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18151.

Question : Describe RNA editing in plants.

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Solution :(i) Chemical modification such as base modification, NUCLEOTIDE insertion or DELETIONS and nucleotide replacements of mRNA results in the alteration of amino acid sequence of protein that is specified is called RNA EDITING. This results in the change in the protein ~oding sequence of RNA following transcription. The coding properties of the RNA transcript is changed. The genetic information encoded in the chloroplast genome is altered by post transcriptional phenomenon which is site - specific (C`rarr` U) in chloroplast of higher plants - RNA editing OCCURS in plant mitochondria and chloroplast.
(ii) In plant cells RNA editing by pyrimidine transitions occurs in mitochondria and plastids (chloroplast). There are two main TYPES of RNA editing.
1. Substitution editing -Alteration of individual nucleotide bases. Mitochondria and chloroplast RNA in plants.
2. Insertion / Deletion editing - Nucleotides are added or deleted from the total number of bases.
Significancc of RNA editing
(a) In higher plant chloroplast, it helps to restore the codons for conserved aminoacids which include initiation and termination codon.
(b) It regulates organellar gene expression in plants.
(c) RNA editing results in the restoration of codons for phylogenetically conserved amino acid residues.
Discussion
18152.

Question : Describe Rivet Popper Hypothesis.

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Solution :Rivet Popper Hypothesis : Rich biodiversity is not only important for ecosystem health but is also essential for survival of humans on the earth.
Since all species on earth are interlinked through various types of RELATIONSHIPS and a loss of few species can have a destabilising effect.
Paul Ehrlich (a Stanford ecologist) explain the importance of diversity by an analogy of .rivet popper hypothesis..
According to this hypothesis, ecosystem is like an aeroplane, all its parts are joined together using thousands of rivets (species).
If every passenger, travelling in it, starts popping a rivet to take home (causing a species to become extinct), it may not affect flight safety (proper functioning of the ecosystem), but as more and more rivets are removed the plane becomes dangerously weak over a period of TIME.
Moreover, which rivet is removed is also crucial eg. loss of rivet on the wings (key species that drive major ecosystem FUNCTIONS) is a more serious threat than loss of rivets on SEATS or WINDOWS inside the plane.
Discussion
18153.

Question : Describe Replication of DNA and Give experimental proof for it.

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Solution :In addition to the double helical structure of DNA, WATSON and CRICK also proposed a scheme of DNA replication.
To quote their original statement that is as follows: (i) It has not ESCAPED our notice that the specific pairing copying mechanism for the genetic material (Watson and Crick, 1953). (ii) The scheme suggested that the two strands would separately and act as template for the synthesis of new complementary strands. (iii) After the completion of replication, each DNA molecule would have one parental and one NEWLY SYNTHESISE strand. (iv) This scheme was termed as semiconservative DNA replication.
Discussion
18154.

Question : Describe process of fertilization.

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SOLUTION :Fertilization : The process of two similar or dissimilar gametes is called as syngamy as a result diploid zygote is formed.
There are two types of fertilization.
EXTERNAL fertilization : In this type of fertilization, syngamy occurs outside the body of organism. Example - in most aquatic organisms LIKE algae, FISH, amphibians.
The major disadvantage of external fertilization is they PRODUCE a large number of offsprings but the offsprings are vulnerable to predators.
Internal fertilization : ..Syngamy occurs inside the body of organism. Example - fungi, birds, mammals.
Male gametes are released close to the female gametes where they fuses to form zygote.
Discussion
18155.

Question : Describe Population Age Distributation.

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Solution :The PROPORTION of the AGE groups (pre-reproductive, reproductive and post reproductive) in a population is its age distribution attribute. This DETERMINES the reproductive status of the population at the given time and is an indicator of the future population size. Usually a rapidly growing population will have larger proportion of young individuals. A stable population will have an even distribution of VARIOUS age classes. A declining population tends to have a larger proportion of older individuals.
Discussion
18156.

Question : Describe pollen-pistil interaction

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Solution :In nature pollination does not guarantee the transfer of the compatible or right type of pollen. Often pollen of the wrong type, either from other species or from the same plant (it is self incompatible) also land on the stigma.
The pistil has the ability to recognise the pollen, whether it is of the right type or of the wrong type. If pollen is not of the right type the pistil accepts the pollen and promotes post-pollination events that leads to fertilisation. If the pollen is of the wrong type of pistil rejects the pollen by preventing pollen germination.
The ability of the pistil to recognise the pollen followed by its ACCEPTANCE or rejection is the result of a continuous dialogues between pollen grain and the pistil. This dialogue is mediated by chemical components and protein elements and releasing various water decomposed enzymes present in it. it is only in recent years that botanists have been able to identify someof the pollen and pistil components and the inter actions leading to the recognition, followed by acceptance or rejection. Compatible pollination, the pollen grain germinates on the stigma to produce a pollen tube through one of the germ pores. The contents of the pollen grain move into the pollen tube. Pollen tube grows through the tissues of the stigma and style and reaches the ovary

In some plants pollen grains are sheded at two celled condition (a vegetative cell and a generative cells). In such plants, the generative cell divides and forms the two male gametes during the growth of pollen tube in the stigma. In plants which sheded pollen in three celled condition pollen tubes carry the two male gametes from the beginning. Pollen tube, after reaching the ovary, enters the ovule through the micropyle and then enters one of the synergids through the filiform apparatus. Filiform apparatus present at the MICROPYLAR part of the synergids guides the entry of the pollen tube.
All these events from pollen deposition on the stigma until pollen tubes enter the ovule are together referred to as pollen -pistil interaction. The knowledge GAINED in this area woudl help the plant breeder in manipulating pollen pistil interaction even in incompatible pollinations, to get desired hybrids.
Experiment: You can easily study pollen germination by dusting some pollen from flowers such as peas, chickpea, crotalaria, balsam and vinca on a glass slide containing a drop of SUGAR solution (about 10 percent). After about 15-30 minutes, observe the slide under the LOW power lens of the microscope. You are likely to see pollen tubes coming out of the pollen grains.
Discussion
18157.

Question : Describe polygenic Inheritance.

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Solution : Mendel.s studies mainly described those traits that have distinct alternate forms such as flower colour which are EITHER purple or white.
= But if you look around you will find that there are many traits which are not so distinct in their occurrence and are spread across a gradient.
= For EXAMPLE in humans we don.t just have tall or short people as two distinct alternatives but a whole range of possible heights such traits are generally controlled by three or more genes and are thus called as polygenic traits.
= Besides the involvement of multiple genes polygenic inheritance also takes into account the influence of environment.
= Human skin colour is another classic example for this.
= In a polygenic trait the phenotype reflects the contribution of each allele i.e. the effect of each allele is additive.
= To understand this better let US assume that three genes A, B, C control skin colour in human with the dominant forms A, B and C responsible for dark skin colour and the recessive forms a, b, and c for light skin colour.
=The genotype with all the dominant alleles [ AA BB CC] will have the darkest skin colour and that with all the dominant alleles (AA, BB, CC) will have the lightest skin colour.
=As expected the genotype with three dominant alleles and three recessive alleles will have an INTERMEDIATE skin colour.
=In this manner the number of each type of allelaes in the genotype would determine the darkness or lightness of the skin in an INDIVIDUAL.
Discussion
18158.

Question : Describe pedigree analysis and draw its symbols.

