Transcript Hardy-Weinberg

Allele Frequency

The first equation looks at the
percentage of alleles in a population
 A+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%)
**These are
the allele
frequencies
So our equation A + a = 1 holds true
(0.649 + 0.351 = 1)
 % dominant alleles + % recessive alleles
= 100%

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=1
 a=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?