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Andhra Pradesh BIEAP AP Inter 2nd Year Botany Study Material 11th Lesson Biotechnology: Principles and Processes Textbook Questions and Answers.

AP Inter 2nd Year Botany Study Material 11th Lesson Biotechnology: Principles and Processes

Very Short Answer Questions

Question 1.
Define Biotechnology.
Answer:
It is a science that utilizes the properties and uses of microorganisms or exploits cells and the cell constituents at the industrial level for generating useful products essential to life and human welfare.

Question 2.
What are molecular scissors? Where are they obtained from?
Answer:
Molecular scissors are the restriction enzymes that cut the DNA at specific locations. Usually, they are obtained from Bacteria.

Question 3.
Name any two artificially restructured plasmids.
Answer:
PBR₃₂₂, PUC_{19 101}.

Question 4.
What is E coRI? How does it function?
Answer:
E.coRI is a restriction enzyme obtained from Escherichia coli. It specifically recognises GAA sites on the DNA and cuts it between G and A.

Question 5.
What are cloning vectors? Give an example.
Answer:
Vectors used for multiplying the foreign DNA sequences are called cloning vectors.
Ex: Plasmids, Bacteriophages, Cosmids.

Question 6.
What is recombinant DNA?
Answer:
The hybrid DNA formed by the fusion of DNA with desirable genes with the vector DNA by DNA ligase is called r-DNA.

Question 7.
What is palindromic sequence?
Answer:
The specific nucleotide sequence of DNA which is recognised by restriction enzyme is called palindrome, which is 4-6 base pairs in length.

Question 8.
What is the full form of PCR? How it is useful in Biotechnology?
Answer:
PCR stands for Polymerase chain reaction. It can be used for the diagnosis of diseases like AIDS, middle ear infection, and tuberculosis.

Question 9.
What is down stream processing?
Answer:
Separation and purification of products before they are ready for marketing is called
downstream processing.

Question 10.
How does one visualize DNA on an agar – gel?
Answer:
To make the DNA visible in the gel, Ethidium bromide is added to the gel solution and the buffer. This positively charged polycyclic compound binds to .DNA inserting itself between the base pairs. Southern blotting may also be used as visualization technique for agarose gels.

Question 11.
How can you differentiate between exonucleases and endonucleases?
Answer:

Short Answer Questions

Question 1.
Write short notes on restriction enzymes.
Answer:
Two enzymes responsible for restricting the growth of Bacteriophage in Escherichia coli were isolated in the year 1963. One of these added methyl groups to DNA and the other cut DNA. The latter was called restriction endonuclease. The first restriction endonuclease – Hind II which cut DNA molecules at a particular point by recognising a specific Sequence of six base pairs, called recognition sequence for Hind II. Today, more than 900 restriction enzymes were isolated from over 200 strains of Bacteria, each of which recognises a different recognition sequence.

E CORI is a restriction enzyme in which, the first letter comes from the Genus (Escherichia), and the second two letters from the species of the Prokaryotic cell [coli], the letter ‘R’ is derived from the name of strain. Roman numbers indicate the order in which the enzymes were isolated from that strain of Bacteria. Restriction enzymes belong to a larger class of enzymes called nucleases. They are of two types.
a) Exonucleases which remove nucleotides from the ends of the DNA.
b) Endonucleases which make cut’s at specific location with in the DNA-

Question 2.
Give an account of amplification of gene of interest using PCR.
Answer:
PCR stands for Polymerase chain reaction. In this reaction, multiple copies of the gene of interest are synthesized in vitro using two sets of primers(oligonucleotides) and the enzyme DNA polymerase. The enzyme extends the primers using the nucleotides provided in the reaction and the. genomic DNA as template. As the process of replication of DNA is repeated many times, the segments of DNA can be amplified to approximately billion times.

Such repeated amplification is achieved by the use of a thermostable DNA polymerase such as Taq polymerase (isolated from thermus aquaticus) which remain active even during high temperature induced denaturation of ds DNA. The amplified fragment, if desired can: now be used to ligate with a vector for further cloning.

Question 3.
What is a bio-reactor? Describe briefly the stirring type of bio-reactor.
Answer:
Bio-reactor is a large vessel which is used for biological conversion of raw material into specific product.

A stirred-tank bio-reactor is usually cylindrical or with a curved base to facilitate the mixing of the reactor contents.

The stirrer facilitates even mixing and oxygen availability throughout the bio-reactor. Alternatively air can be bubbled through the reactor. It has an agitator system, an oxygen delivery system, a foam control system, a temperature .control system, pH control system and sampling ports, so that small volumes of the culture can be with drawn periodically.

Question 4.
What are the different methods of insertion of recombinant DNA into the host cell?
Answer:
There are several methods of introducing the ligated DNA into recipient cells. Recipient cells after making them competent to receive, take up the DNA present in their surrounding. R-DNA can be forced into such cells by incubating the cells with r-DNA on ice followed by placing them briefly at 42eC -(heat shock) and then putting them back on ice. This enables the bacteria to fake up the r-DNA.

In Micro-injection method, r-DNA is directly injected into the nucleus of an animal cell.

In Bio listic of gene gun method – cells are bombarded with high velocity micro particles of gold or tungsten coated with DNA.

In another method, ‘Disarmed pathogen’ vectors are used which when allowed to infect the cell, transfer the r-DNA into the host.

Long Answer Questions

Question 1.
Explain briefly the various processes of recombinant DNA technology.
Answer:
The important methods in recombinant DNA technology are performed through genetic engineering. They are :
i) Isolation of a desired gene,
ii) Insertion of isolated gene into a suitable vector,
iii) Introduction of recombinant vector into a host and
iv) Selection of the transformed host cells.

I) Isolation of a desired gene :

1. The desired gene is isolated from the donor cell. Normally bacteria are the source of desired genes.
2. The cell walls of bacteria are degraded with the help of enzymes.
3. The cell membranes are lysed with the help of detergents.
4. By treating the cellular constituents with phenols and suitable nucleases and by subjecting to gradient centrifugation, pure DNA is isolated.
5. The purified DNA is cut into a number of fragments by restriction endonucleases.
6. The restriction enzymes cleave DNA molecules in two ways.

i) In one way they cut both strands of DNA at exactly opposite points to each other. This results in DNA fragments with blunt ends or flush ends, where two strands end at the same point. Such cut is generally termed as even cut.

ii) But commonly, most enzymes cut the two strands of DNA double helix at different locations. Such a cleavage is generally termed as staggered cut. This generates protruding ends i.e., one strand of DNA double helix extends some bases beyond the other. Since the target site is palindromic in nature, the protruding ends generated by such a cleavage have complimentary base sequence.

As a result, they readily pair with each other and such ends are called cohesive or sticky ends. This stickyness of the ends facilitates the action of the enzyme DNA ligase. When cut by the same restriction enzyme, the resultant DNA fragments have the same kind of ’sticky ends’ and these can be joined together readily by using DNA ligases. E.g.: The restriction enzyme E coRI.
E – The first letter, represents the name of genus Escherichia.
Co – The next two letters, represent the species Escherichia Coli.

The letter R is derived from the name of strain.

Roman numbers following the names indicate the order in which the enzymes were isolated from the strain of bacteria.

This enzyme specifically recognises GAA sites on the DNA and cuts it between G and A (G ↓ A)

7) The resultant fragments are separated from each other by gel electrophoresis.
8) The desired fragments are selected by Southern blotting technique.

II) Insertion of isolated gene into a suitable vector :

1. The selected fragments of DNA are inserted into a suitable vector to produce a large number of copies of genes. This is called gene cloning.
2. There are two major types of vectors, namely plasmids and bacteriophages.
3. Among the two types, plasmids are the ideal cloning vectors.
4. To isolate a plasmid, the Bacterial cell is treated with EDTA (Ethylene diamine tetra acetic acid) along with lysozyme enzyme to digest the cell wall.
5. Then the bacterial cell is subjected to centrifugation in sodium lauryl sulphate to separate the plasmid.
6. The plasmid DNA is cut with the help of restriction endonuclease.
7. The circular plasmid is converted into a linear molecule having sticky ends.
8. The two sticky ends of linear plasmid are joined to the ends of desired gene by DNA ligase.
9. The plasmid containing foreign DNA segments is called recombinant DNA (r DNA) or Chimeric DNA.

III) Introduction of recombinant vector into a suitable host :

1. The r DNA molecule is introduced into suitable bacterial host cell by transfor-mation.
2. The cell containing r DNA is called transformed cell.
3. Bacterial cell walls are not permeable to recombinant vectors, but keeping in dil. Calcium chloride renders the bacterial cell wall permeable to recombinant vectors.
4. The r DNA replicates with in the host cell.
5. The transformed cell grows on the culture medium. Each daughter cell contains r DNA.

IV) Selection of transformed host cells :
1) Selection of transformed cells depends on the nature of gene which is cloned. 2) It can be done in two ways. The are :
a) Without using probes,
b) By using probes.

a) Without using probes:

1. If the gene is cloned for antibiotic resistance, the cells are first incubated on a medium without antibiotic for one hour, to allow the antibiotic resistance gene to be expressed.
2. Then the cells are placed on a medium with an antibiotic for selection of colonies containing rDNA.
3. The cells which have expressed the gene will survive and the others die.

b) By using probes:
When transformed cells are cultured on the nutrient medium, several cells are produced. To select the cells containing the desired gene colony hybridization method is used. In this gene specific probes are used. A probe is a small fragment of single stranded RNA or DNA which is tagged with radioactive, molecule. It can search out complimentary DNA sequences from an organism.

Question 2.
Give a brief account of the tools of recombinant DNA technology.
Answer:
Key tools are :
1) Restriction enzymes :
Two enzymes responsible for restricting the growth of Bacteriophage in Escherichia coli were isolated in the year 1963. One of these added methyl groups to DNA and the other cut DNA. The latter was called restriction endonuclease. The first restriction endonuclease – Hind II which cut.

DNA molecules at a particular point by recognising a specific sequence of six base pairs, called recognition sequence for Hind II. Today, more than 900 restriction enzymes were isolated from over 200 strains of Bacteria, each of which recognises a different recognition sequence.

E CORI is a restriction enzyme in which, the first letter comes from the genus (Escherichia), and the second two letters from the species of the Prokaryotic cell [coli], the letter ‘R’ is derived from the name of strain. Roman numbers indicate the order in which the enzymes were isolated from that strain of Bacteria. Restriction enzymes belong to a larger class of enzymes called nucleases. They are of two types.

a) ExOnucleases which remove nucleotides from the ends of the DNA.
b) Endonucleases which make cuts at specific location with in the DNA.

Most restriction enzymes cut the two strands of DNA double helix at different locations. Such a cleavage is know as staggered cut. E CoRI recognises 5′ GAATT 3′ sites on the DNA and cuts it between G and A results in the formation of sticky ends or cohesive end pieces. This stickyness of the ends facilitates the action of enzyme DNA ligase.

Cloning vectors :
The DNA used as a carrier for transferring a fragment of foreign DNA into a suitable host is called vector. Vectors used for multiplying the foreign DNA sequences are called cloning vectors. Commonly used cloning vectors are plasmids, bacteriophages, cosmids. Plasmids are extrachromosomal circular DNA molecules present in almost all bacterial species. They are inheritable and carry few genes are easy to isolate and reintroduce into the bacterium (Host).

Features required to facilitate cloning into a vector:
a) Origin of replication:
(ori) This is a sequence from where replication starts and any piece of DNA when linked to this sequence can be made to replicate within host cells. It is also responsible for controlling the copy number of the linked DNA.

b) Selectable marker :
In addition to ‘ori’, the vector requires a selectable marker, which helps in identifying and eliminating non-transformants and selectively permitting the growth of the any transformants normally, the genes encoding resistance to antibiotics such as ampicillin, chloramphenicol, tetracycline or kanamycin etc., are useful selectable markers for E.Coli.

c) Cloning sites :
In order to link the alien DNA, the vector needs to have very few, preferably single recognition sites for the restriction enzymes.

d) Molecular weight:
The cloning vector should have low molecular weight.

e) Vectors for cloning genes in plants and animals:
The tumour inducing (Ti) plasmid of Agrobacterium tumifaciens has now been modified into a cloning vector such that it is no more pathogenic to plants. Similarly retroviruses have also been disarmed and are now used to deliver desirable genes into animal cells.

Intext Questions

Question 1.
Do Eukaryotic cells have restriction Endonucleases? Justify your answer.
Answer:
No. only prokaryotic cells have restriction Endonucleases. Today we know 900 or more restriction enzymes have been Isolated from over 230 strains of Bacteria.

Question 2.
Besides better aeration and mixing properties, what other advantages do stirred tank bioreactors have over shake flasks?
Answer:

1. Small volumes of the culture can be withdrawn periodically.
2. Air can be bubbled through the reactor.
3. It has a control system that regulates the temperature and pH.

Question 3.
Can you recall Meiosis and indicate at what stage a recombinant DNA is made?
Answer:
Pachytene of Meiosis – I.

4. Describe briefly the following :
a) Origin of replication b) Bioreactors c) Down stream processing
Answer:
a) Origin of replication :
It is the specific DNA sequence which is responsible for initiating replication.

b) Bioreactors:
They are large vessels which are used for biological conversion of raw material into specific products.

c) Down stream processing :
Separation and purification of products is called down stream processing.

Question 5.
Explain briefly a) PCR b) Restriction enzymes and DNA c) Chitinase
Answer:
a) PCR:
Polymerase chain reaction: Is a biochemical technology in molecular biology to amplify a single or few copies of a piece of DNA generating thousands of copies of a particular DNA sequence.

b) Restriction enzymes and DNA :
Restriction enzymes are DNA cutting enzymes found in Bacteria because they cut with in the molecule.

c) Chitinase :
It is an enzyme obtained from the fungus, used in the digesting the bacterial cell walls.

Question 6.
Discuss with your teacher and find out how to distinguish between
a) Plasmid DNA and Chromosomal DNA
b) RNA and DNA
c) ExonuCleases and Endonucleases

a)

b)

c)

Question 7.
What does ‘H’ in’d’ and HI refer to in the enzyme Hind III?
Answer:
‘H’ – The genes from which it was isolated – Haemophilus
in – Influenza (the species name) .
d – Refers the bacterial strain from which enzyme was isolated.
III – Third isolated enzyme from the strain.

Question 8.
Restriction enzymes should not have more than one site of action in the cloning site of a vector. Comment.
Answer:
The presence of more than one recognition site for the same enzyme with in the vector will generate several fragments which will complicate the gene cloning. So vector needs to have very few, preferably single, recognition site for the commonly used restriction enzymes.

Question 9.
What does ‘competent’ refer to ‘incompetent cells’ used in transformation experiments?
Answer:
DNA is a hydrophilic molecule. It cannot pass through cell membranes. In order to force Bacteria to take up the plasmid, the bacterial cells must be made competent to take up DNA by treating them with a bivalent cation such as calcium which increaes the efficiency with which DNA enters the bacterium.

Question 10.
What is the significance of adding proteases at the time of Isolation of Genetic material (DNA)?
Answer:
Proteins can be removed by treatment with proteases.

Question 11.
While doing a PCR, ‘denaturation’ step is missed. What will be its effect on the process?
Answer:
The segment of DNA cannot be amplified.

Question 12.
What modification is done on the Ti plasmid of Agrobacterium tumefascians to convert it into a cloning vector?
Answer:
It is no more pathogenic to plants but is still able to use the mechanisms to deliver genes of our interest into a variety of plants.

Question 13.
What is meant by gene cloning?
Answer:
It is the replication of DNA fragments by the use of a self-replicating genetic material. It duplicates only individual genes of organism’s DNA.

Question 14.
Decide the ratio between ester bonds and hydrogen bonds that are broken in each palindromic sequence of DNA when treated with E coRI during the formation of sticky ends.
Answer:
1 : 4.

Andhra Pradesh BIEAP AP Inter 2nd Year Botany Study Material 10th Lesson Molecular Basis of Inheritance Textbook Questions and Answers.

AP Inter 2nd Year Botany Study Material 10th Lesson Molecular Basis of Inheritance

Very Short Answer Questions

Question 1.
Distinguish between Heterochromatin and Euchromatin. Which of the two is transcriptionally active?
Answer:
The chromatin that is more densely packed and stains dark is called heterochromatin. The chromatin that is loosely packed and stained light is called Euchromatin. Euchromatin is transcriptionally active chromatin.

