Chapter 4 - Jones & Bartlett Learning

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Gene Linkage and © Jones & Bartlett Learning, LLC Genetic Mapping NOT FOR SALE OR DISTRIBUTION

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CHAPTER ORGANIZATION © Jones & Bartlett Learning, LLC 4.1 SALE LinkedOR alleles tend to stay NOT FOR DISTRIBUTION

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together in meiosis.

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4.4NOT Polymorphic DNA OR sequences are FOR SALE DISTRIBUTION used in human genetic mapping. 128

The degree of linkage is measured by the Single-nucleotide polymorphisms (SNPs) frequency of recombination. 113 are abundant in the human genome. The frequency of recombination is the same SNPs in restriction sites yield restriction for coupling and repulsion heterozygotes. 114 fragment length polymorphisms (RFLPs). © Jones & Bartlett Learning, LLC of recombination differs © Jones & Bartlett Learning, LLC The frequency Simple-sequence repeats (SSRs) often from one gene pair to the next. 114 FOR SALE differ in copy number. NOT FOR SALE OR DISTRIBUTION NOT OR DISTRIBUTION Recombination does not occur in Gene dosage can differ owing to copynumber variation (CNV). Drosophila males. 115 Copy-number variation has helped human 4.2 Recombination results from populations adapt to a high-starch diet.

crossing-over between © Jones & Bartlett linked alleles.

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all Learning, © Jones & Bartlett Learning, LL 116 LLC 4.5 Tetrads contain four products of meiosis. 135 Physical often—but not DISTRIBUTION NOT FOR SALE OR DISTRIBUT NOTdistance FORisSALE OR always—correlated with map distance. One crossover can undo the effects of another.

4.3 Double crossovers are revealed

in three-point crosses. © Jones & Bartlett Learning, LLC Interference the chance of NOT FOR SALE OR decreases DISTRIBUTION multiple crossing-over.

Varieties of maize. [© Katarzyna Citko/ ShutterStock, Inc.]

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Unordered tetrads have no relation to the geometry of meiosis. 136 Tetratype tetrads demonstrate that crossing-over takes place at the fourstrand stage of meiosis and is reciprocal. 137 © Tetrad Jones & Bartlett Learning, LLC analysis affords a convenient test for linkage. 137 NOT FOR SALE OR DISTRIBUTION The geometry of meiosis is revealed in ordered tetrads. 139 Gene conversion suggests a molecular mechanism of recombination. 142

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TXX.XX  A Head Goes Here

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4.6 Recombination is initiated by a

double-stranded break in DNA. 143

Recombination tends to take place at © Jones & Bartlett Learning, LLC preferred positions in the genome. NOT FOR SALE OR DISTRIBUTION the human connection  Starch Contrast

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Chapter Summary Learning Outcomes & Ideas © Issues Jones & Bartlett Learning, LLC Solutions: Step by Step NOT FOR SALE OR DISTRIBUTION Concepts in Action: Problems for Solution

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GENETICS on the web

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tend to remain together in inheritance, a phenomenetic mapping means determining the relaenon known as linkage. Nevertheless, the linkage tive positions of genes along a chromosome. is incomplete. Some gametes are produced that have It is one of the main experimental © Jones & Bartlett Learning, LLC tools in Jones & Bartlett Learning, LLC different ©combinations of the white and miniature genetics. This may seem odd in organisms in which the alleles than those in the parental chromosomes. The NOT FOR SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION DNA sequence of the genome has been determined. If new combinations are produced because homologous every gene in an organism is already sequenced, then chromosomes can exchange segments when they are what is the point of genetic mapping? The answer paired. This process (crossing-over) results in recomis that a gene’s sequence does not always reveal its bination of alleles between the homologous chro­function, nor does a genomic © Jones & Bartlett Learning, LLC DNA sequence reveal © Jones & Bartlett Learning, LLC mosomes. The probabi­ lity of recombination between which genes interact in a complex biological process. any two genesOR serves as a measure of genetic distance NOT FOR SALE OR DISTRIBUTION NOT FOR SALE DISTRIBUTION When a new mutant gene is discovered, the first between the genes and allows the construction of a step in genetic ­analysis is usually genetic mapping to genetic map, which is a diagram of a chromosome determine its position in the genome. It is at this point showing the relative positions of the genes. The linthat the genomic sequence, if known, becomes useful, ear order of genes along a genetic map is consistent because in some cases the position of the mutant gene LLC © Jones & Bartlett Learning, Jones Bartlett Learning, LL with the conclusion that © each gene & occupies a wellcoincides with a gene whose sequence suggests a role defined position, or locus, in the chromosome, with NOT FOR SALE OR DISTRIBUT NOT FOR SALE OR DISTRIBUTION in the biological process being investigated. For examthe alleles of a gene in a heterozygote occupying ple, in the case of flower color, a new mutation may corresponding locations in the pair of homologous map to a region containing a gene whose sequence chromosomes. suggests that it encodes an enzyme in anthocyanin In discussing linked genes, it is necessary to distinsynthesis. But&the functionLearning, of a gene LLC is not always © Jones Bartlett © Jones & Bartlett Learning, guish which alleles are present together in the LLC parenrevealed by its DNA sequence, and so in some cases, tal chromosomes. This is done byDISTRIBUTION means of a slash NOT FOR SALE OR DISTRIBUTION NOT FOR SALE OR further genetic or molecular analysis is necessary to (“/”). The alleles in one chromosome are depicted sort out which one of the genes in a sequenced region to the left of the slash, and those in the homologous corresponds to a mutant gene mapped to that region. chromosome are depicted to the right of the slash. In human genetics, genetic mapping is important For example, in the cross AA BB 3 aa bb, the genobecause it enables genes associated with hereditary type&ofBartlett the doubly heterozygous progeny is denoted © Jones & Bartlett Learning, LLC © Jones Learning, LLC diseases, such as those that predispose to breast canA B/aSALE b because the A and B alleles were inherited NOT FOR SALE OR DISTRIBUTION NOT FOR OR DISTRIBUTION cer, to be localized and correlated with the genomic in one parental chromosome and the alleles a and b sequence in the region. were inherited in the other parental chromosome. In this genotype the A and B alleles are said to be in the coupling or cis ­configuration; likewise, the a and 4.1 Linked alleles tend to stay b alleles are in coupling.©Among four possible © Jones & Bartlett Learning, LLC Jonesthe & Bartlett Learning, LL together in meiosis. types of gametes, the A NOT B andFOR a b types areOR called SALE DISTRIBUT NOT FOR SALE OR DISTRIBUTION parental combinations because the alleles are in the In meiosis, homologous chromosomes form pairs in same configuration as in the parental chromosomes, prophase I by undergoing synapsis, and the individand the A b and a B types are called recombinants ual members of each pair separate from one another (­FIGURE 4.1, part A). at anaphase I. Genes that are close enough together & Bartlettmight Learning, LLC © Jones Bartlett Learning, LLC Another possible& configuration of the A, a and B, b in© theJones same chromosome therefore be expected allele pairs is AFOR b/a B.SALE In this case A and B alleles toNOT be transmitted together. Thomas Hunt Morgan FOR SALE OR DISTRIBUTION NOT OR the DISTRIBUTION are said to be in the repulsion or trans configuration. examined this issue using two genes present in the Now the parental and recombinant gametic types are X chromosome of Drosophila. One was a mutation reversed (Figure 4.1, part B). The A b and a B types are for white eyes, the other a mutation for miniature the parental combinations, and the A B and a b types wings. Morgan found that the white and miniature © Jones & Bartlett Learning, © Jones & Bartlett Learning, LLC are the recombinants. alleles present in each XLLC chromosome of a female do

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(A)–Parental alleles in coupling (A)–or cis configuration

The degree of linkage is measured by the frequency © Jones & Bartlett Learning, LLC b of recombination.

(B)–Parental alleles in repulsion (B)–or trans configuration

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Recombinants

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In his early experiments with Drosophila, Morgan found mutations in each of several X-linked genes that provided ideal materials for studying linkage. One of these genes, © Jones & Bartlett with alleles w1 and w, determines normalLearning, LL red eye color versus whiteFOR eyes; SALE another such NOT OR DISTRIBUT gene, with the alleles m1 and m, determines whether the size of the wings is normal or miniature. The initial cross is shown as Cross 1 in FIGURE 4.2. It was a cross between © Jones Bartlett Learning, LLC females with&white eyes and normal wings and males withSALE red eyesOR and miniature wings: NOT FOR DISTRIBUTION wwmm+ + ww+ +mm × × // ?? YY wwmm+ +

FIGURE 4.1 For any pair of alleles, the gametes produced through meiosis have the alleles either in a parental configuration or in a recombinant configuration. Which types are parental and which recombinant depends In thisLearning, way of writing the genotypes, the © Jones & Bartlett Learning,of LLC © Jones & Bartlett LLC on whether the configuration the alleles in the parent is (A) coupling or horizontal line replaces the slash. Alleles NOT FOR(B) repulsion. SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION

(A) Cross 1

Parents:

Parents: © Jones & Bartlett Learning, LLC Miniature-wing White-eyed and NOT FOR SALE OR DISTRIBUTION

White-eyed females w+ w+

(B)

males

+m Y





miniature-wing females wm  wm



F© 1:

Cross 2

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++ Y



Jones & Bartlett Learning, LLC + wm NOT FOR SALEwOR DISTRIBUTION    Mutant alleles

© Jones & Bartlett Learning, LLC w m OR DISTRIBUTION NOT FOR SALE   w m  Mutant alleles

in homologous chromosomes

in the same chromosomes

F2:

F1:

Y

+m

F2:

Progeny 1

© Jones & Bartlett Learning, LLC 226 White eyes, NOT FOR SALE OR DISTRIBUTION normal wings (maternal gamete: w +)

++

Progeny 2

© Jones & Bartlett Learning, LLC 223 Parental types:NOT FOR Recombinant types: SALE OR DISTRIBUTION 66.5% have parental allele combinations (nonrecombinant).

202 Red eyes, miniature wings (maternal gamete: + m) & Bartlett © Jones

37.7% have nonparental allele combinations (recombinant).

Learning, LLC NOT FOR SALERecombinant OR DISTRIBUTION types: Parental types:

Red eyes, normal wings (maternal gamete: + +)

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White eyes,

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Y

33.5% have nonparental allele combinations (recombinant).

© Jones & Bartlett Learning, LLC miniature wings 644 (maternal gamete: w m) NOT FOR SALE OR DISTRIBUTION

62.3% have parental allele combinations (nonrecombinant).

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FIGURE 4.2 An experiment demonstrating that the frequency of recombination between two mutant alleles is independent of whether they are present in the same chromosome or in homologous chromosomes. (A) Cross 1 produces F1 females with the genotype w 1/1 m, and the w 2 m recombination frequency is 33.5 percent. (B) Cross 2 produces F1 females with the genotype w m/1 1, and the w 2 m & Bartlettfrequency Learning, © range Jones & Bartlett LLC recombination is 37.7LLC percent. These values are within the of variation expectedLearning, to occur by chance.

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written above the line are present in one chromoKEY CONCEPT some, and those written below the line are present in Genes with recombination frequencies smaller than homologous chromosome. & Bartlett Learning, © Jones & the Bartlett Learning, LLC In the females, both ©X Jones50 percent are present in the LLC same c­ hromosome 1 chromosomes carry w and m . In males, the X chroNOT FOR SALE OR DISTRIBUTION (linked). Two genes that undergo independent NOT FOR SALE OR DISTRIBUTION mosome carries the alleles w1 and m. (The Y written assortment, indicated by a recombination f­requency below the line denotes the Y chromosome in the male.) equal to 50 percent, either are in nonhomologous Figure 4.2 illustrates a simplified symbolism, commonly chromosomes or are located far apart in a single used in Drosophila genetics, in which a wildtype allele chromosome. is denoted by a 1 sign in the appropriate position. The LLC © Jones & Bartlett Learning, © Jones & Bartlett Learning, LL 1 symbolism is unambiguous because linked genes NOT FOR SALE OR DISTRIBUT NOT FOR SALEtheOR DISTRIBUTION in a chromosome are always written in the same order. The frequency of recombination is Using the 1 notation,

the same for coupling and repulsion heterozygotes.

w+ w m+ means w Bartlett + w m+ & Learning,

© Jones LLC NOT FOR SALE OR DISTRIBUTION and 1 m Y

means

© Jones Bartlett Learning, LLC Morgan also studied&progeny from the coupling configurationNOT of theFOR w1 and m1 alleles, which results from SALE OR DISTRIBUTION the mating designated as Cross 2 in Figure 4.2. In this case, the original parents had the genotypes

w+ m Y

ww m m

××

1 11 1

Y wm m // Y ?? © Jones & Bartlett Learning, LLC from Cross 1 have the © Jones & Bartlett wLearning, LLC The resulting F1 female progeny genotype 1/1 m (or, equivalently, w m1/w1 m). NOT In NOT FOR SALE ORw DISTRIBUTION FOR SALE OR DISTRIBUTION

The resulting F1 female progeny from Cross 2 have the this genotype, the w1 and m1 alleles are in repulsion. genotype w m/1 1 (equivalently, w m/w1 m1). In this When these females were mated with w m/Y males, case the wildtype alleles are in the same chromosome. the offspring denoted as Progeny 1 in Figure 4.2 were When these F1 female progeny were crossed with obtained. In each class of progeny, the gamete from w m/Y males, they yielded©the types of © Jones & Bartlett Learning, LLC Jones & progeny BartletttabuLearning, LL the female parent is shown in the column at the left, lated as Progeny 2 in Figure 4.2. and the gamete from theFOR maleSALE parent OR carries either NOT FOR SALE OR DISTRIBUT NOT DISTRIBUTION Because the alleles in Cross 2 are in the coupling w m or the Y chromosome. The cross is equivalent configuration, the parental-type gametes carry either to a testcross, and so the phenotype of each class of w m or 1 1, and the recombinant gametes carry ­progeny reveals the alleles present in the gamete from either w 1 or 1 m. The types of gametes are the same the mother. as those 1, but the parental and ©The Jones & Bartlett Learning, LLC ©observed Jonesin& Cross Bartlett Learning, LLC results of Cross 1 show a great departure from recombinant types are opposite. Yet the frequency of theNOT 1 : 1 FOR : 1 : 1 SALE ratio of OR the four male phenotypes that DISTRIBUTION NOT FOR SALE OR DISTRIBUTION recombination is approximately the same: 37.7 percent is expected with independent assortment. If genes versus 33.5 percent. The difference is within the range in the same chromosome tended to remain together expected to result from random variation from experiin ­ inheritance but were not completely linked, this ment to experiment. The consistent finding of equal ­pattern of deviation might be observed. In this case, recombination frequencies in experiments in which the the combinations of phenotypic of Jones © Jones & Bartlett Learning, LLC traits in the parents © & Bartlett Learning, LLC mutant alleles are in the trans or the cis configuration the original cross (parental phenotypes) were present NOT FOR SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION leads to the following conclusion: in 428/644 (66.5 percent) of the F2 males, and nonparental combinations (recombinant phenotypes) of KEY CONCEPT the traits were present in 216/644 (33.5 percent). The 33.5 percent recombinant X chromosomes is called Recombination between linked genes takes place Jones & Bartlett Jones Bartlett the frequency of © recombination, and it Learning, should be LLCwith the same frequency © whether the& alleles of the Learning, LL contrasted with the 50 percent expected genes are in the repulsion (trans)FOR configuration NOT SALE or OR DISTRIBUT NOT FORrecombination SALE OR DISTRIBUTION with independent assortment. in the coupling (cis) configuration; it is the same no The recombinant X chromosomes w1 m1 and w m matter how the alleles are arranged. result from crossing-over in meiosis in F1 females. In this example, the frequency of recombination between ©linked Jones & Bartlett Learning, LLC © Jones &ofBartlett Learning, LLC the w and m genes was 33.5 percent. With other The frequency recombination pairs of linked genes, the of recombination NOT FOR SALE ORfrequency DISTRIBUTION NOT FOR SALE OR DISTRIBUTION differs from one gene pair to ranges from near 0 to 50 percent. Even genes in the the next. same chromosome can undergo independent assortment (frequency of recombination equal to 50 percent) The principle that the frequency of recombination if they are sufficiently far apart. This implies the followdepends on the particular pair of genes may be © Jones & Bartlett Learning, LLC © Jones & Bartlett LLC ing principle: illustrated ­ usingLearning, the recessive allele y of another

