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By Wilmer J. Miller and Willard F. Hollander
BioScience Vol. 45 No 2 Feb. 1995 pp. 98-104
Geneticists now concentrate on the wondrous new molecular techniques. Their promise is being fulfilled in applied as well as theoretical advances. But, while attention is diverted elsewhere, some advances in the classical areas have been neglected. The importance of classical genetics, however, remains undiminished. " it is the simplicity of traditional genetics that has endowed it with such power. Its generalizations do describe most genetic phenomena in most diploid organisms most of the time. Without them, biology could never have progressed to its current understanding of life" (Rennie 1993, p. 132).
Several approaches in classical genetics, alone and in combination, can provide more powerful, consistent, clear, and rapid procedures for analyzing problems. This article focuses on three advances: improved pedigree charting, use of a standard of reference, and calculation of probabilities in complex assortment. Incorporated into the teaching of classical genetics, these methods increase students understanding and problem solving ability.
The diagram of pedigrees and family charts has long followed a right-angle format, with direct connections between parents (figure 1a). An improved method, most likely introduced by Sewall Wright (1921), uses arrows to indicate the genetic contribution of parents to offspring and eliminate repetitions of individuals (figure 1b). J.L. Lush (1945) used it in his popular textbook Animal Breeding Plans. But this method has not been widely applied to genetic problems because freehand or curved lines are difficult to print, and because old habits are hard to break.
The arrow method is needed when calculating coefficients of inbreeding and relationship. Moreover, it clarifies inbreeding complexities and corrects mistaken impressions of the number of generations (emphasized by Hollander 1944, 1972, Miller 1983). It is especially important in analyzing domestic stocks.
While using Figure 1a to represent the partial pedigree of the origin of the Shorthorn breed of cattle, the reader may not immediately note the several repeats of animal names indicating inbreeding. Figure 1b shows the same data reworked in the arrow (network) style in which there are no repetitions. Inbreeding and other relationships are much more readily apparent. In this figure it is obvious that the bull named Favorite has five offspring in this pedigree and that he was mated back to his own mother, Phoenix, and then to his daughter Young Phoenix to produce Comet.
Figure 1. Partial pedigree of the origin of the Shorthorn breed of cattle. . a. In the traditional (bifurcation) style, after Wriedt 1930.. b. Reworked in the arrow or network style, in which there are no repetitions. Dashed lines are used to connect dams and offspring, but solid lines have been used by other authors. Y. Phoenix is Young Phoenix.
Figure 2a shows the pedigree of Racing Homer pigeon in traditional bifurcation format. There seem to be three ancestral generations. In figure 2b, the same data are shown in arrow format, and we see there are actually five ancestral generations and only two great-grandparents.
Figure 2. The pedigree of a Racing Homer pigeon. a. In traditional bifurcation format. b. In arrorw format, which more clearly shows that there are five ancestral generations and only two great-grandparents.
The network pedigree style works well for all traditional types of genetic problems (Miller 1991), and we consider it superior to other styles, especially when inbreeding is involved. Whole families and breeding programs can be charted in larger networks.
Use of a standard of reference
The evolution of genetic symbols has been erratic (Hollander 1953). For some species, committees have established rules, as in Drosophila (Lindsley and Grell 1968). The key advance has been employment of a standard of reference. Using a single phenotype, preferably the modal wild type or normal of the species, other phenotypes are then attributed to single mutants or a combination.
Ordinarily, the wild-type phenotype is more stable than is the phenotype of many mutants. If there is not agreement as to what phenotype is the wild type, as in the horse, a reasonable standard can be chosen. With the use of a standard type there can be economy of description: only phenotype deviations need be mentioned. In turn, the mutant genes responsible for the deviations become the focus for interpretation of gene functions and interactions.
Why did most geneticists not generalize this useful concept? Most textbooks confuse students and teachers because they do not give rules for appropriate gene symbols and do not use gene symbols consistently.
Without using a standard of reference one is likely to confuse nonalleles with alleles and epistasis with dominance. A famous example involves the tortoiseshell cat (Miller and Hollander 1986). Another is exemplified by the black/white plumage colors in chickens (Miller 1985, 1991).
