Final Exam 2008 - WW mam m” Use only words and the sugar...

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Unformatted text preview: WW mam m” Use only words and the sugar names that you learned in class to describe this structure (you can call it Vermatan) in an unambiguous manna (unambiguous means that it could be drawn by one your colleagues from your verbal H OH OH H ’0” description). H0 0 CH2 Draw the sugar on the left side of Vermatan in the g” H H0 furanose form. H H 9H2 H H2C. OH OH M 2 D—gluoo- pyranose 0.6 for D , 0.7 for name 0.7 for cycle 2 linked or. 1, l3 2 0.5 for each anomer, 0.5 for each position 2 D-fructo- furanose 0.6 for D , 0.7 for name 0.7 for cycle If either sugar is not named but they are described as a D-aldo—pyranose with Same- Opposite—Same configurations and a D-keto-furanose with opposite- same configurations , give 0.4 marks for each name If either sugar is not named but described as an D-aldo pyranose with D-L-D configurations and a D—keto furanose with L-D configurations give 0.35 marks for each name. Ifthey say draw a hexagon and then grade-school-style spell out the specifics at each position and all is RIGHT , give 40% of marks. Give 4 (of 6) Iffor all of the above they just say sucrose. D 4marksforthefimosefbrmofD-gluooseltisé “5‘ .3 o H 1.5 forfilranosering(only 0.4 ifcarbon 1 isnotahemiacetal) Ito—CH: H Iforoorrectly showingthatisD "H ‘3”! 1.5 for the correct orientation of the 4 other configurations (0.4 each) Okifthefirranoseisdrawn asaFischerpmjectiomgrade as above. 10= total marks a; Iffuranose is drawn flipped H, 0 forward, etc., yet configurations are 1; .‘ ok give 100%. y 1': . 1;. ff“ 5, OK, if somebody decides to start numbering from the carboxyl. .1 \ ’2} 7 <-. What they have drawn is keto sugar! And shows that they ‘ i‘ = i understand nothing about the reaction. <- Give 1.5!4 for this structure Further breakdown on the grading of this structure: 0.75 for keto furanose ring (Give 0.3 out of .75 if this weird ring closure does not have an 0H 0.35 ifit shows D at carbon 5 0.2 configuration at C4 0.2 configuration at C3 Total 1.5/4 @ ENDFIELD | nxt: specify draw as Hamrth ENDRECORD @@2EXENDFIELD M6503 mdlb ENDFIELD "q Correction of50413ENDFIELD 1) Draw, as 3'-ph05phates, a GC base pair in Watson-Crick DNA. In addition to the hydrogen bonds, your figure should show the configuration at each position of the Sugar and ALL the atoms and double bonds of the bases. Use your figure to explain why an A—T base pair has one less hydrogen bond (no drawing required for this part of the question). Note: the bases are missing RS on this figure But the problem says ALL atoms and the Hs Must be shown for full marks (see below) Student does not have to draw sugars in perspective or give numbering It is not necessary to talk about grooves (the arrows on left) Principle: One of the bases can be drawn in the standard presmtation- the other hm to be flipped over 2 for G 1 for just atoms of base, 0.5 for Hs, 0.5 for dOuble bonds 2 for C 2 for H bonds 0.8leech. only 1 if orientation of H bonds is wrong -1 for am really stupid hydrogen bond give only 1.2 of 2 marks for drawing negligence that shows : O """ NH 2, -0.4 if just 1 bond is like this 1.5 sugar (0.5 ring, 0.5 configurations, 0.5 attachment to base) ~0.5 if C5 of sugar missing, ~0.5 if sugar is ribose 1.0 for phosphate (0.5 for structure, 0.5 attachment point to sugar) we will accept diesters... —O.3 Ifgives both 5' and 3’ p 1.5 1: A is missing a group at position 2. 0.5: group is a hydrogen donor An acceptable answer is that A does not have an acceptor for the hydrogen bond -0.8 if structure of A is wrong, but concept is right; -0.5 if position on A is not given —0.8 ifwrong position on A 10 = total Special cases : total stupidity in terms of putting together as a base pair. Give only 3/10 for whole question even if basic structures are right. H-bonds are perfect but one phosphate joins together sugars of G and C! Give 4110 H (N O’H'b‘ffl Anomeric bond to base is beta / f / N-**' )7“ A drawing like this is 0K 0 N" __ o 0 (3 configurations on sugar are OK) HIM-4+ i 9 2 (9' 43:0 ...
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