Lecture 9 / Chapters 6 and 10 - Bacteria and DNA Discovery Flashcards Preview

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1

In an interrupted mating experiment, the purpose of plating cells on a selective medium is _______.

-to ensure that only recombinant genotypes are recovered
-to determine the genes present in the Hfr
-to eliminate all recipient cells
-to ensure that conjugation has been completed

To ensure that only recombinant genotypes are recovered.

(Selective media in these experiments allow the elimination of parental genotypes and recovery of only those whose genotypes result from the ​transfer of donor genes.)

2

Mapping bacterial genes by conjugation is based on which of the following assumptions?

-F minus bacterial strains are capable of transferring genes to Hfr strains
-Genes are transferred from donor to recipient in a linear fashion
-Bacterial cells are capable of growing in the presence of sodium azide
-Conjugation is not disrupted when a bacterial culture is mixed in a blender

Genes are transferred from donor to recipient in a linear fashion.

(If this were not true, the distance between genes could not be measured as a function of time.)

3

Which of the following statements about mapping bacterial genes by conjugation is NOT true?

-Two genes that are very close together may appear to be transferred at the same time
-The closer a gene to the Hfr origin, the more likely it will be transferred to the recipient during conjugation
-For any two genes transferred from donor to recipient, all genes residing between them have also been transferred
-It is necessary that all Hfr cells be absent from the population of cells recovered for genotyping

The closer a gene to the Hfr origin, the more likely it will be transferred to the recipient during conjugation.

(This is false - it depends on which side of the origin the gene resides.)

4

What is a form of recombination in bacteria that involves the F plasmid?

Conjugation.

5

True or false?
A bacterial strain that is pro + thi + leu − met − will grow on minimal media plus leucine and thiamine.

False.

(The strain is auxotrophic for leucine and methionine, meaning that it cannot synthesize those nutrients. Thus, it requires those two nutrients to be added to minimal media for growth.)

6

Which of the following statements about conjugation is true?
-One strand of DNA from an F+ cell integrates into the chromosome of an F- cell, and the other strand is degraded
-DNA is transferred from an F+ cell to an F- cell
-Only competent cells can undergo conjugation
-The F factor is an element that is found in the chromosome of an F+ cell

DNA is transferred from an F+ cell to an F- cell.

(DNA is transferred through a pilus from an F+ cell to an F- cell.)

7

How is a merozygote formed?
-The F factor is excised from the chromosome of an Hfr strain, causing it to revert to F-
-The F factor and several adjacent genes are excised from the chromosome of an F+ cell and transferred to an F- strain
-The F factor is inserted into the chromosome of an F+ cell, causing it to revert to F-
-The F factor is inserted into the chromosome of an F- cell, causing it to become an Hfr strain

The F factor and several adjacent genes are excised from the chromosome of an F+ cell and transferred to an F- strain.

(A merozygote is formed when the F factor and several adjacent bacterial genes are excised from the bacterial chromosome of an F+ cell and transferred to an F- cell.)

8

Strain 1: arg+ thi- pro- val+
Strain 2: arg- thi+ pro+ val-
Producing Strain 3: arg+ thi- pro- val+ and arg- thi+ pro+ val-

Use the figure to predict the results of Lederberg and Tatum's experiment.

Bacterial cultures from which flask(s) will grow on minimal medium plus pro?

Flask 3.

(Minimal medium only supports the growth of bacteria that are wild type for all growth requirements. When minimal medium is supplemented with proline, it can support the growth of pro - bacteria, but these bacteria must be wild type for other genes. Because strains 1 and 2 have additional mutations, and are auxotrophs, they will not grow under this condition. However, genetic exchange can occur between these two auxotrophic strains when combined together in flask 3, which can produce prototrophs that are wild type for all growth requirements and capable of growing in this condition.)

9

Strain 1: arg+ thi- pro- val+
Strain 2: arg- thi+ pro+ val-
Producing Strain 3: arg+ thi- pro- val+ and arg- thi+ pro+ val-

Use the figure to predict the results of Lederberg and Tatum's experiment.

Bacterial cultures from which flask(s) will grow on complete medium?

Flasks 1, 2, and 3

(Complete medium is extensively supplemented and can support the growth of auxotrophs and phototrophs.)

