Nucleic Acid Biochemistry Flashcards
1
Q
When comparing different species, which of the following best predict the relative number of genes in each species?
a. the size of the organism
b. the complexity of the organism
c. the size of its brain
d. the size of the genome
e. the number of chromosomes
A
d. the size of the genome.
The number of chromosomes is roughly correlated as well.
2
Q
Chromatin in which of the following states is the most euchromatic?
a. in a Barr body
b. in the centromere
c. in a transcribed gene during G1
d. in a silenced gene during M phase
e. in a sperm nucleus
A
c. in a transcribed gene during G1
Genes must be euchromatic (decondensed) in order for transcription factors to have access to them. Chromatin is highly condensed during mitosis, when packed in the sperm head, when part of an inactivated X chromosome (Barr body) and in structural, non-expressed regions of a chromosome such as the centromere.
3
Q
Which of the following would be expected for a genetic disease based on a mutation in mitochondrial DNA?
a. it would be more common in men
b. it would be more common in women
c. it would be inherited maternally
d. it would be easy to diagnose based on symptoms
e. it would affect multiple organ systems
A
c. it would be inherited maternally
e. it would affect multiple organ systems
Mitochondria is maternally inherited but unlike X-linked disorders, will affect both men and women who inherit defective genes. Because of heteroplasmy, different people are affected in different organs and to different extents, making diagnosis tricky. However, since mitochondrial function is essential to so many tissues, symptoms typically arise in multiple organs.
4
Q
Which of the following modifications to histone H3 promote gene silencing?
a. acetylation
b. deacetylation
c. demethylation
d. methylation
e. phosphorylation
A
b. deacetylation
d. methylation
5
Q
A typical eukaryotic gene consists of which of the following elements?
a. internal ribosome binding sites
b. introns
c. exons
d. enhancers
A
b. introns
c. exons
d. enhancers
6
Q
Multiple mRNAs can arise from a primary transcript by use of alternative
a. splicing.
b. poly(A) sites.
c. promoters.
d. ribosome binding sites.
A
a. splicing.
b. poly(A) sites.
c. promoters.
7
Q
Which of the following is (are) encoded by multiple genes present in the human genome?
a. ribosomal RNA
b. transfer RNA
c. histone
d. lysozyme
A
a. ribosomal RNA
b. transfer RNA
c. histone
8
Q
Which of the following statements is (are) true of the human genome?
a. 1.5 percent of the genome corresponds to protein coding sequences.
b. The median length of an intron is 10 kb.
c. 10 percent of the genome is transcribed into pre mRNA precursors.
d. Most human exons contain 500–1000 base pairs.
A
a. 1.5 percent of the genome corresponds to protein coding sequences.
9
Q
DNA fingerprinting is a technique based on differences in the
a. length of introns.
b. number of tandem copies of a simple sequence repeat.
c. number of tandem ribosomal RNA genes.
d. size of protein coding genes.
A
b. number of tandem copies of a simple sequence repeat.
10
Q
Transposition by a bacterial insertion element
a. occurs at a frequency of approximately 1 in 103 cells per generation.
b. can inactivate an essential gene.
c. is mediated through a RNA intermediate.
d. requires the enzyme transposase.
A
b. can inactivate an essential gene.
d. requires the enzyme transposase.
11
Q
Transposition by a retrotransposon requires activity of which of the following enzymes?
a. RNA polymerase
b. reverse transcriptase
c. DNA methylase
d. DNA polymerase
A
a. RNA polymerase
b. reverse transcriptase
12
Q
LINES (long interspersed elements)
a. are retrotransposons that lack LTRs.
b. are approximately 300 base pairs long.
c. are a rare class of mobile elements in mammals.
d. use the enzyme transposase for transposition.
A
a. are retrotransposons that lack LTRs.
13
Q
Human mitochondrial DNA
a. uses the standard genetic code.
b. encodes its own ribosomal RNAs.
c. contains introns like nuclear genes.
d. is larger than yeast mitochondrial DNA.
A
b. encodes its own ribosomal RNAs.
14
Q
Plant mitochondrial DNA
a. is the same size as human mitochondrial DNA.
b. encodes a 5S mitochondrial rRNA.
c. contains multiple copies that recombine with each other.
d. uses the standard genetic code.
A
b. encodes a 5S mitochondrial rRNA.
c. contains multiple copies that recombine with each other.
d. uses the standard genetic code.
15
Q
An open reading frame (ORF) is defined as a DNA sequence that
a. begins with a start codon.
b. ends with a stop codon.
c. contains 50 codons.
d. contains approximately an equal frequency of A, T, G, and C.
