For Reference
KimNgoc Huynh
Bimm101
Section: CO2
TA: Pagkapol Yhew Pongsawakul Bioinformatics Homework#1
Question#1:
The sequence is 8654 nucleotides long.
Question#2:
-10 sequence for the lux operon: 5’-TGTTATA-3’ There are 2 nucleotides different that are G and A.
Question#3:
The lux R, which has its own promoter and is transcribed in the opposite direction from the lux operon, could not be transcribed from the same strand because the RNA polymerase recognizes a promoter sequence only in the direction of 5’ to 3’, and the lux R gets transcribed in the opposite direction from the lux operon. Thus, the transcriptions of luxR and lux operon have to occur on two different strands including coding strand and template strand.
Question#4: -For lux A: (5294-4230) +1= 1065 nt (1065/3)-1= 355 amino acids -For lux B: (6313-5333)+1= 981nt (981/3)-1= 326 amino acids
-The RBS sequence in luxA is AAAGGA. The sequence is 6 nucleotides long (4217-4222).
-Each gene within the lux operon must have its own RBS because the polypeptide synthesis from the lux operon transcript is an independent process, in which each gene within the lux operon such as lux I, C, D, A, B, E, and G genes has a corresponding transcript that has its own ribosome binding site (RBS). Also, we should keep in mind that there is no nucleus to separate the processes of transcription and translation in bacteria. When the lux genes are transcribed, theirs transcripts can immediately be translated. Thus, ribosomes can recognize and bind to the RBS sequences of different genes within lux operon mRNA as a result of different protein products synthesized.
Question#5:
Sal I cuts the sequence at 1st and at 8654th nucleotide of the lux operon’s.
This is relevant to what we are doing in lab because part of the goal of the series of experiment we had done was to isolate the chromosomal DNA of Vibrio Fischeri, and Sal I was used to cleave the DNA because it cuts in way that we could
Bimm101
Section: CO2
TA: Pagkapol Yhew Pongsawakul Bioinformatics Homework#1
Question#1:
The sequence is 8654 nucleotides long.
Question#2:
-10 sequence for the lux operon: 5’-TGTTATA-3’ There are 2 nucleotides different that are G and A.
Question#3:
The lux R, which has its own promoter and is transcribed in the opposite direction from the lux operon, could not be transcribed from the same strand because the RNA polymerase recognizes a promoter sequence only in the direction of 5’ to 3’, and the lux R gets transcribed in the opposite direction from the lux operon. Thus, the transcriptions of luxR and lux operon have to occur on two different strands including coding strand and template strand.
Question#4: -For lux A: (5294-4230) +1= 1065 nt (1065/3)-1= 355 amino acids -For lux B: (6313-5333)+1= 981nt (981/3)-1= 326 amino acids
-The RBS sequence in luxA is AAAGGA. The sequence is 6 nucleotides long (4217-4222).
-Each gene within the lux operon must have its own RBS because the polypeptide synthesis from the lux operon transcript is an independent process, in which each gene within the lux operon such as lux I, C, D, A, B, E, and G genes has a corresponding transcript that has its own ribosome binding site (RBS). Also, we should keep in mind that there is no nucleus to separate the processes of transcription and translation in bacteria. When the lux genes are transcribed, theirs transcripts can immediately be translated. Thus, ribosomes can recognize and bind to the RBS sequences of different genes within lux operon mRNA as a result of different protein products synthesized.
Question#5:
Sal I cuts the sequence at 1st and at 8654th nucleotide of the lux operon’s.
This is relevant to what we are doing in lab because part of the goal of the series of experiment we had done was to isolate the chromosomal DNA of Vibrio Fischeri, and Sal I was used to cleave the DNA because it cuts in way that we could