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BPS 4101 - Human Genome Structure & Function > module 1: genome (dis)organization > Flashcards

module 1: genome (dis)organization Flashcards

1
Q

How would you define the term “genome”?

A

The entire genetic
complement (complete set
of DNA) of a living
organism.

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2
Q

(T/F) Genome only describes the nuclear genomes of eukaryotes.

A

False!

Genome includes both the
nuclear and mitochondrial
genomes of eukaryotes.

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3
Q

Answer the following questions regarding human MITOCHONDRIAL genome:

1) Size (bp)

2) # of different molecules

3) # of DNA molecules per cell

4) Approximate # of genes

5) % of DNA that is protein-coding

A

Size (bp): 16,569

different molecules : 1 (circular)

DNA molecules per cell: 1,000-10,000

Approximate # of genes: 37 (13 protein coding)

% of DNA that is protein-coding: 65%

*mitochondrial DNA is haploid as there is only ONE unique molecule

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4
Q

Answer the following questions regarding human NUCLEAR genome:

1) Size (bp)

2) # of different molecules

3) # of DNA molecules per cell

4) Approximate # of genes

5) % of DNA that is protein-coding

A

Size (bp): 3.05 billion (haploid #)

different molecules: 23 (XX) or 24 (XY)

DNA molecules per cell: 46 (diploid)

Approximate # of genes: ~42000 (~20000 protein coding)

% of DNA that is protein-coding: 1-1.5%

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5
Q

Is the mitochondrial genome similar to eukaryotic or prokaryotic and why?

A

Prokaryotic

  • Single ORI
  • Very few promoters (POLYCISTRONIC)
  • Singular
  • Circular
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6
Q

(T/F) Nuclear chromosomes have a promoter for each gene.

A

True! (MONOCISTRONIC)

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7
Q

Define polycistronic mRNA.

A

mRNA that codes for multiple genes using a single promoter.

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8
Q

Mitochondrial genomes are made of heavy and light strands. Define these strands.

A

Heavy: rich in purines (A/G) - extra rings

Light: rich in pyrimidines (T/C)

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9
Q

What kind of genes are encoded by the mitochondrial genome?

A

22 tRNA, 2rRNA, and 13 protein subunits of mitochondrial respiratory chain

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10
Q

What are the three promoters found in the mitochondrial genome and their functions?

A
  1. HSP1 (heavy strand promoter 1 ): transcription of the two rRNAs
  2. HSP2: transcription of the rest of heavy strand (polycistronic transcript)
  3. LSP (light strand promoter): transcription of the light strand
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11
Q

What are NUMTS?

A

Mitochondrial genomic sequences found in nuclear genomes.

Can be small or the full sequences.

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12
Q

Which one of the statements is true regarding the eukaryotic nuclear genome?

  1. Nuclear genome is split into a set of linear DNA molecules
  2. # of chromosomes is the same between eukaryotes
A

1!

For 2, # of chromosomes is VARIABLE between eukaryotes.

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13
Q

What is the C-value and how is it reported?

A

‘C-value’ means the ‘constant’ value of haploid DNA content per nucleus (total amount of DNA in a gamete)

It is reported in PICOGRAMS (1pg is 1 billion bp)

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14
Q

What is the human C-value?

A

3.05pg

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15
Q

Define the C-value paradox/enigma?

A

Genome size (total # of genes) does not correlate with organismal complexity* (eukaryotes).

*how are the genes turned on? how is protein expression regulated?

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16
Q

What are exons?

A

Sections RETAINED in mature RNA after RNA splicing.

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17
Q

Give two examples of why the statement exons = coding is false.

A
  1. Exons include UTRs that are not part of the mature protein
  2. Non-coding RNAs (RNAs transcribed from the genome are never translated into proteins)
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18
Q

(T/F) There are almost as many non-coding genes as there are coding genes.

A

True!

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19
Q

1) What are pseudogenes?

2) How do pseudogenes arise?

A

Pseudogenes are any genomic sequence that is SIMILAR to another gene and is DEFECTIVE.

They arise from REVERSE TRANSCRIPTION/INTEGRATION of a processed mRNA or from gene duplications that accumulate mutations that prevent translation.

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20
Q

How can you tell the difference between pseudogenes and regular genes?

