Week 1-5 Test Prep

Question 1

A form DNA

Answer 1

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

Question 2

B Form DNA

Answer 2

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

Question 3

Z form DNA

Answer 3

Left handed helix, zig zag backbone

Question 4

Mismatch repair

Answer 4

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

Question 5

What are the causes of mutations (3 types)?

Answer 5

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

Question 6

Nucleotide excision repair

Answer 6

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.

Question 7

Base Excision Repair (BER)

Answer 7

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

Question 8

Double Strand
Breaks

Answer 8

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

Question 9

Non-homologous end joining

Answer 9

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

Question 10

Unique sequences include…

Answer 10

Protein coding and non-protein coding RNA genes

Question 11

Homologous Recombination

Answer 11

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

Question 12

What types of mutations can
occur?

Answer 12

1. Substitution = single nucleotide variant (SNV) is substituted for another
2. Insertion/deletion = indel

Question 13

What are the different classes of repetitive sequences?

Answer 13

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

Question 14

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

Answer 14

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

Question 15

What is satellite DNA and where is it found?

Answer 15

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

Question 16

DNA methylation

Answer 16

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

Question 17

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

Answer 17

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).

Question 18

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

Question 19

DNA Methylation in Prokaryotes

Answer 19

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

Question 20

Deacetylation

Answer 20

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

Question 21

Histones

Answer 21

Positively charged proteins that interact strongly with negatively charged DNA

Question 22

Epigenetics

Answer 22

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

Question 23

How is the epigenome controlled?

Answer 23

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

Question 24

Acetylation

Answer 24

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

Question 25

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

Answer 25

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

Question 26

What is ChIP Seq? And what does it do?

Answer 26

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

Question 27

What are the key steps in the ChIP Seq process?

Answer 27

- 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)

Question 28

What are the key steps in RNA Seq?

Answer 28

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

Question 29

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?

Answer 29

- 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.

Question 30

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

Answer 30

- 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.

Question 31

What is the role of DNA methylation in eukaryotes?

Answer 31

- Methylated regions (hypermethylated) are associated with heterochromatin (closed chromatin) and histone deacetylation

Question 32

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

Answer 32

- 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)

Question 33

What are the two classes of transposons?

Answer 33

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

Question 34

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

Answer 34

- 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

Question 35

How do replicative (class 1) transposons work?

Answer 35

- 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

Question 36

What are nonautonomous transposons and how are they transposed?

Answer 36

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.

Question 37

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

Answer 37

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.

Question 38

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

Answer 38

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