CBSE Class 12 Biology Question Paper (Outside Delhi 2014) with Solutions

Students can use CBSE Previous Year Question Papers Class 12 Biology with Solutions and CBSE Class 12 Biology Question Paper (Delhi 2014) to familiarize themselves with the exam format and marking scheme.

CBSE Class 12 Biology Question Paper (Delhi 2014) with Solutions

Time Allowed: 3 Hours
Maximum Marks: 70

General Instructions:

1. All questions are compulsory.
2. This question paper consists of four Sections A, B, C and D. Section A contains 8 questions of one mark each, Section B is of 10 questions of two marks each, Section C is of 9 questions of three marks each and Section D is of 3 questions of five marks each.
3. There is no overall choice. However, an internal choice has been provided in one question of 2 marks, one question of 3 marks and all the questions of 5 marks weightage. A student has to attempt only one of the alternatives in such questions.
4. Wherever necessary, the diagrams drawn should be neat and properly labelled

Section – A

Question 1.
Name the parts of the flower which the tassels of the corncob represent.
Answer:
The tassels of the corn-cob represent stigma and style of the flower.

Question 2.
Mention any two contrasting traits with respect to seeds in pea plant that were studied by Mendel.
Answer:
The two contrasting traits related to the seeds of pea plant that were studied by Mendel are as follows:
Seed shape- Round seeds and Wrinkled seeds
Seed colour-Yellow and Green

Question 3.
Why is secondary immune response more intense than the primary immune response in humans?
Answer:
The secondary immune response is more intense than primary immune response because the secondary immune response is based on the memory of the first encounter. The secondary immune response takes place when our body encounters the same antigen for second time. Our body have memory of the first encounter that recognize the pathogen quickly on subsequent exposure and initiates the production of antibodies.

Note
The secondary immune response is pathogen specific and is characterised by memory.

Question 4.
Why is it not possible for an alien DNA to become part of a chromosome anywhere along its length and replicate normally?
Answer:
It is not possible for an alien DNA to become part of a chromosome anywhere along its length and replicate normally because the process of replication does not initiate randomly at any place in the DNA as it only initiates at the origin of replication.

Question 5.
State the role of C-peptide in human insulin.
Answer:
The C-peptide in human insulin is a short stretch of amino acids that is present in proinsulin and during the process of maturation of insulin this C-peptide is removed. So, it is involved in the synthesis of mature insulin.

Note
An American company Eli Lilly in 1983 prepared two DNA sequence corresponding to A and B chains of human insulin and introduced them in plasmids of E.coli to produce insulin chains. Chain A & chain B were separately produced & joined together by disulphide bond.

Question 6.
Name the enzymes that are used for the isolation of DNA from bacterial and fungal cells for recombinant DNA technology.
Answer:
Enzymes that can be used for the isolation of DNA from bacterial cell is lysozyme and for fungal cell is chitinase for recombinant DNA technology.

Question 7.
State Gause’s Competitive Exclusion Principle.
Answer:
Gause’s Competitive Exclusion Principle’ states that two closely related species competing for the same resources cannot co-exist indefinitely and the competitively inferior one will be eventually eliminated.

Question 8.
Name the type of association that the genus Glomus exhibits with higher plants.
Answer:
Fungi are able to form symbiotic association with the root nodules of plants and are called mycorrhiza. So many members of the genus Glomus form mycorrhiza. The fungal symbiont in this association absorbs phosphorus from soil and passes it to the plant. Plants having such associations show other benefits such as these are resistance to root-bome pathogens, tolerance against salinity and drought. It also promotes the overall growth and development of plant.

Section – B

Question 9.
Why are the human testes located outside the abdominal cavity? Name the pouch in which they are present.
Answer:
The testes are located outside the abdominal cavity in a pouch like structure called scrotum because the process of spermatogenesis or sperm production requires lower temperature (2-2.5°C) than the normal body temperature (37°C). The scrotum helps in maintaining the low temperature of the testes.

Question 10.
In Snapdragon, A cross between true-breeding red flower (RR) plants and true-breeding white flowered (rr) plants showed a progeny of plants with all pink flowers.
(a) The appearance of pink flowers is not known as blending. Why?
(b) What is the phenomenon known as ?
Answer:
(a) The appearance of pink colour in a Snapdrgon flower or Antirrhinum sp. is not considered as blending because the process of blending involves intermixing of two characters but in Snapdragon flower, the alleles do not mix with each other. So, they tends to maintain their originality and reappear in F2 generation. The development of pink colour is because the dominant allele is not completely dominant over the recessive allele.