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The results of these and other experiments give X-linked gene in Drosophila, which results in yellow support to two general principles of recombination: body color instead of the usual gray color determined The yellow body (y) and white eye © Jones & Bartlett Learning, LLC the y1 allele. © Jones by & Bartlett Learning, LLC n The recombination frequency is a characteristic of (w) genes are linked. The frequency of recombina- NOT FOR SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION a particular pair of genes. tion between the genes is as shown in the data in n Recombination frequencies are the same in cis FIGURE  4.3. The layout of the crosses is like that (coupling) and trans (repulsion) heterozygotes. in Figure 4.2. In Cross 1, the female has y and w in the trans configuration (1 w/y  1); in Cross 2, the Recombination does©not occur alleles are in the cis©configuration (y w/1  1). The Jones & Bartlett Learning, LLC Jones & Bartlett Learning, LL y and w genes exhibit a much lower frequency of in Drosophila males.NOT FOR SALE OR DISTRIBUT NOT FOR SALE OR DISTRIBUTION recombination than that observed with w and m in Early experiments in Drosophila genetics also indicated Figure 4.2. To put it another way, the genes y and w that the organism is unusual in that recombination are more closely linked than are w and m. In Cross does not take place in males. The absence of recom1, the recombinant progeny are 1 1 and y w, and bination in Drosophila males means that all alleles they © account for&130/9027 5 Learning, 1.4 percent ofLLC the total. Jones Bartlett © Jones & Bartlett Learning, LLC located in a particular chromosome show complete In Cross the recombinant are 1 w and NOT2,FOR SALE OR progeny DISTRIBUTION NOT FOR OR linkage in the male. For SALE example, theDISTRIBUTION genes cn (ciny 1, and they account for 94/7838 5 1.2 percent of nabar eyes) and bw (brown eyes) are both in chromothe total. Once again, the parental and recombinant some 2, but they are so far apart that in females, they gametes are reversed in Crosses 1 and 2, because the show 50 percent recombination. Because the genes configuration of alleles in the female parent is trans exhibit 50 percent recombination, the cross in Bartlett Cross 1 but cis in Cross 2, yet the frequency of © Jones & Bartlett Learning, LLC © Jones & Learning, LLC cn bw bw cn cn bw bw between the genes is within experi- NOT FOR SALEcn NOT FORrecombination SALE OR DISTRIBUTION OR DISTRIBUTION ×× // cn cn bw bw?? 11 11 mental error. Cross 1 Parents:

Parents: © Jones & Bartlett Learning, LLC Yellow-body White-eyed and NOT FOR SALE OR DISTRIBUTION

White-eyed females +w +w

Cross 2

males

y+ Y





yellow-body females yw yw



F1© :

Jones & Bartlett Learning, LLC NOT FOR SALE+ wOR DISTRIBUTION  y w  The trans F2:

The cis (coupling) heterozygote

Y

y+

(repulsion) heterozygote

F1:

© Jones & Bartlett Learning, LLC Wildtype body, NOT FOR SALE OR DISTRIBUTION4292 white eyes (maternal gamete: + w)

males





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Yellow body,

44

Y

Progeny 2

98.6% have parental allele combinations (nonrecombinant).

1.2% have nonparental allele combinations (recombinant).

Learning, LLC NOT FOR SALE OR DISTRIBUTION Recombinant types: Parental types: 86



© Jones & Bartlett Learning, LLC 55 Parental types: NOT FOR Recombinant types:DISTRIBUTION SALE OR

4605 Yellow body, red eyes (maternal gamete: y +) & Bartlett © Jones Wildtype body, red eyes (maternal gamete: + +)

++ Y

© Jones & Bartlett Learning, LLC y w OR DISTRIBUTION NOT FOR SALE   y w 

F2:

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© Jones & Bartlett Learning, LL Wildtype NOT FOR SALE OR DISTRIBUT

1.4% have nonparental allele combinations (recombinant).

© Jones Bartlett Learning, LLC white& eyes 9027 (maternal gamete: y w) NOT FOR SALE OR DISTRIBUTION

98.8% have parental allele combinations (nonrecombinant).

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FIGURE 4.3 An experiment demonstrating that the frequency of recombination between two genes depends on the genes. The frequency of recombination between w and y is much less than that between w and m in Figure 4.2. The y2w experiment also confirms the equal frequency of recombination in trans and cis heterozygous genotypes. (A) The trans heterozygous females, 1 w/y 1, yield & Bartlett Learning, LLC © Jones & Bartlett Learning, LLC 1.4 percent recombination. (B) The cis heterozygous females, y w/11, yield 1.2 percent recombination.

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of 3.1 map units therefore equals 3.1 centimorgans and yields progeny of genotype cn bw/cn bw and 1 1/cn bw indicates 3.1 percent recombination between the genes. (the nonrecombinant types) as well as cn 1/cn bw and An example is shown in part ALLC of FIGURE 4.5, which 1 bw/cn bw (the recombinant types) in the proportions © Jones & Bartlett Learning, © Jones & Bartlett Learning, LLC deals with the Drosophila mutants w for white eyes and 1 : 1 : 1 : 1. The outcome of the reciprocal cross is difNOT FOR SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION dm (diminutive) for small body. The female parent in the ferent. Because no crossing-over occurs in males, the testcross is the trans heterozygote, but as we have seen, reciprocal cross this configuration is equivalent in frequency of recomcn cn bw bw cn cn bw bw × × bination to the cis heterozygote. Among 1000 progeny cn cn bw bw // 11 11 ?? there are 31 recombinants.©Using this & estimate, we Learning, can © Jones & Bartlett Learning, LLC Jones Bartlett LL express the genetic distance between w and dm in four NOT FOR SALE OR DISTRIBUT NOT FOR SALE OR DISTRIBUTION yields progeny only of the nonrecombinant genotypes cn completely equivalent ways: bw/cn bw and 11/cn bw in equal proportions. The absence of recombination in Drosophila males is a ­convenience n As the frequency of recombination—in this often exploited in experimental design; as shown in the case 0.031 case of cn and bw, all the alleles ­present in any chromo n As the©percent recombination, 3.1 percent LLC © Jones & Bartlett Learning, LLC Jones & BartlettorLearning, some in a male must be transmitted as a group, without n As the distance in map units—in this example, 3.1 NOT FOR SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION being recombined with alleles p ­ resent in the homologous map units chromosome. The absence of c­ rossing-over in Drosophila n As the distance in centimorgans, or males is atypical; in most other animals and plants, 3.1 ­centimorgans (3.1 cM) recombination takes place in both sexes, though not necessarily with the same frequency.

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© Jones & Bartlett Learning, LLC A genetic map based on these data is shown in ­Figure 4.5, part B.SALE The chromosome is represented as a horizontal NOT FOR OR DISTRIBUTION

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4.2 Recombination results from crossing-over between linked alleles.

line, and each gene is assigned a position on the line according to its genetic distance from other genes. In this example, there are only two genes, w and dm, and they are separated by a distance of 3.1 centimorgans (3.1 cM), Jones Bartlett Learning, © Jones & Bartlett LL or 3.1 map units. Genetic maps are usually truncatedLearning, to The linkage of the©genes in & a chromosome can be LLC show only the genes of interest. The full genetic represented in the NOT form ofFOR a genetic map,OR which shows NOT FOR SALE ORmap DISTRIBUT SALE DISTRIBUTION of the Drosophila X chromosome extends considerably the linear order of the genes along the chromosome farther in both directions than indicated in this figure. spaced so that the distances between adjacent genes is Physically, one map unit corresponds to a length of proportional to the frequency of recombination between the chromosome in which, on the average, one crossthem. A genetic map is also called a linkage map or a © Jones &map. Bartlett Learning, LLCmapping © Jones & Bartlett LLC over is formed in every 50 cells Learning, undergoing meiosis. chromosome The concept of genetic This principle illustrated in FIGURE 4.6. If one meiotic was firstFOR developed Morgan’s student Alfred H. NOT SALEbyOR DISTRIBUTION NOTis FOR SALE OR DISTRIBUTION cell in 50 has a crossover, the frequency of crossingSturtevant in 1913. The early geneticists understood over equals 1/50, or 2 percent. Yet the frequency of that recombination between genes takes place by an recombination between the genes is 1 percent. The corexchange of segments between homologous chromorespondence of 1 percent recombination with 2 percent somes in the process now called crossing-over. Each crossing-over is a Learning, little confusing until you consider that crossover isLearning, manifested physically © Jones & Bartlett LLC as a chiasma, or cross© Jones & Bartlett LLC a crossover in two recombinant chromatids and shapedOR configuration, between homologous chromoNOT FOR SALE DISTRIBUTION NOT FOR SALEresults OR DISTRIBUTION two nonrecombinant chromatids (­Figure 4.6). A crosssomes; chiasmata are observed in prophase I of meiosis. over frequency of 2 percent means that of the 200 chroEach chiasma results from the breaking and rejoining of mosomes that result from meiosis in 50 cells, exactly chromatids during meiosis, with the result that there is 2 chromosomes (those involved in the crossover) are an exchange of corresponding segments between them. recombinant for genetic markers spanning the particuThe theory of crossing-over is that each chiasma results LLC © Jones & Bartlett Learning, © Jones & Bartlett Learning, LL lar chromosome segment. NOT To putFOR the matter in another in a new associationNOT of genetic arkers. This is SALE OR DISTRIBUT FOR­mSALE OR process DISTRIBUTION way, 2 percent crossing-over corresponds to 1 percent illustrated in FIGURE 4.4. When there is no crossingrecombination because only half of the chromatids in over (part A), the alleles present in each homologous each cell with a crossover are actually recombinant. chromosome remain in the same combination. When a In situations in which there are genetic markers crossover does take place (part B), the outermost alleles Jones & Bartlett Jones & Bartlett Learning, along the©chromosome, such as the A, a and B,LLC b pairs in© two of the chromatids areLearning, interchangedLLC (recombined). of allelesNOT in Figure recombination between the The unit distanceOR in aDISTRIBUTION genetic map is called a NOT FORofSALE FOR4.6, SALE OR DISTRIBUTION marker genes takes place only when a crossover occurs map unit; one map unit is equal to 1 percent recombibetween the genes. FIGURE 4.7 illustrates a case in nation. For example, two genes that recombine with a which a crossover takes place between the gene A and frequency of 3.1 percent are said to be located 3.1 map the centromere, rather than between the genes A and units apart. One map unit is also called a centimorgan, © Jones & Bartlett LLC © Jones & Bartlett B. The crossover Learning, does result inLLC the physical exchange abbreviatedLearning, cM, in honor of T. H. Morgan. A distance 116

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(A) No crossing-over

(B) Crossing-over

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© bJonesRecombinant & Bartlett Learning, LLC chromatids NOT FOR SALE OR DISTRIBUTION +

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Crossing-over between these chromatids

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Result: Four & Bartlett Learning, Result: Two recombinant © Jones LLC nonrecombinant and two nonrecombinant NOT FOR SALE OR DISTRIBUTION chromatids chromatids

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FIGURE 4.4 Diagram illustrating crossing-over between two genes. (A) When there is no crossover between two genes, the alleles are not recombined. (B) When there is a crossover between them, the result is two recombinant and two nonrecombinant products, because the exchange takes place between only two of the four chromatids.

© Jones & Bartlett Learning, LLC NOT FOR SALE OR DISTRIBUTION

© Jones & Bartlett Learning, LLC NOT FOR SALE OR DISTRIBUTION

do not result in recombinant chromosomes), there is of segments between the innermost chromatids. Howan important distinction between the distance between ever, because it is located outside the region between A two genes as measured by the recombination frequency and B, all of the resulting gametes must carry either the and as measured in map units: A B or the a b allele combination. These are nonrecombinant chromosomes. The LLC presence of the crossover is © Jones © Jones & Bartlett Learning, & Bartlett Learning, LLC n The map distance between two genes equals onebecause it is not in the region between the NOT FOR NOT FORundetected SALE OR DISTRIBUTION SALE OR DISTRIBUTION half of the average number of crossovers that take genetic markers. place in the region per meiotic cell; it is a measure In some cases, the region between genetic markers of crossing-over. is large enough that two (or even more) crossovers can n The recombination frequency between two genes be formed in a single meiotic cell. One possible conindicates how much recombination is ­actually 4.8. figuration for two crossovers is shown in FIGURE © Jones & Bartlett Learning, LLC © Jones & Bartlett Learning, LL observed in a particular experiment; it is a ­measure In this example, both crossovers areSALE between theDISTRIBUTION same NOT FOR SALE OR DISTRIBUT NOT FOR OR of recombination. pair of chromatids. The result is that there is a physical exchange of a segment of chromosome between the The difference between map distance and recombinamarker genes, but the double crossover remains undetion frequency arises because double crossovers that do tected because the markers themselves are not recomnot yield recombinant gametes, like the one depicted © The Jones & Bartlett Learning, LLC © do Jones & Bartlett Learning, LLC bined. absence of recombination results from the in Figure 4.8, contribute to the map distance but do fact NOT that the second crossover reverses the effect of not contribute to the recombination frequency. The disFOR SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION the first, insofar as recombination between A and B is tinction is important only when the region in question concerned. The resulting chromosomes are either A B is large enough for double crossing-over to occur. If the or a b, both of which are nonrecombinant. region between the genes is short enough that no more Because double crossovers in a region between two than one crossover can occur in the region in any one © Jones & Bartlett Learning, LLC &then Bartlett Learning, LLC genes can remain undetected (this happens when they © Jones meiosis, map units and recombination frequencies

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(A) Cross

(B) Genetic map

Parent:

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© Jones & Bartlett Learning, LLC Diminutive White-eyed NOT FOR SALE OR DISTRIBUTION males females w+ w+ F1:

The trans heterozygote





+ dm Y

3.1 cM



© Jones & Bartlett Learning, LLC w+ dm    wYOR NOT+FOR SALE DISTRIBUTION dm

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F2: Progeny

© Jones & Bartlett Learning, LLC 497 White eyes, NOT FOR SALE normal size OR DISTRIBUTION

Parental types = 969/1000 = 96.9%

(maternal gamete: w +)

Red eyes, diminutive size (maternal gamete: + dm)

© Jones & Bartlett Learning, LLC NOT FOR SALE OR DISTRIBUTION Red eyes,

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19

normal size (maternal gamete: + +) White eyes, 12 diminutive size © Jones & Bartlett 1000 (maternal gamete: w dm)

© Jones & Bartlett Learning, LLC Recombinant types = 31/1000 NOT FOR SALE= 3.1% OR DISTRIBUTION

Genetic distance = frequency of recombination, 0.031 Genetic distance = percent recombination, 3.1% Genetic distance = map distance in map units, 3.1 map units Genetic distance = map distance in centimorgans, 3.1 centimorgans (3.1 cM)

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© Jones & Bartlett Learning, LL NOT FOR SALE OR DISTRIBUT

FIGURE 4.5 An experiment illustrating how the frequency of recombination is used to construct a genetic map. (A) There is 3.1 percent recombination between the genes w and dm. (B) A genetic map with w and dm positioned 3.1 map units (3.1 centimorgans, cM) apart, corresponding to 3.1 percent recombination. The map distance equals frequency of recombination only when the frequency of recombination sufficiently small. © Jonesis& Bartlett Learning, LLC © Jones & Bartlett Learning, LLC

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are the same (because there are no multiple crossovers that can undo each other). This is the basis for defining a map unit as being equal to 1 percent recombination:

frequency between genes y and rb is 7.5 percent, and that between rb and cv is 6.2 percent. The genetic map might be any one of three possibilities, depending on which is in the middle (y, LLC cv, or rb). Map C, which © Jones & Bartlett Learning, LLC © Jones & gene Bartlett Learning, has y SALE in the middle, can be excluded because it implies KEY CONCEPT NOT FOR SALE OR DISTRIBUTION NOT FOR OR DISTRIBUTION that the recombination frequency between rb and cv Over an interval so short that multiple c­ rossovers should be greater than that between rb and y, and this are precluded (typically yielding 10 percent contradicts the observed data. In other words, map C recombination or less), the map distance equals the can be excluded because it implies that the frequency recombination frequency because all crossovers of recombination between© y and cv must be negative. © Jones & Bartlett Learning, LLC Jones & Bartlett Learning, LL result in recombinant gametes. Maps A and B are bothNOT consistent with the observed FOR SALE OR DISTRIBUT NOT FOR SALE OR DISTRIBUTION recombination frequencies. They differ in their predictions regarding the recombination frequency between Furthermore, when adjacent chromosome regions y and cv. Using the principle of additivity of map separating linked genes are so short that multiple crossdistances, the predicted y2cv map distance in A is overs are not formed, the recombination frequencies (and hence the distances) betweenLLC the genes are 13.7 map©units, whereas the predicted y2cv map dis© Jones & map Bartlett Learning, Jones & Bartlett Learning, LLC additive. This important feature of recombination, as tance in BNOT is 1.3 FOR map units. In fact, observed recomNOT FOR SALE OR DISTRIBUTION SALE ORthe DISTRIBUTION well as the logic used in genetic mapping, is illustrated bination frequency between y and cv is 13.3 percent. by the example in FIGURE 4.9. The genes are located Map A is therefore correct. However, there are actually in the X chromosome of Drosophila—y for yellow two genetic maps corresponding to map A. They differ body, rb for ruby eye color, and cv for shortened wing only in whether y is placed at the left or at the right. One crossvein. The experimentally map&is Bartlett y2rb2cv, Learning, which is the one shown in ­Figure 4.9; © Jones & Bartlett Learning, LLC measured recombination © Jones LLC

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r=

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b nonrecombinant © Jones &1Bartlett Learning, LLC NOT FOR SALE OR DISTRIBUTION

a

2 1+1 = 49 + 49 + 49 + 49 + 1 + 1 + 1 + 1 200

= 1 percent = 1 map unit = 1 cM

© Jones & Bartlett Learning, LLC configurations in 50 meiotic cells, © Jones Learning, LLC FIGURE 4.6 Diagram of chromosomal in which 1& hasBartlett a crossover between two genes. (A) The 49 cells without a crossover result in 98 A B and 98 a b chromosomes; these are all nonrecombinant. (B) The cell with a crossover yields chromosomes NOT FOR SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION that are A B, A b, a B, and a b, of which the middle two types are recombinant chromosomes. (C) The recombination frequency equals 2/200, or 1 percent, also called 1 map unit or 1 cM. Hence, 1 percent recombination means that 1 meiotic cell in 50 has a crossover in the region between the genes.

