BIO 315H Lecture Notes - Lecture 3: Dna Replication, Hydrogen Bond, Wild Type

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peachlion181

22 Sep 2017

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If you break phosphodiester bonds you get floating nucleosides. Hydrogen bonding between a and t and between g and c is more stable than mismatched combinations. Active site of dna polymerase is unlikely to form bonds if pairs are mismatched. Dna polymerase can proofread to remove mismatched pairs. Dna polymerase backs up and digests linkages. Outline the steps by which pcr amplifies a specific region of a genome. Pcr polymerase chain reaction is a lab technique that mimics the process of dna replication to amplify specific regions of dna. Pcr is the basis of many different techniques. One prominent use of pcr is the study of genetic polymorphisms (dna changes from wild type{the most common type of the gene} ) A wild type human (or any organism) genome sequence does not exist. The genome sequences of only three people reveal over 5 million dna polymorphisms (sequence differences) Primers (specific to our gene) dntps (to make new dna strands)

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Multiple answers to each question might be possible!

You decide to identify the CFTR mutation by analyzing the genomic DNA of your patients compared to healthy individuals. You specifically are looking to see whether a specific 3' gene truncation has occurred in the patients. You will determine this using hybridization techniques with samples from healthy and CF patients. Which of the following will allow you to accomplish this?

	Using an RNA probe complementary to the region not removed by the truncation.
	Using an RNA probe complementary to the region removed by the truncation.
	Using an DNA probe complementary to the region not removed by the truncation.
	Using an DNA probe complementary to the region removed by the truncation.

To conduct the hybridization experiment, you are trying to decide between using a DNA or RNA probe. Which would be ideal to use and why?

	As both are composed of nucleic acids, using either would result in identical results.
	An RNA probe because RNA has uracil bases.
	An RNA probe because it could also be used in a translation experiment.
	A DNA probe because it is more stable than RNA.
	A DNA probe because RNA cannot bind to DNA.

One step of the Hershey/Chase experiment involved blending the virus/cell mixture before centrifugation and probing the pellet for radioactivity. Why was the blending step necessary?

	To collect the bacteria at the bottom of the tube.
	To break open the bacteria to release the genome.
	To separate the bacteria from the bacteriophages.
	To be able to detect the radioactivity.

Imagine Hershey/Chase had used an RNA virus (genome composed of RNA) instead of a DNA virus in their experiment. Would radioactivity still have been found in the pellet?

	No, because only DNA can be labeled with radioactivity.
	No, because the RNA genome would not enter the bacteria upon infection.
	No, because while DNA and RNA nucleotides are similar, they are not identical.
	Yes, because DNA and RNA nucleotides are similar.
	Yes, because genome in any form (DNA, RNA, protein) would be labeled similarly.

The human genome consists mostly of non-coding DNA. Which of the following are benefits of this?

	Random DNA mutations generally won't affect RNA and protein function.
	It is faster to duplicate the genome when these are present.
	The existence of introns can lead to multiple variations of proteins encoded by a single gene.
	It is unlikely transposons would exist in the genome if there was too much protein coding DNA.

Explain the 5â to 3â directionality of a DNA stand.

	It is due to the fact that the free 5â carbon is on one end and the free 3â carbon is on the other
	It is due to the fact that new nucleotide are added to the 5â carbon of the previous nucleotide
	It is due to the fact that there are 3 phosphate groups attached to the 5â carbon
	It is due to the fact the complementary strand is 3â to 5â
	More than one of the above explain the 5â to 3â directionality

You accidentally add a mutant dNTP (which has an H instead of an OH connected to the 3â carbon) to a reaction where DNA is being replicated. Which of the following is true of this mutant dNTP?

	It can be incorporated into DNA strand but cannot form a phosphodiester bond with an incoming wild type dNTP
	It can be incorporated into a DNA strand but cannot base pair with a complementary nucleotide
	It can be incorporated into a DNA strand and can form a phosphodiester bond with an incoming dNTP, but only if it is another mutant dNTP
	It cannot be incorporated into a DNA strand.

Why does DNA polymerase utilize an RNA primer?