(b) This phenomenon is called incomplete dominance.

Question 11.
With the help of one example, explain the phenomena of co-dominance and multiple allelism in human population.
Answer:
The phenomena of multiple allelism and co-dominance are represented by ABO blood group in human beings. The gene i for the blood group contains three alleles such as I ^A , I ^B and i. These alleles represent the phenomena of multiple allelism as it contain more than two alleles that controls the same characters. The alleles I ^A and I ^B are co-dominant and express themselves independently in a next generation. The presence of both the blood group I ^A and I ^B produces blood group AB and this is an example of co-dominance.

Note
Co-dominance is a form of inheritance in which the alleles of a gene pair in a heterozygote are independently expressed themselves in a hybrid.

Question 12.
Write the scientific name of the fruit-fly. Why did Morgan prefer to work with fruit-flies for his experiments? State any three reasons.
OR
Linkage or crossing-over of genes are alternatives of each other. Justify with the help of an example.
Answer:
The scientific name of the fruit fly is called Drosophila melanogaster. Morgan prefers to work with fruit-flies for his experiment because of the following reasons such as:

• They are easily grown on a simple synthetic medium in a laboratory.
• They complete their life cycle in about two weeks and a single mating could produce a large number of offsprings.
• There was a clear differentiation of sexes as male and female flies are easily distinguishable.

OR

Linkage is defined as the tendency of two or more non-allelic genes to be inherited together because they are located more or less closely on the same chromosome. Whereas crossing over is defined as the process by which two homologous chromosome paired and exchange chromosomal segments of the coiled DNA.

So, if the linkage is high then the crossing over will be low whereas if linkage is less than the crossing over will be more.

For example, in drosophila, a yellow bodied and white eyed female was crossed with brown bodies and red eyed male, F ₁ progeny is produced and then they are inter-crossed. The F ₂ phenotypic ratio of drosophila deviated significantly from Mendel’s 9:3:3:1, the genes for eye colour and body colour are closely located on the ‘X’ chromosome indicates high linkage. So, they are inherited together and the recombinants were formed due to crossing over but at low percentage.

Question 13.
List the symptoms of Ascariasis. How does a healthy person acquire this infection?
Answer:
The symptoms of Ascariasis are as follows:

• Constipation
• Abdominal pain and cramps
• Stools with excess mucous and blood clots

Houseflies act as mechanical carriers and serves to transmit the parasite from faeces of infected person to food and food products. Drinking water and food contaminated by the faecal matter are the main sources of infection.

Question 14.
Explain the significant role of the genus Nucleopolyhedrovirus in an ecological sensitive area.
Answer:
The genus Nucleopolyhedrovirus are baculoviruses that are used as a biological control agents. These viruses are excellent source for species-specific, narrow spectrum insecticidal applications. There will be negative impact on plants, mammals, birds, and fish or even on non-target insects. Such viruses are used for integrated pest management programme (IPM).

Question 15.
How does a restriction nuclease function? Explain.
Answer:
Restriction enzymes belong to a larger class of enzymes called nucleases. They are of two types such as endonuclease and exonucleases. Restriction endonuclease is the most widely used enzyme in recombinant DNA technology and molecular biology. This enzyme is involved in the process of gene modification.

It recognizes specific sequences of nucleotides that is known as recognition site and produces a double strand nick in the desired DNA. The action of restriction endonuclease can be on palindromic sequences. For example, EcoRIcuts the DNA at the following palindromic sequence:
5’ GAATTC 3’
3’CTTAAG 5’
Exonucleases remove nucleotides from the ends of the DNA.

Note
Palindromes are groups of letters that form the same words when read both forward and backward. For example “MALAYALAM”

Question 16.
How have transgenic animals proved to be beneficial in:
(a) Production of biological products
(b) Chemical safety testing
Answer:
(a) The first transgenic cow called Rosie was made in 1997 that produces human-protein-enriched milk. The milk of that cow contains human alpha-lactalbumin and was nutritionally a more balanced diet for human babies than normal natural- cow.

(b) Transgenic animals are made that carry genes which make them more sensitive to toxic substances than non-transgenic animals. They are then exposed to the toxic substances and the effects are studied. Toxicity testing in such animals will allow us to obtain results in less time.

