2. Population Genetics is the study of the genetic
composition of populations rather than individuals
Population geneticists study mendelian populations
of sexually-reproducing organisms in which
there is random mating
A mendelian population may be considered to be a
group of closely-related, sexually reproducing
organisms residing within defined geographical
boundaries within which each member has
an equal chance of mating with any other
member of the population

3. Terminology
A sample is a representative selection from a parent population
A parent population (usually referred to as just population) is the total group from which the sample is selected
Frequency is the total count of a particular variable in a sample or the estimated count for that variable in the population; frequencies are whole numbers, e.g. 105 or 23
Relative Frequency is the frequency of a particular variable divided by the total of all the variables in the sample, also known as the proportion. Relative frequencies (proportions) are always between 0 and 1
Geneticists often use decimals (e.g. p = 0.5) to represent the relative frequency of alleles within populations. The term ‘frequency’ is often used instead of the more accurate term ‘relative frequency’, e.g. “the frequency of allele B is 0.32”
In this presentation, we have chosen the more accurate form of phrasing, e.g. “the relative frequency of allele B is 0.32”

4. This population of 20 fruit flies was bred in
culture bottles in a school laboratory
Some of the flies have normal, long wings governed by the dominant allele vg+ whilst others possess vestigial wings governed by the recessive allele vg

5. The different genotypes within this population are vg+vg+, vg+vg and vgvg
Each individual is described as possessing TWO alleles for wing length and thus the total number of alleles for this characteristic within the population is 40 (20 x 2)

6. time is defined as the GENE POOL
The total number of alleles of a particular gene that are present within a population at a particular
time is defined as the GENE POOL

7. The 20 fruit flies in this population are all heterozygous for normal, long wing length and possess the genotype vg+vg
In this example there is a total of 40 alleles within the population, where 50% of the alleles govern long wing (vg+ ) and 50% govern vestigial wing (vg)

8. In this example, the relative frequency of the dominant long wing allele is 0.5 and the relative frequency of the recessive vestigial wing allele is 0.5

9. Relative frequencies of alleles always provide a total value of ONE, where the value ONE represents the whole population (100%)
In our example, if the relative frequency of the dominant, long wing allele within the population is 0.6 then the relative frequency of the vestigial wing allele is 0.4

10. genetic composition of populations
A general formula is used to represent the relative frequencies of alleles of a single gene within a population:
p + q = 1
where:
p = the relative frequency of the dominant allele and
q = the relative frequency of the recessive allele
Allele frequencies are essential for analysing the genetics of populations but they cannot be observed directly
In 1908, the English mathematician G.H. Hardy and the German geneticist W. Weinberg developed a mathematical approach in which allele frequencies, genotype frequencies and phenotype frequencies could be used to study the
genetic composition of populations

12. Hardy and Weinberg considered a population with a hypothetical gene that has two alleles (A and a) in which the A allele is dominant to the recessive allele a
Given the frequencies of the A and a alleles, we can calculate the EXPECTED frequencies of the progeny genotypes and phenotypes

13. A p a q AA p2 Aa pq Aa pq aa q2 If
p = relative frequency of A alleles in the population
q = relative frequency of a alleles in the population,
then all chance combinations of gametes containing these alleles can be calculated A p a q AA p2 Aa pq Aa pq aa q2

14. The expected genotypic
relative frequencies in the next generation may be summarised as follows:
p pq + q2 = 1
AA Aa aa A p a q AA p2 Aa pq Aa pq aa q2

15. This is the Hardy-Weinberg Equation
p pq + q2 = 1
AA Aa aa
Thus, p2 is the proportion of the next generation EXPECTED to be homozygous dominant (AA)
2pq is the proportion of the next generation EXPECTED to be heterozygous (Aa)
q2 is the proportion of the next generation EXPECTED to be homozygous recessive (aa)
This is the Hardy-Weinberg Equation

16. Within this population there are 891 white sheep and 9 black sheep
Within a population of sheep, white wool is determined by a dominant allele B, and black wool upon its recessive allele b
Within this population there are 891 white sheep
and 9 black sheep
We can use this information to estimate the allele frequencies
and the frequency and number of heterozygous white sheep

