Miss Amlani - population genetics slideshow

1. POPULATION GENETICS
Miss Amlani

2. Lesson Objective
To be able to use the Hardy-Weinberg equation to calculate
allele frequencies in a population.

3. Population definition
A POPULATION is a group of individuals of the same species
that can interbreed. Populations are dynamic - they can expand
or contract due to changes in their birth or death rates or
migration.
The set of genetic information carried by a population is the gene
pool.

4. Darwin
150th Anninversary of ‘The Origin of the Species’
COMPETITIVE STRUGGLE FOR SURVIVAL
VARIATION BETWEEN INDIVIDUALS
SURVIVAL OF THE FITTEST

5. Mendel
PATTERNS OF INHERITANCE

6. Populations rather than
individuals are functional units
of variation
The goal of our previous discussions in this class has been to
understand the inheritance of a single trait, a trait that may be
controlled by one, a few, or many genes. The goal of population
genetics is different. Rather than studying the inheritance of a
trait, population genetics attempts to describe how the frequency
of the alleles which control the trait change over time. To study
frequency changes, we analyse populations rather than
individuals. Furthermore, because changes in gene frequencies
are at the heart of evolution and speciation, population and
evolutionary genetics are often studied together.

7. We observe the phenotype (and not the genotype) of individuals.
To measure the frequency of an allele we need to know:
1. the mechanism of inheritance of a particular trait
2. how many different alleles of that gene there are in the
population.
For traits that show codominance, the frequency of the
heterozygous phenotype is the same as the frequency for the
heterozygous genotype.
Work through blood group example.

8. The Hardy-Weinberg Law of
Genetic Equilibrium
In 1908 G. Hardy and W. Weinberg independently proposed
that the frequency of alleles and genotypes in a population will
remain constant from generation to generation if the population
is stable and in genetic equilibrium. Five conditions are required
in order for a population to remain at Hardy-Weinberg
equilibrium:
1.A large breeding population * 4. No immigration or emigration
2. Random mating *5. No natural selection
3. No mutations

9. A large breeding population
A large breeding population helps to ensure that chance alone
does not disrupt genetic equilibrium. In a small population, only
a few copies of a certain allele may exist. If for some chance
reason the organisms with that allele do not reproduce
successfully, the allelic frequency will change. This random, non
selective change is what happens in genetic drift or a bottleneck
event.

10. Large breeding population

12. Random Mating
In a population at equilibrium, mating must be random. In
assortative mating, individuals tend to choose mates similar to
themselves; for example, large blister beetles tend to choose
mates of large size and small blister beetles tend to choose small
mates. Though this does not alter allelic frequencies, it results in
fewer heterozygous individuals than you would expect in a
population where mating is random.

13. Random Mating

14. No Change in Allelic
Frequency Due to Mutation
For a population to be at Hardy-Weinberg equilibrium, there can
be no change in allelic frequency due to mutation. Any mutation
in a particular gene would change the balance of alleles in the
gene pool. Mutations may remain hidden in large populations for
a number of generations, but may show more quickly in a small
population.

15. No mutations

16. No Immigration or Emigration
For the allelic frequency to remain constant in a population at
equilibrium, no new alleles can come into the population, and no
alleles can be lost. Both immigration and emigration can alter
allelic frequency.

17. No Migration or Emigration

18. No Natural Selection
In a population at equilibrium, no alleles are selected over other
alleles. If selection occurs, those alleles that are selected for will
become more common. For example, if resistance to a particular
herbicide allows weeds to live in an environment that has been
sprayed with that herbicide, the allele for resistance may become
more frequent in the population

19. No natural selection

20. Estimating allelic frequency
If a trait is controlled by two alternate alleles, how can we
calculate the frequency of each allele? For example, let us look at
a sample population of pigs.
The allele for black coat is recessive to the allele for white coat.
Can you count the number of recessive alleles in this population?

21. Estimating allelic frequency of
pigs
Answer: There are 4 individuals with black coat, so it might seem
that there are 8 copies of the recessive allele. In fact, some of the
individuals with white coat may be heterozygous for the trait. So
you cannot estimate the number of recessive alleles simply by
looking at the phenotypes in the population – unless, that is, you
know that the population is at Hardy-Weinberg equilibrium. If
that is the case, then you can determine the frequencies of alleles
and genotypes by using what is called the Hardy-Weinberg
equation.

22. The Hardy-Weinberg equation
To estimate the frequency of alleles in a population, we can use the Hardy-Weinberg
equation. According to this equation:
p = the frequency of the dominant allele (represented here by A)
q = the frequency of the recessive allele (represented here by a)
For a population in genetic equilibrium:
p + q = 1.0 (The sum of the frequencies of both alleles is 100%.)
(p + q)2 = 1
so
p2 + 2pq + q2 = 1
The three terms of this binomial expansion indicate the frequencies of the three genotypes:
p2 = frequency of AA (homozygous dominant)
2pq = frequency of Aa (heterozygous)
q2 = frequency of aa (homozygous recessive)

23. Sample Problem 1
Let's return to our population of pigs. Remember that the allele
for black coat is recessive. We can use the Hardy-Weinberg
equation to determine the percent of the pig population that is
heterozygous for white coat.
Calculate q2
Count the individuals that are homozygous recessive in the
illustration above. Calculate the percent of the total population
they represent. This is q2.