10

Strain 1: arg+ thi- pro- val+
Strain 2: arg- thi+ pro+ val-
Producing Strain 3: arg+ thi- pro- val+ and arg- thi+ pro+ val-

Use the figure to predict the results of Lederberg and Tatum's experiment.

Bacterial cultures from which flask(s) will grow on minimal medium plus val?

Flask 3.

(Minimal medium only supports the growth of bacteria that are wild type for all growth requirements. When minimal medium is supplemented with valine, it can support the growth of val - bacteria, but these bacteria must be wild type for other genes. Because strains 1 and 2 have additional mutations, and are auxotrophs, they will not grow under this condition. Genetic exchange can occur between these strains when combined together in flask 3, which can produce prototrophs which are wild type for all growth requirements and capable of growing in this condition.)

11

Strain 1: arg+ thi- pro- val+
Strain 2: arg- thi+ pro+ val-
Producing Strain 3: arg+ thi- pro- val+ and arg- thi+ pro+ val-

Use the figure to predict the results of Lederberg and Tatum's experiment.

Bacterial cultures from which flask(s) will grow on minimal medium plus pro and thi?

Flasks 1 and 3.

(Minimal medium can only support the growth of bacteria that are wild type for all growth requirements, and if supplemented with proline and thiamine, can support growth of bacteria that are pro - and thi -, if they are wild type for other genes.)

12

Strain 1: arg+ thi- pro- val+
Strain 2: arg- thi+ pro+ val-
Producing Strain 3: arg+ thi- pro- val+ and arg- thi+ pro+ val-

Use the figure to predict the results of Lederberg and Tatum's experiment.

Bacterial cultures from which flask(s) will grow on minimal medium plus val and arg?

Flasks 2 and 3.

(Minimal medium can only support the growth of bacteria that are wild type for all growth requirements. If valine and arginine are added to this medium, it can support growth of bacteria that are val - and arg -, if they are wild type for other genes.)

13

Strain 1: arg+ thi- pro- val+
Strain 2: arg- thi+ pro+ val-
Producing Strain 3: arg+ thi- pro- val+ and arg- thi+ pro+ val-

Use the figure to predict the results of Lederberg and Tatum's experiment.

Bacterial cultures from which flask(s) will grow on minimal medium?

Flask 3.

(The bacteria in strains 1 and 2 are auxotrophs and cannot grow in minimal medium. However, because genetic exchange can occur between these strains when combined together in flask 3, prototrophs which are wild type for all growth requirements can be produced. They are capable of growing in this simple medium.)

14

Name the general category into which double-stranded circular extrachromosomal DNA elements such as F factors, ColE1, and R would fall.

- r-determinant
-partial diploid
-plasmid
-plaque
-capsid

Plasmid.

15

True or False?
An Hfr cell can initiate chromosome transfer from one E. coli to another.

True.

16

With respect to F+ and F- bacterial matings, how was it established that physical contact was necessary?

-By placing a filter​ in a U-tube between bacterial cells. Under this condition, conjugation does not occur.
-By placing bacterial cells in a U-tube. Under this condition, conjugation does not occur.
-By placing a filter in a U-tube between bacterial cells. Under this condition, conjugation occurs.
-By placing bacterial cells in a U-tube. Under this condition, conjugation occurs.

By placing a filter​ in a U-tube between bacterial cells. Under this condition, conjugation does not occur.

17

By treating cells with streptomycin, an ______, it was shown that recombination _______ if one of the two bacterial strains was ______. However, if the other was similarly treated, recombination _______. Thus, directionality was suggested, with one strain being a donor strain and the other being the _______.

Options:
-activator
-antibiotic
-recipient
-inactivator
-would occur
-would not occur
-inactivated
-activated

-antibiotic
-would not occur
-inactivated
-would occur
-recipient

18

What is the genetic basis of a bacterium being F+?

-it lacks the F- plasmid
-it lacks the F- sequence in its genome
-it contains the F+ plasmid
-it contains the F+ sequence in its genome

It contains the F+ plasmid.

19

Describe the origin of F+ bacteria and merozygotes.

The _____ element can enter the host bacterial chromosome, and upon returning to its independent state, it may _____ a piece of a bacterial chromosome. When transferred to a bacterium with a _______, a ______ or merozygote is formed.

