chapter 11 - Gene mutation, DNA repair, homologous recombination Flashcards

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1
Q

What is the role of transposase?

A

Transposase is responsible for making staggered cuts of the DNA seq and carrying the transposable element to the target sequence.

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2
Q

What is an IS element?

A

Insertion element… is transposable.

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3
Q

What is the difference between an IS element and a composite transposon?

A

Composite - a type of transposon that carries genes such as antibiotic resistant genes

IS element - transposable elements, carry genes that code transposase/catalyze transposition activity

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4
Q

What are the steps of transposable element insertion into target sequences?

A

1 - transposase makes staggered cuts and leave DNA strands from target seq, leaving single stranded ends in the target seq.

2- transposable element is inserted into target seq, carried by transposase

3- gaps filled by DNA polymerase

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5
Q

What are transposable elements, also known as “jumping genes”?

A

-DNA sequences found in both eukaryotes and prokaryotes that have multiple copies and are able to move within the genome

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6
Q

What are point mutations?

A

Changes in one or a few base pairs… can be somatic (non heritable mutations, not egg/sperm) or germ-line (heritable, egg/sperm)

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7
Q

What are the two types of BASE PAIR SUBSTITUIONS?

A

Transition - Purine to purine… pyrimidine to pyrimidine

Transversion - Purine to pyrimidine, pyrimidine to purine

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8
Q

What are the types of base-pair substitutions that occur in the coding sequence?

A

Synonymous mutation
missense mutation
nonsense mutation
frameshift deletion
frameshift insertion

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9
Q

What is a wild type sequence also known as?

A

Sense codon

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10
Q

What is a synonymous mutation?

A

-mutation in coding seq
-change in DNA, amino acids stay the same… change in a wobble

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11
Q

What is a missense mutation?

A

-mutation in coding seq
-change in DNA that changes the amino acid

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12
Q

What is a nonsense mutation?

A
  • mutation in coding seq
    -change from a sense codon to a stop codon : UAG, UAA, and UGA
    -shortens seq of proteins
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13
Q

What is a frameshift insertion/deletion?

A
  • mutation in coding seq
    -a single base pair is added or removed, causing all of the amino acids to be off
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14
Q

What are the four regulatory mutations?

A

1- Promotor mutation

2- polyadenylation mutation

3- splice site mutation

4- DNA replication mutation (triple repeat expansion)

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15
Q

Describe the process of strand slippage, and what type of mutation is it?

A

-regulatory, DNA replication mutation

1 - DNA segment contains CAG-triplet repeats
2 - DNA polymerase slips off of strand during synthesis of daughter strand , causing the formation of hairpin loop
3 - in the next replication cycle, CAG repeats number is increased due to the hairpin loop (ex: beginning 6 CAG to end 11 CAG)

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16
Q

What could result from a promotor mutation?

A

Change in timing/amount of transcription

17
Q

What could result from a polyadenylation mutation?

A

Alter sequence of mRNA

18
Q

what may result from a mutation in the splice site?

A

Retaining introns/excluding exons

19
Q

What may result from a DNA replication mutation?

A

increase (sometimes decrease) number of short repeats of DNA

20
Q

What are the three types of reverse mutations?

A

1- true reversion

2- Intragenic reversion

3- second site/intergenic/silence reversion

21
Q

Describe a true reversion?

A

There is a second mutation (revertant) in the same codon that causes the mutated codon to encode for the wild-type amino acid

22
Q

Describe an intragenic reversion?

A

-a mutation that occurs in the same gene, but not the same spot
-two base pairs are deleted, following the insertion of two new base pairs, restoring the reading frame to wild type

23
Q

Describe a second site/intergenic/repressor reversion

A

-a mutation that occurs not in the same gene, occurs in a second gene
- represses certain genes in order to restore to wild-type or near wild-type

24
Q

What is the difference between a spontaneous and induced mutation?

