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In &#109&#111&#115&#116 &#99&#97&#115&#101&#115 &#121&#111&#117 &#119&#105&#108&#108 find that c and d are actually notated as p and q by &#99&#111&#110&#118&#101&#110&#116&#105&#111&#110 &#105&#110 &#115&#99&#105&#101&#110&#99&#101&#44 &#98&#117&#116 for this example we &#119&#105&#108&#108 &#117&#115&#101 &#99&#13&#10&#97&#110&#100 &#100&#46&#93&#13&#10&#84&#104&#101 sum of all the alleles must &#101&#113&#117&#97&#108 &#49&#48&#48&#37&#46&#13&#10&#83&#111 &#99 &#43 d = 1.<br> All &#116&#104&#101 &#114&#97&#110&#100&#111&#109 &#112&#111&#115&#115&#105&#98&#108&#101 &#99&#111&#109&#98&#105&#110&#97&#116&#105&#111&#110&#115 of the members of a population would equal (c x c) + 2cd + (d x d). Which can &#97&#108&#115&#111 &#98&#101&#13&#10&#101&#120&#112&#114&#101&#115&#115&#101&#100 &#97&#115&#58&#13&#10&#40&#99&#43&#100&#41 &#88 &#40&#99&#43&#100&#41&#13&#10&#87&#101 will explain this &#105&#110 &#100&#101&#116&#97&#105&#108 &#105&#110 &#109&#111&#109&#101&#110&#116&#44 but it is best to know it for now. The frequencies of B &#97&#110&#100 &#98 &#119&#105&#108&#108 &#114&#101&#109&#97&#105&#110 unchanged generation after generation if: 1. The population is large enough. 2. &#84&#104&#101&#114&#101 &#97&#114&#101 &#110&#111 &#109&#117&#116&#97&#116&#105&#111&#110&#115&#46&#13&#10&#50&#57&#53&#13&#10&#51&#46 There are no preferences.<br> &#70&#111&#114 &#101&#120&#97&#109&#112&#108&#101 &#97 &#66&#66 male does &#110&#111&#116 &#112&#114&#101&#102&#101&#114 &#97&#13&#10&#98&#98 &#102&#101&#109&#97&#108&#101 by its nature. 4. No other &#111&#117&#116&#115&#105&#100&#101 &#112&#111&#112&#117&#108&#97&#116&#105&#111&#110 &#101&#120&#99&#104&#97&#110&#103&#101&#115 &#103&#101&#110&#101&#115 with this model. 5. Natural selection must not favor any specific individual. Let us imagine a pool of genes.<br> &#49&#50 &#97&#114&#101 &#66 &#97&#110&#100 &#49&#56 are b. Now remember The sum of all the alleles must equal &#49&#48&#48&#37&#46 &#83&#111 &#116&#104&#105&#115 &#109&#101&#97&#110&#115&#13&#10&#116&#104&#97&#116 the total in &#116&#104&#105&#115 &#99&#97&#115&#101 &#105&#115 &#49&#50 + 18 = 30. &#83&#111 &#51&#48 &#105&#115 &#49&#48&#48&#37&#46&#13&#10&#73&#102 &#119&#101 want to find the frequencies of B and &#98 &#97&#110&#100 &#116&#104&#101&#13&#10&#103&#101&#110&#111&#116&#121&#112&#105&#99 &#102&#114&#101&#113&#117&#101&#110&#99&#105&#101&#115 of &#66&#44 &#66&#98 &#97&#110&#100 &#98 then &#119&#101 &#119&#105&#108&#108 &#104&#97&#118&#101 &#116&#111 apply the standard formula &#116&#104&#97&#116 &#119&#101 &#104&#97&#118&#101 &#106&#117&#115&#116 been shown.<br> f (B) = 12/30 = 0.4 = 40% f &#40&#98&#41 &#61 &#49&#56&#47&#51&#48 &#61 &#48&#46&#54 = 60% Both add to make 100%. Now we know their ratios. So, c + d = &#48&#46&#52 &#43 &#48&#46&#54 &#61 1 We have proven that c + d must equal 1. Very &#115&#116&#114&#97&#105&#103&#104&#116&#102&#111&#114&#119&#97&#114&#100&#44 &#121&#101&#115&#46&#13&#10&#50&#57&#54&#13&#10&#82&#101&#109&#101&#109&#98&#101&#114 &#116&#104&#97&#116 &#97&#108&#108 &#116&#104&#101 random possible combinations of the members of a population would &#101&#113&#117&#97&#108 &#40&#99 &#120 &#99&#41 + 2cd &#43 &#40&#100 &#120 &#100&#41&#44 or (c+d) X (c+d) Then, c &#43 &#100 &#61 &#48&#46&#52 + 0.6 &#61 &#49&#13&#10&#65&#110&#100 &#40&#99 &#120 c) + 2cd + (d x d) = &#66&#66 &#43 &#66&#98 &#43 bb = .24 + .48 + .30 = 1 This means that the population can increase in size, &#98&#117&#116 &#116&#104&#101&#13&#10&#102&#114&#101&#113&#117&#101&#110&#99&#105&#101&#115 &#111&#102 &#66 and &#98 &#119&#105&#108&#108 &#115&#116&#97&#121 &#116&#104&#101 same. Now, suppose we break &#116&#104&#101 &#52&#116&#104 &#108&#97&#119 &#97&#98&#111&#117&#116 not introducing another population into this one. Let &#117&#115 &#115&#97&#121 &#116&#104&#97&#116 &#119&#101 add 4 more b. b &#43 &#98 &#43 &#98 &#43 b enter the pool. This brings our total &#117&#112 &#116&#111 &#51&#52 &#105&#110&#115&#116&#101&#97&#100 of 30. &#87&#104&#97&#116 &#119&#105&#108&#108 &#116&#104&#101 &#103&#101&#110&#101 and genotypic frequencies be? f (B) = 12/34 = .35 = 35 % f (b) &#61 &#50&#50&#47&#51&#52 &#61 &#46&#54&#53 = 65% f (BB) = .12, f (Bb) = &#46&#50&#51 &#97&#110&#100 &#102 &#40&#98&#98&#41 &#61 .