Download 10. population genetics

Title: Population Genetics
12th February 2014
Learning question: How can you calculate the frequency
of alleles in a population?
Homework:
Starter
Watch this video!
Population genetics
• A group of individuals of the same species that
can interbreed – a population – can carry a large
number of different genes
• The range of genetic diversity can be measured
with a mathematical equation known as the
Hardy-Weinberg equation
• For this equation to work, certain factors need to
be taken into consideration
Key Considerations
1. There is a stable gene pool
– Because eye colour is not a “key” adaptation
in regards to reproducing, there are no
selection pressures on either characteristic.
You are not more/less likely to reproduce
based on your eye colour. NO SELECTION
TAKING PLACE!
2. There are no mutations taking place.
– A recessive blue allele is not going to
“mutate” and code for brown eyes, vice
versa.
3. There is a large population.
– Large population = no chance of brown or
blue allele randomly disappearing due to
chance.
Key Considerations
4. Mating within the population is random.
– No selective breeding occurring.
5. There is no immigration/emigration.
– That way no alleles leave or enter the population.
– The allele ratio always stays the same!
Measurement of allele and genotype
frequencies
• Observable characteristics of a genotype are
the outward, physical characteristics
(phenotype)
• To measure the allele frequency, we need to
know:
– The mechanism of inheritance of a particular trait
– How many different alleles of the gene for that
trait are in the population
Measurement of allele and genotype
frequencies
•
Co-dominance
– Heterozygous phenotype = heterozygous genotype
– Co-dominant alleles are easy to determine, therefore
the frequency is easy to determine
• Monohybrid inheritance
– If an allele is recessive, it is impossible to determine
the genotype from the phenotype (i.e. Dominant
phenotype HH/Hh).
– The frequency is therefore not as easy to determine
The Hardy-Weinberg principle
• Godfrey Hardy and Wilhelm Weinberg asked
same questions and worked independently
and at same time
• Same conclusion about relation between
allele and genotype frequencies:
• “Hardy-Weinberg law” = mathematical
model
– 1. Allele frequency equation: p + q = 1
– 2. Genotype frequency equation: p2 + 2pq +
q2 = 1
The Hardy-Weinberg principle
• The principle makes the following
assumptions:
– The population is very large (eliminates sampling
error)
– The mating within the population is random
– There is no selective advantage for any genotype
– There is no mutation, migration or genetic drift
Worked example, cystic fibrosis
• Cystic fibrosis, an inherited channel protein disorder of lungs. Heterozygotes
CFcf are symptomless carriers.
• Sufferers are cfcf.
• 1/2000 in a population will suffer from CF. We wish to know how many of the
population are carriers.
•
•
•
•
•
p represents the frequency of the dominant allele CF
q represents the frequency of the recessive allele cf
q2 is the frequency of the genotype cfcf
p2 is the frequency of the genotype CFCF
2pq is the freqency of the genotype CFcf assuming random mating within the
population, where any two individuals of CFcf (pq) mate with each other, the
resulting genotypes of the offspring could be CFCF, CFcf, CFcf, cfcf.
• These translate as p2 + 2pq + q2
• Within a population, p2 + 2pq + q2 adds up to 1 or 100%
Worked example, cystic fibrosis
• We know that q2 is 1/2000, so q2 is 0.0005
• So q is the square root of 0.0005, which is 0.022
• If p + q = 1, then p = 1 – 0.022, which is 0.978
• The frequency of carriers is given by 2pq
• So 2pq = 2 x 0.978 x 0.022
• So 2pq = 0.043
• This means that 4.3 people in 100 are carriers.
• To find out how many people are carriers in our population of 2000,
2000 x 4.3/100 = 86
Genotype frequency
• Frequency of recessive genetic disorder (ie, sickle-cell
anaemia) = q2 frequency
• q = square root of q2 frequency
• p + q = 1 SO p = 1 – q
• With p and q frequencies you can calculate
• frequencies of three genotypes (p2 + 2pq + q2 = 1)
Predict allele frequency
• You can find the frequency of one allele if you
know the frequency of the other:
p+q=1
p = frequency of the
dominant allele
q = frequency of the
recessive allele
Total frequency of
all possible alleles
Predict gene frequency
• You can find the frequency of one genotype if
you know the frequencies of the others:
p2
p2 = frequency
of the
homozygous
dominant
genotype
+ 2pq +
q2=
2pq= frequency of
the heterozygous
genotype
1
Total frequency of
all possible
genotypes
q2 = frequency of
the homozygous
recessive genotype
The Hardy-Weinberg principle
Imagine we live on a planet where there are only 2 possible eye colours.
Brown: B
The only way someone can have blue eyes is if
they inherit two recessive alleles (bb)
There is a stable gene pool in regards to eye colour. E.G. Because
eye colour is not a “key” adaptation in regards to reproducing,
there are no selection pressures on either characteristic. You are
not more/less likely to reproduce based on your eye colour. NO
SELECTION TAKING PLACE!
Lets say we had a population of 2 people.
Allele frequency
Phenotype frequency
Bb Bb
b b bb
If we know that the following assumptions are true
• No selection
• No mutations
• Large population
• Allele frequency is constant
Now the math part!
p + q = 100%
So, if 40% of the alleles are
b
The other 60% must be
B
So we’ve covered allele frequency, lets now look at
genotype frequency.
To get genotype frequency we now need to “square”
the allele frequency (cover all possible genotypes)
2
p
+ 2pq +
2
q
=1
Say we have a population of 100,000 people.
9% of that population have blue eyes (phenotype) of
the genotype (bb). (frequency of bb = 9%)
But is this all the b alleles present in the population?
No! There must be some other b alleles paired with
the dominant B allele.
Back to our equation - p2 + 2pq + q2 = 1
So we know the frequency of having two bb alleles,
how do we work out the total number of b alleles in
a population?
p2 (bb) = 9% so to work out the frequency of
the b allele in the population we must do
another sum.
2
p
2
q
+ 2pq +
=1
bb
bB/Bb
BB
bb (p2)= 9% - to get this you multiply the p by itself.
p = √.09 = .3 or 30% - so if you counted all of the
alleles in the population, you would find that 30% of
them would be b.
To put this another way, there is a 30% chance you
would inherit one b allele from your mother, and a
30% you would inherit one b allele from your father,
making the chances of having blue eyes 9%.
p2 + 2pq + q2 = 1
This means that there is a 70% chance you will
inherit a B allele from either your mother or father,
making the odds of having brown eyes 91%.
That your phenotype would be brown eyes
That your phenotype will be blue eyes
Chance of inheriting B allele
Chance of inheriting b allele
For which phenotype can you always tell the
genotype? Why?
Homozygous recessive genotype, because the trait
needs two recessive alleles, therefore, can only
contain such.
Finish this equation:
p2 +
2pq + q2 = 1
Ok, now answer some of your own . . .
2
p
+ 2pq +
2
q
=1
Ok, so how do we work out the percentage of
people with homozygous dominant alleles?
We know that the brown phenotype is 70%, so we
reverse the equation and square instead of square
root.
BB = 72 = .49 or 49%
bb = 9%
BB = 49%
Bb/bB = 42%