Question 2.
Who proved that DNA is Genetic Material? What is the organism they worked on?
Answer:
Alfred Hershey and Martha chase (1952). They worked with viruses that infect Bacteria, and bacteriophages.

Question 3.
What is the function of DNA polymerase?
Answer:
DNA polymerise is a highly efficient enzyme which catalyse polymerisation of a large number of nucleotides in a very short time. It also catalyse the reaction with a high degree of accuracy.

Question 4.
What are the components of a nucleotide?
Answer:
A nucleotide has three components – a nitrogenous base, a pentose sugar and a phosphate group’.

Question 5.
Given below is the sequence of coding strand of DNA in a transcription unit.
5’AATGCAGCTATTAGG – 3 . Write the sequence of
a) Its complementary strand
b) The mRNA
Answer:
a) 5’TTACGTCGATAATCC-3′
b) 5’ AAUGCAGCUAUUAGG – 3’

Question 6.
Name any three viruses which have RNA as the Genetic Material.
Answer:
Tobacco mosaic virus, QB bacteriophage, HIV.

Question 7.
What are the components of a transcription unit?
Answer:
a) A promotor b) The structural Gene c) A terminator.

Question 8.
What is the difference between exons and Introns?
Answer:

Question 9.
What is meant by capping and tailing?
Answer:
Adding of an unusual nucleotide (methyl guanosine triphosphate) to the 5′ -end of heterogenous nuclear RNA [hnRNA) is called capping.

Adding of adenylate residues to the 3’ – end in a template independent manner is called tailing.

Question 10.
What is meant by point mutation? Give an example.
Answer:
Change of single base pair in the gene for betaglobin chain that results in the change of aminoacid residue glutamate to valine. It results in a diseased condition called sickle cell anaemia.

Question 11.
What is meant by charging of tRNA?
Answer:
Activation of aminoacids in the presence of ATP and linked to their cognate tRNA is known as charging of tRNA or amino acylation of tRNA.

Question 12.
What is the function of the codon AUG?
Answer:
It has dual functions. It codes for methionine and also act as the initiator codon.

Question 13.
Define stop codon. Write the codons.
Answer:
Codons which do not code for any aminoacids are called stop codons. They are UAA, UAG, UGA.

Question 14.
What is the difference between the template strand and a coding strand in a DNA molecule?
Answer:
The two strands have opposite polarity and the DNA-dependant RNA polymerise also catalyses the polymerisation in only one direction that is, 5′ → 3′, the strand that has the polarity 3′ → 5′ acts as a template and is also called template strand. The other strand which has polarity 5′ → 3′ and does not code for anything is called coding strand.

Question 15.
Write any two differences between DNA and RNA.
Answer:

Question 16.
In a typical DNA molecule the proportion of thymine is 30% of the N bases. Find out the percentages of other N bases.
Answer:
Adenine = 30%
Guanine = 20%
Cytosine = 20%

Question 17.
The proportion of nucleotides in a given nucleic acid are Adenine 18%, Guanine 30%, Cytosine 42% and uracil 10%. Name the nucleic acid and mention the number of strands in it.
Answer:
RNA. Only one strand is present.

Short Answer Questions

Question 1.
Define transformation in Griffith’s Experiment. Discuss how it helps in the identifi¬cation of DNA as genetic material.
Answer:
Frederick Griffith (1928) conducted experiments on streptococcus pneumoniae and observed a transformation in bacteria. When streptococcus were grown on a culture plate, some produced smooth shiny colonies (s) while others produced rough colonies (R). Mice injected with ‘s’ shain (mucous coat) die from pneumonia infection but mice injected with R strain do not develop pneumonia.

He injected heat killed ‘s’ strain bacteria to mice, It is healthy. Finally he injected heat killed S and R strains, the mice died. He concluded that the R strain bacteria had been transformed by heat killed ‘s’ strain bacteria. Some transforming principle transferred from heat killed strain to R strain to synthesize a mucous coat and become virulent. This is due to the transfer of genetic material.

Question 2.
Discuss the significance of heavy isotope of Nitrogen in Meselson and Stahl’s experiment.
Answer:
Matthew Meselson and Franklin Stahl, grow E.Coli in a medium containing ¹⁵NH₄Cl and observed 15_N was incorporated in newly synthesized DNA. This heavy DNA molecule could be distinguished from themormal DNA by centrifugation in a cesium chloride density gradient.

They transformed the cells into a medium with ¹⁵NH₄Cl (normal) and took samples at various time intervals and extracted the DNA that remained as double stranded helices. The various samples were separated independently on cesium chloride (cscl) gradients to measure the densities of DNA. Thus the DNA that was extracted from the culture, one generation after the transfer from ¹⁵N to ¹⁴N medium had a hybrid density. DNA extracted from the culture after another generation (40 mts) was composed of equal amounts of this hybrid DNA and of ‘Light’ DNA.

Question 3.
A single base mutation in a gene may not always result in loss or gain of function. Do you think the statement is correct? Defined your answer.
Answer:
A single base mutation in a gene may result in loss or gain of a gene and so a function.
E.g.: A change of single base pair in the gene for betaglobin chain that results in the change of aminoacid residue glutamate to valine. It results in a diseased condition called sickle cell anaemia.

E.g.: 2) Consider a statement that is made like a genetic code is – RAM HAS RED CAP.
If we remove a letter ‘S’ in HAS, it will be RAM HAR EDC AP.

Question 4.
How many types of RNA polymerases exist in cells ? Write their names and functions.
Answer:
Three RNA polymerases exist in Cells. They are :

1. RNA polymerase I – It transcribes r RNAs (28S, 18S, and 5.8S)
2. RNA polymerase II – It transcribes the precursor of RNA, the heterogeneous nuclear RNA (hn RNA)
3. RNA polymerase III – It is responsible for transcription of tRNA 5 Sr RNA and Sn RNAs (small nuclear RNAs)

Question 5.
What are the contributions of George Gamow, H.G. Khorana, Marshall Nirenberg in deciphering the genetic code?
Answer:
George Ganow, suggested that, in order to code for all the 20 Amino’ acids, the code should be made up of three nucleotides. This was a very bold proposition, because a permutation and combination of 4³ would generate 64 codons, generating more codons than required.

H.G. Khorana developed chemical method in synthesising RNA molecules with defined combinations of bases.

Marshall Nirenberg’s cell-free system for protein synthesis finally helped the code to be deciphered. The enzyme polynucleotide phosphorylase was also helpful in polymerising RNA with defined sequences in a template independent manner (enzymatic synthesis of RNA)

Question 6.
On the diagram of the secondary structure of tRNA shown below label the location
of the following parts.
Answer:

a) Anticodon
b) Acceptor stem
c) Anticodon stem
d) 5′ end
e) 3′ end

Question 7.
Draw the schematic / diagrammatic presentation of the lac operon.
Answer:

Question 8.
What are the differences between DNA and RNA.
Answer:

Question 9.
Write the important features of Genetic code?
Answer:

1. The codon is triplet. 61 codons code for aminoacids and 3 codons donot code for any aminoacids called stop codons.
2. One codon codes for only one aminoacid, hence it is unambiguous and specific.
3. Some aminoacids are coded by more than one codon, hence the code is degenerated.
4. The codon is read in mRNA in a contiguous fashion. There are no punctuations.
5. The code is nearly universal: For Ex: UUU code for phenylalanine (Phe) in Bacteria or Humans.
6. AUG has dual functions. It codes for methionine and also acts as initiator codon.

Question 10.
Write briefly on nucleosomes.
Answer:
Nucleosome is a bead like structure of chromosomes. It consists of eight histone molecules and a DNA segment of about 150 base pairs. Each Nucleosome is separated from one another by a linker DNA sequence of about 50 base pairs. Nucleosome helps to fold DNA into a compact form in the interphase nucleus. Otherwise the length of a chromosome, when linear is many orders of magnitude greater than the diameter of the nucleus.

Intext Questions

Question 1.
Group the following as nitrogenous bases and nucleosides : Adenine, Cytidine, Thymine, Guanosine, Uracil and Cytosine.
Answer:
Nitrogenous bases : Adenine, Thymine, Uracil, Cytosine,
Nucleosides : Cytidine, Guanosine.

Question 2.
If a double stranded DNA has 20% of cytosine, calculate the percent of adenine in the DNA.
Answer:
Cytosine = 20%
Guamine = 20%
Adenine = 30%
Thymine = 30%

Question 3.
If the sequence of one strand of DNA is written as follows : Write down the sequence of complementary strand in 3′ → 5′ direction.
5′ – ATGCATGCATGCATGCATGCATGCATGC – 3‘
Answer:
3′ – TACGTACGTACGTACGTACGTACGTACG – 5′

Question 4.
If the sequence of the coding strand in a transcription unit is written as follows.
5 – ATGCATGCATGCATGCATGCATGCATGC – 3 write down the sequence of m RNA
Answer:
3′ AUGC AUGC AUGC AUGC AUGC AUGC AUGC – 5

Question 5.
Which property of DNA double helix led Watson and Crick to hypothesise semi conservative mode of DNA replication? Explain.
Answer:
The two strands would separate and acts as a template for the synthesis of new complementary strands. After completion of replication, each DNA would have one parental and one newly synthesised strand.

Question 6.
Depending upon the chemical nature of the template [DNA or RNA] and the nature of nucleic acids synthesized from it [DNA or RNA], List the types of nucleic acid polymerases.
Answer:
DNA polymerases
RNA polymerases.

Question 7.
How did Hershey and Chase differentiate between DNA and protein in their experiment while proving that DNA is the genetic material?
Answer:
Hershey and chase purified biochemicals (proteins, DNA, RNA etc) from the heat killed ‘S’ cells. They discovered the DNA alone from S bacteria caused R bacteria to become transformed. They also discovered thal protein digesting enzymes and RNA digesting enzymes did not affect transformation. So the transforming principle was not a protein or RNA. Digestron with DNAse did inhibit transformation suggesting that the DNA caused the transformation.

Question 8.
Differentiate between the followings :
a) mRNA and tRNA
b) Template strand and Coding strand
Answer:

a)

b)

Question 9.
List two essential roles of ribosomes during translation.
Answer:

1. Ribosome acts as the site where protein synthesis takes place from individual aminoacids.
2. Ribosome acts a catalyst for forming peptide bond.
   E.g. : 23 S r-RNA in bacteria acts as ribozyme.

Question 10.
In the medium where E. coli was growing, lactose was added. Which induced the Lac operon. Then, why does Lac operon shut down some time after addition of lactose in the medium?
Answer:
Lac operon is a segment of DNA that is made up of three adjacent structural genes namely, an operator gene, a promoter gene and a regulator gene. It works in a coordinated manner to metabolise lactose into glucose and galactose. In Lac operon lactose acts as an inducer. It binds to the repressor and inactivates it. Once the lactose binds to the repressor, RNA polymerase binds to the promotor region.

Hence, three structural genes express their product and respective enzymes are produced. These enzymes act on lactose or metabolise it into glucose and galactose. After sometime, when the level of the inducer decreases, it causes the synthesis of the repressor from regulator gene. The repressor binds to the operator gene and prevents RNA polymerase from transcribing the operon. Hence the transcription is stopped.

Question 11.
Explain (in one or two lines) the functions of the followings :
a) Promoter b) tRNA c) Exons.
Answer:
a) Promoter:
It is a region of DNA that helps in initiating the process of transcription. It serves as the binding site for RNA polymerase. :

b) tRNA :
It is a small RNA that reads the genetic code present on mRNA it carries specific aminoacids to mRNA on ribosome during translation of proteins.

c) Exons:
Exons are coding sequences of DNA in Eukaryotes that transcribe for proteins.

Question 12.
Briefly describe the following :
a) Transcription b) Translation.
Answer:
a) Transcription:
It is the process of synthesis of RNA from DNA template. A segment of DNA gets copied into mRNA during the process. The process of transcription starts at the promotor region of the template DNA and terminates at the terminator region The segment of DNA between these two regions is known as transciption unit. The transcription requires RNA polymerase enzyme, a DNA template, four types of nucleotides, and certain cofactors such as Mg²⁺

During transcription, three events occur. They are : a)Initiations b)Elongation, c) Termination. The DNA dependant RNA polymerase and certain initiation factors bind at the double stranded DNA at the promotor region of the template strand and initiate the process of transcription, RNA polymerase moves along the DNA and leads to the unwinding of DNA duplex in two separate strands.

Then one of the strands, called sense strand acts as template for mRNA synthesis. The epzyme RNA polymerase utilises nucleoside triphosphates as raw material and polymerizes them to form m RNA according to the complimentary bases present on the template DNA. This process of opening of helix and elongation of polynucleotide chain continuous until the enzyme reaches terminator region. As RNA polymerase reaches the terminator region, the newly synthesized mRNA transcripted alpng with enzyme is released. Another factor called terminator factor (p) required for the termination of the transcription.

b) Translation :
It is the process of polymerizing amino acid to form a polypeptide chain. The triplet sequence of base pairs in mRNA defines the order and sequence of amino acids in a polypeptide chain. This process invoivs 3 steps, a) Initiation b) Elongation c) Termination. During the initiation, tRNA gets charged when the aminoacid binds by using ATR The start codon (AUG) present on mRNA is recognized only by the charged tRNA. The ribosome acts as an actual site for the process of translation and contains two separate sites in.a large subunit for the attachment of subsequent aminoacid.

The small subunit of ribosome binds to mRNA at the initiation codon (AUG) followed by the large subunit. Then, it initiates the process of translation. During the elongation process, the ribosome moves one codon dowonstream along with mRNA so as to leave the space for binding of another charged tRNA, The aminoacid brought by tRNA get linked with the previous aminoacid through a peptide bond and this process continues resulting in the formation of a polypeptide chain. When the ribosome reaches one or more STOP codon (UAA, UAG and UGA), the process of translation gets terminated. The polypeptide chain is released and ribosomes get detached from mRNA.

Question 13.
How the polymerization of nucleotides can be prevented in a DNA molecule?
Answer:
Due to Lack of Helicase enzyme, unwounding does not occurs. The single stranded Binding proteins cover the DNA strands preventing them from annealing into a double strand.

Question 14.
In an experiment, DNA is treated with a compound which trends to place itself amongst the stacks of nitrogenous base pairs. As a result of this, the distance between two consecutive base pairs increases. From 0.34 nm to 0.44 nm calculate the length of DNA double helix (which has 2 × 10⁹ bp) in the presence of saturating amount of this compound.
Answer:
2 × 10⁹ × 0.44 × 10^-9 bp.

Question 15.
Recall the experiments done by Frederick Griffith. Where DNA was speculated to be the genetic material. If RNA, instead of DNA was the genetic material, would the heat killed strain of pneumococcus have transformed the R – strain into virulent strain? Explain?
Answer:
RNA is more labile and prone to degradation (owing to the presence of 2’OH group in its ribose). Hence heat killed ‘s’ strain may not have retained its ability to transform the ‘R’ strain into virulent form if RNA was its genetic material.

Question 16.
You are repeating the Hershey – Chase experiment and are provided with two Isotopes : ³²P and ¹⁵N (in place of ³⁵S in the original experiment). How do you expect your results to be different?
Answer:
Use of ¹⁵N will be inappropriate because method of detection of ³⁵P and ¹⁵N is different (³²P being a radioactive Isotope while ¹⁵N is not radioactive but is the heavier Isotope of Nitrogen). Even if ¹⁵N was radioactive, them its presence would have been detected both inside the cell as well as in the supernatant because ¹⁵N would also get incorporated in amino group of aminoacids in proteins. Hence the use of ¹⁵N would not give any conclusive results.

Question 17.
Do you think that the alternate splicing of exons may enable a structural gene to code for several Isoproteins from one and the same gene? If yes, how? If not, why so?
Answer:
Functional m RNA of structural genes need not always include all of its exons. This alternate splicing of exons in sex-specific, tissue – specific, and even developmental stage specific. By such alternate splicing of exons a single gene may encode for several Isoproteins and/or proteins of similar class’. In absence of such a kind of splicing, there should have been new genes for every protein/Isoprotein. Such an extravagency has been avoided in natural phenomena by way of alternate splicing.

