7.05_Problem_Set_3_Solutions

7.05_Problem_Set_3_Solutions - 7.05 Problem Set 3 SOLUTIONS...

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

View Full Document Right Arrow Icon
Background image of page 1

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

View Full Document Right Arrow Icon
Background image of page 2
Background image of page 3

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

View Full Document Right Arrow Icon
Background image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 7.05 Problem Set 3 SOLUTIONS Spring 2008 Question 1 Nucleic Acid Structure (a) Propose a specific mechanism for recognition of an A—T base pair in the major groove of DNA by the side chain of Asn in a regulatory protein, with a structural drawing. Number the atoms in the base(s) that interact with Asn. 3,4, The 6-amin0 and N7 atoms Asparagine M “3 in” H “I L" of Adenine interact with the M“ side chain of Asparagine. {R xx "‘ Si ‘3 7 N a“; I, P is./”'“\1:> \f‘ 5‘ H \ ’ i . a r’ i; m M Thymine g l if . h .,\ "a “I’m. {4’ x x w” A)! “I”? E i K/ ii ‘4 2‘ it / i . \, , . ‘ / Adenme ski “M N. 3 '/ (b) In contrast to DNA, recognition of double—stranded RNA by proteins most often involves the 2—amino and N3 positions of purines, and the 2—keto positions of pyrimidines. Explain, based on your understanding of nucleic acid structure. RNA helices are A-form, which have deep, narrow major grooves and wide, shallow minor grooves. The 2-amino and N3 positions of purines and the 2-keto positions of pyrimidines are accessible through the minor groove. Question 2 DNA Replication (a) When the enzymatic properties of DNA polymerase I were first described, biochemist John Cairns was skeptical that DNA polymerase I was in fact the enzyme that was responsible for DNA replication. List all the activities of DNA polymerase I you learned, and for each explain briefly how these properties can be reconciled with our present understanding of the mechanism of DNA replication. 5’-)3’ DNA-dependent polymerase: accommodates leading strand synthesis, and discontinuous synthesis on the lagging strand. 3’95’ exonuclease activity: provides proofreading mechanism. 5’93’ exonuclease activity: used to remove the RNA primers in lagging strand replication. (b) Cairns isolated a DNA polymerase I mutant which was defective in polymerase activity, although DNA replication occurred normally. Later, another mutant of DNA polymerase I was isolated lacking the small N—terminal fragment of the protein, and these mutants were defective in DNA replication. Explain how two different mutants of this enzyme could have different effects on DNA replication. The first mutant likely had a mutation in the Klenow fragment, affecting polymerase activity, but retained the 5’-)3’ exonuclease activity required to remove RNA primers. Therefore replication could proceed with DNA polymerase III. The mutant lacking the N-terminal fragment, which contains the domain that has 5’93’ exonuclease activity would be defective in DNA replication due to its inability to remove RNA primers used to initiate DNA synthesis. (c) Cairns’ isolation of a mutant that was deficient in polymerase activity of DNA polymerase I allowed other groups to find additional DNA polymerase activities in E. coli cells, including DNA polymerase III. It was noted that there were marked differences in processivity between the different polymerases. Define processivity and compare this property between DNA polymerase I and III. Processivity is a measure of the average number of nucleotides added by a polymerase each time that it associates with template DNA. DNA polymerase III is much more processive, meaning that once it begins synthesis, it incorporates up to 500,000 nucleotides before falling off again. DNA polymerase I is only moderately processive, able to incorporate up to ~200 nucleotides before falling off. (d) For the more processive polymerase, name and describe a structural feature of the enzyme that accounts for its higher processivity. The dimeric f3 subunit of DNA polymerase III functions as a sliding clamp that completely encircles a DNA molecule, permitting the polymerase to slide along DNA, but not dissociate. (6) Name one kinetic property in which DNA polymerases I and III differ substantially. Vmax — DNA polymerase I maxes out at ~20 nucleotides/sec, DNA polymerase III goes up to ~1000 nucleotides/sec. Or kcat (f) On average, fewer than 10 active replication forks are found in an E. coli cell at any time. You are told between DNA polymerases I and III, one can be found at ~10 molecules/cell, where the other is found ~400 