Evolution__3rd_Edition

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Genetic and environmental
influences on a character can be
defined
A simple Mendelian example
illustrates the terms of quantitative
genetics
frequency of A is one-quarter, the population mean would be 0.71875 cm. For aa, P
now is −0.21875 and for AA and Aa, P =+0.28125. In this example the phenotype is
controlled only by genotype. We can symbolize the effect of the genotype by G. G, like
P, is expressed as a deviation from the population mean. In this case, for an individual
with a particular genotype (because the environment has no effect):
Mean of population + P = mean of population + G
The background population mean cancels from the equation, and can be ignored. We
are then left (in this case in which the environment has no effect) with P = G.
The value of a real character will usually be influenced by the individual’s environment
as well as its genotype. If the character under study is something to do with size,
for example, it will probably be influenced by how much food the individual found
during its development, and how many diseases it has suffered. These environmental
effects are measured in the same way as for the genotype, as a deviation from the population
mean. If an individual grew up in an environment causing it to grow a bigger
beak than average, its environmental effect will be positive; and vice versa if it grew up
in an environment giving it a smaller than average beak. The phenotype can then be
expressed as the sum of environmental (E) and genotypic influences:
P = G + E
CHAPTER 9 / Quantitative Genetics 229
This, simple as it is, is the fundamental model of quantitative genetics. For any phenotypic
character, the individual’s value for that character (expressed, remember, as a
deviation from the population mean) is due to the effect of its genes and environment.
We must look further into the genotypic effect. We need to consider both how to
subdivide the genotypic effect, and why the subdivision is necessary. The main point
can be seen in the one-locus example we have already used. The A gene is dominant,
and both AA and Aa birds have 1 cm beaks (P =+0.125). (Because we are investigating
the genotypic effect, it is simplest to ignore environmental effects, so P = G.) Suppose
we take an AA individual and mate it to another bird drawn at random from the
population. The gene frequency is 1 /2 and the random bird is AA with chance 1 /4,
Aa with chance 1 /2, and aa with chance 1 /4; but whatever the mate’s genotype all the
offspring will have beak phenotype P =+0.125 because A is dominant. Now suppose
we take an Aa individual and mate it to a random member of the population. The
average phenotype P of their offspring is 0. (As can be confirmed by working out
( 1 /4 × (+0.125)) + ( 1 /2 × 0) + ( 1 /4 × (−0.125)) for the three offspring genotypes.) A P of
0 means that the average offspring beak size is the same as the population average.
Thus, for two genotypes with the same beak size (P =+0.125), one produces offspring
with beaks like their parent, the other produces offspring with beaks like the population
average.
So some genotypic effects are inherited by the offspring and some are not. The
next step is to divide the genotypic effect into a component that is passed on and a
component that is not. The component that is passed on is called the additive effect (A)
and the component that is not is called (in this case) the dominance effect (D). The full
genotypic effect in an individual is the sum of the two: