Mendelian Transmission and Probability

Every genetics problem you will ever solve, from predicting the coat color of a puppy to estimating a couple's risk of having a child with cystic fibrosis, rests on a handful of rules worked out by a monk counting peas in a monastery garden. Those rules turn inheritance from a mystery into arithmetic: genes come in pairs, the pairs separate when gametes form, different pairs sort independently, and chance decides which gamete meets which. This chapter teaches those rules, the probability tools that make them fast, the chi square test that tells you whether real data fit them, and the pedigree reasoning that lets you apply them to human families.

Section 1Mendel's approach and monohybrid crosses

Why peas, and why Mendel succeeded

Before Mendel, most naturalists believed in blending inheritance: the traits of two parents mixed in their offspring like two colors of paint, so that variation should steadily disappear from a population. Breeders knew that traits sometimes skipped a generation, but no one had a model that could predict how often. Gregor Mendel, working in the 1850s and 1860s and publishing in 1866, succeeded where others failed for reasons of experimental design as much as insight.

The garden pea, Pisum sativum, was an excellent choice. It is cheap, grows quickly, and produces many seeds per plant, so large numbers of offspring can be counted. Its flowers normally self fertilize, because the pollen bearing anthers and the egg containing ovary are enclosed together within the petals, yet a researcher can easily perform cross fertilization by removing the anthers of one flower before they mature and dusting its stigma with pollen from another plant. Commercial seed suppliers already offered many varieties that differed in clear, either or traits.

Mendel's method had four features that remain the backbone of genetic analysis:

• He began with true breeding (pure) lines, plants that produced only one form of a trait generation after generation when allowed to self. This guaranteed that the parents were homozygous.
• He studied traits with two clearly distinct forms, such as round versus wrinkled seeds, rather than continuously varying traits like overall plant weight.
• He followed one or two traits at a time through several generations, keeping the generations separate: the parental generation (P), the first filial generation (F₁) and the second filial generation (F₂), produced by selfing or intercrossing the F₁.
• He counted. Rather than describing offspring as "mostly round," he recorded thousands of individuals and looked for numerical ratios. This quantitative habit, unusual in biology at the time, is what let him see the regularities.

He also performed reciprocal crosses, in which the trait carried by the male parent in one cross is carried by the female parent in the other (round female by wrinkled male, and wrinkled female by round male). The results were the same in both directions, which told him that both parents contribute equally to the offspring for these traits. Reciprocal crosses remain a standard first test: when they give different results, the trait may be sex linked or inherited through the cytoplasm, topics we take up in Chapter 3.

The monohybrid cross and its results

A monohybrid cross follows a single trait. When Mendel crossed a true breeding round seeded plant with a true breeding wrinkled seeded plant, every F₁ seed was round. The wrinkled form had not blended with round; it had vanished. When he let the F₁ plants self fertilize, wrinkled seeds reappeared in the F₂, about one quarter of the total. The table above shows that all seven traits behaved the same way: the F₁ showed only one form, which Mendel called dominant, and the F₂ showed dominant and recessive forms in a ratio close to 3:1.

Mendel reasoned that each plant carries two hereditary "factors" for each trait, that the factors remain discrete rather than blending, and that a plant passes only one of its two factors to each gamete. Today we call his factors genes, and the alternative forms of a gene alleles. The F₁ round plants carried one allele for round (R) and one for wrinkled (r); because R is dominant, they looked round. When these plants made gametes, half carried R and half carried r. Random union of gametes then gives the F₂ genotypes in the proportions 1/4 RR, 1/2 Rr and 1/4 rr.

The vocabulary you will use in every chapter

The genotype is the set of alleles an individual carries; the phenotype is the observable trait, which may be a visible feature, a biochemical property or a disease. An individual with two identical alleles (RR or rr) is homozygous; one with two different alleles (Rr) is heterozygous. The physical position of a gene on a chromosome is its locus (plural loci). By convention the dominant allele is written with an uppercase letter and the recessive with the same letter in lowercase, and gene symbols are italicized. A blank or underscore, as in R_, means "either allele"; it is a compact way of writing "has at least one dominant allele and therefore shows the dominant phenotype."

Dominance describes the relationship between alleles at one locus as seen in the phenotype of the heterozygote. It does not mean that a dominant allele is more common, more beneficial, or that it suppresses the recessive allele. In the case of seed shape, the r allele carries an insertion that disrupts a starch branching enzyme. Seeds with at least one working copy make enough branched starch to stay plump; rr seeds accumulate sugar, take up water during development and wrinkle as they dry. Many recessive alleles are, like this one, loss of function alleles, and the heterozygote is normal because one working copy makes enough product.

The principle of segregation and its chromosomal basis

Mendel's first law, the principle of segregation, states that each individual carries two alleles for a gene, and that these two alleles separate (segregate) when gametes form, so each gamete receives only one allele, with equal probability. Mendel had no knowledge of chromosomes. When cytologists described meiosis around 1900, Walter Sutton and Theodor Boveri noticed that chromosomes behave exactly like Mendel's factors: they come in homologous pairs, one from each parent, and the members of each pair separate into different cells in anaphase I of meiosis. This chromosome theory of inheritance placed genes on chromosomes, and segregation became the visible separation of homologous chromosomes. Because each gamete receives one homolog, a heterozygote produces two types of gametes in equal numbers.

Testcrosses and backcrosses

A plant with the dominant phenotype may be homozygous or heterozygous. To find out, cross it with a homozygous recessive individual, a testcross. The recessive tester contributes only recessive alleles, so the phenotype of each offspring directly reveals which allele the unknown parent contributed. If the unknown is RR, all offspring are round; if it is Rr, round and wrinkled appear in a 1:1 ratio. A backcross is any cross of an F₁ individual with one of its parents (or a genotype identical to a parent); a testcross is the special backcross to the recessive parent. Mendel used testcrosses to confirm his model and obtained the predicted 1:1 ratios.

Ratios to know by heart

The table below collects the monohybrid crosses you will meet most often. Recognizing a ratio in data is often the fastest way to infer the parental genotypes.

Two features of the F₂ deserve emphasis. First, among the dominant phenotype F₂ offspring, the ratio of AA to Aa is 1:2, so two thirds of them are heterozygous. Mendel confirmed this by selfing round F₂ plants: about one third bred true and two thirds again produced wrinkled offspring. This "two thirds" fraction returns in human pedigree problems in Section 5. Second, ratios are expectations about probability, not guarantees. A family of four children from two heterozygous parents may easily have no affected children or two of them; ratios emerge only in large samples, which is why we need the statistical tools of Section 4.

Section 2Dihybrid crosses and independent assortment

Following two traits at once

Having established segregation, Mendel asked whether different traits are inherited together or separately. He crossed a true breeding plant with round, yellow seeds (RRYY) with one that had wrinkled, green seeds (rryy). The F₁ were all round and yellow (RrYy), heterozygous at two loci, so they are called dihybrids. When the F₁ self fertilized, the F₂ contained four phenotypes, two like the parents and two new combinations:

• 315 round, yellow
• 108 round, green
• 101 wrinkled, yellow
• 32 wrinkled, green

That is 556 seeds in a ratio close to 9:3:3:1. The two new combinations, round green and wrinkled yellow, are called recombinant (or nonparental) phenotypes; the other two are parental phenotypes. Notice that if you look at each trait alone, you still see 3:1. Round to wrinkled is (315 + 108) : (101 + 32) = 423 : 133, which is 3.18:1, and yellow to green is (315 + 101) : (108 + 32) = 416 : 140, which is 2.97:1. The 9:3:3:1 ratio is simply two independent 3:1 ratios multiplied together: (3 round : 1 wrinkled) × (3 yellow : 1 green) = 9 round yellow : 3 round green : 3 wrinkled yellow : 1 wrinkled green.

The principle of independent assortment

Mendel's second law, the principle of independent assortment, states that alleles at different loci segregate independently of one another when gametes form. A dihybrid RrYy therefore produces four kinds of gametes, RY, Ry, rY and ry, each with probability 1/2 × 1/2 = 1/4. Combining four kinds of eggs with four kinds of pollen gives a 4 × 4 Punnett square of 16 equally likely cells, containing 9 genotypes and 4 phenotypes.

The chromosomal basis is the random orientation of homologous pairs at metaphase I. If the seed shape and seed color genes lie on different chromosomes, the way one pair lines up at the metaphase plate has no influence on the way the other pair lines up. Half of all meioses place R and Y on the same side, producing RY and ry gametes; the other half place R with y, producing Ry and rY. Averaged over many meioses, all four gamete types are equally common.

Independent assortment holds for genes on different chromosomes and for genes far apart on the same chromosome. Genes that are close together on one chromosome are linked: they tend to travel together, and the dihybrid produces more parental than recombinant gametes. Mendel was fortunate, or perhaps selective, in that the traits he reported in his dihybrid and trihybrid experiments assorted independently. Linkage and its use in mapping are the subject of Chapter 4.

The dihybrid testcross

Crossing a dihybrid with a double recessive (RrYy × rryy) shows independent assortment even more directly. The tester contributes only ry gametes, so each offspring's phenotype reveals the gamete it received from the dihybrid. The four phenotypes appear in a 1:1:1:1 ratio, reflecting the four equally frequent gamete types. When Mendel performed this cross he obtained close to equal numbers of all four classes. In later chapters, a departure of the testcross from 1:1:1:1, with an excess of the parental classes, is the signature of linkage.

Branch diagrams instead of big Punnett squares

A 4 × 4 Punnett square is manageable, but a trihybrid cross needs an 8 × 8 square of 64 cells, and four genes need 256 cells. A branch diagram (also called a forked line method) is faster. Because the genes assort independently, you can work out the outcome for each gene separately and then combine. For the phenotypes of an AaBb × AaBb cross, first split into 3/4 A_ and 1/4 aa; then split each of those into 3/4 B_ and 1/4 bb; multiply along each path. Figure 1.3 in the next section shows exactly this tree. The same approach works for genotypes, with three branches per gene (1/4, 1/2, 1/4) instead of two.

For a trihybrid intercross, AaBbCc × AaBbCc, the phenotypic ratio is (3:1)³ = 27:9:9:9:3:3:3:1. The 27 class is A_B_C_ (3/4 × 3/4 × 3/4 = 27/64); each 9 class is dominant at two loci and recessive at one (3/4 × 3/4 × 1/4 = 9/64); each 3 class is dominant at one locus (3/4 × 1/4 × 1/4 = 3/64); and 1/64 are recessive at all three.

Mixed crosses

Real problems rarely involve two identical heterozygotes. The branch approach works for any combination. In AaBbcc × AabbCc, take each gene separately: Aa × Aa gives 3/4 A_; Bb × bb gives 1/2 Bb and 1/2 bb; cc × Cc gives 1/2 Cc and 1/2 cc. The probability of an offspring showing the dominant phenotype for all three traits is 3/4 × 1/2 × 1/2 = 3/16. The probability of a genotype that matches neither parent can be found by subtraction: P(exactly AaBbcc, like parent 1) = 1/2 (Aa) × 1/2 (Bb) × 1/2 (cc) = 1/8, P(exactly AabbCc, like parent 2) = 1/2 × 1/2 × 1/2 = 1/8, so P(neither parental genotype) = 1 − 1/8 − 1/8 = 3/4.

Section 3Probability rules for genetics problems

Why probability is the language of inheritance

Segregation and fertilization are random events, so every prediction in transmission genetics is a probability. A probability is the expected frequency of an event over many trials, written as a number from 0 (never happens) to 1 (always happens). The probability that a heterozygote Aa transmits a is 1/2, just as the probability that a fair coin lands heads is 1/2. A handful of rules let you solve almost any transmission problem quickly and without drawing Punnett squares.

The product rule

The product rule (multiplication rule) states that the probability of two or more independent events all occurring together is the product of their separate probabilities. Events are independent when the outcome of one does not change the probability of the other. Separate gametes, separate children and separate unlinked genes are all independent. The key word in a question is usually "and": the probability that a child is a girl and has cystic fibrosis, or that one child is affected and the next child is affected.

If two carriers of a recessive disease (Aa × Aa) have a child, the probability the child is aa is the probability that the egg carries a (1/2) times the probability that the sperm carries a (1/2), or 1/4. The probability that their first two children are both affected is 1/4 × 1/4 = 1/16, and the probability that their first child is an affected boy is 1/4 × 1/2 = 1/8.

The sum rule

The sum rule (addition rule) states that the probability of any one of two or more mutually exclusive events occurring is the sum of their separate probabilities. Events are mutually exclusive when only one can happen in a given trial; a single child cannot be both AA and Aa. The key word is "or." In Aa × Aa, the probability that a child has the dominant phenotype is P(AA) + P(Aa) = 1/4 + 1/2 = 3/4. The phenotypic 3/4 you have been using is itself a sum.

Many problems require both rules. In a family of two children from carrier parents, what is the probability that exactly one child is affected? There are two mutually exclusive orders: affected then unaffected (1/4 × 3/4 = 3/16) or unaffected then affected (3/4 × 1/4 = 3/16). The answer is 3/16 + 3/16 = 6/16 = 3/8.

Conditional probability

Sometimes we already know something about an individual, and that knowledge changes the relevant probabilities. A conditional probability is the probability of an event given that another event is known to have occurred, written P(A | B). It is calculated by restricting attention to the outcomes consistent with what we know:

P(A | B) = P(A and B) / P(B)

The most important example in genetics is the unaffected sibling of a person with a recessive disorder. The parents must both be carriers (Aa × Aa). Before anything is known, a child has probability 1/4 of being AA, 1/2 of being Aa and 1/4 of being aa. But we are told the sibling is unaffected, which eliminates aa. Among the remaining 3/4, the chance of being a carrier is (1/2) / (3/4) = 2/3, and the chance of being AA is (1/4) / (3/4) = 1/3. Students often answer 1/2 by reflex; the 2/3 answer is correct because the information "unaffected" has been used.

The binomial expansion

When a question asks for a specific number of outcomes among several independent trials, without specifying the order, use the binomial expansion. If p is the probability of one outcome and q = 1 − p the probability of the other, the probability of exactly s outcomes of the first kind and t of the second in n = s + t trials is

P = [n! / (s! t!)] p^s q^t

The factor n! / (s! t!) counts the number of different orders in which the outcomes can occur. (Remember that 0! = 1.) For small n, you can also read the coefficients from Pascal's triangle: for n = 4 they are 1, 4, 6, 4, 1.

"At least one" problems

Questions that ask for "at least one" are easiest by subtraction. The probability of at least one affected child among four from carrier parents is 1 minus the probability that none is affected: 1 − (3/4)⁴ = 1 − 81/256 = 175/256, about 0.68. Doing this by addition would require summing the probabilities of exactly 1, 2, 3 and 4 affected children.

Section 4Chi square testing of genetic ratios

The problem of chance deviation

Real data never match expected ratios exactly. Mendel's 5,474 round and 1,850 wrinkled seeds were not exactly 3:1, and a student who counts 72 dominant and 48 recessive offspring from a supposed testcross must decide whether 72:48 is "close enough" to 1:1. The deviation could be due to chance alone (sampling error), or it could mean the hypothesis is wrong, perhaps because the parent is not the genotype assumed, the genes are linked, or one class has lower survival. The chi square (χ²) goodness of fit test is the standard tool for deciding.

Hypotheses and the logic of the test

The test begins with a null hypothesis (H₀): there is no real difference between the observed and expected numbers, and any deviation is due to chance. In genetics, the null hypothesis is usually a specific ratio, such as "the offspring segregate 3:1." The test asks: if H₀ were true, how often would chance alone produce a deviation at least as large as the one observed? That probability is the P value. If it is small, conventionally less than 0.05, we reject the null hypothesis. If it is larger, we fail to reject it: the data are consistent with the hypothesis, though that does not prove the hypothesis true.

Calculating chi square

The statistic is

χ² = Σ (O − E)² / E

where O is the observed number in each class, E the expected number, and the sum runs over all phenotypic classes. Each term squares the deviation (so positive and negative deviations do not cancel) and scales it by the expected number (so a deviation of 10 matters more when 20 were expected than when 2,000 were). The steps are always the same:

1. State the null hypothesis and the ratio it predicts.
2. Find the total number of offspring, and calculate the expected number in each class by multiplying the total by the expected fraction. Expected numbers may be decimals; do not round them to whole numbers.
3. For each class compute (O − E)² / E, then add the terms.
4. Determine the degrees of freedom (df). For a genetic ratio test, df = number of phenotypic classes − 1. Once the total and all but one class are fixed, the last class is determined, so it is not free to vary.
5. Compare the χ² value with the critical value for that df at the chosen significance level (usually 0.05) in a chi square table, and state the conclusion.

Read the table this way: with 3 degrees of freedom, a chi square of 7.815 or more would arise by chance only 5 percent of the time if the null hypothesis were true. Larger chi square values mean larger deviations and smaller P values.

Rules and pitfalls

• Use counts, never percentages or ratios. The chi square value depends on sample size. 60:40 out of 100 gives χ² = 4.0, but 6:4 out of 10 gives only 0.4, even though the percentages are identical. Larger samples give the test more power to detect a real departure.
• Expected numbers should be at least about 5 in every class. With very small expected values, the chi square approximation becomes unreliable; combine classes or collect more data.
• Degrees of freedom come from the number of classes, not the number of offspring. A 9:3:3:1 test has df = 3 whether you counted 50 or 5,000 offspring. (In some tests, such as Hardy Weinberg tests in Chapter 12, an additional degree of freedom is lost for each parameter estimated from the data.)
• Failing to reject is not proof. A small chi square says the data are compatible with the hypothesis, but other hypotheses may fit as well.
• Suspiciously good fits deserve attention too. The statistician R. A. Fisher pointed out in 1936 that Mendel's data, taken together, fit his expectations more closely than chance would usually allow. Historians still debate whether this reflects unconscious bias in scoring, selective reporting or something else, but no one doubts that the underlying principles are correct.

Section 5Pedigree analysis basics

Why human genetics needs pedigrees

Geneticists working with peas or fruit flies design the crosses they want and count hundreds of offspring. Human geneticists can do neither. Matings are not controlled, families are small, generations are long, and many traits of interest are rare. Instead we reconstruct inheritance from a pedigree, a standardized family tree that records who is related to whom and who shows the trait. By asking which modes of inheritance are consistent with the pattern, we can usually identify the most likely mode and then calculate risks for relatives.

Pedigree conventions

Males are drawn as squares and females as circles; a diamond is used when sex is unknown or not relevant. Affected individuals are filled in. A horizontal line between a male and a female is a mating, and a vertical line drops from it to a horizontal sibship line from which the children hang in birth order, left to right. A double mating line marks a consanguineous mating, one between relatives such as first cousins. Generations are numbered with Roman numerals and individuals within a generation with Arabic numerals, so II-3 is the third person in the second generation. The person through whom the family came to attention, the proband (or propositus), is marked with an arrow. Heterozygous carriers, when known, may be shown by a half filled symbol or a dot in the center. A diagonal slash marks a deceased person. Twins hang from a common point; identical (monozygotic) twins are joined by a short horizontal bar.