A
a. begins with a start codon.
b. ends with a stop codon.
16
Q
Two related genes that are derived from a gene duplication event are considered to be
a. homologous.
b. paralogous.
c. orthologous.
d. members of a gene family.
A
a. homologous.
b. paralogous.
d. members of a gene family.
17
Q
A transcriptionally active gene compared with a transcriptionally inactive gene would be expected to
a. contain acetylated histones.
b. contain unacetylated histones.
c. be sensitive to DNase I.
d. be resistant to DNase I.
A
a. contain acetylated histones.
c. be sensitive to DNase I.
18
Q
Scaffold-associated regions
a. are the chromosome attachment points for the mitotic spindle.
b. can insulate transcription units from each other.
c. are the points at which DNA interacts with histone proteins.
d. are found between transcription units.
A
b. can insulate transcription units from each other.
d. are found between transcription units.
19
Q
Metaphase chromosomes can be identified
a. by shape.
b. by the size and number of introns.
c. by banding patterns with Giemsa reagent.
d. by chromosome painting.
A
a. by shape.
c. by banding patterns with Giemsa reagent.
d. by chromosome painting.
20
Q
Telomerase
a. extends DNA strands during DNA synthesis.
b. has reverse transcriptase activity.
c. is a protein-RNA complex.
d. replicates repetitious DNA located at the centromere.
A
b. has reverse transcriptase activity.
c. is a protein-RNA complex.
21
Q
Compare and contrast simple transcription units and complex transcription units.
A
A simple transcription unit produces a single monocistronic mRNA, which is translated into a single protein. A complex transcription unit produces primary transcripts that can be processed in alternative ways and translated into multiple proteins.
22
Q
Describe the three different ways that multiple mRNAs can arise from a complex transcription unit in eukaryotes.
A
A complex transcription unit is transcribed into multiple mRNAs by using alternative promoters, poly(A) sites, and splicing. The use of alternative promoters produces mRNAs with different 5´ exons but common 3´ exons. The use of alternative poly(A) sites produces mRNAs with common 5´ exons but different 3´ exons. The use of alternative splice sites produces mRNAs with common 5´ and 3´ exons but different combinations of internal exons.
23
Q
Describe the general organization of the human genome in terms of protein-coding and functional RNA genes, repetitious DNA, and spacer DNA.
A
The human genome consists of protein-coding and functional RNA genes, repetitious DNA, and spacer DNA. Protein-coding genes and functional RNA genes make up approximately 30 percent of the genome. Protein-coding genes can exist as solitary genes or duplicated genes. Functional RNAs such as tRNA, rRNA, and snRNA are present as tandemly repeated genes. Repetitious DNA makes up almost half of the human genome and is usually concentrated at specific chromosomal locations, e.g., at the centromere. The remainder of the genome consists of spacer or intergenic DNA.
24
Q
Describe the molecular basis for the DNA fingerprinting technique. How can this be used to differentiate between two individuals?
A
The DNA fingerprinting technique is based on differences in the length of simple-sequence DNAs. Simple-sequence DNA usually occurs in tandem arrays. The number of simple-sequence repeat units at a given genetic locus varies between individuals, and thus the total length of the tandem array differs. These differences in the length of tandem arrays throughout the genome are unique to an individual and is the basis for an individual’s unique DNA fingerprint.
25
Describe the structural features of bacterial insertion sequences and transposons.
Bacterial insertion sequences (IS) are about 1–2 kilobases long. At each end of the IS element is an approximately 50-base-pair inverted repeat. Between the inverted repeats is a region that encodes the enzyme transposase—which is required for transposition or the "cut and paste" operation of the element—to a new site in the genome.
26
Describe the possible role that mobile DNA elements played in evolution.
Mobile DNA elements are hypothesized to have had a profound effect on the evolution of organisms. Insertion of a mobile DNA element into or near a gene can cause a mutation in the gene. Some evidence suggests that recombination between repeat sequences (e.g., mobile elements) of two separate genes can generate a novel combination of exons, a process known as exon shuffling. Unequal crossing over between repeat sequences can result in gene duplication. Furthermore, transposition of DNA adjacent to mobile DNA elements can move transcriptional control elements to new regions of the genome.
27
How does plant mitochondrial DNA differ from mammalian mitochondrial DNA?
Plant mitochondrial and animal mitochondrial DNA are circular molecules that encode for tRNA, rRNA, and essential mitochondrial proteins. In contrast to mammals, plants contain multiple mitochondrial DNAs that appear to recombine with one another. Plant mitochondrial DNAs are much larger and more variable in size than those of other organisms. Plant mitochondrial DNA also encode a 5S mitochondrial rRNA, which is present only in plant mitochondrial ribosomes, and the α subunit of the F1 ATPase. Furthermore, plant mitochondrial DNA uses the standard genetic code, whereas mammalian mitochondria use a modified code.
