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MBG*2040 > Exam Review > Flashcards

Exam Review Flashcards

1
Q

Watson and Crick suggested that the double helical DNA model suggests a _________ copying mechanism which means…

A

Semi conservative

after one round of DNA replication, each DNA contains 1 parental and 1 newly syntheized strand

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2
Q

What are the 3 models of DNA copying mechanisms

A
  1. Conservation replication
  2. Dispersive Replication
  3. Semi-conservative Replication
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3
Q

What are the 3 modes of DNA replication

A
  1. Theta replication
  2. Rolling circle replication
  3. Linear chromosome replication
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4
Q

Describe theta replication

A
  • Replication that occurs in mostly circular DNA
  1. double stranded DNA unwinds at the replication origin
  2. This produces single stranded templates DNA synthesis and replication bubble forms with a replication fork at each end
  3. the fork then proceeds around the circle bidirectionally
  4. Eventually two circular DNA molecules are produced
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5
Q

Rolling Circle Replication

A
  • Specialized form of replication that occurs in the F factor and some viruses only
  1. replication is initiated by a break in one of the nucleotide strands
  2. synthesis begins at the 3’ end of the broken strand; the inner strand is then used as a template. the 5’ en d of the broken strand is displaced
  3. cleavage releases a single stranded linear DNA and a double-stranded circular DNA. Replication is unidirectional.
  4. the linear DNA may circularize and sere as a template for the synthesis of a complementary strand
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6
Q

linear replication

A
  • Linear chromosome replication occurs in the linear chromosomes of eukaryotic cells.
  • This DNA replication is bidirectional.
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7
Q

What are the requirements for DNA replication (4)

A
  1. DNA dependant DNA polymerase
  2. Four deoxyribonucleoside triphosphates (dNTPs)
  3. A single stranded template of DNA to be copied
  4. An RNA primer (provides the 3’ OH end to initiate DNA synthesis by DNA polymerase)
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8
Q

DNA is ALWAYS synthesized in what direction?

A

5’ -> 3’

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9
Q

DNA synthesis is ____ on the leading strand and ___ in the lagging strand

A

continuous, discontinuous

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10
Q

Fragments formed during synthesis of the lagging strand is called

A

Okazaki fragments

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11
Q

What are DnaA proteins involved in prokaryotic DNA replication

A
  • They are initiator proteins that bind to oriC (the origin)
  • This causes a short stretch of DNA to unwind
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12
Q

DNA Helicase (Prokaryotes)

A
  • Unwinds the DNA in the 5’ to the 3’ direction
  • Only travels on the lagging strand
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13
Q

Single Strand binding proteins (Prok.)

A

Coats unwound single stranded DNA to protect and keep DNA single-stranded (prevents hairpins)

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14
Q

DNA Gyrase (prok.)

A
  • A topoisomerase that makes double stranded breaks in DNA to relive the torsional strain that builds up from the activity of the DNA helicase.
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15
Q

____ requires a primer to initiate synthesis ___ does not

A

DNA polymerization, RNA polymerase

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16
Q

Primase (RNA Polymerase) (prok.)

A

Synthesizes short RNA primer that provides the 3’ OH end for DNA polymerase to begin DNA synthesis

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17
Q

Primase binds to ___ to initiate synthesis

A

Helicase

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18
Q

DNA Polymerase (prok.)

A

Elongates the new polynucleotide strand by catalyzing DNA polymerisation.

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19
Q

T or F: DNA polymerase requires a primer

A

True

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20
Q

DNA Pol III (prok.)

A
  • The main DNA Pol
  • has 5’-> 3’ polymerase activity and 3’ to 5’ exonuclease activity.
  • can add dNTP on the 3’ end and also back up to remove a nucleotide that has been mis incorporated
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21
Q

DNA Pol I (prok.)

A
  • Similar to DNA Pol III, but has 5’ -> 3’ exonuclease activity so it can remove the RNA primer and replace it with DNA
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22
Q

DNA Ligase (prok.)

A

Makes phosphodiester bonds between 5’ phosphate and the 3’ OH froup and seals the nicks in the lagging strand

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23
Q

Helicase (Euk.)

A
  • Binds to initiator protein on double streanded DNA
  • involves the MCM complex of proteins
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24
Q

In eukaryotes DNA replication only occurs during ___ phase

A

S

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25
Q

Polymerase Alpha (euk)

A
  • Has primase activity
  • generates the RNA primer
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26
Q

Polymerase Epsilon (euk)

A

preforms leading strand replicaiton

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27
Q

Polymerase delta (euk)

A

Preforms lagging strand replicaition

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28
Q

when DNA is doubled in euk. histones must be ___ and the modifications on them need to be replicated

A

doubled

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29
Q

The telomere problem

A
  • The chromosome end will be degraded causeing shortening every round of replication.
  • leaves gaps in DNA that are not replicated causing shortening
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30
Q

Telomerase

A
  • Activity extends eukaryotic chromosome ends in replicating cells
  • can extent the 3’ end without use of complementary DNA template
  • results in shortening telomeres over time
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31
Q

Shorter telomeres is associated with…

A

Cellular senescence and death

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32
Q

What makes RNA less stable than DNA?

