Allele Frequency

The first equation looks at the percentage of alleles in a populationA + a = 1

○ A is the percentage of dominant alleles○ a is the percentage of recessive alleles

There are 30 million people in Canada, so how many alleles for eye colour should there be?60 million –two for each person

If 38.9 million of those alleles are brown, what does that mean for the A and a values?A = 38.9/60 = 0.649 (64.9%)a = 21.1/60 = 0.351 (35.1%)

So our equation A + a = 1 holds true (0.649 + 0.351 = 1)

% dominant alleles + % recessive alleles = 100%

**These are the allele frequencies

Say in New Brunswick (population 750 000), the dominant allele frequency is 71.6%

How many dominant alleles should there be?0.716 x 1 500 000 = 1 074 000

Recessive alleles?A + a = 1a = 1 – A

= 1 – 0.716

= 0.284 (28.4%)

0.284 x 1 500 000 = 426 000

How does this relate to population dynamics?

Probability… If the dominant allele frequency is 64.9%,

then we have a 64.9% chance of selecting that allele

We would have a (0.649)(0.649) of selecting it twice -> (0.649)(0.649) = (0.649)2 = 0.421

So, what is A2? You may recognize it as AA -> homozygous

dominant

This means two things:1. A random individual in Canada has a 42.1%

chance of being homozygous dominant for brown eyes

2. 42.1% of individuals in Canada should be homozygous dominant

This holds true for homozygous recessive a = 0.351

a2 = (0.351)2

= 0.123

12.3% of individuals should be homozygous recessive (blue eyes)

What about the carriers (heterozygotes?) Aa = (0.649)(0.351) = 0.228 But, there are two ways to become

heterozygous -> your father’s allele is dominant, your mother’s recessive OR you mother’s is dominant, your father’s recessive

So, we multiply by 2 2Aa = 2(0.649)(0.351) = 0.456 45.6% are heterozygous

These combine into the equation:A2 + 2Aa + a2 = 1 (look familiar?)

This means percentage of individuals that are homozygous dominant plus heterozygous plus homozygous recessive equals 100% -> everyone!

Please note that you’ll never use this equation as a whole ->you will always merely take out terms from it to do individual calculations

Does it work backwards?

Yes! If you know the percentage of people that

are homozygous recessive (usually the easiest thing to identify), you can make assumptions about the allele frequency

If 15.4% of people are homozygous recessive for PTC paper tasting ability, what percentage are homozygous dominant, and what percentage are carriers?

a2 = 0.154

a = (0.154)0.5 (this means square root of 0.154, but I couldn’t show that on PPT ’07. Thanks Microsoft…)

a = 0.392

A + a = 1

A = 1 – a

= 1 – 0.392

= 0.608 With this info, the values are easy to

calculate

A2 = (0.609)2

= 0.370

37.0 % are homozygous dominant 2Aa = 2(0.609)(0.392)

= 0.477

47.7 % are carriers All percentages add up to 100 (ok, 100.1,

sig figs…)

Example

Batten disease is a rare recessive neurodegenerative disease, affecting 3 out of every 100 000 people in North America. Based on this knowledge, what percentage of people are carriers and could pass it onto their offspring?

Answer

We define the dominant, normal allele as B, and the recessive as b

Since occurrence is 3 out of 100 000, b2=0.00003

So, frequency of recessive allele is b=√0.00003 = 0.005

The frequency of the dominant allele is B = 1-b = 1-0.005 = 0.995

The frequency of carriers would be 2Bb=2(0.995)(0.005) = 0.0095

So, approximately 1% of the population are carriers for this disease

Example (try on your own)It is believed that approximately 4% of Canadians of South American decent are carriers for the recessive condition sickle cell anemia. If 98% of the alleles in this population are dominant, what should the prevalence of sickle cell anemia be?