28
What is a BLAST search and how can it be used to help determine the function of an unknown cloned gene?
The BLAST (basic local alignment and search tool) program searches the DNA/protein databases for significant matches between a query sequence and stored sequences. The search program then assigns a score based on the extent of the match. For an unknown cloned gene, the DNA sequence or predicted amino acid sequence can be used as the query sequence for the BLAST search. If a known gene or protein in the database shows significant similarity to the query sequence, then it is likely that the unknown cloned gene is functionally similar to the known gene.
29
What are the applications of DNA microarrays?
A DNA microarray consists of thousands of individual, closely packed, gene-specific sequences attached to a surface such as glass or a nylon membrane. The power of the microarray is its ability to examine the expression of thousands of genes simultaneously. It is the functional equivalent of performing thousands of Northern blots at one time.
30
Describe the role that nonhistone proteins play in organizing chromosome structure.
Nonhistone proteins provide a structural scaffold for organizing chromatin loops. The chromosome scaffold, which has the shape of a metaphase chromosome, consists of nonhistone proteins that serve as binding sites for chromatin loops.
31
How does chromatin differ between transcriptionally inactive and transcriptionally active regions of DNA?
Experimental evidence indicates that transcriptionally inactive DNA has a condensed chromatin structure. This structure makes the DNA inaccessible to RNA polymerase. In contrast, actively transcribed DNA is present in a more extended, "beads-on-a-string" form of chromatin. DNA with this more open conformation is more accessible to RNA polymerase and other proteins required for transcription.
32
What three functional elements are required for replication and stabilization of chromosomes?
Eukaryotic chromosomes require origins of replication to initiate DNA replication, a centromere for proper segregation of chromosomes during mitosis, and telomeres to stabilize the ends.
33
```
How many base pairs make up the human genome?
A. 3.3 million
B. 5.1 billion
C. 3.3 billion
D. 5.1 million
```
C. 3.3 billion
34
When was Sanger sequencing developed and what is another name for it?
1975, chain terminator
35
What is the typical read length for a Sanger sequencing product?
A. 300-800 bps
B. 500-1,000 bps
C. 100-300 bps
A. 300-800 bps
36
What are the major differences between Sanger sequencing and massively parallel sequencing?
1. Longer read length with Sanger
2. Single template read for Sanger, many template reads with MPS
3. Sanger cannot be used for gene expression or quantitative assays
4. No prior knowledge of the genome is required for MPS
37
Which of the following mutations can result in a reduction of β-galactosidase?
a. a mutation in adenylate cyclase
b. a mutation in catabolite activator protein (CAP)
c. a mutation in the CAP site in the lac control region
d. a mutation in the repressor binding site in the operator
a. a mutation in adenylate cyclase
38
Specific DNA control elements in promoters can
a. interact with general transcription factors.
b. interact with repressor proteins.
c. interact with activator proteins.
d. remain unavailable because of condensed chromatin.
a. interact with general transcription factors.b. interact with repressor proteins.c. interact with activator proteins.d. remain unavailable because of condensed chromatin.
39
Reporter genes are used to
a. express enzymes that are not easily assayed in cell extracts.
b. express enzymes that are easily assayed in cell extracts.
c. characterize DNA control elements.
d. characterize reporter plasmids.
b. express enzymes that are easily assayed in cell extracts.c. characterize DNA control elements.
40
The three eukaryotic RNA polymerases can be distinguished by
a. the types of genes they transcribe.
b. the number and types of large subunits.
c. their differential sensitivities to cycloheximide.
d. their differential sensitivities to α-amanitin
a. the types of genes they transcribe.andd. their differential sensitivities to α-amanitin.
41
"Which of the following can be identified using a series of promoter linker scanning mutations?
a. areas of the promoter that are non-essential
b. areas of the promoter that are essential
c. the presence of separate transcriptional control regions
a. areas of the promoter that are non-essentialb. areas of the promoter that are essentialc. the presence of separate transcriptional control regions
42
An enhancer
a. can be located upstream of a promoter.
b. can be located downstream of a promoter.
c. can be located a variable distance from the promoter.
d. is always located within 1 kb of the promoter.
e. can be cell-type-specific.
a. can be located upstream of a promoter.b. can be located downstream of a promoter.c. can be located a variable distance from the promoter.e. can be cell-type-specific.
