📖 ContentsChapter 4
Genetics · Chapter 4

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.

By the end of this chapter you can

  • Explain linkage and distinguish genes in coupling (cis) from genes in repulsion (trans).
  • Describe how crossing over between nonsister chromatids at the four strand stage produces recombinant gametes, and why recombination frequency cannot exceed 50 percent.
  • Calculate recombination frequency from testcross data and convert it to map units (centimorgans).
  • Use a chi square test to decide whether two genes assort independently.
  • Analyze a three point testcross to determine gene order and the distances between genes.
  • Calculate the coefficient of coincidence and interference, and convert recombination frequencies to map distances with the Haldane and Kosambi mapping functions.
  • Calculate and interpret a LOD score for linkage in human pedigrees.
  • Compare molecular markers (RFLPs, microsatellites and SNPs) and explain how they made human gene mapping possible.

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:

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.

Prophase I: one crossover Four products of meiosis AAaa BbBb sister chromatids of one homolog (blue) and of the other homolog (red) crossover (chiasma) point ABAb aBab parentalrecombinantrecombinantparental A crossover involves only two of the four chromatids, so a meiosis with one crossover in the interval gives 2 parental and 2 recombinant products (50 percent). Meioses with no crossover give 0 percent. Therefore recombination frequency = 1/2 × fraction of meioses with a crossover in the interval
Figure 4.1 Crossing over happens after DNA replication, when each homolog consists of two sister chromatids. A single exchange between nonsister chromatids converts two of the four chromatids into recombinants, so even a meiosis with a crossover between two genes yields only half recombinant gametes.

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.

Situation (testcross of AaBb to aabb)Gametes from the heterozygoteTestcross phenotype ratio
Genes on different chromosomes (independent assortment)AB, Ab, aB, ab each 25 percent1 : 1 : 1 : 1
Same chromosome, complete linkage, coupling (AB/ab)AB and ab, 50 percent each1 : 1, parental types only
Same chromosome, incomplete linkage, couplingAB and ab more than 25 percent each; Ab and aB fewerTwo large parental classes, two small recombinant classes
Same chromosome, incomplete linkage, repulsion (Ab/aB)Ab and aB more than 25 percent each; AB and ab fewerThe large classes are Ab and aB
Same chromosome but far apart (50 cM or more)All four near 25 percentAbout 1 : 1 : 1 : 1; looks unlinked
Coupling (cis): AB / abRepulsion (trans): Ab / aB ABab AbaB genes 20 map units apart genes 20 map units apart AB 0.40 parentalab 0.40 parental Ab 0.10 recombinantaB 0.10 recombinant Ab 0.40 parentalaB 0.40 parental AB 0.10 recombinantab 0.10 recombinant recombinants total 0.20recombinants total 0.20
Figure 4.2 Linkage phase decides which gametes are parental. The same two genes, 20 map units apart, give opposite gamete frequencies in coupling and repulsion heterozygotes. The total frequency of recombinants (20 percent) is the same in both.
Study tip. Worked problem: A true breeding AAbb plant is crossed with an aaBB plant. The F1 is testcrossed to aabb, and 500 offspring are scored. Genes A and B are 20 map units apart. How many of each phenotype do you expect? Step 1. Find the phase. The F1 received Ab from one parent and aB from the other, so it is Ab/aB, in repulsion. Step 2. Parental gametes are Ab and aB; recombinant gametes are AB and ab. Step 3. 20 map units means 20 percent recombinants, split equally: AB 10 percent and ab 10 percent. The parentals share the remaining 80 percent: Ab 40 percent and aB 40 percent. Step 4. Multiply by 500: Ab 0.40 × 500 = 200; aB 200; AB 0.10 × 500 = 50; ab 50. Step 5. Check: 200 + 200 + 50 + 50 = 500. Answer: 200 A_bb, 200 aaB_, 50 A_B_ and 50 aabb. The common error is to assume coupling and call AB and ab the parental types.
Study tip. Worked problem: Cytological study shows a chiasma between genes D and E in 30 percent of meioses (and never more than one). What is the recombination frequency? Step 1. A meiosis with one crossover in the interval gives 50 percent recombinant chromatids. Step 2. Meioses without a crossover give none. Step 3. RF = 0.30 × 0.50 + 0.70 × 0 = 0.15. Answer: 15 percent, or 15 map units. Remember this factor of one half; it is the reason the maximum recombination frequency is 50 percent.
Key point

Linked genes travel together because they share a chromosome. Crossing over between nonsister chromatids in prophase I, visible as chiasmata, exchanges chromosome segments and creates recombinant gametes. Because one crossover affects only two of the four chromatids, recombinants make up at most half of the products. Always determine the linkage phase, coupling or repulsion, before deciding which offspring are parental and which are recombinant.

