Unit 2 - M/C Flashcards
Based on Chargaff’s rules, which of the following
is TRUE of double stranded DNA molecules?
A. (C+T) = (A+G)
B. (A+T)/(C+G) = 1
C. (A+G)/(U+C) =1
D. C/T = 1
E. A/U = G/C
B. (A+T)/(C+G) = 1
If a DNA molecule contains 27% cytosine bases (C),
then what percentage of thymine bases will it have?
A. 10%
B. 27%
C. 46%
D. 23%
E. 52%
D. 23%
A DNA molecule of 50 base pairs contains 15 cytosine
bases (C). How many thymine bases will it have?
A. 10
B. 15
C. 30
D. 35
E. 60
D. 35
How did Chargaff’s rules contribute to Watson and
Crick’s elucidation of the structure of DNA?
A. The rules suggested an equal concentration
of sugars and phosphates.
B. The rules suggested the amounts of all four
bases were equal.
C. The rules suggested the base-pairing
combinations of adenine with thymine and
guanine with cytosine.
D. The rules suggested that each base
corresponds to an amino acid.
C. The rules suggested the base-pairing
combinations of adenine with thymine and
guanine with cytosine.
Consider two molecules of DNA. They are the same length, and they are both double-stranded. At 75°C, the first one will completely denature; however, at the same
temperature the second one is still double-stranded. What factors might allow for these different melting
temperatures?
A. The first molecule has a higher C+G content than
the second molecule.
B. The second molecule has a higher C+G content
than the first molecule.
C. The first molecule has a secondary structure of B
form, while the second molecule is an A form.
D. The second molecule is methylated more than the
first molecule.
B. The second molecule has a higher C+G content
than the first molecule.
Which of the following DNA strands would
have the LOWEST melting temperature?
A. 10% AT and 90% GC
B. 30% AT and 70% GC
C. 50% AT and 50% GC
D. 70% AT and 30% GC
E. 90% AT and 10% GC
E. 90% AT and 10% GC
You’ve grown an overnight culture of bacteria in media
that is enriched with radioactive nucleotides, so that
both strands of every DNA molecule found in the
bacteria will be radioactive. You harvest the cells and
plan to grow them in new media with nucleotides that
are NOT radioactive, so any DNA synthesized in this
culture will not be radioactive. After the cells grow for
two rounds of replication in the non-radioactive media,
what proportion of DNA molecules will be radioactive?
A. 0
B. 1/8
C. 1/4
D. 1/3
E. 1/2
B. 1/8
GIVEN THE FOLLOWING STATEMENTS:
1) binding to replication origin and
separating the double stranded
DNA to initiate replication.
2) binding to single stranded DNA to
prevent the duplex from
reforming.
3) unwinding the DNA double helix
at the replication fork.
4) sealing nicks in the sugar–
phosphate backbone of newly
synthesized DNA.
5) creating and resealing double
strand breaks to remove coiling.
6) synthesize primers for initiation of
DNA synthesis.
MATCH WITH THE FOLLOWING EVENTS:
A. DNA gyrase
B. DNA ligase
C. Initiator protein
D. DNA primase
E. Single stranded binding protein
F. DNA helicase
Initiator protein: Binding to replication origin and separating the double-stranded DNA to initiate replication.
Single-stranded binding protein: Binding to single-stranded DNA to prevent the duplex from reforming.
DNA helicase: Unwinding the DNA double helix at the replication fork.
DNA ligase: Sealing nicks in the sugar–phosphate backbone of newly synthesized DNA.
DNA gyrase: Creating and resealing double-strand breaks to remove coiling
DNA primase: Synthesize primers for initiation of DNA synthesis.
