Chapter 22 - Quantitative Genetics

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Quantitative inheritanceQuantitative traits exhibit continuous variation in their phenotypic expression, e.g. height, weight, IQ, or skin color in humans. These traits are considered to have phenotypic expressions that are due to the additive influence of three or more genes that govern the same trait. These complexes of genes are referred to as multiple genes or polygenes (Figure 18.1). Refer back to amonohybrid crossin which the alleles are incompletely dominant (figure 22.1 and 22.2 and table 22.1). e.g. snapdragons (red) RR x rr (white) F_{1}x F_{1}Rr (pink) F_{2}RR Rr rr -- genotype (red) (pink) (white) -- phenotype 1 2 1 -- genotypic ratio 2 1 0 -- units of red colorDihybrid, incomplete dominance(Figure 18.3 and Figure 18.5). (red) AABB x aabb (white) F_{1}AaBb (pink) F_{1}x F_{1}red reddish pink pink whitish pink white 1 4 6 4 1 1 AABB (red, 4 additive alleles) 2 AaBB, 2 AABb (reddish pink, 3 additive alleles) 1 aaBB, 4 AaBb, 1 AAbb (pink, 2 additive alleles) 2 aaBb, 2 Aabb (whitish pink, 1 additive alleles) 1 aabb (white, 0 additive alleles) Genotypic ratios AABB AaBB aaBB AABb AaBb aaBb aabb Aabb AAbb 1 2 1 2 4 2 1 2 13 genes(table 22.1 and figure 22.5) AABBCC x aabbcc F_{1}AaBbCc F_{2}yields a phenotypic ratio of . . . red pink white 1 6 15 20 15 6 1 This produces a curve in which most individuals are pink with a very small number of individuals that are either red or white. With more genes, the curve becomes smoother until it approaches the standard bell shaped curve. Skin color in humans is aquantitative trait. The children (F_{1}) of black-skinned and white-skinned parents have skin color that is intermediate between the parents. By examining the children (F_{2}) from a F_{1}x F_{1}marriage, the F_{2}s are on average similar to the F_{1}s, but with greater variance. Remember that in a 4-hybrid cross, the completely homozygous recessive individual should occur 1 out of (2^{4})^{2}times (Table 2.3). In marriages between individuals who are F_{1}s, the most extreme phenotypes (black or white) occur about 1 in every 256 births. By solving the equation 1 1 ----- = ------ 256 (2^{n})^{2}for n, we can see that n equals 4. Thus, the data are consistent with a model of four loci, each segregating two alleles. At each locus, one allele adds a measure of color whereas the other does not. The additive genetic effects at a locus are what determines that an offspring will, in general, have a phenotypic value intermediate between its parents.Additive Genetic VarianceUltimately, we would like to determine the effect of genes on the expression of traits in a population. This influence will be due to genetics and to the environment. In evaluating the effects of genes, we need to examine such parameters as the heritability of a trait and a population's response to selection. Both of these phenomenon depend upon an understanding of the additive genetic variance and the additive genetic variance depends upon the concept of allele frequencies. As response to selection and heritability are population phenomenon, we need to introduce a little population genetics.Frequencies of alleles in a populationThe phenotypic distribution of flower color among a sample of 300 flowers. Red flowers (RR) = 120 f(RR) = 120/300 = .40 Pink flowers (Rr) = 150 f(Rr) = 150/300 = .50 White flowers (rr)= 30 f(rr) = 30/300 = .10 # of r alleles f(r) = ----------------- total # of alleles (It is easier to think in terms of alleles rather than genotypes.) 2(120) + 150 304 p = f(R) = ------------- = --- = .65 2(300) 400 2(30) + 150 96 q = f(r) = ------------ = --- = .35 2(300) 400 As there are only two alleles in this example, p + q = 1 and q = 1 - p. (See pages 671 - 672)Phenotypic Mean and Variance of a PopulationWhat is the mean flower color of the population above? mean(xbar) = sigma (sum across all classes of) f_{i}*x_{i}where f_{i}is the frequency of the ith class and x_{i}is the phenotypic value of ith class. This equation is derived from equation 22.1. For the population above, where red is assigned a phenotypic value of 1, pink is assigned a value of 0, and white is assigned a value of -1, the equation for the mean phenotypic value becomes mean(xbar) = .40*1 + .50*0 + .10*(-1) = .30, which means that the average color of a flower is reddish-pink, though no flower in the population will have this exact color. Now we need to derive a more general formula for the mean and then one for the variance. In our 1 locus problem, the mean phenotype in the population is a weighted average of the three phenotypic values times the genotypic frequencies. The genotypic frequencies in a population are determined by the allele frequencies. The relationship is called the Hardy-Weinberg relationship, which we will cover more thoroughly later. RR Rr rr -- genotype p^{2}2pq q^{2}--expected genotypic ratio a 0 -a -- phenotypic value where p^{2}is the frequency of the red flowers, 2pq is the frequency of the pink flowers, q^{2}is the frequency of the white flowers, and a is a unit of red color produced by a single additive allele mean (xbar) x = p^{2}(a) + 2pq(0) + q^{2}(-a) remember that p + q = 1 and simplify the equation to mean (xbar) = 2ap - a variance V = sigma f_{i}(x_{i}- xbar)^{2}This equation is derived from equation 18.2. V = p^{2}[a - (2ap - a)]^{2}+ 2pq[0 - (2ap - a)]^{2}+ q^{2}[-a - (2ap - a)]^{2}remember that p + q = 1 and simplify the equation to V = 2pqa^{2}Note that V is directly proportional to p*q, and p*q is a maximum when p = q = .5 We now have an equation to determine the additive genetic variance for a population, if we know the allelic frequencies in the population and the contribution to the phenotype of each additive allele. For the vast majority of cases, the inheritance of traits is too complicated to