MCB1RecitationWorkshop6

BIOL-UA 21 Molecular and Cell Biology I Fall 2023 Recitation Workshop 6: Transcriptional regulation 1. Consensus DNA-binding sequences On the next two pages, you will find a list of DNA sites from different genes that are bound by the E. coli CAP transcriptional activator. The WebLogo web site takes these binding sites, aligns the sequences to one another and finds the nucleotides in common between the sequences. The WebLogo web site makes a visual representation of the most likely nucleotides at each position, with the height of each nucleotide proportional to the importance of that nucleotide at that site. This is known as a position weight matrix (PWM). Go to http://weblogo.berkeley.edu/logo.cgi and paste the text into the Multiple Sequence Alignment box at the top. Leave all other settings as they are, and then click on the Create Logo button. It may be helpful to paste the PWM here: (a) From the motif, what can you predict about how many molecules of CAP bind a single binding-site? Explain your answer. (b) What do you predict would happen to DNA-binding of CAP if you changed the nucleotides in positions 9 to 14 of the PWM to ATATT? What do you predict would happen to DNA-binding of CAP if you changed the nucleotides in positions 4 to 8 of the PWM to ATATT? Describe one experiment to test these predictions? (c) What do you predict would happen to expression of the lac operon if the CAP binding site upstream of the promoter could not bind cAMP-bound CAP? Reduced binding of RNA polymerase to lac operon promoter, decreased transcription, decreased expression of lac operon. The nucleotides in position 9 to14 are not as conserved as nucleotides in 4-8. Change in nucleotides would have more negative effect for 4-8 because these positions are crucial for CAP protein recognition. You perform gel shift assay. You can radioactively label DNA fragments with modified sequences and incubate them with CAP protein. Then, you run the gel. If protein binds to DNA, there would be a sift in migration of DNA. By comparing degree of shift of original sequence and modified sequence, you can determine effect of nucleotide changes on CAP binding. 2. There is region of symmetry. TGTGA and TCACA are palindromic. 2 CAP complexes close to each other bind to single binding site, forming dimer.

BIOL-UA 21 Molecular and Cell Biology I Fall 2023 >aldB -18->4 attcgtgatagctgtcgtaaag >ansB 103->125 ttttgttacctgcctctaactt >araB1 109->131 aagtgtgacgccgtgcaaataa >araB2 147->169 tgccgtgattatagacactttt >cdd 1 107->129 atttgcgatgcgtcgcgcattt >cdd 2 57->79 taatgagattcagatcacatat >crp 1 115->137 taatgtgacgtcctttgcatac >crp 2 gaaggcgacctgggtcatgctg >cya 151->173 aggtgttaaattgatcacgttt >cytR 1 125->147 cgatgcgaggcggatcgaaaaa >cytR 2 106->128 aaattcaatattcatcacactt >dadAX 1 95->117 agatgtgagccagctcaccata >dadAX 2 32->54 agatgtgattagattattattc >deoP2 1 75->97 aattgtgatgtgtatcgaagtg >deoP2 2 128->150 ttatttgaaccagatcgcatta >fur 136->158 aaatgtaagctgtgccacgttt >gal 56->78 aagtgtgacatggaataaatta >glpACB (glpTQ) 1 54->76 ttgtttgatttcgcgcatattc >glpACB (glpTQ) 2 94->116 aaacgtgatttcatgcgtcatt >glpACB (glpTQ) 144->166 atgtgtgcggcaattcacattt >glpD (glpE) 95->117 taatgttatacatatcactcta >glpFK 1 120->142 ttttatgacgaggcacacacat >glpFK 2 95->117 aagttcgatatttctcgttttt >gut (srlA) 72->94 ttttgcgatcaaaataacactt >ilvB 87->109 aaacgtgatcaacccctcaatt >lac 1 (lacZ) 88->110

BIOL-UA 21 Molecular and Cell Biology I Fall 2023 2. Analyzing regulatory DNA to identify the relevant control regions You are studying a simple organism that has 3 different types of taste receptor neurons – one type for sweet taste, a second type for salty taste, and a third type for bitter taste. Each type of taste receptor neuron expresses one receptor gene. You are particularly interested in the regulation of Sweet1 , which encodes the sweet taste receptor gene. Sweet 1 is normally only expressed in neurons that respond to sweet-tasting food, and not in neurons that respond to salty or bitter tastes. You build transcriptional reporter genes with different lengths of the Sweet 1 regulatory region upstream of GFP as shown below. (a) Name two reasons why GFP is a good reporter gene. Reporter 1 is only expressed in neurons that respond to sweet food. Reporter 2 is expressed in neurons that respond to sweet food and neurons that respond to salty food. Reporter 3 is expressed in all 3 types of taste receptor neurons. Reporter 4 is not expressed at all. (b) Using these data, draw a model of the 1kb region upstream of Sweet1 that explains how Sweet1 expression is normally restricted to neurons that respond to sweet-tasting food. [ Hint: one of your essential skills is to understand the normal function of something - such as a DNA regulatory element - by finding out what happens when it is missing.] Reporter 1 -750 GFP -1000 GFP GFP GFP Reporter 2 Reporter 3 Reporter 4 -450 -150 Sweet1 regulatory region 器 Tibi ii a.int i an3 Repressor Repressor Activator iii 1000 750 450 150 easily visible, not normally in cell, not affecting normal functioning sweet sweet, salty sweet, salty, bitter none

