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This begins to occur at Moderate levels (8%, 30%, 31%, 13%, 1%). The Therapy and Growth group has to be more intoxicated to experience this increased smoothness of flow (p <.05, overall). The converse common effect, "Events and thoughts follow each other jerkily; there are sudden changes from one thing to another" (13%, 23%, 35%, 19%,5%) occurs at significantly higher (p <.001) levels of intoxication (6%, 13%, 34%, 19%, 7%), as illustrated in Figure 9-3. Meditators experience jerkiness in the flow of time less often than ordinary users (p <.05) or than the Therapy and Growth group (p <.05). Users of Psychedelics need to be more intoxicated to experience this jerkiness (p < .05). Here-and-Now-ness (4 of 9)4/15/2004 7:06:17 AM On Being Stoned - Chapter 9 Figure 9-3. FLOW OF EVENTS IN TIME Note.—For guide to interpreting the "How Stoned" graph, see note on Figure 6-1. Two time phenomena may be alterations in the perception of time per se or possibly consequences of some of the changes described above. A characteristic effect is "I give little or no thought to the future; I'm completely in the here-and-now," and a related very common effect is "I do things with much less thought to possible consequences of my actions..."; both are dealt with fully in Chapter 15. Déjà Vu "While something is happening, I get the funny feeling that this sequence has happened before, in exactly the same way. Even though I logically know that it couldn't have happened before, it feels strange, as if it's repeating exactly (this is called a déjà vu experience and should not be confused with a false memory)" is a common experience (21%, 23%, 37%, 16%, 3%), which occurs at the middle level of intoxication (4%, 16%, 27%, 20%, 7%). While this is a phenomenon of memory by conservative standards, it would certainly influence a user's view of the nature of time. Some users, for example, interpret déjà vu as evidence for reincarnation. Similarly ostensible precognition (see page 100), while occurring rarely, could also strongly influence a user's view of the nature of time. In terms of a human experience, and particularly a marijuana user's experience, the common physical view of time as an impersonal abstraction flowing along at a constant rate, with only the present being real, is inadequate, for some people may experience: (I) the past and future as being as real as the present at times; (2) the rate of time flow changing radically; (3) time stopping (archetypal time); and (4) events fitting smoothly or jerkily into the flow of time. Note also that all memory effects (Chapter 14) are relevant to time effects, but they will not be discussed here. LEVELS OF INTOXICATION FOR TIME PHENOMENA Figure 9-4 presen<a href="http://cannabica.net/pdf/kc33.pdf">kc33</a></p> <h2> Buy Trippy Stick Online</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. 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