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:

Drosophila X chromosome (map positions in cM) y 0.0w 1.5v 33.0m 36.1f 56.7 yellowwhitevermilionminiatureforked w to v: 31.5 cM by adding short intervals y to f: 56.7 cM apart, yet a direct cross shows under 50 percent recombination
Figure 4.3 Part of the standard genetic map of the Drosophila X chromosome. Map positions are built by adding short, accurately measured intervals, so distant genes such as yellow and forked can be more than 50 map units apart even though no cross can show more than 50 percent recombination between them.

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.

Degrees of freedomCritical χ2 at P = 0.05Critical χ2 at P = 0.01
13.8416.635
25.9919.210
37.81511.345
49.48813.277
Study tip. Worked problem: A dihybrid AaBb is testcrossed to aabb. The 400 offspring are 140 AB, 132 ab, 62 Ab and 66 aB (named by the alleles received from the dihybrid). Are the genes linked, and if so how far apart? Step 1. Null hypothesis: independent assortment, so each class is expected at 400 ÷ 4 = 100. Step 2. Compute (O − E)2/E for each class: AB (140 − 100)2/100 = 1600/100 = 16.00; ab (132 − 100)2/100 = 1024/100 = 10.24; Ab (62 − 100)2/100 = 1444/100 = 14.44; aB (66 − 100)2/100 = 1156/100 = 11.56. Step 3. Sum: χ2 = 16.00 + 10.24 + 14.44 + 11.56 = 52.24. Step 4. Four classes give 3 degrees of freedom; the critical value at P = 0.05 is 7.815. Since 52.24 is far larger, reject independent assortment: the genes are linked. Step 5. The large classes, AB and ab, are parental, so the dihybrid was in coupling. Recombinants = 62 + 66 = 128. RF = 128 ÷ 400 = 0.32. Answer: linked, 32 map units apart. A caution: at 32 map units some double crossovers are being missed, so the true distance is probably a little larger.

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:

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

Key point

Recombination frequency, the fraction of offspring that are recombinant, measures distance: 1 percent recombination equals 1 map unit or centimorgan. It cannot exceed 50 percent, and it underestimates large distances because double crossovers go unseen, so maps are built by adding short intervals. A chi square test against the 1:1:1:1 expectation tells whether a departure from independent assortment is real. Genetic distance tracks gene order faithfully but its relation to base pairs varies with hotspots, sex and region.

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:

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:

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.

