Oakley_PracticeFinalKey - . h‘fl l. TW'p is the process...

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Unformatted text preview: . h‘fl l. TW'p is the process of making RNA from a DNA template, where 17‘ m is the process of making protein by reading the mRNA code. 2. Splicing of enkaryofic RNA molecules removes the fleMS and links together the 2x131} 5 within the genetic sequence. . Which enzyme catalyzes transcription in prokaryotes? Poll! was!“ . During transcription, nucleotides are added (a 5’ to 3’ b. 3' to 5’ c. from N terminus to C terminus d. from C terminus to N terminus e. a and c f. b and d . What is one major role of the “Sigma factor” in prokaryotic transcription? Imam promotr Spatfi‘crfij . Name 2 ways eukaryotic mRNA may be processed afier transcription and before translation. polv'fi \‘m‘l addl‘hbn, Tum“ spirale . Ignoring the difference between uracil and thymine (RNA’s have U’s DNA’s have T’s) circle all molecules that are identical in sequence to the transcribed mRNA for a given gene: 0 RNA-like strand of DNA . Template strand of DNA 0. First-strand cDNA d. Processed mRNA . In prokaryotes, a search for genes in a DNA sequence involves scanning the DNA for long reading frames uninterrupted by stop codons. What is a potential problem for this approach in eukaryotes? Evan's Wm Cams may/NA, Wot/t (mud how 3"“? coJMS. I 9. (Genetic code will be provided — you don’t have to memorize it). The sequence of a segment of mRNA, beginning with the start codon, is given below along with t ,- several mutant sequences: a. Normal AUGACACAUGGG Translation: b. Mutant] AUGACACAUGGA Translation: 77 c. Mutant2 AUGATACAUGGG Translation: d. Mutant3 UGACACAUGGG Translation: Type of mutation (e.g. missense, nonsense, etc) 9AM (U) A W b: c: d: W“ 10. You find a new form of life that uses a completely new genetic code, which you aim to crack with an in vitro translation system. You create artificial mRNA with the sequence ...UCUCUCUCUC. .. and discover that the protein generated is a string of Glycine amino acids. What is the codon for Glycine in the new form of life? a. UCU b. CUC c. AGA d. GAG @ a or b r c or d g. none of the above I 1- I: ll: L l' .iI'. .Il -. W: - Eh I .I. II' .' 'l .| I:l1r:: :11 DNA molecules in a reaction (Y-axis) as a function of time in number of cycles (X-axis). Why does the number increase exponentially in PCR, but only linearly in sequencing? 34W M (WP/wt,” aapr 05m 0AM QM wot WW4 mould. (le HWA {) a real)- { H) 0! lol"\ se, Mom 91% *- IZrCo'iisiclg' fifripdifferent kings of h arr: librapies: genoni, brain c SVA fibrary, liver cDNA library, and a genomic library that lacks repetitive DNA. Assuming inserts of about equal size, which would contain the greatest number of different clones‘7Why? 61 “H (616 Hr My” ‘2 m my ANAdrw/rsdy, mm AffillCllt/H‘i CW will maul/:2 5- ATCGACTGATCGATCGATCGTAGCTAGCTAGCATCG — 3 3- tagctgactagctagctagcatcgatcgatcgtagc — 5 13. For the double-stranded DNA sequence above, design a sequencing primer that is ten bases in length and that will sequence the DNA strand indicated in lowercase. 5’fihfiIQQIéQQ' 3’ 14. Besides protein coding genes, Wme two other types of genes: I. Q9399 2. {EN W 15. Give two different reasons for the much higher ratio of total DNA to DNA that encodes proteins in the human genome compared to bacterial genomes. IHWS, NowCoqu WHHVC. DNA 16. What sequence information about a gene is lacking in a cDNA library? an5 17. Assemble the following fragments, which were shotgun sequenced into a contiguous sequence (assume the same strand is sequenced in each clone): I z 3 ‘i z tgggtaatgtgccccca acgtaaaa taaaagatggg ccata a .-——, ..__.____, W 4 ___39 -—7 p42)» M, W7 K, swat «m. ‘1‘: 1 8. microarray experiment a“ (- h (£14 oh M 19. A constitutive phenotype will be produced by mutations in which (circle all that apply): Operator Repressor c. Promoter d. RNA polymerase er Sigma factor 20. Which proteins would be bound to the lac operon in the presence of only lactose (Circle all that apply) a RNA polymerase @ CRP c. Lacl (Repressor) d. Permease e. Beta-galactosidase i 21. You fused different fragments of the upstream DNA of a gene of interest to a GFP gene (reporter gene) that lacks a promoter. From the results below, which region contains the enhancer(s) and which contains the promotor(S)? Regulatory Region 1 3 4 Egression 0 8 l“ “f 8 go [-3 IA hum aw in l 8 90 22. What is the major difference between a LCR and an operon? 6W «’6 am uc (L Kant, M W Vimwcohyy. awn-no Rowe l pmmohfif 23. Circle the processes below that occur in prokaryotes, underline those that occur in _I|-.r.'.'-.'I_I...II':;I 24. A mutation in which protein would cause constitutive expression in yeast? 25. The regulation of gap genes and the distribution of their gene products are regulated in part by: a. pair rule genes. b. segment polarity genes. maternal genes. . all of the above 26. Which of the following is not a property of the hunchback gene? 1 Hunchback (hb) mRNA is uniformly distributed in the egg )( Hunchback transcription is enhanced by Bicoid ad Translation of hb mRNA is inhibited by Nanos I Hunchback protein is distributed in a gradient in the egg Huchback protein directs the distribution of bicoid <32 am 3% 3.3 waa ~ #1 f 3%.9 ” sumac 2: mo 2:: 35902“. 2: Sons 332.8 :0» on «E3 63% page 05 E Sun 2: 93 J83 31.8.2: 5 mEE=mm< .3:2o&% 853% $2 a 2.: v5 860% mac EEuEv 025 Eoc :EoEoEo: 969:8 =8» :2...st 8v 8m 8m 8. I .3 DUB .E. mum u .250 .H. muOan <9 BU m uouoco: .H. H moxfiuo ¢ . _ . . a p / .AvfiooEzv ooh msoioEEan fl 0 $2: 2: BED .00.: unocowoifi 933mg :03 Sm Roam banana—go mo REE—E W3 05 23:28 SnoEfiHn 95:33. .30533 <ZD EBA 12.0583 93 320m .wn ...
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This note was uploaded on 10/16/2010 for the course EEMB 129 taught by Professor Ajnarivera during the Spring '09 term at UCSB.

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Oakley_PracticeFinalKey - . h‘fl l. TW'p is the process...

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