A

Pseudogenes are usually intron-less.

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21
Q

_________ sequences make up the bulk of our genome (more than 50%). They are referred to as _____ DNA.

Give examples.

A

Repetitive; Junk

  1. Tandem repeats
  2. Telomeres
  3. Centromeres
  4. Transposable elements
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22
Q

What are tandem repeats? What are the two types found?

A

Tandem repeats are sequences of DNA bases that are repeated one after the other.

  1. Microsatellites
  2. Minisatellites
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23
Q

Microsatellites are __-__ bp motif, repeated multiple times (total length up to ___bp).

Found in:

Aka:

A

2-10 (very short), 100bp

coding as well as non-coding regions

SHORT TANDEM REPEATS (STRs)

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24
Q

Minisatellites are __-__ bp motif, repeated multiple times (up to __kb).

Found in:

Aka:

A

10-100 (5kb)

subtelomeric regions (adjacent to the telomeres)

VARIABLE NUMBER TANDEM REPEATS (VNTRs)

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25
Q

What are telomeres?

A

Conserved tandem repeats (5’-TTAGGG-3’) at the ends of a chromosome (3,000x).

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26
Q

(T/F) Telomeric DNA contains double-stranded base pairs and single-stranded G-overhang.

A

True.

9-15,000 ds base pairs and 150-300nt G-overhangs.

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27
Q

What is the T-loop? What helps stabilize it?

A

T-loop forms to protect the single-stranded ends of telomeres from the DNA repair machinery.

Binding of the SHELTERIN complex (6 proteins) helps stabilize it.

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28
Q

What is the “end-replication” problem?

What is the role of telomeres?

A

At the very 5’ end of the DNA, there is an RNA primer that gives a 3’-OH to initiate synthesis in the 5’—>3’ direction. This primer can not be replaced with DNA because there is nothing to prime the synthesis reaction, leaving a gap at the 5’ end.

This causes the telomeres of chromosomes to get shorter and shorter with each round of replication.

Telomeres protect against the loss of genetic material.

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29
Q

What happens when telomeres get really short?

A

When telomeres get so short, they withdraw from the cell cycle and enter the G0 phase.

They are in a SENESCENCE state - not dormant as they are metabolically active but no longer dividing.

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30
Q

(T/F) Tumours have to overcome the problem of senescence.

A

True!

Senescence is a barrier to proliferation. Tumours have to overcome this problem by re-expressing the telomerase enzyme that allows them to extend these telomeres which causes the cell division process to start over again.

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31
Q

What are centromeres?

A

Repetitive DNA sequences (tandem repeats; AT-rich).

*Repetitive DNA found in centromeres is often called alpha-satellite DNA.

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32
Q

(T/F) Between two monomers in centromeres, the sequence can vary ~60%.

A

False!

Between two monomers in centromeres, the sequence can vary ~40% in sequences.

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33
Q

Monomers (171bp) in centromeres are arranged head-to-tail and are repeated to form _____ ____ ____, which are organized in tandem to form _________.

A

Higher Order Repeat (HOR); homogenous higher order alpha satellite array

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34
Q

Within the 171bp monomers, there is a ______ box (17 box motif) that directs the binding of protein ________, which recruits other proteins. Together they form the structure called ___________.

A

Within the 171bp monomers, there is a CENP-B box (17 box motif) that directs the binding of protein CENP-B, which recruits other proteins. Together they form the structure called KINETOCHORE.

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35
Q

1) What is CENP-A?

2) What is the function of the kinetochore?

3) What % of centromeres make up the human genome?

A
  1. CENP-A is a histone H3 variant ONLY FOUND in centromeres, part of the kinetochore.
  2. Chromosome segregation (attachment of microtubules) and stability
  3. 3%
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36
Q

(T/F) Transposable elements, another form of repetitive DNA sequences, make up to 45% of the genome.

A

True!

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37
Q

What are the two types of transposable elements?

Briefly describe each.

A
  1. Retrotransposons: move through the genome using RNAs (DNA - RNA - DNA). Flanked by long terminal repeats (LTRs)
  2. DNA transposons: “cut-and-paste” (no RNA intermediate)
38
Q

What are the two classes of retrotransposons?