Options:
-pick up
-partial diploid
-F+
- F-
-reduced chromosome
-complete chromosome
-leave
-Hfr

-F+
-pick up
-complete chromosome
-partial diploid

20

How do we know that bacteria undergo genetic recombination, allowing the transfer of genes from one organism to another?

-From experiments involving conjugation
-From experiments involving transduction
-From experiments involving transformation
-From a variety of experiments involving transformation, conjugation, and transduction

From a variety of experiments involving transformation, conjugation, and transduction.

21

How do we know that conjugation leading to genetic recombination between bacteria involves cell contact, which precedes the transfer of genes from one bacterium to another?

-involving a Davis U-tube and a filter: when the filter separated two auxotrophic strains, genetic recombination occurred
-involving a Davis U-tube and a filter: when the filter separated two auxotrophic strains, NO genetic recombination occurred
-involving a Davis U-tube: when the tube separated two auxotrophic strains, NO genetic recombination occurred
-involving a Davis U-tube: when the tube separated two auxotrophic strains, genetic recombination occurred

Involving a Davis U-tube and a filter: when the filter separated two auxotrophic strains, NO genetic recombination occurred.

22

How do we know that during transduction bacterial cell-to-cell contact is not essential?

-Bacterial contact can be eliminated with a U-tube
-Bacterial contact can be eliminated with a filter
-This is a hypothesis

Bacterial contact can be eliminated with a filter.

23

How do we know that intergenic exchange occurs in bacteriophages?

-It was demonstrated by using a Davis U-tube without a filter
-It was demonstrated by using a Davis U-tube and a filter
-It was demonstrated by mixed infections that yielded recombinants
-It was demonstrated by a single infection that yielded recombinants

It was demonstrated by using a Davis U-tube and a filter.

24

Why are the recombinants produced from an Hfr × F−
cross rarely, if ever, F+?

-Because the F factor is the last element to be transferred and the conjugation tube is fragile, the liklihood for complete transfer is low
-Because the F factor is the last element to be transferred, the likelihood for complete transfer is low
-Because the conjugation tube is fragile, the likelihood for complete transfer is low
-Because the F factor is the first element to be transferred, the likelihood for complete transfer is high

Because the F factor is the last element to be transferred and the conjugation tube is fragile, the liklihood for complete transfer is low.

25

All EXCEPT which of the following are characteristics of the genetic material?

-It is composed of protein
-It must be replicated accurately
-It contains all the information needed for growth, development, and the reproduction of the organism
-It must be capable of change

It is composed of protein.

(Although early observations favored protein as the genetic material, subsequent experiments demonstrated that the genetic material was nucleic acid.)

26

In 1928, Frederick Griffith established that _______.

-mouse DNA could be transferred into bacterial cells
-proteases have no effect on DNA
-heat-killed bacteria harbor the constituents necessary to convey genetic properties to living bacteria
-mice could be infected with bacteria

Heat-killed bacteria harbor the constituents necessary to convey genetic properties to living bacteria.

(Because some of the nonvirulent bacteria acquired properties of the virulent bacteria, instructions for this transformation must be carried by the virulent bacteria.)

27

To be certain that the extract prepared from virulent cells still contained the transforming principle that was present prior to lysis, Avery _______.

-injected mice with the extract
-incubated virulent cells with complete​ extract
-destroyed proteins, polysaccharides, DNA, and RNA contained in the extract
-incubated nonvirulent cells with the complete extract

Incubated nonvirulent cells with the complete extract.

(The complete extract possessed the same ability to induce transformation in IIR bacteria as whole heat-killed IIS bacteria.)

28

If Avery had observed transformation using only the extracts containing degraded DNA, degraded RNA, and degraded protein, but NOT the extract containing degraded polysaccharides, he would have concluded that _______.

-polysaccharides are the genetic material
-RNA is the genetic material
-the preparations were contaminated
-mice with diets rich​ in polysaccharides are resistant to bacterial infection

Polysaccharides are the genetic material.

(Failure to transform suggests that the chemical degraded in that preparation is the one responsible for transformation, in this case polysaccharides.)

29

The Hershey and Chase experiments involved the preparation of two different types of radioactively labeled phage. Which of the following best explains why two preparations were required?

