Microbiology with Diseases by Taxonomy Chapter 7 Answers

Microbiology with Diseases by Taxonomy Chapter 7 Answers

Microbiology with Diseases by Taxonomy Answers

Microbiology with Diseases by Taxonomy Chapter 7 Answers 1CM

Microbiology with Diseases by Taxonomy Chapter 7 Answers 1CT
DNA of nucleotides used as template for replication and transcription, with radioactive deoxyribose as template:
Explanation
Deoxyribonucleic acid (DNA) encodes genetic instructions in all living organisms. It is a nucleic acid, composed of major macro molecules, containing double-stranded helices. It organized into long structures known as chromosomes. Moreover, it is used in genetic engineering, forensics, bioinformatics, and DNA nanotechnology.
Similarly, ribonucleic acid (RNA) is a biological molecule, which plays an important role in coding, decoding regulation, and expression of genes. However, it is usually single-stranded and plays an important role in the cells by catalyzing reactions.
The percentage of DNA and RNA strands that are radioactive after DNA replication are:
Explanation
After completion of DNA replication cycles, DNA strand contain 12.5 % of radioactive sugars. Therefore, in RNA molecules do not contain radioactive sugars, because it uses only ribose sugar, whereas in experiment only deoxyribose is radioactive.

Microbiology with Diseases by Taxonomy Chapter 7 Answers 1FB
The three steps in RNA transcription are initiation, elongation, and termination.
Cells first make an RNA (ribonucleic acid) copy of the gene; this process is known as transcription. In RNA transcription three steps are involved, they are initiation of transcription, elongation, and termination.
RNA polymerase is an enzyme, which produces RNA. It attaches to the promoters (specific nucleotide sequences) that is located in the beginning of the gene and initiates the transcription. It transcribes the DNA template in to RNA. RNA is produced in 5’-3’ direction.
In elongation, RNA polymerase moves down the DNA template to synthesize RNA.
Finally RNA polymerase falls off on the 5’ end of the DNA template. It leads to the termination of the transcription.
Hence, the correct answers are initiation, elongation, and termination.

Microbiology with Diseases by Taxonomy Chapter 7 Answers 1L
The given diagram is showing the discontinuous synthesis of a lagging strand during DNA replication.
The parts a, b, c, d, e, f, g, h, i, j, k and l can be identified as listed below:
a. Replication fork:
It is the spot at which both strands of DNA get separated to allow replication of individual strands.
b. Stabilising proteins:
They help in unwinding and parting of the two strands of the DNA molecule.
c. Nucleotide trisphosphate:
These are the main components of nucleic acids. They not only provide energy but also give phosphate groups during phosphorylation reactions.
d. Leading strand:
This is strand produced in continuation in 5’ to 3’ direction due to polymerization.
e. Helicase (enzyme):
It is an enzyme which is responsible for separation of double helical strands of DNA.
f. Primase (enzyme):
It is the enzyme that builds short RNA molecules complementary to the DNA template.
g. DNA polymerase III (enzyme):
This is the enzyme involved in the proof reading function when nucleotides are being added to the new strand.
h. RNA primer:
It is a small strand composed of nucleic acids and is the starting point to begin DNA synthesis.
i. Okazaki fragments:
The short segments of DNA on the lagging strand during replication. These are joined as one by an enzyme ligase.
j. DNA polymerase I (enzyme):
It is the enzyme which removes the RNA primer. RNA primer is replaced by the DNA. It also plugs in the required nucleotides amid the Okazaki fragments.
k. Lagging strand:
The newly formed discontinuous strand opposite to replication fork in leading strand in 3’ – 5’ direction.
l. Ligase (enzyme):
This enzyme is responsible for filling the gaps in sugar phosphate back bone of the DNA. It also helps in joining the Okazaki fragments.

Microbiology with Diseases by Taxonomy Chapter 7 Answers 1MC
An organized structure of DNA (deoxyribonucleic acid), protein and RNA (ribonucleic acid) found in cells is known as chromosome. It contains DNA-bound proteins which serve packaging the DNA and controls it functions.
a) 4,000,000 base pairs:
Each cell contains more than one type of chromosome. Chromosomes are existed as either duplicated (two identical copies) or unduplicated (single duplicated strands).
In eukaryotes linear chromosome is present whereas in prokaryotes small circular chromosome is present.
In eukaryotes chromosomes are packed by proteins into a condensed structure is known as chromatin. In bacterial chromosomes the 4,000,000 base pairs are present. Hence, this option is correct.
b) 4000 base pairs:
Each cell contains more than one type of chromosome. Chromosomes are existed as either duplicated (two identical copies) or unduplicated (single duplicated strands).
In eukaryotes linear chromosome is present whereas in prokaryotes small circular chromosome is present.
In eukaryotes chromosomes are packed by proteins into a condensed structure is known as chromatin. In bacterial chromosomes the 4,000,000 base pairs are present. Hence, this option is incorrect.
c) 400 base pairs:
Each cell contains more than one type of chromosome. Chromosomes are existed as either duplicated (two identical copies) or unduplicated (single duplicated strands).
In eukaryotes linear chromosome is present whereas in prokaryotes small circular chromosome is present.
In eukaryotes chromosomes are packed by proteins into a condensed structure is known as chromatin. In bacterial chromosomes the 4,000,000 base pairs are present. Hence, this option is incorrect.
d) 40 base pairs:
Each cell contains more than one type of chromosome. Chromosomes are existed as either duplicated (two identical copies) or unduplicated (single duplicated strands).
In eukaryotes linear chromosome is present whereas in prokaryotes small circular chromosome is present.
In eukaryotes chromosomes are packed by proteins into a condensed structure is known as chromatin. In bacterial chromosomes the 4,000,000 base pairs are present. Hence, this option is incorrect.
Hence, the correct option is (a) 4,000,000 base pairs which is the present in the bacterial chromosome.

Microbiology with Diseases by Taxonomy Chapter 7 Answers 1SA
Genotype of bacterium determines its phenotype:
Explanation
Set of genes in the genome is known as genotype of an organism. Physical features and function traits of an organism is known as phenotype. Examples are antibiotic resistance, morphology, structure, and metabolism. Moreover, it is mainly used in determining the difference in genetic makeup of an individual by observing the DNA sequence of biological assays.
The following methods of genotyping, which includes are:

1. Restriction fragment length polymorphism identification (RFLPI).
2. Amplified fragment length polymorphism detection (AFLPD).
3. Random amplified polymorphic detection (RAPD).
4. DNA sequencing.
5. Polymerase chain reaction (PCR).
6. Allele specific oligonucleotide (ASO).

Similarly, genotype of bacteria is encoded by gene in DNA, which is transcribed into mRNA. The mRNA is translated into protein (polypeptide). In this translation, ribosomes play an important role.
Thus, a bacterium contains different genotypes with different proteins and determines different phenotype.

Microbiology with Diseases by Taxonomy Chapter 7 Answers 2CT
The amino acid sequences in polypeptide are synthesized by eukaryotic ribosomes:
Explanation

1. Messenger RNA is large family of molecules. They transmit genetic information to DNA and to ribosome.
   Transcription of mRNA is followed by the RNA polymerase, and then it translated to polymers of amino acids and proteins.
2. In DNA mRNA genetic information is determined in nucleotide sequence.
3. If mRNA have the nucleotide base sequences, then sequence of amino acids in polypeptides are synthesized by eukaryotic ribosomes.

The sequence of amino acids in polypeptides is synthesized by eukaryotic ribosomes by the following below.

• AUGGGGAUACGCUACCCC
  Met-Gly-Ile-Arg-Tyr-Pro
• CCGUACAUGCUAAUCCCU
  Pro-Tyr-Met-Leu-Ile-Pro
• CCGAUGUAACCUCGAUCC
  Pro-Met (stop)
• AUGCGGUCAGCCCCGUGA
  Met-Arg-Ser-Ala-Pro (stop

Microbiology with Diseases by Taxonomy Chapter 7 Answers 2FB
A triplet nucleotide sequence that specifies a particular amino acid is known as codon.
The mRNA (messenger RNA) is large family of the RNA and it transmits genetic information from DNA (deoxyribonucleic acid) to the ribosome.
It mainly involves in transcription, transport and translation. Important function of mRNA carries genetic information.
Genetic codon is a set of rules by which the information stored in DNA or RNA is translated into proteins. Generally the codon is a sequence of three nucleotides called triplet. Each triplet specifies one amino acid during protein synthesis.
Hence, the correct answer is codon.

