{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

MT2key_1 - BIO320 spring 2010 Midterm 2 Answer Key 1 2 3 4...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
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)
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
BLUE Short answer question 1. 5'..GTGCGATCTA…… more intron ……GTCACGAGCGCATTCGA..3' 3'..CACGCTAGAT……… sequence ………CAGTGCTCGCGTA AGCT..5' (a) The DNA sequence includes the translation start site, which will appear in DNA as 5’ATG3’. 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…… more intron ……UAGAUCGCAC..3’ (b) To derive the messenger RNA sequence, the intron must be removed. The splice donor (5’GU3’) and acceptor (5’AG3’) sequences each appear only once (in bold ) in this part of the transcript: 5’..UCGAAUG CGCUC GU GAC…… more intron ……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…… more intron …… 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…… more intron …… (NH 2 )- Met Arg Ser STOP
Background image of page 2
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. Short answer question 2b Even though repressor is normally bound to the operator in an uninduced cell and prevents almost all transcription, a very occasional transcription event can occur. This was experimentally demonstrated to happen about once every few cell cycles in the absence of inducer.
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}