From DNA to Protein
Multiple
Choice
1. During the process of translation:
A. the peptide is
‘passed’ from the tRNA in the P-site to the tRNA in the A-site.
B. incoming tRNAs must
first bind to the E-site.
C. initiation begins with
the binding of the ribosomal SSU to the poly-A tail of the mRNA.
D. the mRNA is translated
by one ribosome at a time.
2. The presence of
a poly-A tail on a RNA molecule indicates that:
A. there are exons
present that must be removed.
B. this RNA molecule does
not contain introns.
C. the transcript should
be immediately degraded.
D. this is a rRNA
molecule.
E. None of the above
answers is correct
3. A ‘proteasome’ is a large structure in the
cytoplasm that:
A. translates mRNA into protein C. supercoils DNA
B. processes RNA D. enzymatically degrades proteins
4. During the
processing of introns, a single snRNP complex catalyzes both the cleavage of
the RNA and the joining of the cut ends.
What would be the consequence if these two processes were catalyzed by
separate enzymes not associated in a single complex?
A. the rate of RNA
processing would be much faster.
B. the cell would be
unable to identify the correct cleavage sites.
C. the exons might not be
joined in the correct sequence.
D. exons instead of
introns would be cleaved from the RNA molecule.
5. The nucleolus
of the nucleus is the site where:
A. RNA processing occurs
B. rRNA is transcribed
and ribosomal subunits are assembled
C. tRNA are charged with
amino acids
D. mRNA is translated
into protein
6. During
"RNA processing"
A. all of the exons are
removed and discarded
B. the RNA molecule is
made from a DNA template.
C. introns are cut from
the RNA and the exons are spliced together.
D.
the RNA molecule is translated into a protein molecule.
7. How does the
cell ‘mark’ the positions of introns in an immature RNA?
A. There is a special
snRNP for each type of intron.
B. Codons called ‘cut’
and ‘paste’ (copyrighted by Microsoft) are present within the RNA.
C. It doesn’t need to, since the boundary
between an intron and exon alternates frequently.
D. Special sequences are
located near the splicing sites which are recognized by ribozymes.
8. “Alternative
splicing” refers to:
A. the use of introns as
exons, or vice versa, during RNA processing
B. splicing out of
damaged DNA by DNA repair enzymes.
C. joining of RNA from
two different genes to form a new mRNA.
D. the use of alternative
reading frames when translating an mRNA.
E. a new dance for people
with alternative life styles.
9. During transcription of DNA to RNA:
A. an RNA polymerase
moves along the DNA in the 5’ to the 3’ direction .
B. the 3’ end of the RNA
molecule is produced first.
C. an RNA polymerase must
first bind to a promoter sequence.
D. transcription is
always initiated at a “start codon.”
10. During the ‘elongation’ stage of translation,
after the arrival of each new tRNA:
A. the amino acid is
‘passed’ from the tRNA in the A-site to the tRNA in the P-site.
B. newly arriving tRNAs
must first bind to the E-site.
C. the peptide is
‘passed’ from the tRNA in the P-site to the tRNA in the A-site.
D. the new tRNA must
first bind to the P-site of the ribozome.
11. During transcription of a particular gene, the RNA
polymerase will transcribe:
A. both strands, but only
one of RNA molecule will be used.
B. only one of the DNA strands,
moving in a 3’ to 5’ direction along the template.
C. both strands, but
moving 3’ to 5’ for one and 5’ to 3’ along the other.
D. only the exons of the
gene while skipping over the introns.
12. Since the two
strands of the DNA molecule are complementary, for any given gene:
A. The RNA polymerase can bind to either strand.
B. Only one strand actually carries the genetic code for a
particular gene.
C. Each gene possesses an exact replica that can be used
should a mutation occur.
D. A gene transcribed in the 5’ to 3’ direction on one
strand can be transcribed in the 3’ to
5’ direction on the other strand.
13. In the genetic
code there are:
A. more tRNAs than
codons. C. more nucleotides than codons.
B. more codons than amino
acids. D. the same number of
codons and amino acids
14. Initiation of
translation begins when the:
A. large and small
subunits link together, then bind to the mRNA.
B. ribosomal small
subunit holding an initiator tRNA binds to the 5’ end of the mRNA.
C. ribosome binds to the
start codon and an initiator tRNA enters the ribosome.
D. initiator tRNA binds
to the start codon, followed by binding of the ribosome large subunit.
15. A “TATA box” is
A. the translation termination sequence.
B. an important base sequence in the promoters of
bacteria.
C. the site where the RNA polymerase II binding
complex is assembled.