	DNA polymerase is unable to initiate strand synthesis but RNA polymerase can
	DNA polymerase can only add a dNTP to an rNTP
	DNA synthesis proceeds in the 3â to 5â when initiating strand synthesis
	Chromosomal DNA contains interspersed RNA fragments
	The RNA primer increases stability of the newly synthesized strand

rosehippopotamus950

QUESTION 1

A mutation caused by exposure to gamma rays is called a spontaneous mutation

	a.	True
	b.	False
	c.	Sometimes, depending on the type of base change
	d.	Gamma rays do not cause mutations

QUESTION 2

A recessive mutation affecting an essential biochemical pathway can be detected________

	a.	when the organism is heterozygous for that mutation.
	b.	when the organism is homozygous for that mutation.
	c.	in either homozygous or heterozygous conditions.
	d.	only by sequencing (recessive mutations never show a phenotype).

QUESTION 3

During DNA replication, the newly synthesized chain grows by adding

	a.	nitrogenous bases
	b.	the sugar-phosphate backbone
	c.	nucleosides
	d.	dNTPs

QUESTION 4

During replication, the DNA strand 5â- AAGTCTAGCCTAG -3â will serve as a template for the polymerization of:

	a.	5â- CTAGGCTAGACTT -3â
	b.	5â - TTCAGATCGGATC -3â
	c.	5â - GATCCGATCTGAA -3â
	d.	3â - AAGTCTAGCCTAG -5â

QUESTIOn 5

Given the following DNA sequences, select which statement(s) is(are) correct.

Sequence 1:

5â-TGGACGCTAA-3â

3â-ACCTGCGATT-5â

Sequence 2:

5â-AATCGCAGGT-3â

3â-TTAGCGTCCA-5â

Sequence 3:

5â-ACCTGCGATT-3â

3â-TGGACGCTAA-5â

	a.	Sequences 1 and 2 are the same
	b.	Sequences 1 and 3 are the same
	c.	Sequences 2 and 3 are the same
	d.	Sequences 1, 2 and 3 are all different

QUESTION 6

Repetitive sequences in the genome are hotspots for:

	a.	Deamination
	b.	Depurination
	c.	Thymine dimer formation
	d.	Replication errors

QUESTION 7

Select which statement(s) about the Ames test is(are) correct

	a.	it has been designed to understand mutation repair systems in Salmonella typhimurium
	b.	it uses mammalian liver extract
	c.	it allows to study whether chemical compounds or their enzymatic breakdown products are mutagenic
	d.	it is based on whether a chemical compound causes reversion mutations from his+ to his-
	e.	two of the above are correct
	f.	three of the above are correct

QUESTION 8

Select which statement(s) is(are) correct.

	a.	All DNA strands have a direction, and it is specified by the carbons in the sugar backbone.
	b.	All DNA strands have a direction, and it is specified by hydrogen bonds between nucleotides.
	c.	In order to form proper base pairs in a double stranded DNA molecule, the two strands must run in opposite directions.
	d.	A and C are correct.
	e.	B and C are correct.

QUESTION 9

Studies of gene mutation frequencies have shown that:

	a.	mutations are rare, and genomes are generally stable.
	b.	mutation frequencies differ among organisms and also between genes, suggesting certain genes are more susceptible to mutation.
	c.	mutation frequencies are consistent between organisms, and each region of DNA is equally susceptible to random mutations.
	d.	Both A and B are correct.
	e.	Both A and C are correct.

QUESTION 10

The compound 5-bromodeoxyuridine (BrdU) is a derivative of uracil, and if BrdU becomes incorporated during DNA replication, it pairs with adenine. This compound is best classified as which type of mutagen?

	a.	base analog
	b.	base inducer
	c.	intercalating agent
	d.	oxidative agent
	e.	alkylating agent

QUESTION 11

The rate of mutation of the fruit fly is higher than the rate of mutation of Algae

True

False

QUESTION 12

Thymine dimers are most commonly caused by which of the following?

	a.	X-rays
	b.	Alkylating agents
	c.	U.V. irradiation
	d.	DNA intercalating agents

QUESTION 13

What chemical group is found in the 3â end of a DNA strand?

	a.	an alcohol
	b.	a hydroxyl
	c.	a methyl
	d.	a phosphate

QUESTION 14

What chemical group is found in the 5â end of a DNA strand?

	a.	an alcohol
	b.	a hydroxyl
	c.	a methyl
	d.	a phosphate

QUESTION 15

Which is the correct order of molecules binding to DNA during DNA replication?

