{"id":9508,"date":"2019-12-03T10:58:01","date_gmt":"2019-12-03T05:28:01","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=9508"},"modified":"2021-07-02T15:44:17","modified_gmt":"2021-07-02T10:14:17","slug":"2nd-puc-biology-question-bank-chapter-6","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-biology-question-bank-chapter-6\/","title":{"rendered":"2nd PUC Biology Question Bank Chapter 6 Molecular Basis of Inheritance"},"content":{"rendered":"

You can Download Chapter 6 Molecular Basis of Inheritance Questions and Answers, 2nd PUC Biology Question Bank with Answers<\/a>, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 2nd PUC Biology Question Bank Chapter 6 Molecular Basis of Inheritance<\/h2>\n

2nd PUC Biology Molecular Basis of Inheritance Ncert Text Book Questions and Answers<\/h3>\n

Question 1.
\nGroup the following as nitrogenous bases and nucleosides:
\nAdenine, Cytidine, Thymine, Guanosine, Uracil and Cytosine.
\nAnswer:
\nNitrogenous bases Adenine, Thymine, Uracil, Cytosine. NucleosidesCytidine, Guanosine.<\/p>\n

Question 2.
\nIf a double stranded DNA has 20 percent of cytosine, calculate the percent of adenine in the DNA.
\nAnswer:
\n\"2nd
\nQuestion 3.
\nIf the sequence of one strand of DNA is written as follows:
\n5ATGCATGCATGC ATGCATGCATGC ATGC-3′
\nWrite down the sequence of complementary strand in 5’\u2014>3\u2019 direction.
\nAnswer:
\n5′ – ATGCATGC ATGC ATGC ATGC ATGC ATGC – 3′ (In question)
\ncomplement 3′ – TACG TACG TACG TACG TACG TACG – 5′
\ncomplement\u00a0in 5′ – 3′ direction 5′ – GCAT GCAT GCAT GCAT GCAT GCAT GC AT – 3′.<\/p>\n

Question 4.
\nIf coding strand in a transcription unit is written as follows:
\n5′-ATGCATGCATGCATGCATGCATGCATGC – 3′ Write down the sequence of m-RNA.
\nAnswer:
\nm RNA carries same code as coding strand except T is replaced by U.
\n(Question) transcription strand 5′ – ATGC ATGC ATGC ATGC ATGC ATGC ATGC – 3′ m RNA \u2192 5′ – AUGC AUGC AUGC AUGC AUGC AUGC AUGC – 3′.<\/p>\n

Question 5.
\nWhich property of DNA double helix led Watson and Crick to hypothesise semi-conservative mode of DNA replication? Explain.
\nAnswer:
\n2 strands of DNA has complementary base pairs which run in opposite direction. Property was that during DNA replication one strand was conserved while the complementary is formed new.<\/p>\n

\"KSEEB<\/p>\n

Question 6.
\nDepending upon the chemical nature of the template (DNA or RNA) and the nature of nucleic acids synthesised from it (DNA or RNA), list the types of nucleic acid polymerases.
\nAnswer:
\nDNA Template<\/p>\n

    \n
  • DNA polymerases for DNA replication.<\/li>\n
  • RNA polymerases for RNA synthesis or transcription<\/li>\n<\/ul>\n

    RNA Template<\/p>\n

      \n
    • RNA dependent RNA polymerases for synthesis of RNA is some RNA viruses.<\/li>\n
    • Reverse transcriptase to synthesize DNA over RNA template.<\/li>\n<\/ul>\n

      Question 7.
      \nHow did Hershey and Chase differentiate between DNA and protein in their experiment while proving that DNA is the genetic material?
      \nAnswer:
      \nThey raised 2 types of bacteriophages<\/p>\n

        \n
      • On radioactive phosphorous (32<\/sup>P)<\/li>\n
      • On radioactive sulphur (35<\/sup>S).<\/li>\n<\/ul>\n

        35<\/sup>S gets into protein and 32<\/sup>P into DNA When both bacteriophages infected bacteria differently and by shaking them, viral protein coat was separated<\/p>\n

        After raising these bacteria it was found that those infected with 32<\/sup>P bacteriophage \u2192 radioactivity was found. But with 35<\/sup>S \u2192 no radioactivity was found.<\/p>\n

        Question 8.
        \nDifferentiate between the followings:
        \n(a) Repetitive DNA and Satellite DNA
        \nAnswer:<\/p>\n\n\n\n\n\n\n
        Repetitive DNA<\/strong><\/td>\nSatellite DNA<\/strong><\/td>\n<\/tr>\n
        (i) sequence of N2\u00a0<\/sub>bases present in more than 1 copy in a genome.<\/td>\n(i) part of DNA \u00a0having repeated short sequences of N2 <\/sub>bases<\/td>\n<\/tr>\n
        (ii) Repeated DNA sequence may or may not be present in text.<\/td>\n(ii) Repeated sequence occur in tandem<\/td>\n<\/tr>\n
        (iii) Variability may or may not be present<\/td>\n(ii) Variability occurs<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

