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About 60% of the base pairs in a human DNA molecule are A+T. If the human genome has 3.2 billion base pairs of DNA, about how many times will the following restriction sites be present? (Total 3 points)(i)BamHI (restriction site is 5? ? GGATCC ? 3? )? (1 point)(ii)EcoRI (restriction site is 5? ? GAATTC ? 3? )? (1 point)(iii)HaeIII (restriction site is 5? ? GGCC ? 3? )? (1 point)

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Bunny GreenfelderLv2
28 Sep 2019
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Related questions

1.) You measure the OD600 of your culture of chloramphenicol-treated XL1-Blue/pTrc99A cells as described in Step 1 of Experiment 3A and obtain a value of 0.265. Assuming a plasmid copy number of 570 plasmids per cell, and a 100% recovery of plasmid DNA from 3.00 mL of the cells, what is the theoretical yield of plasmid DNA in micrograms (ug)? 2 points (Note: the size of pTrc99A is 4176 base pairs and the average molecular weight of double-stranded DNA is 660 g/mole of base pairs; report your answer with three significant figures) Tries 0/3

2.) Would you expect your actual yield of DNA to be more or less than the theoretical yield? 1 point Tries 0/1

3.) Assuming 100% recovery of DNA from the phage DNA isolation (Experiment 3B) and considering that not all the phage in your stock are able to infect E. coli successfully, would you expect your actual yield of DNA to be more or less than the theoretical yield calculated from your titer of the phage stock? 1 point Tries 0/1

4. Choose (True or False) whether each of the following parameters effects the rate of migration of DNA through agarose gels. 2 points Length of time in electric field DNA shape (supercoiled vs. relaxed circular vs. linear)

Volume of sample loaded (assuming all of the sample is contained within the loading well and enters the gel)

Electric field strength (volts per centimeter)

DNA size (molecular weight)

Physical length of the agarose gel

Polarity of the electric field Porosity of gel (% (w/v) agarose) Tries 0/3

5.) Ethidium bromide stains DNA by intercalating between adjacent base pairs, resulting in an increase in fluorescence, and can be used to quantify the mass of DNA present in a band of DNA on an agarose gel. Will the staining intensity of 100 ng of supercoiled plasmid DNA be the same as 100 ng of the plasmid that has been linearized by a single-site digestion with a restriction endonuclease?1 point Tries 0/1

6.) Choose (True or False) whether each of the following parameters effects the distance of migration of DNA through agarose gels. 1 point Polarity of the electric field

DNA size (molecular weight)

Length of time in electric field

Electric field strength (volts per centimeter)

Volume of sample loaded (assuming all of the sample is contained within the loading well and enters the gel)

Physical length of the agarose gel

Porosity of gel (% (w/v) agarose)

DNA shape (supercoiled vs. relaxed circular vs. linear)

2

a) You also wish to synthesize a degenerate oligonucleotide primer that could be used eventually to clone DNA that encodes your mutant protein. Design this degenerate oligo, keeping in mind the rules we talked about in class. More information about designing degenerate primers and primers in general are located on the following page.

Write out the sequence of this oligo here

b) If you synthesized the above oligonucleotide primer, how many different oligos would be present in the mixture?

RULES TO FOLLOW for creating primers (Modified from the Thermo Fisher web site):

Designing degenerate primers:

Write out your amino acid sequence, label amino and carboxy termini

For each amino acid, use the genetic code to predict the various nucleic acid codons that can encode this amino acid. For all but Met, you should have redundancy present.

Now you need think about 5’ and 3’ considerations, and keep in mind that DNA within genes is double-stranded. It is helpful to keep in mind that the 5’ end of a gene corresponds to the amino terminus of a protein. Using 5’ and 3’ labels write out a single strand DNA sequence corresponding to the your amino seq of interest. You may choose to start with only possible codon for each amino acid, or you can do all possibilities using the notation I showed you in class to account for the wobble position of codons.

Now make your DNA double stranded by filling in the opposite DNA strand; label the 5’ and 3’ ends of this second strand and make sure you have an anti-parallel arrangement of DNA strands.