Phenotype of progeny (alleles from trihybrid)NumberClass
+ + c405Parental
a b +401Parental
+ b +58Single crossover, region I (a to c)
a + c56Single crossover, region I (a to c)
+ b c38Single crossover, region II (c to b)
a + +36Single crossover, region II (c to b)
+ + +3Double crossover
a b c3Double crossover
Total1,000
Study tip. Worked problem: A female heterozygous for recessive alleles a, b and c is crossed to an a b c / a b c male. The 1,000 progeny are listed in the table above, written in alphabetical order (+ means the wild type allele). Find the gene order and the map distances. Step 1. Identify the parentals. The two largest classes are + + c (405) and a b + (401), so the female's chromosomes were + + c and a b +. Step 2. Identify the double crossovers. The two smallest classes are + + + (3) and a b c (3). Step 3. Find the middle gene. Compare parental + + c with DCO + + +: they differ only at c. So c is in the middle, and the true order is a, c, b. Step 4. Rewrite the parentals in order: + c + and a + b. Step 5. Assign the singles. A crossover between a and c turns + c + into + + b and a + b into a c +; in alphabetical notation these are + b + (58) and a + c (56), region I. A crossover between c and b gives + c b and a + +, written + b c (38) and a + + (36), region II. Step 6. Region I distance: (58 + 56 + 3 + 3) ÷ 1,000 = 120 ÷ 1,000 = 0.120, so a to c is 12.0 map units. Step 7. Region II distance: (38 + 36 + 3 + 3) ÷ 1,000 = 80 ÷ 1,000 = 0.080, so c to b is 8.0 map units. Step 8. Outside distance: 12.0 + 8.0 = 20.0 map units. Check: counting only offspring recombinant for a and b gives (58 + 56 + 38 + 36) ÷ 1,000 = 188 ÷ 1,000 = 18.8 map units. The difference, 20.0 − 18.8 = 1.2, equals 2 × 6/1,000 = 0.012, the double crossovers counted twice. Answer: a, 12.0 cM, c, 8.0 cM, b.
A double crossover swaps only the middle gene +c+ a+b +++ acb parental chromosomesdouble crossover products acbacb Compare parental + c + with DCO + + + : only c differs, so c is the middle gene. Order: a, c, b acb 12.0 cM (region I)8.0 cM (region II) a to b = 20.0 cM by summing; 18.8 cM if measured directly
Figure 4.4 Determining gene order in a three point cross. A double crossover exchanges the middle segment, so the middle gene is the one that differs between a parental class and the double crossover class that matches it at the other two genes. The bottom bar shows the resulting map, drawn to scale.

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).
Study tip. If the heterozygote's chromosomes are not stated, never assume they are all wild type on one homolog and all mutant on the other. In the example above, the parental classes were + + c and a b +, a mixed arrangement. Always let the two largest classes tell you the phase. A second quick check: the two parental classes must be exact complements of each other, as must each single crossover pair and the DCO pair. If they are not, you have paired the classes wrongly.
Key point

A three point testcross gives eight classes: two large parental classes, two pairs of single crossovers and two rare double crossovers. The gene that differs between the parental and the matching double crossover class is the middle gene. Each interval distance counts its single crossovers plus the double crossovers, and the outside distance is the sum of the two intervals, which is always larger than a direct two point estimate.

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:

Value of c.o.c.Interference IMeaning
01Complete interference: no double crossovers at all in the interval
Between 0 and 1Between 0 and 1Positive interference: fewer DCOs than expected; the common case
10No interference: crossovers occur independently
Greater than 1NegativeNegative interference: more DCOs than expected; seen in some fungi, bacteriophages and very short intervals
Study tip. Worked problem: Use the three point cross from Section 3 (a to c 12.0 cM, c to b 8.0 cM, 6 DCOs among 1,000 offspring) to calculate the coefficient of coincidence and interference. Step 1. Expected DCO frequency = 0.120 × 0.080 = 0.0096. Step 2. Expected DCO number = 0.0096 × 1,000 = 9.6. Step 3. Observed DCOs = 3 + 3 = 6. Step 4. c.o.c. = 6 ÷ 9.6 = 0.625. Step 5. I = 1 − 0.625 = 0.375. Answer: c.o.c. = 0.625 and I = 0.375, meaning that 37.5 percent of the expected double crossovers did not occur. A frequent slip is to use numbers in one place and frequencies in another; keep both observed and expected as counts, or both as frequencies.

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.

r = 0.5 limit 00.10.20.30.40.5 050100150 map distance d (cM) recombination frequency r r = d (no correction)KosambiHaldane (no interference)
Figure 4.5 Mapping functions. For short distances recombination frequency equals map distance, but as distance grows undetected multiple crossovers make r level off toward 0.5. Kosambi's function, which allows for interference, rises more steeply than Haldane's before leveling off.
Study tip. Worked problem: Two genes show 30 percent recombination. Estimate the map distance with each mapping function, and convert 12 percent recombination with Haldane's function. Step 1. Haldane, r = 0.30: 1 − 2r = 1 − 0.60 = 0.40; ln 0.40 = −0.916; d = −½ × (−0.916) = 0.458 Morgans, or 45.8 cM. Step 2. Kosambi, r = 0.30: (1 + 0.60) ÷ (1 − 0.60) = 1.60 ÷ 0.40 = 4.00; ln 4.00 = 1.386; d = ¼ × 1.386 = 0.347 Morgans, or 34.7 cM. Step 3. Haldane, r = 0.12: 1 − 0.24 = 0.76; ln 0.76 = −0.274; d = 0.137 Morgans, or 13.7 cM. Answer: 45.8 cM (Haldane) and 34.7 cM (Kosambi) for r = 0.30; 13.7 cM for r = 0.12. Notice that the correction is small at 12 percent and large at 30 percent, and that Kosambi corrects less than Haldane because it assumes interference has already removed some double crossovers.
Key point