        (b) mRNA and tRNA
        \nAnswer:<\/p>\n\n\n\n\n
        m-RNA<\/strong><\/td>\nt – RNA<\/strong><\/td>\n<\/tr>\n
        (i) Large sized RNA with cap and tail.
        \n(ii) Carries codon information<\/td>\n        		(i)\u00a0\u00a0 Small sized with 3-4 loops and a limb
        \n(ii) Carries infor\u00admation for association of AA with anti Condon for incor\u00adporation<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

        (c) Template strand and Coding strand
        \nAnswer:<\/p>\n\n\n\n\n
        Template strand<\/strong><\/td>\nCoding strand<\/strong><\/td>\n<\/tr>\n
        (i) The strand of DNA which takes part in transcription.<\/td>\n(i) It does not takes part in transcription.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

        Question 9.
        \nList two essential roles of ribosome during translation.
        \nAnswer:<\/p>\n

          \n
        • Provides sites for attachment of m-RNA and charged t – RNA for polypeptide synthesis.<\/li>\n
        • One of its r – RNA functions as peptidyl transferase for peptide bond formation.<\/li>\n<\/ul>\n

          Question 10.
          \nIn the medium where E. coli was growing, lactose was added, which induced the lac operon. Then, why does lac operon shut down some time after the addition of lactose in the medium?
          \nAnswer:
          \nDue to decrease in lactose substrate concentration, lac operon shuts down as it is inducible operon.<\/p>\n

          Question 11.
          \nExplain (in one or two lines) the function of the followings:<\/p>\n

          (a) Promoter
          \nAnswer:
          \nIt is a gene that lies near operator which functions as binding site structural genes if operator allows.<\/p>\n

          (b) tRNA
          \nAnswer:
          \nFunctions as adapter molecule that picks\u00a0 up a particular amino – acid from cellular pool and takes same over A site, m – RNA for incorporation into polypeptide<\/p>\n

          (c) Exons
          \nAnswer:
          \nCoding segments present in primary transcript which on splicing by snRNP’s get joined to form functional m-RNA.<\/p>\n

          \"KSEEB<\/p>\n

          Question 12.
          \nWhy is the Human Genome project called a mega project?
          \nAnswer:
          \nBecause it involves<\/p>\n

            \n
          • Sequencing of over 3 \u00d7 109 bp<\/li>\n
          • Identification of all genes and alleles in human genome and their function.<\/li>\n
          • Storage of data requires storage space equal to 3300 books 1000 pages, each page with 1000 letters.<\/li>\n<\/ul>\n

            Question 13.
            \nWhat is DNA fingerprinting? Mention its applications.
            \nAnswer:
            \nDNA fingerprinting is technique of determing similarity and dissimilarity of VNTR’s between 2 samples of DNA so as to bring out relationship if any VNTR’s are specific for each individual. They are derived from parents in 50: 50 ratio.
            \nApplications<\/p>\n

              \n
            • Identification – of criminals<\/li>\n
            • Paternity and Maternity disputes.<\/li>\n
            • Migrations<\/li>\n
            • Human Lineage.<\/li>\n<\/ul>\n

              Question 14.
              \nBriefly describe the following:
              \n(a) Transcription
              \nAnswer:
              \nTranscription – Formation of-RNA over template of DNA.
              \n\"2nd
              \nThe snRNA formed has codon information similar to sense or coding strand of DNA with just U replacing T.<\/p>\n

                \n
              • The DNA strand which function as a template is template or anti-sense strand.<\/li>\n
              • Only one of the DNA strands is transcribed<\/li>\n
              • The main enzyme taking part in transcription is RNA polymerase.<\/li>\n<\/ul>\n

                \"2nd<\/p>\n

                (b) Polymorphism
                \nAnswer:
                \nPolymorphism It is the occurrence of more than one form of genetic material
                \nTypes \u2192<\/p>\n

                  \n
                • Allelic polymorphism – multiple alleles in a gene. So it alters structure and function of protein formed.<\/li>\n
                • SNP or single nucleotide Polymorphism – Unique in every human being. Useful in locating specific alleles, disease associated sequences.<\/li>\n
                • RFLP or Restriction Fragment Length polymorphism – Different sized fragments are formed by cleavage with same enzyme most RFPL have no effect on phenotypes.<\/li>\n<\/ul>\n

                  (c) Translation
                  \nAnswer:
                  \nTranslation (Biosynthesis of proteins) Coded genetic message brought by m – RNA from DNA is charged into polypeptide chain (proteins).
                  \n\"2nd
                  \nDNA transcription mRNA translation Proteins Materials required for translation are ribosomes, AA, t RNA’s, aninoacylt – RNA synthetase, m – RNA and same factors Step in Translation . Activation of Amino Acids \u2192 charging of t RNA\u2192 Initiation \u2192 Elongation \u2192 Termination \u2192 Modification.<\/p>\n

                  (d) Bioinformatics
                  \nAnswer:
                  \nScience of handling, storing as databases, analysing, modelling and providing access to various aspects of biological information, especially molecules connected with genomies and proteomics
                  \nApplications<\/p>\n

                    \n
                  • Organisation of Biological Data<\/li>\n
                  • Functional Genomics<\/li>\n
                  • Proteomics, Pharmacogenomics<\/li>\n
                  • Medical informatics, chemoinformatics<\/li>\n
                  • Faster drey research.<\/li>\n<\/ul>\n