For general gene symbol usage, one should be able to tell at a glance whether the gene is mutant or normal, whether it has some degree of dominance or is recessive, and whether it is allelic to particular other genes. Rules for creating appropriate gene symbols may be stated as follows:
Because consistency helps in learning and using symbols, these rules aid students and promote accurate communication among scientists. There are many historical exceptions to one or more of these rules. A major exception to this usage is in physiological genetics such as blood grouping, in part because a standard type has not been recognized despite thousands of publications. (Human ABO blood grouping is still confusing because of this lack of a standard.) However, often in spite of the lack of a standard type, many genetic combinations of such characters have been clearly identifiable. To further confuse the matter, in immunology the + symbol signifies a reaction of antibodies with antigens.
For species crosses, genetic analyses and gene symbols pose special difficulties (e.g., see Miller and Weber 1969). One species may be arbitrarily chosen as a standard of reference.
Classical testing for allelism of mutants
The classical determination of allelism and avoiding false allelism next needs consideration. We need not only to use a standard type in comparison with mutant or variant alternatives, but we must be sure which are single mutant forms. There may be other complications such as individual mutants being pleiotropic.
Given two or more single mutants, we need to know whether they are independent or allelic. It is essential to know the phenotypes of the single mutants. Figure 3a provides the simplest evidence for identification of four such mutants in rabbit coat colors. We assume for this example that the P1 (parental) stocks are homozygous and that the ratios is the F2 (second filial generation) are reliable.
Agouti is our standard of reference. The chinchilla, himalayan, and yellow mutants act as simple recessives--in crosses with the standard they do not show in the F1 and occur as the 1/4 class in the F2. Black is a dominant mutant-- it shows in the F1 and is the 3/4 class in the F2 generation.
Each of the mutant colors is allelic to agouti. Are they therefore allelic to each other? Figure 3b shows data that test possible allelism of the recessive mutants with each other. Reversion to wild type in the F1 generation is the rule in crosses of nonallelic recessives (such as himalayan x yellow) and certainly is not reverse mutation, because it would require two extremely unlikely events. It is complementation--evidence that the parental types are not allelic.
The appearance of a mutant phenotype, light chinchilla in F1 from chinchilla x himalayan is evidence that these parental mutants are allelic. We can use the symbol c for chinchilla, ch for himalayan, and y for yellow. We could go on to the F2 generation for further evidence.
For dominant mutants, the F1 results are inconclusive. In figure 3c, data are presented for testing allelism of the dominant black and two of the recessive mutants. Black and yellow are found to be allelic mutants, because the F2 results show a monohybrid ratio (single-gene pair F2 segregation) and no wild type. Therefore, if yellow is given the symbol y, black should have the symbol Y.
Black and chinchilla cannot be alleles because the F2 in the second family includes recovery of wild type and exhibits a dihybrid ratio (with epistasis of black over chinchilla). Therefore, genotype of the F1 should be C+c YY+. Had we not known from the data in Figure 3a that yellow and black were single mutants forms in the first F2 family, in Figure 3c, an alternative explanation to allelism would have been that the stocks carried a mutant in common and that mutant was homozygous in each stock.
What are the consequences of not using proven single mutants in testing for allelism? As shown in the breeding data of Figure 3d, from the frequency 3/4:1/4 of the first F2 set, one might conclude that white is recessive to yellow. This conclusion is erroneous, however, because the wild type was not included. In the second F2 set, the proportion is again 3/4:1/4. Because it includes the wild type and only one other phenotype, the conclusion that yellow is recessive to agouti is correct.
The F2 ratio 9:3:3:1 is called a dihybrid ratio. Therefore, white results from the combination effect of the yellow mutant with the chinchilla mutant. In the upper F2 group (figure 3d), the white phenotype was not identifiable as a double mutant. It can best be explained by hypothesizing that the yellow mutant blocks synthesis of the pigment eumelanin and the chinchilla mutant blocks synthesis of the pigment phaeomelanin. Their action together blocks both pigments, leaving the hair unpigmented, or white.
Figure 3. a. Evidence identifying four mutants in rabbit coat colors. Agouti, the wild type, has hairs that are banded black and yellow. Chinchilla has no yellow in the bands. Himalayan has whitening of the fur of the warm parts of the body, yellow has entirely yellow hair, and the black mutant's hair is entirely black. b. Data that test recessive mutants for allelism with each other. c. Testing allelism of the dominant black and two of the recessive mutants. d. The consequences of not using proven single mutants in testing for allelism. From the frequency of the first F2 set, one might conclude erroneously that white is recessive to yellow. In the second F2 set, the conclusion that yellow is recessive to agouti is correct.
