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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