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Molecular Basis of Inheritance Test - 99

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Molecular Basis of Inheritance Test - 99
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  • Question 1
    1 / -0
    Select the correct option that correctly fill the blanks i - iv
    I. Less than (i) of genome represents structural genes that code for proteins.
    II. Chemical substance that binds with repressor and convert it into a non-DNA binding State is (ii).
    III. In prokaryotes, during replication RNA primer is removed by (iii) whereas in eukaryotes it is removed by (iv).
    Solution
    The human genome is made of 3 billion DNA base pairs and out of them 2% only are responsible for coding for proteins and the rest may have regulatory functions. In an inducible operon such as the Lac Operon the inducer binds to the repressor to inactivate it such that the repressor can no longer bind to the operator and the gene expression can take place. DNA polymerase I (or Pol I) is an enzyme that participates in the process of prokaryotic DNA replication. This is responsible for proof reading and excising or removing off the RNA primers required for initiation of replication. DNA polymerase, beta, also known as POLB, is an enzyme present in eukaryotes involved in base excision and repair, also called gap-filling DNA synthesis.
    So, the correct option is '(i) - 2%, (ii) inducer, (iii) - DNA polymerase I, (iv) - DNA polymerase  β '.

  • Question 2
    1 / -0
    Match column I with column II and select the correct option from the-given codes.
    Column I
    (Codons)
    Column II
    (Translated amino acid)
    A.UUU(i)Serine
    B.GGG(ii)Methionine
    C.UCU(iii)Phenylalanine
    D.CCC(iv)Glycine
    E.AUG(v)Proline
    Solution
    The genetic codes is a degenerate code that arises from the sequence of mRNA and three bases give rise to a single type of amino acid. As there can be about 64 such combinations of the bases A,U,C and G and there are 20 major amino acids more than one code can represent a single amino acid. According to the universal genetic code UUU codes for Phenylalanine, GGG codes for Glycine, UCU codes for Serine, CCC codes for Proline and AUG which is also the start codon of translation codes for methionine. 
    So, the correct option is 'A-(iii), B-(iv),C-(i), D-(v), E-(ii)'.
  • Question 3
    1 / -0
    The promoter site and the terminator site for transcription are located at
    Solution
    The promoter sequence needs to lie in front of the start site or upstream to it in the 5' end and the terminator should lie in the 3' end downstream as transcription proceeds in the 5' to 3' direction. 
    So, the correct answer is, '5 (upstream) end and 3 (downstream) end, respectively of the transcription unit'.