A

The presence of 2’OH groups in ribose causes it to react intramolecularly with the 3’OH site resulting in phosphate breakage.

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33
Q

What is transcription

A

Synthesis of RNA from DNA

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34
Q

In what Direction is RNA synthesis

A

RNA is synthesized in the 5’ to 3’ direction using the 3’ to 5’ DNA template strand.

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35
Q

The precursors to RNA syntheis are..

A

Ribonucleoside triphosphates (rNTPs)

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36
Q

RNA Polymerase Holoenzyme Subunits

A
  • alpha
  • beta
  • beta-prime
  • omega
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37
Q

What is the function of the alpha subunit?

A

Involved in the assembly of the tetrametric core

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38
Q

what is the function of the beta subunit?

A

Contains the ribonucleoside triphosphate (rNTP) binding site

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39
Q

What is the function of the beta prime subunit?

A

Contains the DNA binding region

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40
Q

What is the function of the omega subunit?

A

It helps to stabilize the tetrameric core

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41
Q

What is the function of the sigma subunit

A
  • it binds to the RNA polymerase tetrametric core and assists in the correct initiation of transcription specifically at the promotor region of the prokaryotic gene. Many types of sigma factors -> allow for specificity.
  • recognizes and binds to the -35 and -10 consensus sequances in the promotor region, properly positioning the RNA polymerase to begin transcription
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42
Q

Why is the -10 consensus sequence prone to unwinding?

A

Due to its AT rich content

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43
Q

Describe initiation of transcription in prokaryotes

A
  1. holoenzyme binds to the promoter and RNA polymerase is positioned over the transcription start site and has unwound the DNA to produce a single stranded template
  2. the orientation and spacing of the consensus sequence on a DNA strand determine which strand will be the template for transcription and therby determine the direction of transcription
  3. RNA polymerase binds, unwinds and joins first 2 nucleotides
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44
Q

Does RNA synthesis require a primer

A

no

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45
Q

Elongation

A
  1. Occurs when sigma factor is released and RNA polymerase begins to move along the 3’ to 5’ DNA template strand.
  2. Complementary nucleotides are added
  3. Localized DNA unwinding ahead of RNA polymerase and generates a “transcription bubble”
  4. Transcription bubble moves along with the unwound DNA rewinds behind the bubble
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46
Q

RNA polymerase has both helix ___ and ___ activities

A

unwinding, rewinding

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47
Q

Termination

A
  • stop when RNA polymerase reaches the terminator region of the gene
  • occurs upstream of where the actual termination takes place
  • the newly synthesized RNA together with the RNA polymerase are released
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48
Q

What are the two major types of terminators in bacterial cells?

A

Rho-dependant (requires the Rho factor) and Rho-independant (aka intrinsic terminators)

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49
Q

Describe Rho-dependant Termination

A

Sequence Features:

  1. DNA sequence of terminator site causes polymerase to pause
  2. DNA sequance upstream of terminator encodes a stretch of RNA that is C rich and devoid of secondary structure

a) called the rho utilization

b) rho binds to rut site

  • Rho moves alonf RNA towards paused polymerase
  • rho factor has helicase activity
  • unwinds the RNA-DNA hybrid
  • ends transcription
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50
Q

Describe Rho-independent termination

A

Common features

  1. contains inverted repeats
  2. a string of 6-9 A’s follows the inverted repeats
  • poly A sequance is transcribed into a poly U tail after hairpin is transcribed
  • causes polymerase pause
  • hairpin forms and destabilizes the DNA-RNA hybrid
  • assisted by weak A-U base pairing
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51
Q

true or false: Pol I, II, and III are inly recruited to their promoter specific accessory proteins

A

true

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52
Q

Initiation in Eukaryotes

A

Involves a step wise assembly of general Transcription Factors of Pol II

  • Tata box is the fist to assemble then followed by the remaining general TFs and Pol II
  • assembly of the TFs and Pol II causes 11-15 bp of surrounding DNA to unwind
  • template strand positioned within active site of RNA pol
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53
Q

Is the pre-initiation complex sufficient to initiate transcription?

