Download - Ftt1033 7 population genetics-2013
POPULATION GENETICS
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.
Estimating Allelic Frequencies Example: blood type locus◊ two alleles: LM or LN, ◊ three genotypes: LMLM, LMLN, LNLN
Blood type Genotype Number of
individualsM LMLM 1787
MN LMLN 3039N LNLN 1303
Total 6129
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
Genotype Number of individuals Allele LM Allele LN
LMLM 1,787 3,574 0
LMLN 3,039 3,039 3,039
LNLN 1,303 0 2,606
Total 6,129 6,613 5,645Total alleles 12,258
f(LM) = (3,574 + 3,039)/12,258 = 0.5395
f(LN) = (3,039 + 2,606)/12,258 = 0.4605.
Genotype Number of individuals Allele LM Allele LN
LMLM 1,787 3,574 0
LMLN 3,039 3,039 3,039
LNLN 1,303 0 2,606
Total 6,129 6,613 5,645Total alleles 12,258
Estimating Allelic Frequencies
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
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.
The Hardy-Weinberg Law For one gene with two alleles
where:p2 is frequency for the AA genotype2pq is frequency for the Aa genotype, andq2 is frequency for the aa genotype.
(p + q)2 = p2 + 2pq + q2
p + q = 1
and
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.
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
Male gametesp (A) q (a)
p (A) p2(AA) pq(Aa)q (a) pq(Aa) q2(aa)
Some examples1. 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?
Some examplesSolution 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.
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?
Some examples2. 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 bb. What are the genotype frequency of BB
and Bb?
Some examplesSolution 2:
a. q2(bb) = 0.04
q(b) = 0.04 = 0.2
p(B) = 1 – 0.2 = 0.8
b. p2(BB) = (0.8)2 = 0.64
2pq(Bb) = 2 x 0.8 x 0.2 = 0.32.
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 bb. What are the genotype
frequency of BB and Bb?
Frequencies of multiple allelesFor one gene with two alleles
where:p2 is frequency for the AA genotype2pq is frequency for the Aa genotype, andq2 is frequency for the aa genotype.
(p + q)2 = p2 + 2pq + q2
p + q = 1
and
Frequencies of multiple alleles For one gene with three alleles:
Example of one gene with three alleles: ABO blood group:◊ IA : produce antigen A◊ IB : produce antigen B◊ i : does not produce any antigen.
(p + q + r)2 = p2 + q2+ r2 + 2pq + 2pr + 2qr
p + q + r = 1and
Frequencies of multiple alleles For ABO blood group:
Blood type
Genotype Frequency Total
AIAIA p2
p2 + 2prIAi 2pr
BIBIB q2
q2 +2qrIBi 2qr
AB IAIB 2pq 2pqO ii r2 r2
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?
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
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
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
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
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
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.
Genotype AA Aa aa TotalFrequency p2 2pq q2 1.00Fitness 1 1 1-sProportion after selection
p2 2pq q2(1-s) 1-sq2
Table formulating selection:
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 AA Aa aa
Relative fitness 1 1 1-0.1 = 0.9
Frequency(at fertilization)
p2 = 0.25 2pq = 0.50 q2= 0.25
Selection against the recessive
In forming the next generation, each genotype will contribute gametes in proportion to its frequency and relative fitness
Genotype AA Aa aa
Relative contribution to next generation
(0.25) x 1 = 0.25
(0.50) x 1 = 0.50
(0.25) x 0.9 = 0.225
Selection against the recessive
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 AA Aa aa
Proportional contribution to next generation
0.256 0.513 0.231
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
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.
Selection against the recessive
20 40 60 80 100 120 140 160 180 200 220 240 260 280
Selection against the recessive
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.
Try these1. 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.
Try these2. 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)?
Try these3. 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
Try these4. You sample 1,000 individuals from a large
population for the MN blood group:
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.
Blood type Genotype Number of individuals
Resulting frequency
M MM 490 0.49
MN MN 420 0.42
N NN 90 0.09
ANY QUESTION?
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