43
The fact that a specific protein leaves a footprint on a DNA molecule is indicative of
a. a lack of interaction between the specific protein and DNA.
b. protection from DNAse by the specific protein.
c. binding of the specific protein to all types of DNA.
d. binding of the specific protein to a specific sequence of DNA.
b. protection from DNAse by the specific proteinandd. binding of the specific protein to a specific sequence of DNA.
44
The C-terminal activation domain of transcriptional activators is capable of
a. binding to DNA.
b. stimulating transcription.
c. interaction with other transcriptional machinery.
d. functioning in a fusion with a DNA-binding domain from an unrelated transcriptional activator.
a. binding to DNA.
b. stimulating transcription.
c. interaction with other transcriptional machinery.
d. functioning in a fusion with a DNA-binding domain from an unrelated transcriptional activator.
45
Which of the following are not found in DNA-binding proteins?
a. homeodomains
b. zinc fingers
c. leucine zippers
d. CpG islands
d. CpG islands
46
Transcription factors can:
a. exhibit cooperative binding.
b. exist as heterodimers.
c. act to repress transcription of transcription factor genes.
d. undergo conformational changes which alter activity.
e. never interact with co-repressors.
a. exhibit cooperative binding.
b. exist as heterodimers.
c. act to repress transcription of transcription factor genes.
d. undergo conformational changes which alter activity.
47
Which is the first factor to bind at the promoter of eukaryotic genes?
a. RNA polymerase
b. TFIIA
c. TFIIB
d. TFIID
e. TATA box binding protein
d. TFIID,
| e. TATA box binding protein
48
Which of the following occur(s) during Pol II transcription preinitiation complex formation?
a. TFIIA binds to TFIIB.
b. TFIIB unwinds the DNA.
c. TFIIB contacts both TATA box binding factor and DNA.
d. DNA bends.
c. TFIIB contacts both TATA box binding factor and DNA.
| d. DNA bends.
49
Chromatin-mediated repression of transcription involves
a. modification of lysine residues in histones.
b. large, multiprotein complexes.
c. acetylation of histone tails.
d. deacetylation of histone tails.
a. modification of lysine residues in histones.
b. large, multiprotein complexes.
c. acetylation of histone tails.
50
Which of the following does (do) not require a DNA helicase activity?
a. SWI/SNF function
b. Pol II open-complex formation
c. transcription-factor binding to DNA
d. deacetylation of histone tails
c. transcription-factor binding to DNA
| d. deacetylation of histone tails
51
The yeast two-hybrid system can be used to identify
a. proteins that interact with a known or unknown protein.
b. proteins that interact with a kinase domain.
c. cDNAs that encode interacting proteins.
d. co-activators and co-repressors.
a. proteins that interact with a known or unknown protein.
c. cDNAs that encode interacting proteins.
d. co-activators and co-repressors.
52
The expression of which gene(s) is (are) regulated by promoter proximal pausing?
a. hsp70
b. lac operon
c. tat
d. none of the above
a. hsp70
53
The promoter sequences of RNA polymerase III-transcribed genes are located
a. within the transcribed sequences of genes.
b. upstream of the transcribed sequences of genes.
c. downstream of the transcribed sequences of genes.
d. within the first intron.
a. within the transcribed sequences of genes.
54
Regulation of the lac operon is under both negative and positive control. Describe how these two transcriptional control mechanisms regulate expression of the lac operon in the presence of glucose and/or lactose.
In the presence of glucose and the absence of lactose, cAMP levels are low. The lac repressor is bound to the lac operator, which blocks the synthesis of lac mRNA. In the presence of glucose and lactose, the lac repressor is bound to lactose and is unable to bind to the lac operator. However, because cAMP is low in the presence of glucose, very little lac mRNA is synthesized. In the absence of glucose and the presence of lactose, the lac repressor is again unable to bind to the lac operator because it is bound to lactose. However, cAMP levels are high in the absence of glucose. A cAMP-CAP complex forms, which binds to the CAP site and stimulates transcription of lac mRNA.
55
Compare the structure and function of the three eukaryotic RNA polymerases.
In eukaryotic cells, three different RNA polymerases catalyze the formation of different RNAs. RNA polymerase I is located in the nucleolus and synthesizes pre-ribosomal RNA (the precursor to 28S, 5.8S, and 18S rRNAs). RNA polymerase II synthesizes messenger RNAs. RNA polymerase III synthesizes tRNAs, 5S rRNAs, and a number of small RNAs. These small RNAs include an RNA involved in RNA splicing and the 7S RNA of the signal-recognition particle. All three RNA polymerases contain two large sub-units and 15 smaller subunits. Some of these subunits are shared among the different polymerases and others are unique.
