(repetition from last time)
Mendel’s first law: the principle of
segregation
For any particular trait, the pair of
alleles of each parent separate and
only one allele passes from each
parent on to an offspring. Which
allele in a parent's pair of alleles is
inherited is a matter of chance.
We now know that this segregation of
alleles occurs during the process of sex
cell formation (meiosis).
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From alleles to genotypes
Allele & genotype frequencies
(repetition from last time)
A – is the normal allele (wild type),
A – is the normal allele (wild type),
a – is the mutant allele
IF frequency of A allele is 50%
= 0.5
IF frequency of a allele is 50%
= 0.5
a – is the mutant allele
AA is the homozygous wild genotype
aa is the homozygous mutant genotype
Aa is the heterozygous genotype
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From alleles to phenotypes
The Hardy-Weinberg equilibrium
a – is the recessive allele for non-twistable tongue
Frequency of A allele = 70%
= 0.7
Frequency of a allele = 30%
= 0.3
What is the frequency of the following phenotypes:
twistable tongue
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Basic rule for population genetics: frequencies of alleles and
genotypes do not change across generations unless forces,
such as natural selection or migration change them!
A – is the dominant allele for twistable tongue,
What is the frequency of the following genotypes:
AA = 0.5 x 0.5 = 0.25
Aa = (both Aa & aA) = 2 x 0.5 x 0.5 = 0.5
aa = 0.5 x 0.5 = 0.25
=(AA or Aa)= 0.49 + 0.42 = 0.91
A – normal allele, a – mutant allele
Frequency of A allele = p
%
Frequency of a allele = q
Genotypes: AA Aa aA aa
A
Sper
p2 + 2pq + q2 = 1
m a
p+q=1
A Eggs a
p2
pq
pq
q2
AA = 0.7 x 0.7 = 0.49 Aa = 2 x 0.3 x 0.7 = 0.42
non-twistable tongue = (aa genotype) = 0.3 x 0.3 = 0.09
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With this information, you can figure out rate of
disorders & carriers in the population!
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1
Is it true, that 1 in every 50 people are
carriers for PKU?
from last time: Prevalence of PKU is 1 in every 10,000!!!
A – normal allele, a – mutant (recessive) allele
Carriers have heterozygous Aa genotypes!
What is the frequency of the Aa (or aA) genotype?
aa (affected) = 1/10000 q2=0.0001
Frequency of a allele q=0.01
Frequency of A allele p=1-0.01=0.99
2pq (carriers) =0.99 x 0.01 x 2=0.0198=1/50.5
In humans, free ear lobes are
dominant to attached ear lobes.
After randomly sampling a
population of 1000 people you
find that 650 of them had free
earlobes. Estimate the
frequencies of the dominant
and recessive alleles.
We seek: p = ?, q = ?
We know: p2 + 2pq= 650/1000 = 0.65
Since p2 + 2pq + q2 = 1 we can calculate that q2 = 1 – 0.65 = 0.35
q = square root of 0.35 = 0.5916, about 59%
Since p + q = 1 we can calculate that p = 1 – 0.59 = 0.41 41%
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How many people carry the recessive
allele of albinism in heterozygous form?
•Frequency of Albinism in North America = 1/20,000
A – normal allele, a – mutant allele
• What is the genotype for
Frequency of A allele = p
Aa, aA
heterozygous carriers? Frequency of a allele = q
Genotypes: AA Aa aA aa
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• Frequency of the aaqgenotype: p + 2pq + q = 1
p+q=1
• q2 = 1/20,000
• q= √1/20,000 = .007
• What is the frequency of the normal allele: p
• p=1-q
• p= 1 - 0.007 = 0.993
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• The Hardy-Weinberg equilibrium (frequencies of alleles
and genotypes do not change across generations unless forces,
such as natural selection or migration change them)
p2 + 2pq + q2 = 1
p+q=1
• Using the Hardy-Weinberg equilibrium to calculate rate
of disorders & carriers in the population
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Selected readings
allele frequencies (based on Mendel’s first law, dominant recessive traits)
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• Frequency of heterozygotes (carriers): 2pq
•2 x 0.007 x 0.993 = 0.014 (1.4% or about 1/71)
• Calculating genotype and phenotype frequencies from
An example of possible exam questions:
Textbook: Ch 2 page 13 (Box 2.2).
The outline from this lecture presentation
will be available at the course website!
Frequency of a rare autosomal recessive trait
is = 15/10.000. What is the frequency of
carriers of his trait?
9985/10.000
0.003
7.4%
1/27
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A – normal allele, a – mutant allele
Frequency of A allele = p
Frequency of a allele = q
Genotypes: AA Aa aA aa
p2 + 2pq + q2 = 1
p+q=1
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