BIOL 110H Lecture Notes - Lecture 3: Plasmid, Restriction Enzyme, Methyl Group

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maroonfox199

28 Aug 2015

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Bio110h- lecture 3- prokaryotic cells: cellular and genetic organization. Negative- loss of function; cell would die. Benefit of sexual reproduction- if there"s a problem on one strand of dna, it may be fixed by using the other strand. Each time a mutation is produced in the dna, it"s stuck and more mutations can be introduced. In eukaryotes, genomic errors can be corrected via recombination of maternal and paternal genes. In bacteria, these errors can be fixed if it accepts dna around it into its plasmid and transforms. If there"s no origin of replication, the dna circles can"t replicate. Restriction enzymes- enzymes that recognize certain sequences of dna and cut the. ( ) = the frequency of a restriction enzyme cutting along a random genome. There can be modification (ex. introducing a methyl group) A bacteriophage infects bacteria with its dna and kills it (transduction) Sometimes host dna can enter the head of the bacteriophage (generalized.

Related Questions

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.

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%

peachmarten804

Need help with evolution questions:

1.) A haplotype is best defined as the ________________.

	haploid genotypes of all the gametes produced by a diploid individual

ABO blood type conferred by an individual gamete

genotype of either the paternal or maternal chromosomal complement

multilocus genotype of a chromosome or gamete

2.) Which of the following statements regarding linkage disequilibrium is true?

Exists when D is less than zero, but not when it is greater than zero.

Is reduced by sexual reproduction.

Is increased by crossing-over during meiosis.

Is increased by any random sampling error that happens to create or destroy certain chromosome genotypes but not others.

Is reduced by selection that favors certain combinations of genotypes.

3.) Selection on multilocus genotypes in random-mating populations leads to linkage disequilibrium when _______________.

it reduces mutation rates

it eliminates all haplotypes from the population

it prevents crossing-over during meiosis

some allele combinations confer greater fitness than do others

4.) When two loci are in linkage equilibrium, the allele carried on a chromosome at one locus is independent of the allele carried on the chromosome at another locus. When this is the case, _________________.

allele frequencies will not change over time

the two loci (or, more specifically, their protein products) do not interact

the frequency of each haplotype can be calculated by multiplying the frequency of its two alleles

all possible haplotypes occur at equal frequency

5.) Which of the following accurately characterize(s) Muller's ratchet model for the selective advantage of sexual reproduction?

Drift establishes linkage disequilibrium because multilocus genotypes with few mutations are eliminated by chance events.

In populations without sexual reproduction, deleterious mutations accumulate, eventually imposing such a burden that the population goes extinct.

Sex is favored selectively because it recreates the zero-mutation genotypes lost due to drift.

All of the above correctly characterize Muller's ratchet model for the evolution of sexual reproduction.

6.) The genetic signature of recent positive selection on a locus is _______________.

high frequency and low linkage disequilibrium

low frequency and low linkage disequilibrium

low frequency and high linkage disequilibrium

high frequency and high linkage disequilibrium

7.) Sexual reproduction is advantageous both in changing environments and in constant environments, compared to asexual reproduction.

True

False

8.) Sexual reproduction causes genetic recombination via crossing-over, which combines genotypes from different parents.

True

False

9.) One reason sex may be maintained in natural populations is proposed by the "Red Queen Hypothesis", which states that.....

genotypes created by sex are created by natural selection over and over

recombination creating novel genotypes that can resist infection by parasites

mutation load is reduced by novel adaptive mutations

the ability of sex to fix multiple

Question 1

1 pts

Cystic fibrosis is caused by nonsense and missense mutations in the CFTR gene, which encodes for a chloride channel. You are studying cystic fibrosis patients to determine what mutation they possess in the CFTR gene. The difference between the mutant and wild type CFTR genes can be uncovered by examining the CFTR:

	DNA
	mRNA
	protein
	tRNA

Question 2

1 pts

	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.

Question 3

1 pts

You would like to ensure that this experiment (to determine whether patients have a specific CFTR gene truncation using hybridization) is properly controlled. Which of the following samples must you test?

	The genomic DNA of a healthy individual who does not have cystic fibrosis.
	The genome of a CFTR patient known to have the specific truncation you are trying to identify.
	The genome of a CFTR patient with a missense mutation but full length gene.
	The genome of a healthy individual married to a CFTR patient with the specific truncation you are trying to identify.
	The genome of a patient with muscular dystrophy, which can be due to a trucation in the dystrophin gene.

Question 4

1 pts

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.

Question 5

1 pts

Which of the following will lower the Tm of a given DNA strand?

	Increasing the percentage of GC base pairs.
	Raising the pH of the solution from neutral to basic.
	Decreasing the buffer concentration from 50mM NaCl to 5mM NaCl.
	None of the above.

Question 6

1 pts

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.

Question 7

1 pts

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.

Question 8

1 pts

Griffith and Avery's transformation experiments allowed us to identify that DNA is our genetic information. Which of the following scenarios would result in bacterial cells that are capable of killing mice upon injection?

	Heat killed non-virulent bacteria is added to a live virulent bacteria strain.
	Heat killed virulent bacteria is added to a heat killed non-virulent bacteria strain.
	A heat killed virulent bacteria that is treated with a nuclease, is then added to a non-virulent bacteria strain.
	Heat killed mouse cells are added to a non-virulent bacteria.

Question 9

1 pts

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.

Question 10

1 pts

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

tealcockroach755

BIOL 110H Lecture Notes - Lecture 3: Plasmid, Restriction Enzyme, Methyl Group

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