A

yes

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54
Q

what is a mediator

A

involves multi-subunit that permits interactions with other activator/repressor proteins bound to upstream/downstream regulatory regions or enhancer sequences

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55
Q

Elongation in Eukaryotes

A
  • Polymerase moves along the template strand leaving transcription factor intact at the promoter for reinitiation of transcrption with new polymerase
  • RNA pol maintains 8nt transcption bubble
  • DNA-RNA hybrid bends at right angle
  • positions the OH group at the active site where nucleotides are added to the 3’ end of the growing chain
  • newly sythesized RNA is seperated from DNA and exits through another cleft
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56
Q

Termination in RNA Pol I

A

Requires a termination factor similar to the rho factor in prokaryotes

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57
Q

Termination in RNA Pol III

A

Ends after transcribing a terminator sequence that produces a string of Us that is downstream of hairpin. Similar to rho-independant termination

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58
Q

Termination in Pol II

A

Transcription continues past termination sequence

  • RNA is cleaved at consensus sequence in the RNA
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59
Q

Termination in Eukaryotes

A
  • Cleaved RNA results in 2 RNAs: one that will encode a protein and the other with its 5’ trailing out of the RNA polymerase
  • Rat 1 (5’ -> 3’ exonuclease) attaches to the 5’ end and chews up the RNA
  • rat 1 does not bind to the 5’ end of the preotein coding mRNA because it is protected by a 5’ cap
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60
Q

What is Colinearity

A
  • the coding region in prokaryotes that is uninterupted
  • The sequance of the gene directly corresponds to the amino acid sequance

the number of nucleotides is proportional to the number of amino acids (3:1)

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61
Q

The shine delgarno sequance is found in ___ (prok or euk)

A

Prokaryotes

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62
Q

What are the three primary regions of a mature mRNA

A

5’ Untranslated Region (5’UTR): Does not code for amino acids, binds the ribosomal complex

Protein Coding Region: Comprises the codons that specify the amino acids begins with a start codon and ends with a stop codon

3’ UTR: does not code for amino acids. Affects the stability of the mRNA and regulates its translation

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63
Q

Exons are…

A

Protein coding segments

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64
Q

Introns are…

A

inervating (non-coding) segments

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65
Q

removal of ___ along with additional RNA processing steps are required to form the mRNA that will be translation into ___

A

introns, polypeptide

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66
Q

What is required to remove the intron?

A

the splicosomal complex which contains small nuclear RNAs and proteins

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67
Q

Where does splicing take place?

A

In the nucleus

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68
Q

What are the 3 main processing steps in eukaryotic nuclear pre-mRNA

A
  1. the addition of 7-methyl guanosine cap. linked betwen the 5’ phosphate of the 7 MG and the 5’ phophphate of the first RNA
  2. addition of a poly A tail to the 3’ end by poly A polymerase
  3. Removal of introns from pre-mRNA called splicing
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69
Q

Describe the 5’ cap

A

Methyl groups are added to:

  • 7’ position of the terminal guanine
  • 2’ OH group of the sugar of the first and sometimes second nucleotide
  • to adenine if it is the second nucleotide
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70
Q

Descibe the 3’ poly A tail

A

50-250 As added to the 3’ end

  • A are added without a template (polyadenylation)
  • Poly A consensus sequance (AAUAAA) is the polyadenylation signal located 11 to 30 nucleotides upstream of the cleavage site
  • affects stability of mRNA, facilitates attachment of the ribosome and export out of the nucleous
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71
Q

What is the polyadenylation signal

A

AAUAAA

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72
Q

Descibe RNA splicing

A
  • removal of introns from pre-RNA
  • Removal of introns must be precise in order to properly fuse the 3’ end of one exon to the 5’ end of the next exon
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73
Q

How does the eukarotic cell ensure that introns are properly removed?

A

Every intron has 3 conserved sequances thata re required for its precise removal:

  1. 5’ splice sequances containing the junction seequances GU (5’ end of the intron)
  2. 3’ splice sequence containing the junction sequence AG
  3. Intron branch point: a conserved A residue is located 18-40 nt upstream of 3’ splice site
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74
Q

What are the Steps of RNA splicing

A
  1. Pre-mRNA is cut at the 5’ splice site and the 5’ end of this intron attaches to the branch point (Lariat)
  • 5’ phosphate of the G nucleotide binds with the 2’ OH group of the A nucleotide at the branch point
  1. Cut is made at the 3’ splice site and simultaneously the 3’ end of exon 1 becomes covalently attached to the 5’ end of exon 2
  • intron is released as a lariat and degraded
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75
Q

Describe a Spliceosome

A
  • RNA/ pritein structure
  • contains 5 small nuclear RNAs
  • The snRNAs associated with about 300 small proteins to form small nuclear SNURPS
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76
Q

What is alternative splicing?

