Molecular Genetics chapter 05

• Except for viruses, which may have ssDNA, dsDNA, ssRNA, or dsRNA genomes, the arrangement of all genetic information in an organism takes the form of double-stranded DNA
• A single linear or circular molecule makes up the genome in many viruses and prokaryotes
• Genome size often grows with organism complexity
• Mean weight of a NT pair is 1.0858264521 x 10-9 pg
• One NT pair mean weight is 1.0855 × 10-9 pg
• One pg of DNA correspond to 0.921 × 109 base pairs (bp)
• Genome size (bp)


  0.921 x 10^9 x DNA content (pg)
• DNA content (pg)


  Genome size (bp) / 0.921 x 10^9
• 1 pg = 921 Mb
• Megabases (Mbs) are units of measurement used to indicate the size of DNA
• 1 Mb = 1 x 10^6 base pairs
• According to one research, the haploid genome size of humans is predicted to be 3.055 x 10^9 bp (Nurk S. et al. 2022), or 1.9 x 10^12 Da
• The amount of DNA in a haploid cell of an organism is not related to its evolutionary complexity, and this in known as the C-value paradox
• There is no correlation between the presumed evolutionary complexity and the number of chromosomes
• Cot Curve


  The DNA has the ability for denaturation (heat or denaturant agents) and renaturation or re-annealing (cool down or removal of the denaturant) and this ability is used to estimate the complexity of the sequence of a genome
• Key factors for the rate of hybridization

  • Concentration of the DNA molecule (Co)
  • Time allotted for renaturation (t)
  • Amount of homology between the molecules to be hybridized
• Steps required in DNA denaturation and renaturation experiments

  1. Shear the DNA to a size of around 400 bp

  2. Heat the DNA to 100 °C to denature it

  3. Cool gradually while taking samples at periodic intervals

  4. Calculate the percentage of DNA that is single-stranded at each time point
• Cot curve


  1. axis: Fraction of the DNA that has re-associated (1 – C/Co), representing the % of DNA that remains single stranded
  2. axis: log of Cot, showing the product of the starting [DNA] (moles/L) multiplied by the duration (time) of the reaction progress (seconds)
• The size or complexity of the genome and the quantity of repetitive DNA present in the genome both influence the form of a Cot curve for a specific species
• Cot curves of the genomes for Enterobacteria phage lambda (bacteriophage lambda), Escherichia coli, and Saccharomyces cerevisiae (yeast), will all have the same shape when plotted, but the yeast's Cot½ will be the greatest, E. coli's next, and lambda's lowest

  
• It will take longer for any two complementary sequences to encounter each other and reanneal if the genome is more complex (there are more unique sequences available)
• A more complex species will thus take longer to reach Cot½ at similar DNA concentrations in solution
• Cot analysis of large-sized eukaryotic genomes shows repeated DNA sequences that, due to their nature, anneal fast when compared with unique (non-repeated) sequences that anneal slowly

  
• Genomes containing repeated and non-repeated sequences reanneal in a different way than genomes containing solely single-copy sequences

  
• The Cot value of a particular sequence will be half as large as that of a sequence represented only once in the genome if it is represented twice in the genome because it will have two complementary sequences to pair with

  
• Eukaryotic genome components classified by abundance

  • Unique sequence (single-copy, low-copy, non-repetitive DNA)

  Moderately repetitive DNA

  Highly repetitive DNA

  
• Eukaryotic genome components classified by function

  • Genic DNA

  Regulatory DNA

  Intergenic DNA, spacer DNA

  Satellite DNA

  Selfish DNA

  Junk DNA

  
• As optical density (O.D.) decreases over time, it may be utilized to calculate how much double-stranded DNA was produced

  
• Human DNA comprises of highly repetitive, moderately repetitive, and non-repetitive DNA sequences, its renaturation curve differs from that of Escherichia coli
• Hydroxyapatite column


  Used to run a DNA sample through, which only allows ssDNA to pass through and keeps only dsDNA
• It is simple to determine how much DNA has been retained
• Quantifying the amount of double-stranded DNA created during renaturation

  1. Samples taken at regular intervals

  2. S1 nuclease treatment breaks down ssDNA, and quantity of dsDNA can be determined in the sample

  3. Renaturation will take a longer time for fragment of larger sizes

  
• DNA-DNA denaturation, renaturation, and hybridization procedures

  Can be used to examine the sequence complexity of eukaryotic DNA
• dsDNA


  Unwinds into ssDNA by heating
• The rate of renaturation depends on the concentration of nucleotide strands and their sequences
• DNA renaturation is governed by second order kinetics and is dependent on a chance collision of the complementary strands
• Cot1/2


  The product of the concentration and time needed to proceed halfway through a reaction
• A higher Cot1/2 indicates a slower reaction
• In terms of unineme structure and semi-conservative DNA replication, prokaryotic and eukaryotic chromosomes are similar

  
• Unineme structure


  A chromosome is formed of a single continuous length of DNA double helix