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Microbial Physiology > Lecture 1 - Exam 2 > Flashcards

Lecture 1 - Exam 2 Flashcards

1
Q

Who discovered DNA? What did he specifically discover?

A

Friedrich Miescher. He discovered “nuclein” in nuclei of human white blood cells.

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2
Q

What are nucleotides in DNA composed of?

A
  1. A phosphate group
  2. A pentose sugar (ribose or deoxyribose)
  3. A single nitrogen-containing base
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3
Q

What are the two basic categories of nitrogenous bases?

A

Purines: Adenine (A) and Guanine (G)
Pyrimidines: Cytosine (C), Thymine (T), and Uracil (U)

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4
Q

What are DNA’s nitrogenous bases?

A

A, G, C and T

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5
Q

What are RNA’s nitrogenous bases?

A

A, G, C, and U

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6
Q

Cytosine can spontaneously deaminate _____? If this base was used routinely in DNA, then could the DNA repair mechanisms detect this spontaneous mutation?

A

Uracil
No, it couldn’t detect this spontaneous mutation.

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7
Q

What allows the repair mechanism in DNA to identify a spontaneous Cytosine to Uracil mutation?

A

Thymine is (basically) methylated Uracil, so using Thymine in DNA allows the repair mechanism to identify a spontaneous Cytosine to Uracil mutation, and replace U with a C.

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8
Q

What is a nucleoside?
Adding what would make it a nucleotide?

A

A pentose sugar plus nitrogenous base
The addition of a phosphate group to the 5’ C of a nucleoside makes it a nucleotide.

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9
Q

________ are joined together to make one of the strands of DNA.

A

Nucleotides

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10
Q

The ______ of one nucleotide can form a bond with the _______ group of another nucleotide. Joined together by __________ bonds.

A

3’ OH ; 5’ Phosphate ; Phosphodiester

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11
Q

The 3’ OH of one nucleotide can form a bond with the 5’ phosphate group of another nucleotide. What does this mean for the polynucleotide chain?

A

This means that the polynucleotide chain has biological polarity or directionality, meaning that the strands always grow from 5’ to 3’.

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12
Q

What are the bonds between the 3’ OH and the 5’ phosphate group?

A

Phosphodiester bonds and are very strong.

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13
Q

What are the bonds between nitrogenous bases?

A

Hydrogen bonds and are much weaker than phosphodiester bonds.

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14
Q

Why is it important that the hydrogen bonds are weaker between the nitrogen bases than the phosphodiester bonds?

A

The hydrogen bonds have to be weaker in order for easy separation for replication.

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15
Q

Nucleic acids are ______ sugars.

A

Pentose

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16
Q

What is the pentose sugar in RNA?
What about DNA?

A

Ribose is the pentose sugar in RNA.
Deoxyribose is the pentose sugar in DNA.

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17
Q

What is the structural difference between Deoxyribose and Ribose?

A

There is a 2’ Hydroxyl group in Ribose, but a 2’ Hydrogen in Deoxyribose.

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18
Q

What are the two classes of nitrogenous bases?

A

Pyrimidines: Contains one carbon-nitrogen ring and two nitrogen atoms
-Cytosine
-Thymine
-Uracil
Purines: A pyrimidine ring fused to an imidazole ring. Contains two carbon-nitrogen rings and four nitrogen atoms.
-Adenine
-Guanine

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19
Q

Where is the nitrogenous base bound to the deoxyribose sugar?
Where is the phosphate group attached to of the deoxyribose?

A

A nitrogenous base is bound to the 1’ C of the deoxyribose sugar.
The phosphate group is attached at the 5’ position of the deoxyribose sugar.

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20
Q

When is DNA double helix formed?

A

When nitrogenous bases on one DNA strand form hydrogen bonds with a complementary base on a second strand of DNA.

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21
Q

When is a single strand of DNA made?

A

When nucleotides joined together by phosphodiester bond between he 5’ phosphate of one nucleotide and the 3’ OH of a neighboring nucleotide.

