Your Name: _______________________________________TA:______________ 7.012 Quiz 2 Answers A≥85 ~12% of test takers B≥72 ~31.2% of test takers C≥60 ~34.1% of test takers D≥50 ~16.3% of test takers F≥49 ~6.2% of test takers Regrade requests (with a note attached indicating the problem and part you want looked at) accepted until Thursday November 4 th , 5pm. Question Value Score 1 17 2 16 3 30 4 17 5 20 100 MIT Biology Department 7.012: Introductory Biology - Fall 2004 Instructors: Professor Eric Lander, Professor Robert A. Weinberg, Dr. Claudette Gardel 2 Question 1 In the bacterium Funditus fabricatus, the metabolism of the sugar ridiculose is dependent on the rid operon shown below. The ridiculose operon encodes the enzymes shown in the following pathway. Rid U Rid V Rid X Ridiculose Ridiculose Ridiculose-5P Seriose-5P (outside cell) (inside cell) (intermediate) (used by cell) The ridW gene is constitutively expressed. The expression of ridU, ridV, and ridX genes is off in the absence of ridiculose and is activated by the product of the ridW gene in the presence of ridiculose. There is an artificial inducer of ridUVX expression, called GIG-L, and an artificial substrate for Rid X, called STRN that turns red in the presence of active Rid X protein. a) Several specific Rid- mutants of F. fabricatus are shown below. Predict their phenotypes when grown in the presence of STRN, with and without the addition of the inducer, GIG-L. 10 pts (Fill in the chart with either RED or WHITE) Strain of F. Fabricatus + GIG-L - GIG-L _______________________________________________________________________ WT RED WHITE _______________________________________________________________________ M1 (deletion of P UVX ) White White _______________________________________________________________________ M2 (control region that can’t bind RidW protein) White White _______________________________________________________________________ M3 (RidW protein that can’t bind ridiculose) White White _______________________________________________________________________ M4 (nonsense mutation early in ridX) White White _______________________________________________________________________ M1337 (RidW protein that always binds control region) Red Red ________________________________________________________________________ ridWridXridVridU P W P UVX Control region Name:___________________________________ TA:_______________________ 3 b) Predict whether the following F. fabricatus strains, that are merodiploid for the ridiculose operon, will grow on minimal media with or without ridiculose as the only carbon source. Fill in the chart with YES if the merodiploid will GROW or NO if the merodiploid will NOT GROW 7 pts (5 points for the first column, 2 pts for entire last column) Merodiploid +Ridiculose - Ridiculose _______________________________________________________________ WT / WT YES NO _______________________________________________________________________ M1 / M2 (deletion of P UVX ) NO NO (control region that can’t bind RidW) _______________________________________________________________________ M2/ M3 (control region that can’t bind RidW) YES NO (RidW protein that can’t bind ridiculose) _______________________________________________________________________ M3 / M1337 (RidW protein that can’t bind ridiculose) YES NO (RidW protein that always binds control region) _______________________________________________________________________ M4 / M 1 (nonsense mutation early in ridX) NO NO (deletion of P UVX ) _______________________________________________________________________ M1337 / M4 (RidW protein that always binds control region) YES NO (nonsense mutation early in ridX) _______________________________________________________________ 4 Question 2 You believe that a disruption in a gene, sokS, may contribute to an interesting disease phenotype in cardinals. You wish to PCR amplify and sequence sokS from both wild type and diseased cardinals. The sokS gene is shown below. Exons are represented by numbered boxes and the terminal sequences are depicted in bold. 5'CTCGAGGTAT (1) (3) GTAATCGATA 3' 3'GAGCTCCATA (2) (4) CATTAGCTAT 5' a) For PCR amplification, the primers should be identical to which of the following sequences? (Circle one.) 3 pts 1 and 3 2 and 4 1 and 4 2 and 3 b) Which primer(s) should you use in one sequencing reaction to obtain sequence that looks most similar to the mRNA (the coding strand)? 