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Solution :The idea that disorders are inherited has been prevailing in the human society since long.
= This was based on the heritability of certain characteristic features in families.
=After the rediscovery of Mendel.s work the practice of analysing inheritance pattern of traits in human beings began.
= Since it is EVIDENT that control crosses that can be performed in pea PLANT or some other ORGANISMS are not possible in case of human being study of the family HISTORY about inheritance of a particular trait provides an alternative.
= Such an analysis of traits in a several of generation of family is called pedigree analysis.
=In the pedigree analysis the inheritance of a particular trait is represented in the family tree over generations.
=In human genetics, pedigree study provides a strong tool, which is utilised to trace the inheritance of specific trait, abnormality or disease.
= Some of the important standard SYMBOLS used in the pedigree analysis have been shown in figure.
=Inheritance pattern of traits in human beings cannot be studied by crosses as in case of other organisms due to following reasons : (i) The progeny produced in very small (usually one) and therefore, take long time. (ii) Controlled crosses cannot be performed.
Discussion
18159.

Question : Describe one gene inheritance.

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Solution :The study of inheritance of a single pair of alleles or factors of a trait at a time (monohybrid cross) is called one gene inheritance.
= Based on his observations on monohybrid crosses, Mendel proposed two general rules in order to consolidate his understanding of inheritance in monohybrid crosses.
= These rules are called the principles or laws of inheritance.
= They are :
=Law of Dominance (First Law) : (i) Characters are controlled by discrete units called FACTOR. (ii) Factors occur in pairs. (iii) In a DISSIMILAR pair of factors, one member of the pair is dominant while the other is recessive.
= This law is used to explain the expression of only one of the parental characters in a monohybrid cross in the `F_1` generation and the expression of both in the `F_2` generation. It ALSO explains the proportion of 3 : 1 ratio obtained in the `F_2` generation.
=Law of Segregation (Second Law) :This law STATES that though the parents contain two alleles during gamete formation the factors or alleles of a pair SEGREGATE from each other, such that gamete receives only one of the two factors.
= Hence, the alleles do not show any blending and both the characters are recovered as such in the `F_2` generation, though one of these is not seen in the `F_1` generation
Discussion
18160.

Question : Describe organs/parts of a typical flower

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Solution :A TYPICAL flower possesses organs of four GROUPS. In which two groups of outer side are sterile and two groups of inside are fertile organs.
Sterile organs: In sterile organs, sepals unite and make calyx.
Fertile organs: In two types of fertile organs stamens (microspores) unite and form Androecium and carpels (megaspores) unite and form cynoecium.
Stamen: Stamen is a thin organ and has three clear parts LIKE proximal sterile.It is CALLED filament and terminal fertile parts is called another. The part CONNECTING anther and filament is called connective

Pistil/Carpel: The proximal part of end/ of pistil, consisting ovules is called ovary and terminal end accepting pollen grains is called stigma and the sterile part between ovary and stigma is called style.
Discussion
18161.

Question : Describe one example of adaptive radiation.

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Solution : The ground finches of Galapagos Islands show the phenomenon of adaptive radiation. In general, they POSSESS STOUT, conical BEAK adapted for crushing seeds but by undergoing great diversification in their feeding habits, they show variations in the shape and SIZE of their beak.
Discussion
18162.

Question : Describe mutualism in detail.

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Solution :• Interaction conferring benefits on both the interacting species.
• Lichens-mutualistic relationship between fungus and photosynthesising algae or cyanobacteria.
• MYCORRHIZAE- mutualistic relationship between fungus and roots of higher plants. Fungi absorbs essential nutrients from soil and in return plant provides energy yielding carbohydrates to fungi.
• Most spectacular and evolutionary fascinating example is plant-animal interaction.
• Plant needs animals for POLLINATION, dispersal of SEEDS.
• Animals are rewarded for their services in the form of pollens, nectars, juices, fruits and seeds.
• In many species of fig trees, there is a tight one-to-one relationship with pollinator species of wasp.
• Fig species is pollinated by its partner wasp and no other species.
• Female wasp uses fruit not only as an OVIPOSITION (egg-laying) site but uses the developing seeds within fruit for nourshing its larvae.
• The wasp pollinates the fig inflorescence while searching for suitable egg-laying SITES.
Discussion
18163.

Question : Explain Stanley - Miller’s spark discharge apparatus experiment on the origin of life with a neat labelled diagram.

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Solution :In 1953, S.L. Miller, an American scientist created similar conditions in a laboratory scale relevance to early earth.s ATMOSPHERE. He created electric discharge in a closed flask containing `CH_(4), NH_(3), H_(2)` and water vapour at `800^(@)C`.
The flask was completely evacuated, boiling water led to the formation of water vapour. Water vapour was passed across spark discharge chamber CONNECTED to electrodes and filled with MIXTURE of gases. STEAM carrying the mixture of gases were allowed to condense and water containing organic compounds were allowed to accumulate in trap. Under such simulated conditions similar to early earth.s atmosphere, Miller observed the formation of amino acids.
Discussion
18164.

Question : Describe Mendel.s Monohybrid experiment.

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Solution : If we use alphabetical symbols for each gene then the capital letter is used for the trait expressed at the `F_1` stage and the small alphabet for the other trait.
= For EXAMPLE, in case of the character of height "T" is used for the Tall trait and .t. for the .dwarf. and .T. and .t. are alleles of each other.
= Hence in plants the pair of alleles for height would be TT, Tt or tt.
= Mendel also proposed that in a true breeding tall or dwarf pea variety the allelic pair of genes for height are identical or homozygous, TT and tt respectively.
= TT and tt are called genotype of the plant while the descriptive terms tall and dwarf are the phenotyped.

= As Mendel found the PHENOTYPE of the `F_1` heterozygote Tt to be exactly like TT parent in appearance he proposed that in a pair of dissimilar factors, one Parental dominates the other (as in the `F_1`) and hence is called the dominant factor while the other characters/trait PAIRS that he studied.
=It is convenient and logical to use the capital and lower case of an alphabetical symbol to remember this concept of dominance and fi generation
=Alleles can be similar as in the case of homozygotes TT and tt or can be dissimilar as in the case of heterozygote Tt।
=Since the Tt plant is heterozygous for genes controlling one character (height) it is a monohybrid and the cross between TT and tt is a monohybrid cross.
=From the observation that the recessive parental trait is expressed without any blending in the `F_2`
=generation, we can infer that when the tall and dwarf plant produce gametes by the PROCESS of meiosis the alleles of the parental pair separate or segregate from each other and only one allele is transmitted to gametes.
=This segregation of alleles is a random process and so there is a 50 percent chance of a gamete containing either allele as has been verified by the results of the crossings.
=In this way the gametes of the tall TT plants have the alleles .T. and the gametes of the dwarf tt plants have the allele t.
=During fertilisation the two alleles .T. from one parent say, through the pollen and .t. from the other parent then through the egg are united to produce zygote that have one .T. allele and one l. allele.
=In other words the HYBRIDS have Tt.
=Since these hybrids contain alleles which express contrasting traits the plants are heterozygous.
=The production of gametes by the parents, the formation of zygotes, the `F_1 and F_2` plants can be understood from diagram called punnett square.
Discussion
18165.

Question : Describe Mendelian Disorder.

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Solution :Mendelian disorders are mainly determined by alternation or mutation in the single gene.
=These disorders are transmitted to the offspring on the same lines as the principle of inheritance.
= The pattern of inheritance of Mendelian disorders can be traced in a family by the pedigree analysis.