-Each scientist had his own method for labeling phages, so each conducted the same experiment using a different isotope.
-The bacteriophage used in the experiments was a T2 phage.
-Establishing the identity of the genetic material required observation of two phage generations.
-It was necessary that each of the two phage components​, DNA and protein, be identifiable upon recovery at the end of the experiment.

It was necessary that each of the two phage components, DNA and protein, be identifiable upon recovery at the end of the experiment.

(Because it was concluded that the component associated with bacteria at the end of the experiment must be the genetic material, it was critical that the component be identifiable as either DNA or protein.)

30

Which of the following statements best represents the central conclusion of the Hershey-Chase experiments?

-Phage T2 is capable of replicating within a bacterial host
-Some viruses can infect bacteria
-When radioactive sulfur is supplied in a growth medium, it is primarily DNA that incorporates radioactive label
-DNA is the identity of the hereditary material in phage T2

DNA is the identity of the hereditary material in phage T2.

(Because phage DNA and not protein were​ associated with bacteria at the end of the experiment, it could be concluded that DNA - not protein - must be the genetic material.)

31

Which of the following outcomes would be most likely if the Hershey-Chase experiments were repeated without the step involving the blender?

-The phage would fail to infect bacteria
-Neither preparation of infected bacteria would exhibit radioactivity
-Both preparations of infected bacteria would contain both P32 and S35
-Both preparations of infected bacteria would exhibit radioactivity

Both preparations of infected bacteria would exhibit radioactivity.

(Instead of being removed from the preparation, the "ghosts" would be retained. Because both bacterial preparations would include ghosts as well as viral DNA, both would be radioactive, one with P32, one with S35.)

32

The classic Hershey and Chase (1952) experiment that offered evidence in support of DNA being the genetic material in bacteriophages made use of which of the following labeled component(s)?

-hydrogen
-tritium
-phosphorus and sulfur
-nitrogen and oxygen
-none of the answers listed is correct

Phosphorus and Sulfur

33

What results did Avery, McLeod, and McCarty obtain in their experiments with virulent bacteria?

-Protease destroyed the transforming activity
-The transforming principle was too complex and difficult to be purified
-DNase destroyed the transforming activity
-RNase destroyed the transforming activity

DNase destroyed the transforming activity.

(Treatment of the transforming principle with DNase destroyed the DNA and thus its ability to transform bacteria.)

34

True or False?

Deoxyribonuclease is an enzyme that adds 3'-hydroxyl groups to RNA.

False

35

True or False?

Hershey and Chase used labeled DNA and protein to determine that DNA is the genetic material in bacteria.

False

36

True or False?

The transforming principle discovered by Griffith is RNA.

False

37

What are the three main parts of the structure of adenosine triphosphate (ATP)?

-triphosphate, ribose, adenine
-ribose, triphosphate, adenine
-triphosphate, deoxyribose, adenine
-adenine, ribose, triphosphate
-deoxyribose, adenine, triphosphate

Triphosphate, Ribose, Adenine

(Adenosine-5-triphosphate (ATP) is a nucleoside triphosphate composed of three phosphate groups, a ribose sugar,​ and the nitrogenous base adenine.
Triphosphate consists of three phosphate groups (PO4) linked together (A, orange).
Ribose is an aldopentose, a five-carbon sugar molecule (B).
Adenine is a purine nucleobase, consisting of a pyrimidine ring linked to an imidazole ring (C).)

38

What is the structural difference between ATP and dATP?

-ATP has a 2' H and 3' OH while dATP has a 2' OH and a 3'OH
-ATP has a 2' OH and 3' OH while dATP has a 2' OH and 3' H
-ATP has a 2' OH and 3' OH while dATP has a 2' H and 3' OH
-ATP has a 2' OH and 3' H while dATP has a 2' OH and 3' OH

ATP has a 2' OH and 3' OH while dATP has a 2' H and 3' OH

(ATP and dATP both consist of three phosphate groups, a sugar molecule, and adenine.
The difference is in the number of hydroxyl (OH) groups attached to the sugar.
The sugar in ATP is ribose, which has an OH group on both the 2' and 3' carbons.
The sugar in dATP is deoxyribose, which only has an OH group on the 3' carbon. The full name of dATP is 2'-deoxyadenosine triphosphate. The prefix de- means "away from" or "without", indicating that the 2' carbon is without an oxygen atom.)