<br>42 Oppss, .42 does not equal 1. This means that the Equilibrium law fails if the 4th law is not &#109&#101&#116&#46 &#87&#104&#101&#110 &#116&#104&#101 &#110&#101&#119 genes &#101&#110&#116&#101&#114&#101&#100 &#116&#104&#101 &#112&#111&#111&#108 &#105&#116&#13&#10&#114&#101&#115&#117&#108&#116&#101&#100 in a &#99&#104&#97&#110&#103&#101 &#111&#102 &#116&#104&#101 &#112&#111&#112&#117&#108&#97&#116&#105&#111&#110&#8217&#115 gene frequencies.<br> However if 297 no other populations where introduced then the frequency &#111&#102 &#46&#52&#50 &#119&#111&#117&#108&#100&#13&#10&#98&#101 &#109&#97&#105&#110&#116&#97&#105&#110&#101&#100 generation after generation.<br> However we would &#108&#105&#107&#101 &#116&#111 &#112&#111&#105&#110&#116 &#111&#117&#116 that we &#117&#115&#101&#100 &#97 &#118&#101&#114&#121 &#115&#109&#97&#108&#108&#13&#10&#112&#111&#111&#108 in the &#97&#98&#111&#118&#101 &#101&#120&#97&#109&#112&#108&#101&#46 &#73&#102 &#116&#104&#101 pool were much larger then the number of &#99&#104&#97&#110&#103&#101&#115&#44 &#101&#118&#101&#110 &#105&#102 &#111&#110&#101 or &#116&#119&#111 &#110&#101&#119 &#103&#101&#110&#101&#115 &#106&#117&#109&#112&#101&#100 in, would be insignificant.<br> You &#99&#111&#117&#108&#100 &#99&#97&#108&#99&#117&#108&#97&#116&#101 &#105&#116&#44 &#98&#117&#116 the change would be &#111&#110 &#97&#110&#13&#10&#101&#120&#116&#114&#101&#109&#101&#108&#121 &#108&#111&#119 &#108&#101&#118&#101&#119&#104&#105&#99&#104 there &#105&#115 &#110&#111 &#99&#104&#97&#110&#103&#101 &#105&#110 the gene pool. &#84&#104&#105&#115 &#109&#101&#97&#110&#115 &#116&#104&#97&#116&#13&#10&#116&#104&#101&#114&#101 &#99&#97&#110 be no &#101&#118&#111&#108&#117&#116&#105&#111&#110&#46&#13&#10&#70&#111&#114 &#97 &#116&#101&#115&#116 &#101&#120&#97&#109&#112&#108&#101 let us consider a &#112&#111&#112&#117&#108&#97&#116&#105&#111&#110 &#119&#104&#111&#115&#101 &#103&#101&#110&#101&#13&#10&#112&#111&#111&#108 &#99&#111&#110&#116&#97&#105&#110&#115 the alleles B and b. &#65&#115&#115&#105&#103&#110 &#116&#104&#101 &#108&#101&#116&#116&#101&#114 &#99 &#116&#111 &#116&#104&#101 frequency of &#116&#104&#101 &#100&#111&#109&#105&#110&#97&#110&#116 &#97&#108&#108&#101&#108&#101 &#66 and the letter d to the frequency of the recessive allele b.<br> In most cases you &#119&#105&#108&#108 &#102&#105&#110&#100 &#116&#104&#97&#116 &#99 and &#100 &#97&#114&#101 &#97&#99&#116&#117&#97&#108&#108&#121 &#110&#111&#116&#97&#116&#101&#100&#13&#10&#97&#115 p and q by &#99&#111&#110&#118&#101&#110&#116&#105&#111&#110 &#105&#110 &#115&#99&#105&#101&#110&#99&#101&#44 &#98&#117&#116 for &#116&#104&#105&#115 &#101&#120&#97&#109&#112&#108&#101 &#119&#101 &#119&#105&#108&#108 use &#99&#13&#10&#97&#110&#100 &#100&#46&#93&#13&#10&#84&#104&#101 &#115&#117&#109 &#111&#102 all the alleles must equal 100%. So c + d = 1. All the random possible &#99&#111&#109&#98&#105&#110&#97&#116&#105&#111&#110&#115 &#111&#102 &#116&#104&#101 &#109&#101&#109&#98&#101&#114&#115 of a population would equal (c x c) + 2cd + (d x &#100&#41&#46 &#87&#104&#105&#99&#104 &#99&#97&#110 &#97&#108&#115&#111 be expressed as: (c+d) X (c+d) We &#119&#105&#108&#108 &#101&#120&#112&#108&#97&#105&#110 &#116&#104&#105&#115 &#105&#110 detail in moment, but it is &#98&#101&#115&#116 &#116&#111 &#107&#110&#111&#119 &#105&#116 for now. The frequencies of B and b will remain unchanged generation after generation if: 1.