This preview shows page 1. Sign up to view the full content.
Unformatted text preview: Problem 1 (20 points, 5 points each) A) Gel should show 7 bands but it showed 6. B) One possibility is that the cleavage sites for the two enzymes are so close that the smaller band cannot be resolved on the gel. The other possibility is the sizes of some fragments are similar and only one band is seen. C) If you add up all the size of the fragments and it is similar to that of the plasmid, it is the first possibility. If the total size is smaller than that of the plasmid, it is the second possibility which can be checked by running the gels separately and comparing. Also running a high resolution gel can help. D) The situation would be normal for bacteriophage lambda, which has linear genome and would have one more fragment than circular one and leave no discrepancy. Problem 2 (20 points) Problem 3 (20 points) A) ¼ x ½ = 1/8 Note: all other 18 bases have a probability of 1, since they are fixed. B) a) Add Mn2+ (MnCl2) change the ratio of dNTPs added from 1:1:1:1 to something else to create a bias, use DNA polymerase without proofreading acticity b) Transform the PCR reaction products into the desired cell line. Select at least 10 clones positive for the insert and sequence the gene insert. c) 5 amino acids d) Simple probability: 20 amino acids in each position of the 500 amino acid protein: 20500. e) Transformation. It is finite. Typical transformations will maximize a library at 1010 for bacterial and yeast libraries, up to 1011 for phage, and 1015 for ribosome display. Problem 4 (20 points) 5’ CATTCCAATTGCAG 3’ => 5’ CATTCCAAGTGCAG 3’ Diseased Healthy Problem 5 (20 points, 10 points each) A) Since this is non‐coding strand so it would need to be reverse complemented for the synthesizing RNA. Therefore, this will be read from 3’ ‐> 5’. Hence for AUG (start codon) on coding strand, it would be presented on the non‐coding strand as 5’‐CAT‐ 3’. Likewise, stop codon on non‐coding strand would be 5’‐TTA‐3’, 5’‐CTA‐3’, or 5’‐TCA‐3’. 5’‐CATACTTACTACTAGATTACGATTAGACGATTAGGATGGCCGACTCGTGCAGTAAC AGCATGACCGAGGCCTAGACCAGATTAGGAGCCGGACCAGGACGGACCAGCGACT‐3’ 12 mino acids would be encoded B) For Transcription: Promoter For Translation: Ribosome binding site (RBS) ...
View Full Document
This note was uploaded on 12/13/2011 for the course BME 339 taught by Professor Georgiou during the Spring '11 term at University of Texas at Austin.
- Spring '11