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22
Q

What is the nature of the two strands of DNA that form the double helix? Why is it this way?

A

Antiparallel.
The antiparallel nature allows for the hydrogen bond between the nitrogenous bases, it makes the nitrogenous bases closer.

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23
Q

The sugars in the backbone have directionality. What does this mean?
How do the sugars in the backbone elongate?

A

The 5’ phosphate is first and the 3’ hydroxyl is last.
The sugars elongate by the addition of dNTPs to the 3’ OH.

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24
Q

Without the antiparallel nature of DNA, would replication be possible?

A

NO!

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25
Q

How is RNA different than DNA?

A

Ribose sugar instead of deoxyribose, uracil instead of thymine, the complementary strand is not present (this doesn’t mean it is not doubled stranded or structured, instead it forms complementary strand bonds with itself), and it is much more sensitive to degradation and enzymes than DNA is.
Structurally, OH on the 2’ C (whereas DNA has a H on 2’C)

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26
Q

What is Chargaff’s Rule?

A

A purine (A, G) can only pair with a pyrimidine (C,T).

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27
Q

If you think of DNA as a spiral staircase, the nitrogenous base pairs would be?

A

“Steps”

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28
Q

All “steps” (hydrogen bond) must be the same size. What happens if they aren’t?
How many angstroms (A) are there between two complementary strands of DNA? This only allows space for what?

A

The spiral will be distorted.
20 A.
This only allows space for a purine to pair with a pyrimidine.

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29
Q

How are the hydrogen bonds broken between the purines and pyrimidines?

A

Easily broken with DNA helicases during replication and can be easily reformed after synthesis of new strands.

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30
Q

How many hydrogen bonds are formed between A & T?
How about G & C?
Which one takes more energy to break? Do you want to start the template somewhere where it is harder or easier to break apart?

A

A & T: Two
G & C: Three
G & C bonds are harder to break (b/c there are more hydrogen bonds than A & T).
You want to start the template somewhere where it is harder to break apart.

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31
Q

What are the main DNA double helix conformations?

A

B form: Watson-Crick structure, right-handed helix with 10bp per turn. The most common form.
A form: right-handed, more compact, 11 bp per turn. Shorter, wider form. Rarely found under normal physiological consequences.
Z form: left-handed, 12 bp per turn, zig-zag pattern in the phosphodiester backbone, found in GCGCGC (harder to break this run) sequences. Is a transient form of DNA. Seems that is plays an important biological role in protection against viral disease.

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32
Q

The Meselson - Stahl experiment proved what?

A

That DNA is replicated semiconservatively.

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33
Q

What is semiconservative replication?

A

Each strand of DNA acts as a template for the synthesis of a daughter strand.

34
Q

Describe the Meselson - Stahl Experiment.

A

E. coli cells were grown in 15N for several generations so that all the nitrogen in the DNA is heavy.
-After centrifugation in CsCl density gradient, all the DNA settles near the bottom.
Next, the ‘heavy’ E. coli are grown in 14N (light) media.
-After 1 generation, all of the DNA was ‘hybrid’ 15N-14N (heavy-light)
After another generation in 14N (light) media, both 15N-14N (heavy-light) and 14N-14N (light) DNA are present.
This proved semi-conservative replication because one strand of DNA acted as a template for each replication.

35
Q

In DNA replication, a single DNA strand can act as a _______ for the synthesis of a _______ in which each base forms a hydrogen-bonded pair with one on the template strand.

A

Template ; new nucleic acid strand

36
Q

In DNA replication, a single DNA strand can act as a template for the synthesis of a new nucleic acid strand in which each base forms a hydrogen-bonded pair with one on the template strand. The new sequence is thus _________ to the template strand.

A

Complementary.

37
Q

Define replication.

A

The copying of DNA molecules to produce more DNA is known as replication.

38
Q

Define transcription.

A

The synthesis of RNA using a DNA template is called transcription.