= Most common and prevalent Mendelian disorders can be traced in a family by the pedigree analysis.
= Most common and prevalent Mendelian disorder are Haemophilia, Cystic fibroses, Sickle cell anaemia, Colour blindness, PHENYLKETONURIA, Thalassemia etc.
=Such Mendelian disorder may be dominant or recessive.
=By pedigree analysis one can easily understand whether the trait in question is dominant or recessive.
=Similarly the trait may also be linked to the sex chromosome as in case of haemophilia.
=It is evident that this X-linked recessive trait shows transmission from carrier female to male progeny.
= A representative pedigree is shown in figure for dominant and recessive traits.
= Sub Que : Write short note on Colour Blindness.
= Colour blindness is sex linked recessive disorder due to defect in either red or green cone of eye resulting in failure to discriminate between red and green colour.
=This defect is due to mutation in certain genes present in the x-chromosome. It occurs in about 8% of males and only about 0.4% of females. This is because the genes that leads to red-green colour blindness are on the x-chromosome. Males have only one x-chromosome and females have two. The son of a woman who carries the gene has a 50% chance of being colour blind.
=The mother is not herself colourblind because the gene is recessive.
= That MEANS that its effect is suppressed by her matching dominant normal gene. A daughter will not NORMALLY be colour blind, unless her mother is a carrier and her father is colourblind.
=Sub Que : Write short note on Haemophilia.
=Haemophilia is sex linked recessive disease, which show its transmission from unaffected carrier female to some of the male progeny has been widely studied.
=In this disease, a single protein that is a part of the cascade of proteins involved in the clotting of blood is affected.
=Due to this, in an affected individual a simple cut will result in non stop bleeding.
= The heterozygous female (carrier) for haemophilia may transmit the disease to sons.
= The possibility of a female becoming a haemophilic is extremely rare because mother of such a female has to be atleast carrier and the father should be haemophilic (Unviable in the later stage of life).
= The family pedigree of Queen Victoria shows a number of haemophilic descendents as she was a carrier of the disease.
= Sub Que : Describe cystic fibroses.
= A recessive mutant allele on an autosome (chromosome 7) causes this disease.
=A unique glycoprotein is produced by the gene which leads to the formation of mucus of abnormally high viscosity.
= This mucus interferes with the functioning of many exocrine glands like sweatgland, liver, pancreas and lungs.
=There is accumulation of mucus in long tubules which makes it susceptible to infection leading to bronchitis.
=Sub Que : Write short note on sickle-cell anaemia.
= Sickle-cell anaemia is a autosomal linked recessive trait that can be transmitted from parents to the OFFSPRINGS when both the parents are carrier for the gene (or heterozygous). The disease is controlled by a single pair of allele, `Hb^A and Hb^S`.
=Out of the three possible genotypes only homozygous individuals for `Hb^S [Hb^S Hb^S]` show the diseased phenotype.
=Heterozygous `[Hb^A Hb^S]` individuals appear apparently unaffected but they are carrier of the disease as there is 50 percent probability of transmission of the mutant gene to the progeny thus exhibiting sickle-cell trait.
=The defect is caused by the substitution of Glutamic acid (Glu) by Valine (Val) at the sixth position of beta globin chain of the haemoglobin molecule. The substitution of amino acid in the globin protein results due to the single base substitution at the sixth codon of the beta globin gene from GAG to GUG.
=The mutant haemoglobin molecule undergoes polymerisation under low oxygen tension causing the change in the shape of the RBC from biconcave disc to elongated sickle like structure.
=
=Micrograph of the red blood cells and the amino acid composition of the relevant portion of `beta` -chain of haemoglobin: (a) From a normal individual (b) From an individual with sickle-cell anaemia
=Sub Que : Write short note on Phenylketonuria.
=Phenylketonuria is inborn error of metabolism is also INHERITED as the autosomal recessive trait.
The affected individual lacks an enzyme that converts the amino acid phenylalanine into tyrosine.
=As a result of this phenylalanine is accumulated and converted into phenylpyruvic acid and other derivatives.
= Accumulation of these in brain result in mental retardation. These are also excreted through urine because of its poor absorption by kidney.
= Sub Que : Write short note on Thalassemia.
=Thalassemia is autosome linked recessive blood disease transmitted from parents to the offspring when both the partness are unaffected carrier for the gene (or heterozygous).
=The defect could be due to either mutation or deletion which ultimately results in reduced rate of synthesis of one of the globin chains (a and b chains) that makes up haemoglobin.
=This causes the formation of abnormal haemoglobin molecules resulting into anaemia which is characteristic of the disease.
=Thalassemia can be classified according to which is characteristic of the disease.
=Thalassemia can be classified according to which chain of the haemoglobin molecule is affected.
=In a Thalassemia, production of a globin chain is affected while in `beta` Thalassemia production of `beta ` globin chain is affected.
=`alpha` Thalassemia is controlled by two closely linked genes HBA 1 and HBA 2 on chromosome 16 of each parent and it is observed due to mutation or deletion of one or more of the four genes.
=The more genes affected, the less alpha globin molecules produced.
=While `beta` Thalassemia is controlled by a single gene HBB on chromosome 11 of each parent and occurs due to mutation of one or both the genes.
=Thalasemia differes from sickle-cell anaemia in that the former is a quantitative problem of synthesizing too few globin molcules while the latter is a qualitative problem of synthesizing an incorrectly functioning globin.
Discussion
18166.

Question : Describe location and external structure of ovaries.

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Solution :Ovaries are LOCATED one on each side of the lower abdomen. Ovaries maintain their condition/ position by successive tendons.
Ovaries are in a pair and almond like glands in size and shape. It is 2-4 cm long. 2 cm wide and 1 cm in height.
Each ovary has Hilus. It is an entrance of blood vessels and nerves.
Each ovary is COVERED by a thin epithelium which enclose the OVARIAN stroma. The stroma is divided into two zones - a peripheral CORTEX and an inner medulla.
Discussion
18167.

Question : Describe linkage and recombination in detail.

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Solution :Morgan and his group observed in Drosophila that when the two genes in a dihybrid cross were situated on the same chromosome, the proportion of parental gene combination were much higher than the non parental type.
= They attributed this due to physical association of the two genes and coined the term "linkage. to describe this physical association of genes on a chromosome and the term recombination to describe the generation of non parental gene combinations.
= Thus linkage is a phenomenon of genic inheritance in which gene of a particular chromosome show their tendency to inherit together.
= Morgan and his group also found that even when gene were GROUPED on the same chromosome, some genes were tightly linked i.e. linkage is stronger between two genes if the FREQUENCY of recombination is low.
=Whereas the frequency of recombination is higher if genes are loosely linked i.e. linkage is weak between two genes.
=Recombination of linked genes is by crossing over (exchange of corresponding parts between the chromatids of homologous chromosomes).
= All the genes linked together in a SINGLE chromosome constitute a linkage group.
= - The number of linkage group in an organism is equal to there haploid number of chromosomes.
=This hypothesis was proved by TH Morgan by his experiments on Drosophila.
= Morgan and his group hybridised yellow bodies and white eyed females with brown bodies and red eyed males (wild type) and intercrossed their Fi-progeny (cross-A).
=It was observed that the two genes did not segregate independently of each other and the `F_2` ratios deviated significantly from 9: 3:3:1 ratio.
=In `F_2` generation parental combinations were 98.7 % and recombinant were 1.3%.
= In another cross (cross-B), between white bodies fly with miniature wing and a male fly with yellow body and normal wing, parental combinations were 62.8% and recombinants were 37.2% in` F_2` generation.
=Thus it was proved from the CROSSES that the linkage between genes for yellow body and white eyes is stronger than the linkage between the white body and miniature wing.
= His student Alfred Sturtevant used the frequency of recombination between gene pairs on the same chromosome as a measure of the distance between genes and .mapped their position on the chromosome.
= Today genetic maps are extensively used as a starting point in the sequencing of whole GENOME as was done in the case of Human Genome Sequencing Project.
Discussion
18168.

Question : Describe Lac operon concept with labelled sketch.