© Jones & Bartlett Learning, LLC A B SALE OR DISTRIBUTION A FOR NOT

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© Jones & Bartlett Learning, LLC is not NOTCrossing-over FOR SALE OR DISTRIBUTION detected because no recombination between A and B genes occurred.

outside of the region between the A and B genes.

FIGURE 4.7 Crossing-over outside the region between two genes is not detectable through recombination. Although a segment of © Jones & Bartlett Learning, LLC © Jones & Bartlett Learning, LLC chromosome is exchanged, the genetic markers outside the region of the crossovers stay in the nonrecombinant configurations, in this NOT FORcase SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION A B and a b.

B

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© Jones & Bartlett Learning, LL OR DISTRIBUT

Double& crossover is not © Jones Bartlett Learning, LLC detected because it does NOTnot FOR SALE OR DISTRIBUTION result in recombination

same pair of chromatids.

between A and B genes.

FIGURE 4.8 If two crossovers take place between marker genes A and B, and both involve the same pair of chromatids, then neither

© Jones & Bartlett Learning, Jones & Bartlett crossover is detected because allLLC of the resulting chromosomes are©nonrecombinant A B or a b. Learning, LLC NOT FOR SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION

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(A)

y

rb 7.5 cM

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6.2 cM rb

y

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© Jonescv & Bartlett Learning, LLC NOT FOR SALE ORis the DISTRIBUTION If this correct genetic map,

rb

cv

© Jones & Bartlett Learning, LLC y rb NOT FOR SALE OR DISTRIBUTION 7.5 cM cv

then the distance y to cv should be 7.5 cM + 6.2 cM = 13.7 cM.

© Jones & Bartlett Learning, LL NOT FOR SALE OR DISTRIBUT If this is the correct genetic map, then the distance y to cv should be 7.5 cM – 6.2 cM = 1.3 cM.

rb 6.2 cM

© Jones & Bartlett Learning, LLC y cv rb NOT FOR SALE OR DISTRIBUTION (C)

rb

© Jones & Bartlett Learning, LLC This genetic map is not consistent NOT FOR SALE OR DISTRIBUTION

cv

y

7.5 cM

© Jones & Bartlett Learning, LLC NOT FOR SALE OR DISTRIBUTION

6.2 cM

with the data; the map implies that the frequency of recombination between y and cv is negative (because 6.2 cM – 7.5 cM = –1.3 cM); however, a negative frequency of Bartlett Learning, recombination is not LLC possible.

© Jones & NOT FOR SALE OR DISTRIBUTION

FIGURE 4.9 In Drosophila, the genes y (yellow body) and rb (ruby eyes) have a recombination frequency of 7.5 percent, and rb and cv (shortened wing crossvein) have a recombination frequency of 6.2 percent. There are three possible genetic maps, depending on whether rb is in the middle (part A), cv is in the middle (part B), or y is in the middle (part C). Map (C) can be excluded because it implies that rb and y should be closer than rb and cv, whereas the observed recombination frequency between rb and y is actually greater than that between rb and cv. Maps and (B) & are Bartlett compatible with the data given. © (A) Jones Learning, LLC © Jones & Bartlett Learning,

LL NOT FOR SALE OR DISTRIBUT

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Courtesy of M. G. Neuffer, College of Agriculture, Food, and Natural Resources, University of Missouri.

Courtesy of M. G. Neuffer, College of Agriculture, Food, and Natural Resources, University of Missouri.

R1-mb in a cross between a heterozygous genotype and the other is cv2rb2y. The two ways of depicting the a homozygous wildtype. genetic map are completely equivalent because there is no way of knowing from the recombination data © Jones LLC © Jones & Bartlett Learning, LLC whether y or & cv Bartlett is closer toLearning, the telomere. (Other data NOT that FOR SALE NOT FOR SALE OR DISTRIBUTION indicate y is, in fact,OR nearDISTRIBUTION the telomere.) A genetic map can be expanded by this type of reasoning to include all of the known genes in a ­chromosome; these genes constitute a linkage group. The number of linkage groups is the same as the haploid © Jones & Bartlett LLC © Jones & Bartlett Learning, LLC number ofLearning, chromosomes of the species. For example, Physical distance is often—but not cultivated (Zea mays) has ten pairs of chromosomes NOT FOR SALE ORcorn DISTRIBUTION NOT FOR SALE OR DISTRIBUTION always—correlated with map distance. and ten linkage groups. A partial genetic map of chroGenerally speaking, the greater the physical separamosome 10 is shown in FIGURE  4.10, along with the tion between genes along a chromosome, the greater dramatic phenotypes caused by some of the mutations. the map distance between them. Physical distance and The ears of corn shown in parts C and F demonstrate genetic map distance are©usually correlated, because © Jones & Bartlett Jones & Bartlett Learning, LL the result of Mendelian segregation. The earLearning, in part C LLC a greater distance between genetic markers a shows a 3 : 1 segregation yellow : orange kernels proNOT FOR SALEaffords OR DISTRIBUT NOT ofFOR SALE OR DISTRIBUTION greater chance for a crossover to take place; crossingduced by the recessive orange pericarp-2 (orp-2) allele in over is a physical exchange between the chromatids of a cross between two heterozygous genotypes. paired homologous chromosomes. On the other hand, the general correlation between © Jones & Bartlett Learning, LLC Jonesand & genetic Bartlett Learning, physical © distance map distance isLLC by no means absolute. We have already noted that the freNOT FOR SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION quency of recombination between genes may differ in males and females. An unequal frequency of recombination means that the sexes can have different map distances in their genetic maps, although the physical The ear in part F shows a 1 : 1 segregation of © Jones & Bartlett Learning, LLC by the dominant allele © Jones & Bartlett chromosomes of Learning, the two sexesLLC are the same and the marbled:white kernels produced

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Short arm (10) Centromere

(C)

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© Jones & Bartlett Learning, LLC NOT FOR SALE OR DISTRIBUTION (B)

php1 glu1

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zn1 du1 li1

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sr2

135 & Bartlett Learning, LL © Jones NOT FOR SALE OR DISTRIBUT Telomere

FIGURE 4.10 Genetic map of chromosome 10 of corn, Zea mays. The map distance to each gene is given in standard map units (centimorgans) relative to a position 0 for the telomere of the short arm (lower left). (A) Mutations in the gene oil yellow-1 (oy1) result in a yellow-green plant. The plant in the foreground is heterozygous for the dominant allele Oy1; behind is a normal plant. (B) Mutations in the gene lesion-16 (les16) in many small to medium-sized, on the leaf blade and sheath. The photograph © Jones &result Bartlett Learning, LLCirregularly spaced, discolored©spots Jones & Bartlett Learning, LLC shows the phenotype of a heterozygote for Les16, a dominant allele. (C) The orp2 allele is a recessive expressed as orange pericarp, a maternal NOT FOR SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION tissue that surrounds the kernels. The ear shows the segregation of orp2 in a cross between two heterozygous genotypes, yielding a 3 : 1 ratio of yellow : orange seeds. (D) The gene zn1 is zebra necrotic-1, in which dying tissue appears in longitudinal leaf bands. The leaf on the left is homozygous zn1, that on the right is wildtype. (E) Mutations in the gene teopod-2 (tp2) result in many small, partially podded ears and a simple tassel. An ear from a plant heterozygous for the dominant allele Tp2 is shown. (F) The mutation R1-mb is an allele of the r1 gene, resulting in red or purple color in the aleurone layer of the seed. Note the marbled color in kernels of an ear segregating for R1-mb. [Adapted from an illustration by E. H. Coe. Photos courtesy of M. G. Neuffer, College of Agriculture, Food, and Natural Resources, University of Missouri.]

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genes must have the same linear order. Distance in map units on genetic map For example, because there is no recom54.5 3.0 49.5 male Drosophila, © Jones & Bartlett Learning, LLC © Jones & bination BartlettinLearning, LLCthe map distance between any pair of genes located in PhysicalNOT map FOR SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION the same chromosome, when measured in Euchromatin Heterochromatin Euchromatin the male, is 0. (On the other hand, genes on different chromosomes do undergo independent assortment in males.) Genetic map The general correlation between phys- Learning, LLC © Jones & Bartlett © Jones & Bartlett Learning, LL pr cn sp net ical distance and genetic distance can NOT FOR SALE OR DISTRIBUT NOTmap FOR SALE OR DISTRIBUTION even break down in a single chromosome. 54.5 57.5 107.0 0.0 For example, crossing-over is much less Map position frequent in heterochromatin, which consists primarily of gene-poor regions near Very little recombination the© centromeres, than in Learning, euchromatin.LLC Jones & Bartlett © Jones & takes Bartlett LLC place Learning, in heterochromatin; Consequently, given OR length of heteroa small distance in the genetic NOT FOR aSALE DISTRIBUTION NOT FOR SALE OR DISTRIBUTION chromatin will appear much shorter in map corresponds to a large distance on the chromosome. the genetic map than an equal length of euchromatin. In heterochromatic regions, therefore, the genetic map gives a disFIGURE 4.11 Chromosome 2 in Drosophila as it appears in metaphase of torted picture of the physical mitosis (physical map, & top)Bartlett and in the genetic map (bottom). © Jones & Bartlett Learning, LLC map. An © Jones Learning, LLCThe genes pr and cn are actually in euchromatin but are located near the junction with heterochromatin. example of such distortion is illustrated in NOT FOR SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION The total map length is 54.5 1 49.5 1 3.0 5 107.0 map units. The heterochromatin ­ IGURE 4.11, which compares the physical F accounts for 3.0/107.0 5 2.8 percent of the total map length but constitutes map and the genetic map of chromosome approximately 25 percent of the physical length of the metaphase chromosome. 2 in ­Drosophila. The physical map depicts the appearance of the chromosome in metaphase of mitosis. Two genes the tips and two LLC crossover can be canceled © byJones another & crossover further © Jones & near Bartlett Learning, Bartlett Learning, LL near the euchromatin–­ h eterochromatin junction are along the way. If two exchanges between the same NOT FOR SALE OR DISTRIBUT NOT FOR SALE OR DISTRIBUTION indicated in the genetic map. The map distances across two chromatids take place between the genes A and the euchromatic arms are 54.5 and 49.5 map units, B, then their net effect will be that all chromosomes respectively, for a total euchromatic map distance of are nonrecombinant, either A B or a b. Two of the 104.0 map units. However, the heterochromatin, which products of this meiosis have an interchange of their constitutes 25 percent LLC of the entire middle segments, but&the chromosomes are not LLC recom© Jonesapproximately & Bartlett Learning, © Jones Bartlett Learning, chromosome, has a genetic length in map units of binant for the genetic markers and so are genetically NOT FOR SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION only 3.0 percent. The distorted length of the heteroindistinguishable from noncrossover chromosomes. ­ chromatin in the genetic map results from the reduced The possibility of such canceling events means that the frequency of crossing-over in the heterochromatin. In observed recombination value is an underestimate of the spite of the distortion of the genetic map across the true exchange frequency and the map distance between heterochromatin, in the regions of euchromatin there the genes. In higher organisms,LLC double crossing-over is © Jones & Bartlett Learning, LLC © Jones & Bartlett Learning, is a good correlation between the physical d ­ istance effectively precluded in chromosome segments that are NOT FOR SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION between genes and their ­distance, in map units, in the sufficiently short, usually about 10 map units or less. genetic map. Therefore, multiple crossovers that cancel each other’s effects can be avoided by using recombination data for closely linked genes to build up genetic ­linkage maps. One crossover can undo the effects The minimum recombination frequency between © Jones & Bartlett Learning, LL of another. © Jones & Bartlett Learning, LLC two genes is 0. The recombination frequency NOT FOR SALEalso ORhas DISTRIBUT NOT FOR SALE OR DISTRIBUTION a maximum: When two genes are located far apart along a chromosome, more than one crossover can be formed between them in a single meiosis, and this complicates the KEY CONCEPT interpretation of recombination data. The probabilNo matter how far apart two genes may be, the Bartlett increases Learning, © Jones & Bartlett Learning, LLC ity©ofJones multiple&crossovers withLLC the distance maximum frequency of recombination between any between the genes. Multiple crossing-over complicates NOT FOR SALE OR DISTRIBUTION NOTis 50 FOR SALE OR DISTRIBUTION two genes percent. genetic mapping because map distance is based on the number of physical exchanges that are formed, and Fifty percent recombination is the same value that some of the multiple exchanges between two genes do would be observed if the genes were on nonhomolonot result in recombination of the genes and hence are © Jones & Bartlett Learning, © Jones Bartlett Learning, LLC gous&chromosomes and assorted independently. The not detected. As we sawLLC in Figure 4.8, the effect of one

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Chromosomes Recovered Parental Recombinant

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© Jones & Bartlett Learning, LLC © Jones & Bartlett Learning, LLC NOT FOR SALE OR NOT OR DISTRIBUTION FIGURE 4.12 Diagram showing thatDISTRIBUTION the result of two crossovers in the interval between twoFOR genes isSALE indistinguishable from independent assortment of the genes, provided that the chromatids participate at random in the crossovers. (A) A two-strand double crossover. (B) and (C) The two types of three-strand double crossovers. (D) A four-strand double crossover.

which & three chromatids participate. maximum of recombination is observed © Jones © Jones & Bartlettfrequency Learning, LLC Bartlett Learning, LLCThe final possiis that the OR second exchange connects the chrotheOR genes are so far apart in the chromosome NOTbility NOT FORwhen SALE DISTRIBUTION FOR SALE DISTRIBUTION

matids that did not participate in the first exchange that at least one crossover is almost always formed (four-strand double crossover, part D), in which case between them. In part B of Figure 4.6, it can be seen all four products are recombinant. that a single exchange in every meiosis would result In most organisms, when double crossovers are in half of the products having parental combinations take part& in the twoLearning, LL and the other half ­having recombinant ­combinations © Jones & Bartlett Learning, formed, LLC the chromatids that © Jones Bartlett exchange events are selected random. In OR this DISTRIBUT of the genes. The NOT occurrence two OR exchanges NOTatFOR SALE FOR of SALE DISTRIBUTION case, the expected proportions of the three types of between two genes has the same effect, as shown double exchanges are 1/4 four-strand doubles, 1/2 in FIGURE  4.12. Part A shows a two-strand double three-strand doubles, and 1/4 two-strand doubles. crossover, in which the same chromatids participate This means that on the average, (1/4)(0) 1 (1/2)(2) in both exchanges; no recombination of the marker © isJones & Bartlett Learning, LLC have ©2Jones & Bartlett Learning, LLC 1 (1/4)(4) 5 recombinant chromatids will be found genes detectable. When the two exchanges among the NOT 4 chromatids produced from meioses with one NOT chromatid common (three-strand double FOR in SALE OR DISTRIBUTION FOR SALE OR DISTRIBUTION two exchanges between a pair of genes. This is the crossover, parts B and C), the result is indistinguishsame proportion obtained with a single exchange able from that of a single exchange; two products between the genes. Moreover, a maximum of 50 perwith parental combinations and two with recomcent recombination is obtained for any number of binant combinations are produced. Note that there & LLCdoubles, depending on © Jones & Bartlett Learning, LLC exchanges. areBartlett two typesLearning, of three-strand

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a

c c

a © Jones & Bartlett Learning, LLC + + NOT FOR SALE OR DISTRIBUTION +

+

simple method for determining the order of the three genes, as we will see in the next section.

b b

© Jones & Bartlett Learning, LLC NOT FOR OR crossovers DISTRIBUTION 4.3 SALE Double are revealed

+ +

in three-point crosses.

