Natural Selection on semi- & incompletely dominant phenotypes with Additive & Genic fitness
In classical genetics, if the phenotype of the AB genotype
is intermediate between AA & BB, but
closer to that of the AA than the AB genotype,
A is described as incompletely
dominant to B. If the AB phenotype
is precisely intermediate between that of the two homozygous
genotypes, A and B are described as semi-dominant.
With respect to fitness phenotypes, semi-dominance occurs for example when genotypes AA, AB, & BB are assigned fitness values of W0 = 0.2, W1 = 0.4, and W2 = 1.0 : with the notation of selection coefficients, the fitness values would be written as WAA = (1 - 2s), WAB = (1 - s), and WBB = (1), where s = 0.4. That is, each A allele contributes an additive selective disadvantage of s = 0.4, so that an AA homozygote is at twice the disadvantage of the AB heterozygote.
In the table below for Additive fitness, let initial q = f(B) = 0.001. Note that if s > 0.5, the additive fitness of AA homozygotes W2 < 0 and therefore undefined. At s = 0.5, the alternative B allele is semi-lethal, and for s < 0.5, fitness is positive though initially low.
Note once again that A and B are semi-dominant, not because one has superior fitness (and might be said to "dominate" the other allele), but because the AB phenotype is intermediate between that of the AA and BB. Genetic dominance is a genotypic, not a phenotypic, relationship.
Compare this model with that for Genic (Multiplicative) fitness. Again, let initial q = f(B) = 0.001. Using the notation of selection coefficients with s = 0.4 as above, WBB = (1), WAB = (1 - s), and WAA = (1 - s)(1 - s) = (1 - s)2 , so W2 = 1.0 , W1 = 0.4, and W0 = 0.36. That is, each A allele reduces fitness by a factor of (1 - s). The fitness effect of a single allele is (1 - s) in either model. However, the two models make very different predictions about dq over the range 0.1 ~ s ~ 0.5. At smaller values of s, the expected difference between models becomes negligible and too small to be measured. This is because genic fitness (1 - s)2 = 1 - 2s + s2 ~ (1 - 2s) as in additive fitness, when s2 << 2s or s << 2.
Simple additive dominance may be typical at many gene loci, where the phenotype is a consequence of equal expression by both alleles. For example, each allele at a protein-coding locus may contribute half the total amount of gene product. This might explain so-called "null alleles", in which one allele has entirely lost its function. One functional allele that produces 50% of the expected gene product may (or may not) be sufficient for standard phenotypic expression. Incomplete genic dominance may be typical at other gene loci, where the phenotype is (much) more strongly influenced by one allele than the other. For example, given a null allele that produces no gene product, the standard allele may be "up-regulated" so that the amount of gene product in the AB heterozygote is (much) closer to that of the AA homozygote.
It remains a major point of contention what fraction of heterozygous allelic variation detected originally by protein electrophoresis and (or) nowadays by DNA sequencing has any measurable effect on the observed phenotype relative to that of the homozygotes, as is clear from the math above. The so-called "Neutralist - Selectionist" controversy will be discussed elsewhere in the course.
HOMEWORK:
(1) For an initial f(A) = 0.01 and s≼
0.5
(2) For an initial f(A) = 0.01 and s = 1/10 of the values below [i.e., shift the decimal in s one place to the right for s≼ 0.5
(3) How do you interpret the values obtained for s > 0.5.
With respect to fitness phenotypes, semi-dominance occurs for example when genotypes AA, AB, & BB are assigned fitness values of W0 = 0.2, W1 = 0.4, and W2 = 1.0 : with the notation of selection coefficients, the fitness values would be written as WAA = (1 - 2s), WAB = (1 - s), and WBB = (1), where s = 0.4. That is, each A allele contributes an additive selective disadvantage of s = 0.4, so that an AA homozygote is at twice the disadvantage of the AB heterozygote.
In the table below for Additive fitness, let initial q = f(B) = 0.001. Note that if s > 0.5, the additive fitness of AA homozygotes W2 < 0 and therefore undefined. At s = 0.5, the alternative B allele is semi-lethal, and for s < 0.5, fitness is positive though initially low.
Note once again that A and B are semi-dominant, not because one has superior fitness (and might be said to "dominate" the other allele), but because the AB phenotype is intermediate between that of the AA and BB. Genetic dominance is a genotypic, not a phenotypic, relationship.
Compare this model with that for Genic (Multiplicative) fitness. Again, let initial q = f(B) = 0.001. Using the notation of selection coefficients with s = 0.4 as above, WBB = (1), WAB = (1 - s), and WAA = (1 - s)(1 - s) = (1 - s)2 , so W2 = 1.0 , W1 = 0.4, and W0 = 0.36. That is, each A allele reduces fitness by a factor of (1 - s). The fitness effect of a single allele is (1 - s) in either model. However, the two models make very different predictions about dq over the range 0.1 ~ s ~ 0.5. At smaller values of s, the expected difference between models becomes negligible and too small to be measured. This is because genic fitness (1 - s)2 = 1 - 2s + s2 ~ (1 - 2s) as in additive fitness, when s2 << 2s or s << 2.
Simple additive dominance may be typical at many gene loci, where the phenotype is a consequence of equal expression by both alleles. For example, each allele at a protein-coding locus may contribute half the total amount of gene product. This might explain so-called "null alleles", in which one allele has entirely lost its function. One functional allele that produces 50% of the expected gene product may (or may not) be sufficient for standard phenotypic expression. Incomplete genic dominance may be typical at other gene loci, where the phenotype is (much) more strongly influenced by one allele than the other. For example, given a null allele that produces no gene product, the standard allele may be "up-regulated" so that the amount of gene product in the AB heterozygote is (much) closer to that of the AA homozygote.
It remains a major point of contention what fraction of heterozygous allelic variation detected originally by protein electrophoresis and (or) nowadays by DNA sequencing has any measurable effect on the observed phenotype relative to that of the homozygotes, as is clear from the math above. The so-called "Neutralist - Selectionist" controversy will be discussed elsewhere in the course.
HOMEWORK:
(1) For an initial f(A) = 0.01 and s
(2) For an initial f(A) = 0.01 and s = 1/10 of the values below [i.e., shift the decimal in s one place to the right for s
(3) How do you interpret the values obtained for s > 0.5.
[Note: in this table only, the left-to-right order of W0 W1 W2 is reversed]
Table & text material © 2024 by Steven M. Carr