56
Describe the three types of eukaryotic promoters.
1. contains a highly conserved sequence located approximately 25-35 base pairs upstream of the transcription start site known as the TATA box. The TATA box acts similarly to a prokaryotic promoter to position RNA polyeramerase II for transcription initiation.
2. lack a TATA box and instead contain an alternative promoter element called an initiator. The initiator sequence is not highly conserved.
3. contains neither a TATA box nor an initiator element. These genes contain a CG-rich stretch of 20-50 nucleotides (CpG islands) approximately 100 base pairs upstream of the start site.
57
How can linker scanning mutation analysis be used to map the location of transcription-control elements?
In linker scanning mutation analysis, a region of DNA that contains putative regulatory elements is cloned into a plasmid upstream of a reporter gene. A series of overlapping linker scanning mutations are introduced into the DNA containing the regulatory elements. The mutations are introduced by scrambling the nucleotide sequence in a short stretch of DNA. The effects of these mutations are evaluated by assaying reporter gene expression.
58
Describe the sequence of events that occurs during formation of a transcription initiation complex on a TATA-box promoter.
A number of sequential protein binding steps occurs during the formation of a transcription initiation complex. The TATA box-binding protein (TBP) included within TFIID first binds to a TATA box in the promoter. Once TBP has bound to the TATA box, TFIIB can bind. The next step is the binding of a pre-formed TFIIF and PolII complex. Then TFIIE and TFIIH bind, completing assembly of the transcription initiation complex.
59
Describe transcription initiation for RNA polymerase I and III transcribed genes.
Transcription initiation by RNA polymerase I is mediated by a core element, which includes the start site of the pre-rRNA gene and an upstream control element, located approximately 100 base pairs upstream. Two transcription factors up-stream binding factor (UBF) and selectivity factor 1 (SL1) bind to and help stabilize the initiation complex. Transcription initiation by RNA polymerase III is mediated by internal promoter elements, termed the A box and B box, present in all tRNA genes. Two transcription factors, TFIIIC and TFIIIB, are required for transcription initiation of tRNA genes. An additional factor (TFIIIA) is required for transcription initiation of 5S rRNA genes.
60
Regulation of the cell cycle is controlled by:
a. telomerase
b. protein kinases
c. ATP
d. growth factors
b. protein kinases
61
What is the function of DNA in the cell?
DNA is a storage system. The function of DNA is to store genetic information.
62
Compare the structure of nitrogen bases. How do purines and pyrimidines differ?
Purines have a double ring, while pyrimidines have a single ring.
63
Write the complementary sequence to the following:
5' AGG TCA CGT CTA GCT AGC TAG A '3
5' T CTA GCT AGC TAG ACG TGA CCT '3
64
Which of the ribose carbons participate in the phosphodiester bond?
The 5' ribose carbon carries the phosphate group that forms a phosphodiester bond with the hydroxyl group on the 3' ribose carbon.
65
Which of the ribose carbons carries the nitrogen base?
The 1' carbon of the ribose carries the nitrogen base.
66
Why does DNA polymerase require primase activity?
DNA polymerase cannot begin synthesis with a 3' OH group.
67
What is the covalent bond between nucleotides catalyzed by DNA polymerase?
The covalent bond between nucleotides is a phosphodiester bond.
68
Is DNA replication conservative or semi-conservative?
DNA replication is semi-conservative.
69
What is the lagging strand in DNA synthesis?
During DNA replication, the lagging strand is the strand positions 5' to 3' with respect to the direction of synthesis, requiring the replication assembly to jump ahead and read 3' to 5' back toward the moving replication fork. The leading strand is read continuously 3' to 5'. Synthesis thus proceeds 5' to 3' on both strands.
70
Short pieces of DNA involved in DNA synthesis in replicating cells are called what type of fragments?
Short pieces of DNA involved in DNA synthesis in replicating cells are called Okazaki fragments.
71
Name the function of the following enzyme: polymerase
Polymerase catalyze formation of a phosphodiester bond between nucleotides.
72
Name the function of the following enzyme: helicase.
helicase unwinds nucleic acids, relieving stress on the double helix by breaking and re-attaching phosphodiester bonds.
73
Name the function of the following enzyme: primase
Primase is an RNA polymerase that starts replication in the absence of a 3' OH group.
74
Name the function of the following enzyme: exonuclease
Exonuclease digests phosphodiester bonds from the ends of nucleic acid molecules.
75
Name the function of the following enzyme: endonuclease
Endonuclease breaks the phosphodiester bonds in the nucleic acids within the polymer chains, rather than from the ends.