D. an example of one of the translation stop codons.
16. According to
the RNA-world theory:
A. RNA molecules were the first
organic molecules formed on earth.
B. Life evolved on another planet
called the “RNA World.”
C. All RNA molecules in cells are
“ribozymes.”
D. Primitive RNA molecules evolved
before protein and DNA
17. It seems
probable that DNA contains thymine instead of uracil because:
A. thymine is chemically much more
stable than uracil.
B. when uracil is chemically
deaminated, thymine is produced.
C. thymine was one of the first four
nucleotides in primitive RNA molecules.
D. if cytosine is deaminated, the
altered base can be detected and removed.
True or False
1. Each of the 3
potential reading frames of an mRNA can produce a functional, but different,
protein.
2. Transcription is
terminated when the RNA polymerase encounters a poly-U sequence.
3. Translation
ends when a ‘release factor’ protein binds to a stop codon.
4. Initiation of
translation in prokaryotes involves binding of the sigma factor to a promoter.
5. Only rRNAs are
polyadenylated.
6. Because genes
can be coded on either strand of the DNA double helix, the coding regions of
different genes can overlap.
7. Some
antibiotics have selective effects upon prokaryotic ribosomes.
8. The promoter is
located downstream from the coding region of a gene.
9. RNA processing,
common in eukaryotes, does not occur in prokaryotes.
10. General transcription factors are regulatory proteins that bind to
eukaryotic RNA polymerases.
11. Ribozymes are
primitive forms of RNA molecules that no longer exist in cells.
Fill in, etc
1. In a ribosome, the formation of the peptide bonds
of the new peptide chain occurs in the __________________ subunit, whereas
matching the codons of the mRNA are exposed on the surface of the
_______________ subunit. During the
peptide elongation stage of translation, each in-coming aminoacyl-tRNA binds to
the ___-site of the ribosome, where as the growing peptide chain is held on the
tRNA in the ___-site.
2. The end of translation is signaled by a ________
codon, which binds a protein called the _____________________.
3. In bacteria, the protein called the
__________________ associated with the RNA polymerase is principally
responsible for binding to the promoter.
4. Place the following events in their correct
sequence:
___
Translation
___
Transcription
___
Polyadenylation
___
Capping
___
RNA processing
___
Nuclear export
6. RNA molecules possess both genotype and phenotype:
A. The
genotype of an RNA molecule is held in its ________________________.
B. What are
two examples of phenotype properties of an RNA molecule?
7. Show and
briefly explain how complementary base pairing can lead to replication of a
strand of RNA with the following sequence:
AUCGCGUUAACCGUA
8. ‘Wobble base
pairing’ will occur for which one of the following pairs codons?
A. AUG and UGG C. GGA and GGC
B. AAA and UUU D. UAG and UGA
9. The codon for
methionine is _____, the anticodon is _____, and the DNA code is ______.
10. Based upon the
results of DNA sequencing for the Human Genome Project, the number of promoters
suggests that there are around 25,000 genes in the human genome. However, the number of different types of
proteins may actually be much higher than this.
Why?
11. Introns are
‘junk’ DNA that create a burden on the species.
Give at least two reasons why this statement is incorrect?
12. A research lab purified the messenger RNA for a mouse
protein called ‘GFI’ and reverse transcribed it into cDNA. They then used the cDNA as a probe to locate
the gene on the mouse chromosome, which they isolated and cloned. To their surprise the coding region of the GFI
gene was 5500 bases longer than that of the cDNA. In an attempt to mass-produce the GFI
protein, they transfected E. coli with
double-stranded DNA derived from the cDNA, hoping that it could be produced in
the bacteria, but the gene was not transcribed.
A. How would you explain the difference
in size between the cDNA and genomic forms of the GFI gene?
B. What would you do differently to
achieve transcription of the GFI gene in bacteria?
13. The following
is a segment of DNA containing the beginning of a gene
3¢- GGCATACTTCAGTCAAGAGACATAG
-5¢
5¢- CCGTATGAAGTCAGTTCTCTGTATC -3¢
A. If an RNA polymerase were to transcribe the
gene from left to right, is the top or the bottom strand serving as the template?
B. What will be the sequence of the mRNA produced
(be sure to label the 5¢ and 3¢ ends of your RNA molecule)?
C. What is the amino acid sequence of the
peptide that would be translated from the mRNA.
Label the N- and the C-terminus amino acids.
D. Which one of the following mutations
would have the greatest effect on the structure of the protein? Explain.
a.
deletion of the underlined ‘GC’ pair
b.
substitution of TA pair for the underlined GC pair
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