Question: Minor allele frequency calculation
2
gravatar for Tark
3.9 years ago by
Tark30
japan
Tark30 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 

please explain with calculation

Regards

snp next-gen • 35k views
ADD COMMENTlink modified 12 months ago by gpatil0 • written 3.9 years ago by Tark30
17
gravatar for Maxime Lamontagne
3.9 years ago by
Québec
Maxime Lamontagne2.0k 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.

ADD COMMENTlink written 3.9 years ago by Maxime Lamontagne2.0k

Thank you for the clear explanation.

ADD REPLYlink written 2.2 years ago by Jonathan King0

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

ADD REPLYlink written 5 months ago by Amy0
2
gravatar for Danielson
3.5 years ago by
Danielson20
Netherlands
Danielson20 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)."

ADD COMMENTlink written 3.5 years ago by Danielson20
0
gravatar for gpatil
12 months ago by
gpatil0
gpatil0 wrote:

=(2Ref_allele)/(2(Ref_allele + alternate_allele)) e.g. Population size 108, reference allele “A” = 70 and Alternate_allele “G”=38 =(270)/(2(70+38) =0.648 =1-0.648=0.352 MAF of alternate_allele

ADD COMMENTlink written 12 months ago by gpatil0
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