve.
(b) Briefly describe nature’s carrying capacity for a species in a particular habitat.
Answer : (a) 

(b) In nature, a given habitat has resources to support a certain number of individuals of a population, beyond which no further growth is possible. This limit is called nature’s carrying capacity (K) for that species in that habitat. Nature’s limited resources limit the carrying capacity for a species in a particular habitat.

29. (a) What do you mean by the term biopiracy?
(b) What measures are being taken by the Indian Government to prevent biopiracy? 
Answer : (a) When bioresources of one country are exploited or patented by some multinational companies or organisations of other countries without proper authorisation from concerned country and compensation is referred to as biopiracy. (b) To prevent unauthorised exploitation of bioresources and traditional knowledge, the Indian Parliament has recently cleared the second amendment of the Indian Patents Bill, that takes biopiracy issues into consideration, including patent terms, emergency provisions, research and development initiatives.


(a) Identify the labelled part ‘A’ and ‘B’.
(b) What does the diagram illustrate?
(c) Mention the type of immune response provided by the given structure.
Answer : (a) ‘A’ – Antigen binding site
‘B’ – Variable region of heavy chain
(b) This diagram illustrates the structure of an
antibody molecule.
(c) Humoral immune response is an antibody
mediated immune response, which is regulated by
antibodies in body fluids.


(a) Write the scientific name of the source plant of following drugs:
(i) Marijuana (ii) Opium (iii) Cocaine (iv) LSD
(b) Name any two natural hallucinogens and mention their effects on human body.
Answer : (a) (i) Cannabis sativa
(ii) Papaver somniferum
(iii) Erythroxylum coca
(iv) Claviceps purpurea
(b) The two natural hallucinogens are: (i) LSD (Lysergic acid diethylamide): It causes horrible dreams, emotional outbursts, hallucination, chronic psychosis. It also brings about chromosomal and fetal abnormalities. (ii) Cannabinoids: These raise the blood sugar level and increase the frequency of urination. They bring about a state of well being, excitement, sometimes uncontrolled laughter and dilation of pupil of eyes.


31. (a) Spleen acts as a lymphoid organ. Justify the statement.
(b) Differentiate between the following:
(i) Innate and acquired immunity
(ii) B – lymphocytes and T – lymphocytes
(c) What are antigen presenting cells? Mention its types.
Answer :  (a) Spleen acts as a secondary lymphoid organ where mature B and T lymphocytes undergo proliferation and differentiation. In spleen, lymphocytes develop immune response and become effector cells.
(b) (i) Differences between innate and acquired immunity are as follows:

(ii) Differences between the B-lymphocytes (B-cells) and T-lymphocytes (T-cells) are :

(c) The cells that engulf antigens and present their fragments to T-cells are called antigen presenting cells (APCs).
There are three types of APCs found in the body : (i) Macrophages, (ii) Dendritic cells, (iii) B-cells.


How sewage water is treated before it could be discharged into natural water bodies?
Answer : Sewage water can be purified by passing it through sewage treatment plants with the action of microorganisms. A sewage treatment plant separates solids from liquids by physical processes and purifies the liquid by biological processes. There are three stages of this treatment; primary, secondary and tertiary. Primary treatment is physical, secondary biological and tertiary chemical. Primary treatment phase of sewage treatment removes floating and suspended solids from sewage through two processes of filtration and sedimentation. First floating matter is removed through sequential filtration. The filtrate is kept in large open settling tanks where grit settles down. Aluminium or iron sulphate is added in certain places to flocculation and settling down of solids. The sediment is called primary sludge while the supernatant is called effluent. The primary sludge traps a lot of microbes and debris. It is subjected to composting or land fill where anaerobic digestion removes the organic matter. During secondary treatment, the primary effluent is taken to aeration tanks. A large number of aerobic heterotrophic microbes grow in the aeration tank. They form flocs are masses of bacteria held together by slime and fungal filaments to form mesh-like structures. The microbes digest a lot of organic matter, converting it into microbial biomass and releasing a lot of minerals. As a result the BOD of the waste matter is reduced to 10-15% of raw sewage, it is passed into settling tank. In settling tank, the bacterial flocs are allowed to undergo sedimentation. The effluent or supernatant is generally passed into natural water bodies and sediment of settling tank is called activated sludge.

32. (a) Draw a diagram of an enlarged view of T.S. of one microsporangium of an angiosperm and label the following parts:
(i) Tapetum
(ii) Middle layer
(iii) Endothecium
(iv) Microspore mother cells
(b) Mention the characteristic features and function of tapetum.
(c) Explain the following with reasons:
(i) Pollen grains are well preserved as fossils.
(ii) Pollen tablets are in use by people these days.
Answer : (a) Sectional view of microsporangium of an angiosperm is as follows: 

(b) The characteristics of tapetum are as follows:
(i) Tapetal cells are filled with protoplasmic contents as well as nutrients.
(ii) They are either multinucleate or their nucleus becomes polyploid due to endoploidy.
(iii) Tapetum is of two types – amoeboid and secretory.
(iv) In amoeboid type, the tapetal cells fuse to form a plasmodium or periplasmodium because it passes in between the sporogenous cells to nourish them.
(v) The cells of secretory tapetum pass out substances over the sporogenous cells for their growth and differentiation.
Tapetum has a number of functions:
(i) Nourishment of the developing microspore mother cells and pollen grains.
(ii) It produces lipid rich ubisch granules containing sporopollenin for exine formation, pollenkitt, special proteins for the pollen grains which recognise compatibility and hormone IAA.
(iii) It secretes enzyme callase responsible for the degradation of callose wall around pollen tetrad.
(c) (i) Sporopollenin is present in exine layer of pollen grains. Sporopollenin is highly resistant fatty substance not degraded by any enzyme and not affected by high temperature, strong acid or strong alkali therefore, pollen grains can be well preserved as microfossils.
(ii) Pollen tablets are used as food supplement by people to improve health.


