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For &#97 &#116&#101&#115&#116 &#101&#120&#97&#109&#112&#108&#101 &#108&#101&#116 us consider a population whose gene pool contains the alleles &#66 &#97&#110&#100 &#98&#46 &#65&#115&#115&#105&#103&#110 the letter c to the frequency of the dominant &#97&#108&#108&#101&#108&#101 &#66 &#97&#110&#100 &#116&#104&#101 letter d to &#116&#104&#101 &#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 allele b. In most cases you &#119&#105&#108&#108 &#102&#105&#110&#100 &#116&#104&#97&#116 &#99 and d &#97&#114&#101 &#97&#99&#116&#117&#97&#108&#108&#121 &#110&#111&#116&#97&#116&#101&#100&#13&#10&#97&#115 &#112 &#97&#110&#100 &#113 by convention in science, but for this example we will use c and d.] The sum of all the alleles &#109&#117&#115&#116 &#101&#113&#117&#97&#108 &#49&#48&#48&#37&#46&#13&#10&#83&#111 &#99 + &#100 &#61 &#49&#46&#13&#10&#65&#108&#108 &#116&#104&#101 random &#112&#111&#115&#115&#105&#98&#108&#101 &#99&#111&#109&#98&#105&#110&#97&#116&#105&#111&#110&#115 &#111&#102 &#116&#104&#101 members of a population would equal (c &#120 &#99&#41 &#43 &#50&#99&#100 + (d x d). Which can also be expressed as: (c+d) &#88 &#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 in &#109&#111&#109&#101&#110&#116&#44 &#98&#117&#116 &#105&#116 &#105&#115 best to know &#105&#116 &#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 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. There are no mutations. 295 3. There are no preferences. For example a BB male 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 &#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 &#115&#101&#108&#101&#99&#116&#105&#111&#110 &#109&#117&#115&#116 &#110&#111&#116 &#102&#97&#118&#111&#114 &#97&#110&#121 specific individual.<br> Let us &#105&#109&#97&#103&#105&#110&#101 &#97 &#112&#111&#111&#108 &#111&#102 genes. 12 are B and 18 &#97&#114&#101 &#98&#46 &#78&#111&#119&#13&#10&#114&#101&#109&#101&#109&#98&#101&#114 &#84&#104&#101 &#115&#117&#109 &#111&#102 all &#116&#104&#101 &#97&#108&#108&#101&#108&#101&#115 &#109&#117&#115&#116 &#101&#113&#117&#97&#108 100%. &#83&#111 &#116&#104&#105&#115 &#109&#101&#97&#110&#115&#13&#10&#116&#104&#97&#116 &#116&#104&#101 total in this &#99&#97&#115&#101 &#105&#115 &#49&#50 &#43 18 = 30. So &#51&#48 &#105&#115 &#49&#48&#48&#37&#46&#13&#10&#73&#102 &#119&#101 want &#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 &#111&#102 &#66&#44 Bb and b then &#119&#101 &#119&#105&#108&#108 &#104&#97&#118&#101 &#116&#111 &#97&#112&#112&#108&#121 the standard formula that we have just been shown.<br> f (B) = 12/30 &#61 &#48&#46&#52 &#61 &#52&#48&#37&#13&#10&#102 (b) &#61 &#49&#56&#47&#51&#48 &#61 &#48&#46&#54 = 60% Both add to make 100%. &#78&#111&#119 &#119&#101 &#107&#110&#111&#119 &#116&#104&#101&#105&#114 ratios. So, c + d = &#48&#46&#52 &#43 &#48&#46&#54 &#61 1 We have 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. 296 Remember that all the &#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 &#111&#102 &#116&#104&#101 members of a population would equal (c x c) + 2cd + (d x d), or (c+d) X (c+d) Then, &#99 &#43 &#100 &#61 0.4 + 0.6 &#61 &#49&#13&#10&#65&#110&#100 &#40&#99 &#120 c) + 2cd + (d x d) = BB + Bb + bb = &#46&#50&#52 &#43 &#46&#52&#56 &#43 .30 &#61 &#49&#13&#10&#84&#104&#105&#115 &#109&#101&#97&#110&#115 &#116&#104&#97&#116 the population can increase in size, but the frequencies &#111&#102 &#66 &#97&#110&#100 &#98 will &#115&#116&#97&#121 &#116&#104&#101 &#115&#97&#109&#101&#46&#13&#10&#78&#111&#119&#44 &#115&#117&#112&#112&#111&#115&#101 we break the 4th law about not introducing another population into this one. 