                    \"KSEEB<\/p>\n

                    2nd PUC Biology Molecular Basis of Inheritance Additional Questions and Answers<\/h3>\n

                    2nd PUC Biology Molecular Basis of Inheritance One Mark Questions<\/strong><\/p>\n

                    Question 1.
                    \nExpand the term DNA, RNA and tRNA.
                    \nAnswer:
                    \nDNA – Deoxy ribonucleic acid RNA – Ribo nucleic acid tRNA – transfer – RNA<\/p>\n

                    Question 2.
                    \nDefine genetic code.
                    \nAnswer:
                    \nRelationship between the sequence of nucleotides or bases on mRNA and the sequence of amino acids in the polypeptide.<\/p>\n

                    Question 3.
                    \nWhat are Okazaki fragments?
                    \nAnswer:
                    \nThe small stretches of DNA formed due to the opposite running of DNA template at the time of replication is called Okazaki fragments.<\/p>\n

                    Question 4.
                    \nWhat is codon?
                    \nAnswer:
                    \nIt is the triplet nitrogenous base sequence which code for one amino acid. It lies on the m- RNA.<\/p>\n

                    Question 5.
                    \nWhat is anti codon?
                    \nAnswer:
                    \nIt in the triplet nitrogenous bases in the t-RNA complementary to an m-RNA codon. It identifies a particular codon on mRNA.<\/p>\n

                    Question 6.
                    \nWhat is a translation?
                    \nAnswer:
                    \nIt is the process of the formation of protein directed by a m-RNA molecule.<\/p>\n

                    Question 7.
                    \nDefine Transcription?
                    \nAnswer:
                    \nIt is the process of the formation of m-RNA on a DNA template.<\/p>\n

                    Question 8.
                    \nName the enzyme that catalyses
                    \n(a) Replication of DNA and
                    \n(b) Formation of RNA. (CBSE 1995)
                    \nAnswer:
                    \n(a) Topoisomerase.
                    \n(b) RNA polymerase.<\/p>\n

                    Question 9.
                    \nWhat is the genetic material of tobacco mosaic virus (TMV)
                    \nAnswer:
                    \nRNA.<\/p>\n

                    Question 10.
                    \nName the bond present between two adjacent nucleotides
                    \nAnswer:
                    \nPhosphodiester bond.<\/p>\n

                    Question 11.
                    \nDue to mistake during transcription, ATG forms UAG in m-RNA. What change would occur in polypeptide chain translated by this m-RNA? (Cbse 1996)
                    \nAnswer:
                    \nUAG is a termination codon and so at point protein synthesis will get stopped.<\/p>\n

                    \"KSEEB<\/p>\n

                    Question 12.
                    \nName three different non – sense codons.
                    \nAnswer:
                    \nUAA, UAG and UGA.<\/p>\n

                    Question 13.
                    \nIf the base sequence of one strand of DNA is CAT TAG TAC GAC, what will be the base sequence
                    \n(a) Of complementary DNA strand, and
                    \n(b) Of its complementary RNA strand? (CBSE 1991)
                    \nAnswer:
                    \n(a) GTA ATC ATG CTG
                    \n(b) GUA AUCAUGCUG.<\/p>\n

                    Question 14.
                    \nName the scientist who proposed one gene one enzyme hypothesis. (CBSE 1997)
                    \nAnswer:
                    \nBeadle and Tatum.<\/p>\n

                    Question 15.
                    \nName the enzyme that joints the short pieces in the lagging strand during synthesis of DNA.
                    \nAnswer:
                    \nDNA legase. (CBSE 1998)<\/p>\n

                    Question 16.
                    \nIn which direction 5′ – 3′ or 3′ -5′ are the new strands of DNA formed during replication ? (CBSE 1992, 2K)
                    \nAnswer:
                    \n5′ – 3′ direction<\/p>\n

                    Question 17.
                    \nGive the present-day representation of central dogma.
                    \nAnswer:
                    \n\"2nd<\/p>\n

                    Question 18.
                    \nState Chargaff s base complementary rule.
                    \nAnswer:
                    \nThe total molar amount of adenine in any specimen of DNA is always equal to that of thymine. In a given DNA A = T and G = C.<\/p>\n

                    Question 19.
                    \nDefine mutation
                    \nAnswer:
                    \nIt is an abrupt and distinct change in the structure of base pair. It is a discontinuous inheritable and sudden change in an organism.<\/p>\n

                    Question 20.
                    \nWhat do you call the kind of mutation in which a single base is added to a base strand : (CBSE 2K)
                    \nAnswer:
                    \nFrame shift mutation.<\/p>\n

                    Question 21.
                    \nSickle cell anaemia is caused due to abnormal haemoglobin. Which chain of haemoglobin is responsible for this disease?
                    \nAnswer:
                    \n\u03b2 – chain of haemoglobin<\/p>\n

                    Question 22.
                    \nWhat do the triplets AUG and UGA respectively code for during proteins synthesis?
                    \nAnswer:
                    \nAUG – Methionine UGA – Termination codon (Nonsense codon)<\/p>\n

                    Question 23.
                    \nName the technique used by Watson and Crick to propose the double-helical structure of DNA molecule?
                    \nAnswer:<\/p>\n

