07exam1key - \1 Gene Action Exam 1 Wm». . lame: I 2 Q I \....

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Unformatted text preview: \1 Gene Action Exam 1 Wm». . lame: I 2 Q I \. 5 U) K 1. (17 pts) Imagine that the following DNA sequence comes from the middle of the wild- gpe Iac/ open reading frame, as indicated (the arrow shows the direction of transcription and translation of the gene). 5’ A"?GATATGGAGTTAACGCTCGCATACACTAGCSA 3’ 3’ TACT ACCTCAATTGCGAGCGTATGTGATCGT 5’ a. What ami acid sequence does this DNA sequence encode? (Diagram the codonsa d the corresponding amino acid sequence below.) ’5 arrectreadinj GM) @cxho UUA (3m one «(,0 630) mm" at A11: *9? “1" 6‘“ u” M“ 1"" Tm ser czcraA‘T L; 6461C “pr-40.0“ (J’le n -. (3'. :9 (:16 Tue—96L fl what ooclms b. On your sequence above, circle the codons which can be converted into codons for arginine by single base pair changes in the DNA, and write the base pair change necessary below each. 1; SWA “,an Second Position ‘03 flUA UW mol- IIL ’"1" ‘J .. mace 'M6;9,—)(:1L “NRA—mu .7611 (puwmj «(he fl askibuhofi as (of ber‘ec‘f FCaGiURj First Position (5' end) (pua .c) uomsod Pllul L—Ofl‘l‘y 2. (15 pts) Imagine a bacterial cell with a linear douole-stranded DNA molecule that is half-replicated. The cell is incubated just long enough to allow replication to be completed, generating two daughter DNA molecules, as shown here: RNAL. primers Starting with the same half-replicated DNA molecule, imagine that the incubation takes place in the absence of DNA ligase, but all other replication enzymes function normally. The daughter DNA products might be predicted to be as shown here (on the right): RNAL primas For each of the following functions, predict how the daughter DNA products would look in the absence of the function listed (Draw the products you would expect). In each case, except for the one function listed, assume that all the other replication proteins are resent and can act normall . - p y l. Hahn-4e unuMJ‘S shunts \ ft- tvr Elma 3n (P0113 2. DNA palm (or-A3‘lh} l (0,. h". Sfavms (L‘\fiégz\) a. DNA primase Hm.“ sh“; W LONA Pu I emu: RNA . flaw; _ ,w mAfi-lum wlr‘tDNA \M\\\~H\\>\W Ll “3"” “'45 “NI” ‘9‘? for N: flcl-(oh‘o m “U‘C (or C((Afhf Show :L'" from“. (5:!ofiiomh.‘ Pol 1 deaf-511$ “M—S \ ‘ _\ \ W} .y“ rtrJH A B'QthanS—me‘nkugg # 1‘ lp‘l‘S (no unwino‘.‘43\ 1, DNA Pot Inner: (204A 9+ 1%: ‘F-Ifi‘mlal/x (90‘ S\ ' l b. DNA helicase {W Ms M Wk BNA 18‘ 9* hr “‘3”: 2. Lax-tr! S‘cnls nidc‘ I] W \\‘\\\l_t\\| CO VA ‘ «Hund- r’r (PN;AS \\ gh—Skom» I I \/ c A fur-k (no ua-u-AAA‘D l, Helms: Mania/15 stasis WM. kw: 0. DNA polymerasel 2. PM}: wth ‘ (“1:1 Wm! M RNA 3. DNA PrMME [zrx'mes Pea-mg“ ea L K L {t \ \ \ \ ‘ “' .— i;_\..‘.AM LAM ‘1‘ DNA P" m “WAS mnzlnMurA DNA Map-SM P chunk Fm‘MMH 09A ‘0 + ____,____,_... __a_l___k.4- ‘\>iuhn\~sl\\\‘ (RNA PQMACQS; NJ“ mucky!“ 7__A§__é__ Nip” l m MAJ M3 3" rematm“) \l swam 3. (20 pts) In E. coli, metabolism of the sugar galac...se is carried out by the enzymes encoded by the galK, galT and galE genes, as shown below. The letters X, Y and Z represent intermediate metabolites in the pathway. (further metabolism) _—“—‘—> CO; Cells with mutations in galE lyse (i.e., cannot grow) in the presence of galactose because the absense of functional GaIE enzyme permits accumulation of intermediates X and Y, which are toxic to cells in high concentratiOns. Imagine you have isolated a mutant strain with an amber mutation in galE. The strain has a complete wild-type lac operon (i.e., its lac genotype is Iac"). Recall that [3- galactosidase cleaves lactose to yield galactose and glucose. You mutagenize the strain and plate it on media that contains lactose and glycerol (an alternative carbon source which does not require lac or gal genes and which does not affect their expression). Colonies appear. a. Name Mg mutations in the lac region that could lead to the observed colonies. WW“) LacE' Lac‘l ' laco— LALIS 6r?) or (+7) or (+7) or L42} (+4 told) . Describe one type of mutation within the gaIE gene that could lead to the observed colonies. (<10 word Wston h) N“; type of shale. lap GM 0’; Mica (Odom (suofegmm) . Describe one type of mutation outside of both the lac