76
A plasmid was digested with the enzyme HpaII. On agarose gel electrophoresis, you observe three bands: 100, 230, and 500 bp. How many HpaII sites are present in this plasmid?
There are three HpaII sites in this plasmid.
77
A plasmid was digested with the enzyme HpaII. On agarose gel electrophoresis, you observe three bands: 100, 230, and 500 bp. What are the distances between each site?
The distances between the sites are 100, 230, and 500 bp.
78
A plasmid was digested with the enzyme HpaII. On agarose gel electrophoresis, you observe three bands: 100, 230, and 500 bp. What is the size of the plasmid?
The plasmid is 100 + 230 + 500 = 830 bp.
79
A plasmid was digested with the enzyme HpaII. On agarose gel electrophoresis, you observe three bands: 100, 230, and 500 bp. A second cut of the plasmid with BamHi yields two pieces, 80 and x bp. How many BamHI sites are in the plasmid?
There are two BamHI sites in the plasmid.
80
A plasmid was digested with the enzyme HpaII. On agarose gel electrophoresis, you observe three bands: 100, 230, and 500 bp. A second cut of the plasmid with BamHi yields two pieces, 80 and x bp. What is the length of x in base pairs (bp)?
x = 830 - 80 = 750 bp
81
A plasmid was digested with the enzyme HpaII. On agarose gel electrophoresis, you observe three bands: 100, 230, and 500 bp. A second cut of the plasmid with BamHi yields two pieces of 80 and 750 bp. How would you determine where the BamHI sites are in relation to the HpaII sites?
Cut the plasmid with both the BamHi and HpaII restriction enzymes in the same reaction and note the fragment sizes that are produced.
82
A plasmid has one EcoR1 site into which you want to clone a blunt-ended fragment. What type of enzyme could turn an EcoR1 sticky end into a blunt end?
A single-strand exonuclease can digest the EcoR1 overhang, forming a blunt end. Alternatively, a polymerase could fill in the overhang to make a blunt end.
83
Compare how DNA moves from one cell to another by: conjugation.
For conjugation, DNA moves from cell to cell through physical contact.
84
Compare how DNA moves from one cell to another by: transduction.
DNA moves from cell to cell through intermediary viruses or bacteriophages in transduction.
85
Compare how DNA moves from one cell to another by: transformation
In transformation, unprotected DNA moves from cell to cell without physical contact or viral carriers.
86
Bacteria with phenotype A+ are mixed with bacteria with phenotype A- in a culture. Some of the A- bacteria become A+. When A+ bacteria are removed from the culture and A- bacteria are grown in the A+ cell culture medium only, they all remain A-. What type of transfer occurred in the mixed culture?
Conjugation occurred in the mixed culture. If the A-bacteria became positive through transformation or transduction, the cell culture medium from the A+ bacteria would contain DNA or phage, respectively, capable of transforming A- bacteria to A+.
87
What was the rationale for labeling bacteriophage with 35S and 32P in the Hershey and Chase 'blender' experiment?
35S labels protein, specifically since there is no S in nucleic acis. 32P labels nucleic acid specifically, since there is no P in the phage proteins.
88
A plasmid that carries genes for its own transfer and propagations is called _____.
A plasmid that carries genes for its own transfer and propagation is called self-transmissible
89
What enzymes can convert a supercoiled plasmid to a relaxed circle?
Helicases (topoisomerases) can digest the phosphodiester bond in one strand of the double-stranded DNA, allowing the twisted strands to unwind.
90
Name three processing steps undergone by messenger RNA.
Three processing steps of mRNA are polyadenylation, capping, and splicing.
91
What is the function of polyadenylate polymerase?
Polyadenylate polymerase digests RNA and adds adenines to the 3' end of the transcript.
92
What is unusual about the phosphodiester bond formed between mRNA and its 5' cap?
The 5' cap is attached by an unusual 5' to 5' pyrophosphate linkage.
93
The parts of the hnRNA that are translated into protein (open reading frames) are the ______.
a. exons
b. introns
c. enhancers
d. miRNAs
a. exons
94
The intervening sequences that interrupt protein coding sequences in the hnRNA are _____.
a. exons
b. introns
c. enhancers
d. miRNAs
b. introns
95
Proteins that bind to DNA to control gene expression are called ______ .
a. restriction enzymes
b. cis factors
c. enhancers
d. trans factors
d. trans factors
96
The binding sites on DNA for proteins that control gene expression are ______.
a. restriction enzymes
b. cis factors
c. enhancers
d. trans factors
b. cis factors
97
How might a single mRNA produce more than one protein product?
A single RNA may be alternatively spliced to produce more than one product.