Microbiology with Diseases by Taxonomy Chapter 7 Answers 2MC
Prokaryotic chromosomes are circular and they are present in the cytosol.
a) They typically have two or three origins of replication:
In prokaryotes genome is made by the plasmid and chromosome. They contain only one chromosome.
They contain circular chromosome and introns are absent. In prokaryotes nucleus is absent and chromosome is present in the cytosol.
Their DNA (deoxyribonucleic acid) is naked that means nucleosomes are absent. They are haploid and their chromosome is unpaired. They contains single molecule of circular DNA. Hence, this option is incorrect.
b) They contain single-strand DNA:
In prokaryotes genome is made by the plasmid and chromosome. They contain only one chromosome.
They contain circular chromosome and introns are absent. In prokaryotes nucleus is absent and chromosome is present in the cytosol.
Their DNA is naked that means nucleosomes are absent. They are haploid and their chromosome is unpaired. They contains single molecule of circular DNA. Hence, this option is incorrect.
c) They are located in the cytosol:
In prokaryotes genome is made by the plasmid and chromosome. They contain only one chromosome.
They contain circular chromosome and introns are absent. In prokaryotes nucleus is absent and chromosome is present in the cytosol.
Their DNA is naked that means nucleosomes are absent. They are haploid and their chromosome is unpaired. They contains single molecule of circular DNA. Hence, this option is correct.
d) They are associated in linear pairs:
In prokaryotes genome is made by the plasmid and chromosome. They contain only one chromosome.
They contain circular chromosome and introns are absent. In prokaryotes nucleus is absent and chromosome is present in the cytosol.
Their DNA is naked that means nucleosomes are absent. They are haploid and their chromosome is unpaired. They contains single molecule of circular DNA. Hence, this option is incorrect.
Hence, the correct option is (c) they are located in the cytosol which is the location prokaryotic chromosomes are present.

Microbiology with Diseases by Taxonomy Chapter 7 Answers 2SA
Eukaryotic messenger RNA differs from prokaryotic mRNA:
Explanation
In prokaryotes, the site of transcription and translation of mRNA is cytoplasm. In eukaryotes, transcription occurs in the nucleus and translation of mRNA occurs in the cytoplasm. In addition, prokaryotic mRNA is not modified in transcription, whereas in eukaryotes mRNA is modified in transcription by RNA splicing. Eukaryotic mRNA has only one site for initiation and termination of the protein synthesis, whereas in eukaryotes mRNA is monocistronic.
Furthermore, in eukaryotes mRNA is synthesized from DNA (deoxyribonucleic acid) template. Therefore, in prokaryotes mRNA is being transcribed from the DNA molecule.
Prokaryotic mRNA lifespan is very short, whereas eukaryotic mRNA has longer life span and they are metabolically stable. In eukaryotes transcribed mRNA involves major post transcriptional modifications, whereas prokaryotes undergoes little post transcriptional modifications
Thus, these are the main differences between prokaryotic mRNA and eukaryotic mRNA.

Microbiology with Diseases by Taxonomy Chapter 7 Answers 3CT
Drugs ddC and AZT are used in AIDS treatment:
Explanation
Deoxyribonucleic acid (DNA) encodes genetic instructions in all living organisms. It is a nucleic acid, composed of major macro molecules. Moreover, it contains double-stranded helices and organized into long structures known as chromosomes. Additionally, it is used in genetic engineering, forensics, bioinformatics, and DNA nanotechnology.
Similarly, the drugs ddC (dideoxycytidine) and ATZ (azidothymidine) are mainly used in the treatment of the AIDS (acquired immunodeficiency syndrome). Besides, ATZ is a nucleoside analog, reverse transcriptase inhibitor. It is an antiretroviral drug used for treatment of AIDS.
Therefore, the drugs ddC and ATZ are nucleotide analog bases, in DNA. These drugs are incorporated in the strand of DNA; they bring to an end in the replication of new DNA strand.

Microbiology with Diseases by Taxonomy Chapter 7 Answers 3FB
Three effects of point mutations are silence, missense, and non-sense.
Point mutations replace the single base nucleotide with another nucleotide of the genetic material DNA (deoxyribonucleic acid) or RNA (ribonucleic acid). It occurs at the single point during DNA replication.
They are arising from the spontaneous mutations. The rate of mutations is increased by mutagens like UV rays and X-rays. Three effects of point mutations are silence, missence and nonsense mutations.
Silence mutations code for the same amino acid, and hence it does not show any effect on the protein functioning. A single nucleotide change in this mutation, but the new codon codes for the same amino acid as wild type.
Missense mutations code for entirely different amino acid.
Nonsense mutations code for a stop codon that terminates the protein synthesis. It converts a functional codon into a termination codon.
Hence, the correct answers are silence, missense, and non-sense.

Microbiology with Diseases by Taxonomy Chapter 7 Answers 3MC
a) A molecule of RNA (ribonucleic acid ) found in bacterial cells:
An extra chromosomal genetic material is known as plasmid. They are circular deoxyribonucleic acids which are replicate individually of the bacterial chromosome.
They are commonly found as small circular and double stranded DNA (deoxyribonucleic acid) molecules in bacteria. They also present in eukaryotes and archaea.
They are also considering as replicons because they replicate autonomously with in suitable host. They are mainly used as vectors for molecular cloning. Hence, this option is incorrect.
b) A distinguished from a chromosome by being circular:
An extra chromosomal genetic material is known as plasmid. They are circular deoxyribonucleic acids which are replicate individually of the bacterial chromosome.
They are commonly found as small circular and double stranded DNA molecules in bacteria. They also present in eukaryotes and archaea.
They are also considering as replicons because they replicate autonomously with in suitable host. They are mainly used as vectors for molecular cloning. Hence, this option is incorrect.
c) A structure in bacterial cells formed from plasma membrane:
An extra chromosomal genetic material is known as plasmid. They are circular deoxyribonucleic acids which are replicate individually of the bacterial chromosome.
They are commonly found as small circular and double stranded DNA molecules in bacteria. They also present in eukaryotes and archaea.
They are also considering as replicons because they replicate autonomously with in suitable host. They are mainly used as vectors for molecular cloning. Hence, this option is incorrect.
d) Extra chromosomal DNA:
An extra chromosomal genetic material is known as plasmid. They are circular deoxyribonucleic acids which are replicate individually of the bacterial chromosome.
They are commonly found as small circular and double stranded DNA molecules in bacteria. They also present in eukaryotes and archaea.
They are also considering as replicons because they replicate autonomously with in suitable host. They are mainly used as vectors for molecular cloning. Hence, this option is correct.
Hence, the correct option is (d) extra chromosomal DNA which is known as plasmid.

Microbiology with Diseases by Taxonomy Chapter 7 Answers 3SA
Contrast and comparison of introns and exons:
Explanation
Introns are the nucleotide sequence and exons are the nucleic acid sequence, which are related to genes. Moreover, introns are placed between the exons, whereas exons contain code for the complete protein.
Additionally, introns are non-coding sequence and do not carry genetic information and they don’t involve with the coding of proteins, whereas exons are coding DNA sequence and code for proteins.
Furthermore, introns are spliced out by the RNA (ribonucleic acid) splicing mechanism after transcription, whereas exons join them. Introns do not contain any open reading frame (ORF). They have only stop codons and random sequences.
Thus, exons are named as DNA (deoxyribonucleic acid) bases that are translated into mRNA. Introns are present in the genome of the higher vertebrates.

Microbiology with Diseases by Taxonomy Chapter 7 Answers 4CT
Insertion of three nucleotides is less likely result in deleterious effect than an insertion of single nucleotide:
Explanation
Nucleotides are organic molecules. They are building blocks of the nucleotides (DNA and RNA). They carry energy within the cell and also involve in cell signaling. They are composed of nucleobase, a five carbon sugar, and one or more phosphate groups. They play important role in the metabolism.
Additionally, insertion of three nucleotides gives complete codon to a sequence. It results in incorporation of an amino acid into polypeptide product. Polypeptides have regions, which are flexible for extra amino acid without force.
Furthermore, insertion of a single nucleotide into gene sequence produces a shift of the entire reading frame. It results in the changes in the all codons of the downstream insertion.
Thus, the gene product has an entirely unusual amino acids sequence downstream of the insertion and is non-functional.

