Download LS1a ICE 4

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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).
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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.
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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.
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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.
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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.
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The Genetic Code
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