molecules/cell. Specify which polymerase you think is found at which concentration and briefly explain why. Due to its high processivity and role as the major DNA polymerase of the cell, one would not expect a cell to require hundreds of copies of DNA polymerase 111 working on fewer than 10 replication forks. Therefore, expect to find DNA polymerase III closer to ~10 molecules/cell. One may expect to find DNA polymerase I at closer to ~400 molecules/cell due to it lower processivity and role of filling in gaps of DNA generated from Okazaki fragments on the lagging strand. Question 3* (from Question 5 Exam 1 2005) Imagine that you found an organism on Mars that synthesized proteins using the same 20 amino acids as life here on earth, but it had a different triplet code that you already knew was non- overlapping and nonpunctuated. You made an in Vitro translation system from this organism, then set out to determine its code and made the following observations: poly(AUU) coded for poly—Ala, poly—Asp, and poly-Thr poly(CUAU) coded for a mixed polymer containing Ala, Phe, and Gly What are the new conclusions that you could draw about the code of this Martian organism based on these two observations? Codon UAU codes for Ala. Asp is coded for by either UUA or AUU, and Thr will be coded for by the other. Phe is coded for by one of CUA, AUC, or UCU. Gly is coded for by another of CUA, AUC, or UCU. The code is degenerate-the last of CUA, AUC, or UCU codes for either Phe, Gly, or Ala. Question 4* (from Question 5 Exam 2 2006) You isolate a protein (Protein X) from an uncharacterized virus and then submit it for protein sequencing and manage to obtain the following sequence from the N-terminus: MITISGREAT. , Comparison to the databases reveals that this 10—aa sequence has similarity to two very different proteins (A and B), which have folds very different from each other. The fact that Proteins A and B are so different from each other suggests that at least one of them is not evolutionarily related to your protein. Proteins A and B have peptides with sequence MFHISMCLAT and MLAVKGQQAT, respectively. (a) Considering the alignments shown below, which protein (A or B) do you think is related to your protein X? Justify your answer. BLOSUM62 Score 8 6 —3 6 6 —4 —5 —4 6 7 :23 Protein X peptide M I T I S G R E A T Protein A peptide M I H I S M C L A T Protein X peptide M I T I S G R E A T Protein B peptide M L A V K G Q Q A T BLOSUM62 Score 8 2 0 4 0 8 1 3 6 7 =39 _—__—_——.—_—————_—_ Protein B is related to Protein X. Scoring the sequence alignments with the BLOSUM62 matrix shows that the sum for the A-X alignment is less than the B-X alignment. Since a higher score indicates that amino acid substitutions are observed more, B is more related to X. (b) As sequences change during evolution, Glutamines sometimes change to other amino acids. When considering changes from Gln to Glu, Lys or Asn (i .e. Gln 9 Glu or Gln 9 Lys or Gln 9 Asn), which of these three possibilities is least common? How did you arrive at your answer? Using BLOSUM62 scores: Q9N score=0 Q9K score=2 Q9E score=3 The Q9N change is the least common possibility because it has the lowest score. (0) Provide an explanation for why your choice in part (b) would be less common than the other two choices. Amino Acid Codons CAA, CAG GAA, GAG AAA, AAG AAU, AAC Changing Q to E or K requires changing one base in the codons. Changing Q to N requires changing two bases in the codons. Therefore a Q to N change is a less likely alteration. Question 5 Translation Which steps in bacterial protein synthesis are dependent on ATP? Which are dependent on GTP? Aminoacylation of tRNA by an Aminoacyl tRNA synthetase requires ATP. During Initiation, IF2 requires GTP binding to associate with initiator tRNA leading to formation of the ternary complex. Hydrolysis of this GTP leads to transition from the 308 initiation complex to the 708 initiation complex. During Elongation, EF-Tu requires GTP to bind aminoacyl-tRNA and to bind to the ribosome at the A-site. Hydrolysis of this GTP occurs when an appropriate interaction between the EF-Tu-aminoacyl-tRNA complex and ribosome has formed (proper pairing between codon and anticodon). Hydrolysis also causes release of EF-Tu from the ribosome. During Elongation, EF-G requires GTP to facilitate translocation. On GTP hydrolysis, EF-G undergoes a conformational change that displaces the peptidyl-tRNA in the A site to the P site. ...
View Full Document

{[ snackBarMessage ]}

Page1 / 4

7.05_Problem_Set_3_Solutions - 7.05 Problem Set 3 SOLUTIONS...

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