Microbiology with Diseases by Taxonomy Chapter 7 Answers 4FB
Insertion and deletions in the genetic code are also called frame shift mutations.
Point mutations replace the single base nucleotide with another nucleotide of the genetic material DNA (deoxyribonucleic acid) or RNA (ribonucleic acid). It occurs at the single point during DNA replication. It includes insertion, deletions and substitutions.
They are arising from the spontaneous mutations. The rate of mutations is increased by mutagens like UV rays and X-rays.
Insertion and deletion of a base pair is called frame shift mutations. In this mutation nucleotide triplets are displaced and creates new sequence of codons, it results in altered polypeptide sequences that mainly affect proteins.
A frameshift mutation is caused by deletion or insertion of a nucleotide in a codon. Due to the triplet nature of a reading frame, any insertion or deletion results in a complete different translation. This affects the protein being synthesized.
Hence, the correct answer is frame shift mutation.

Microbiology with Diseases by Taxonomy Chapter 7 Answers 4MC
a) Chromatin:
it is combination of DNA (deoxyribonucleic acid) and protein which makes the content of the nucleus of the cell. Their main functions are packaging DNA in to a small volume to fit in the cell, to strength the DNA to allow mitosis and DNA damage prevention. It also controls gene expression and DNA replication. Hence, this option is incorrect.
b) Bacteriocin:
a proteinaceous toxin produced from the bacteria is known as bacteriocin. It inhibits the growth of similar bacterial strains. It is narrow spectrum antibiotic. Hence, this option is incorrect.
c) Histone:
Histones are many small, basic proteins most commonly found in association with the DNA in the chromatin of eukaryotes.
They are chief protein component of chromatin. They play very important role in the gene regulation.
They undergo post transcriptional modifications which alter their interaction with DNA and nuclear proteins.
They are basic proteins and their positive charge forms ionic bonds with eukaryotic DNA and stabilizes it. Hence, this option is correct.
d) Nucleosome:
it is a basic unit of DNA packaging in eukaryotic cells. They form fundamental repeating units of eukaryotic chromatin. Hence, this option is incorrect.
Hence, the correct option is (c) histone which is basic protein and their positive charge forms ionic bonds with eukaryotic DNA and stabilizes it.

Microbiology with Diseases by Taxonomy Chapter 7 Answers 4SA
Regulation synthesizes to conserve energy in cells:
Explanation
Polypeptide synthesis is also known as protein synthesis. Biological cells synthesis new proteins, known as polypeptide synthesis. Moreover, it balances by loss of cellular proteins through degradation, assembly of the proteins by ribosomes. It involves in two steps; transcription and translation and differs in prokaryotes and eukaryotes. For example, photosynthesis is best example for polypeptide synthesis. In photosynthesis, the light energy is converted into chemical energy and depends on the light, carbon dioxide, and temperature.

Microbiology with Diseases by Taxonomy Chapter 7 Answers 5CT
The lac operon is an operon, which requires transport and metabolism of the lactose in Escherichia coli and some enteric bacteria. Moreover, it contains lacZ, lacY, and lacA structural genes. These genes are instructed by the enzymes, β-galactosidase, lactose permease, and thiogalactoside transacetylase.
Similarly, E. coli withstands mutation in the gene for lac operon repressor. The mutation in lac operon prevents repressor binding to the promoter, which results in the continuous transcription of the lac operon genes.
Additionally, the continuous production of enzymes is bacterium is catabolized to lactose. However, continuous production of wild type is disadvantage to cells the lac enzymes is. This is because in continuous production of enzyme if no lactose is available, it leads to the waste cellular resources.

Microbiology with Diseases by Taxonomy Chapter 7 Answers 5FB
An operon consists of promoter, operator and series of genes and is associated with a regulatory gene.
Functioning unit of genomic DNA (deoxyribonucleic acid) is known as operon. It consists of promoter, operator and series of genes. It controls the cluster of genes under the promoter.
Promoter is a specific nucleotide sequence on DNA (deoxyribonucleic acid) that initiates and allows gene transcription. It is located on the same DNA strand that will be transcribed and is situated upstream of the sequence towards the 3’ region. The promoter sequence will be recognized by RNA (ribonucleic acid) polymerases and by transcription factors. These sequences are usually 100 to 1000 base pairs in length.
Operator is a segment of the DNA, which is regulator binding. It is defined as a segment between promoter and the series of genes. It is the site where transcription factors bind. The main function of the operator is to regulate the expression of the gene. The function of the operator is explained very clearly in lac operon.
Series of genes are co-regulated by the operon. These genes code for the enzymes that are necessary for metabolic pathway. For instance, the series of genes present in lac operon are lacZ, lacY, and lacA. They code for different proteins and enzymes that are required in lactose metabolism.
Hence, the correct answers are promoter, operator, and series of genes.

Microbiology with Diseases by Taxonomy Chapter 7 Answers 5MC
Nucleotides are organic molecules which forms basic building blocks of the nucleic acids like DNA (deoxyribonucleic acid) and RNA (ribonucleic acid).
a) Carry energy:
In DNA replication process two identical copies are produced from the original DNA molecule. DNA replication is semi conservative model because each strand of the DNA serves as template for the production of complementary strand.
In this process DNA polymerase produce new DNA strand by adding complimentary nucleotides to the template strand. The nucleotide is carry energy in this process.
Nucleotides are found in four forms, each with a deoxyribose sugar, a phosphate and a base. They are present in the cytosol of the cells as triphosphate nucleotides. Hence, this option is correct.
b) Are found in four forms, each with a deoxyribose sugar, a phosphate and a base:
In DNA replication process two identical copies are produced from the original DNA molecule. DNA replication is semi conservative model because each strand of the DNA serves as template for the production of complementary strand.
In this process DNA polymerase produce new DNA strand by adding complimentary nucleotides to the template strand. The nucleotide is carry energy in this process.
Nucleotides are found in four forms, each with a deoxyribose sugar, a phosphate and a base. They are present in the cytosol of the cells as triphosphate nucleotides. Hence, this option is correct.
c) Are present in the cytosol of the cells as triphosphate nucleotides:
In DNA replication process two identical copies are produced from the original DNA molecule. DNA replication is semi conservative model because each strand of the DNA serves as template for the production of complementary strand.
In this process DNA polymerase produce new DNA strand by adding complimentary nucleotides to the template strand. The nucleotide is carry energy in this process.
Nucleotides are found in four forms, each with a deoxyribose sugar, a phosphate and a base. They are present in the cytosol of the cells as triphosphate nucleotides. Hence, this option is correct.
d) All of the above:
In DNA replication process two identical copies are produced from the original DNA molecule. DNA replication is semi conservative model because each strand of the DNA serves as template for the production of complementary strand.
In this process DNA polymerase produce new DNA strand by adding complimentary nucleotides to the template strand. The nucleotide is carry energy in this process.
Nucleotides are found in four forms, each with a deoxyribose sugar, a phosphate and a base. They are present in the cytosol of the cells as triphosphate nucleotides. Hence, this option is correct.
Hence, the correct option is (d) all of the above which is nucleotides are used in the replication of DNA for caring energy. They are found in four forms, each with a deoxyribose sugar, a phosphate and a base and they are present in the cytosol of the cells as triphosphate nucleotides.

Microbiology with Diseases by Taxonomy Chapter 7 Answers 5SA
Operon model in gene regulation:
Explanation
Operon is functioning unit of the genomic DNA. It is related to regulons, stimulons and modulons. It exists in both prokaryotes and eukaryotes. They are also found in viruses like bacteriophages. Moreover, it was discovered by Jacob and Jacques Monod. Operon has three parts operator, promoter, and repressor.
In addition, during stress circumstances the repressor is bound to the RNA (ribonucleic acid) promoter. Moreover, it increases the RNA polymerase to bind the operator and structural genes of enzymes, which are important for breakdown of the sugars. Thus, in this way operon regulates the gene expression.

Microbiology with Diseases by Taxonomy Chapter 7 Answers 6CT
Nucleotide analogs can be carcinogenic:
Explanation
Nucleotide analogs resemble nucleotide like sequences of DNA (deoxyribonucleic acid). They are mainly used in the treatment of cancer by inhibiting the enzyme reverse transcriptase. Moreover, they are used as antiviral drug; for instance, nucleotides analogs are AZT (azidothymidine) and ddI.
Additionally, nucleotide analogs are carcinogenic; they are incorporated into DNA molecule and cause base pair mismatches, which will synthesis mutations in the new DNA strand sequences.
Therefore, they block the DNA replication by preventing further elongation of strand that contains them. It results in inability to replicate and prevent the continued growth of cancer.

