Human Genome Karyotype

Question 1

C-value

Answer 1

Amount of DNA in one copy of the genome

• 3.2 x 10^9 bp of DNA
• 22,000 genes
• 22 pairs of autosomes and 2 sex chromosomes

Question 2

Tandem repeats

Answer 2

1. Ancient repeats
   
   1. Have diverged in nucleotide sequence over time
2. Recent repeats
   
   1. Have greater than 90% sequence identity
   2. The repeats occured recently and therefore have not diverged in nucleotide sequence yet.

Question 3

Recombination between repeated sequences

Answer 3

1. Repeats are substrates for recombination because they are similar to or identical in nucleotide sequence
2. If sequence identity exists in more than 2 places, recombination can occur between these regions
3. Recombination between them may cause inversion, duplication, or deletion.
4. Examples
   
   1. Red-Green color blindness
      
      1. Genes for red and green receptor misalign on X chromosome
      2. One X gets only red and the other X gets duplicated red receptor
   2. Rh-factor
   3. Di George or Velocardiofacial syndrome and Prader Willi and Angelman’s syndrome
      
      1. Recombination between ADJACENT genes can cause deletion of a block of DNA that contains multiple genes
      2. Failure of pharyngeal pouches to develop
         
         1. Parathyroid, thymus, cardiac defects, failure of pharyngeal pouches to develop
   4. Hemophilia A (X-linked)

Question 4

Short repeats

Answer 4

1. Satellite sequences
   
   1. Tandem repeats of sequences of a few hundred bp long
   2. Mostly seen at centromeres and telomeres
2. Micro-satellites
   
   1. Repeats of a few nucleotides
      
      1. i.e (CA)n dinucleotide
      2. Copy number n is highly variable

Question 5

How are retrotransposons incorporated into DNA

Answer 5

1. Transcription by RNA polymerase from DNA to RNA
2. Reverse transcription of RNA to cDNA via reverse transcriptase
3. Incorporation of cDNA into genome
4. I.E. LINES and SINES
   
   1. LINES –> encode reverse transcriptase
   2. SINE –> short cellular RNA
   3. Pseudogenes –> copies of cellular mRNA that are not transcribed due to lacking promoter

Question 6

G-banding

Answer 6

• Creates a pattern of dark and light bands unique to each chromosome
  
  • Dark bands pick up the stain more and are called G-bands
• Can be used to determine a particular chromosome
  
  • Size
  • Centrosomere position
  • Banding pattern
• Technique
  
  • Cells incubated with colchcine which binds microtubules and arrests cell in metaphase
  • Chromosomes condense steadily during prolonged metaphase
  • Stained with giemsa dye–> 500-800 bands per haploid set oc chromosomes is seen
• Why use it?
  
  • Can detect only relatively LARGE changes in chromosome structure

Question 7

Under what context should you consider a diagnosis of chromosome abnormality?

Answer 7

1. Problems with physical or mental development in fetus or child
2. Infertility/miscarriage
3. Pregnancy in women 35 yrs or older
4. Cancer

Question 8

Fluorescent in situ hybridization

Answer 8

• Technique
  
  • Chromatin or chromosomes are fixed to slide
  • Probe binds to DNA of complementary sequence
• Interphase FISH
  
  • Faster because it is done directly on clinical samples
  • Lower resolution because DNA is not condensed
• Metaphase FISH
  
  • Requires culture to amplify cell number and then incubated with colchcine to cause chromosomes to dense
  • Higher resolution
• Why is it used?
  
  • Detects changes in chromosome structure too small to see in G-banding such as INDELs
  • Only detects presence/absence/position of DNA to which the probe binds
  • Location of the mutation must be known to design a probe
  • Cannot detect single nucleotide changes
  • Resolution decreases as number of probes increases so whole chromosome FISH lacks the resolution

Question 9

Comparitive Genomic Hybridization

Answer 9

• Technique?
  
  • Array of oligonucleotides immobilized at different positions on glass slide complementary to sequences on genome
  • Comparison of PCR amplified genome DNA with reference genome (control) in ability to hybridize with oligonucleotides
  • Plot of CGH hybridization
    
    • Left side is p arm and right side is q arm
• Why is it used?
  
  • Can also be used to see small changes in chromosome structure such as deletions or duplications but unlike FISH, you don’t need to know location of structural change
  • Cannot detect inversions or translocations but you can see translocations in GIEMSA and FISH
  • Can detect increase/decrease in copy number

Question 10

Pericentric

Answer 10

Break points are in different arms and involves an inversion around the centromere

Question 11

Paracentric

Answer 11

Both breakpoints are in the same arm and do not involve centromere.