Population Genetics Hardy Weinberg Equilibrium. 6.1 Mendelian Genetics in Populations: The...

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Transcript of Population Genetics Hardy Weinberg Equilibrium. 6.1 Mendelian Genetics in Populations: The...

  • Slide 1
  • Population Genetics Hardy Weinberg Equilibrium
  • Slide 2
  • 6.1 Mendelian Genetics in Populations: The Hardy-Weinberg Equilibrium Principle
  • Slide 3
  • Population Genetics Population genetics is concerned with the question of whether a particular allele or genotype will become more common or less common over time in a population, and Why. Example: Given that the CCR5- 32 allele confers immunity to HIV, will it become more frequent in the human population over time?
  • Slide 4
  • Predicting Allele Frequencies Populations in Hardy-Weinberg equilibrium
  • Slide 5
  • Yule vs. Hardy What are the characteristics of a population that is in equilibrium or another words, not evolving. Yule thought that allele frequencies had to be 0.5 and 0.5. for a population to be in equilibrium. Hardy proved him wrong by developing the Hardy-Weinburg equation.
  • Slide 6
  • Punnett square 60 % of the eggs carry allele A and 40% carry allele a 60% of sperm carry allele A and 40% carry allele a.
  • Slide 7
  • Sample problem AA Aa, aa.In a population of 100 people, we know that 36% are AA, 48% are Aa, and 16% are aa. A aDetermine how many alleles in the gene pool are A or a. Each individual makes two gametes.... A How many A alleles are in this populations gene pool? _____ a How many a alleles? _____ 120 80 (36*(2)+48) (16*(2) +48)
  • Slide 8
  • A a What percent of the alleles are A or a ? A 120 / 200 =.6 or 60% A ; or A.6 = frequency of allele A a ; 80 / 200 =.4 or 40% a ; or a.4 = frequency of allele a
  • Slide 9
  • Creating the Hardy- Wienburg equation is a matter of combining probabilities found in the Punnett square.
  • Slide 10
  • Combining Probabilities The combined probability of two independent events will occur together is equal to the product of their individual probabilities. What is the probability of tossing a nickel and a penny at the same time and having them both come up heads? x =
  • Slide 11
  • Combining Probabilities The combined probability that either of two mutually exclusive events will occur is the sum of their individual probabilities. When rolling a die we can get a one or a two (among other possibilities), but we cannot get both at once. Thus, the probability of getting either a one or a two is 1/6 + 1/6 = 1/3
  • Slide 12
  • Calculating Genotype Frequencies We can predict the genotype frequencies by multiplying probabilities.
  • Slide 13
  • Hardy-Weinburg equation Genotype Frequencies
  • Slide 14
  • ZygotesAllelic frequencyGenotype frequency AA(p)(p)p2p2 Aa(p)(q) 2pq aA(q)(p) aa(q)(q)q2q2 Genotype frequencies described by p 2 +2pq+q 2 =1.0
  • Slide 15
  • The relationship between allele and genotype frequency A aLet original A frequency be represented by p and original a frequency be represented by q AaSince there are only two alleles possible for this gene locus, The frequencies of A and a must equal 1.0 Therefore, p + q =1.0
  • Slide 16
  • Sample: calculating genotype frequencies from allele frequencies? If a given population had the following allele frequencies: A allele frequency (p) for A of 0.8 a allele frequency (q) for a of 0.2 Determine the genotype frequencies of this population? AAAaaa AA Aa aa 0.640.32 0.04 AAAaaa AA = p 2 ; Aa = 2pq ; and aa = q 2 as follows
  • Slide 17
  • We can also calculate the frequency of alleles from the genotype frequencies. When a population is in equilibrium the genotype frequencies are represented as.. P 2 + 2pq +q 2 The allele frequency can therefore be calculated as follows. A = p 2 + (2pq) and a = q 2 + (2pq)
  • Slide 18
  • Examining our example again we see that if we use the frequencies we calculated for each genotype. p 2 2pq q 2 0.64 AA.32 Aa.04 aa A = p 2 + (2pq) A=.64 + (.32) A = 0.8 and since q = 1-p ; then a = 1-(0.8 ) a = 0.2
  • Slide 19
  • These rules hold as long as a population is in equilibrium. Hardy Weinberg Equilibrium describes the conclusions and assumptions that must be present to consider a population in equilibrium.