Question 17.
Describe the mutual relationship between Fig tree and wasp and comment on the phenomenon that operates in their relationship.
Answer:
The relationship between the fig tree and wasp shows mutualism, the wasp lays its egg and also pollinates the fig. On the other hand, the fig not only provides shelter (fruit) for oviposition to wasp but also the insect its larva can feed on seeds.

Note
Mutualism is a type of population interaction that confers benefits on both the interacting species.

Question 18.
Construct an age pyramid which reflect an expanding growth status of human population.
Answer:
Diagrammatic Representation of Expanding age pyramid:

Expanding age pyramid indicates that the population is growing.

Note
Age pyramids represent age distribution of males and females in a combined diagram. As, the shape of the pyramid reflects the growth status of the population.

Section – C

Question 19.
Make a list of any three outbreeding devices that flowering plants have developed and explain how they help to encourage cross-pollination.
OR
Why are angiosperm anthers called dithecous? Describe the structure of its microsporangium.
Answer:
The outbreeding devices developed by plants are as follows:

• Some species of plants, pollen release an receptivity of stigma are not synchronised. Either the pollen is released before the stigma becomes receptive or stigma becomes receptive much before the release of pollen.
• Some species of plants, the anther and stigma are placed at different positions so that the pollen cannot come in contact with the stigma of the same flower. Both these devices prevent autogamy.
• The third device helps to prevent inbreeding is self incompatibility. It is genetic mechanism that prevents self-pollen (from the same flower or other flowers of the same flower) from fertilising the ovules by inhibiting pollen germination or pollen tube growth in the pistil.

OR

An angiosperm anther is bilobed with each lobe having two theca and because of this is called dithecous.

Structure of microsporangium:

A microsporangium is surrounded by four wall layers such as epidermis, endothecium, middle layer and tapetum. The outer three wall layers provide protection to the microsporangium and also help in dehiscence of anther to release the pollen. The innermost wall layer is the tapetum. It provide nourishment to the developing pollen grains. Cells of the tapetum possess dense cytoplasm and contain more than one nucleus.

Note
The outer hard layer called the exine is made up of sporopollenin which is one of the most resistant organic material. It can tolerate high temperature as well as all biochemical and enzymatic degeradation.

Question 20.
If implementation of better techniques and new strategies are required to provide more efficient care and assistance to people, then why is there a statutory ban on amniocentesis? Write the use of this technique and give reason to justify the ban.
Answer:
Aminocentesis is a foetal-sex determination test based on the chromosomal pattern in the amniotic fluid surrounding the developing embryo. This technique is used for the determination of sex and other metabolic disorders of the developing embryo. It is also used for the determination of genetic disorders in the developing foetus. Their is statutory ban on amniocentesis for sex-determination of developing foetus that increases female foeticides.

Question 21.
Why is pedigree analysis done in the study of human genetics? State the conclusions that can be drawn from it.
Answer:
Pedigree analysis is done to study the human genetics because it provides all essential information that can be utilised to trace the inheritance of a specific trait such as abnormality or disease. Several disorders are inheritable and depend on the genetics of the families. They inherit the genes from such families. So pedigree analysis is done to trace such inheritance patterns.

The conclusions that can be drawn from the pedigree analysis are:

• The pattern of inheritance and tracing of Mendelian disorders.
• The trait in question is dominant or recessive.
• The trait is linked to the sex chromosomes or autosomes.

Note
In the pedigree analysis the inheritance of a specific trait is represented in the family tree over generations.

Question 22.
Identify ‘a’, ‘b’, ‘c’ ‘d’, ‘e’ and ‘f’ in the table given below :

Answer:

Note
The chromosomal disorders are caused due to absence or excessive or abnormal arrangement of one or more chromosomes.

Question 23.
Community Service department of your school plans a visit to a slum area near the school with an objective to educate the slum dwellers with respect to health and hygiene.
(a) Why is there a need to organize such visits?
(b) Write the steps you will highlight, as a member of this department, in your interaction with them to enable them to lead a healthy life.
Answer:
(a) There is need to Organize Community Service
Department to visit a slum area is to create awareness about disease and their effects on the body and health and hygiene.