Expected double crossovers equal the product of the two interval frequencies times the number of offspring. The coefficient of coincidence (observed ÷ expected) is usually less than 1, and interference (1 − c.o.c.) is positive because one crossover inhibits others nearby. Mapping functions correct recombination frequencies for unseen multiple crossovers: Haldane's assumes no interference, d = −½ ln(1 − 2r), while Kosambi's allows for it.

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.

IIIIII 1/13/31/32/2 1/23/21/21/23/23/23/21/23/23/2 R Disease travels with marker 1 in the affected father. 9 nonrecombinant, 1 recombinant
Figure 4.6 A phase known family for an autosomal dominant disease. The affected grandfather is homozygous for marker allele 1, so in the affected father the disease allele must sit on the chromosome carrying allele 1. Children who received allele 1 should be affected; one child who received allele 3 is affected, a recombinant. Filled = affected.

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) = θR(1 − θ)NR and L(0.5) = 0.5R + 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:

Study tip. Worked problem: In the family of Figure 4.6 there are 10 informative, phase known meioses with 1 recombinant. Calculate the LOD score at θ = 0.1. Step 1. L(0.1) = 0.11 × 0.99. 0.99 = 0.3874, so L(0.1) = 0.03874. Step 2. L(0.5) = 0.510 = 0.0009766. Step 3. Odds = 0.03874 ÷ 0.0009766 = 39.7. Step 4. LOD = log10 39.7 = 1.60. Answer: 1.60, suggestive but short of 3. If a second, similar family also gives 1.60 at θ = 0.1, the combined LOD is 1.60 + 1.60 = 3.20, which crosses the threshold. Note that LOD at θ = 0 is minus infinity here, because a single recombinant is impossible if the loci are completely linked. Also note that 0.1 is the best estimate because it equals R ÷ total = 1/10.
LOD 3: linkage accepted 01234 00.10.20.30.40.5 recombination fraction θ LOD score one family: max 1.60two families: max 3.20 best θ = 0.1
Figure 4.7 LOD score curves. The curve for one family of ten phase known meioses with one recombinant peaks at θ = 0.1 with LOD 1.60. Because LOD scores from independent families add, two such families give a peak of 3.20, enough to declare linkage.

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[θR(1 − θ)NR + θNR(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:

MarkerWhat variesHow it is detectedStrengths and limits
RFLP (restriction fragment length polymorphism)A base change that creates or destroys a restriction enzyme siteDigestion and Southern blot, later PCR plus digestionFirst DNA markers (1980); usually only two alleles, so often uninformative
VNTR (minisatellite)Number of tandem copies of a repeat unit of about 10 to 60 base pairsSouthern blot or PCRHighly polymorphic; used in early DNA fingerprinting; unevenly spread
Microsatellite (STR)Number of copies of a 1 to 6 base pair repeat, such as (CA)nPCR and fragment sizing by electrophoresisMany alleles, very informative, found every few tens of kilobases; the workhorse of 1990s linkage maps and forensic identification
SNP (single nucleotide polymorphism)One base pair differenceDNA microarrays or sequencingUsually two alleles each, but millions exist, about 1 per 1,000 base pairs between two genomes; typed by the hundreds of thousands at once

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.