The data in TABLE 4.1 result from a testcross in corn with three genes in a single chromosome. The a c& Bartlett b © Jones Learning, LLC © Jonesto& interpreting Bartlett Learning, LL analysis illustrates the approach ­ a three-point cross. The recessive allelesSALE of the genes in NOT FOR OR DISTRIBUT NOTa FOR SALE OR DISTRIBUTION + b this cross are lz (for lazy or prostrate growth habit), gl (for glossy leaf), and su (for sugary endosperm), and + + c the multiply heterozygous parent in the cross had the genotype + + +

© Jones & Bartlett Learning, LLC NOT FOR SALE OR DISTRIBUTION FIGURE 4.13 Diagram showing that two crossovers that occur

© Jones & Bartlett Learning, LLC Lz Gl SuOR DISTRIBUTION NOT FOR SALE lz gl su

between the same chromatids and span the middle pair of alleles in a triple heterozygote will result in a reciprocal exchange of the middle pair of alleles between the two chromatids.

where each symbol with an initial capital letter represents the dominant allele. (The use of this type of symbolism is customary in corn genetics.) © Jones & Bartlett Learning, LLC © Jones & Bartlett Learning, LLC The two classes of progeny that inherit noncrossover Double crossing-over is detectable in recombinaNOT FOR SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION (parental-type) gametes are therefore the wildtype plants and those tion experiments that employ three-point crosses, with the lazy-glossy-sugary phenotype. The number of which include three pairs of alleles. If a third pair of progeny in these classes is far larger than the number alleles, c1 and c, is located between the outermost in any of the crossover classes. Because the frequency genetic markers (FIGURE 4.13), double exchanges in of recombination is never © greater than percent, Learning, the © Jones & Bartlett Learning, Jones &50 Bartlett LL the region can be detected when the crossovers flank LLC very fact that these progeny are the most numerous the c gene. The twoNOT crossovers, which in this example NOT FOR SALE OR DISTRIBUT FOR SALE OR DISTRIBUTION indicates that the gametes that gave rise to them have take place between the same pair of chromatids, would the parental allele configurations, in this case Lz Gl Su result in a reciprocal exchange of the c1 and c alleles and lz gl su. Using this principle, we could have inferred between the chromatids. A three-point cross is an the genotype of the heterozygous parent even if the efficient way to obtain recombination data; it is also a

© Jones & Bartlett Learning, LLC NOT FOR SALE OR DISTRIBUTION TABLE

© Jones & Bartlett Learning, LLC NOT FOR SALE OR DISTRIBUTION

4.1

Interpreting a Three-Point Cross © Jones & Bartlett Learning, LLC Phenotype of DISTRIBUTION Genotype of gamete NOT FOR SALE OR from hybrid parent

Number of progeny

Wildtype

Lz Gl Su

286

Lazy

lz Gl Su

33

testcross progeny

Glossy

© Jones & Bartlett Learning, LLC NOT FOR SALE OR DISTRIBUTION The two most frequent classes identify the nonrecombinant gametes.

Lz gl Su & Jones

59 © Bartlett Learning, LLC Lz Gl su SALE OR DISTRIBUTION 4 NOT FOR The two rarest classes iden-

Sugary Lazy, glossy

lz gl Su

2

Lazy, sugary

lz Gl su

44

Glossy, sugary

Lz gl su

40

© Jones & Lazy, glossy, sugary

tify the double-recombinant gametes.

BartlettlzLearning, LLC272 gl su

© Jones & Bartlett Learning, LLC NOT FOR SALE OR DISTRIBUTION

NOT FOR SALE OR DISTRIBUTION 740 These reciprocal classes result from single recombination between one pair of adjacent genes.

© Jones & Bartlett Learning, LLC NOT FOR SALE OR DISTRIBUTION 124

© Jones & Bartlett Learning, LL NOT FOR SALE OR DISTRIBUT

These reciprocal classes result from single recombination between the other pair of adjacent genes.

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genotype had not been stated. This is a point important enough to state more generally:

the parental gamete lz gl su except for the allele Su. The middle gene can be identified because the “odd man out” in the comparisons—in this case, the alleles © Jones & Bartlett Learning, LLC © Jones & Bartlett Learning, LLC of Su—is always the gene in the middle. CONCEPT NOT FOR SALE OR DISTRIBUTIONThe reason is NOT FORKEY SALE OR DISTRIBUTION that only the middle pair of alleles is interchanged by In any genetic cross, no matter how complex, the double crossing-over. two most frequent types of gametes with respect to Taking the correct gene order into account, the any pair of genes are nonrecombinant; these progenotype of the heterozygous parent in the cross yieldvide the linkage phase (cis versus trans) of the alleles ing the progeny in Table 4.1 should be written as © Jones & Bartlett Learning, LLC © Jones & Bartlett Learning, of the genes in the multiply heterozygous parent.

LL NOT FOR SALE OR DISTRIBUT Lz Su Gl

NOT FOR SALE OR DISTRIBUTION

lz su gl In mapping experiments, the gene sequence is ­ sually not known. In this example, the order in which u The consequences of single crossing-over in this genothe three genes are shown is entirely arbitrary. H ­ owever, type are shown in FIGURE 4.15. A single crossover in there©is Jones an easy&way to determine the ­ c orrect order Bartlett Learning, LLC © Jones & Bartlett Learning, LLC the lz2su region (part A) yields the reciprocal recomfrom NOT three-point data. The gene order can be deduced FOR SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION binants Lz su gl and lz Su Gl, and a single crossover in by identifying the genotypes of the d ­ ouble-crossover the su2gl region (part B) yields the reciprocal recomgametes produced by the heterozygous p ­ ­arent and binants Lz Su gl and lz su Gl. The consequences of comparing these with the non­ recombinant gametes. double crossing-over are illustrated in FIGURE 4.16. Because the probability of two simultaneous exchanges is There are four different types of LLC double crossovers: a considerably than that of either single exchange, © Jones © Jones & Bartlett smaller Learning, LLC & Bartlett Learning, two-strand double (part A), two types of three-strand the double-crossover gametes will be the least frequent NOT FOR SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION doubles (parts B and C), and a four-strand double (part types. It is clear in Table 4.1 that the classes composed D). These types were illustrated earlier in Figure 4.12, of four plants with the sugary phenotype and two plants where the main point was that with two genetic markwith the lazy-glossy phenotype (products of the Lz Gl su ers flanking the crossovers, the occurrence of double and lz gl Su gametes, respectively) are the least frequent and therefore constitute the double-crossover © Jones & Bartlett Learning, LLC © Jones & Bartlett Learning, LL progeny. Now we apply another principle: NOT FOR SALE OR DISTRIBUT NOT FOR SALE OR DISTRIBUTION (A)

KEY CONCEPT

Lz

The effect of double crossing-over is to interchange the members of the middle pair of © Jones & Bartlett Learning, LLC alleles between the chromosomes.

NOT FOR SALE OR DISTRIBUTION

Lz

Gl Gl

lz

gl

lz

gl

Su Su

Lz

Gl

Su

Lz

gl

Su

su

su lz Gl © Jones & Bartlett Learning, LLC su gl NOT FOR SALE OR lzDISTRIBUTION su

This principle is illustrated in FIGURE 4.14. With (B) three genes there are three possible orders, Gl Lz Su depending on which gene is in the middle. If gl Gl Su Lz were in the middle (part A),LLC the double-recom© Jones & Bartlett Learning, © Jones & Bartlett Learning, LLCLz su Su Gl Gl Lz gametes would be Lz gl Su and lz Gl su, NOT FORbinant SALE OR DISTRIBUTION NOTlz FOR SALE OR DISTRIBUTION gl su gl lz which is inconsistent with the data. Likewise, Su lz if lz were in the middle (part C), the doublegl su su gl lz recombinant gametes would be Gl lz Su and gl Lz su, which is also inconsistent with the data. (C) The correct order © of Jones the genes, lz2su2gl,Learning, & Bartlett LLC © Jones & Bartlett Learning, LL Su Gl is given in part B, because in this case, OR the DISTRIBUTION Lz Lz NOT FOR SALE OR DISTRIBUT NOT FOR SALE Su Gl double-recombinant gametes are Lz su Gl and Lz Gl lz Su Su Gl lz Su gl, which Table 4.1 indicates is actually su gl the case. Although one can always infer which lz su gl Lz gene is in the middle by going through all three gl su lz lz su gl possibilities, there is a shortcut. Each double-­ © Jones & Bartlett Learning, LLC © Jones & Bartlett Learning, LLC recombinant gamete will always match one of NOT FOR SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION FIGURE 4.14 The order of genes in a three-point testcross may be the parental gametes in two of the alleles. In deduced from the principle that double recombination interchanges the Table 4.1, for example, the double-recombinant middle pair of alleles. For the genes Lu, Gl, and Su, there are three possible gamete Lz Gl su matches the parental gamete orders (parts A, B, and C), each of which predicts a different pair of gametes as Lz Gl Su except for the allele su. Similarly, the the result of double recombination. Only the order in part B is consistent with double-recombinant gamete lz gl Su matches © Jones & Bartlett Learning, LLC © Jones &suBartlett LLC the finding that Lz Gl and lz gl SuLearning, are the double-recombinant gametes.

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125

(A) Single crossover in lz–su region Lz

Su

Gl

© Jones & Bartlett Learning, LLC Gl Su Lz NOT FOR SALE OR DISTRIBUTION lz su gl

lz

su

gl

su–gl region (B) Single crossover ©inJones & Bartlett Lz

Su

crossovers cannot be detected genetically. The difference in the present case is that, here, the genetic marker su is located in the middle © & Bartlett Learning, LLC gl su Jones Lz between the two crossovers, NOT FOR SALE OR DISTRIBUTIONso some of the Gl Su lz double crossovers can be detected genetically. On the right in F ­ igure 4.16, the asterisks mark su gl lz the sites of crossing-over between nonsister chromatids. In terms of recombination, the result is that Learning, LLC © Jones & Bartlett Learning, Lz

Gl

Su

Lz Gl Su Gl FOR SALE OR DISTRIBUTION NOT

LL NOT FOR SALE OR DISTRIBUT n  A two-strand double crossover (part A)

yields the reciprocal double-recombinant products Lz su Gl and lz Su gl. lz gl Gl su lz n One three-strand double crossover (part B) lz su gl su gl lz the double-recombinant ­product © Jones & Bartlett Learning, LLC ©yields Jones & Bartlett Learning, LLC Lz su Gl and two single-­recombinant NOT FOR SALE OR DISTRIBUTION NOT FOR ORlz Su DISTRIBUTION ­products, LzSALE Su gl and Gl. FIGURE 4.15 Result of single crossovers in a triple heterozygote, using the Lz2Su2Gl region as an example. (A) A crossover between Lz and Su n The other three-strand double crossover results in two gametes that show recombination between Lz and Su and (part C) yields the double-recombinant two gametes that are nonrecombinant. (B) A crossover between Su and Gl product lz Su gl and two single-­recombinant results in two gametes that show recombination between Su and Gl and products, Lz su gl and lz su Gl. two gametesLearning, that are nonrecombinant. © Jones & Bartlett LLC © Jones & Bartlett Learning, LLC n T  he four-strand double crossover (part C) NOT FOR SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION (A) Two-strand double crossover yields reciprocal single recombinants Lz in the lz2su region, namely Lz su gl and Gl Su Lz Su Gl lz Su Gl, and reciprocal single recombi* * su Su Lz Gl Gl Lz nants in the su2gl region, namely Lz Su gl lz * gl * LLC and lz su Gl. © Jones & Bartlett Learning, & Bartlett Learning, su © Jones gl lz Su Lz

Su su

lz

su

Lz

Gl

gl

Su

LL gl FOR SALE OR DISTRIBUTION Note that the products NOTofFOR SALE OR DISTRIBUT NOT recombination in the su gl lz

three-strand double crossovers (parts B and C) are the reciprocals of each other. Because these (B) Three-strand double crossover two types of double crossovers are equally fre* gl Lz Su Gl Lz Su quent, the reciprocal products of recombination * © Jones & Bartlett Learning, LLC © expected Jones to & appear Bartlett Learning, LLC * Su Gl are in equal numbers. su Lz Gl Lz NOT lzFOR suSALEgl OR DISTRIBUTION * NOT FOR OR the DISTRIBUTION We can nowSALE summarize data in Table 4.1 Gl Su lz in a more informative way by writing the genes lz su gl in the correct order and grouping reciprocal su gl lz gametic genotypes together. This grouping is shown in TABLE 4.2. Note that each class of (C) Three-strand double crossover © Jones & Bartlett Learning, LLC © Jones & Bartlett Learning,consists LLC of two reciprocal Lz Gl Su single recombinants Su Gl Lz products and that these are found in approxiNOT FOR SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION * gl Lz su Su Gl Lz mately equal frequencies (40 versus 33 and * lz su 59 versus 44). This observation illustrates an gl gl lz Su * important principle: su lz gl * su

lz

(D)

Gl

© Jones & Bartlett Learning, LLC KEY CONCEPT © Jones & Bartlett Learning, LL Four-strand double crossover NOT FOR SALE NOT FOR SALE OR DISTRIBUTION The two reciprocal products resulting fromOR any DISTRIBUT Lz

Su

Gl

Lz

Su

su

*

Su

Lz

Su

Gl

* Lz

lz

su

gl

lz

© Jones & Bartlett Learning, LLC lz su gl lz NOT FOR SALE OR DISTRIBUTION

su

*

gl

gl Gl

*

Gl

FIGURE 4.16 Result of double crossovers in a triple heterozygote, using the Lz2Su2Gl region as an example. Note that chromosomes showing double recombination derive from the two-strand double crossover (A) or from either type of three-strand double crossover (B and C). The four-strand Bartlett Learning, LLC © Jones double crossover (D) results only in single-recombinant chromosomes.

© Jones & NOT FOR SALE OR DISTRIBUTION 126

crossover, or any combination of crossovers, are expected to appear in approximately equal frequencies among the progeny.

© Jones & Bartlett Learning, LLC NOT FOR SALE DISTRIBUTION In calculating theOR frequency of recombination from the data, remember that the ­double-recombinant chromosomes result from two exchanges, one in each of the chromosome regions defined by the three genes. Therefore, Bartlett Learning, LLC chromosomes that are recombinant between lz

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and su are represented by the following c­ hromosome types:

© Jones & Bartlett Learning, LLC Lz su gl 40 NOT FOR SALE OR DISTRIBUTION lz Su Gl 33

Lz su Gl 4 lz Su gl 2 79

f­requency between adjacent genes. You can keep from falling into this trap by remembering that the doublerecombinant chromosomes have LLC single recombination © Jones & Bartlett Learning, in both regions. NOT FOR SALE OR DISTRIBUTION

Interference decreases the chance of multiple crossing-over.