Autosomal recessive inheritance

A trait caused by a recessive allele on an autosome typically shows these features:

• Affected children are usually born to unaffected parents, who are both heterozygous carriers. The trait often appears to "skip" generations.
• Males and females are affected in roughly equal numbers, and either parent can transmit the allele.
• Among the children of two carriers, about one quarter are affected. (In real pedigrees the observed fraction is often higher, because families with no affected child are never ascertained.)
• The trait is more common among the children of consanguineous matings, because relatives are more likely to carry the same rare allele inherited from a common ancestor.
• Two affected parents (aa × aa) have only affected children, unless the parents carry mutations in different genes (see complementation, Chapter 2).

Examples include cystic fibrosis, sickle cell disease, phenylketonuria, Tay Sachs disease and albinism.

Autosomal dominant inheritance

A trait caused by a dominant allele on an autosome typically shows:

• Every affected person has at least one affected parent (unless the allele arose by new mutation or is not fully penetrant). The trait appears in every generation, a vertical pattern.
• Males and females are affected about equally, and both sexes transmit the trait, including male to male transmission, which rules out X linked inheritance.
• An affected heterozygote mated with an unaffected person has, on average, half affected children.
• Two unaffected parents do not have affected children (again barring new mutation or incomplete penetrance).
• Two affected parents can have an unaffected child (Aa × Aa gives 1/4 aa); this pattern is the single best clue that a trait is dominant.

Most people affected by rare dominant disorders are heterozygous, because homozygotes require two affected parents and are often more severely affected. Examples include Huntington disease, Marfan syndrome, neurofibromatosis type 1, achondroplasia and familial hypercholesterolemia.

Reasoning through a pedigree

A systematic approach avoids most errors. First, look for two unaffected parents with an affected child: if present, the trait must be recessive. Next, look for two affected parents with an unaffected child: if present, the trait must be dominant. Then check for sex bias and male to male transmission to decide between autosomal and X linked modes (Chapter 3). For each hypothesis that remains, try to assign genotypes to every person; a hypothesis is ruled out if any individual cannot be given a consistent genotype. Finally, if more than one mode fits, choose the one that requires the fewest assumptions, such as the fewest unrelated carriers marrying into the family. For a rare trait, assume that people marrying into the family are homozygous for the common allele unless the pedigree forces otherwise.

Ascertainment and the limits of pedigrees

Pedigrees are small, so the ratios within any one family seldom match Mendelian expectations, and the way families are found (ascertainment) can bias the data: a family is usually studied because it has at least one affected child, which inflates the observed proportion of affected children. Other complications include new mutations, which can produce an isolated affected child with a dominant disorder; incomplete penetrance, in which a carrier of a dominant allele is unaffected; locus heterogeneity, in which mutations in different genes cause the same phenotype; and nonpaternity or adoption, which break the assumed genetic relationships. Chapter 2 discusses penetrance and expressivity, and Chapter 14 returns to risk calculation in genetic counseling.

Words worth knowing

Extensions of Mendel

Mendel's peas gave clean 3:1 and 9:3:3:1 ratios, but most real genes are not so tidy. Heterozygotes can look intermediate, both alleles can show at once, a gene can have dozens of alleles, some genotypes never survive to be counted, and one gene can hide the effects of another. This chapter shows that none of these complications overturn Mendel: they are his rules operating on the biochemistry of real gene products, and once you see the pathway behind a strange ratio, the ratio stops being strange.

Section 1Incomplete dominance, codominance and multiple alleles

Dominance is a property of the heterozygote's phenotype

In Chapter 1, dominance meant that a heterozygote looked like one homozygote. That is complete dominance. But dominance is not built into an allele; it describes how the phenotype of the heterozygote compares with the two homozygotes, and the answer depends on the trait and on how closely you look. Three relationships are recognized:

• Complete dominance: the heterozygote is indistinguishable from one homozygote. This usually happens when one functional allele makes enough product for a normal phenotype (the gene is haplosufficient).
• Incomplete dominance (partial dominance): the heterozygote phenotype is intermediate between the two homozygotes, often because the amount of product, and thus the trait, depends on the dose of a functional allele.
• Codominance: the heterozygote shows the phenotypes of both alleles simultaneously, because each allele makes its own distinct, detectable product.

Incomplete dominance

The classic example is flower color in snapdragons (Antirrhinum) and four o'clocks (Mirabilis). A true breeding red line crossed with a true breeding white line gives pink F₁ plants. Intercrossing the pink plants gives an F₂ of 1 red : 2 pink : 1 white. This looks at first like blending, but it is not: red and white reappear unchanged in the F₂, proving that the alleles stayed discrete. The red allele C^R encodes an enzyme in anthocyanin pigment synthesis, while C^W makes no functional enzyme. Two doses give full red pigment, one dose gives half as much, which we see as pink, and none gives white.

With incomplete dominance the phenotypic ratio equals the genotypic ratio, 1:2:1, because each genotype has its own phenotype. Symbols change accordingly: since neither allele is dominant, geneticists use superscripts on a common base letter rather than upper and lower case letters.

Codominance

In codominance both alleles are expressed in the heterozygote, each detectable in its own right. The human MN blood group is a clean example. The GYPA gene encodes glycophorin A, a red cell surface protein, and two common alleles, L^M and L^N, make forms that differ at two amino acids. People with genotype L^ML^M have type M cells, L^NL^N have type N, and heterozygotes L^ML^N have type MN, with both antigens present on their cells. Two MN parents have children in the ratio 1 M : 2 MN : 1 N.

Whether a pair of alleles looks dominant, incompletely dominant or codominant often depends on the level of analysis. Sickle cell disease illustrates this. At the level of the whole person under ordinary conditions, the sickle allele Hb^S is recessive: heterozygotes (sickle cell trait) are generally healthy. At the level of red cell shape under low oxygen, heterozygote cells sickle somewhat, so the allele is incompletely dominant. At the level of the protein, heterozygotes make both normal hemoglobin A and hemoglobin S, which can be separated by electrophoresis, so the alleles are codominant. Same alleles, three answers.

Multiple alleles

Any one diploid individual carries at most two alleles of a gene, but a population can carry many. A gene with three or more alleles in the population has multiple alleles, and the alleles form an allelic series. With n alleles, the number of possible genotypes is n(n + 1)/2: n homozygotes plus n(n − 1)/2 heterozygotes. Three alleles give 6 genotypes; four give 10.

The ABO blood group combines multiple alleles with both dominance and codominance. The ABO gene encodes a glycosyltransferase that adds a sugar to a cell surface precursor called the H antigen. The I^A allele's enzyme adds N acetylgalactosamine, making the A antigen; the I^B enzyme adds galactose, making the B antigen; the i allele carries a frameshift and makes no active enzyme, so the H antigen is left unmodified (type O). I^A and I^B are each dominant to i and codominant with each other.

Another kind of allelic series is a dominance hierarchy, in which each allele is dominant to those below it. Rabbit coat color is the textbook case, with four alleles of the C gene: C (full color, agouti) > c^ch (chinchilla, silvery gray) > c^h (Himalayan, white with dark extremities) > c (albino). A c^chc^h rabbit is chinchilla, a c^hc rabbit is Himalayan, and so on. The Himalayan allele is itself temperature sensitive, a point we return to in Section 4. Plant self incompatibility systems carry the idea to an extreme: the S locus of some species has dozens of alleles, and pollen cannot fertilize a plant that carries the same S allele, which enforces outcrossing.

Section 2Lethal alleles and gene interaction

The yellow mouse puzzle

Soon after Mendel's work was rediscovered, the French geneticist Lucien Cuénot studied coat color in mice and hit a ratio that made no sense. Yellow mice never bred true. When yellow was crossed with yellow, the offspring were always about 2 yellow : 1 nonyellow (agouti), never 3:1. And every yellow mouse, when testcrossed, proved to be heterozygous. Where were the homozygous yellow mice?

The answer is that they die as embryos. The yellow allele, A^Y, is an allele of the agouti gene. It has a dominant effect on coat color (one copy makes the coat yellow) but a recessive lethal effect: A^YA^Y embryos die shortly after implantation. A yellow by yellow cross (A^YA × A^YA) is expected to give 1/4 A^YA^Y (dead), 1/2 A^YA (yellow) and 1/4 AA (agouti). Among the survivors, yellow : agouti = 1/2 : 1/4 = 2:1. The mechanism was later found: A^Y carries a large deletion that fuses the agouti coding sequence to the promoter of a neighboring gene, Raly, which is essential for early development. The deletion destroys Raly function (lethal when homozygous) while driving agouti protein expression everywhere (yellow coat, plus obesity and diabetes in adult yellow mice).

Kinds of lethal alleles

A lethal allele causes death of the organism, usually before reproductive age. Most are recessive lethals: heterozygotes survive (often appearing normal), and homozygotes die. Many essential genes, whose complete loss is incompatible with life, can mutate to recessive lethal alleles, and every human is estimated to carry a small number of them in the heterozygous state. A cross of two heterozygous carriers of a recessive lethal with no visible heterozygous phenotype simply produces about one quarter fewer offspring, with all survivors normal in appearance; only when the heterozygote has a visible phenotype, as with yellow mice, does the 2:1 ratio appear.

Other examples: the Manx allele in cats (M) causes taillessness in heterozygotes and is lethal early in development in homozygotes, so Manx by Manx crosses give 2 Manx : 1 tailed. Creeper chickens, with short legs, behave the same way. In humans, achondroplasia (caused by an activating mutation in FGFR3) is dominant for short limbed dwarfism, while homozygotes have a severe skeletal dysplasia that is usually lethal in infancy.

Dominant lethal alleles kill even in one copy. Those that act before reproduction are removed from the population in the same generation they arise, so they are seen only as new mutations. A dominant lethal can persist only if its effect appears after reproduction has begun. Huntington disease is the textbook example: symptoms typically begin between 30 and 50 years of age, after many affected people have had children, and the disease is eventually fatal.

Conditional lethal alleles are lethal only under certain conditions. Temperature sensitive (ts) mutants of bacteria, yeast and flies survive at a permissive temperature but die at a restrictive one, usually because the mutant protein misfolds at higher temperature. Geneticists prize ts alleles because they allow the study of essential genes: grow the organism at the permissive temperature, then shift it and see what fails. In humans, glucose 6 phosphate dehydrogenase deficiency is conditional in a different sense: carriers are usually healthy but can suffer severe hemolysis after eating fava beans or taking certain drugs.

Gene interaction that produces novel phenotypes

So far each trait has been controlled by one gene. But most traits are the products of several genes whose products interact, and when two genes affect the same trait, the dihybrid ratio may be modified or new phenotypes may appear. The term gene interaction covers all such cases; the simplest are those in which the 9:3:3:1 ratio is preserved but the four classes are four distinct forms of one trait.

The classic example, analyzed by William Bateson and Reginald Punnett in the early 1900s, is comb shape in chickens. Two genes, R (rose) and P (pea), act together:

• R_P_: walnut comb
• R_pp: rose comb
• rrP_: pea comb
• rrpp: single comb

A cross of true breeding rose (RRpp) with true breeding pea (rrPP) gives an F₁ with walnut combs, a phenotype seen in neither parent. The F₂ shows 9 walnut : 3 rose : 3 pea : 1 single. The ratio is Mendelian, but the walnut and single classes are novel phenotypes created by the combination of alleles at two loci. Similar interactions produce the red, brown, scarlet and white eyes that segregate when Drosophila carrying the brown and scarlet mutations are crossed and bell pepper colors (red, peach, brown, green, from two genes).

How can you tell gene interaction for a single trait from two separate traits? Look at the parents and the F₁. If one trait (comb shape) appears in four forms in the F₂, and the F₁ shows a form not present in either parent, two genes are acting on that one trait. Testcrossing the F₁ to rrpp confirms it: the four comb types appear 1:1:1:1.

Section 3Epistasis and modified ratios

When one gene masks another

Epistasis is a form of gene interaction in which the genotype at one locus masks or modifies the phenotypic expression of a genotype at another locus. The gene doing the masking is epistatic; the gene whose effect is hidden is hypostatic. The word comes from Greek for "standing upon," and the image is apt: one gene stands on top of another. Epistasis is different from dominance. Dominance is an interaction between alleles at the same locus; epistasis is an interaction between different loci.

In a dihybrid intercross with epistasis, the 16 Punnett cells still occur in the usual proportions of 9 A_B_ : 3 A_bb : 3 aaB_ : 1 aabb. Independent assortment has not changed. What changes is that some of these genotype classes share a phenotype, so the four classes merge into three or two phenotypic classes. Every epistatic ratio therefore still adds up to 16, and recognizing a ratio with a total of 16 is the first clue that two genes are involved.

Thinking in pathways

Epistasis is easiest to understand through biochemical pathways. Many traits are the end product of a series of enzyme catalyzed steps, each enzyme encoded by a different gene. A loss of function mutation in any step blocks everything downstream. If a mutation blocks an early step, it does not matter what alleles are present at genes acting later; the later enzymes have no substrate. The early acting gene is therefore epistatic to the later one.

Recessive epistasis: 9:3:4

In recessive epistasis, the homozygous recessive genotype at one locus masks the other locus. Labrador retriever coat color is the standard example. The B gene (TYRP1) determines whether the dark pigment eumelanin is black (B_) or brown (bb). The E gene (MC1R, the melanocortin 1 receptor) determines whether eumelanin is deposited in the hair at all; ee dogs make little eumelanin in the coat and are yellow, although their noses and lips still show the B genotype. A cross of two dogs heterozygous at both loci (BbEe × BbEe) gives 9 black (B_E_) : 3 chocolate (bbE_) : 4 yellow (3 B_ee + 1 bbee). Albinism in many animals works the same way: cc individuals lacking tyrosinase are white whatever their other coat color genes say, which is why albino mice are a frequent textbook example of 9:3:4.

The Bombay phenotype in humans is recessive epistasis too. The FUT1 gene (the H locus) makes the fucosyltransferase that builds the H antigen, the substrate on which the ABO enzymes act. People homozygous for a null allele (hh) make no H antigen, so even if they carry I^A or I^B, no A or B antigen can be made, and they type as O. The famous clue is a pedigree in which a "type O" mother has an AB child: she must carry an I^B or I^A allele masked by hh.

Dominant epistasis: 12:3:1

In dominant epistasis, a dominant allele at one locus masks the other locus. Fruit color in summer squash shows it. A dominant allele W blocks pigment formation, giving white fruit regardless of the second gene. Among ww plants, Y_ gives yellow and yy gives green. The F₂ of WwYy × WwYy is 12 white (9 W_Y_ + 3 W_yy) : 3 yellow (wwY_) : 1 green (wwyy).

Duplicate recessive epistasis (complementary genes): 9:7

When two genes act in series in one pathway and a recessive loss of function at either one gives the same mutant phenotype, both genes must have a dominant allele for the wild type phenotype. Bateson and Punnett found this in sweet peas: two true breeding white flowered strains crossed to give purple F₁ plants, and the F₂ was 9 purple : 7 white. Purple requires C_P_; every genotype with cc or pp is white (Figure 2.3A). This is also called complementary gene action, because the two white parents each supply what the other lacks, which is exactly the logic of the complementation test in Section 5.

Duplicate dominant epistasis: 15:1

When either of two genes can supply the same function, as with duplicated genes, a dominant allele at either locus is enough for the wild type phenotype, and only the double homozygous recessive is mutant. Seed capsule shape in shepherd's purse (Capsella) segregates 15 triangular (any genotype with at least one A or one B) : 1 ovoid (aabb). Duplicate genes are common in plants with polyploid ancestry, and 15:1 ratios are frequent in crops such as wheat.

Dominant and recessive epistasis: 13:3

Sometimes the dominant allele of one gene and the recessive genotype of the other both produce the same phenotype. In chickens, C is required for color, and a dominant inhibitor I suppresses color. Only iiC_ birds are colored. Crossing two white breeds (white Leghorn, IICC, and white Wyandotte, iicc) gives white F₁ (IiCc) and an F₂ of 13 white (9 I_C_ + 3 I_cc + 1 iicc) : 3 colored (iiC_).

Duplicate genes with cumulative effect: 9:6:1

Fruit shape in another summer squash cross gives 9 disc (A_B_) : 6 sphere (A_bb or aaB_) : 1 long (aabb). Here a dominant allele at either locus has the same effect and the effects add: two loci with dominant alleles give disc, one gives sphere, and none gives long. Strictly this is gene interaction with additive effects rather than masking, but it is usually taught alongside epistatic ratios.

Section 4Penetrance, expressivity and the environment

Same genotype, different phenotypes

Mendel's traits behaved as if each genotype produced exactly one phenotype. Many genes are less predictable. Two people can carry the identical disease allele, yet one is severely affected, one mildly, and one not at all. Geneticists describe this variability with two terms that students often confuse: penetrance and expressivity.

Penetrance

Penetrance is the percentage of individuals with a particular genotype who show the expected phenotype at all. It answers a yes or no question across a group: does the person with the genotype have the trait? If 80 of 100 people carrying a dominant allele show the trait, penetrance is 80 percent. When every individual with the genotype shows the phenotype, the allele is fully penetrant; otherwise it shows incomplete (or reduced) penetrance. Incompletely penetrant dominant traits can appear to skip a generation: an unaffected parent may transmit the allele to an affected child, creating a pedigree that looks recessive at first glance.

Familiar examples include polydactyly (extra fingers or toes), which in some families shows clearly reduced penetrance, and inherited cancer predispositions. Women who carry a harmful BRCA1 variant have a lifetime breast cancer risk commonly cited at about 55 to 70 percent, far above the population risk of roughly 12 to 13 percent, but not 100 percent. Retinoblastoma, caused by RB1 mutations, shows about 90 percent penetrance in carriers of typical mutations.

Penetrance is often age dependent. Huntington disease alleles with 40 or more CAG repeats are essentially fully penetrant over a normal lifespan, but the probability of having symptoms rises with age, so a 25 year old carrier usually shows no signs. When counseling relatives, geneticists use age specific penetrance curves rather than a single number.

Expressivity

Expressivity is the degree or intensity with which a trait is expressed in those individuals who do show it. It answers a "how much" question about each affected person. An allele with variable expressivity produces a range of severity. Neurofibromatosis type 1 (NF1) is the standard example: almost all carriers show some sign by adulthood (penetrance near 100 percent), but one affected person may have only a few café au lait spots while a relative has hundreds of neurofibromas, learning difficulties and skeletal problems. Marfan syndrome likewise ranges from mild tall stature and joint laxity to life threatening aortic dilation within the same family.

Why penetrance and expressivity vary

Variation in penetrance and expressivity has three main sources. Modifier genes elsewhere in the genome alter the effect of the primary gene; differences in genetic background are why a mutation can produce a severe phenotype in one inbred mouse strain and a mild one in another. The environment, including diet, temperature, drugs, infections and toxins, changes how gene products work or how much they are needed. And chance events during development, such as which cells happen to acquire a second hit in a tumor suppressor gene or random variation in gene expression, can make even genetically identical twins differ.

Environmental effects on gene expression

Some of the clearest examples involve temperature. Himalayan rabbits and Siamese cats carry temperature sensitive alleles of the tyrosinase gene. The enzyme works at the cooler temperatures of the ears, nose, paws and tail, but is inactive at core body temperature, so pigment forms only at the extremities. Shave a patch of a Himalayan rabbit's back and keep an ice pack on it, and the new fur grows in dark. The genotype has not changed; its expression depends on the temperature.