                      \n
                    • X-ray crystallography<\/li>\n
                    • X-ray diffraction method.<\/li>\n<\/ul>\n

                      Question 24.
                      \nWho discovered nucleic acid DNA? What was it called them?
                      \nAnswer:
                      \nFredrich Meischer. It was called as nuclein.<\/p>\n

                      Question 25.
                      \n(a) What is the length of the pitch of helix?
                      \n(b) What is the distance between 2 base pairs is a stand of DNA?
                      \nAnswer:
                      \n(a) 3.4 nm
                      \n(b) 34 nm.<\/p>\n

                      Question 26.
                      \nWhy is the distance between the 2 nucleotide chains in a DNA maintained almost constant?
                      \nAnswer:
                      \nThe complementary base pairing between the 2 strands i.e. adenine and thymine (double bond) and Guanine and cytosine (triple bond) is responsible for the uniform distance between the 2 strands.<\/p>\n

                      Question 27.
                      \nName the process which occurs in virus where the formation of DNA occurs from RNA.
                      \nAnswer:
                      \nReverse transcription.<\/p>\n

                      Question 28.
                      \nName the components ‘a’ and ‘b’ in the nucleotides with a parent given below. (CBSE, Delhi 2008)
                      \nAnswer:
                      \n\"2nd<\/p>\n

                      Question 29.
                      \nWhat are histones?
                      \nAnswer:
                      \nHistones are positively charges proteins found in association with DNA in eukaryotic cell.<\/p>\n

                      Question 30.
                      \nName 2 organisms where RNA as the genetic material.
                      \nAnswer:
                      \nQB Bacteriophage, Tobacco mosaic virus.<\/p>\n

                      Question 31.
                      \nName the main enzyme involved in the replication of DNA.
                      \nAnswer:
                      \nDNA dependent DNA polymerase enzyme.<\/p>\n

                      \"KSEEB<\/p>\n

                      Question 32.
                      \nName the types of synthesis ‘a’ and ‘b’ occurring in the replication of DNA as shown below (CBSE 2008)
                      \nAnswer:
                      \n\"2nd<\/p>\n

                        \n
                      • a – continuous synthesis<\/li>\n
                      • b – discontinuous synthesis.<\/li>\n<\/ul>\n

                        Question 33.
                        \nWhat is replication fork in DNA?
                        \nAnswer:
                        \nDuring DNA replication the unwinding of DNA leads to the formation of ‘Y’ shaped structure to the 2 strands of DNA duplex. This is known as the replication fork.<\/p>\n

                        Question 34.
                        \nDefine a cistron
                        \nAnswer:
                        \nA cistron is defined as the length of mRNA, that codes for a polypeptide.<\/p>\n

                        Question 35.
                        \nWhat is meant by hnRNA?
                        \nAnswer:
                        \nThe hnRNA is the precursor of mRNA transcribed by RNA polymerase II in eukaryotic cells hn represents heteronuclear RNA.<\/p>\n

                        Question 36.
                        \nWhen and at what end does the ‘tailing’ of hnRNA take place? (AI 2009)
                        \nAnswer:
                        \nAfter splicing, tailing occurs at the 3′ end of hn RNA.<\/p>\n

                        Question 37.
                        \nWhy is hn RNA required to undergo splicing?
                        \nAnswer:
                        \nSince hn RNA contain both the coding sequences (exon) and the non-coding sequences (introns), hn RNA has to undergo splicing for the removal of introns.<\/p>\n

                        Question 38.
                        \nWho proposed the operon concept?
                        \nAnswer:
                        \nFrancois Jacob and Jacque Monod proposed the operon concept.<\/p>\n

                        Question 39.
                        \nName the induce of lac operon in E. Coli?
                        \nAnswer:
                        \nLactose.<\/p>\n

                        \"KSEEB<\/p>\n

                        Question 40.
                        \nGiven below is a schematic representation of a lac operon in the absence of an inducer. Identify ‘a’ and ‘b’ in it
                        \nAnswer:
                        \n\"2nd
                        \na – Repressor
                        \nb – operator.<\/p>\n

                        Question 41.
                        \nWhat term is given to a single base DNA differences?
                        \nAnswer:
                        \nWhen the repressor binds to the operator, the operon is switched off and transcription is stopped.<\/p>\n

                        Question 42.
                        \nExpand VNTR.
                        \nAnswer:
                        \nVariable Number of Tandem Repeats.<\/p>\n

                        Question 43.
                        \nRegulation of lac operon by repressor is referred to as negative why is it so?
                        \nAnswer:
                        \nWhen the repressor binds to the operator, the operon is switched off and transcription is stopped.<\/p>\n

                        Question 44.
                        \nName 2 plants whose genome have been sequenced.
                        \nAnswer:
                        \nRice and Arabidopsis.<\/p>\n

                        Question 45.
                        \nDefine DNA polymorphism
                        \nAnswer:
                        \nInheritable mutations at high frequency in a population. It refers to the variation at genetic level.<\/p>\n

                        Question 46.
                        \nWho discovered the techniques of DNA fingerprinting
                        \nAnswer:
                        \nAlec Jeffreys.<\/p>\n

                        Question 47.
                        \nWhat is a probe in DNA-finger printing?
                        \nAnswer:
                        \nA probe is a short stretch of DNA, with the nucleotide sequence that is complementary to that of VNTR sequence.<\/p>\n

