Explain the outcomes of crosses involving incomplete dominance, codominance, sex-linked traits, gene interactions and gene linkage; read pedigrees; compare observed data with predictions statistically; and use chromosome maps to identify linkage groups

Learning goal: Explain the outcomes of a series of genetic crosses using Mendelian and non-Mendelian genetics, including incomplete dominance, codominance, sex-linked traits and dihybrid crosses; discuss gene interactions, gene linkage and the sources of phenotypic variation; identify genotypes from pedigrees and from the phenotypes of offspring; use the principles of statistics to compare real-world data with predicted outcomes; and evaluate chromosome maps to identify linkage groups.

Beyond simple dominance. Mendel's peas showed complete dominance, but many traits do not. In incomplete dominance the heterozygote has a phenotype between the two homozygotes: in snapdragons, a red-flowered plant (RR) crossed with a white one (WW) gives all pink (RW) offspring, and pink × pink gives 1 red : 2 pink : 1 white. The alleles have not blended; cross two pinks and the red and white reappear. In codominance both alleles are fully expressed in the heterozygote. Human ABO blood type is the standard example: the alleles I-A and I-B are codominant with each other, so an I-A I-B person has type AB blood with both markers on the red cells, and both are dominant to i, so I-A i is type A and ii is type O. ABO also shows multiple alleles: three alleles exist in the population although any one person carries only two.

Sex-linked traits. In humans, females have two X chromosomes and males have one X and one Y. Genes on the X chromosome are sex-linked, and because a male has only one X, whatever allele it carries is expressed, with no second copy to mask it. Red-green color blindness and hemophilia are X-linked recessive conditions. Write the alleles on the chromosome: X-H for normal clotting and X-h for hemophilia. A carrier mother (X-H X-h) and an unaffected father (X-H Y) have four equally likely children: X-H X-H daughter, X-H X-h carrier daughter, X-H Y unaffected son and X-h Y affected son. So half the sons are expected to be affected, half the daughters to be carriers, and a quarter of all children to be affected. This is why such conditions appear far more often in males and why an affected son usually has an unaffected carrier mother.

Gene interactions. Most phenotypes depend on more than one gene. In epistasis, one gene's alleles alter or hide the effect of another gene. Coat color in Labrador retrievers is the classic case: gene B determines black (B) or chocolate (b) pigment, and a second gene E determines whether pigment is deposited in the hair at all; dogs with ee are yellow regardless of their B alleles. Cross two dogs BbEe × BbEe: of 16 Punnett cells, 9 are B_E_ black, 3 are bbE_ chocolate, and 4 are ee yellow, a 9 : 3 : 4 ratio in place of 9 : 3 : 3 : 1. Polygenic traits such as human height and skin color depend on many genes of small effect and show continuous variation. The environment interacts too: the dark points of a Siamese cat come from a pigment enzyme that works only in the cooler parts of the body, so the same genotype gives a paler cat raised in a warm room.

Pedigrees. A pedigree is a family diagram: squares for males, circles for females, shaded shapes for affected individuals, a horizontal line joining mates and vertical lines to their children. Reading one is hypothesis testing. Assuming a single gene, full penetrance and no new mutation, if two unaffected parents have an affected child the trait is recessive: for an autosomal gene both parents are carriers, and for an X-linked gene the mother is a carrier (a father cannot carry an X-linked recessive allele without showing it). If every affected person has an affected parent, dominance is likely. If affected individuals are mostly male and the trait skips generations through unaffected mothers, X-linked recessive inheritance is likely. With the mode settled, genotypes of individuals can be inferred, at least partly, and probabilities for future children calculated.

