本期的內(nèi)容為哈迪-溫伯格定律。本文集的這一部分是遺傳、進化與生態(tài)學(xué) Genetics, Evolution, and Ecology. 這門課理論上建議在閱讀完文集的第一部分的內(nèi)容之后再開始學(xué)習(xí)，但基礎(chǔ)不足的朋友也可以嘗試閱讀喔~

這一部分的主要內(nèi)容均來自 Prof. Angela J. Roles 的 BIOL 200 課程，因此本文集的這一部分均不會標(biāo)記為原創(chuàng)。但由于文本來源不清晰，UP主還是一個字一個字碼出來的文章，本文禁止非授權(quán)的轉(zhuǎn)載，謝謝！

Lesson 15: Population Genetics: Hardy-Weinberg

[1] Populations

Important vocabulary for genetics of populations:

????Locus:?refers to a particular region of DNA of a chromosome; may or may not be coding DNA; plural is loci.

????Gene:?a functional unit of DNA coding for a protein (or an active RNA). A gene is also a locus.

????Allele:?a variant form of a locus (e.g., purple vs. white alleles for flower color OR 18 vs. 20 repeat units for microsatellite loci).

????Allele frequency:?the proportion of total alleles in the population that are represented by a particular allele. (The total number of alleles is 2 times the population size for an autosomal locus in a diploid.).

????Genotype frequency:?the proportion of individuals in the entire population that possess a particular genotype

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Genome-wide association studies (GWAS)

- From individuals to populations

How can we use linkage to find genes underlying phenotypes?

Genome-Wide Association Studies (GWAS): Sample populations, look for an association between the phenotype and a genetic location.

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- Correlating phenotype to genotype to find genes involved in disease

(1)?Sample people with and without a disease phenotype (ex: indicators of cardiovascular disease);

(2)?Genotype: for each person, sequence thousands of SNP loci (single nucleotide polymorphisms) across the genome;

(3)?Correlate phenotype and genotype to identify loci that are linked to potential disease-related genes;

????Additional work needed to find potential causal genes and test for effects on phenotype (each SNP locus might be linked to hundreds of candidate genes).

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Identify candidate loci: Compare genotypes between those with vs. without the

????SNP1 (a single base pair in the DNA sequence) is more often a G in individuals with the disease phenotype than in individuals without the disease phenotype.

????Thus, there is linkage between SNP1 and coronary artery disease.

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[2] Frequencies

Imagine a population of plants...

?????Let’s consider a population of monkey flower plants, which have flowers in 3 colors: pink, orange, and red (as below). This color difference is related to the production of carotenoid pigments (more carotenoids = more orange/red color).

?????In a population in nature, we find all three flower colors (any given plant has flowers of a single color). To start, we want to determine how common plants of each flower color are.

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Calculating observed (obs)?genotype frequencies

?????Consider the CAR locus with 2 alleles: L = encodes no carotenoids and C = encodes carotenoid production:

?????Imagine you count 40 pink, 40 orange, and 20 red-flowered plants. N = 100 total.

How common is each allele in the population?

????- Allele Frequencies Method 1: Counting up alleles

????????How many total alleles? N=100 individuals, 2 alleles each so 2N = 200 alleles;

????????How many L alleles? (40 pink × 2 each) + (40 orange × 1 each) = 120 L alleles;

????????How many C alleles? (20 red × 2 each) + (40 orange × 1 each) = 80 C alleles;

????????Allele frequencies are: p = frequency of L = 120/200 = 0.6,?q = frequency of C = 80/200 = 0.4.

????- Allele Frequencies Method 2: Use observed genotype frequencies

????????Recall: P(obs)?= freq of LL = 0.4,?H(obs)?= freq of LC = 0.4,?Q(obs)?= freq of CC = 0.2;

????????Thus: p = P(obs)?+ H(obs)/2 =?0.4 + 0.4/2 = 0.6,?q = Q(obs)?+ H(obs)/2 = 0.2 + 0.4/2 =?0.4.?

*Remember that the sum of all allele frequencies must be 1.0.*

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Calculating expected (exp)?genotype frequencies

????The same ideas that underlie Punnett squares as predictors of offspring ratios in a genetic cross can be used to calculate the genotype frequencies we’d expect BY CHANCE given the allele freuqencies that we have in our sample.

? ? What is the probability that the allele?from the male parent is an L allele? And?from the female parent?

Thus, the probability that two L alleles are randomly joined to make a?homozygous L zygote is:

For other genotypes,?we expect:

[3] The Hardy Weinberg (HW) Principle:

Hardy Weinberg Principle: Modeling stasis

?????In the absence of forces causing change, allele and genotype frequencies?will remain the same over time.

?????Assumptions for no genetic change at a locus:

????????(1)?Mating is random with respect to genotype for this locus;

????????(2)?No mutation at this locus;

????????(3)?No natural selection?(all genotypes have equal rate of survival and reproductive success);

????????(4)?No gene flow?(no gametes or individuals are entering or leaving the population);

????????(5)?The population is large?(genetic drift, random change in frequency, is not occurring).

????? Assumptions for no change at a locus: random mating; no mutation; no?selection; no gene flow; no genetic drift

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If the assumptions are met, then we predict:

????(1)?Allele frequencies do not change over generations, p of time 1 = p of time 2;

????(2) Genotype frequencies within a generation are a simple function of allele frequencies:

HW Expected: equilibrium for any value of p

????Expected genotype frequencies for any value of p, when in HW

?????x-axis shows allele?frequencies;

?????y-axis shows expected?genotype frequencies;

?????Expected heterozygosity is?maximized when?p?=?q?=?0.5;

?????When an allele is rare, we?expect most copies of it to?be found in heterozygotes.