8.5
Starter
 What is the term for all the alleles of all the genes in a
population?
 Gene Pool
 What do you think the term would be for the number of
times an allele occurs within a gene pool?
 Allelic Frequency
Learning Objectives
 Describe the terms ‘gene pool’ and ‘allelic frequency’
 Explain what the Hardy-Weinberg principle is
 Explain how the Hardy-Weinberg principle can be used to
calculate allele, genotype and phenotype frequencies
Allelic Frequencies
 Every person has 2 alleles of a gene (e.g. TT, Tt or tt)
 If a population has 5000 people, then there are 10,000
alleles of each gene
 The total number of alleles in a population is said to be
1.0
Allelic Frequencies
 Frequencies of the 2 alleles must add up to 1.0
 A recessive/dominant situation
 If everyone in a population was TT, then what would the
frequency of the T allele be? 1.0
 If everyone in a population was Tt, then what would the
frequency of the T allele be? 0.5 The t allele? 0.5
 However, realistically you see a mixture of genotypes in a
population, so this way of working out allele frequencies
cannot be used. We must use the Hardy-Weinberg Principle
The Hardy-Weinberg Principle
 An equation to work out the frequencies of the alleles in
a population
 To use the equation the following 5 things must be true:
 No mutations
 Population is isolated
 No selection
 Large population
 Mating is random
The Hardy-Weinberg Principle
 The equation = p2 + 2pq + q2
 Frequency of allele T = p
 Frequency of allele t = q
 So, p + q must equal 1.0
 Four possible arrangements of the alleles:
TT, Tt, tT, tt which must equal 1.0
 Times them together to create the equation:
p2 + 2pq + q2
Example
 Work out, using the Hardy-Weinberg equation, the allele
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frequencies of cystic fibrosis, a recessive condition
affecting the lungs. In a population of 15,000 people, 1
person suffers from the disease.
Recessive, so the frequency of tt = 1/15000
So, q2 = 1/15000 = 0.000067
So, q = square root of 0.000067 = 0.0081854 = 0.0082
p + q = 1.0
So, 1.0 – q = p
p = 1.0 – 0.0082 = 0.9918 (This is the frequency of allele T)
Example
 Now that we know p and q we can work out the
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frequency of heterozygous individuals in the population
Heterozygous = 2pq
= 2 x 0.9918 x 0.0082
= 0.0163
This means 163 individuals in 10000 are carriers for the
recessive allele and could potentially pass on cystic
fibrosis. This is equivalent to 244 people in our 15000.
Task
 Complete application questions on page 127
1. Not sex linked as approximately equal numbers of males and
females in each colour.
2. 562/2215 = 0.254
3. a) q = Square root of 0.254 = 0.504
b) p + q = 1.0. p = 1.0 – 0.504 = 0.496
c) Heterozygotes = 2pq = 2 x 0.496 x 0.504 = 0.5, so
heterozygotes = 50%
4. Collect a sample of moths. Mark them unobtrusively.
Release back into the population and allow time to mix.
Later, randomly catch a sample and count the number of
marked and unmarked moths. Calculate by: (total number in
first sample x total number of moths in second sample) /
number of marked moths recaptured
Exam Question on H-W
 1.
(a)
(b)
(q2 = 0.52 / q = 0.72)
(p = 1 – 0.72 = 0.28)
p + q = 1 / p2 + 2pq + q2 = 1 ;
Answer = 2pq / use of appropriate numbers;
Answer = 40%;
3
Any three from: (MARK AS A WHOLE)
Small founder population / common ancestor;
Genetic isolation / small gene pool / no immigration /
no migration / in-breeding;
High probability of mating with person having H-allele;
Reproduction occurs before symptoms of disease are apparent;
Genetic argument – Hh x hh ® 50% / Hh x Hh ® 75% affected
offspring;
No survival / selective disadvantage;
3 max
Ignore ‘survival of the fittest’
[6]
Task
 You will be given a selection of sweets to represent alleles
in a population
 The type with fewer present will be the recessive allele
and the other the dominant
 Work out q, then p
 Then calculate the frequency of heterozygous individuals
in the population