Chromosome Structure Flashcards

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1
Q

Facultative heterochromatin

A

Heterochromatin that can be both active (decondensed) or inactive (condensed)

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2
Q

Constitutive heterochromatin

A

Highly condensed heterochromatin that shows little or no evidence of active gene expression.

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3
Q

Alpha satellite DNA

A

171 monomer unit tandemly repeated to form a large array. Role in centromere development.

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4
Q

Beta satellite

A

68bp monomer repeat unit

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5
Q

Gamma satellite

A

Mostly observed on X chromosome. Prevents spread of heterochromatin.

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6
Q

Inversion variants

A
inv(2)(p11.2q13)
inv(3)(p11q12)
inv(3)(p13q12)
inv(5)(p13q13)
inv(10)(p11.2q21.2)
inv(Y)(p11.2q11.2)
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7
Q

What is a fragile site?

A

Segments of uncoiled chromatin likely to show gaps and chromosome breaks. Line in gene content and high in LINE content. Sites where genes can be silenced by hypermethylation. Disruption of the DNA due to a) DNA elongation inhibition by folate sensitivity b) DNA polymerase inhibition by BrdU c) starving cells of thymidine.

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8
Q

What type of culture conditions induce fragile sites?

A

Folate sensitivity, thymidine starvation, BrdU or distamycin A exposure.

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9
Q

What is the incidence of a rare fragile site?

A

1/several hundred individuals (less than 5% of pop), Rare ones tend to be folate sensitive.

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10
Q

Intermediate fragile sites (1-5% of population)

A

fra(10)(q25)

fra(16)(q22)

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11
Q

What is FRA11B?

A

A fragile site at 11q23.3 within proto-oncogene CBL2. CCG repeat which behaves similar to FRAXA. Expansion of this repeat in mothers can be a vulnerable point for de novo del (11q) - Jacobsen syndrome.

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12
Q

Euchromatic variants

A

var(4)(p16.1p16.1) - resembles a dup

var(8)(p23.1 p23.1) - can appear cytogenetically indistinguishable from 8p23.1 dup syndrome which includes dup of GATA4. Distinguish by FISH or MLPA

var(9)(p12p12) - can appear as dup or trp

var(15)(q11.2q11.2) - difficult to distinguish between PWS/AS critical region.

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13
Q

Why do some euchromatic variants show no phenotypic effect?

A
  • low gene content
  • absence of dosage sensitive loci
  • psuedogenes (copy of a gene that lacks introns or other essential DNA sequences essential for function)
  • functional compensation elsewhere in the genome
  • positional effect e.g. inactivation of a duplicated gene
  • may be a small phenotypic effect but patient within the normal range so difficult to detect
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14
Q

Transposable elements

A

Units of DNA that move within the genome. e.g. LINES (long interspersed nuclear elements) and SINES (short interspersed nuclear elements),
Reterotranspons are transposons that have done a copy and paste.
As transposons accumulate over evolution they can result in repetitive sequences of DNA interspersed throughout the genome.
Transposons can be associated with disease if they insert within a gene or disrupt homologoius recombination and/or mismatch repair resulting in structural rearrangements.

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15
Q

VNTRs

A

variable number tandem repeats aka satellite DNA

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16
Q

Where do you find minisatellites?

A

Cluster at telomere, but can be found throughout the genome. (10-50bp)

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17
Q

How long are microsatellites (STRs)?

A
18
Q

What is restriction lenth polymorphism analysis?

A

The cleavage of DNA is done by restriction enzymes. Individuals can have different lengths of DNA due to the existence of VNTR or SNP mutations at the cleavage site, which can be separated by electrophoresis. This is one of the first fingerprinting techniques.

19
Q

Synonymous SNP

A

Does not change the amino acid sequence

20
Q

Non-synonymous SNP

A

Changes the amino acid sequence

21
Q

What is contained within a nucleotide unit?