39

How many OH groups are attached to the sugar of ddATP?
-none
-one
-two
-three

None.

(Adenosine-5'-triphosphate (ATP) has two hydroxyl groups (OH) on the 2' and 3' carbons of the sugar molecule (ribose).
2'-Deoxyadenosine (dATP) has only one hydroxyl group (OH) on the 3' carbon of the sugar molecule (deoxyribose).
2',3'-Dideoxyadenosine-5'-triphosphate (ddATP) is lacking hydroxyl groups (OH) on both the 2' and 3' carbons of the sugar (dideoxyribose).)

40

Which of the following molecules is a nucleotide precursor that is incorporated into the newly synthesized DNA strand during normal DNA replication?

-ATP
-dATP
-ddATP

dATP

(DNA is short for deoxyribonucleic acid.
The sugar in DNA is 2'-deoxyribose, meaning that it does not have a hydroxyl group attached to the 2' carbon of the sugar.
In order for a nucleotide to be added to a growing DNA strand during DNA synthesis, it must have a free 3' OH group.)

41

If you were to set up a PCR reaction (in vitro DNA synthesis) with a DNA template, primers, DNA polymerase, dATP, dGTP, dCTP, dTTP,​ and a small amount of ddATP, what would be the result?

-DNA synthesis would happen normally. All DNA molecules produced would be the same length as the template.
-DNA synthesis might be terminated after the addition of any adenine base (at random). DNA molecules of many different lengths would be produced.
-DNA synthesis would be terminated after the first adenine base is added. All DNA molecules produced would be the same length, shorter than the template.

DNA synthesis might be terminated after the addition of any adenine base (at random). DNA molecules of many different lengths would be produced.

(To link the next nucleotide to the growing DNA strand, the DNA polymerase requires a free 3'OH group.
2'-Deoxyadenosine triphosphate (dATP) has a 3' OH group, and 2',3'-dideoxyadenosine triphosphate (ddATP) does not. If a dATP is added to the chain, DNA synthesis continues normally. If a ddATP is added, DNA synthesis terminates. This is why ddNTPs are sometimes called “chain terminating nucleotides” or “chain terminators.”
Because there is more dATP than ddATP, DNA synthesis does not always terminate at the first A that's added. However if a dATP is added, synthesis continues to the next nucleotide normally. If however a ddATP is added, synthesis stops.
As a result, DNA molecules of different sizes are produced, depending on the position of adenines and whether a dATP is added or a ddATP is added.)

42

The nucleic acids DNA and RNA are made from chains of nucleotides. Nucleotides consist of three components: a five-carbon sugar (either ribose or deoxyribose), a nitrogenous base attached to the sugar’s 1'-carbon, and a phosphate group attached to the sugar’s 5'-carbon.

Identify three possible components of a DNA nucleotide.

-deoxyribose, phosphate group, thymine
-adenine, phosphate group, ribose
-deoxyribose, phosphate group, uracil
-cytidine, phosphate group, ribose
-guanine, phosphate group, ribose
-cytosine, phosphate group, ribose

Deoxyribose, Phosphate group, Thymine

(DNA and RNA have similar structures: a pentose sugar with a nitrogenous base and a phosphate group. DNA and RNA differ in the type of pentose sugar each possesses (DNA has deoxyribose; RNA has ribose) and in one base (DNA has thymine; RNA has uracil).

43

The nucleic acids DNA and RNA are made from chains of nucleotides. Nucleotides consist of three components: a five-carbon sugar (either ribose or deoxyribose), a nitrogenous base attached to the sugar’s 1'-carbon, and a phosphate group attached to the sugar’s 5'-carbon.

Identify three possible components of a DNA nucleotide.

-deoxyribose, phosphate group, thymine
-adenine, phosphate group, ribose
-deoxyribose, phosphate group, uracil
-cytidine, phosphate group, ribose
-guanine, phosphate group, ribose
-cytosine, phosphate group, ribose

Deoxyribose, Phosphate group, Thymine

(DNA and RNA have similar structures: a pentose sugar with a nitrogenous base and a phosphate group. DNA and RNA differ in the type of pentose sugar each possesses (DNA has deoxyribose; RNA has ribose) and in one base (DNA has thymine; RNA has uracil).

