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BSCI 222 Spring 2012 Exam II Study Guide - Prof. David A. O'Brochta, Exams of Genetics

Answers and explanations for exam ii of the bsci 222 spring 2012 course. It includes questions related to chromatin remodeling complexes, cpg methylation, and dna replication. The document also includes calculations related to the total number of nucleotides in the genome and the number of origins per chromosome.

Typology: Exams

2012/2013

Uploaded on 05/03/2013

phungt1118
phungt1118 🇺🇸

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Download BSCI 222 Spring 2012 Exam II Study Guide - Prof. David A. O'Brochta and more Exams Genetics in PDF only on Docsity!   BSCI 222 Spring 2012 Master Key to Exam II question white (w) pink (p) tan (t) 1 B   A   C   2 C   C   B   3 D   C   D   4 C   D   C   5 A   B   A   6 D   E   E   7 E   B   D   8 B   D   B   9 E   D   C   10 B   D   B   11 C   B   D   12 B   B   D   13 C   C   C   14 D   E   E   15 D   C   B   16a A   A   A   16b C   C   C   17 A   B   B   18 C   D   D   19 D   A   C   20 B   C   A   21 CpG   increases,  decreases   A   22 increases,  decreases   CpG   –10   23 –10   –10   chromatin-­‐ remodeling   complexes   24 3'   chromatin-­‐ remodeling   complexes   CpG   25 chromatin-­‐remodeling  complexes   3'   small   26 64   II   3'   27 II   A   II   28 small   small   64   29 A   64   increases,  decreases   w p t 30 31 30 a)    2(6.6  x109)  bp  =  1.32  x  1010  total  nucleotides     1.32  x  1010  total  nucleotides  x  (50  nt/sec)–1  =  2.64  x  108  sec     2.64  x  108  sec  x  (60  sec/min)–1  =  4.4  x  106  min     4.4  x  106  min  /  5  min  =  8.8  x    105  or  880,000  origins     b)  There  are  46  chromosomes  in  a  normal  human  cell.  Therefore,   880,000  origins/46  human  chromosomes  =  19,130   origins/human  chromosome.   31 30 33 The CTD is phosphorylated 32 32 31   a)  The  top  strand,  which  has  5'  to  3'  synthesis  in  the  same  direction  as   movement  of  the  replication  fork.    All  DNA  synthesis  is  5’à3’,   addition  to  3’OH  can  occur  continuously  when  using  the  top  strand   as  template     b)  The  bottom  strand,  which  experiences  discontinuous  synthesis   because  the  direction  of  synthesis  is  opposite  that  of  the   movement  of  the  replication  fork.  Because  DNA  synthesis  occurs   5’à3’  further  unwinding  must  occur  before  there  is  adequate   amounts  of  bottom  strand  template  for  priming  and  then   extension.   33 34 34
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