Bioinformatics Lab 9
Question 1:
The sequence in the entry that was obtained from sequencing a piece of DNA from Vibrio fischeri genomic DNA digested with Sal I is 8654 bp long.
Question 2:
The -10 sequence for the lux operon is tgttata. 2 nucleotides are different when compared to the ideal sequence.
Question 3:
The lux R has its own promoter and is transcribed in the opposite direction from the lux operon, so it cannot be transcribed from the same strand because the RNA Polymerase recognizes promoter sequences only in the 5’ 3’ direction (lux R is transcribed in the opposite direction). Therefore, luxR has to be transcribed on a different strand.
Question 4:
The coding sequence for the luxA and luxB genes lie within this sequence: * luxA cds 4230-5294 * (5294-4230)+1 = 1065 nt long * (1064+1)/3 = 355 AA long protein expected * luxB cds 5333-6313 * (6313-5333)+1 = 981 nt long * (980+1)/3 = 327 AA long protein expected * lux A RBS: luxA rbs 4217-4222 (aaag ga) * Each gene within the lux operon must have its own RBS because the polypeptide synthesis from the lux operon transcript is an independent process where each gene within the lux operon has its own corresponding transcript and RBS. Also, in bacteria, lux genes are transcribed and immediately translated because there is no nucleus. Therefore, ribosomes can bind to different RBS sequences within he lux operon mRNA, synthesizing different proteins.
Question 5:
- Sal 1 cuts the sequence at:
5’ …G I TCGAC…3’
3’…CAGCT I G…5’ * This is relevant because we isolated the chromosomal DNA of Vibrio fischeri, and Sal I was used to cleave the DNA. * There are 6 Hind III sites in the lux sequence. * There are more sites for Hind III than Sal I because HindII has less G-C content in the cutting sites (SalI has 4GC’s while HindIII has 2 GC’s). Based on the total content of 3926 GC and 6690 AT of the lux sequence, there are more AT and less GC content so it is more
The sequence in the entry that was obtained from sequencing a piece of DNA from Vibrio fischeri genomic DNA digested with Sal I is 8654 bp long.
Question 2:
The -10 sequence for the lux operon is tgttata. 2 nucleotides are different when compared to the ideal sequence.
Question 3:
The lux R has its own promoter and is transcribed in the opposite direction from the lux operon, so it cannot be transcribed from the same strand because the RNA Polymerase recognizes promoter sequences only in the 5’ 3’ direction (lux R is transcribed in the opposite direction). Therefore, luxR has to be transcribed on a different strand.
Question 4:
The coding sequence for the luxA and luxB genes lie within this sequence: * luxA cds 4230-5294 * (5294-4230)+1 = 1065 nt long * (1064+1)/3 = 355 AA long protein expected * luxB cds 5333-6313 * (6313-5333)+1 = 981 nt long * (980+1)/3 = 327 AA long protein expected * lux A RBS: luxA rbs 4217-4222 (aaag ga) * Each gene within the lux operon must have its own RBS because the polypeptide synthesis from the lux operon transcript is an independent process where each gene within the lux operon has its own corresponding transcript and RBS. Also, in bacteria, lux genes are transcribed and immediately translated because there is no nucleus. Therefore, ribosomes can bind to different RBS sequences within he lux operon mRNA, synthesizing different proteins.
Question 5:
- Sal 1 cuts the sequence at:
5’ …G I TCGAC…3’
3’…CAGCT I G…5’ * This is relevant because we isolated the chromosomal DNA of Vibrio fischeri, and Sal I was used to cleave the DNA. * There are 6 Hind III sites in the lux sequence. * There are more sites for Hind III than Sal I because HindII has less G-C content in the cutting sites (SalI has 4GC’s while HindIII has 2 GC’s). Based on the total content of 3926 GC and 6690 AT of the lux sequence, there are more AT and less GC content so it is more