A
  1. Long interspersed nuclear elements (LINEs) - 3-7kb
  2. Short interspersed nuclear elements (SINEs) - 100-300bp

*interspersed means the opposite of tandem

39
Q

(T/F) Along with exons, non-protein coding genes, pseudogenes, repetitive sequences (tandem repeats, telomeres, centromere, transposable elements), the human genome also contains regulatory elements.

A

True!

40
Q

Match the terms to their definitions:

  1. Metacentric
  2. Submetacentric
  3. Acrocentric
  4. Telocentric

a) one arm of the chromosome is shorter

b) centromere is at the telomere

c) centromere is in the middle, both arms are equal

d) centromere is closer to the telomere. there is also a 2˚ point of constriction

A

Metacentric: centromere is in the middle, both arms are equal

Submetacentric: one arm of the chromosome is shorter

Acrocentric: centromere is closer to the telomere. there is also a 2˚ point of constriction

Telocentric: centromere is at the telomere

41
Q

Define:

1) Karyotype

2) Karyogram

3) Ideogram

A

Karyotype: an individual’s chromosome collection (#, size, shape). separated into 7 groups based on size and relative position of the chromosome. ambiguous identification

Karyogram: photo of an individual’s chromosomes (sorted & arranged by size). position of centromere aligned. exception 21 & 22.

Ideogram: diagrammatic representation of the karyotype. each chromosome displays a banding pattern, which are shared between individuals of a species.

42
Q

Because structural details are difficult to detect under a light microscope, DNA-binding stains are used, giving each chromosome a unique “bar-code.”

What kind of method is most common?

A

G-banding. It is stained with Giesma (mix of 3 dyes).

43
Q

Which of the statements is true regarding G-banding?

1) It preferentially stains GC-rich regions.

2) Pre-treatment consists of mild proteolysis (trypsin)

3) Chromosomes must be in G0 phase.

A

2!

  1. It preferentially stains AT-rich regions (heterochromatin regions like centromeres)
  2. Chromosomes must be in mitosis.
44
Q

The __ is the short arm, while the __ is the long arm.

Which arm is towards the top?

A

p; q

p arm

45
Q

(T/F) When performing G-banding using Geimsa, more condensed chromosomes (metaphase of mitosis) will yield fewer bands (lower resolution) than less condensed chromosomes (prophase of mitosis).

A

True.

Metaphase has less # of bands than prophase. DNA is more accessible when it is less condensed and more spots are able to be stained.

46
Q

(T/F) In an ideogram, each band represents thousands of base pairs, so small changes can be seen.

A

False!

Each band represents millions of base pairs, so small changes can’t be seen!

47
Q

In 12p15.31, what is the
1) chromosome
2) arm
3) region
4) band
5) sub-band
6) sub-sub-band

A

1) chromosome: 23
2) arm: p
3) region: 1
4) band: 5
5) sub-band: 3
6) sub-sub-band: 1

48
Q

(T/F) Structural and numerical abnormalities can be inherited or can occur spontaneously.

A

True!

Can occur spontaneously after fertilization during embryonic development.

49
Q

What are the abnormalities that affect some cells but not others called?

A

Chimeric/mosaic

50
Q
  1. Describe ANEUPLOIDY.
  2. What kinds can there be?
  3. When are aneuploid gametes produced and why are most incompatible with life?
A
  1. Gain or loss of complete chromosomes
  2. Monosomy (loss) or trisomy (gain)
  3. Produced during MEIOSIS, but most are incompatible with life due to GENE DOSAGE being unbalanced. Quantity of protein made in embryonic development is tightly regulated; imbalance leads to dramatic results.
51
Q

How do numerical abnormalities arise?

A

Due to erros in meiosis.

Failure of separation of homologous chromosomes in meiosis I (NON-DISJUNCTION).

or

Failure of separation of sister chromatids in meiosis II.

52
Q

What are the outcomes of non-disjunction in meiosis I and II (only one pair of sister chromatids fails to separate)?

A

Meiosis I: When fertilized, there are 3 copies (3n) of chromosomes in 2 cells, 1 copy (1n) in 2 cells.

Meiosis II: n, 3n, 2n, 2n (50% are euploid gametes)

53
Q

Though non-disjunction can occur in egg or sperm, and during meiosis I or II, which is more common?