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Solution :It is an inducible operon, where lactase is the inducer, it is the substrate for the enzyme B- galactasidase the Lac - operon has three stractural gene
Gene Z codes for B- galactosidase enzyme which brakes lactose into galactose and glucose
Gene Y codes for permease which increase the permeability of the cell to lactose
Gene A codes for transacetylase, that catalyses the transacetylation of lactose in active form
When lactose is absent, the gene regulates and produces repressor MRNA which produces the repressor protein. This protein binds to the operator region of the operòn and as a RESULT PREVENTS RNA polymerase to bind to the operon, the operon is SWITCHED off.
When lactose is PRESENT, it acts as an inducer which binds to the repressor and form an inactive repressor, it fails to bind to operator gene, the RNA polymerase binds to the polycistronie structural gene, it produces all the enzymes B- galactosidase, permease and transacetylase, the lac operon is switched on.
Discussion
18169.

Question : Describe International Initiatives for Conservation of Biodiversity.

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Solution :BIODIVERSITY conservation is a collective responsibility of all the nation.
Convention on BIOLOGICAL DIVERSITY (The Earth Summit) held in Rio de Janerio in 1992 called upon all the nations to take INITIATIVE in biodiversity conservation.
In a follow up World Summit on Sustainable Development was held in 2002 in Johannesburg, South Africa. In this 190 countries MADE commitments to significantly reduce the current rate of biodiversity loss at global, regional and local levels by 2010.
In 2012, the United Nations conference on Sustainable Development was again held at Rio and is called Rio + 20 or Rio Earth Summit 2012.
Discussion
18170.

Question : Describe J - shape and S - shape growth curves.

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Solution :Populations show characteristic growth patterns or forms. These patterns can be plotted and termed as J - SHAPED growth form and S - shaped growth form (Sigmoid form). Their growth is represented by S shaped growth cure.J shaped growth form : When a population increases rapidly in an exponential fashion and then stops abruptly due to environmental resistance or due to sudden appearance of a LIMITING factor, they are said exhibit J - shaped growth form. Many insects show explosive increase in number during the rainy season FOLLOWED by their disappearance at the end of the season. S- Shaped growth form (sigmoid growth) Some populations, as in a population of small marmmals, increase slowly at first then more rapidly and gradually slow down as environmental resistance increases whereby equilibrium is reached and MAINTAINED. Their growth is represented by S shaped growth curve.
Discussion
18171.

Question : Describe internal structure of microsporangium

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Solution :A typical microsporangium appears near circular in outline
It is generally surrounded by four wall layers. They are epidermis, endothecium, middle layers and the tapetum
The outer three wall layers perform the function of PROTECTION and help in dehiscence of anther to release the pollen.
The innermost wall layer is the tapetum. It nourishes the developing pollen GRAINS. Cells of the tapetum possess dense cytoplasm and generally have more than one NUCLEUS
Tapetum secretes enzyme and hormone both and special protein. It decides correspondence of pollen GRAIN
The cells of epidermis are stretched or SCATTERED or flat

(a) Transverse section of a young anther (b) Enlarged view of one microsporangium showing wall layers (c ) A mature dehisced anther.
Endothecium is fibroius layer
Discussion
18172.

Question : Describe incomplete dominance.

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Solution : When experiments on peas were repeated using other traits in other plants it was found that sometimes the `F_1` had a phenotype that did not resemble either of the two parents and was in between the two.
= The inheritance of flower colour in the dog flower (Snapdragon or Antirrhinum SP.) is a GOOD EXAMPLE to understand incomplete dominance.
=In a cross between true - breeding red flowered (RR) and true - breeding white flowered plants (rr) the `F_1` [Rr] was pink.
= When the `F,_1` was self POLLINATED the `F_2` resulted in the following ratio 1 (RR) Red : 2[Rr] Pink : 1 [rr] White.
= Here the genotype ratios were exactly as we would expect in any mendelian monohybrid cross, but the phenotype ratio had changed from the 3:1 dominant: recessive ratio.
= .R. was not completely dominant over r and this made it POSSIBLE to distinguish Rr as pink from RR [red] and rr (white).
Discussion
18173.

Question : Differentiate incomplete dominance and codominance.

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SOLUTION :The German Botainst Carl Correns.s (1905) , Experiment - In 4 O. clock plant, Mirabilis Jalapa when the pure breeding homozygous red `(R^1R^1)`parent is crossed with homozygous white `(R^2R^2)` , the phenotype of the `F_1` hybrid is heterozygous pink `(R^1R^2)` . The `F_1` heterozygous phenotype differs from both the parental homozygous phenotype. This cross did not exhibit the character of the dominant parent but an intermediate of the dominant parent but an intermediate colour pink. When one allele is not completely dominant to another . allel it shows incomplete dominance. Such ALLELIC interaction is known as incomplete dominance . `F_1` generation produces intermediate phenotype pink coloured flower.

When pink coloured plant of `F_1` generation were interbred in `F_2` both phenotypic and genotypic ratios were found to be identical as `1:2:1` (1 red : 2 pink : 1 white) . Genotypic ratio is `1R^1R^1 : 2 R^1R^2: 1R^2R^2`From this we conclude that the alleles themselves remain discrete and unltered proving the Mendel .s Law of Segregation . The phenotypic and genotypeic ratios are the same . There is not blending of genes . In the `F_2` generation `R^1 and R^2` genes segregate and recombine to produce red, pink and white in the ratio of 1 : 2 : 1. `R^1` allele codes for an enzyme responsible for the formation of red pigment . `R^2` allele codes for deceive enzyme . `R^1 and R^2` genotypes produce only enough to make the flower pink. Two `R^1R^1` are needed for producing red flowers. Two `R^2R^2` genes are needed for white flowers. If blending had taken place , the ORIGINAL pure traits would not have appeared and all `F_2` plants would have pink flowers . It is very clear that Mendel.s particulate INHERITANCE takes place in this cross which is confirmed by the reappearance of original phenotype in `F_2`.
Discussion
18174.

Question : Describe inheritance of two gene.

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Solution :Mendel also worked with and crossed pea PLANTS that differed in two characters as is seen in the cross between a pea plant that has seeds with yellow colour and round shape and one that had seeds of green colour and wrinkled shape.
= Mendel found that the seeds resulting from the crossing of the parents had yellow coloured and round shaped seeds.
= Thus yellow colour was dominant over green and round shape dominant over wrinkled.
= These results were IDENTICAL to those that he got when he made separate MONOHYBRID crosses between yellow and green seeded plants and between round and wrinkled seeded plants.
= Let us use the genotypic symbols .Y. for dominant yellow seed colour and .y. for recessive green seed colour, R for round shaped seeds and r for wrinkled seed shape.
= The genotype of the parents can then be written as RRYY and rryy.
=The cross between the two plants can be written down as in the given figure SHOWING the genotypes of the parent plants.
= The gametes RY and ry unite on fertilisation to produce the `F_1`, hybrid RrYy.
=When Mendel self hybridised the `F_1` plants he found that `3//4^(th)` of `F_2` plants had yellow seeds and 1/`4^(th)` had green.
= The yellow and green colour segregated in a 3:1 ratio.
= Round and wrinkled seed shape also segregated in a 3:1 ratio, just like in a monohybrid cross.
Discussion
18175.

Question : Describe in short about RNA.

