BICD100 MID 2 KEY

BICD100 MID 2 KEY - MIDTERM 2 ANSWER KEY BICD100 GENETICS...

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BICD100 GENETICS Prof. Chisholm MIDTERM 2 KEY Approximate grade scheme: > 199.2 = A > 139.6 = B > 80 = C > 20 = D 0-20 = F Descriptive statistics (as of 2/25/08) (158 exams) Minimum 5.0 Median 136.5 Maximum 260.0 Mean 139.589 Std. Deviation 59.5784 Histogram of Midterm 2:Freq. dist. (histogram) 0 2 4 6 8 1 0 5 10 15 20 25 Bin Center MIDTERM 2 ANSWER KEY 1
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Question 1: You are using the Ames test to examine some exciting new hair dye mutagens. You use two ‘tester’ strains of Salmonella, TA1 and TA2, each of which is unable to synthesize histidine due to different mutations in the hisD gene. Below is the amino acid sequence of the part of HisD protein mutated in the tester strains: Wild type HisD protein: Met Phe Tyr Trp Glu Gln Leu Lys TA1 hisD: Met Phe Tyr STOP TA2 hisD Met Phe Tyr Trp Gly Thr STOP (a) What changes in the DNA sequence of hisD are in the TA1 and TA2 strains, compared to the wild type? State whether the mutations are transitions, transversions, or frameshifts (insertions or deletions). (25 points) TA1 has changed Trp to STOP. From genetic code this is TGG to TAG or TGA. Either change is a G-to-A transition. (10 points) TA2 has a frameshift in the codon after the Trp, leading to two missense residues and a premature stop codon. The wild type amino acid is Glu, so the WT codon must be GAA or GAG. The mutant amino acid is Gly which is GG-. So the frameshift must be a single G insertion. (15 points) You now expose the tester strains to each mutagen and plate on minimal medium to select for His+ revertants. You find the following (average # colonies per plate): Tester strain Mutagen A Mutagen B no mutagen TA1 240 16 18 TA2 17 178 15 (b) Explain why mutagen A does not revert TA1 and why mutagen B does not revert TA2. You do not have to draw out chemical structures, just explain the difference between the mutagens (25 points) Mutagen A must be reverting A back to G. Mutagen A causes base pair substitutions. (10 points) Mutagen B must be a frameshifting mutagen (10 points). Frameshifting mutagens cannot revert base pair substitutions nor vice versa (5
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BICD100 MID 2 KEY - MIDTERM 2 ANSWER KEY BICD100 GENETICS...

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