Microbiology with Diseases by Taxonomy Chapter 7 Answers 6FB
In general, inducible operons are inactive until the substrate of their genes polypeptides is present.
Operons can be repressible or inducible. Repressible operon is the operon that is active unless a repressor molecule inhibits it. Inducible operon is the operon that is switched off, but is induced by a small molecule.
In inducible operon, regulatory repressor protein is bound to the operator that ceases the transcription. It is switched off when inducer (chemical) is present.
If inducer molecule is present it binds to the repressor, so repressor does not bind to the operator. It leads to the expression of the operon. Lac operon is example of inducible operon.
Hence, the correct answer is inducible operon.

Microbiology with Diseases by Taxonomy Chapter 7 Answers 6MC
DNA polymerase III is functions as a proofreader for newly replicated strand of DNA (deoxyribonucleic acid) .
a) DNA polymerase III:
DNA polymerase III is a holoenzyme and it is involved in the prokaryotic replication. It contains two DNA polymerase embedded in a particle with nine other subunits.
The main function of DNA polymerase III is proofreader for newly replicated strand of DNA. Hence, this option is correct.
b) Primase:
It is an enzyme involved in DNA replication. It plays very important role in the DNA replication. It produces short RNA (ribonucleic acid) or DNA segment in some organisms is known as primer. In bacteria it binds to the DNA helicase and forms primosome. Hence, this option is incorrect.
c) Helicase:
It is a class of enzyme which is important to all living organisms. They play important function is to unpackaged the genes of organisms. Hence, this option is incorrect.
d) Ligase:
it is an enzyme which catalyzes the joining of two large molecules by forming a chemical bond. Hence, this option is incorrect.
Hence, the correct option is (a) DNA polymerase which functions as a proof reader for newly replicated strand of DNA.

Microbiology with Diseases by Taxonomy Chapter 7 Answers 6SA
Components and structure of DNA and RNA in prokaryotes:
Explanation
Prokaryotic cells lack nucleus and do not having organelles. The DNA (deoxyribonucleic acid) contains only one chromosome. Moreover, the cells lack membrane bound nucleus and most of them are unicellular organisms. In prokaryotes DNA is located in the nucleoid and is spread throughout the entire cell.
Additionally, the transfer of DNA in prokaryotes is through transduction, conjugation, and transformation. Prokaryotic chromosomes have circular shape and have few proteins. They contain small circular DNA that replicate independently, known as plasmids.
Furthermore, prokaryotes contain large RNA (protein). RNA (mRNA) is immediately translated by ribosomes in to proteins. Larger subunit of the ribosomal RNA has 50 Svedberg units (50S) in molecular size and smaller subunit has 30 S in molecular size. These subunits join together to form protein factory. Thus, these are the main difference between prokaryotic DNA and RNA.

Microbiology with Diseases by Taxonomy Chapter 7 Answers 7CT
DNA polymerase:
Explanation
DNA polymerase is a cellular polymerase enzyme. They produce DNA (deoxyribonucleic acid) molecules from the nucleotide building blocks. Moreover, plays an important role in the DNA replication, repair, genetic recombination, and reverse transcriptase. It is mainly used in molecular biology, DNA sequencing, and molecular cloning. Besides, it reads DNA strand in replication process and create another strand. It also helps in the production and catalyzes the bonds between nucleic acids in DNA.
Thus, DNA refers to deoxyribonucleic acid. It denotes the genetic makeup and -ase refers to enzyme or protein. Hence, DNA is named as DNA polymerase.

Microbiology with Diseases by Taxonomy Chapter 7 Answers 7FB
A daughter DNA molecule is composed of one original strand and one new strand because DNA replication is semi conservative.
In DNA replication two strands of complementary DNA separates in to two original strands.
Each strand uses as a template for the synthesis of new complementary strand, so the DNA replication is semi conservative. It contains half original strand and half new strand.
Semi conservative replication was confirmed by the Meselson-Stahl experiment.
Hence, the correct answer is semi conservative.

Microbiology with Diseases by Taxonomy Chapter 7 Answers 7MC
The methyl group addition to cytosine of DNA (deoxyribonucleic acid) is known as methylation.
a) Methylation:
Addition of the methyl group to cytosine of deoxyribonucleic acid is known as methylation. It modifies the expression of gene in the cells.
It removed in the zygote formation and reestablished in the cell divisions. It modifies which regulate gene expression. Moreover, it suppresses harmful stretches of deoxyribonucleic acid and the expression of endogenous retroviral genes. It forms bases of chromatin structure.
It plays very important role in the development of all types of cancer. Hence, this option is correct.
b) Restriction:
Restriction enzyme is used for the cutting DNA at specific recognition nucleotide sequence. It is also known as restriction endonuclease. They are mainly used for the DNA modification and manipulation. Hence, this option is incorrect.
c) Transcription:
It is first step of the gene expression. In this process DNA segment is copied into RNA (ribonucleic acid) polymerase. Hence, this option is incorrect.
d) Transversion:
The substation of purines for pyrimidines is known as transversion. It can be caused by the ionizing radiation. Hence, this option is incorrect.
Hence, the correct option is (a) methylation addition of the methyl group to the cytosine of the deoxyribonucleic acid.

Microbiology with Diseases by Taxonomy Chapter 7 Answers 7SA
Difference in function of RNA polymerase with DNA polymerase:
Explanation
DNA (deoxyribonucleic acid) polymerase plays an important role in the replication process, but it is not capable of initiating, whereas RNA (ribonucleic acid) polymerase has the capability of initiation. Thymine base is present in DNA, whereas uracil base is present in RNA. RNA is single stranded, whereas DNA is double stranded. In addition, DNA polymerase requires RNA primer for initiating replication, but RNA polymerase does not require any primer.
RNA polymerase does not require helicase enzyme to unwind the DNA strands, whereas DNA require helicase enzyme for replication process. Moreover, RNA polymerase is capable of many functions on comparing with DNA polymerase. DNA polymerase starts function at 3’ end. RNA polymerase functions anywhere of the DNA strand, 3’ end to 5’ end direction.

Microbiology with Diseases by Taxonomy Chapter 7 Answers 8CT
Effect of Corynebacterium diphtheria on cellular metabolism:
Explanation
Corynebacterium diphtheriae is gram positive bacteria. They mainly cause disease known as diphtheria. The bacteria produce a toxin called as diphtheria toxin. The toxin supports in the alteration of protein function in the host by inactivating the elongation factor (EF-2). Moreover, it slowly shows immediate toxic effect on the cell, because cell requires the productions of proteins continuously. Thus, there is no long term effect because cell is already dead.
Hence, Corynebacterium diphtheria secretes a toxin, which inactivates elongation factors of eukaryotic cell.

Microbiology with Diseases by Taxonomy Chapter 7 Answers 8FB
A gene for antibiotic resistance can move horizontally among bacterial cells by transformation, transduction and bacterial conjugation.
Insertion of genetic material into the non-bacterial cells is known as transformation. It was demonstrated by the Frederick Griffith in 1928 in Streptococcus pneumoniae.
Injection of foreign DNA (deoxyribonucleic acid) by bacteriophage virus in to the host bacterium is known as transduction.
Transfer of genetic material between two bacterial cells through direct contact is known as conjugation.
By the process of transformation, transduction and bacterial conjugation antibiotic resistance can move among bacterial cells.
Hence, the correct answer is transformation, transduction and bacterial conjugation.