Answer»

Solution :Ribonucleic Acid (RNA) was the first genetic material to be discovered.
It is evident through VARIOUS scientific researches that essential life PROCESSES such as metabolism, translation and splicing etc. evolved around RNA.
RNA used to act as a genetic material as WELL as a catalyst.
But RNA being a catalyst was reactive and hence unstable.
Therefore DNA has evolved from RNA with chemical modifications that make it more stable.
DNA being double stranded and having complementary strand further resists changes by evolving a process of repair.
The nitrogenous base in RNA are of two types : (i) Purines (adenine and guanine) (ii) Pyrimidines (Cytosine and Uracil) Ribose sugar `+` Nitrogenous base `rarr` Ribonucleoside
Ribonucleoside `+` phosphate group `rarr` Ribonucleotide.
Discussion
18176.

Question : Describe in detail about asexual reproduction in animals.

Answer»

SOLUTION :Asexual reproduction is a type of reproduction seen in single celled organism.
In these organisms, cell DIVISION is itself a mode of reproduction.
These organisms REPRODUCE by binary fission.
Binary fission : In this type of cell division cell DIVIDES into two halves and each rapidly grows into an adult.
E.g. Amoeba, Paramoecium.
Clones : Binary fission INVOLVES mitosis only and consequently the resultant offsprings are genetically identical to the parent and to each other.
Budding : In yeast, the division is unequal and small buds are produced which remain attached initially to the parent cell which, eventually gets separated and mature into new yeast organisms or cells.
Discussion
18177.

Question : Describe HSK pathway of photosynthesis.

Answer»

Solution :(1) In tropical plants like maize, sugar cane,Jowar, Amaranthus, Atriplex, etc, there exists an alternative pathway of photosynthesis. This pathway is called HSK pathway of photosynthesis named after its discoverers, viz.,Hatch, Slack and Kortschak.
(2) HSK pathway occurs in low concentration of `CO_(2)` as low as 10 ppm (parts per million) in the atmosphere.
(3) In HSK pathway, the initial product is a 4-carbon compound, i.e., oxaloacetic acid(oxalocetate). Therefore, this pathway of photosynthesis is also known as `C_(4)` pathway.
(4) The various reactions that occur during HSK pathway are completed in two parts at two different sites which are as follows :
(i) Reactions in MESOPHYLL chloroplasts :
(1) Initially, the atmospheric carbon dioxidewhich enters the mesophyll chloroplasts is accepted by PEPA (phosphoenol pyruvic acid) a 3-carbon compound to form a 4-C compound called oxaloacetic acid in the presence of water and PEP carboxylase enzyme as shown below :
`underset((C_(3)))(PEPA)+CO_(2)+H_(2)O overset("PEP carboxylase")to underset((C_(4)))(OA A+H_(3)PO_(4))`
(2) Oxaloacetic acid is reduced to malic acid (MALATE) by the enzyme malate dehydrogenase in the presence of `NADPH_(2)` or changed aspartic acid by amination in the presence of `NADPH_(2)` and the enzyme transminase as shown below :
`OA A+NADPH_(2) overset("Malate dehydrogenase")tounderset(("Malate"))("Malic acid")+NADP`
`OA A+NADPH_(2)+NH_(3) overset("Transaminase")to underset(("Malate"))("Aspartic acid")+NADP+H_(2)O`
(ii) Reactions in bundle sheath chloroplasts :
(1) Malic or aspartic acid is then transported to the bundle sheath chloroplasts where it undergoes decarboxylation to form pyruvic acid (pyruvate) with the release of `CO_(2)` as shown below :
`underset((4C))("Malic acid") +NADP to underset(("Pyruvate"))("Pyruvic acid") +underset((3C))(CO_(2)+)NADPH_(2)`
(2) `CO_(2)` released in the above reaction is accepted by RuBP (second carbon acceptor). It enters the Calvin cycle and eventually forms glucose which is stored as starch in the bundle sheath chloroplasts.
(3) Pyruvic acid which is formed due to decarboxylation of malic acid is transported to mesophyll cells where it undergoes phosphorylation to regenerate PEPA as shown below :
`underset((3C))("Pyruvic")+ATP overset("Phosphorylation") to underset((3C)+iP)( PEPA +ADP )`
(4) From the above account it is abivous that in `C_(4)` plants, `CO_(4)` is FIXED twice in two different cells during the daytime.

`CO_(2)` fixation in `C_(4) ` plants can occur at very low concentrantions of `CO_(2)` i.e., as low as 1-2 parts per million. `C_(4)` plants can carry on photosynthesis enven if the light intensity is high, TEMPERATURE is high and the amount of available water is less.
Discussion
18178.

Question : Describe how deforestation might contribute to global warming.

Answer»

Solution : Deforestation refers to CUTTING down of trees. Trees are one of the major sources of `CO_2`, uptake. `CO_2`, is a major green house GAS. If there is a LARGE scale destruction of FORESTS, the level of `CO_2`, will be INCREASED leading to global warming.
Discussion
18179.

Question : Describe how oxygen and chemical composition of detritus control decomposition do.

Answer»

Solution :Decomposition is largely an oxygen-requiring PROCESS. The rate of decomposition is controlled by chemical composition of detritus and climatic factors. In a particular climatic condition, decomposition rate is slower if detritusis RICH in lignin and chitin, and quicker, if detritus is rich in NITROGEN and water-soluble substances like sugars. Temperature and soil moisture are the most IMPORTANT climaticfactors that regulate decomposition through their effects on the activities of soil microbes. Warm and moist environment favor decompositionwhereas low temperature andanaerobiosisinhibit decomposition resulting in buildup of organic materials.
Discussion
18180.

Question : Describe how Haryana Kisan Welfare Club was created ?

Answer»

Solution :Integrated organic farming is a cyclical, zero-waste procedure in which waste products from one process are cycled in as nutrients for other processes to allow the maximum utilization of resource and increase the efficiency of production.
Ramesh Chandra Dagar, a farmer in Sonipat, Haryana has created this club with a current membership of 5000 farmers.
It includes bee-keeping, DAIRY management, water harvesting, composting and agriculture in a chain of processes which support each other and allow an extremely ECONOMICAL and sustainable venture.
Crop waste is used to create compost, which can be used as a natural fertilizer or can be used to GENERATE gas for SATISFYING the energy needs of the farm.
No chemical fertilizer is used in this process.
Discussion
18181.

Question : Describe Hershey - Chase experiment or How it was proved the DNA is genetic material ?

Answer»

Solution :The unequivocal proof that DNA is the genetic material CAME from the experiments of Alfred Hershey and Martha Chase (1952).
They worked with viruses that infect bacteria called bacteriophages.
The bacteriophage attaches to the bacteria and its genetic material then enters the bacterial cell.
The bacterial cell treats the viral genetic material as if it was its own and SUBSEQUENTLY manufactures more virus particles.
Hershey and chase worked to discover whether it was protein or DNA from the viruses that entered the bacteria.
They grew some viruses on a medium that contained radioactive phosphorus and some others on medium that contained radioactive sulfur.
Viruses grow in the presence of radioactive phosphorous contained radioactive DNA but not radioactive protein because DNA contains phosphorus but protein does not.
Similarly viruses grown on radioactive sulfur contained radioactive protein but not radioactive DNA because DNA does not contain sulfur.
Radioactive PHAGES were allowed to attach to E. coli bacteria.
Then, as the infection proceeded, the viral coats were removed from the bacteria by agitating them in a blender.
The virus particles were separated from the bacteria by spinning them in a centrifuge.
Bacteria which was infected with viruses that had radioactive DNA were radioactive, indicating that DNA was the material that passed from the virus to the bacteria.
Bacteria that were infected with viruses that had radioactive proteins were not radioactive.
This INDICATES that proteins did not ENTER the bacteria from the viruses.
DNA is therefore the genetic material that is passed from virus to bacteria.
Discussion
18182.

