MIDTERM_ANSWERS

MIDTERM_ANSWERS - / MCB124 Fall 2011 page3 NAME 1....

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Unformatted text preview: / MCB124 Fall 2011 page3 NAME 1. VOCABULARY AND CONCEPTS (10 pts) Match the words with the definition or property. Only one correct answer i secondary structure a. double-stranded DNA or RNA ‘ loop b. classification of the 3-D path of a polypeptide or RNA j- A-forrn c. a four way junction created from two (nearly)-identical DNA duplexes _ base step (1. provides the widest access to the basepairs in B-DNA helices Q Holliday junction e. Conserved sequence patterns that indicate a specific fold or function _5__ alpha helix f. regular structures characterized by repeating backbone torsion bequence motif angles and maximal backbone H— bonds r" orthologs g. where DNA duplexes cross in a loop or supercoil major groove h. result of London dispersion forces between planar bases node i. contains the cytosine 02 and adenine N3 atoms, narrow in A-tracts item j. protein segment that does not adopt compact “regular” 2“ structure h stacking interactions k. 3-D contacts between residues that are far apart in sequence 11“ .1elical repeat 1. two consecutive base pairs L tertiary interactions m.cortnect two polypeptide sequences with minimal loss of hydrogen bonds multimer n. a protein sequence pattern that places residues on One side of an OL-helix ~ i fold 0'. most common geometry of RNA helix gr duplex p. binds to GNRA loops in folded RNAs tetraloop receptor q. contains almost 2 residues/turn and extends 3.2-3.4 Alresidue beta strand r. proteins from a different organisms with a similar functions 3; minor groove 5. 3.6 residuesiturn and extends 1.5 X's/residue t. a protein complex, contains more than one subunit II. SECONDARY STRUCTURE (8 pts) Answer each question BREIFLY (USE FULL SENTENCES) I A.(4 pts) Describe and Explain energetic contributions that favor formation of specific secondary structures (oi-helices, B—strands, etc) in proteins. Secondary structures are the campacied conformation for a polypeptide in which backbone H- bonds are satisfied while simultaneous minimizing backbone torsional energies. The driving force for compactness arises from the twin influences of excluded volume and surface ‘ tension/hydrophobic eflect. T he local context in the 3—D siruclure ultimately defermines which secondary structure is (or is not) preseni.‘ B. m the important energetic factor(s) that drive duplex formation in B-DNA . and pgplafl how they do so (4 pts) I Base siacking, which dominates DNA pairing energetics, increases by aboui 50% upon formation of a dupiexfl‘om single strands. Base-pair H—bona’s minimize the cost of desolvaiing I polar groups on bases while optimizing stacking inferaciions, which provides the basepairing specy’icziy. MCBlZ4 Fall 2011 page4 NAME III. BUILDING BLOCKS (27 pts) A. Amino acidffbase properties— match the factoid with the BEST amino acid OR base (one-letter code). Some amino acids may have more than one match, but only a single way gets ALL of points. (9 pts)Factoid One letter code M first amino acid in newly translated proteins E side chain is a benzyl group Lites a basic side chain that contains a primary amino group LL side chain pKa is near physiological pH fl disrupts backbone hydrogen bonding when in helices and sheets 1 lsosteric (same shape as) to valine fi often found in turns with phi—psi angles near (90,0) 1 phenolic side chain found in topoisomerasc active sites fl side chain is the amide ofD di- forrns cyclic Hoogstien quartets in telomeric DNA Q side chain is -CH2COOH m a fluorophore that absorbs light maximallly at 280 mm C under Oxidizing conditions, cross-links polypeptides containing it _& side chain contains three nitrogens, pKa ~12 2 deamination product of cytidine _& is most stabilizing for alpha helices sidechain is -CH2CH2COOH, , pKa is ~4 defining buried residue in alpha-helical coiled-coil proteins agorm<xcdawozmmwg> B. Basepairs (9 pts) i. Draw an RNA U-A base pair below. The sugar-phosPhate can be abbreviated as Show the Watson-Crick H-bonds with dotted lines (5 pts), ii. Label the “MAJOR” and “MINOR” grooves and circle the H—bonding atoms that protrude into them (2 pts), Indicate the atom(s) susceptible to alklyation with electrophiles with an (1 pt), iv. indicate the atom(s) that are susceptible to spontaneous deamination with a “D” (1 pt). MCB124 Fall 2011 pages NAB/LE C. Peptide and nucleic acids (9 pts) Serine recombinases can catalyze recombination at specific DNA sequences by a cleavage and rejoining mechanism not unlike topoisomeraSes. They cleave the DNA backbone, forming a temporary 5’ phosphodiester bond between the cleaved DNA 5’ end and an active-site SERINE during strand exchange. The DNA-linked covalent intermediates can be isolated and digested with nucleases and proteases to give the peptide S-H-P (one-letter-code) with the S’phospho-dinucleotide (pCpG) attached. a) Draw the SHP tripeptide in a covalent intermediate with a 5’-phosphodeoxycytidyl-deoxyguanosine dinucleotide: S(pdedG)—H-P. (three amino acids, two nucleotides, 1 pt each residue, no partial) (5 ms) b) write and circle the predominant charges next to evem atom that is ionized at pH 5. (3 pts) c) Indicate the peptide bond that is most likely to be cis” with a “*” (1 pts) can 9; /‘?\ WV Q“ J I r‘ \I 0 8 Na. I” \\_.e (.926 W (9’ :1: t” IV TERTIARY AND QUARTERNARY STRUCTURE (16 pts) A. (10 pts) A protein deduced from the human genome sequence GOTNA, has an unknown fold and structure, and is involved in detecting oxidative stress. The gene predicts a small 56 amino acid protein. Dr. Gleek agrees to study this protein. 1) What is the approximate molecular weight of a GOTNA monomer? é I [:2 (2 (1 pt) (hint—see the “cheat sheet”) MCB124 Fall 2011 page6 NAME Circular dichroism measurements suggest that it has about 48% alpha helix and 42% beta strand, the remainder being turns or loops. 2) Given the secondary structure composition how many residues are in alpha helices and beta strands ? (2 pts) 3) What are the total lengths of the helical segments ( SE 2 A) and strand segments (—35:30— A) 7 (2 M) In the absence of oxygen, or the presence of reducing agents like beta-mercaptoethanol, gel-filtration chromatography and ultracentrifiigation indicate the protein is a disk-shaped cylinder, 40 A x 60 A x 60 A. The apparent molecular mass is 24-25,000 Da (24-25 kD). In the presence of strong oxidizing agents like peroxide, it has a different shape and an approximate size of 40 A x 25 A x 25 A. 4) What is the agfiegation state ofthe oxidized {‘00 "GE g [ and reduced i MMU—fonns? (2 P‘s) qué 0n “file, uolomelor wé— icons) 5) Noting the overall dimensions of the oxidized protein draw % plausible architecture and topology. Remember to clearly diagram any potential strand-strand, helix-helix, or helix-strand contacts. ' CLEARLY label the N- and C-termini of each chain AND secondary structural element to get CREDIT. (Use the symbols from the cheat sheet for clarity). (4 pts) 7 C A . N C .. N \ 01w c ' \péflumlo n . (Cook‘ng om where) (ghost 3W1, w/ 5WJ2 w in Pickup 6) Briefly describe an diagram at pla 1e structure for reduced form of th protein. CLEAREY label the N- and C—termini of each chain AND secondary structural element to get CREDIT. (3 pts) NH Jrfi‘rrotwer 'i’lec'i‘ show; 1) pat/UM Upfind or 222 £014 ‘ 2’) LiO db ‘3‘“? Pfirpmdiculour +0 Dfle‘ which 4m 3} Some 80(4- 05 {eqsomtola