The Structural Basis of Cellular Information: DNA, Chromosomes, and the Nucleus

What was the main conclusion that could be drawn from Frederick Griffith's key experiment that utilized R (rough) and S (smooth) bacteria?

DNA and/or RNA was likely the genetic material.

A "transforming principle" was present in the mixture of heat-killed S bacteria and living R bacteria.

DNA was likely the genetic material.

S (smooth) bacteria were toxic to mice.

A "transforming principle" was present in the mixture of heat-killed R bacteria and living S bacteria.

A "transforming principle" was present in the mixture of heat-killed S bacteria and living R bacteria.

Why did living S (smooth) bacteria cause the mice to die when they were injected into mice, while the living R (rough) bacteria did not?

S bacteria produce a deadly toxin that the R strain does not.

S bacteria have a capsule that protects them from the host immune system, which the R strain lacks.

S bacteria induce auto-antibodies that the R strain does not.

S bacteria are more likely than the R strain to infect the lungs.

S bacteria cause a more active immune response than the R strain causes.

S bacteria have a capsule that protects them from the host immune system, which the R strain lacks.

Frederick Griffith used a series of bacterial mixtures in order to interrogate the source of heredity. Various cultures containing R (rough) bacteria and/or S (smooth) bacteria were injected into mice. Sort the mixtures below based on whether the mice would be expected to live or die after injection.

heat-killed R bacteria 

Mice live

Frederick Griffith used a series of bacterial mixtures in order to interrogate the source of heredity. Various cultures containing R (rough) bacteria and/or S (smooth) bacteria were injected into mice. Sort the mixtures below based on whether the mice would be expected to live or die after injection.

living S bacteria 

Mice dead

Frederick Griffith used a series of bacterial mixtures in order to interrogate the source of heredity. Various cultures containing R (rough) bacteria and/or S (smooth) bacteria were injected into mice. Sort the mixtures below based on whether the mice would be expected to live or die after injection.

heat-killed R bacteria mixed with living S bacteria 

Mice dead 

Frederick Griffith used a series of bacterial mixtures in order to interrogate the source of heredity. Various cultures containing R (rough) bacteria and/or S (smooth) bacteria were injected into mice. Sort the mixtures below based on whether the mice would be expected to live or die after injection

living R bacteria 

Mice living

Frederick Griffith used a series of bacterial mixtures in order to interrogate the source of heredity. Various cultures containing R (rough) bacteria and/or S (smooth) bacteria were injected into mice. Sort the mixtures below based on whether the mice would be expected to live or die after injection.

heat-killed S bacteria mixed with living R bacteria 

Mice dead 

Frederick Griffith used a series of bacterial mixtures in order to interrogate the source of heredity. Various cultures containing R (rough) bacteria and/or S (smooth) bacteria were injected into mice. Sort the mixtures below based on whether the mice would be expected to live or die after injection.

heat-killed R bacteria

Mice living

What was the main conclusion that could be drawn from the key experiment done by Oswald Avery and colleagues?

DNA and/or RNA was likely the genetic material.

DNase treatment destroys DNA.

It was impossible to determine what the genetic material was.

A "transforming principle" was present in the mixture of heat-killed S bacteria and living R bacteria.

DNA was likely the genetic material.

DNA was likely the genetic material

What would the predicted result be if Griffith injected mice with a mixture of heat-killed R bacteria and living S bacteria?

All mice would live, but only if DNAse was added to the mixture prior to injection.

All mice would die, but only if DNAse was added to the mixture prior to injection.

All mice would live.

All mice would die.

All mice would die.

The lytic cycle of bacteriophage infection ends with the _____.    

replication of viral DNA

entry of the phage protein coat into the host cell

assembly of viral particles into phages

the injection of phage DNA into a bacterium

rupture of the bacterium

rupture of the bacterium  

As a result of the lytic cycle, _____.    

the host cell is not destroyed

the host cell's DNA is destroyed

viral ribosomes are produced

viral DNA is incorporated into host cell DNA

a prophage is created

the host cell's DNA is destroyed

Who demonstrated that DNA is the genetic material of the T2 phage?    

Hershey and Chase

Meselson and Stahl

Darwin and Wallace

Franklin

Watson and Crick

Hershey and Chase

The radioactive isotope ³²P labels the T2 phage's _____.    