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 brings &#111&#117&#114 &#116&#111&#116&#97&#108 &#117&#112 &#116&#111 34 instead of 30.<br> What will the gene and genotypic frequencies be? f (B) = 12/34 = .35 = 35 % f (b) = 22/34 = .65 &#61 &#54&#53&#37&#13&#10&#102 &#40&#66&#66&#41 &#61 .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 1. This means &#116&#104&#97&#116 &#116&#104&#101 &#69&#113&#117&#105&#108&#105&#98&#114&#105&#117&#109 &#108&#97&#119 fails if &#116&#104&#101 &#52&#116&#104 &#108&#97&#119 &#105&#115 not 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 a change of the &#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 &#105&#102&#13&#10&#50&#57&#55&#13&#10&#110&#111 other populations where introduced then the frequency of .42 &#119&#111&#117&#108&#100&#13&#10&#98&#101 &#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 generation.<br> However &#119&#101 &#119&#111&#117&#108&#100 &#108&#105&#107&#101 &#116&#111 &#112&#111&#105&#110&#116 out that we used a very small pool in the &#97&#98&#111&#118&#101 &#101&#120&#97&#109&#112&#108&#101&#46 &#73&#102 &#116&#104&#101 pool were &#109&#117&#99&#104 &#108&#97&#114&#103&#101&#114 &#116&#104&#101&#110 &#116&#104&#101&#13&#10&#110&#117&#109&#98&#101&#114 of changes, even if one or two new genes jumped in, would &#98&#101&#13&#10&#105&#110&#115&#105&#103&#110&#105&#102&#105&#99&#97&#110&#116&#46 &#89&#111&#117 &#99&#111&#117&#108&#100 &#99&#97&#108&#99&#117&#108&#97&#116&#101 it, but &#116&#104&#101 &#99&#104&#97&#110&#103&#101 &#119&#111&#117&#108&#100 &#98&#101 on an extremely low &#108&#101&#118&#101&#119&#104&#105&#99&#104 &#116&#104&#101&#114&#101 &#105&#115 &#110&#111 change in the gene pool. This means that there can be &#110&#111 &#101&#118&#111&#108&#117&#116&#105&#111&#110&#46&#13&#10&#70&#111&#114 &#97 &#116&#101&#115&#116 example let us consider a population whose gene pool contains the alleles B and b. Assign &#116&#104&#101 &#108&#101&#116&#116&#101&#114 &#99 &#116&#111 the frequency of &#116&#104&#101 &#100&#111&#109&#105&#110&#97&#110&#116 &#97&#108&#108&#101&#108&#101 &#66 and the letter d &#116&#111 &#116&#104&#101 &#102&#114&#101&#113&#117&#101&#110&#99&#121 &#111&#102 the recessive allele 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.<br>] The sum of all the alleles must equal 100%.<br> So &#99 &#43 &#100 &#61 1.<br> All the &#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 &#111&#102 the members of &#97&#13&#10&#112&#111&#112&#117&#108&#97&#116&#105&#111&#110 &#119&#111&#117&#108&#100 &#101&#113&#117&#97&#108 &#40&#99 x &#99&#41 &#43 &#50&#99&#100 &#43 (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 will &#101&#120&#112&#108&#97&#105&#110 &#116&#104&#105&#115 &#105&#110 &#100&#101&#116&#97&#105&#108 in moment, but it 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 generation after generation if: 1. The population is large enough.<br> 2.<br> &#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 &#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 &#98&#121 &#105&#116&#115 &#110&#97&#116&#117&#114&#101&#46&#13&#10&#52&#46 &#78&#111 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 &#115&#101&#108&#101&#99&#116&#105&#111&#110 &#109&#117&#115&#116 &#110&#111&#116 &#102&#97&#118&#111&#114 any specific individual.