Prominent erroneous textbook examples of pigment-gene allelism include: Shorthorn cattlered, roan, and white hair; horses chestnut, palomino, and cremello hair; catsblack, tortoiseshell, and yellow fur; chickensblack blue, and whitish plumage; and chickenswhite, black, and buff plumage. In each of these examples at least two of the mutants are nonallelic according to analyses using suitable standard types as in rabbits: blackish-brown for cattle (Olson 1975), bay for horses and tabby for cats (Searle 1968), and black-and-red = red jungle fowl for chickens (Jaap and Hollander 1954). The superficially confusing question"Are things allelic to the same thing allelic to each other?" was encountered in considering that agouti, the wild-type color in rabbits (and many other animals), is allelic to chinchilla, himalayan, yellow, yellow, and black. But yellow and black are not allelic to chinchilla or himalayan (Searle 1968). The answer to the question is that the identification of mutants implies as standard type that does not describe a single locus or several loci as does the word mutant. Hollander (1959) long ago pointed out the obvious: there is no single gene for wild type. All loci must act normally to produce this standard phenotype. For example, one cannot refer to "the gene" for normal (red) eye color in Drosophila. But a mutant phenotype results from some kind of failure or change at just one or a few loci. At least 50 loci are known to control coat color in mice (Silvers 1979). Single mutants are allelic to each other if they are both allelic to another single mutant.
Mendel was fortunate in not being faced with such questions. Mendel did not even differentiate between mutants and wild type (Hollander 1975). Research after 1900 revealed great complexities.
Table 1. Non-Mendelian complexities:
Pleiotropism Mimics, duplicates Variable dominance Reduced penetrance Lethality Early Late Sterility Sex-linkage Sex-influence Mutable (unstable) Multiple alleles Linkage Epistasis-hypostasis Dominant Recessive Collaboration Additive Multiplicative Novelty Duplicates, silent singly Modifiers
Inheritance of even single mutants is now known to be often far from simple. It may involve such perplexing phenomena as pleiotropic effects, variable expression as affected by environment and modifier genes, reduced penetrance, irregular dominance, lethal effects, and instability. A mutations precise chromosomal basis is seldom known.
In his law of dominance, Mendel did not accommodate different degrees of dominance. As such examples were discovered (Bateson 1913), various new terms were introduced. Within and between textbooks of genetics definitions are inconsistent. Various names have been used: partial dominance, incomplete dominance, codominance, lack or absence of dominance, intermediate dominance, imperfect dominance, egalitarian dominance, and transdominance. The definitions vary from text to text and depend on interpretation of allelic function, although an alleles function is seldom known and often must be assumed. In all these usages there is one consistent aspect; each genotype has a distinguishable phenotype, and the genotype may be inferred from the phenotype. Perhaps none of the terms that have been used are all-inclusive, but some such term is desirable for teaching purposes. We have chosen the term codominance as simplest, shortest, and adequately inclusive. One can still use specialized sub-definitions for well-analyzed cases. In our more that 20 years of teaching this method has worked well.
When we are dealing with more than one mutant there may be further complexities: allelism, linkage, and various interaction effects. The complications that may be encountered regarding mutants and their interrelations are summarized in Table 1.
Examples of some of these complications cannot be found in all areas of geneticssuch as plants, animal, and human. Duplicates called silent singly are known in maize, for example the phenotype called lemon-white (Robertson 1975). But such duplicates have not been found in animals or humans. In additions, environmental variability may confuse some phenotypic classification.
Many of these examples are unlikely to be used in an introductory course unless actual data make them necessary. Teachers tend to present the students with exercise problems that are not too difficult. The students should be taught to use parsimony (often called "Occams razor") in making hypothesesthat is, the simplest hypotheses that is likely to explain the data is preferred. For example, a peculiar F2 ratio involving two mutants may have various hypothetical explanations, such as inadequate numbers, epistasis, linkage, or tampering by the Devil.
Calculation of probabilities
The third procedure is called the arithmetic or product method of combining probabilities. It was suggested by Mendel (1865). Its use in textbooks is occasional and incomplete. Basically this method is simply to multiply the expected frequencies for events at independent loci to get the overall probability or frequency. For many questions it is much faster and simpler than the Punnett square or so-called checkerboard method. Further, one can easily deal with complex mongrel matings, where one may have a combination of backcross, F2, and other matings. It can also deal with both phenotypic and genotypic results. A disadvantage of the arithmetic method for teaching, however, is that unlike the more graphic Punnett square it does not show how gametes combine.
