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<body><DIV class=Contentbg> <DIV class=Topborder></Div> <DIV id=BigImage> ivapor for ivapor </DIV> <DIV class=Bottomshadow></DIV> <DIV class=Bottomshadow2><h1> trippy stick for sale in canada Blue Moonshine Yumbolt</h1></DIV> </DIV> <DIV class=Contentbg> <DIV class=Topborder></Div> <DIV id=Content> <A name="content"></a> <DIV id=SideInfo> <div id=othernavcontainer> <div id=othernav> <H3> Order Marijuana Seeds</H3> <ul class=navlist> <li><li><a href="http://cannabica.net/wherecanyoubuytrippystickcartridgesonline.shtml" title="Wherecanyoubuytrippystickcartridgesonline" alt="Wherecanyoubuytrippystickcartridgesonline" >Wherecanyoubuytrippystickcartridgesonline</a></li><li><a href="http://cannabica.net/cannabis paypal.shtml" title="Cannabis Paypal" alt="Cannabis Paypal" >Cannabis Paypal</a></li><li><a href="http://cannabica.net/buytrippystick.shtml" title="Buytrippystick" alt="Buytrippystick" >Buytrippystick</a></li><li><a href="http://cannabica.net/trippystickforsaleincanada.shtml" title="Trippystickforsaleincanada" alt="Trippystickforsaleincanada" >Trippystickforsaleincanada</a></li><li><a href="http://cannabica.net/shishkaberry outdoor.shtml" title="Outdoor Trippy Stick Buy Online Outdoor" alt="Outdoor Trippy Stick Buy Online Outdoor" >Outdoor Trippy Stick Buy Online Outdoor</a></li><li><a href="http://cannabica.net/growing marijuana seeds/" title="Marijuana Growing Growing" alt="Marijuana Growing Growing" >Marijuana Growing Growing</a></li><li><a href="http://cannabica.net/aeroponicsmarihuana.shtml" title="Aeroponicsmarihuana" alt="Aeroponicsmarihuana" >Aeroponicsmarihuana</a></li><li><a href="http://cannabica.net/trippystickcartridges.shtml" title="Trippystickcartridges" alt="Trippystickcartridges" >Trippystickcartridges</a></li><li><a href="http://cannabica.net/ivaportrippystickforsaleonline.shtml" title="Ivaportrippystickforsaleonline" alt="Ivaportrippystickforsaleonline" >Ivaportrippystickforsaleonline</a></li><li><a href="http://cannabica.net/wheretobuytrippystick.shtml" title="Wheretobuytrippystick" alt="Wheretobuytrippystick" >Wheretobuytrippystick</a></li><li><a href="http://cannabica.net/sitemap.aspx">Sitemap</a></li><li><a href="http://cannabica.net/sitemap.xml">Sitemap XML</a></li><li><a href="http://cannabica.net/rss/">RSS</a></li><li><a href="http://cannabica.net/mobile.aspx">Mobile version</a></li></li> <li></li> </ul> <H3>seeds canada canada</H3> <ul class=navlist> <li></li> </ul> </div> </div> </DIV> <DIV class=Contentarea> <h1> order marijuana seeds</h1> <h3> ivapor for sale</h3> <p><img border="1" src="/images/big-bud-super-skunk.jpg" ></p> <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. 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