Question : Hershey and Chase experiment with bacteriophage showed that

Answer»

Solution :Alftred Hershey and Martha Chase (1952) conducted experiments on bacteriophages that infect bacteria. PHAGE `T_(2)`, is a virus that infects the bacterium Escherichia coli. When PHAGES (Virus) are added to bacteria, they adsorb to the outer surface, sóme material enters the bacterium, and then later each bacterium lyses to release a large number of progeny phage. Hershey and Chase wanted to observe whether it was DNA or protein that entered the bacteria. All nucleic acids contain phosphorus, and contain sulphur (in the amino acid cysteine and methionine). Hershey and Chase designed an experiment using radioactive isotopes of Sulphur `""^(35)S`and phosphorus `""^(32)P` to keep SEPARATE track of the viral protein and nucleic acids during the infection process. The phages were allowed to infect bacteria in culture medium which containing the radioactive isotopes `""^(35)S` or `""^(32)P`. The bacteriophage that grew in the presence of `""^(35)S` had labelled proteins and bacteriophages grown in the presence of P had labelled DNA. The differential labelling thus enabled them to identify DNA and proteins of the phage. Hershey and Chase mixed the labelled phages with unlabeled E. coli and allowed bacteriophages to attack and inject their genetic material. Soon after infection before lysis of bacteria), the bacterial cells were gently agitated in a blender to loosen the adhering PHASE particles. It was observed that only `""^(32)P` was found associated with bacterial cells and `""^(35)S` was in the surrounding medium and not in the bacterial cells. When phage progeny was studied for radioactivity, it was found that it carried only `""^(32)P` and not `""^(35)S`. These results clearly indicate that only DNA and not protein coat entered the bacterial cells. Hershey and Chase thus conclusively proved that it was DNA, not protein, which carries thehereditary information from virus to bacteria.
Discussion
18183.

Question :Describe haploid diploid sex determination system?

Answer»

Solution :The sex determination in honey bee is based on the number of sets of chromosomes an individual receives .
=An offspring formed from the union of a sperm and an egg develops as a FEMALE (QUEEN or worker) and an unfertilized egg develops as a male (drone) by means of parthenogenesis.
= This means that the males have half the number of chromosomes than that of a female.
= The females are diploid having 32 chromosomes and males are haploid i.e. having 16 chromosomes.
= This is called haploid diploid sex determination system and has SPECIAL characteristic features such as the males produce sperms by mitosis, they do not have father and thus cannot have sons but have grandfather and can have grandsons.
Discussion
18184.

Question : Describe Haplodiploid sex determination system in Honey Bees.

Answer»

Solution :The off springs from the union of sperm and egg develops female or queen. Unfertilized egg develops male (drone) by-parthenogenesis.
FEMALES are diploid having 32 chromosomes, PRODUCE egg by meiosis, where as males are haploid having 16 chromosomes, SPERMS are produced by MITOSIS.
Discussion
18185.

Question : Describe Growth "Models"//"Curves."

Answer»

Solution :Populations show characteristic growth patterns or forms. These patterns can be plotted and termed as J-shaped potential CAPACITY growth form and S-shaped growth form (Sigmoid form). Their growth is represented by S shaped growth CURVE. Jshaped growth form: When a population increases rapidly in an exponential fashion and then stops abruptly due to ENVIRONMENTAL resistance or due to sudden appearance of a limiting factor, they are said to exhibit growth form. Many INSECTS show explosive

increase in number during the rainy SEASON followed by their disappearance at the end of the season. S-Shaped growth form (sigmoid growth) Some populations, as in a population of small mammals, increase slowly at first then more rapidly and gradually slow down as environmental resistance increases whereby equilibrium is reached and maintained. Their growth is represented by S-shaped growth curve.
Discussion
18186.

Question : Describe functionalareasof cerebrumwith the help of neat and labelleddiagram .

Answer»

Solution :1. Structureof cerebrum :
(1) Cerebrum is the largest part of the brain. It forms 80-85 % of the brain.
(2) A median longitudinal fissure dividesthe cerebruminto twocerebral hemispheres. These hemispheresare interspheres are interconnected by athick band of transversenervefibrescalledcorpuscallosum.
(3) The outerpart ofcerebrumis called cortex while theinner part is called medulla.

(4) The roof of cerebrumis called pallium. Pallium is highly folded formingsulci and gyri. Sulci are depressionswhile gyri are ridges. The gyriincrease the surfacearea of cerebal cortex.
(5) The ventrolateral walls of cerebrum arethickenedand are called corporastriata.
(6) The cerebral cortexhas three deep sulci, the central , lateraland parieto-oocipital.
(7) These sulci dividethe cerebral hemisphere into FOUR lobes. These are frontal, partietal, occipital and temporal lobes.
(8) The central sulcus SEPARATES frontaland parietal lobes. the lateral sulcusseparatesparietaland temporal lobes and the parieto-occipital sulcusseparatesparietaland occipital lobe.
(9) There are three functionalareas in cerebral cortex. viz., sensory,associationand motor area. Sensoryarea analyzesthe sensoryinputs. Association area processesthe information .Motorarea sendsthe motor outputs.
2. Functional AREASOF cerebrum are :
(1) General motor area and general sensory area.
(2) Association area
(3) Broca's area
(4) Sensory speech areaor Wernick's area.
The functionsgiven below are associatedwith these areas .
(6) Sensory visual area
(7) Auditory area.
3. Functions of cerebrum :
(1) The cerebrum controls the voluntary activities.
(2) The cerebrumperceives various sensorystimuli receivedthroughvision. taste , smell , sound , touch, speech , etc.
(3) It isthe centre of memory, willpower , intelligence resoning and learning.
(4) The cerebrumis the centre for EMOTIONS, thoughtsand fellings, PAIN , plessure, fear, fatigue,pressure,temperature , etc.
(5) It is also the centre for micturition, defecation , weeping , laughing etc.
Discussion
18187.

Question : Describe fission and its types.

Answer»

Solution :Asexual reproduction is a type of reproduction SEEN in single celled organism.
In these organisms, CELL division is itself a mode of reproduction.
These organisms reproduce by binary fission.
Binary fission : In this type of cell division cell divides into two halves and each RAPIDLY grows into an adult.
E.g. Amoeba, Paramoecium.
Clones : Binary fission involves mitosis only and consequently the resultant offsprings are genetically identical to the parent and to each other.
Budding : In yeast, the division is unequal and small buds are produced which remain attached initially to the parent cell which, eventually gets separated and mature into new yeast organisms or CELLS.
Discussion
18188.

Question : Describefemalehetrogamyand itstypes. HeterogameticFemales:

Answer»

Solution :In thismethodof sex determination, the homogameticmalepossessestwo .X. chromosomesas in certaininsectsandcertainvertebrateslikefishes, repliesand birdsproducinga singletypeof gamete ,whilefemalesproducedissimilargametes .The femalesexconsistsof a single.X.chromosomeor one.X. and one.Y . CHROMOSOME . Thusthe femalesareheterogametic an producetwo typesof eggs.Heterogameticfemales are of twotypes, ZO - ZZ typeandZW - ZZ type .
ZO - ZZ type.
This methodof SEXDETERMINATION isseenin certainmoths ,butterfliesand domesticchickens . In thistype ,the femaleposses sessingle.Z.chromosomein itsbody cellsand is hetrogametic(ZO) PRODUCING two kindsof eggssome with.Z. chromosomeand somewithout.Z.chromosome, while the malepossessestwo.Z. chromosomes and ishomogametic(ZZ) .