98
The type of transcription producing RNA that is continually required and relatively abundant in the cell is called _____ transcription.
The type of transcription producing RNA that is continually required and relative abundant in the cell is called constitutive transcription.
99
A set of structural genes transcribed together on one polycistronic mRNA is called _____.
A set of structural genes transcribed together on one polycistronic mRNA is called an operon.
100
Would the methylation of cytosine bases 5' to the gene increase or decrease expression of the gene?
Methylation of cytosine bases 5' to the gene will decrease expression of the gene.
101
Would histone acetylation close to the gene increase or decrease expression of the gene?
Histone acetylation close to the gene will increase expression
102
What two factors are responsible for regulation of RNA synthesis?
cis factors, which are DNA sequences that mark places on DNA involved in the initiation and control of RNA synthesis
and
trans factors, which are proteins that bind to the cis sequences and direct the assembly of transcription complexes at the proper gene.
103
Define Operon.
a series of structural genes transcribed together on one mRNA and subsequently separated into individual proteins. (e.g. in a small genome an operon might encode the enzymes of a metabolic pathway).
104
What is encoded by the lac operon?
three structural genes, LacZ, LacY, and LacA, which are all required for the metabolism of lactose.
LacZ - beta-galactosidase (hydrolyzes lactose into glucose and galactose)
LacY - lactose permease (transports lactose into the cell)
LacA - thiogalactoside transacetylase (transacetylates galactosides)
105
Name three modes of regulation for mRNA transcription in prokaryotes.
1. induction (e.g. lac operon)
2. repression (e.g. arg operon)
3. activation (e.g. mal operon)
106
Name two distal regulatory elements for transcription.
enhancers and silencers
107
Define star activity.
Under non-standard conditions, restriction enzymes can find and cut sequences that differ from their recognition sequences.
108
Name five reaction conditions that an induce star activity.
1. suboptimal buffer
2. contamination with solvents or high concentrations of glycerol
3. prolonged reaction time
4. high concentration of enzymes
5. divalent cation imbalance
109
Briefly explain Restriction Fragment Length Polymorphisms (RFLPs).
Given that inherited and somatic differences in human DNA, the number and location of restriction sites for a given restriction enzyme are not the same in all individuals. These differences in size or number of restriction fragments are called RFLPs.
110
Name two uses for RFLPs (Restriction Fragment Length Polymorphisms).
Molecular-based human identification and analysis of structural changes in chromosomes (translocations, deletions, and insertions).
111
Northern blots are used to determine:
a. gene structure
b. transcript structure
c. protein processing
d. DNA binding proteins
b. transcript structure
explanation: Northern blots target RNA using a nucleic acid probe and can determine transcript structure, processing, and gene expression.
112
The Southern blot is used to determine:
a. gene structure
b. transcript structure
c. gene expression
d. protein processing
a. gene structure
| Explanation: The Southern blot targets DNA using a nucleic acid probe to determine gene structure.
113
Briefly explain the method of Southern blot.
DNA is isolated and cut with restriction enzymes. The fragments are separated by gel electrophoresis, denatured, and then transferred to a solid support (nitrocelluluose). DNA fragments are exposed to a labeled DNA or RNA probe that is complementary to the region of interest. The signal of the probe is detected and the presence of absence of the signal indicates whether the sequence in question is detected.
114
A Southwestern blot:
a. Targets DNA with a nucleic acid probe
b. Targets RNA with a nucleic acid probe
c. Targets protein with a DNA probe
d. Targets protein with a protein probe
c. Targets protein with a DNA probe
| A Southwestern blot targets Protein with a DNA probe to determine DNA binding proteins and gene regulation.
115
A Western blot:
a. Targets DNA with a nucleic acid probe
b. Targets RNA with a nucleic acid probe
c. Targets protein with a DNA probe
d. Targets protein with a protein probe
D. Targets protein with a protein probe.
A Western is used to target protein with a protein probe (often an antibody) to determine protein processing and gene expression.
116
A Northern blot:
a. Targets DNA with a nucleic acid probe
b. Targets RNA with a nucleic acid probe
c. Targets protein with a DNA probe
d. Targets protein with a protein probe
b. Targets RNA with a nucleic acid probe
A Northern blot targets RNA with a nucleic acid probe in order to determine transcript structure, processing, and gene expression.
117
A Southern blot:
a. Targets DNA with a nucleic acid probe
b. Targets RNA with a nucleic acid probe
c. Targets protein with a DNA probe
d. Targets protein with a protein probe
a. targets DNA with a nucleic acid probe
| A Southern blot targets DNA with a nucleic acid probe to determine gene structure.