17. Total Population = 900 sheep
891 white sheep
9 black sheep
Total Population = 900 sheep
Let p = relative frequency of dominant allele (B)
Let q = relative frequency of recessive allele (b)
According to the Hardy-Weinberg Equation:
p pq + q2 = 1
BB Bb bb genotypes
All the black sheep are homozygous recessive with the genotype bb; it is not possible to determine from the phenotype which of the white sheep are homozygous or heterozygous
We can calculate the frequency of the black wool allele by first determining the frequency of homozygous recessive black sheep within the population

18. Relative frequencies of alleles: p = 0.9 q = 0.1
891 white sheep
9 black sheep
Total Population = 900 sheep
q2 = relative frequency of homozygous recessive individuals, i.e. bb genotypes
Therefore, q2 = 9/900 = 1/100 = 0.01
Therefore, q = = 0.1
Therefore the relative frequency of the b allele is 0.1 (q)
Since p + q = 1, then p = 0.9
Therefore the relative frequency of the B allele is 0.9 (p)
Relative frequencies of alleles: p = 0.9 q = 0.1

19. p2 + 2pq + q2 = 1 p2 = relative frequency of homozygous dominants
Allele relative frequencies for wool colour
p = 0.9
q = 0.1
p pq + q2 = 1
p2 = relative frequency of homozygous dominants
2pq = relative frequency of heterozygotes
q2 = relative frequency of homozygous recessives
According to the Hardy-Weinberg equation, the relative frequency of heterozygotes is 2pq
Therefore the relative frequency of Bb individuals is 2 x 0.9 x 0.1 = 0.18
The estimated NUMBER of heterozygotes (Bb) is therefore
0.18 x 900 = 162

20. The Hardy-Weinberg Law states that:
In a large randomly mating population, the allele and genotype frequencies remain constant from generation to generation
If a population conforms to the expectations predicted
by the law, then the population is said to be in
Hardy-Weinberg Equilibrium
However, Hardy and Weinberg based their law on a
number of conditions; if any of these conditions are
not met then allele and genotype frequencies will
change over the generations – the process of evolution

21. The Hardy-Weinberg equilibrium is maintained if all the following conditions are met:
The population is large
There is random mating within the population
No selection pressures are operating – each genotype has equal fitness and there is no differential mortality
The population is closed – there is no immigration or emigration of individuals from or into another population
There is no mutation from one allele into another or, if mutation does occur, the forward and back mutations are equal
There is no genetic drift, i.e. changes in allele frequencies resulting from random, chance processes (most likely to affect small populations)
The character being studied is not sex linked

22. In the example of the sheep population, the allele frequencies were determined by estimating the frequency of the recessive allele from the number of homozygous recessive individuals
In cases of co-dominance, the heterozygotes may be phenotypically distinguishable from the homozygotes, e.g. both red and white snapdragons are homozygous (RR & rr) whereas only pink snapdragons are heterozygous (Rr)
In these cases, the frequencies of the genotypes can be determined by counting and thus provide observed rather than estimated values

23. In a species of bean, seed colour is determined by a pair of alleles C and c, where the allele C determines golden seeds and the allele c determines dark green seeds - there is co-dominance
The heterozygote (Cc) has light green seeds and is thus distinguishable from the other two genotypes (CC and cc)
500 seeds of this species were randomly selected ‘in the field’ and the following phenotypes were recorded:
golden 100
light green 250
dark green 150

24. golden CC
light green Cc
dark green cc
In this situation let: p = the relative frequency of the C allele and q = the relative frequency of the c allele
In this situation the ACTUAL allele frequencies can be determined by counting the alleles, as the heterozygotes (Cc) are distinguishable from both homozygotes (CC & cc)
Each individual is described as possessing TWO alleles for seed colour such that the total number of alleles for seed colour in this sample is ( ) x 2 = 1000 alleles

25. golden CC
light green Cc
dark green cc
Total number of alleles in the sample = 1000
100 homozygous golden seeds (CC) represent 200 C alleles
250 heterozygous seeds represent 250 C alleles as the genotype is Cc and each individual carries only ONE C allele
Therefore the total number of C alleles in the sample is = 450
The proportion (relative frequency) of C alleles in the sample is therefore 450/1000 = 0.45
Therefore p = 0.45

26. golden CC
light green Cc
dark green cc
Total number of alleles in the sample = 1000
150 homozygous dark green seeds (cc) represent 300 c alleles
250 heterozygous seeds (Cc) represent 250 c alleles
Therefore the total number of c alleles in the sample is = 550
The frequency of c alleles in the sample is therefore 550/1000 = 0.55
Therefore q = (N.B. this simply confirms p + q = 1)