A
  • an alternative processing pathway
  • pre-mRNA is processed in different ways to produce alternative types of mRNA
  • alternative exons are used
  • results in different proteins from the same DNA sequance (unless in the UTR region)
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77
Q

Describe the concept of multiple 3’ cleavage sites?

A
  • alternative procssing pathway

-Can generate longer or shorter exons

  • may or may not produce a different protein. depends if upstream or downstream of stop codon
  • note that this is not the same as the 3’ splice site
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78
Q

What is Calcitonin

A

an alternative processing pathway

regulates calcium concentration in the blood

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79
Q

what is CGRP

A

an alternative processing pathway

Causes dialation of blood vessels and the transmission of pain

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80
Q

How is RNA editing carried out?

A
  • Carried out by guide RNAs that can be made in the cell
  • gRNAs direct the insertion of urodine bases into the mRNA by a repair polymerase
  • This perminatly modifies the mRNA by making new codons that specify new amino acids
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81
Q

What is the editing of Apoplipoproteins B mRNA

A
  • Apoplipoproteins are blood proteins that carry lipids
  • RNA editing by cytidine deaminase changes C to U converting a normal glutamine codon to a termination codon which truncates the protein and gives it a different function with respect to lipid binding
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82
Q

Describe tRNA modifications

A
  • the anticodon of the tRNA base pairs with the codon of mRNA
  • The amino acid is covalently attached to the 3’ end of the tRNA
  • tRNAs often contain modified ribonucleotides through the action of tRNA modifying enzymes
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83
Q

What processing events do tRNAs undergo

A
  • removal of extra 5’ and 3’ sequances
  • removal of introns
  • nucleotides 5’-CCA-3’ are added to the 3’ end of the tRNA for all tRNAs
  • modification of several nucleotides
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84
Q

What is a ribosome

A
  • comprised of 2 subunits and has a key role in protein synthesis including the formation of peptide bonds between amino acids
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85
Q

Where does rRNA sytheisis occur in both prok. and euk.

A

Prokaryotes

  • No nuclear envelope so rRNA synthesis and ribosome assembly occur in the cytoplasm

Eukaryotes

  • nucleus is the site of rRNA synthesis and ribosome assembly
86
Q

snRNAs

A

Splices introns from genetic transcripts

87
Q

snoRNAs

A
  • act in complezes with proteins
  • in euk., guide the enzymatic chemical modifications of ribosomal RNAs, transfer RNAs and small nuclear RNAs
88
Q

siRNA, miRNA

A
  • found in euk.
  • act as short, single stranded RNAs that bind to complementary sequances in mRNA
89
Q

Crispr RNA

A
  • found in prok.
  • encoded by DNA sequances found in prokaryotic genomes

works in association with the prokaryotic cas 9 nuclease to cleave foreign DNA that might happen to enter a host cell

90
Q

LncRNA

A
  • known to function in eukaypotic cells
  • Longer than miRNAs
  • regulate and control gene expression at the level transcription or translation by binding mRNA or sequestering micro-RNAs that control gene expression
  • bind or recruit proteins involved in DNA modification
91
Q

What is sickle cell anemia?

A

It is caused by single acid change in one of the protein chains that make up hemoglobin red blood cells

92
Q

true or false: transcription and translation often occur simultaneously in prokaryotes

93
Q

where do transcription and translation occur in eukaryotes?

A

the nucleolus and cytoplasm respectively

94
Q

What is the one gene, one colinear polypeptide theory

A

The sequance of base pair triplets in the coding region of a gene specify a colinear sequance of amino acids in its polypeptide product

95
Q

Structure of Amino acids

A
  • Proteins are made of polypeptides
  • amino acids has a free amino group (N-term), a free carboxyl group (C-term), and side group R
96
Q

How are amino acids joined

A

they are joined covalently by peptide bonds

97
Q

Why is the genetic code a triplet code?

A

4^3 cause that is the only way there was enough codons to encode for 20 amino acids

4^3 = 64 therefore there is degeneracy in the code

98
Q

How was the genetic code determined

A
  1. Synthetic mRNAs of repeated sequance (homopolymers) were tested in vitro for protein synthesis: produced homopolypeptides
  2. Each individual amino acid was tested until the polypeptide set off the geiger counter
  3. this confirmed UUU specifies phenylalinine
  4. they then moved on to the next homopolymer and repeat the process
99
Q

What is the wobble position

A

Flexibility in the binding at the 3rd codon position (1st anticodon.

  • This permits the tRNA anticodon to bind codons that bind the same amino acid even if the sequance isnt complementary to the tRNA
100
Q

What feature in the genetic code explains degeneracy

A

Oftentimes the 3rd codon position can be changed and still specify the same amino acid

101
Q

What are the 4 Macromolecules of Translation

A
  1. Ribosomes/ ribosomal RNA
  2. Amino acid activating enzymes
  3. tRNA molecules
  4. soluble proteins involved in polypeptide chain initiation, elongation and termination
102
Q

What is a Ribosome?