An example of the power of the arithmetic method may be taken from laboratory mouse genetics. Consider the independent mutants designated by cc for albino (colorless), hh for hairloss, PP+ Pintail, PP for very short tail, RR+ for roan, RR for lethal embryo, and ss for short ear. What is the expected frequency of normal progeny from crossing a pintail roan heterozygous for hairloss and albino with a mouse that is hairloss pintail roan short ear and heterozygous for albino? Notice that this example mixes F2 mice with a testcross and with P1 parental types and mixes a recessive with partial dominants and semidominant lethals and phenotypes. We calculate the expectation for each locus.
Figure 4. An example of the use of the arithmetic method in laboratory mouse genetics. The abbreviations are: cc for albino (Colorless), hh for hairloss, PP+ for pintail, PP for very short tail, RR+ for roan, RR for lethal embryo, and ss for short ear. The blank in the symbols C+_ indicates that the second allele can be either C+ or c.
PARENT #1 C+c H+h PP RR+ S+S+
PARENT #2 C+c h h PP+ RR+ s s
NORMAL C+_ H+h P+P+ R+R+ S+s PROGENY
PROBABLILITY 3/4 x 1/2 x 1/4 x 1/3 x 1 = 3/96 = 1/32
Figure 4 gives the problem and the solution. (The blank in the symbols C+_ indicates that the second allele can be either C+ or c but either case yields the same phenotype. This convention is a convenient shorthand symbolization.) Who now would want to use the Punnett square approach?
Linkage complicates the calculations but can be accommodated by the arithmetic method. Consider this simple example in chickens (Figure 5). The dominant mutant controlling pea comb (P) and that controlling blue egg shell (O, the presence of oöporphyrin) are approximately five crossover units apart on an autosome (Bitgood 1985). The dominant mutant rose comb (R) is independent of the other two. What fraction of the female offspring would be normal from the mating of a parent heterozygous for P, O, and R (as shown in Figure 5) and a parent heterozygous only for R?
The probability of getting the wild-type allele at the P locus is 1/2, and the probability of getting it with the wild-type allele for the O locus by crossover is 1/20. The probability of getting the wild-type allele at the R locus from both parents is 1/4. The product of these probabilities is the answer, 1/160.
Figure 5. Accommodating linkage in calculations using the arithmetic method to examine chicken inheritance of the dominant mutant controlling pea comb (P) and that controlling blue egg shell (O, the presence of oöporphyrin).
Integration of the three procedures
Where interaction of mutants seems likely, integration of the above three procedures helps solve the difficulties. These procedures can be used together in a pedigree chart of interacting characters, such as comb types in chickens, feather color in ducks, and bulb color in onions. As teaching tool, such exercise pedigree problems can provide an additional step toward situations encountered in research. Currently such problems are found in only one text (Miller 1991).
Figure 6 illustrates a problem of intermediate difficulty concerning coat colors of the guinea pig, Cavia. The three advances in classical genetics allow a logical, straightforward solution. For simplicity, we exclude sex linkage, multiple allelism, and linkage, and the data are assumed to be statistically reliable. The wild type is agouti coloration, the hairs being banded black and yellow. The student should identify the single mutant phenotypes and fill in the missing phenotypes, genotypes, and frequencies.
Normal Wild Type = Agouti (hairs banded black and yellow) Identify the mutant genes and fill in the missing genotypes, phenotypes, and frequencies. Use the shorthand symbolization (see second procedure) appropriately in the last generation and show how frequencies are obtained. Sex-linkage and multiple alleles are excluded in this problem to obviate the need for additional crosses.
The problems solution does not necessarily involve a rigid series of steps. We offer a reasonable analysis of the problem and finished answer sheet (Figure 7).
Examine the layout of the information given. Does any cross between purebred P1 give more than one phenotype in F1 ? If so, despite purebred stocks, heterozygosity, sex-linkage, or some unusual condition would be implied. In this example, all the P1 types to be homozygous.
Next note where the standard of reference (wild type, agouti, normal) occurs. Does any nonstandard P1 stock produce only standard type F1 in a particular cross? If so, that P1 stock, cannot have any dominant or codominant mutants. That is the case here for all the nonstandard P1 stocks, so we can conclude that all the mutants are recessives.
Does any cross of two nonstandard P1 stocks fail to give standard F1? If so, those two P1 stocks must have at least one recessive mutant in common. We see this situation occurring for chocolate x tricolor, chocolate x cinnamon, and chocolate x pink-eyed lavender but not for albino x pink-eyed lavender, although these last two both have pink eyes.