44

Sort the parts of a nucleic acid according to whether each occurs exclusively in DNA, exclusively in RNA, or in both types of nucleic acid.

Exclusively DNA: Thymine, Deoxyribose
Exclusively RNA: Uracil, Ribose
Both DNA and RNA: Cytosine, Guanine, Adenine, Phosphate

(DNA is used for storage of genetic information. The presence of deoxyribose as the sugar in DNA makes the molecule more stable and less susceptible to hydrolysis. The 2'-oxygen on the ribose found in RNA makes RNA much more susceptible to breakdown. It is important that mRNA be easily broken down, to ensure that the correct levels of protein are maintained in the cell.)

45

True or False?

Guanine and adenine are purines found in DNA.

True

(Guanine and adenine are indeed purines found in DNA; thymine and cytosine are the pyrimidines found in DNA.)

46

Which of the following statements about DNA structure is true?
-Nucleic acids are formed through phosphodiester bonds that link nucleosides together.
-The pentose sugar in DNA is ribose.
-Hydrogen bonds formed between the sugar-phosphate backbones of the two DNA chains help to stabilize DNA structure.
-The nucleic acid strands in a DNA molecule are oriented antiparallel to each other, meaning they run in opposite directions.

The nucleic acid strands in a DNA molecule are oriented antiparallel to each other, meaning they run in opposite directions.

(This statement is true; the 5′–3′ orientation of each chain runs in opposite directions.)

47

What is the complementary DNA sequence to 5′ ATGCTTGACTG 3′?
-5' ACTCTACGTAG 3'
-5' CAGTCAAGCAT 3'
-5' TACGAACTGAC 3'
-5' ATGCTTGACTG 3'

5' CAGTCAAGCAT 3'

(This sequence is complementary and in the correct orientation.)

48

True or False?

Adenine and thymine are complementary and join via two hydrogen bonds. Guanine and cytosine are complementary and join via three hydrogen bonds.

True

49

What is the complementary DNA sequence for the DNA strand: 3' ATTGCAGTACC 5'?

5' TAACGTCATGG 3'

(This sequence is complementary and in the correct orientation.)

50

The basic structure of a nucleotide includes ______.

-amino acids
-phosphorus and sulfur
-base, sugar, and phosphate
-mRNA, rRNA, tRNA
-tryptophan and leucine

base, sugar, and phosphate

51

Regarding the structure of DNA, the covalently arranged combination of a deoxyribose and a nitrogenous base would be called a(n) ________.

-oligonucleotide
-ribonucleotide
-nucleotide
-nucleoside
-monophosphate nucleoside

Nucleoside

52

Which statement about the polarity of DNA strands is true?

-The 3' end has a free phosphate group
-The 3' end has a free OH group
-The 5' end has a free OH group

The 3' end has a free OH group.

(Nucleotides are joined together by a phosphodiester linkage, which requires a free 3’ OH group on the sugar molecule, and an alpha phosphate of an entering dNTP.)

52

Which statement about the polarity of DNA strands is true?

-The 3' end has a free phosphate group
-The 3' end has a free OH group
-The 5' end has a free OH group

The 3' end has a free OH group.

(Nucleotides are joined together by a phosphodiester linkage, which requires a free 3’ OH group on the sugar molecule, and an alpha phosphate of an entering dNTP.)

53

To help achieve proper base pairing and hence form a double helix, which condition must be met?

-A purine base must pair with a purine base.
-A purine base must pair with a pyrimidine base.
-A pyrimidine base must pair with a pyrimidine base.

A purine base must pair with a pyrimidine base.

(Chargaff’s rules state that there is a 1:1 ratio of purines to pyrimidines. More specifically the amount of adenine is equal to the amount of thymine, and the amount of cytosine is equal to the amount of guanine. Two purines would place the bases too far apart for hydrogen bonds to form. Two pyrimidines would place the bases too close together, making the bond unstable due to steric repulsion.)

54

If you were to try and pair a thymine with a cytosine (a non Watson-Crick base pairing), then would you expect to see any stability with respect to the hydrogen bonding (assuming the geometrical configurations of both bases were favorable to each other)? If yes, then how many hydrogen bonds could form between these two bases?

