cx-ncert-exercise-class12-chap05-molecular-basis-of-inheritance-en

<h6 id="successmolecular-basis-of-inheritancesuccess-3"><Success>Molecular basis of inheritance</Success></h6>
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<p><strong>Solution:-</strong> As per Chargaff’s rule, DNA molecules are required to have an equal ratio of purine (adenine and guanine) and pyridine (cytosine and thymine). This is to say that the number of adenine molecules is equivalent to the cytosine molecule.</p>
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<p>Percentage of adenosine = percentage of thymine</p>
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<p>Percentage of guanine = percentage of cytosine</p>
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<p>Hence, according to the law, if the double-stranded DNA has 20% of cytosine, it should have 20% of guanine. Therefore, the percentage of G + percentage of C = 40%</p>
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<p>The other 60% indicates both A + T percentage molecules. As adenine and thymine are always found in equal numbers, the adenine content is 30%.</p>
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<p><strong>Solution:-</strong> Watson and Crick noted that the two DNA strands are anti-parallel, complementary to each other with regards to their base sequences which facilitates each strand to serve as a template to synthesise a new strand. This organisation of the DNA molecule furthered the hypothesis that the replication of DNA is semi-conservative.</p>
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<p>In other words, the double-stranded DNA molecule splits, and in turn, each of the separated strands serves as a template to synthesise a new complementary strand. Subsequently, each DNA molecule will have one parental strand and a newly synthesised daughter strand.</p>
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<p>As only one parental strand is conserved in each of the daughter molecules, the mode of replication is termed as semi-conservative.</p>
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<p><strong>Solution:-</strong> To prove DNA is the genetic material, Hershey and Chase studied and worked on bacteriophage and E.Coli. To label the protein coat and DNA of the bacteriophage, they used different radioactive isotopes. In a medium containing radioactive phosphorous (32P), they cultivated some bacteriophages to detect DNA and a few more on a medium comprising radioactive sulphur (35S) to detect protein. The labelled radioactive phages were then made to infect the bacteria – E.coli. Once infected, the protein coat of the bacteriophage was segregated from the bacterial cell by mixing and then subject to the centrifugation process.</p>
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<li>It was observed that, in the supernatant, as the protein coat was lighter, the infected bacteria got settled at the bottom of the centrifuge tube. In case I – the supernatant was observed to be radioactive, indicating that the protein did not enter the bacterial cell when infected. But in case II – the bacterial cells were radioactive as they possess radioactive DNA. Thus, it was proved that DNA is a genetic material as it was transferred from viruses to bacteria.</li>
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<h6 id="successmolecular-basis-of-inheritancesuccess-22"><Success>Molecular basis of inheritance</Success></h6>
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<p><strong>Solution:-</strong> A segment of DNA which comprises three adjacent structural genes – a promoter gene, an operator gene, and a regulator gene is known as a lac operon.</p>
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<p>It is functional in a coordinated manner in order to metabolise lactose into galactose and glucose. Lactose serves as an inducer in the lac operon, which binds to the repressor, inactivating it.</p>
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<p>Once the lactose is bound to the repressor, RNA polymerase binds to the area of the promoter. Therefore, the structural genes express their product, and corresponding enzymes are yielded.</p>
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<p>Lactose is metabolised due to the action of enzymes into galactose and glucose. Later on, the level of inducer decreases as enzymes completely metabolise them, which results in the synthesis of the repressor from the regulator gene.</p>
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<p>Thus, the repressor is bound to the operator gene, restricting RNA polymerase from transcribing the operon; subsequently, transcription is ceased. This kind of regulation is termed negative regulation.</p>
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<p><strong>Solution:-</strong> (a) Transcription</p>
<p>The process of transcription is initiated when the DNA-dependent RNA polymerase and several initiation factors bind at the promoter region of the template strand at the double-stranded DNA. The RNA polymerase passes along the DNA, causing the DNA duplex to unwind into two distinct strands. One of the strands is known as the sense strand, which serves as a template for mRNA synthesis.</p>
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<p>The RNA polymerase enzyme uses nucleoside triphosphates (dNTPs) as raw material to polymerize them to form mRNA as per the complementary bases positioned in the template DNA. This phenomenon of the opening of the helix and elongation of the polynucleotide chain lasts until the enzyme arrives at the terminator region.</p>
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<p>Once the RNA polymerase arrives at the terminator section, the freshly produced mRNA transcripted in addition to the enzyme is liberated. Additionally, another factor known as the terminator factor is essential for the transcription process to end.</p>
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<p>(b) Polymorphism</p>
<p>It is a form of genetic variation wherein different nucleotide sequences can be present at different sites in a molecule of DNA. There is a high frequency of this heritable mutation to be observed in a population which emerges as a result of mutation either in the germ cells or somatic cells. The germ cell mutation can be passed from parents to offspring, which leads to the accumulation of different mutations in a population, causing variation and polymorphism in the population. This has a crucial role to play in the evolution and speciation process.</p>
<p>(c) Translation</p>
<p>It is the process wherein amino acids are polymerized for the formation of a polypeptide chain, a ribosome, by reading an mRNA molecule. In mRNA, the triplet sequence of base pairs defines the order and sequence of amino acids in a polypeptide chain.</p>
<p>This phenomenon includes three steps:</p>
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<li>Initiation  2. Elongation  3. Termination</li>
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<h6 id="successmolecular-basis-of-inheritancesuccess-35"><Success>Molecular basis of inheritance</Success></h6>
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<p>The tRNA gets charged during the initiation of the translation when the amino acids bind to it with the help of ATP. The initiation code – AUG that is located on the mRNA is identified only by the charged tRNA. For the process of translation, the ribosome acts as an actual site, containing two distinct sites in a large subunit to attach to the following amino acids.</p>
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<p>The smaller subunit of the ribosome is bound to the mRNA at AUG (initiation code), subsequently by the large subunit. The process of translation is then said to be initiated.</p>
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<p>In the elongation process, the ribosome passes one codon downstream with the mRNA in order to leave some space to bind another charged tRNA. Amino acid brings tRNA, which gets linked with the former amino acid via a peptide bond.</p>
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<p>The process continues leading to the formation of a polypeptide chain. When the ribosome arrives, one or more STOP codons (UAA, UGA, and UAG). The phenomena of translation end, polypeptide chain releases and the ribosomes get detached from mRNA.</p>
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<p>(d) Bioinformatics</p>
<p>It is the application of statistical and computational methodologies to the molecular biology stream. It explains practical issues that arise from the analysis and management of biological data. The stream of bioinformatics developed after accomplishing the human genome project as a huge amount of data was produced during the process, which has to be stored and managed for easy access and analysis for later use.</p>
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<li>Thus, bioinformatics includes creating biological databases which store huge amounts of data on biology. It comes up with a few tools for efficient and easy access to data which can be used. Also, bioinformatics brought in new algorithms and statistical techniques to figure out the dynamics between data, to predict the structure of protein and their functionalities and to group protein sequences into their associated families.</li>
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