39
Q

Describe bacterial chromosomes.

A

Most (not all) of the bacterial chromosomes are closed circles.
Causes topological problems during DNA replication and transcription.

40
Q

Why is it a problem that bacterial double helix DNA are closed circles?

A

The DNA in front of it starts to tangle up.

41
Q

What is the nature of DNA polymerases?

A

DNApol synthesize DNA by using a single strand as a template.
DNApol can ONLY extend a DNA chain, they cannot initiate the extension.
DNApol requires a free 3’ OH to add the phosphate group of the next nucleotide to the growing chain.
Needs a primer.

42
Q

E. coli chromosomes have about _______ base pairs which results in a length of _____.

A

4.7 x 10^6
1.6 mm

43
Q

An E. coli cell is only ______ long.

A

0.002 mm

44
Q

How do bacteria solve the problem of their chromosomes being longer than their actual cells?

A

Bacteria solve this problem by compacting their DNA in the nucleoid.

45
Q

What is the bacterial nuceloid?

A

The bacterial nucleoid is an amorphous mass of DNA, unbounded by a membrane, lying in the center of the cell. It is the site of DNA and RNA synthesis.
The DNA is very tightly coiled and folded into supercoiled loops.

46
Q

What happens if one of the loops in the nucleoid breaks?

A

The whole thing comes apart, so they have histone-like proteins to help.

47
Q

Discuss the tightly coiled circular DNA in bacteria.

A

Tightly coiled circular DNA causes problems during replication and transcription. As the DNA strands are separated, the DNA in front of the replication fork becomes so supercoiled that the replication and transcription and machinery cannot continue.

48
Q

What does supercoiling refer to?

A

The twisting of the DNA double helix round its central axis, making it more compact.

49
Q

As the duplex is pulled apart at the replication fork in the closed circle, the unreplicated portion ahead of the replication fork twists so that the helix becomes…?

A

Overwound into a positive supercoil.

50
Q

Unless something happens to prevent overwinding, the positive supercoils would cause the DNA to bunch up, and further unwinding (and replication) would stop. How is this solved?

A

DNA gyrase (or topoisomerase II) relaxes the supercoiling by nicking both strands of DNA and converts them into underwound (negative supercoiled) DNA.

51
Q

Topoisomerases change what?

A

Topoisomerases change DNA topology.

52
Q

What are DNA topoisomerases?

A

Introduce (or remove) supercoils from DNA by controlled breaking and rejoining of DNA strands.

53
Q

What are type I topoisomerases?

A

Relax DNA by breaking one strand of DNA and passing the other strand through the gap, then reseals the nicked DNA.
The degree of negative supercoiling is reduced, the DNA becomes relaxed.

54
Q

What are type II topoisomerases?

A

Break both strands and pass double stranded DNA through the gap, which adds negative supercoils into the DNA.
Ex: DNA gyrase and topoisomerase IV

55
Q

Even in bacteria, with their smaller genomes, DNA replication involves…?

A

An incredibly sophisticated, highly coordinated series of molecular events. These events are divided into four major stages:
1. Initiation
2. Unwinding
3. Primer synthesis
4. Elongation

56
Q

Replication is unidirectional. T or F.

A

False, bidirectional - replicated in both directions, simultaneously.

57
Q

Where does the initiation of replication begin in prokaryotes?

A

At a precise location on the bacterial chromosome called the origin of replication.

58
Q

Most bacteria have _______ closed circular chromosomes.

A

covalently

59
Q

Most bacterial chromosomes have one origin of replication, which is?

A

oriC

60
Q

What is the origin of replication sequence (oriC) is recognized by…?

A

an initiator protein, DnaA.

61
Q

What is the function of DnaA?

A

To assemble the replication fork machinery at oriC. Within oriC, there are several DnaA boxes, DNA motifs that are DnaA recognition sites.

62
Q

Each DnaA box has a __nucleotide sequence.

A

9.