                        Question 48.
                        \nName the branch of science that HGP is closely associated with.
                        \nAnswer:
                        \nBioinformatics.<\/p>\n

                        \"KSEEB<\/p>\n

                        Question 49.
                        \nGiven below is the sequence of steps of transcription in eukaryotic cell. Fill up the blank 1, 2, 3, 4 left in the sequence.
                        \nAnswer:
                        \n\"2nd
                        \nSingle nucleotide polymorphism (SNPS).
                        \n1 – RNA polymerase
                        \n2 – hn
                        \n3 – processed RNA
                        \n4 – tail.<\/p>\n

                        Question 50.
                        \nWrite any 3 unusual bases present in Yeast’s alanine tRNA with their sources.
                        \n(CBSE 2004)
                        \nAnswer:
                        \nAll tRNA have 2 unusual bases – dihydro uridine (derived from uracil) and pseudo uridine (from uraicl. The third common unusual base is hypoxanthine (from adenine).<\/p>\n

                        2nd PUC Biology Molecular Basis of Inheritance Two Marks Questions<\/strong><\/p>\n

                        Question 1.
                        \nWhat are 5′ end, 3′ end of a polynucleotide chain?
                        \nAnswer:<\/p>\n

                          \n
                        • The poly nucleotide chain has at one end a free phosphate at the 5′ end of the pentose sugar. Which is referred to as 5′ end.<\/li>\n
                        • The other end of polynucleotide chain has a free. 3′ – OH group at the 3rd end of the pentose sugar which is referred to as 3′ end of polynucleotide chain.<\/li>\n<\/ul>\n

                          Question 2.
                          \nLabel the diagram 1, 2, 3, 4, 5, 6. (CBSE 2004)
                          \nAnswer:
                          \n\"2nd<\/p>\n

                            \n
                          • 5′ end<\/li>\n
                          • Ribosome binding site<\/li>\n
                          • Start<\/li>\n
                          • Stop signal<\/li>\n
                          • Open reading frames (ORFs)<\/li>\n
                          • 3′ end.<\/li>\n<\/ul>\n

                            Question 3.
                            \nName the components (parts) A and B of the transcription unit \u2018given below.
                            \nAnswer:
                            \n\"2nd
                            \nA – Promoter
                            \nB – Coding strand.<\/p>\n

                            Question 4.
                            \n(a) Who first proposed semi-conservative mode of replication of DNA.
                            \n(b) Which organisms is used in this experiment?
                            \n(c) Name the techniques used
                            \n(d) What is the result of first, second and 3rd generation?
                            \nAnswer:
                            \n(a) Watson and Crick (1953)<\/p>\n

                              \n
                            • Eschirisia Coli<\/li>\n<\/ul>\n

                              (i) Use heavy isotope 15<\/sup>N instead of normal 14
                              \n<\/sup>(ii) Use of Cesium chloride based density gradient configuration with ethidium bromide as flarochrome.
                              \n(iii) First generation DNA is hybrid as with intermediate heaviness due to presence of both 14<\/sup>N and 15<\/sup>N strands. The parent generation is heaviest.
                              \nSecond generation 50% light (14<\/sup>N\/14<\/sup>N) and 50% intermediate heavy (14<\/sup>N\/15<\/sup>N). Third generation75% light (14<\/sup>N\/14<\/sup>N) and 25% of intermediate heaviness (14<\/sup>N\/15<\/sup>N).\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 This is possible only if DNA is double stranded with semi conservative replication. (one parent strand and other new strand) UACGAG AGAUUUi<\/p>\n

                              Question 5.
                              \n\"2nd
                              \nStudy the messenger RNA segment given above which is complete to be translated into a polypeptide chain<\/p>\n

                                \n
                              • Write the codons and ‘a’ and ‘b’<\/li>\n
                              • What do they code for?<\/li>\n
                              • How is peptide bond formed between 2 amino acids in the ribosome (CBSE 2008)<\/li>\n<\/ul>\n

                                Answer:
                                \n(i) a – AUG
                                \nb – UAA | UAG | UGA<\/p>\n

                                (ii) AUG codes for methionine.
                                \nUAA\/ UAG\/ UGA is stop \/ nonsense codon.<\/p>\n

                                (iii) Peptide bond is formed between -COOH group of P- site amino acid and NH2<\/sub> – group of A – site amino acid with the help of ribozine peptidyl transferase provided by ribosome.<\/p>\n

                                Question 6.
                                \nState any one reason to explain why RNA virus mutate and evolve faster than other viruses.
                                \nAnswer:
                                \nRNA is an unstable highly reactive molecule due to its single-stranded structure and exposure of its nitrogen bases. But DNA is stable, the molecule, as its, nitrogen bases are not exposed, became its double-helical nature.<\/p>\n

                                Repressor binds to the operator region
                                \n(o)\u00a0 and prevent RNA polymerase from transcribing the given.
                                \nLook at the figure above depicting lac operon in E-Coii.
                                \n(a) What could be a series of events when an inducer is present in the medium in which E. Coli is growing.
                                \n(b) Name the inducer.
                                \nAnswer:
                                \n(a) When the inducer is present in the medium it is absorbed at first slowly into the bacterium. The inducer binds with the repressor attached to the operator gene the repressor leaves the operator gene and allows the RNA polymerase to pass from promotor to the structural genes for transcription or formation of a polycistronic mRNA.
                                \n(b) Lactose or galactoside.<\/p>\n