DNA

tail

base plate

head

protein coat

DNA        

Hershey and Chase used _____ to radioactively label the T2 phage's proteins.    

222Ra

32P

35S

92U

14C

³⁵S      

After allowing phages grown with bacteria in a medium that contained ³²P and ³⁵S, Hershey and Chase used a centrifuge to separate the phage ghosts from the infected cell. They then examined the infected cells and found that they contained _____, which demonstrated that _____ is the phage's genetic material.    

labeled protein .... protein

labeled DNA .... protein

labeled DNA ... DNA

labeled DNA ... labeled protein

labeled protein ... DNA

labeled DNA ... DNA    

In an experiment like that of the 1952 Hershey and Chase experiments, ³⁵S was added to a phage replicating within its bacterial host. The new phage particles were carefully isolated and used to infect fresh bacterial cells in the absence of any radioisotopes. Where would you expect to find the ³⁵S radioisotope immediately after infection?

inside the bacterial cells, but separate from the bacterial proteins

incorporated into the bacterial DNA

in the phage ghosts outside the bacterial cells

incorporated into the bacterial matrix proteins

inside the bacterial cells, but separate from the bacterial DNA

in the phage ghosts outside the bacterial cells

Avery, MacLeod, and McCarty (1944) demonstrated that DNA was the genetic material by treating heat-killed Streptococcus pneumoniae S (smooth) strain with DNase, thereby preventing transfer of the "transforming substance" from the killed S strain to the live R (rough) strain when the two were mixed together and injected into mice. Another way that one could demonstrate transformation in bacteria would be to extract DNA from

both S and R strains and mix it to allow recombination to take place.

an R strain and mix it with cells of an S strain.

an ampicillin-sensitive strain and mix it with cells of an ampicillin-resistant strain.

an S strain and mix it with cells of an R strain.

None of the above are correct.

an S strain and mix it with cells of an R strain.

True or False:

Viruses contain DNA, proteins, and RNA

True

True or False:

Viruses contain ribosomes

False

Virus components

capsid, capsomere (core protein), and envelope with glycoproteins 

True or False:

Both the lytic and lysogenic cycle have viral genes replicated

True 

Lysogenic cycle components

1.cell reproduces normally 2.the viral DNA integrates into the chromosome of the host cell

Lytic cycle components

1.New phages are assembled from viral DNA and proteins 2.the cell is lysed (broken apart) 3.the host is destroyed

Which statements about viruses are true?

HIV contains reverse transcriptase.

A retrovirus contains RNA.

The capsid enters the host cell if the virus is enveloped.

All RNA-containing viruses are retroviruses.

Some viruses contain DNA and RNA.

Enveloped viruses bud from the host cell.

HIV contains two identical strands of DNA.

1.HIV contains reverse transcriptase. 2.A retrovirus contains RNA. 3.The capsid enters the host cell if the virus is enveloped 4. Enveloped viruses bud from the host cell.

Which of the following accurately identifies the chemical basis of Chargaff's rule?

the presence of deoxyribose instead of ribose

the ionic interactions of the negatively charged phosphates with positively charged histone proteins

the covalent bonding of the sugar-phosphate backbone

the hydrogen bonding properties of the DNA bases

the hydrogen bonding properties of the DNA bases

A researcher heats a solution of DNA and notices that as the temperature increases, the absorbance of ultraviolet light at 260 nm increases dramatically. This is because __________.

DNA gyrase relaxes supercoiling of the DNA

the double strands of DNA are separating

the DNA becomes supercoiled

there are more base pairs synthesized

the double strands of DNA are separating

One strand of a region of DNA has the sequence 5’-ATTCCG-3’. The complementary strand for this one is __________.

5’-ATTCCG-3’

5’-ACCTTA-3’

5’-CGGAAT-3’

5’-TAAGGC-3’

5’-CGGAAT-3’

Which of the following are components of Chargaff's rules of bases?

% purines = % pyrimidines

%A = %T

%C + %T = %A + %G

%G = %C

All of these are true.

All of these are true.

You are given two samples of DNA, each of which melts at 92∘C during thermal denaturation. After denaturing the DNA, you mix the two samples together and then cool the mixture to allow the DNA strands to reassociate. When the newly reassociated DNA is denatured a second time, the sample now melts at 85∘C.

