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Contents : MIT Biology Department 7.012: Introductory Biology - Fall 2004 Instructors: Professor Eric Lander Professor Robert A. Weinberg Dr. Claudette Gardel Name: TA: 7.013 Problem Set 3 FRIDAY October 8th 2004 Problem set answers must be inserted into the box outside Problem sets will NOT be accepted late. Solutions will be posted on the web. Question 1 a) Below is a schematic of an inset electron micrograph showing simultaneous transcription and translation of a gene in E. coli. Select the names from the following list of structures depicted in the schematic and write them on the lines adjacent to their symbol. (Use each term only once you do not need to use all of them.) Some of the terms are explained in the book i.e. page 265. 1. 2. 3. 4. N-terminal of polypeptide C-terminal of polypeptide 3 end of mRNA Shine-Delgarno sequence 5. 6. 7. 8. 5 end of mRNA ribosome polypeptide RNA polymerase 9. 10. 11. 12. messenger RNA GTP cap 3 end of gene promoter 13. 14. 15. 16. DNA 5 end of DNA DNA polymerase stop codon A B C D E F G Image removed due to copyright reasons. b) According to the schematic in which direction are the structures symbolized by the checkered ovals moving up down not moving left right c) In which direction are the structures symbolized by the striped boxes moving up down not moving left right 1 Question 2 Assume mRNA is being transcribed starting from the far left side of the following double stranded DNA template. 5 GTGCTAGCGGGAATGAGCTGGGATACTAGTAGGGCT3 3 CACGATCGCCCTTACTCGACCCTATGATCATCCCGA5 a) What are the first five nucleotides of the mRNA sequence b) What are the first 5 amino acids encoded c) The following sequences show (in bold) different mutations affecting the above DNA sequence. Assume none affect the expression of the mRNA synthesis. A. 5 GTGCTGAGCGGGAATGAGCTGGGATACTAGTAGGGCT3 3 CACGACTCGCCCTTACTCGACCCTATGATCATCCCGA5 B. 5 GTGCTAGCGGGAATGAGCTGCGGATACTAGTAGGGCT3 3 CACGATCGCCCTTACTCGACGCCTATGATCATCCCGA5 C. 5 GTGCTAGCGGGAATGAGCTGAGATACTAGTAGGGCT3 3 CACGATCGCCCTTACTCGACTCTATGATCATCCCGA5 D. 5 GTGCTAGCGGGAATGAGCTGGGAAACTAGTAGGGCT3 3 CACGATCGCCCTTACTCGACCCTTTGATCATCCCGA5 E. 5 GTGCTAGCGGGAATGAGCTGGGACACTAGTAGGGCT3 3 CACGATCGCCCTTACTCGACCCTGTGATCATCCCGA5 F. 5 GTGCTAGCGGGAATGAGCTGGCATACTAGTAGGGCT3 3 CACGATCGCCCTTACTCGACCGTATGATCATCCCGA5 WT 5 GTGCTAGCGGGAATGAGCTGGGATACTAGTAGGGCT3 3 CACGATCGCCCTTACTCGACCCTATGATCATCCCGA5 2 i) For the above mutations fill in the following box. Sequence Type of mutation Effect on protein Choose from insertion deletion substitution. Choose from missense nonsense frameshift silent. A B C D E F ii) Order the mutations according to the likelihood that they will result in an inactive protein from most likely to least likely. If you think two mutations have an equal likelihood of resulting in an inactive protein write an equal sign between them. Your answer should be in the format X Y Z V WT(wild type). The Genetic Code U U C A G UUU UUC UUA UUG CUU CUC CUA CUG AUU AUC AUA AUG GUU GUC GUA GUG phe (F) phe (F) leu (L) leu (L) leu (L) leu (L) leu (L) leu (L) ile (I) ile (I) ile (I) met (M) val (V) val (V) val (V) val (V) C UCU UCC UCA UCG CCU CCC CCA CCG ACU ACC ACA ACG GCU GCC GCA GCG ser (S) ser (S) ser (S) ser (S) pro (P) pro (P) pro (P) pro (P) thr (T) thr (T) thr (T) thr (T) ala (A) ala (A) ala (A) ala (A) A UAU UAC UAA UAG CAU CAC CAA CAG AAU AAC AAA AAG GAU GAC GAA GAG tyr (Y) tyr (Y) STOP STOP his (H) his (H) gln (Q) gln (Q) asn (N) asn (N) lys (K) lys (K) asp (D) asp (D) glu (E) glu (E) G UGU UGC UGA UGG CGU CGC CGA CGG AGU AGC AGA AGG GGU GGC GGA GGG cys (C) cys (C) STOP trp (W) arg (R) arg (R) arg (R) arg (R) ser (S) ser (S) arg (R) arg (R) gly (G) gly (G) gly (G) gly (G) U C A G U C A G U C A G U C A G 3 Question 3 Part I In E. coli when glucose is present galactose is used to make a component of the bacterial cell wall. When glucose is absent galactose is MOSTLY used to make energy but a little is used to make the cell wall. The enzymes for the conversion of galactose into a cell wall component are transcribed as an operon. The operon is transcribed at a high level when glucose is present and at a low level when glucose is absent. In many organisms this type of regulation is accomplished by having two promoters in front of the operon one that is strong meaning it causes a high level of transcription and one that is weak meaning it causes a low level of transcription. In this problem we will be learning how the transcriptional regulation of this type of operon occurs. You isolate mutants that do not respond ap

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