mid1_2007_blue_key

mid1_2007_blue_key - Name Last, First MCB121, Page 1 of 8...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Name MCB121, Page 1 of 8 Last, First Harmer Student Authorization: I, , authorize the University to distribute publicly this graded exam ( e. g. , handed out in class, or left in a bin for me to pick up). Signature Date MCB121 First Midterm, April 24, 2007 Instructions: There are eight pages in this exam including the cover sheet, please count them before you start to make sure all are present. Write your name on each page of the exam. Write your answers in the space provided below each question. If you need more space use the back of the page and indicate clearly that you have continued your answer on the back. Do not use additional paper. Question Value Score 1 20 2 12 3 26 4 12 5 12 6 25 7 25 8 18 Total 150 Tm (°C) = 81 + 16.6 (log10 ci) + 0.4 [%(G+C)] - 0.6 (% formamide) -1.5(%mismatch) - 600/n (ci = cation concentration in M; n = length of DNA in base pairs)
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Name MCB121, Page 2 of 8 Last, First Harmer 1. (20) Mark the single most correct answer to each of the following. All nucleic acid sequences are written 5’ to 3’. a) (5) Liz Blackburn isolated the enzyme telomerase from the protozoan Tetrahymena . Tetrahymena telomeres consist of the repeated sequence TTGGGG. Consistent with telomerase being a reverse transcriptase, she isolated an RNA associated with the telomerase protein that contained the sequence CAACCCCAA. She next identified the gene encoding this RNA and altered the sequence to CG ACCCCG A. She found that Tetrahymena expressing this altered gene had telomeres consisting of the following repeated sequence: CGACCC TTGGGG TTGGGGTTG TCGGGGTCG X TCGGGG b) (5) You generate a yeast mutant that does not have a functional telomerase gene. Now, the yeast chromosomes will become shorter with increasing rounds of replication. Consider one such round of replication. Which of the following statements best describes the status of the two daughter chromosomes relative to the parent chromosome? One daughter chromosome will be shorter at one end; the other daughter chromosome will be normal at both ends. One daughter chromosome will be shorter at both ends; the other daughter chromosome will be normal at both ends. X Both daughter chromosomes will be shorter at one end, which is the same end in the two chromosomes (i.e, both 5’or both 3’ends). Both daughter chromosomes will be shorter at one end, which is the opposite end in the two chromosomes (i.e, the 5’end on one and the 3’ end on the other). (Problem 1 is continued on the next page.)
Background image of page 2
Name MCB121, Page 3 of 8 Last, First Harmer c) (5) In E. coli , the clamp loader helps regulate the action of the replisome by directly binding to which of the below proteins? (Indicate all correct answers; there may be more than one.) X DNA polymerase DNA ligase X sliding clamp topoisomerase X helicase 2 4 7 1 3 5 6 d) (5) Write in the number corresponding to the atom that is subject to nucleophilic attack by the 3’ OH of the growing nucleic acid chain during DNA replication: 5 2. (12) You are studying the mode of action of Rad51, the eukaryotic homolog of RecA.
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 04/07/2010 for the course MCB 102,103,12 taught by Professor Segel during the Spring '10 term at UC Davis.

Page1 / 8

mid1_2007_blue_key - Name Last, First MCB121, Page 1 of 8...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online