Chapter 10: DNA: Structure, Replication, and Variation Flashcards Preview

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Flashcards in Chapter 10: DNA: Structure, Replication, and Variation Deck (64):
1

In 1928, Frederick Griffith established that _______.

heat-killed bacteria harbor the constituent(s) 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.

2

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

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.

3

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

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

4

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?

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.

5

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

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

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

6

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

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.

7

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

DNase destroyed the transforming activity.

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

8

What observation did Griffith make in his experiments with Streptococcus pneumoniae?

The mouse did not survive when injected with a mixture of live, avirulent (rough) Streptococcus pneumoniae and heat-killed, virulent Streptococcus pneumoniae.

Something in the heat-killed preparation was able to transform the avirulent strain to a virulent form.

9

Guanine and adenine are ____ found in DNA.

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

10

The following statements about DNA structure are true?

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

The 5′–3′ orientation of each chain runs in opposite directions.

11

What is the complementary DNA sequence to 5′ ATGCTTGACTG 3′?

5′ CAGTCAAGCAT 3′

12

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

nucleoside

13

If 15% of the nitrogenous bases in a sample of DNA from a particular organism is thymine, what percentage should be cytosine?

35%

14

Identify three possible components of a DNA nucleotide.

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).

15

List 3 main differences between DNA and RNA.

RNA often occurs as both single- and partially double-stranded forms, whereas DNA most often occurs in a double- stranded form.

Ribose in RNA replaces deoxyribose in DNA.

Uracil in RNA replaces thymine in DNA.

16

One of the most common spontaneous lesions that occurs in DNA under physiological conditions is the hydrolysis of the amino group of cytosine, converting the cytosine to uracil. What would be the effect on DNA structure of a uracil group replacing cytosine?

A base substitution of G:C to A:T after two rounds of replication.

17

In 1953, Watson and Crick published a paper that described the ____.

In 1953, Watson and Crick published a paper that described the structure of DNA.

18

RNA differs from DNA EXCEPT what way?

the 5'-3' orientation of the polynucleotide strand

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

19

All EXCEPT which of the following statements are evidence that DNA, and not protein, is the genetic material in eukaryotes as well as bacteria?

DNA has four nucleotides.

DNA is located only where the primary genetic function is known to occur.

UV light is most mutagenic at a wavelength at which DNA and RNA strongly absorb.

Introduction of a cloned DNA into another organism results in the production of the corresponding protein product.

DNA has four nucleotides, and proteins have 20 amino acids. Protein was thought to have the chemical diversity and complexity necessary for the genetic material and DNA was not.

20

Which technique can be used to identify the location of genes on a chromosome?

FISH

Fluorescent in situ hybridization is a technique used to identify the location of a gene or a specific DNA sequence on a chromosome. The technique involves hybridizing single-stranded DNA probes, labeled with a fluorescent dye, to their complementary DNA strand in the genome. Their location on the chromosome can be viewed under a fluorescent microscope.

21

___ has no sulfur, and ____ have no phosphorus.

DNA has no sulfur, and proteins have no phosphorus.

22

In a DNA sequence, the purine ____ always pairs with the pyrimidine ____ , and the purine ____ always pairs with the pyrimidine ____.

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

23

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

5′−CGATATTGAGCTAAGCTT−3′

3′−GCTATAACTCGATTCGAA−5′

24

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.

How does the number of hydrogen bonds between the two bases affect the stability of a base pair?

25

All other factors being equal, the renaturation of the three classes of complementary nucleic acid sequences occurs in what order, from fastest to slowest?
A) unique sequences
B) moderately repetitive sequences
C) highly repetitive sequences

C,B,A

26

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.

27

Genetic material qualities

Replication
Storage
Expression
Variation by mutations

28

DNA to Protien

DNA >>> TRANSCRIPTION >>> RNA

mRNA, tRNA, rRNA >>> Ribosome

Ribosome >>> TRANSLATION >>> Protien

29

What is the genetic material?
History

Proteins and nucleic acids were considered possibilities.