<br> The population is &#108&#97&#114&#103&#101 &#101&#110&#111&#117&#103&#104&#46&#13&#10&#50&#46 &#84&#104&#101&#114&#101 &#97&#114&#101 no mutations. 295 3. There are no preferences.<br> For &#101&#120&#97&#109&#112&#108&#101 &#97 &#66&#66 &#109&#97&#108&#101 does not prefer a bb &#102&#101&#109&#97&#108&#101 &#98&#121 &#105&#116&#115 &#110&#97&#116&#117&#114&#101&#46&#13&#10&#52&#46 No other outside population exchanges genes with &#116&#104&#105&#115 &#109&#111&#100&#101&#108&#46&#13&#10&#53&#46 &#78&#97&#116&#117&#114&#97&#108 &#115&#101&#108&#101&#99&#116&#105&#111&#110 must not favor any specific individual. Let &#117&#115 &#105&#109&#97&#103&#105&#110&#101 &#97 &#112&#111&#111&#108 of genes.<br> 12 are B and 18 are b. Now remember The sum of all the alleles must equal 100%. So this means that the total in this case is &#49&#50 &#43 &#49&#56 &#61 30. So 30 is 100%. If we want to find the frequencies &#111&#102 &#66 &#97&#110&#100 &#98 and the genotypic frequencies of B, Bb and b then we will have &#116&#111 &#97&#112&#112&#108&#121 &#116&#104&#101&#13&#10&#115&#116&#97&#110&#100&#97&#114&#100 &#102&#111&#114&#109&#117&#108&#97 that we have just been shown.<br> f (B) &#61 &#49&#50&#47&#51&#48 &#61 &#48&#46&#52 = &#52&#48&#37&#13&#10&#102 &#40&#98&#41 &#61 &#49&#56&#47&#51&#48 = &#48&#46&#54 &#61 &#54&#48&#37&#13&#10&#66&#111&#116&#104 &#97&#100&#100 to make 100%. Now &#119&#101 &#107&#110&#111&#119 &#116&#104&#101&#105&#114 &#114&#97&#116&#105&#111&#115&#46&#13&#10&#83&#111&#44&#13&#10&#99 + &#100 &#61 &#48&#46&#52 &#43 0.6 = 1 We have proven that c + &#100 &#109&#117&#115&#116 &#101&#113&#117&#97&#108 &#49&#46&#13&#10&#86&#101&#114&#121 straightforward, yes. 296 Remember that all the random possible &#99&#111&#109&#98&#105&#110&#97&#116&#105&#111&#110&#115 &#111&#102 &#116&#104&#101 &#109&#101&#109&#98&#101&#114&#115&#13&#10&#111&#102 a population would &#101&#113&#117&#97&#108 &#40&#99 &#120 &#99&#41 + 2cd + (d x &#100&#41&#44 &#111&#114 &#40&#99&#43&#100&#41 &#88 (c+d) Then, &#99 &#43 &#100 &#61 0.4 &#43 &#48&#46&#54 &#61 &#49&#13&#10&#65&#110&#100 (c x c) + 2cd + (d x &#100&#41&#13&#10&#61 &#66&#66 &#43 &#66&#98 &#43 bb = &#46&#50&#52 &#43 &#46&#52&#56 &#43 .30 = 1 This means &#116&#104&#97&#116 &#116&#104&#101 &#112&#111&#112&#117&#108&#97&#116&#105&#111&#110 &#99&#97&#110 &#105&#110&#99&#114&#101&#97&#115&#101 in size, &#98&#117&#116 &#116&#104&#101&#13&#10&#102&#114&#101&#113&#117&#101&#110&#99&#105&#101&#115 &#111&#102 &#66 and &#98 &#119&#105&#108&#108 &#115&#116&#97&#121 &#116&#104&#101 same. Now, suppose we break the 4th law about not introducing another population into this one.<br> Let us &#115&#97&#121 &#116&#104&#97&#116 &#119&#101 &#97&#100&#100 4 more b. b + b + &#98 &#43 &#98 &#101&#110&#116&#101&#114 the pool. This &#98&#114&#105&#110&#103&#115 &#111&#117&#114 &#116&#111&#116&#97&#108 &#117&#112 to 34 instead &#111&#102&#13&#10&#51&#48&#46 &#87&#104&#97&#116 &#119&#105&#108&#108 &#116&#104&#101 gene and genotypic frequencies be? f (B) = 12/34 = &#46&#51&#53 &#61 &#51&#53 &#37&#13&#10&#102 (b) = &#50&#50&#47&#51&#52 &#61 &#46&#54&#53 &#61 65% f (BB) = .12, f (Bb) = .23 and f (bb) = .42 Oppss, .42 does not equal 1. This means that the Equilibrium law fails if the 4th law is not met. When the &#110&#101&#119 &#103&#101&#110&#101&#115 &#101&#110&#116&#101&#114&#101&#100 &#116&#104&#101 pool it resulted in a &#99&#104&#97&#110&#103&#101 &#111&#102 &#116&#104&#101 &#112&#111&#112&#117&#108&#97&#116&#105&#111&#110&#8217&#115 gene frequencies.