in m 5'- 3'- Shown below is a schematic diagram illustrating a very short gene with 3000 bp region of an unknown Escherichia coli genome. (Note: Transcription starts at Transcription Start Site (TSS).) TSS -3' -5' +1 (i) Name the specific regions that can be recognized by sigma factor and indicate the locations in the diagram above. (ii) How does Sigma factor trigger the initiation of transcription?

1. (a) By binding one L-tryptophan molecule/monomer, the trp repressor binds to DNA to sup- press synthesis of L-tryptophan in E. coli. Below is the amino acid sequence of the helix - reverse turn - helix region of the trp repressor that binds to DNA compared to the sequence of the corresponding DNA binding motif of the Prl protein. A diagram of the trp repressor dimer is also shown. Trp Prl Trp Prl 80 -Gly-Glu-Met-Ser-Gln-Arg-Glu-Leu-Lys-Asn-Glu-Leu-Gly-Ala-Gly-Ile- -Ser-Glu-Glu-Ala-Lys-Glu-Glu-Leu-Ala-Lys-Lys-Cys-Gly-Ile-Thr-Val- trp helix 5 70 trp helix 4 Prl helix 80 Prl helix Ala-Thr-Ile-Thr-Arg-Gly-Ser-Asn-Ser-Leu-Lys-Ala-Ala- Ser-Gln-Val-Ser-Asn-Trp-Phe-Gly-Asn-Lys-Arg-Ile-Arg- reverse turn 90 Comparing the two protein sequences above, identify all amino acid pairs that differ in electrostatic charge due to proton dissociable groups (assume pH 7). Indicate the charge of both residues for each such pair. (b) Circle the pair of residues for which the electrostatic charge due to…

RNA polymerases generally require a primer to begin transcription. (T) (F)   The Death Cap Mushroom Amanita phalloides is toxic because of its ability to produce alpha-amanitin, which is an inhibitor of RNA Polymerases I and III. (T) (F)   In bacteria, transcription and translation can occur simultaneously. (T) (F)   In eukaryotes, transcription and translation can occur simultaneously (T) (F)   RNA polymerase II has no form of proofreading activity. (T) (F)   Sigma factors specify binding of bacterial RNA Polymerases to specific promoters (T) (F)   An E. coli strain with mutations in genes encoding both the dam methylase and the RecA protein would likely be inviable (dead) (T) (F)   An E. coli culture grown in a pure (100%) N2 atmosphere would likely have a lower rate of mutations than a culture grown under normal conditions (~30% O2 and 70% N2) (T) (F)   Non-homologous end joining repairs double strand DNA breaks with no loss of information, restoring the original…

Consider the mechanism of the enzyme RNase:   What would happen to the Km (i.e., would it increase, decrease, or stay the same) if the his12 was mutated to a lysine? Explain.   What would happen to the Kcat (i.e., would it increase, decrease, or stay the same) if the his12 was mutated to a valine? Explain.

Searching the yeast Saccharomyces cerevisiae genome, researchers found approximately 4,000 DNA sites with a sequence which could potentially bind the yeast transcription factor GAL4.  GAL4 activates the transcription of galactose genes.  Yet there are only 10 GAL4-binding sites which control the genes necessary for galactose metabolism.  The GAL4 binding sequence is CGGAT#AGAAGC*GCCG, where # is T, C or G, and * is C or T. In one chromatin immunoprecipitation experiment (ChIP), yeast growing on galactose were lysed, and subjected to cross-linking reagents which cross-linked transcription factors and activators to DNA. Next the DNA was sheared into small fragments, and antibodies to GAL4 were added.  These antibodies coprecipitated the GAL4 and the DNA it was cross-linked to.  The cross-linking was then chemically reversed, and the DNA was isolated, cloned into a library of plasmids and sequenced.  Results showed that only 10 different DNA sequences had GAL4 bound.  Since the…

A novel, RNA-containing splicing factor is purified on the basis of its physical interaction with a previously known subunit of the spliceosome. You sequence the RNA within the new factor, and identify a region that is perfectly complementary to a stretch of the RNA from the previously known subunit with which it interacts. You suspect that the interaction between the new factor and the subunit is mediated by hybridization between these complementary RNA sequences, but would like to be sure. How could you test this experimentally?

DNA: 5’-CTCTACTATAAACTCAATAGGTCC-3’ Draw a box around the sequence where RNA polymerase will bind to the DNA. What is this sequence called? Will transcription start at this sequence, to the left of this sequence (“upstream”) or, to the right of this sequence (“downstream”)? Draw a small arrow above the DNA strand where transcription will begin. Which DNA strand will RNA polymerase transcribe? Highlight this strand with your highlighter. (Hint: RNA pol is similar to DNA pol because it can only make new RNA in the 5’ to 3’ direction. Draw in an arrow to show the direction that RNA polymerase will move along the DNA strand.

Given the following diagram of how protein AWESOME1 binds to it's target DNA, describe the potential effects of each of the 5 mutations shown below. The wild-type sequence of a helix #1 is also shown in the blue box, and all the mutations are in helix #1 (see numbers for identifying particular residues). a helix #1 R(1)-V-I-L-Y-F-W-I-M-Y-F-S-H-Y-W-R(16) #1 Predict the consequence of the following mutations: 1) Arg(1) to Glu 2) Arg(1) to Ala 3) Phe(6) to lle 4) Trp(7) to Phe 5) Met(9) to Pro in