	a.	Helicase, SSB, primase, DNA pol III, DNA pol I, ligase
	b.	SSB, DNA pol I, ligase, helicase, DNA pol I, primase
	c.	primase, helicase, DNA pol III, ligase, DNA pol I, SSB
	d.	SSB, helicase, primase, DNA pol III, DNA pol I, ligase

QUESTION 16

Which statement(s) is(are) correct about strand slippage?

	a.	It can cause disorders such as Huntingon disease.
	b.	It is a process that causes mutations altering the number of DNA repeats.
	c.	It is a process that incorporates nucleotide base analogs and trinucleotide repeats.
	d.	a and b are correct.
	e.	b and c are correct.
	f.	a, b and c are correct.

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chocolatesea-lion434

	Using an RNA probe complementary to the region not removed by the truncation.
	Using an RNA probe complementary to the region removed by the truncation.
	Using an DNA probe complementary to the region not removed by the truncation.
	Using an DNA probe complementary to the region removed by the truncation.

To conduct the hybridization experiment, you are trying to decide between using a DNA or RNA probe. Which would be ideal to use and why?

	As both are composed of nucleic acids, using either would result in identical results.
	An RNA probe because RNA has uracil bases.
	An RNA probe because it could also be used in a translation experiment.
	A DNA probe because it is more stable than RNA.
	A DNA probe because RNA cannot bind to DNA.

Imagine Hershey/Chase had used an RNA virus (genome composed of RNA) instead of a DNA virus in their experiment. Would radioactivity still have been found in the pellet?

	No, because only DNA can be labeled with radioactivity.
	No, because the RNA genome would not enter the bacteria upon infection.
	No, because while DNA and RNA nucleotides are similar, they are not identical.
	Yes, because DNA and RNA nucleotides are similar.
	Yes, because genome in any form (DNA, RNA, protein) would be labeled similarly.

The human genome consists mostly of non-coding DNA. Which of the following are benefits of this?

	Random DNA mutations generally won't affect RNA and protein function.
	It is faster to duplicate the genome when these are present.
	The existence of introns can lead to multiple variations of proteins encoded by a single gene.
	It is unlikely transposons would exist in the genome if there was too much protein coding DNA.

You accidentally add a mutant dNTP (which has an H instead of an OH connected to the 3â carbon) to a reaction where DNA is being replicated. Which of the following is true of this mutant dNTP?

	It can be incorporated into DNA strand but cannot form a phosphodiester bond with an incoming wild type dNTP
	It can be incorporated into a DNA strand but cannot base pair with a complementary nucleotide
	It can be incorporated into a DNA strand and can form a phosphodiester bond with an incoming dNTP, but only if it is another mutant dNTP
	It cannot be incorporated into a DNA strand.

Andrew Murray's sister, Andrea, is adding to her brother's work on chromosomes. She is using cells that are unable to synthesize adenine (âade) and histidine (âhis). The plasmid she is currently working with consists of an origin of replication and the Ade gene.

Following her transformation of the plasmid into her yeast, what media will the cells be plated on to select for cells that have picked up the plasmid?

	Media containing histidine
	Media containing adenine
	Media lacking adenine
	Media lacking histidine

She starts by attempting to add the centromere DNA into a plasmid containing the origin of replication. Unfortunately, when adding the centromere sequence, the origin of replication is removed, thus leaving a plasmid with only a centromere and selection marker. Following plasmid transformation, what growth result will she see on her plates?

	No colonies
	Little colonies
	Big colonies

She fixes the mistakes from the previous experiment and now has a complete plasmid (selection marker, origin of replication, centromere). She then inserts telomere sequences into the plasmid. How will this impact her transformation?

	It will not impact her transformation
	Her transformation will no longer work because plasmids donât require telomeres
	She will now see much larger colonies
	She will now see fewer but larger colonies
	She will now see smaller colonies

In the Meselson/Stahl experiment, E. coli were first grown in media containing heavy nitrogen, 15N, and then transferred to light nitrogen, 14N, at the beginning of the experiment.

Imagine that their data showed that replication occurs in a conservative manner instead of semi-conservative. What fraction of the DNA helices will consist of mixed DNA after 4 rounds of replication in this case?

	None
	More than 75%
	25-50%
	51-75%
	Less than 25%

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BIO 315H Lecture Notes - Lecture 3: Dna Replication, Hydrogen Bond, Wild Type

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