A

5 carbon deoxyribose sugar, a phosphate and one of four nitrogenous bases (A,T,G,C)

22
Q

Purines

A

A and G (two interlocked rings)

23
Q

Pyrimidines

A

C and T (single ring)

24
Q

Length of a single pitch (complete turn of DNA)

A

3.6nm

25
Q

Hydrogen bonds for DNA

A

2 for A joining to T

3 for G joining to C

26
Q

Define A, B, H and Z DNA

A

A-DNA: right handed helix with 11bp per turn. Not found in vivo.
B-DNA: right handed helix with 10bp per turn. Most DNA in bacterial and eukaryotic cells adopt this.
Z-DNA: left handed, 12bp per turn. HIgh GD rich favours this form. Conformation occurs during transcription of genes.
H-DNA - triple helix structure caused by inverted repeats of polypurine/polypyrimidine DNA stretches. May have a role of functional regulation of gene exp.

27
Q

Hairpin structure

A

inverted repeats of polypurine and polypyrimidine DNA stretches that form hairpin structures through intra strand pairing (DNA folds back on itself).

28
Q

Cruciform

A

Two hairpin loops arranged in a 4 way junction (such as holliday junction formed during recombination)

29
Q

Promotor region of DNA

A

TATA box (so called because of its 8bp conserved sequence made solely of AT pairs). It lies 30bp up from the mRNA start site. This sequence is recognised by the transcription factor TATA binding protein (TBP).

30
Q

5’ untranslated region (5UTR)

A

The region of a gene from the transcription start site to the nucleotide before the start codon (usually 100 to 220bp, but vary across species). High GC content (approx 60%) is highly conserved. The 5’UTR is an important binding site for ribosomal RNA that initiates mRNA translocation and as such this region is important in regulation of protein expression.

31
Q

Translation initiation codon

A

ATG (codes for methionine)

32
Q

Majority of introns start and end with what base sequence?

A
Start GT (becoming GU in intronic RNA)
End AG
33
Q

Largest gene is what?

A

Dystrophin gene, which is involved in muscular dystrophy. 99% of it is intronic, with 79 introns.

34
Q

Translation stop site codons

A

TAA, TAG or TGA

35
Q

3’ untranslated region

A

Regulator elements within the 3UTR can influence mRNA polyadenylation, translation efficiency, localisation and stability and thus ultimately control protein expression. 3UTR contains the polyadenylation primary signal for the poly A tail to be added to the mRNA to control nuclear export, translation and stability of mRNA.

36
Q

Mitochondrial gene

A

Circular DNA containing 37 genes. Maternally inherited.

37
Q

Mutations can arise by?

A

1) DNA damage
2) failure to to DNA damage
3) errors in DNA replication
4) recombination

38
Q

MMR

A

Mismatch repair. Carried out by MutSa and MutSB, which recognise mismatched base pairs. Recurits MutL complex which excises the mismatched DNA. When MMR is lost there is a decrease in apoptosis and a 100 to 1000 fold increase in mutation rates.

39
Q

BER

A

Base excision repair. Principal repair pathway for oxidative damage. DNA glycosylases recognise and remove damaged bases and inserts new ones. Either short patch (single nucleotide) or long patch (multiple)

40
Q

NER

A

Nucleotide excision repair. Removes pyrimidine dimers caused by UV radiation.

41
Q

DSBs - how are they repaired?

A

1) homologous recombination repair (HRR): a single strand from the homologous chromosome invades the damaged DNA and acts as a template for accurate repair. Involves Rad51.
2) non homologous end joining (NHEJ): Ku and DNA dependent protein kinase complexes bind to the end of the DSB. Ku unwinds the ends, by chance revealing short homologous sequences in the two DNAs, which base pair to yield a region of microhomology. The unpaired single stranded 5’ends are removed and the two double stranded molecules ligate together. As a result the DNA is repaired, but several base pairs at the site of breakage are removed. Leads to permanent change in DNA.