A

Egg; Meiosis I

54
Q

Briefly describe premature sister chromatid separation (PSCS) and when/why it happens.

What is the outcome of one sister chromatid separation out of the 2 homologous chromosomes?

A

PSCS occurs when sister chromatids separate during MEIOSIS I.

Sister chromatids are held together by proteins called COHESINS. As we age, these proteins decrease in our cell, causing less proteins to hold the chromatids together. Then, they prematurely separate in meiosis I instead of meiosis II.

This is the MAIN CAUSE of monosomy and trisomy, especially in older women.

Outcome: 2n, 3n, 2n, n

55
Q

(T/F) After 40 years of age in women, there is a 40-60% chance of aneuploidy.

A

True!

Results in miscarriages.

56
Q

Which one of the statements is true regarding trisomies:

1) Down syndrome is the second most common trisomy that is viable.

2) Down syndrome is denoted as 48, XY, +21

3) The other aneuploidy of autosomes is trisomy of 13 and 18, but they are short survival.

A

3!

Down syndrome is the MOST common trisomy that is viable.

Down syndrome is denoted as 47 (chromosomes), XY (doesn’t matter), +21 (gain of chromosome 21).

57
Q

Why are aneuploidies of other chromosomes (not 13, 18, and 21) non-viable?

A

GENE DOSAGE!

13, 18, and 21 have the fewest # of genes; excess of these genes will be tolerated more.

*Y chromosome has the fewest # of genes.

58
Q

Why is trisomy of sex chromosomes tolerated better than autosomal chromosomes?

A

+Y: easily tolerated as it has the lowest # of genes

+X: X-inactivation (one is active, rest are all silent)

59
Q

What is the klinefelter syndrome?

A

Extra X chromosome

47, XXY

60
Q

Aside from the mosaic, what is the only example of viable monosomy in humans?

A

Monosomy of the sex chromosome (X)

45, X

61
Q

(T/F) Humans can tolerate loss of chromosomes than gain.

A

False!

Humans can tolerate gain of chromosomes more than loss.

*monosomy of x chromosome has a lower % than the other gain examples.

62
Q

1) Why do structural abnormalities occur?

2) How long can they span?

3) What is balanced vs unbalanced structural abnormalities?

A

1) They are often the result of INCORRECT JOINING TOGETHER of two broken chromosome ends

2) Abnormalities can be 1000-millions of NTs.

3) Balanced (no change to the total amount of genetic material) vs Unbalanced (duplication/deletions)

63
Q
  1. What is duplication?
  2. What are the two kinds possible?
  3. What is 46,XX,dup(8)(pter p22) represent?
A
  1. extra copies of a chromosomal region
  2. tandem (x-x) vs displaced (x-y-x)
  3. There is a duplication occurring on chromosome 8 on the terminal region of the p arm up to p22.
64
Q

(T/F) Duplications affect gene dosage and thus are always harmful to humans.

A

False!

Though it affects gene dosage, duplication can provide material for evolution!

65
Q

Describe Charcot-Marie-Tooth disease.

A

Duplication maps to 17p11.1

3 copies of the gene that codes for the synthesis of the PMP-22, that is involved in myelin sheath formation (results in muscular weakness of legs + hands)

66
Q

How can duplications arise (two ways)?

A
  1. Unequal crossing over during meiosis I (aligns incorrectly; one gamete has a duplication, while the other has a deletion)
  2. Slippage of DNA polymerase during transcription (rejoins & synthesizes same region)
67
Q
  1. What are the two different types of deletions?
  2. What do they cause?
  3. What are hCONDELs?
A
  1. TERMINAL (deletions at the end; doesn’t affect telomeres) and INTERSTITIAL deletions (deletions within chromosomes)
  2. Haploinsufficiency (gene dosage)
  3. hCONDELs: human conserved deletions. ~510 deletions near genes that are involved in steroid and neural signaling which have contributed to human EVOLUTION.
68
Q

Match the following diseases to their definitions:

  1. Cri du chat
  2. Prader Willi

A. XY, del(15)(q11q12) - deletion of regions 11 to 12 of p arm of the chromosome 15.

B. (46,XX,5p-) can be a small deletion or full loss of the p arm; size of deletion correlates with severity of intellectual development and other symptoms.