Human disease offers equally clear cases. Phenylketonuria (PKU) results from homozygosity for loss of function alleles of phenylalanine hydroxylase. On an ordinary diet, phenylalanine accumulates and causes severe intellectual disability; on a low phenylalanine diet begun in the newborn period, affected children develop with normal or near normal intelligence. Newborn screening exists precisely so that the environment can be changed early enough to alter the phenotype. Lactase persistence and lactose intolerance, G6PD deficiency with fava bean exposure, and the effect of smoking on lung disease in people with alpha 1 antitrypsin deficiency are further gene by environment examples.

Plants show strong environmental effects too. Hydrangea flower color depends on soil pH, which controls how much aluminum is available to bind the pigment: acidic soil gives blue flowers and alkaline soil pink ones, in plants of the same genotype.

Phenocopies and norms of reaction

A phenocopy is a phenotype produced by environmental factors that mimics a phenotype normally produced by a specific genotype. Phenocopies are not inherited. Thalidomide, a sedative given to pregnant women around 1960, caused limb malformations that resembled those of rare inherited syndromes such as Roberts syndrome. Heat shock of Drosophila pupae can produce wing defects that look like specific mutations. In a pedigree, a phenocopy can mislead an analyst into seeing an inherited trait where there is none, or into assigning the wrong genotype.

A norm of reaction is the range of phenotypes a single genotype produces across a range of environments. It is usually drawn as a graph with environment on the horizontal axis and phenotype on the vertical axis, one line per genotype. When the lines are parallel, genotype and environment act additively; when they cross, the best genotype in one environment is not the best in another, a genotype by environment interaction. Classic studies of Drosophila eye facet number at different temperatures showed that genotypes respond to the same environmental change by different amounts, and studies of yarrow clones grown at different elevations showed that the ranking of genotypes can even reverse from one environment to another. Norms of reaction are a reminder that a phenotype belongs to a genotype in an environment, never to a genotype alone.

Section 5Complementation tests

The question complementation answers

Suppose a geneticist screens thousands of fruit flies and recovers ten independent recessive mutants that all have white eyes, or a plant breeder finds five dwarf lines, or a clinic sees several families with the same recessive form of deafness. A basic question comes first: are these mutations in the same gene or in different genes? The phenotype alone cannot say, because, as Section 3 showed, mutations that block different steps of one pathway produce the same phenotype. The complementation test answers the question with a single cross.

How the test works

Cross two homozygous recessive mutants that have the same phenotype, and examine the F₁:

• If the F₁ is wild type, the mutations complement each other, and they are in different genes. Each parent was homozygous mutant for a different gene and supplied a working copy of the gene the other lacked. Mutant 1 is a¹a¹ BB, mutant 2 is AA b²b², and the F₁, Aa¹ Bb², has one working copy of each gene.
• If the F₁ is mutant, the mutations fail to complement, and they are (with rare exceptions) in the same gene. The F₁ carries two different mutant alleles of one gene, a¹a², and has no working copy.

The logic depends on the mutations being recessive loss of function alleles. In the F₁ the two mutations sit on different homologs, a configuration called trans. Seymour Benzer, working on bacteriophage T4 in the 1950s, formalized this as the cis trans test and used the word cistron for a unit of function defined by complementation. A cistron corresponds, in nearly all cases, to what we now call a gene.

Notice how closely this mirrors the 9:7 ratio of Section 3. Two white sweet pea strains crossed to give purple F₁ plants: that is complementation, and it told Bateson and Punnett that the strains carried mutations in different genes, C and P.

Complementation groups

With many mutants, every pair is crossed and the results are arranged in a complementation matrix. Mutants that fail to complement one another are placed in the same complementation group, and each group corresponds to one gene. The number of groups is therefore an estimate of the number of genes in which mutations can produce the phenotype. Complementation analysis of yeast cell division cycle (cdc) mutants, many of them temperature sensitive, identified dozens of genes needed for cell cycle progression, work honored with a Nobel Prize in 2001. Similar analyses defined the genes of the T4 phage rII region, the yeast mating pathway and many metabolic pathways.

The matrix above is the same data in the conventional half matrix form. The diagonal is always −, because a mutant crossed with itself cannot complement.

Complementation in human families

Humans cannot be crossed deliberately, but nature sometimes performs the test. Congenital deafness is often autosomal recessive, and mutations in well over 50 different genes can cause it. Deaf people frequently marry one another, and the outcomes are informative. When two deaf parents with recessive deafness have only deaf children, they likely carry mutations in the same gene. When two deaf parents have children who all hear normally, the parents' mutations must be in different genes, and each child is heterozygous at both loci. The same reasoning applies to recessive forms of blindness such as retinitis pigmentosa, which can be caused by mutations in dozens of genes. Clinicians call this phenomenon locus heterogeneity: the same phenotype arising from mutations at different loci. Complementation can also be tested in cultured cells by fusing cells from two patients; if the hybrid cell recovers normal function, the patients have defects in different genes. This was how the complementation groups of xeroderma pigmentosum, a DNA repair disorder, were defined.

Limits and exceptions

• Dominant mutations cannot be tested this way. The F₁ will show the mutant phenotype whether or not the mutations are in the same gene, because one dominant allele is enough. Complementation tests require recessive mutations.
• Intragenic (interallelic) complementation is a rare exception in which two mutations in the same gene complement. It occurs when the gene product is a multimer: two differently defective subunits can sometimes assemble into a partly functional protein. The partial activity is usually weaker than wild type, a useful clue.
• Nonallelic noncomplementation is the opposite exception: mutations in two different genes fail to complement, typically because their products physically interact, so being heterozygous for both is enough to reduce function below a threshold.
• Polar mutations in bacterial operons can block expression of downstream genes in the same transcript, making mutations in separate genes appear to fail to complement.
• Complementation is not recombination. Complementation is assessed in the F₁ and asks about function. Recombination between two mutations is assessed in the next generation and asks about physical position (Chapter 4). Two mutations in one gene can recombine to give wild type gametes at low frequency, yet still fail to complement.

Words worth knowing

Sex Linkage and Extranuclear Inheritance

Mendel's peas gave the same results no matter which parent carried which allele, but many real traits refuse to behave that way. A son inherits color blindness from his mother, and a deletion on chromosome 15 causes one disease when it comes from the father and another when it comes from the mother. This chapter explains why the sex of the parent, and the sex of the child, can change the outcome of a cross.

Section 1Sex determination systems

In most animals and in a minority of flowering plants, individuals come in two sexes that make different gametes: females make large, nutrient rich eggs and males make small, mobile sperm or pollen. Sex determination is the mechanism that decides which developmental path an embryo takes. It matters here because when sex is chromosomal, the chromosomes that decide sex also carry ordinary genes whose inheritance is tied to sex.

Chromosomal sex determination

In chromosomal systems, the two sexes differ in their sex chromosomes, and every other chromosome is an autosome. The sex that carries two different sex chromosomes makes two kinds of gametes and is called the heterogametic sex; the sex with two matching sex chromosomes makes one kind of gamete and is the homogametic sex.

• XX/XY. Females are XX (homogametic) and males are XY (heterogametic). This is the system of mammals, of Drosophila and of a few dioecious plants such as white campion. Because each sperm carries either an X or a Y, the father's gamete decides the sex of each child, and the expected sex ratio is 1:1.
• XX/XO. Females are XX and males have a single X and no partner chromosome (written XO, where O means absence). Many grasshoppers and crickets use this system. Males make two kinds of sperm, those with an X and those with no sex chromosome. The nematode Caenorhabditis elegans is a variant: XX individuals are self fertilizing hermaphrodites and XO individuals are males.
• ZZ/ZW. Here the female is heterogametic (ZW) and the male is homogametic (ZZ). Birds, butterflies and moths, and some fish and reptiles use it. The letters Z and W are used precisely so that no one confuses this system with XX/XY. In birds, the egg (not the sperm) decides the sex of the offspring. Evidence suggests that two doses of the Z linked gene DMRT1 push the bird gonad toward a testis.

Haplodiploidy

Bees, wasps and ants (the order Hymenoptera) have no sex chromosomes at all. Fertilized eggs develop into diploid females and unfertilized eggs develop into haploid males, a system called haplodiploidy. A queen honeybee controls whether each egg she lays is fertilized, and so its sex. Because a male is haploid, he makes sperm by mitosis, and every sperm he produces is genetically identical to every other.

In many hymenopterans the real switch is a single locus called complementary sex determiner (csd). Individuals heterozygous at csd become female; individuals that are hemizygous (haploid) become male. Inbreeding produces some eggs homozygous at csd, which develop as sterile diploid males.

Mammals: SRY decides

The study of people with unusual numbers of sex chromosomes showed that in humans it is the presence of a Y chromosome, not the number of X chromosomes, that makes an embryo male. A person with 45,X (one X, no second sex chromosome) has Turner syndrome and develops as a female, usually with short stature and ovaries that fail early. A person with 47,XXY has Klinefelter syndrome and develops as a male, typically tall with small testes and reduced fertility. People with 47,XXX are female and those with 47,XYY are male.

The critical gene on the Y is SRY (sex determining region Y), located on the short arm near the boundary of the pseudoautosomal region. SRY encodes a transcription factor with an HMG box, a DNA binding domain that bends DNA. In the bipotential gonad of a six to seven week embryo, SRY switches on the autosomal gene SOX9. SOX9 makes gonad cells become Sertoli cells, which secrete anti Müllerian hormone to remove the female duct system, while Leydig cells make testosterone to masculinize the rest of the body. Without SRY, the gonad follows the ovarian pathway, supported by genes such as WNT4, RSPO1 and FOXL2.

Two rare conditions confirm SRY's role. In XX males, an abnormal crossover during the father's meiosis has carried SRY from the Y onto the X, so the child has two X chromosomes and testes. In XY females (Swyer syndrome in some cases), the Y is present but SRY is deleted or carries a loss of function mutation, and the gonads develop as nonfunctional streaks.

The pseudoautosomal regions

The human X and Y are very different in size and gene content: the X carries roughly 800 to 900 protein coding genes while the Y carries only a few dozen. Yet in male meiosis the X and Y must pair and segregate. They do so through short regions of shared sequence at their tips, the pseudoautosomal regions (PAR1 at the end of the short arms and PAR2 at the end of the long arms). In male meiosis at least one crossover occurs in PAR1, which is why genes there, such as SHOX, are inherited as though they were autosomal. Outside the PARs, the male specific region of the Y does not recombine and passes intact from father to son.

Drosophila: the X to autosome ratio

Fruit flies have XX females and XY males, but their sex is decided differently. In flies the Y carries no sex determining gene; instead, sex depends on the ratio of X chromosomes to sets of autosomes, the X:A ratio. Calvin Bridges worked this out in the 1920s by studying flies with unusual chromosome sets. A ratio of 1.0 or higher produces a female, 0.5 or lower produces a male, and values in between produce intersexes. Molecularly, a double dose of X linked signal genes switches on Sex lethal (Sxl), which starts a cascade of alternative RNA splicing (Sxl to transformer to doublesex) that produces female development. The Y is still needed in male flies for fertility, because it carries genes for sperm motility, but an XO fly is a sterile male and an XXY fly is a fertile female.

Environmental sex determination

In some animals, sex is not decided by genotype at all. Many turtles, all crocodilians and some lizards show temperature dependent sex determination (TSD): the temperature at which an egg incubates during a sensitive window of development decides the sex. In many turtles warm nests produce females and cool nests produce males; in the American alligator the pattern is reversed, with cool incubation giving females. Temperature acts partly through aromatase, the enzyme that converts testosterone to estrogen. Some fish, such as clownfish, even change sex as adults in response to social cues.

Section 2X linked inheritance and pedigrees

Genes on the X chromosome are X linked. A male has one X, so for X linked genes he is neither homozygous nor heterozygous but hemizygous: whatever allele his single X carries is expressed. A female has two X chromosomes and can be homozygous or heterozygous, just as for autosomal genes. This asymmetry gives X linked traits their signature: results that change when the sexes of the parents are swapped.

Morgan's white eyed fly

In 1910 Thomas Hunt Morgan found a single male fruit fly with white eyes in a bottle of normal, red eyed flies. He crossed it to a red eyed female. All the F1 flies had red eyes, so white was recessive. When he intercrossed the F1, the F2 contained 3 red to 1 white, a Mendelian ratio, but with a twist: every white eyed fly was male. Half the F2 males had white eyes and none of the females did.

Morgan proposed that the gene lies on the X and that the Y has no copy. Using the convention X⁺ for the wild type (red) allele and X^w for white, the original male was X^wY and the female X⁺X⁺. The F1 females were X⁺X^w and the F1 males X⁺Y. In the F2, daughters received X⁺ from their father and so were all red eyed, while sons received an X from their heterozygous mother and were X⁺Y or X^wY in equal numbers.

The decisive test was the reciprocal cross: a white eyed female (X^wX^w) crossed to a red eyed male (X⁺Y). For an autosomal gene the reciprocal cross gives the same result as the original. Here it did not. All F1 daughters were red eyed (X⁺X^w) and all F1 sons were white eyed (X^wY). This pattern, in which sons resemble their mothers and daughters resemble their fathers, is called crisscross inheritance. Intercrossing these F1 flies gives an F2 of 1 red female : 1 white female : 1 red male : 1 white male.

Bridges and nondisjunction: proof of the chromosome theory

Morgan's student Calvin Bridges pushed the argument to its conclusion. When he crossed white eyed females to red eyed males, about 1 in 2,000 offspring broke the crisscross rule: a few daughters were white eyed and a few sons were red eyed. Bridges predicted that these exceptions came from nondisjunction, the failure of the two X chromosomes to separate in the mother's meiosis. An egg carrying both X^w chromosomes, fertilized by a Y bearing sperm, would give an X^wX^wY fly: female (X:A = 1.0) and white eyed. An egg with no X, fertilized by an X⁺ sperm, would give an X⁺O fly: a red eyed, sterile male. Under the microscope, the exceptional females had a Y and the exceptional males lacked one, exactly as predicted: the most convincing proof yet that genes are carried on chromosomes.

Reading X linked recessive pedigrees

In human families we cannot set up crosses, so we read pedigrees. An X linked recessive trait leaves a characteristic set of fingerprints:

• Many more males than females are affected. A male needs only one copy of the allele; a female needs two. If the allele frequency is q, the expected frequency of affected males is q and of affected females is q². For red green color vision defects, which affect about 8 percent of males of northern European ancestry, the expected frequency in females is about 0.08² = 0.0064, or roughly 0.6 percent, close to what is observed.
• No male to male transmission. A father gives his X only to daughters and his Y to sons, so an affected father cannot pass the trait to his sons.
• All daughters of an affected male are carriers. Each daughter receives her father's only X.
• The trait often skips a generation, passing from an affected grandfather through a carrier daughter to an affected grandson.
• Sons of carrier mothers have a 1/2 chance of being affected, and daughters of carrier mothers have a 1/2 chance of being carriers.
• An affected female usually has an affected father and a mother who is a carrier or affected.

X linked dominant and Y linked patterns

An X linked dominant trait appears in anyone with at least one copy of the allele. Its pedigree features are different: an affected father passes the trait to all of his daughters and none of his sons, and an affected heterozygous mother passes it to half of her children of either sex. Because females have two chances to inherit the allele, affected females outnumber affected males, often about two to one. Examples include X linked hypophosphatemic rickets, in which a defect in the PHEX gene causes the kidney to waste phosphate, and incontinentia pigmenti and Rett syndrome, in which the mutation is usually lethal or far more severe in hemizygous males, so most patients are female.

Y linked (holandric) traits pass only from father to son, to every son, and never to daughters. Nearly all known Y linked traits concern male fertility, such as deletions that block sperm production.

Section 3Dosage compensation and X inactivation

Gene dosage matters. Extra or missing copies of an autosome are usually lethal or cause serious disease because hundreds of genes are suddenly expressed at the wrong level. Yet in XX/XY species, females have two copies of every X linked gene and males have one, with no harm to either sex. The reason is dosage compensation, a set of mechanisms that equalize the expression of X linked genes between the sexes. Remarkably, three well studied animals solve the problem in three different ways.

Barr bodies and the Lyon hypothesis

In 1949, Murray Barr and Ewart Bertram noticed a small, darkly staining mass at the edge of the nucleus in nerve cells of female cats that was missing in males. This Barr body turned out to be a condensed, transcriptionally silent X chromosome, a form of facultative heterochromatin. In 1961 Mary Lyon proposed what is now called the Lyon hypothesis:

1. In each somatic cell of a female mammal, one X chromosome is inactivated and becomes a Barr body.
2. Inactivation happens early in embryonic development.
3. In the embryo proper, the choice of which X to silence, maternal or paternal, is random and is made independently in each cell.
4. Once made, the choice is stable and clonally inherited: all descendants of a cell keep the same X inactive.

A female mammal is therefore a mosaic of two cell populations, one expressing her maternal X and one expressing her paternal X. The rule extends to people with extra X chromosomes: cells keep exactly one X active and inactivate all the others, so the number of Barr bodies equals the number of X chromosomes minus one (the n minus 1 rule).

How an X is silenced

X inactivation is controlled by a region of the X called the X inactivation center (XIC). Its key gene is XIST (X inactive specific transcript), which encodes a long noncoding RNA of about 17,000 nucleotides that is never translated. The process can be pictured in steps:

1. Counting. The cell senses how many X chromosomes it has relative to its autosome sets. If there is more than one X, inactivation proceeds.
2. Choice. One X is chosen to stay active. On the future inactive X, XIST is switched on; on the future active X, XIST is kept off. In mice, an antisense transcript called Tsix, made from the opposite DNA strand across Xist, helps repress Xist on the active X.
3. Spreading. XIST RNA coats the chromosome that makes it, working only in cis (on its own chromosome). It spreads outward from the XIC, using repetitive sequences such as LINE elements as way stations.
4. Silencing and maintenance. XIST recruits proteins that remove active histone marks and add repressive ones, especially trimethylation of histone H3 at lysine 27 by Polycomb repressive complex 2. The inactive X gains the histone variant macroH2A, its CpG island promoters become methylated, and it replicates late in S phase. DNA methylation makes the silent state heritable through cell division.

In the germ line the process is reversed. In the developing oocyte, the inactive X is reactivated so that each egg carries a working X. In mouse extraembryonic tissues and in all tissues of marsupials, inactivation is imprinted: the paternal X is always the one silenced.

Escape from inactivation

Inactivation is not complete. In humans about 15 percent of X linked genes escape inactivation and are expressed from both X chromosomes, and another 10 percent or so escape in some individuals or tissues. The escapees include genes in the pseudoautosomal regions, which have partners on the Y and so are present in two active copies in both sexes. Escape explains a puzzle: if one X is inactivated anyway, why are 45,X and 47,XXY not harmless? The answer is that people with Turner syndrome have only one copy of the escape genes instead of two, and people with Klinefelter syndrome have an extra copy. Loss of one copy of the pseudoautosomal gene SHOX, for example, accounts for much of the short stature in Turner syndrome.