If you haven’t already, now consider the redundancy present on both strands, and make sure you have indicated sequences that account for all possible redundancies.

Realize that when a researched synthesizes a degenerate oligo that the final synthesis contains oligos with every substitution possible. The amount of degeneracy is defined by the number of different primer combinations in the mix. You can determine the degree of degeneracy by multiplying the number of changes present at each position together. If a position has no degeneracy then you multiple by 1 for that position.

Keep in mind that the trade-off between primer specificity and efficiency can be modified by altering the degeneracy of the primer. For example, the more degenerate the primers, the less specific annealing will be; however, decreased degeneracy will allow more potential to identify unknown variants. Anything over 1024 fold degeneracy is at the upper limit of potential usefulness.

Designing Primers:

Good primer design is essential for a successful PCR reaction. There are many factors to take into account when designing the optimal primers for your gene of interest. This information is taken from the Fisher Scientific website and are used when actually designing an experiment. Here are some tips to consider when designing primers.

In general, a length of 18 nucleotides is the minimum necessary for efficient primer binding.

Try to make the melting temperature (Tmm) of the primers between 65°C and 75°C, and within 5°C of each other.

If the Tmm of your primer is very low, try to find a sequence with more GC content, or extend the length of the primer a little.

Aim for the GC content to be between 40 and 60%, with the 3' of a primer ending in C or G to promote binding. This is called a GC clamp.

Typically, 3 to 4 nucleotides are added 5’ of the restriction enzyme site in the primer to allow for efficient cutting. Restriction enzymes are ENDO nucleases, so they cut better when their recognition site is not on the end of DNA.

Try to avoid regions of secondary structure, and have a balanced distribution of GC-rich and AT-rich domains.

Avoid intra-primer homology (more than 3 bases that complement within the primer) or inter-primer homology (forward and reverse primers having complementary sequences). These circumstances can lead to self-dimers or primer-dimers instead of annealing to the desired DNA sequences.


1. In addition to identifying the genetic material, the experiments of Avery, MacLeod, and McCarty with different strains of Streptococcus pneumoniae demonstrated that
DNA is a double helix.
DNA may be taken up by bacterial cells and be active.
DNA is present in bacteriophage T2.
eukaryotic cells may be genetically transformed.
DNA binds to the cell wall of the bacterium.
2. In order to show that DNA in cell extracts is responsible for genetic transformation in Streptococcus pneumoniae, important corroborating evidence should indicate that _______ also destroy transforming activity.
temperatures high enough to kill cells
enzymes that hydrolyze proteins
enzymes that hydrolyze DNA
enzymes that hydrolyze polysaccharides
enzymes that hydrolyze RNA
3. Based on what you have learned about the experiments conducted by Griffith and Avery and colleagues with bacteria, which of the following would result in transformation of living R cells?
Pure RNA from S cells
Cell-membrane extracts from S cells
A mixture of pure RNA and protein from S cells
Cell-free extracts from S cells
Pure protein from S cells
4. A-T base pairs in a DNA double helix
form three hydrogen bonds with each other.
are more heat-stable than G-C base pairs.
are of a substantially different length than G-C base pairs.
are not accessible to DNA binding proteins.
are chemically distinct from G-C base pairs.
5. If 23 percent of the bases in a sample of double-stranded DNA are adenine, what percentage of the bases are uracil?
27
50
0
73
23
6. The uniform diameter of the DNA structure provides evidence for
antiparallel DNA strands.
a right-handed helix.
a double-stranded helix.
3′ end phosphorylation.
purine–pyrimidine pairing.
7. If a sequence of one strand of DNA is 5′-TGACTATC-3′, what is the complementary strand?
5′-ACTGATAG-3′
3′-TGACTATC-5′
3′-ACTGATAG-5′
3′-ACTGATAG-3′
5′-TGACTATC-3′
8. What structural aspect of the DNA facilitates dissociation of the two DNA strands for replication?
3′ end phosphorylation
Purine–pyrimidine pairing
Antiparallel DNA strands
Covalent bonds between sugars and phosphate groups
Hydrogen bonding between paired bases
9. If the Meselson–Stahl density gradient experiment had resulted in two bands of DNA molecules after only one round of replication, one containing only 15N and the second only 14N, this result would have indicated that replication was
unidirectional.
dispersive.
conservative.
bidirectional.