Connection. Before the Huntington disease gene was cloned, linkage to nearby markers allowed the first predictive testing: in families where the phase could be worked out, a person's risk could be estimated from which grandparental marker allele he or she had inherited. The test was accurate only to within the recombination fraction between marker and gene, a few percent, and it required samples from several relatives. Direct testing for the CAG repeat expansion replaced it in 1993, but the same linkage logic is still used in preimplantation diagnosis to confirm results with flanking markers.
Connection. Plant and animal breeders use marker assisted selection: once a marker is tightly linked to a valuable gene, such as disease resistance in wheat or rice, seedlings can be screened by a DNA test long before the trait can be measured. Linkage also explains linkage drag, the unwanted genes from a wild relative that come along with a desired resistance gene because they lie close to it on the chromosome. Breaking that association requires many generations of backcrossing and selection for rare recombinants.
Key point

Human linkage is measured in pedigrees by counting recombinants in informative meioses and summarizing the evidence as a LOD score, the log10 of the odds of linkage at θ versus no linkage. Scores from families add; 3 or more establishes linkage and −2 or less excludes it. Polymorphic DNA markers, from RFLPs to microsatellites to SNPs, made mapping human disease genes routine, and linkage disequilibrium extends the method to whole populations in association studies.

The chapter in one breath

  • Genes on the same chromosome are linked and tend to be inherited together; the number of linkage groups equals the haploid chromosome number.
  • The linkage phase, coupling (AB/ab) or repulsion (Ab/aB), decides which gametes are parental and which are recombinant.
  • Crossing over between nonsister chromatids in prophase I, seen as chiasmata, creates recombinants; one crossover affects two of four chromatids.
  • Recombination frequency = recombinants ÷ total; 1 percent equals 1 map unit or centimorgan, and RF never exceeds 50 percent.
  • A chi square test against 1:1:1:1 (3 degrees of freedom, critical value 7.815) tests for independent assortment.
  • In a three point testcross, the parental classes are largest, the double crossovers smallest, and the gene that differs between them is in the middle.
  • Interval distances include the double crossovers; the outside distance is the sum of the intervals and exceeds any direct two point estimate.
  • Coefficient of coincidence = observed ÷ expected DCOs; interference = 1 − c.o.c., usually positive.
  • Mapping functions correct for hidden crossovers: Haldane, d = −½ ln(1 − 2r), assumes no interference; Kosambi allows for it.
  • Human linkage is assessed with the LOD score; LOD 3 or more supports linkage, −2 or less excludes it, and scores add across families.
  • Molecular markers (RFLPs, microsatellites, SNPs) and linkage disequilibrium made positional cloning and genome wide association studies possible.

Words worth knowing

TermWhat it means
LinkageThe tendency of genes close together on the same chromosome to be inherited together.
Linkage groupA set of genes that are all linked to one another; in a well mapped species, one per chromosome.
Coupling (cis)The arrangement in which the two dominant alleles of a double heterozygote are on one homolog and the two recessives on the other (AB/ab).
Repulsion (trans)The arrangement in which each homolog carries one dominant and one recessive allele (Ab/aB).
Parental gameteA gamete carrying the same combination of linked alleles that the individual inherited from one parent.
Recombinant gameteA gamete carrying a new combination of alleles produced by crossing over or independent assortment.
Crossing overReciprocal exchange of segments between nonsister chromatids of homologous chromosomes during prophase I.
ChiasmaThe visible X shaped connection between homologs at the site of a crossover (plural chiasmata).
Synaptonemal complexThe protein structure that holds homologous chromosomes closely paired during prophase I.
Recombination frequencyThe proportion of progeny or gametes that are recombinant for two genes; symbol RF or θ.
Map unit (centimorgan)The genetic distance giving 1 percent recombination between two genes.
SyntenicLocated on the same chromosome, whether or not the genes show measurable linkage.
TestcrossA cross to a homozygous recessive individual, so that progeny phenotypes reveal the gametes of the other parent.
Chi square testA statistical test that compares observed with expected numbers to judge whether a deviation is due to chance.
Three point testcrossA testcross of an individual heterozygous for three linked genes, used to find gene order and two distances at once.
Double crossoverTwo crossovers in adjacent intervals on the same chromatid pair, which swap the middle gene relative to the outside genes.
Coefficient of coincidenceThe ratio of observed to expected double crossovers.
InterferenceThe reduction in the chance of a second crossover near a first one; equal to 1 minus the coefficient of coincidence.
Mapping functionA formula that converts recombination frequency to map distance by correcting for undetected multiple crossovers.
Informative meiosisA meiosis in a parent heterozygous at both loci, in which recombinants can be recognized.
LOD scoreThe base 10 logarithm of the odds that two loci are linked at a given recombination fraction rather than unlinked.
MicrosatelliteA short tandem repeat of a 1 to 6 base pair unit whose copy number varies, making a highly polymorphic marker.
Single nucleotide polymorphismA position in the genome where people differ by a single base; the most common type of marker.
Positional cloningFinding a gene by first mapping it near linked markers and then searching that region of DNA.
Linkage disequilibriumThe nonrandom association of alleles at nearby loci in a population.