Visible mosaics

The most famous mosaic is the calico or tortoiseshell cat. An X linked gene, Orange, has an allele X^O that makes orange fur and an allele X^o that allows black fur. A heterozygous female (X^OX^o) has patches of orange where the X^o chromosome was inactivated and patches of black where X^O was inactivated. White spotting from an unrelated autosomal gene adds the white areas of a calico. Because the patches depend on two different X chromosomes, calico cats are nearly always female. The rare male calico is usually XXY, and like an XXY man, he is sterile.

Human females show the same mosaicism, though usually less visibly. Women heterozygous for X linked anhidrotic ectodermal dysplasia have patches of skin without sweat glands, mapped by spraying the skin with a starch iodine indicator. Women heterozygous for G6PD deficiency have two populations of red cells, one normal and one deficient. Because inactivation is random, the proportion of cells using each X varies from woman to woman. When most cells happen to inactivate the normal X, called skewed X inactivation, a carrier can show symptoms. Such women are called manifesting heterozygotes; for example, some carriers of Duchenne muscular dystrophy develop muscle weakness or heart muscle disease.

Section 4Sex influenced and sex limited traits

Not every trait that differs between males and females is caused by a gene on a sex chromosome. Many autosomal genes are expressed differently in the two sexes because males and females have different hormonal environments. Two categories are worth distinguishing, because they produce patterns that can be mistaken for sex linkage.

Sex influenced traits

In a sex influenced trait, both sexes can express the trait, but the heterozygote's phenotype depends on its sex: the allele behaves as dominant in one sex and recessive in the other. Homozygotes have the same phenotype in both sexes.

The standard example is horns in certain sheep breeds. Let H be the allele for horns and h the allele for no horns. HH sheep of both sexes are horned and hh sheep of both sexes are hornless. Hh rams are horned, while Hh ewes are hornless. In effect H is dominant in males and recessive in females, a difference driven by testosterone. Beards in goats follow the same logic. Pattern baldness is often taught the same way, dominant in men and recessive in women. In reality many genes contribute, including variants near the X linked androgen receptor gene, so treat the one gene model as a practice simplification.

Sex limited traits

In a sex limited trait, the phenotype is expressed in only one sex, although both sexes carry and transmit the genes. Milk yield in cattle is the classic example: bulls carry genes that influence milk production and pass them to their daughters, but bulls themselves give no milk. Breeders judge a bull by his daughters' yields. Other examples include genes affecting egg production in hens, the timing of puberty in boys with familial male limited precocious puberty (an autosomal dominant activating mutation in the luteinizing hormone receptor that has no effect in carrier females), and the cock feathering trait of chickens.

In chickens, males usually have long, curved, pointed tail and neck feathers (cock feathering) and females have short, rounded feathers (hen feathering). An autosomal allele H produces hen feathering in both sexes. Females are hen feathered no matter what their genotype is, so the trait of cock feathering is limited to hh males.

Telling them apart from sex linkage

Sex influenced traits can mimic X linkage in a single generation. Cross a hornless ram (hh) to a horned ewe (HH): every lamb is Hh, so sons are horned like their mother and daughters hornless like their father, which looks like crisscross inheritance. Two tests reveal the truth. An autosomal gene can pass from father to son, which an X linked gene cannot, and the F2 of a sex influenced gene shows 3:1 in one sex and 1:3 in the other.

Section 5Mitochondrial and chloroplast inheritance, maternal effect and genomic imprinting

This section covers situations in which the two parents do not contribute equally. In extranuclear inheritance, genes in mitochondria or chloroplasts come mostly from one parent. In maternal effect, the mother's nuclear genotype, not the offspring's, decides the offspring's phenotype. In genomic imprinting, a gene inherited from one parent is silenced. They are easily confused, so learn to tell them apart.

The mitochondrial genome

Mitochondria descend from bacteria that were engulfed by an ancestral cell more than a billion and a half years ago, and they still carry a small genome of their own. The human mitochondrial DNA (mtDNA) is a circular molecule of 16,569 base pairs containing 37 genes: 13 encode polypeptides of the oxidative phosphorylation system (subunits of complexes I, III, IV and ATP synthase, but none of complex II), 22 encode transfer RNAs and 2 encode ribosomal RNAs. The genome is compact, with almost no noncoding DNA except the D loop control region, and it uses a slightly different genetic code (for example, UGA codes for tryptophan instead of stop). The other 1,000 or more proteins that a mitochondrion needs are encoded in the nucleus, made in the cytosol and imported. For that reason, many mitochondrial diseases are inherited in ordinary Mendelian patterns.

A cell holds hundreds to thousands of copies of mtDNA. Its mutation rate is much higher than that of nuclear DNA, and with no recombination this makes mtDNA useful for tracing maternal ancestry.

Maternal inheritance, heteroplasmy and the threshold

In humans and most animals, mtDNA is inherited from the mother. An egg contains roughly 100,000 or more copies of mtDNA, while the few mitochondria in the sperm midpiece are tagged with ubiquitin and destroyed in the early embryo. A mother with an mtDNA mutation therefore can pass it to all of her children, sons and daughters alike, but an affected father passes it to none.

When all mtDNA copies in a cell are identical, the cell is homoplasmic; when a cell has a mixture of normal and mutant copies, it is heteroplasmic. Three ideas explain why mitochondrial diseases are so variable, even among siblings:

• The bottleneck. During oogenesis, only a small number of mtDNA molecules from each primordial germ cell repopulate the mature egg. Random sampling means a heteroplasmic mother can produce eggs with very different proportions of mutant mtDNA, so one child may be severely affected and another barely at all.
• Replicative segregation. When a cell divides, its mitochondria are distributed to daughter cells at random, so the proportion of mutant mtDNA drifts between cell lineages and tissues over a lifetime.
• The threshold effect. A tissue shows dysfunction only when the fraction of mutant mtDNA exceeds a threshold, often 60 to 90 percent depending on the mutation. Tissues with high energy demand, such as brain, retina, skeletal muscle and heart, reach their limits first.

Chloroplast inheritance

Chloroplasts also carry their own DNA, typically a circle of 120,000 to 160,000 base pairs with roughly 100 to 130 genes for photosynthesis proteins, ribosomal RNAs and transfer RNAs. In 1909 Carl Correns studied the four o'clock plant Mirabilis jalapa, whose branches can be green, white (lacking chlorophyll) or variegated. He found that the phenotype of seedlings depended only on the branch that bore the seed, the female parent, and not at all on the pollen source. Seeds from green branches gave green seedlings, seeds from white branches gave white seedlings (which die once their seed reserves run out), and seeds from variegated branches gave a mix of green, white and variegated seedlings. The variegated branches carried a mixture of normal and defective plastids, and the egg cells received different mixtures, the plant equivalent of heteroplasmy.

Maternal effect

A maternal effect is not the same as maternal inheritance. In a maternal effect, the gene is an ordinary nuclear gene that follows Mendel's laws, but the offspring's phenotype is decided by the mother's genotype, because the mother loads the egg with mRNA and proteins made from her own genes before fertilization. Those products direct the earliest stages of development, before the embryo's own genes are switched on.

The classic example is the direction of shell coiling in the pond snail Lymnaea peregra. The allele D for right handed (dextral) coiling is dominant to d for left handed (sinistral) coiling, but the direction is set by the orientation of the spindle in the first cleavage divisions, which depends on products the mother deposited in the egg. Snails are hermaphrodites, so a single snail can be crossed or self fertilized. The key results are:

• A dd female crossed to a DD male gives F1 offspring that are all Dd, yet all are sinistral, because their mother is dd.
• Selfing the F1 gives F2 offspring in a ratio of 1 DD : 2 Dd : 1 dd, but all are dextral, because their mother (an F1 snail) is Dd.
• Selfing each F2 snail reveals the 3:1 ratio one generation late: the offspring of DD and Dd F2 snails are dextral and the offspring of dd F2 snails are sinistral. So 3/4 of F2 snails produce dextral broods and 1/4 produce sinistral broods.

Maternal effect genes are central to early development. In Drosophila, the mother deposits bicoid mRNA at the front end of the egg. Its protein gradient specifies head and thorax, and a homozygous mutant mother lays eggs that develop without heads even when the embryo carries a normal paternal allele.

Genomic imprinting

For most genes, the copy from the mother and the copy from the father are expressed equally. For a few hundred mammalian genes (estimates for humans run from about 100 to more than 200), expression depends on the parent of origin: one copy is silenced according to which parent it came from. This is genomic imprinting. An imprinted gene is "maternally imprinted" if the maternal copy is silenced (so only the paternal copy is expressed) and "paternally imprinted" if the paternal copy is silenced. Because these terms are easy to flip, "paternally expressed" or "maternally expressed" is clearer.

Imprints are set by DNA methylation at imprinting control regions (ICRs) during gamete formation. In the primordial germ cells of the embryo, old imprints are erased, and new ones are written according to the sex of the individual: a male writes paternal imprints on every chromosome he will pass on, and a female writes maternal imprints. After fertilization, imprinted methylation is protected from the wave of demethylation that resets most of the genome, and it is maintained in somatic cells for life.

The Igf2 and H19 genes on mouse chromosome 7 (human 11p15) show how an imprint works. Igf2 encodes insulin like growth factor 2, a fetal growth factor, and is expressed only from the paternal chromosome. H19, a noncoding RNA gene, is expressed only from the maternal chromosome. Between them lies an ICR. On the maternal chromosome the ICR is unmethylated and binds the insulator protein CTCF, which blocks shared downstream enhancers from reaching Igf2, so the enhancers activate H19 instead. On the paternal chromosome the ICR is methylated, CTCF cannot bind, the enhancers reach Igf2, and methylation spreading into the H19 promoter silences it. A mouse that inherits a disrupted Igf2 gene from its father is small, but one that inherits the same disrupted gene from its mother is normal, because the maternal copy was silent anyway.

Prader Willi and Angelman syndromes

The clearest human example of imprinting is a cluster of genes on chromosome 15q11 to q13. Some genes in the region, including the small nucleolar RNA cluster SNORD116, are expressed only from the paternal chromosome; one gene, UBE3A, is expressed only from the maternal chromosome in neurons. The same deletion of about 5 million base pairs therefore causes two different disorders depending on its parent of origin:

• Prader Willi syndrome results from loss of the paternal contribution: poor muscle tone and feeding problems in infancy, followed by insatiable appetite and obesity, short stature, small hands and feet, hypogonadism and mild to moderate intellectual disability. About 70 percent of cases are paternal deletions and most of the rest are maternal uniparental disomy, in which the child received two copies of chromosome 15 from the mother and none from the father.
• Angelman syndrome results from loss of the maternal contribution: severe intellectual disability, absent speech, seizures, a jerky gait and a frequently happy demeanor with easily provoked laughter. About 70 percent are maternal deletions; others arise from paternal uniparental disomy, imprinting defects or point mutations in the maternal UBE3A.

Imprinting defects at 11p15 cause Beckwith Wiedemann syndrome (overgrowth and embryonal tumor risk, often from excess IGF2) and its mirror image, Silver Russell syndrome (growth restriction). The parental conflict hypothesis proposes that imprinting evolved as a tug of war over fetal growth: paternally expressed genes such as Igf2 tend to promote it and maternally expressed genes tend to restrain it.

Words worth knowing

Linkage and Genetic Mapping

Mendel's second law says that genes assort independently, but a human cell has about 20,000 genes and only 23 pairs of chromosomes, so most genes must share a chromosome with hundreds of others. Genes that travel together break Mendel's ratios in a way that turns out to be enormously useful: how often they are separated tells us how far apart they are. This chapter shows how to turn counts of offspring into maps of chromosomes, from fruit flies to the human disease genes found by following families.

Section 1Linkage and crossing over

Independent assortment works because nonhomologous chromosomes line up independently at metaphase I. Two genes on different chromosomes therefore sort independently. Two genes on the same chromosome are physically connected, and unless something breaks that connection, the alleles that entered meiosis together on one chromosome will leave together in the same gamete. Genes that tend to be inherited together because they lie near each other on a chromosome are linked, and a group of genes that are all linked to one another is a linkage group. In a well mapped species the number of linkage groups equals the haploid chromosome number: four in Drosophila melanogaster, ten in maize, and 23 in humans (counting the X; 24 if the Y is counted separately).

The first hints

The earliest sign of linkage came in 1905, when William Bateson, Edith Saunders and Reginald Punnett crossed sweet peas differing in flower color (purple, dominant, versus red) and pollen shape (long, dominant, versus round). The F2 did not give 9:3:3:1. Instead, the two parental combinations, purple long and red round, were much more common than expected, and the new combinations, purple round and red long, were rarer. Bateson and Punnett called this "coupling," but they had no chromosomal explanation. In 1911 Thomas Hunt Morgan, studying X linked genes in Drosophila such as white eyes and miniature wings, saw the same excess of parental combinations and proposed the explanation we still use: the genes lie on the same chromosome, and the new combinations arise when homologous chromosomes exchange segments during meiosis. He called the exchange crossing over, and he suggested that the farther apart two genes are, the more often a crossover will fall between them.

Parental and recombinant gametes

Consider an individual heterozygous for two linked genes, with A and B on one homolog and a and b on the other. Its gametes fall into two classes:

• Parental (nonrecombinant) gametes carry the same allele combinations that the individual received from its parents: AB and ab.
• Recombinant gametes carry new combinations: Ab and aB.

The arrangement of alleles on the two homologs is called the linkage phase. When the two dominant alleles are on one chromosome and the two recessive alleles on the other (AB/ab), the genes are in coupling, also called the cis configuration. When each chromosome carries one dominant and one recessive allele (Ab/aB), the genes are in repulsion, or the trans configuration. The phase determines which gametes are parental. For a repulsion heterozygote, Ab and aB are the parental gametes and AB and ab are the recombinants. Students often forget this, so always identify the phase before labeling classes.

The slash notation is worth learning. Writing AB/ab means one homolog carries A and B and the other carries a and b. Writing the genotype as AaBb tells you only that the individual is a double heterozygote and says nothing about phase.

Crossing over happens at the four strand stage

Crossing over takes place in prophase I of meiosis, after DNA replication, when homologous chromosomes are tightly paired by the synaptonemal complex. The process begins when the protein Spo11 makes deliberate double strand breaks in the DNA; some of these breaks are repaired using a nonsister chromatid as a template and resolved as crossovers, which reciprocally exchange the chromosome arms beyond the break. When the homologs begin to separate in diplotene, the crossover sites remain visible as chiasmata, X shaped connections that hold the homologs together until anaphase I. Each pair of homologs normally forms at least one crossover, the obligate crossover, which is necessary for the pair to orient and segregate properly. Failure of crossing over raises the risk of nondisjunction.

Three facts follow from the timing. First, crossing over involves two of the four chromatids at any one site, so one crossover between two genes yields two recombinant and two parental chromatids. Second, for that reason the frequency of recombinant gametes is half the frequency of meioses in which a crossover occurs in the interval. Third, because a crossover can also involve different pairs of chromatids at different sites, double crossovers can restore the parental combination of outside markers, a point we will return to in Section 3.

Proof that crossing over is physical exchange

In 1931 Harriet Creighton and Barbara McClintock in maize, and independently Curt Stern in Drosophila, showed that genetic recombination is accompanied by physical exchange of chromosome parts. Creighton and McClintock used a maize chromosome 9 that carried two visible landmarks, a dark knob at one end and an extra piece of another chromosome at the other, along with alleles for kernel color and starch type. Plants that were recombinant for the alleles had also exchanged the landmarks, exactly as the crossing over model predicts. Genetic maps were from then on understood as maps of real physical objects.

Complete and incomplete linkage

If two genes are so close together that a crossover essentially never occurs between them, or if the species does not cross over in one sex, linkage is complete and only parental gametes are made. Drosophila males have no meiotic crossing over at all, and neither do females of the silkworm moth. More often linkage is incomplete: parental gametes are in the majority but some recombinants appear. The best way to see the difference is a testcross to a homozygous recessive, in which each offspring's phenotype directly reveals the gamete it received from the heterozygous parent.

Section 2Recombination frequency and map units

In 1911 Alfred Sturtevant, then an undergraduate in Morgan's laboratory, realized that the frequency of recombination between linked genes could serve as a measure of the distance between them. That night, as he later recalled, he neglected his homework and produced the first genetic map, ordering five X linked genes in Drosophila. The idea rests on a simple assumption: crossovers occur more or less at random along a chromosome, so the chance that one falls between two genes is proportional to the distance separating them.

Calculating recombination frequency

The recombination frequency (RF) between two genes is the proportion of gametes, or of testcross progeny, that are recombinant:

RF = number of recombinant offspring ÷ total number of offspring

One percent recombination defines one map unit (m.u.), also called a centimorgan (cM) in honor of Morgan. So genes with 8 percent recombination are 8 map units apart. One hundred centimorgans make one Morgan, a unit used mainly in mapping functions.

Why recombination frequency cannot exceed 50 percent

As two genes get farther apart, more than one crossover can occur between them. Each crossover switches which homolog the chromatid is copying, so an odd number of crossovers on a chromatid between two genes gives a recombinant, and an even number (two, four and so on) restores the parental combination. When genes are very far apart, crossovers are so frequent that the chance of an odd number approaches the chance of an even number, and recombinants approach, but never exceed, 50 percent. That is exactly the value for genes on different chromosomes. So:

• An RF near 50 percent means the genes are either on different chromosomes or far apart on the same chromosome. Such genes are syntenic (on the same chromosome) but genetically unlinked.
• RF underestimates distance once double crossovers become common, roughly above 10 to 15 map units, because double crossovers between two markers are counted as parentals.
• Map distances are therefore best measured over short intervals and added up. Added in this way, a chromosome's total map length can be much longer than 50 cM. The Drosophila X chromosome is about 66 cM long, and a large human chromosome such as chromosome 1 is more than 250 cM long.

Testing for linkage with chi square

Data from real crosses never match expectations exactly, so we need a way to decide whether a deviation from 1:1:1:1 in a testcross is due to chance or to linkage. The chi square test compares observed numbers (O) with the numbers expected (E) under a null hypothesis; for linkage questions the null hypothesis is that the genes assort independently. The statistic is

χ² = Σ (O − E)² / E

summed over all classes. The degrees of freedom are the number of classes minus one. The calculated value is compared with a critical value from a chi square table at the chosen significance level, usually P = 0.05. If the calculated value is larger than the critical value, the deviation is too large to blame on chance and the null hypothesis is rejected.

Genetic distance versus physical distance

Map units measure crossover frequency, not DNA length, and the two do not convert at a fixed rate. Averaged over the human genome, about 1 cM corresponds to roughly 1 million base pairs (1 Mb), but in baker's yeast, which has a tiny genome and very high crossover rates, 1 cM is only about 3,000 base pairs. Even within one genome, the relationship varies widely:

• Hotspots and coldspots. In humans and mice, most crossovers cluster in short hotspots one or two thousand base pairs long, many of them marked by binding sites for the protein PRDM9, which places histone marks that attract Spo11. Regions near centromeres and in heterochromatin have very little crossing over, so genes there look closer on the genetic map than they really are.
• Sex differences. The human female map is roughly 1.6 times longer than the male map, meaning women make more crossovers per meiosis. In Drosophila the difference is absolute: males make none.
• Structural changes. In an individual heterozygous for an inversion, crossovers within the inverted region produce unbalanced gametes that are not recovered, so recombination appears suppressed (Chapter 6).