The total implies that©79/740, percent,Learning, of the Jonesor&10.7 Bartlett LLC © Jones & Bartlett The detection of double crossing-over makes it pos-Learning, LL chromosomes recovered in the progeny are recombisible to determine whether exchanges in two different NOT FOR SALE OR DISTRIBUT NOT FOR SALE OR DISTRIBUTION nant between the lz and su genes, so the map distance regions of a pair of chromosomes are formed indepenbetween these genes is 10.7 map units, or 10.7 centidently of each other. Using the information from the morgans. Similarly, the chromosomes that are recomexample with corn, we know from the recombination binant between su and gl are represented by frequencies that the probability of recombination is © Jones & Bartlett Learning, LLC © Jones Bartlett Learning, LLC If 0.107 between lz and su & and 0.147 between su and gl. Lz Su gl 59 recombination is independent in the two regions (which NOT FOR SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION lz su Gl 44 means that the formation of one crossover does not Lz su Gl 4 alter the probability of the second crossover), the prob lz Su gl 2 ability of a single recombination in both regions is the 109 product of these separate probabilities, or 0.107 3 0.147 In Bartlett this case the recombination © Jones & Learning, LLC frequency between su © Jones & Bartlett Learning, LLCthat in a sample 5 0.0157 (1.57 percent). This implies and gl is 109/740, or 14.7 percent, so the map distance 740 gametes, number of double recomNOT FOR SALE OR DISTRIBUTION NOTofFOR SALE the ORexpected DISTRIBUTION between these genes is 14.7 map units, or 14.7 centibinants would be 740 3 0.0157, or 11.6, whereas the morgans. The genetic map of the chromosome segment number actually observed was only 6 (Table 4.2). Such in which the three genes are located is therefore deficiencies in the observed number of double recombinants are common; they reflect a phenomenon called lz su© Jones & Bartlett Learning, gl LLC Jones & Bartlett chromosome i­ nterference, in©which a crossover in oneLearning, LL region of a chromosome reduces probability of a DISTRIBUT NOT the FOR SALE OR FOR SALE 10.7 map units NOT 14.7 map units OR DISTRIBUTION second crossover in a nearby region. Over short genetic distances, chromosome interference is nearly complete. The most common error in learning how to interpret The coefficient of coincidence is the observed three-point crosses is to forget to include the dounumber of double-recombinant chromosomes divided ble recombinants when calculating the recombination © Jones & Bartlett Learning, LLC © Jones &Its Bartlett Learning, LLC by the expected number. value provides a quantitative measure of the degree of interference, which is NOT FOR SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION defined as TABLE



4.2

© Jones & Comparing Bartlett Learning, LLC Reciprocal Products in a Three-Point Cross NOT FOR SALE OR DISTRIBUTION Genotype of gamete from hybrid parent

Number of progeny

Intervals showing recombination

Lz Su Gl 286 lz su gl 272

i 5 Interference 5 1 2 (Coefficient of coincidence)

© Jones & Bartlett Learning, LLC From the data in the corn example, the coefficient of NOTcoincidence FOR SALE OR DISTRIBUTION is calculated as follows: n Observed number of double recombinants 5 6 n Expected number of double recombinants 5 0.107

3 0.147 3 740 5 11.6

© Jones & Bartlett Learning, n LLC © Jones & Bartlett Learning, LL Coefficient of coincidence 5 6/11.6 5 0.52 Lz su gl 40 NOT FOR SALE OR DISTRIBUT NOT FOR SALE OR DISTRIBUTION lz2su lz Su Gl 33

The 0.52 means that the observed number of double recombinants was only about half of the number expected if crossing-over in the two regions were independent. The value of the interference depends on Lz su Gl 4 lz2su  1  su2gl lz Su gl 2 the distance©between genetic markers and onLLC the © Jones & Bartlett Learning, LLC Jonesthe & Bartlett Learning, ­species. In some the interference increases as NOT FOR SALE 740OR DISTRIBUTION NOT species, FOR SALE OR DISTRIBUTION the ­distance between the two outside markers becomes smaller, until Total number of recombinants in Total number of recombinants in a point is reached at which double lz2su region: su2gl region: ­crossing-over is ­eliminated; that is, 40 1 33 1 4 1 2 5 79 59 1 44 1 4 1 2 5 109 double recombinants are found, & Bartlett Learning, LLC © Jones & BartlettnoLearning, LLC Lz Su gl 59 su2gl lz su Gl 44

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Recombination frequency (r )

0.5

© Jones & Bartlett Learning, LLC No interference NOT FOR SALE OR DISTRIBUTION (i = 0)

© Jones & Bartlett Learning, LLC 0.4 NOT FOR SALE OR DISTRIBUTION 0.3

Interference decreasing with distance (i = 1 – 2r)

0.2

© Jones & Bartlett CompleteLearning, interference LLC (i = 1) NOT FOR SALE OR DISTRIBUTION

0.1 0

0

20

40

60

© Jones & Bartlett Learning, LLC

© Jones & Bartlett Learning, LL NOT FOR SALE OR DISTRIBUT

80 100 120 Distance in map units (cM)

140

160

180

200

© Jones & Bartlett Learning, LLC

FIGURE mapping function is the relation between genetic map distance across intervalSALE and the observed frequency of NOT 4.17 FORA SALE OR DISTRIBUTION NOTanFOR OR DISTRIBUTION recombination across the interval. Map distance is defined as one-half the average number of crossovers converted into a percentage. The three mapping functions correspond to different assumptions about interference, i.

mapping function,Learning, as shown by the three examples in and the coefficient of coincidence equals 0 (or, to say © Jones & Bartlett Learning, LLC © Jones & Bartlett LLC Figure 4.17. the same thing, the interference equals 1). In Drosophila NOT FOR SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION this distance is about 10 map units.

The effect of interference on the relationship 4.4 Polymorphic DNA sequences between genetic map distance and the frequency of are used in human genetic recombination is illustrated in FIGURE 4.17. Each mapping. © Jones & Bartlett Learning, LL curve is an example©ofJones a mapping function, which is the LLC & Bartlett Learning, mathematical relation between the genetic distance Until quite recently, mapping human OR beings NOTgenes FORinSALE DISTRIBUT NOT FOR SALE OR DISTRIBUTION across an interval in map units (centimorgans) and the was very tedious and slow. Numerous practical obstacles observed frequency of recombination across the intercomplicated genetic mapping in human pedigrees: val. In other words, a mapping function tells one how 1. Most genes that cause genetic diseases are rare, so to convert a map distance between genetic markers into they are observed in only a small number of families. a ­r© ecombination frequency between the markers. As we Jones & Bartlett Learning, LLC © Jones & Bartlett Learning, LLC have seen, when the map distance between the mark 2. Many mutant genes of interest in human ­genetics NOT FOR SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION ers is small, the recombination frequency equals the are recessive, so they are not detected in heterozymap distance. This principle is reflected in the curves gous genotypes. in Figure 4.17 in the region in which the map distance 3. The number of offspring per human family is is smaller than about 10 cM. At less than this distance, relatively small, so segregation cannot usually be all of the curves are nearly straight lines, which means in single sibships. LLC © Jones & Bartlett Learning, LLC © Jones detected & Bartlett Learning, that map distance and recombination frequency are 4. The human geneticist cannot perform testcrosses NOT FOR SALE NOT FOR SALE OR DISTRIBUTION equal; 1OR mapDISTRIBUTION unit equals 1 percent recombination, and or backcrosses, because human matings are not 10 map units equals 10 percent recombination. For dictated by an experimenter. distances greater than 10 map units, the recombinaHuman genetics has been revolutionized by the tion frequency becomes smaller than the map distance use of techniques for manipulating DNA. These techaccording to the pattern of interference along the chro© Jones & Bartlett Learning, LLC © Jones Bartlett Learning, LL niques have enabled investigators to & carry out genetic mosome. Each pattern of interference yields a different

NOT FOR SALE OR DISTRIBUTION

Q

NOT FOR SALE OR DISTRIBUT

A MOMENT TO THINK

Problem: In his pioneering 1913 studies in Drosophila that resulted in the first genetic linkage map, A. H. Sturtevant included three © Jones &now Bartlett & Bartlett Learning, genetic markers known to Learning, cover almost theLLC entirety of the euchromatin of the©X Jones chromosome. The marker w (white eyes) is LLC near the tip, mFOR (miniature body)OR near the middle, and r (rudimentary wings) near the centromere. In two-point Sturtevant obtained NOT SALE DISTRIBUTION NOT FOR SALEcrosses, OR DISTRIBUTION recombination frequencies of 0.32 for the interval w2m, 0.25 for the interval m2r, and 0.45 for the interval w2r. He noted that 0.45 is smaller than the sum of 0.32 1 0.25 5 0.57, which is the value expected if the frequencies of recombination over such large distances were additive. He commented that the discrepancy “is probably due to the occurrence of two breaks in the same chromosome, or double crossing-over.” From Sturtevant’s data, calculate the coincidence and the interference across the region. (The answer can be found at the end of the chapter.)

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­ apping in human pedigrees primarily by using m Identifying the particular nucleotide present at each genetic markers present in the DNA itself, rather than of a million SNPs is made possible through the use the phenotypes composedLLC of about 20 m ­ illion of DNA microarrays & Bartlett Learning, © Jones through & Bartlett Learning,produced LLC by mutant genes. © Jones There are many minor differences in DNA sequence infinitesimal spots on a glass slide about the size of a NOT FOR SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION from one person to the next. On the average, the postage stamp. Each tiny spot contains a unique DNA DNA sequences at corresponding positions in any oligonucleotide sequence present in millions of copies two c­ hromosomes, taken from any two people, differ synthesized by microchemistry when the microarray at approximately one in every thousand base pairs. is manufactured. Each oligonucleotide sequence is A genetic ­difference © that is relatively common in a designed to hybridize specifically with small fragmentsLearning, LL Jones & Bartlett Learning, LLC © Jones & Bartlett population is called a NOT ­polymorphism. Most polymorof genomic DNA that include one or the other the DISTRIBUT NOT FOR SALEofOR FOR SALE OR DISTRIBUTION phisms in DNA sequence are not associated with any nucleotide pairs present in a SNP. The microarrays inherited disease or disability; many occur in DNA also include numerous controls for each hybridizasequences that do not code for proteins. Neverthetion. The controls consist of oligonucleotides containless, each of the p ­ olymorphisms serves as a conveing ­deliberate mismatches that are intended to guard nient©genetic marker, and those genetically linked to against being by & particular nucleotide sequences Jones & Bartlett Learning, LLC ©misled Jones Bartlett Learning, LLC genesNOT that FOR cause SALE hereditary d ­ iseases are particularly that are particularly “sticky” and hybridize too ­readily OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION important. with genomic fragments and other sequences that form structures that h ­ ybridize poorly or not at all. Such Single-nucleotide polymorphisms microarrays, sometimes called SNP chips, enable the SNP genotype of an individual to be determined with (SNPs) are abundant in the nearly & 100Bartlett percent accuracy. © Jones & Bartlett Learning, LLC © Jones Learning, LLC human genome. The principles behind oligonucleotide hybridizaNOT FOR SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION tion are illustrated in FIGURE 4.18. Here the length A single-nucleotide polymorphism, or SNP (pronounced “snip”), is present at a particular nucleotide of each oligonucleotide is seven nucleotides, but in site if the DNA molecules in the population often differ practice this is too short. Typical SNP chips consist of in the identity of the nucleotide pair that occupies the ­oligonucleotides at least 25 nucleotides in length. Part site. For example, some DNA molecules mayLearning, have a A shows the two types of DNA duplexes that mightLearning, LL © Jones & Bartlett LLC © Jones & Bartlett T–A base pair at a particular nucleotide site, whereas form a SNP. In this example, some chromosomes NOT FOR SALE carry OR DISTRIBUT NOT FOR SALE OR DISTRIBUTION other DNA molecules in the same population may a DNA molecule with a T–A base pair at the position have a C–G base pair at the same site. This difference shown in red, while the DNA molecule in other chroconstitutes a SNP. The SNP defines two “alleles” for mosomes has a C–G base pair at the corresponding which there could be three genotypes among indiposition. Short fragments of genomic DNA are labeled viduals the population: homozygous withLLC T–A at the with a fluorescent tag,&and then the single strands © in Jones & Bartlett Learning, © Jones Bartlett Learning, LLC corresponding site in both homologous chromosomes, h ­ ybridized with a SNP chip containing the compleNOT FOR SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION homozygous with C–G at the corresponding site in both mentary oligonucleotides as well as the numerous homologous chromosomes, or heterozygous with T–A controls. The duplex containing the T–A hybridizes in one chromosome and C–G in the homologous chroonly with the two oligonucleotides on the left, and that mosome. The word allele is in quotation marks above containing the C–G hybridizes only with the two oligobecause the SNP need not be in a coding sequence, or nucleotides on the right. © Jones & Bartlett Learning, LLC © Jones & Bartlett Learning, LLC even in a gene. In the human genome, any two ranAfter the hybridization takes place, the SNP chip NOT FOR SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION domly chosen DNA molecules are likely to differ at one is examined with fluorescence microscopy to detect SNP site about every 1000 bp in noncoding DNA and the spots that fluoresce as a result of the tag on the at about one SNP site every 3000 bp in protein-coding genomic DNA. The possible patterns are shown in part B. Genomic DNA from an individual whose chroDNA. In the definition of a SNP, the DNA molecules contain two copies©ofJones the T–A of theLearning, LL must differ at the nucleotide site “often.” This provision © Jones & Bartlett Learning, mosomes LLC & form Bartlett duplex (homozygous TA/TA) causes the two leftmost excludes rare genetic variation of the sort found in less NOT FOR SALE OR DISTRIBUT NOT FOR SALE OR DISTRIBUTION spots to fluoresce, but the two rightmost spots remain than 1 percent of the DNA molecules in a population. unlabeled. Similarly, genomic DNA from a homozyThe reason for the exclusion is that genetic variants that gous CG/CG individual causes the two rightmost spots are too rare are not generally as useful in genetic analyto fluoresce but not the two on the left. Finally, genomic sis as the more common variants. SNPs are the most DNA from © a Jones heterozygous TA/CG Learning, individual causes common form of genetic differences among people. © Jones & Bartlett Learning, LLC & Bartlett LLC fluorescence of all four spots because the TA duplex About 10 million SNPs have been identified in which NOT FOR SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION labels the two leftmost spots and the CG duplex labels the alternative nucleotides are each relatively common the two rightmost spots. in the human population, and about half a million of Use of SNP chips or other available technologies these are typically used in a search for SNPs that might for high-throughput genotyping of millions of SNPs be associated with complex diseases such as diabetes or in thousands of individuals allows genetic risk factors high blood pressure. © Jones & Bartlett Learning, LLC © Jones & Bartlett Learning, LLC

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4.4  Polymorphic DNA Sequences Are Used in Human Genetic Mapping

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In this example the SNP consists of a TA base pair in some DNA duplexes and a CG base pair in others.

(A)

© Jones & Bartlett Learning, LLC NOT FOR SALE OR DISTRIBUTION

© Jones & Bartlett Learning, LLC Fragments of genomic NOT FOR SALE OR DISTRIBUTION DNA are labeled with a fluorescent molecule.

A C T G C AG TGACGT C

© Jones & Bartlett Learning, LLC NOT FOR SALE OR DISTRIBUTION GC AT CG GC TA in Each oligonucleotide © Jones & Bartlett Learning, LLC CG a SNP chip is attached A T slide. to a glass NOT FOR SALE OR DISTRIBUTION

(B)

© Jones & Bartlett Learning, LLC Hybridization signal NOT FOR SALE OR DISTRIBUTION

TA GC AT CG GC TA CG

Strands from each duplex will hybridize only with their complementary oligonucleotides.

A C C G C AG TGGCGT C

© Jones & Bartlett Learning, LL NOT FOR SALE OR DISTRIBUT

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from homozygous AT/AT

Hybridization signal from heterozygous AT/CG

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FIGURE 4.18 (A) Oligonucleotides attached to a glass slide in a SNP chip can be used to identify duplex DNA molecules containing base pairs for a SNP; in this example, a TA base pair versus a CG base pair. (B) The SNP genotype of an individual can be determined by hybridization because DNA from genotypes that are homozygous TA/TA, homozygous CG/CG, or heterozygous TA/CG each gives a © Jones & Bartlett Learning, LLC © Jones & Bartlett Learning, LLC different pattern of fluorescence.

