Other DNA stuff

Question 1

How are pyrimidine bases numbered?

Answer 1

Number counter-clockwise
Get to the second nitrogen as quickly as possible
Thymine has additional methyl group (CH3) at 5’

Question 2

What is the difference between an RNA nucleotide and a DNA nucleotide?

Answer 2

RNA has ribose sugar instead of deoxyribose sugar in DNA. 2’ position of sugar has hydrogen atom in DNA while RNA has an OH- group
Uracil replaces Thymine in RNA

Question 3

Describe the structure of DNA

Answer 3

Nucleotide consists of ribose sugar, phosphate group and a nitrogenous base (adenine, thymine, cytosine, guanine).
Glycosidic bond between 1’C on base and 1’C on sugar
Phosphodiester bond between base and phosphates - makes up backbone
2 H bonds between TA, 3 between GC
Strands have antiparallel arrangement and wrap around each other in plectonemic coil

Question 4

Why is DNA a good molecule to encode genetic information and make stable genomes?

Answer 4

1. Presence of 2’ OH in RNA, makes it alkali lable (breaks up at OH acts as catalyst for hydrolysis)
2. Cytidine is unstable and presence of T allows products of cytidine instability to be identified, removed and repaired
3. dsDNA doesn’t fall apart when phosphodiester backbone nicked since stabilised by H bonds and by hydrophobic interactions between faces of individual bps
4. Nucleotide bases protected from aqueous phase by phosphodiester backbone
5. Double helix can be melted under physiological conditions

Question 5

Describe the process of Cytosine deamination

Answer 5

Cytosine + H20 –> Uracil + NH3
Hydrolytic Reaction
Potentially mutagenic but sometimes biologically exploited

Question 6

How is Uracil removed from DNA?

Answer 6

Base Excision Repair (BER)
Uracil glycosylase breaks bond between sugar and uracil
AP endonuclease removed part of backbone at point where uracil was
Restores cytosine

Question 7

What evidence supports the suggestion of RNA coming before DNA?

Answer 7

1. Both can assemble on basis of complementarity and store information but only RNA is catalytically active
2. Idea of RNA world where RNA performs catalysis as starting point for life so must come first
3. DNA precursors made from RNA precursors by ribonucleotide reductase
4. Rwo different thymidylate synthetases that methylate dUMP in many organisms suggested T evolved twice
5. PBS1 and PBS2 phage of B. subtilis contain U instead of T in their DNA

Question 8

What are the three types of double helices?

Answer 8

B - dsDNA under most conditions, major (can see bases) and minor grooves (can’t see bases) important for sequence-specific protein binding
A - found in RNA-DNA duplexes and in dsRNA
Z - formed by alternating purines and pyrimidines, left handed, function unproved

Question 9

How is the double helix stabilised?

Answer 9

1. Hydrogen bonds between bases
2. Hydrophobic stacking interactions between faces of the bases (stronger at high salt conc, so DNA more stable at higher salt conc)

Question 10

Explain Cytidine deaminases as an important role in RNA Editing.

Answer 10

e.g. Apolipoprotein B (ApoB) - 2 versions have same mRNA but different proteins due to RNA editing - Glutamine (CAA) 2153 converted to in frame stop (UAA) by APOBEC1 acting on ssDNA
ApoB100 - liver, transports lipids from liver
ApoB48 - small intestine, synthesis of chylomicrons by gut
9/11 human APOBECs convert cytidine to deoxyuridine (dU) in polynucleic acids
Target cytidines need to be in ssDNA/RNA (during replication, transcription, repair after damage)
Most target TC with CC being less efficient

Question 11

Explain Cytidine deaminases as an important role in Innate Immunity To Retroviruses and Transposable Elements.

Answer 11

Some APOBECs restrict HIV and ssRNA virus propagation
Enzyme hacks away at ssDNA produced from viral RNA, deaminating C to U - endonucleases severe it and less likely to be functional in host cell
Cytidine deaminases don’t normally get into viruses (viruses have method of ‘punching back’ against deaminases)

Question 12

Explain Cytidine deaminases as an important role in Adaptive Immunity.