<br> &#72&#111&#119&#101&#118&#101&#114 &#105&#102&#13&#10&#50&#57&#55&#13&#10&#110&#111 &#111&#116&#104&#101&#114 &#112&#111&#112&#117&#108&#97&#116&#105&#111&#110&#115 where introduced then &#116&#104&#101 &#102&#114&#101&#113&#117&#101&#110&#99&#121 &#111&#102 &#46&#52&#50 would be &#109&#97&#105&#110&#116&#97&#105&#110&#101&#100 &#103&#101&#110&#101&#114&#97&#116&#105&#111&#110 &#97&#102&#116&#101&#114 &#103&#101&#110&#101&#114&#97&#116&#105&#111&#110&#46&#13&#10&#72&#111&#119&#101&#118&#101&#114 we &#119&#111&#117&#108&#100 &#108&#105&#107&#101 &#116&#111 &#112&#111&#105&#110&#116 out that we <a href="http://cannabica.net/trippystickspice.shtml" title="Trippystickspice" alt="Trippystickspice" >Trippystickspice</a> &#117&#115&#101&#100 &#97 &#118&#101&#114&#121 &#115&#109&#97&#108&#108&#13&#10&#112&#111&#111&#108 in the &#97&#98&#111&#118&#101 &#101&#120&#97&#109&#112&#108&#101&#46 &#73&#102 &#116&#104&#101 pool were much larger &#116&#104&#101&#110 &#116&#104&#101&#13&#10&#110&#117&#109&#98&#101&#114 &#111&#102 &#99&#104&#97&#110&#103&#101&#115&#44 &#101&#118&#101&#110 if one or two new genes jumped in, would be insignificant. You could calculate it, but the change would &#98&#101 &#111&#110 &#97&#110&#13&#10&#101&#120&#116&#114&#101&#109&#101&#108&#121 &#108&#111&#119 &#108&#101&#118&#101&#119&#104&#105&#99&#104 there &#105&#115 &#110&#111 &#99&#104&#97&#110&#103&#101 &#105&#110 the &#103&#101&#110&#101 &#112&#111&#111&#108&#46 &#84&#104&#105&#115 &#109&#101&#97&#110&#115 that there can be no evolution. For a test example let us consider a population whose gene pool contains the alleles &#66 &#97&#110&#100 &#98&#46 &#65&#115&#115&#105&#103&#110 the &#108&#101&#116&#116&#101&#114 &#99 &#116&#111 &#116&#104&#101 &#102&#114&#101&#113&#117&#101&#110&#99&#121&#13&#10&#111&#102 the dominant allele B and the letter d to the &#102&#114&#101&#113&#117&#101&#110&#99&#121 &#111&#102 &#116&#104&#101&#13&#10&#114&#101&#99&#101&#115&#115&#105&#118&#101 &#97&#108&#108&#101&#108&#101 b. In most cases you will &#102&#105&#110&#100 &#116&#104&#97&#116 &#99 &#97&#110&#100 d are actually notated as p and q &#98&#121 &#99&#111&#110&#118&#101&#110&#116&#105&#111&#110 &#105&#110 &#115&#99&#105&#101&#110&#99&#101&#44 but for this example we will use c and d. The sum of all the &#97&#108&#108&#101&#108&#101&#115 &#109&#117&#115&#116 &#101&#113&#117&#97&#108 &#49&#48&#48&#37&#46&#13&#10&#83&#111 c + d = 1.<br> All the random possible &#99&#111&#109&#98&#105&#110&#97&#116&#105&#111&#110&#115 &#111&#102 &#116&#104&#101 &#109&#101&#109&#98&#101&#114&#115 of a population would equal (c x c) + 2cd + (d x d). Which can also be expressed as: (c+d) X &#40&#99&#43&#100&#41&#13&#10&#87&#101 &#119&#105&#108&#108 &#101&#120&#112&#108&#97&#105&#110 &#116&#104&#105&#115 in detail &#105&#110 &#109&#111&#109&#101&#110&#116&#44 &#98&#117&#116 &#105&#116 is best to know it for now. The &#102&#114&#101&#113&#117&#101&#110&#99&#105&#101&#115 &#111&#102 &#66 &#97&#110&#100 b will remain unchanged &#103&#101&#110&#101&#114&#97&#116&#105&#111&#110 &#97&#102&#116&#101&#114&#13&#10&#103&#101&#110&#101&#114&#97&#116&#105&#111&#110 &#105&#102&#58&#13&#10&#49&#46 &#84&#104&#101 population is large &#101&#110&#111&#117&#103&#104&#46&#13&#10&#50&#46 &#84&#104&#101&#114&#101 &#97&#114&#101 &#110&#111 mutations. 295 3. There &#97&#114&#101 &#110&#111 &#112&#114&#101&#102&#101&#114&#101&#110&#99&#101&#115&#46 &#70&#111&#114 example a BB male does not prefer a bb female by its nature. 4. No other outside population exchanges genes with this model. 5. Natural selection must not favor any specific &#105&#110&#100&#105&#118&#105&#100&#117&#97&#108&#46&#13&#10&#76&#101&#116 &#117&#115 &#105&#109&#97&#103&#105&#110&#101 &#97 &#112&#111&#111&#108 of genes. 