A

Cri du chat: (46,XX,5p-) can be a small deletion or full loss of the p arm; size of deletion correlates with severity of intellectual development and other symptoms.

Prader Willi: XY, del(15)(q11q12) - deletion of regions 11 to 12 of q arm of the chromosome 15.

69
Q

Briefly describe how Prader Willi occurs:

A

Due to MATERNAL IMPRINTING (specific pattern of METHYLATION), genes found in regions q11q12 of the MATERNAL allele not expressed (heterochromatin, repressed environment). They are only expressed from the paternal allele.

Prader Willi can occur because of:
1) deletion of the same region occurs from the PATERNAL allele
2) UNIPARENTAL DISOMY (both alleles coming from the mom due to an error in meiosis)

*if deletion occurred on the maternal allele, no impact on individuals as it is already silenced

70
Q
  1. What are the two different types of inversion?
  2. Are inversions balanced or unbalanced?
  3. Why can’t individuals be homozygous for inversion?
A
  1. PARACENTRIC inversion (inversion does not include the centromere), PERICENTRIC (inversion that includes the centromere).
  2. Balanced: no net loss of genetic material (usually viable with LITTLE or NO phenotypic consequences)
  3. The other allele can not compensate if it is a homozygous inversion.
71
Q

(T/F) For an inversion to occur, there needs to be one double stranded DNA break required to allow rotation and ligation.

A

False!

For an inversion to occur, there needs to be TWO double stranded DNA breaks required to allow rotation and ligation.

72
Q

(T/F) Inversions can cause problems if the breakpoints occur within the coding/non-coding region of a gene sequence of importance.

A

True!

73
Q

How can heterozygous inversion reduce fertility by 50%?

A

Chromosome with inversion creates a loop to pair up homologous region in “normal” chromosome, which causes two inviable/deleted products (chromosome), and two normal products.

74
Q

1) What is a translocation and what are the two types of translocations?

2) In which translocation type is there a net loss of genetic material?

3) What is 46,XY,t(4;20)(q28.2;q12) telling us?

A

1) Translocation is the transfer of genetic material ENDS from one chromosome to a non-homologous chromosome.

Reciprocal and non-reciprocal.

2) Both translocations result in NO NET LOSS of genetic material.

3) translocation; region from q28.2 all the way to the end of chromosome 4 has traded places with region q12 of chromosome 20.

75
Q

1) What are Robertsonian translocations?

2) What kinds of chromosomes are involved?

3) How many chromosomes do the carriers have?

4) What are the phenotypic consequences?

A

1) One normal, non-homologous chromosome breaking and rejoining with another chromosome.

2) ACROCENTRIC; the long arms are joining to be metacentric/submetacentric.

3) 45 chromosomes but all genes are present. The short arms of the chromosomes that fuse together include repetitive sequences mostly.

4) Carriers are normal but they have issues during fertilization.

76
Q

Robertsonian translocation occurs in which chromosomes to cause 5% of down syndrome? How?

A

Chromosomes 14 and 21.

The parent would have a regular chromosome 14, a fused 14/21 chromosome, and a regular chromosome 21.

When there is fertilization by A NORMAL GAMETE, it results in 2n, balanced carrier, trisomy 14 (NV), monosomy 14 (NV), monosomy 21 (NV) and trisomy 21.

*draw it out

77
Q

(T/F) Chromosomal deletions are used to diagnose the different stages of blood cancer.

A

False!

Chromosomal translocations are usually found in many cancers (especially blood). They can be used to diagnose different stages of the cancer depending on the GENETIC CHANGES.

78
Q

What is the philadelphia chromosome? How does it cause Chronic Myelogenous Leukemia (CML)?

A

Philadelphia chromosome occurs when there is a RECIPROCAL TRANSLOCATION of chromosomes 9 and 22.

Derivative 9 is longer, while derivate 22 is shorter. t(9;22)(q34;q11)

In normal chromosome 9, there is a tyrosine kinase (ABL) that targets cell cycle progression and division. This kinase is highly regulated.

When this translocation occurs, ABL fuses with a protein (BCR) found in chromosome 22. This fusion protein retains the phosphorylating ability but loses its ability to be regulated. This causes it to be always in the cytosol and activating proteins involved in cell cycle division, causing CML.