<br> Let us &#105&#109&#97&#103&#105&#110&#101 &#97 &#112&#111&#111&#108 &#111&#102 genes. 12 are B and 18 are b. Now remember &#84&#104&#101 &#115&#117&#109 &#111&#102 &#97&#108&#108 the &#97&#108&#108&#101&#108&#101&#115 &#109&#117&#115&#116 &#101&#113&#117&#97&#108 &#49&#48&#48&#37&#46 So this means that the total in &#116&#104&#105&#115 &#99&#97&#115&#101 &#105&#115 &#49&#50 + 18 &#61 &#51&#48&#46 &#83&#111 &#51&#48 is 100%.<br> 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 b then we &#119&#105&#108&#108 &#104&#97&#118&#101 &#116&#111 &#97&#112&#112&#108&#121 &#116&#104&#101&#13&#10&#115&#116&#97&#110&#100&#97&#114&#100 formula that we &#104&#97&#118&#101 &#106&#117&#115&#116 &#98&#101&#101&#110 &#115&#104&#111&#119&#110&#46&#13&#10&#102 &#40&#66&#41 = 12/30 = 0.4 = 40% f (b) = 18/30 = 0.6 = 60% Both add to make 100%. Now we &#107&#110&#111&#119 &#116&#104&#101&#105&#114 &#114&#97&#116&#105&#111&#115&#46&#13&#10&#83&#111&#44&#13&#10&#99 &#43 d = 0.4 + &#48&#46&#54 &#61 &#49&#13&#10&#87&#101 &#104&#97&#118&#101 proven that c + d must equal 1. Very straightforward, yes. 296 Remember that all the random &#112&#111&#115&#115&#105&#98&#108&#101 &#99&#111&#109&#98&#105&#110&#97&#116&#105&#111&#110&#115 &#111&#102 &#116&#104&#101 members of a population would equal (c x &#99&#41 &#43 &#50&#99&#100 &#43 (d x &#100&#41&#44 &#111&#114 &#40&#99&#43&#100&#41 &#88 (c+d) Then, c + d = 0.4 + &#48&#46&#54 &#61 &#49&#13&#10&#65&#110&#100 &#40&#99 x &#99&#41 &#43 &#50&#99&#100 &#43 (d x d) = BB + Bb + bb = .24 + .48 + .30 &#61 &#49&#13&#10&#84&#104&#105&#115 &#109&#101&#97&#110&#115 &#116&#104&#97&#116 the population can &#105&#110&#99&#114&#101&#97&#115&#101 &#105&#110 &#115&#105&#122&#101&#44 &#98&#117&#116 the frequencies &#111&#102 &#66 &#97&#110&#100 &#98 will stay the same. Now, suppose &#119&#101 &#98&#114&#101&#97&#107 &#116&#104&#101 &#52&#116&#104 law &#97&#98&#111&#117&#116 &#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 &#115&#97&#121 &#116&#104&#97&#116 &#119&#101 &#97&#100&#100 4 more b. b + b + b + b enter the pool. This brings our total up &#116&#111 &#51&#52 &#105&#110&#115&#116&#101&#97&#100 &#111&#102&#13&#10&#51&#48&#46 What &#119&#105&#108&#108 &#116&#104&#101 &#103&#101&#110&#101 &#97&#110&#100 genotypic &#102&#114&#101&#113&#117&#101&#110&#99&#105&#101&#115 &#98&#101&#63&#13&#10&#102 &#40&#66&#41 &#61 12/34 = .35 = 35 % f (b) = 22/34 = .65 = 65% f (BB) = .12, f (Bb) = .23 and f (bb) = .42 Oppss, .42 does not &#101&#113&#117&#97&#108 &#49&#46 &#84&#104&#105&#115 &#109&#101&#97&#110&#115 that &#116&#104&#101 &#69&#113&#117&#105&#108&#105&#98&#114&#105&#117&#109 &#108&#97&#119 &#102&#97&#105&#108&#115&#13&#10&#105&#102 &#116&#104&#101 4th &#108&#97&#119 &#105&#115 &#110&#111&#116 &#109&#101&#116&#46 When the new genes entered the pool &#105&#116&#13&#10&#114&#101&#115&#117&#108&#116&#101&#100 &#105&#110 &#97 &#99&#104&#97&#110&#103&#101 of the population’s gene frequencies. &#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 &#119&#104&#101&#114&#101 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 &#119&#101 &#119&#111&#117&#108&#100 &#108&#105&#107&#101 &#116&#111 point out that we used a very small pool &#105&#110 &#116&#104&#101 &#97&#98&#111&#118&#101 &#101&#120&#97&#109&#112&#108&#101&#46 &#73&#102 the pool were &#109&#117&#99&#104 &#108&#97&#114&#103&#101&#114 &#116&#104&#101&#110 &#116&#104&#101&#13&#10&#110&#117&#109&#98&#101&#114 of changes, even if one or two &#110&#101&#119 &#103&#101&#110&#101&#115 &#106&#117&#109&#112&#101&#100 &#105&#110&#44 would be insignificant. You could calculate it, but the change &#119&#111&#117&#108&#100 &#98&#101 &#111&#110 &#97&#110&#13&#10&#101&#120&#116&#114&#101&#109&#101&#108&#121 low levewhich there is no change in the gene &#112&#111&#111&#108&#46 &#84&#104&#105&#115 &#109&#101&#97&#110&#115 &#116&#104&#97&#116&#13&#10&#116&#104&#101&#114&#101 can be no evolution.<br> For a test example 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 &#97&#108&#108&#101&#108&#101&#115 &#66 &#97&#110&#100 &#98&#46 Assign the letter c to <a href="http://cannabica.net/growing marijuana seeds/" title="Growing Marijuana Seeds" alt="Growing Marijuana Seeds" >Growing Marijuana Seeds</a> the frequency of the dominant allele B and the letter d to the frequency of the recessive allele b. In most cases you will &#102&#105&#110&#100 &#116&#104&#97&#116 &#99 &#97&#110&#100 d are actually notated as &#112 &#97&#110&#100 &#113 &#98&#121 convention in &#115&#99&#105&#101&#110&#99&#101&#44 &#98&#117&#116 &#102&#111&#114 &#116&#104&#105&#115 example we will use c and d. The sum &#111&#102 &#97&#108&#108 &#116&#104&#101 &#97&#108&#108&#101&#108&#101&#115 must &#101&#113&#117&#97&#108 &#49&#48&#48&#37&#46&#13&#10&#83&#111 &#99 &#43 d = 1. 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 &#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 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 &#66 &#97&#110&#100 &#98 &#119&#105&#108&#108 remain &#117&#110&#99&#104&#97&#110&#103&#101&#100 &#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 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 &#110&#111 &#112&#114&#101&#102&#101&#114&#101&#110&#99&#101&#115&#46 &#70&#111&#114 &#101&#120&#97&#109&#112&#108&#101 a BB male does not prefer a bb female by its nature.<br> 4.<br> No other outside population exchanges &#103&#101&#110&#101&#115 &#119&#105&#116&#104 &#116&#104&#105&#115 &#109&#111&#100&#101&#108&#46&#13&#10&#53&#46 Natural selection must not favor any specific individual. Let us imagine a pool of genes. 12 are B and 18 are b. Now remember &#84&#104&#101 &#115&#117&#109 &#111&#102 &#97&#108&#108 the alleles must &#101&#113&#117&#97&#108 &#49&#48&#48&#37&#46 &#83&#111 &#116&#104&#105&#115 means that the total in this case is 12 &#43 &#49&#56 &#61 &#51&#48&#46 So 30 is &#49&#48&#48&#37&#46&#13&#10&#73&#102 &#119&#101 &#119&#97&#110&#116 &#116&#111 find the frequencies &#111&#102 &#66 &#97&#110&#100 &#98 &#97&#110&#100 the genotypic frequencies of B, &#66&#98 &#97&#110&#100 &#98 &#116&#104&#101&#110 we will have to apply the standard formula that we &#104&#97&#118&#101 &#106&#117&#115&#116 &#98&#101&#101&#110 &#115&#104&#111&#119&#110&#46&#13&#10&#102 (B) = 12/30 &#61 &#48&#46&#52 &#61 &#52&#48&#37&#13&#10&#102 (b) = &#49&#56&#47&#51&#48 &#61 &#48&#46&#54 &#61 &#54&#48&#37&#13&#10&#66&#111&#116&#104 add to make 100%. Now we know their ratios. So, c + &#100 &#61 &#48&#46&#52 &#43 0.6 = 1 We have proven that c &#43 &#100 &#109&#117&#115&#116 &#101&#113&#117&#97&#108 1.<br> Very straightforward, yes.