How might you explain the lowering of the melting temperature from 92∘C to 85∘C?

During denaturation, strands could be damaged and would not be able to associate completely. Because the resulting hybrids would not be complementary, you would get a lower melting temperature.

Strands from one sample might destabilize strands from the other sample so that some bases would dissociate to monomers. Because there would be a lower number of the resulting hybrids, you would get a lower melting temperature.

Strands from one sample might hybridize to strands from the other not identical sample. Because the resulting hybrids would not be exactly complementary, you would get a lower melting temperature.

Since the strands could orientate differently to each other when mixed and the resulting hybrids would not be complementary, you would get a lower melting temperature.

Strands from one sample might hybridize to strands from the other not identical sample. Because the resulting hybrids would not be exactly complementary, you would get a lower melting temperature.

If the newly reassociated DNA had melted at 92∘C instead of 85∘C, what conclusions might you have drawn concerning the base sequences of the two initial DNA samples?

Select all that apply.

The two initial samples were related in a sequence.

The two initial samples of DNA might have been identical.

The relative proportion of GC base pairs in both samples was similar.

The two initial samples may have been totally unrelated to each other in the nucleotide sequence.

1.The two initial samples of DNA might have been identical. 2.The relative proportion of GC base pairs in both samples was similar. 3.The two initial samples may have been totally unrelated to each other in the nucleotide sequence

DNA isolated from Aspergillus has an adenine content of 25%. Based upon this information, what is the %G + %C within the Aspergillus DNA?

0%

25%

50%

75%

The answer cannot be determined from this information.

50%

Which of the following shows Chargaff's equivalence?

double-stranded DNA

RNA

single-stranded DNA

histones

deoxyribose

double-stranded DNA

Which of the following statements regarding the Watson - Crick model of DNA is false?

The bases are "stacked" in the center of the molecule.

The backbone consists of alternating sugars and phosphates.

The two DNA strands form a left-handed helix.

It is the most common biological form of DNA.

The two DNA strands are antiparallel.

The two DNA strands form a left-handed helix

DNA is different from RNA in that

RNA is made up five bases, whereas DNA is made up of four.

in general, RNA molecules are longer than DNA molecules.

RNA contains an additional oxygen atom on the ribose sugar.

RNA cannot exist as a double helix.

All of these statements are true.

RNA contains an additional oxygen atom on the ribose sugar.

Circular DNA molecules found in nature are

not supercoiled, due to their small size.

positively supercoiled.

always oriented in a triple helix to conserve space.

negatively supercoiled.

always in the Z-DNA formation.

negatively supercoiled.

Based on their chemical properties, what groups of amino acids would you least expect to find within the histone tails?

nonpolar amino acids

hydrophobic amino acids

polar, uncharged amino acids

hydrophilic amino acids

acidic amino acids

basic amino acids

1.nonpolar amino acids 2.hydrophobic amino acids 3.acidic amino acids

Which of the following is the correct order of the levels of DNA packaging in eukaryotic chromosomes?

nucleosome → chromatin fiber → looped domains → heterochromatin

chromatin fiber → heterochromatin → nucleosome → looped domains

heterochromatin → nucleosome → chromatin fiber → looped domains

nucleosome → looped domains → chromatin fiber → heterochromatin

chromatin fiber → nucleosome → looped domains → heterochromati

nucleosome → chromatin fiber → looped domains → heterochromatin

Which of the following statements is true regarding mitochondrial and chloroplast genomes?

They only encode for ribosomal and transfer RNAs.

They are able to replicate independently of the nuclear DNA.

They contain all the genes necessary for aerobic respiration and photosynthesis, respectively.

They contain mostly interspersed and tandemly repeated DNA.

They are degenerate bacterial genomes that don't contain important protein-coding genes.

They are able to replicate independently of the nuclear DNA

Packaging of prokaryotic chromosomes

is essentially the same as eukaryotic DNA packaging.

doesn't involve histone or histone-like proteins.

involves large regions of positively supercoiled DNA but lacks histones or histone-like proteins.

is very simple and lacks higher-level structure, such as looped domains.

involves structural RNA molecules that contribute to the formation of looped domains.

involves structural RNA molecules that contribute to the formation of looped domains.