A
  • Ribosomes are composed of proteins and several diff rRNAs
  • Ribosomes are composed of both a large and small subunit
  • a ribosome is an RNA machine with key roles in protein synthesis including the formtion of peptide bonds between amino acids
103
Q

What is Aminoacyl tRNA syntheses?

A
  • this enzyme charges the tRNA with its specific amino acid
  • an amino acid becomes attached to the appropriate tRNA by the amino-acyl tRNA synthetade in a two step rxn:
  1. amino acid reacts with ATP
  2. the amino acid is tranfered to tRNA and AMP is released
104
Q

What are tRNAs

A
  • tRNA are adapters between amino acids and the codons in mRNA
  • the anticodon of the tRNA base pair with the codon of mRNA
  • The amino acid is covalently attached to the 3’ end of the tRNA
105
Q

Translation initiation (prok)

A
  1. 16S of the 30S small ribosomal sunbunit contains the complement to the shine-delgarno sequance in the mRNA. Pairing of the two sequances positions the ribosome near the AUG start codon
  2. IF-3 inhibits the large subunit from binding (30S) small subunit
  3. IF-1 and IF-2 position fMet-tRNA over the start codon
  4. IF-3 dissociated allowing the 50S subunit to bind
  5. GTP is hydorlyzed to GDP and the remaining IF factors dissociate
  6. complete comllex is now called 70S initation complex
106
Q

What. is the difference between prok. and euk. translation initiation

A
  1. fmet in prok. just met in euk.
  2. no shine delgarno sequance in euk.
  3. Kozak sequance is used in place of the shine delgarno sequance
  4. poly A tail of the euk mRNA interacts with the mRNA 5’ 7-MG cap structure via the cap binding protein complex to promote translation initiation
107
Q

Translation elongation

A
  1. A charged tRNA binds to the A-site of the ribosome
  2. once in the A site, GTP is hydrolyzed to GDP and is released along with EF-Tu
  3. rRNA in large subunit catalyzes formation of a peptide bond
  4. chanin is transfered to the aa in the A-site
  5. rRNA subunit catalyzes formation of the peptide bond
  6. releases the tRNA in the P site from its AA
108
Q

what site does the F-Met or first Met in a protein strat in during protein synthesis

A

the P-site

109
Q

what direction does the ribosome move does the mRNA

A

5’ -> 3’

110
Q

What direction does the polypeptide chain grow

A

N-term -> C-term

111
Q

Translation Termination

A

Termination occurs when stop codon enters the A-site of the ribosome?

  1. when stop codon is encountered a release factor (RF) binds to the A-site
  2. RF-1 or 2 binding alters activity of the peptidyl transferase, releasing the polypeptide and leading to the termination.
  3. the binding of RF 3 and GTP to the ribosome assists in the dismantling of the entire complex
112
Q

What stop codons do RF-1 recognize

113
Q

what stop codons does RF 2 recognize

A

UAA, and UGA

114
Q

what does Chloromycetin inhibit?

A

Formation of peptide bonds

115
Q

what does Erythromycin inhibit

A

translocation of mRNA along the ribosome

116
Q

what does Neomyacin inhibit

A

Interactions between tRNA and mRNA

117
Q

What does Streptomycin inhibit

A

initiation of translation

118
Q

what does tetracycline inhibit

A

binding of tRNA to ribosome

119
Q

what does Paromomycin inhibit

A

Validation of mRNA-tRNA match

120
Q

Define a Mutant

A

An organism that carries one or more mutations in its genetic material

121
Q

True or False: Mutations are the source of all genetic variation

122
Q

During meiosis, recombination between __ chromosomes rearranges genetic variability into a new combination

A

homologous

123
Q

Somatic Mutations

A

Occur in somatic cells; it will occur only in the descendants of that cell and will not be transmitted to the progeny.

124
Q

Germinal Mutations

A

Occur in germ-line cells and will be transmitted through the gametes to the progeny

125
Q

What are the 4 main types of Gene Mutations?

A
  1. Base Substitutions
  2. Insertions and Deletions
  3. Tautomeric Shift
  4. Expanding nucleotide repeats
126
Q

Describe a transition Mutation

A

Replaces a pyrimidine with another pyrimidine, or purine with purine.

127
Q

Describe a transversion mutation

A

replaces a pyrimidine with a purine or vise versa

128
Q

Describe insertion and deletion mutations

A

Add one or two base pairs that alter the reading frame of the gene distal to the site of the mutation.

129
Q

What is a tautomeric shoft mutation?