  • Question 4
    1 / -0
    Transcription unit
    Solution
    Transcription unit is the distance between sites of initiation and termination by RNA polymerase. It may include more than one gene. RNA polymerase produces transcription unit that extends from the promoter to the termination sequences.
  • Question 5
    1 / -0
    Which of the following statements regarding human genome is incorrect?
    Solution
    Repeated or repetitive sequences make up large portion of human genome. Repetitive sequences are nucleotide sequences that are repeated many times, sometimes hundred to thousand times. They have no direct coding function but provide information as to chromosome structure, dynamics and evolution. Approximately 1 million copies of short 5-8 base pair repeated sequences are clustered around centromeres and near the ends of chromosomes. They represent junk DNA. 
    So, the correct answer is 'Repeated (or repetitive) sequences are not present in human genome'.
  • Question 6
    1 / -0
    Choose the correct answer from the alternatives given :
    A DNA template plus primer with the structure
    3 P - TG C GAATTAG C GACAT - P5
    5 P - ATCGGTACGACGCTTAAC OH 3
    (where P = a phosphate group) is placed in an in vitro DNA synthesis system containing $$Mg^{2+}$$, an excess of the four deoxyribonucleoside triphosphates, etc. and a mutant form of E. Coli DNA polymerase I that lacks 5 $$\rightarrow$$ 3 exonuclease activity. The 5 $$\rightarrow$$ 3 polymerase and. 3 $$\rightarrow$$ 5 exonuclease activities of this aberrant enzyme are identical to those of normal E. Coli DNA polymerase I. It simply has no 5 $$\rightarrow$$ 3 exonuclease activity. What will be the structure of the final product?
    Solution
    The DNA synthesis will not occur on the left end because all DNA polymerases require a free 3 OH terminus, but here the 3 terminus of potential primer strand is blocked with a phosphate group.
  • Question 7
    1 / -0
    Choose the correct answer from the alternatives given :
    If Meselson and Stahls experiment is continued for four generations in bacteria, the ratio of $$N^{15}$$- $$N^{15}$$, $$N^{14}$$ - $$N^{15}$$, $$N^{14}$$ - $$N^{14}$$  containing DNA in the fourth generation would be
    Solution
    As per Messelson and Stahl's experiment, parent DNA was first isolated from E.coli grown in heavy $$^{15}\texterm{N}$$ medium. It was then put in light $$^{14}\texterm{N}$$ medium. DNA shows semi conservative nature, hence, in fourth generation, the ratio of $$^{15}\texterm{N}/^{15}\texterm{N} \, : \, ^{15}\texterm{N}/^{14}\texterm{N} \, : \, ^{14}\texterm{N}/^{14}\texterm{N}$$ will be 0 : 1 : 7.
  • Question 8
    1 / -0
    DNA is a polymer of nucleotides which are linked to each other by 3-5 phosphodiester bond. To prevent polymerisation of nucleotides, which of the following modifications would you choose?
    Solution
    If 3 OH group is removed/replaced in deoxyribose, there will be no formation of pliosphodiester bonds and hence polymerisation of nucleotides will be prevented.
  • Question 9
    1 / -0
    Which of the following statements are correct?
    (i) RNA polymerase I transcribes rRNAs.
    (ii) RNA polymerase II transcibes snRNAs.
    (iii) RNA polymerase III transcibes hnRNA,
    (iv) RNA polymerase II transcribes hnRNA.
    Solution
    RNA polymerase 1 (also known as Pol I) is, in higher eukaryotes, the polymerase that only transcribes ribosomal RNA (but not 5S rRNA, which is synthesized by RNA polymerase III), a type of RNA that accounts for over 50% of the total RNA synthesized in a cell. Pol I is a 590 kDa enzyme that consists of 14 protein subunits (polypeptides). Twelve of its subunits have identical or related counterparts in RNA polymerase II (Pol II) and RNA polymerase III (Pol III). The other two subunits are related to Pol II initiation factors and have structural homologues in Pol III. Ribosomal DNA transcription is confined to the nucleolus, where about 400 copies of the 42.9-kb rDNA gene are present, arranged as tandem repeats in nucleolus organizer regions. Each copy contains a 13.3 kb sequence encoding the 18S, the 5.8S, and the 28S RNA molecules, interlaced with two internal transcribed spacers, ITS1 and ITS2, and flanked upstream by a 5' external transcribed spacer and a downstream 3' external transcribed spacer. Because of the simplicity of Pol I transcription, it is the fastest-acting polymerase and contributes up to 60% of cellular transcription levels in exponentially growing cells.
    RNA polymerase II (RNAP II and Pol II) is a multiprotein complex. It is one of the three RNAP enzymes found in the nucleus of eukaryotic cells. It catalyzes the transcription of DNA to synthesize precursors of mRNA and most snRNA and microRNA. A 550 kDa complex of 12 subunits, RNAP II is the most studied type of RNA polymerase. A wide range of transcription factors are required for it to bind to upstream gene promoters and begin transcription. It catalyzes the transcription of DNA to synthesize precursors of mRNA and most snRNA and microRNA. A 550 kDa complex of 12 subunits, RNAP II is the most studied type of RNA polymerase. A wide range of transcription factors are required for it to bind to upstream gene promoters and begin transcription.
    So the correct option is ' (i) and (ii)'.
  • Question 10
    1 / -0
    Choose the correct answer from the alternatives given :
    Discontinuous synthesis of DNA occurs in one strand, because
    Solution
    The DNA-dependent DNA polymerases catalyze polymerization only in the one direction, that is 5  3. This creates some additional complications at the replicating fork. Consequently, on one strand (the template with polarity 3  5), the replication is continuous, while on the other (the template with polarity 5  3), it is discontinuous. The discontinuously synthesizing fragments are later joined by the enzyme DNA ligase.
    So, the correct answer is 'DNA dependent DNA polymerase catalyses polymerisation only in one direction (5 → 3)'.
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