One of these is NOT required for DNA polymerases
to initiate DNA replication. Which one is it?
a) DNA template
b) RNA primer
c) 3’ to 5’ polymerase activity
d) Initiator proteins
e) Deoxyribonucleotides
d) Initiator proteins
When replicating a chromosome, the synthesis of
DNA is initiated from:
a) RNA template
b) DNA primer
c) 3’ OH
d) Deoxyribonucleotides
e) 5’ phosphate
b) DNA primer
Consider this statement about the synthesis of a molecule. A polymerase enzyme forms phosphodiester bonds using nucleoside triphosphates. Which of the following are described by this statement?
a) RNA only
b) DNA only
c) both RNA and DNA
d) neither RNA nor DNA
c) both RNA and DNA
Both strands of a DNA molecule are used simultaneously as a template when which of the following molecules is synthesized?
a) RNA only
b) DNA only
c) both RNA and DNA
d) neither RNA nor DNA
b) DNA only
If the sequence of an RNA molecule is
5ʹ–GGCAUCGACG–3ʹ
what is the sequence of the nontemplate strand of DNA?
a) 5ʹ–GGCATCGACG–3ʹ
b) 3ʹ–GGCATCGACG–5ʹ
c) 5ʹ–CCGTAGCTGC–3ʹ
d) 3ʹ–CCGTAGCTGC–5ʹ
e) 3ʹ–CGTCGATGCG–5ʹ
a) 5ʹ–GGCATCGACG–3ʹ
If the sequence of an RNA molecule is
5ʹ–GGCAUCGACG–3ʹ
what is the sequence of the template strand of DNA?
a) 5ʹ–GGCATCGACG–3ʹ
b) 3ʹ–GGCATCGACG–5ʹ
c) 5ʹ–CCGTAGCTGC–3ʹ
d) 3ʹ–CCGTAGCTGC–5ʹ
e) 3ʹ–CGTCGATGCC–5ʹ
d) 3ʹ–CCGTAGCTGC–5ʹ
You are sequencing an RNA molecule. Which
of the following sequences would be the
template strand for this RNA molecule:
5ʹ–UUCAUCGACG–3ʹ
a) 3ʹ–AAGUAGCUGC–5ʹ
b) 5ʹ–TTCATCGACG–3ʹ
c) 5ʹ–AAGTAGCTGC–3ʹ
d) 3ʹ–AAGTAGCTGC–5ʹ
e) 3ʹ–TTCATCGACG–5ʹ
d) 3ʹ–AAGTAGCTGC–5ʹ
Which of the following statements about RNA
polymerase is NOT true?
A. RNA polymerase has many subunits.
B. During transcription of a gene, RNA
polymerase reads only one of two strands of
DNA.
C. RNA polymerase reads a template strand of
DNA 3ʹ to 5ʹ.
D.RNA polymerase adds ribonucleotides to the 3ʹ
end of a growing RNA molecule.
E. RNA polymerase binds to an enhancer
sequence to initiate transcription
C. RNA polymerase reads a template strand of
DNA 3ʹ to 5ʹ.
Modification is necessary for efficient initiation of _____________ , __________ of mRNA from ________ and protects mRNA from _______________, and enhances ______ _______.
Modification is necessary for efficient initiation of translation, export of mRNA from nucleus, and protects mRNA from degradation, and enhances translation efficiency.
The spliceosome is large ___________ complex,
which contains five _______ molecules and
functions in the _________ of the cell.
a) Protein; rRNAs; nucleus.
b) Protein; snRNAs; cytoplasm.
c) Ribonucleoprotein; snRNAs; cytoplasm.
d) Ribonucleoprotein; snRNAs; nucleus.
e) Ribonucleoprotein; rRNAs; nucleus.
d) Ribonucleoprotein; snRNAs; nucleus.
During pre-mRNA splicing, the branch point is
recognized and bound by which spliceosome
subunit?
a) U1 snRNP
b) U2 snRNP
c) U2, U5 and U6 snRNPs
d) U1 and U2 snRNPs
e) U2 and U6 snRNPs
b) U2 snRNP
What is the CORRECT order for the process of pre-
mRNA processing by the spliceosome?
