Quiz 3 Flashcards
A 20 bp duplex with __________ content at __________ will have the highest absorbance at 260 nm
a) 60% A-T; 50C
b) 60% G-C; 50C
c) 60% G-C; 50C and 50 mM NaCl
d) 40% G-C; 50C and 50 mM NaCl
a) 60% A-T; 50*C
NOTES: The absorbance of a nucleic acid strand is greater for single-stranded DNA than a duplex due to hyperchromicity. A higher G-C content leads to a more stable duplex, as does the addition of salt. A duplex with a higher A-T content will denature at a lower temperature than the same length duplex with a higher G-C content.
for a circular, unbroken piece of DNA, which of the following is true?
a) increasing the twist also increases the writhe
b) increasing the twist also increases the linking number
c) decreasing the writhe also decreases the twist
d) decreasing the writhe increases the twist to keep the linking number constant
d) decreasing the writhe increases the twist to keep the linking number constant
NOTE: The linking number is the sum of the twist and the writhe. For a circular piece of DNA whose backbone is unbroken, the linking number remains constant, so changes to the twist require the opposite changes in writhe.
for the figure shown below, which of the following is correct?
a) the minor groove is labeled B
b) the major groove is labeled A
c) the blue and yellow strands have a parallel orientation
d) the blue and yellow strands have an antiparallel orientation
d) the blue and yellow strands have an antiparallel orientation
NOTE: The two strands that make up a DNA double helix run in an antiparallel direction. When the strands twist into the double helix, a larger major groove and a smaller minor groove are formed.
imagine isolated DNA from an organism for which it is determined that 15% of the nucleotides are T. what percentage of the nucleotides are G?
a) 15%
b) 25%
c) 35%
d) 45%
c) 35%
NOTE: Chargaff’s rule is that for an organism, A = T and G = C. This is because of the base pairing of A to T and C to G in DNA. If you know the amount of one of the nucleotides, you can determine the amount of the other nucleotides. If T = 15 percent, then A is also 15 percent. The remaining 70 percent would be split between G and C, resulting in 35 percent of each.
RNA-seq is considered to be complementary to gene expression microarrays because RNA-seq
a) probes mRNA, while gene expression microarrays probe DNA
b) is based on a predetermined set of DNA sequences
c) is able to identify low-abundance, alternatively spliced transcripts
d) does not require the formation of cDNA
c) is able to identify low-abundance, alternatively spliced transcripts
NOTE: Both RNA-seq and gene expression microarrays probe mRNA to gain knowledge about the transcriptome. Gene expression microarrays are based on a pre-determined set of DNA sequence, while RNA-seq is unbiased and allows one to survey the entire transcriptome. Both processes require the formation of cDNA. RNA-seq has an additional advantage of allowing for the identification of low-abundance, alternatively spliced transcripts.
the figure below shows a nucleosome core particle. which of the following is INCORRECT?
a) the histone protein core consists of eight histone proteins
b) H1 histone protein is not part of the histone protein core
c) DNA wraps around the histone protein core twice
d) the interactions between the DNA and the histone protein core are sequence specific
d) the interactions between the DNA and the histone protein core are sequence specific
NOTE: The histone protein core is made up of a total of eight proteins, two of each histone proteins H2A, H2B, H3, and H4. Histone protein H1 is located outside of the protein core. The DNA wraps around the core twice, which involves 147 bp. The interactions between histones and DNA are not sequence specific. The exterior histone protein surfaces are positively charged, which allows them to interact with the negatively charged DNA backbone, regardless of the sequence of DNA.
The figure below shows the temperature cycling used for a typical PCR reaction. Which of the following is occurring at step B?
a) annealing of primers to target DNA
b) denaturation of target DNA
c) denaturation of primers
d) extension of primers
a) annealing of primers to target DNA
NOTE: PCR is a three-step process. In the first step, the template DNA and the primers are heated to 95 degrees C to disrupt hydrogen bonding, thus denaturing the strands. In the next step, labeled B in the figure, the reaction temperature is lowered to allow the primers to anneal to the target DNA. This temperature will depend on the Tm, or melting temperature, of the primers. The final step, C, is elongation, extension, or synthesis. DNA polymerase will extend the primer DNA and produce the complement to the template DNA
Which component of a common plasmid cloning vector is necessary for blue-white screening?
a) origin of replication
b) lacZ gene
c) antibiotic-resistance gene
d) multiple cloning site (MCS)
b) lacZ gene
NOTE: The product of the lacZ gene, β-galactosidase, cleaves the chromogenic substrate X-gal and produces blue color colonies, indicating a functional gene on transformed plasmid. Disruption of the lacZ gene by insertion of a DNA fragment of interest, stops production of β-galactosidase and leads to white colonies. These white colonies have been transformed with plasmids containing the DNA insert.
