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26 Oct 2018

Consider the structure of the DNA double helix.

A. You and a friend want to split a double-stranded DNA molecule so you each have half. Is it better to cut the length of DNA in half so each person has a shorter length, or to separate the strands and each take one strand? Explain.

B. In the original 1953 publication describing the discovery of the structure of DNA, Watson and Crick wrote, âIt has not escaped our notice that the specific pairings we have postulated immediately suggest a possible copying mechanism for the genetic material.â What did they mean?

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Beverley SmithLv2

29 Oct 2018

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Related questions

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.

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

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

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.

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

The uniform diameter of the DNA structure provides evidence for

	antiparallel DNA strands.
	a right-handed helix.
	a double-stranded helix.
	3â² end phosphorylation.
	purineâpyrimidine pairing.

If a sequence of one strand of DNA is 5â²-TGACTATC-3â², what is the complementary strand?

	5â²-ACTGATAG-3â²
	3â²-TGACTATC-5â²
	3â²-ACTGATAG-5â²
	3â²-ACTGATAG-3â²
	5â²-TGACTATC-3â²

What structural aspect of the DNA facilitates dissociation of the two DNA strands for replication?

	3â² end phosphorylation
	Purineâpyrimidine pairing
	Antiparallel DNA strands
	Covalent bonds between sugars and phosphate groups
	Hydrogen bonding between paired bases

If the MeselsonâStahl 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 3â² hydroxyl group (âOH). 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 90Â°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.

magine that the year is 1928 and the scientific world is in an uproar trying to determine what component of the cell isresponsible for genetic inheritance. Is it DNA, protein, perhaps even RNA? You are busy in your lab, trying to help solve this crucial mystery that still plagues (then) modern day biology. Your work focuses on a microorganism, a bacterium named Streptococcus pneumoniae, and its effects on mice. There are two strains of S. pneumoniae (which is responsible for pneumonia), the âRoughâ strain (R) and âSmoothâ strain (S). The strains are very similar except the S strain has a smooth polysaccharide coat that enables the microorganism to avoid the organismâs immune system and cause illness (and death in mice). The R strain has a rough coat which is detected and dealt with by the immune system (and as a result causes no death in mice). You decide to perform a series of experiments using the R and S strain of S. pneumoniae involving injections of these strains in mice. Your results are documented in the figure below: living S strain of S. pneumoniae mouse dies of infection living R strain of S. pneumoniae mouse lives mouse lives S strain heat-killed living R strain mouse dies of infection living, pathogenic S strain recovered S strain heat-killed From these results, you figure out that something from the heat killed S strain is transforming the R strain into the virulent strain. What is this âTransforming principleâ? Is it related to the cellular component that was passed from one generation to the next? As you know, this was the initial experiment performed by Dr. Frederick Griffith that eventually led to the discovery of the true cellular component that is the genetic inheritance molecule as well as the âtransforming principleâ (Hint: Itâs DNA!).

1. Several experiments were required to demonstrate how traits are inherited. Which scientist or team of scientists first demonstrated that cells contain some component that can be transferred to a new population of cells and permanently cause changes in the new cells?

a. Griffith b. Watson and Crick c. Avery, MacLeod, and McCarty d. Hershey and Chase

2. From your results, which one of the cellular components can be determined NOT to be the component responsible for genetic inheritance? Why?

a. Proteins because these macromolecules become denatured with heat. b. DNA because these macromolecules form a single helix in solution. c. RNA because it's double stranded 100% of the time d. Sugars because glycolysis happens.

3. Which one of the following experimental trials below are considered to be controls for this experiment? a. S Strain grown in 10% ethanol medium. b. Living S strain; Living R strain; Heat Killed S Strain c. Living R Strain with heat killed S strain; Living S Strain d. Heat Killed S strain; Living R strain; R strain grown in 10% ethanol medium

4. Which scientist or team of scientists obtained definitive results demonstrating that DNA is the genetic molecule? a. Griffith b. Watson c. Crick d. Hershey and Chase

5. Hershey and Chase used radiolabeled macromolecules to identify the material that contains heritable information. What radioactive material was used to track DNA during this experiment?

a. 3H b. 14C c. 35S d. 32P

6. DNA molecules, like proteins, consist of a single, long polymeric chain that is assembled from small monomeric subunits.

a. True b. False

7. The polarity of a DNA strand results from the polarity of the nucleotide subunits.

a. True b. False

8. There are five different nucleotides that become incorporates into a DNA strand.

a. True b. False

9. Hydrogen bonds between each nucleotide hold individual DNA strands together.

a. True b. False

brownpanda894

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 5â and 3â considerations, and keep in mind that DNA within genes is double-stranded. It is helpful to keep in mind that the 5â end of a gene corresponds to the amino terminus of a protein. Using 5â and 3â 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 5â and 3â ends of this second strand and make sure you have an anti-parallel arrangement of DNA strands.

If you havenât 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 65Â°C and 75Â°C, and within 5Â°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 5â 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.