Microbiology with Diseases by Taxonomy Chapter 7 Answers 8MC
In translation the binding site through which tRNA (transfer ribonucleic acid) molecules leave is known as E site.
a) A site:
Translation is a process that cellular ribosomes produce proteins and also synthesis polypeptides. It is a part of gene expression.
It complete in four process they are; initiation, elongation, translocation and termination. In this process the binding site through which tRNA (transfer ribonucleic acid) molecules leave is known as E site.
A site is accommodates a tRNA delivering an amino acid. Hence, this option is incorrect.
b) X site:
Translation is a process that cellular ribosomes produce proteins and also synthesis polypeptides. It is a part of gene expression.
It complete in four process they are; initiation, elongation, translocation and termination. In this process the binding site through which tRNA (transfer ribonucleic acid) molecules leave is known as E site. Hence, this option is incorrect.
c) P site:
Translation is a process that cellular ribosomes produce proteins and also synthesis polypeptides. It is a part of gene expression.
It complete in four process they are; initiation, elongation, translocation and termination. In this process the binding site through which tRNA (transfer ribonucleic acid) molecules leave is known as E site.
P site is holds a tRNA and the growing polypeptide. Hence, this option is incorrect.
d) E site:
Translation is a process that cellular ribosomes produce proteins and also synthesis polypeptides. It is a part of gene expression.
It complete in four process they are; initiation, elongation, translocation and termination. In this process the binding site through which tRNA (transfer ribonucleic acid) molecules leave is known as E site. In this site tRNA is exist. Hence, this option is correct.
Hence, the correct option is (d) E site which is the binding site of the tRNA molecules.

Microbiology with Diseases by Taxonomy Chapter 7 Answers 8SA
Function and formation of mRNA, rRNA, and tRNA in eukaryotes and prokaryotes:
Explanation
The mRNA (messenger RNA) is large family of the RNA and it transmits genetic information from DNA (deoxyribonucleic acid) to the ribosome. In eukaryotes, translation occurs in ribosomes. Translation is not directly coupled to the transcription. In eukaryotes mRNA contains only one polypeptide.
Similarly, in prokaryotes mRNA is translated into proteins. The mRNA does not require processing or transported. Transcription immediately begins after the end of transcription. In prokaryotes translation is coupled to the transcription. Besides, eukaryotic and prokaryotic mRNA involves in transcription, transport, and translation. Main function of mRNA is to carry genetic information.
Additionally, rRNA (ribosomal RNA) is RNA, component of the ribosome. It is important in the protein synthesis of living organism. Prokaryotes contain 30S small subunit and 50S large subunit. In eukaryotes 28S, 5.8S, 18S, and 5S subunits of four types of cytoplasmic rRNA and 12S and 16S two types of subunits in mitochondria.
Furthermore, tRNA plays important role to translate the code words in mRNA. Each specific tRNA molecule contains three base sequences, which base pair with complimentary mRNA. It is specific to amino acid specific, scans and detects mRNA. Its structure resembles with clover leaf and has several extended loops. Thus, acts as tool in the translation process of mRNA.

Microbiology with Diseases by Taxonomy Chapter 7 Answers 9CT
Use of siRNA to turn off a cancer-inducing gene:
Explanation
Small or short interfering RNA (siRNA) belong to a class of double stranded RNA molecules. It plays important role in the RNA (ribonucleic acid) interference path way and post transcriptional gene silencing. Moreover, they are used as a tool in validating gene function and drug targeting of post genome.
In addition, they are used in the turn off of cancerous cells, to produce specific proteins that make them differ from the normal cells. Scientists use this technique to cease the process, which converts a normal cell to cancerous cell from the beginning.

Microbiology with Diseases by Taxonomy Chapter 7 Answers 9FB
Transposons are nucleotide sequences containing palindromes and genes for protein that cut DNA (deoxyribonucleic acid) strands.
Transposons are DNA segments of 700-40,000 bp in length that can move from one place to another place within the genome.
They were discovered by the American scientist McClintock in 1902-1992. They are also known as jumping genes. The action of the transposons is known as transposition. It occurs between the chromosomes and the plasmids. They transfer transposons from one cell to another cell.
Hence, the correct answer is transposons.

Microbiology with Diseases by Taxonomy Chapter 7 Answers 9MC
Ames test is used for the auxotroph and liver extract to reveal potential mutagens.
a) Used auxotroph and liver extract to reveal potential mutagens:
Ames test is mainly used for strains of bacteria such as Salmonella typhimurium which carries mutations in gene involved in production of histidine.
These are autotrophic mutants they require histidine for growth. This test is mainly used for the auxotroph and liver extract to reveal potential mutagens.
A positive test indicates that the chemical is mutagenic and carcinogenic. Hence, this option is correct.
b) Is time intensive and costly:
Ames test is mainly used for strains of bacteria such as Salmonella typhimurium which carries mutations in gene involved in production of histidine. Hence, this option is incorrect.
c) Involves the isolation of a mutant by eliminating wild-type phenotypes with specific media:
Ames test is mainly used for strains of bacteria such as Salmonella typhimurium which carries mutations in gene involved in production of histidine. Hence, this option is incorrect.
d) Proves that suspected chemicals are carcinogenic:
Ames test is mainly used for strains of bacteria such as Salmonella typhimurium which carries mutations in gene involved in production of histidine. Hence, this option is incorrect.
Hence, the correct option is (a) uses auxotroph and liver extract to reveal potential mutagens which is known as Ames test.

Microbiology with Diseases by Taxonomy Chapter 7 Answers 9SA
Package of DNA in prokaryotes and eukaryotes:
Explanation
DNA (deoxyribonucleic acid) is a double stranded helical structure. Moreover, it is a molecule that encodes genetic instructions for developing and functioning.
Similarly, it was identified by the Friedrich Miescher and its structure was predicted by the James Watson and Francis Crick. The structure of DNA contains four bases, which contains are adenine (A), cytosine (C), guanine (G), and thymine (T). It is used in the genetic engineering, bioinformatics, and DNA nanotechnology.
Additionally, eukaryotic, DNA is wrapped around the proteins known as histones. DNA is present within the nucleus. In this genetic material is condensed as linear and linear chromosome.
In prokaryotes DNA is compacted by the basic proteins known as histone like proteins. Genetic material is found in covalently circular form.

Microbiology with Diseases by Taxonomy Chapter 7 Answers 10CT
Nucleotide analogs are useful to cancer research:
Explanation
Nucleotide analogs resemble nucleotide like sequences of DNA (deoxyribonucleic acid). They are mainly used in the treatment of cancer by inhibiting the enzyme reverse transcriptase. Moreover, they are used as antiviral drug; for instance, nucleotides analogs are AZT (azidothymidine) and ddI. Additionally, nucleotide analogs are carcinogenic; they are incorporated into DNA molecule and cause base pair mismatches, which will synthesis mutations in the new DNA strand sequences.
Additionally, they block DNA replication by preventing elongation of strand. It results in inability to replicate and prevent the continued growth of cancer. In contrast, they are active in the native form and work in the cells to activate phosphorylation by the drug. It clearly shows that they can act against HIV in infected cells.
Furthermore, they break DNA replication cells undergoing mitosis, which is a character of cancer cells. It leads to the cell to undergo apoptosis and later die. It is a useful method to kill cancer cells to avoid damage to normal cells.
Thus, in this way nucleotide analogs are useful to cancer researcher.

Microbiology with Diseases by Taxonomy Chapter 7 Answers 10FB
Crossing over is a recombination event that occurs during gamete formation in eukaryotes.
Exchange of chromosomal segments between two non-sisters chromatids is known as crossing over. It is a reciprocal recombination that that involves in breaking and exchange between two nonsister chromatids. It occurs in prophase-I of meiosis. It provides genetic variation during meiosis.
Hence, the correct answer is crossing over.

Microbiology with Diseases by Taxonomy Chapter 7 Answers 10MC
DNA (deoxyribonucleic acid) repair involves enzymes that recognize and a correct nucleotide error in unmethylated strands of DNA is known as mismatch repair.
a) Light repair of thymine dimers:
A cell identifies and corrects damage of DNA molecule which codes genome if known as DNA repair. U.V light and radiation of environmental factors cause DNA repair and some other lesions also cause mutations in cell.
Thymine dimers repaired by the two mechanisms photoreactivation repair and excision repair. Hence, this option is incorrect.
b) Dark repair of pyrimidines dimers:
A cell identifies and corrects damage of DNA molecule which codes genome if known as DNA repair. U.V light and radiation of environmental factors cause DNA repair and some other lesions also cause mutations in cell.
Pyrimidines dimers can be repaired by the photoreactivation or nucleotide excision repair. This type of repair mainly causes the melanomas in humans. Hence, this option is incorrect.
c) Mismatch repair:
A cell identifies and corrects damage of DNA molecule which codes genome if known as DNA repair. U.V light and radiation of environmental factors cause DNA repair and some other lesions also cause mutations in cell.
DNA repair involves enzymes that recognize and a correct nucleotide error in unmethylated strands of DNA is known as mismatch repair. Hence, this option is correct.
d) SOS response:
A cell identifies and corrects damage of DNA molecule which codes genome if known as DNA repair. U.V light and radiation of environmental factors cause DNA repair and some other lesions also cause mutations in cell.
The cell cycle is arrested and DNA repair and mutagens are induced is known as SOS response. In this process RecA protein is involved, it is produced by the single stranded DNA. Hence, this option is incorrect.
Hence, the correct option is (c) mismatch repair which is involves enzymes that recognize and a correct nucleotide error in unmethylated strands of DNA.

