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Name: TF: Section Time: LS1a ICE 4 Practice ICE Version A Name: ____________________________ 1. (10 points) The development of a eukaryotic embryo is dependent upon the concentration gradients of transcription factors and the binding affinity of the transcription factors for their DNA binding sites. a. (5 points) Transcription factor Lsck is an activator (i.e., Lsck positively-regulates genes) and it binds two enhancer sites with different affinity. Multiple Lsck-DNA binding reactions were performed with both enhancer binding sites: each reaction included the same concentration of binding-site DNA (as indicated in the flat lines in the diagrams below) but with a variable concentration of Lsck protein (as indicated by the triangle, showing increasing protein concentration from left to right) to determine which site Lsck binds most tightly, with the following results: Does Lsck have a higher Keq of binding for Binding Site 1 or Binding Site 2? Briefly explain your answer. Lsck has a higher Keq of binding for Binding Site 2 because it takes a higher relative concentration of Lsck protein to induce the shift indicative of binding for Binding Site 1 (i.e., it takes a lower relative concentration of Lsck protein to induce the shift indicative of binding for Binding Site 2). 2 Name: ____________________________ b. (5 points) The concentration gradient of transcription factor Lsck is highest at the anterior of the embryo and lowest at the posterior, as diagrammed below: At the posterior, will there be more expression of genes under the control of Binding Site 1 or Binding Site 2? I would expect more genes under the control of Binding Site 2 to be expressed at the posterior than Binding Site 1 because Lsck is an activator and since it binds more tightly to Binding Site 2, expression of Binding Site 2 genes should extend further in a posterior direction. 3 Name: ____________________________ 2. (12 points) Shown below is the DNA sequence of a prokaryotic gene. The start and stop codons are shown in bold. A portion of the sequence, represented as “…”, is not shown. 1 921 981 AATCTTCAGC GGACCGGCGA GTATACCTGA AGAAAGGACG TCAGATGCTT TTAGCTACGG… … …CGCTCCCCGC AACCGGGCCG TGGTGTAGGG GTATTATTAA CTGGCGGATT TATCGTATGG CTGGCGATGT TGTACTGGTT ATCGCCAATA CTCGTTGAAT AACTGGAAAC GCATTACTCT… Several experiments were conducted to test how specific changes to the DNA sequence affect translation of the gene. Shown below is a gel showing the sizes of the proteins produced from this gene. Lane 2 shows the size of the original protein, and Lanes 3-5 show the sizes of the proteins that result from making changes 1-3 to the gene. Match the DNA sequence changes (a) through (c) listed below to changes 1 through 3 using the gel data above. Each of the changes (1-3) corresponds to one, and only one, of the changes to the DNA sequence (a-c). Briefly explain why each change results in the observed gel data. A codon table is provided at the end of the exam. a. (4 points) G979 is changed to A This change corresponds with Change 2. An earlier stop codon is created. As a result, the protein is slightly shorter than usual. The codon TGG (tryptophan) is converted to TAG, inserting a premature stop codon, losing 14 amino acids. Since the resulting protein is only slightly smaller than the original protein, it will be Change 2 rather than Change 1. 4 Name: ____________________________ b. (4 points) T51 is changed to A This change corresponds with Change 3. This causes a codon to change from TTA (leucine) to ATA (isoleucine). This changes the protein, but not in a way that is observable on the protein gel. c. (4 points) A new “C” is inserted between G47 and C48 This change corresponds to Change 1. This insertion causes a reading frame shift such that a totally different codon sequence is translated. The new reading frame that must contain a MUCH earlier stop codon, causing the translated protein to be MUCH shorter than the normal protein. 5 Name: ____________________________ The Genetic Code 6