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lacking the cleavage site will be cleaved into one larger for disease to be identified. A typical study compares fragment instead of two smaller ones (FIGURE  4.20). the genotypes of patients with particular diseases with healthy people matched with the patients for such facMore rarely, a mutation in the DNA sequence will cretors as sex,Learning, age, and ethnic group. Comparing the SNP ate a&new site rather than destroy one already present. © Jones & Bartlett LLC © Jones Bartlett Learning, LLC genotypes these groups often reveals which The main point is that any difference in DNA sequence NOT FOR SALE OR among DISTRIBUTION NOT FOR SALE OR DISTRIBUTION SNPs are in the genome. that alters a cleavage site also changes the length of the DNA fragments produced by cleavage with the SNPs in restriction sites corresponding restriction enzyme. The different DNA fragments can be separated by size by an electric field yield restriction fragment length in a supporting gel and © detected means. © Jones & Bartlett Learning, LLC Jonesby&various Bartlett Learning, LL polymorphisms (RFLPs). Differences in DNA fragment length produced presNOT FOR SALEbyOR DISTRIBUT NOT FOR SALE OR DISTRIBUTION Some polymorphisms in DNA sequence are detected by ence or absence of the cleavage sites in DNA molecules means of a type of enzyme called a restriction endoare known as restriction fragment length polymornuclease, which cleaves double-stranded DNA molphisms (RFLPs). ecules wherever a particular, short sequence of bases RFLPs are typically formed in one of two ways. JonesFor&example, Bartlettthe Learning, © Jones & Bartlett Learning, is © present. restriction LLC enzyme EcoRI A mutation that changes a base sequence mayLLC result cleaves GAATTC appears in loss orNOT gain of a cleavage is recognized by NOTDNA FORwherever SALE the ORsequence DISTRIBUTION FOR SALEsite ORthat DISTRIBUTION in either strand, as illustrated in FIGURE 4.19. Restricthe restriction endonuclease in use. FIGURE 4.21, part tion enzymes are considered in detail in C ­ hapter 6. A, gives an example. On the left is shown the relevant For present purposes, their significance is related to the region in the homologous DNA molecules in a person fact that a difference in DNA sequence that eliminates who is heterozygous for such a sequence polymor© Jones & Bartlett © Jones & Bartlett Learning, LLC a cleavageLearning, site can be LLC detected because the region phism. The homologous chromosomes in the person

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EcoRI cleavage sites A A T T C Learning, G ALLC AT TC © Jones & GBartlett CT TAAG CT TAAG NOT FOR SALE OR DISTRIBUTION 5’ 3’

and b types of chromosomes (genotype ab) would yield three bands in a gel. Similarly, DNA from homozygous GAAT © T C Jones & Bartlett Learning, LLCaa would yield CT TAAG one band, and that from homozygous bb NOT FOR SALE OR DISTRIBUTION would yield two bands. 3’ 5’

Treatment of DNA with EcoRI

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Simple-sequence repeats (SSRs) often©differ Jonesin & copy Bartlett Learning, LL number. NOT FOR SALE OR DISTRIBUT

Yet another type of DNA polymorphism results from differences in the number of Cleavage Cleavage Cleavage copies of a short DNA sequence that may be repeated many times in ­tandem at a par© Jones & Bartlett Learning, LLC © Jones & aBartlett Learning, LLC DNA fragments resulting ticular site in chromosome (Figure 4.21, from Eco RI cleavage NOT FOR SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION part B). In a particular chromosome, the tandem repeats may contain any number of copies, typically ranging from ten to a few hundred. When a DNA molecule is cleaved with a restriction endonuclease that cleaves © Jones & Bartlett Learning, LLC © Jones & Bartlett Learning, LLC repeat, the size at sites flanking the tandem NOT FORFIGURE SALE4.19 ORThe DISTRIBUTION NOT FOR SALE OR DISTRIBUTION restriction enzyme EcoRI cleaves double-stranded DNA of the DNA fragment produced is deterwherever the sequence 59-GAATTC-39 is present. In the example shown here, the mined by the number of repeats present in DNA molecule contains three EcoRI cleavage sites, and it is cleaved at each site, the molecule. ­Figure 4.21, part B, illustrates producing a number of fragments. homologous DNA sequences in a heterozygous person containing one chromosome a are distinguished by the a and b. In the region of © letters Jones & Bartlett Learning, LLCtwo copies of the repeat© Jones Bartlett Learning, LL with and another&chromosome interest, chromosome NOT a contains two cleavage sites and NOT FOR SALE OR DISTRIBUT FOR SALE OR DISTRIBUTION b with five copies of the repeat. When cleaved and sepachromosome b contains three. On the right is shown rated in a gel, chromosome a yields a shorter fragment the position of the DNA fragments produced by cleavthan that from chromosome b, because a contains fewer age after separation in an electric field. Each fragment copies of the repeat. A genetic polymorphism resulting appears as a discrete band in the gel. The fragment from a tandemly repeated short DNA sequence is called from © chromosome migrates Learning, more slowly LLC than those Jones & aBartlett © Jones & Bartlett Learning, a simple sequence repeat (SSR). An example ofLLC an from NOT chromosome b because it is longer, and longer FOR SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION SSR is the repeating sequence fragments move more slowly through the gel. In this 59-...TGTGTGTGTGTG...-39 example, DNA from a person heterozygous for the a

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FIGURE 4.20 A minor difference in the DNA sequence of two molecules can be detected if the difference eliminates a restriction site. (A) This molecule contains three restriction sites for EcoRI, including one at each end. It is cleaved into two fragments by the enzyme. (B) This molecule has a mutant base sequence in the EcoRI site in the middle. It changes 59-GAATTC-39 into 59-GAACTC-39, which is no & Bartlett Learning, LLC © Jones & Bartlett longer cleaved by EcoRI. Treatment of this molecule with EcoRI results in one larger fragment. Learning, LLC

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(A) DNA in chromosomes Longer DNA

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FIGURE 4.21 Two types of genetic&variation that are widespread inLLC most natural populations of animals and plants.& (A)Bartlett RFLP (restriction © Jones Bartlett Learning, © Jones Learning, LL fragment length polymorphism), in which alleles differ in the presence or absence of a cleavage site in the DNA. The different alleles NOT FOR SALE OR DISTRIBUTION yield different fragmentNOT lengthsFOR (shownSALE in the gelOR pattern at the right) when the molecules are cleaved with a restriction enzyme. (B) SSRDISTRIBUT (simple sequence repeat), in which alleles differ in the number of repeating units present between two cleavage sites.

© the Jones & Bartlett Learning, LLC in the © Jones & Bartlett Learning, LLC and polymorphism consists of differences TABLE 4.3 number TG repeats. ­particular “allele” of the SSR NOT of FOR SALE AOR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION is defined by the number of TG repeats it includes. Some Simple Sequence Repeats in the One source of the utility of SSRs in human genetic Human Genome mapping is the high density of SSRs across the genome. There is an average of one SSR per 2 kb of human SSR repeat unit Number of SSRs in the human genome DNA. Some examples are shown in TABLE 4.3. The © Jones & Bartlett Learning, LLC © Jones & Bartlett Learning, LLC 59-AC-39 80,330 prevalence different SSRs differs. Some dinucleoNOT FOR SALE OR of DISTRIBUTION NOT FOR SALE OR DISTRIBUTION 59-AT-39 56,260 tide repeats, such as 59-AC-39 and 59-AT-39, are far 59-AG-39 23,780 more frequent than others, such as 59-GC-39. Overall, dinucleotide repeats are much more abundant than 59-GC-39   290 trinucleotide repeats. The second source of utility of 59-AAT-39 11,890 SSRs in genetic mapping is the&large numberLearning, of alleles LLC © Jones Bartlett © Jones & Bartlett Learning, LL 59-AAC-39  7,540 that can be present in any human population. The NOT FOR SALE OR DISTRIBUT NOT FOR SALE OR DISTRIBUTION 59-AGG-39  4,350 large number of alleles also implies that most people will be heterozygous, and so their DNA will yield two 59-AAG-39  4,060 bands upon cleavage with the appropriate restriction 59-ATG-39  2,030 endonuclease. Because of their high degree of variation 59-CGG-39  1,740 © Jones & DNA Bartlett Learning,are LLC © Jones & Bartlett Learning, LLC among people, polymorphisms also widely 59-ACC-39  1,160 used in DNA in criminal investigations. NOT FORtyping SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION In genetic mapping, the phenotype of a person with 59-AGC-39   870 respect to a DNA polymorphism is a pattern of bands 59-ACT-39   580 in a gel. As with any other type of genetic marker, the Data from International Human Genome Sequencing Consortium, Nature 409 (2001): genotype of a person with respect to the polymorphism 860–921. © Jones & Bartlett © Jones & Bartlett Learning, LLC is inferred, Learning, insofar as it isLLC possible, from the ­phenotype. NOT FOR SALE OR DISTRIBUTION 132

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FIGURE 4.22 Human pedigree showing segregation of SSR alleles. Six alleles (1–6) are present in the pedigree, but any one person © only Jones & Bartlett Learning, LLC © Jones & Bartlett Learning, LLC can have one allele (if homozygous) or two alleles (if heterozygous).

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Linkage between different polymorphic loci is detected human genome can be duplicated or deleted in much through lack of independent assortment of the alleles in larger but still submicroscopic chunks ranging from © Jones & Bartlett LLC Learning, LLC pairs, or one pedigrees, andLearning, recombination and genetic mapping are © Jones 1 kb to&1 Bartlett Mb (Mb stands for megabase outOR using the same principles as apply in other NOT­mFOR illion base pairs). type of variation is known as NOT FORcarried SALE DISTRIBUTION SALE ORThis DISTRIBUTION organisms, except that in human beings, because of the copy-number variation (CNV). The extra or misssmall family size, different pedigrees are pooled for analing copies of the genome in CNVs can be detected ysis. Primarily through the use of DNA polymorphisms, by means of hybridization with oligonucleotides in genetic mapping in humans has progressed rapidly. DNA microarrays. Since each spot on the microarray © Jones & Bartlett Learning, LLC of millions of identical © Jones & aBartlett To give an example of the type of data used in consists copies of particularLearning, LL human genetic mapping, a three-generation pedigree of oligonucleotide sequence, the number these OR that DISTRIBUT NOT FORofSALE NOT FOR SALE OR DISTRIBUTION a family segregating for several alleles of an SSR is illusundergo hybridization depends on how many coptrated in FIGURE 4.22. In this example, each of the paries of the complementary sequence are present in ents is heterozygous, as are all of the children. Yet every genomic DNA. A typical region is present in two person can be assigned his or her genotype because the copies (one inherited from the mother and the other © Jones & BartlettAtLearning, LLC © Jones & Bartlett LLC SSR alleles are codominant. present, DNA polymorfrom the father). If an individual hasLearning, an extra copy of phisms are FOR the principal genetic markers used the region, NOT the ratio of SALE hybridization and, therefore, NOT SALEtypes OR ofDISTRIBUTION FOR OR DISTRIBUTION in genetic mapping in human pedigrees. Such polymorfluorescence intensity is 3 : 2, while if an individual phisms are prevalent, are located in virtually all regions has a missing copy, the ratio of hybridization and, of the chromosome set, and have multiple alleles and therefore, fluorescence intensity is 1 : 2. These differso yield a high proportion of heterozygous genotypes. ences can readily be detected with DNA microarrays. Furthermore, only a small amount Moreoever, since CNVs are r­ elatively © Jones & Bartlett Learning, LLC of biological material © Jones & Bartlett Learning, LLClarge, a microarneeded OR to perform the necessary tests. typically has OR many different oligonucleotides that NOT FORisSALE DISTRIBUTION NOTrayFOR SALE DISTRIBUTION One feature of the human genetic map is that there are complementary to sequences at intervals across the is about 60 percent more recombination in females than CNV; hence, the CNV results in a increase or decrease in males, so the female and male genetic maps differ in in signal intensity of all the oligonucleotides included length. The female map is about 4400 cM, the male map in the CNV. Current SNP chips also include about about 2700 cM. Averaged over both the length of aLLC million oligonucleotide probes designed to detectLearning, LL © Jones &sexes, Bartlett Learning, © Jones & Bartlett the human genetic map for allFOR 23 pairs of chromosomes known CNVs. NOT FOR SALE OR DISTRIBUT NOT SALE OR DISTRIBUTION is about 3500 cM. Because the total DNA content per CNVs by definition exceed 1 kb in size, but haploid set of human chromosomes is 3200 million base many are much larger. In one study of about 300 pairs, there is, very roughly, 1 cM per million base pairs individuals with their ancestry in Africa, Europe, in the human genome. or Asia, approximately 1500 CNVs were discovered © Jones & Bartlett Learning, LLC © Jones Bartlett Learning, LLC by hybridization with µarrays. These averaged 200 to 300 NOT kb in length. In the aggregate, the CNVs NOT FOR SALE OR DISTRIBUTION FOR SALE OR DISTRIBUTION Gene dosage can differ owing to included 300 to 450 ­million base pairs, or 10 to 15 copy-number variation (CNV). percent of the nucleotides in the entire genome. Many of the CNVs were located in regions near In addition to the small-scale variation in copy numknown mutant genes associated with hereditary ber represented by such genomic features as simple© Jones & Bartlett Learning, & Bartlett diseases. CNVs inLearning, alpha andLLC beta hemoglobin sequence repeats (SSRs), aLLC substantial portion of the © Jones

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© Jones & Bartlett Learning, LLC © Jones & Bartlett LLCconnection the Learning, human NOT FOR SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION

Starch Contrast

­ roteinaceous resources (for examp ple, meats and blood) and simple saccharides (for example, from fruit, honey,© and milk). .&. .Bartlett We estimated © Jones & Bartlett Learning, LLC Jones Learning, LL includes macaques and mangabeys, George H. Perry1,2, Nathaniel J. [salivary amylase gene] AMY1 copy NOT FOR SALE OR DISTRIBUT NOT FOR SALE OR DISTRIBUTION produces even more salivary amylase Dominy3, Katrina G. Claw1, Arthur number in three high-starch and S. Lee2, Heike Fiegler4, Richard than humans. Cercopithecines are four low-starch population samRedon4, John Werner1, and 6 other unique among primates in storing ples. . . . Notably, the proportion authors 2007 starchy foods, such as the seeds of of individuals from the combined 1 Arizona State University, Tempe, AZ; unripe fruits, in a cheek pouch, and 2 Brigham and Women’s Hospital, Boston, MA; with atLLC least © Jones & Bartlett Learning, LLC © Joneshigh-starch & Bartlettsample Learning, 3 University of California, Santa Cruz, CA; it is a plausible hypothesis that the six AMY1 copies (70 ­ p ercent) was 4 FOR SALE ORHinxton DISTRIBUTION NOT FOR SALE OR DISTRIBUTION TheNOT Wellcome Trust Sanger Institute, UK. increased amylase facilitates digesnearly two times greater than that Diet and the Evolution of Human Amylase Gene tion of the starch. It is not known for low-starch populations (37 Copy Number Variation whether the increased amylase propercent). . . . Diet more strongly duction in cercopithecines is due to predicts AMY1 copy number than copy-number ©variation some Learning, geographic LLC proximity. . . . We favor © Jones & Bartlett Learning, LLC Jonesor& to Bartlett other mechanism. a model in which AMY1 copy numWe OR favorDISTRIBUTION a model in which NOT FOR SALE NOT FOR SALE OR DISTRIBUTION ber has been subject to positive or AMY1 copy number has directional selection in at least some been subject to positive high-starch populations but has or directional selection in Hominin evolution is characterized evolved neutrally (that is, through at least some high-starch by significant dietary shifts, faciligenetic drift) in low-starch popula© Jones & Bartlett Learning, LLC © Jones & Bartlett Learning, LL populations . . . tated in part by the development tions. . . . Comparisons with other NOT FOR SALE OR DISTRIBUT NOT FOR SALEofOR DISTRIBUTION stone tool technology, the congreat apes suggest that AMY1 copy Evolution of increased enzyme activtrol of fire and, most recently, the number was probably gained in ity can occur through regulatory domestication of plants and anithe human lineage. . . . The initial mutations that increase transcription mals. . . . Starch, for instance, has human-specific increase in AMY1 of a single gene or through increases become an increasingly prominent copy number may have been coin© Jones & Bartlett Learning, LLC © Jones & Bartlett Learning, LLC in gene copy number. This study component of the human diet, parcident with a dietary shift early in NOTaFOR ORbetween DISTRIBUTION reports strong SALE correlation ticularly among agricultural NOT socie- FOR SALE OR DISTRIBUTION hominin evolutionary history. For amount of starch in the diets of ties. . . . A distinction can be made example, it is hypothesized that human populations and the number between “high-starch” populations starch-rich plant underground of copies of a gene encoding salivary for which starchy food resources storage organs such as bulbs, amylase, a starch-degrading enzyme. comprise a substantial portion of corms and LLC tubers were a critical © Jones & Bartlett LLC Jones & Bartlett Humans areLearning, not alone among primates the diet and© “low-starch” popu- Learning, food resource for early hominins. in having higher amylase lations with NOT traditional thatOR DISTRIBUTION NOT FOR SALE ORevolved DISTRIBUTION FORdiets SALE activity. A group of Old World Monincorporate relatively few starchy keys called ­ cercopithecines, which foods [but] instead emphasize Source: Science 328: 710–722.

