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BMS2.02 Human Genome and Karyotype (MATISE) > Human Genome and Karyotype > Flashcards

Human Genome and Karyotype Flashcards

1
Q

Genome size/C-value:

A
  • amount of DNA in one copy of the genome.
  • Humans have 3.2 x 109 base pairs of DNA in each somatic cell of our bodies.
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2
Q

Humans have _____ base pairs of DNA in each somatic cell of our bodies.

A

3.2 x 109 base pairs

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3
Q

How many genes are in the human genome?

A

about 22,000

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4
Q

Humans have ____ pairs of autosomal chromosomes and ____ sex chromosomes (X or Y).

A
  • 22 pairs of autosomal chromosomes
  • 2 sex chromosomes (X or Y).
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5
Q

What cell phase does condensation of chromatin into chromosomes occur?

A

prophase of mitotic cell division

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6
Q

Genome size generally ______ with an organisms complexity.

A

increases

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7
Q

The C-value enigma:

A
  • Wide variations exist between genome size and organism complexity.
    • e.g., some single-celled protists have genomes much larger than that of humans.
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8
Q

Ploidy is:

A

the number of chromosomes in an organism

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9
Q

Does chromosome number (ploidy) increase with organism complexity?

A

NO

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10
Q

The two basic mechanisms by which genome complexity/size arises:

A
  1. duplication: partial or whole
  2. incorporation​: of other specie’s DNA
    • “lateral transfer”
    • mitochondria
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11
Q

The three models of human genome make-up:

A
  1. Mostly junk, some useful parts.
    • due to incorporation; “littered”
  2. Highly complicated, each part has specific function and everything works together.
  3. Highly complicated machine, with a simple function.
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12
Q

Five regions of genomic DNA:

A
  • Protein coding
  • Regulatory (promoters, enhancers, etc).
  • Long non-coding (RNA; no protein)
  • Short non-coding (RNA; no protein)
  • Unknown function (RNA; no protein)
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13
Q

Goal of ENCODE project:

A
  • map the human genome in about 80 different human cell types
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14
Q

What criteria were used by the ENCODE project to determine what was “functional” DNA?

A
  1. Transcripts and protein-encoding exons.
  2. Chromatin (histone) modification and DNA methylation.
  3. DNAse hypersensitivity.
  4. Binding of ~100 known transcription factors, RNA Pol II/III, and other proteins.
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15
Q

The ENCODE project determined that chromatin exists in how many functional states?

A

7

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16
Q

The ENCODE project determined that ____% of our DNA is transcribed into RNA?

A

60-75%

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17
Q

What was the main conclusion of the ENCODE project?

A
  • At least 80% of the genome is likely to be functional, implying thatnon-coding regions may be as, or more important than, protein-encoding regions (as determinants of health and disease).
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18
Q

Single-nucleotide polymorphisms (SNPs) definition and prevalence:

A
  • A:T vs G:C (single nucleotide swap out)
    • about SNPs 3,000,000 per genome
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19
Q

Insertions and deletions definition and prevalence:

A
  • GCATT vs. GT
    • about 800,000 indels per genome
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20
Q

Block substitutions definition and prevalence:

A
  • GCATT vs. GTTAT
    • about 50,000 per genome
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21
Q

Inversions definition and prevalence:

A
  • GCATT vs GTACT (sequence reversed)
    • about 100 per genome
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22
Q

Copy number variations definition and prevalence:

A
  • GCATT vs. GCATCATT (sequence repeated)
    • about 50-100 per genome; small and large
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23
Q

Three types of repetitive sequences present in the human genome:

A
  1. tandem repeats (of genes of blocks of genes)
  2. short repeats
  3. retrotransposons
    • ​​repeats that are the products of reverse transcriptases
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24
Q

How to determine the age of a tandem repeat:

A
  • the more recent the incorporation of the repeat, the more sequence identity it will share with the original sequence
    • more divergent = older
    • older = much different encoded proteins
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25
Q

What kinds of genome changes form “hot spots” for recombination?

A

tandem repeats

  • increases the chance of structural change in chromosomes and the frequency of some genetic conditions.
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26
Q

Tandom repeats are substrates for recombination because:

A
  • they are similar or identical in nucleotide sequence.
  • If sequence identity exist in more than 2 places, recombination can occur between these regions.
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27
Q

Recombination between repeats may cause:

A
  • inversion, duplication, or deletion, depending on the position and orientation of the repeats.
  • occurs during meiosis.
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28
Q

Red-green color blindness is caused by:

A
  • recombination between duplicated genes with almost identical sequence identity on the X chromosome (tandem repeats).
  • occurs during meiosis - misalignment of chromosomes.
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29
Q

Contiguous gene syndromes are due to:

(a.k.a. microdeletions)

A
  • Recombination occurs between large repeats deletes a block of DNA which contains multiple genes.
    • ex:
      • diGeorge syndrome
      • prader-willis
      • angelman
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30
Q

Satellite sequences:

A
  • short repeat
  • tandem repeats of sequences of a few hundred base pairs long
  • hundreds to thousands of copies, mostly at centromeres and telomeres.
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31
Q

Microsatellites:

A
  • short repeats of a few nucleotides.
  • relatively common.
  • copy number highly variable.
  • widely used to identify specific chromosomes in genetic counseling, because often each of the four parental copies will be different.
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32
Q

The two types of short repeat sequences:

A
  • satellites
  • microsatellites
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33
Q

Where are satellite sequences normally found?