Question 18.
Can you recall what centrifugal force is, and think why a molecule with higher mass/ Density would sediment faster?
Answer:
Proteins have lower density when compared to others. So a molecule with higher mass would sediment faster than molecules with light weight (density).

Question 19.
Do Retroviruses follow central Dogma? Give one example.
Answer:
Genetic material of Retroviruses in RNA. At the time of synthesis of protein RNA is reverse transcribed to its complementary DNA first/which is opposite to the central Dogma. Hence Retroviruses are not known to follow central Dogma.

Question 20.
If there are 2.9 × 10⁹ complete turns in a DNA molecule. Estimate the length of the molecule (1 angstrom = 10^-8 cm).
Answer:
1 turn of DNA = 3.4 nm.
Number of turns are = 2.9 × 10⁹
The length of the DNA molecule = 2.9 × 10⁹ × 3.4 nm.

Andhra Pradesh BIEAP AP Inter 2nd Year Botany Study Material 9th Lesson Principles of Inheritance and Variation Textbook Questions and Answers.

AP Inter 2nd Year Botany Study Material 9th Lesson Principles of Inheritance and Variation

Very Short Answer Questions

Question 1.
What is the cross between the F₁ Progeny and the Homozygous recessive parent called? How is it useful?
Answer:
Test cross. It is used to test whether an Individual is Homozygous (pure) or Heterozygous (Hybrid).

Question 2.
Do you think Mendel’s laws of inheritance would have been different if the characters that he chose were located on the same chromosome?
Answer:
Yes, it depends on the linkage of genes on the same chromosome.

Question 3.
Who proposed the Chromosome theory of inheritance?
Answer:
Walter Sutton and Theodore Boveri.

Question 4.
Define true breeding. Mention its significance.
Answer:
All offsprings has the same phenotype as its parents is called the breeding. They express characters for several generations.

Question 5.
Explain the terms Phenotype and Genotype.
Answer:
The Physical or external appearance of a character is called Phenotype. The genetic makeup of an individual is called Genotype.

Question 6.
What is point mutation? Give an example.
Answer:
Mutation which occurs in a single base pair of DNA is called point mutation.
E.g.: Sickle cell anemia.

Question 7.
What is the genotype of wrinkled phenotype of pea seeds?
Answer:
rr is the genetic nature of the phenotypic wrinkled pea seeds.

Question 8.
What will be the phenotypic ratio in the offsprings obtained from the following crosses.
a) Aa x aa b) AA x aa c) Aa x Aa d) Aa x AA
Note : Gene A is dominant over gene a.
Answer:

Question 9.
In garden pea, the gene T for tall is dominant over its allele for dwarf. Give the genotypes of the parents in the following crosses.
a) Tall x Dwarf producing all tall plants.
b) tall x tall producing 3 tall and 1 dwarf plants.
Answer:

Short Answer Questions

Question 1.
Mention the advantages of selecting pea plant for experiment by Mendel.
Answer:

1. It is an annual plant that has well defined characteristics.
2. It can be grown and crossed easily.
3. It has bisexual flowers containing both male and female parts.
4. It can be self fertilized conveniently.
5. It has a short life cycle and produces large number of offsprings.

Question 2.
Differentiate between the following.
a) Dominant and Recessive b) Homozygous and Heterozygous
Answer:

Question 3.
Explain the law of Dominance using a Monohybrid cross.
Answer:
Characters are controlled by discrete units called factors. Factors occur in pairs. In a dissimilar pair of factors pertaining to a character one number of the pair dominates the other. The law of Dominance is used to explain the expression of only one of the parental characters in a Monohybrid cross in the F₁ generation and the expression of both in F₂ generation in the ratio of 3 : 1.

Question 4.
Define and design a test cross.
Answer:
Crossing between Fx indivi – duals with the recessive parent is called test cross. It is used to test whether an individual is homozygous or heterozygous. A mono hybrid test cross gives a phenotype ratio of 1 : 1 and a dihybrid test cross gives a ratio of 1 : 1 : 1 : 1.

Question 5.
Explain the co-dominance with example.
Answer:

a) Co-dominance:
The phenomenon where heterozygotes have features of both the homozygotes, that is, an allele is neither dominant nor recessive to the other. 1) Different types of red blood cells that determine ABO blood grouping in humans 2) Seed coat pattern size in lentil plants. A cross between pure breeding spotted lentils with pure breeding dotted lentils produce heterozygotes. These Fx hybrids show the phenotypic features of both parents which means that neither the spotted nor the dotted, allele is dominant or recessive to the other. Because both traits show up equally in the heterozygote’s phenotype, the alleles are termed co-dominant. F₁ plants are self pollinated produce F₂ progeny in the ratio of (lspotted : 2 spotted and dotted : 1 dotted).

Question 6.
Explain the Incomplete Dominance with example.
Answer:

Incomplete Dominance: It is the condition where one allele of a gene is not completely dominant over the other allele and results in the heterozygous having phenotype different from the dominant and recessive homozygotes.

In a cross between true beeding red flowered plants (RR) and true white flowered plants, (rr) the F₁ was pink (Rr). When the F₁ was self pollinated, the F₂ resulted in the ratio of 1(RR) : 2(Rr) : 1(rr)

Red Pink White

Here the genotype ratios were as in monohybrid cross of Mendel but the phenotypic ratio had changed from the 3 : 1 because, R was not completely dominant over r and is possible to distinguish Rr as pink from RR and rr. Thus the phenotypic and genotypic ratios in F₂ progeny are the same, that is 1 : 2 : 1.

Question 7.
Write a brief note on chromosomal mutations and gene mutations.
Answer:
a) Chromosomal mutations :
Any change in the number or structure of chromosomes are called chromosomal mutations. They may be

a) Structural changes :
During prophase I of Meiosis, homologous regions of chromosomes show many changes in structure inorder to form pairs or bivalents. They are :
1) Deletion : A part of chromosome is lost or broken.
2) Duplication : A particular part of chromosome is repeated.

b) Gene Mutations :
Mutations which occur due to change in a single base pair of DNA are called gene mutations or point mutations.

Question 8.
Define Law of Segregation and Law of Independent Assortment.
Answer:
Law of segregation :
The two alleles of a gene when present together in a heterozygous state, they do not fuse or blend in any way but remain distinct and segregate during meiosis or in the formation of gametes, so that each meiotic product of gamete will carry only one of them.

Law of Independent Assortment:
When two pairs of traits are combined in a hybrid, seggregation of one pair of characters is independent of the other pair of characters.

Long Answer Question

Question 1.
Explain the dihybrid cross with the help of punnel square board by taking contrasting traits, seed colour and seed shape.
Answer:
The cross made between individuals differing in two characters is known as Dihybrid cross. Mendel studied the inheritance of two characters at a time. Mendel crossed a homozygous pea plant having yellow and round seeds with another pea plant having green and wrinkled seeds. The F₁ hybrids were found to have yellow and round seeds. This shows that are dominant.

When F₁ plants allowed to cross among themselves, four distinct types of seeds appeared in F₂ generation. Two of these were similar to parental combinations, while the other two were new combinations. The phenotypic ratio of 9 : 3 : 3 : 1 can be derived as a combination series of 9 round yellow, 3 round green, 3 wrinkled yellow, 1 wrinkled green.

Intext Questions

Question 1.
Mendel crossed pea plants producing round seeds with those producing wrinkled seeds. From a total of 7324 F₂ seeds, 5474 were round and 1850 were wrinkled. Using the symbols R and r for genes, predict the a) the parental (p) genetypes, b) the gametes, c) F₁ progeny, d) the cross between F₁ hybrids, e) genotypes, phenotypes, genotypic frequency, phenotypic ratio of F₂ progeny.
Answer:
a)

d) Cross between F₁ hybrids

e) 3 : 1 – phenotypic ratio of F₂ progeny

Question 2.
The following data was obtained from an experiment on peas. The grey coloured seed is dominant over white coloured seed. Use the Letter G for grey and g for white traits. Predict genotypes of the parents in each of the following crosses.

Answer:

Question 3.
In tomatoes red fruit colour (R) is dominant yellow (r). Suppose a tomato plant homozygous for red is crossed with one homozygous for yellow. Determine the appearence of following.
a) The F₁, b) The F₂, c) The offspring of a cross of the F₁ back to the red parent, d) The offspring of a cross of the F₁ back to the yellow parent.
Answer:

Question 4.
In pea, axillary position of flowers T is dominant over its terminal position (t). Coloured flowers (C) are dominant to white flowers (c). A true breeding plant with coloured flowers in axils is crossed to one with white terminal flowers. Give the phenotypes, genotypes and expected ratios of F₁, F₂ back cross and test cross progenies. What genotype ratio is expected in the F₂ progeny?
Answer:

In Back cross 1 : 1 : 1 : 1
Test cross 1 : 1 : 1 : 1
Genotypic ratio of this cross = 1 : 2 : 2 : 4 : 1 : 2 : 1: 2 : 1
Phenotypic ratio of this cross = 9 : 3 : 3 : 1

Question 5.
In summer squash, a plant with white flowers and disc shaped fruits is crossed to a plant with yellow flowers and sphere shaped fruits. The F₁ hybrids had white flowers and disc shaped fruits. Which phenotypes are obminant? Give the genotypes of the parents and the hybrids. If these hybrids were selfed and 256 progeny were obtained, what would be the frequency of the various phenotypes?
Answer:

In dihybrid cross 9 : 3 : 3 : 1 phenotypic ratio is obtained.
The hybrids were selfed gives 256 progeny. Among them
White disc shaped fruits are 144 [\(\frac{256}{19}\) × 9
White sphere shaped fruits = 48 \(\frac{256}{10}\) × 3
Yellow; disc shaped fruits = 48 \(\frac{256}{16}\) × 3
Yellow sphere shaped fruits = 16 \(\frac{256}{16}\) × 1]

Question 6.
Give the ratios of the following
a) Monohybrid test cross
b) Dihybrid test cross
c) F₂ Phenotypic ratio of mono-hybrid cross
d) F₂ Phenotypic ratio of dihybrid cross
e) F₂ Genotypic ratio of monohybrid cross
f) F₂ Genotypic ratio of dihybrid cross.
Answer:
a) 1 : 1
b) 1 : 1 : 1 : 1
c) 3 : 1
d) 9 : 3 : 3 : 1
e) 1 : 2 : 1
f) 1 : 2 : 2 : 4 : 1 : 2 : 1 : 2 : 1

Question 7.
A diploid organism is heterozygous for 4 loci. How many types of gametes can it produce?
Answer:
Locus is a fixed position on a chromosome which is occupied by a single or more genes. A diploid organisms which is heterozygous at four loci, will have over different contratling characters at four different loci. For example the organisms have Aa, Bb, Cc, Dd then during meiosis, it will seggregate to form 8 separate gametes.

If the genes are not linked, then the diploid organism will produce 16 different gametes. However if the genes are linked, the gametes will reduce their number as the genes might be linked and the linked genes will be inherited togethers during the process of meiosis.

Question 8.
What is crossing over? In which stage of cell division crossing over occurs? What is its significance?
Answer:
Exchange of chromatid segments between two homologous chromosomes is called crossing over. It ccurs in pachytene sub stage of meiosis. It leads to the evolution of new genetic recombinations.

Question 9.
“Genes contain the information that is required to express a particular trait”. Explain.
Answer:
Genes are the functional segments of DNA, which is a part of chromosome, controls traits. If proteins work efficiently then trait get expressed in a better way, height depends on the release of growth hormones by proteins. Thus through proteins present in genes, genes controls the traits.

Question 10.
For the expression of traits, genes provide only the potentiality and the environment provides the opportunity. Comment on the varaeity of the statement.
Answer:
Phenotype = Genotype + Environment
Trait = (Potentiality) (Opportunity)

Question 11.
Two heterozygolis parents are crossed. If the two loci are linked what would be the distribution of phenotypic features in F₁ generation for a dihybrid cross?
Answer:

The above condition explained by using dihybrid crosses in drosophila were carried out by Morgan. He crossed yellow bodied (y) and white eyed (w) female with brown bodied (y⁺) and red eyed (w⁺) male, which produced F₁ with brown bodied red eyed female and yellow bodied white eyed male.

Question 12.
In peas, tallness is dominant over dwarfness and violet colour of flowers is dominant over the white colour. When a tall plant bearing violet flowers was pollinated with a dwarf plant bearing white flowers, different phenotypic groups were obtained in the progeny in numbers mentioned against them.
Tall, violet = 138
Tall, white = 132
Dwarf, violet = 136
Dwarf, white = 128
Mention the genotypes of the two parents and of the four offsprings.
Answer:
Tall plant violet flowers
Dwarf plant white flowers
Test cross:

Question 13.
How do genes and chromosomes share similarities from the view point of genetical studies?
Answer:
The genes are located on chromosomes so therefore they both contain genetic information.

Gene :
A gene is an ordered sequence of nucleotides located in a particular position on a particular chromosome that encodes a specific functional product.

Chromosome components in a cell that contain genetic information. Each chromosome contains genes. Chromosomes occur in pairs, one from mother and one from father.

Andhra Pradesh BIEAP AP Inter 2nd Year Botany Study Material 8th Lesson Viruses Textbook Questions and Answers.

AP Inter 2nd Year Botany Study Material 8th Lesson Viruses

Very Short Answer Questions

Question 1.
What is the shape of the T₄ phage? What is its genetic material?
Answer:
Tad pole-shape. Its Genetic material is double-stranded DNA.

Question 2.
What are virulent phages? Give an example.
Answer:
T-even phages that attack the bacterium E.coli cause lysis of the cells and are called virulent phages Eg: Bacteriophage.

Question 3.
What is lysozyme and what is its function?
Answer:
Lysozyme is an viral enzyme, which breaks or lysed the bacterial cell wall to release their newly produced phage particles.

Question 4.
Define ‘lysis’ and ‘burst size’ with reference to viruses and their effects on host cells.
Answer:
The cycle in which the host cell wall ruptures to release the virions is called lysis. The number of newly synthesized bacteriophage particles released from a single affected host cell is called burst size.

Question 5.
What is a prophage?
Answer:
The phage DNA which is incorporated into bacterial DNA is called prophage.

Question 6.
What are temperate phages? Give an example.
Answer:
The phage DNA gets integrated into the circular bacterial DNA, becomes part of it and remains latent. Such phages are called temperate phages. Eg : A [Lambda] phage.

Question 7.
Mention the difference between virulent phages and temperate phages.
Answer:

Question 8.
What is the shape of Tmv? What is its genetic material?
Answer:
The shape of Tmv is rod and genetic material is single stranded RNA consisting of 6500 nucleotides.

Short Answer Questions

Question 1.
What is ICTV? How are viruses named?
Answer:
International Committee on Taxonomy of Viruses [ICTV] regulates the norms of classification and nomenclature of viruses. The ICTV has only three hierarchial levels, The family, Genus and Species. The family names end with the suffix viridae’ while the Genus name ends with virus and the Species names are common English expressions . Viruses are named after the disease they cause.
Eg : Polio virus.

Question 2.
Explain the chemical structure of viruses.
Ans. All viruses consist of two basic components, a core of nucleic acid that forms the genome and the surrounding coat of protein known as capsid. The capsid gives shape to virus and also protects the genome. Capsid is made up of protein sub-units called capsomeres.

The nucleic acid may be either a double stranded DNA (ds DNA) or single stranded DNA [ss DNA]. In general, viruses that infect plants have ss RNA and viruses that infect animals have ds DNA. Most viruses have a single nucleic acid molecule but a few have more than one. Eg: HIV has two identical RNA molecules.

Question 3.
Explain the structure of TMV.
Answer:

1. It is a rod shaped virus. It is about’ 300 nm long and 18 or 19 nm in diameter with a molecular weight of 39 × 10⁶ daltons.
2. The capsid is made up of 2,130 sub-units called capsomeres.
3. Capsomeres are arranged in a helical manner around a central core of 4 nm.
4. Each protein sub-unit is made up of a single polypeptide chain with 158 amino acids.
5. Single stranded RNA is present inside the capsid and is spirally coiled.
6. RNA of TMV consists of 6,500 nucleotides.