Despite these quirks, the order of genes on a genetic map matches their physical order on the chromosome. Only the spacing differs.

Section 3Three point crosses and gene order

Mapping genes two at a time works, but it is slow and it leaves the order of genes uncertain when distances are similar. If A and B are 10 map units apart and B and C are 4 apart, C could lie between A and B or outside them. A three point testcross solves both problems at once. A female (or other individual) heterozygous for three linked genes is crossed to a tester that is homozygous recessive for all three. Every offspring then reveals, gene by gene, which alleles were on the chromatid it received, and a single cross gives the order of the genes, the two distances and a direct count of double crossovers.

The eight classes

A trihybrid makes 2³ = 8 kinds of gametes, so a three point testcross gives eight phenotypic classes. They pair up into four reciprocal pairs, and the pairs come in characteristic sizes:

• Parental (noncrossover) classes: the two largest classes, carrying the allele combinations of the two chromosomes in the heterozygote.
• Single crossover classes in region I: a pair produced by a crossover between the first and middle genes.
• Single crossover classes in region II: a pair produced by a crossover between the middle and last genes.
• Double crossover (DCO) classes: the two smallest classes, produced by one crossover in each region. Since the probability of two independent events is the product of their probabilities, double crossovers are rarer than either kind of single crossover.

Each reciprocal pair should contain roughly equal numbers, because each crossover produces both reciprocal products. If the numbers in a pair are wildly different, suspect reduced viability of one phenotype.

Finding the middle gene

The key insight of three point mapping is that a double crossover swaps only the middle gene relative to the two outside genes. Compare a parental class with a double crossover class that shares the same alleles at two genes: the one gene that differs is the gene in the middle. Once the middle gene is known, rewrite every class in the correct order, and the single crossover classes sort themselves into region I and region II.

Calculating the distances

The distance between two adjacent genes is the frequency of all offspring recombinant in that interval. Double crossovers are recombinant in both intervals, so they must be added to each:

• Distance in region I = (single crossovers in region I + double crossovers) ÷ total
• Distance in region II = (single crossovers in region II + double crossovers) ÷ total

The distance between the two outside genes is the sum of the two intervals. If you instead calculate it directly by counting only offspring recombinant for the two outside genes, you miss the double crossovers, which look parental for those genes. The direct value is smaller than the sum by exactly twice the DCO frequency, the numerical proof that two point mapping over long distances underestimates the truth.

A systematic routine

1. Pair up the eight classes as reciprocals and add each pair.
2. Call the largest pair parental and the smallest pair double crossovers.
3. Find the middle gene by comparing the parentals with the DCOs.
4. Rewrite all classes in the correct gene order.
5. Assign the remaining two pairs to region I or region II.
6. Compute each interval, remembering to add the DCOs to both.
7. Check that the outer distance equals the sum of the intervals, then compute interference (Section 4).

Section 4Interference, coefficient of coincidence and mapping functions

Expected versus observed double crossovers

If crossovers in two adjacent regions occurred independently, the probability of a double crossover would equal the product of the two single interval recombination frequencies. In the three point example, that expectation is 0.12 × 0.08 = 0.0096, or 9.6 double crossovers per 1,000 offspring. Only 6 were observed. Fewer double crossovers than expected is the usual finding, and it means that one crossover reduces the chance of another nearby. This phenomenon is called crossover interference (or chiasma interference).

Interference is measured in two related ways:

• The coefficient of coincidence (c.o.c.) is the ratio of observed to expected double crossovers: c.o.c. = observed DCO ÷ expected DCO.
• Interference (I) is its complement: I = 1 − c.o.c.

Interference is strongest over short distances and fades as intervals get longer. In Drosophila and in humans, two crossovers almost never occur within about 10 cM of each other on the same chromatid pair. The mechanism is still being worked out, but the leading models propose that the formation of one crossover sends a signal, possibly mechanical stress relieved along the chromosome axis, that inhibits others nearby. Most organisms have two classes of crossovers: an interfering class, which accounts for the majority, and a smaller noninterfering class made by a different resolution pathway. Interference also helps ensure that each chromosome pair gets its obligate crossover without wasting crossovers close together.

Mapping functions

Because double crossovers hide, recombination frequency is a biased estimate of true map distance, especially above 10 to 15 cM. A mapping function is a formula that converts an observed recombination frequency (r) into an estimated map distance (d) by correcting for undetected multiple crossovers.

Haldane's mapping function assumes no interference: crossovers fall at random, so the number per chromatid in an interval follows a Poisson distribution. If d is the map distance in Morgans, the mean number of crossovers per meiosis in the interval is 2d (each crossover involves only two of the four chromatids). The probability of no crossover is e^−2d, and every meiosis with at least one crossover produces 50 percent recombinant chromatids on average. Therefore

r = ½(1 − e^−2d), and rearranged, d = −½ ln(1 − 2r)

As d becomes large, e^−2d approaches zero and r approaches 1/2, which is why recombination frequency levels off at 50 percent.

Kosambi's mapping function allows for interference that is strong over short distances and weakens with distance, which fits most real data better:

d = ¼ ln[(1 + 2r) ÷ (1 − 2r)]

For small values of r, both functions give almost exactly d = r; they diverge as r grows. Human linkage maps are usually reported in Kosambi or Haldane centimorgans, and the choice should always be stated.

Section 5Mapping in humans: LOD scores and molecular markers

Humans cannot be testcrossed, families are small, and for most of the twentieth century there were few genes whose alleles could be followed through a family. Human gene mapping became practical when two things came together: a statistical method for combining evidence from many small families, and DNA molecular markers, harmless sequence differences scattered by the hundreds of thousands across the genome.

Informative meioses and phase

Linkage can be measured only in meioses from a parent who is heterozygous at both loci: the disease gene and the marker. Such a meiosis is informative. To count recombinants we also need to know the phase, which marker allele sits on the same chromosome as the disease allele. In a phase known family, the grandparents' genotypes reveal the phase. In a phase unknown family, both phases must be considered, which weakens the evidence from each child.

The LOD score

In 1955 Newton Morton introduced the LOD score (logarithm of the odds) to weigh the evidence for linkage. For a chosen recombination fraction θ (theta, the human geneticist's symbol for RF), the LOD score compares the probability of the observed family data if the loci are linked at θ with the probability if they are unlinked (θ = 0.5):

LOD(θ) = log₁₀ [ L(θ) ÷ L(0.5) ]

For phase known meioses with R recombinants and NR nonrecombinants, L(θ) = θ^R(1 − θ)^NR and L(0.5) = 0.5^{R + NR}. LOD scores are calculated for a series of θ values, and the value of θ that gives the highest LOD is the best estimate of the recombination fraction. Because logarithms of independent likelihoods add, LOD scores from different families can simply be summed at each θ, which is the method's great strength.

The conventional thresholds are:

• LOD of 3 or more (odds of 1,000 to 1 in favor of linkage) is accepted as evidence of linkage. The bar is high because any two loci picked at random are unlikely to be linked; with that low prior probability, a LOD of 3 corresponds to roughly a 5 percent chance that the linkage is false.
• LOD of −2 or less (odds of 100 to 1 against) excludes linkage at that θ.
• Values between −2 and 3 are inconclusive; more families are needed.

For phase unknown families, the likelihood is averaged over the two possible phases, each with probability 1/2. For example, for a father whose phase is unknown and whose children show R of one kind and NR of the other, L(θ) = ½[θ^R(1 − θ)^NR + θ^NR(1 − θ)^R]. In practice, computer programs calculate LOD scores for complex pedigrees, taking account of penetrance, missing individuals and marker allele frequencies.

Molecular markers

A good marker is polymorphic (many people are heterozygous, so meioses are informative), common across the genome, easy to type and inherited codominantly, so both alleles are visible. Several generations of markers have been used:

Markers transformed medical genetics through positional cloning: find a marker linked to the disease, narrow the interval with more markers and more families, then search the interval for the gene. Huntington disease was linked to a marker on chromosome 4 in 1983, using one of the first RFLPs, and the gene itself, HTT, was identified ten years later. Cystic fibrosis was mapped to chromosome 7 and its gene, CFTR, cloned in 1989. Today whole genome sequencing often finds rare disease genes directly, but linkage still narrows the search.

Linkage disequilibrium and association studies

Linkage analysis in families follows a handful of meioses. Population studies take advantage of thousands of past meioses. When a new mutation arises, it sits on a particular chromosome with a particular set of nearby marker alleles, a haplotype. Over many generations, recombination breaks the association with distant markers, but very close markers can remain associated for thousands of years. This nonrandom association of alleles at nearby loci is called linkage disequilibrium. Genome wide association studies (GWAS) type hundreds of thousands of SNPs in thousands of cases and controls and look for SNPs whose allele frequencies differ; because of linkage disequilibrium, a SNP that is not itself causal can flag a nearby causal variant. GWAS has found thousands of common variants with small effects on traits such as height, type 2 diabetes and coronary disease, complementing family linkage studies, which are best at finding rare variants of large effect.

Words worth knowing

Bacterial and Viral Genetics

Bacteria and viruses have no meiosis, yet they exchange genes constantly, and that exchange is why antibiotic resistance spreads through a hospital in months and why a new influenza strain can appear almost overnight. The same tiny genomes gave geneticists their sharpest tools: billions of offspring in a single tube, selection for one rare recombinant among millions, and maps fine enough to resolve neighboring base pairs. This chapter shows how bacteria and phages move DNA, how each route of transfer becomes a mapping method, and how viral genomes, including retroviruses, are built and copied.

Section 1Bacterial genomes, plasmids and mutants

Why bacteria are such powerful genetic organisms

A culture of Escherichia coli in rich medium doubles about every 20 minutes, so a single cell becomes more than a billion cells overnight. Bacteria are haploid, which means every mutation shows its phenotype at once, with no dominant allele hiding a recessive one. And because you can spread a hundred million cells on one Petri plate and allow only a rare type to grow, you can find events that happen once in a million or once in a billion cells.

The bacterial chromosome

Most bacteria carry a single, circular, double stranded DNA chromosome. The E. coli K-12 chromosome is about 4.6 million base pairs (4.6 Mb) and encodes roughly 4,300 genes. It is compacted into an irregular region called the nucleoid. There is no nuclear membrane. Compaction comes from negative supercoiling, introduced by DNA gyrase (a type II topoisomerase), and from nucleoid associated proteins such as HU, H-NS and IHF that bend and bridge DNA.

Bacterial genomes are gene dense: about 88 percent of the E. coli chromosome codes for protein, introns are essentially absent, and functionally related genes are often clustered into operons that are transcribed as one messenger RNA. Replication begins at a single origin, oriC, and proceeds in both directions until the two forks meet in the terminus region opposite the origin.

Plasmids

Plasmids are small DNA molecules, usually circular and double stranded, that replicate independently of the chromosome. Each has its own origin of replication, which also sets its copy number: some plasmids are kept at one or two copies per cell under tight control, while others reach dozens or hundreds. Plasmids are generally not essential for growth under ordinary conditions, but the genes they carry can be decisive when conditions change. An episome is a plasmid that can either replicate on its own or integrate into the chromosome; the F factor is the classic example.

Notation for genotypes and phenotypes

Bacterial geneticists use a compact notation. Genotypes are written as three lowercase italic letters followed by an uppercase letter for the specific gene: leuA, lacZ, trpE. A superscript plus means the wild type allele (leu⁺, able to make leucine) and a superscript minus means a mutant allele (leu⁻, unable to make leucine). str^r means streptomycin resistant. Phenotypes are written in roman type with the first letter capitalized: Leu⁺, Lac⁻, Str^r.

Kinds of bacterial mutants

A prototroph is a wild type strain that grows on minimal medium: water, salts, a carbon source such as glucose and nothing else. From those raw materials it builds every amino acid, nucleotide and vitamin it needs. An auxotroph has lost the ability to make one or more of these and grows only if the missing nutrient is supplied. A his⁻ auxotroph, for example, grows on minimal medium plus histidine but not on minimal medium alone. Complete medium contains everything, so both prototrophs and auxotrophs grow on it.

• Nutritional (auxotrophic) mutants cannot synthesize a nutrient.
• Carbon source mutants cannot use a particular sugar; a Lac⁻ mutant cannot grow when lactose is the only carbon source.
• Resistance mutants survive an antibiotic or a phage that kills the wild type.
• Conditional mutants show the mutant phenotype only under restrictive conditions. A temperature sensitive mutant makes a protein that works at 30 °C (permissive) but unfolds at 42 °C (restrictive). They let geneticists study essential genes.

Finding mutants: selection, screening and replica plating

Resistance mutants are easy to find by direct selection: spread 10⁸ cells on a plate containing the antibiotic, and only resistant cells form colonies. Auxotrophs are harder, because the mutant grows on fewer media than the wild type, not more. Joshua and Esther Lederberg solved this in 1952 with replica plating. Cells are spread on a complete medium master plate, giving a few hundred colonies. A sterile velvet pad pressed onto the master picks up cells from every colony and can then print that exact pattern onto a series of fresh plates. Any colony present on the master and on complete medium but absent from minimal medium is an auxotroph, and its position tells you where to pick it from the master. Printing onto minimal medium supplemented with single nutrients then identifies exactly which nutrient it needs.

Do mutations arise before or after selection? The fluctuation test

In the early 1940s it was an open question whether phage T1 caused resistance in the few cells that survived it, or whether resistant mutants arose spontaneously and the phage merely revealed them. Salvador Luria and Max Delbrück answered with the fluctuation test (1943). They grew many small, independent cultures from a few cells each, then plated each entire culture with phage.

The logic is simple. If the phage induced resistance at the moment of plating, every culture would have about the same small chance per cell, and the counts would vary little from culture to culture, like samples from one flask. If mutations arose at random during growth, a mutation early in the history of one culture would leave a huge resistant clone (a "jackpot"), while most cultures would have few or none. The independent cultures showed enormous fluctuation, with many zeros and a few jackpots, while the samples from the single flask were uniform. Mutations are spontaneous and random; selection only reveals them.

Section 2Conjugation and mapping by interrupted mating

The discovery of gene exchange

In 1946 Joshua Lederberg and Edward Tatum mixed two E. coli auxotrophs. One strain was met⁻ bio⁻ thr⁺ leu⁺; the other was met⁺ bio⁺ thr⁻ leu⁻. Neither grew on minimal medium alone, but the mixture produced prototrophic colonies at about 1 in 10⁷ cells. Double reversion could not explain them. Bernard Davis then placed the two strains on opposite sides of a U-tube separated by a filter that let medium and large molecules through but not cells. No prototrophs formed, showing that exchange required cell to cell contact. This contact dependent transfer is conjugation.

The F factor and the three mating types

Conjugation depends on the F (fertility) factor, a plasmid of about 100 kb. A cell carrying F is F⁺; a cell lacking it is F⁻. The tra genes of F encode the sex pilus, which pulls an F⁻ cell close, plus the transfer machinery. Transfer begins when a relaxase nicks one strand at the origin of transfer (oriT). That single strand, 5′ end first, threads into the recipient while the donor replaces it by rolling circle replication; the recipient synthesizes the complementary strand. The result of an F⁺ × F⁻ mating is two F⁺ cells. F spreads rapidly, yet F⁺ cells rarely transfer chromosomal genes.

About once in 10⁴ to 10⁵ F⁺ cells, the F factor integrates into the chromosome by recombination between short homologous sequences (often insertion sequences) on F and on the chromosome. The result is an Hfr (high frequency of recombination) cell. When an Hfr cell conjugates, transfer still starts at oriT, but now the strand that follows is chromosomal DNA. Genes beside the integration site enter first, then genes farther away, in a fixed order and direction set by where and in which orientation F inserted. Transferring the entire 4.6 Mb chromosome takes about 100 minutes at 37 °C, and the mating pair almost always breaks apart before that. The part of F that lies behind oriT is transferred last, so the recipient of an Hfr mating nearly always remains F⁻.

Occasionally F excises from the Hfr chromosome imprecisely and carries a piece of neighboring chromosome with it. This plasmid is an F′ (F prime) factor. An F′ transfers efficiently like F, and the recipient becomes a partial diploid, or merozygote, carrying two copies of the captured genes.

Recombination in the recipient

The donor fragment that enters the recipient is linear and cannot replicate on its own. To be inherited, its genes must be recombined into the recipient's circular chromosome, and that requires an even number of crossovers, at least two, one on each side of the gene. A single crossover would open the circle into a linear molecule that cannot survive. The donor genes that do not recombine in are degraded. The recipient therefore needs a functional RecA protein.

Interrupted mating

In the mid 1950s Élie Wollman and François Jacob turned the ordered transfer of Hfr strains into a mapping method. They mixed an Hfr strain sensitive to streptomycin (str^s) with an F⁻ recipient that carried several auxotrophic markers and was streptomycin resistant (str^r). At intervals they removed samples and broke the mating pairs apart in a kitchen blender, then plated the cells on media containing streptomycin to kill the donors (counterselection) and lacking a particular nutrient to select for recipients that had received a donor allele.

Each donor gene first appeared among recombinants at a characteristic time of entry, and the frequency of recombinants for that gene then rose to a plateau. Genes that entered early reached higher plateaus than genes that entered late, because the longer the transfer runs, the more likely the pair is to break apart spontaneously. This decline is called the gradient of transfer. The time of entry measures position: the E. coli chromosome is mapped in minutes, from 0 to 100, with the thr locus placed at 0. One minute corresponds to roughly 46 kb.

Building a circular map from several Hfr strains

Different Hfr strains have F inserted at different places and in different orientations, so each transfers the chromosome starting from a different point and in one of two directions. No single Hfr can map the whole chromosome, because pairs break apart long before 100 minutes. But the orders from several strains overlap, and when you align them the only arrangement consistent with all of them is a circle.

Section 3Transformation and transduction

Transformation: taking up naked DNA

Transformation is the uptake of free DNA from the surroundings and its incorporation into the recipient's genome. It was the first form of bacterial gene transfer discovered. In 1928 Frederick Griffith found that heat killed virulent (smooth, encapsulated) Streptococcus pneumoniae could convert living harmless (rough) cells into virulent ones inside a mouse. In 1944 Oswald Avery, Colin MacLeod and Maclyn McCarty purified the "transforming principle" and showed that it was destroyed by DNase but not by protease or RNase. That experiment identified DNA as the genetic material.

Some species are naturally competent: they switch on a set of genes, usually in response to crowding or stress, that builds a DNA uptake machine. Streptococcus pneumoniae, Bacillus subtilis and Neisseria gonorrhoeae are classic examples. One strand of bound DNA enters and pairs with the homologous region of the chromosome with the help of RecA, forming a heteroduplex, and mismatch repair or the next round of replication resolves it into one recipient chromosome and one transformed chromosome. Because the entering strand is linear, at least two crossovers are again needed, one on each side of the marker.

E. coli is not naturally competent, so laboratories make it artificially competent, either by chilling cells in calcium chloride and giving a brief heat shock, or by electroporation, a short high voltage pulse that opens transient pores.