© Jones & Bartlett Learning, LLC NOT FOR SALE OR DISTRIBUTION genes are known to be associated with resistance to malaria, and CNVs in an HIV-1 receptor gene CCL3 are resistance to AIDS. The preva© associated Jones & with Bartlett Learning, LLC lence of CNVs has prompted the ­ i nclusion of CNV NOT FOR SALE OR DISTRIBUTION ­oligonucleotides onto DNA microarrays to be able to assess what effects CNVs may have on the risk of complex diseases such as autism or Alzheimer’s disease.

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Copy-number variation has helped human populations adapt to © Jones & Bartlett Learning, LLC a high-starch diet.

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Increased copy number is sometimes beneficial, with the result that natural selection operates to gradually increase copy number throughout an entire population of organisms. An excellent example in human evolution

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Average amylase-gene copy number per © Jones individual differs & Bartlett Learning, LL according to the amount NOT FOR SALE OR DISTRIBUT of starch in the diet.

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Amylase copy number is greater in populations with a high-starch diet even when the populations are © Jones & Bartlett Learning, LLC geographically close.NOT FOR SALE OR DISTRIBUTION

FIGURE 4.23 Amylase copy number varies with the amount of starch in human diets. [Data from G. H. Perry, et al. Nat. Genet. 39 (2007): 1256–1260.]

© Jones & Bartlett Learning, LLC © Jones & Bartlett Learning, LL is copy-number variation the gene for OR the starchamylase results from ­copy-number The results NOTvariation. FOR SALE OR DISTRIBUT NOTinFOR SALE DISTRIBUTION

degrading enzyme amylase, which is produced in the of one study of amylase gene copy number are shown in salivary gland and pancreas. ­FIGURE 4.23. Seven populations were assayed for amyAs anatomically modern humans left their original lase gene copy number, four with low-starch diets (blue) homes in Southern Africa about 200,000 years ago and and three with high-starch diets (red). On average, © Jones & Bartlett Learning, © Jones & Bartlett Learning, migrated first north to populate all of AfricaLLC and then, individuals in populations with a low-starch diet LLC had aboutNOT 60,000 yearsSALE ago, into theDISTRIBUTION Middle East and then an average of 5.4 copies the amylase gene, whereas FOR OR NOT FOR of SALE OR DISTRIBUTION other parts of the world, the migrating populations those with a high-starch diet had an average of 6.7 cophad to cope with different climates, soil types, water ies of the gene. The comparisons exclude geography as availability, and native plants and animals. Some of the an explanation for the difference. Note, for example, the novel conditions were accommodated by changes in contrasting populations in Tanzania in East Africa at the diet. Traditional hunter-gatherer lower right and the Learning, contrasting populations in Siberia © Jones & Bartlett Learning, LLC populations in humid © Jones & Bartlett LLC tropical forests typically eat a diet rich NOTand Japan at the OR upperDISTRIBUTION right. Our nearest primate relaNOT FORclimates SALEsuch ORasDISTRIBUTION FOR SALE in meat or fish and relatively low in starch. Pastoralists, tives, chimpanzees and gorillas, have only one copy of who follow herds like reindeer or bison as they migrate the amylase gene; therefore, amylase gene copy number during the year, also have diets rich in meat and relaapparently has increased generally in human evolution, tively low in starch. In contrast, hunter-gatherer popuperhaps beginning at a time when our ancestors left the lations in arid climates©rely more& onBartlett roots andLearning, tubers humid and&began to eatLearning, LL Jones LLC forests for the arid savannahs © Jones Bartlett for food and increase their starch. Populations more starchy roots and tubers. NOT FOR SALE OR DISTRIBUT NOTintake FORof SALE OR DISTRIBUTION that switch to agriculture also have a dramatic increase in dietary starch because cultivated staple foods like 4.5 Tetrads contain all four wheat, rice, potatoes, and corn are rich in starch. products of meiosis. One consequence of copy-number variation is that © Jones & Bartlett Learning, LLC © Jones & Bartlett Learning, the amount of protein produced from any gene included In some species of fungi, each meiotic tetrad is LLC conin theNOT regionFOR is usually proportional to the copy number. tained in a sac-like structure, calledOR an ascus, and can be SALE OR DISTRIBUTION NOT FOR SALE DISTRIBUTION Increased gene copy number usually implies increased recovered as an intact group. Each product of meiosis is protein level. In human populations with high-starch included in a reproductive cell called an ascospore, and diets, it might be expected that increased amylase is all of the ascospores formed from one meiotic cell remain beneficial because more calories can be absorbed as together in the ascus FIGURE 4.24. The a­ dvantage of © Jones & Learning, LLC populations, increased © Jones & Bartlett theBartlett starch is digested. In human these organisms for Learning, the study of LLC ­recombination is the

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The 2 A : 2 a segregation means that the Mendelian ratio of 1 : 1 is realized in the a products of each individual © Jones & Bartlett Learning, LLC meiotic division © Jones & Bartlett Learning, LLC and is not merely an average over a large Chromosome NOT FOR SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION duplication number of meioses. Two other features of ascus-producing A organisms are especially useful for genetic analysis: (1) They are haploid, so dominance A is not a complicating because Learning, the © Jones & Bartlett Learning, LLC © Jonesfactor & Bartlett LL Meiosis I genotype is expressed directly in the phenoa NOT FOR SALE OR DISTRIBUT NOT FOR SALE OR DISTRIBUTION type. (2) They produce very large numbers of progeny, making it possible to detect rare a events and to estimate their f­requencies accurately. Furthermore, the life cycles of Segregation of chromosomes in meiosis I the organisms tend toLearning, be short. The only © Jones & Bartlett Learning, LLC © Jones & Bartlett LLC diploid stage is the zygote, which undergoes NOT FOR SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION meiosis soon after it is formed; the resulting a A haploid meiotic products (which form the ascospores) germinate to regenerate the a A vegetative stage FIGURE 4.25. In most of the organisms, the LLC meiotic products, or their © Jones & Bartlett Learning, LLC © Jones & Bartlett Learning, Segregation of chromosomes in meiosis II derivatives, are not arranged in any particuNOT FOR SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION lar order in the ascus. However, bread molds of the genus Neurospora and related organA a isms have the useful characteristic that the Meiosis II a A meiotic products are arranged in a definite order directly©related to the planes of Learning, the © Jones & BartlettSpore Learning, LLC Jones & Bartlett LL meiotic divisions. In Neurospora, each of the formation NOT FOR SALE OR DISTRIBUT NOT FOR SALE OR DISTRIBUTION four products of meiosis also undergoes a mitotic division, with the result that each member of the tetrad yields a pair of genetiAscospores cally identical ascospores. We will examine the ordered system after first looking © Jones & Bartlett Learning, LLC © Jones & Bartlett Learning, LLCat A unordered tetrads. NOT FOR SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION A

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genotypes are represented, and their alleles FIGURE 4.24 Formation of an ascus containing all of the four products of a have the same combisingle meiosis. Each ascospore present in the ascus is a reproductive cell formed nations found in the from one of the products of meiosis. parents. © Jones & Bartlett Learning, LLC © Jones & Bartlett Learning, LLC Ab Ab aB aB rOR eferred to as nonpaNOT FOR SALE OR DISTRIBUTION NOT FOR SALE DISTRIBUTION potential for analyzing all of the products from each rental ditype, or NPD. meiotic division. For example, one can see immediately Only two genotypes are from the diagram in Figure 4.24 that a tetrad containing represented, but their the products of a single meiosis in a heterozygous Aa alleles have nonparental Bartlett Learning, LLC © Jones & Bartlett Learning,combinations. LLC organism contains 2 A ascospores and 2 a ascospores. Ascus

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Diploid cells (2 n )

a a© Jones

& Bartlett Learning, LLC NOT FOR SALE OR DISTRIBUTION α α

α

Ascus with four

FIGURE 4.25

© Jones & Bartlett Learning, LLC NOT FOR SALE OR DISTRIBUTION αa

α

αa

Meiosis and sporulation

Ascospore germination a

© Jones & Bartlett Learning, LLC NOT FOR SALE OR DISTRIBUTION

© Jones & Bartlett Learning, LL NOT FOR SALE OR DISTRIBUT

Meiosis and sporulation produce haploid spores.

ascospores (n) © Jones & Bartlett Learning, LLC © Jones & Bartlett Learning, LL SALE FOR SALE OR DISTRIBUTION Life cycle NOT of the yeast Saccharomyces cerevisiae. Mating type is determined by the alleles NOT a and a.FOR Both haploid andOR DISTRIBUT

diploid cells normally multiply by mitosis (budding). Depletion of nutrients in the growth medium induces meiosis and sporulation of cells in the diploid state. Diploid nuclei are shown in red, haploid nuclei in yellow.

© Jones & Bartlett Learning, LLC as tetratype, or TT. All NOT FOR SALE to OR DISTRIBUTION

AB Ab aB ab referred

four of the possible genotypes are present.

Tetratype tetrads demonstrate that

© Jones & Bartlett Learning, crossing-over takesLLC place at the NOT FOR SALE OR DISTRIBUTION

© Jones & Bartlett Learning, LLC

2. The exchange consistsOR of the breaking and NOT process FOR SALE DISTRIBUTION rejoining of the two chromatids, ­resulting in the reciprocal exchange of equal and ­corresponding segments. Tetratype tetrads ­demonstrate this point because they contain the reciprocal products (A b and&a Bartlett B if the parental alleles were Jones Learning, LLCin coupling, A B and a b if they were in repulsion).

© NOT FOR SALE OR DISTRIBUTION

f­ our-strand stage of meiosis and is reciprocal.

Tetrad analysis affords a convenient test for linkage.

We noted earlier that tetrads from heterozygous organisms regularly contain 2 A and 2 a ascospores, way to determine whetherLearning, LL © Jones & Bartlett Learning, Tetrad LLC analysis is an effective © Jones & Bartlett which implies that Mendelian segregation takes place two genes are linked, because NOT FOR SALE OR DISTRIBUT FORofSALE ORtetrads DISTRIBUTION in each meiosis. TheNOT existence tetratype for linked genes demonstrates two features about KEY CONCEPT ­crossing-over that we have assumed, so far without When genes are unlinked, the parental ditype proof.

© Jones LLCchro 1. The exchange&ofBartlett segmentsLearning, between parental matids in the first meiotic prophase, NOT takes FORplace SALE OR DISTRIBUTION after the chromosomes have duplicated. Tetratype tetrads demonstrate this assertion because only two of the four products of meiosis show recombination. This would not be possible unless crossing& Bartlett LLC over took Learning, place at the four-strand stage.

© Jones NOT FOR SALE OR DISTRIBUTION

tetrads and the nonparental ditype tetrads are & Bartlett expected© in Jones equal frequencies (PD 5Learning, NPD).

LLC NOT FOR SALE OR DISTRIBUTION

The reason for the equality PD 5 NPD for unlinked genes is shown in part A of FIGURE 4.26, where the two pairs of alleles A, a and B, b are located in different Jones & Bartlett LLC chromosomes. In theLearning, absence of crossing-over between

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(A) No crossing-over

A

© Jones & Bartlett Learning, LLC A NOT FOR SALE OR DISTRIBUTION a

B B b

a

b

Ascospore

genes, meiotic cells with no crossovers always ­outnumber those with four-strand double crossTherefore, Jonesovers. & Bartlett Learning, LLC Ascus

© AB NOT FOR SALE OR DISTRIBUTION

AB

ab

ab Parental ditype tetrad

KEY CONCEPT

Linkage is indicated when nonparental ditype tetrads appear with a much lower frequency than parental ditype tetrads (NPD ,, PD).

© Jones & Bartlett Learning, LLC b NOT FOR SALE OR DISTRIBUTION

A

© Jones & Bartlett Learning, LL NOT FOR SALE OR DISTRIBUT

The relative frequencies of the different types of tetrads can be used to determine the map disa B aB tance between two linked genes. The simplest case is one in which the genes are sufficiently close Nonparental a B ditype that double and higher levels of crossing-over can © Jones & Bartlett Learning, LLC © Jones & Bartlett Learning, LLC tetrad be neglected. In this case, tetratype tetrads arise NOT FOR SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION (B) Crossing-over between one of the genes and its centromere only from meiotic cells in which a single crossover occurs between the genes (Figure 4.27, part A and B A part B). As we saw in Figure 4.6, the genetic map AB b A distance across an interval is defined as one-half B aB Ab the proportion of cells with © Jones & Bartlett Learning, LLC © Jones & Bartlett Learning, LLCa crossover in the a b ab i ­ nterval, so the map distance implied by the tetrads NOT FOR SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION a Tetratype b is given by tetrad A

b

A

Ab

Ab

aB

Map distance 5 Number of tetratype tetrads 1 × 100 (4.1) × 2 Total number tetrads & Bartlett Learning, ©ofJones

b

A

b

Ab

AB ab © Jones & Bartlett Learning, LLC LL a B aB To take a specific example, 100 tetrads NOTsuppose FOR SALE ORare DISTRIBUT NOT FOR SALE OR DISTRIBUTION B

a

Tetratype tetrad

FIGURE 4.26 Types of unordered asci produced with two genes in different chromosomes. (A) In the absence of a crossover, © Jones & Bartlett Learning, LLC random arrangement of chromosome pairs at metaphase I results in two different NOT FOR SALE OR DISTRIBUTION combinations of chromatids, one yielding PD tetrads and the other NPD tetrads. (B) When a crossover takes place between one gene and its centromere, the two chromosome arrangements yield TT tetrads. If both genes are closely linked to their centromeres (so that crossing-over is rare), few TT tetrads are produced.

analyzed from the cross A B 3 a b, and the result is that 91 are PD and 9 TT. The finding that NPD ,, PD means that the genes are linked, and the fact that NPD 5 0 means that the genes are so closely Jones & Bartlett Learning, linked©that double crossing-over does notLLC occur between them. TheSALE map distance between A and B NOT FOR OR DISTRIBUTION is calculated as follows: Map distance =

1 9 × × 100 = 4 . 5 cM 2 100

© Jones & Bartlett Learning, LLC NOT FOR SALE OR DISTRIBUTION

We & must emphasize that Equation © Jones Bartlett Learning, LLC (4.1) is valid only whenSALE NPD 5OR 0, soDISTRIBUTION that interference across the region NOT FOR

© Jones & NOT FOR SALE OR DISTRIBUTION

NOT FOR SALE OR DISTRIBUTION

prevents the occurrence of double crossing-over. When double crossovers do take place in the interval, then either gene and its centromere, the two chromosomal NPD  0, and the formula for map distance has to be configurations are equally likely at metaphase I, and so modified to take the double crossovers into account. PD 5 NPD. When there is a crossover between either tetrads differs from gene and its centromere (Figure part B), a tetra- LLC The mapping procedure © Jones & 4.26, Bartlett Learning, © using Jones & Bartlett Learning, LL that presented earlier in the chapter that the map type tetrad results,NOT but this doesSALE not change the fact NOT FORinSALE OR DISTRIBUT FOR OR DISTRIBUTION distance is not calculated directly from the number of that PD 5 NPD. recombinant and nonrecombinant chromatids. Instead, In contrast, when genes are linked, parental ditypes the map distance is calculated directly from the tetrads are far more frequent than nonparental ditypes. To see and the inferred crossovers that give rise to each type why, assume that the genes are linked and consider © events Jonesrequired & Bartlett Learning, Jones it&isBartlett Learning, LLC of tetrad.© However, not necessary to carry out a full the for the productionLLC of the three tetrad analysis for estimating linkage. The alternative types tetrads. FIGURE shows that when no NOTof FOR SALE OR4.27 DISTRIBUTION NOT FOR SALE OR DISTRIBUTION is to examine spores chosen at random after allowcrossing-over takes place between the genes, a PD ing the tetrads to break open and disseminate their tetrad is formed. Single crossover between the genes spores. This procedure is called random-spore analysis, results in a TT tetrad. The formation of a two-strand, and the linkage relationships are determined exactly as three-strand, or four-strand double crossover results Bartlett Learning, LLC respectively. With linked © Jones & Bartlett Learning, described earlier for DrosophilaLLC and corn. In particular, in a PD, TT, or NPD tetrad, 138