(b) The steps we will highlight, as a member of this department in our interaction with them to enable them to lead a healthy life will be:

• Explain them the importance of healthy life and healthy people more efficient at work.
• Health also increases productivity, economy, longevity and also reduces infants as well as maternal mortality.
• To make them aware of the various diseases and their effect.
• Teach people about the proper disposal of waste, control of vectors like mosquitoes, importance of hygienic food and drinking, balanced diet.
• Maintenance of hygienic environment, proper disposal of waste, and control of vector.

Question 24.
The following graph shows the species area relationship. Answer the following questions as directed.

(a) Name the naturalist who studied the kind of relationship shown in the graph. Write the observation made by him.
(b) Write the situations as discovered by the ecologists when the value of ‘Z’ (slope of the line) lies between
(i) 0.1 and 0.2
(ii) 0.6 and 1.2
What does ‘Z’ stand for ?
(c) When would the slope of the line ‘b’ become steeper?
Answer:
(a) Alexander Von Humboldt was the naturalist who studied the relationship shown in the graph. He observed that within a region the species richness increases with the increase in explored area it only upto a limit.

(b) When the value of z lies between
(i) z = 0.1 to 0.2 for small or average area.
(ii) z = 0.6 to 1.2 for large area for example continent.
Z stands for slope of line.

(c) The slope of the line ‘b’ becomes steeper when very large areas such as continents are considered for species-area relationship.

Question 25.
Name and describe the technique that helps in separating the DNA fragments formed by the use of restriction endonuclease.
Answer:
The technique that is used for separating DNA fragments in the lab is called gel electrophoresis. In this technique, restriction endonuclease is used for cutting DNA fragments.

• The DNA fragments are negatively charged molecules so they can be separated by forcing them to move towards the anode under an electric field through a medium or matix.
• The most commonly used matrix is called agarose that is a natural polymer extracted from sea weeds.
• The DNA fragments separate according to their size through sieving effect provided by the agarose gel
• So, the smaller the fragment size, farther, it moves.
• The separated DNA fragments can be visualised 7 only after staining the DNA with a compound known
  as ethidium bromide followed by exposure to UV radiation.
• A DNA marker with fragments of known lengths is usually run through the gel at the same time as the samples.
• By comparing the bands of the DNA samples with those from the DNA marker one can work out at the approximate length of the DNA fragments in the samples.

Question 26.
State the function of a reservoir in a nutrient cycle. Explain the simplified model of carbon cycle in nature.
Answer:
The movement of nutrient elements through the various components of an ecosystem is called nutrient cycling. It is also called biogeochemical cycles. It is of two types such as gaseous and sedimentary. The reservoir for gaseous type of nutrient cycle such as nitrogen and carbon cycle exists in the atmosphere and for the sedimentary cycle such as sulphur and phosphorus cycle. This reservoir is located in Earth’s crust.

Environmental factors such as soil, moisture, pH, temperature and so on tends to release the nutrients in the atmosphere.

The main function of the reservoir is to meet with the deficit that occurs due to imbalance in the rate of influx and efflux.

Question 27.
Since the origin of life on earth, there were five episodes of mass extinction of species.
(i) How is the ‘Sixth Extinction’, presently in progress, different from the previous episodes ?
(ii) Who is mainly responsible for the ‘Sixth Extinction’?
(iii) List any four points that can help to overcome this disaster.
Answer:
(i) The sixth extinction is different from previous extinctions in several ways:
The sixth extinction occurs rapidly such as the reduction in a number of species per unit area per unit time. It is accelerated by human activities such as deforestation, industrialization and so on.

(ii) Human activities that lead to global warming and disruption of environmental as well as ecological balance are responsible for the sixth extinction.

(iii) The disaster can be overcome by following ways:

• Afforestation
• Reduction in over-exploitation of natural resources
• Conservation of species and their natural habitats in order to minimize the losses.
• Create awareness among people regarding global warming and their consequences.

Section – D

Question 28.
(a) Where does fertilisation occur in humans? Explain the events that occur during this process.
(b) A couple where both husband and wife are producing functional gametes, but the wife is still unable to conceive, is seeking medical aid. Describe any one method that you can suggest to this couple to become happy parents.
OR
(a) Explain the different ways apomictic seeds can develop. Give an example of each.
(b) Mention one advantage of apomictic seeds to farmers.
(c) Draw a labelled mature stage of a dicotyledonous embryo.
Answer:
(a) The process of fertilisation takes place only if the ovum and sperms are transported simultaneously to the ampullary-isthmic junction. The process of fusion of a sperm with an ovum at this junction is ampullary- isthmic junction of fallopian tube.