A

The movement of H atoms from one position in a purine or pyrimidine base to another.
- This is a rare mutation that can occur spontaneously during DNA replication where they alter DNA base pairing

130
Q

Can tautomeric shifts cause spontaneous mutations?

131
Q

Can a tautomeric shift generate A:C and G:T pairs?

132
Q

Describe expanding nucleotide repeats

A
  • Expansion of triple repeats
  • involves DNA replication
  • Dynamic mutation because the repeat copy number is in flux with each round of replication.
  • Mutations can be outside of the coding region that affects its expression
133
Q

Forward Mutation

A

wildtype -> mutant type

134
Q

Reverse Mutation

A

Mutant type -> wild type

135
Q

Missense Mutation

A

amino acid -> different amino acid

136
Q

Nonsense Mutation

A

Sense codon -> nonsense codon (stop)

137
Q

Silent Mutation

A

codon -> synonoomous codon still specifying the same amino acid

138
Q

Neutral Mutation

A

amino acid change with no observable change in protein function. The amino acid changes to one of similar chemical type

139
Q

Loss of function mutation

A

Result of mutations that cause complete or partial loss of normal protein function

140
Q

Gain of function mutation

A

is the result of a mutation that causes the cell to produce a protein or gene product whose function is not normally present

141
Q

Conditional Mutation

A

Is expressed only under certain conditions

142
Q

lethal mutation

A

causes pre mature cell death

143
Q

Temperature sensitive allele

A

An allele whose functional product is conditional on the temperature

144
Q

Supressor Mutation

A

A second site mutation that hides or suppresses the effect of the first mutation

145
Q

Intragenic Supressor

A

Supressor mutations within the same gene

146
Q

Intergenic suppressor

A

A suppressor mutation can be present within a different gene

147
Q

Induced mutations

A

Mutations that occur as a result of exposure to external factors

148
Q

Spontaneous mutations

A

Mutations that occur under normal circumstances as a result of internal factors

149
Q

Spontaneous Replication Errors

A
  • Tautomeric shifts
  • Mispairing due to other structures
  • Incorporated errors and replicated errors
  • causes deletions and insertions
  • spontaneous chemical changes
150
Q

What causes deletions and insertions (2)

A
  1. Strand slippage: can be insertion or deletion
  2. Unequal crossing over during meiosis
151
Q

Describe Depurination

A
  • a spontaneous chemical change
  • breakage of the covalent bond between the purine base and the 1’ carbon atom of the deoxyribose sugar. creates an apurinic site
152
Q

Describe Deamination

A
  • spontaneous chemical change
  • Deamination is a loss of amino group, typically from cytosine. This gives rise to uricil, which will pair with adenine during replication. In the next cycle it will pair with T. End result is C -> T
153
Q

Mutagen

A

An environmental agent that increases the mutation rate above the spontaneous rate

154
Q

Base Analogue (Mutagen)

A

Chemical with structures similar to any of the standard 4 mucleotides

155
Q

What is 5-bromouracil an anolgue of

A

thyamine which causes G -> transition

156
Q

what is 2-aminopurine an analogue of

A

Adenine which can mispair eith cytosine

157
Q

What are Alkylating Agents (mutagens)

A

Mutagens that react with DNA bases and add methyl or ethyl groups
- EMA and mustard gas

158
Q

Describe Deaminating Chemicals (mutagens)

A
  • This is in addition to spontaneous deamination noted for spontaneous chemical changes
  • Nitrous acid causes c->T and G->A transitions
159
Q

Describe Hydroxylamine (mutagen)

A
  • Vary specific base-modifying mutagen
  • adds hydroxyl group to cytosine -> hydroxylaminocytosine
  • increases occurence of rare tautomer that pairs with adenine
  • only affects cytosine
  • C -> T transition
160
Q

Oxidative Radicles (mutagens)

A
  • Reactive forms of oxygen
  • producef through normal anerobic metabolism
  • can also be produced by chemicals and radiation
  • can affect different bases
161
Q

Intercalating Agents (mutagens)

A
  • sandwich themsekces (intercalate) between adjacent base pairs
  • distorts DNA helix
  • causes insetions and deletions -> frameshift mutations
162
Q

Radiation (mutagen), 2 types of radiation

A

Ionizing radiation
- dislofge electrons from the atoms they encounter
- Stable atom -> freee radicles, reactive ions
- Breaks phosphodiester bonds leading to double stranded breaks

UV Radtion
- less intense than ionizing
- Pyrimidine bases absorb UV light
- induces chemical bonds between two adjacent pyrimidine molecules on the same strand of DNA
- thymidine dimers most frequant

163
Q

What are the two general features of DNA Repair

A
  1. Most DNA repair mechanisms require double stranded DNA: As one nucleotide is being repaired, the complementary strand is acting as a template for the repair
  2. DNA repair has built in redundancy. Many types of DNA damage can be repaired by more than one repair system, helps ensure survival.
164
Q