-Yes, one hydrogen bond could form between thymine and cytosine.
-No, hydrogen bonds cannot form between thymine and cytosine.
-Yes, two hydrogen bonds could form between thymine and cytosine
-Yes, three hydrogen bonds could form between thymine and cytosine

Yes, two hydrogen bonds could form between thymine and cytosine.

(One hydrogen bond could form between the C4 carbonyl group on thymine (a hydrogen bond acceptor) and the C4 amino group on cytosine (a hydrogen bond donor). Another hydrogen bond could form between N3 of thymine (a hydrogen bond donor) and the N3 of cytosine (a hydrogen bond acceptor). Note that the C2 carbonyl groups found on both bases are both hydrogen-bond​ acceptors and therefore a hydrogen bond cannot be formed between them.)

55

If a DNA-binding protein “reads” a short stretch of DNA and detects the following “second” genetic code provided by the functional groups located on each base as H-HD-CH3-HA-HA-HA-HA-HD, then what is the corresponding sequence of bases?

*C-T-A-G
*C-G-G-A
*C-T-G-A
*C-A-G-A

C-T-G-A

(The first two functional groups (nonpolar hydrogen and exocyclic amino group) indicate cytosine, the second two (methyl and carbonyl groups) indicate thymine, the next two (N7 and carbonyl group) indicate guanine and the last two functional groups (N7 and exocyclic amino group) indicate the presence of ​adenine.)

56

The minor groove contains less information about the identity of base pairs than the major groove because of the

-glycosidic bond angles and structure of purine bases
-geometry of base pairing and structure of ribose
-glycosidic bond angles and geometry of base pairings
-glycosidic bond angles and structure of pyrimidine bases

Glycosidic bond angles and structure of pyrimidine bases.

(The geometry of a base pair is such that the two glycosidic bonds each point outward from their respective bases, thereby creating a narrow (120°) angle on one side of the base pair and a broad (240°) angle on the other side. Moreover, because the glycosidic bonds are found on the N1 of each pyrimidine, only three functional groups can be found in each minor (narrow angle) groove, whereas four functional groups are found in the corresponding major (broad angle) groove.)

57

True or False?

In a DNA sequence, the purine adenine always pairs with the pyrimidine thymine, and the purine guanine always pairs with the pyrimidine cytosine.

True

58

Write the complementary sequence for the following DNA sequence, in order from 3' to 5':
5′−CGATATTGAGCTAAGCTT−3′

3' GCTATAACTCGATTCGAA 5'

59

True or False?

The base pair adenine-cytosine occurs very rarely in nature. It only happens during a mutation event. When the DNA is replicated, one of the two daughters will contain a guanine-cytosine base pair in the location of the mutation, and the other daughter will contain an adenine-thymine base pair.

True

60

The GC content of a DNA molecule is 60%.
What is the molar percentage of guanine (G)?

30%

61

The GC content of a DNA molecule is 60%.
What is the molar percentage of adenine (A)?

20%

62

The GC content of a DNA molecule is 60%.
What is the molar percentage of thymine (T)?

20%

63

The GC content of a DNA molecule is 60%.
What is the molar percentage of cytosine (C)?

30%

64

Which of the following is a purine in DNA?

-adenine
-thymine
-guanine
-cytosine
-uracil

Adenine

65

Which of the following is a pyrimidine in DNA?

-cytosine
-adenine
-uracil
-thymine
-guanine

Thymine

66

During the polymerization of nucleic acids, covalent bonds are formed between neighboring nucleotides. Which carbons are involved in such bonds?
Select the two carbons involved in the bond.

-C-1'
-C-2'
-C-3'
-C-4'
-C-5'

C-3'
C-5'

67

True or False?

When considering the structure of DNA, we would say that complementary strands are antiparallel.

True

68

RNA differs from DNA in all EXCEPT which of the following ways?

-the presence of uracil
-the number of different functions performed
-the sugar molecule
-the 5'-3' orientation of the polynucleotide strand

The 5'-3' orientation of the polynucleotide strand

(Both RNA and DNA have the same 5' amino group and 3' hydroxyl group chemical orientation.)

69

When Avery and his colleagues had obtained the transforming factor from the IIIS virulent cells, they treated the fraction with proteases, ribonuclease, and deoxyribonuclease, followed by the assay for retention or loss of transforming ability.

What was the purpose of these experiments?