Question 4.
Explain the structure of T-even Bacteriophages.
Answer:

1. The viruses which attack bacteria are called Bacteriophages. They were discovered by Twort (1915).
2. Felix’d’ Herelle (1917) coined the term Bacteriophage.
3. Bacteriophages are tadpole-shaped with a large head and a tail.
4. The head is hexagonal and is capped by hexagonal pyramid, measures about 65 × 95 nm.
5. The head is formed with several cap-someres, each of which is a single protein.
6. The head protein forms a semipermeable membrane enclosing the folded double stranded DNA which is 1000 times longer than the phage.
7. The tail is composed of several parts present around central core.
8. The tail core is 95 nm long and 8 nm in diameter. It is surrounded by tail sheath composed of 144 sub-units which are arranged in 24 rings of 6 each.
9. The head and tail are joined by collar whose function is not known.
10. At the tip of the tail, hexagonal tail plate is present with six tail pins and tail fibres.
11. The tail fibres help in attachment of the phage on to the host cell.

Question 5.
Explain the lytic cycle with reference to certain viruses.
Answer:
T-even Bacteriophages that attack the bacterium E.coli cause lysis of the cells and are called virulent phages. They show lytic cycle, which is a five step process involving (a) attachment (b) penetration (c) biosynthesis (d) maturation and (e) release

a) Attachment:
The tail fibres of phages help in attachment to the complementary receptor sites on the bacterial cell wall.

b) Penetration:
The tail sheath of phage contracts and the tail core is driven in through the bacterial cell wall. When the tip of the core reaches the plasma membrane, the DNA from the bacteriophage head passes through the tail Core through the plasma membrane and enters the bacterial cell. The capsid remains outside the bacterial cell and is called ghost. Thus the phage particle functions like hypodermic syringe and injects its DNA into the bacterial cell.

c) Biosynthesis:
When the phage DNA reaches the cytoplasm of the host cell, many copies of phage DNA, enzymes, and capsid proteins are synthesized using the cellular machinery of the host cell.

d) Maturation:
Bacteriophage DNA and capsids are assembled into complete virions. This period of time between the infection by a virus and the appearance of mature virions is called Eclipse period.

e) Release:
The plasma membrane of the host cell gets dissolved or lysed due to viral enzyme, lysozyme. The bacterial cell wall breaks, releasing the newly produced phage particles or virions.

Long Answer Questions

Question 1.
Write about the discovery and structural organization of viruses.
Answer:
In 1892, the Russian Pathologist Dmitri Iwanowski, while studying Tobacco mosaic disease, he filtered the sap of diseased tobacco leaf through filters and is injected into a healthy plant. He found the symptoms of Mosaic disease in it. Finally he reported that, a filterable agent was responsible for the disease. Later Martinus Beijerinck repeated Iwanowski’s experiments and concluded that, the disease causing agent was a contagious living fluid’ (contagium vivium fluidum).

W.M Stanley (1935) purified the sap and announced that the virus causing mosaic disease in tobacco could be crystallized and was named TMV. Fralnke conrael (1956) confirmed that the genetid material of TMV is RNA.

Structure:-
Viruses range in size from 300 nm as in TMV to 20 nm as in Parvoviruses. All viruses consist of two basic components, a core of nucleic acid that forms the genome and the surrounding coat of protein known as capsid. The capsid gives shape to the virus and provides a protective covering for the genome. It is made up of protein sub-units called, capsomeres. A virus contains its genetic information in either a double stranded (ds) DNA or single stranded (ss) DNA. In general, phytophages have ss RNA and zoophages have ds DNA. Bacteriophages are usually ds DNA viruses.

Shape:-
a) Helical viruses resemble long rods.
Eg: Rabies Virus, Tobacco Mosaic Virus.

b) Polyhedral viruses. They resemble polyhedral shape (many sided)
Eg : Herpes simplex and polio viruses.

c) Enveloped viruses: The capsid is covered by an envelope which are roughly spherical.
Eg : Influenza virus.

d) Complex viruses: Viruses which infect bacteria have complicated structures.
Eg : Bacteriophages have polyhedral symmetry in the head and helical symmetry in the tail sheath.

Question 2.
Describe the process of multiplication of viruses.
Answer:
T-even Bacteriophages that attack the bacterium E.coli cause lysis of the cells and are called virulent phages. They show lytic cycle, which is a five step process involving (a) attachment (b) penetration (c) biosynthesis (d) maturation and (e) release

a) Attachment:
The tail fibres of phages help in attachment to the complementary receptor sites on the bacterial cell wall.

b) Penetration:
The tail sheath of phage contracts and the tail core is driven in through the bacterial cell wall. When the tip of the core reaches the plasma membrane, the DNA from the bacteriophage head passes through the tail core through the plasma membrane and enters the bacterial cell. The capsid remains outside the bacterial cell and is called ghost. Thus the phage particle functions like hypodermic syringe and injects its DNA into the bacterial cell.

c) Biosynthesis:
When the phage DNA reaches the cytoplasm of the host cell, many copies of phage DNA, enzymes, and capsid proteins are synthesized using the cellular machinery of the host cell.

d) Maturation:
Bacteriophage DNA and capsids are assembled into complete virions. This period of time between the infection by a virus and the appearance of mature virions is called Eclipse period.

e) Release:-
The plasma membrane of the host cell gets dissolved hr lysed due to viral enzyme, lysozyme. The bacterial cell wall breaks, releasing the newly produced phage particles or varions.

Some Bacteriophages such as λ(Lambda) phages do not cause lysis and death of host cell when they multiply. Instead, the phage DNA, upon penetration into an E.coli cell gets integrated in to the circular bacterial DNA, becomes part of it and remains latent. Such phages are called temperate phages. The inserted phage DNA is now called prophages.

The prophages replicate along with the bacterial genetic material. The prophage remains latent with in the progny cells. In some rare spontaneous events, or when the host cell is exposed to UV light or some chemicals the phage DNA separates from the bacterial genetic material leading to the initiation of the lytic cycle. This lysogenic cycle facilitates the phenomenon of transduction.

Intext Questions

Question 1.
When discussing the multiplication of viruses, virologists prefer to call the process as replication, rather than reproduction. Why?
Answer:
Viruses must invade a host cell and take over the hosts metabolic machinery. So it is called replication.

Question 2.
In dealing with Public Health, the approach to deal with bacterial diseases is treatment. Canyou guess the nature of the general Public Health approach to viral diseases? What example do you cite to support your answer?
Answer:
“Prevention is better than cure” – in AIDS.

Andhra Pradesh BIEAP AP Inter 2nd Year Botany Study Material 7th Lesson Bacteria Textbook Questions and Answers.

AP Inter 2nd Year Botany Study Material 7th Lesson Bacteria

Very Short Answer Questions

Question 1.
Write briefly on the occurrence of microorganisms.
Answer:
Microorganisms are omnipresent in vast numbers and are described as ubiquitous. Among them, Bacteria are found in soil, water, air and on or inside living organisms. They also occur in a variety of foods, can withstand extreme cold, heat and drought conditions.

Question 2.
Define Microbiology.
Answer:
Microbiology is a branch of biology that deals with the scientific study of microorganisms. That are too small to be seen with the naked eye.

Question 3.
Name the Bacteria which is a common inhabitant of human intestine. How is it used in biotechnology?
Answer:
Escherichia coli (E.coli). It is used in biotechnology for the production of Insulin hormone.

Question 4.
What are Pleomorphic Bacteria? Give an example.
Answer:
The Bacteria which change their shape depending up on the type of environment and nutrients available are called Pleomorphic Bacteria.
Eg : Acetobacter.

Question 5.
What is sex pilus? What is its function?
Answer:
Sex pilus is a conjugation apparatus that pulls two cells together prior to DNA transfer. It help in binding two conjugants and also help in passing F plasmid from F⁺ cell to F^– cell.

Question 6.
What is a genophore?
Answer:
Genophore is the main genetic material of Bacteria which is circular, double stranded DNA without histone proteins. It is also called Bacterial chromosome.

Question 7.
What is a plasmid? What is its significance?
Answer:
“Plasmid is a small self-replicating circular, double stranded DNA molecule present in a Bacteria in addition to main genetic material”. They are used as agents (vectors) in modem Genetic Engineering techniques.

Question 8.
What is conjugation? Who discovered it and in which organism?
Answer:
Transfer of DNA from one Bacterium (Donor) to the recipient Bacterium through direct contact is called conjugation. It was first observed in Escherichia coli by Lederberg and Tatum (1946).

Question 9.
What is transformation? Who discovered it and in which organism?
Answer:
Uptake of naked DNA fragments from the surrounding environment and their incorporation into the recipient cell is called transformation. It was discovered by Frederick Griffith in Streptococcus pneumoniae (1928).

Question 10.
What is transduction? Who discovered it and in which organism?
Answer:
The transfer of genetic material from one bacterium to another through bacteriophage is known as transduction. It was observed by Lederberg and Zinder in Salmonella typhimurium (1951).

Short Answer Questions

Question 1.
How are bacteria classified on the basis of Morphology?
Answer:
Bacteria are classified into four types based on their shape. They are
I) Cocci : Spherical Bacteria.
Based on the number of cells and their arrangement, the Cocci are classified into
(a) Monococcus : A single cell
(b) Diplococcus : A pair of cells
(c) Tetracocci : A group of four cells
(d) Streptococcus : A linear chain of cells
(e) Staphylococci : A bunch of cells
(f) Sarcina : A group of 8 cells, arranged in cubes

II) Bacilli :
Rod-shaped Bacteria : The Bacillus forms exist as
a) Monobacillus : A single rod-shaped ceil.
b) Diplobacilli : A pair of rod-shaped cells.
c) Streptobacilli : A chain of rod-shaped cells.

III) Spirillum Helical rod-shaped cell.
It may be a distinct helical shape (spirillum) or Slender long and cork-screw shaped (Spirochete).

IV) Vibrio : Comma-shaped Bacteria.

Question 2.
How are Bacteria classified on the basis of number and distribution of flagella?
Answer:
Bacteria are classified into four types based on the number and arrangement of flagella.
They are
1) Monotrichous : A single polar flagellum is present at one end of Bacterial cell.
2) Lophotrichous : A tuft of flagella at one pole of the cell.
3) Amphitrichous : A single flagellum at each end of the cell.
4) Peritrichous : Flagella distributed over the entire cell.

Question 3.
What are the nutritional groups of Bacteria based on their source of .energy and carbon?
Answer:
1) Photoautotrophs :
They are photosynthetic i.e., they capture light energy and transform it into chemical energy and obtain carbon from atmospheric CO₂.
Eg: Chlorobium, Chromatium,

2) Chemoautotrophs :
They derive energy from the oxidation of inorganic substances and carbon from atmospheric CO₂.
Eg: Nitrosomonas, Nitrobacter, Beggiaotoaand Methanogens.

3) Photoheterotrophs :
They capture light energy and carbon from other organic sources.
Eg : Rhodospirillum and Rhodopseudomonas.

4) Chemoheterotrophs:
They derive both carbon and energy from organic compounds.
Eg: Xanthomonas, Salmonella.

Question 4.
Explain the conjugation in Bacteria.
Answer:
Transfer of Genetic material between two live Bacteria is called conjugation. It was first observed in 1946 by Lederberg and Tatum in Escherichia coli.

In E.coli, a small circular DNA strand occurs in the cytoplasm in addition to nucleoid called on F plasmid. The cell with F plasmid is called F⁺ cell and without F plasmid is called F^– cell. The F⁺ cell or Donor cell produces the sex pilus that makes contact with the recipient cell or F^– cell. During conjugation, F⁺ and F^– cells bind with each other with the help of sex pilus forms a bridge between them. The F plasmid replicates and the replicated DNA passes through bridge to the F^– cell. The F^– cell becomes F⁺ cell as it receives the F plasmid. After conjugation, the two cells separate from each other.

Long Answer Questions

Question 1.
Explain different methods of sexual reproduction in Bacteria.
Answer:
True sexual reproduction is absent in Bacteria. However, the exchange of genetic material takes place in three ways. They are 1) Conjugation 2) Transformation 3) Transduction

1) Conjugation:-
Transfer of Genetic material between two live Bacteria is called conjugation. It was first observed in 1946 by Lederberg and Tatum in Escherichia coli.

In E.coli, a small circular DNA strand occurs in the cytoplasm in addition to nucleoid called on F plasmid. The cell with F plasmid is called F⁺ cell and without F plasmid is called F^– cell. The F⁺ cell or Donor cell produces the sex pilus that makes contact with the recipient cell or F cell. During conjugation, F⁺ and F^– cells bind with each other with the help of sex pilus forms a bridge between them. The F plasmid replicates and the replicated DNA passes through bridge to the F^– cell. The F^– cell becomes F⁺ cell as it receives the F plasmid. After conjugation, the two cells separate from each other.

2) Transformation:-
It is uptake of naked DNA fragments from the surrounding environment and their incorporation into the recipient chromosome in a heritable form is known as transformation. It was discovered by Frederick Griffith (1928) in Streptococcus pneumoniae.

3) Transduction:-
The transfer of genetic material from one Bacterium to another through Bacteriophage is known as transduction. It was discovered by Lederberg and Zinder in Salmonella typhimurium.

Question 2.
“Bacteria are friends and foes of man” – Discuss.
Answer:
Many Bacteria are directly or indirectly beneficial and harmful to humans. Hence they are considered both as friends and foes of man.
1) Beneficial activities:-
Some Bacteria are important to man in the field of agriculture, industry, medicine and biotechnology.

I) Agriculture :
Bacteria play a significant role in maintaining the fertility of the soil.

They are :
a) Ammonifying Bacteria:
These Bacteria convert the proteins, and amino acids of the dead bodies into Ammonia. This process is called Ammonification.
Eg: Bacillus
b) Ammonia is oxidised to nitrates, by nitrifying bacteria, This process is called Nitrification Eg: Nitrosomonas, Nitrobacter.
c) Symbiotic Bacteria like Rhizobium, non-symbiotic Bacteria like Clostridium and photosynthetic Bacteria like Rhodospirillum, Rhodomicrobium and Chlorobacterium fix the atmospheric gaseous nitrogen and enrich the soil.
d) Bacillus thuringiensis is used as an agent against insect pests. Hence popularly used as a bioinsecticide.

II) Industry:
Industrially, Bacteria are employed in a number of processes. They are :
a) Clostridium butyricum and C. felcinium are used in retting of sunhemp and flax respectively to extract the fibre.
b) Some Bacteria are used in tanning industry.
c) Bacteria such as Bacillus megaterium and Micrococcus are used in curing of tobacco and tea respectively.
d) Some Bacterm are used in fermentation process, Eg: Lactobacillus.
e) Bacteria like Methanococcus and Methanobacillus ferment the dung anaerobically and produce methane that is commonly referred to as gobar gas.
f) In various chemical industries, Bacteria are extensively used for the production of chemicals. Few, examples are given below.

III) Medicine:
a) Corynebacterium glutamicum produces lysine, an essential amino acid.
b) A large number of antibiotics are produced by Bacteria. Species of Streptomyces and Bacillus produce important antibiotics as mentioned below.

IV) Biotechnology:
a) With the help of recombinant DNA technology, it was made possible to produce insulin hormone from Escherichia coli.
b) Some Bacteria store large quantities of proteins inside their cells. These are used as a source of single cell protein (SCP).
Eg : Brevibatiterium.
c) Agrobacterium tumefaciens is used as vector in genetic engineering.

2) Harmful activities :
A few saprophytic and all parasitic Bacteria carry on some processes which are harmful to man.

i) Spoilage of food materials :
Bacteria grow on different types of food materials and render them unsuitable for human consumption. Some of these Bacteria produce powerful toxins while growing on the food materials. For instance, Clostridium botulinum releases a very potential toxin, ‘botulin’ which causes ’botulism’, a type of food poisoning.

ii) Plant diseases :
A number of species of Bacteria are reported to cause different plant diseases as mentioned below.

Some important crop diseases caused by Bacteria

Some Bacteria cause Human diseases. They are :

Intext Questions

Question 1.
Many people believe that bacteria do little more than cause human illness and infectious diseases. How does the information in the chapter help you correct that misconception?
Answer:
Bacteria are directly or indirectly beneficial to Humans. Hence they are considered both as “friends and foes of man”.

Question 2.
An organism is described as a peritrichous bacillus. How might you translate this bacteriological language into a description of the organism.
Answer:
It is Rod shaped Bacteria.

Andhra Pradesh BIEAP AP Inter 2nd Year Botany Study Material 6th Lesson Plant Growth and Development Textbook Questions and Answers.