(a) Explain any three outbreeding devices in flowering plants.
(b) Suggest the advantages to a farmer of using apomictic seeds of hybrid varieties.
Answer : 

(a) The three outbreeding devices that flowering plants have developed are as follows:
(i) Dicliny (Unisexuality) : Flowers are unisexual so that self pollination is not possible. The plants may be monoecious (bearing both male and female flowers, e.g., maize) or dioecious (bearing male and female flowers on different plants, e.g., mulberry, papaya). (ii) Dichogamy : Anthers and stigmas mature at different times in a bisexual flower so as to prevent self pollination.
(a) Protandry : Anthers mature earlier than stigma of the same flower. Their pollen grains become available to stigmas of the older flowers, e.g., sunflower, Salvia.
(b) Protogyny : Stigmas mature earlier so that they get pollinated before the anthers of the same flower develop pollen grains, e.g., Mirabilis jalapa, Gloriosa, Plantago.
(iii) Heterostyly : There are 2 or 3 types of flowers with different heights of styles (and stamens), e.g., in diheterostyly (dimorphic heterostyly), there are two types of flowers, pin eyed (long style and short stamens) and thrum eyed (short style and long stamens), e.g., Primula (primrose), jasmine. Pollination occurs between anthers and stigmas of the same height present in different flowers.
(b) Advantages of using apomictic seeds of hybrid varieties are as follows:
(i) Apomixis can be used to preserve desired characters.
(ii) It will reduce the cost of purchasing hybrid seeds every year. (iii) Embryos formed through apomixis are generally free from infections.

33. Explain the process of protein synthesis.
Answer : The process of decoding of the message from mRNA to protein with the help of tRNA, ribosome and enzymes is called protein synthesis or translation. Protein synthesis occurs over ribosomes. The 4 main steps in protein synthesis are: activation, initiation, elongation and termination of polypeptide chain. The newly synthesised mRNA joins the smaller subunit of ribosome at 5′ end. mRNAs carry the codon and tRNAs carry the anticodon for the same codon. Activation of amino acid is catalysed by the enzyme aminoacyl tRNA synthetase in the presence of ATP. In presence of ATP an amino acid combines with its specific aminoacyl-tRNA synthetase to produce aminoacyl adenylate enzyme complex. This reacts with tRNA to form aminoacyl-tRNA complex. Activated tRNA is taken to ribosome mRNA complex for initiation of protein synthesis. Initiation of protein synthesis is accomplished with the help of 3 initiation factors (IF3, IF2, IF1) in prokaryotes and 9 in eukaryotes (eIF2, eIF3, eIF1, eIF4A, eIF4B, eIF4C, eIF4D, eIF5, eIF6). The ribosome binds to the mRNA at the start codon (AUG) that is recognised only by the initiator tRNA. A polypeptide chain forms as tRNAs deliver amino acids to the ribosome. Large ribosomal subunit binds the initiation complex forming two (A and P) binding sites for tRNA molecules. The first site is P site or peptidyl site which is occupied by tRNAmet. The second site is A or aminoacyl site and is positioned over the second codon. Now, the ribosome proceeds to the elongation phase of protein synthesis. During this stage, complexes composed of an amino acid linked to tRNA, sequentially bind to the appropriate codon in mRNA by forming complementary base pairs with the tRNA anticodon. The ribosome moves from codon to codon along the mRNA. Amino acids are added one by one, translated into polypeptide sequences dictated by DNA and represented by mRNA. The enzyme peptidyl synthetase catalyses the formation of peptide bond between the carboxylic group of amino acid at P site and amino group of amino acid at A site. Enzyme translocase brings about the movement of mRNA by one codon. The termination of protein synthesis occur when a non-sense codon reaches at A site of ribosome. The chain detaches from the ribosome. A release factor binds to the stop codon, terminating translation and releasing the complete polypeptide from the ribosome. Two subunits of ribosomes dissociate with the help of dissociation factor.


State three principles of Mendel’s law of inheritance. Using Punnett square demonstrate the law of independent assortment in a dihybrid cross involving two heterozygous parents.
Answer : The three principles of Mendel’s law of inheritance are :
(i) Law of dominance
: This law explains that when two individuals of a species, differing in a pair of contrasting forms of a trait are crossed, the form of the trait that appears in the F1 hybrid is dominant and the alternate form that remains hidden, is called recessive.
(ii) Law of segregation : This law states that the members of the allelic pair that remained together in the parent, segregate during gamete formation and only one factor enters a gamete.
(iii) Law of independent assortment : This states that in the inheritance of two pairs of contrasting characters, the factors of each pair of characters segregate independently of the factors of the other pair of characters. The principle or law of independent assortment can be studied by means of dihybrid cross between heterozygous parents having YyRr genotype.
This can be demonstrated through Punnett square as follows:

hus, the phenotypic ratio of a dihybrid cross is 9 : 3 : 3 : 1. The occurrence of four types of plants (parental types) in the F2 generation of dihybrid cross shows that the factors of each of the two characters assort independently of the other as if the other pair of factor are not present.