2 pts 1 2 3 4 You use the dideoxy sequencing method to sequence your PCR products. You see the following pattern in the sequencing gel representing the sequence spanning the end of the first intron and the beginning of exon 2. Wild Type (WT) Diseased (D) ddA ddC ddG ddT ddA ddC ddG ddT c) These gels correspond to which WT and diseased sequences respectively? (WT, D) (Circle i, ii, iii, or iv.) 4 pts i) 5’-ACTGGAAC-3’, 5’-ACTTGAAC-3’ ii) 5’-TGACCTTG-3’, 5’-TGAACTTG-3’ iii) 5’-CAAGGTCA-3’, 5’-CAAGTTCA-3’ iv) 5’-GTTCCAGT-3’, 5’-5’-GTTCAAGT-3’ 1 2 3 Start of gene End of gene Name:___________________________________ TA:_______________________ 5 To determine how this mutated sequence might affect the mRNA and the protein product, you obtain two cDNA libraries: one derived from wild type cardinal RNA and one derived from diseased cardinal RNA. d) You choose to use a radioactively labeled DNA probe to screen these cDNA libraries for clones containing sokS. Which region of sokS would make the best probe for screening the available cDNA libraries? (Circle either Region A, Region B, or Region C.) 4 pts Region Region B Region C 100 100 40 200 400 50 Distances in base pairs Your probe hybridizes to cDNA clones in both cDNA libraries. You purify the plasmids from single clones and cut them with the restriction enzyme used for cloning to verify insert size. The gel is shown below. Ladder WT Diseased e) Based on all evidence above, what is the most likely explanation for the difference in restriction enzyme digestion patterns of the sokS cDNA clones? 3 pts i) A mutation in the sokS cDNA from the diseased cardinal cDNA library gives rise to an additional restriction enzyme site. ii) The mRNA encoded by the sokS gene from the diseased cardinal is incorrectly spliced. iii) There is likely a problem with the gel and it should be rerun. iv) The sokS gene from the diseased cardinal acquired a spontaneous insertion. v) The mRNA encoded by the sokS gene from the diseased cardinal has no poly A tail. 1000 bp 700 bp 500 bp 300 bp 50 bp Vector 1 2 3 6 Question 3 a) Match the following. Choose only one answer for each blank below. 10 pts A. Unwinds DNA B. Where Okazaki fragments are synthesized C. Where DNA can be replicated continuously D. Synthesizes RNA primers on DNA E. Synthesizes DNA primers on RNA F. Catalyzes the addition of dNTPs to lipids G. Relieves tension in DNA caused by unwinding H. 5’ to 3’ proofreading activity I. 3’ to 5’ proofreading activity __B__ lagging strand __I__ DNA polymerase __G__ topoisomerase __D__ primase __A__ helicase b) Where in the eukaryotic cell does the following processes of the central dogma occur? One word answer for each, and please write legibly. 6 pts DNA RNA Protein 1 2 3 Process 1____nucleus_______________ Process 2_______nucleus____________ Process 3______cytoplasm___________ Name:___________________________________ TA:_______________________ 7 c) Below are schematics of transcription, splicing and translation. Match the numbered boxes with the following terms. Use each number only once. It’s okay to leave blanks. 8 pts __3_ 5’ cap _____3’ Cap __6_ Amino acid _7__ Anti-codon __1__ Coding (Non-template) strand __8_ Codon ____ Exon __4__Intron __5_ mature mRNA ____ PolyA tail ____ Pre-mature mRNA __2__ Template strand ____ tRNA d) In the box below, write the sequence that would be in box 7 above. Designate the 5’ and 3’ orientations. 3 pts e) Circle the specific name of the structure that is in box 6 above. 3 pts. Alanine Arginine Cysteine Cytosine Cyanide Methionine Threonine Uracil 5’ – ACG – 3’ H 5 8 Question 4 Below is a diagram of a transmembrane protein called Soxwin. Normally Soxwin is found embedded in the plasma membrane. You obtain a mutant in which Soxwin is mislocalized. The mutation resides in the DNA that encodes the transmembrane domain. a) What type of mutation occurred in the soxwin gene? Circle your answer(s). 2pts deletion frameshift insertion missense nonsense silent b) How has this mutation changed the chemical property of the transmembrane domain? Fill in each blank with one word. 2pts From __hydrophobic, non-polar_ in WT to _hydrophilic, polar, or charged__ in mutant c) Where do you expect the majority of the mutant Soxwin to accumulate? Circle your answer. 