                                \"KSEEB<\/p>\n

                                Question 7.
                                \nDraw schematically a single polynucleotide strand (with at least three nucleotides)
                                \nAnswer:
                                \n\"2nd<\/p>\n

                                Question 8.
                                \nDifferentiate between euchromatin and heterochromatin.
                                \nAnswer:
                                \nEuchromation<\/p>\n

                                  \n
                                • These are the regions where chromatin is loosely packed.<\/li>\n
                                • Euchromation stains lighter<\/li>\n
                                • This is transcriptionally more active.<\/li>\n<\/ul>\n

                                  Hetero chromatin<\/p>\n

                                    \n
                                  • These are the regions where chromatin is tightly packed<\/li>\n
                                  • Hetero chromatin stains darker<\/li>\n
                                  • This is transcriptionally less active or inert.<\/li>\n<\/ul>\n

                                    Question 9.
                                    \nWrite the functions of DNA polymerase.
                                    \nAnswer:<\/p>\n

                                      \n
                                    • DNA polymerase catalyses the polymerisation of nucleotides into a polynucleotide strand of DNA.<\/li>\n
                                    • It carries out proof reacting in prokaryotes; it moves back and removes any wrong bases added before polymerisation continues.<\/li>\n<\/ul>\n

                                      Question 10.
                                      \nCompare the roles of the enzymes DNA polymerase and DNA ligase in the replication fork of DNA.
                                      \nAnswer:
                                      \nDNA polymerases the nucleotides in the 5\u2019 \u2014> 3′ direction, as a continuous stretch on the template strand with 3′ \u2014> 5′ polarity and short stretches on the template strand with 5′ \u2014> 3′ polarity. \u2014>\u00a0DNA ligase joins the short stretches of DNA formed on the template strand with 5′-3′ polarity.<\/p>\n

                                      Question 11.
                                      \nWrite the differences between mono-cistronic and polycistronic mRNAs.
                                      \nAnswer:<\/p>\n\n\n\n\n
                                      Monocistromic mRNA<\/td>\nPolycistromic mRNA<\/td>\n<\/tr>\n
                                      (1) It is the mRNA that can code for only one polypeptide i.e. it has one cistron.
                                      \n(2) it is normally found in eukaryotic cell.<\/td>\n                                      		(1) It is the mRNA that can code for more than one polypeptide i.e. it has more than one cistron.
                                      \n(2) It is found in prokaryotic cell.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                                      Question 12.
                                      \nDifferentiate between Exons and Introns.
                                      \nAnswer:
                                      \nExons (HOTS)<\/p>\n

                                        \n
                                      • They are the coding sequence of DNA\/RNA transcript, that form parts of mRNA and code for different regions of the polypeptide.<\/li>\n
                                      • They are joined together during splicing to make the information continuous.<\/li>\n<\/ul>\n

                                        Introns<\/p>\n

                                          \n
                                        • Introns are non coding sequences of DNA\/ RNA transcript that do not become part of mRNA.<\/li>\n
                                        • They are removed during splicing.<\/li>\n<\/ul>\n

                                          Question 13.
                                          \nGenetic code is specific and nearly universal justify. (Delhi 2008)
                                          \nAnswer:
                                          \nSince a codon codes only for a particular amino acid, genetic code is specific or unambiguous. A codon codes for one particular amino acid in all living, be it a bacterium or a human, hence it is universal.<\/p>\n

                                          \"KSEEB<\/p>\n

                                          Question 14.
                                          \nWhy is that transcription and translation can be coupled in prokaryotic cell but not in eukaryotic cells?
                                          \nAnswer:
                                          \n(a) In prokaryotes, mRNA does not require any processing to become active<\/p>\n

                                          (b) Transcription and translation occur in the same compartment cytosol, as there is no well defined nuclear membrane. There for it can be coupled.<\/p>\n

                                          In eukaryotic m-RNA has to be processed (splicing) before it become active. Since RNA is synthesised inside the nucleus and translation occurs in the cytoplasm, coupling of transcription and translation is not possible.<\/p>\n

                                          Question 15.
                                          \nWhy is it essential that RNA binds to both an amino acid and an mRNA codon, during protein synthesis?
                                          \nAnswer:
                                          \n(a) By binding an amino acid at its 3′ end the tRNA transports the amino acid to the site of protein synthesis.<\/p>\n

                                          (b) It binds to the codon on mRNA through hydrogen bonds with its anticodon as it is recognised by the codon of the amino acid it carries, it is to release the amino acid at the correct site for polypeptide elongation.<\/p>\n

                                          Question 16.
                                          \nWhat is an operon? How does lactose act as an inducer in the lac operon? (Al 2008)
                                          \nAnswer:
                                          \nAll the genes controlling a metabolic pathway, collectively constitute an operon. Lactose binds to the repressor and inactivates it, and prevents it from binding to the operator. As a result, the RNA polymerase gets access to the promoter and transcription proceeds, ie., the operon is induced to function.<\/p>\n

