 Population Genetics. Macrophage CCR5 CCR5-  32.

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Population Genetics

Macrophage

CCR5 CCR5-  32

What accounts for this variation? Random? Past epidemics (plague, smallpox)? What will happen to this variation in the future? Will  32 allele increase in frequency?

These are the questions that “population genetics” is designed to address

Hardy-Weinberg Principle 1.Allele frequencies remain constant from generation to generation unless some outside force is acting to change them 2.When an allele is rare, there are many more heterozygotes than homozygotes (if p is small, then ____ is very small)

Assumptions of H-W 1)Mating is random across the entire population. 2)All genotypes have equal viability and fertility (no selection). 3)Migration into the population can be ignored. 4)Mutation does not occur, or is so rare it can be ignored. 5)Population is large enough that the allele frequencies do not change from generation to generation due to chance (random genetic drift). 6)Allele frequencies are the same in females and males.

Usefulness of H-W If you know the allele frequencies, you can predict the genotype frequencies: Q: In S. France, the frequency of the  32 allele is 10% (i.e., q=0.10). What proportion of individuals will be homozygous for the allele? What proportion will be heterozygous? AAAaaa p2p2 2pqq2q2

Usefulness of H-W If you know the frequency of one of the homozygous genotypes, you can estimate allele frequencies, and predict the frequencies of the other genotypes. Q: Among individuals of European descent, 1/1700 newborns have cystic fibrosis (a recessive genetic disorder). What proportion of this population are heterozygous carriers? Hint: q 2 = 1/1700 = 0.00059 A:

Multiple Alleles: ABO blood types p = freq of A allele q = freq of B allele r = freq of O allele Expansion of [p + q + r] 2 = p 2 + q 2 + r 2 + 2pq + 2pr + 2qr

Assumptions of H-W 1)Mating is random across the entire population. 2)All genotypes have equal viability and fertility (no selection). 3)Migration into the population can be ignored. 4)Mutation does not occur, or is so rare it can be ignored. 5)Population is large enough that the allele frequencies do not change from generation to generation due to chance (random genetic drift). 6)Allele frequencies are the same in females and males.

What happens when any of these assumptions are violated? Selection Mutation Non-random mating _______________ If any of these processes are occurring, will tend to get ____________ from H-W expected proportions

How can we detect deviation from H-W expectations? Do observed genotype frequencies match HW expectations?

p=.5425 q=.4575

Test for H-W Genotype Frequencies

Importance of H-W H-W is an important tool for population genetics. If assumptions are met, we can use it to estimate allele and genotype frequencies that would otherwise be difficult to measure. If assumptions are not met (can be tested statistically), then we know that some outside force is perturbing allele or genotype frequencies.

Change in allele frequencies over generations Evolution is defined as a change in allele (or genotype) frequencies over generations, and evolution will be caused by violation of any of the assumptions of H-W.

Forces that cause deviation from H-W (evolution) 1.Selection 2.Mutation 3.Genetic Drift 4.Nonrandom Mating 5.Gene Flow (Migration)

Genotype A has a constitutive mutation for enzyme production in the lactose operon. B is the normal inducible lactose operon. A and B grown together in environment with limited lactose.

p = 0.5, q = 0.5 Genotypes Number: AA 25 Aa 50 aa 25 Survival to reproduction 25 100% = 1 50 100% = 1 20 80% = 0.8 Gamete contribution 25/95 A 25/95 A; 25/95 a 20/95 a New allele frequencies? New genotype frequencies (assume random mating): p = = ____ q = = ____ AA Aa aa 0.28 0.50 0.22

Consistent differences in survival or reproduction between genotypes = genotypic-specific differences in fitness When fitness values are expressed on a scale such that highest fitness=1, then the values are called relative fitness To conveniently calculate change in allele frequency due to selection, need concept of average fitness

Change in allele frequency Genotype AA Aaaa Genotype Frequencyp 2 2pqq 2 Relative Fitness W AA W Aa W aa W=average fitness= (p 2 W AA )+ (2pqW Aa )+ (q 2 W Aa ) Freq of A after one gen. of selection: p' = p 2 W AA /W + pqW Aa /W Freq of a after one gen. of selection: (1-p’) or: q'= q 2 W aa /W + pqW Aa /W

CCR5 Example; p (+) =0.9; q (  32) =0.1 Genotype frequency: +/+ p 2 =0.81 +/  32 2pq=0.18  32/  32 q 2 =0.01 Relative Fitness W +/+ = 0.99 W +/  32 = 0.99 W  32/  32 = 1.0 Average fitness W = 0. 81*0.99 + 0.18*0.99 + 0.01*1 = 0.9901 q'=q 2 W  32/  32 /W + pqW +/  32 /W=0.01009 +0.089991=0.100091 p’= 1-q’ = 0.89999 Next generation genotype freq. p 2 0.80998 2pq 0.18016 q 2 0.01002

q q2q2

Selection will increase the frequency of  32 allele Selection is relatively weak The favored allele is recessive and the favored genotype is very rare The change in allele frequency (response to selection) will be relatively slow

Response to selection can be fast! Selection is strong Favored allele is partially dominant Both alleles are common

Selection is not always “Directional” Heterozygote advantage Frequency dependence Selection varying in space or time

Heterozygote advantage Fitnes s A aa A

Hb A /Hb A Hb A /Hb S Hb S /Hb S Relative Fitness0.881.00.14 Fitness (in symbols)1-t11-s Selection coefficientst=0.12s=0.86 Relative fitness of hemoglobin genotypes in Yorubans Equilibrium frequencies: p eq = s/(s+t) = 0.86/(0.12+0.86) = 0.88 q eq = t/(s+t) = 0.12/(0.12+0.86) = 0.12 Predict the genotype frequencies (at birth): HW proportions0.774 0.211 0.0144

Variable selection: genotypes have different fitness effects in different environments Fitness

Frequency-dependent selection

Other Examples of Freq-dep. Selection