Transcript Hardy-Weinberg Equation Uses

D.4 High Level Only
• D.4 The Hardy-Weinberg Principle
– D.4.1 Explain how the Hardy-Weinberg
equation is derived
– D.4.2 Calculate allele, genotype and
phenotype frequencies for two alleles of a
gene using the Hardy Weinberg Equation
– D.4.3 State the Assumptions made when the
Hardy-Weinberg Equation is used
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Hardy-Weinberg Equation Uses
• Useful in determining how fast a population is
changing (allele frequency is changing)
• Predicting outcomes of mating crosses
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
D.4.1 Explain how the Hardy-Weinberg
equation is derived
• p = frequency of DOMNANT allele in a population
• q = frequency of RECESSIVE allele in a
population
• Frequencies of the alleles on a chromosome must
add up to 1
• THUS
• p+q=1
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
D.4.1 Explain how the Hardy-Weinberg
equation is derived
• p = frequency of DOMNANT allele in a population
– EXAMPLE: T frequency is 0.25 or 25%
• q = frequency of RECESSIVE allele in a
population
– Example: t frequency is 0.75 or 75%
• Frequencies of the alleles on a chromosome must
add up to 1 or 100%
• p+q=1
.75 + .25 = 1
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
D.4.1 Explain how the Hardy-Weinberg
equation is derived
• Apply p + q = 1 to a diploid situation
• Because we are all diploid
( p + q )2 =1
• If you remember your mathematics about
polynomials ( p + q )2 =1 can be changed to
– p2 + 2pq + q2 = 1
• Now you know how the Hardy-weinberg equation
was derived
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Hardy-Weinberg Equations---What it means..
• If p and q represent the relative frequencies of the
only two possible alleles in a population at a
particular locus, then
– p2 + 2pq + q2 = 1
– And p2 and q2 represent the frequencies of the
homozygous genotypes. Examples:
• p2 = p x p = TT (homozygous dominant)
• and 2pq represents the frequency of the
heterozygous genotype
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Hardy-Weinberg Equations---What it means..
• p2 + 2pq + q2 = 1
• p2 = p x p = TT (homozygous dominant)
• q2 = q x q = tt ( homozygous recessive)
• 2 pq Heterozygote
T
t
T
TT
Tt
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t
Tt
tt
Hardy-Weinberg Equation Uses
• Useful in determining how fast a population is
changing (allele frequency is changing)
• Predicting outcomes of mating crosses
Allele
Recessive t
Frequencies
Dominant T
q
p
Genotype
Homozygous Recessive q2
Frequencies
Heterozygote
2pq
Homozygous Dominant p2
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Example Problems: p2 + 2pq + q2 = 1
One Square
Two Square
One square
Genotypes
TT
2 Tt
tt
Phenotypes
¼
½
¼
•
•
•
•
•
Frequency of TT = p2 = ¼
Frequency of Tt = 2pq = ½
Frequency tt = q2 = ¼
¼+½+¼=1
0.25 + 0.50 + 0.25 = 1
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Problem 1 calculating allele frequency
• Recessive allele t is 10% of a given population.
Calculate the percentage of the dominant allele
• q = 0.10 or 10%
• p + q =1
• So…….p = 1 - 0.10
• p= 0.90 or 90%
• Remember this is allele frequency NOT
genotype frequency.
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Problem 2 calculating allele frequency
•
In a study 989 members of the population from example 1, it was found
that 11 people had showed the recessive phenotype (t). Calculate the
frequency of of the recessive allele (t).
•
1st calculate the percentage of people who have the recessive
phenotype (tt)
•
11/ 989 = 0.011 ----thus 1.1 % of the population have this phenotype
tt)
•
Hence q2 = 0.011
•
To calculate q (frequency of recessive allele) just take the square root
of q2 = 0.011
•
√ q2 =√ 0.011 = 0.105
•
This means that the frequency of this recessive allele is 10.5 % of the
population
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Problem 3 calculating genotype frequency
• Use the information from the previous problems to fill in the
charter below:
Allele
Recessive t
Frequencies
Dominant T
q
p
Genotype
Homozygous Recessive q2
Frequencies
Heterozygote
2pq
Homozygous Dominant p2
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Problem 3 calculating genotype frequency
• We know from problem 1, q= 0.10 so q2 = 0.01
• we know from problem 1, p = 0.9 so p2 = 0.81
• So 2pq = 2 x 0.10 x 0.9 = 0.18
Allele
Frequencies
Recessive t
q
Dominant T
p
0.1
0.9
Genotype Frequencies
Homozygous Recessive
q2
0.01
Heterozygote
2pq
0.18
Homozygous Dominant
p2
0.81
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Practice Problem
In a randomly breeding population of mice, 640 had black fur
and 360 brown fur. Black fur is dominant to brown fur. The
Hardy-Weinberg Principle (p2 + 2pq + q2 =1) can be used
to calculate allele and phenotype frequencies.
• (a) Calculate the frequency of the recessive allele (1
point).
• Solve for q
• Calculate q2 frequency of homozygous recessive genotype
• q2 = 360/640 = 0.5625
• q = √q2 = √0.5625 = 0.75 or 75%
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
The Hardy-Weinberg Theorem
• The Hardy-Weinberg theorem describes a
population that is not evolving
• It states that frequencies of alleles and
genotypes in a population’s gene pool remain
constant from generation to generation, provided
that only Mendelian segregation and
recombination of alleles are at work
• Mendelian inheritance preserves genetic
variation in a population
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Preservation of Allele Frequencies
• In a given population where gametes contribute
to the next generation randomly, allele
frequencies will not change
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Hardy-Weinberg Equilibrium
• Hardy-Weinberg equilibrium describes a
population in which random mating occurs
• It describes a population where allele
frequencies do not change
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Conditions for Hardy-Weinberg Equilibrium
• The Hardy-Weinberg theorem describes a
hypothetical population
• In real populations, allele and genotype
frequencies do change over time
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• The five conditions for non-evolving populations
are rarely met in nature:
– Extremely large population size
– No gene flow
– No mutations
– Random mating
– No natural selection
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
• More Help check out this tutorial
• http://www.youtube.com/watch?v=xPkOAnK20kw
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings