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MCDB 3500

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F third position is wobble. 5. An amber codon is recognized by a stop tRNA. F release factor ... Wobble position (or hypothesis) Pab1p binds poly(A) tail. eIF4E ... – PowerPoint PPT presentation

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Title: MCDB 3500


1
MCDB 3500
Exam 4
Spring, 2004
50 minutes close everything be to the point!
Name
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ID
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Q1 (20 points)
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Q2 (20 points)
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Q3 (20 points)
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Q4 (20 points)
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Q5 (20 points)
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Total (100)
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Grade
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2
  • Question 1. Please indicate whether the following
    statements are true or false. If false, correct
    it. (20 points)
  • 1. Protein synthesis occurs in a 5 to 3
    direction.
  • F N to C terminus
  • 2. EF-Tu and EF-G hydrolyze ATP for their
    functions in translation.
  • F GTP
  • 3. A ribosomal protein in the large subunit most
    likely plays a direct role to catalyze peptide
    bond formation.
  • F RNA
  • 4. All three positions of a codon are equally
    important for translation.
  • F third position is wobble
  • 5. An amber codon is recognized by a stop tRNA.
  • F release factor
  • 6. U1 snRNA transiently appears in the cytoplasm
    during its life cycle.

Question 2. A. Describe an experiment that would
show that ribosome subunits dissociate and
reassociate with each round of translation (use
schematic). B. What experimental outcome would
make you conclude that they do not dissociate and
reassociate. (20 points)
HH and LL combination means no dissociation
New Heavy-Light size suggest Exchange, when the
ribosomes Were grown in Heavy (C13 and N15) ,
shifted to Light (C12 and N14) medium, and
separated by sedimentation
3
Question 3A. The following constructs (1 to 4)
were incubated in a cell-free translation system
in the presence of 35 S-methionine. The
resulting products were separated on a
polyacrylimide gel (lanes 1 to 6). 1) What
three important conclusions can you draw from
this experiment? (7 points) 2) What conclusion(s)
cannot be drawn if lanes 4 and 6 are missing,
and why? (3 points)
1)
1. Stem-loop too close to cap inhibits
translation. 2. Stable Stem-loop
inhibits translation. 3. Constructs 1 and 4 dont
poison the translation system. Conclusion 3
cannot be drawn if lanes 4 and 6 missing because
construct 3 is translated, when mixed with
either 1 or 4..
2)
4
Question 3B. Label proteins 1-5, and explain why
you chose those labels. (10 points)
Pab1p binds poly(A) tail eIF4E binds cap eIF3
binds 40S eIF4A helicase to unwind RNA ahead of
40S eIF4G connects 4E and pab1p
Question 4. Explain the following. (20 points
766)
tRNA suppression or suppressor tRNA
A mutation in tRNA antiocodon can suppress an
amber mutation or trnaslation stop codon because
this tRNA can base-pair with the amber codon.
RNA editing
Many RNAs get modified posttranscriptionally, by
deamination, insertion/deletion of uridines.
Wobble position (or hypothesis)
The third position in codon can be recognized
through non-watson-crick base-pairing.,like G-U,
A-I, and I-U.
5
Question 5A. The Shine-Dalgarno sequence is
important for translation initiation in
bacteria. It hybridizes to a sequence 5 ACCUCCU
3 in the 16S rRNA. Using the given genetic
codon table, indicate the amino acid sequence of
the polypeptide that will be translated from the
following mRNA sequence, and explain why. (10
points) 5 pppACCGCCAUGGCCUAGGAGGUUUUGACUAUGCGAGC
UUAGUGGACC 3 (10 points) SD Because it uses
the first AUG downstream of SD Question 5B. If
the following mRNA was translated in a eukaryotic
cell, what will be the most likely polypeptide
sequence, and explain why? (10 points) m7GpppACCG
CCAUGGCCUAGGAGGUUUUGACUAUGCGAGCUUAGUGGACC
MetArg
MetAla
Because it uses the first AUG from cap, which is
also a good kozak consensus
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