Microbiology with Diseases by Taxonomy Chapter 7 Answers 10SA
Central dogma of genetics:
Explanation
Central dogma of the genetics is first coined by the Francis Crick in 1958. It explains flow of genetic information in biological system, which involves from DNA (deoxyribonucleic acid) to RNA (ribonucleic acid) and finally to protein.
Moreover, DNA carries genetic information and transcript to RNA. Then forms mRNA molecules carry amino acid sequence for protein synthesis.
Additionally, it forms the basis of storage, transmission, and expression of hereditary information. It is mainly divided into three groups, which are replication, translation and transcription.
Therefore, DNA copied into DNA is known as replication. DNA can be copied into mRNA, mediated by the process of transcription. In translation proteins are produced by using mRNA, as a template.

Microbiology with Diseases by Taxonomy Chapter 7 Answers 11FB
Transfer RNA, its conformation studies:
Explanation
An adaptor molecule is composed of RNA (ribonucleic acid) and is known as tRNA. It contains 73 to 94 nucleotides in length. They are small RNA molecules. Moreover, they are important component in protein translation and production of new proteins. However, contains primary, secondary, and tertiary structure.
Similarly, tRNA is visualized as cloverleaf structure (3-D L shaped structure), which allows to P and A sites of the ribosomes. Whereas, in eukaryotes tRNA as transcribed by the RNA polymerase-3. They carry amino acids to organelles of the cell by ribosomes.
Thus, transfer RNA carries amino acids.

Microbiology with Diseases by Taxonomy Chapter 7 Answers 11MC
The natural mechanism of genetic transfer and recombination is transduction, transformation and conjugation.
a) Transduction:
Insertion of foreign DNA (deoxyribonucleic acid) through a bacteriophage virus into the host bacterium is known as transduction. It needs physical contact between cells.
It plays important role in introducing foreign DNA in to host cell genome. Viruses which infect bacteria are known as bacteriophage. When bacteriophages to the bacterial cell.
It explains mechanism of antibiotic drugs become infective by the transfer of antibiotic resistance genes of bacteria. Hence, this option is incorrect.
b) Transformation:
Genetic alteration of a cell. It results from direct uptake, incorporation and expression of exogenous genetic material from its surrounding is known as transformation.
Some species of bacteria transformation occurs. It is also known as insertion of genetic material in to nonbacterial cells. It also used for the animals and plant cells.
It was demonstrated by the Frederick Griffith in 1928. It main principle is heat killed strain was responsible for producing harm less strain virulent. Hence, this option is incorrect.
c) Transcription:
It is first step of the gene expression. In this process DNA segment is copied into RNA (ribonucleic acid) polymerase. It is not a natural genetic mechanism and recombination mechanism. Hence, this option is correct.
d) Conjugation:
Transfer of genetic material between two bacterial cells through direct contact is known as conjugation. It was discovered by Joshua Lederberg and Edward Tatum in 1946
It is a horizontal gene transfer mechanism. In this process genetic element is plasmid or transposon
It was very important to transfer of antibiotic resistance and xenobiotic tolerance. Hence, this option is incorrect.
Hence, the correct option is (c) transcription which is not a mechanism of natural genetic transfer and recombination.

Microbiology with Diseases by Taxonomy Chapter 7 Answers 11SA
Comparison and contrast between transformation, transduction, and conjugation:
Explanation
Transformation is the process of alteration of genetic information in the cell. It includes the uptake, integration, and exogenous genetic material from the surrounding; for instance, in few species of bacteria transformation. It includes the incorporation of genetic material into non-bacterial cells. It is used for the animals and plant cells. Moreover, it was demonstrated by the Frederick Griffith in 1928. The main principle lying behind is the strains of heat killed that are responsible in producing virulent, harmless strains.
Transduction is the insertion of foreign DNA into the host bacterium, through bacteriophage virus. It needs physical contact between cells. It plays an important role in introducing foreign DNA into host cell genome. Viruses, which infect bacteria, are known as bacteriophage. It explains mechanism of antibiotic drugs become infective by the transfer of antibiotic resistance genes of bacteria
Conjugation is the process of transferring genetic material between the two cells of bacteria through direct contact. It was discovered by Joshua Lederberg and Edward Tatum in 1946. Moreover, it the process of gene transfers mechanism that takes place horizontally. Thus, in this process genetic element is plasmid or transposon.
Hence, it is very important to transfer antibiotic resistance and xenobiotic tolerance.

Microbiology with Diseases by Taxonomy Chapter 7 Answers 12FB
Sense and antisense RNA are complementary to nucleic acid molecule:
Explanation
Small interfering ribonucleic acid is also known as short interfering ribonucleic acid. In addition, small interfering ribonucleic acid is also called as silencing RNA, which plays an important role in post-transcriptional gene silencing of plants.
It encourages RNA interference in mammalian cells and plays a vital role in authorizing gene function and drug targeting of post genomic era. In addition, micro RNA is small non-coding macro molecule, which was found in plants and animals. It mainly involves in the transcriptional and post transcriptional gene expression.
Furthermore, the use of miRNA is complementary to a part of one or more messenger RNAs. It functions mainly in emerging of gene regulation.
Small interfering RNA (siRNA) and micro RNA (miRNA) are antisense; that is, they are complementary to another nucleic acid molecule.

Microbiology with Diseases by Taxonomy Chapter 7 Answers 12MC
Cell which has the ability to take up extra cellular DNA (deoxyribonucleic acid) from their environment is known as competent cell.
a) Hfr cell:
High frequency recombinant cell (Hfr cell) is a bacterium. It is integrated with conjugative plasmid in to its genomic DNA. It was first discovered by the Luca Cavalla Sforza. Hence, this option is incorrect.
b) Genomic:
Genome of the entire organism is known as genomic cell. It encodes DNA or RNA. It includes both gene and RNA (ribonucleic acid)\ DNA non coding sequence. Hence, this option is incorrect.
c) Transposing:
DNA sequence change their position within the genome is known as transposing cells. It creates reversing mutations and changes the genome size of cell. Hence, this option is incorrect.
d) Competent:
Cell which has the ability to take up extra cellular DNA from their environment is known as competent cell. It was discovered by Frederick Griffith. It can be divided into natural competence and artificial competence. Hence, this option is correct.
Hence, the correct option is (d) competent which has ability to take up DNA from their environment.

Microbiology with Diseases by Taxonomy Chapter 7 Answers 12SA
Replication is the process of DNA (deoxyribonucleic acid) by synthesizing two identical copies from the original DNA molecule.
Transcription is a gene expression. It is a particular segment of DNA that is copied into RNA (ribonucleic acid) by the RNA polymerase.
Translation is the process in cellular ribosomes to create proteins, which is responsible in producing polypeptides.
The following table depicts the filled blanks:

Microbiology with Diseases by Taxonomy Chapter 7 Answers 13MC
Transfer of genetic material between two bacterial cells through direct contact is known as conjugation.
a) Conjugation requires a sex pilus extending from the surface of a cell:
Transfer of genetic material between two bacterial cells through direct contact is known as conjugation. It was discovered by Joshua Lederberg and Edward Tatum in 1946
It is a horizontal gene transfer mechanism. In this process genetic element is plasmid or transposon
It requires a sex pilus extending from the surface of a cell. It was very important to transfer of antibiotic resistance and xenobiotic tolerance. Hence, this option is correct.
b) Conjugation involves a factor:
Transfer of genetic material between two bacterial cells through direct contact is known as conjugation. It was discovered by Joshua Lederberg and Edward Tatum in 1946
It is a horizontal gene transfer mechanism. In this process genetic element is plasmid or transposon
It requires a sex pilus extending from the surface of a cell. It was very important to transfer of antibiotic resistance and xenobiotic tolerance. Hence, this option is incorrect.
c) Conjugation is an artificial genetic engineering technique:
Transfer of genetic material between two bacterial cells through direct contact is known as conjugation. It was discovered by Joshua Lederberg and Edward Tatum in 1946
It is a horizontal gene transfer mechanism. In this process genetic element is plasmid or transposon
It requires a sex pilus extending from the surface of a cell. It was very important to transfer of antibiotic resistance and xenobiotic tolerance. Hence, this option is incorrect.
d) Conjugation involves DNA that has been released into the environment from dead organisms:
Transfer of genetic material between two bacterial cells through direct contact is known as conjugation. It was discovered by Joshua Lederberg and Edward Tatum in 1946
It is a horizontal gene transfer mechanism. In this process genetic element is plasmid or transposon
It requires a sex pilus extending from the surface of a cell. It was very important to transfer of antibiotic resistance and xenobiotic tolerance. Hence, this option is incorrect.
Hence, the correct option is (a) conjugation requires a sex pilus extending from the surface of a cell is known as conjugation.