ZW - ZZ type
Thismethodof sexdetermination occursincertaininsects(gypsy moth ) and in vertebratessuch as fishes, reptiles and birds. In this methodthe femalehas one.Z.and one.W.chromosome(ZW) producingtwotypes of eggs , somecarrying theZchromosomesand somecarrythe W chromosome. The malesexhas two .Z. chromosome and is homogametic(ZZ)producinga singletype of SPERM .
Discussion
18189.

Question : Describe double helix structure of DNA and how are the sequences of other strand predicted.

Answer»

Solution :The base pairing confers a very unique property to the polynucleotide chains.
They are said to be complementary to each other and therefore if the SEQUENCE of bases in one strand is known then the sequence in other strand can be predicted.
Also, if each strand from DNA acts as a template for synthesis of a new strand, the two double stranded DNA (daughter DNA) thus produced would be identical to the parental DNA molecule.
Because of this, the genetic implications of the structure of DNA become very clear.
The salient features of the double helix structure of DNA are as follows:
(i) It is made of two polynucleotide chain where the backbone is constituted by SUGAR PHOSPHATE and bases project inside.
(ii) The two chains have anti-parallel polarity. It means, if one chain has the polarity `5. rarr 3.` the other has `3.rarr 5.`.
(iii) The bases in two strands are paired through hydrogen bond (H-bonds) forming base pairs (bp).
Adenine forms two hydrogen bonds with Thymine from opposite strand and vice-versa.

Similarly Guanine is bounded with Cytosine with three H-bonds.
As a result always a purine comes opposite to a pyrimidine.
This generates approximately uniform distance between the two strands of the helix.
(IV) The two chains are coiled in a right handed fashion.
The pitch of the helix is 3.4 nm (a nanometer is one billionth of a metre, that is 10-9 m) and there are roughly 10 bp in each turn.
Consequently the distance between a bp in a helix is approximately equal to 0.34 nm.
(v) The plane of one base pair stacks over the other in double helix.
This, in addition of H-bond, confers stability of the helical structure.
Discussion
18190.

Question : Describe dominant epistasis with an example.

Answer»

Solution :Dominant EPISTASIS - It is a gene interaction in which two alleles of a gene at one locus interfere and suppress or mask the phenotypic expression of a different PAIR of alleles of another gene at another locus. The gene that suppresses or masks the phenotypic expression of gene at another locus is KNOWN as epistatic. The gene whose expression is interfered by non - allelic genes and prevents from exhibiting its character is known as hypostatic . When both the genes are present together , the phenotype is determined by the epistatic gene and not by the hypostatic gene .
It the summer squash the fruit colour locus has a dominant allel .W. for white colour and a recessive allel.w. for coloured fruit . .W. allele is dominant that masks the expression of any colour . In another locus hypostatic allel .G. is for yellow fruit and its recessive allel. .g. for green fruit. In the first locus the white is dominant to colour where as in the second locus yellow is dominant to green. When the white fruit with genotypeWWgg is crossed with yellow fruit with genotype wwGG , the `F_1` plants have white fruit and are heterozygous (WwGg). When `F_1` heterozygous plants are crossed they give rise to `F_2` with the phenotypic ratio of 12 white : 3 : yellow : 1 green Since W is epistatic to the allels .G. and .g. . the white which is dominant , masks the effect of yellow or green. Homozygous recessive ww genotypes only can give the coloured fruits (4/16) . DOUBLE recessive .wwgg. will given green fruit (1/16) . The plants having only .G. in its genotype (wwGg o wwGG) will give the yellow fruit (3/16) .
Discussion
18191.

Question : Describe dispersal of fruit and seeds by animals.

Answer»

Solution :Birds and mammals, including HUMAN beings play an efficient and important role in the dispersal of FRUIT and seeds. They have the following devices.
(i) Hooked fruit: The surface of the fruit or seeds have HOOKS (Xanthium), barbs (Andropogon), spines (Aristida) by means or which they ADHERE to the body of animals or clothes of human beings and get dispersed.
(ii) Sticky fruits and seeds:
a. Some fruits have sticky glandular hairs by which they adhere to the fur of grazing animals. a. EXAMPLE: Boerhaavia and Cleome.
b. Some fruits have viscid layer which adhere to the beak of the bird which eat them and when they rub them on to the branch of the tree, they disperse and germinate. Example:
Cordia and Alangium.
(iii) Fleshy fruits: Some fleshy fruits with conspicuous colours are dispersed by human beings to distant places after consumption.
Discussion
18192.

Question : Describe diagrammatic representation of age structure showing declining population.

Answer»

Solution :(1) The relative proportion of the INDIVIDUALS of various age groups in the population is called the age structure of that population.
(2) There are three age groups, viz. young or pre-reproductive age group (0 to 14 years), adults or reproductive group (15 to 59 years) and old or post-reproductive group (60 years and above).
(3) The age structure of a population decides the trend of a population.
(4) Based on the distribution of the age groups the populations can be either, growing population, steady population or declining population.
(i) Growing population : Population having larger number of individuals of the pre-reproductive age groups GROWS at a very rapid rate. Therefore such a population is called a growing population.
(ii) Steady population: When the pre-reproductive age group is larger, the population remains steady.
(iii) Declining population : Larger number of old people and make the number of young people make the population DECLINE.
Discussion
18193.

Question : Describe different experiment which proves that DNA is a Genetic material.

Answer»

Solution :Even though the discovery of nuclein by Meischer and the PROPOSITION for principles of inheritance by MENDEL were almost at the same time but that the DNA acts as a geneticmaterial took long to be discovered and PROVEN.
By 1926, the quest to DETERMINE the mechanism for genetic inheritance had reached the molecular level.
Previous discoveries by Gregor Mendel, Walter Sutton, Thomas Hunt Morgan an numerous other scientists had narrowed the search to the chromosomes located in the nucleus of most cells.
Discussion
18194.

Question : Describe the Darwin's theory of natural selection

Answer»

Solution :1. Darwin's theory of Natural selection : Darwin's theory of Natural selection is based on the following principles :
(1) Over production or enormous fertility or prodigality of production : Every living organism has a great POTENTIALITY of reproduction. Therefore, they produce many YOUNG ones but only few can survive up to maturity. Any population tends to increase in geometrical proportion causing over production of its members.
(2) Struggle for existence: Organisms multiply in geometric ratio but the space and food remain constant leading to competition for survival. This competition is called struggle for existence. The struggle may be intraspecific, intersp'ecific or environmental.
(3) Variations and heredity: Variations are the differences between individuals of the same species. Variations may be harmful. i.e. UNFAVOURABLE, neutral or useful i.e. favourable. The useful variations are preserved and passed on to their offsprings.
(4) Survival of the fittest or Natural selection : The organisms having useful and favourable variations succeed in the struggle for existence. These organisms are the fittest for survival. While the organisms with unfavourable variations are unfit to survive and thus nature does not select them. This is called survival of the fittest. Since the nature selects the organisms which are provided with favourable variations hence the theory is known as the theory of Natural selection.
(5) Origin of new species : A new species originates by the gradual accumulation of favourable variations for a number of generations. Nature selects organisms with favourable variations which have BETTER CHANCE for survival. Due to gradual chap.ges in the organisms after many generations the descendants appear to be different from their ancestors and are identified as new species.
Discussion
18195.

Question : Describe contribution of microbes in production of biogas.