Answer 12

Activation-induced deaminase AID necessary for production of high affinity IgG in mammals and birds and for class switching in all verts
1. Somatic hyper-mutation (SHM) necessary for high-affinity antibodies
2. Class switch recombination (CSR) - Low affinity IgM is produced, genes encoding variable region gets rearranged at immunoglobulin locus stitching that part of gene onto one coding IgG molecule
3. Gene conversion in birds

Question 13

Explain Cytidine deaminases as an important role in Cancer.

Answer 13

Unregulated expression generates general mutator phenotype in tumour subclones
Enzymes mutate DNA and contribute to development of tumours and tumour drug resistance
C to T/G substitutions in TC dinucleotides; strand coordinated in localised regions (kataegis)

Question 14

What is Cytosine Methylation and its consequences?

Answer 14

5-methyl cytosine is deaminated to thymine which is a natural base so the enzyme doesn’t recognise this and the mutations don’t get repaired as well

Question 15

What are CpG islands?

Answer 15

Two adjacent nucleotides in DNA (palindromic in opposite way) - CytosinepGuanine
C of CpG 70-80% methylated in mammals
Generally silences transcription, X inactivation, genomic imprinting, repression of TEs
5-methyl C hypermutable and CpG found at only 25% levels - TpG and CpA elevated (CpG suppression)
CpG islands are 200bp-1kb stretches that are unmethylated and show no suppression with elevated C+G
Often found around promoters of genes, thought to be binding site for TFs

Question 16

What is epigenetics?

Answer 16

The heritable change in phenotype that doesn’t involve a change in DNA sequence
Inheritance of methylation pattern DNA methylation is one mechanism but there are others including small RNA expression
Tortoiseshell cat is an example of epigenetic inheritance

Question 17

What is the C value paradox?

Answer 17

A term used to describe the variation in the size of eukaryotic genomes. It refers to the observation that genome size doesn’t correlate with organismal complexity.
C = number of genes in haploid genome
E.coli - 4.5Mb, 4300 genes
Paramecium - 96Mb, 40000 genes
Human - 3000Mb, 19-25000 genes
Outliers include Lilies and lung fish

Question 18

What are the two approaches into working out how much of the genome is functional?

Answer 18

Biochemistry - What binds, what is transcribed (considered less powerful approach)
Bioinformatics - What is conserved or constrained and mutates less frequently than expected - one method is to look for conserved sequences (divide genome into windows and identify those with excess of conserved residues as compared to local ancestral repeat)
This approach suggests:
8.2% constrained by purifying selection
~5% shows evidence of lineage-specific constraints
2.2% constrained in human and mouse
Human rule of thumb: 10% functional, 1.5% coding

Question 19

Why are circular chromosomes small?

Answer 19

1. Large circular chromosomes are unstable
2. Bacteria are small and have very large effective population sizes and so their genomes are not going to grow as extra non functional DNA will be efficiently removed by selection

Question 20

Why is sister chromatid exchange so dangerous for circular chromosomes?

Answer 20

Can cause the formation of a dicentric chromosome that undergoes a breakage fusion bridge cycle (BFB)
Chromosome replicates
Uneven Sister Chromatid Exchange
Bridge forms (now 2 are in 1 big circle)
Centromeres pulled apart - breakage
Each side replicates
Tries to repair by fusing back into circle
This can also occur in vertebrates if telomeres are lost and broken ends are fixed by forming dicentric chromosomes

Question 21

How do Sister Chromatid Exchanges effect linear chromosomes?

Answer 21

Do not affect the structure (still leaves two distinct molecules)
Unequal SCE leads to duplications and deletions (aren’t lined up perfectly when SCE occurs) - can accumulate tandemly repeated sequences (head to tail repeats)
One loses and other gains - if it keeps occuring, chromosome will eventually look like ABABAB

Question 22

What are the theories for explanation of extra genomic DNA?

Answer 22

1. Selfish DNA (doesn’t explain why tandem repeats are also bigger)
2. Bulk DNA (adaptive change to increase nuclear volume and cell size; not tested at population level)
3. Metabolic cost (no correlation between cell division rate and genome size; bacterial cells often contain C present in nested set of many copies; DNA only constitutes 2-5% of dry weight of cell)
4. Competition between genome growth (mal-adaptive) and selection
5. Power of natural selection is proportional to the effective population size (Ne)

Question 23

What are the consequences of having linear chromosomes?