AP Inter 2nd Year Botany Study Material 6th Lesson Plant Growth and Development

Very Short Answer Questions

Question 1.
Define plasticity. Give an example.
Answer:
Ability of plants to follow different pathways in response to the environment or phases of life to form different kinds of structures in called plasticity.
Ex : Heterophily.

Question 2.
What is the disease that formed the basis for the identification of gibberellins in plants? Name the causative fungus of the disease.
Answer:
Bakane (foolish seedling) disease in rice seedlings. It is caused by a fungal pathogen Gibberella fujikuroi.

Question 3.
What is apical dominance? Name the growth hormone that causes it.
Answer:
Growing Apical bud inhibits, the growth of Axillary buds is called Apical dominance. It is caused by Auxins.

Question 4.
What is meant by bolting ? Which hormone causes bolting?
Answer:
Sudden elongation of internodes prior to flowering is called bolting. It is caused by Gibberellins.

Question 5.
Define respiratory climactic. Name the PGR associated with it.
Answer:
The rise in the rate of respiration during the ripening of fruits is known as respiratory climatic. It is associated by ethylene.

Question 6.
What is Ethephon? Write its role in agricultural practices.
Answer:
It is an Ethylene releasing chemical formulation. It hastens fruit ripening in tomatoes and apples and accelerates abscission in flowers and fruits. It promotes female flowers in cucumbers, there by increasing the yield.

Question 7.
Which of the PGRs is called stress hormone and why?
Answer:
ABC [Abscisic acid] is called stress hormone. ABA stimulates the closure of stomata in the epidermis and increases the tolerance of plants to various kinds of stresses.

Question 8.
What do you understand by vernalisation?
Answer:
The method of inducing flowering quantitatively or qualitatively on exposure to low temperature is called vernalisation. It prevents precocious reproductive development late in the growing season, and enables the plant to have sufficient time to reach maturity. It specially refers to the promotion of flowering. It also stimulates a subsequent photoperiodic flowering response in biennials, [cabbage, carrot]

Question 9.
Define the terms quiescence and dormancy.
Answer:
The condition of a seed when it is unable to germinate only because favourable external condition normally required for growth are not present is called quiescence.

The condition of a seed when it fails to germinate because of internal conditions even though external conditions are suitable is called dormancy.

Short Answer Questions

Question 1.
Write a note on agricultural / horticultural applications of auxins.
Answer:

1. IBA, NAA and IAA help to initiate rooting in stem cuttings, widely used for plant propagation in horticulture.
2. Auxins like 2,4-D, 2,4,§-T acts as herbicides and kills broad leaved dicot weeds to prepare weed – free lawns.
3. Auxins stimulates fruit growth. Ex : Tomatoes.
4. Auxins induces flowering in pineapple.
5. Prevents pre harvest fruit drop.

Question 2.
Write the physiological responses of Gibberellins in plants.
Answer:

1. Gibberellins delay senescence. Thus fruits can be left on the tree longer so as to extend the market period. .
2. Spraying of Gibberellins on sugarcane crop, increases the length of the stem, thus increasing the yield as much as 20 tonns per acre.
3. GA hastens the maturity period of conifers thus leading to early seed production.
4. GA also promotes bolting in cabbages, beet etc.,
5. They also produce parthenocarpic fruits in grapes and tomato.
6. Gibberellins favour the formation of male flowers in cucurbita.

Question 3.
Write any four physiological effects of cytokinins in plants.
Answer:

1. Cytokinins induces cell division.
2. They help to produce new leaves, chloroplasts in leaves, lateral shoot growth and adventitious shoot formation.
3. Cytokinins help to overcome apical dominance.
4. They promote nutrient mobilisation which helps in the delay of senescence.
5. Cytokinins help in the opening of stomata by increasing the concentration of k+ ions in guard cells.

Question 4.
What are the physiological prodesses that are regulated by ethylene in plants?
Answer:

1. Ethylene promotes the ripening of fruits.
2. Ethylene promotes the senescence and abscission of leaves and flowers.
3. Ethylene breaks seed and bud dormancy, initiates germination in peanut seeds and sprouting of potato tubers.
4. Ethylene promotes rapid intemode/petiole elongation in deep water rice plants.
5. It also promotes root growth and root hair formation, thus helping plants to increase their absorption surface.
6. Ethylene is used to initiate flowering (mango) and for synchronising fruit set in pineapples.
7. It promotes female flowers in cucumbers, thereby increasing the yield.

Question 5.
Write short notes on seed dormancy.
Answer:
The inability of seed to germinate or grow is called dormancy. It may be due to either external factors or internal factors.

Internal factors:
1) Immature embryo :
The embryo has not reached morphological maturity to germinate.
Ex : Ranunculus.

2) Hard seed coat:
In fabacea members, seeds have hard seed coats which prevent uptake of oxygen or water. It can be broken by scarification in which the hard seed coat is ruptured or weakened.

3) Chemicals :
Seeds of some plants (tomato) contain chemical compounds which inhibit their germination.

External factors:
1) Low temperature treatment :
Many seeds (polygonum) with not germinate until they have been exposed to low temperatures in moist conditions in the presence of oxygen for weeks to months. The practice of layering the seeds during winter in moist sand and peat is called stratification or prechilling.

2) Seeds of many domestic plants may be limited only by lack of moisture or warm temperature.

Question 6.
Which one of the plant growth regulators would you use if you are asked to.
a) Induce rooting in a twig
Answer:
Auxins

b) Quickly ripen a fruit
Answer:
Ethylene

c) Delay leaf senescence
Answer:
Cytokinins

d) Induce growth in axillary buds
Answer:
Cytokinins

e) ‘Bolt’ a rosette plant
Answer:
Gibberellins

f) Induce immediate stomatal closure in leaves
Answer:
Abscisic acid

g) Overcome apical dominance
Answer:
Cytokinins

h) Kill dicotyledonous weeds
Answer:
Auxins – 2, 4 – D

Question 7.
Describe briefly, a) Sigmoid growth curve, b) Absolute and relative growth rates.
Answer:

a) Sigmoid Growth Curve : It consists of 3 phases namely
1) lag phase
2) log pase
3) stationary phase.

1) Lag phase :
Growth is slow

2) Log phase :
Growth increases rapidly at an exponential rate

3) Stationary phase :
Limited nutrient supply, slows down the growth leading to a stationary phase. If we plot the parameter of growth against time, we get a typical sigmoid or S-curve.

b) Absolute and relative growth rates :
Measurement and comparision of the total growth per unit time is called absolute growth rate. The growth of the given system per unit time expressed on a common basis, e.g., per unit initial parameter is called relative growth rate.

Long Answer Question

Question 1.
List five natural plant growth regulators. Write a note on discovery, physiological functions and agricultural / horticultural applications of any one of them.
Answer:
Auxins, Gibberellins, Cytokinins, Abscisic acid, Ethylene

Auxins:
Discovery :
Observation of Charles Darwin and his son Francis Darwin, that the coleoptiles of canary grass responded to unilateral illumination by growing towards the light. It was concluded that the tip of the coleoptile was the site of transmittable influence that caused the bending of the entire coleoptile. Auxin was isolated by F.W. Went from the tips of coleoptiles of oat seedlings.

Physiological functions :

1. Auxins help of initiate rooting in stem cuttings.
2. Auxins promote flowering in pineapples.
3. They help to prevent fruit and leaf drop at early stages but promote the abscission of older mature leaves and fruits.
4. Auxins promote apical dominance.
5. Auxins also induce parthenocarpy in tomatoes.
6. They are widely used as herbicides 2,4-D widely used to kill dicotyledonous weeds.
7. Auxins also controls zylem differentiation and help in cell division.

Agricultural / Horticultural applications ;

1. Auxins help to initiate rooting in stem cuttings, widely used for plant propagation in Horticulture.
2. Auxins, 2, 4-D widely used as herbicide ie, kills dicotyledonous weeds to prepare a weed free-lawns.

Intext Questions

Question 1.
Fill in the blanks with appropriate word/words.
a) The phase in which growth is most rapid is ………… .
b) Apical dominance as expressed in dicotyledonous plants is due to the presence of more ………… in the apical bud than in the lateral ones.
c) In addition to ouxin, a …………. must be supplied to the culture medium to obtain a good callus in plant tissue culture.
d) ………….. of vegetative plants are the sites of photoperiodic perception.
Answer:
a) LOg phase or exponential phase
b) Auxins
c) Cytokinin
d)Shootapices of plants

Question 2.
Gibberellins promote the formation of Male flowers on genetically Modified (dwarf) plants in cannabis whereas Ethylene promotes formation of Female flowers on genetically Modified (dwarf) plants.

Question 3.
Classify the following plants into long day plants(LDP), short day plants(SDP) and Day Neutral plants(DNP). Xanthium, Spinach, Henbane (Hyoscyamus), Rice, Strawberry, Bryophyllum, niger, Sunflower, Tomato, Maize.
Answer:
Long day plants – Henbane, Spinach,
Short day plants – Xanthium, Sunflower, Rice, Tomato
Day nutral plants : Maize, Bryophyllum

Question 4.
A farmer grows cucumber plants in his field. He wants to increase the number of female flowers. Which plant growth regulator can be applied to achieve this?
Answer:
Ethyline.

Question 5.
Where are the following hormones synthesized in plants?
a) IAA b) Gibberellins c) Cytokinins.
Answer:
a) IAA – growing apices of the stems and roots.
b) Gibberellins : Fungal pathogen, Gibberella fujikuroi.
c) Cytokinins – roots apices, shoot…, young fruits.

Question 6.
Light plays an important role in the life of all organisms. Name any three physiological processes in plants which are influenced by light.
Answer:
Growth, differentiation and development.

Question 7.
Growth is one of the characteristics of all living organisms. Do unicellular organisms also grow? If so, what all the parameters?
Answer:
Yes, unicellar organisms also grow, some parameters are increase in fresh weight, length, area, volume and cell number.

Question 8.
Rice seedlings infected with fungus Gibberella fujikuroi are called Foolish seedlings. What is the reason?
Answer:
The fungus cause foolish seeding disease in rice which show, plant grow. Very tall, become pale, produce less tillers and less yield.

Question 9.
Why isn’t any one parameter good enough to demonstrate growth throughout the life of a flowing plant?
Answer:
One parameter is not good enough to demonstrate growth because various parts show growth… is estimated in different parameters Eg: Growth of the pollentube is measured interms of the length, increase in surface are dexoles growth in a dorsiventral leaf.

Question 10.
‘Both growth and differentiation in higher plants are open’ Comment.
Answer:
Both growth and diffentitation in higher plants are open because, cells/tissues arising out of the same meristem have different structures of maturity and their location. Eg. : Cells away from root apical meristem differentiate as root cap cells, while those pushed to the periphery nature as epidermis.

Question 11.
Both a short day plant and a long day plant can produce flowers simutaeneously in a given place’. Explain.
Answer:
5 me plants require the exposure to light for a period exceeding critical duration and some require a light for a period less than critical duration for flowing.

Question 12.
Would a defoliated plant respond to photoperiodic cycle? Why?
Answer:
No, because some hormonal substance migraterwom leaver to shoot apices for induce flowing.

Question 13.
What would be expected to happen if
a) GAj is applied to rice seedlings.
b) Dividing cells stop differentiating.
c) A rotten fruit gets mixed with unripe fruits.
d) You forget to add cytokinnin to culture medium.
Answer:
a) Appearance of disease symptoms (Bakane disease) in uninfected rice seedings.
b) Growth of the plant body in stopped.
c) Unripe fruits became ripened due to ethylene.
d) Initiation of heafy shoots is inhibited.

Andhra Pradesh BIEAP AP Inter 2nd Year Botany Study Material 5th Lesson Respiration in Plants Textbook Questions and Answers.

AP Inter 2nd Year Botany Study Material 5th Lesson Respiration in Plants

Very Short Answer Questions

Question 1.
Different substrates get oxidised during respiration. How does respiratory quotient (RQ) indicate which type of substrate (i.e) carbohydrate, fat or protein is getting oxidised?
RQ = A/B. What do A & B stand for?
What type of substrates have RQ of 1, <1; > 1?
Answer:

carbohydrates have RQ = 1
Fats have RQ value = < 1
Organic acids have RQ = > 1

Question 2.
What is the specific role of F₀ – F₁ particles in Respiration?
Answer:
The F₁ headpiece is a peripheral membrane protein complex and contains the site for the synthesis of ATP from ADP and inorganic phosphate. F₀ is an intergral membrane protein complex that forms the channel through which protons cross the inner membrane. The passage of protons through the channel is coupled to the catalytic site of the F₁ component for the production of ATP

Question 3.
When does anaerobic respiration occur in man and yeast?
Answer:
When oxygen is inadequate for cellular respiration, anaerobic respiration occurs.

Question 4.
What is the common pathway for aerobic and anaerobic respirations? Where does it take place?
Answer:
Glycolysis. It occurs in cytoplasm of the cell.

Question 5.
What cellular organic substances are never used as respiratory substrates?
Answer:
Pure proteins or fats are never used as respiratory substrates.

Question 6.
Why is the respiratory quotient (RQ) of fats less than that of carbohydrates?
Answer:
During the oxidation of fats, the amount of CO₂ evolved is less than the amount of O₂ consumed so respiratory quotient value is less than one. Where as in oxidation of carbohydrates, the amount of CO₂ evolved is equal to the amount of O₂ consumed. So the respiratory quotient value is one.

Question 7.
What is meant by Amphibolic pathway?
Answer:
Respiratory pathway involves both anabolism and catabolism so it is considered as amphibolic pathway. In this, the breakdown and the synthesis of fatty acids and proteins took place.

Question 8.
Name the mobile electron carriers of the respiratory electron transport chain in the inner mitochondrial membrane.
Answer:
Ubiquinone and cytochrome – C.

Question 9.
What is the final acceptor of electrons in aerobic respiration? From which complex does it receive electrons?
Answer:
Oxygen. It receives electrons from complex IV

Question 10.
Do you know of any step in Kreb’s cycle where there is a substrate level phosphorylation? Explain.
Answer:
In Kreb’s cycle during the conversion of succinyl – CoA to succinic acid a molecule of GTP is synthesised. This step is a substrate level phosphorylation.

Short Answer Questions

Question 1.
Why is the respiratory pathway referred to as amphibolic pathway? Explain.
Answer:
Respiration involves the breakdown of substrates so traditionally called catabolic process and the respiratory pathway as a catabolic pathway. In respiration, different substrates enter into respiratory pathway at different points. If fatty acids were respired, j they would be degraded to acetyl CoA and enter the pathway Glycerol would enter the pathway after converted to PGAL. The proteins and the aminoacids would enter the pathway at pyruvate.

Fatty acids would be brokendown to acetyl CoA before entering into respiratory pathway. Rut when the organism needs to synthesis fatty acids, Acetyl CoA would be withdrawn from the respiratory pathway for it. Hence the respiratory pathway comes into the picture both during the breakdown and the synthesis of fatty acids. In this issue, respiratory pathway is involved in both anabolism and catabolism, it would be better to consider it as amphibolic pathway rather than as a catabolic one.

Question 2.
Write about two ATP yielding reactions of glycolysis.
Answer:
1) 1, 3 BPGA (bisphosphoglyceric acid) looses phosphate group in the presence of phosphoglycerokinase to form 3-phosphoglyceric acid. ADP accepts phosphate group and gets converted to ATP.

2) Phosphoenol pyruvic acid undergoes dephosphorylation in the presence of pyruvic kinase results in the formation of pyruvic acid. ADP accepts phosphate group and gets converted to ATP.

