Genetic Constants In DNA and RNA - 4

Fig. 1
Link

This post analyzes the quality of sequences of ancient human DNA and compares those results with modern human DNA analyses.

The comparison technique is to see how each sequence matches up to the "perfect DNA sequence" using the DNA constants method (Genetic Constants In DNA and RNA, 2, 3).

There are three appendices to today's post: One details the DNA of ancient Eurasian humans which was reported on in a recent Nature paper (The spatiotemporal distribution of human pathogens in ancient Eurasia).

Another one details the DNA of ancient Egyptians that was detailed in another recent Nature paper (First human genome from ancient Egypt sequenced).

And the third appendix details modern human chromosome DNA that has been reported in various Nature papers.

In the first post of this series the fundamentals of this way of testing the veracity of a DNA or RNA sequence was featured. 

It is simple arithmetic to count the number of A, T, C, and G nucleotides there are in one or more DNA molecules (Fig. 1).

And it is also simple arithmetic to count the number of A, T, C, and G nucleotides there are in a DNA genome sequence.

That makes it easy to determine if the researchers were able to perfectly acquire the DNA in a sample and then to perfectly determine the number of A, T, C, and G's in that sample.

For example, there are two A's and  T's,  and two C's and G's in the Fig. 1 graphic, so using the data supplied in the first post of this series.

Let's review a portion of that post:

DNA:

(Adenine)  'A' = "C₅H₅N₅" (5 Carbon, 5 Hydrogen, 5 Nitrogen)

(Cytocine) 'C' = "C₄H₅N₃O₁" (4 Carbon, 5 Hydrogen, 3 Nitrogen, 1 Oxygen)

(Thymine) 'T' = "C₅H₆N₂O₂" (5 Carbon, 6 Hydrogen, 2 Nitrogen, 2 Oxygen)

(Guanine)  'G' = "C₅H₅N₅O₁" (5 Carbon, 5 Hydrogen, 5 Nitrogen, 1 Oxygen)

DNA molecules contain 59 atoms

Fig. 2

(Genetic Constants In DNA and RNA).  So,  there would be (59 x 2) 118 total atoms in that Fig.1 segment; (5+5+5) + (4+5+3+1) + (5+6+2+2) + (5+5+5+1) = 59, and 59 x 2 = 118.

If a larger segment had 597 instead of two, you multiply each numeral by 597 to derive the totals (597 x 118 = 70,446). 

The simple arithmetic applies no matter how many A, T, C, and G's are in a genome.

Likewise, the constants as percentages are also simple arithmetic no matter the size of a genome.

It is only addition, subtraction, multiplication, or division.

No algebra, trigonometry, or calculus is involved. 

When, as in Fig. 2, some nucleotides are unknown because of a damaged or incomplete sample in a sequence, the letter 'N' is a filler designation.

Closing Comments

Fig. 3

Finally, in terms of counts, the graphic at Fig. 3 indicates that the 'A' count must match the 'T' count.

It also indicates that the 'C' count must match the 'G' count.

The HTML tables in the appendices contain the counts of those nucleotides in those Base pairs.

Another thing that came to mind is that when one of those nucleotides in a Base pair (A-T) (C-G) is missing we can change the 'N' there to what it must be because the two are bonded pairs and only one is missing.

Check out the graphs in the appendices while remembering that DNA naturally degrades, so the ancient mummy DNA and the ancient Eurasian DNA would be expected to be of less quality than the modern DNA in chromosomes would be.

That brings us to some large genome examples shown in the appendices (Appendix Eurasia, Appendix Egypt, and Appendix Chromosomes).

They help show why explanations in the first post of this series are a much needed help for grading the sequence sample to ascertain its quality, and should remain so.

The previous post in this series is here.