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