Brainscape's Knowledge GenomeTM

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


See full index

BIOL241 - Genetics > Week 1-5 Test Prep > Flashcards

Week 1-5 Test Prep Flashcards

You may prefer our related Brainscape-certified flashcards:
Biology 101 flashcards
1
Q

A form DNA

A

Right handed double helix with a wider major groove
different conformation of nucleotides that allow for closer packing

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

B Form DNA

A

Right Handed double helix with both major and minor
grooves and a smooth backbone

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

Z form DNA

A

Left handed helix, zig zag backbone

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

Mismatch repair

A

After the replication fork has passed, the base mismatch is recognised by the MMR system and excised, and is filled in by DNA polymerase and ligated back together

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

What are the causes of mutations (3 types)?

A
  1. Replication errors - Misincorporation of nucleotide
  2. Chemical modification - deamination (loss of amine
    group from nucleotide), depurination/depyrimidatio, base analogous, base modifiying agents (mutagens, chemical
    warfare etc.)
  3. Radiation - x-rays break chromosomes, UV in sunlight creates thymine dimers
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
6
Q

Nucleotide excision repair

A

Repair system in which DNA is unwound into a bubble and a large stretch of bases is removed around the damaged base. DNA polymerase inserts the correct base and DNA
ligase ligates the strand.

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

Base Excision Repair (BER)

A

Repair system in which a single damaged base precisely is removed, DNA polymerase inserts the correct base and DNA ligase ligates the strand.

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

Double Strand
Breaks

A

Damage that occurs naturally through processes like meiosis or exposure to radiation. Can be repaired by
Homologous Recombination or Non - Homologous End Joining

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

Non-homologous end joining

A

Quick repair mechanism in which loose strands of DNA
are jammed together.

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

Unique sequences include…

A

Protein coding and non-protein coding RNA genes

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

Homologous Recombination

A

Repair process by which a cell replaces a stretch of DNA with a segment that has a similar nucleotide sequence

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
12
Q

What types of mutations can
occur?

A
  1. Substitution = single nucleotide variant (SNV) is substituted for another
  2. Insertion/deletion = indel
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
13
Q

What are the different classes of repetitive sequences?

A
  1. Moderately repetitive (few to 100,000 copies per
    genome) - tandem repeats (micro-satellites) and interspersed retrotransposons (SINEs & LINEs)
  2. Highly repetitive (100,000 to 10,000,000 copies per
    genome) - Satellite DNA
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
14
Q

Who has more repetitive sequences compared to unique sequences, Eukaryotes or Prokaryotes?

A

Eukaryotes! Differences in genome size is due to these
repetitive DNA regions, although in eukaryotes this does not correlate with how many code for proteins and the overall complexity of the organism

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

What is satellite DNA and where is it found?

A

They are large areas of highly repetitive sequences organised in a tandem array, and are found at telomeres and centromeres - structural

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

DNA methylation

A

The addition of methyl groups to bases of DNA after DNA synthesis; may serve as a long-term control of gene expression.

17
Q

What are minisatellites (VNTRs)? And why are combined VNTR sequences more useful in identifying an individual?

A

Variable number tandem repeats are 7-100 bps (repeated many times), and are considered to be polymorphic
(different between individuals of a species), and can be used as a DNA fingerprint. Combined VNTR sequences use lots of different sections, rather than just a few in order to more accurately work out an individuals DNA fingerprint (CODIS).

18
Q

What are multiple copy genes? And how do they differ in prokaryotes and eukaryotes?

A
19
Q

DNA Methylation in Prokaryotes

A

Methylation is written onto each new strand after replication by a DNA methyltransferase enzyme - part of the bacterial immune defense. Also useful in repair - identifies the original strand

20
Q

Deacetylation

A

Histone deacetylase (HDAC) erases acetyl mark on the
histone which become positively charged causing chromatin to wind tighter

21
Q

Histones

A

Positively charged proteins that interact strongly with negatively charged DNA

22
Q

Epigenetics

A

The study of environmental influences on gene expression that occur without a DNA change

23
Q

How is the epigenome controlled?