Question 3.
The net gain of ATP for the complete aerobic oxidation of glucose is 36. Explain.
Answer:
Balance sheet of ATP production in aerobic oxidation of Glucose.
1) Glycolysis :
1. ATP produced by substrate level phosphorylation
Bisphosphoglyceric acid to phosphoglyceric acid : 2 × 1 = 2ATP
phosphoenol pyruvic acid to pyruvic acid : 2 × 1 = 2ATP
ATP consumed : for the phosphorylation of glucose
and fructose-6-phosphate : -2 ATP
Net gain of ATP : +2 ATP

2. ATP from NADH generated in glycolysis :
G-3-P to BPGA (2NADH, each worth 2ATP) 2 × 2 = 4ATP
Total ATP gain from glycolysis in the presence of O₂ : ^(a)6 ATP

2) Oxidative decarboxylation of pyruvic acid
Pyruvic acid to acetyl COA
(2 NADH, each worth 3 ATP) : ^(b)2 × 1 = 6 ATP

3) Krebs cycle
1. ATP produced in substrate level phosphorylation:
Succinyl CoA to succinic acid : 2 × 1 = 2 ATP
2. ATP from NADH: Isocitric acid to Oxalosuccinic acid : 2 × 3 = 6 ATP
a-Ketoglutaric acid to succinyl CoA : 2 × 3 = 6 ATP
Malic acid to Oxalocaetic acid : 2 × 3 = 6 ATP
3. ATP from FADH2 : Succinic acid to fumaric acid : 2 × 2 = 4 ATP
Total ATP value of krebs cycle : ^(a)24 ATP
Net gain of ATP in aerobic respiration per mole glucose
(a + b + c) : 36 ATP

Question 4.
Define RQ. Write a short note on RQ.
Answer:
The ratio of the volume of CO₂ evolved to the volume of O₂ cosumed in respiration is called the Respiratory Quotient (RQ).

The respiratory quotient depends upon the type of respiratory substrate used during respiration.
1) When carbohydrates are used as substrate, the RQ is 1, because equal amounts of CO₂ and O₂ are evolved and consumed.
Ex : C₆H₁₂O₆ + 6 O₂ → 6 CO₂ + 6 H₂O + Energy
RQ = \(\frac{6CO_2}{6O_2}\) = 0.1

2) When fats are used in respiration the RQ is less than 1.
Ex : 2(C₅₁ H₉₈ O₆) + 145 O₂ → 1 O₂ CO₂ + 98 H₂O + Energy Tripalmitin
RQ = \(\frac{102CO_2}{145O_2}\) = 0.7

3) When protein are respiratory substrates the ratio would be about 0.9

Question 5.
Describe briefly the process of fermentation.
Answer:
In fermentation, the incomplete oxidation of glucose is achieved under anaerobic conditions by set of reactions where pyruvic acid is converted to CO₂ & Ethanol. The ezymes, pyruvic acid decarboxylase, and alcohol dehydrogenase catalyse these reactions. Two types of fermetations are :

1. Alcoholic fermentation
2. Lactic acid fermentation

1) Alcoholic fermentation :
This is most common type of fermentation process. In alcoholic fermentation, pyruvic acid is broken to ethyl alcohol & CO₂ in two steps.
i) The pyruvic acid is decarboxylated to acetaldehyde in the presence of enzyme pyruvic decarboxylase.

ii) Acetaldehyde is then reduced to ethyl alcohol by NADH⁺ + H⁺ produced in glycolysis acetaldehyde is final hydrogen acceptor. The reaction is catalysed by alcohol dehydrogenase.

2) Lactic acid fermentation:
Pyruvic acid formed at the end of glycolysis is converted to lactic acid by lactobacillus lactibacteria.

Question 6.
Explain various complexes involved in electron transport system of respiration.
Answer:
Electron flow through the mitochondrial electron transport (Aerobic respiration) is carried out by five enzyme complexes. These complexes are the integral proteins of the inner mitochondrial membrane. They are complex I, II, III, IV & V out of these first four complexes involved in Electron transport chain.
i) Complex I :
NADH – dehydrogenase (or) NADH-Q-reductase. This complex I transfers electrons from NADH to ubiquinone.

ii) Complex II :
Succinate dehydrogenase (or) ubiquinone.This complex trasfers electrons from succinate to ubiquinone via Fe – S – centres.

iii) Complex III :
Cytochrome ‘c’ Reductase or cyt-b-c, complex. This complex contains cytochrome ‘b’ and cytochrome ’c’ cyt-c(mobile carrier). The ultimate electron acceptor of complex-III. This complex works through Q-cycle mechanism.

iv) Complex IV :
Cytochrome -c- oxidase. It contains two heme proteins called cyta and cyta₃ and two copper proteins. This complex transfers electrons to oxygen.

v) Complex V :
ATP synthase or F₀ – F₁ complex. This complex involved for the production of ATP from ADP & inorganic phosphate.

Question 7.
Describe the structure of Complex-V and explain the process of oxidative phospho-rylation as explained by chemosmotic hypothesis.
Answer:

Complex – V is an ATP synthase. This complex-V consists of two major components, F₁ and F₀. The F₁ head piece is a peripheral membrane protein complex and contains the site for synthesis of ATP from ADP and inorganic phosphate. F₀ is an integral membrane protein complex that froms the channel through which protons cross the inner membrane. The passage of protons through the channel is coupled to the catalytic site of the F₁ component for the production of ATP For each ATP produced, 3H⁺ passes through F₀ from the intermembrane space to the matrix down the electrochemical proton gradient.

In photophosphorylation light energy is utilised for the production of proton gradient required for phosphorylation, but in respiration it is the energy of oxidation reduction utilized for the proton gradient. It is for this reason that the process is called oxidative phosphorylation. Oxidation of one molecule of NADH gives rise to 3 molecules of ATP, while that one molecule of FADH₂ produces 2 molecules of ATP.

Long Answer Questions

Question 1.
Give an account of glycolysis. Where does it occur? What are the end products? Trace the fate of these products in both aerobic and anaerobic respiration.
Answer:
Glucose is broken down into 2 molecules of pyruvic acid is called glycolysis. It was given by gustav embden, Otto Mayerhof and J.Parnas so called EMP pathway. It occurs in the cytoplasm of the cell and takes place in all living orgnisms. In this, 4 ATP are formed of which two are utilised and 2 NADPH⁺ K⁺ are formed. A⁺ the end of glycolysis, 2PA, 2ATP and 2NADPH⁺ H⁺ are formed as end products. The ATP and NADPH⁺ H⁺ are utilised for fixation of CO₂.

Glycolysis occur in cytoplasm. Pyruvic acid, 2ATP, 2NADPH⁺H are the end products. In aerobic respiration, pyruvic acid, 2 NADPH + H⁺ are completely oxidised through TCA cycle, ETS pathway and produce 36 ATP molecules. In Anaerobic respiration, pyruvic acid is partially oxidised results in the formation of Ethyol alcohol and CO₂.

Question 2.
Explain the reactions of Krebs cycle.
Answer:

The acetyl CoA enters into the [mito chondrial matrix] a cyclic pathway tricarboxylic acid cycle, more commonly called krebs cycle after the scientist Hans Krebs who first elucidated it.

1) Condensation :
In this acetyl CoA condenses with oxaloacetic acid and water to yield citric acid in the presence of citrate synthetase and CoA is released.

2) Dehydration :
Citric acid looses water molecule to yield cisaconitic acid in the presence of aconitase.

3) Hydration :
A water molecule is added to cis aconic acid to yield isocitric acid in the presence of a conitase.

4) Oxidation I:
Isocitric acid undergoes oxidation in the presence of dehydrogenase to yield succinic acid

5) Decarboxylation :
Oxalosuccinic acid undergoes decarboxylation in the presence ofdecarboxylase to form a-keto glutaric acid.

6) Oxidation II, decarboxylation :
α – keto glutaric acid undergoes oxidation and decarboxylation in the presence of dehydrogenase and condenses with co.A to form succinyl co. A.

7) Cleavage :
Succinyl co.A splits into succinic acid and co.A in the presence of thiokinase to form succinic acid. The energy released is utilised to from ATP from ADP and PI.

8) Oxidation – III:
Succinic acid undergoes oxidation and forms Fumaric acid in the presence of succinic dehydrogenase.

9) Hydration :
A water molecule is alcohol to Fumaric acid in the presence of Fumarase to form Malic acid.

10) Oxidation IV :
Malic acid undergoes oxidation in the presence of malic dehydrogenase to form oxaloacetic acid.

In TCA cycle, for every 2 molecules of Acetyl co.A undergoing oxidation, 2 ATP, 8 NADPH+ H⁺, 2FADH₂ molecules are formed.

Intext Questions

Question 1.
Differentiate between
a) Respiration and Combustion
b) Glycolysis and Krebs cycle
c) Aerobic respiration and Fermentation
Answer:
a)

b)

c)

Question 2.
What are respiratory substrates? Name the most common respiratory substrate.
Answer:
Substances which oxidises in respiration are called respiratory substrates. Ex: Glucose, Fats, Proteins and Organic acids. Among them glucose is the common respiratory substrate.

Question 3.
Give the schematic representation of glycolysis.
Answer:

Question 4.
What are the main steps in aerobic respiration? Where does it take place?
Answer:
Glycolysis = Cytoplasm
PAOD = Matiix (Mitochondria)
Krebs cycle = Matrix (Mitochondria)
Electron transport system = Matrix (Mitochondria)

Question 5.
Give the schematic representation of an overall view of Krebs cycle.
Answer:

Question 6.
Explain ETS
Answer:
NADPH + H⁺ or FADH₂ formed in glycolysis and Mitochondrial matrix are oxidised through the electron transport system and the electrons are passed on to O₂ resulting in the formation of H₂O. The metabolic pathway through which an electron passes from one carrier to another is called electron transport system.

NADPH + H⁺ is oxidised and releases electrons in the presence of NADH dehydrogenase (complex I), which are transferred to ubiquinone located with in the inner membrane. Ubiquinone also receives reducing equivalents via FADH₂ (Complex II). The reduced ubiquinone (Ubiquinol) is oxidised and releases electrons which are accepted by cytochrome c₁ Cytochrome_c (Complex III) cytochrome C is a small protein, acts as a mobile carrier for the transfer of electrons between complex III and TV to cytochromes a and a₃, finally reaches \(\frac{1}{2}\)O₂ along with 2H⁺, produce 1 H₂O molecule.

During this electron flow, 10H⁺ moves from matrix to inner mitochondrial membrane [4 H⁺ at complex I, 4H⁺ at complex III, 2 at complex IV]. As a result H⁺ concentration increases towards the inner membrane of mitochondria. So the H⁺ comes back to the matrix side through ATPage [F₀, F₁], involves in the synthesis ATP.

Question 7.
Distinguish between the following
a) Aerobic respiration and Anaerobic respiration
b) Glycolysis and Fermentation
c) Glycolysis and Citric acid cycle
Answer:
a)

b)

c)

Question 8.
What are the assumptions made during the calculations of net gain of ATP?
Answer:
It is possible to make calculations of the net gain of ATP for every glucose is oxidised.
These calculations can be made only on certain assumptions that:

1. There is a sequential, orderly pathway functioning with one substrate forming the next and with glycolysis, TCA cycle, ETS pathway following one after another.
2. The NADH synthesised in glycolysis is transferred into the mitochondria and undergoes oxidative phosphorylation.
3. None of the intermediates in the pathway are utilised to synthesize any other compound.
4. Only glucose is being transferred – no other alternative substrates are entering in the pathway at any of the intermediary stages.

But these kind of assumption are not really valid in a living system all pathways work simultaneously and do not take place one after another.

Differences

Question 9.
Discuss “The respiratory pathway is an amplibolic pathway”.
Answer:
Respiration involves the breakdown of substrates so traditionally called catabolic process and the respiratory pathway as a catabolic pathway. In respiration, different substrates enter into respiratory pathway at different points. If fatty acids were respired, they would be degraded to acetyl CoA and enter the pathway Glycerol would enter the 1 pathway after converted to PGAL. The proteins and the aminoacids would enter the pathway at pyruvate.

Fatty acids would be brokendown to acetyl CoA before entering into respiratory pathway. Rut when the organism needs to synthesis fatty acids, Acetyl CoA would be withdrawn from the respiratory pathway for it. Hence the respiratory pathway comes into the picture both during the breakdown and the synthesis of fatty acids. In this issue, respiratory pathway is involved in both anabolism and catabolism, it would be better to consider it as amphibolic pathway rather than as a catabolic one.

Question 10.
Define respiratory quotient (RQ). What is its value for fats?
Answer:
The ratio of the volume of CO₂ evolved to the volume of O₂ consumed in respiration is called respiratory quotient.
Respiratory quotient value of fats is less than one
2C₅₁ H₉₈ O₆ + 145 O₂ → 1 02 CO₂ + 98 H₂O + energy
(Tripalmitin) Respiratory quotient = \(\frac{102}{145}\) = 0.7.

Question 11.
What is oxidative phosphorylation?
Answer:
Synthesis of ATP from ADP and Pi coupled to electron transport from substrate to molecular oxygen is called oxidative phosphorylation.

Question 12.
What is the significance of step-wise release of energy in respiration?
Answer:
The key is to oxidise glucose not in one step but in several small steps enabling some steps to be just large enough such that the energy is released can be coupled to ATP synthesis. All the energy entained in a substrate is not released free into the cell or in a single step. It is release in a series of slow step wise reactions controlled by enzymes, and it is trapped as chemical energy in the form of ATP.

13. Find the correct ascending sequence of the following, on the basis of energy released in respiratory oxidation.
a) 1 gm pf fat
b) 1 gm of protein
c) 1 gm of glucose
d) 0.5 gm of protein + 0.5 gm of glucose
Answer:
c → b → d → a

Question 14.
Name the products, respectively in aerobic glycolysis in skeletal muscle and anaerobic fermentation in yeast.
Answer:
In aerobic glycolysis :
Pyruvic acid, ATP, NADPH + H⁺

In skeletal muscle :
Lactic acid

In anaerobic fermentation :
CO₂ and Ethanol

Question 15.
If a person is feeling dizzy, glucose or fruit juice is given immediately but not a cheese sandwich, which might have more energy. Why?
Answer:
Glucose or fruit juice. They contain more sugars which oxidise in the presence of O₂ and releases more energy.

Question 16.
In a way green plants and cyanobacteria have synthesised all the food on earth. Comment.
Answer:
Only green plants and cyanobacteria can prepare their own food but not all food on earth.

Question 17.
It is known that red muscle fibres in animals can work for longer periods of time continuously. How is this possible?
Answer:
Red muscle fibres utilise fats or carbohydrates as their food. They can work for long time but with little force.

Andhra Pradesh BIEAP AP Inter 2nd Year Botany Study Material 3rd Lesson Enzymes Textbook Questions and Answers.

AP Inter 2nd Year Botany Study Material 3rd Lesson Enzymes

Very Short Answer Questions

Question 1.
How are prosthetic groups different from co-factors?
Answer:
Prosthetic groups are organic compounds that are tightly bound to the apoenzyme (protein part of the enzyme) whereas cofactors are non-protein parts of the holoenzyme.

Question 2.
What is meant by feedback inhibition?
Answer:
The end product of a chain of enzyme-catalyzed reactions inhibits the enzyme of the first reaction as a part of homostatic control of metabolism is called feedback inhibition.

Question 3.
W hy are ‘oxido reductases’ so named?
Answer:
Enzyme which catalyse oxidation and reduction of substrates usually involving hydrogen transfer are called oxido reductases.

Question 4.
Distinguish between apoenzyme and cofactor.
Answer:
The protein part of a holoenzyme is called apoenzyme the non-protein part of a holoenzyme is called cofactor.

Question 5.
What are competitive enzyme inhibitors? Mention one example.
Answer:
Substances which are closely resembles the substrate molecules and inhibits the activity of the enzyme are called competitive inhibitors.
Eg : Inhibition of succinic dehydrogenase by malonate which closely resembles the substrate succinate.

Question 6.
What are non-competitive enzyme inhibitors? Mention one example.
Answer:
The inhibitor has no structural similarity with the substrate and bind to an enzyme of locations other than the active sites so that the globular structure of the enzyme is changed are called non-Competitive enzyme inhibitors.
Eg : Metal ions of Copper, Mercury.

Question 7.
What do the four digits of an enzyme code indicate?
Answer:
In enzyme code, the first digit indicates the major class of the enzyme the second digit and the third digit indicate sub class and sub-sub class respectively. The last digit of the code is the serial number of the enzyme.

Question 8.
Who proposed ‘Lock and Key hypotheses’ and ‘Induced fit hypothesis’?
Answer:
Lock and key hypothesis was proposed by Emil Fisher (1884). Induced fit hypothesis was proposed by Daniel E. Koshland (1973).

Question 9.
Define Michael’s constant?
Answer:
Substrate cencentration required to cause half the maximal reaction rate is termed as michalis menten constant (km).

