Genetics

1. POPULATION GENETICS

2. Gene (or Allelic) Frequencies
 Genetic data for a population can be
expressed as gene or allelic frequencies
 All genes have at least two alleles
 Frequencies can vary widely among the
alleles in a population
 Two populations of the same species
do not have to have the same allelic
frequencies.

3. Estimating Allelic Frequencies
 Example: blood type locus
◊ two alleles: LM or LN,
◊ three genotypes: LMLM, LMLN, LNLN
Blood
type
M
LMLM
Number of
individuals
1787
MN
LMLN
3039
N
LNLN
1303
Genotype
Total
6129

4. Estimating Allelic Frequencies
 To determine the allelic frequencies we
simply count the number of LM or LN
alleles and divide by the total number
of alleles
Number of
individuals
Allele LM
Allele LN
LMLM
LMLN
LNLN
1,787
3,574
0
3,039
3,039
3,039
1,303
0
2,606
Total
6,129
6,613
5,645
Genotype
Total alleles
12,258

5. Estimating Allelic Frequencies
Number of
individuals
Allele LM
Allele LN
LMLM
LMLN
LNLN
1,787
3,574
0
3,039
3,039
3,039
1,303
0
2,606
Total
Total alleles
6,129
6,613
12,258
5,645
Genotype
 f(LM) = (3,574 + 3,039)/12,258 = 0.5395
 f(LN) = (3,039 + 2,606)/12,258 = 0.4605.

6. Estimating Allelic Frequencies
 By convention one of the alleles is
given the designation p and the other q
 Also p + q = 1
 p (LM) = 0.5395 and q (LN) = 0.4605

7. The Hardy-Weinberg Law
 The unifying concept of population genetics
 Named after the two scientists who
simultaneously discovered the law
 The law predicts how gene frequencies will
be transmitted from generation to
generation with some assumptions:
◊
◊
◊
◊
◊
Population large
Random mating population
No mutation
No migration
No natural selection.

8. The Hardy-Weinberg Law
For one gene with two alleles
(p + q)2 = p2 + 2pq + q2
and
p+q=1
where:
p2 is frequency for the AA genotype
2pq is frequency for the Aa genotype, and
q2 is frequency for the aa genotype.

9. The Hardy-Weinberg Law
 the gene frequencies will not change
over time, and the frequencies in the
next generation will be:
◊ p2 for the AA genotype
◊ 2pq for the Aa genotype, and
◊ q2 for the aa genotype.

10. The Hardy-Weinberg Law
 If p equals the frequency of allele A in a
population and q is the frequency of
allele a in the same population, union
of gametes would occur with the
following genotypic frequencies:
Female
gametes
p (A)
q (a)
Male gametes
p (A)
q (a)
p2(AA)
pq(Aa)
pq(Aa)
q2(aa)

11. Some examples
1. Assume that a community of 10,000
people on an island is in HardyWeinberg equilibrium and there are
100 sickle cell individuals
(homozygous recessives).
a. What are the frequencies of the alleles
(sickle cell and normal)?
b. What is expected number of
heterozygous carriers in the
community?

12. Some examples
Assume that a community
of 10,000 people on an
island is in Hardy-Weinberg
equilibrium and there are
100 sickle cell individuals
(homozygous recessives).
a. What are the
frequencies of the
alleles (sickle cell and
normal)?
b. What is expected
number of
heterozygous carriers
in the community?
Solution 1:
a..q2(aa) = 100/10,000 = 0.01
q(a) = 0.01 = 0.1
p(A) = 1 – 0.1 = 0.9
b. Frequencies heterozygous:
2pq(Aa) = 2 x 0.9 x 0.1 = 0.18
Number of heterozygous carriers =
0.18 x 10,000 = 1800 people.

13. Some examples
2. In a randomly mating laboratory
population of Drosophila melanogaster,
4 percent of the flies have black body
(black is the autosomal recessive, b)
and 96 percent have brown bodies (the
natural color, B). If this population is
assumed to be in Hardy-Weinberg
equilibrium:
a. What are the allelic frequency of B and b
b. What are the genotype frequency of BB
and Bb?

14. Some examples
Solution 2:
a. q2(bb) = 0.04
q(b) = 0.04 = 0.2
p(B) = 1 – 0.2 = 0.8
In a randomly mating
laboratory population of
Drosophila melanogaster, 4
percent of the flies have black
body (black is the autosomal
recessive, b) and 96 percent
have brown bodies (the natural
color, B). If this population is
assumed to be in HardyWeinberg equilibrium:
a. What are the allelic
frequency of B and b
b. What are the genotype
frequency of BB and Bb?
b. p2(BB) = (0.8)2 = 0.64
2pq(Bb) = 2 x 0.8 x 0.2 = 0.32.

15. Frequencies of multiple alleles
For one gene with two alleles
(p + q)2 = p2 + 2pq + q2
and
p+q=1
where:
p2 is frequency for the AA genotype
2pq is frequency for the Aa genotype, and
q2 is frequency for the aa genotype.

16. Frequencies of multiple alleles
 For one gene with three alleles:
(p + q + r)2 = p2 + q2+ r2 + 2pq + 2pr + 2qr
and
p+q+r=1
 Example of one gene with three alleles:
ABO blood group:
◊ IA : produce antigen A
◊ IB : produce antigen B
◊ i : does not produce any antigen.

