17 lab ex1 sum07 key

17 lab ex1 sum07 key - Biol o gy 1 A La b Ex am#1 J uly 26...

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Unformatted text preview: Biol o gy 1 A - La b Ex am #1 J uly 26 th , 2 0 07 Answers Exam 1 Bio 1A, Summer 2007 1 B 6 C 11 A 16 C 21 D 26 B 2 E 7 E 12 E 17 C 22 B 27 E 3 D 8 B 13 E 18 A 23 B 28 D 4 B 9 A 14 E 19 B 24 C 29 B 5 A 10 D 15 C 20 E 25 D 30 Mean =60.75, STDEV = 13.3 . A (some form) 70-100, B (some form) 59-69, C (some form) 47-58, D (some form) 39- 46, F X < 39. Short answers are on the back. 2) From the curve you determine the original conc. Is 50 mM, too dilute to make a 100 mM unless boil away, etc. Units matter! 5) 1N = 2, 2N = 4. For mitosis 4 chromsomes with sister chromatids line up. No 1N cell in animals undergo mitosis. 6) In portion 3 additional enzyme does not change the O.D. which means all S converted to P in the time course of the experiment (substrate is limiting, enzymes are idle). 8) You don't change the ploidy but instead change the amount of DNA, C value increases. 9) G2 = 4C and zygote = 2C. Determine #bp for 4.8 X 10-11 grams. 4.8 X 10-11 grams /6.0 X 102 g/mole bp = 0.8 X 10-13 moles X 6.0 X 1023 (Avogadros #) = 4.8 X 1010. 10) Note units, 400 cells/ul there need to do 1:4 dilution and then 2 serial dilutions of 1:10. 12) Isoenzymes catalyze the same reaction but they don't have the same primary sequence because if they did they would be the same enzyme, not isoenzyme. 13) Double crossover events between d&e yield a gamete that looks like a parental, should be large number. 19) From A you know it is linear. All the bands have the same amount of DNA. D must be smaller than B. 20) Genes versus alleles. In a diploid individual there are 2 alleles per gene (genetic locus). 23) Membranes are hydrophobic, everything is pretty much hydrophilic. Like dissolves like. From lab you know that the more hydrophobic the compound the farther it migrates. 25) Since Alice had a child with disease K you know she must be heterozygous. A heterozygote crossed with a hetero yields a genotypic ratio of 1:2:1 (phenotypic ratio 3:1) but you know Tony isn't homo recessive so there is a 2/3 chance Tony is heterozygous. Only of the hetero w/hetero display the recessive phenotype. 2/3 X = 2/12. 26) Since there is complementation the alleles must be in different genes. 27) dATP has no 2'OH (ribose does) and a 3' OH. 28) 6 base pair cutter. 29) Different genetic loci implies a heterozygote should be wild type. Short Answer 2. a) 256 (1/( X X X ) b) 256 (1/( X X X ) c) 128 the size of 256, 1/(1/256 + 1/256). 3. a) b) Color is sex-linked, allergenicity is autosomal. B+ = white, B = black. ha+ = allergenic, ha = non-allergenic. #3 + XB XB + XB + White XB + XB + Black XB XB Y White ha ha+ ha+ ha ha+ ha ha ha ha ha+ ha #4 0.4 DE 0.4 DE + No 0.4 D + E No 0.1 DE No 0.1 D+E + No Black ha + 0.4 D + E + 0.1 D + E 0.1 DE + No No Yes = 0.16 Yes = 0.04 No No No No No Yes = 0.01 No Note = black, = white. For allergenicity induce it, do not. 4. X 0.21. 1 pt for 1/2 , 1 pt for multiplication and 2 pts for 0.21. See the box, upper right. Summer 2007 Lab exam 1 Key 5. Template = 5'--- CGTACGTACGTACGTGTC- 3' , Radioactive primer = 3'-CAG- 5' Each lane = 1 pt. ddG ddA ddC ddT All 4 6. a) 4 b) isolated blue colonies c) Pick a blue colony but it most likely will not have a plasmid containing pBio1B ligated to pBIo 1A because that is an unlikely event compared to the chance that pBio1A will ligate back to itself of that pBio 1B will ligate back to itself. 7. 1 pt for order, 1 pt for each complementation group. Mutant 4 Fo r mat Ex a mpl e: No # 8 9. The DNA must be linear from lane 1. From Eco R1 you know You get three peices, each of 1,000 bp long (total = 3,000) From Pst I you know Case 1 0 0 1,000 2,000 Q Mutants 2 & 3 S P Mutants 1 & 5 R R1 R1 Pst1 1,500 Pst1 Pst1 1,500 Pst1 2,500 or Case 2 0 2,000 Either Pst1 map works. Case 1 yields pieces of 4 X 500 = 1 X 1,000. Case 2 yields pieces of 2 X 500 and 1 X 1,000. If you had intensity to look at you could distinguish between the two possibilities. You were given credit for either answer but you had to include the location of cuts and the sizes in your final map. Case 1 0 R1 Pst1 R1 & Pst1 2,000 Case 2 0 R1 Pst1 R1 Pst1 2,000 2,500 1,000 1,500 1,000 1,500 10) From the data in the F1s you can see that the heterozygotes have gray wings (incomplete dominance). Clear veins and wide abdomen, wild type traits, are dominant to the mutant alleles (dark vein and narrow abdomen). From the F2 you see there is a difference in veins and abdomen between males and females and therefore both traits are linked to the X chromosome. + + + + + + Ge n otyp e F1 Fe m ale B ; X dv n a Ge noty pe F1 M ale B ; X dv n a B X dv na B Y Le gend: Since White is the wild type trait and back wing is mutant you should use B for mutant and B+ for wild type, note that both are uppercase. dv is mutant, dv+ is wild type for veins and na is mutant and na+ is wild type for abdomen. Summer 2007 Lab exam 1 Key ...
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