- Attachment of the U1 snRNP to the 5’ splice site
- Transcription of the DNA template into the pre-
mRNA molecule - Release of lariat structure
- Splicing together of exons
- Transesterification reaction at the branch point
adenine
a) 1, 2, 3, 4, 5
b) 4, 1, 3, 5, 2
c) 2, 1, 5, 3, 4
d) 3, 5, 1, 2, 4
e) 5, 3, 4, 1, 2
b) 4, 1, 3, 5, 2
Which end of the polypeptide chain emerges from the
Ribosome first? amino (N) or carboxyl (C)
The amino (N)-terminal end of the polypeptide chain emerges from the ribosome first during translation.
Synonymous codons always match with the same tRNA . T/F
False
Synonymous codons always code for the same amino acid. T/F
True
The anticodon region helps aminoacyl-tRNA synthetase match the tRNA to the correct amino acid. T/F
True
How many different aminoacyl-tRNA synthetases are there?
20 different aminoacyl-tRNA synthetases in total, one for each amino acid. Each aminoacyl-tRNA synthetase is specific to its corresponding amino acid and is responsible for attaching that amino acid to its corresponding tRNA molecule.
What do you call the proteins that help with the Initiation of Translation
initiation factors. These proteins play crucial roles in the initiation phase of protein synthesis, facilitating the assembly of the ribosome and the binding of the initiator tRNA to the start codon on the mRNA molecule.
There are 3 sites in the Ribosome used by tRNAs, list them in order (5’ to 3’ on mRNA)
A site (aminoacyl site): This site is where the incoming aminoacyl-tRNA binds to the ribosome.
P site (peptidyl site): This site is where the peptidyl-tRNA, carrying the growing polypeptide chain, is located.
E site (exit site): This site is where the deacylated tRNA exits the ribosome after releasing its amino acid.
What is the full name of the site that usually holds the polypeptide chain?
The full name of the site that usually holds the polypeptide chain during translation is the “peptidyl site,” often abbreviated as the “P site.” This site is where the peptidyl-tRNA, carrying the growing polypeptide chain, is located within the ribosome during protein synthesis.
According to the central dogma, double-stranded DNA serves as the template for the production of RNA during transcription. Which of the two DNA strands serves as the template for transcription?
a) Either DNA strand may be used as a template by RNA polymerase, but a single DNA strand oriented in the 3-5” direction is used as a template each time transcription occurs.
b) The DNA strand oriented in the 5’-3’ direction is the only template because only this strand contains the promoter sequences necessary to initiate transcription.
c) The DNA strand oriented in the 3’-5’ direction is the only template because RNA polymerase synthesizes only in the 3’-5’ direction.
d) Both DNA strands serve as templates simultaneously for RNA polymerase, with continuous transcription on the leading strand and discontinuous transcription on the lagging strand.
a) Either DNA strand may be used as a template by RNA polymerase, but a single DNA strand oriented in the 3-5” direction is used as a template each time transcription occurs.
The descriptions pertain to either prokaryotic or eukaryotic transcription. Match each description to the appropriate category.
I. Prokaryotic transcription
II. Eukaryotic transcription
Description:
- can be terminated by rho helicase
- includes spliceosome processing
- requires TFIID
- promoter includes - 10 consensus sequence
- promoter includes a TATA box
I. Prokaryotic transcription
- can be terminated by rho helicase
- promoter includes - 10 consensus sequence
II. Eukaryotic transcription
- includes spliceosome processing
- requires TFIID
- promoter includes a TATA box
Match each function to the appropriate type of RNA.