Pre 1940’s, protein was leading contender:
-Protein is abundant in cell
-There is a wide variety of proteins in cells
-DNA was known to be made up of 4 nucleotides, nucleic acids were “too simple” to be genetic material
-Proteins contained 20 different amino acids, thus more potential variation

30

Model systems and differential labeling

Choice of the correct organism to perform an experiment

Methodology to deferentially label or separate molecules of interest

31

Evidence of DNA as Genetic Material:
Early work (1927) by Frederick Griffith

He analyzed virulence of various strains of Diplococcus pneumoniae

Some strains caused pneumonia in humans and mice (VIRULENT)

Some did NOT cause pneumonia (AVIRULENT)

Virulence of Diplococcus pneumoniae caused by polysaccharide capsule of bacterium

Those with capsule virulent & grow as smooth, shiny (S) colonies

Those without capsule are avirulent & grow as rough (R) colonies

Therefore, you can distinguish virulent from avirulent by growth characteristics

32

Griffith’s “Transforming principle” experiments

Model system: D. pneumoniae

Many sereotypes of rough and smooth colonies, eg. type I, type II, type III

Differential identification: Rough colonies= aviurlent; Smooth colonies= virulent

Griffith used seroytpes II and III of rough and smooth strains in his experiments (IIR and IIIS)

33

Griffith’s critical experiment Results:

Lastly also known, if you inject heat killed virulent bacteria:
LIVE MOUSE

Inject heat-killed virulent IIIS bacteria & living avirulent IIR bacteria into mouse:
DEAD MOUSE, Also recover living virulent IIIS bacteria from dead mouse tissue!

34

Griffith’s conclusion

Somehow the heat-killed IIIS bacteria were transforming the live avirulent IIR bacteria into virulent IIIS bacteria!

Griffith suggested the transforming principle might be something in the capsule, however the capsule alone did not cause pneumonia

35

Further transformation studies

Work extended in the 1930’s showed that transformation could happen in vitro

Soluble filtrate of heat-killed S cells was sufficient to cause transformation

Experimental evidence suggested that a chemical substance was responsible for the observed transformation phenomena

36

What is the “transforming” molecule?

Experiments of Avery, McLeod, and McCarty

Classic paper, published in 1944 after 10 years of work!

Conclusion: DNA was the transforming molecule

37

Avery, McLeod, and McCarty experiment , which demonstrated that DNA is the transforming principle.

First, they transformed with crude extract, but still got transformation

Next: treat remaining extract with protease to remove any remaining protein. Still got transformation

Next: treat remaining extract with RNase, degrades RNA: Transformation still occurs

Next: treat remaining extract with deoxyribonuclease, degrades DNA: Transformation does NOT OCCUR, therefore DNA must be transforming material!!!

38

Summary of Avery, MacLeod, and McCarty’s experiment

Differential separation of cell molecules from virulent strain

Test these separate components for ability to transform avirulent into virulent

Conclusion: The only molecule capable of transformation is DNA

39

Hershey-Chase Experiments

Provided convincing evidence of DNA as genetic material

Key was the model system used (E. coli and its viruses) and differential labeling of proteins and nucleic acids
-Provides method to “differentiate one thing from another thing”

40

Hershey-Chase Experiments: Model system

E. coli: bacteria easily grown and analyzed

T2 bacteriophage: a virus of E. coli that replicates within E. coli following infection

Proteins labeled with radioactive 35S, DNA labeled with radioactive 32P. This is Differential Labeling!

41

State of knowledge of bacteriophage T2 in early 1950’s

T2 phages ~50% protein, 50% DNA

Infection of E. coli is initiated by adsorption of phage to bacterial cell
The production of new viruses occurs within the E. coli cell

42

Summary of Hershey–Chase experiment. demonstrating that DNA, and not protein, is responsible for directing the reproduction of phage T2 during the infection of E. coli.

Demonstrated that DNA, and not protein, is responsible for directing the reproduction of phage T2 during the infection of E. coli.

Differentially labeled phages,
DNA with 32P; proteins labeled with 35S

Infect non-radioactive E. coli with labeled phage:
32P DNA, 35S proteins

Add phage to E. coli 32P medium.
Phage progeny labeled.
32P labeled phages infect unlabeled bacteria.

Add phage to E. coli 35S medium.
Phage progeny labeled.
35S labeled phages infect unlabeled bacteria.

43

KEY RESULTS of the Hershey–Chase experiment

Removed phage ghosts, only radioactive phage ghosts were those labeled with 35S

Viable phage were radioactive only from those that were labeled with 32P, therefore it was DNA that served as the genetic material!

44

Hershey-Chase conclusion

DNA was acting as the genetic material in T2 bacteriophage

Additional evidence followed in subsequent years that supported this conclusion

45

DNA as genetic material in eukaroytes

Indirect evidence: DNA found only where primary genetic function occurs, correlation of ploidy and the quantity of genetic material molecule

Indirect evidence: UV light is most mutagenic at wavelength that RNA and DNA absorb UV light the most

46

action spectrum

determines the most effective mutagenic UV wavelength

47

absorption spectrum

shows the range of wavelength where nucleic acids and proteins absorb UV light

48

The structure of DNA?