Microbiology with Diseases by Taxonomy Chapter 7 Answers 14MC
Segments of DNA (deoxyribonucleic acid) are known as transposons. It was discovered by the American scientist McClintock in 1902-1992. They are also known as jumping genes.
a) Hfr cell:
High frequency recombinant cell (Hfr cell) is a bacterium. It is integrated with conjugative plasmid in to its genomic DNA. It was first discovered by the Luca Cavalla Sforza. Hence, this option is incorrect.
b) Transducing phages:
They can able to bind bacterial cell and inject their contents. They have segment of bacterial DNA integrated into their genome. Hence, this option is incorrect.
c) Palindromic sequences:
It is a nucleotide sequence of DNA or RNA (ribonucleic acid). It read 5’-3’ on one strand and 3’-5’ direction of complementary strand. Hence, this option is incorrect.
d) Transposons:
Segments of DNA are known as transposons. They are 700-40,000 bp in length. They can move from one place to another place on the same molecule.
It was discovered by the American scientist McClintock in 1902-1992. They are also known as jumping genes. The action of the transposons is known as transposition.
Transposition was occurs between the chromosomes and the plasmids. They transfer transposons from one cell to another cell. Hence, this option is correct.
Hence, the correct option is (d) transposons which are also known as jumping genes.

Microbiology with Diseases by Taxonomy Chapter 7 Answers 15MC
a) Crossing over of P and Q:
Exchange of genetic material between two homologous chromosomes is known as crossing over. It results in recombinant chromosomes. It was discovered by Thomas Hunt Morgan. Hence, this option is incorrect.
b) Vertical gene transfer:
The passing of the genes from the parent to the offspring is known as vertical gene transfer. It can be done by sexual or asexual. Hence, this option is incorrect.
c) Horizontal gene transfer:
Transfer of genes between organisms by using non-reproductive methods is known as horizontal gene transfer. In this process donor cell donate genome the recipient cell that varies species from the donor. It was rarely occur in the 1% of prokaryotes. Hence, this option is correct.
d) Transposition:
Transfer of genetic material between two organisms other than by vertical gene transfer is known as transposition. Hence, this option is incorrect.
Hence, the correct option is (c) horizontal gene transfer which is a non-reproductive method.

Microbiology with Diseases by Taxonomy Chapter 7 Answers 16MC
A triplet nucleotide sequence in RNA (ribonucleic acid) is known as codon. It is a part of each molecule of mRNA (messenger RNA).
a) Palindrome:
Palindrome is word, number, phrase or element. Palindromic sequence is a nucleotide sequence of DNA (deoxyribonucleic acid) or RNA. It read 5’-3’ on one strand and 3’-5’ direction of complementary strand. Hence, this option is incorrect.
b) Anticodon:
It is a region of tRNA (transfer RNA). It is a sequence of three bases that are complementary to a codon in the mRNA (messenger RNA). Hence, this option is incorrect.
c) Codon:
The mRNA (messenger RNA) is large family of the RNA and it transmits genetic information from DNA to the ribosome.
It mainly involves in transcription, transport and translation. Important function of mRNA is carries genetic information.
The mRNA contains three groups of nucleotides; codon specifies the amino acid which makes up the protein. Hence, this option is correct.
d) Base pair:
Building blocks of the DNA double helix is known as base pair. It donates folded structure to the DNA and RNA. It maintains the helical structure of the DNA. Hence, this option is incorrect.
Hence, the correct option is (c) codon which is a part of each molecule of mRNA.

Microbiology with Diseases by Taxonomy Chapter 7 Answers 17MC
Nucleotides are organic molecules which forms basic building blocks of the nucleic acids like DNA (deoxyribonucleic acid) and RNA (ribonucleic acid)
a) A five carbon sugar:
Nucleotides are used in the replication of DNA for caring energy. They are found in four forms, each with a deoxyribose sugar, a phosphate and a base.
They are present in the cytosol of the cells as triphosphate nucleotides. They are made up of three parts they are; five carbon sugar, nitrogenous base and phosphate. Hence, this option is correct.
b) Phosphate:
Nucleotides are used in the replication of DNA for caring energy. They are found in four forms, each with a deoxyribose sugar, a phosphate and a base.
They are present in the cytosol of the cells as triphosphate nucleotides. Nucleotides are made up of three parts they are; five carbon sugar, nitrogenous base and phosphate. Hence, this option is correct.
c) A nitrogenous base:
Nucleotides are used in the replication of DNA for caring energy. They are found in four forms, each with a deoxyribose sugar, a phosphate and a base.
They are present in the cytosol of the cells as triphosphate nucleotides. They are made up of three parts they are; five carbon sugar, nitrogenous base and phosphate. Hence, this option is correct.
d) All of the above:
Nucleotides are used in the replication of DNA for caring energy. They are found in four forms, each with a deoxyribose sugar, a phosphate and a base.
They are present in the cytosol of the cells as triphosphate nucleotides. They are made up of three parts they are; five carbon sugar, nitrogenous base and phosphate. Hence, this option is correct.
Hence, the correct option is (d) all of the above which are made up of three parts they are; five carbon sugar, nitrogenous base and phosphate

Microbiology with Diseases by Taxonomy Chapter 7 Answers 18MC
In DNA (deoxyribonucleic acid) adenine forms two hydrogen bonds with thymine.
a) Three/uracil:
Hydrogen bonding is a chemical reaction. In DNA adenine, guanine, cytosine and thymine are nitrogen bases.
They are bonded with various number of hydrogen. In DNA adenine forms two hydrogen bonds with thymine. Hence, this option is incorrect.
b) Two/uracil:
Hydrogen bonding is a chemical reaction. In DNA adenine, guanine, cytosine and thymine are nitrogen bases.
They are bonded with various number of hydrogen. In DNA adenine forms two hydrogen bonds with thymine. Hence, this option is incorrect.
c) Two/thymine:
Hydrogen bonding is a chemical reaction. In DNA adenine, guanine, cytosine and thymine are nitrogen bases.
They are bonded with various number of hydrogen. In DNA adenine forms two hydrogen bonds with thymine. Hence, this option is correct.
d) Tree/ thymine:
Hydrogen bonding is a chemical reaction. In DNA adenine, guanine, cytosine and thymine are nitrogen bases.
They are bonded with various number of hydrogen. In DNA adenine forms two hydrogen bonds with thymine. Hence, this option is incorrect.
Hence, the correct option is (c) two/thymine which is formed by the DNA, adenine.

Microbiology with Diseases by Taxonomy Chapter 7 Answers 19MC
A sequence of nucleotides formed during replication of the lagging DNA (deoxyribonucleic acid) strand is known as an Okazaki fragments.
a) A palindrome:
Palindrome is word, number, phrase or element. Palindromic sequence is a nucleotide sequence of DNA or RNA (ribonucleic acid). It read 5’-3’ on one strand and 3’-5’ direction of complementary strand. Hence, this option is incorrect.
b) An Okazaki fragment:
They are short, newly produced DNA fragments. They formed lagging template strand in DNA replication. They are complimentary to the lagging template strand. They are 1000 to 2000 nucleotides long. Hence, this option is correct.
c)A template strand:
It is mainly used in the synthesis mRNA. It is used to describe DNA strands. It is a sequence of the DNA that is copied in mRNA (messenger RNA) synthesis.
Hence, this option is incorrect.
d) An operon:
Operon is functioning unit of the genomic DNA. It is related to regulons, stimulons and modulons.
It was discovered by the Jacob and Jacques Monod. Operon has three parts operator, promoter and repressor.
They exist in both prokaryotes and eukaryotes. They are also found in viruses like bacteriophages. It regulates the gene expression. Hence, this option is incorrect.
Hence, the correct option is (b) an Okazaki fragment which is a sequence of nucleotides formed during replication of the lagging DNA strand.