A

Spontaneous - no external reason for mutation outside of cell

induced - can detect a reason outside of the cell for mutation

25
Q

Describe chemical mutagen: Hydroxylating agent

A
  • addition of hydroxyl (OH) group to a recipient compound by a donor known as hydroxylating agent
  • hydroxylamine is a hydroxylating agent that adds a hydroxyl group to cytosine by displacing an H2, creating hydroxylaminocytosine.
    -Hydroxylaminocytosine pairs with guanine but frequently mispairs with adenine, leading to transition mutations
26
Q

What is the purpose of the Ames test?

A
  • Uses bacteria to detect carcinogens
  • determine if chemicals can cause mutations in the cell
  • chemicals that have mutagenic effect on cells is considered carcinogenic
  • detects a reversion from his- to his+
27
Q

Why add s9 activation enzymes from rat liver in the Ames test?

A

The S9 extract allows you to see if a mutagenic substance can be created by digestive reactions.

28
Q

How does his- strains differ with base-pair substitutions and frameshift mutations?

A

Base pair sub - there is consistent growth in revertant colonies as medium increases

frameshift - there is no growth at all ]

conclusion: his- strains only revert under base-pair substitution

29
Q

What are UV photoproducts?

A

formation of photoproducts cause by UV irradiation

formed from adjacent pyrimidines (TT)

distorts double helix, blocks replication

30
Q

How does UV radiation damage DNA?

A
  • mutagenic properties of UV radiation derives from legions it creates in DNA
  • alters DNA nucleotides by inciting formation of additional bonds known as photoproducts (TT bonds)
31
Q

Describe photoreactive repair?

A

1- UV light causes a thymine dimer to form (TT)

2- photolyase binds to the dimer

3- visible light (300-500 nm) energized photolyase

4- photolyase released, DNA repaired

32
Q

Describe Nucleotide excision repair (NER)?

A

1- UVR A binds to UVR B and helps it get to the damage site, binding opposite of the thymine dimer

2- UVR A leaves, and UVR C binds to UVR B, catalyzing 3’ and 5’ cuts.

3 - UVR D (helicase) releases damaged strand, DNA poly + ligase fill gaps, restoring DNA

33
Q

What is Xeroderma Pigmentosum?

A

a skin condition caused by a mutation in the NER system, individuals are homozyous recessive and cannot go into the sun

  • freckles
    -photophobia
  • skin cancer
    -tumor in eye
    -premature skin aging
  • neurological problems
34
Q

Describe Mismatch repair?

A

1 - mutH binds to hemi-methylated CTAG recognition site

2 - MutS binds to the mismatch site

3- MutL is attracted by MutS, forms the bed which contacts MutH

4 - MutH cleabes from pre mismatch to site of methylation, causing single strand gap

5 - gap filled in by ligase and polymerase

35
Q

What are the three types of DNA repair systems?

A

1 - photoreactive repair

2 - nucleotide excision repair (NER)

3 - Mismatch repair

36
Q

Describe Meiotic recombination?

A

Crossing over occurs during meiosis between two homologous non sister chromatids….

1 - protein binds to one of the strands, causing a double strand break, making the strands offset

2 - Enzymes bind to the strands, digesting them 5-3 creating single strand segments

3 - proteins bind to exposed single strands, top strand falls between the two bottom strands (strand invasion), causing bottom strand to form a D Loop/holliday junction/heteroduplex region

4 - strands are extended by polymerase, displaced D loop, pairs with complementary single strand dna to form second heteroduplex region

5 - dna poly extension fills gaps in strand paired with d loop, forming a second d loop in top strands

6 - double hollidays junctions form

37
Q

What are the two possible outcomes of meiotic recombination?

A

Opposite sense resolution:
- EW and NS cuts : common, results in recombination of flanking genes, creates offset heteroduplex regions.

Same sense resolution:
- EW and EW cuts : not common, no recombination of flanking genes, creates offset heteroduplex region