Mapping by cotransformation

A transforming fragment is only a small piece of the chromosome, typically a few tens of kilobases. Two genes can be transferred together on one fragment, an event called cotransformation, only if they lie close together. The frequency of cotransformation therefore measures linkage: the closer two genes, the more often they travel together. Two tests are used.

• Compare double transformants with the product of single rates. If genes a and b are far apart, a cell must take up two separate fragments, and the double transformation frequency is roughly the product of the single frequencies. If they are close, double transformants are far more common than that product.
• Order three genes by pairwise cotransformation. The two genes that are cotransformed least often are the outside genes; the third lies between them.

Transduction: phages as couriers

Transduction is gene transfer from one bacterium to another carried inside a phage particle. Zinder and Lederberg discovered it in 1952 with phage P22 in Salmonella: transfer crossed a U-tube filter and resisted DNase. There are two types, and they differ in what the phage packages.

Generalized transduction

In generalized transduction, any part of the host chromosome can be transferred. Phages such as P1 (E. coli) and P22 (Salmonella) degrade the host chromosome into fragments during a lytic infection and package DNA into heads by a headful mechanism: they fill the head with as much DNA as fits. Occasionally, about once in every thousand or so particles, a head is filled with a piece of host DNA instead of phage DNA. Such a transducing particle contains no phage genes. It injects its bacterial DNA into the next cell it infects, and that cell survives because no phage genes enter. If the donor fragment recombines into the recipient chromosome by two crossovers, the result is a stable transductant (complete transduction). A fragment that persists without recombining or replicating gives abortive transduction.

A P1 head holds about 100 kb, roughly 2 percent of the E. coli chromosome, or about 2 minutes of the map. Two genes can be cotransduced only if they lie within that distance, and the closer they are, the more often they are.

Specialized transduction

In specialized transduction, only genes next to a prophage's integration site can be transferred. The classic example is phage lambda (λ), which integrates at a single att site between the gal and bio genes of E. coli. When a lysogen is induced, the prophage normally excises precisely. Rarely (about once in 10⁵ to 10⁶ excisions), the excision cut is in the wrong place, and the circle that loops out contains some host DNA on one side (gal or bio) and lacks a corresponding piece of phage DNA on the other. The resulting particle, written λdgal (defective, carrying gal) or λdbio, can inject the host gene into a new cell. A lysate that contains such particles transduces gal or bio at high frequency, but never transduces genes elsewhere on the chromosome.

Mapping by cotransduction

Cotransduction frequency is used exactly like cotransformation. You select for one donor marker and score how many of the selected transductants also inherited a second, unselected marker. Close genes are cotransduced often; genes more than one transducing fragment apart are never cotransduced. For P1, Wu derived a useful approximation relating cotransduction frequency C to map distance d (in minutes) when the fragment length L is 2 minutes:

C = (1 − d / L)³

Three factor transduction crosses resolve gene order without any distance formula. Select for one marker and score the other two among the transductants. The class that requires four crossovers rather than two will be by far the rarest, and it identifies which gene is in the middle.

Section 4Bacteriophage genetics and fine structure mapping

Phage life cycles

Bacteriophages (phages) are viruses that infect bacteria. A phage such as T4 has a protein head holding its double stranded DNA genome (about 169 kb) and a tail that attaches to receptors on the cell surface and injects the DNA. In the lytic cycle, the phage takes over the cell: early genes shut down host functions and degrade host DNA, middle and late genes replicate phage DNA and build heads and tails, the DNA is packaged, and a lysozyme-like enzyme bursts the cell. At 37 °C, a T4 infection lasts about 25 minutes and releases a burst of roughly 100 to 200 new phages.

Temperate phages such as lambda can choose. After infection, lambda either enters the lytic cycle or enters the lysogenic cycle, in which its 48.5 kb genome integrates into the host chromosome at att by site specific recombination, catalyzed by phage integrase. The integrated DNA is a prophage, and the host cell, now a lysogen, divides normally and copies the prophage along with its own DNA. The lambda cI repressor keeps the lytic genes silent and also makes the lysogen immune to superinfection by another lambda. When the host DNA is damaged (for example by ultraviolet light), activated RecA stimulates self cleavage of the cI repressor, the prophage excises (integrase plus excisionase), and the lytic cycle begins: this is induction.

Plaques: how phage genetics is scored

Phages are counted by the plaque assay. Phages are mixed with a large excess of bacteria in soft agar and poured over a plate. Bacteria grow into a cloudy lawn, but wherever a single phage infected a cell, successive rounds of infection and lysis clear a small circular zone, a plaque. Each plaque comes from one phage, so plaques measure the titer:

titer (plaque forming units per mL) = number of plaques / (volume plated in mL × dilution)

Phage mutants change plaque appearance or host range. Rapid lysis (r) mutants of T2 and T4 make large, sharp edged plaques, while wild type (r⁺) makes small, fuzzy plaques. Host range (h) mutants can infect bacterial strains that resist the wild type. Phage crosses use mixed infection: each cell receives both genotypes, which recombine as they replicate.

Benzer and the fine structure of the rII region

In the 1950s, Seymour Benzer asked how finely recombination could divide a gene. He used the rII mutants of phage T4. The trick was a pair of hosts:

• On E. coli B, rII mutants grow well and form large r plaques.
• On E. coli K12(λ), a strain lysogenic for lambda, rII mutants cannot grow at all, while wild type T4 grows normally.

That asymmetry created an extraordinarily sensitive selection. After a cross between two rII mutants, plating on K12(λ) reveals only the rare wild type recombinants, while plating on B counts every progeny phage. Because one plate can hold progeny from a whole lysate, Benzer could detect one wild type recombinant among hundreds of millions of progeny, enough to measure recombination between mutations only one or a few base pairs apart.

Two different tests: complementation and recombination

Benzer used two tests that are easy to confuse, so keep them separate.

The complementation test asks whether two mutations are in the same functional unit. Cells of K12(λ) are coinfected with two rII mutants. If the two mutations are in different genes, each phage supplies the product the other lacks, the infection succeeds and the culture lyses: they complement. If both mutations are in the same gene, neither phage makes a working product and nothing happens. This is a trans test (the mutations are on different DNA molecules). Its control is the cis arrangement (both mutations on one molecule, the other molecule wild type), which always gives wild type function. Benzer called the unit defined by the cis-trans test a cistron, and found that the rII region contains two: rIIA and rIIB.

The recombination test asks whether two mutations are at different sites, and how far apart. Cells of E. coli B (on which all rII mutants grow) are coinfected, the progeny are collected, and the lysate is plated on both hosts. Wild type recombinants are one of two reciprocal products; the other (the double mutant) cannot be seen on K12(λ), so the frequency of wild type is doubled:

RF = 2 × (titer on K12(λ)) / (titer on B)

Mutations at the identical site never give wild type recombinants, and mutations in the same cistron can still recombine if they are at different sites.

Deletion mapping

Rather than crossing thousands of mutants pairwise, Benzer used deletion mutants, which are recognizable because they never revert. A point mutation crossed with a deletion can give wild type recombinants only if the point mutation lies outside the deleted segment; if the site is missing from the deletion phage, there is no wild type copy of that stretch to recombine in. A panel of overlapping deletions therefore divides the region into segments, and a single spot test against the panel assigns each new mutation to a segment. Fine mapping by recombination is then needed only among the few mutations within that segment.

What the rII work showed

Benzer mapped more than 2,000 rII mutations to about 300 sites. Recombination could separate mutations at neighboring sites, so the gene is not indivisible: the units of mutation and recombination are much smaller than the gene, ultimately single base pairs. He also found mutational hotspots. The map was linear, like DNA.

Section 5Viral genomes and retroviruses

What a virus is, genetically

A virus is a genome wrapped in a protein coat (the capsid), sometimes surrounded by a lipid envelope taken from the host membrane. Viruses have no ribosomes of their own, but their genomes vary enormously. Viral genomes can be DNA or RNA, single stranded or double stranded, linear or circular, in one piece or split into segments. They range from about 2 kb to more than a megabase.

The Baltimore classification

David Baltimore grouped viruses by the route their genome takes to messenger RNA (mRNA, defined as the plus strand), which predicts which enzymes each virus must carry.

Note that the naked genome of a plus strand RNA virus is infectious by itself, because host ribosomes can translate it directly, whereas minus strand and double stranded RNA viruses must carry their polymerase inside the virion. RNA dependent RNA polymerases and reverse transcriptases generally lack proofreading, so RNA viruses mutate far faster than DNA organisms, roughly 10⁻⁴ to 10⁻⁵ errors per nucleotide per replication compared with about 10⁻⁹ to 10⁻¹⁰ for cellular DNA replication. Coronaviruses (about 30 kb) are an exception: they encode a proofreading exonuclease.

Viral recombination and reassortment

Viruses exchange genes in two ways. When two related viruses infect the same cell, their polymerases can switch templates, producing recombinant genomes. Viruses with segmented genomes can also swap whole segments, a process called reassortment. Influenza A has eight minus strand RNA segments, including the ones encoding the surface proteins hemagglutinin (HA) and neuraminidase (NA). Point mutations in HA and NA cause antigenic drift (hence yearly vaccines); reassortment between, say, human and avian strains gives a new HA or NA, antigenic shift, the source of pandemics.

The retrovirus genome

Retroviruses reverse the usual flow of genetic information: their RNA genome is copied into DNA, which integrates permanently into the host chromosome. Each virion carries two copies of a plus strand RNA genome of about 7 to 10 kb (HIV-1 is about 9.7 kb), so retroviruses are effectively diploid. Every retrovirus has three core genes:

• gag (group specific antigen): structural proteins of the matrix, capsid and nucleocapsid.
• pol: the enzymes reverse transcriptase, integrase and protease.
• env: the envelope glycoproteins that bind receptors and fuse membranes (gp120 and gp41 in HIV).

HIV also carries regulatory and accessory genes such as tat, rev and nef.

The retroviral life cycle

1. Entry. Env binds the receptor (CD4 plus the coreceptor CCR5 or CXCR4 for HIV), the envelope fuses with the cell membrane, and the core enters the cytoplasm.
2. Reverse transcription. Reverse transcriptase, an RNA dependent and DNA dependent DNA polymerase with an RNase H domain that degrades the RNA in RNA DNA hybrids, makes a double stranded DNA copy of the genome. It is primed by a host transfer RNA (tRNA^Lys3 for HIV). Two template "jumps" generate identical long terminal repeats (LTRs) at the two ends of the DNA. Switching between the two RNA copies also causes recombination.
3. Integration. The viral DNA enters the nucleus, and integrase cuts the host DNA with a staggered cut and joins the viral ends to it. Repair of the gaps creates a short target site duplication (5 bp for HIV) flanking the provirus. The provirus is now a permanent part of that cell's genome and of all its descendants.
4. Expression. Host RNA polymerase II transcribes the provirus from a promoter in the 5′ LTR; the 3′ LTR supplies the polyadenylation signal. Spliced transcripts make Env; full length RNA makes Gag and Gag-Pol and serves as new genome.
5. Assembly and maturation. Two genome RNAs, Gag and Gag-Pol assemble at the membrane and bud off with an envelope. The viral protease then cleaves the polyproteins into mature proteins, converting the particle into an infectious virion.

Retroviruses, oncogenes and our own genome

Retroviruses gave cancer genetics its first oncogene. Rous sarcoma virus, which causes tumors in chickens, carries v-src, a gene that the virus captured long ago from a normal chicken gene, c-src, a proto-oncogene that encodes a tyrosine kinase. The viral copy is mutated and overexpressed, so it drives uncontrolled growth. Other retroviruses carry no oncogene but cause cancer by insertional activation, integrating beside a proto-oncogene such as MYC so that the strong LTR promoter or enhancer drives it.

When a retrovirus infects a germ line cell, the provirus can be inherited like any other gene. About 8 percent of the human genome consists of such endogenous retroviruses, almost all inactivated by mutation. Related LTR retrotransposons move within the genome by the same copy, reverse transcribe and integrate mechanism.

Words worth knowing

Chromosome Mutations

A single base change can alter one protein, but a chromosome mutation can add, remove or rearrange hundreds of genes at once. These large scale changes cause about half of early miscarriages, Down syndrome and many developmental disorders, drive certain leukemias, and built some of the most important crops on Earth. This chapter shows how chromosomes are described, how they break and rejoin, how whole chromosomes or whole sets are gained or lost, and what each change does in meiosis and in the clinic.

Section 1Chromosome structure and karyotypes

Parts of a chromosome

Every eukaryotic chromosome is one long DNA molecule packaged with proteins into chromatin. After replication, and until anaphase, each chromosome consists of two identical sister chromatids held together at the centromere. Three elements are essential for a chromosome to be copied and passed on faithfully:

• Origins of replication, many per chromosome, so that a long molecule can be copied in the hours of S phase.
• A centromere, where the kinetochore assembles and attaches to spindle microtubules. Human centromeres are built on long arrays of a 171 bp repeat called alpha satellite DNA, but they are defined epigenetically by nucleosomes containing the histone H3 variant CENP-A.
• Two telomeres, the protective ends. Human telomeres are thousands of tandem copies of TTAGGG bound by the shelterin protein complex. Telomeres keep chromosome ends from being mistaken for broken DNA and joined together.

The importance of these elements explains much of this chapter. A chromosome fragment with no centromere (acentric) cannot attach to the spindle and is lost at division. A chromosome with two centromeres (dicentric) may be pulled toward both poles and break. A broken end without a telomere is "sticky" and tends to fuse with other broken ends, which is how many rearrangements form.

Centromere position and chromosome shape

The centromere divides a chromosome into two arms. By convention the short arm is p (for petit) and the long arm is q. Chromosomes are classified by where the centromere lies, which is measured by the arm ratio (q length divided by p length) or the centromeric index (p length divided by total length).

The tiny short arms of the five human acrocentric chromosomes carry the tandem ribosomal RNA genes, the nucleolus organizer regions. Because the same genes are repeated on all five, losing the short arms of two of them, as happens in a Robertsonian translocation, has no effect on the person.

Making and reading a karyotype

A karyotype is the complete set of chromosomes of a cell, arranged in pairs by size, centromere position and banding pattern. Dividing cells (commonly cultured white blood cells, amniocytes or bone marrow) are treated with a spindle poison such as colchicine or colcemid, which arrests them in metaphase when chromosomes are most condensed. A hypotonic solution swells the cells, they are fixed and dropped onto slides, and the chromosomes are stained.

The standard stain is G-banding: a brief trypsin treatment followed by Giemsa dye. Dark G bands tend to be AT rich, gene poor and late replicating; light bands are GC rich and gene dense. Each chromosome has a reproducible pattern of about 400 to 550 bands per haploid set in a routine preparation, and up to about 850 in high resolution prometaphase spreads. Bands are numbered outward from the centromere: 7q31.2 means chromosome 7, long arm, region 3, band 1, sub-band 2.

A band at routine resolution contains roughly 5 to 10 Mb of DNA, so a G-banded karyotype detects gains, losses and rearrangements of about that size or larger. Molecular methods see far smaller changes:

• Fluorescence in situ hybridization (FISH) uses a fluorescent DNA probe for a specific region. The number and location of signals reveal deletions, duplications and translocations, even in nondividing (interphase) cells.
• Spectral karyotyping paints each chromosome a different color, making complex translocations obvious.
• Chromosomal microarray compares the amount of patient DNA with a reference at hundreds of thousands of points across the genome. It detects copy number variants as small as about 50 to 100 kb, but it cannot detect balanced rearrangements, which change position without changing amount.

Karyotype notation

The international system for human cytogenomic nomenclature (ISCN) writes the total chromosome number, then the sex chromosomes, then any abnormality. A normal female is 46,XX and a normal male 46,XY. A plus or minus sign before a chromosome number means a whole extra or missing chromosome. Structural changes use abbreviations followed by the chromosome(s) in one set of parentheses and the breakpoints in the next.

Section 2Deletions, duplications, inversions and translocations

How rearrangements arise

Structural rearrangements begin with double strand breaks, caused by ionizing radiation, some chemicals, replication errors or the cell's own programmed cuts. If the broken ends are rejoined wrongly, or if crossing over occurs between similar sequences at the wrong positions (nonallelic homologous recombination), the gene order changes. Repeated sequences, such as segmental duplications and transposable elements, are hotspots because they provide misleading homology.

Deletions

A deletion (deficiency) removes a segment, either from the end (terminal) or from within (interstitial). In a deletion heterozygote, the normal homolog has nothing to pair with across the missing region, so it buckles out as a deletion loop during synapsis, visible in Drosophila polytene chromosomes. Two genetic effects reveal deletions. First, a deletion uncovers recessive alleles on the other homolog: a recessive allele opposite a deletion is expressed, a phenomenon called pseudodominance, which makes deletions excellent tools for mapping genes to small regions (just as Benzer used them in phage). Second, many genes are haploinsufficient: one copy does not make enough product for a normal phenotype. Large deletions are usually lethal when homozygous and often harmful when heterozygous. Deletion of part of the short arm of human chromosome 5 causes cri du chat syndrome, named for the high pitched, catlike cry of affected infants, together with intellectual disability and a small head.

Duplications

A duplication repeats a segment. In a tandem duplication the copies are adjacent and in the same orientation; in a reverse duplication they are in opposite orientation; a duplicated segment may also be inserted elsewhere. Duplications commonly arise by unequal crossing over between misaligned repeats, which creates a duplication on one chromatid and a reciprocal deletion on the other.

The Bar mutation of Drosophila is the classic case. Bar is a tandem duplication of a small region of the X chromosome that reduces the number of facets in the eye. Females with one Bar region on each X (B/B) have four copies of the region and small eyes, but females with a triple copy on one X and a single copy on the other (double Bar over wild type) also have four copies and even smaller eyes. Same dose, different phenotype: the arrangement mattered, an example of a position effect.

Duplications are also the raw material of evolution. Once a gene is duplicated, one copy can keep the original function while the other accumulates mutations and may gain a new role. The globin genes, olfactory receptor genes and Hox clusters are families that arose this way.

Inversions

An inversion cuts out a segment and reinserts it in reverse. If the segment does not include the centromere, it is a paracentric inversion; if it does, it is a pericentric inversion, which can change the arm ratio and therefore the shape of the chromosome. An inversion homozygote usually has a normal phenotype, because no genes are lost or gained, unless a breakpoint disrupts a gene or moves it next to heterochromatin. The genetic action happens in heterozygotes. To pair gene for gene, one homolog must twist into an inversion loop.

A single crossover within the loop produces unbalanced chromatids:

• In a paracentric inversion heterozygote, the crossover produces a dicentric chromatid (two centromeres) and an acentric fragment (none). At anaphase I the dicentric stretches between the poles as a dicentric bridge and breaks, while the acentric fragment is lost. Both recombinant products are missing genes.
• In a pericentric inversion heterozygote, each recombinant chromatid has one centromere but carries a duplication of one end and a deficiency of the other.

In both cases the two nonrecombinant chromatids (one normal, one inverted) are balanced. Gametes carrying the unbalanced recombinant chromatids are inviable in plants or give inviable zygotes in animals, so recombinant progeny are not recovered for genes within the inversion. Inversions therefore act as crossover suppressors, keeping groups of alleles together. Geneticists exploit this in Drosophila balancer chromosomes, which carry multiple overlapping inversions, a dominant visible marker and a recessive lethal, to maintain a mutation in a stock without it being lost by recombination.