A

centromeres and telomeres

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34
Q

Retrotransposon process:

A
  1. ds-DNA transcribed into mRNA by RNA polymerase.
  2. mRNA transcribed back into ds-cDNA by reverse transcriptase.
  3. ds-cDNA randomly incorporated back into ds-DNA. This reintegration can disrupt normal gene function.
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35
Q

The three types of retrotransposons:

A
  1. LINES
  2. SINES
  3. pseudogenes
36
Q

LINES retrotransposons:

A
  • “long interspersed nuclear elements”
  • mRNA-encoding reverse transcriptase
  • functional
37
Q

SINE retrotransposons:

A
  • “short interspersed nuclear elements”
  • copies of a short cellular RNA.
    • Most abundant are Alu sequences
  • functional
38
Q

Pseudogene retrotransposons:

A
  • copies of cellular mRNAs.
  • Not transcribed because they lack promoter sequences.
  • non-functional
39
Q

Karyotype refers to:

A
  • the number and structure of chromosomes
  • humans have 22 autosomal chromosomes and 2 sex chromosomes
40
Q

In what 4 contexts should you consider a diagnosis of chromosome abnormality?

A
  1. Problems with physical or mental development of a fetus or child.
  2. Infertility, spontaneous abortion, or stillbirth, especially if repeated.
  3. Pregnancy in a woman 35 years of age or older.
  4. Cancer
41
Q

Three techniques used for visual identification of chromosomes and to detect changes in their structure:

A
  • G-banding
  • Fluorescent in situ hybridization (FISH)
  • Comparative genomic hybridization (CGH)
42
Q

G-banding:

A
  • Giemsa staining creates pattern of dark and light bands unique to each chromosome.
  • indiscriminant stain.
43
Q

Fluorescent in situ hybridization (FISH):

A
  • detects changes in chromosome structure too small to see by G-banding.
    • uses small probes to identify specific genes.
  • The location must be known (to design probe).
44
Q

Comparative genomic hybridization (CGH):

A
  • detects deletions or duplications in chromosomes even if their location is not known.
  • usually microarray approaches
45
Q

G-banding steps:

A
  1. Cells incubated with colchicine – binds tubulin, prevents spindle function, arrests cells in metaphase.
  2. Chromosomes condense steadily during prolonged metaphase with time, the number of cells in mitosis increases but the number of visible bands decreases.
  3. Stained with Giemsa dye.
46
Q

Colchicine:

A
  • used in G-banding
  • binds tubulin, prevents spindle function
  • arrests cells in metaphase
47
Q

How are G-bands utilized?

A
  • G-banding stains chromosomes with a specific and reproducible pattern. Each chromosome has a unique staining pattern.
  • Compare patien’ts G-banding pattern to a normal G-banding pattern.
48
Q

Centromere position in this G-band stained chromosome:

A
  • metacentric
    • meta = middle
    • ex: chromosome 3
49
Q

Centromere position in this G-band stained chromosome:

A
  • sub-metacentric
    • ex: chromosome 18
50
Q

Centromere position in this G-band stained chromosome:

A
  • acrocentric
    • (acro = high)
    • ex: chromosome 22
51
Q

Q-arm and P-arm of G-band stained chromosome:

A
  • P-arm = short arm (‘p’ for “petite”)
  • Q-arm = long arm
52
Q

Telocentric:

A
  • describes location of centromere when the centromere is at the end of a chromosome
  • not seen in humans, but in other vertebrates
53
Q

How to name a chromosome location:

6q31.1

A

6q31.1

  • chromosome 6
  • Q-arm
  • position 31.1 on the Q-arm
54
Q

What changes in a chromosome can you detect with G-banding?

A
  • can only detect large changes in chromosome structure
    • ex: big deletions
55
Q

How large is one G-band on a chromosome?