use this formula to determine the additive genetic variance and we must use other methods. However, it is important to understand that the additive genetic variance is the variance among the phenotypes in a population that is due to the additive effects of the alleles affecting the trait in question.Partitioning the Phenotypic VarianceThe total phenotypic variance can be partitioned (split) into that due to genetics and that due to the environment. V_{P}= V_{G}+ V_{E}+ V_{GE}(equation 22.11) V_{P}= total phenotypic variance in the population V_{G}= contribution to the variance from genetics V_{E}= contribution to the variance from the environment V_{GE}= contribution to the variance from a genotype by environment interaction The genetic variance V_{G}can be further partitioned into the additive genetic variance, the dominance genetic variance, and the epistatic (genic interaction) genetic variance, where V_{G}= V_{a}(additive) + V_{d}(dominance) + V_{e}(epistatic) Thus V_{P}= V_{a}+ V_{d}+ V_{e}+ V_{E}+ V_{GE}(equation 22.13)heritability in the narrow sense= h_{N}^{2}= V_{a}/V_{P}Heritability in the narrow sense represents the proportion of the total phenotypic variance in a population that is due to the additive genetic variance. One of several ways to estimate h_{N}^{2}is to plot the offspring phenotypic value versus the midparent phenotypic value for a large number of families in a population. The slope of the regression line through the data is h_{N}^{2}. (Figure 22.17). Another way to determine h_{N}^{2}is to calculate the correlation for a phenotypic trait bwtween pairs of dizygotic twins and then divide that by the expected correlation coeficient based on the degree of relatedness. That ratio is an estimate of h_{N}^{2}. If we set up careful experiments using regression against other relatives, we can estimate V_{d}and V_{e}. However, these estimates require large sample sizes if we are striving for precision.Response to SelectionPractically any character you could examine or measure of a species would vary within a population. This includes not only morphological traits, but physiological traits (heart rate), behavioral traits (aggression), biochemical traits (resistance to toxins), etc. In addition, many of the characters are inherited in some fashion. That is h_{N}^{2}is positive and there is a positive V_{a}. Because of this, for most characters that have been examined, the mean of a population can be shifted via selection (selective breeding regimens) (figures 22.21 and 22.22). For a character that has no additive genetic variance (V_{a}= 2pqa^{2}= 0), a change due to selection is not possible.Selectioncan only cause a change in a population when V_{a}> 0, and the rate and extent of change is proportional to V_{a}. V_{a}V_{a}h_{N}^{2}= ----------------- = -------- V_{a}+ V_{d}+ V_{e}+ V_{E}V_{P}The response to selection is directly proportional to h_{N}^{2}, where R = h_{N}^{2}* S, where S is called the selection differential and R is the response to selection If we have a population whose mean is Ybar before selection and we select a few individual with one extreme phenotype for breeders for the next generation, then the mean of the offspring (Ybar_{o}) minus Ybar is the response to selection. The mean of the breeders (Ybar_{b}) minus Ybar is the selection differential. R/S equals h_{N}^{2}(equation 22.22).Twin studiesMonozygotic twinsresult from one fertilized egg splitting into two embryos Pages 141 - 145. Monozygotic rates are relatively constant when considering race and maternal age.Dizygotic twinsresult from two eggs being fertilized. They occur because of multiple ovulations and are more common in older women. This may be under some genetic control as it tends to run in families. Also, some ethnic groups have higher rates.Monozygotic twins reared togetherare genetically identical (G) and have the same familial environment (E_{2}) and they only differ on their individual environments (E_{1}). If you plot the phenotypic value of one twin versus the other twin, the correlation between them is reduced from 1.0 due to the effects of their different individual environments (E_{1}).Monozygotic twins reared apartare genetically identical (G), but have different familial and individual environments (E_{1}+ E_{2}). The correlation between monozygotic twins reared apart is due to the common genotype that they share. Thus that correlation is G. The difference between the correlations for monozygotic twins reared apart and monozygotic twins reared together is E_{2}, the familial environment. In one study on intelligence, the correlation between monozygotic twins reared together was 0.87, which means that E_{1}is equal to 0.13. The correlation between monozygotic twins reared apart was 0.69, which means that G is equal to 0.69. The difference between the two correlation coefficients (0.87 - 0.69) is 0.18, which is equal to E_{2}. Thus genetics plays a major role in determining the variance in intelligence in this population.Multiple gene hypothesishas 6 assumptions concerning the behavior of the alleles and their expression. These six assumptions basically summarize the behavior of multiple loci and their effects on a particluar trait. 1) no allelic pairs exhibit dominance of one allele over another 2) each contributing allele in a series has an equal effect 3) the effect of each contributing allele is additive 4) no genetic interaction, or epistasis, occurs among the alleles at different loci in a polygenic series 5) no genetic linkage is exhibited between the genes in a polygenic series, and therefore they assort independently 6) there are no environmental effects The last two assumptions are violated most often. The multiple gene hypothesis forms a satisfactory place from which to start when analyzing a quantitative trait.

Last update on 24 October 2004