Translocations

A translocation moves a segment to a nonhomologous chromosome. In a reciprocal translocation, two nonhomologous chromosomes exchange segments. A carrier with a balanced reciprocal translocation has all its genes, usually in normal amounts, and is typically healthy. In meiosis, however, the two normal and two translocated chromosomes pair as a cross shaped quadrivalent, and how the four chromosomes segregate determines whether gametes are balanced:

• Alternate segregation: the two normal chromosomes go to one pole and the two translocated chromosomes to the other. Both gametes are balanced (one normal, one balanced carrier).
• Adjacent-1 segregation: each pole receives one normal and one translocated chromosome, with nonhomologous centromeres going together. Both gametes are unbalanced, carrying a duplication of one segment and a deficiency of another.
• Adjacent-2 segregation: homologous centromeres go to the same pole. Both gametes are unbalanced; this is rarer.

Because adjacent and alternate segregation occur at similar frequencies in many plants, a translocation heterozygote is often semisterile, with about half its pollen or ovules aborted. In humans, balanced translocation carriers may have recurrent miscarriages or a child with an unbalanced karyotype. Translocations also produce pseudolinkage: genes near the breakpoints on the two different chromosomes behave as if linked, because only the alternate segregation products survive.

Robertsonian translocations

A Robertsonian translocation fuses the long arms of two acrocentric chromosomes at or near their centromeres, and the two tiny short arms are lost. The carrier has 45 chromosomes but is healthy, because the lost short arms carry only redundant ribosomal RNA genes. About 1 in 1,000 people carry one; rob(13;14) and rob(14;21) are the most common. The danger is to offspring.

Section 3Aneuploidy and nondisjunction

Terms for changes in chromosome number

Changes in chromosome number come in two kinds. Euploidy changes involve whole sets (the subject of Section 4). Aneuploidy is the gain or loss of individual chromosomes, so that the number is not an exact multiple of the haploid number. Using 2n for the normal diploid number:

• Nullisomy (2n − 2): both members of a pair missing.
• Monosomy (2n − 1): one chromosome missing, for example 45,X.
• Trisomy (2n + 1): one extra chromosome, for example 47,XX,+21.
• Tetrasomy (2n + 2): two extra copies of the same chromosome.
• Double trisomy (2n + 1 + 1): one extra copy each of two different chromosomes.

Aneuploidy is usually far more harmful than polyploidy, even though it changes fewer genes. The reason is gene balance: many proteins work in complexes or pathways whose parts must be made in the right proportions. Adding a whole set keeps every ratio at 1:1; adding one chromosome raises the dose of hundreds of genes by 50 percent relative to all the others. In humans, every autosomal monosomy and most autosomal trisomies are lethal before birth.

Nondisjunction

Most aneuploidy arises from nondisjunction, the failure of chromosomes to separate properly. In meiosis I nondisjunction, a pair of homologs goes to the same pole; in meiosis II nondisjunction, a pair of sister chromatids goes to the same pole. The two produce different gamete sets:

• After nondisjunction in meiosis I, all four products are abnormal: two are n + 1, carrying one copy of each homolog, and two are n − 1.
• After nondisjunction in meiosis II (in one of the two secondary cells), half the products are abnormal: one n + 1 carrying two copies of the same chromatid, one n − 1 and two normal.

That difference is the key to working out where nondisjunction occurred. If an extra chromosome carries two different alleles from one parent (both of that parent's homologs), the error was in meiosis I. If it carries two identical copies of one parental homolog, the error was most likely in meiosis II. (Crossing over complicates this for loci far from the centromere, so markers near the centromere give the cleanest answer.) Related mechanisms include anaphase lag, in which a chromosome is left behind and lost, and premature separation of sister chromatids in meiosis I.

Human aneuploidies

Only three autosomal trisomies commonly survive to birth, those of chromosomes 21, 18 and 13, which are among the most gene poor autosomes. Sex chromosome aneuploidies are far better tolerated, for two reasons: all X chromosomes beyond one are inactivated (Chapter 3), and the Y carries few genes.

About 95 percent of Down syndrome is free trisomy 21, and in roughly 90 percent of these the extra chromosome comes from the mother, most often from a meiosis I error. About 3 to 4 percent of cases are due to a Robertsonian translocation (Section 2), which matters because it can recur in a family, and 1 to 2 percent are mosaic. The risk of free trisomy rises steeply with maternal age: approximately 1 in 1,250 births at age 25, 1 in 1,000 at 30, 1 in 350 at 35, 1 in 100 at 40 and 1 in 30 at 45. A leading explanation is that human oocytes begin meiosis I before birth and remain arrested for decades; the cohesin complexes that hold homologs and sisters together slowly deteriorate, and with fewer or poorly placed crossovers, homologs separate prematurely or not at all.

Mosaicism and uniparental disomy

Nondisjunction in a mitotic division after fertilization produces a mosaic, a person with two or more cell lines of different karyotype, such as 46,XX/47,XX,+21. Mosaics are usually more mildly affected than full trisomics, in proportion to how many cells, and which tissues, are abnormal. Mosaicism confined to the placenta can confuse prenatal tests that sample chorionic villi.

A trisomic embryo can sometimes "rescue" itself by losing one of the three copies in an early mitosis. If the lost copy is the one from the parent who contributed only one, the embryo ends up with both copies from the same parent, which is uniparental disomy (UPD). UPD is harmless for most chromosomes but causes disease when the region contains imprinted genes, or when the doubled chromosome carries a recessive mutation (isodisomy). Maternal UPD of chromosome 15 causes about a quarter of cases of Prader Willi syndrome, because the paternally expressed genes in 15q11-q13 are missing.

Section 4Polyploidy

Whole extra sets

Polyploids carry more than two complete chromosome sets. Geneticists use x for the number of chromosomes in one basic set and n for the number in a gamete. In a diploid these are the same, but in polyploids they differ. Bread wheat, for instance, is 2n = 6x = 42: each body cell has six sets of 7 chromosomes, and each gamete (n) has 21. Triploids (3x), tetraploids (4x), hexaploids (6x) and octoploids (8x) are all known.

Polyploidy is common in plants (most flowering plant lineages have undergone at least one whole genome duplication in their history) and in some fish, amphibians and insects, but rare in mammals. Plants tolerate it better for several reasons: many reproduce asexually or self fertilize, they lack chromosomal sex determination that would be upset by extra sets, and their development is more flexible. Polyploid plants often have larger cells and larger organs (fruits, flowers, leaves), the "gigas" effect, which made them attractive in agriculture.

Autopolyploidy

An autopolyploid has multiple sets from the same species. It can arise when a diploid gamete (from a failure of meiosis) fuses with a normal gamete, giving a triploid, or with another diploid gamete, giving a tetraploid. In the laboratory, colchicine, which binds tubulin and prevents spindle formation, blocks anaphase so that the chromosomes of a dividing cell are not separated, doubling the chromosome number in that cell and its descendants.

Autotetraploids are fertile but segregate in unusual ratios, because each chromosome has four homologs that can pair as a quadrivalent or as two bivalents, and any two of the four go to each gamete. Triploids are almost always highly sterile. With three homologs for each chromosome, one pole gets two and the other one, at random for each chromosome type. A gamete is balanced only if it receives exactly one copy of every chromosome (n = x) or two copies of every chromosome (2x). Seedless watermelons and the Cavendish banana (a triploid, 2n = 3x = 33) owe their seedlessness to this sterility.

Allopolyploidy

An allopolyploid combines chromosome sets from different species. A hybrid between two diploid species is usually sterile: its two sets are not homologous, so the chromosomes cannot pair and meiosis produces unbalanced gametes. If the hybrid's chromosome number doubles, however, every chromosome now has a partner, meiosis proceeds normally, and the result is a fertile amphidiploid (also called an allotetraploid). Because it cannot produce fertile offspring with either parent species, it is, in effect, a new species formed in a single generation.

The first deliberately made example was Georgii Karpechenko's 1928 cross of radish (Raphanus, 2n = 18) with cabbage (Brassica, 2n = 18). The F₁ had 9 radish and 9 cabbage chromosomes and was sterile; a rare doubled form with 36 chromosomes was fertile. (Unfortunately for gardeners, it had the root of a cabbage and the leaves of a radish.)

Polyploidy in crops and in animals

Many crops are polyploid, and their histories can be read in their genomes. Bread wheat (Triticum aestivum) arose from two hybridizations: a diploid AA wheat crossed with a BB goat grass to form tetraploid AABB (the ancestor of durum wheat), which later crossed with a third species, DD goat grass, giving hexaploid AABBDD. Plant breeders have made new allopolyploids too, such as triticale, a wheat by rye hybrid doubled with colchicine.

In animals, whole genome duplication occurred twice early in vertebrate evolution, which is why we have four Hox clusters where an insect has one, and again in the ancestors of teleost fish and of salmon. Living polyploid animals include some frogs, lizards and fish, often reproducing asexually. In humans, triploidy (69 chromosomes) occurs in about 1 to 2 percent of conceptions, usually from fertilization of an egg by two sperm, and almost all are miscarried. Many human tissues contain endopolyploid cells, in which DNA has replicated without division: liver cells are often tetraploid, and megakaryocytes reach 16n or more. The giant polytene chromosomes of Drosophila salivary glands are an extreme case, with about 1,000 aligned copies of each chromatid, which makes deletions and inversions directly visible.

Section 5Chromosome mutations in human disease and cancer

How common are chromosome abnormalities?

Chromosome errors are strikingly common at conception and much rarer at birth, because most are lethal. About half of first trimester miscarriages are chromosomally abnormal. Among these, autosomal trisomies make up roughly half (trisomy 16, never seen at birth, is the single most common), followed by 45,X and triploidy. Among liveborn infants, about 1 in 150 (roughly 0.6 percent) has a chromosome abnormality, divided among aneuploidies, balanced rearrangements and unbalanced structural changes.

Microdeletion and microduplication syndromes

Some deletions are too small to see on a routine karyotype (under about 5 Mb) yet remove a group of neighboring genes, causing a recognizable contiguous gene syndrome. Many recur because they are flanked by low copy repeats that promote nonallelic homologous recombination; the same mechanism produces reciprocal duplications. They are diagnosed by FISH or chromosomal microarray.

Chromosome changes in cancer

Cancer cells are genetically unstable, and chromosome mutations are among their most characteristic features. Most solid tumors are aneuploid, often with dozens of numerical and structural changes. But some cancers are driven by one specific rearrangement, and those have taught us how chromosome mutations activate genes. Rearrangements cause cancer in three main ways.

1. Fusion genes. A translocation joins parts of two genes into a new gene encoding a hybrid protein with abnormal activity.
2. Promoter or enhancer swapping. A translocation moves a proto-oncogene next to a strongly active regulatory region, so it is overexpressed even though its protein is normal.
3. Copy number change. Amplification of an oncogene raises its dose, and deletion removes a tumor suppressor gene.

The first consistent cancer rearrangement found was the Philadelphia chromosome, a small chromosome 22 seen in about 95 percent of cases of chronic myeloid leukemia (CML). It results from the reciprocal translocation t(9;22)(q34;q11). The breakpoints fall within the ABL1 gene on 9q34 and the BCR gene on 22q11, and the derivative chromosome 22 carries a BCR-ABL1 fusion gene. Its protein is a tyrosine kinase that is always switched on, driving proliferation of myeloid cells. That discovery led to imatinib, a drug that blocks the ATP binding site of the BCR-ABL1 kinase and transformed CML from a fatal disease into a manageable one.

Amplified genes appear cytologically in two forms: double minutes, small paired acentric chromatin bodies, and homogeneously staining regions, long stretches of a chromosome that lack normal bands because they consist of repeated copies of the amplicon. Some tumors also show chromothripsis, in which one or a few chromosomes shatter into many pieces in a single catastrophic event and are stitched back together in random order.

Words worth knowing

DNA Replication, Recombination, Mutation and Repair

Each dividing cell copies billions of base pairs with fewer than one error per billion, and repairs thousands of DNA injuries every day. When copying, recombination or repair goes wrong, the result is a mutation, the raw material of evolution and the root of many inherited diseases and cancers. This chapter shows how the molecular machinery works and, just as important, how geneticists reasoned their way to knowing it.

Section 1Semiconservative replication and its enzymes

Three models and one decisive experiment

Watson and Crick noted in 1953 that base pairing suggests a copying mechanism. Semiconservative replication predicts that each daughter molecule keeps one old strand and gains one new strand. Two alternatives were also on the table. In conservative replication the parental duplex would stay intact and a wholly new duplex would be built beside it. In dispersive replication each strand of every daughter molecule would be a patchwork of old and new segments.

In 1958 Matthew Meselson and Franklin Stahl distinguished these models with a density label. They grew Escherichia coli for many generations in medium whose only nitrogen source was the heavy isotope ¹⁵N, so every DNA molecule was "heavy." They then moved the cells to medium containing ordinary ¹⁴N and took samples after each round of division. DNA from each sample was spun in a cesium chloride (CsCl) density gradient, where molecules settle into a band at the position matching their own buoyant density. Heavy DNA bands lower in the tube, light DNA higher, and a hybrid of one heavy and one light strand exactly in between.

After one generation in ¹⁴N, all the DNA formed a single band at the hybrid position. That result alone ruled out conservative replication, which would have given one heavy and one light band. After two generations they saw two bands of equal intensity, one hybrid and one light. A dispersive mechanism would have produced a single band that drifted steadily lighter, never a discrete light band, so the second generation ruled out dispersive replication. Only the semiconservative model predicts both observations.

Origins, forks and the direction of synthesis

Replication begins at specific sites called origins of replication. The circular E. coli chromosome, about 4.6 million base pairs (4.6 Mb), has a single origin named oriC, about 245 bp long. It contains repeated 9 bp binding sites for the initiator protein DnaA and a cluster of AT-rich 13 bp repeats. DnaA bound to ATP wraps the origin DNA and melts the AT-rich region, which opens easily because A:T pairs have only two hydrogen bonds. The helicase loader DnaC then places two hexameric DnaB helicases on the single strands, one facing each direction. Replication is therefore bidirectional: two replication forks move away from the origin around the circle until they meet roughly opposite, in a terminus region flanked by Ter sites bound by the Tus protein, which act as one-way traps for forks. Because the replicating circle looks like the Greek letter theta, this is called theta replication; some plasmids and phages instead use rolling circle replication.

Every DNA polymerase obeys two rules that shape the fork. First, it adds nucleotides only to a free 3'-OH, so synthesis always runs 5' to 3' and the template is read 3' to 5'. Second, no DNA polymerase can start a chain from nothing; it needs a primer. In cells the primer is a short stretch of RNA, about 10 nucleotides in bacteria, laid down by primase (DnaG in E. coli). Each dNTP is joined by its innermost phosphate, releasing pyrophosphate.

Because the two strands of the helix are antiparallel, these rules force an asymmetry. On one template the fork opens in the same direction that the polymerase travels, so a single primer is enough and the new strand, the leading strand, grows continuously toward the fork. On the other template the polymerase must move away from the fork. That strand, the lagging strand, is made discontinuously as a series of short pieces, each started by its own primer. These pieces are Okazaki fragments, about 1,000 to 2,000 nucleotides long in bacteria and only 100 to 200 nucleotides in eukaryotes.

The cast of enzymes at the fork

Unwinding the helix creates two problems. The separated strands want to snap back together, and the duplex ahead of the fork becomes overwound. Single-strand binding proteins (SSB) coat exposed single strands, keeping them extended and protected. DNA gyrase, a type II topoisomerase, works ahead of the fork, cutting both strands, passing a segment of duplex through the break and resealing it to introduce negative supercoils that relieve the strain. When replication of the circle is finished, the two daughter circles are interlocked like chain links, and topoisomerase IV separates (decatenates) them.

The main replicative enzyme in E. coli is DNA polymerase III holoenzyme, a multi-subunit machine. Its alpha subunit performs polymerization; its epsilon subunit is a 3' to 5' exonuclease that removes a mismatched nucleotide just added, a process called proofreading; and its beta clamp, a ring that encircles the DNA, holds the polymerase on the template so it can add thousands of nucleotides without falling off. This property, called processivity, lets pol III synthesize at about 1,000 nucleotides per second. A clamp loader complex opens the beta ring and closes it around each primer. One holoenzyme replicates both strands, the lagging template looping out so both cores move together (the trombone model).

DNA polymerase I finishes the lagging strand. It has a 5' to 3' exonuclease activity that removes the RNA primer of the fragment ahead while its polymerase activity fills the gap with DNA, a combined process called nick translation. Finally, DNA ligase seals the remaining nick by forming the last phosphodiester bond. Bacterial ligase uses NAD⁺; eukaryotic ligases use ATP.

Fidelity: three layers of accuracy

Replication is astonishingly accurate, and the accuracy comes in layers. Base selection by the polymerase active site, which fits only a correctly shaped Watson and Crick pair, gives roughly one error per 10⁴ to 10⁵ nucleotides. Proofreading by the 3' to 5' exonuclease improves this about 100-fold, to roughly 10^-7. Mismatch repair (Section 5) catches most of the remaining errors after the fork has passed, bringing the final rate to about one error per 10⁹ to 10¹⁰ nucleotides replicated.

Section 2Replication in eukaryotes and telomeres

Many origins, licensed once per cycle

Eukaryotic chromosomes are linear, far larger than bacterial chromosomes, packaged in chromatin, and replicated by forks that move more slowly, roughly 50 nucleotides per second compared with about 1,000 in E. coli. A single origin per chromosome would never finish in time, so each chromosome uses many origins. The stretch of DNA replicated from one origin is a replicon. Yeast origins are short defined sequences called autonomously replicating sequences (ARS); human origins lack a simple consensus, and tens of thousands fire in each S phase.

A genome with thousands of origins faces a bookkeeping problem: every segment must be copied exactly once per cell cycle. Over-replication creates extra copies of genes and broken chromosomes. Cells solve this by separating two steps in time. During G1, when cyclin-dependent kinase (CDK) activity is low, the origin recognition complex (ORC) binds origins and, with the helpers Cdc6 and Cdt1, loads inactive double rings of the MCM2 to 7 helicase. This loaded pre-replicative complex marks the origin as licensed. When S phase begins, rising CDK and the kinase DDK (Dbf4-dependent kinase) activate licensed origins by recruiting Cdc45 and the GINS complex, which together with MCM form the active CMG helicase. The same high CDK activity that fires origins also prevents new licensing (Cdc6 is degraded or exported and Cdt1 is inhibited by the protein geminin), so no origin can be loaded again until the cell passes through mitosis and returns to G1. Active euchromatin generally replicates early in S phase and heterochromatin late.

Eukaryotic polymerases and lagging strand maturation

Eukaryotes divide the work among several polymerases. Pol α-primase starts every chain: its primase subunit makes about 10 nucleotides of RNA and pol α then adds about 20 nucleotides of DNA. Pol α lacks proofreading and has low processivity, so it is quickly replaced. Pol ε synthesizes most of the leading strand and pol δ most of the lagging strand; both proofread. The ring-shaped clamp PCNA (proliferating cell nuclear antigen), loaded by RFC, plays the role of the bacterial beta clamp. Because eukaryotic polymerases lack the 5' to 3' exonuclease of bacterial pol I, primers are removed differently: pol δ runs into the previous fragment and displaces its 5' end as a single-stranded flap, which the flap endonuclease FEN1 cuts off (with help from RNase H, which degrades RNA in RNA:DNA hybrids). DNA ligase I then seals the nick. Mitochondrial DNA is copied by pol γ.

Chromatin must also be copied. Nucleosomes ahead of the fork are disassembled and parental histones are passed to both daughter duplexes, while new histones are deposited by the chaperone CAF-1, which is recruited to PCNA. Parental histones keep their chemical marks, which helps pass on gene activity patterns (Chapter 9).

The end replication problem

Linear chromosomes have a weakness that circular ones do not. At the very end of a chromosome, the lagging strand template has no place for a final primer that can be replaced with DNA: when the last RNA primer is removed, nothing upstream can supply a 3'-OH to fill the gap. In addition, nucleases process the ends to leave a single-stranded overhang. The result is that each daughter molecule would be a little shorter than its parent, and after enough divisions genes near the ends would be lost. This is the end replication problem.

Chromosome ends are protected by telomeres, long arrays of a short repeat. In humans and all other vertebrates the repeat is 5'-TTAGGG-3' on the strand running 5' to 3' toward the chromosome end, repeated to a total length of roughly 10 to 15 kb at birth. The G-rich strand ends in a 3' single-stranded overhang of about 50 to 300 nucleotides. That overhang tucks back and invades the double-stranded repeats to form a protective lasso called the t-loop, held in place by a six-protein complex called shelterin (including TRF1, TRF2 and POT1). The t-loop hides the end so the cell does not mistake it for a broken chromosome and try to repair it by joining chromosomes together.

Telomerase

The enzyme that restores telomere length is telomerase, discovered by Elizabeth Blackburn and Carol Greider in the ciliate Tetrahymena. Telomerase is a ribonucleoprotein with two essential parts: a protein subunit, TERT (telomerase reverse transcriptase), and an RNA, TERC (also called TR), which carries a short template complementary to the telomere repeat. In humans the template region reads 3'-CAAUCCCAAUC-5'. Telomerase binds the 3' overhang so that the end of the DNA pairs with the 3' part of the template, then TERT copies the rest of the template, adding one GGTTAG unit (completing a TTAGGG repeat). It then slides forward and repeats. Telomerase is thus a reverse transcriptase carrying its own template. Once the G-rich strand has been lengthened, pol α-primase fills in the complementary C-rich strand by ordinary lagging strand synthesis.

Telomeres, aging and cancer

Most human somatic cells express little or no TERT, so their telomeres shorten by roughly 50 to 200 bp with each division. When telomeres become critically short, the exposed ends trigger a DNA damage response and the cells enter permanent growth arrest, called replicative senescence. This is the molecular basis of the Hayflick limit, the observation that normal human fibroblasts divide only about 40 to 60 times in culture. Germ cells and many stem cells do express telomerase. About 85 to 90 percent of human cancers reactivate telomerase, often through mutations in the TERT promoter, and most of the rest maintain telomeres by recombination between telomeres, a pathway called alternative lengthening of telomeres (ALT). Inherited telomerase defects cause dyskeratosis congenita, with bone marrow failure.

Section 3Homologous recombination

What recombination is for

Homologous recombination is the exchange of genetic information between two DNA molecules that share a long stretch of identical or nearly identical sequence. You already know one of its products: the crossovers between homologous chromosomes in meiosis that reshuffle alleles and that geneticists use to build linkage maps (Chapter 4). Its most ancient job, though, is repair. A double-strand break severs a chromosome completely, and the only way to restore the lost or damaged sequence without error is to copy it from an intact homologous template, usually the sister chromatid after replication or the homolog in meiosis. It also restarts collapsed forks and, in bacteria, integrates DNA brought in by transformation, transduction and conjugation (Chapter 5).

Homologous recombination should be distinguished from two other kinds of DNA rearrangement. Site-specific recombination joins two short, specific sequences recognized by a dedicated enzyme (phage lambda integration, or Cre recombinase acting on loxP sites). Transposition moves a mobile element into a new site that needs no homology at all (Section 5). Only homologous recombination depends on base pairing between long matching sequences.

The Holliday model and the Holliday junction

In 1964 Robin Holliday proposed that single-strand nicks at the same position in two homologous duplexes let the nicked strands swap partners, each pairing with the complementary strand of the other duplex. The result is a four-armed, cross-shaped structure now called a Holliday junction. Wherever a strand from one parent pairs with a strand from the other, the duplex is a heteroduplex; if the two parents differ at a site in that region, the heteroduplex contains a mismatch. The junction can slide along the DNA, a process called branch migration, lengthening the heteroduplex. Finally, the junction is cut by a resolvase. A Holliday junction can be cut in either of two planes. Cutting the two strands that were exchanged restores the original partners at flanking markers, a noncrossover that leaves only a patch of heteroduplex. Cutting the two strands that were not exchanged joins the left part of one chromosome to the right part of the other, a crossover.

The double-strand break repair model

Yeast experiments later showed that meiotic recombination begins with a double-strand break and that the broken chromosome receives information. The resulting double-strand break repair (DSBR) model is now the textbook framework. Its steps are:

1. Break. In meiosis, the topoisomerase-like protein Spo11 cuts both strands deliberately at hundreds of sites; in mitotic cells breaks come from radiation, chemicals or collapsed forks.
2. Resection. Nucleases (the MRN complex with CtIP, then Exo1 and Dna2 in eukaryotes) degrade the 5' ends on both sides of the break, leaving long 3' single-stranded tails. RPA coats the tails at first.
3. Strand invasion. The recombinase Rad51 (and in meiosis also its relative Dmc1) replaces RPA, forming a helical nucleoprotein filament on a 3' tail. Loading Rad51 requires the mediator protein BRCA2. The filament finds a matching sequence and invades it, displacing one strand as a D-loop.
4. Synthesis. A DNA polymerase extends the invading 3' end, using the homolog as template. This copies any sequence lost at the break.
5. Second end capture. The enlarged D-loop pairs with the other 3' tail, which is also extended. Ligation of the ends produces two Holliday junctions, a double Holliday junction.
6. Resolution. The two junctions are cut by resolvases or, in somatic cells, dissolved by a helicase and topoisomerase complex. If the two junctions are cut in the same orientation the outcome is a noncrossover; if they are cut in opposite orientations the outcome is a crossover.

Many breaks never form a double Holliday junction. In synthesis-dependent strand annealing (SDSA), the invading strand is extended, released from the D-loop and annealed back to the other tail on its own chromosome. SDSA always gives a noncrossover, which is why most double-strand breaks in mitotic cells are repaired without exchanging chromosome arms. This matters because a mitotic crossover can cause loss of heterozygosity, unmasking a recessive tumor suppressor mutation (Chapter 13).

The bacterial machinery

Much of what we know about recombination enzymes was worked out in E. coli. The RecBCD complex binds a double-stranded end and travels along it, unwinding and degrading DNA, until it meets a short sequence called Chi (5'-GCTGGTGG-3'). At Chi, RecBCD stops degrading the 3' strand and loads RecA onto it. RecA is the bacterial counterpart of Rad51: it polymerizes on single-stranded DNA, searches for homology and catalyzes strand invasion. The RuvA and RuvB proteins bind Holliday junctions and drive branch migration, and RuvC resolves them by symmetric nicks.

Gene conversion

Mendel's first law predicts that a heterozygote A/a produces meiotic products in a 2:2 ratio of A to a. In fungi such as Neurospora and Sordaria, where all products of a single meiosis stay together in an ascus, and where a postmeiotic mitosis turns the four products into eight spores (an octad), the normal ratio is 4:4. Occasionally, however, an ascus shows 6:2 or 2:6, meaning one allele was converted into the other, or 5:3 and 3:5, meaning one spore pair contains both alleles. This nonreciprocal transfer of information is gene conversion.

Recombination explains both kinds of aberrant ascus. When a heteroduplex forms across a site where the parents differ, the two strands of that region carry different alleles and form a mismatch. If mismatch repair corrects the mismatch, it may restore the original allele (giving a normal 4:4) or convert it to match the invading strand (giving 6:2 or 2:6). If the mismatch escapes repair, the two strands separate at the next DNA replication, the postmeiotic mitosis, and one spore pair becomes a non-identical pair, giving 5:3 or 3:5, an event called postmeiotic segregation. Gap repair in the DSBR model also converts: sequence lost by resection at the break is replaced by copying from the homolog.

Section 4Mutation types, causes and rates

Kinds of mutation

A mutation is any heritable change in DNA sequence. Mutations in germ-line cells can be passed to offspring; somatic mutations arise in body cells and are passed only to that cell's descendants, as in cancer. Here we treat changes of one or a few nucleotides; larger changes were covered in Chapter 6.

The simplest gene mutation is a base substitution. A transition replaces a purine with the other purine or a pyrimidine with the other pyrimidine (A:T to G:C or G:C to A:T). A transversion replaces a purine with a pyrimidine or the reverse (for example G:C to T:A). Although there are twice as many possible transversions, transitions are more common, because deamination and tautomeric mispairing produce transitions. Insertions and deletions (indels) add or remove nucleotides. If the number added or removed is not a multiple of three in a coding region, the reading frame shifts and every downstream codon is misread; this is a frameshift mutation, and it nearly always destroys protein function, usually by reaching a premature stop codon soon after the shift.

Mutations are also classified by their effect on the protein:

• A synonymous (silent) mutation changes a codon to another codon for the same amino acid, most often at the third position.
• A missense mutation changes one amino acid to another. A conservative missense change swaps similar amino acids (leucine to isoleucine) and may be harmless; a nonconservative change swaps dissimilar ones and is more likely to disrupt structure.
• A nonsense mutation changes an amino acid codon into a stop codon, truncating the protein. Messenger RNAs with premature stop codons are often destroyed by nonsense-mediated decay (Chapter 8).
• A splice-site mutation alters the signals for intron removal, often causing exon skipping or intron retention.
• A regulatory mutation in a promoter or enhancer changes how much protein is made rather than what kind.

A forward mutation changes the wild type to a mutant; a reverse mutation (reversion) restores the wild-type phenotype, either by restoring the original sequence or by a second change. A suppressor mutation is a second mutation at a different site that hides the effect of the first; it may be intragenic (a compensating change in the same gene, such as a +1 frameshift near a −1 frameshift) or intergenic (for example, a mutant tRNA that reads a stop codon as an amino acid). Loss-of-function mutations are usually recessive; gain-of-function mutations, which give a protein a new or excessive activity, are usually dominant.

Expanding repeats

In trinucleotide repeat expansions, a run of a three-base repeat becomes unstable past a threshold length and tends to grow in each generation. In Huntington disease, a CAG repeat encoding polyglutamine in the huntingtin gene causes disease at 40 or more copies (36 to 39 copies give incomplete penetrance). In fragile X syndrome, a CGG repeat in the 5' untranslated region of FMR1 is normal below about 45 copies, a premutation from about 55 to 200, and a full mutation above 200, at which point the gene is methylated and silenced. Myotonic dystrophy type 1 involves a CTG repeat in a 3' untranslated region. Because longer repeats are more unstable and cause earlier, more severe disease, these disorders show anticipation: symptoms appear earlier or more severely in successive generations.

Spontaneous mutation

Many mutations arise without any outside agent:

• Replication errors. Despite proofreading, polymerases occasionally insert the wrong base. Rare tautomeric shifts (the amino to imino form of adenine or cytosine, the keto to enol form of guanine or thymine) and wobble pairing allow mispairs such as G:T and A:C, which become transitions after the next round of replication.
• Strand slippage. In runs of a repeated base or short repeat, the new or template strand can loop out, producing small insertions or deletions. These create mutational hot spots in repetitive sequences.
• Depurination. The bond joining a purine base to deoxyribose breaks spontaneously, roughly 10,000 times per human cell per day, leaving an apurinic (AP) site with no base to guide replication.
• Deamination. Cytosine loses its amino group to become uracil, which pairs with adenine; if unrepaired, G:C becomes A:T. Cells recognize uracil as foreign in DNA and remove it efficiently. But deamination of 5-methylcytosine yields thymine, a normal DNA base, so the resulting G:T mismatch is repaired less reliably. That is why methylated CpG sites are the strongest mutational hot spots in mammalian genomes, mutating 10 to 50 times faster than other sites.
• Oxidative damage. Reactive oxygen species from normal metabolism convert guanine to 8-oxoguanine, which pairs with adenine and so causes G:C to T:A transversions.

Induced mutation and mutagens

Mutagens raise the mutation rate, and knowing how each acts predicts the kind of change it causes.

The Ames test, developed by Bruce Ames in the 1970s, screens chemicals for mutagenic potential. It uses strains of Salmonella carrying mutations in a histidine synthesis gene (his⁻), so they cannot grow without added histidine. Tester strains carry substitution or frameshift alleles and lack excision repair. Because many compounds become mutagenic only after liver metabolism, the chemical is mixed with a liver extract (S9 fraction). The bacteria are plated on medium without histidine, and only cells in which a new mutation has reverted the his⁻ allele can form colonies. More revertant colonies than on a control plate indicates a mutagen, and most chemical carcinogens score as mutagens.

Mutations are random: the fluctuation test

Do mutations arise at random, or are they induced by the selective condition that reveals them? In 1943 Salvador Luria and Max Delbrück addressed this with bacteria and phage T1. They grew many small, separate cultures of E. coli from a few cells each, then plated each entire culture on phage-coated plates and counted resistant colonies. If resistance were induced by contact with phage, every culture would give a similar small number, with a variance about equal to the mean (a Poisson distribution). Instead, the counts fluctuated wildly: most cultures had zero or a few resistant cells, while a few had hundreds. This "jackpot" pattern is expected if mutations occur randomly during growth: an early mutation leaves many resistant descendants, a late one only a few. Joshua and Esther Lederberg confirmed this with replica plating, showing resistant colonies existed before exposure to phage.

The fluctuation test also gives a way to measure the mutation rate, the probability of a mutation per cell per division (or per gene per generation). The simplest version, the P₀ method, uses the fraction of cultures with no mutants. If mutations occur at random, the number of mutational events per culture follows a Poisson distribution with mean m, so the fraction of cultures with no mutational events is P₀ = e^−m, and m = −ln P₀. Dividing m by the number of cell divisions in a culture (approximately the final number of cells) gives the rate.

How often mutations happen

Mutation rates are usually quoted at two scales. Per gene, forward mutation rates typically fall between 10^-5 and 10^-7 per generation. Per nucleotide, the human germ-line rate is about 1.2 × 10^-8 per base pair per generation. With a diploid genome of about 6.2 × 10⁹ bp, that predicts 1.2 × 10^-8 × 6.2 × 10⁹ = about 74 new point mutations in each child, in line with the 60 to 75 measured by sequencing parents and children. Most arise in the father, whose sperm precursors keep dividing throughout life. RNA viruses, whose polymerases lack proofreading, mutate on the order of 10^-4 per nucleotide per replication, which is why influenza and HIV evolve so rapidly.

Section 5DNA repair systems and transposable elements

Direct reversal

The simplest repair undoes the damage chemically without removing anything. Photolyase uses the energy of visible light to split cyclobutane pyrimidine dimers back into separate bases, a process called photoreactivation; it works in bacteria, fungi, plants and many animals but is absent from placental mammals. O⁶-methylguanine DNA methyltransferase (MGMT) transfers the methyl group from O⁶-methylguanine onto one of its own cysteines. The enzyme is inactivated by the reaction, so it is called a suicide enzyme.

Excision repair

Base excision repair (BER) handles small lesions that do not distort the helix: uracil, 8-oxoguanine, alkylated and deaminated bases. A DNA glycosylase specific for one kind of damage flips the base out and cuts the bond to its sugar, leaving an AP site. AP endonuclease nicks the backbone, DNA polymerase (pol β in mammals) inserts the correct nucleotide, and ligase seals the strand. Using thymine rather than uracil in DNA lets cells treat any uracil as deaminated cytosine.

Nucleotide excision repair (NER) handles bulky lesions that distort the helix, chiefly UV dimers and chemical adducts. In E. coli, UvrA and UvrB find the distortion, UvrC cuts the damaged strand on both sides, and UvrD helicase removes an oligonucleotide of 12 to 13 nucleotides; pol I and ligase fill the gap. In humans the same logic uses about 30 proteins, including the XP proteins and the helicases of TFIIH, and removes 24 to 32 nucleotides. Transcription-coupled NER repairs lesions that stall RNA polymerase faster than global genome NER.

Mismatch repair

Mismatch repair corrects replication errors that escaped proofreading. The key problem is deciding which strand is wrong. In E. coli the answer comes from methylation: the Dam methylase adds methyl groups to the A of every GATC sequence, but only after a delay, so just behind the fork the new strand is briefly unmethylated. MutS recognizes the mismatch, MutL links it to MutH, and MutH nicks the unmethylated (new) strand at a nearby GATC. An exonuclease removes the new strand past the mismatch and pol III resynthesizes it. Eukaryotes use MutS and MutL homologs (MSH2, MLH1 and others) and recognize the new strand by its nicks.

Double-strand breaks and damage tolerance

Double-strand breaks are repaired by two main routes. Homologous recombination (Section 3) is accurate but needs a sister chromatid, so it operates in S and G2. Nonhomologous end joining (NHEJ) works throughout the cycle: the Ku70/Ku80 heterodimer binds the broken ends, recruits DNA-PKcs, and the ends are trimmed and joined by DNA ligase IV. NHEJ often adds or loses a few nucleotides, so it is error prone. Lesions that block replication can also be bypassed rather than repaired. In E. coli, single-stranded DNA at stalled forks activates RecA, which stimulates self-cleavage of the LexA repressor and induces more than 40 genes of the SOS response, including the error-prone translesion polymerases pol IV and pol V. Humans have several translesion polymerases; pol η correctly inserts two A's opposite a thymine dimer.

Transposable elements

Transposable elements are DNA sequences that move from one place in the genome to another. Barbara McClintock discovered them in maize in the 1940s, when the Ds (Dissociation) element moved only in the presence of Ac (Activator), which supplies the enzyme; Ds is a defective copy. Two main classes exist.

• Class II, DNA transposons, move as DNA. Each carries short terminal inverted repeats recognized by the transposase it encodes. Bacterial insertion sequences (IS elements), 0.7 to 2.5 kb, carry only a transposase gene. Composite transposons such as Tn10 are two IS elements flanking other genes, often antibiotic resistance genes. Most DNA transposons move by cut-and-paste (conservative) transposition; others, such as Tn3, use replicative transposition, leaving a copy behind.
• Class I, retrotransposons, move through an RNA intermediate that is reverse transcribed, so they copy and paste. LTR retrotransposons (yeast Ty1) resemble retroviruses. LINEs such as human L1 (about 6 kb, roughly 17 percent of the human genome) encode their own reverse transcriptase. SINEs such as Alu (about 300 bp, over one million copies, about 10 percent of the genome) encode nothing and borrow L1 enzymes.