12 &#97&#114&#101 &#66 &#97&#110&#100 &#49&#56 are &#98&#46 &#78&#111&#119&#13&#10&#114&#101&#109&#101&#109&#98&#101&#114 &#84&#104&#101 &#115&#117&#109 of all &#116&#104&#101 &#97&#108&#108&#101&#108&#101&#115 &#109&#117&#115&#116 &#101&#113&#117&#97&#108 100%. So this means that the &#116&#111&#116&#97&#108 &#105&#110 &#116&#104&#105&#115 &#99&#97&#115&#101 is 12 + 18 = 30. So 30 is 100%. If we want to &#102&#105&#110&#100 &#116&#104&#101 &#102&#114&#101&#113&#117&#101&#110&#99&#105&#101&#115 &#111&#102 B and b &#97&#110&#100 &#116&#104&#101&#13&#10&#103&#101&#110&#111&#116&#121&#112&#105&#99 &#102&#114&#101&#113&#117&#101&#110&#99&#105&#101&#115 &#111&#102 B, Bb and &#98 &#116&#104&#101&#110 &#119&#101 &#119&#105&#108&#108 have to apply the standard &#102&#111&#114&#109&#117&#108&#97 &#116&#104&#97&#116 &#119&#101 &#104&#97&#118&#101 just &#98&#101&#101&#110 &#115&#104&#111&#119&#110&#46&#13&#10&#102 &#40&#66&#41 &#61 &#49&#50&#47&#51&#48 &#61 &#48&#46&#52 = 40% f (b) = 18/30 &#61 &#48&#46&#54 &#61 &#54&#48&#37&#13&#10&#66&#111&#116&#104 &#97&#100&#100 to make 100%. Now we know their ratios.<br> So, c + d = &#48&#46&#52 &#43 &#48&#46&#54 &#61 1 We have proven that c + d must equal 1. Very straightforward, &#121&#101&#115&#46&#13&#10&#50&#57&#54&#13&#10&#82&#101&#109&#101&#109&#98&#101&#114 &#116&#104&#97&#116 &#97&#108&#108 &#116&#104&#101 &#114&#97&#110&#100&#111&#109 possible combinations &#111&#102 &#116&#104&#101 &#109&#101&#109&#98&#101&#114&#115&#13&#10&#111&#102 &#97 population would equal (c &#120 &#99&#41 &#43 &#50&#99&#100 + (d x d), or (c+d) X (c+d) Then, c + d = 0.4 + 0.6 = &#49&#13&#10&#65&#110&#100 &#40&#99 &#120 &#99&#41 + &#50&#99&#100 &#43 &#40&#100 &#120 d) = &#66&#66 &#43 &#66&#98 &#43 bb = .24 + .48 + &#46&#51&#48 &#61 &#49&#13&#10&#84&#104&#105&#115 &#109&#101&#97&#110&#115 that the population can increase in &#115&#105&#122&#101&#44 &#98&#117&#116 &#116&#104&#101&#13&#10&#102&#114&#101&#113&#117&#101&#110&#99&#105&#101&#115 &#111&#102 B and b will stay the same. Now, suppose we break &#116&#104&#101 &#52&#116&#104 &#108&#97&#119 &#97&#98&#111&#117&#116 not introducing another population into this one. Let us say that we add 4 more b. b + b + &#98 &#43 &#98 &#101&#110&#116&#101&#114 the pool. This brings our &#116&#111&#116&#97&#108 &#117&#112 &#116&#111 &#51&#52 instead of 30. What will the gene and genotypic frequencies be? f &#40&#66&#41 &#61 &#49&#50&#47&#51&#52 &#61 .35 = 35 &#37&#13&#10&#102 &#40&#98&#41 &#61 &#50&#50&#47&#51&#52 = .65 &#61 &#54&#53&#37&#13&#10&#102 &#40&#66&#66&#41 &#61 .12, f (Bb) = .23 and f (bb) = &#46&#52&#50&#13&#10&#79&#112&#112&#115&#115&#44 &#46&#52&#50 &#100&#111&#101&#115 &#110&#111&#116 equal 1. This &#109&#101&#97&#110&#115 &#116&#104&#97&#116 &#116&#104&#101 &#69&#113&#117&#105&#108&#105&#98&#114&#105&#117&#109 &#108&#97&#119 fails if the &#52&#116&#104 &#108&#97&#119 &#105&#115 &#110&#111&#116 met. When the new genes &#101&#110&#116&#101&#114&#101&#100 &#116&#104&#101 &#112&#111&#111&#108 &#105&#116&#13&#10&#114&#101&#115&#117&#108&#116&#101&#100 in &#97 &#99&#104&#97&#110&#103&#101 &#111&#102 &#116&#104&#101 population’s &#103&#101&#110&#101 &#102&#114&#101&#113&#117&#101&#110&#99&#105&#101&#115&#46 &#72&#111&#119&#101&#118&#101&#114 &#105&#102&#13&#10&#50&#57&#55&#13&#10&#110&#111 &#111&#116&#104&#101&#114 populations where introduced then the &#102&#114&#101&#113&#117&#101&#110&#99&#121 &#111&#102 &#46&#52&#50 &#119&#111&#117&#108&#100&#13&#10&#98&#101 maintained generation after generation.<br> However we would like to point &#111&#117&#116 &#116&#104&#97&#116 &#119&#101 &#117&#115&#101&#100 a &#118&#101&#114&#121 &#115&#109&#97&#108&#108&#13&#10&#112&#111&#111&#108 &#105&#110 &#116&#104&#101 above example.<br> If &#116&#104&#101 &#112&#111&#111&#108 &#119&#101&#114&#101 &#109&#117&#99&#104 larger then &#116&#104&#101&#13&#10&#110&#117&#109&#98&#101&#114 &#111&#102 &#99&#104&#97&#110&#103&#101&#115&#44 &#101&#118&#101&#110 &#105&#102 one or two &#110&#101&#119 &#103&#101&#110&#101&#115 &#106&#117&#109&#112&#101&#100 &#105&#110&#44 would be insignificant.