MT2key_1 - BIO320 spring 2010 Midterm 2 Answer Key # BLUE...

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Unformatted text preview: BIO320 spring 2010 Midterm 2 Answer Key # BLUE ORANGE RED GREEN 1 D E D D 2 D E C D 3 B D A B 4 C D A E 5 D E C B 6 A B D C 7 B D A C 8 B E C B 9 C C D E 10 E B B C 11 B D A A 12 C C E D 13 B D C A 14 A C A B 15 B C D B 16 C A A B 17 B A A D 18 E D D C 19 C C A B 20 D C D C (SHORT answers on following pages) BLUE Short answer question 1. 5'..GTGCGATCTA moreintron GTCACGAGCGCATTCGA..3' 3'..CACGCTAGAT sequence CAGTGCTCGCGTA AGCT..5' (a) The DNA sequence includes the translation start site, which will appear in DNA as 5ATG3. This occurs only once (underlined), in the bottom strand (reading 5 to 3, right to left). Therefore this strand must be the sense strand. The primary transcript, which will include the intron , will be: 5..UCGAAUG CGCUCGUGAC moreintron UAGAUCGCAC..3 (b) To derive the messenger RNA sequence, the intron must be removed. The splice donor (5GU3) and acceptor (5AG3) sequences each appear only once (in bold ) in this part of the transcript: 5..UCGAAUG CGCUC GU GAC moreintron U AG AUCGCAC..3 Removing the intron sequence (bounded by and including these sites) gives the mRNA sequence: 5..UCGAAUG CGCUCAUCGCAC..3 (c) The START codon sets the reading frame for translation: 5..UCGA AUG CGC UCA UCG CAC..3 From the code table: (NH 2 )- Met Arg Ser Ser His (d) The mutation changes the splice donor site, which will now not be recognized as a splice site by the spiceosome. The intron, or at least this part of it, will be retained in the mRNA: * 5..UCGAAUG CGCUC AU GAC moreintron Translation will begin as before, but it will now proceed into the intron sequence, and encounter a STOP codon (underlined) * 5..UCGA AUG CGC UC A U GA C moreintron (NH 2 )- Met Arg Ser STOP BLUE Short answer question 2a Complete the following partial diploid genotype by filling in the blank allele designations [ ], to make a cell that produces galactosidase inducibly and permease constitutively, but never produces transacetylase. Choose from [+], [], [c], or [S]. lac I[] P[+] O[+] Z[+] Y [ ] A[ ] / F lac I[+] P[+] O[c] Z[] Y [+] A[ ] Galactosidase is produced from the lacZ[+] gene on the chromosomal (left) copy of the operon, its expression is regulated by the repressor protein encoded on the F copy. Permease is produced constitutively from the F copy, which has a mutant operator. Both copies of the lacA gene are mutant, so transacetylase is not produced....
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This note was uploaded on 04/21/2010 for the course BIO 89329 taught by Professor Hollingsworth during the Spring '10 term at SUNY Stony Brook.

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MT2key_1 - BIO320 spring 2010 Midterm 2 Answer Key # BLUE...

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