MIT Department of Biology 7.014 Introductory Biology, Spring 2005 7.014 Quiz II Handout **This will be a closed book exam**Question 1 Shown below is a representation of an origin of replication. 5' 3' 5' 3' A C B D GGAAC CAAGG GGAAC CAAGG fork 1 fork 2 template 4 template 3template 1 template 2 ORI a) For the following, use sites A and B with respect to fork 1 and sites C and D with respect tofork 2. i) On which strand(s) will replication be continuous? template 1 template 2 template 3 template 4 ii) To which site or sites (A, B, C, or D) can the primer 5'-GUUCC-3' bind to initiatereplication? iii) When DNA ligase is inhibited, it differentially affects the synthesis from the leadingand the lagging strands. Explain which strand (leading or lagging) is more affected bythe lack of DNA ligase and why. 2Question 1, continued b) The next nucleotide to be added to a growing DNA strand is dCTP (shown). • Circle the part of the growing DNA chain to which the next base is attached. • Circle the part of the dCTP that is incorporated into the growing DNA chain. NN NH 2 Growing DNA chain N N dCTP CH3 O O O O -P OCH2OO P O OP O O -O -O -O C C C CH2 C C C OC OH H C OH H c) DNA Replication involves many different enzymatic activities. Match each enzyme activitylisted below with the function(s) that it has in the replication process. The first one is done for you. Enzyme Activity Function(s) Topoiosmerase k Primase (synthesizes primer) DNA polymerase to elongate newDNA strand Helicase to unwind DNA DNA polymerase to replace RNAwith DNA Processivity factor Choose From: a) 3’ 5’ growth of new DNA strand b) 5’ 3’ growth of new DNA strand c) 3’ 5’ exonuclease d) 5’ 3’ exonuclease e) Makes RNA primer complementaryto the lagging strand f) Makes RNA primer complementaryto the leading strand g) Makes peptide bonds h) Separates the two DNA strands i) Maintains DNA polymerase on template j) Provides 3’ hydroxyl for initiation of DNA polymerization k) Untangles super-coiled DNA 3___ Question 2 Below is the partial sequence of the sevenohwunforin (7014in) gene, hypothesized to be mutantin students who take 7.013 Introductory Biology and in those students at the other school upthe river. The promoter is underlined and transcription begins at and includes the bold G/C base pair. 5’ TGCCA TCCGA TTGGT GTTCC TTCCA TGAAG GATGC ACAAC GCAAA 3’ 3’ ACGGT AGGCT AACCA CAAGG AAGGT ACTTC CTACG TGTTG CGTTT 5’ 5’ TACAC GCTTA GCTGA CTATA AGGAC GAATC GCTAC AACGA TGCGA 3’ 3’ ATGTG CGAAT CGACT GATAT TCCTG CTTAG CGATG TTGCT ACGCT 5’ 5’ TGCCA TCCGA TTGGT GTTCC TTCCA TGAAG GATGC ACAAC GCAAA 3’ 3’ ACGGT AGGCT AACCA CAAGG AAGGT ACTTC CTACG TGTTG CGTTT 5’ a) What are the first 12 nucleotides of the transcript encoded by the 7014in gene? Label the 5’and 3’ ends. __’--__’ b) On the DNA sequence above, circle the DNA bases that encode the first amino acid of the protein. c) What are the first four amino acids encoded by the 7014in transcript? Label the N- and C-terminus ____--d) You want to create a system to translate a specific mRNA in a test tube. To an appropriatewater and salt solution you add many copies of this mRNA and ATP (energy). What other key components must you add? You succeed in translating the mRNA in your test tube. You repeat the experiment with twoidentical test tubes. You add limiting amounts of the antibiotic puromycin to test tube 2 only.Puromycin is a molecule that has structural similarities to the 3’ end of a charged tRNA. It can enter the ribosome and be incorporated into the growing protein. When puromycin isincorporated into the polypeptide, it stalls the ribosome and the polypeptide is released. You do not know if puromycin recognizes a specific codon or not. e) What effect would puromycin have on transcription? f) What effect would puromycin have on translation? 4Question 2, continued g) You examine the length of the polypeptide produced in both test tubes. i) In test tube 1 (no puromycin) you get a polypeptide that is 100 amino acids long. At least how many bases was the mRNA that you added? ii) Which of the following would you find in test tube 2 (has limiting amounts ofpuromycin) if puromycin does NOT recognize a specific codon. Only a single type of polypeptide Only 2 types of polypeptides that are each different lengths Only 3 types of polypeptides that are each different lengths Only 4 types of polypeptides that are each different lengths Polypeptides of all sizes, i.e., dipeptides, tripeptides, … a polypeptide that is 100amino acids long iii) Which of the following would you find in test tube 2 (has limiting amounts ofpuromycin) if puromycin recognizes a specific codon that occurs three times in themRNA. Only a single type of polypeptide Only 2 types of polypeptides that are each different lengths Only 3 types of polypeptides that are each different lengths Only 4 types of polypeptides that are each different lengths Polypeptides of all sizes, i.e., dipeptides, tripeptides, … a polypeptide that is 100amino acids long 5Question 2, continued The 7014in gene encodes a protein (7014IN) that binds the neurotransmitter serotonin, asshown below. The five amino acids 7014IN involved in binding serotonin are shown. To understand the difference between introductory biology students, you have determined theDNA sequence for the 7014in gene in a group of 7.013 and 7.014 students. Below is the 7014IN protein and the DNA sequence that encodes it. The amino acids depicted in the picture above are underlined. From a 7.014 student: DNA 5’ ACC AAT GGA CCA GCA GGA AGC GGG GTA GCT GAG TAC 3’ 3’ TGG TTA CCT GGT CGT CCT TCG CCC CAT CGA CTC ATG 5’ Protein N- Thr Asn Gly Pro Ala Gly Ser Gly Val Ala Glu Tyr –C h) You find that 7.013 student 1 has the following DNA sequence for the 7014IN: 5’ ACC AAT GGA CCA GCA GGA TAG CGG GGT AGC TGA GTAC 3’ 3’ TGG TTA CCT GGT CGT CCT ATC GCC CCA TCG ACT CATG 5’ i) Indicate (circle/underline) the site of the mutation on the sequence directly above. ii) Does student 1 have an insertion, deletion, or substitution mutation? iii) Would you expect this DNA sequence to encode a protein that binds serotonin?Why or why not? A chart of the amino acids is found on page 10. i) You find that 7.013 student 2 has the following DNA sequence for the 7014IN: 5’ ACC AAT GGA CCA GCA GGA AGC GGG GTA GCT GAT TAC 3’ 3’ TGG TTA CCT GGT CGT CCT TCG CCC CAT CGA CTA ATG 5’ i)
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