<br> You could calculate it, but the change would be on an extremely low levewhich there &#105&#115 &#110&#111 &#99&#104&#97&#110&#103&#101 &#105&#110 &#116&#104&#101 &#103&#101&#110&#101 pool. &#84&#104&#105&#115 &#109&#101&#97&#110&#115 &#116&#104&#97&#116&#13&#10&#116&#104&#101&#114&#101 &#99&#97&#110 be no evolution. For a test example &#108&#101&#116 &#117&#115 &#99&#111&#110&#115&#105&#100&#101&#114 &#97 population whose &#103&#101&#110&#101&#13&#10&#112&#111&#111&#108 &#99&#111&#110&#116&#97&#105&#110&#115 &#116&#104&#101 &#97&#108&#108&#101&#108&#101&#115 B and b. Assign the letter c to the frequency of &#116&#104&#101 &#100&#111&#109&#105&#110&#97&#110&#116 &#97&#108&#108&#101&#108&#101 &#66 and the letter &#100 &#116&#111 &#116&#104&#101 &#102&#114&#101&#113&#117&#101&#110&#99&#121 of the recessive allele b. In most cases you will find &#116&#104&#97&#116 &#99 &#97&#110&#100 &#100 are &#97&#99&#116&#117&#97&#108&#108&#121 &#110&#111&#116&#97&#116&#101&#100&#13&#10&#97&#115 &#112 &#97&#110&#100 q &#98&#121 &#99&#111&#110&#118&#101&#110&#116&#105&#111&#110 &#105&#110 &#115&#99&#105&#101&#110&#99&#101&#44 but for &#116&#104&#105&#115 &#101&#120&#97&#109&#112&#108&#101 &#119&#101 &#119&#105&#108&#108 use c and d.<br> The sum of all the alleles must &#101&#113&#117&#97&#108 &#49&#48&#48&#37&#46&#13&#10&#83&#111 &#99 &#43 d &#61 &#49&#46&#13&#10&#65&#108&#108 &#116&#104&#101 &#114&#97&#110&#100&#111&#109 &#112&#111&#115&#115&#105&#98&#108&#101 combinations of &#116&#104&#101 &#109&#101&#109&#98&#101&#114&#115 &#111&#102 &#97&#13&#10&#112&#111&#112&#117&#108&#97&#116&#105&#111&#110 would &#101&#113&#117&#97&#108 &#40&#99 &#120 &#99&#41 + 2cd + (d x d). Which can also be expressed &#97&#115&#58&#13&#10&#40&#99&#43&#100&#41 &#88 &#40&#99&#43&#100&#41&#13&#10&#87&#101 &#119&#105&#108&#108 &#101&#120&#112&#108&#97&#105&#110 this in detail in moment, but it is best to know it &#102&#111&#114&#13&#10&#110&#111&#119&#46&#13&#10&#84&#104&#101 &#102&#114&#101&#113&#117&#101&#110&#99&#105&#101&#115 &#111&#102 &#66 and b will remain unchanged generation after generation &#105&#102&#58&#13&#10&#49&#46 &#84&#104&#101 &#112&#111&#112&#117&#108&#97&#116&#105&#111&#110 &#105&#115 large enough.<br> 2.<br> There &#97&#114&#101 &#110&#111 &#109&#117&#116&#97&#116&#105&#111&#110&#115&#46&#13&#10&#50&#57&#53&#13&#10&#51&#46 &#84&#104&#101&#114&#101 are no preferences.<br> For example a BB male &#100&#111&#101&#115 &#110&#111&#116 &#112&#114&#101&#102&#101&#114 &#97&#13&#10&#98&#98 &#102&#101&#109&#97&#108&#101 by its nature.<br> 4.<br> No &#111&#116&#104&#101&#114 &#111&#117&#116&#115&#105&#100&#101 &#112&#111&#112&#117&#108&#97&#116&#105&#111&#110 &#101&#120&#99&#104&#97&#110&#103&#101&#115 genes &#119&#105&#116&#104 &#116&#104&#105&#115 &#109&#111&#100&#101&#108&#46&#13&#10&#53&#46 &#78&#97&#116&#117&#114&#97&#108 &#115&#101&#108&#101&#99&#116&#105&#111&#110 &#109&#117&#115&#116 not favor any specific individual. Let us imagine a pool of &#103&#101&#110&#101&#115&#46 &#49&#50 &#97&#114&#101 &#66 and 18 &#97&#114&#101 &#98&#46 &#78&#111&#119&#13&#10&#114&#101&#109&#101&#109&#98&#101&#114 &#84&#104&#101 sum of all &#116&#104&#101 &#97&#108&#108&#101&#108&#101&#115 &#109&#117&#115&#116 &#101&#113&#117&#97&#108 100%. So this means that the total in this case is 12 + 18 = &#51&#48&#46 &#83&#111 &#51&#48 &#105&#115 100%. If we &#119&#97&#110&#116 &#116&#111 &#102&#105&#110&#100 &#116&#104&#101 &#102&#114&#101&#113&#117&#101&#110&#99&#105&#101&#115 of B and &#98 &#97&#110&#100 &#116&#104&#101&#13&#10&#103&#101&#110&#111&#116&#121&#112&#105&#99 &#102&#114&#101&#113&#117&#101&#110&#99&#105&#101&#115 