Enter a DNA or RNA sequence into the calculator to determine the percentage of each nucleotide (A, T/U, C, G), along with GC content (and AT/AU content). You can also generate a DNA reverse complement and compute GC% across a sequence using a sliding window.
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DNA Nucleotide Percentage Formula
The following formulas are used to calculate nucleotide percentages (base composition) within a single DNA (or RNA) sequence.
\%X = (N_X / N_{total}) * 100Common composition summaries:
\%GC = ((N_G + N_C) / N_{total}) * 100For DNA, AT content is:
\%AT = ((N_A + N_T) / N_{total}) * 100For RNA, AU content is:
\%AU = ((N_A + N_U) / N_{total}) * 100Variables:
- %X is the percentage of a specific nucleotide X (A, T/U, C, or G) (%)
- NX is the count of nucleotide X in the sequence
- Ntotal is the total number of nucleotides in the sequence (sequence length, in nt)
To calculate a nucleotide percentage, divide the count of that nucleotide by the total sequence length, then multiply by 100. GC%, AT%, and AU% are calculated the same way using the combined counts.
What is a DNA Percentage?
On this page, “DNA percentage” refers to base composition: the percentage of each nucleotide (A, T, C, and G) present in a DNA sequence, along with summary measures like GC content (G+C) and AT content (A+T). These percentages describe a sequence itself (or windows within a sequence), not how much DNA two people share. Shared-DNA percentages used in ancestry/relationship testing come from population genetics methods and are a different concept.
How to Calculate DNA Percentage?
The following steps outline how to calculate nucleotide percentages (base composition) for a DNA sequence.
- Write down the DNA sequence and remove any spaces or line breaks.
- Count how many A, T, C, and G bases are present.
- Compute the total length Ntotal (in nucleotides, nt).
- Calculate each nucleotide percentage using %X = (NX / Ntotal) × 100.
- Optionally compute GC% = ((NG + NC) / Ntotal) × 100 and AT% = ((NA + NT) / Ntotal) × 100.
Example Problem:
Use the following variables as an example problem to test your knowledge.
DNA sequence = ATGCGCATTAACGTTAGC (18 nt)
A = 5 (27.78%), T = 5 (27.78%), C = 4 (22.22%), G = 4 (22.22%), GC% = 44.44%