BIOL-UA 21 Molecular and Cell Biology I Fall 2023 3. Using gel shifts to determine binding specificity. You are studying the transcriptional repressor Rep1, which is 300 amino acids long. You identify a binding site X1 in the regulatory region of gene X and want to test if Rep1 can bind to this site X1 and to similar sequences in vitro. (a) You decide to perform a gel shift assay. What kind of gel do you use and why? You synthesize 5 short 10 base pair double-stranded DNA sequences and label them each with radioactivity to use in a gel shift with Rep1. The sequences are shown on the left, with the changes from Sequence A in bold and underlined. Sequences B and C differ from Sequence A at positions 5 & 6, while Sequences D and E differ from Sequence A at positions 4 & 7. You run a gel shift with equal amounts of radioactively-labeled DNA in each of the 6 lanes. You use Sequence A in lanes 1 and 2, Sequence B in lane 3, Sequence C in lane 4, Sequence D in lane 5 and Sequence E in lane 6. Equal amounts of Rep1 were added in each of lanes 2-6. The shifted bands in lanes 2, 5 and 6 are much darker than lanes 3 and 4. (b) What can you conclude about the sequences that Rep1 interacts with to bind DNA? Be as specific as possible, referring back to the sequences above. Position number: 1 2 3 4 5 6 7 8 9 Sequence A: A A C A C G T G T T T T G T G C A C A A Sequence B: A A C A G C T G T T T T G T C G A C A A Sequence C: A A C A T A T G T T T T G T A T A C A A Sequence D: A A C C C G G G T T T T G G G C C C A A Sequence E: A A C G C G C G T T T T G C G C G C A A 10 Rep1 5’ - 5’ - 5’ - 5’ - 5’ - - 3’ - 3’ - 3’ - 3’ - 3’ Polyacrylamide gel, because it can separate small DNA-protein complexes from free DNA, and shows shifted protein bound bands and unshifted DNA bands. Rep1 has strong binding affinity for sequence A, D, E. Positions 5, 6 are crucial for Rep1 binding, since sequence B, C show lighter shifted bands and have reduced binding affinity. Change at positions 4&7 in sequence D&E does not significantly affect Rep1 binding affinity, because sequence D(lane 5) and sequence E(lane 6) have dark shifted bands similar to sequence A. Changes at positions 4&7 are not as critical for Rep1 binding as positions 5&6. Changes in positions 5&6 reduce binding affinity of Rep1.

BIOL-UA 21 Molecular and Cell Biology I Fall 2023 You have 2 different antibodies that recognize Rep1: Antibody 1 recognizes the N-terminal 100 amino acids, and Antibody 2 recognizes the C-terminal 100 amino acids of Rep1. You add them to the gel shift reaction (Antibody 1 in Lane 3 and Antibody 2 in Lane 4) and see the following results. (d) Why do the antibodies have such different effects on the gel shift? What can you conclude about the N-terminal and C-terminal regions of Rep1? Rep1 added - + + + Antibody added - - 1 2 Lane 1 2 3 4 Lane 3 suggests that antibody1 has bound to Rep1-DNA complex, and increased molecular size, which causes it to migrate slowly. Binding of antibody1 does not inhibit DNA binding ability of Rep1. It means N-terminal region is likely not directly involved in DNA binding. Lane 4 suggests binding of antibody2 to Rep1 protein inhibits Rep1 binding to DNA. It means C-terminal region of Rep1 is crucial for DNA binding.

BIOL-UA 21 Molecular and Cell Biology I Fall 2023 4. Deletion analysis of a protein You discover a new human protein and call it NHP1 (New Human Protein 1). NHP1 has domains that look similar to those of a well-studied transcriptional activator from Drosophila . You decide to test if NHP1 can bind DNA using a band shift (gel shift) assay. You make a set of NHP1 deletion constructs, where the missing domain is shown in the schematic below and the rest of the protein is fused together. You purify each protein and test each one’s ability to bind to a radioactively labeled DNA probe. The proteins are named by  followed by a number showing the domain deleted. The results are shown below. (a) Which domain of NHP1 is necessary to bind to DNA? (b) To test if the domain you wrote in (a) above is sufficient to bind DNA, you purify this domain on its own and run it alongside the other deletion proteins. If it binds DNA, where would you expect it to run on the gel? Show your answer with an arrow next to the gel above. (c) The sequence of Domain 3 makes you suspect that it is a transcriptional activation domain. Describe an experiment to test if Domain 3 is capable of activating transcription without any other parts of NHP1. What control experiments will be important to include? NHP1 NHP1 D 1 NHP1 D 2 NHP1 D 3 Domain 2 Domain 1 Domain 3 Domain 2 Domain 3 Domain 1 Domain 3 Domain 2 Domain 1 á domain 1 Experiment: 1.Clone domain3 fused to DNA binding domain(DBD) upstream of a reporter gene, whose promoter contains binding sites for DBD 2. Transfect cells 3. Measure reporter gene expression (increased reporter gene expression if domain 3 is a transcriptional activation domain) Control experiment: Transfect cells with DBD alone(without domain3). Transfect cells with full NHP1 protein fused to DBD. Transfect cells without any construct.