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2

Just need to solve the equation. p1 = X11 + X12; q1 = X11 + X21; 1 = X11 + X12 + X21 + X22. D = X11 - (X11 + X12) * (X11 + X21) D = X11 - (X11X11 + X11X21 + X11X12 + X12X21) D = X11 - X11X11 - X11X21 - X11X12 - X12X21 D = X11 * (1 - X11) - X11X21 - X11X12 - X12X21 D = X11 * (X11 + X12 + X21 + X22 - X11) - X11X21 - X11X12 - X12X21 D = X11 * (X12 + X21 + ...


3

The breeder's equation as you wrote it: $$R = h^2S$$ The heritability that is the ratio of additive genetic variance over the total phenotypic variance is called the heritability in the narrow sense and is noted $h_N^2 = \frac{V_a}{V_P}$, where $V_A$ and $V_P$ are the additive genetic and phenotypic variance respectively. In contrast the heritability that ...


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It's simplest to think about a haploid population with size $N$. Say there are two alleles, $A$ and $a$, with $A$ having frequency $p$ in this generation. Each $A$ individual will have some realized fitness (i.e., number of offspring) in this generation; let's call this number $w_A^{(i)}$ for the $i^\text{th}$ individual. The total number of $A$ individuals ...


1

If the fitness of a heterozygote is $(1+s/2)$ and of a homozygote is $(1-s/2)$ then why is the probability for a given state $(1+s/2)^j(1-s/2)^{m-k}$ $$\binom mj (1/2)^j(1/2)^{m-j}= \binom mj (1/2)^m~~ ?$$ As you pointed out earlier, in the general case it need not be true that $p = q = 1/2$ but that is what the form of the probability above implies. So ...


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@Remi.b's list is excellent, but it should also include Gillespie's Population Genetics: A Concise Guide.


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Not in general -- there can be linkage disequilibrium among the loci. For instance, say that there are two di-allelic loci, $A/a$ and $B/b$, and that the frequencies of the $A$ and $B$ alleles are both $1/2$ and that they have the same effect on the trait, with no dominance. If all haplotypes in the population are either $Ab$ or $aB$ (with no $AB$ or $ab$ ...


1

To derive it, first use that $E[x(1-x)]= E[x-x^2]=E[x]-E[x^2]$ and that $E[x^2]=\text{Var}[x]+E[x]^2$ to rewrite the left-hand side: $$E\left[x_{t+1}(1-x_{t+1})\right] = E\left[x_{t+1}\right](1-E\left[x_{t+1}\right])-\text{Var}\left[x_{t+1}\right].$$ The equation for $p_{ij}$ is just saying that $2Nx_{t+1}$ is binomially distributed with $2N$ trials with ...


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The notation at this site resembles that in your question but preserves the $\frac{x_t}{2N}$ notation for probability of selecting an allele. $$E[\frac{x_{t+1}}{2N})(1 - \frac{x_{t+1}}{2N} )|x_t] = (\frac{x_{t}}{2N})(1 - \frac{x_{t}}{2N}) (1 - \frac{1}{2N}) $$ The expression $(\frac{x_{t}}{2N})(1 - \frac{x_{t}}{2N}) $ is the probability of heterozygosity ...


2

A Biologist's guide to mathematical modeling in evolution and ecology (Otto) is a very good book that is presented for people that have highschool level in mathematics (It makes a good review in linear algebra for example). It is highly accessible and in the meantime it goes pretty far as it ends up talking about the application of diffusion equation in ...


0

Well, the total genetic variance is just, by the definition of the variance, $$ \sigma^2 =\sum_{i,j} f_i f_j (w_{ij}-\bar{w})^2 $$ (using $f_i$ and $w_{ij}$ for frequency and fitness, respectively), and $$\bar{w} = \sum_{i,j} f_i f_j w_{ij}$$ is just the average fitness. You can calculate the additive genetic variance for different loci by simply assuming ...



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