Brainscape's Knowledge GenomeTM

Browse over 1 million classes created by top students, professors, publishers, and experts.


See full index

BLGY1232 > DNA > Flashcards

DNA Flashcards

1
Q

What must the genetic material be capable of doing?

A
  • replication with sufficient fidelity so organisms can pass on their characteristics to their offspring
  • undergoing some change or evolution could not occur
  • encoding vast amounts of information in order to programme the complexity of living organisms
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
2
Q

Griffiths - Discovers Genetic Transformation

Experiment Description

A
  • type IIR living, non virulent bacteria injected into mouse, the mouse survives and no bacteria are recovered
  • IIIS bacteria with polysaccharide capsule, living and virulent injected into mouse, mouse dies and IIIS bacteria recovered
  • IIIS with capsule but heat killed, injected into mouse, mouse survives, no bacteria recovered
  • IIR living, non virulent and IIIS heat killed, nonvirulent both injected into mouse, mouse dies, live IIIS recovered even though it was live IIR and dead IIIS injected
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
3
Q

Griffiths - Discover Genetic Transformation

Conclusions

A
  • Griffith’s thought that the live IIR were taking up something from the dead IIIS enabling them to become smooth
  • this was called the transforming principle
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
4
Q

1944 - Avery

Showing DNA was the transforming principle using rough and smooth bacteria

A
  • start with RNA and DNA from S (smooth) bacteria
  • treat with RNase
  • only DNA remains
  • mix with R (rough) bacteria
  • plate on growth medium
  • S transformants produced, some colonies had smooth phenotype instead of rough
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
5
Q

1944 - Avery

Showing that RNA was not the transforming principle using rough and smooth bacteria

A
  • start with RNA and DNA from smooth bacteria
  • treat with DNase
  • only RNA remains
  • mix with rough bacteria
  • plate on growth medium
  • no S transformants produced, all colonies have rough phenotype
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
6
Q

1944 - Avery

Conclusions

A

-the DNA must contain the instructions for the production of the smooth phenotype

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
7
Q

1952 - Hershey & Chase - Used bacteriophage T2 to demonstrate that DNA was the genetic material
-preparation of radioactively labelled T2

A

i) T2 phage mixed with E.coli and grown on medium containing radioactive phosphorous (P32), phosphorous is only in the DNA not the proteins so only DNA is labelled. This produces T2 phages with radioactively labelled DNA
ii) T2 phage mixed with E.coli and grown on medium containing radioactive sulphur (S35), sulphur only found in protein not DNA. This produces T2 phages with radioactively labelled head, sheath and tail fibres but not the DNA.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
8
Q

1952 - Hershey & Chase - Used bacteriophage T2 to demonstrate that DNA was the genetic material
-experiment to show that DNA is the genetic material

A

i) P32 labelled phages mixed with e.coil, blended to separate phage ghosts and E.coli infected with phage DNA, centrifuged, radioactivity only recovered in the host and not in the phage ghosts
ii) S35 labelled phages mixed with E.coli, blended briefly to separate phage ghosts and infected bacteria, centrifuged, radioactivity only recovered in the phage ghosts and not in the infected bacterial cells

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
9
Q

1952 - Hershey & Chase - Used bacteriophage T2 to demonstrate that DNA was the genetic material
-conclusions

A

-the genetic material injected into the bacteria by the phages must be DNA

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
10
Q

Charggaff’s Rules

A

-the percentage of the genome that is cytosine is equal to the percentage that is guanine
-the percentage of the genome that is thymine is equal to the percentage that is adenine
-bases are always found in the same ratio such that
C=G and A=T

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
11
Q

Franklin & Wilkes - X-ray Diffraction

A
  • provided evidence that DNA was a double helix
  • X ray beam directed at DNA sample, diffracted, projected onto screen
  • X pattern characteristic of a helix
  • height of the x corresponds to the 0.34nm gap between each base pair
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
12
Q

Structure of DNA

A
  • antiparallel double helix
  • hydrogen bonding between bases
  • sugar phosphate back bone
  • molecule has polarity
  • alternate minor and major grooves in the helix
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
13
Q

What does the structure of DNA explain about its function?

A
  • it can be replicated with high fidelity due to the complimentary base pairing
  • base pairing also explains why DNA obeys Chargaff’s rules
  • the linear base sequence provides means by which genetic information can be encoded
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
14
Q

Hydrogen Bonds Between Bases

A

G-C -> 3 hydrogen bonds
A-T -> 2 hydrogen bonds
-this means that a higher temperature would be required to break a molecule with only G-C base pairs than one with only A-T base pairs

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
15
Q

How do proteins access the base sequence of DNA?

A

the DNA sequence is accessible to the proteins that regulate gene expression via the MAJOR groove of the double helix

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
16
Q

What else can be the genetic material? (apart from DNA)

A

in certain organisms, it is RNA that is the genetic material

17
Q

Tobacco Mosaic Virus

-showing that RNA is the genetic material

A
  • starts with two different versions of the virus TMVA and TMVB
  • degrade both samples to remove the protein coat leaving only their RNA
  • coat the TMVA RNA with TMVB proteins and coat the TMVB RNA with TMVA proteins
  • infect two different leaves, one with each recombinant virus
  • the TMVA RNA with TMVB coat will produce TMVA progeny (RNA and protein)
  • the TMVB RNA with TMVA coat will produce TMVB progeny (RNA and protein)
  • this is because only the genetic material is passed onto the progeny an then encodes for the protein coat, the changes made to the original viruses were not changes to the genetic material therefore they were not passed on to the progeny
  • the genetic material of the TMV must be RNA
18
Q

Which end of a DNA strand carries the phosphate group?

A

5’ end

19
Q

Which end of a DNA strand carries the -OH group?

A

3’ end

20
Q

Which direction do polynucleotides grow in?

A

polynucleotides grow when nucleotides are added to their 3’ ends

21
Q

The Central Dogma of Molecular Biology

A
  • DNA can be copied to produce more DNA by replication
  • DNA can be converted into RNA by transcription
  • RNA can be converted into DNA by reverse transcription
  • RNA can be converted into a protein by translation
22
Q

DNA Replication

A
  • hydrogen bonds between the base pairs are broken
  • the two strands separate
  • a new complementary strand is synthesised opposite each strand by DNA polymerase
23
Q

DNA Polymerase

A
  • one enzyme with three activities
    1) extends new chain 5’ to 3’ by copying the template strand from its 3’ to its 5’
    2) 5’ to 3’ exonuclease activity: chews away ahead of the polymerase
    3) 3’ to 5’ exonuclease activity: checks for mismatches and removes them
24
Q

Transcription

A
  • hydrogen bonds between the template and coding strand are broken
  • RNA polymerase travels along the template strand from the 3’ end of the template strand to the 5’ end of the template strand
  • this allows the new RNA chain to grow by extension of its 3’ end
  • the RNA chain will have the same sequence as the coding strand
25
Q

Translation

A
  • the ribosome moves from the 5’ end of the mRNA to the 3’ end of the mRNA
  • the N terminal amino acid is added first, and the C terminal amino acid is the last to be added
  • each codon, sequence of three bases, codes for one amino acid
26
Q

Start Codons

A

AUG (usually)

GUG (sometimes)

27
Q

Terminator Codons

A

UGA
UAA
UAG

28
Q

Reverse Transcription

A

-catalysed by reverse transcriptase, an RNA dependent DNA polymerase

BLGY1232 (19 decks)

Brainscape helps you reach your goals faster, through stronger study habits.
© 2024 Bold Learning Solutions. Terms and Conditions