{"id":23718,"date":"2018-05-25T06:13:23","date_gmt":"2018-05-25T06:13:23","guid":{"rendered":"https:\/\/cbselibrary.com\/?p=23718"},"modified":"2018-05-25T06:13:23","modified_gmt":"2018-05-25T06:13:23","slug":"microbiology-with-diseases-by-taxonomy-chapter-7-answers","status":"publish","type":"post","link":"https:\/\/cbselibrary.com\/microbiology-with-diseases-by-taxonomy-chapter-7-answers\/","title":{"rendered":"Microbiology with Diseases by Taxonomy Chapter 7 Answers"},"content":{"rendered":"

Microbiology with Diseases by Taxonomy Chapter 7 Answers<\/strong><\/span><\/h2>\n

Microbiology with Diseases by Taxonomy Answers<\/a><\/p>\n

Microbiology with Diseases by Taxonomy Chapter 7 Answers 1CM<\/strong><\/span>
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Microbiology with Diseases by Taxonomy Chapter 7 Answers 1CT<\/strong><\/span>
\nDNA of nucleotides used as template for replication and transcription, with radioactive deoxyribose as template:<\/strong>
\nExplanation<\/strong>
\nDeoxyribonucleic 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.
\nSimilarly, 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.
\nThe percentage of DNA and RNA strands that are radioactive after DNA replication are:<\/strong>
\nExplanation<\/strong>
\nAfter 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.<\/p>\n

Microbiology with Diseases by Taxonomy Chapter 7 Answers 1FB<\/strong><\/span>
\nThe three steps in RNA transcription are initiation<\/strong>, elongation<\/strong>, and termination<\/strong>.
\nCells 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.
\nRNA 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\u2019-3\u2019 direction.
\nIn elongation, RNA polymerase moves down the DNA template to synthesize RNA.
\nFinally RNA polymerase falls off on the 5\u2019 end of the DNA template. It leads to the termination of the transcription.
\nHence, the correct answers are initiation<\/strong>, elongation<\/strong>, and termination<\/strong>.<\/p>\n

Microbiology with Diseases by Taxonomy Chapter 7 Answers 1L<\/strong><\/span>
\nThe given diagram is showing the discontinuous synthesis of a lagging strand during DNA replication.
\nThe parts a, b, c, d, e, f, g, h, i, j, k and l can be identified as listed below:
\na. Replication fork: <\/strong>
\nIt is the spot at which both strands of DNA get separated to allow replication of individual strands.
\nb. Stabilising proteins:<\/strong>
\nThey help in unwinding and parting of the two strands of the DNA molecule.
\nc. Nucleotide trisphosphate:<\/strong>
\nThese are the main components of nucleic acids. They not only provide energy but also give phosphate groups during phosphorylation reactions.
\nd. Leading strand:<\/strong>
\nThis is strand produced in continuation in 5\u2019 to 3\u2019 direction due to polymerization.
\ne. Helicase (enzyme):<\/strong>
\nIt is an enzyme which is responsible for separation of double helical strands of DNA.
\nf. Primase (enzyme):<\/strong>
\nIt is the enzyme that builds short RNA molecules complementary to the DNA template.
\ng. DNA polymerase III (enzyme):<\/strong>
\nThis is the enzyme involved in the proof reading function when nucleotides are being added to the new strand.
\nh. RNA primer:<\/strong>
\nIt is a small strand composed of nucleic acids and is the starting point to begin DNA synthesis.
\ni. Okazaki fragments:<\/strong>
\nThe short segments of DNA on the lagging strand during replication. These are joined as one by an enzyme ligase.
\nj. DNA polymerase I (enzyme):<\/strong>
\nIt 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.
\nk. Lagging strand: <\/strong>
\nThe newly formed discontinuous strand opposite to replication fork in leading strand in 3\u2019 \u2013 5\u2019 direction.
\nl. Ligase (enzyme):<\/strong>
\nThis enzyme is responsible for filling the gaps in sugar phosphate back bone of the DNA. It also helps in joining the Okazaki fragments.<\/p>\n

Microbiology with Diseases by Taxonomy Chapter 7 Answers 1MC<\/strong><\/span>
\nAn 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.
\na) 4,000,000 base pairs:<\/strong>
\nEach cell contains more than one type of chromosome. Chromosomes are existed as either duplicated (two identical copies) or unduplicated (single duplicated strands).
\nIn eukaryotes linear chromosome is present whereas in prokaryotes small circular chromosome is present.
\nIn 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.
\nb) 4000 base pairs:<\/strong>
\nEach cell contains more than one type of chromosome. Chromosomes are existed as either duplicated (two identical copies) or unduplicated (single duplicated strands).
\nIn eukaryotes linear chromosome is present whereas in prokaryotes small circular chromosome is present.
\nIn 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.
\nc) 400 base pairs:<\/strong>
\nEach cell contains more than one type of chromosome. Chromosomes are existed as either duplicated (two identical copies) or unduplicated (single duplicated strands).
\nIn eukaryotes linear chromosome is present whereas in prokaryotes small circular chromosome is present.
\nIn 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.
\nd) 40 base pairs:<\/strong>
\nEach cell contains more than one type of chromosome. Chromosomes are existed as either duplicated (two identical copies) or unduplicated (single duplicated strands).
\nIn eukaryotes linear chromosome is present whereas in prokaryotes small circular chromosome is present.
\nIn 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.
\nHence, the correct option is (a) 4,000,000 base pairs <\/strong>which is the present in the bacterial chromosome.<\/p>\n

Microbiology with Diseases by Taxonomy Chapter 7 Answers 1SA<\/strong><\/span>
\nGenotype of bacterium determines its phenotype:<\/strong>
\nExplanation<\/strong>
\nSet 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.
\nThe following methods of genotyping, which includes are:<\/strong><\/p>\n

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

    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.
    \nThus, a bacterium contains different genotypes with different proteins and determines different phenotype.<\/p>\n

    Microbiology with Diseases by Taxonomy Chapter 7 Answers 2CT<\/strong><\/span>
    \nThe amino acid sequences in polypeptide are synthesized by eukaryotic ribosomes:<\/strong>
    \nExplanation<\/strong><\/p>\n

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

      The sequence of amino acids in polypeptides is synthesized by eukaryotic ribosomes by the following below.<\/p>\n