Revised Key - ANSWER KEY LAB EXAM 1 Summer 2010 Mean =...

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ANSWER KEY LAB EXAM 1, Summer 2010 Mean = 80.5 (56.2), Stdev = 13.4, Median score = 56. RANGE 22-90. A = 75-100, A- = 71-75, B+ = 67-70, B = 62-66, B- =59-61, C+ =56-58, C = 47-55, C- = 42-46 D+ = 41- D = 38-40 D- = 35-37 F = 34 or less. 1 C 6 D 1 1 D 1 6 A 2 1 A 2 6 B 3 1 E 3 6 A or B or D. 4 1 B 4 6 D 2 E 7 C 1 2 B 1 7 B 2 2 A 2 7 C 3 2 B 3 7 C 4 2 B 4 7 C 3 C 8 E 1 3 C 1 8 A 2 3 B 2 8 D 3 3 C 3 8 C 4 3 B 4 8 B 4 B 9 A 1 4 E 1 9 B 2 4 A 2 9 E 3 4 D 3 9 B 4 4 B 4 9 5 E 1 0 B 1 5 C 2 0 D 2 5 D 3 0 C 3 5 A 4 0 E 4 5 C 5 0 1) Since 25 and 30 is greater than 50 and 50 is maximum randomness, the two outer loci assort independently and you expect 4 types of gametes- each at 25%. 2) 1 X 10 -16 L X 1 X 10 -4 H H+/L = 1 X 10 -20 Moles X 6.0 X 10 23 molecules/mole = 6,0000. 3) When pigments are extracted they can’t transfer their energy and as a result will have more fluorescence than intact or lysed chlroplasts. This integrated the demo in lab. 4) NH 2 + H + > NH 3 + This equation indicates that at a pH of 9 there are equal amounts (the definition of pKa. If you are at pH 10 then there are fewer protons than at pH 9, and as a result you shift the reaction to the left, meaning you increase the amount of NH 2 and decrease the amount of NH 3 + 5) Annealing involves base pairing which involves the formation of H bonds (A with T, C with G). 6) Simple identification. 7 & 8 ) The processed mRNA is shown below (and the codon and corresponding amino acids are shown below and below that are shown the number of codons that yield the same amino acid. The first AUG is in bold, and the codons are alternating italics and normal font. There are 1 X 2 X 1 X 2 X 4 X 3 possible mRNAs that would yield the same penta-peptide (in the question it specifically states you should include the stop codon when determining the number). C G G U A U G C A U A U G U U C G C A U A G G A MET HIS MET PHE ALA STOP 1 2 1 2 4 3 9) Bacilloid means rod-shaped and gram + bacteria stain purple. 10) Since sperm are 1C the G1 cell would twice as much DNA. 1.2 X 10 -13 g DNA X 6.0 X 10 23 = 7.2 X 10 10 grams DNA. 7.2 X 10 10 grams DNA/6 X 10 2 g Mole/bp = 1.2 X 10 8 bp but remember it is twich as much DNA in the G1 cell versus the sperm. 11) Rate should be double but since both started with the same amount of starch and went to completion they should yield the same O.D. 12) A high dN/dS ratio implies many changes in amino acid composition of the protein. 13) The Km experiment should show leveling off as you increase the pecentage of starch. There should have been multiple blanks, each corresponding to the initial starch concentration used for the Km determination. 14)
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This note was uploaded on 09/25/2010 for the course BIO biology taught by Professor Meighan during the Spring '09 term at Berkeley.

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Revised Key - ANSWER KEY LAB EXAM 1 Summer 2010 Mean =...

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