Short Answer Questions

Question 1.
Write briefly about enzyme inhibitors.
Answer:
The chemicals that can shut off enzyme activity are called inhibitors. They are of 3 types.
1) Competitive inhibitors:
Substances which closely resemble the substrate molecules and inhibits the activity of the enzyme are called competitive inhibitors.
Eg : Inhibition of succinic dehydrogenase by malonate which closely resembles the substrate succinate.

2) Non-competitive inhibitors :
The inhibitors which have no structural similarity with the substrate and bind to an enzyme at locations other than the active sites, so that the globular structure of the enzyme is changed are called non competitive inhibitors.
Eg: Metal ions of Copper, Mercury.

3) Feedback inhibitors :
The end product of a chain of enzyme catalysed reactions inhibit the enzyme of the first reaction as a part of homoeostatic control of metabolism are called feed back inhibitors.
Eg : During respiration (Glycolysis) accumalation of Glucose-6 Phosphate occurs, it inhibits the Hexokinase.

Question 2.
Explain different types of co-factors.
Answer:
Cofactors are two types.
1) Metal ion cofactor :
Metallic cations get tightly attached to the apoenzyme are called metallo enzymes.
Eg : Cu⁺² cytochrome oxidase

2) Organic cofactors :
They are two types.
a) Coenzyme :
They are small organic molecules which are loosely associated with the apoenzyme.
Eg : Thiamine pyrophosphate, Vitamin B.

b) Prosthetic group :
They are the organic cofactors which are tightly bounded to the apoenzyme.
Eg : Haeme is the prosthetic group of enzyme peroxidase.

Question 3.
Explain the mechanism of enzyme action.
Answer:
The substrate ‘S’ has to bind the enzyme of its active site with in a given cleft. The substrate has to diffuse towards the active site leads to the formation of ES complex. This is called transition state structure. Very Soon, after the expected bond breaking /making is completed, the product is released from the active site. In other words, the structure of substrate gets transformed into the structure of products.

When we represent this pictorially through a graph by taking potential energy on y-axis and progress of reaction on x-axis, we notice the energy level difference between ‘S’ and ‘P’. If ‘P’ is at lower level than S, it is an exothermic reaction. One need flot supply energy to form the product. However, whether it is an exothermic or an endothermic reaction (Energy requiring reaction) the ‘S’ has to go through a much higher energy state or transition state.

The difference between average energy content of ‘S’ and that of this transition state is called activation energy. –

Each enzyme [E] has a substrate [S] binding site in its molecule so that a highly reactive enzyme substrate complex [ES] is produced. This complex is short lived and dissociates into its products [P] and the unchanged enzyme with an intermediate formation of the enzyme product complex [EP].
E+S → ES → EP → E+P

Formation of ES complex has been explained with lock and key hypothesis by Emil Fisher and much later with induced fit hypothesis by Daniel E. Koshland.
The catalytic cycle of an enzyme action ‘is

1. The substrate binds to the active site of the enzyme.
2. The binding of the substrate induces the enzyme to alter its shape, fitting more tightly around the substrate.
3. The active site of the enzyme breaks the chemical bonds of the substrate and the new enzyme product complex is formed.
4. The enzyme releases the products of the reaction and the free enzyme is ready to bind to another molecule of the substrate.

Intext Questions

Question 1.
Explain how pH effects enzyme activity with the help of a graphical representation.
Answer:

Enzymes generally function in a narrow range of pH. Each enzyme shows its highest activity at a particular pH called optimum pH.
Activity declines both below and above the optimum level.

Question 2.
Explain the importance of [ES] complex formation.
Answer:
Each enzyme [E] has a substrate [S] binding site in its molecule so that a highly reactive enzyme substrate complex [ES] is produced. This complex is short lived and dissociates into its products [P] and the unchanged enzyme with an intermediate formation of the enzyme product complex [EP].
E+S → ES → EP → E+P

Andhra Pradesh BIEAP AP Inter 2nd Year Botany Study Material 2nd Lesson Mineral Nutrition Textbook Questions and Answers.

AP Inter 2nd Year Botany Study Material 2nd Lesson Mineral Nutrition

Very Short Answer Questions

Question 1.
Define hydroponics.
Answer:
The technique of growing plants in a specified nutrient solution in the complete absence of soil is known as hydroponics.

Question 2.
How do you categorize a particularly essential element as a macro or micronutrient?
Answer:
The elements which are present in large amounts i.e., in excess of 10m mole kg^-1 of dry matter are called Macronutrients. The elements which are needed in small amounts i.e., less than 10m mole kg^-1 of dry matter are called micronutrients.

Question 3.
Give two examples of essential elements that act as activators for enzymes.
Answer:
Molebdinum, Magnesium and Zn⁺².

Question 4.
Name the. essential mineral element that play an important role in photolysis of water.
Answer:
Calcium and Manganese.

Question 5.
Out of the 17 essential elements, which elements are called non-mineral essential elements?
Answer:
Carbon, Oxygen, Hydrogen and Nitrogen.

Question 6.
Name two amino acids in which sulphur is present.
Answer:
Cysteine and Methionine.

Question 7.
When is an essential element said to be deficient?
Answer:
The element said to be deficient when present below the critical concentration [The Cone, of the essential elements below which plant growth is retarded is termed as critical concentration].

Question 8.
Name two elements whose symptoms of deficiency first appear in younger leaves.
Answer:
Sulphur and Calcium.

Question 9.
Explain the role of the pink colour pigment in the root nodule of legume plants. What is it called?
Answer:
The enzyme nitrogenase is highly sensitive to the molecular oxygen. It is protected from oxygen by pink colour pigment cailed an oxygen scavenger, leg haemoglobin.

Question 10.
Which element is regarded as the 17^th essential element? Name the disease caused by its deficiency.
Answer:
Nickel. It’s deficiency cause Mouse ear in pecan. [Slightly wrinkled leaves round or blunt leaflets].

Question 11.
Name the essential elements present in nitrogenase enzyme. What type of essential elements are they?
Answer:
Nitrogenase enzyme contains Mo and Fe elements. They are Micronutrients.

Question 12.
Write the balanced equation of nitrogen fixation.
Answer:
N₂ + 8H⁺ + 8e^– + 16 ATP → 2NH₃ + H₂ + 16ADP + 16 Pi

Question 13.
Name any two essential elements and the deficiency diseases caused by them.
Answer:
Chlorosis occurs due to the deficiency of N, K, Mg, S, Fe, Mn, Zn and Mo.
Neurosis occurs due to the deficiency of Ca, Mg, Cu, K.

Short Answer Questions

Question 1.
Explain the steps involved in the formation of root nodule.
Answer:

1. Roots of legumes release sugars, amino acids which attracted Rhizobium. They get attached to epidermal and root hair cells of the host.
2. The root hair curl and the bacteria invade the root hair.
3. An infection thread is produced, carrying the bacteria into cortex of the root.
4. Bacteria initiate nodule formation in the cortex of the root. Then the bacteria released from the thread into the cortical cells of the host and stimulate the host cells to divide. Thus leads to the differentiation of specialized nitrogen fixing cells.
5. The nodule thus formed establishes a direct vascular connection with the host for exchange of nutrients.

Question 2.
Write in brief, how plants synthesize amino acids.
Answer:
Amino acids are synthesized into two ways. They are
1) Reductive Amination :
In this ammonia reacts with a-ketoglutaric acid and forms glutamic acid.

2) Transamination :
In this, transfer of an amino group from an amino acid to the keto group of a keto acid. Glutamic acid is the main amino acid from which the transfer of NH₂ takes place and other amino acids are formed in the presence of transaminase.

Question 3.
Explain in brief, how plants absorb essential elements.
Answer:
The process of absorption can be demarcated into two main phases. In the 1^st phase, there is an initial uptake of ions into the free space or outer space of cells – the apoplast. It is a passive process. In the second phase of uptake, the ions are taken in slowly into the inner space – the symplast.

The passive movement of ions into the apoplast from the cell along the concentration gradient usually occurs through ion-channels. The entry or exit of ions to and from the symplast against the concentration gradient requires metabolic energy which is an active process.

Question 4.
Explain the Nitrogen cycle, giving relevant examples.
Answer:
The cyclic movement of nitrogen from the atmosphere to soil and from soil back into the atmosphere through plants, animals and micro-organisms is termed as nitrogen cycle. Nitrogen cycle is a continuous pathway and it involves five steps.

1. Nitrogen Fixation
2. Nitrogen Assimilation
3. Ammonification
4. Nitrification
5. Denitrification

1) Nitrogen Fixation :
The gaseous dinitrogen is fixed into inorganic nitrogenous substances is called nitrogen fixation. It occurs by two methods.
a) Abiological or physical method,
b) Biological method – Dia2otrophy.

a) Abiological or physical method :

1. It occurs in atmosphere. Due to thunders and lightening, dinitrogen is converted into nitric acid which is further oxidised to nitrogen dioxide.
2. These oxides dissolves in rain water and reaches soil as nitrous and nitric acids.
3. These acids react with alkali radicles of soil and form nitrates.
4. The soluble nitrates are directly absorbed by plants. The reactions of physical nitrogen fixation are as follows.
   1) N₂ + O₂ → 2NO
   2) 2NO + O₂ → 2NO₂
   3) 2NO₂ + H₂O → HNO₂ + HNO₃
   4) HNO₃ + Ca or K salts → Ca or K nitrates.
   5) A biological nitrogen fixation is carried put on industrial scale by Haber Bosch process at 0°C with 1000°C bars pressure.

b) Biological method – Diazotrophy :
1) Conversion of dinitrogen into nitrogenous compounds by prokaryotes is called biological nitrogen fixation. Such microbes are called as diazotrophs or nitrogen fixers.
Eg : Free-living bacteria – Azotobacter, Clostridium.
Symbiotic bacteria – Rhizobium in the root nodules of Fabaceae members.
Blue green algae – Nostoc and Anabaena.

2) Nitrogen Assimilation :

1. The process of absorbing nitrates, Ammonia to produce organic nitrogen constituents is called nitrogen assimilation. It is the second step of nitrogen cycle.
2. Nitrates and Ammonia formed in the first step are absorbed by plants and converted and constitute the organic nitrogen.
   3) When plants are eaten by animals, this Organic nitrogen is passed on into the animal body.

3) Ammonification :

1. The process of conversion of organic nitrogenous compounds from the dead bodies of plants and animals into ammonia is called Ammonification. It is mineralization process.
2. The bacteria responsible for this are called Ammonifying bacteria.
   Eg: Bacillus ramosus, B. vulgaris, B. mycoides.

4) Nitrification :

1. This is fourth step of nitrogen cycle. It is an oxidative and an exergonic process.
2. The conversion of ammonia into nitrites and nitrates by bacteria is called nitrification. Sucfybacteria are called nitrifying bacteria. It occurs in two steps.

A) In the step, Ammonia is converted into nitrites by bacteria like Nitrosomonas and Nitrococcus.
2NH₃ + 3O₂ → 2NO₂^– + 2H⁺ + 2H₂O

B) In the next step, the nitrites are further oxidised to nitrates by Nitrobacter.
2NO₂^– + O₂ → 2NO₃^–

5) Denitrification :
This is final step in nitrogen cycle. It is also exergonic process released around 11,000 cal of energy. Conversion of nitrates of soil into molecular dinitrogen is called denitrification. Denitrification occurs in four steps
NO₃^– → NO₂^– → NO → N₂

This process is brought about by denitrifying bacteria like Thiobacillus denitrificans, Pseudomonas denitrificans and Micrococcus denitrificans.

Intext Questions

Question 1.
Who should be credited for initiation of Hydroponics?
Answer:
Julius von sachs (1860).

Question 2.
Are all the essential elements required by plants mineral elements? Explain.
Answer:
No. based upon the criteria of essentiality, only a few elements have been found to be essential for plant growth and metabolism. They are micro elements and microelements.

Question 3.
Which essential element is needed to activate the enzymes required for CO₂ fixatin?
Answer:
Mg⁺².

Question 4.
Name a cation and anion that maintain osmotic balance in cells?
Answer:
K⁺ and Cl^–

Question 5.
Which element is required for the formation of mitotic spidle?
Answer:
Calcium

Question 6.
What is the role of sulphur in plant life?
Answer:
Sulphur is present in two amino acids, cysteine and methionine. It is the main constituent of several coenzymes, vitamins (Thiamine, biotin, coenzyme A) and ferredoxin. It forms disulphide bridges which help in stabilizing protein structure.

Question 7.
Which micro element is required in more quantify than die other micronutrients?
Answer:
Iron.

Question 8.
Which element is necessary for the synthesis of the chief photosynthetic pigment without being its structural component?
Answer:
Iron, Magnesium

Question 9.
Which micronutrient necessary for photolysis of water is absorbed by plants in anionic form?
Answer:
Chlorine (Cl^–)

Question 10.
Which enzyme is activated by the 17^th essential element?
Answer:
Nickel acts as an activator for urease.

Question 11.
When is an element considered to be toxic?
Answer:
Any mineral ion concentration in tissues that reduces the dry weight of tissues by about 10 percent is considered toxic.

Question 12.
Which element when supplied in excess leads to appearence of brown spots surrounded by chlorotic venis?
Answer:
Magnesium.

Question 13.
Name an anaerobic, free living, photo-hetrotrophic nitrogen fixing bacterium.
Answer:
Rhodospirillum.

Question 14.
Which microorganism produces nitrogen-fixing nodules in Alnus?
Answer:
Frankia.

Question 15.
When the cross section of root nodules of ground nut plants are observed under microscope, they appear pinkish, why?
Answer:
This is due to the presence of leguminous haemoglobin or leg-haemoglobin which is a pink coloured pigment and is also called oxygen scavenger.

Question 16.
Apart from the cortical cells, which other cells are stimulated to divide by the bacteroids inside the root nodules?
Answer:
Inner cortex and pericycle cells.

Question 17.
What is the ratio of electrons and protons required for the fixation of atmospheric molecular nitrogen through biological mode?
Ans. 8H⁺ + 8e^– = 1 : 1.

Question 18.
What acts as oxygen scavenger in the legume-root nodule combination?
Answer:
Leg-hemoglobin.

Question 19.
In what why does aspargine differ from aspartic acid?
Answer:
Aspargine contains more nitrogen than aspartic acid.

Question 20.
Through which tissue the amino acids are transported inside the plant body?
Answer:
Xylem vessels.

Question 21.
Plants like the Picture and Venesfly trap have special nutritional adaptations. Name the essential element and its source for which they show such adaptations.
Answer:
Nitrogen.

Question 22.
Excess ‘Mn’ in soils leads to deficiency of Ca, Mg and Fe. Justify.
Answer:
Manganese competes with iron and Magnesium for uptake and with magnesium for binding with enzymes. Mn also inhibits calcium translocation in the shoot apex. Therefore excess of Mn may infact, induce deficiencies of Ca, Mg and Fe.

Question 23.
What acts as a reservoir of essential elements for plants? By what process is it formed?
Answer:
Soil, it is formed due to weathering and breakdown of Rocks.

Question 24.
Nitrogen fixation is shown by prokaryotes only. Why not by eukaryotes?
Answer:
In prokaryotes, nitrogenase enzyme which is a Mo-Fe protein is present and is capable of nitrogen reduction. It is absent in Eukaryotes.

Andhra Pradesh BIEAP AP Inter 2nd Year Botany Study Material 1st Lesson Transport in Plants Textbook Questions and Answers.

AP Inter 2nd Year Botany Study Material 1st Lesson Transport in Plants

Very Short Answer Questions

Question 1.
What are porins? What role do they play in diffusion?
Answer:
Porins are proteins that form huge pores in the outer membranes of the Plastids, mitochondria, and some bacteria, allowing molecules upto the size of small proteins to pass through.

Question 2.
Define water potential. What is the value of the water potential of pure water?
Answer:
Water potential is defined as the chemical potential of water and a measure of energy for reaction or movement. The value of the water potential of pure water is zero.

Question 3.
Differentiate osmosis from diffusion.
Answer:

Question 4.
What are apoplast and symplast?
Answer:

Question 5.
How does ffuttation differ from transniration?
Answer:

Question 6.
What are the physical properties of water responsible for the ascent of sap through xylem in plants?
Answer:
1) Cohesion :
Mutual attraction between water molecules.

2) Adhesion :
Attraction of water molecules to polar surface (such as the surface of trachery elements).