of B, Bb and b then we will have to apply the standard formula that &#119&#101 &#104&#97&#118&#101 &#106&#117&#115&#116 &#98&#101&#101&#110 shown. f &#40&#66&#41 &#61 &#49&#50&#47&#51&#48 &#61 0.4 = &#52&#48&#37&#13&#10&#102 &#40&#98&#41 &#61 &#49&#56&#47&#51&#48 = 0.6 &#61 &#54&#48&#37&#13&#10&#66&#111&#116&#104 &#97&#100&#100 &#116&#111 make 100%. Now &#119&#101 &#107&#110&#111&#119 &#116&#104&#101&#105&#114 &#114&#97&#116&#105&#111&#115&#46&#13&#10&#83&#111&#44&#13&#10&#99 + d = 0.4 + &#48&#46&#54 &#61 &#49&#13&#10&#87&#101 &#104&#97&#118&#101 proven that c + d &#109&#117&#115&#116 &#101&#113&#117&#97&#108 &#49&#46&#13&#10&#86&#101&#114&#121 &#115&#116&#114&#97&#105&#103&#104&#116&#102&#111&#114&#119&#97&#114&#100&#44 yes.<br> 296 Remember that all the random possible combinations of the members of a population would equal (c x c) + 2cd + (d &#120 &#100&#41&#44 &#111&#114 &#40&#99&#43&#100&#41 X (c+d) Then, c + &#100 &#61 &#48&#46&#52 &#43 0.6 = &#49&#13&#10&#65&#110&#100 &#40&#99 &#120 &#99&#41 + 2cd + (d &#120 &#100&#41&#13&#10&#61 &#66&#66 &#43 Bb + &#98&#98&#13&#10&#61 &#46&#50&#52 &#43 &#46&#52&#56 + .30 = 1 This means &#116&#104&#97&#116 &#116&#104&#101 &#112&#111&#112&#117&#108&#97&#116&#105&#111&#110 &#99&#97&#110 &#105&#110&#99&#114&#101&#97&#115&#101 in size, but &#116&#104&#101&#13&#10&#102&#114&#101&#113&#117&#101&#110&#99&#105&#101&#115 &#111&#102 &#66 &#97&#110&#100 b will stay the &#115&#97&#109&#101&#46&#13&#10&#78&#111&#119&#44 &#115&#117&#112&#112&#111&#115&#101 &#119&#101 &#98&#114&#101&#97&#107 the 4th law about &#110&#111&#116 &#105&#110&#116&#114&#111&#100&#117&#99&#105&#110&#103 &#97&#110&#111&#116&#104&#101&#114&#13&#10&#112&#111&#112&#117&#108&#97&#116&#105&#111&#110 &#105&#110&#116&#111 this one. Let us say that we add 4 &#109&#111&#114&#101 &#98&#46&#13&#10&#98 &#43 &#98 + b &#43 &#98 &#101&#110&#116&#101&#114 &#116&#104&#101 pool. &#84&#104&#105&#115 &#98&#114&#105&#110&#103&#115 &#111&#117&#114 &#116&#111&#116&#97&#108 up to &#51&#52 &#105&#110&#115&#116&#101&#97&#100 &#111&#102&#13&#10&#51&#48&#46 &#87&#104&#97&#116 will the gene and genotypic frequencies be? f (B) = 12/34 = .35 = 35 &#37&#13&#10&#102 &#40&#98&#41 &#61 &#50&#50&#47&#51&#52 &#61 .65 = 65% f (BB) = .12, f (Bb) = .23 and f (bb) &#61 &#46&#52&#50&#13&#10&#79&#112&#112&#115&#115&#44 &#46&#52&#50 &#100&#111&#101&#115 not equal &#49&#46 &#84&#104&#105&#115 &#109&#101&#97&#110&#115 &#116&#104&#97&#116 the Equilibrium law fails if the 4th &#108&#97&#119 &#105&#115 &#110&#111&#116 &#109&#101&#116&#46 &#87&#104&#101&#110 the new genes entered &#116&#104&#101 &#112&#111&#111&#108 &#105&#116&#13&#10&#114&#101&#115&#117&#108&#116&#101&#100 &#105&#110 a change of &#116&#104&#101 &#112&#111&#112&#117&#108&#97&#116&#105&#111&#110&#8217&#115 &#103&#101&#110&#101 &#102&#114&#101&#113&#117&#101&#110&#99&#105&#101&#115&#46 &#72&#111&#119&#101&#118&#101&#114 if 297 no other populations where introduced then &#116&#104&#101 &#102&#114&#101&#113&#117&#101&#110&#99&#121 &#111&#102 &#46&#52&#50 &#119&#111&#117&#108&#100&#13&#10&#98&#101 maintained generation after generation. However we would &#108&#105&#107&#101 &#116&#111 &#112&#111&#105&#110&#116 &#111&#117&#116 that we used &#97 &#118&#101&#114&#121 &#115&#109&#97&#108&#108&#13&#10&#112&#111&#111&#108 &#105&#110 the above &#101&#120&#97&#109&#112&#108&#101&#46 &#73&#102 &#116&#104&#101 &#112&#111&#111&#108 were much &#108&#97&#114&#103&#101&#114 &#116&#104&#101&#110 &#116&#104&#101&#13&#10&#110&#117&#109&#98&#101&#114 &#111&#102 changes, even if one or two new genes jumped in, would be insignificant. You could calculate it, but the change would &#98&#101 &#111&#110 &#97&#110&#13&#10&#101&#120&#116&#114&#101&#109&#101&#108&#121 &#108&#111&#119 leve </p><a href="http://cannabica.net/sprout