24. q2 is:
Four of the sixteen individuals show the recessive phenotype, so
the correct answer is 25% or 0.25.

25. Find q
Find q
Take the square root of q2 to obtain q, the frequency of the
recessive allele.

26. q is:
q = 0.5

27. Find p
The sum of the frequencies of both alleles = 100%, p + q = l. You
know q, so what is p, the frequency of the dominant allele?

28. p is:
p = 1 - q, so p = 0.5

29. Find 2pq
Find 2pq
The frequency of the heterozygotes is represented by 2pq. This
gives you the percent of the population that is heterozygous for
white coat:

30. 2pq is:
2pq = 2(0.5) (0.5) = 0.5 , so 50% of the population is
heterozygous.

31. Sample Problem 2
In a certain population of 1000 fruit flies, 640 have red eyes
while the remainder have sepia eyes. The sepia eye trait is
recessive to red eyes. How many individuals would you expect to
be homozygous for red eye colour?
Hint: The first step is always to calculate q2! Start by
determining the number of fruit flies that are homozygous
recessive. If you need help doing the calculation, look back at the
Hardy-Weinberg equation.

32. Solution
You should expect 160 to be homozygous dominant.
Calculations:
q2 for this population is 360/1000 = 0.36
q = = 0.6
p = 1 - q = 1 - 0.6 = 0.4
The homozygous dominant frequency = p2 = (0.4)(0.4) = 0.16.
Therefore, you can expect 16% of 1000, or 160 individuals, to
be homozygous dominant.

33. Sample Problem 3
The Hardy-Weinberg equation is useful for predicting the
percent of a human population that may be heterozygous carriers
of recessive alleles for certain genetic diseases. Phenylketonuria
(PKU) is a human metabolic disorder that results in mental
retardation if it is untreated in infancy. In the United States, one
out of approximately 10,000 babies is born with the disorder.
Approximately what percent of the population are heterozygous
carriers of the recessive PKU allele?

34. Solution
Answer:Approximately 2% of the U.S. population carries the
PKU allele.
Calculation:
q2= 1/10,000 = 0.0001
q = = 0.01
p = 1 - q = 1 - 0.01 = 0.99
The carriers are heterozygous. Therefore, 2pq = 2 (0.99) (0.01)
= 0.0198= 1.98%

35. Allelic frequency vs. Genotypic
frequency
Allelic Frequency
If you are told that the frequency of a recessive allele in a
population is 10%, you are directly given q, since by definition q
is the frequency of the recessive allele. This comprises all the
copies of the recessive allele that are present in heterozygotes as
well as all the copies of the allele in individuals that show the
recessive phenotype. What is q for this population?

36. Answer
q = 0.1

37. Allelic frequency vs. Genotypic
frequency
Genotypic Frequency
Genotypic frequency is the frequency of a genotype —
homozygous recessive, homozygous dominant, or heterozygous
— in a population. If you don't know the frequency of the
recessive allele, you can calculate it if you know the frequency of
individuals with the recessive phenotype (their genotype must be
homozygous recessive).

38. Sample Problem
If you observe a population and find that 16% show the recessive
trait, you know the frequency of the aa genotype. This means you
know q2. What is q for this population?

39. Answer
q is the square root of 0.16 = 0.4

40. Class Quiz
1(c)The question tells you that p = 0.9 and q = 0.1. From this, you can calculate the heterozygotes: 2pq = 2 (0.9) (0.1) = 0.18. If you
selected e as your response, you may have confused the allelic frequency with genotypic frequency. This problem gives you the allelic
frequency of a, which is 10%.
2(b) The conditions described all contribute to genetic equilibrium, where it would be expected for initial gene frequencies to remain
constant generation after generation. If you chose e, remember that genetic equilibrium does not mean that the frequency of A = the
frequency of a.
3(d) Like question 2, this question is intended to emphasise the point that the initial frequency of alleles has nothing to do with genetic
equilibrium.
4(d) Where q2 = 0.09, so q = 0.3. p = 1 - q, so p = 1 - 0.3 = 0.7 AA = q2 = 0.49
5(d) Where q2 = 0.16; q = 0.4 p = 1 - q, so p = 0.6 = 60%

41. Cohen Syndrome
is a developmental disorder inherited as an autosomal recessive
trait.
http://www.cbsnews.com/video/watch/?
id=700552n&tag=related;photovideo

42. Ellis-Van Creveld Syndrome
Ellis-van Creveld is passed down through families (inherited). It
is caused by defects in one of two Ellis van Creveld syndrome
genes (EVC and EVC2) that are next to each other.
The disease is autosomal-recessive
The severity of the disease varies from person to person. The
highest rate of the condition is seen among the Old Order Amish
population of Lancaster County, Pennsylvania. It is fairly rare in
the general population.

43. Consanguinity
Consanguinity means descent from a common ancestor; a
consanguineous couple is usually defined as being related as
second cousins or closer. The word derives from ‘con’+
‘sanguine’ – from the Latin, meaning ‘of the same blood’.
Consanguinuous marriage today is most prevalent in
communities originating from North Africa, the Middle East, and
large parts of Asia. In the British Pakistani community it is
estimated that 50-60% of marriages are consanguineous, and
there is evidence that this proportion is rising. Geographical or
social isolation of migrant groups may play a part in this.
http://www.youtube.com/watch?v=Swadss8D8zw