Answer 23

1. Need special machinery to protect and replicate ends of chromosomes - telomeres
2. Repeated sequences are common in eukaryotic genome - tandem and interspersed

Question 24

What are Ploidy, Euploidy, Aneuploidy and Polyploidy?

Answer 24

Ploidy = the chromosome number within a cell or with the cells of an organism
Euploidy = variation in the number of complete sets of a chromosome
Aneuploidy = variation in the number of particular chromosomes within a set (causes disease)
Polyploidy = alteration in the number of chromosome sets per cell

Question 25

What are examples of polyploid organisms?

Answer 25

Strawberry - 8N, 2N
Xenopus laevis (4N) and tropicalis (2N)
Bread Wheat (6N)

Question 26

What is the nomenclature for polyploidy

Answer 26

x = monoploid number of chromosomes in a basic, ancestral set (human = 23)
n = haploid number/in gamete (human = 23)
2x = diploid, 3x = triploid, 3x = tetraploid, 6x = hexaploid
Example bread wheat, hexaploid, 42 C per cell
6x; x=7; six sets of 7
Gametes contain 3 sets so n = 21

Question 27

How can Polyploids form?

Answer 27

Autopolyploids (all from same species; 4x=AAAA) form by spontaneous premeiotic endoreduplication

Allopolyploids (chromosome sets from different species; 4x=AA BB) form by interspecific hybridisation and endoreduplication (2 copies of each)
A and B sets of partially homologous (homeologous)

Question 28

What are the consequences of polyploidy?

Answer 28

Advantages:
- Increased cell size leads to larger fruits, flowers and organism
- Advantageous characteristics from multiple species (allopolyploids)
- Whole genome duplications allow long term evolution of genetic novelty

Disadvantages:
- Sterility (problems forming balanced gametes during meiosis), especially for odd-numbered polyploids.

Question 29

Describe segregation at meiosis 1 in triploids.

Answer 29

Two chromosomes pair correctly
3rd chromosome can’t pair so segregates randomly
Therefore the products of the first meiotic division aren’t balanced e.g. One cell has 2 blue and 2 red, other cell has opposite
For proper development you need equal number of copies of all chromosomes, otherwise products of genes encoded by the 1 blue C aren’t in balance with the products encoded by the 2 red C - won’t be viable

Question 30

What are the types of programmed changes in number of chromosome/genome sets per cell and what is their function?

Answer 30

Usually to provide increased RNA synthesis rate/cell; usually incompatibly with normal chromosome segregation and maintenance of intact genomes
1. Endopolyploidy: Repeated chromosome replication without cell division e.g. nurse cells of Drosophila egg chamber
2. Polytenization: A kind of endopolyploidy with in register chromatid pairing
3. Multinucleate cells: temporary endopolyploidy, nuclear multiplication without cell division, germ line genome conserved
4. Chromosome diminution: some somatic cell chromosomes break down before segregation leaving altered somatic chromosomes in daughter cells
5. Chromosome elimination: e.g. sciarid flies where paternal chromosomes are selectively eliminated at 3 stages of the life cycle
6. Gene amplification in whole organisms:
a. rDNA amplification in Xenopus oocytes involves extrachromosomal rolling circle
b. Maturation of macronucleus in ciliated protozoa e.g. Tetrahymena
c. Chorion gene amplification in Drosophila nurse cell

Question 31

What are the two types of programmed gene amplification?

Answer 31

1. Extrachromosomal - doesn’t permanently affect genome of dividing cells or germ line
   Rolling circle - replication disrupts other strand which is then copied too
2. Intrachromosomal - alters genome; incompatible with chromosome segregation i.e. dead end cells do not divide but are specialised for high rates of chorion RNA/protein production

Question 32

What is meant by the term “tandemly repeated DNA”?

Answer 32

A sequence that is repeated many times head to tail along the DNA in an array
The unit repeats are similar but not necessarily identical and may vary in length between 2bp and >10kb
Are almost always polymorphic in copy number

Question 33

How can rounds of successive unequal SCE can homogenise an array and rapidly spread a mutation through the array?