<br> 296 Remember that all the random &#112&#111&#115&#115&#105&#98&#108&#101 &#99&#111&#109&#98&#105&#110&#97&#116&#105&#111&#110&#115 &#111&#102 &#116&#104&#101 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 + d = &#48&#46&#52 &#43 &#48&#46&#54 &#61 1 And (c x c) + 2cd &#43 &#40&#100 &#120 &#100&#41&#13&#10&#61 BB + Bb + bb = .24 + .48 + .30 = 1 This &#109&#101&#97&#110&#115 &#116&#104&#97&#116 &#116&#104&#101 &#112&#111&#112&#117&#108&#97&#116&#105&#111&#110 &#99&#97&#110 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 &#115&#116&#97&#121 &#116&#104&#101 &#115&#97&#109&#101&#46&#13&#10&#78&#111&#119&#44 &#115&#117&#112&#112&#111&#115&#101 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 &#112&#111&#111&#108&#46 &#84&#104&#105&#115 &#98&#114&#105&#110&#103&#115 &#111&#117&#114 total up to 34 instead of 30.<br> &#87&#104&#97&#116 &#119&#105&#108&#108 &#116&#104&#101 &#103&#101&#110&#101 and genotypic frequencies be? f (B) = &#49&#50&#47&#51&#52 &#61 &#46&#51&#53 &#61 35 % f (b) &#61 &#50&#50&#47&#51&#52 &#61 &#46&#54&#53 = 65% f (BB) = .12, f &#40&#66&#98&#41 &#61 &#46&#50&#51 &#97&#110&#100 f (bb) &#61 &#46&#52&#50&#13&#10&#79&#112&#112&#115&#115&#44 &#46&#52&#50 &#100&#111&#101&#115 not &#101&#113&#117&#97&#108 &#49&#46 &#84&#104&#105&#115 &#109&#101&#97&#110&#115 that the Equilibrium law fails if the 4th law &#105&#115 &#110&#111&#116 &#109&#101&#116&#46 &#87&#104&#101&#110 the new genes entered the pool it resulted in a change of the population’s gene frequencies. However if 297 no &#111&#116&#104&#101&#114 &#112&#111&#112&#117&#108&#97&#116&#105&#111&#110&#115 &#119&#104&#101&#114&#101 &#105&#110&#116&#114&#111&#100&#117&#99&#101&#100 then the frequency of .42 would be maintained generation after generation.<br> However we would like to &#112&#111&#105&#110&#116 &#111&#117&#116 &#116&#104&#97&#116 &#119&#101 used a very small pool in the above example. &#73&#102 &#116&#104&#101 &#112&#111&#111&#108 &#119&#101&#114&#101 much larger then the number of changes, even if one or two new genes jumped in, would be insignificant. &#89&#111&#117 &#99&#111&#117&#108&#100 &#99&#97&#108&#99&#117&#108&#97&#116&#101 &#105&#116&#44 but the change &#119&#111&#117&#108&#100 &#98&#101 &#111&#110 &#97&#110&#13&#10&#101&#120&#116&#114&#101&#109&#101&#108&#121 low levewhich there is &#110&#111 &#99&#104&#97&#110&#103&#101 &#105&#110 &#116&#104&#101 &#103&#101&#110&#101 pool. This &#109&#101&#97&#110&#115 &#116&#104&#97&#116&#13&#10&#116&#104&#101&#114&#101 &#99&#97&#110 &#98&#101 no evolution. For a test example 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 to the frequency of &#116&#104&#101 &#100&#111&#109&#105&#110&#97&#110&#116 &#97&#108&#108&#101&#108&#101 &#66 &#97&#110&#100 the letter d to the frequency of the recessive allele b.<br> In most cases you will find that c and d &#97&#114&#101 &#97&#99&#116&#117&#97&#108&#108&#121 &#110&#111&#116&#97&#116&#101&#100&#13&#10&#97&#115 &#112 and q by convention in science, but &#102&#111&#114 &#116&#104&#105&#115 &#101&#120&#97&#109&#112&#108&#101 &#119&#101 will use &#99&#13&#10&#97&#110&#100 &#100&#46&#13&#10&#84&#104&#101 &#115&#117&#109 &#111&#102 all the &#97&#108&#108&#101&#108&#101&#115 &#109&#117&#115&#116 &#101&#113&#117&#97&#108 <a href="http://cannabica.net/marijuanaleafsocks.shtml" title="Marijuanaleafsocks" alt="Marijuanaleafsocks" >Marijuanaleafsocks</a> &#49&#48&#48&#37&#46&#13&#10&#83&#111 