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# Btt2153 7 population genetics_print1

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### Btt2153 7 population genetics_print1

1. 1. POPULATION GENETICS
2. 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. 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 LN LN 1303 Genotype Total 6129
4. 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 LN LN 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. 5. Estimating Allelic Frequencies Number of individuals Allele LM Allele LN LMLM LMLN LN LN 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  f(LM) = (3,574 + 3,039)/12,258 = 0.5395  f(LN) = (3,039 + 2,606)/12,258 = 0.4605.
6. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 17. Frequencies of multiple alleles  For ABO blood group: Blood type A B AB O Genotype IAI A I Ai IBIB IBi I AIB ii Frequency p2 2pr q2 2qr 2pq r2
18. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 31. Selection against the recessive 20 40 60 80 100 120 140 160 180 200 220 240 260 280
32. 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. 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. 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. 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. 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. 37. ANY QUESTION?
38. 38. THANK YOU