63
Q

After DnaA binds to the DnaA boxes, the _________ will be ___________.

A

AT-rich 13-mer
Unwound by a histone-like protein (HU) to form an open complex.

64
Q

After DnaA binds to the DnaA boxes, the AT-rich 13-mer will be unwound by a histone-like protein (HU). Why is it important that it is AT-rich?

A

Takes less energy to break the hydrogen bonds with AT-rich sites due to their only being two hydrogen bonds, and not three (like in GC sites).

65
Q

Before DNA synthesis can begin, what must be formed?

A

A prepriming complex.

66
Q

How is the prepriming complex made?

A

To make the prepriming complex so that two replication forks can be initiated, the two DNA strands must first unwind so that each strand can act as a template.

67
Q

The DNA-binding protein DnaA binds to the DnaA box. A histone-like protein, HU, then aids in the unwinding the AT-rich 13-mer region to form an open complex. To prevent this open complex from re-forming the duplex, what happens?

A

Single stranded binding (SBB) proteins coat the unwound strands. SBB also prevents degradation from enzymes.

68
Q

Once the open complex is formed, what happens?

A

The double stranded DNA at each replication fork must be unwound by a protein called a DnaB.

69
Q

DnaB is a…?

A

Helicase that separates complementary DNA strands.

70
Q

The helicase DnaB requires what in order to bind to DNA?

A

The helicase DnaB cannot bind to the DNA by itself, it depends on another protein, DnaC, to be transferred to the open complex.

71
Q

How do bacteria prevent the single strands of DNA (ssDNA) from coming together and forming double-stranded DNA (dsDNA)?

A

The single-stranded DNA binding protein (SSB) binds to ssDNA.

72
Q

Once the prepriming complex has been formed, what needs to happen in order for replication to happen?

A

The two strands need to be separated. This is achieved by the helicase DnaB which binds to the template strand and moves long it, separating the two strands. The separated strands are prevented from re-associating by the single-stranded DNA -binding protein SBB.

73
Q

As DnaB separates the two strands of DNA, what happens?

A

Positive supercoils form in front of the replication fork.

74
Q

As DnaB separates the two strands of DNA, positive supercoils form in front of the replication fork. What removes the positive supercoils ahead of the replication fork?

A

Type II topoisomerase, DNA gyrase.

75
Q

Describe DNA replication.

A

During replication, one of the new strands (the leading strand) can be synthesized in the 5’ to 3’ direction. The enzyme responsible for synthesis of the leading strand is DNA polymerase III.
Since nucleic acids can only be synthesized in the 5’ to 3’ direction, the 3’ to 5’ strand (the lagging strand) must be made backwards. This is done by making new strands in short fragments (Okazaki fragments). DNA polymerase III adds nucleotides in short fragments. DNA ligase fills in the gaps between Okazaki fragments, joining all the pieces together into a complete strand.

76
Q

What kind of polymerases are required for replication?

A

Both DNA polymerases and RNA polymerases.

77
Q

Discuss the roles of DNA polymerases and RNA polymerases required for DNA replication.

A

DNA polymerases can only extend a previously existing molecule - a DNA template.
DNA polymerase must have a 3’-OH to add a dNTP to.
RNA polymerases are able to initiate synthesis of new nucleic acids without a primer. Each fragment on leading and lagging strand is started with a short piece of RNA, produced by the action of special RNA polymerase (Primase). Primase synthesizes short RNA primer from DNA template.
This RNA primer can then be extended by DNA polymerase II.
The primer is subsequently removed and the gap filled in by DNA polymerase I. After the gap has been filled, the fragments that have been produced are joined together by DNA ligase.

78
Q

DNA is replicated _________.

A

Bidirectionally, minimizing replication time.

79
Q

DNA polymerase III catalyzes polymerization at ~ _________ nt/s in vivo.

A

1000

80
Q

Multiple rounds of DNA replication can occur at _____________ in actively growing cells.

A

The same time.

Microbial Physiology (22 decks)

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