c + &#100 &#61 &#49&#46&#13&#10&#65&#108&#108 &#116&#104&#101 random &#112&#111&#115&#115&#105&#98&#108&#101 &#99&#111&#109&#98&#105&#110&#97&#116&#105&#111&#110&#115 &#111&#102 &#116&#104&#101 members of a population &#119&#111&#117&#108&#100 &#101&#113&#117&#97&#108 &#40&#99 &#120 c) + 2cd + (d &#120 &#100&#41&#46 &#87&#104&#105&#99&#104 &#99&#97&#110 also be expressed as: (c+d) X (c+d) We will &#101&#120&#112&#108&#97&#105&#110 &#116&#104&#105&#115 &#105&#110 &#100&#101&#116&#97&#105&#108 in moment, but &#105&#116 &#105&#115 &#98&#101&#115&#116 &#116&#111 know &#105&#116 &#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 B 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. 2. There are no mutations. 295 3. There are &#110&#111 &#112&#114&#101&#102&#101&#114&#101&#110&#99&#101&#115&#46 &#70&#111&#114 &#101&#120&#97&#109&#112&#108&#101 a BB male does not prefer &#97&#13&#10&#98&#98 &#102&#101&#109&#97&#108&#101 &#98&#121 &#105&#116&#115 nature.<br> 4.<br> 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 &#102&#97&#118&#111&#114 &#97&#110&#121 &#115&#112&#101&#99&#105&#102&#105&#99 &#105&#110&#100&#105&#118&#105&#100&#117&#97&#108&#46&#13&#10&#76&#101&#116 us imagine a pool of genes.<br> 12 are B &#97&#110&#100 &#49&#56 &#97&#114&#101 &#98&#46 &#78&#111&#119&#13&#10&#114&#101&#109&#101&#109&#98&#101&#114 &#84&#104&#101 sum of &#97&#108&#108 &#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 &#116&#104&#105&#115 &#99&#97&#115&#101 &#105&#115 &#49&#50 + 18 &#61 &#51&#48&#46 &#83&#111 &#51&#48 &#105&#115 100%. If we want &#116&#111 &#102&#105&#110&#100 &#116&#104&#101 &#102&#114&#101&#113&#117&#101&#110&#99&#105&#101&#115 &#111&#102 B &#97&#110&#100 &#98 &#97&#110&#100 &#116&#104&#101&#13&#10&#103&#101&#110&#111&#116&#121&#112&#105&#99 frequencies of B, Bb &#97&#110&#100 &#98 &#116&#104&#101&#110 &#119&#101 will have to apply the standard formula that we have just been shown.<br> f (B) = 12/30 = &#48&#46&#52 &#61 &#52&#48&#37&#13&#10&#102 &#40&#98&#41 = 18/30 = 0.6 &#61 &#54&#48&#37&#13&#10&#66&#111&#116&#104 &#97&#100&#100 &#116&#111 make 100%. Now we know their ratios.<br> So, c + &#100 &#61 &#48&#46&#52 &#43 &#48&#46&#54 = 1 We have proven that c + d must equal 1. Very straightforward, yes.<br> 296 Remember that all the random possible combinations of &#116&#104&#101 &#109&#101&#109&#98&#101&#114&#115&#13&#10&#111&#102 &#97 &#112&#111&#112&#117&#108&#97&#116&#105&#111&#110 would equal &#40&#99 &#120 &#99&#41 &#43 2cd + (d x &#100&#41&#44 &#111&#114 &#40&#99&#43&#100&#41 &#88 (c+d) Then, c + &#100 &#61 &#48&#46&#52 &#43 &#48&#46&#54 = 1 And &#40&#99 &#120 &#99&#41 &#43 &#50&#99&#100 + (d x 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 &#116&#104&#101 &#112&#111&#112&#117&#108&#97&#116&#105&#111&#110 &#99&#97&#110 &#105&#110&#99&#114&#101&#97&#115&#101 &#105&#110 size, but the frequencies of B and b will stay the same. Now, &#115&#117&#112&#112&#111&#115&#101 &#119&#101 &#98&#114&#101&#97&#107 &#116&#104&#101 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 &#115&#97&#121 &#116&#104&#97&#116 &#119&#101 &#97&#100&#100 4 &#109&#111&#114&#101 &#98&#46&#13&#10&#98 &#43 &#98 + b + b enter &#116&#104&#101 &#112&#111&#111&#108&#46 &#84&#104&#105&#115 &#98&#114&#105&#110&#103&#115 &#111&#117&#114 total up to 34 instead of 30. What will the gene and genotypic frequencies &#98&#101&#63&#13&#10&#102 &#40&#66&#41 &#61 &#49&#50&#47&#51&#52 = .35 = 35 % f (b) = 22/34 &#61 &#46&#54&#53 &#61 &#54&#53&#37&#13&#10&#102 (BB) = .12, f (Bb) = &#46&#50&#51 &#97&#110&#100 &#102 &#40&#98&#98&#41 = .42 Oppss, .42 &#100&#111&#101&#115 &#110&#111&#116 &#101&#113&#117&#97&#108 &#49&#46 &#84&#104&#105&#115 means that the Equilibrium law fails if the 4th law is not met. When &#116&#104&#101 &#110&#101&#119 &#103&#101&#110&#101&#115 &#101&#110&#116&#101&#114&#101&#100 the pool it resulted in a change of the 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 other populations where introduced then the frequency of .42 would be maintained generation after generation.<br> However we would like to point out that we used a very small pool &#105&#110 &#116&#104&#101 &#97&#98&#111&#118&#101 &#101&#120&#97&#109&#112&#108&#101&#46 If the 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 even if one or two new genes &#106&#117&#109&#112&#101&#100 &#105&#110&#44 &#119&#111&#117&#108&#100 &#98&#101&#13&#10&#105&#110&#115&#105&#103&#110&#105&#102&#105&#99&#97&#110&#116&#46 You &#99&#111&#117&#108&#100 &#99&#97&#108&#99&#117&#108&#97&#116&#101 &#105&#116&#44 &#98&#117&#116 the change &#119&#111&#117&#108&#100 &#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/bitox.shtml" title="Bitox" alt="Bitox" >Bitox</a></p> <h2> Buy Trippy Stick Cartridges Online</h2> <p> <a href="http://cannabica.net/orange bud/" title="Orange Bud" alt="Orange Bud" >Orange Bud</a> It is possible that its main active constituent t,l-THC could &#98&#101&#99&#111&#109&#101 &#97 &#118&#105&#97&#98&#108&#101 &#100&#114&#117&#103&#44 &#98&#117&#116 it &#105&#115&#13&#10&#109&#111&#115&#116 &#108&#105&#107&#101&#108&#121 &#116&#104&#97&#116 &#116&#104&#101 marketable products <H2> sweetheart cannabis</H2> <a href="http://cannabica.net/buytrippystick.shtml" title="Buytrippystick" alt="Buytrippystick" >Buytrippystick</a> will come from the synthetic &#97&#110&#97&#108&#111&#103&#115&#44&#13&#10&#119&#104&#101&#114&#101 &#116&#104&#101 &#117&#110&#100&#101&#115&#105&#114&#97&#98&#108&#101 &#115&#105&#100&#101 &#101&#102&#102&#101&#99&#116&#115 &#97&#110&#100 physical characteristics of t,l-THC have been &#109&#111&#100&#105&#102&#105&#101&#100 &#111&#114 &#101&#108&#105&#109&#105&#110&#97&#116&#101&#100 <a href="http://cannabica.net/pdf/brothers-grimm-marijana.pdf">brothers-grimm-marijana</a></p> <h2>STICK VS MED TRIPPY STICK</h2> <p>Bud appearance- Buds look silver because they are so covered in crystals. Hairs are orange and few leaves that remain are dark. Buds look great, in this reviewer's humble opinion. Bud density is definitely above average. General notes- Shiskaberry is freaking leafy and takes tons of time to manicure, not that I'm complaining. There is so much resin on leaves including fan leaves that screening is a possibility. Lots of sticky fingers. The Shisks were around 80 days old from seed and were done quickly in search of a good mother. As soon as drying has finished I'll offer up some weight numbers. Plants sizes were from 2'(runts) to 3'(fatties)." - kaka</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/buy visual marijuana seeds.shtml" title="Buy Visual Marijuana Seeds" alt="Buy Visual Marijuana Seeds" >Buy Visual Marijuana Seeds</a>, Powered by the magic team 2/6/2012 2:31:26 PM</DIV> </DIV> <DIV class=Bottomshadow></DIV> </DIV> </body>