                                          Question 17.
                                          \nExplain VNTR as the basis of DNA finger printing.
                                          \nAnswer:<\/p>\n

                                            \n
                                          • Variable number of tandem repeats belong to the class of satellite DNA referred to as mini satellite.<\/li>\n
                                          • The number of repeats shows very high degree of polymorphism as a result the size of VNTR varies.<\/li>\n
                                          • After hybridisation of DNA sample with VNTR probe, an autoradiogram developed which gives many bands give the characteristic pattern of an individual DNA.<\/li>\n<\/ul>\n

                                            Question 18.
                                            \nGive the applications of DNA fingerprinting?
                                            \nAnswer:<\/p>\n

                                              \n
                                            • To identify criminals in the forensic labs.<\/li>\n
                                            • To determine the biological parent in case of dispute.<\/li>\n
                                            • To verify whether an immigrant is really a close relative of the mentioned resident.<\/li>\n
                                            • To identify racial groups to rewrite the biological evolution.<\/li>\n<\/ul>\n

                                              Question 19.
                                              \nHow is nucleosome formed? Draw a diagram of the nucleosome.
                                              \nAnswer:
                                              \n\"2nd
                                              \nIn eukaryotes, histones which are positively charged proteins, become organised as a unit of 8 molecules, called histone octamer. The negatively charged DNA is wraped around the positively charged histone octamer, to form the structure, called nucleosome.<\/p>\n

                                              \"KSEEB<\/p>\n

                                              Question 20.
                                              \nWhat is meant by R – cells and S – cell with which Frederick Griffith carried out his experiments on Diplococcus pneumonial? What did he prove from these experiments?
                                              \nAnswer:
                                              \nR – cells are those bacterial cells that form rough colonies, without a capsule and are non-virulent. S – cells are those cells, that form smooth colonies, with a capsule and are virulent. He proved that virulence of S – cells had some how been transferred into R – cells, which became transformed into S- cells, it is the genetic material that had effected transformation.<\/p>\n

                                              Question 21.
                                              \nDraw the schematic representation to show continuous or discontinuous synthesis of DNA (replication of DNA) and label it.
                                              \nAnswer:
                                              \n\"2nd<\/p>\n

                                              Question 22.
                                              \nExplain frames shift mutation.
                                              \nAnswer:
                                              \n(a) It is the type of mutation where insertion or deletion of one or two bases changes the reacting frame from the point of insertion or deletion.<\/p>\n

                                              (b) When three or multiples of three bases are added there is addition of one or more amino acids and the reacting of the frame remains unaltered after that. This provides the genetic proof that codons are triplets.<\/p>\n

                                              Question 23.
                                              \nA t-RNA is charged with the amino acid phenylalanine.
                                              \n(a) At what end of the tRNA is the amino acid attached ?
                                              \n(b) Give the mRNA codon that codes for phenylalanine.
                                              \n(c) Which enzyme is responsible for this attachment ?
                                              \nAnswer:
                                              \n(a) Amino acid is attached to the 3′ end of t-RNA.
                                              \n(b) UUC or UUU are the codes for phenylalanine.
                                              \n(c) Amino Acyl tRNA synthetase enzyme is responsible for this attachment.<\/p>\n

                                              Question 24.
                                              \n(a) Draw the schematic diagram of tRNA showing the following.
                                              \n(i) Methionine attached to the amino acid accept site.
                                              \n(ii) Correct base sequence at the anticodon loop
                                              \n(b) Write the role of “untranslated regions” on mRNA segment play in protein synthesis?
                                              \nAnswer:
                                              \n(a)
                                              \n\"2nd
                                              \n(b) They are needed for efficient translation<\/p>\n

                                              Question 25.
                                              \nDraw a labelled diagram depiciting schematically the process of elongation in translation
                                              \nAnswer:
                                              \n\"2nd
                                              \nQuestion 26.
                                              \nDescribe the goals of Human genoine project.
                                              \nAnswer:
                                              \nThe major goals of Human genome project are<\/p>\n

                                                \n
                                              • Determine the sequence of the 3 billion base pairs present in Human DNA.<\/li>\n
                                              • Identify the genes in the human DNA.<\/li>\n
                                              • Store the information in data bases.<\/li>\n
                                              • Improve the tool for data analyses.<\/li>\n
                                              • Address ELSI (ethical, leagal, social issues) that arise from project.<\/li>\n
                                              • Transfer the technologies to other sectors.<\/li>\n<\/ul>\n

                                                2nd PUC Biology Molecular Basis of Inheritance Five Marks Questions<\/strong><\/p>\n

                                                Question 1.
                                                \nDescribe the features of double helical model of DNA.
                                                \nAnswer:
                                                \nThe features of double helical DNA are<\/p>\n

                                                  \n
                                                • It is made up of 2 polynucleotide of sugar phosphate and the nitrogen bases inside.<\/li>\n
                                                • The two chains have antiparallel polarity ie., one has 5′ \u2014> 3\u2019 polarity and the other with 3′ \u2014> 5′ polarity.<\/li>\n
                                                • The bases of 2 strands are joined by double hydrogen bond between adenine and thymine and triple hydrogen bond between guanine cytosine.<\/li>\n
                                                • The distance between 2 base pairs is 0.34nm and the distance between each turn is 3.4nm. Each turn consists of 10 base pairs.<\/li>\n
                                                • The plane of one base pair stacks over the other in double helen this gives the stability of double helical structure.<\/li>\n<\/ul>\n