Statistics: comparing observed data with predictions. A Punnett square predicts ratios; real crosses produce counts that never match exactly. The chi-square test asks whether the gap between observed and expected counts is small enough to be chance. For each class compute (observed minus expected) squared, divided by expected, and add the results. The test is sound only under conditions: it is run on counts, never on percentages; each offspring is counted once and independently of the others; and the expected count in every class is reasonably large (a common guide is at least five). Worked in the prose: a monohybrid cross yields 100 plants, 68 purple and 32 white, where 3 : 1 predicts 75 and 25. Purple: (68 − 75)² ÷ 75 = 49 ÷ 75 = 0.65. White: (32 − 25)² ÷ 25 = 49 ÷ 25 = 1.96. Chi-square = 2.61. When the expected ratio is fixed in advance by the hypothesis, as it is here, the degrees of freedom are the number of classes minus one, so with two classes there is one degree of freedom (if the expected values had themselves been estimated from the data, further degrees of freedom would be lost, a case for later courses), and the conventional threshold at the 5 percent level is 3.84. Because 2.61 is below 3.84, the deviation is the kind chance produces at least 5 percent of the time, so the data are consistent with the 3 : 1 hypothesis. Had the count been 60 : 40, chi-square would be 3 + 9 = 12, far above the threshold, and the hypothesis would be rejected in favor of some other explanation, perhaps reduced survival of one class.

Gene linkage and chromosome maps. Independent assortment holds for genes on different chromosomes. Genes close together on the same chromosome tend to be inherited together; they are linked, and all the genes on one chromosome form a linkage group. The number of linkage groups in a species equals its number of chromosome pairs: fruit flies have four, garden peas have seven. Linked genes are not inherited together every time, because crossing over between them separates them at a rate that depends on how far apart they lie. In a test cross of a doubly heterozygous parent carrying A and B on one chromosome and a and b on the other with a recessive aabb mate, the parental combinations AB and ab dominate, and the recombinant combinations Ab and aB appear at a frequency set by the distance. Geneticists define one map unit as one percent recombination. Suppose 1,000 offspring give AB 412, ab 408, Ab 92, aB 88: recombinants are 180 of 1,000, so the recombination frequency is 18 percent and the genes are placed about 18 map units apart. A map distance found this way is an estimate from a simple model that counts every recombinant as one crossover; that is close to true for genes near each other, while over longer distances double crossovers go uncounted and the percentage underestimates the true distance. Adding a third gene C with B–C at 7 units and A–C at 25 units fixes the order A, B, C, since 18 + 7 = 25. A chromosome map is that line of genes with the distances marked; two genes on the same map belong to the same linkage group, while a gene whose recombination with every gene on the map is 50 percent, the same as independent assortment, shows no detectable linkage to them: most often it lies on a different chromosome, though a gene very far away on the same chromosome gives the same result, because so many crossovers occur between them that the observed recombination levels off near 50 percent; only mapping genes in between can tell the two apart.

Sources of phenotypic variation, gathered. Offspring vary because meiosis shuffles alleles (independent assortment, crossing over, random fertilization), because many alleles show incomplete dominance, codominance or interaction with other genes, because new alleles arise by mutation, and because the environment shapes how a genotype is expressed. A complete explanation of a population's variation names all of these rather than dominance alone.

Misconceptions. "Incomplete dominance means the alleles blend" is wrong, because the parental phenotypes reappear in the next generation. "Sex-linked means the trait is only in males" is wrong; females can be affected when they inherit two recessive alleles, just less often. "Observed 68 : 32 does not fit 3 : 1" mistakes any deviation for a contradiction; statistics measures whether the deviation is larger than chance allows.

Where the rest of the standards is taught. B.H.3 is shared with the meiosis lesson, which models crossing over and independent assortment. From B.H.4 this lesson teaches evaluating chromosome maps to identify linkage groups; analyzing and classifying mutations is taught in the mutations lesson.