Which of the following interactions it NOT likely to be found between the lac repressor protein and its DNA binding partner?
a) covalent linkage
b) hydrogen bonding
c) ionic interactions
d) van der Waals interactions
a) covalent linkage
NOTE: The interactions between proteins and DNA are dynamic and rely on weak interactions, including hydrogen bonding, ionic interactions, and van der Waals interactions. Covalent linkages are stronger and would prevent release of the binding partners, which could lead to inactive proteins and could inactivate genes. A combination of these weak interactions leads to the specific interaction between a DNA sequence and a protein, such as that between the lac repressor and its DNA binding site.
Which of the following is INCORRECT regarding eukaryotic genes?
a) the 5’ UTR and 3’ UTR are not translated, but play important regulatory roles
b) exon sequences are removed from the primary transcript prior to translation
c) mature RNAs have a 5’ cap and a 3’ poly(A) tail
d) alternative splicing allows for one gene to produce several different protein products
b) exon sequences are removed from the primary transcript prior to translation
NOTE: Introns, or intervening sequences, are removed from primary transcripts prior to translation. Exon sequences are the coding regions that need to be spliced together to produce the mature RNA for translation. The inclusion and exclusion of exons during this process, known as alternative splicing, leads to different mature RNAs from a single gene, which ultimately results in different protein products. The untranslated regions (UTR) at the 5′ and 3′ end are important for processing the RNA.
Which of the following is NOT true for Sanger DNA sequencing compared to Maxam-Gilbert DNA sequencing?
a) sanger sequencing is also called chain termination sequencing
b) sanger DNA sequencing uses ddNTPs
c) sanger DNA sequencing is a high efficiency method of DNA sequencing
d) sanger DNA sequencing relies on chemical degradation of DNA
d) sanger DNA sequencing relies on chemical degradation of DNA
NOTE: Maxam-Gilbert DNA sequencing is based on the chemical degradation of DNA. Multiple different reactions are required to fully determine the sequence of a segment of DNA. This low throughput, low efficiency technique is generally not used for large scale sequencing efforts. Sanger DNA sequencing, or chain termination sequencing, is higher efficiency and can be automated, allowing for higher throughput.
Which of the following is considered gene-rich chromatin?
a) centromeres
b) euchromatin
c) heterochromatin
d) telomeres
b) euchromatin
NOTE: Euchromatin is gene-rich and less condensed, which makes it more accessible for protein binding. Centromeres and telomeres are types of heterochromatin, which is more condensed and contains mostly noncoding DNA. Centromeres are important during chromosomal division. Telomeres are located at the ends of chromosomes and maintain the length of chromosomes after replication.
Which of the following is correct about restriction enzymes?
a) restriction enzymes degrade foreign DNA by cleaving methylated DNA
b) Type II and Type III restriction enzymes cut outside of their recognition sequences
c) type II restriction enzymes recognize palindrome sequences
d) all restriction enzymes produce the same type of cleaved ends
c) type II restriction enzymes recognize palindrome sequences
NOTE: In general, restriction enzymes do not cleave methylated DNA because the bacteria’s own DNA will be methylated to protect it. Type I and Type III restriction enzymes cleave outside of their recognition sites, while Type II enzymes cleave inside their recognition sites. This makes Type II enzymes the most useful restriction enzymes for cloning. The recognition sites are palindromic as many restriction enzymes are dimers that recognize each side of the DNA duplex. Restriction enzymes can produce a blunt end if the cut occurs in the middle of the recognition site. 5′ or 3′ overhangs are produced when a restriction enzyme cleaves after the first or before the last nucleotide.
Which of the following is correct about topoisomerases?
a) type I topoisomerases cleave both strands of DNA and reduce supercoiling by one turn
b) type II topoisomerases cleave one strand of DNA and reduce supercoiling by two turns
c) type I topoisomerases are generally dimeric
d) Type II topoisomerases reduce positive supercoiling generated during replication or transcription
d) Type II topoisomerases reduce positive supercoiling generated during replication or transcription
NOTE: Type I topoisomerases are generally monomeric and cleave just one strand of DNA to reduce supercoiling by one turn. Type II topoisomerases are generally dimeric. They cleave both strands of DNA to reduce supercoiling by two turns. This processing is important for replication and transcription to be successful. Without relief of the supercoiling, polymerases would be unable to pass through the structured region.
Which of the following is the correct order for DNA condensation in eukaryotes?
a) nucleosomes, condensed coils, loops, chromatin, chromosome
b) nucleosomes, chromatin, loops, condensed coils, chromosome
c) chromatin, nucleosomes, loops, condensed coils, chromosome
d) chromatin, nucleosomes, condensed coils, loops, chromosome
b) nucleosomes, chromatin, loops, condensed coils, chromosome
NOTE: DNA is first wrapped around histones to form nucleosomes, or the classic “beads on a string” form. Next, the nucleosomes pack to form chromatin fiber, which then begins to form loops. These loops then condense into coils and ultimately form the chromosomes.