A
  • about 4-7 Mb, 45 or so genes.
  • to detect smaller changes, need FISH or CGH
56
Q

Fluorescent in situ hybridization (FISH) process:

A
  • Chromatin or Chromosomes are fixed to slide
  • Probe binds to DNA of complementary sequence
    • probe has flourescent dye
  • resolution of FISH is determined by probe length
    • shorter probe = higher resolution
57
Q

The two types of FISH:

A
  • metaphase FISH - higher resolution
    • DNA condensed
  • interphase FISH - lower resolution
    • DNA not condensed
58
Q

FISH detects only:

A
  • presence/absence/ position of DNA to which the probe binds. So, you must know the sequence or chromosome you are looking for.
  • A normal result with one probe does not rule out a genetic defect elsewhere.
59
Q

Comparative Genome Hybridization (CGH) process:

A
  • array of oligonucleotides immobilized at different positions on a glass slide (microarray), complementary to sequences spaced across the genome.
  • Compare PCR-amplified patient genome (test) DNA with reference genome (control) in ability to hybridize with oligonucleotides.
60
Q

Yellow signal in Comparative Genome Hybridization (CGH) indicates:

A
  • equal amount of test and reference signal
  • patient is normal for that gene
61
Q

CGH G:R Signal Ratio:

A

1:1

62
Q

CGH G:R Signal-Ratio:

A

0.5:1

63
Q

CGH G:R Signal Ratio:

A

3: 2
1. 5:1

64
Q

CGH detects only:

A
  • only increases/decreases in copy number.
    • very small changes anywhere in genome.
    • Do not need to know where to look.
  • Cannot detect rearrangements without gain or loss, e.g. inversion or translocation.
65
Q

Haploid cells:

A

1N

  • germ cells (sperm and egg)
  • contain 1 copy of each chromosome
66
Q

Diploid cells:

A

2N

  • zygotes and somatic cells
  • contain 2 copies of each chromosome
    • one mother
    • one father
67
Q

In diploid cells the two copies of each chromosome are homologues and form a:

A

homologous pair

  • one homologue maternal
  • one homologue paternal
68
Q

Euploidy:

A
  • normal number of chromosomes:
    • 22 pairs of autosomes and one pair of sex chromosomes:
    • Karyotype designation: 46, XX or 46, XY
69
Q

Aneuploidy:

A
  • extra or missing chromosomes:
    • Missing chromosome: monosomy (2N – 1)
    • Extra chromosome: trisomy (2N + 1)
70
Q

Monosomy:

A

missing one chromosome

71
Q

Trisomy:

A

extra chromosome

72
Q

Triploidy:

A

3N

  • two sperm fertilize one egg: not viable
  • 3 complete chromosome sets; lethal.
73
Q

Male with Trisomy 21 (Down syndrome) karyotype designation:

A

47, XY, +21

74
Q

Female with Trisomy 13 karyotype designation:

A

47, XX, +13

75
Q

Male with two X chromosomes karyotype designation:

A

47, XXY

“Kleinfelter syndrome”

76
Q

Female with Monosomy X karyotype designation:

A

45, X

“Turner syndrome”

77
Q

Types of chromosomal abnormalities:

A
  1. translocation
    • reciprocal or non-reciprocal
  2. inversion
  3. deletion
  4. duplication
78
Q

Translocation:

A
  • genetic material moved from one chromosome to another.
  • Two types:
    • Reciprocal (two chromosomes exchange segments)
    • Non-reciprocal (movement of DNA from one chromosome to another).
79
Q

Karyotype designation:

A

46, XY, t(1;3)(q31;q24)

80
Q

How chromosomes are altered:

A
  • double strand breaks
  • important for carcinogenesis
  • frequency increased by radiation
81
Q

How to identify a chromosome abnormality:

A
  • number of centromeres
    • tells us how many chromosomes are present in a karyotype
  • identity of centromere
    • i.e. if a chromosome has the centromere of chromosome 22 but an arm of chromosome 21, it is considered an altered chromosome 22
82
Q

The most common transclocations occur between what types of chromosomes?

A
  • acrocentric autosomal chromosomes:
    • group D (13-15)
    • group G (21-22)
  • the p-arms of these chromosomes contain only genes for rRNA
83
Q

Robertsonian Translocations:

A
  • Breakpoints occur within the centromeres of D- and G-group chromosomes, with fusion of chromosomes and loss of p arms.
  • carrier has normal phenotype
  • no loss of DNA from Q-arms; the Q-arms are fused together
84
Q

Isochrome 21

A
  • both arms from 21q
  • viable since diploid from 21q
  • Fertilization will result in either:
    • trisomy 21 (often lethal, Down syndrome if viable)
    • monosomy 21 (lethal).
85
Q

Pericentric inversions:

A
  • both breakpoints are in the same arm
  • broken fragments relocate to a different location on the same chromosome
  • 46, XY, inv(6)(p23;q21)
86
Q

Duplication diagram and karyotype:

A
  • segmental repeats
  • 46, XX, dup8(q13 → qter)
87
Q

Deletion diagram and karyotype designation:

A

46, XY, del19(q13.1;q13.3)

BMS2.02 Human Genome and Karyotype (MATISE) (2 decks)

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