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Lecture 10

BIOL 359 Lecture 10: Population Genetics
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Department
Biology
Course
BIOL 359
Professor
Kristen Muller
Semester
Winter

Description
Population Genetics: Change in allele frequencies means that evolution has occurred Genetic Variation Hypotheses: β€’ Classical: selection maintaining best allele for population o Heterozygotes are rate and deleterious and eliminated by selection o Very little variation in population β€’ Balance: heterozygous at many loci o Maintains variability within population β€’ Heterozygotes provide variation in population for example, sickle cell Measuring Heterozygosity: # π‘œπ‘“ π»π‘’π‘‘π‘’π‘Ÿπ‘œπ‘§π‘¦π‘”π‘œπ‘‘π‘’π‘  π»π‘’π‘‘π‘’π‘Ÿπ‘œπ‘§π‘¦π‘”π‘œπ‘ π‘–π‘‘π‘¦ (𝐻) = π‘‡π‘œπ‘‘π‘Žπ‘™ π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ πΌπ‘›π‘‘π‘–π‘£π‘–π‘‘π‘’π‘Žπ‘™π‘  Example: There are 100 individuals and we have the following results in genotype: AA = 25 AB = 50 BB = 25 50 (𝐻) = 100 = 0.5 β€’ Diversity of populations used to be measured by protein electrophoresis; today DNA sequencing is used. β€’ Large amounts of genetic variation in populations in organisms o Most populations are highly variable o Classical hypothesis was incorrect as shown by protein electrophoresis o Balance hypothesis was supported o At the DNA level, variation is even greater High Variability led to new Debates/Hypothesis: β€’ Selectionist: Heterozygotes have higher fitness o Maintenance of high genetic variability as a result of balancing hypothesis β€’ Neutral: Most alleles in natural populations are neutral, doesn’t affect fitness Calculating Genotype Frequency: # πΌπ‘›π‘‘π‘–π‘£π‘–π‘‘π‘’π‘Žπ‘™π‘  π‘œπ‘“ π‘”π‘’π‘›π‘œπ‘‘π‘¦π‘π‘’ 𝐴𝐴 𝐹𝐴𝐴 = π‘‡π‘œπ‘‘π‘Žπ‘™ # π‘œπ‘“ πΌπ‘›π‘‘π‘–π‘£π‘–π‘‘π‘’π‘Žπ‘™π‘  (𝑁) In a population of 200, AA= 50 AB = 100 and BB = 50 N= 200 𝐹 = 50 = 0.25 𝐴𝐴 200 Calculating Allele Frequency: In a population of 200, AA= 50 AB = 100 and BB = 50 N= 200 so 2N = 400 2 𝐴𝐴 + (𝐴𝐡) 𝐹𝐴= 2𝑁 2 50 + (100) 𝐹𝐴= 400 𝐹𝐴= 0.50 YoucanfindF bBsubtractingfrom1orfromdoingsameprocessforF alwaBscheckthat the frequencies add up to 1. Estimating Allele Frequencies from Genotype Frequencies: If you were given genotype frequencies, you can also find allele frequencies: Example: F(AA) = 0.16 F(AB)= 0.48 and F(BB)= 0.36 Find frequencies of allele A and B 𝐹(𝐴𝐡) 𝐹 𝐴 = 𝐹 𝐴𝐴 +) 2 0.48 𝐹 𝐴 = 0.16 + 2 𝐹 𝐴 = 0.40 You can find F(B) by subtracting F(A) by 1. Hardy-Weinberg Principle β€’ Shows how allele and genotype frequencies behave in natural populations o No external forces can apply and has to be random mating o Calculate genotype frequency from allele frequency o Null hypothesis for how population evolves (Starting point in population genetics) β€’ Panmictic Principle: o Sexually reproducing population here each male has equal probability of mating w/ each female (vice versa) o Random mating o Each sperm has equal probability of fusing with each egg (vice versa) o Not realistic, null hypothesis of haploid gametes, so