118
An Eastern blot:
a. Targets DNA with a nucleic acid probe
b. Targets RNA with a nucleic acid probe
c. Targets protein with a DNA probe
d. Targets protein with a protein probe
d. Targets protein with a protein probe
An Eastern blot targets protein with a protein probe as a modification of a Western blot. The assay uses an enzymatic detection (PathHunter (tm)); also, an Eastern blot can be used for the detection of specific agriculturally important proteins.
119
What range of DNA input is commonly used for Southern blot analysis?
a. 5-10 ug
b. 10-20 ug
c. 10-50 ug
d. 30-100 ug
c. 10-50 ug
More or less DNA may be required depending on the sensitivity of the detection system, the volume and configuration of the wells, and the abundance of the target DNA. Also, in the clinical laboratory, specimen availability may limit the input DNA.
120
Name three scenarios that might result in a complicated interpretation of a Southern blot.
a. degraded DNA (smear present)
b. incomplete cutting of restriction enzyme (anomalous patterns)
c. impurity in the DNA may result in a resistance to restriction digest
121
In a Southern blot, if there is variation in the brightness of genomic DNA smears between lanes which is the best explanation?
a. not all DNA lanes were loaded with an equal amount of DNA
b. restriction enzyme activity was incomplete.
c. DNA is degraded.
a. not all DNA lanes were loaded with an equal amount of DNA.
The brightness of the DNA smears resulting from digestion with restriction enzymes digestion of genomic DNA should be similar from lane to lane, ensuring that equal amounts of starting DNA were added to each lane and reaction.
122
In genomic DNA cut with restriction enzymes, what is the explanation if a large aggregate of DNA remains near the top of the lane when run on an agarose gel?
a. not all DNA lanes were loaded with an equal amount of DNA
b. restriction enzyme activity was incomplete.
c. DNA is degraded.
b. restriction enzyme activity was incomplete.
123
In genomic DNA cut with restriction enzymes, what is the explanation if a smear is located primarily in the lower region of the lane on an agarose gel?
a. not all DNA lanes were loaded with an equal amount of DNA
b. restriction enzyme activity was incomplete.
c. DNA is degraded.
c. DNA is degraded.
124
What is the difference between the two results obtained from a Southern blot: the smeared appearance and the banded fragments?
The smeared agarose gel is viewed with UV to determine if the restriction digestion was successful. The banded or resolved appearance follows transfer of the agarose gel to a membrane that is then probed with a known single-stranded DNA or RNA labeled probe.
125
Describe briefly the process of depurination of a Southern blot and why it is important.
before moving DNA fragments to a membane, the double-stranded DNA must be denatured into single-stranded DNA to bind to the probe. Short fragments (>500 bp) are denatured by exposing the gel to a strong base (sodium hydroxide), which promotes the breakage of the hydrogen bonds holding the DNA strands together. Large fragments are more efficiently denatured if they are depurinated first. Depurination requires soaking the gel in hydrogen chloride (HCL), which removed purine bases from the sugar-phosphate backbone. This loosens the larger fragments for denaturation with a strong base (sodium hydroxide).
126
Why is nitrocellulose the membrane of choice for Southern blots?
ss-DNA binds to nitrocellulose membranes with a noncovalent, but irreversible, connection. The binding interaction is hydrophobic and electrostatic between the negatively charged DNA and the positive charge on the membrane.
127
Why is it possible to reprobe a Southern blot?
Because the bond between the membrane and the DNA is much stronger than the hydrogen bonds that hold the complementary probe to the DNA.
128
Briefly describe capillary transfer for Southern blots.
Capillary transfer is a simple and relatively inexpensive. The buffer moves from the bottom of the reservoir by capillary action through the gel, through the membrane and then through dry paper on top of the membrane. When the DNA comes in contact with the membrane it will bind to it. Issues with this method are bubbles, crystals, or other particulates that might be between the membrane and the gel will cause staining artifacts or loss of information. Also, this procedure is slow and usually takes overnight.
129
What is the purpose of the prehybridization step in Southern blot analysis?
The purpose of the prehybridization step is to prevent the probe from binding to nonspecific sites on the membrane surface, which will cause high background noise.
130
Name at two blocking agents for a Southern blot.
Denhardt solution (Ficoll, polyvinyl pyrrolidane, bovine serum albumin) and salmon sperm DNA.
131
What is the best gel type to resolve small nucleic acid fragments?
a. Acrylamide
b. Agarose
c. Poly-acrylamide
c. poly-acrylamide
Poly-acrylamide is a combination of acrylamide and the cross-linker methylene bisacrylamide that gives consistent resolution characteristics.
