GCD GROUP PROBLEMS TEST QUESTIONS WITH CORRECT DETAILED ANSWERS
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GCD GROUP PROBLEMS
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GCD GROUP PROBLEMS
GCD GROUP PROBLEMS TEST QUESTIONS WITH CORRECT DETAILED ANSWERS
What is the consensus sequence of the following DNA sequences?
TGACTGCCAT TGTCAGCCAT AGTCACGCAT TGGGAGCAAA TGTCAGCCTA
TGTCAGGCAA - Answer-Consensus sequence: TGTCAGCCA A/T
A mutation causes sigma factor to bind more tightly to...
A mutation causes sigma factor to bind more tightly to the core polymerase enzyme
compared to a normal sigma factor. How would this mutation affect the rate of
transcription? Explain. - Answer-Release of sigma factor marks the transition to the
elongation stage of transcription. During transcription only small parts of the DNA
molecule (single gene or just a few) are transcribed into RNA. Sigma binding is required
only for promoter binding and initiation. After it's done with its job, the sigma factor is
supposed to detach from the core enzyme and allow it to slide down the DNA strand,
synthesizing RNA along the way - this step is known as the elongation phase. In a
normal routine, a terminator is reached that causes the polymerase and RNA transcript
to dissociate from DNA (termination phase). If a mutation causes sigma to bind more
tightly to the core polymerase, the mutation may affect the ability of the sigma factor
protein to detach from the core enzyme, which may slow or disrupt the rate of
transcription.
A gene encodes a pre-mRNA with 9 exons and 8 introns. A mutation eliminates the 3'
splice site from intron 5. What exons would be found in the mRNA after splicing was
completed? - Answer-If the 3' splice site in intron 5 is deleted, a spliceosome will
connect to the next available 3' splice cite, which would be at the end of intron 6.
Assuming exon 6 is in between intron 5 and intron 6, exon 6 will not be found in the
mRNA after splicing is completed. Exons 1-5 and 7-9 will be found in the mRNA.
What is a difference between endonuclease and exonuclease? Also, describe how
these activities are involved in the processing of tRNA molecules. - Answer-An
endonuclease is an enzyme that can cut in the middle of an RNA strand. This enzyme
will break down the RNA strand into two or more shorter chains by cleaving the internal
covalent bonds linking the nucleotides of a RNA strand.
An exonuclease is an enzyme that will digest an RNA or DNA strand from the end. This
involves a DNA polymerase that will digest nucleotides at the end of an RNA or DNA
strand.
Endonuclease will remove the intron of pre-tRNA and catalyze the last stage in the
maturation of tRNA molecules.
, During the initiation of translation, why does a bacterial mRNA bind specifically to the
small ribosomal subunit? - Answer-Bacterial mRNA will bind specifically to the small
ribosomal subunit because of a complementarity between a short region of the mRNA
and the 30S ribosomal subunit. A sequence within bacterial mRNA known as ribsomal-
binding site or Shine-Dalgarno sequence is the short region that will bind to the 30S
ribosomal subunit. The ribosomal-binding site is complementary to a short sequence
within the 16S rRNA, which will promote hydrogen bonding of the bacterial mRNA to the
30S ribosomal subunit.
A mutation occurs within a bacterial tRNA gene that encodes a tRNA that carries
tyrosine (tRNATyr). The normal anticodon for this tRNA is 3'-AUA-5'. The mutant gene
encodes a tRNATyr in which the anticodon is 3'-AUU-5'. In spite of this change, the
tRNATyr is still recognized by the correct aminoacyl-tRNA synthetase and is charged
with tyrosine. Explain how this mutation would affect cellular translation. Explain
whether you think it would be detrimental to the growth of the bacterial cells. - Answer-I
don't think that this would be detrimental to the growth of the bacterial cells because the
mutation did not effect the amino acid produced, tyrosine. However, when being
translated, it could possibly produce a different amino acid. The anticodon 3' - AUU - 5'
corresponds to the STOP codon (5' - UAA - 3'). So, this mutation would place tyrosine
at the end of a polypeptide where it shouldn't be. This would also allow the polypeptide
to grow until the mRNA codes for a different stop codon (such as 5' - UAG - 3'). This
could alter the affect of polypeptides and may be very detrimental to the growth of
bacterial cells.
In which of the ribosomal sites, the A site, P site, and/or E site, could the following be
found? - Answer-A. A tRNA without an amino acid attached: E site
B. A tRNA with a polypeptide attached: P site or A site
C. A tRNA with a single amino acid attached: A site or P site (if it is the initiator tRNA)
A mutation in the lacI gene alters the lac repressor in a way that causes it to have a two-
fold lower affinity for allolactose. How would this affect the expression of the lac operon?
- Answer-If the lac repressor has a lower affinity for allolactose than normal, then it will
require lower cytoplasmic amounts of allolactose for a conformational change to occur
in the lac repressor. In other words, the lac repressor will stop inhibiting RNA
polymerase earlier than normal, resulting in more, unneeded proteins for lactose uptake
and metabolism.
A gene is under positive control. Which of the following could explain the action of a
small effector molecule? - Answer-The answer can be A or B. Positive control refers to
transcriptional regulation by an activator protein. Activator proteins bind to the DNA and
promote gene expression. Effector molecules that work on activator proteins include
inducers and inhibitors.
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