region and the ga/E gene that could lead to the observed colonies. (<10 words) God K, Of“ Armor tlZMlhswere‘oeDr (+4 loll-4‘) m) (at) . If you plate the mutagenized strain on media that contains IPTG, P-Gal (phenyl- B—D-galactoside) and glycerol (no lactose), name the most likely mutation within (*4 the lac region that will lead to growth. (<10 words) Lac? " (r4) Lac?’ or LacIS are less Mal and war—Wt -l— l\ . If the lac operon used a positive “logic” of regulatio , and if the strain as plated as in question (d), name the tw_o most likely types of mutations within the lac region that would lead to growth. (<10 words) LacO or is \ac. Lac:— Lac—i, 6”; less \{Mlo an!) er?) Li?) [WW Q: We (+4 tole 4. (13 pts) Imagine a growth condition In w...c.. \ is no spontaneous deamination of cytosine residues. Under this condition, the most common type of spontaneous DNA change is conversion of adenine in DNA into a non-standard base (call it “X") which base—pairs with cytosine. .I. L] a. What would the most common type of spontaneous base pair change leading to mutation be (write the predicted base pair change)? A:T——> 61:5 4-th 47-» )(C @+:»§+2> 4. 2/ b. Is the base-pair change listed a transition or a transversion? WW 4. 7 c. You isolate a mutant in which the frequency of mutations due to the A9X change is greatly increased at all adenine residues. What (hypothetical) function might be defective in the mutant? (<20 words) @wam )MW‘ X W391" Mama/25. m4m— Name” 5. (20 pts) The table below lists the diploid genotypes for stable F' strains that carry two copies of the lac operon — one on the chromosome and one on the F' factor. For each diploid genotype, indicate in the first column whether active B-galactosidase can be made if lactose is present (write Yes or No); in the second column, indicate whether synthesis of fi-galactosidase activity is inducible, constitutive or neither (write I, C or N); in the third column, indicate whether synthesis of lactose permease activity is inducible, constitutive or neither (write I, C or N); and, in the fourth column, indicate whether or not the strain could grow with lactose as the sole carbon source (write Yes or No). Assume that no recombination or additional mutation takes place. Four answers are provided as examples. I, 0, Z and Yare used for [ac], lacO, lacZ, lacY,»for simplicity. ls indicates lac superrepressor, which is non-responsive to lactose. Can active Is synthesis of Is synthesis of B-galactosidase fi—galactosidase lac pennease Can the cell be made if activity activity grow on lactose lactose is inducible (I), inducible (I), as the sole present constitutive (C) constitutive (C) carbon source (Yes or No ? or neither (N)? or neither (N)? (Yes or No)? rz-xm rmrow In rocmw 3 lo h‘c‘? livle 1P 6. (15 pts) Purified DNA polymerase I from various E. coli strains is incubated in the presence of the following DNA substrate in which some of the T and C residues are radioactively labeled, as shown: radioactively labeled residues CCCCC 5’ . . .TTTTTTTTTTTTTTTTTTTTTTTTTTTC 3’... You measure the loss of labeled dT and d0 residues over time either without any dTTP present (no DNA synthesis is possible), or with dTTP present (DNA synthesis can occur). Enzyme purified from wild-type cells yields the following results: dTTP Absent dTl‘P Present 8 % label remaining “la label remaining 0 8 8 8 8 0 time (minutes) time (minutes) / Enzymes purified from two different mutant strains, however, yield these re ults: Mutant A: dTTP Absent dTTP Present % label remaining 8 8 8 8 % label remaining 0 time (minutes) time (minutes) a. What activity is affected in mutant A and how is it affected ‘ “NP; “Pf-00%rtuJ-Mj' (“In 3‘45‘ Mflvthek as w: I) l- b. What activity is affected in fT-Zf’tglt B and how is it affected? (510 words) DNA Pol 1: rue—«Kiln r1 Mow-L (+00") c ,‘U‘tn‘i’ (twain) lanai/Ll mm (,fl-FGMH7I (m (upcd’ an”) 0. What growth phenotype would you predict for mutant B? (510 words) S(0v~’£f~ {yin/TR )V'L +0 +W/{mr37 JKA) 3n “{AdUm/Q’r PaafixxJ-b ...
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07exam1key - \1 Gene Action Exam 1 Wm». . lame: I 2 Q I \....

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