Microbiology with Diseases by Taxonomy Chapter 7 Answers 20MC
Operon is functioning unit of the genomic DNA (deoxyribonucleic acid). It is related to regulons, stimulons and modulons. It was discovered by the Jacob and Jacques Monod. Operon has three parts operator, promoter and repressor.
a) Operator:
Operator is a segment of the DNA, which is regulator binding. It is defined as segment between promoter and the series of genes. Hence, this option is incorrect.
b) Promoter:
Promoter is a nucleotide sequence. It allows the gene transcription. It was recognized by RNA (ribonucleic acid) polymerase. It indicates which gene used for the mRNA (messenger RNA) in RNA production. Hence, this option is incorrect.
c) Origin:
It is a point of where is something starts or created. It is not a part of operon. Hence, this option is correct.
d) Gene:
Gene is co-regulated by the operon. It codes the enzymes in metabolic pathway. Hence, this option is incorrect.
Hence, the correct option is (c) origin which is a not a part of operon.

Microbiology with Diseases by Taxonomy Chapter 7 Answers 21MC
RNA (ribonucleic acid) transcription is responsible for the operons are important in regulation prokaryotic.
a) DNA replication:
It is a process of synthesizing two identical copies from original strand of DNA (deoxyribonucleic acid) is knows as DNA replication. Hence, this option is incorrect.
b) RNA transcription:
It is first step of gene expression. It is a segment of DNA that is copied into the RNA by the enzyme RNA polymerase. In RNA transcription operons are important in regulation. Hence, this option is correct.
c) Ribosomal RNA (rRNA) processing:
It is synthesized from large transcript (pre rRNA) which under goes cleavage of both ends of the transcript. Hence, this option is incorrect.
d) Sugar metabolism:
Sugar is metabolized by the all organisms. They are stored as long polymers of the glucose molecules with glucosidic bonds. It gives structural support and energy storage. Hence, this option is incorrect.
Hence, the correct option is (b) RNA transcription which is responsible for the operons are important in regulation prokaryotic.

Microbiology with Diseases by Taxonomy Chapter 7 Answers 22MC
Transcription is first step of the gene expression. It produces RNA (ribonucleic acid) molecules.
a) DNA molecules:
DNA (deoxyribonucleic acid) is molecule encodes genetic instructions for developing and functioning. It is double stranded helical structure.
It was identified by the Friedrich Miescher and its structure was predicted by the James Watson and Francis Crick.
It contains four bases; they are adenine (A), guanine (G), cytosine (C) and thymine (T). It was used in the genetic engineering, bioinformatics and DNA nanotechnology. Hence, this option is incorrect.
b) RNA molecules:
They are large biological molecule. They plays important role in coding, decoding, regulation and gene expression.
They catalyze biological reactions and control gene expression. They are produced by the transcription process.
Transcription is first step of the gene expression. In this process DNA segment is copied into RNA polymerase. Hence, this option is correct.
c) Polypeptides:
Polypeptides are short chains of amino acids that are linked by the peptide bonds. They are long, continuous and unbranched peptide chain. Hence, this option is incorrect.
d) Palindromes:
Palindrome is word, number, phrase or element. Palindromic sequence is a nucleotide sequence of DNA or RNA. It read 5’-3’ on one strand and 3’-5’ direction of complementary strand. Hence, this option is incorrect.
Hence, the correct option is (b) RNA molecules which are produced by the transcription process.

Microbiology with Diseases by Taxonomy Chapter 7 Answers 23MC
Ligase plays a major role in the lagging strand replication.
a) Lagging strand replication:
Ligase is a specific type of enzyme. It plays important role in the joining of DNA (deoxyribonucleic acid) strands together by forming phosphodiester bonds. It plays very important role in the lagging strand replication.
It plays important role in repairing single stranded breaks in the DNA. It also use in gene cloning. Hence, this option is correct.
b) Messenger RNA (mRNA) processing in eukaryotes:
Eukaryotic mRNA has only one site for initiation and termination of the protein synthesis, where as in eukaryotes mRNA is monocistronic.
In eukaryotes mRNA is synthesized from DNA template. Therefore in prokaryotes mRNA is still being transcribed from the DNA molecule. Hence, this option is incorrect.
c) Polypeptide synthesis by ribosome:
Polypeptides are short chains of amino acids that are linked by the peptide bonds. They are long, continuous and unbranched peptide chain. Hence, this option is incorrect.
d) RNA transcription:
It is first step of gene expression. It is a segment of DNA that is copied into the RNA (ribonucleic acid) by the enzyme RNA polymerase. In RNA transcription operons are important in regulation. Hence, this option is incorrect.
Hence, the correct option is (a) lagging strand replication which is an important role of ligase.

Microbiology with Diseases by Taxonomy Chapter 7 Answers 24MC
Change of genome of nucleotide sequence of genome of an organism is known as mutation. They cause unrepaired damage to the DNA (deoxyribonucleic acid) and RNA (ribonucleic acid) genomes.
a) Lasting:
Change of genome of nucleotide sequence of genome of an organism is known as mutation. They cause unrepaired damage to the DNA and RNA genomes.
They play vital role in the normal and abnormal biological process. They can the sequences. They also occur in the nongenic regions.
Before mutations can affect the population permanently they must be lasting, inheritable and beneficial. Hence, this option is correct.
b) Inheritable:
Change of genome of nucleotide sequence of genome of an organism is known as mutation. They cause unrepaired damage to the DNA and RNA genomes.
They play vital role in the normal and abnormal biological process. They can the sequences. They also occur in the nongenic regions.
Before mutations can affect the population permanently they must be lasting, inheritable and beneficial. Hence, this option is correct.
c) Beneficial:
Change of genome of nucleotide sequence of genome of an organism is known as mutation. They cause unrepaired damage to the DNA and RNA genomes.
They play vital role in the normal and abnormal biological process. They can the sequences. They also occur in the nongenic regions.
Before mutations can affect the population permanently they must be lasting, inheritable and beneficial. Hence, this option is correct.
d) All of the above:
Change of genome of nucleotide sequence of genome of an organism is known as mutation. They cause unrepaired damage to the DNA and RNA genomes.
They play vital role in the normal and abnormal biological process. They can the sequences. They also occur in the nongenic regions.
Before mutations can affect the population permanently they must be lasting, inheritable and beneficial. Hence, this option is correct.
Hence, the correct option is (c) all of the above which is before mutations can affect the population permanently they must be lasting, inheritable and beneficial.

Microbiology with Diseases by Taxonomy Chapter 7 Answers 25MC
A group of genes which are used together for synthesis of tryptophan is known as tryptophan operon. It was present in the bacteria. It regulates when tryptophan is present in the environment.
a) Active/an inducer:
A group of genes which are used together for synthesis of tryptophan is known as tryptophan operon. It was discovered by the Jacques Monod in 1953.
It was present in the bacteria. It regulates when tryptophan is present in the environment.
They are repressible that means they are active and directly controlled by a repressor. Hence, this option is incorrect.
b) Active/ a repressor:
A group of genes which are used together for synthesis of tryptophan is known as tryptophan operon. It was discovered by the Jacques Monod in 1953.
It was present in the bacteria. It regulates when tryptophan is present in the environment.
They are repressible that means they are active and directly controlled by a repressor. Hence, this option is correct.
c) Inactive/an inducer:
A group of genes which are used together for synthesis of tryptophan is known as tryptophan operon. It was discovered by the Jacques Monod in 1953.
It was present in the bacteria. It regulates when tryptophan is present in the environment.
They are repressible that means they are active and directly controlled by a repressor. Hence, this option is incorrect.
d) Inactive/a repressor:
A group of genes which are used together for synthesis of tryptophan is known as tryptophan operon. It was discovered by the Jacques Monod in 1953.
It was present in the bacteria. It regulates when tryptophan is present in the environment.
They are repressible that means they are active and directly controlled by a repressor. Hence, this option is incorrect.
Hence, the correct option is (b) active/a repressor which is function of the tryptophan operon.