The events involved in the process of fertilisation are as follows:

• During the process of fertilisation, a sperm comes in contact with the zona pellucida layer of the ovum and also induces changes in the membrane that block the entry of additional sperms.
• It ensures that only one sperm can fertilise an ovum.
• The secretions of the acrosome help the sperm enter into the cytoplasm of the ovum through the zona pellucida and the plasma membrane.
• This induces the completion of the meiotic division of the secondary oocyte.
• The second meiotic division is also unequal and results in the formation of second polar body and a haploid ovum.
• The haploid nucleus of the sperms and that of the ovum fuse together to form a diploid zygote.

Note
The acrosome of the sperm secretes hyaluronidase enzyme to peneterate the corona radiata layer of the ovum.

(b) The technique suggested to the couples who are not able to conceive is IVF (Invitro fertilisation). This method involves embryo transfer. In this method, ova from the wife or donor female and sperms from the husband or donor male are collected. The ova and sperm are induced to form zygote under stimulated conditions in the laboratory.

Note
IVF is the method of fertilisation that takes place outside the body in almost the conditions similar to that of body. This technique is also called test tube programme.

OR

(a) There are two ways by which the apomitic seeds are produced:
Agamospermy: In this method, the seeds are produced from diploid cells without meiosis and fertilisation. For example: Apple.
Adentive embryony: In this method, the nucellus and integuments extends into the embryo sac and develops into embryo. It involves the formation of more than one embryo. For example: citrus fruit.

(b) The main advantage of apomictic seeds is that there is no segregation of characters in the hybrid progeny. The apomictic seeds are cost effective and high yielding.

(c) Diagrammatic Representation of mature stage of dicotyledonous embryo:

Question 29.
(a) Describe the various steps of Griffith’s experiment that led to the conclusion of the ‘Transforming Principle’.
(b) How did the chemical nature of the ‘Transforming Principle’ get established?
OR
Describe how the lac operates, both in the presence and absence of an inducer in E.coli.
Answer:
(a) The transforming principle was proposed by Frederick Griffith in 1928. He performed his experiment with Streptococcus pneumonia a bacteria which is responsible for causing pneumonia.

The following steps are involved in his experiment:

• When Streptococcus pneumonia bacteria are grown on a culture plate, some bacteria produces smooth shiny colonies (s) whereas other produce rough colonies (R).
• It is because the S strain bacteria have a mucous polysaccharide coating whereas the R strain bacteria lack this coating.
• When the mice infected with the S strain or virulent strain, the mice die because of pneumonia infection.
• When the mice infected with the R strain do not develop pneumonia.
• Then, Griffith was able to kill bacteria by heating. He observed that heat-killed S strain bacteria injected into mice did not kill them.
• When he injected a mixture of heat-killed S and live R bacteria, the mice died. He recovered living S bacteria from the dead mice.

After this experiment, he concluded that the R strain bacterium has been transformed by the heat-killed S strain bacteria. As some ‘transforming principle’ was transferred from the heat-killed S strain and enabled the R strain to synthesise a smooth polysaccharide coating and become virulent. This is because of the transfer of the genetic material.

(b) The biochemical characterisation of Transforming principle was determined by Oswald Avery, Colin Macleod and Maclyn McCarty. Prior it was thought that the genetic material was protein.

The following steps are involved in his experiment:

• They purified biochemicals such as proteins, DNA and RNA from the heat-killed S cells to determine which one could transform live R cells into S cells.
• They discovered that DNA alone from S bacteria caused R bacteria to become transformed.
• They also discovered that protein-digesting enzymes such as proteases and RNA-digesting enzymes such as RNases did not affect the transformation.
• Hence, it was proved that the transforming substance was not protein and RNA.
• Digestion with DNase did inhibit transformation and the DNA caused the transformation.
• So, they concluded that DNA is the hereditary material. (2 Marks)

OR

Lac operon was proposed by Jacob and Monad in 1961. It contains following components such as:

1. Structural gene: There are three types of structural genes that codes for different enzymes and facilitates the process of transcription in the presence of inducer (lactose).
   • The z gene codes for enzyme beta-galactosidase that regulates the switching on and is responsible for the hydrolysis of disaccharide, lactose into its monomeric unit’s glucose.
   • Y gene codes for enzyme permease that increases the permeability of the cell to beta-galactosides.
   • a gene codes for enzyme transacetylase.
2. Promoter: It is the sequence of DNA at which the RNA polymerase enzyme get binds and initiates the process of transcription.
3. Operator: It is sequence of DNA that is adjacent to promoter.
4. Regulator gene: A gene that codes for repressor protein and binds with the operator and because of it operon is switched “off’.
5. Inducer: Lactose is inducer that helps in switching “on” of operon.

