Exam 3: 32 DNA Recombination and Technology (complete?) Flashcards
Clicker:
Compare ZFN, TALEN, and CRISPR/Cas9/Cas12 a
which statement is correct ?
A. They all have a nuclease domain/component
B. ZFN and TALEN use DNA-binding protein/domain as a guide for nuclease cleavage site
C. CRISPR/Cas9 uses complementary RNA as a guide for determining the cleave site
D. Cas12a generate sticky end DNA
E. All of the above
E. All of the above
Clicker:
Which of the following proteins bind to the single-strand DNA and is important for DNA recombination during the pairing of two chromosomes?
A. RuvC proteins
B. RecA protein
C. Recombinases
D. Topisomerases
B. RecA protein
What does RecA protein do?
A) resolve holliday junctions
B) interwine single stranded DNA and DNA duplex
C) Regulates DNA supercoiling
B) interwine single stranded DNA and DNA duplex
- In each cycle of PCR, often there is one step of “95°C, 30 seconds”. What purpose does this step serve?
A) Allowing the double-stranded DNA to melt into two single-strands
B) Allowing the primers to form base-pairs with the template
C) Allowing DNA to be synthesized
D) Allowing researchers to add DNA polymerase into the reaction
A) Allowing the double-stranded DNA to melt into two single-strands
(Denaturing)
The purpose of the “95°C, 30 seconds” step in PCR is to allow the double-stranded DNA to melt into two single strands. This step is called denaturation.
Each cycle of PCR has three steps:
- Denaturation (95°C for 30 seconds)
- Annealing primers (50-60°C for 30 seconds)
- Elongation (72°C for 1kb/minute)
Multiple cycles of these reactions lead to massive amplification of the DNA fragment.
Which of the following describes denaturing?
A) occurs at 95°C, 30 seconds; allows for double stranded DNA to melt to 2 single stranded DNA
B) occurs at 50-60°C, 30 seconds; allows primers to form base pairs with the template
C) occurs at 72°C, 30 seconds; allows DNA polymerase to add nucleotides to the primers to form base pairs with the template as a guide
A) occurs at 95°C, 30 seconds; allows for double stranded DNA to melt to 2 single stranded DNA
Which of the following describes Elongation?
A) occurs at 95°C, 30 seconds; allows for double stranded DNA to melt to 2 single stranded DNA
B) occurs at 50-60°C, 30 seconds; allows primers to form base pairs with the template
C) occurs at 72°C, 30 seconds; allows DNA polymerase to add nucleotides to the primers to form base pairs with the template as a guide
C) occurs at 72°C, 30 seconds; allows DNA polymerase to add nucleotides to the primers to form base pairs with the template as a guide
Which of the following describes Annealing Primers?
A) occurs at 95°C, 30 seconds; allows for double stranded DNA to melt to 2 single stranded DNA
B) occurs at 50-60°C, 30 seconds; allows primers to form base pairs with the template
C) occurs at 72°C, 30 seconds; allows DNA polymerase to add nucleotides to the primers to form base pairs with the template as a guide
B) occurs at 50-60°C, 30 seconds; allows primers to form base pairs with the template
Clicker:
Which of the following reagents is essential
for making cDNA?
A. Helicase
B. Clamp loader
C. Reverse transcriptase
D. Ligase
C. Reverse transcriptase
The essential reagent for making cDNA is C. Reverse transcriptase.
Here’s why:
- cDNA stands for complementary DNA, which is a DNA segment produced using mRNA as a template.
- Reverse transcriptase is an enzyme that catalyzes the synthesis of DNA from an RNA template, hence its name.
The other options are not essential for making cDNA:
- Helicase is an enzyme that unwinds double-stranded DNA during replication. While it is important for DNA replication, it is not directly involved in cDNA synthesis.
- Clamp loader is a protein complex that loads the DNA polymerase clamp onto the DNA template during replication. Like helicase, it is important for DNA replication but not cDNA synthesis.
- Ligase is an enzyme that joins DNA fragments together. While it might be used in some downstream applications of cDNA, it is not required for the initial synthesis of cDNA from an mRNA template.
Clicker:
In each cycle of PCR, often there is one step
of “55°C, 30 seconds”. What purpose does
this step serve?
A. Allowing the double-stranded DNA to melt into two single-strands
B. Allowing the primers to form base-pairs with the template
C. Allowing DNA to be synthesized
D. Allowing researchers to add DNA polymerase
into the reaction
B. Allowing the primers to form base-pairs with the template
(Annealing)
Each cycle of PCR involves three steps:
- Denaturation: This step occurs at 95°C for 30 seconds and allows the double-stranded DNA to melt into two single strands.
- Annealing: This step occurs at 50–60°C for 30 seconds and allows the primers to attach to the template DNA.
- Elongation: This step occurs at 72°C for 1 kb/minute and allows DNA polymerase to synthesize new DNA, starting from the primers.
Multiple cycles of these reactions lead to a massive amplification of the DNA fragment.
(Question from discussion)
Suppose that you want to clone your favorite gene into the vector pSAP. Sequence analysis revealed that your favorite gene contains the following restriction enzymes sites:
Xhol, Ndel, Hindlll, Smal, and Sall.
The sequence upstream of your favorite gene contains the sites for EcoRI and Hindlll
The sequence downstream of your favorite gene contains the sites Xhol and Kpnl.
Which of the following enzyme(s) will you choose for your cloning project?
A) Hindlll and Smal
B) EcoRI and Kpnl
C) EcoRI and Xhol
D) Hindlll and Ndel
B) EcoRI and Kpnl
Upstream: cut first one
Down stream: cut the last one
- To make cDNA (complementary DNA), reverse transcriptase is used.
