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Miscellaneous question discussion series i (molecular biology)

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  • 1. Miscellaneous Question Discussion series-I Topic:- Molecular Biology © www.cteck.org © www.cteck.co.in
  • 2. © www.cteck.org © www.cteck.co.inQuestion No1 Why Chargaff’s Rule (A+G = C+T) is not a universal rule? a) Because genetic material of some marine bacteria doesn’t obey it b) Because genetics material of Φx174 doesn’t obey it c) Because genetic material of Pneuococcus doesn’t obey it d) None of the above The answer is b. Since the mole percent of A is 24, C is 22, G is 23 & T is 31 in Φx174 which is not in accordance to Chargaff’s rule. This is the only exception found in nature.
  • 3. © www.cteck.org © www.cteck.co.in Question No.2 What is the level of packing observed in 30nm fibre? 0.27 fold 2.7 fold 27 fold Answer 0.027 fold Explanation • Since 20 nucleosomes are found in 30nm fiber & each nucleosome is associated with 200 bases (roughly) while the distance between each base is 0.34nm • The total length of DNA duplex in 50nm (length) of the fiber = 20 x 200 x 0.34 = 1360nm • Since 1360nm fiber is reduced to 50 nm, hence the level of packing is 1360/50 = 27.2
  • 4. © www.cteck.org Question No. 3 © www.cteck.co.inA dicentric chromosome is not observed in nature 1. Because the two centromeres will interfere with each other 2. Because the chromosome will become unstable during meiosis 3. Because the chromosome will become unstable during cell division 4. Because the chromosomes will attach to spindle fiber at two places which will break the chromosome Which of the above statements are correct? a) Only 1 b) 1&2 c) Only 3 d) 3&4 Answer
  • 5. Question No.4 © www.cteck.org © www.cteck.co.in The base composition of phage M13 DNA is A, 23%; T, 36%; G, 21%;C, 20%. This leads to one of the following conclusion. 1. The phage contains dsDNA as the genetic material 2. The phage doesn’t contain dsDNA 3. The phage contains dsDNA only during replication 4. The phage contains ssRNA as the genetic material– Which of the following combination of statements is correct? a) Only 1 b) 1&2  The answer is b as the virus doesn’t c) Only 3 seems to follow Chargaff’s rule, whose d) 4&3 only exception is Φx174
  • 6. Question No 5 © www.cteck.org © www.cteck.co.in The complete genome of the simplest bacterium known, Mycoplasma genitalium, is a circular DNA molecule with 580,070 bp. What is the molecular weight and contour length (when relaxed) of this molecule? 1. 1.8 x 10^8 & 100 microns 2. 2.8 x 10^8 & 150 microns 3. 3.8 x 10^8 & 200 microns 4. 4.8 x 10^8 & 250 microns The molecular weight of a single nucleotide pair is 650, so the approximate molecular weight of the Mycoplasma DNA molecule is (580,070 bp)(650/bp) = 3.8 x10^8 Given a length of 3.4 Å per base pair, the contour length is (580,070 bp)(3.4 Å/bp) = 2.0x10^6 Å = 200 mm
  • 7. Question No.6 © www.cteck.org © www.cteck.co.in An enzyme isolated from rat liver has 192 amino acid residues and is coded by a gene with 1,440 bp. Which of the following statement is correct according to the above mentioned data? 1. The enzyme is not fully synthesized 2. The enzyme is coded by 576 bases out of 1440 bases 3. The coding sequence of enzyme contains non coding sequence of the length 764 bases 4. The extra size of the gene is because of the presence of untranslated region present on the genea) 1&2b) 2&3 The answer is c.c) 2&4  Each amino acid is encoded by a triplet of 3 bp, so the 192d) 1&4 amino acids are encoded by 576 bp. The gene is in fact longer (1,440 bp). The additional 864 bp could be in introns(noncoding DNA, interrupting a coding segment) or could code for a signal sequence (or leader peptide). In addition, eukaryotic mRNAs have untranslated segments before and after the region coding for the polypeptide chain, which also contribute to the “extra” size of genes
  • 8. Question No7 © www.cteck.org © www.cteck.co.in Bacteriophage Lambda infects E. coli by integrating its DNA into the bacterial chromosome. The success of this recombination depends on the topology of the E. coli DNA. When the superhelical density (sigma) of the E. coli DNA is greater than -0.045, the probability of integration is 20%; when sigma is less than -0.06, the probability is >70%. Plasmid DNA isolated from an E. coli culture is found to have a length of 13,800 bp and an Lk of 1,222. Calculate sigma for this DNA and predict the likelihood that bacteriophage Lambda will be able to infect this culture. 1. -0.067 & the phage will integrate into the E.coli genome Answer 2. -0.077 & the phage will not integrate into the E.coli genome 3. -0.087 & the phage will integrate into the E.coli genome 4. -0.097 & the phage will not integrate into the E.coli genomeSuperhelical density, also known as specific linking difference, is given by sigma = (Lk - Lko)/LkoIn this example, Lko = (13,800 bp)/(10.5 bp) = 1,310, andSigma = (1,222 -1,310)/1,310 = -0.067Because 0.067 is less than 0.06, there is a greater than 70% probability that the phage DNA will be incorporated into the E. coli DNA
  • 9. Question No 8 © www.cteck.org The E. coli chromosome contains 4,639,221 bp.www.cteck.co.in © How many turns of the double helix must be unwound during replication of the E. coli chromosome? 1. 4.42 x 10^5 turns 2. 5.52 x 10^5 turns 3. 6.52 x 10^5 turns 4. 7.52 x 10^5 turns  During DNA replication, the complementary strands must unwind completely to allow the synthesis of a new strand on each template. Given the 10.5 bp/turn in B-DNA, and approximating the E. coli chromosome as 4.64 x10^6 bp  The no of helical turns = no of base pairs / no of base pairs per helical turn = 4.64 x 10^6 bp / 10.5 bp/turn = 4.42 x 10^5 turns
  • 10. © www.cteck.org © www.cteck.co.inQuestion no.9 Which of the following is the most common mismatches in the DNA of eukaryotes? A=T G=C  Spontaneous deamination of 5-methylcytosine produces thymine, and thus a G–T A=C mismatched pair. Such G–T pairs are among the most common mismatches in the DNA of G=A eukaryotes
  • 11. © www.cteck.org © www.cteck.co.inQuestion No 10  Determine the minimum energy cost, in terms of ATP equivalents expended, required for the biosynthesis of the beta-globin chain of hemoglobin (146 residues), starting from a pool including all necessary amino acids, ATP, and GTP. 1. 500 • The total number of ATP equivalents 2. 583 required for the synthesis of one molecule 3. 600 of beta-globin is (4 x 146) - 1 = 583 (since 4. 683 the formula for polypeptide synthesis is 4n-1, where n = no of amino acids present in the polypeptide chain)