Lactose acts as the substrate for enzyme beta-galactosidase. This enzyme regulates the switching on and off the operon because of this it is termed as inducer. So, in the absence of glucose (carbon source), if lactose is added in the growth medium of the bacteria. The lactose is transported into the cells by the action of permease enzyme that increases permeability of the cell to beta-galactosides.

Lactose induces the operon in following manner:

• In a lac operon, the repressor protein is synthesised from the i gene.
• This repressor protein gets bind with the operator region of the operon and prevents RNA polymerase enzyme from transcribing the operon.
• In the absence of lactose, the repressor gene produces repressor protein and get binds with the operator gene. It prevents the RNA polymerase enzyme to get binds with the operon.

Diagrammatic Representation of Lac operon in the absence of lactose:

In the presence of lactose as an inducer, the repressor protein is inactivated. It allows RNA polymerase enzyme to activate the promoter and initiates the process of transcription by structural genes.

Diagrammatic Representation of Lac operon in the presence of inducer:

Question 30.
With advancement in genetics, molecular biology and tissue culture, new traits have been incorporated into crop plants.
Explain the main steps in breeding a new genetic variety of crop.
OR
(a) State the objective of animal breeding.
(b) List the importance and limitation of inbreeding. How can the limitations can be overcome.
(c) Give an example of new breed of cattle and poultry.
Answer:
Plant breeding refers to a technique that involves the manipulation of plant species in order to produce desired genotypes and phenotypes. The main steps that involves in plant breeding for the production of genetic variety of crop are as follows:

(i) Collection of variability: The genetic variability is the root of any crop breeding programme. In many crops, the pre-existing genetic variability is available from the wild relatives of the crop. It involves the collection and preservation of all the different wild varieties, species and relatives of the cultivated species. The entire collection of plants or seeds that contains all the diverse alleles for all genes in a given crop is called germplasm collection.

(ii) Evaluation and selection of parents: The germplasm is evaluated to identify the plants with desirable
combination of characters. The selected plants are multiplied and used in the process of hybri’disation. Purelines arc created wherever desirable and possible.

(iii) Cross bybridisation among the selected parents: The desired characters are formed by the combination of two different plants. For example high protein quality of one parent may need to be combined with disease resistance from another parent. It can be achieved by cross hybridising the two parents to produce hybrids that genetically combine the desired characters in one plant.

(iv) Selection and testing of superior recombinants: This step involves the selection of the progeny of the
hybrids. The plants that have the desired character combination. The selection process involves the
success of the breeding objective and requires careful scientific evaluation of the progeny. This step yields plants that are superior to both of the parents.

(v) Testing, release and commercialisation of new cultivars: The newly selected lines are evaluated for their yield and other agronomie traits of quality, disease resistance and so on. This evaluation is done by growing these in the research field and recording their performance under ideal fertiliser application irrigation and other crop management practices. The evaluation in research fields is followed by testing the materials in farmer’s field for atleast three growing seasons at several locations in the country. Then the material is evaluated in comparison to the best available local crop cultivar to check or reference cultivar.

OR

(a) Animal breeding aims to increase the yield of animals and also improving the desirable qualities of the animals.
(b) Inbreeding refers to the mating of more closely related individuals within the same breed for 4-6 generations.

Importance of Inbreeding:

• Inbreeding increases homozygosity.
• It helps in accumulation of superior genes
• It helps in the elimination of desirable genes.

Limitations of inbreeding:

• inbreeding leads to inbreeding depression.
• It reduces fertility and productivity.

The limitations of inbreeding can be overcome by techniques such as outbreeding, outerossing, cross breeding and interspecific hybridisation. All these techniques involve the breeding of selected animals with unrelated superior animals of the same breed.

(c) Example of new breed of cattle is Hisardale and Example of new breed of poultry is Hampshire.