Question: Minor allele frequency calculation
4
Tark50 wrote:

hi everyone

I need to understand Minor allele frequency

a SNP with a allele (G) frequency of 0.40 implies that 40% of a population has G allele versus the most common allele (major allele), which is found in 60% of the population.

if a SNP allele frequency is 0.05 how will be calculate its percentage  in the population

Regards

snp next-gen • 41k views
modified 3 months ago by pltbiotech_tkarthi170 • written 4.6 years ago by Tark50

The statement "which is found in 60% of the population" is incorrect, and should be corrected. It would be more correct to say "makes up 60% of the alleles in the population"

21
Maxime Lamontagne2.1k wrote:

If a SNP allele frequency is 0.05 in a population of 100 people:

Total number of alleles for each SNP: 100 * 2 = 200  (each individual has two alleles)

Total number of alleles for your SNP with a MAF of 0.05: 200 * 0.05 = 10

The minor allele is present ten times in your population. However, you don't know how many heterozygote and homozygote.

Thank you for the clear explanation.

What could be the solution if you want to know frequencies for homozygote and heterozygote ? Any databses that provide those infos ? Thanks !

4
Danielson40 wrote:

An explanation on MAF is giving on ncbi: http://www.ncbi.nlm.nih.gov/projects/SNP/docs/rs_attributes.html#gmaf

Which states: "...In other words, if there are 3 alleles, with frequencies of 0.50, 0.49, and 0.01, the MAF will be reported as 0.49"

An example given: ""MAF/MinorAlleleCount:G=0.249/542". This means that for rs222, minor allele is 'G' and has a frequency of 24.9% in the 1000Genome phase 1 population and that 'G' is observed 542 times in the sample population of 1088 people (or 2176 chromosomes)."

0
gpatil0 wrote:
``````=(2*Ref_allele)/(2*(Ref_allele + alternate_allele))
e.g. Population size 108, reference allele “A” = 70 and Alternate_allele “G”=38
=(2*70)/(2*(70+38)
=0.648
=1-0.648=0.352 MAF of alternate_allele
``````