Explore Britten's DNA reassociation rates for different organisms.
Britten’s Cot curve analysis compares DNA complexity.
A useful number in this type of analysis is Cot1/2. This is the value when 50% of the DNA has reassociated. For example, the Cot1/2 of E. coli is 9.
What is the approximate Cot1/2 of bacteriophage T4 DNA?
9 (No, this is the Cot1/2 of E. coli.)
1 (No, this is not correct.)
10 (No, this is close to the Cot1/2 of E. coli.)
10 -1 (That is correct)
The Cot1/2 of T4 DNA is 3 X 10-1.
The Cot1/2 value is derived from the log of the DNA concentration X time. A greater Cot1/2 value means a slower reaction. It takes longer for E. coli DNA to completely reassociate than bacteriophage T4.
The Cot1/2 value is also related to the amount of DNA present in the genome — the concentration of complementary sequences. A larger genome has more DNA, but by comparing Cot1/2 values with E. coli, the "complexity" of a genome can be calculated.
The complexity is the total length of different sequences present in a genome and is usually expressed in base pairs. The E. coli genome has only unique sequences; the total length of different sequences is 4.2 X 106bp. What is the complexity of bacteriophage T4 DNA?
4.2 X 10 6 bp (No, this is the size of the E. coli genome.)
4.2 X 10 5bp (No, this is not correct.)
1.4 X 10 6 bp (No, this is not correct.)
1.4 X 10 5 bp (That is correct)
Bacteriophage T4 has a Cot1/2 of 3 X 10 -1, and the Cot1/2 of E. coli is 9. The complexity of the bacteriophage T4 genome is 1.4 X 10 5bp.
Now, let’s look at eukaryotic DNA.
Since there are three components to this reassociation curve, each component has its own Cot1/2value. What are the Cot1/2values of the fast, medium and slow components?
0.004, 3, 1000 (That is correct)
10, 10, 10 (No, the Cot1/2 values aren’t all the same.)
0.004, 10, 1000 (No, one of the values is incorrect.)
1, 10, 1000 (No, two of the values are incorrect.)
The Cot1/2values of the fast, medium and slow components of this curve are 0.004, 3 and 1000.
The fast component of this curve represents 25% of the total DNA, the medium component is 30% and the slow component is 45%. If we could isolate these fractions and run them as separate reassociation reactions, then the Cot1/2values would be:
0.004, 3, 1000 (No, these are the original Cot values.
0.25, 0.3, 0.45 (No, these are the % DNA in each component.)
1 X 10 -3, 0.9, 450 (That is correct)
If we treat each of the components as separate reactions, then the Cot1/2values for the fast, medium and slow curves would be 1 X 10 -3, 0.9, 450.
What is the complexity of the slow fraction of this eukaryotic DNA?
1.47 X 10 8 bp. (No, that is not correct.)
4.7 X 10 8 bp. (That is correct)
5.95 X 10 8 bp. (No, that is not correct.)
6.62 X 10 8 bp. (No, that is not correct.)
Since you are comparing the Cot1/2value of 45% of the eukaryotic DNA, the E. coli Cot1/2 value needs to be factored for 45%. This means that:450 / (9 X 0.45) = complexity of the slow fraction / 4.2 X 10 6= 4.7 X 10 8.
The complexity of the slow fraction is the total length of unique sequence in this genome. Since the slow fraction is only 45% of the DNA, what is the estimated size of the whole genome?
4.7 x 10 8 bp. (No, this is not correct)
1.47 x 10 8 bp. (No, this is not correct)
1.0 x 10 9 bp. (That is correct)
If 4.7 X 10 8 bp represents 45% of the genome, then 1.0 X 10 9 bp represents 100% of the genome.
YOU'RE SO SMART!
dna reassociation, dna complexity, dna concentration, complementary sequences, britten, cot curve analysis