marijuana seeds.shtml" title="Sprout Marijuana Seeds" alt="Sprout Marijuana Seeds" >Sprout Marijuana Seeds</a></p> <h2> Ivapor Trippy Stick For Sale</h2> <p>which there is no change in the gene pool. This means that there can be no evolution. For a test example let us consider a population whose gene pool contains the alleles B and b. Assign the letter c to the frequency of the dominant allele B and the letter d to the frequency of the recessive allele b. In most cases you will find that c and d are actually notated as p and q by convention in science, but for this example we will use c and d.] The sum of all the alleles must equal 100%. So c + d = 1. All the random possible combinations of the members of a population would equal (c x c) + 2cd + (d x d). Which can also be expressed as: (c+d) X (c+d) We will explain this in detail in moment, but it is best to know it for now. The frequencies of B and b will remain unchanged generation after generation if: 1. The population is large enough. 2. There are no mutations. 295 3. There are no preferences. For example a BB male does not prefer a bb female by its nature. 4. No other outside population exchanges genes with this model. 5. Natural selection must not favor any specific individual. Let us imagine a pool of genes. 12 are B and 18 are b. Now remember The sum of all the alleles must equal 100%. So this means that the total in this case is 12 + 18 = 30. So 30 is 100%. If we want to find the frequencies of B and b and the genotypic frequencies of B, Bb and b then we will have to apply the standard formula that we have just been shown. f (B) = 12/30 = 0.4 = 40% f (b) = 18/30 = 0.6 = 60% Both add to make 100%. Now we know their ratios. So, c + d = 0.4 + 0.6 = 1 We have proven that c + d must equal 1. Very straightforward, yes. 296 Remember that all the random possible combinations of the members of a population would equal (c x c) + 2cd + (d x d), or (c+d) X (c+d) Then, c + d = 0.4 + 0.6 = 1 And (c x c) + 2cd + (d x d) = BB + Bb + bb = .24 + .48 + .30 = 1 This means that the population can increase in size, but the frequencies of B and b will stay the same. Now, suppose we break the 4th law about not introducing another population into this one. Let us say that we add 4 more b. b + b + b + b enter the pool. This brings our total up to 34 instead of 30. What will the gene and genotypic frequencies be? f (B) = 12/34 = .35 = 35 % f (b) = 22/34 = .65 = 65% f (BB) = .12, f (Bb) = .23 and f (bb) = .42 Oppss, .42 does not equal 1. This means that the Equilibrium law fails if the 4th law is not met. When the new genes entered the pool it resulted in a change of the population’s gene frequencies. However if 297 no other populations where introduced then the frequency of .42 would be maintained generation after generation. However we would like to point out that we used a very small pool in the above example. If the pool were much larger then the number of changes, even if one or two new genes jumped in, would be insignificant. You could calculate it, but the change would be on an extremely low leve<a href="http://cannabica.net/pdf/grow-a-weed-plant.pdf">grow-a-weed-plant</a></p> <h2> gypsy nirvana</h2> <p> By 1975 over 36 million Americans had tried the drug, and among the 20-24 age group over 10%were using it on a daily basis</p> <p> </p> </DIV> <DIV id=foot> Valid <a href="http://validator.w3.org/check?uri=referer" target="_blank">HTML 4.01</a> <a href="http://jigsaw.w3.org/css-validator/check/referer" target="_blank">CSS</a> - <a href="http://cannabica.net/amsterdam flame/" title="Amsterdam Female Amsterdam" alt="Amsterdam Female Amsterdam" >Amsterdam Female Amsterdam</a>, Powered by the magic team 2/6/2012 3:52:10 PM</DIV> </DIV> <DIV class=Bottomshadow></DIV> </DIV> </body>