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1
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- Change of allele abundance/frequency is the basic evolutionary change.
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2
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- Chapter 24 Evolutionary Processes is the reading assigned for this
lecture.
- Frequency is used in the sense of a proportion, frequency takes values
between 0 and 1.
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3
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- The increase in abundance is a true measure of fitness.
- Most of population models deal with frequency, ranging from 0 to 1, of
two or more types, i.e., relative abundance.
- We will continue in that tradition.
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4
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5
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6
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- If the A locus has classic dominance, there are only two phenotypes,
which we can symbolize as A_ and aa.
- The frequency of the A_ (the dominant phenotype) is 0.24 + 0.41 = 0.65;
the frequency of the homozygous recessive phenotype is the same as the
homozygous recessive genotype = 0.35
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7
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8
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9
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- Random mating means that mating pairs are drawn independently from the
available pool of individuals.
- Independent means
- Pr(AA x aa mating) = Pr(AA)•Pr(aa), etc.
- With some algebra I will not show, it can be shown that random mating is
equal to a random pick of alleles from a gene pool.
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10
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11
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- Expected genotype frequencies can be predicted from allele frequencies.
- e.g. freq(IB i) = 2•freq(IB)•freq(i)
- If genotype frequencies can be predicted then phenotype frequencies are
also predicted.
- e.g. freq(B blood type) = 2•freq(IB)•freq(i)
- + freq(IB)• freq(IB)
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12
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- Given freq(IA) = 0.3, freq(IB) = 0.1 so freq(i) =
1 -0.3 -0.1 = 0.6 (as they must add to 1)
- THEN the expected frequency of the IBi genotype (using
Hardy-Weinberg) is 2•.1•.6 = 0.12
- The expected frequency of the B phenotype is expected freq(IBi)
+ freq(IBIB) = 2•.1•.6 + .1•.1 = 0.13 or 13%
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13
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- Knowing the allele frequencies in the current generation and that mating
is random with respect to the genotype of interest allows one to
calculate the expected genotype frequencies in the next generation
(Hardy-Weinberg).
- The observed frequencies will rarely equal the expected frequencies.
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14
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- If you are told that p = 0.6 (and therefore q must equal 0.4) and you
observe the following numbers of AA, Aa & aa, respectively 30, 40,
10
- You note that 80 individuals were observed and calculate the expected
numbers as N•p2, … = 28.8, 38.4, 12.8
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15
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- Χ2 = 1.22/28.8
+ 1.62/38.4 + 2.82/12.8
- Χ2 = 0.05 +
0.067 + 0.6125
- Χ2 = 0.73
- Χ2 is less than
the critical value, 5.99, with 2 degrees of freedom so hypothesis is
accepted.
- The degrees of freedom (df) in this problem are 2, because you were
given the gene frequencies and the observed numbers. Normally one has to
estimate allele frequency from data. If that is done there is one less
df.
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16
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17
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18
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- Species are typically divided into many populations which are clumped in
space.
- Allele frequencies are likely to be different among populations of a
species.
- Movement of individuals between populations is known as gene flow (also
called migration.)
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19
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20
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- If an allele increases in frequency no matter what its initial
frequency, it is considered to be directionally selected.
- The most clear examples of directional selection are examples of DDT
resistance and antibiotic resistance.
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21
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- Directional selection at phenotypic level results when survival
increases (or decreases) in increasing phenotypic value.
- Stabilizing selection is the pattern where the middle phenotypic values
have the greatest survival and the extremes have the least.
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22
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- The calculation of expected genotype frequencies given random mating
assumes that p does not change with time. That is not possible with
finite populations.
- All real populations have a finite number of members, and even if mating
is independent of genotype, the actual numbers change slightly from the
expected due to sampling and the population size, N, becomes an
important variable.
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23
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- Allele frequency
- Genotype frequency
- Phenotype frequency
- Random mating
- Expected genotype frequency
- Hardy-Weinberg
- Directional selection
- Stabilizing selection
- Finite
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