3 pts cytoplasm endoplasmic reticulum golgi apparatus mitochondria nucleus outside the cell peroxisomes plasma membrane ribosomes Name:___________________________________ TA:_______________________ 9 d) There’s another soxwin mutant in which a missense mutation abolishes the function of the signal sequence. Where would you expect the majority of this mutant Soxwin protein to accumulate? Circle your answer. 3 pts cytoplasm endoplasmic reticulum golgi apparatus mitochondria nucleus outside the cell peroxisomes plasma membrane ribosomes e) Match the following. (Multiple answers may be chosen for each blank.) 7 pts Destinations of proteins Molecular events ____C___ cytoplasm A. co-translational transport ____B, C___ mitochondria B. post-translational transport ____B,C___nucleus C. entire protein synthesized on a free ribosome ____A,D___ extracellular space D. signal sequence recognized by SRP 10 Question 5 a) A plasmid is a… (Circle your answer(s).) 2pts bacterium circular piece of DNA cell multipurpose enzyme petri plate vesicle b) Match each vector feature with its function. Not all answers need be used. 8 pts ____B____Restriction site A) Required for expression of insert B) Allows for insertion of DNA into vector ____E____Origin of replication C) Encodes an enzyme to cut DNA D) Enables selectability for strain that has taken up the vector ____A____Promoter E) Required for duplication of vector F) Required for SRP to bind ____D____Drug resistance G) Site for ribosome to bind c) To obtain the gene that rescues a tryptophan biosynthesis E.coli mutant strain named, NY-trp-Zup, you construct a genomic library from wild-type E.coli by cutting the genome with BamHI and inserting the fragments into pGoSOX!, a plasmid which has been very successful in the lab. pGoSox! contains the genes for tetracycline and kanamycin resistances and has a unique Bam HI restriction site that maps to the kanamycin resistance gene. You transform the library into NY-trp-Zup and plate the transformants onto rich agar medium. You replica plate the colonies onto different media shown below. Below are the plates shown in the same orientation after colonies form. Rich Medium Minimal medium Rich +Tetracycline Rich +Kanamycin (lacks tryptophan) i) Which colony (ies) contain the original pGoSOX!? 3 pts 1@ None 1 2 3 4 5 6 ii) Which colony (ies) carry pGoSOX! containing an insert? 4 pts 2@ None 1 2 3 4 5 6 iii) Which colony (ies) would you choose to further study the gene encoding the tryptophan biosynthetic enzyme that is deficient in NY-trp-Zup? 3 pts 3@ None 1 2 3 4 5 6 1 2 3 4 5 6 Name:___________________________________ TA:_______________________ 11 C C OO H NH 3 CH 2 CH 2 CH 2 CH 2 NH 3 + ALANINE (ala) ARGININE (arg) ASPARAGINE (asn) + - ASPARTIC ACID (asp) CYSTEINE (cys) - + C N CH 2 CH 2 CH 2 H H H - C OO + GLUTAMIC ACID (glu) GLUTAMINE (gln) GLYCINE (gly) HISTIDINE (his) ISOLEUCINE (ile) LEUCINE (leu) LYSINE (lys) METHIONINE (met) PHENYLALANINE (phe) PROLINE (pro) SERINE (ser) THREONINE (thr) TRYPTOPHAN (trp) TYROSINE (tyr) VALINE (val) STRUCTURES OF AMINO ACIDS at pH 7.0 C N H H H H H H C C OO H NH 3 CH 3 - + C C OO H NH 3 CH 2 CH 2 CH 2 N H C NH 2 NH 2 - + C C OO H NH 3 CH 2 C O NH 2 - + C C OO H NH 3 CH 2 C O O - + C C OO H NH 3 CH 2 SH - + C C OO H NH 3 CH 2 CH 2 O O C - + C C OO H NH 3 CH 2 CH 2 O C NH 2 - + C C OO H NH 3 H - + C C OO H NH 3 CH 2 C N C N H H H H - + C C OO H NH 3 C H CH 3 CH 2 CH 3 - + C C OO H NH 3 CH 2 C H CH 3 CH 3 - + - + C C OO H NH 3 CH 2 CH 2 S CH 3 - + C C OO H NH 3 CH 2 H H H H H - + C C OO H NH 3 CH 2 OH - + C C OO H NH 3 C H OH CH 3 - + C C OO H NH 3 CH 2 - + C C OO H NH 3 CH 2 OH H H H H - + C C OO H NH 3 C H CH 3 CH 3 - + 12 The Genetic Code U C A G U UUU phe (F) UUC phe (F) UUA leu (L) UUG leu (L) UCU ser (S) UCC ser (S) UCA ser (S) UCG ser (S) UAU tyr (Y) UAC tyr (Y) UAA STOP UAG STOP UGU cys (C) UGC cys (C) UGA STOP UGG trp (W) U C A G C CUU leu (L) CUC leu (L) CUA leu (L) CUG leu (L) CCU pro (P) CCC pro (P) CCA pro (P) CCG pro (P) CAU his (H) CAC his (H) CAA gln (Q) CAG gln (Q) CGU arg (R) CGC arg (R) CGA arg (R) CGG arg (R) U C A G A AUU ile (I) AUC ile (I) AUA ile (I) AUG met (M) ACU thr (T) ACC thr (T) ACA thr (T) ACG thr (T) AAU asn (N) AAC asn (N) AAA lys (K) AAG lys (K) AGU ser (S) AGC ser (S) AGA arg (R) AGG arg (R) U C A G G GUU val (V) GUC val (V) GUA val (V) GUG val (V) GCU ala (A) GCC ala (A) GCA ala (A) GCG ala (A) GAU asp (D) GAC asp (D) GAA glu (E) GAG glu (E) GGU gly (G) GGC gly (G) GGA gly (G) GGG gly (G) U C A G