Answer 33

Starting with 1 mutation in row of 4 TRs
UEC results in gain of mutation repeat (2/5 mutant TRs)
UEC can result in loss of normal repeat (2/4 mutant TRs)
Can keep gaining and losing until 4 repeats all have mutant
evolve rapidly through evolution

Question 34

What is NAHR

Answer 34

Non-allelic homologous recombination e.g. unequal sister chromatid exchange
Can be intra-chromosomal (resulting in a ring chromosome and a deletion) or inter-chromosomal (resulting in a duplication and a deletion)
May occur in meiosis 1 if pairing occurs between different low copy repeats that are out of register

Question 35

Describe Micro-satellite DNA as an example of Tandemly repeated sequences in the eukaryotic genome.

Answer 35

Simple repeat extending for a few hundred bp
Value as markers in that exact number of repeats can be determined
Present throughout genome (100,000)
Used to create early linkage map of genome
Can be used as polymorphic markers - size fractionate of PCR products to resolve sizes within single base
Microsatellite polymorphism arises by slippage during replication inconsistent with stepwise mutation model

Question 36

Describe Mini-satellite DNA as an example of Tandemly repeated sequences in the eukaryotic genome.

Answer 36

Unit repeat 5-100bp, array length 0.5-30kb
Tend to be located near telomeres
Contain core of GGGCAGGACG
Early use as DNA fingerprinting reagents but needs gel electrophoresis, southern blotting and filter hybridisation to detect polymorphism
New alleles likely arise by gene conversion and not SCE

Question 37

Describe Satellite DNA as an example of Tandemly repeated sequences in the eukaryotic genome.

Answer 37

Arrays of TRs, 5bp-400bp
Can occupy significant proportion of genome
Can be separated on basis of buoyant density - bound DNA to equilibrium in centrifuge in heavy salt solution (typically CsCl) at high speed over a few days - forms density gradient with DNA sedimenting at same level as salt solution but some other less dense (rich in A+T)
Often found at centromeres and telomeres
Often stains differently from the bulk of genomic DNA

Question 38

Describe Telomeric DNA as an example of Tandemly repeated sequences in the eukaryotic genome.

Answer 38

Often tandemly repeated
Short repeats in most organisms - 5-7bp
In humans, one strand in TTAGGG and in another one is AATCCC

Question 39

Describe Tandemly repeated gene families as an example of Tandemly repeated sequences in the eukaryotic genome.

Answer 39

Classical example is rDNA; encodes rRNA which cell needs many copies of
Acrocentric chromosomes
rRNA Tandem repeats form nucleolar organising regions (NORs)

Question 40

Describe Clustered gene families as an example of Tandemly repeated sequences in the eukaryotic genome.

Answer 40

Many genes found clustered in genome with homologues in repeated sequences (head to tail, tandemly) and often perform similar functions in different contexts
E.g. Globin genes - 2 main clusters
Direction of trans matches direction of expresh
Human Hb change during development and stay consistent after birth
Organisation of clustered gene families rapid in evolution due to NAHR: beta-globin cluster; variation in copy number at different loci corresponding to beta and alpha genes
Sequence duplication is important because it allows evolution of new genes without loss of old ones; may be responsible for what some have termed ‘genetic redundancy’; often source of polymorphism due to variation in number of repeats; important mutational mechanism
Unequal exchange as a result of NAHR at meiosis is mutational mechanism that reduces gene copy number a-globin locus - reduced a-globin chain synthesis causes a-thalasemia

Question 41

Describe Segmental Duplications/LCRs as an example of a type of repetitive DNA where unequal exchange as a result of NAHR is important in understanding variation.

Answer 41

Segment 10-100kb repeated within genome or at dispersed positions along chromosome
Head/tail or head/head
NAHR between segmental duplications give rise to many structural rearrangements
e.g. Charcot Marie Tooth Syndrome, Hereditary neuropathy with pressure palsies
Some repeats complex - human C22q11
Head/head duplications and intra-chromosomal recombination is associated with inversion polymorphism - associated with dicentric inversion duplications if they involve interchromosomal sister chromatid or meiotic exchange

Question 42

Describe Copy Number Variation as an example of a type of repetitive DNA where unequal exchange as a result of NAHR is important in understanding variation.

Answer 42

In humans, CNS’s covering 115Mb (3%) and so very important source of human genetic diversity
Segmental duplications frequently associated with CNVs