Hardy-Weinberg (AQA A Level Biology): Revision Note

Exam code: 7402

Jenna Quinn

Written by: Jenna Quinn

Reviewed by: Lára Marie McIvor

Updated on

Natural selection: Hardy-Weinberg principle

  • The Hardy-Weinberg principle states that allele frequencies in a population will remain constant from one generation to the next if specific conditions are met

  • The equation can be used to:

    • Calculate allele and genotype frequencies

    • Predict how these frequencies will change across generations

Conditions for the Hardy-Weinberg principle

  • For the Hardy-Weinberg principle to be correctly applied to a population a series of conditions, or assumptions, need to be met

    • Organisms are diploid

    • Organisms reproduce by sexual reproduction only

    • There is no overlap between generations (parents do not mate with offspring)

    • Mating is random

    • The population is large

    • There is no migration, mutation, or selection

      • This would mean no individuals entering the population (immigration) or leaving (emigration)

      • Selection refers to both natural and artificial selection

    • Allele frequencies are equal in both sexes

  • The Hardy-Weinberg principle can be useful when building models and making predictions, but the assumptions listed are very rarely, if ever, all present in nature

Hardy-Weinberg equations

  • If the phenotype of a trait in a population is determined by a single gene with only two alleles (we will use B / b as examples throughout this section) then the population will consist of individuals with three possible genotypes:

    • Homozygous dominant (BB)

    • Heterozygous (Bb)

    • Homozygous recessive (bb)

  •  All frequencies are expressed as proportions (values between 0 and 1)

  • For example:

    • if every individual in the population has the homozygous dominant genotype (BB), its frequency will be 1

    • if half of the population show this genotype, then the frequency will be 0.5

Allele frequencies

  • For a gene with two alleles (B and b):

    • The letter p represents the frequency of the dominant allele (B)

    • The letter q represents the frequency of the recessive allele (b)

    • As there are only two alleles for this gene:

p + q = 1

  • For example:

    • In a population of 100 individuals, there are 200 alleles

    • If there are 120 dominant (B) alleles, the frequency of the dominant allele = 120/200

    • Therefore:

      • p = 120 ÷ 200 = 0.6

      • q = 1 - 0.6 = 0.4

Genotype frequencies

  • Frequency of genotypes can also be represented; this is the proportion of all of the individuals with a particular genotype

  • For a gene with two alleles (B and b):

    • The homozygous dominant (BB) genotype is represented by p2

    • The heterozygous genotype (Bb) is represented by 2pq

    • The homozygous recessive genotype (bb) is represented by q2

  •  As these are all the possible genotypes of individuals in the population, the following equation can be constructed:

p2 + q2 + 2pq = 1

Worked Example

In a population of birds 10% of the individuals exhibit the recessive phenotype of white feathers. Calculate the frequencies of all genotypes.

Solution:

  • We will use F / f to represent dominant and recessive alleles for feather colour

  • Those with the recessive phenotype must have the homozygous recessive genotype, ff

  • Therefore q2 = 0.10 (as 10% of the individuals have the recessive phenotype and q2 represents this)

To calculate the frequencies of the homozygous dominant ( p2 ) and heterozygous ( 2pq ):

Step 1: Find q

Natural Selection Hardy-Weinberg Principle Worked Example equation

Step 2: Find p (the frequency of the dominant allele F). If q = 0.32, and p + q = 1

p + q = 1

p = 1 - 0.32

p = 0.68

Step 3: Find p2 (the frequency of homozygous dominant genotype)

0.682 = 0.46

p2 = 0.46

Step 4: Find 2pq = 2 x (p) x (q)

2 x (0.68) x (0.32) = 0.44

Step 5: Check calculations by substituting the values for the three frequencies into the equation; they should add up to 1

p2 + 2pq + q2 = 1

0.46 + 0.44 + 0.10 = 1

In summary:

  • Allele frequencies:

    • p = F = 0.68

    • q = f = 0.32

  • Genotype frequencies:

    • p2 = FF = 0.46

    • q2 = ff = 0.10

    • 2pq = Ff = 0.44

Examiner Tips and Tricks

Tips for approaching a Hardy-Weinberg calculation question

  1. Start with q²: Always begin by identifying the frequency of the recessive phenotype

    • This is the only genotype you can determine directly (must be homozygous recessive, bb)

  2. Identify known and unknown values:

    • Check what the question gives you (e.g. or p)

    • Work out what it asks for (e.g. 2pq)

    • Use the appropriate equation(s) to connect them

  3. Don’t confuse the equations with the principle:

    • The equations estimate allele and genotype frequencies

    • The principle assumes no change in allele frequencies between generations (i.e. genetic equilibrium)

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Jenna Quinn

Author: Jenna Quinn

Expertise: Head of Humanities & Social Science

Jenna studied at Cardiff University before training to become a science teacher at the University of Bath specialising in Biology (although she loves teaching all three sciences at GCSE level!). Teaching is her passion, and with 10 years experience teaching across a wide range of specifications – from GCSE and A Level Biology in the UK to IGCSE and IB Biology internationally – she knows what is required to pass those Biology exams.

Lára Marie McIvor

Reviewer: Lára Marie McIvor

Expertise: Biology, Psychology & Sociology Subject Lead

Lára graduated from Oxford University in Biological Sciences and has now been a science tutor working in the UK for several years. Lára has a particular interest in the area of infectious disease and epidemiology, and enjoys creating original educational materials that develop confidence and facilitate learning.