Which activity of the reverse transcriptase (RT) is used for synthesizing the DNA in the first round.
A) 3’-5’ DNA directed RNA polymerase
B) 3’-5’ DNA directed DNA polymerase
C) 5’-3’ RNA directed DNA polymerase
D) 5’-3’ RNA directed RNA polymerase
C) 5’-3’ RNA directed DNA polymerase
- Reverse transcriptase is used to make cDNA, which is a DNA segment made using mRNA as a template.
- Reverse transcriptase is an RNA-directed DNA polymerase, meaning it uses RNA as a template to synthesize DNA. It does this by adding nucleotides to the 3’ end of a growing DNA strand.
- All DNA polymerases synthesize DNA in the 5’ to 3’ direction, meaning they add nucleotides to the 3’ end of a growing DNA strand.
Let’s look at why the other options are incorrect:
- A) 3’-5’ DNA directed RNA polymerase: This describes an RNA polymerase that uses DNA as a template to synthesize RNA. RNA polymerase is involved in transcription, not the synthesis of cDNA.
- B) 3’-5’ DNA directed DNA polymerase: This describes a DNA polymerase that uses DNA as a template to synthesize DNA and that synthesizes DNA in the 3’ to 5’ direction. No known DNA polymerase synthesizes DNA in the 3’ to 5’ direction. DNA polymerase is involved in DNA replication, not cDNA synthesis.
- D) 5’-3’ RNA directed RNA polymerase: This describes an RNA polymerase that uses RNA as a template to synthesize RNA. RNA polymerase is involved in transcription, not cDNA synthesis.
- In order to clone the intact ABC gene into a vector with the given linker (Hindlll, Kpnl, Xbal, BamHI), which enzyme(s) should be used?
A) XbaI and BamHI
B) HindIII and EcoRI
C) HindIII and BamHI
D) HindII and KpnI
C) HindIII and BamHI
(HW 32)
What reagents are needed for a typical polymerase chain reaction (PCR)?
- Two primers
- Four nucleotides (A,T,C, and G)
- DNA polymerase from E. coli
- Taq polymerase
- Template DNA strand
-Four nucleotides (A,U,C and G)
-DNA ligase
-Two primers
- four nucleotides (A, T, C, and G)
- Taq polymerase
-Template DNA strand
Polymerase chain reaction (PCR) is used to amplify the quantity of a sequence in a sample of DNA. PCR requires a template DNA strand, two DNA primers, Taq polymerase, and DNA nucleotides (A, T, C, and G).
The template DNA strand contains the target DNA sequence that will be amplified. The two DNA primers are short, single‑stranded DNA molecules that are complementary to the template DNA. Primers define the segment of the DNA that will be amplified. Taq polymerase is a special polymerase, which was isolated from the bacterium Thermus aquaticus and maintains functionality at high temperatures.
A PCR cycle is a three‑step process that is repeated many times to quickly amplify the target sequence of the template DNA. First, the temperature of the reaction is raised in order to denature, or separate, the strands of the template DNA. Next, the reaction is allowed to cool so that the DNA primers may anneal to the DNA template. Finally, Taq polymerase extends the primer by adding nucleotides that are complementary to the template DNA strand, replicating the DNA sequence.
DNA ligase is an enzyme that joins fragments of DNA and is not involved in PCR. Taq polymerase must be used rather than DNA polymerase from E. coli because DNA polymerase from E. coli would be denatured at the high temperature at which PCR takes place. Only the nucleotides found in DNA (A, T, C, and G) are used in the experiment; uracil (U) found in RNA is not involved.
PCR has many applications, including such things as analysis of crime scene evidence, comparison of DNA of different organisms, and DNA sequencing.
(HW 32)
Targeting a DNA sequence for editing by the CRISPR-Cas9 system requires that a protoscpacer-adjacent motif (PAM) sequence appear just downstream of the target sequence. For S. pyogenes Cas9, this PAM is the trinucleotide 5’-NGG-3’, where N represents any nucleotide.
Assuming a fragment of DNA includes equal amounts of A,C, G, and T bases, how far apart would one expect to find these PAMS?
Average PAM distances:___ bases.
16 bases
PAM Spacing Calculation:
Given the PAM sequence 5’-NGG-3’, where N can be any of the four nucleotides (A, C, G, or T), and assuming equal proportions of each nucleotide in a DNA fragment, the probability of finding a specific nucleotide at any given position is 1/4.
- To encounter the ‘GG’ dinucleotide, we need two consecutive ‘G’ nucleotides.
- The probability of finding a ‘G’ at one position is 1/4.
- The probability of finding another ‘G’ immediately after is also 1/4.
- Therefore, the probability of encountering the ‘GG’ dinucleotide is (1/4) * (1/4) = 1/16.
This means, on average, one would expect to find the ‘GG’ dinucleotide once every 16 base pairs. Since ‘N’ can be any nucleotide, the PAM sequence ‘NGG’ would also occur, on average, every 16 bases.
Important note: This calculation is based on the assumption of equal nucleotide distribution, which may not always hold true in real DNA sequences. Genomic DNA often has regions with varying nucleotide compositions.
(HW 32)
Suppose you have a DNA fragment you would like to insert into pSAP. The fragment has PstI and EcoRI restriction endonuclease sites near the 5’ end, and Hindlll and SmaI restriction endonuclease sites near the 3’ end.
What are the best restriction enzymes to use to digest both the DNA fragment and pSAP?
A) EcoRI and HindIII
B) PstI and Hindlll
C) EcoRI and Smal
D) Pstl and Smal
A) EcoRI and HindIII
Upstream(3’): cut first one
Downstream(5’): cut the last one