Nucleic acid chemistry helped to work out the structure of DNA

DNA is a nucleic acid, nucleotides are building blocks of nucleic acids

3 components: nitrogenous base, pentose (5-carbon) sugar, and a phosphate group

49

Nitrogenous bases

2 types:

Purines: 9 member double ring.
-Adenine (A)
-Guanine (G)

Pyrimidines: 6 member single ring.
-Cytosine (C)
-Thymine (T)
-Uracil (U)

DNA has A, C, G, T
RNA has A, C, G, U

Trick to remember:
Short name (purine), BIG (double ring) structure.
Long name (pyrimidine): SMALL (single ring) structure.

50

RNA

join Ribose to purine or pyrimidine in specific manner; result is ribonucleosides.

2 -OH groups in ribonucloeside

Structure resembles DNA, with some exceptions

Ribose sugar replaces deoxyribose

Uracil replaces Thymine

Most RNA is single stranded, but can base pair with itself

51

DNA

join 2-Deoxyribose to purine or pyrimidine in specific manner; result is deoxyribonucleosides

1 -OH groups
Deoxy= the difference in deoxyribonucleotide vs. ribonucleotide

52

DNA & RNA

Nucleotides are the building blocks of each

NMP: nucleoside monophosphate

NDP: nucloeside diphosphate

NTP: nucleoside triphosphate

53

Polynunucleotides

formed between mononucleotides with phosphodiester bond, linkage between C-3’ to C-5’

Linkage of two nucleotides by the formation of a 5' phosphodiester bond, produce a dinucleotide.

54

Watson-Crick DNA model

Previous studies were integrated into their model

Built models based on the known constraints experimentally determined

The ribbonlike strands constitute the sugar-phosphate backbones, and the horizontal rungs constitute the nitrogenous base pairs, of which there are 10 per complete turn. Major and minor grooves are apparent.

The antiparallel nature of the helix and the horizontal stacking of the bases.

55

The Watson-Crick DNA model features:

2 polynucleotide chains coiled around a central axis, forming right-handed helix

The 2 chains are anti-parallel: 5’-->3’ runs in opposite directions on each strand

The bases lay flat, perpendicular to the axis, stacked on one another 3.4 Angstroms apart

Each complete turn of the helix is 34 Angstroms

In the chain, there are major grooves alternating with minor grooves

The double helix is 20 Angstroms in diameter

56

Alternative DNA Forms

Using different isolation conditions, different forms of DNA have been obtained

2 forms known at the time of Watson & Crick model: A & B

B-DNA is the “normal” form of DNA in vivo

Other forms now found: C, D, E, P, & Z DNA (roles in vivo still not understood)

57

RNA & DNA analysis

Both RNA and DNA absorb UV light most strongly at 260nm

The amount of UV light absorbed is related to the concentration of RNA or DNA in a sample

This can be used to quantify RNA or DNA

A260 = 1 = 50ug/ml

58

Nucleic Acid Centrifugation

RNA’s differentiated by their sedimentation behavior

Measure is called Svedberg coefficient

Higher S, in general, larger molecule

Ribosomal RNAs the largest, eg. eukaryotic 18S rRNA

59

Gradient centrifugation of nucleic acids
and
Collection of fractions

Separation of a mixture of 2 types of nucleic acid by gradient centrifugation.

To fractionate the gradient, successive samples are eluted from the bottom of the tube.

Each is measured for absorbance of ultraviolet light at 260 nm, producing a profile of the sample in graphic form.

60

Nucleic Acid Separation

Another method: gel electrophoresis

61

DNA denaturation/hybridization

Hydrogen bonds of DNA can be broken with thermal energy (heat)

Lower temperature, DNA reanneals

Basis of molecular hybridiztion techniques (like FISH)

62

DNA reassociation/ Cot curves

Shear/sonicate DNA

Renature (denature?)

Slow initial association
Once association begins, then goes faster

Last association takes the longest, unique sequences finding one another

Cot1/2 = point where 50% of the DNA is reassociated

In general, smaller genomes reassociate faster than larger genomes

Linear relationship between Cot1/2 and genome size

63

The ideal time course for reassociation of DNA (C/C0) when, at time zero, all DNA consists of unique fragments of single-stranded complements.

C0: initial single stranded DNA in moles per liter of nucleotides

C: single stranded DNA concentration after time t has elapsed

The abscissa (C0t) is plotted logarithmically.

We can use comparisons of unknown DNA reassociation to known curves to characterize the unknown

64

Eukaryotic DNA reassociation

Initial rapid reassociation: repetitive DNA

Next, unique sequences take longer to find their match (curve flattens)

Lastly, the remaining sequences come together

Take home point: DNA reassociation curves reflect the complexity of the genome