I. Messenger RNA (mRNA)
II. Ribosomal RNA (rRNA)
III. Transfer RNA (tRNA)
Function:
- contains the codons for the polypeptide sequence
- enzymatic amide bond synthesis
- hydrogen bonds with codon
- transports amino acids to the ribosome
I. Messenger RNA (mRNA)
- contains the codons for the polypeptide sequence
II. Ribosomal RNA (rRNA)
- enzymatic amide bond synthesis
III. Transfer RNA (tRNA)
- hydrogen bonds with codon
- transports amino acids to the ribosome
In eukaryotic gene regulation, RNA interference occurs through
a) the accumulation of mRNA that blocks transcription of the target gene by feedback inhibition.
b) the action of microRNAs that block translation of specific mRNA molecules.
c) the action of RNA-protein complexes that inhibit translation by altering the three dimensional configuration of rRNA molecules.
d) the accumulation of mRNA that blocks transcription of nearby genes.
b) the action of microRNAs that block translation of specific mRNA molecules.
As DNA is replicated, both continuous and discontinuous replication occur. Discontinuous replication is the result of which specific feature of DNA?
a) action of helicase at the replication fork
b) antiparallel strands
c) purine and pyrimidine nucleotides
d) major groove
e) bonds between base pairs
b) antiparallel strands
How can microRNAs (miNAs) regulate gene expression?
a) prevent translation by binding to mRNA and degrading the mRNA strand
b) prevent translation by binding to tRNA and interfering with protein synthesis
c) prevent transcription by binding to RNA polymerase and denaturing the enzyme
d) prevent transcription by binding to DNA and removing transcription factors
a) prevent translation by binding to mRNA and degrading the mRNA strand
Select the gene regulatory mechanisms that are different between prokaryotes and eukaryotes.
a) Multiple basal transcription factors are found in eukaryotes.
b) Nuclear export of RNA occurs in prokaryotes.
c) The structure of thymine is different in eukaryotic DNA.
d) Bukaryotic DNA contains enhancers.
e) Chromatin packaging occurs in eukaryotes.
a) Multiple basal transcription factors are found in eukaryotes.
e) Chromatin packaging occurs in eukaryotes.
A mutant strain of E. coli produces $-galactosidase in the presence and in the absence of lactose.
Where in the operon might the mutation in this strain occur, and why?
a) in the CAP binding site, where the mutation leads to the inefficiency of RNA polymerase activity
b) in the lac gene, which leads to an inactive lac repressor
c) in the promoter region of the operon, where the mutation leads to the failure of the RNA polymerase to bind to the promoter
d) in the operator region, where the mutation leads to the failure of the operator to normally bind the repressor
e) near the lac gene, where the mutation leads to increased levels of lac repressor being made
f) in the operator region, which leads to increased binding of the lac repressor to the operator
b) in the lac gene, which leads to an inactive lac repressor
d) in the operator region, where the mutation leads to the failure of the operator to normally bind the repressor
Napoli et al. (1990) produced genetically-modified petunias by inserting numerous copies of the petunia CHS gene into ndividual genomes. CHS codes for a key enzyme in the synthesis of anthocyanin, a pigment that produces light purple petals in wild type petunias. The investigators hypothesized that additional copies of CHS would increase expression of anthocyanin, producing a phenotype of dark purple petals. Unexpectedly, 42% of modified plants yielded petals that were white or white with patches of purple. Contrary to initial predictions, results showed that overexpression of CHS had silenced expression of CHS.
Select the best explanation for how overexpression of CHS silences expression of CHS.
a) CHS overexpression increases concentration of CHS mRNA in the cytosol, which activates an RNA-induced silencing complex that degrades mRNA or suppresses its translation.
b) CHS overexpression interferes with tRNA secondary structure formation, preventing amino acid incorporation into
protein products by ribosomes.
c) CHS overexpression causes formation of CHS introns, which interfere with translation by forming double-stranded molecules with tRNA molecules.
d) CHS overexpression induces mutation in the promoter, which decreases template affinity for RNA polymerase and prevents transcription of CHS.
e) CHS overexpression induces addition of 5’ caps and 3’ poly (A) tails to mRNA molecules, modifications which prevent ribosomes from binding to the mRNA.
a) CHS overexpression increases concentration of CHS mRNA in the cytosol, which activates an RNA-induced silencing complex that degrades mRNA or suppresses its translation.