  • Slide 20
  • Hardy Weinberg Conclusions 1.The allele frequencies in a population will not change from generation to generation. You would need at least 2 generations of data to demonstrate this. 2.If the allele frequencies in a population are given by p and q then the genotype frequencies will be equal to p 2 ; 2pq ; q 2. AA Aa aa Therefore if AA can not be predicted by p 2 Aa cannot be predicted by 2pq and aa cannot be predicted by q 2 then the population is not in equilibrium
  • Slide 21
  • There are 5 assumptions which must be met in order to have a population in equilibrium 1.There is no selection. In other words there is no survival for one genotype over another 2.There is no mutation. This means that none of the alleles in a population will change over time. No alleles get converted into other forms already existing and no new alleles are formed 3.There is no migration (gene flow)New individuals may not enter or leave the population. If movement into or out of the population occurred in a way that certain allele frequencies were changed then the equilibrium would be lost
  • Slide 22
  • 4.There are no chance events (genetic drift) This can only occur if the population is sufficiently large to ensure that the chance of an offspring getting one allele or the other is purely random. When populations are small the principle of genetic drift enters and the equilibrium is not established or will be lost as population size dwindles due to the effects of some outside influence 5.There is no sexual selection or mate choice Who mates with whom must be totally random with no preferential selection involved. Exceptions to Hardy Weinberg cont.
  • Slide 23
  • Problem #6 on page 219 Go to your text page 219 and answer question number 6.
  • Slide 24
  • In humans, the COL1A1 locus codes for a certain collagen protein found in bone. The normal allele at the locus is denoted with S. A recessive allele s is associated with reduced bone mineral density and increased risk of fractures in both Ss and ss women. A recent study of 1,778 women showed that 1,194 were SS, 526 were Ss, and 58 were ss. Are these two alleles in Hardy-Weinberg equilibrium in this population? What information would you need to determine whether the alleles will be in Hardy-Weinberg equilibrium in the next generation?
  • Slide 25
  • Problem approach Check that conclusion #2 holds First figure genotype frequencies from the data (percentages) A a Then from the data, count the actual A alleles in the population and the actual a alleles in the population. What are their frequencies? Then calculate the predicted genotype frequencies of Hardy Weinberg and compare to actual numbers.
  • Slide 26
  • The genotype frequencies are: SS =1194/1778 =.67 Ss= 526/1778 =.30 ss= 58/1778 =.03 2914/ 3556 S alleles = 0.82 S frequency or.67+1/2(.30) = 0.82 642 / 3556 s alleles = 0.18 s frequency or.03 + (.30)=0.18 If the population is in Hardy-Weinberg equilibrium, the allele frequencies should predict the genotype frequencies. SS genotype frequency would be (0.82 )2, = 0.67; Ss frequency would be 2 (.82) (.18), = 0.30; ss frequency would be (0.18 )2, = 0.03. These numbers almost exactly match the measured genotype frequencies - so this population may be in Hardy- Weinberg equilibrium Calculate allele frequencies from genotype frequencies
  • Slide 27
  • However, what also must we know to be sure? We would need to check in future generations to make sure that the allele frequencies are not changing. So here we confirmed conclusion #2 but have not yet verified conclusion#1.
  • Slide 28
  • Example Initial frequencies 15 B 1 B 1 50 B 1 B 2 15 B 2 B 2 80 total
  • Slide 29
  • Example Initial frequencies 15 B 1 B 1 50 B 1 B 2 15 B 2 B 2 80 total Calculate genotype frequencies 15/80 =.187550/80 =.62515/80 =.188
  • Slide 30
  • Example Initial frequencies 15 B 1 B 1 50 B 1 B 2 15 B 2 B 2 80 total Calculate genotype frequencies 15/80 =.187550/80 =.62515/80 =.188 Calculate allele frequencies in the population B 1 = 15+1/2(50)/80 =.5 B 2 = 15+1/2(50)/80 =.5
  • Slide 31
  • Example Initial frequencies 15 B 1 B 1 50 B 1 B 2 15 B 2 B 2 80 total Calculate genotype frequencies 15/80 =.187550/80 =.62515/80 =.188 Calculate allele frequencies in the population B 1 = 15+1/2(50)/80 =.5 B 2 = 15+1/2(50)/80 =.5 Can we predict the genotype frequency from the allele frequency? (Frequency of B 1 ) 2 (0.5) 2 =.25 2( B 1 B 2 ) 2(.5)(.5) =.5 (Frequency of B 1 ) 2 (0.5) 2 =.25
  • Slide 32
  • Example Initial frequencies 15 B 1 B 1 50 B 1 B 2 15 B 2 B 2 80 total Calculate genotype frequencies 15/80 =.187550/80 =.62515/80 =.188 Calculate allele frequencies in the population B 1 = 15+1/2(50)/80 =.5 B 2 = 15+1/2(50)/80 =.5 Can we predict the genotype frequency from the allele frequency? (Frequency of B 1 ) 2 (0.5) 2 =.25 2( B 1 B 2 ) 2(.5)(.5) =.5 (Frequency of B 1 ) 2 (0.5) 2 =.25 Allele frequency does not predict genotype frequency. Population is not in Hardy-Weinberg equilibrium because it violates conclusion 2
  • Slide 33
  • Using the Hardy-Weinberg equilibrium with more th