1. THE HARDY-WEINBERG PRINCIPLE
AHL Option D.4
IB Biology
Miss Werba

2. OPTION D - EVOLUTION D.1 ORIGIN OF LIFE ON EARTH D.2
SPECIES AND SPECIATION
D.3
HUMAN EVOLUTION
AHL D.4
THE HARDY–WEINBERG PRINCIPLE
AHL D.5
PHYLOGENY AND SYSTEMATICS
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3. THINGS TO COVER Derivation of the H-W equation
Calculations using the H-W equation
Assumptions made when using the H-W equation
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4. THE HARDY–WEINBERG EQUATION Command term = EXPLAIN
It is a mathematical model for determining the allele frequencies for a gene with 2 alleles.
Dominant allele is A and its frequency is p (a number between 0-1)
Recessive allele is a and its frequency is q (a number between 0-1)
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5. THE HARDY–WEINBERG EQUATION Command term = EXPLAIN
A gene must have alleles.
The only options are A or a.
p + q = 1
If only A is present (homozygous dominant):
This makes the frequency of A 100% - ie. p=1.0
This means that the frequency of a must be 0% - ie. q=0.0
If A is not present (homozygous recessive):
p=0.0 and q=1.0
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6. THE HARDY–WEINBERG EQUATION Command term = EXPLAIN
If the frequency of A is p, then the frequency of AA is….
p x p = p2
If the frequency of a is q, then the frequency of aa is….
q x q = q2
The frequency of having one of each (Aa or aA) is therefore….
(p x q) + (q x p) = 2pq
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7. THE HARDY–WEINBERG EQUATION Command term = EXPLAIN
Since any genotype is AA or Aa or aa, it means that:
p2 + 2pq + q2 = 1
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8. USING THE HARDY-WEINBERG EQUATION Command term = CALCULATE
Example 1:
A particular population of Drosophila fruit flies contained 64 individuals with red eyes (wild-type / DOMINANT) and 36 individuals are found to have white eyes.
Find the allele frequency for each allele and the genotype and phenotype frequencies.
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9. USING THE HARDY-WEINBERG EQUATION Command term = CALCULATE
100
Total population = __________
Let’s calculate the allele frequencies first:
There are 36 white eyed flies. These must be homozygous recessive so we can find q:
Now we have q, we can find p:
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10. USING THE HARDY-WEINBERG EQUATION Command term = CALCULATE
Now we can calculate the genotype and phenotype frequencies:
Phenotypes:
red = 64/100 = 0.64
white = 36/100 = 0.36
Genotypes:
AA = p2 = 0.4 x 0.4 = 0.16
Aa = 2pq = 2 x 0.4 x 0.6 = 0.48
aa = q2 = 0.6 x 0.6 = 0.36
 Use INFO FROM QUN OR p/q VALUES
 Use INFO FROM QUN OR p/q VALUES
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11. D.4.3
ASSUMPTIONS MADE WHEN USING THE HARDY-WEINBERG EQUATION Command term = STATE
The H-W equation can be used when the following conditions are met:
large population
random mating
no migration / immigration / emigration
no mutations
no natural (or directional) selection
no allele specific mortality
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12. Sample questions Q1
Using the Hardy-Weinberg principle, calculate the percentage of carriers in a population where the occurrence of the condition cystic fibrosis is 1 in births (2)
Predict the effect on a gene pool in succeeding generations if the conditions of the Hardy-Weinberg principle are not met. (1)
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13. Sample questions Q2
In a randomly breeding population of mice, 640 had black fur and 360 brown fur. Black fur is dominant to brown fur.
a) Calculate the frequency of the recessive allele (1)
b) Calculate the number of homozygous black mice in the sample. (2)
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14. Sample questions Q3
Tay-Sachs disease is an autosomal recessive disorder of the enzyme hexosaminodase. The disorder causes a build-up of fatty deposits in the brain. A child affected by the disease usually dies by the age of four. The frequency of Tay-Sachs disease (tt) in a Mediterranean population is
a) Calculate the frequencies in the population of allele t and genotype Tt (2)
b) State two conditions required for the Hardy-Weinberg equation to be valid (2)
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15. Sample questions A1 means:
q2 = , q = 0.02 and p = pq = 2(0.98)(0.02) = = 3.9 % carriers (accept 4 %) (2)
gene pool will change / evolution will occur; allele frequency will change / genotype frequency will change; (1)
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16. Sample questions a) 0.6 / 60 % (1)
b) Apply error carried forward - indicate by ECF dominant allele is = 0.4 / p2 = 0.16; 160; (2)
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17. Sample questions Q2
a) q/frequency of allele t: 0.02 / 0.017; 2pq/frequency of genotype Tt: 0.04 / or 0.03 / 0033; (2)
b) no natural selection / no allele specific mortality; random mating; large population; no mutation; no immigration / no emigration / no migration; constant allele frequency over time; (1)
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18. D.4
NEED MORE???
/hwetutorial.html#Q13B
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