the o Genotype AA: only produce A gametes o Genotype AB: produce A and B gametes (equally) o Genotype BB: only produce B gametes β–ͺ Gametes carrying specific allele are produced in same proportion as frequency of allele in population β€’ Hardy-Weinberg Principle: p +2pq+q = 1 2 o If no external forces/factors, this would be frequency (Allele pool) o No allele frequency change from 1 generation to the next, if no external factors acting on it o Assumptions: 1. No Selection acting on population 2. No Mutation occurring 3. No Migration of population (in or out) 4. Large population with random mating 5. Population is diploid 6. Population reproduces sexually HW principle provide null hypothesis to predict genotype frequencies from allele frequencies β€’ How we see change (expected frequency) and (observed frequency) and their difference o If change occurred, you must see why o If no change occurred, then population is in HW equilibrium o Allele and genotype frequencies remain unchanged and no evolution occurs o Next generation would have given frequency if no change occurred. o Everyone has same chance of survival, mutation doesn’t have huge impact on frequency, & according to HW-equilibrium, no mutation occurs o If population already passed on a certain disease, it cannot be selected for out of population (disease that occurs after reproductive age) Null Hypothesis (HW principle) used to test Alternate Hypotheses (accept/reject HW principle) (π‘Άπ’ƒπ’”π’†π’“π’—π’†π’…βˆ’π‘¬π’™π’‘π’†π’„π’•π’†π’…) 𝟐 β€’ 𝑿 = βˆ‘ 𝑬𝒙𝒑𝒆𝒄𝒕𝒆𝒅 π‘¨π’π’˜π’‚π’šπ’” 𝒑𝒖𝒕 π’π’–π’Žπ’ƒπ’†π’“π’π’ π’Šπ’π’…π’Šπ’—π’Šπ’…π’–π’‚π’π’” 𝒂𝒏𝒅 𝒏𝒐𝒕 𝒕𝒉𝒆 π’‡π’“π’†π’’π’–π’†π’π’„π’š π’‚π’”π’”π’π’„π’Šπ’‚π’•π’†π’… Example: AA= 40, AB = 47 and BB = 13 Population: 100 1) Estimate allele frequencies 𝟐 𝑨𝑨 + 𝑨𝑩 𝐅(𝐀) = 𝟐𝟎𝟎 = 𝟎.πŸ”πŸ‘πŸ“ 𝑭 𝑩 = 𝟏 βˆ’ 𝟎.πŸ”πŸ‘πŸ“ = 𝟎.πŸ‘πŸ”πŸ“ 2) Find expected numbers π„π±π©πžπœπ­πžπ 𝐀𝐀 = (𝑨 ) 𝟏𝟎𝟎 = 𝟎.πŸ”πŸ‘πŸ“ π‘ΏπŸπŸŽπŸŽ = πŸ’πŸŽ.πŸ‘ ) 𝑬𝒙𝒑𝒆𝒄𝒕𝒆𝒅 𝑨𝑩 = πŸπ‘¨π‘© = 𝟐𝟐𝟎.πŸ”πŸ‘πŸ“ 𝟎.πŸ‘πŸ”πŸ“ 𝟏𝟎𝟎 = πŸ’πŸ”.πŸ‘πŸ“ ) 𝑬𝒙𝒑𝒆𝒄𝒕𝒆𝒅 𝑩𝑩 = 𝑩) ( )(𝟏𝟎𝟎 = 𝟎.πŸ‘πŸ”πŸ“ 𝟏𝟎𝟎 = πŸπŸ‘.πŸ‘πŸ 3) Plug it into Chi-Square Eq: 𝑿 = 𝟎.πŸŽπŸπŸ—πŸ‘ 4) Degrees of Freedom Df: β€’ K= number of genotypes (AA,AB,BB) =3 β€’ K-1, minus number of independent parameters =1 β€’ Df= 1 If you check this value with the chi square table, you will see that we do not have statistical evidence to reject null hypothesis. Observed versus expected: β€’ Determine why population is out of HW equilibrium β€’ Compare observed and expected heterozygosity β€’ If Heterozygosity is less than expected, it’s result of heterozygote defici HW Disequilibrium β€’ HW disequilibrium will always result as a consequence of a heterozygote deficit or excess β€’ Shift in allele frequency β€’ Both evolutionary and non-evolutionary forces can result in this deviation Forces that can change genotype/allele frequencies β€’ Natural selection o Selection before reproduction is important β€’ Mutation o Doesn’t shift allele frequency a lot o Important for variat
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