Investigation you can do. Analyze a set of virtual fruit-fly crosses or a published class data set, or generate one: put 60 red beads and 20 white beads in a bag to model a 3 : 1 cross, draw one with replacement 100 times, record the counts, and run the chi-square test against 75 : 25. Repeat with 50 : 30 in the bag and 100 draws, test again against 75 : 25 and see whether the test rejects. A result in which the first sample passes and the second usually fails supports the claim that the test separates chance deviation from a real difference in the underlying ratio. What the model cannot show: beads have no genes, so the model cannot show why a real cross deviates, such as one genotype surviving less well, and a single test at the 5 percent level will wrongly reject a true hypothesis about one time in twenty.

Worked X-linked cross: A woman who is a carrier for red-green color blindness (X-B X-b) has children with a color-blind man (X-b Y). Her gametes: X-B or X-b; his: X-b or Y. Children: X-B X-b daughter, carrier, normal vision; X-b X-b daughter, color-blind; X-B Y son, normal; X-b Y son, color-blind. So half the daughters and half the sons are expected to be color-blind, which shows that females can be affected when the father is affected and the mother carries the allele.

Worked pedigree inference: Two unaffected parents have three children: an affected son, an unaffected daughter and an unaffected son. Assuming one gene, full penetrance and no new mutation, since neither parent is affected but a child is, the trait is recessive and both parents are carriers (Aa × Aa if autosomal). The affected son is aa. Each unaffected child is AA with probability 1/3 or Aa with probability 2/3, because among the three unaffected genotype outcomes of the cross (AA, Aa, Aa) two are carriers. The probability that the next child is affected is 1/4. If the trait were X-linked recessive, the affected son would be X-a Y and the mother a carrier; the pedigree as given cannot distinguish the two modes, and more relatives, especially affected females, would be needed.

Worked chi-square for a dihybrid cross: A cross predicted to give 9 : 3 : 3 : 1 yields 160 offspring: 95, 30, 28, 7. Expected: 90, 30, 30, 10. Terms: (95 − 90)² ÷ 90 = 25 ÷ 90 = 0.28; (30 − 30)² ÷ 30 = 0; (28 − 30)² ÷ 30 = 4 ÷ 30 = 0.13; (7 − 10)² ÷ 10 = 9 ÷ 10 = 0.90. Sum 1.31. With four classes there are three degrees of freedom, and the 5 percent threshold is 7.81. Because 1.31 is well below it, the data are consistent with independent assortment of the two genes.

Worked chromosome map: Three linked genes give recombination frequencies D–E 12 percent, E–F 9 percent and D–F 21 percent. Because 12 + 9 = 21, E lies between D and F, and the map reads D at 0, E at 12 and F at 21 map units. A fourth gene G shows 50 percent recombination with D, E and F alike, which is what independent assortment produces, so G shows no linkage to them; most likely it is on another chromosome, although a gene very far away on the same chromosome, where multiple crossovers push observed recombination up to about 50 percent, would give the same result. If instead D–F had measured 19 percent rather than 21, the shortfall would be explained by double crossovers between D and F, which restore the parental combination and go uncounted.

Supported record: Draw a pedigree of your own invention for a hypothetical family of three generations with an X-linked recessive trait, showing at least one carrier female, one affected male and one unaffected male, and write the genotype under every individual (use a question mark where the pedigree cannot decide). Beside it, draw a chromosome map for the genes D, E, F and G of the worked example and label each distance.

Independent performance: Using a virtual cross simulator, a published data set or the bead model, obtain results for one monohybrid and one dihybrid cross of at least 100 offspring each, and write an analysis. Success criteria: (a) the hypothesis about the mode of inheritance and the predicted ratio are stated before the data are analyzed; (b) the conditions for the test are checked (counts, independent offspring, expected counts of at least about five), the chi-square statistic is computed correctly, the degrees of freedom and the 5 percent threshold are named, and the conclusion is stated as consistent or not consistent with the hypothesis; (c) the analysis explains one non-Mendelian pattern (incomplete dominance, codominance, sex linkage or epistasis) with a worked cross; (d) it explains what a recombination frequency below 50 percent between two genes would mean and how it is turned into a map distance; (e) it lists at least three sources of phenotypic variation in a population; (f) it states what the data cannot decide.