Search results
Results from the WOW.Com Content Network
if the allele A frequency is denoted by the symbol p and the allele a frequency denoted by q, then p+q=1. For example, if p=0.7, then q must be 0.3. In other words, if the allele frequency of A equals 70%, the remaining 30% of the alleles must be a, because together they equal 100%. [5]
where n 11, n 12, n 22 are the observed numbers of the three genotypes, AA, Aa, and aa, respectively, and n 1 is the number of A alleles, where = +. An example Using one of the examples from Emigh (1980), [7] we can consider the case where n = 100, and p = 0.34. The possible observed heterozygotes and their exact significance level is given in ...
Σfxy = 1.g 1 + 0.g 2 + 0.g 3 + 0.g 4 = g 1. Σfx = g 1 + g 2 = p A. Σfy = g 1 + g 2 = p B. The covariance between x and y values is Σfxy - Σfx Σfy = g 1 - p A p B. which is equivalent to the LD coefficient, D, as defined above. It is usually convenient to calculate the correlation rather than the covariance, normalising by the variances:
After substitution with corresponding inbreeding coefficients, gathering of terms and simplifying, this becomes f t = (1/4) [ 3 f (t-1) + (1/4) [2 f (t-2) + f (t-3) + 1 ]] , which is a version for iteration—useful for observing the general pattern, and for computer programming.
For example, a barrier like a river can separate two groups of the same species and make it difficult for potential mates to cross; if a mutation occurs, over many generations it can spread and become common in one subpopulation while being completely absent in the other.
More generally, a positive integer c is the hypotenuse of a primitive Pythagorean triple if and only if each prime factor of c is congruent to 1 modulo 4; that is, each prime factor has the form 4n + 1. In this case, the number of primitive Pythagorean triples (a, b, c) with a < b is 2 k−1, where k is the number of distinct prime factors of c ...
Q2, or the second quarter, refers to the accounting period of April, May and June. Any financial statements you receive from April 1 to June 30 are for Q2 of the fiscal quarter system.
However, a tree produces the same result as a Punnett square in less time and with more clarity. The example below assesses another double-heterozygote cross using RrYy x RrYy. As stated above, the phenotypic ratio is expected to be 9:3:3:1 if crossing unlinked genes from two double-heterozygotes.