Which type of DNA makes up the largest portion of the human genome?

unique noncoding DNA

tandemly repeated DNA

interspersed repeated DNA

introns

exons

interspersed repeated DNA

A student is working in a DNA analysis laboratory and performs numerous DNA isolations from eukaryotic cells. However, the student realizes at the end of the day that he forgot to label whether or not the DNA samples he isolated were from nuclei or mitochondria. Which of the following would NOT be a strategy the student could use to determine the origin of his DNA isolates?

The student could compare whether the DNA in each sample was circular or linear.

The student could determine whether histones were present or not in each sample.

The student could measure the average size of the DNA isolated in each sample.

The student could determine if the sample absorbed light of 260 nm.

The student could determine if the sample absorbed light of 260 nm.

Alteration of histones by addition of methyl and acetyl groups to particular amino acids

changes B-DNA to Z-DNA.

helps preserve telomeres during DNA replication.

creates G bands seen after Giemsa staining of chromosomes.

results in the remodeling of chromatin that can activate or inhibit gene expression.

alters the denaturation and renaturation of DNA.

results in the remodeling of chromatin that can activate or inhibit gene expression.

Fluorescent in situ hybridization (FISH) would be helpful in determining

the position of particular chromosomes or sequences inside the nucleus.

the location of B-DNA and Z-DNA.

the position and structure of nucleosomes.

the base composition of a DNA sample.

the number of nuclear pores on the nuclear envelope

the position of particular chromosomes or sequences inside the nucleus.

Based on your knowledge of glycolysis and energy metabolism, would you expect to find a nuclear localization signal (NLS) within any protein encoded in the mitochondrial genome? Why or why not?

You would expect some of them to contain an NLS. Most of the genes in the mitochondrial genome are involved in the production of ATP in the nucleus, so these proteins should have an NLS to traffic into the nucleus.

You would not expect any of them to contain an NLS. Most mitochondrial-encoded proteins function in the endoplasmic reticulum, so an NLS would mistakenly route them to the nucleus.

You would expect some of them to contain an NLS. Several genes in the mitochondrial genome are involved in the transcription of nuclear genes, so these proteins should have an NLS to traffic into the nucleus.

You would not expect any of them to contain an NLS. Most mitochondrial-encoded proteins function in mitochondria, so an NLS would mistakenly route them to the nucleus.

There is no way to predict whether any mitochondrial-encoded genes would contain an NLS.

You would not expect any of them to contain an NLS. Most mitochondrial-encoded proteins function in mitochondria, so an NLS would mistakenly route them to the nucleus.

Which of these statements is NOT true of the nucleolus?

It is connected to the nucleus via nuclear pores.

It synthesizes rRNA.

It contains a nucleolus organizer region.

It contains fibrils of rDNA.

It is connected to the nucleus via nuclear pores.

Sucrose crosses the nuclear envelope so rapidly that its rate of movement cannot be accurately measured.

Unlike most membranes, the nuclear envelope appears to be freely permeable to a polar organic molecule.

The nuclear envelope has pores that are designed to transport sucrose molecules so that they can cross it without any resistance.

Sucrose can dissolve the nuclear envelope because of its polarity. After sucrose molecules have passed, nuclear envelope goes back to its initial state.

There is a small concentration difference between two sides of the nuclear envelope that forces sucrose to move very fast.

Unlike most membranes, the nuclear envelope appears to be freely permeable to a polar organic molecule.

Colloidal gold particles with a diameter of 5.5 nm equilibrate rapidly between the nucleus and cytosol when injected into an amoeba, but gold particles with a diameter of 15 nm do not.

Large gold particles are very slow and unwieldy, preventing them from passing through polar membranes

The aqueous channels in nuclear pore complexes have diameters of at least 5.5 nm, but not as great as 15 nm.

The aqueous channels in nuclear pore complexes have diameters of about 15-20 nm, but only particles with a diameter of one third could pass through them.

Large gold particles have a larger charge which prevents them from passing through polar membrane.

The aqueous channels in nuclear pore complexes have diameters of at least 5.5 nm, but not as great as 15 nm.

Nuclear pore complexes sometimes stain heavily for RNA and protein.

The stained pores contain complexes of RNA and protein, probably ribonucleo-protein particles caught in transit.