Answer»

Solution :Biogas is a mixture of gases (containing predominantly methane) PRODUCED by the microbial activity and which may be used as fuel. Microbes produce different types of gaseous end products during growth and metabolism. The type of a gas produced depends upon the microbes and the organic substrates they utilise. In the examples cited in relation to fermentation of dough, chees making and production of beverage, the main gas produced was `CO_2`. HOWEVER certain bacteria which grow anaerobically on cellulose material, produce large amount of methane along with `CO_2` and `H_2` . These bacteria are collectively called methanogens and one such common bacterium is Methano-bacterium. These bacteria are present in the rumen part of cattle. A lot of cellulosic material present in the food of cattle is ALSO present in the rumen. In rumen, these bacteria help in the breakdown of cellulose and play an important role in the nutrition of cattle. Thus the excreta (dung) of cattle is called Gobar also. They possess high level of these bacteria. Dung can be used for generation of biogas commonly called gobar gas. The biogas plant consists of a concrete tank (10-15 feet deep) in which bio-wastes are collected and a slurry of dung is fed. A floating cover is placed over the slurry. Which keeps on rising as the gas is produced in the tank due to the microbial activity.

The biogas plant has an outlet which is connected to a pipe to supply biogas to NEARBY houses. The spent slurry is removed through another outlet and may be used as fertilisers Cattle dung is available in large quantities in rural areas where cattle are used for a variety of purposes. The biogas thus produced is used for cooking and lighting. The technology of biogas production was developed in India mainly due to efforts of Indian Agricultural Research Institute (IARI) and KHADI and Village Industries Commission (KVIC).
Discussion
18196.

Question : Describe commensalism

Answer»

Solution :• The interaction in which one species benefits and other is neither harmed nor benefited.
• An orchid growing as an EPIPHYTE on mango branch.
• Barnacles growing on the back of whale benefit while neither the mango tree nor the whale DERIVES any apparent benefit.
• The cattle erget and grazing cattle in close association, the erget ALWAYS forage close to where cattle are grazing because the cattle, as they move, stir up and flush out insects from vegetation that otherwise might be difficult for the ergets to FIND and catch.
• Sea anemone stinging tentacles and clown fish, fish gets PROTECTION from predators which stay away from the stinging tentacles, anemone does not get any benefit or is not harmed.
Discussion
18197.

Question : Describe chromosomal theory of inheritance.

Answer»

Solution :Mendel published his work on inheritance of characters in 1865 but for several reasons it REMAINED unrecognised till 1900.
=Firstly communication was not easy in those days and his work could not be widely publicised.
= Secondly, his concept of genes (factors) as stable and discrete units that controlled the expression of traits and of the pair of alleles which did not .blend. with each other was not accepted by his contemporaries as an explanation for the apparently continuous variation seen in nature.
= Thirdly, Mendel.s approach of using mathematics to explain biological phenomena was totally new and unacceptable to many of the biologists of his time.
=Finally, though Mendel.s work suggested that factors (gene) were discrete units he could not provide any physical proof for the existence of factors or say what they were made of.
=In 1900 three scientists (de Vries, Correns and Von Tschermak) independently rediscovered Mendel.s results on the inheritance of characters.
=Also by this time due to advancements in microscopy that were taking place, Scientists were able to carefully observe cell division.
= This led to the discovery of structures in the nucleus that appeared to double and divide just before each cell division. These were called chromosomes (COLOURED bodies, as they were visualised by staining).
=By 1902, the chromosome movement during meiosis had been worked out.
= Walter Sutton and Theodore Boveri noted that the BEHAVIOUR of chromosomes was parallel to the behaviour of genes and used chromosome movement to explain Mendel.s laws.
= They united the knowledge of chromosomal segregation with Mendelian principles and called it chromosomal theory of inheritance.
= According to this theory : (i) All hereditary characters must be with sperms and EGG cell as they provide bridge from one generation to the other. (ii) The hereditary factors must be CARRIED by the nuclear material. (iii) Chromosomes are also found in pairs like the Mendelian alleles. (iv) The two alleles of gene pair are located on homolous sites on the homologous chromosomes. (v) The sperm and eggs have haploid sets of chromosomes which fuses to re-establish the diploid state. (vi) The genes are carried onto the chromosomes. (vii) Homologous chromosomes synapse during meiosis and get separated to pass into different cells.
= This is the basis for segregation and independent assortment.
Discussion
18198.

Question : Describe chromosomal disorder and describe its examples.

Answer»

Solution :The chromosomal disorders are caused due to absence or excess or abnormal arrangement of one or more chromosomes.
= Failure of segregation of chromatids during cell division cycle results in the gain or loss of a chromosome(s) called aneuploidy.
=For example Down.s syndrome results in the gain of extra copy of chromosome 21.
= Similarly, Turner.s syndrome results due to loss of an X-chromosome in human females.
=Failure to cytokinesis after telophase stage of cell division result in an increase in a WHOLE set chromosomes in an organism and this phenomenon is known as polyploidy.
=This condition is often seen in plants. The total number of chromosomes in a normal human cell is 46 (23 pairs).
=Out of these 22 pairs are autosomes and one pair of chromosomes are sex chromosome.
= Sometimes, though rarely either an additional copy of a chromosome MAY be included in an individual or an individual may lack one of any one pair of chromosomes. These SITUATIONS are known as trisomy or monosomy of a chromosome respectively.
= Such a situation leads to very SERIOUS consequences in the individual. Down.s syndrome, Turner.s syndrome, Klinefelter.s syndrome are common examples of chromosomal disorders.
= Sub Que : Describe Down.s Syndrome.
= The cause of this genetic disorder is the presence of an additional copy of the chromosome number 21 (trisomy of 21).
= The disorder was first described by Langdon Down (1866). The affected individual is short statured with SMALL round head, furrowed tongue and partially open mouth.

= Sub Que : Describe Klinefelter.s Syndrome.
=Klinefelter.s syndrome is caused due to the presence of an additional copy of Xchromosome resulting into a Karyotype of 47, XXY.

=Diagrammatic representation of genetic disorders due to sex chromo some composition in humans: (a) Klinefelter Syndrome (b) Turner.s Syndrome
=Such an individual has overall masculine development however the feminine development (development of breast, i.e. Gynaecomastia) is also expressed. - Such individuals are sterile.
=Sub Que : Describe Turner.s Syndrome.
=Turner syndrome is caused due to the absence of the X-chromosomes i.e. 45 with XO.
=Such females are sterile as ovaries are rudimentary besides other features including lack of other secondary sexual characters.
Discussion
18199.

Question : Describe briefly the steps involved in the breeding of new genetic variety of crops.

Answer»

Solution :Breeding of new genetic variety of crops involve-
a) Collection of variability The entire collection ( of plants and seeds ) having all diverse alleles for all GENES in a given crop is called germplasm collection.
b) EVALUATION and selection of parents -The germplasm is evaluated in order to identify plants with desirable combination of characters and selection of parents.
c) Cross hybridisation among the selected parents- Involves cross hybridizing the two parents to produce hybrids that genetically combine the desired characters in one plant.
d) Selection and testing of superior recombinants It involves the selection of superior plants than both the parents among the progeny of the hybrids. Plants with desired character of combination are self pollinated for SEVERAL generators, so that the characters do not segregate in the progeny.
e) Testing, release and COMMERCIALISATION of new cultivars - The newly selected lines are tested and evaluated by growing in research fields. Then later, the materials are tested in farmers field for atleast THREE growing seasons at several agro climatic zones where the crop is usually grown. The material is evaluated in comparison to the best available local crop cultivar- a check or reference cultivar and commercialised.
Discussion
18200.

Question : Describe briefly the following: Origin of replication

Answer»

Solution :It is the sequences of bases on the DNA from where the replication starts and when a segment of foreign DNA linked to this segments, the forein DNA will replicate along with the rest of the vector DNA.
This sequence is ALSO responsible for CONTROLLING the number of COPIES of the foreign DNA.
To obtain a LARGE number of copies of the foreign DNA the cloning should be done in a vector which has an origing supporting HIGH copy production.
Discussion