3) Transpiration pull:
Driving force for upward movement of water.

Question 7.
With reference to transportation of food within a plant cells, what are the sources and sink?
Answer:
Source is the part of the plant which synthesise the food. Sink is the part of the plant which utilizes depending on the season or the plants needs.

Question 8.
Does transpiration occurs at night? Give an example.
Answer:
Yes. In succulent plants, stomata opens during the night and remain closed during the day time. (Scotoactive stomata) Eg: Bryophyllum, Cacti.

Question 9.
Compare the pH of guard cells during the opening and closing of stomata.
Answer:
Increase in the pH of guard cells leads to the opening of stomata. Decrease in the pH of guard cells leads to closure of stomata.

Question 10.
In the wake of transpirational loss, why do the C₄ plants are of more efficient than C₃ plant?
Answer:
C₄ plants minimise water loss and fixes more CO₂ than C₃ plants.

Question 11.
What is meant by transport saturation how does it influence facilitated diffusion.
Answer:
Transport rate reaches a maximum when all of the protein transporters are being used is called transport saturation. It allows cell to select substances for uptake.

Question 12.
How does ABA bring about the closure of stomata under water stress conditions?
Answer:
Under water stress condition, ABA, a natural anti-transpirant drive the K⁺ ion put of the guard cells making them close.

Question 13.
Compare imbibing capacities of pea and wheat seeds.
Answer:
Proteinaceous pea seeds swell more on imbibition than starchy wheat seeds, because proteins have very high imbibing capacities compared to carbohydrates.

Short Answer Questions

Question 1.
Define and explain water potential
Answer:
“The measure of the relative tendency of water to move from one area to another is called water potential”. It is denoted by the Greek symbol Psi or Ψ and is expressed in pascals (Pa). It has two components namely solute potential and pressure potential.

a) Solute potential:
If some solute is added to pure water, the solution has fewer free water molecules and the concentration of water decreases reducing its water potential. The magnitude of this lowering due to dissolution of a solute is called solute potential. It is denoted as Ψ_s. It is always negative.

b) Pressure potential :
When water enters a plant ceil due to diffusion, causing a pressure to build up against the cell wall. This makes the cell turgid. The magnitude of increase in water potential in such turgid cell is called pressure potential It is usually positive and is denoted as Ψ_p.
Ψ = Ψ_S + Ψ_P

Question 2.
Write short note on facilitated diffusion.
Answer:
Facilitated diffusion :
The diffusion rate depends on the size of the substances. Obviously smaller substances diffuse faster. The diffusion of any substance across a membrane also depends on its solubility in lipids, which also move faster. Substances that have a hydrophilic moiety find it difficult to move through the membrane. Their movement has to be facilitated by membrane proteins without utilising metabolic energy and there must be concentration gradient.

This is called facilitated diffusion. It cannot cause net transport of molecules from low to high concentration, this would require input of energy. Facilitated diffusion is very specific, it allows cell to select substances for uptake. It is sensitive to inhibitors which react with protein side chains.

Some proteins allow diffusion only if two types of molecules move together. In asymport both molecules cross the membrane in the same direction, in an antiport, they move in opposite directions. When a molecule moves across a membrane independent of other molecules, the process is called uniport.

Question 3.
What is meant by plasmolysis? How is it practically useful to us?
Answer:
Plasmolysis is the process in plant cells where the cytoplasm pulls away from the cell wall due to the loss of water through osmosis. This occurs when cell is placed in a hypotonic solution, water moves out causes the shrinkage of protoplast leading to the seperation of plasma membrane from the cell wall in the comers called incipient plasmolysis. The salting of pickles and preserving of fish and meat in salt are good examples of practical applications.

Question 4.
How does ascent of sap occur in tall trees?
Answer:
Upward movement of water through xylem against gravitational force is called ascent of sap. The transpiration driven ascent of xylem sap depends on (a) Cohesion – mutual – attraction between water molecules (b) Adhesion – attraction of water molecules to polar surfaces (c) transpiration pull – driving force for upward movement of water. These properties give water high tensile strength and high capillarity.

In plants capillarity is aided by the small diameter of the tracheary elements. As water evaporates through the stomata, Since the thin film of water over the cells is continuous, it results in pulling of water, molecule by molecule into the leaf from the xylem. Also, because of lower concentration of water vapour in the atmosphere, water diffuses into the surrounding air. This creats transpiration pull. The forces generated by the transpiration can create pressure sufficient to lift a xylem sized column of water over 130 metres high.

Question 5.
Explain pressure flow hypothesis of translocation of sugars in plants.
Answer:
Glucose is prepared at the source is converted to sucrose, then moved into the companion cells and then into the living phloem sieve tube cells by active transport. Water in the adjacent xylem moves into the phloem by osmosis. As osmotic pressure builds up, the phloem sap moves to the cells which will use the sugar converting it into energy, starch or cellulose. As sugars are removed, the osmotic pressure decreases and water moves out of the phloem.

To explain this, Munch conducted one experiment. In this, he took two osmometers A and B. In ‘B’ he took concentrated sugar solution and in A, he took pure water. These two bulbs A and B are connected by a tube, ‘C’. He then, placed A and B bulbs in X and Y water tubs which are connected by tube ‘Z’. Due to osmosis, water moves from A to B, B to A through ‘C and then A-X and finally X-Y through Z occurs until the solution becomes isotonic. In this he compared ‘B’ bulb as source, A’ bulb as sink, ‘C’ tube as phloem, X apd Y tubs are xylem. .

Question 6.
Transpiration is necessary evil’. Explain.
Answer:
Transpiration has both beneficial and harmful effects. They are

Beneficial effects :

1. It helps in passive absorption of water.
2. It also helps in passive absorption of mineral salts by Mass flow mechanism.
3. It is the main force for ascent of sap.
4. It regulates the temperature of plant body and provides cooling effect.

Harmful effects :

1. Excessive transpiration makes the cells flaccid which retards growth.
2. Excessive transpiration leads to closure of stomata thus obstructing gaseous exchange.
   Hence transpiration is considered to be a necessary evil.

Question 7.
Transpiration and photosynthesis – a compromise. Explain.
Answer:
Transpiration has more than one purpose; it

1. Creates transpiration pull for absorption and transportation in plants
2. Supplies water for photosynthesis
3. Transports minerals from the soil to all parts of the plant
4. Cools leaf surface
5. Maintains the shape arid structure of the plants by keeping the cells turgid.

An actively photosynthesising plant has an insatiable need for water. Photosynthesis is limited by available water which can be swiftly depleted by transpiration. C₄ photo- synthetic system is one of the strategies for maximising the availability of CO₂ and minimizing water loss. C₄ plants are twice efficient than C₃ plants in fixing carbon and also water loss.

Question 8.
Explain the mechanism of opening and closing of stomata.
Answer:

Mechanism of opening and closing of stomata :
Levitt (1974) proposed K⁺ pump theory to explain the opening and closing of stomata. According to this, K⁺ ions are accumulated into guard cells from the subsidiary cells in the presence of light. This coupled efflux of protons leads to increase in pH of the guard cells. K⁺ ion accumulation is associated with influx of Cl^– ions to decrease the water potential of guard cells. Water enters into guard cells making them turgid. The outer walls of the guard cells are thin, expand out- wardly with the help of microfibrils in the cell walls of guard cells resulting in opening of stomata.

At night, the K⁺ and Cl^– ions move out of the guard cells due to which the water potential of guard cells increases and water stands moving out of guard cells leading to the closure of stomata.

Under water stress conditions ABA (Abscisic acid), a natural antitranspirant drives the K⁺ ions out of guard cell making them close.

Intext Questions

Question 1.
Differentiate uphill and down till transport.
Answer:

Question 2.
Compare facilitated diffusion and simple diffusion.
Answer:

Question 3.
What happens when two solutions of different concentrations are separated by an egg membrane? State the reason.
Answer:
If two solutions are separated by an egg membrane, the solution moves from low concentrated solution to a high concentrated solution through egg membrane (semipermeable membrane) until the two solutions become equally concentrated. It is called osmosis.

Question 4.
In general in a plant which path of water movement is more and why?
Answer:
Transpiration pull. Generally due to transpiration, tension is created in mesophyll cells which withdraws water from stem → root → Root hair and finally from the soil. Thus water goes upwards in the form of a continuous column.

Question 5.
Why pinus seeds fail to germinate in the absence of mycorrhizae?
Answer:
Pinus plants have an obligate association with the mycorrhizae. The fungus only provides water and minerals. That is why seeds of pinus cannot germinate in the absence of mycorrhiza.

Question 6.
Why do stomata close under water stress conditions?
Answer:
In water stress conditions, ABA (Abscisie acid), a natural antitranspirants drives thfe K⁺ ions out of guard cells making the stomata close.

Question 7.
How are stomata distributed in a typical monocot plant?
Answer:
Stomata are equally distributed on both the surfaces in monocot also plant.

Question 8.
In what form the sugars are transported through phloem?
Answer:
Sucrose, other sugars, hormones and amino acids.

Question 9.
Why does the root endodermis transport ions in one direction only?
Answer:
The walls of the endodermis are suberised (casparian strips) and is impervious to water. So water is directed to wall regions that are not suberised into the cells proper through membranes. So because of suberin, it has the ability to transport water in one direction only.

Question 10.
If a ring of bark is removed from an actively growing plant, what will happen and why?
Answer:
In the absence of downward movement of food, the portion of the bark above the ring on the stem becomes swollen after a few weeks. This shows that, phloem is the tissue responsible for translocation of food.

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Differential Equations Solutions Exercise 8(e) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2B Differential Equations Solutions Exercise 8(e)

I. Find the I.F. of the following differential equations by transforming them into a linear form.

Question 1.
x\(\frac{dy}{dx}\) – y = 2x² sec² 2x
Solution:
x\(\frac{dy}{dx}\) – y = 2x² sec² 2x
\(\frac{dy}{dx}\) – \(\frac{1}{x}\) . y = 2x. sec² 2x
This is linear in x
I.F = e^{∫-\(\frac{1}{x}\)dx} = e^{-log x} = e^{log 1/x} = \(\frac{1}{x}\)

Question 2.
y\(\frac{dy}{dx}\) – x = 2y³
Solution:
y\(\frac{dy}{dx}\) – x = 2y³
\(\frac{dy}{dx}\) – \(\frac{1}{y}\).x = 2y²
I.F = e^{∫-\(\frac{1}{x}\)dx} = e^{-log y} = e^{log 1/y} = \(\frac{1}{y}\)

II. Solve the following differential equations.

Question 1.
\(\frac{dy}{dx}\) + y tan x = cos³ x
Solution:

Solution is 2y = x cos x + sin x. cos² x + c. cos x

Question 2.
\(\frac{dy}{dx}\) + y sec x = tan x
Solution:
I.F. = e^{∫sec x dx} = e^{log(secx + tan x)} = sec x + tan x
y. (sec x + tan x)
= ∫tan x (sec x + tan x)dx
= ∫(sec x. tan x + tan² x)dx
= ∫(sec x tan x + sec² x – 1)dx
Solution is
y(sec x + tan x) = sec x + tan x – x + c

Question 3.
\(\frac{dy}{dx}\) – y tan x = e sec x. dx
Solution:
I.F. = e^{-∫tan x dx} = e^{log cos x} = cos x
y. cos x = ∫e^x.sec x. cos x dx = ∫ e^x dx
= e^x + c

Question 4.
x\(\frac{dy}{dx}\) + 2y = log x.
Solution:
I.F. = e^{∫\(\frac{2}{x}\)dx} = e^{2log x} = e^{log x²} = x²
Solution is

Question 5.
(1 + x²)\(\frac{dy}{dx}\) + y = e^{tan-1 x}
Solution:

Question 6.
\(\frac{dy}{dx}+\frac{2y}{x}\) = 2x².
Solution:
I.F. = e^{∫\(\frac{2}{x}\)dx} = e^{2log x} = e^{log x²} = x²
y. x² = ∫2x⁴ dx = \(\frac{2x^5}{5}\) + c

Question 7.
\(\frac{dy}{dx}+\frac{4x}{1+x^2}y=\frac{1}{(1+x^2)^2}\)
Solution:

Question 8.
x\(\frac{dy}{dx}\) + y = (1 + x)e^x
Solution:

Question 9.
\(\frac{dy}{dx}+\frac{3x^2}{1+x^3}y=\frac{1+x^2}{1+x^3}\).
Solution:

Question 10.
\(\frac{dy}{dx}\) – y = -2e^-x.
Solution:
I.F = e^∫-dx = e^-x
y. e^-x = -2∫e^-2x dx = e^-2x + c
y = e^-x + c. e^x

Question 11.
(1 + x²)\(\frac{dy}{dx}\) + y = tan^-1 x.
Solution:

Put t = tan^-1 x so that dt = \(\frac{dx}{1+x^2}\)
R.H.S = ∫t. e^tdt = t. e^t – ∫e^tdt
= t. e^t – e^t = e^t(t – 1)
Solution is y. e^{tan-1 x} = e^{tan-1 x} (tan^-11 x – 1) + c
y = tan^-1 x – 1 + c. e^{-tan-1 x}

Question 12.
\(\frac{dy}{dx}\) + y tan x = sin x.
Solution:
I.F. =e^{∫tan x dx} = e^{log sec x} = sec x
y. sec x = ∫ sin x. sec x dx = ∫tan x dx
= log sec x + c

III. Solve the following differential equations.

Question 1.
cos x. \(\frac{dy}{dx}\) + y sin x = sec² x
Solution:
\(\frac{dy}{dx}\) + tan x. y = sec³ x
I.F. = e^{∫tan x dx} = e^{log sec x} = sec x
y. sec x = ∫sec⁴ x dx = ∫(1 + tan² x) sec² x
dx = tan x + \(\frac{\tan^3x}{3}\) + c

Question 2.
sec x. dy = (y + sin x) dx.
Solution:
\(\frac{dy}{dx}=\frac{y+sin x}{sec x}\) = y cos x + sin x. cos x
\(\frac{dy}{dx}\) – y. cos x = sin x. cos x
I.F. = e^{-∫cos x dx} = e^{– sin x}
= y. e^{-sin x} = ∫ e^{-sin x}. sin x. cos x dx ……….. (1)
Consider ∫e^{-sin x}. sin x. cos x dx
t = – sin x ⇒ dt = -cos x dx
∫e^{-sin x}. sin x. cos x dx = + ∫e^t t dt
= t. e^t – e^t + c.
= e^{-sin x} (- sin x – 1) + c
y. e^{-sin x} = – e^{-sin x} (sin x + 1) + c
or y = – (sin x + 1) + c. e^{sin x}.

Question 3.
x log x.\(\frac{dy}{dx}\) + y = 2 log x.
Solution:

Question 4.
(x + y + 1)\(\frac{dy}{dx}\) = 1.
Solution:
\(\frac{dy}{dx}\) = x + y + 1
\(\frac{dy}{dx}\) – x = y + 1
I.F. = e^∫-dy e^-y
x . e^-y = ∫e^-y (y + 1) dy
= -(y + 1). e^-y + ∫e^-y. dy
= -(y + 1) e^-y – e^-y
= -(y + 2) e^-y+ c
x = -(y + 2) + c. e^-y

Question 5.
x(x – 1)\(\frac{dy}{dx}\) – y = x³(x – 1)³
Solution:
\(\frac{dy}{dx}-\frac{1}{x(x-1)}\)y = x²(x – 1)²

Question 6.
(x + 2y³)\(\frac{dy}{dx}\) = y
Solution:

Solution is x = y(y² + c)

Question 7.
(1 – x²)\(\frac{dy}{dx}\) + 2xy = x\(\sqrt{1-x^2}\)
Solution:

Question 8.
x(x – 1)\(\frac{dy}{dx}\) – (x – 2)y = x³(2x – 1)
Solution:


Solution is y(x – 1) = x²(x² – x + c)

Question 9.
\(\frac{dy}{dx}\)(x²y³ + xy) = 1
Solution:
\(\frac{dy}{dx}\) = xy + x²y³
This is Bernoulli’s equation

Question 10.
\(\frac{dy}{dx}\) + x sin 2y = x³ cos² y
Solution:

Question 11.
y² + (1 – \(\frac{1}{y}\)).\(\frac{dy}{dx}\) = 0
Solution:


Solution xy = 1 + y + cy.e^1/y