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Buzz- The first time I Trippystickvaporizerforsale smoked some Shiskaberry yesterday it really didn’t stone me. Today it has whipped me
:) A
shishkaberry seeds for sale few hits on the way to school were a few too many, it hit and I became unmotivated. This evening a friend
and I finished off two bowls and it was quite the experience. Fair amount of visual distortion, lots of laughs
which lead to tears and falling on floor. Totally a fun indica. I place it a little below the NL x shiva in power
level, but still above average. I'd put the Shisk in the social indica category, with Trippystickvaporizerforsale the nlxshiva toward the
unsociable side. I did have to pull myself from the couch also. Differenttypesoftrippystick
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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.
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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.
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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
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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