semiconservative.
10. The nucleoside analogue acyclovir, which is used to treat herpes simplex virus (HSV) infections, lacks a 3′ hydroxyl group (–OH). Predict what will happen if the host cell DNA polymerase incorporates a molecule of acyclovir into an elongating strand of HSV DNA.
The replication complex will no longer bind to origins of replication.
The DNA polymerase will only incorporate acyclovir molecules.
Acyclovir will bind single-strand binding proteins.
DNA polymerase will not be able to link a successive nucleotide.
Acyclovir will block the activity of the DNA helicase.
11. Which of the following does not demonstrate the stability of the DNA double helix?
Single-strand binding proteins are required to keep the strands apart.
Paired bases form hydrogen bonds.
High temperatures are required to denature it.
DNA synthesis requires energy.
DNA helicases require ATP.
12. What effect would a primase inhibitor have on DNA replication?
DNA polymerase would not be able to add bases to the DNA.
Primase would be blocked from joining the DNA replication complex.
Primase would not be able to provide primers for DNA polymerases.
DNA polymerase would not be able to use DNA as a template.
There would be no effect on DNA replication.
13. To replicate their DNA in a timely manner, most eukaryotic chromosomes
normally have uncoiled DNA.
have multiple origins of replication.
contain linear molecules of single-stranded DNA.
have two replication forks that move in the same direction.
are composed entirely of DNA.
14. Which statement about DNA replication is false?
Okazaki fragments are synthesized as parts of the leading strand.
Error rates for DNA replication are reduced by proofreading of the DNA polymerase.
The sliding clamp increases the rate of DNA synthesis.
Replication forks represent areas of active DNA synthesis on the chromosomes.
Ligases and polymerases function in the vicinity of replication forks.
15. In many eukaryotes, there are repetitive sequences called telomeres at the ends of chromosomes. After successive rounds of DNA replication, the _______ strand becomes shorter. In some cells, an enzyme called _______ repairs the shortened strand.
leading; telomerase
lagging; telomerase
lagging; DNA polymerase I
leading; DNA polymerase I
leading; replicase
16. A researcher studies normal human fibroblast cells. They can be maintained in culture but die off after about 30 cell generations. Unexpectedly, a colony of cells continues to survive and divide past 30 generations. Which scenario is most likely true for these cells?
DNA polymerase is constantly active.
Primase is able to extend the telomeres.
The mismatch repair system is extra efficient.
The telomerase gene is being expressed.
The cells do not need the ends of their chromosomes.
17. If DNA polymerase III introduces an incorrect nucleotide, what is the first corrective action made by the DNA repair system?
The replication complex excises the incorrect nucleotide.
The excision repair proteins remove the incorrect nucleotide.
The mismatch repair system scans for base-pairing mismatches.
There is no correction, because nucleotide errors by DNA polymerases are not repaired.
DNA ligase connects the correct nucleotide.
18. Choose the correct order of the following four events in the excision repair of DNA:
(1) Base-paired DNA is made complementary to the template.
(2) Damaged bases are recognized.
(3) DNA ligase seals the new strand to existing DNA.
(4) Part of a single strand is excised.
1, 2, 3, 4
2, 1, 3, 4
3, 4, 2, 1
2, 4, 1, 3
4, 2, 3, 1
19. Six complete cycles of PCR should result in a _______-fold increase in the amount of DNA.
64
32
12
128
6
20. When double-stranded DNA is heated to temperatures above 90°C, it denatures. Denaturation is a process that
breaks covalent bonds.
hydrolyzes DNA.
separates double-stranded DNA into two single strands.
causes new hydrogen bonds to form.
irreversibly destroys the double helical structure.

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