79
Q

What are some of the limitations of G-banding?

A

Resolution; anything less than 5 million base pairs can not be seen; anomalies have to be VERY large to be seen.

80
Q

1) What does FISH stand for?

2) Briefly describe the process.

3) What kinds of anomalies can it detect?

A

1) Fluorescent In-Situ Hybridization

2) a short sequence (50-60 NTs) of a gene of interest is used as a ds DNA Probe, which is labelled with fluorescent dye. Gene of interest is denatured and DNA probe binds to complementary sequence.

3) Numerical, deletions, & duplications

81
Q

Which statement is true?

1) FISH is done on interphase chromosomes.

2) Novel translocations and inversions can be seen using FISH.

3) Tandem duplication may not be seen using FISH.

A

3!

For 1, FISH is usually done on METAPHASE chromosomes as we can tell chromosomes apart. Chromosomes have not replicated and DNA is more spread out and less organized in interphase.

For 2, a specific area has to be translocated with FISH. It is not useful for novel translocations as we don’t know where to create the DNA binding probe for. It is very helpful to confirm translocations!

Inversions are hard to see using regular FISH.

82
Q

(T/F) Fish can detect inversions using orientation-specific single-stranded DNA probes.

A

True!

*look up notes on how it works

83
Q

Describe Spectral karyotyping (SKY) aka chromosome painting.

A

All chromosomes are broken down into fragments and are labelled with different fluorophores.

Probes spanning all metaphase chromosomes are mixed in a DNA PROBE COCKTAIL and hybridized to a sample (ie. patient’s chromosomes).

Unlike G-banding, there are no patterns seen in the chromosomes. These are just colourful.

Deletions, duplications and inversions are hard to see but translocation can be seen!

84
Q

Briefly describe how Comparative Genomic Hybridization (CGH) works.

A

Isolation/labeling of entire genome DNA from references (untreated/healthy) cells one colour.

Isolation/labeling of entire genomic DNA from test (experimental) cells another colour.

Competitive hybridization to differentially labelled whole-genome probes to a MICROARRAY.

*labelled with Cy3 or Cy5

85
Q

What are microarrays?

A

Microarrays are glass slides with thousands to millions of distinct DNA probes (single-stranded DNA) immobilized on a solid surface that represent different regions of the genome.

There are thousands of copies of the same sequence per spot. In a different spot, there are thousands of copies of another sequence.

86
Q

(T/F) A SurePrint G3 Human CGH microarray 1x1M contains 60-mer probes that correspond to 1 million different regions across the genome (esp gene enriched regions) targeting regions that are ~2.1 kilobases apart.

A

True!

60-mer probes: 60 NT long ss DNA probes

87
Q

Normal gDNA is labelled with Cy5 (red) and patient gDNA is labelled with Cy3 (green).

If the patient has a type of cancer caused by a deletion of a large portion of the p-arm of chromosome 3. What would you expect to see for that corresponding region when analyzing the CGH results?

  1. more green fluorescence
  2. more red fluorescence
  3. each amounts of red and green (yellowish)
A

2!

88
Q

Match the following scenarios to their description:

1) Heterozygous deletion

2) Heterozygous duplication

3) No change

A) Log2(1) = 0

B) Log2(0.5) = -1

C) Log2(1.5) = 0.584

A

Experimental:Control Ratio

Heterozygous deletion:
1 experimental:2 control = 1:2
Log2(0.5) = -1

Heterozygous duplication:
3 experimental:2control = 3:2
Log2(1.5) = 0.584

No change:
2 experimental:2control = 1:1
Log2(1) = 0

89
Q

(T/F) It is possible to get exactly zero fluorescence for a homozygous deletion.

A

False!

0 experimental : 2 control = 0
Log2(0) is not possible!

Thus, we can’t get zero fluorescence!

We have to put a very small number: log2(0.01) = -6.0.

90
Q

Though CGH is the most powerful tool to detect chromosomal abnormalities due to its very high resolution, ability to detect denovo changes, and looks at the entire genome, what are some of its dis-advantages?

A
  • Limited to COPY NUMBER CHANGES ONLY (deletions and duplications)!
  • Can’t detect translocations and inversions.
  • Expensive

BPS 4101 - Human Genome Structure & Function (4 decks)

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