The stained pores contain complexes of RNA and protein, probably ribonucleo-protein particles that are part of the nuclear pore complex.

The stained pores contain complexes of RNA and protein, probably ribonucleo-protein particles attached to the membrane by hydrogen bonds.

The stained pores contain complexes of RNA and protein, probably ribonucleo-protein particles caught in transit

If gold particles up to 26 nm in diameter are coated with a polypeptide containing a nuclear localization signal (NLS) and are then injected into the cytosol of a living cell, they are transported into the nucleus. If they are injected into the nucleus, however, they remain there.

The NLS only triggers transport from cytosol to nucleus, not from nucleus to cytosol, suggesting that gold particles have a large charge that prevents their moving away from nucleus.

The NLS only triggers transport from nucleus to cytosol, not from cytosol to nucleus, suggesting that NLS receptor proteins (importins) function only in the nucleus.

The NLS only triggers transport from cytosol to nucleus, not from nucleus to cytosol, suggesting that NLS receptor proteins (importins) function only in the cytosol.

The NLS only triggers transport from cytosol to nucleus, not from nucleus to cytosol, suggesting that NLS receptor proteins (importins) are attached to the outer side of the nucleus.

The NLS only triggers transport from cytosol to nucleus, not from nucleus to cytosol, suggesting that NLS receptor proteins (importins) function only in the cytosol.

Many of the proteins of the nuclear envelope appear from electrophoretic analysis to be the same as those found in the endoplasmic reticulum.

The nuclear membranes are changed during electrophoretic analysis so that they become similar to those of  the endoplasmic reticulum.

The proteins in the nuclear envelope that are probably caught in transit are the same as those found in the endoplasmic reticulum..

The nuclear membranes and the endoplasmic reticulum are likely to have a common origin.

The nuclear membranes and the endoplasmic reticulum are likely to have a common origin.

Ribosomal proteins are synthesized in the cytosol but are packaged with rRNA into ribosomal subunits in the nucleus.

Ribosomal proteins must pass inward from the cytosol to the nucleus at a rate adequate to sustain ribosomal subunit assembly; ribosomal subunits must move outward from the nucleus to the cytosol at a rate commensurate with their rate of assembly.

Ribosomal proteins must pass outward from the nucleus to the cytosol to the nucleus at a rate adequate to sustain ribosomal subunit assembly; ribosomal subunits must move inward from the cytosol to the cytosol at a rate commensurate with their rate of assembly.

Ribosomal proteins must pass inward from the cytosol to the nucleus at a rate adequate to prevent proteins from dissolving; ribosomal subunits must move outward from the nucleus to the cytosol at a rate commensurate with their rate of dissolving.

Ribosomal proteins must pass inward from the cytosol to the nucleus at a rate adequate to sustain ribosomal subunit assembly; ribosomal subunits must move outward from the nucleus to the cytosol at a rate commensurate with their rate of assembly.

Treatment of nuclei with the nonionic detergent Triton X-100 dissolves the nuclear envelope but leaves an otherwise intact nucleus.

The nucleus seems to produce a new insoluble envelope following the dissolution of the previous one.

While dissolving the nuclear envelope, Triton X-100 produces a new envelope that protects the nucleus.

The integrity of the nucleus does not seem to depend entirely on the envelope.

The integrity of the nucleus does not seem to depend entirely on the envelope.

The proteins that compose the nuclear matrix and are implicated in a number of degenerative diseases are called ________.

lamins

In the Ran/importin pathway ________ provides the energy for transport of proteins into the nucleus.

GTP hydrolysis

Each of the following structures is found within the nucleus of a eukaryotic cell except

plasmids.

the nucleolus.

transfer RNA.

histones.

All of these are found in the nucleus.

plasmids.

You are using genetic engineering to design a protein that needs to be synthesized in the cytoplasm and then localized in the nucleus of the cell. To accomplish this, you could add

a nucleotide or few nucleotides to the DNA sequence for the protein.

the protein during mitosis, at a time when the nuclear membrane is dispersed.

inverted repeats to the DNA sequence for the protein.

a series of specific polysaccharides to the carboxyterminal end of the protein.

a DNA sequence for a nuclear localization signal into the DNA that will be present in the mature protein.

a DNA sequence for a nuclear localization signal into the DNA that will be present in the mature protein.