DNA Mismatch repair

A
  • Old DNA has methyl groups attached to it, new DNA does not
  • Allows the MMR machinery to differentiate the new versus original strand
  • MMR degrades new strand between nick and mismatched bases
  • Polymerases and DNA ligases fill the gap using the old strand as the template
  • Similar processes between prok. and euk., but euk. dont have methylated DNA, so unknown how MMR recognizes old and new strands
  • Mutations in MMR leads to elevated somatic mutations -> cancer
165
Q

Direct DNA Repair

A

Retores the original structure:
1. photoreactivation of UV-induced pyrimidine dimers. Breaks the covalent bond in the dimers
2. Enzyme call O-methylguanine DNA methyltransferase removed the methyl grouo from the O methylguanine, restoring th ebase to guanine

166
Q

Base Excision DNA Repair

A
  • DNA glycosylases excises the damaged base or bases
  • these enzymes are specific for each modified base
  • just the base is removed not the deoxyribose sugar
  • AP endonuclease cleaves the phosphodiester bond
  • other enzymes remove the deoxyribise sugar
  • A DNA polymerase fills in the gap using the undamaged complementaey strand of DNA as the template
  • DNA ligase seals the brak left by DNA polymerase
167
Q

Nucleotide-Excision DNA Repair

A
  • NER is complex of enzymes that scans the DNA for distortions in the DNA
  • DNA is seperated and stabilized by single stranded DNA Binding proteins
  • Phosphodiester bonds on either side of mutation are cleaved
  • damaged strand is removed by helicase
  • gap is filled in with DNA polymerase
  • ends are sealed by DNA ligase
168
Q

Repair of Double stranded DNA breaks

A
  • Caused by ionizing radiation, oxidative radicles and other DNA damaging agents
  • stalls DNA replication and could lead to chromosomal rearrangements which in turn can cause deletions, duplications, inversions and translocations

Two paths repair these
1. Homology directed repair
2. nonhomologous end joing

169
Q

Homology dependant double stranded DNA repair

A
  • homologous recombination
  • repairs a briken DNA molecule by using the genetic information contained in the sister chromatid
  • Employs the same system as in homologous recombination that occurs during crossing over during meiosis
170
Q

Non-homologous end joining in double stranded breaks

A
  • Repairs double stranded breaks
  • independant of homologous template
  • proteins recognize broken ends of DNA and bind them together
  • leads to deletions and insertions and translocations
  • important in somatic recombination
  • involves DNA polymerases, nucleases and ligases
171
Q

The enzyme with a built-in RNA strand template, which enables DNA synthesis at the end of the eukaryotic chromosome, is:

A

Telomerase

172
Q

What are transposable elements

A

Segments of DNA capable from moving from one location in a chromosome to another, or even to a different chromosome

173
Q

General Characteristics of Transposable Elements

A
  1. Create staggered breaks in target DNA
  2. Transposable element attaches to single stranded ends of DNA
  3. DNA is replicated at the gaps
    - flanking direct repeats
    - generated in the process of transposition
  4. Terminal inverted repeats associated with the transposable elements
    - inverted and complimentary
174
Q

Terminal inverted repeats are recognized by …. that catalyze transposition

A

Transposases

175
Q

There are ___ classes of transposable elements based on ___

A

2, DNA or RNA

176
Q

Class I. Retrotransposons (RNA intermediate)

A
  • DNA copy of element made by reverse transcription from its RNA and then inserted into a new chromosomal site
  • requires a reverse transcriptase - reverse flow of genetic information from RNA to DNA

Two Types;
1. Retrovirus or Retrovirus like elements
2. Retroposons or retrotransposons

Both use replicative transposition

177
Q

Class II. DNA transposons-catalyzed by transposase

A
  1. Non-replicative transposons - element is physically cut out of one site in a chromosome or plasmid and pasted into a new site (cut and paste)
  2. Replicative Transposons - element is replicated with one copy and inserted at a new site and one remains at the og. site (copy and paste)
178
Q

How are transposable elements controlled

A
  1. DNA is methylated where transposons are common
  2. alterations in chromatin structure prevent transcription
  3. Control of transposase translation by piwi-interacting RNAs (piRNAs). Bind to Piwi proteins and inhibit tranlation of transposase mRNA
179
Q

what is the mutagenic effect (transposable elements)

A
  • Transposition entails the exchange of DNA sequances and recombination
  • Leads to rearrangements such as duplications, deletions, inversions, and translocations
180
Q

mutanagenic effect can occur by pairing of direct repeat which leads to ___

181
Q

mutanagenic effect can occur by pairing of inverted repeat which leads to ___

182
Q

What are the transposable element groups in bacteria

A
  1. Simple transposable elements
  2. Complex transposable elements
183
Q

Simple Transposable Elements

A
  • only carry the information required for movement
  • contain the transposase gene
  • contain both terminal inverted repeats and direct repeats
184
Q

