Short Exam 3, February 18 Name True False. (3 points for each correct answer) 5.

Question

Short Exam 3, February 18 Name True False. (3 points for each correct answer) 5. The RNA Pol II general transcription factors: a. were originally isolated as a single complex. b. are not required for transcription of all human genes. c. are not required during activated gene transcription. d. join the core promoter in a specific order lac repressor CAP 6. Chromatin: -a. olves proteins called "nuclcosomes" b. does not regulate gene expression in bacteria. c. plays a role in gene regulation in eukaryotes. d. does not change during the cell cycle. ac CBS laco WT E. coli 7a. Low glucose => CAP-CBS binding Fili in the biamk. I poind eachAnswers can be purchased for I point 7c*. You transform WT E. coli with a vector containing the following reporter cassette. You grow the cells on lactose-> high Lac repressor-Lac ) binding +glucose+lactose/+X-Gal media. What color will the colonies be? Explain why, including a discussion of the molecular interactions involved. (Note: GFP = green fluorescent protein) (8 points) promoler 7d*. Next, you transform the same vector into a E. coli strain with a mutation in the CBS (CAP binding sequence) that is no longer recognized by CAP. You grow the cells on -glucose +lactose/+X-Gal media What color will the colonies be? Explain why, including a discussion of the molecular interactions involved t8 points) 8. You want the use PCR to amplify the shaded region in the sequence below and include a fluorescent tag at the 3' end of the fragment. Assuming 15 bp is sufficient, what primers will you order? Note: Fluorescein (aka FAM) is a dye that can be chemically linked to either end of an oligonucleotide. 6 points) 5'-caggtggagctgggeggeccggcgacctgcaggtggggcactgtccttctgccatggccctgtggatgcgcctcctgecc ctgctggcggggggctggtacacacccaagacccgccgggaggcagagatcagaagaggccatcaagcacat-3

Explanation / Answer

7a. high
7b. Low

7C. In glucose medium lac operon is inhibited. So, the repressor will binds to LacO in the vector containing the reporter gene. The colorless strain will form.

In lactose medium lac operon is ON. So repressor binds with allolactose and becomes inhibited. So, the repressor will not binds to LacO in the vector containing the reporter gene. The green flourescence is activated and strain will become green.

In X-gal media, beta galactosidase hydrolyzed to form 5-bromo-4chloro-indoxyl which form blue color.

7.d - If no gulcose means CAP binds to CBS then cAMP binds and activates the lac operon. Mutation in CBS no longer to binding of CAP so no lac operon will activates.

In gulcose medium repressor binds to LacO in the vector containing the reporter gene. The colorless strain will form.

In lactose medium no binding of CAP-CBS so no activation of LAC operon. But in vector carrying reporter will become activates due to their own strong promoter and repressor will leave from LacO. So it looks green in color.

In X-gal media no activity of Lac operon so no beta galactosidase. The colonies looks like white.

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