17. Frequencies of multiple alleles
 For ABO blood group:
Blood
type
A
B
AB
O
Genotype
Frequency
IAIA
IAi
IBIB
IBi
IAIB
ii
p2
2pr
q2
2qr
2pq
r2
Total
p2 + 2pr
q2 +2qr
2pq
r2

18. Example
 In the population of 1000 people, there
are 42 persons having blood type of A,
672 of B, 36 of AB and 250 of O.
◊ What is the frequency of IA?
◊ What is the frequency of IB?
◊ What is the frequency of i?
◊ How many persons from 42 of A type are
A heterozygote?
◊ How many persons are B homozygote?

19. In the population of 1000 people, there are 42 persons having blood
type of A, 672 of B, 36 of AB and 250 of O.
 Solution:
◊ From that data, the frequency of allele
that can directly be calculated is of i
◊ From 1000 people, there are 250 of O
blood type
◊ r2(ii) = 250/1000 = 0.25
◊ r(i) = 0.25 = 0.5

20. In the population of 1000 people, there are 42 persons having blood
type of A, 672 of B, 36 of AB and 250 of O.
 Now, we add A and O blood types, and
we will have
◊ A + O = 42 + 250 = 292
◊ A = p2 + 2pr and O = r2
◊ p2 + 2pr + r2 = 0.292
◊ (p + r)2 = 0.292
◊ p + r = 0.54
◊ Since r(i) = 0.5 then p(IA) = 0.54 – 0.50 =
0.04

21. In the population of 1000 people, there are 42 persons having blood
type of A, 672 of B, 36 of AB and 250 of O..
◊ What is the frequency of IB?
◊ p+q+r=1
◊ q(IB) = 1 – 0.04 – 0.50 = 0.46

22. In the population of 1000 people, there are 42 persons having blood
type of A, 672 of B, 36 of AB and 250 of O..
 How many persons from 42 of A type
are A heterozygote?
◊ The frequency of heterozygous A is 2pr
◊ 2 x 0.04 x 0.5 x 1000 = 40 persons

23. In the population of 1000 people, there are 42 persons having blood
type of A, 672 of B, 36 of AB and 250 of O..
 How many persons are B homozygote?
◊ The frequency of homozygous B is q2
◊ 0.462 x 1000 = 212 persons

24. Selection against the recessive
 Selection (s) against the recessive is
relative compared to the dominant
types
 The proportion selected of a given
genotype is given the symbol s, which
do not reproduce in every generation
 Therefore, the fitness is equal to 1-s.

25. Selection against the recessive
Table formulating selection:
Genotype
Frequency
Fitness
Proportion
after selection
AA
p2
1
p2
Aa
2pq
1
2pq
aa
q2
1-s
q2(1-s)
Total
1.00
1-sq2

26. Selection against the recessive
 Let’s assume that initially
◊ the frequency of A is p = 0.5,
◊ the frequency of a is q = 0.5 and
◊ s1 = 0.1
Genotype
Relative fitness
Frequency
(at fertilization)
AA
Aa
aa
1
p2 = 0.25
1
2pq = 0.50
1-0.1 = 0.9
q2= 0.25

27. Selection against the recessive
 In forming the next generation, each
genotype will contribute gametes in
proportion to its frequency and relative
fitness
Genotype
Relative
contribution to
next generation
AA
Aa
aa
(0.25) x 1 =
0.25
(0.50) x 1 =
0.50
(0.25) x 0.9
= 0.225

28. Selection against the recessive
 If we divide each of these relative
contribution by their sum (0.25 + 0.50 +
0.225 = 0.975) we obtain
Genotype
Proportional
contribution to
next generation
AA
Aa
aa
0.256
0.513
0.231

29. Selection against the recessive
 The frequency of the a allele after one
generation of selection is from
homozygote aa and from half of
heterozygote Aa:
q‘(a) = 0.231 + (1/2)(0.513) = 0.487

30. Selection against the recessive
 The frequency q' represents the genes
which survive and therefore
corresponds to the gene frequency in
the next generation before selection.
 The formula can be applied repeatedly
generation after generation.
 In the right side of the formula q' is
calculated in the preceding generation
and so forth.

31. Selection against the recessive
20
40
60
80
100 120 140 160 180 200 220 240 260 280

32. Selection against the recessive
 The figure shows such an
application. By strong
selection (s=1) the gene
frequencies change very
rapidly at high gene frequencies.
 If the gene frequency in contrasts is low, the
selection will hardly affect the frequency.
 by weak selection pressure the changes in
the gene frequency are always very slow.

33. Try these
1. The ability to taste the compound
PTC is controlled by a dominant allele
T, while the individuals homozygous
for the recessive allele t are unable to
taste this compound. In a genetics
class of 125 students, 88 were able to
taste PTC, 37 could not.
a. Calculate the frequency of the T and t
allele in this population.
b. Calculate the frequency of the
genotypes.

34. Try these
2. In a given population, only the "A" and
"B" alleles are present in the ABO
system; there are no individuals with
type "O" blood or with O alleles in this
particular population. If 200 people
have type A blood, 75 have type AB
blood, and 25 have type B blood, what
are the alleleic frequencies of this
population (i.e., what are p and q)?

35. Try these
3. Cystic fibrosis is a recessive condition
that affects about 1 in 2,500 babies in
the Caucasian population of the
United States. Please calculate the
following.
a. The frequency of the recessive allele in
the population.
b. The frequency of the dominant allele in
the population.
c. The percentage of heterozygous
individuals (carriers) in the population

36. Try these
4. You sample 1,000 individuals from a large
population for the MN blood group:
Blood type
Genotype
Number of
individuals
Resulting
frequency
M
MM
490
0.49
MN
MN
420
0.42
N
NN
90
0.09
Calculate the following:
a. The frequency of each allele in the population.
b. Supposing the matings are random, the
frequencies of the matings.
c. The probability of each genotype resulting from
each potential cross.

37. ANY QUESTION?

38. THANK YOU