Complex transposable elements

A

Composite transposons
- any sequence that is flanked by two insertion sequences

Noncomposite transposons
- lack insertion sequances
- posses the transposase gene
- have terminal inverted repeats

185
Q

Ac-Activator Gene (transposons)

A
  • Contains terminal inverted repeats
  • has transposase gene
  • autonomous transposition
186
Q

Ds Dissociation gene (transposon)

A
  • Similar to Ac
  • Posses inactivated transposase gene
  • Requires transposase from Ac to transpose
  • Noautonomous transposition
187
Q

Ac stimulates chromosome ___ at site of Ds

188
Q

Transposable elements in Eukaryotes

A

p elements
- possess terminal inverted repeats
- DNA transposons
- contains both a transposase and repressor of transposition
- creates phenomenon of hybrid dysgenesis

189
Q

What is Hybrid Dysgenesis?

A
  • Occurs when p elements are introduced into a cell that does not have them in its genome
  • sudden appearance of numerous mutations
  • chromosome aberrations
  • sterility
190
Q

Describe the process of Hybrid Dysgenesis

A
  • Repressor protein is a cytoplasmic protein that inhibits transposition
  • inhibitor is in the egg of the female (P+)
  • if this P+ is crossed with P- or P+ male, the inhibitor in the egg will prevent transposition and the fly will be normal or wild type

if a P- female is crossed with a P+ male the lack of inhibitor in the egg will alow for rapid transposition, leading to mutations and sterility

191
Q

Sines - Transposable Elements

A
  • not identicle
  • have short flanking direct repeats when inserted into DNA
  • arose through an RNA intermediate
  • Alu is a short interpersed elements
192
Q

Lines - Transposable Elements

A

Long interspersed elements

193
Q

Constitutively expressed Genes

A
  • Essential components of almost all living cells
  • Genes that specify products of this type are continuously expressed in most cells
194
Q

Inducible and Repressible Genes

A
  • Some gene products are needed for cell growth only under certain environmental conditions
  • Regulatory mechanisms allow the synthesis of these gene products only when they are needed (inducible) or shut down genes when they are no longer need (repressable)
195
Q

Transcription in bacteria is regulated by ___

196
Q

Negative regulatory proteins ___ transcription
Positive regulator proteins ___ transcription

A

inhibit, activate

197
Q

Inducible operon

A

When transcription of the operon is normally off and something happens that turns it on.

198
Q

Repressible Operon

A

Normally turned on, and something happens that it needs to be turned off

199
Q

In a negative inducible operon:
Regulator protein is a ___ and blocks the binding if the RNA polymerase to the ___

A

inhibitor, promoter

200
Q

Negative Inducible Operon

A
  • Transcription is normally turned off and must be turned on
  • Repressor needs to be relived of its duties in order for transcription to proceed
  • inducer activates it
  • repressor can no longer bind to DNA

Negative: Regulator protein is an inhibitor or repressor
Inducible: something inactivates repressor which then induces transcription

201
Q

Negative Repressible Operon

A
  • Transcription normally takes place and must be turned off
  • inactive in the absence of a co-factor
  • co-repressor binds to the repressor which allows it to bind to the operator and inhibit transcription

Negative: regulator protein is an inhibitor or repressor
Repressible: Something activates the repressor which. then represses transcription

202
Q

Positive Control - Gene expression

A
  • regulatory protein is an activator
  • Activator binds operator and induces txn
  • Binds just upstream of the promoter and enhances the binding of RNA polymerase to the promoter
203
Q

Describe the lac operon

A

Negative inducible:
- regulator is an inhibitor
- allolactose inactivates inhibitor

Positive Control:
- CAP+cAMP enhances transcription

204
Q

what is the role of permase (lacY)

A

To transport lactose across the bacterial cell membrane

205
Q

Role of b-galactosidase (lacZ)

A

catalyzes the breakdown of lactose to glucose and galactose

also converts lactose into allolactose

206
Q

Cis regulation of expression

A

control of gene expression of genes only on the same piece of DNA

207
Q

Trans regulation of expression

A

Control gene expression on other DNA molecules

208
Q

non functional mustations are designted by

209
Q

Catabolite Repression

A

Bacteria will actively turn off the other metobolic pathways when glucose is present

209
Q

Positive Control

A

Positive glucose metabolism results in repression of other pathways

209
Q

levels of cAMP are ___ proportional to glucose

A

inversely

glucose low = cAMP high

MBG*2040 (1 decks)

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