soot/Ath (mi-qu f Parka fl J. (DI/Fa {Jot claim? The GOTNA sequence contains a single Cys residue. What do you think the effect of peroxide is and how does it cause the change in size? (2 pts) OH o H o F" W (3.4.3 t§ Otgid‘vz‘ecf +0 % 65F. orb'ffio 01ml (ands/*5 lQ‘OM rte/v po[o4. \ +0 polqu—fiacs wtgh’i‘ diSfuP’l' q SeBUni-l- “TR—Qfl—brag MCBIZ4 Fall 2011 page 7 NAME V. FUN WITH NUCLEIC ACIDS (19 pts) 1. In the spaces above nucleic acid 2° structures named below, use lines and arrOWS to sketch the following nucleic acid structures (the duplex is give as an example). Arrows point 5’=>3’ -USE THEM FOR canon. (3 pts) I : j‘T 4—' - l Duplex Stem-loop D-loop Hollillay junction 2. The following ssDNA sequence can form ONE OR MORE of the 4 above structures. Circle ALL of the ones that can be formed from this sequence. 5’-AGC1"AGCC I'I'I‘I'I'AGGCTAGCI‘ (1 pt) 3. 2100 bp plasmid is subjected to the several treatments. For all the products below, calculate the linking number (Lk), twist (T) and writhe (W). For step 4, indicate'the treatment that caused the change(s). In step 2, ethidium from step 1 is still present, but it is removed in Step 3. (6 pts, 1 each, N0 PARTIAL) Lk @U 290 l a; @c l“) 1? T 100 | 99 I 39 1032 9.00 VV ._CL___ __Jtl__fi 4:;§___ ___£1___ __:jL__ €24%~@—vl§>*? 1. Emidium 2. DNA Gyrase 3. anngQ <= WHAT TREATMENT? Bromide ATP 4. I‘CmOVB EtBrfprotein 5. A transcription factor binds to a double-stranded DNA sequence GTI‘CA. Two different chromosomal Sequences are its target (below). In a crystal structure of the protein-DNA complex, the DNA duplex looks like as in (3) protein not shown. (4 pts) (1) ATCGTTCAATGCTGAACAAT (3) V (2) GGAGTTCAGAATTGAACCTG a) Underline all of the Sequences bound in sequences ( 1) and (2) (Remember there is a complimentary strand) (1 pt) ' b) How many protein molecules bind to each sequence? 3L (1 pt) c) Which sequence is probably bound better, (1) or (2)? 1- (1 pt) :1) Briefly explain why? (1 pts) l c ' {AAA} h . 6. Nucleosomes compact dsDNA by wrapping it around them in 150 bp loops. Which of the below sequences would likely bind BEST? (circle) (2 pts) a) (GCGCGCGCGCLS b) (CI'ITI‘AAAAGLS c) 0111mm)” apt) Briefly explain why? (1 pt) - . AWqu Rumor, refine/Ice \ More. Casilq be wrapped QS‘ow‘c“ Q nudeoscwe- ob AA- M03124 Fall 2011 page 8 NAME ' _ . 7. (3 pts) At low concentrations, ethidium bromide can decrease nucleosome binding affinity to closed- circular DNAs but not linear ones, without directly interacting with the nucleosome-bound DNA. Explain why this happens? fl lntercalation by EtBr requires DNA unwinding which removes twist. To maintain Lk in a closed circular DNA, a Iplositive supercoil will be created for every 10-11 EtBr molcules bound. Linear D A is not topologically constrained and does not accumulate supercoils. For similar reasons, nucleosome wrapping of DNA in a negative toroid also creates positive supercoils in the unbound region. Since the cost of adding supercoils increases with the square of thenumber of existing supercoils, EtBr increases the cost of nucleosome binding. Question: Do nucleosomes weaken EtBr binding? EXTRA CREDIT (10 pts): Name TWO non-B-form but regular DNA structures that have potential medical I implications. Describe the structure of each, its biological role and what this has to do With health and/or disease. gpts each) Bren/L A Q.qu Wm .Loolc {Jr 99‘. ...
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This note was uploaded on 11/10/2011 for the course MCB 124 taught by Professor Baldwin during the Spring '09 term at UC Davis.

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MIDTERM_ANSWERS - / MCB124 Fall 2011 page3 NAME 1....

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