-to prove the stability of the "transforming principle" in a typical cell medium
-to determine the exact molecular species of the "transforming principle"
-to prove that transforming ability does not depend on the presence of any enzymes
-to determine the enzyme that can suppress virulence

To determine the exact molecular species of the "transforming principle".

70

When Avery and his colleagues had obtained the transforming factor from the IIIS virulent cells, they treated the fraction with proteases, ribonuclease, and deoxyribonuclease, followed by the assay for retention or loss of transforming ability.

What was the result? What conclusions were drawn?

-The DNA treatment was proven to be the "transforming principle". Treatment with deoxyribonuclease resulted in the loss of transforming ability.
-Protein was proven to be the "transforming principle". Treatment with protease resulted in the loss of transforming ability.
-The RNA was proven to be the "transforming principle". Treatment with ribonuclease resulted in the loss of transforming ability.
-The deoxyribonuclease was proven to be the "transforming principle". Treatment with deoxyribonuclease resulted in the retention of transforming ability.

The DNA treatment was proven to be the "transforming principle". Treatment with deoxyribonuclease resulted in the loss of transforming ability.

71

Describe the various characteristics of the Watson-Crick double helix model for DNA.
Which of the following are true about nitrogenous bases and base pairing?
Check all that apply.

-The base composition is such that A=T and G=C
-The base composition is such that T=C and A=G
-There are 3 hydrogen bonds forming the A to T pair and 2 forming the G to C pair
-There are 2 hydrogen bonds forming the A to T pair and 3 forming the G to C pair
-Bases are stacked 0.34 nm apart
-Bases are stacked 0.50 nm apart

-The base composition is such that A=T and G=C
-There are 2 hydrogen bonds forming the A to T pair and 3 forming the G to C pair
-Bases are stacked 0.34 nm apart

72

Describe the various characteristics of the Watson-Crick double helix model for DNA.
Which of the following are true about the two strands of the double helix?
Check all that apply.

-Phosphodiester bonds hold the 2 polynucleotide chains together
-Hydrogen bonds hold the 2 polynucleotide chains together
-The hydrophobic phosphodiester backbone is located in the center of the molecule, whereas the hydrophilic nitrogenous​ bases are on the outside.
-The hydrophobic nitrogenous bases are located in the center of the molecule, whereas the hydrophilic phosphodiester​ backbone is on the outside.
-The 2 strands are antiparallel
-The 2 strands are parallel

-Hydrogen bonds hold the 2 polynucleotide chains together.
-The hydrophobic nitrogenous bases are located in the center of the molecule, whereas the hydrophilic phosphodiester​ backbone is on the outside.
-The 2 strands are antiparallel

73

Describe the various characteristics of the Watson-Crick double helix model for DNA.
Which of the following is true about the overall structure of the DNA helix?
Check all that apply.

-There is one complete turn for each 34 nm, which constitutes 46 bases per turn
-There is one complete turn for each 3.4 nm, which constitutes 10 bases per turn
-The double helix is right-handed
-The double helix is left-handed
-The double helix is approximately 20 Angstroms in diameter
-The double helix is approximately 34 Angstroms in diameter

-There is one complete turn for each 3.4 nm, which constitutes 10 bases per turn
-The double helix is right-handed
-The double helix is approximately 20 Angstroms in diameter

74

______ showed that a chemical extract of the virulent cells was sufficient to cause transformation of avirulent cells, further reinforcing the hypothesis that the transforming factor has a chemical basis.

-Alloway
-Griffith
-Avery

Alloway

75

______ performed experiments with different strains of Diplococcus pneumoniae in which a heat-killed pathogen, when injected into a mouse with a live nonpathogenic strain, eventually led to the mouse's death. A summary of this experiment is provided in the text. Examination of the dead mouse revealed living pathogenic bacteria. The scientist suggested that heat-killed virulent (pathogenic) bacteria transformed the avirulent (nonpathogenic) strain into a virulent strain.

-Alloway
-Griffith
-Avery

Griffith

76

______ and coworkers systematically searched for the transforming principle originating from the heat-killed pathogenic strain and determined it to be DNA. Others showed that transformed bacteria are capable of serving as donors of transforming DNA, indicating that the process of transformation involves a stable alteration in the genetic material (DNA).

-Alloway
-Griffith
-Avery

Avery