339 HW 5 Solutions - Fall 2010 BME339/CHE339/BIO335 HW#5...

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Fall 2010 BME339/CHE339/BIO335 HW#5 Solutions Problem 1 Circular genome size = 2.8 x 10 6 bp 0.65 for G/C ; 0.35 for A/T ; At any given base: p(G) = p(C) = 0.65/2 = 0.325 p(A) = p(T) = 0.35/2 = 0.175 For NcoI site: p(CCATGG) = 0.325 * 0.325 * 0.175 * 0.175 * 0.325 * 0.325 = 3.41 x 10 -4 Number of fragments = number of NcoI sites n = 2.8 x 10 6 * 3.41 x 10 -4 n = 956 Estimate to generate 956 fragments Problem 2
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Fall 2010 BME339/CHE339/BIO335 HW#5 Solutions Problem 3 EcoRI: 2400,2200,1800,800,600 HindIII: 2400,2000,900,200 Fragments for each partial digestions: EcoRI P32 end HindIII P32 end Double * * * * * * * * * * * * * * * * * * * * * * * * Note: each * is 100 bp fragment longest fragment is 900 bps Problem 4 a) Since this is non-coding strand so it would need to be reverse complemented for the synthesizing RNA. Therefore, this will be read from 3’ -> 5’. Hence for AUG (start codon) on coding strand, it would be presented on the non-coding strand as 5’-CAT- 3’. Likewise, stop codon on non-coding strand would be 5’-TTA-3’, 5’-CTA-
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This note was uploaded on 12/13/2011 for the course BME 339 taught by Professor Georgiou during the Spring '11 term at University of Texas at Austin.

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339 HW 5 Solutions - Fall 2010 BME339/CHE339/BIO335 HW#5...

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