A
  1. DNA Methylation
  2. Histone Modification
  3. Chromatin Remodelling
  4. Non-Coding RNA
24
Q

Acetylation

A

Histone Acetyl transferase (HAT) writes an acetyl onto the histone tail causing nucleosomes to loosen and spread apart

25
Q

What are microsatellites (STRs)? And why are they prone to evolution?

A

Short, tandem repeats (STRs - 2-6 bps) of varying length
across species/individuals
- The DNA polymerase tends to slip over short repeat sequences, either forwards resulting in deletion, or backward resulting in insertion of sequences

26
Q

What is ChIP Seq? And what does it do?

A
  • Chromatin Immunoprecipitation combined with high-throughput sequencing
  • Identifies the locations in the genome bound by proteins
27
Q

What are the key steps in the ChIP Seq process?

A
  • Formaldehyde (or analogous substance) used to glue all of the proteins bound to the DNA together
  • DNA cut into small fragments (approx. 300 bp)
  • Isolate the protein we are interested in by using an antibody, and wash away other proteins
  • Reverse formaldehyde glue to DNA by heating, then wash away the proteins (including histones) to isolate the DNA
  • Then follow RNA Seq process (PCR amplify library, check library concentration, sequence, filter out garbage reads)
28
Q

What are the key steps in RNA Seq?

A
  1. PCR amplify library
  2. Check library concentration
  3. Sequence
  4. Filter out garbage reads
29
Q

What is the resulting output from a ChIP Seq experiment? And what is a ‘track’ and why is it created from our ChIP Seq reads?

A
  • A long list of genomic coordinates for all the reads (across many chromosomes)
  • The control track uses some of the same input chromatin for the ChIP Seq experiment, but doesn’t try to enrich for any particular protein binding. This is used to verify that a high concentration of reads in the ChIP-seq track is due to a protein binding there and not because a lot of reads mapped to a repetitive region.
30
Q

How can we identify the functional role of a specific protein based on the ChIP Seq output?

A
  • By looking at where it binds relative to the genes, if it binds to the start of the gene, for example, it may regulate the expression of that gene.
31
Q

What is the role of DNA methylation in eukaryotes?

A
  • Methylated regions (hypermethylated) are associated with heterochromatin (closed chromatin) and histone deacetylation
32
Q

How does the cell know which is the correct sequence and which is the mutant one in both prokaryotes and eukaryotes?

A
  • Prokaryotes: DNA is methylated, which is written onto the new strand after replication
  • Eukaryotes: related to nicks found in newly synthesised strand (both lagging and leading strands have nicked DNA)
33
Q

What are the two classes of transposons?

A

Class 1 - Replicative (RNA intermediate = retrotransposon)
Class 2 - Conservative - DNA transposon = Inverted Terminal repeat Element (ITE) OR Terminal Inverted Repeat (TIR)

34
Q

What kind of process occurs in Conservative DNA transposons, and what are the mechanics of “jumping”?

A
  • Cut and paste process (piece of DNA excised and inserted somewhere else in the genome)
  • Full length/fully functional TEs are autonomous - encode for the transposase needed for jumping
  • The transcribed Transposase recognises terminal inverted repeats (TIR) to be able to be excised/”jump”
  • Transposase dimerizes (DNA loops around) and is excised from donor site, to be inserted at new site in the genome with terminal inverted repeats
35
Q

How do replicative (class 1) transposons work?

A
  • Copy and paste process: Transposons can be transcribed to make RNA, then using a reverse transcriptase enzyme (encoded by the transposon itself) the RNA is converted back to DNA
36
Q

What are nonautonomous transposons and how are they transposed?

A

Miniature ITE (MITEs) do not encode for transposase, due to mutational decay of the sequences over time. They can still be transposed using transposase encoded from other sites due to still have the TIRs to attach to.

37
Q

Why would a cell (prokaryote) initiate gene expression? And what happens when the signal is received?

A

Due to an environmental signal, which will be picked up by a specific transcription factor. This TF then binds to a DNA sequence upstream of the transcription start site.

38
Q

What are the three main classes of transcription factors in prokaryotes?

A
  1. Arrangement of DNA - Nulceotide associated proteins (NAPs), open chromatin
  2. Activator - bind to operator sequences, assist RNA pol
  3. Repressor - bind to operator sequences, block RNA pol
39
Q
A

BIOL241 - Genetics (2 decks)

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