The first F2 family is from F1 that must be heterozygous in at least two loci (dihybrid). We expect the F2 ratio to be 9/16 agouti, 3/16 yellow, 3/16 black, 1/16 yellow-black combination. To account for the given 1/4 yellow, the simplest hypothesis is that the yellow mutant is epistatic in the combination. If we choose the gene symbols b for the black mutant and y for the yellow mutant, the to fill in the genotypes, and calculate the probability for agouti, we have B+_Y+_ = 3/4 x 3/4 = 9/16.
Let us hypothesize that the black F1 from chocolate x pink-eyed lavender is showing the same mutant b as in the black P1 stock. Then the F2 family from this F1 was heterozygous for two other mutants. We should expect the F2 ratio to be 9/16 black, 3/16 chocolate, 3/16 pink-eyed lavender, and 1/16 combination of chocolate with pink-eyed lavender. The only gene difference between pink-eyed lavender and black is a lightening or diluting change. The pink-eyed tan given seems a reasonable phenotype for the combination of this diluting effect with the gene difference of black and chocolate. Using as tentative symbols p for pink-eyed-dilute and c for chocolate, we can fill in genotypes and probabilities
(for example: b b C+ _ P+_ = 1 x 3/4 x 3/4 = 9/16 black).
Chocolate x cinnamon gave cinnamon F1. Because cinnamon differs from standard in having brown instead of black bands in the hair, and chocolate is all brown, the mutant in common most probably is c, and cinnamon is probably the single mutant effect.
Chocolate x tricolor gave black F1. Assuming that this black is homozygous for b, then tricolor must have this mutant but not c. Tricolor also gave black F1 in the cross with pink-eyed lavender, so it cannot have p. Tricolor x albino gave agouti F1, so albino cannot have b, and we already deduced that albino does not have p. There for, we can use the symbol a for albino.
The backcross of the agouti F1 from tricolor x albino to tricolor is stated to give 1/8 tortoiseshell, which suggests a trihybrid phenotypic segregation. We would not expect albino to appear; therefore, the agouti F1 should be heterozygous for three more mutants from the tricolor P1. Because no yellows appeared, the yellow patches in tortoiseshell are apparently not produced by y. The only difference between tortoiseshell and wild type is that yellow replaces the black banding in patches. This simple difference seems likely to be a single-mutant effect, so let us use t for tortoiseshell.
The white patches in tricolor seem to be the effect of still another mutant. Let us use s (for spots). We have already used the symbols p (for pink-eyed dilute) and a for albino, and so we should not use them for white patches. Then we can fill in the genotypes and probabilities expected, for example:
A+ _ B+b S+ s T+t = 1 x 1/2 x 1/2 x 1/2 = 1/8 agouti.
The remaining second-generation family is a cross of the black F1 from chocolate x pink-eyed lavender with the agouti F1 from cinnamon x pink-eyed lavender. The frequencies given are multiples of 1/32 (1/32, 3/32, ), so we can expect at least loci to segregating. Because we have already worked out genotypes for the P1 stocks, we can work out this family. The 3/32 pink-eyed-dilute fits the single-mutant phenotype for the p gene.
By listing the mutant genes and the actions deduced from their single affects, one should be able to predict new phenotypes. These include bb ss, which is black with spots (piebald), and bb pp tt, which is lavender with yellow patches.
Identify the mutant genes and fill in the missing genotypes, phenotypes, and frequencies. Use the shorthand symbolization (see second procedure) appropriately in the last generation and show how frequencies are obtained. Sex-linkage and multiple alleles arc excluded in this problem to obviate the need for additional crosses.
Single Mutants and Their Effects
tt - Tortoiseshell = yellow patches
Interaction Effects of Mutants
pp cc bb - pink-eyed tan
Figure 7. The solution of the problem shown in Figure 6. The terms with all letters capitalized represent the problem as presented to a class. The genotypes and the terms in lower case letters represent the correct solution.
Three important but widely unappreciated approaches to classical genetics are the arrow pedigree format, use of a standard of reference, and the arithmetic method of combining probabilities. A combination of these methods gives clear, powerful, consistent, rapid, and logical handling of classical genetics results. Their synergy can provide an impressive step forward for genetics teaching and research.
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Wilmer J. Miller and Willard F. Hollander are professors emeritus in Department of Zoology and Genetics, Iowa State University, Ames, Iowa 50011-3223. ? 1995 American Institute of Biological Sciences.
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