Check yourself

A testcross of a double heterozygote gives 45 percent AB, 45 percent ab, 5 percent Ab and 5 percent aB. What was the phase of the heterozygote, and how far apart are the genes?
Answer. Coupling (AB/ab), because AB and ab are the large parental classes. The recombinants total 5 + 5 = 10 percent, so the genes are 10 map units apart.
Why can recombination frequency between two genes never exceed 50 percent, even when they are far apart on a long chromosome?
Answer. Each crossover involves only two of the four chromatids, and an even number of crossovers between two genes restores the parental combination. When genes are far apart, odd and even numbers of crossovers become equally likely, so recombinants approach but cannot pass 50 percent, the same value as for unlinked genes.
An Ab/aB individual with the genes 10 map units apart is testcrossed. What fraction of offspring will be aabb?
Answer. 5 percent. In repulsion, ab is a recombinant gamete. Recombinants total 10 percent and are split equally between AB and ab, so ab gametes are 0.10 ÷ 2 = 0.05.
In Drosophila, a dihybrid female testcrossed gives four classes with two large and two small, but a dihybrid male of the same genotype testcrossed gives only the two parental classes. Explain.
Answer. Drosophila males do not undergo meiotic crossing over, so linked genes show complete linkage in males. Recombination must be measured in females.
In a three point testcross, the parental classes are + + + and a b c, and the double crossover classes are a + c and + b +. Which gene is in the middle?
Answer. b. Compare parental + + + with the DCO + b +: they differ only at b, so b was swapped by the double crossover and lies between a and c.
Genes x and y are 10 cM apart and y and z are 20 cM apart, with y in the middle. Among 1,000 testcross offspring, 10 double crossovers are observed. Calculate the coefficient of coincidence and interference.
Answer. Expected DCOs = 0.10 × 0.20 × 1,000 = 20. c.o.c. = 10 ÷ 20 = 0.5, and I = 1 − 0.5 = 0.5. Half of the expected double crossovers were prevented by interference.
A testcross of 200 offspring gives 60, 56, 44 and 40 in the four classes, with the two larger classes parental. Is there significant evidence of linkage at P = 0.05?
Answer. No. Expected is 50 per class. χ2 = (100 + 36 + 36 + 100) ÷ 50 = 272 ÷ 50 = 5.44, which is below the critical value of 7.815 for 3 degrees of freedom. The deviation could be due to chance, so independent assortment is not rejected, although more data might reveal loose linkage.
Use Haldane's mapping function to convert a recombination frequency of 0.20 into map distance.
Answer. d = −½ ln(1 − 0.40) = −½ ln 0.60 = −½ × (−0.511) = 0.255 Morgans, or about 25.5 cM, slightly more than the 20 cM suggested by the raw frequency.
A phase known family has 8 informative meioses with no recombinants between a disease gene and a marker. What is the maximum LOD score, and is linkage established?
Answer. The maximum is at θ = 0: LOD = log10(1 ÷ 0.58) = log10 256 = 2.41. That is below 3, so linkage is suggested but not established; more families are needed.
Why was a microsatellite with 12 alleles more useful for family linkage studies than a single SNP, and why do SNPs dominate today?
Answer. With many alleles, most parents are heterozygous for a microsatellite and the two parents usually carry different alleles, so most meioses are informative. A SNP has only two alleles and is often homozygous. SNPs dominate now because there are millions of them and microarrays and sequencing can type hundreds of thousands at once, so many neighboring SNPs together form highly informative haplotypes.