                                                  \"KSEEB<\/p>\n

                                                  Question 2.
                                                  \n(1) Represent diagramatically the Watson Crick model for semi conservation replication of DNA.
                                                  \n(2) Differentiate between continuous and discontinuous synthesis of DNA.
                                                  \nAnswer:
                                                  \n\"2nd
                                                  \n(2)<\/p>\n\n\n\n\n\n\n\n
                                                  Continuous<\/strong><\/td>\nDiscontinuous<\/strong><\/td>\n<\/tr>\n
                                                  (a) One stand of DNA is synthesised as a continuous strech in the
                                                  \n5′ \u2014>3′ direction.<\/td>\n                                                  		(a) Short stretches \u00a0are synthesised in the 5’\u2014>3′ direction from replication fork<\/td>\n<\/tr>\n
                                                  (b) The template strand of DNA strand is with 3’\u2014>5′ polarity.<\/td>\n(b) The DNA strand with 5′ \u2014> 3′ polarity is the template strand \u00a0for this.<\/td>\n<\/tr>\n
                                                  (c) No need of enzyme ligase (for joining)<\/td>\n(c) DNA ligase \u00a0enzyme is required \u00a0for joining short stretches.<\/td>\n<\/tr>\n
                                                  (d) There is no need for primers.<\/td>\n(d) There is need for primers.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                                                  Question 3.
                                                  \nWrite short notes on different types of RNAs.
                                                  \nAnswer:
                                                  \nThe RNAs are of 3 types<\/p>\n

                                                    \n
                                                  • Messenger RNA (m – RNA)<\/li>\n
                                                  • Transfer RNA (t – RNA)<\/li>\n
                                                  • Ribosomal RNA (r – RNA)<\/li>\n<\/ul>\n

                                                    m – RNA It provides the template for polypeptide synthesis, ie., it decides the sequence of amino acids in the polypeptide through the sequence of bases on it. t – RNA It has the shape of clover leaf in two dimensional structure It transports the amino acids to the site of protein synthesis it recognises the codon on m RNA. r – RNA It plays the structural and catalytic role during translation.<\/p>\n

                                                    Question 4.
                                                    \nDefine genetic code and write its salient features.
                                                    \nAnswer:
                                                    \nGenetic code is the relationship between the sequence of nucleotides on mRNA and sequence of amino acids in the polypeptide.
                                                    \nIts features are<\/p>\n

                                                      \n
                                                    • The genetic code is universal ie., the codons code for one amino acid is same in all organism.<\/li>\n
                                                    • Codons are triplet codons and there are 64 codons. 61 codons code for twenty different amino acids while the other three (UAA, UAG, UGA) are termination codons which do not code for any amino acids.<\/li>\n
                                                    • Each codon codes for only one particular amino acid. Therefore it is unambiguous.<\/li>\n
                                                    • Some amino acids are coded by more than one codon there for it is said to be degenerate.<\/li>\n
                                                    • The codons are read in a continuous manner, without any punctuation ie., codons are comma less.<\/li>\n
                                                    • AUG has dual functions of coding for metheonine as well as functioning as initiation codon.<\/li>\n<\/ul>\n

                                                      \"KSEEB<\/p>\n

                                                      Question 5.
                                                      \n(1) Represent schematically the process of transcription in eukaryotic cells.
                                                      \n(2) How does DNA polymerase function in the replication fork of DNA?
                                                      \nAnswer:
                                                      \n\"2nd
                                                      \n(2) The DNA polymerase can catalase the polymerisation of nucleotides only in one direction is 5′ \u2014> 3′ direction.
                                                      \nOn the template stand with 3’\u2014>5′ polarity the polymerisation occurs continuously and on the template strand with 5′ \u2014> 3′ polarity, polymerisation occurs in short stretches (discontinuous synthesis).<\/p>\n

                                                      Question 6.
                                                      \n(a) Explain with the help of schematic representation, the lac operon in E-Coli.
                                                      \n(b) Mention the role of lactose in this operon.
                                                      \nAnswer:
                                                      \n(a) Jacob and Monod explained that lactose inducer the expression of genes leading to its catabolism.
                                                      \n\"2nd
                                                      \n(b) Role of lactose<\/p>\n

                                                        \n
                                                      • Lactose is a substrate for the enzyme (3- galactosidase.<\/li>\n
                                                      • It function as the inducer and regulates the switching on and off of the operon.<\/li>\n
                                                      • When lactose is present, it combines with the repressor protein which otherwise has a high affinity for the operator.<\/li>\n
                                                      • This inactivates the repressor from binding to the operator and hence transcription continuous ie., the operon is switched on.<\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"

                                                        You can Download Chapter 6 Molecular Basis of Inheritance Questions and Answers, 2nd PUC Biology Question Bank with Answers, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations. Karnataka 2nd PUC Biology Question Bank Chapter 6 Molecular Basis of Inheritance 2nd PUC Biology Molecular Basis of Inheritance …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[82],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/9508"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=9508"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/9508\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=9508"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=9508"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=9508"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}