27. These actual values of p and q can be applied to the
golden CC
light green Cc
dark green cc
p = 0.45 q = 0.55
These values of p and q are ACTUAL values determined from the sample and not ESTIMATES of the parent population as are determined by the Hardy-Weinberg equation
These actual values of p and q can be applied to the
Hardy-Weinberg equation in order to determine whether the parent population is in Hardy-Weinberg equilibrium

28. According to the Hardy-Weinberg Law;
p = 0.45 q = 0.55
According to the Hardy-Weinberg Law;
p2 represents the EXPECTED proportion of gold homozygotes (CC)
2pq represents the EXPECTED proportion of light green heterozygotes (Cc)
q2 represents the EXPECTED proportion of dark green homozygotes (cc)
Determine the expected frequencies and hence the expected numbers of the different genotypes in the population

29. According to the Hardy-Weinberg Law;
p = 0.45 q = 0.55
According to the Hardy-Weinberg Law;
p2 = (0.45)2 = = expected proportion of CC individuals
Therefore expected number of CC individuals = x 500 = (500 = total number of seeds in the sample)
2pq = 2 x 0.45 x 0.55 = expected proportion of Cc individuals
Therefore expected number of Cc individuals = x 500 = 247.5
q2 = (0.55)2 = = expected proportion of cc individuals
Therefore expected number of cc individuals = x 500 =

30. The observed results can be compared with those expected from the Hardy-Weinberg equation in order to determine whether genotype frequencies (CC, Cc and cc) are in equilibrium
The significance of differences between the OBSERVED and EXPECTED results can be assessed using the Chi-squared test (χ²)
The basis of the Chi-squared test is the difference between observed results (O) and the expected results (E) predicted by the ‘null hypothesis’

31. State the ‘null hypothesis’ and ‘alternative hypothesis’
Genotype
Observed Result (O)
Expected Result (E)
O – E
(O – E)2
(O – E)2 /E CC
100
101.25 Cc
250
247.5 cc
150
151.25
Σ =
Σ =
151.25
150 cc
247.5
250 Cc
101.25
100 CC
(O – E)2 /E
(O – E)2
O - E
Expected Result (E)
Observed Result (O)
Genotype
State the ‘null hypothesis’ and ‘alternative hypothesis’
Complete the table above to obtain the c2 value
Use the c2 value and critical values table to determine c2crit, at the 5% significance level for one degree of freedom (number of alleles - 1)
Comment on the result

32. Σ =
1.5625
-1.25
151.25
150 cc
6.25
+2.5
247.5
250 Cc
101.25
100 CC
(O – E)2 /E
(O – E)2
O – E
Expected Result (E)
Observed Result (O)
Genotype
Null Hypothesis; there is no difference between the observed results and those expected from the Hardy-Weinberg equation
Alternative Hypothesis; there is a difference between the observed results and those expected from the Hardy-Weinberg equation
The c2 value is 0.05

33. For a c2 value of 0.05 and 1 degree of freedom there is between an 80% and 90% probability that chance alone has caused the difference between the observed values and those expected if the null hypothesis is true, i.e. p lies between 0.80 and 0.90; there is NO significant difference between observed and expected results
The c2 value (0.05) is less than the critical value, ccrit2 (3.84) for a 5% (0.05) level of significance and thus we accept the null hypothesis and reject the alternative hypothesis
The observed and expected results are NOT significantly different and therefore this bean population is in Hardy-Weinberg equilibrium

34. If the observed and expected results had shown a significant difference, then factors other than chance would have caused this difference and the population would not be in Hardy-Weinberg equilibrium
In such a case, one or more of the factors considered as assumptions of the law must be operating in the population, e.g. there is mutation from one allele into the another

35. Past Question
A species of beetle, only found on guano island, has a characteristic controlled by a pair of codominant alleles, CM and CN.
Define codominance
If there are 500 beetles in the total population and 300 individuals have the genotype CM CM , 150 had the genotype
CM CN , and 50 had the genotype CN CN . Calculate the actual frequency of the allele CN. Show your working.

36. answer
0.25 or 25% 2 marks
CN= 250/ mark

37. Cont..
Use your answer and the Hardy-Weinberg equation to calculate the number of beetles that you would expect to have the genotype CN CN

38. answer
p2 = (0.25)2 / / square of calculated figure for CN marks.
Formula = mark