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AGGGGGATTCACCGTGCGAAATTTTTTTAACAACTGCTCAGTCTGACAGGCAACTGTCAA
CTGACTGAATTGTGACACAGATTACACTTGTTACCCACGTACCACGAATCAGGTTATGCC
TCAGTCAATATTAAACTGCACTTCAGCAAATCCGGAGCCTGATTCACAGGTACTGGATTT
GATTGTGACAGTCATTCCTGTCAACTGAGCACTTTGCAGTAGCGGTTGCAGATTCCAGCG
ACTGGTCCAGTATTCTTTCCCGGCTACTTTTACTGTGAATGTATCGTTCTCATTATACTT
GGAAAACTCAATTTTACCTTTAGCACAATCTGCCGCCATTGCATTAACAGAAACTAATGC
AAATAAAACCGCCATAAACATCTTCTTCATACTTAACTCCTTTATTCACCCGTTGTATAT
AAAGACTGTGACTTTCTGTTCAGAAACGCTGCTGCTGTATTACTTTCCGATAATCTATTG
TTTATTTTTA

1. From 5' to 3', list the DNA sequences of all possible ORFs that are encoded by this piece of double stranded DNA.

2. What protein-coding gene or genes does this piece of double stranded DNA encode?

3. E. coli O26 can utilize glucose through glycolysis. (TRUE/FALSE)

4. In the absence of glucose, E. coli O26 can utilize arabinose, a five-carbon monosaccharide. Which transport mechanism is used to import arabinose by E. coli O26?

5. E. coli O26 can use trimethylamine N-oxide (TMAO) as an electron acceptor for anaerobic respiration. However, E. coli does not make TMAO reductase unless TMAO is present in the environment. How does E. coli O26 know TMAO is present or not?

6. Bacteria can evolve much faster than animals and plants because they grow much faster and have larger population sizes. To put this in a context, do you think it is possible statistically that every single base pair in the E. coligenome has experienced a mutation in a 5 ml E. coli overnight culture? We know that there are ~1010 cells in the 5 ml overnight culture and the mutation rate of E. coli is 10-9 per base pair per DNA replication. Justify you conclusion.

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Patrina Schowalter
Patrina SchowalterLv2
28 Sep 2019

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