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the frequency of recombination equals the number of spores that are recombinant for the genetic markers divided&byBartlett the total number of spores. Jones Learning, LLC

(A) No crossing-over

A

B

© Jones & Bartlett Learning, LLC A B NOT FOR SALE OR DISTRIBUTION

(B)

a

b

a

b

AB AB

ab ab

Parental ditype (PD)

© NOT FOR SALE OR DISTRIBUTION

The geometry of meiosis is revealed in ordered tetrads. In the bread mold Neurospora crassa, the products of

© Jones & Bartlett Learning, LLC © Jones meiosis are contained in an ordered array&ofBartlett ascosporesLearning, LL Single crossover (FIGURE 4.28). A zygote nucleus, contained in a sac-like NOT FOR SALE OR DISTRIBUT NOT FOR SALE OR DISTRIBUTION

ascus, undergoes meiosis almost immediately after it is formed. The four nuclei produced by meiosis are in a b B A linear, ordered sequence in the ascus, and each of them Ab aB a b undergoes a mitotic division to form two genetically idenab © Jones &aBartlett Learning, LLC © Jones & Bartlett Learning, LLC tical and adjacent ascospores. Each mature ascus contains b Tetratype (TT) eight ascospores arranged in four pairs, pair derived NOT FOR SALE OR DISTRIBUTION NOT FOR SALE OR each DISTRIBUTION from one of the products of meiosis. The ascospores can (C) 2-strand double crossover be removed one by one from an ascus and each germinated in a culture tube to determine its genotype. A B Ordered asci also can be classified as PD, NPD, or TT AB B A © Jones & Bartlett Learning, LLC © Jones & Bartlett Learning, with respect to two pairs of alleles,LLC which makes it posAB ab a b sible to assess the degree of linkage between the genes. NOT FOR SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION ab The fact that the arrangement of meiotic products is a b Parental ordered also makes it possible to determine the recomditype (PD) bination frequency between any particular gene and its (D) 3-strand double crossover centromere. The logic of the mapping technique is based & Bartlett Learning, on LLC © Jones & 4.29 Bartlett Learning, . the feature of meiosis shown in FIGURE A © Jones B A

B

AB

A a a

NOT FOR SALEAbOR DISTRIBUTION B b b

AB

aB

ab Tetratype (TT)

© Jones & Bartlett Learning, LLC crossover (E) 3-strand NOT FORdouble SALE OR DISTRIBUTION A

B

A a

B

KEY CONCEPT

LL NOT FOR SALE OR DISTRIBUT

Homologous centromeres of parental chromosomes separate at the first meiotic division; the centromeres of sister chromatids separate at the © Jones & Bartlett Learning, second meiotic division.

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Thus, in the absence of crossing-over between a gene and its centromere, the alleles of the gene (for example, A Ab ab and a) must separate in the first meiotic division; this sepb aB aration& is called first-division segregation. © Jones & Bartlett Learning, LLC © Jones Bartlett Learning, LLC If, instead, a b a is formed between the gene and its centromere, NOT FOR SALE OR DISTRIBUTION Tetratype (TT) NOTcrossover FOR SALE OR DISTRIBUTION the A and a alleles do not become separated until the second meiotic division; this separation is called second(F) 4-strand double crossover division segregation. The distinction between first-diviA B sion and second-division segregation is shown in Figure Ab B & Bartlett Learning, 4.29. A © Jones two possible arrange-Learning, LL LLCAs shown in part A, only © Jones & Bartlett Ab aB a ments of the products of meiosis canFOR yield SALE first-division b NOT OR DISTRIBUT NOT FOR SALEaBOR DISTRIBUTION segregation—A A a a or a a A A. However, four patterns a b Nonparental of second-division segregation are possible because of the ditype (NPD) random arrangement of homologous chromosomes at metaphase I and of the chromatids at metaphase II. These FIGURE 4.27 Types of tetrads produced with two linked & Bartlett Learning, © Jones &are Bartlett four arrangements, which shown inLearning, part B, are LLC genes.©(A)Jones In the absence of a crossover, a PD tetradLLC is produced. AB

(B) With a single crossover between genes, a TT tetrad is NOT FOR SALE ORthe DISTRIBUTION produced. (C2F) Among the four possible types of double crossovers between the genes, only the four-strand double crossover in part F yields an NPD tetrad.

© Jones & Bartlett Learning, LLC NOT FOR SALE OR DISTRIBUTION

NOT OR DISTRIBUTION A a A a,  a AFOR a A,  SALE A a a A, and a AAa

The percentage of asci with second-division segregation patterns for a gene can be used to map the gene with respect to its centromere. For example, let us assume Jones Bartlett that 30&percent of a Learning, sample of asciLLC from a cross have a

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Cross fertilization

Asexual cycle

Conidia (n)

Asexual cycle

Conidia (n)

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Haploid nuclei fuse to form diploid nucleus.

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Diploid cell forms in the protoperithecium.

Hyphae of mating-type A

Hyphae of mating-type a

© Jones & Bartlett Learning, LLC NOT FOR SALE OR DISTRIBUTION Zygote nucleus (2n)

© Jones & Bartlett Learning, LLC NOT FOR SALE OR DISTRIBUTION

© Jones & Bartlett Learning, LLC NOT FOR SALE OR DISTRIBUTION

A a

Zygote nucleus (2n)

A a First meiotic division

A

© Jones & Bartlett Learning, LLC NOT FOR SALE AORa DISTRIBUTION

a

Second meiotic division

© Jones &ABartlett Learning, LLC a a A NOT FOR SALE OR DISTRIBUTION

A

A

a



Jones & Bartlett Learning, LL NOT FOR SALE OR DISTRIBUT

Mitotic divisions

a a a A A A A LLC © Jones & Bartlett Learning, NOT FOR SALE OR DISTRIBUTION Ascus with eight

A A A a a& a a © AJones Bartlett

Learning, LLC NOTwith FOR SALE OR DISTRIBUTION Ascus eight

a

ascospores (n) in ordered array

© Jones & Bartlett Learning, LLC NOT FOR SALE OR DISTRIBUTION

ascospores (n) in ordered array

© Jones & Bartlett Learning, LLC NOT FOR SALE OR DISTRIBUTION

FIGURE 4.28 The life cycle of Neurospora crassa. The vegetative body consists of partly segmented filaments called hyphae. Conidia are asexual spores that function in the fertilization of organisms of the opposite mating type. A protoperithecium develops into a structure in which numerous cells undergo meiosis.

© Jones & Bartlett Learning, LLC

© Jones & Bartlett Learning, LL

Equation (4.2) is valid asNOT long FOR as theSALE gene isOR close second-division segregation pattern for the A andDISTRIBUTION a alleles. DISTRIBUT NOT FOR SALE OR enough to the centromere for us to neglect multiple This means that 30 percent of the cells undergoing meiosis crossovers. Reliable linkage values are best deterhad a crossover between the A gene and its centromere. mined for genes that are near the centromere. The Because the map distance between two genes is, by defilocation of more distant genes is then accomplished nition, equal to one-half times the proportion of cells with by mapping these genes relative Learning, to genes nearer © Jones & Bartlett Learning, LLC between © Jones & Bartlett LLCthe a crossover between the genes, the map distance centromere. a gene its centromere is given by the equation NOTand FOR SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION If a gene is far from its centromere, crossing-over Map distance 5 between the gene and its centromere will be so freNumber of asci with quent that the A and a alleles become randomized with 1 second division segregation respect to the four chromatids. The result is that the six × 100 × (4.2) possible spore arrangements Bartlett Learning, LLC © Jones & Bartlett Learning,shown LLC in Figure 4.29 are Total number of asci 2

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(A) First-division segregation

(B) Second-division segregation

A

© Jones & Bartlett Learning, LLCA A FOR SALE OR DISTRIBUTION b A NOT

© Jones & Bartlett Learning, LLC Homologous NOT FOR SALE OR DISTRIBUTION centromeres of parental chromosomes

a

a

a

a

No crossing-over © Jones & Bartlett Learning, LLC Single crossover © Jones & Bartlett Learning, LL between gene and between gene and Meiosis I Meiosis I NOT FOR SALE OR DISTRIBUT NOTcentromere FOR SALE OR DISTRIBUTION occurs. centromere occurs. A

A

© Jones & Bartlett Learning, LLC NOT FOR SALE OR DISTRIBUTION Homologous centromeres separate.

A

A and a separated

© Jones & Bartlett Learning, LLC a NOT FOR SALE OR DISTRIBUTION A and a not yet separated A

a

© Jones & Bartlett Learning, LLC NOT FOR SALE OR DISTRIBUTION

© Jones & Bartlett Learning, LLC a NOT a FOR SALE OR DISTRIBUTION Meiosis II

A

© Jones & Bartlett Learning, LLC NOT FOR SALE OR DISTRIBUTION Centromeres split and separate.

A

a

a

A

a

Mitosis

A A A A a a a a

segregation

A

© Jones & Bartlett Learning, LL NOT FOR SALE OR DISTRIBUT

A and a separated

A and a separated

© Jones & Bartlett Learning, LLC NOT FOR SALE OR DISTRIBUTION

© Jones & BartlettPossible Learning, LLC spore patterns with first-division NOT FOR SALE OR DISTRIBUTION

Meiosis II

© Jones & Bartlett Learning, LLC a NOT FOR SALE OR DISTRIBUTION Mitosis

A A a a A A a a

© Jones & Bartlett Learning, LLC or NOT FOR SALE OR DISTRIBUTION

or

a a a a A A A A

a a A A A A a a or

© Jones & Bartlett Learning, LLC spore patterns NOT FOR SALEPossible OR DISTRIBUTION with second-division segregation

a a A A a a A A

© Jones & Bartlett Learning, LL FOR SALE OR DISTRIBUT

NOT or

A A a a a a A A

FIGURE 4.29 First- and second-division segregation in Neurospora. (A) First-division segregation patterns are found in the ascus when a crossover between&the gene and centromere doesLLC not take place. The alleles separate in meiosis I. TwoLearning, spore patternsLLC © Jones Bartlett Learning, © (segregate) Jones & Bartlett are possible, depending on the orientation of the pair of chromosomes on the first-division spindle. The orientation shown results in NOT FOR SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION the pattern in the upper ascus. (B) Second-division segregation patterns are found in the ascus when a crossover between the gene and the centromere delays separation of A from a until meiosis II. Four patterns of spores are possible, depending on the orientation of the pair of chromosomes on the first-division spindle and that of the chromatids of each chromosome on the second-division spindle. The orientation shown results in the pattern in the top ascus.

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141

all equally frequent. Therefore, when the chromatids participating in each cross­over are chosen at random,

© Jones & Bartlett Learning, LLC KEY CONCEPT NOT FOR SALE OR DISTRIBUTION

The maximum frequency of second-division segregation asci is 2/3.

meiosis can be recovered in a four-spore (yeast) or eight-spore (Neurospora) ascus. As we have noted, most asci & from heterozygous Aa diploids Jones Bartlett Learning, LLCcontain ratios of

© NOT FOR SALE OR DISTRIBUTION 2 A : 2 a in four-spored asci, or 4 A : 4 a in eight-spored asci

© Jones & Bartlett Learning,

demonstrating normal Mendelian segregation. Occasionally, however, aberrant©ratios are also found, such as LLC Jones & Bartlett Learning,

LL

Gene conversion a molecular NOT FOR SALE OR DISTRIBUT NOTsuggests FOR SALE OR DISTRIBUTION 3 A : 1 a or 1 A : 3 a in four-spored asci, and mechanism of recombination. Genetic recombination may be regarded as a process of breakage and repair between two DNA molecules. In © eukaryotes, process takes place early in meiosis Jones &theBartlett Learning, LLC after each molecule has replicated, and with respect NOT FOR SALE OR DISTRIBUTION to genetic markers, it results in two molecules of the parental type and two recombinants. For genetic studies of recombination, fungi such as yeast or Neurospora are particularly useful, because all four products of any

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5 A : 3 a or 3 A : 5 a in eight-spored asci

Different types of aberrant ratios can also occur. The aberrant © asciJones are said&toBartlett result from gene conversion Learning, LLC because it appears as if one allele has “converted” the NOT FOR SALE OR DISTRIBUTION other allele into a form like itself. Gene conversion is frequently accompanied by recombination between genetic markers on either side of the conversion event, even when the flanking markers are tightly linked. This

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Region containing a mismatched base pair GAGT CGAAT C

C T C A G LLC TT TA G © Jones & Bartlett Learning, NOT FOR SALE OR DISTRIBUTION

© Jones & Bartlett Learning, LL NOT FOR SALE OR DISTRIBUT

One or the other

© Jones & Bartlett Learning, LLC strand is chosen GA GT C GAA T C to be repaired. C T OR C AG NOT FOR SALE DISTRIBUTION AGTT T

© Jones & Bartlett Learning, LLC NOT FOR SALE OR DISTRIBUTION

A segment of the strand to be repaired is excised © Jones & and discarded.

© Jones & Bartlett Learning, LLC TCGAA NOT FOR SALE OR DISTRIBUTION GAG T C CTCAGT T TAG

Bartlett Learning, LLC NOT FOR SALE OR DISTRIBUTION GAG T C CTCAGT T TAG

G A GT C GA A T C C T C AG

The gap remaining © Jones & Bartlett Learning, LLC is filled in by repair NOT FOR SALE OR DISTRIBUTION synthesis using the other

© Jones & Bartlett Learning, LL NOT FOR SALE OR DISTRIBUT

strand as a template.

© Jones & Bartlett Learning, LLCThe two possible types of© Jones & Bartlett Learning, LLC GAGTCGAAT C GAGTCAAAT C repair result in different NOT FOR NOT FOR SALE C T OR C A GDISTRIBUTION C T CSALE A G T T TOR A G DISTRIBUTION CT TAG DNA sequences.

FIGURE 4.30 Mismatch repair consists of the excision of a segment of a DNA strand containing a base mismatch followed by repair

© Jones & Bartlett Learning, LLC © Jones & mismatch Bartlett Learning, LLC synthesis. Either strand can be excised and corrected. In this example, the G2T is corrected to either G2C (left) or A2T (right). NOT FOR SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION 142

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implies that gene conversion can be one conseA a quence of the recombination process. Gene conversion results from a normal DNA © JonesG & Bartlett Learning, LLCG T C A A A T © Jones & Bartlett Learning, LLC TCGAAT repair process in the cell known as mismatch CAG C T T A OR DISTRIBUTION CAGT T TA NOT FOR SALE NOT FOR SALE OR DISTRIBUTION repair. In this ­process, an enzyme recognizes any base pair in a DNA duplex in which the paired bases are mismatched—for example, G paired with T, or A paired with C. When such a mismatch is found in a molecule of a small seg© duplex JonesDNA, & Bartlett Learning, LLC © Jones & Bartlett Learning, LL ment of one strand isNOT excised and replaced with NOT FOR SALE OR DISTRIBUT FOR SALE OR DISTRIBUTION a new segment synthesized using the remaining GTCGAAT GTCAAAT strand as a template. In this manner the misCAGT T TA CAGCT TA matched base pair is replaced. ­FIGURE 4.30 shows an example in which a mismatched G2T pair is being©repaired. strand thatLearning, is excised could JonesThe & Bartlett LLCbe © Jones & Bartlett Learning, LLC Heteroduplex Heteroduplex eitherNOT the strand containing T or the one containFOR SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION ing G, and the newly synthesized (repaired) segment, shown in red, would contain either a C or an A, respectively. The two possible products of repair TCGAAT T CGAAT A A T C G A G G CAGCT TA CAGCT TA differ in DNA sequence. The role of mismatch LLC repair in gene conver© Jones & Bartlett Learning, © Jones & Bartlett Learning, LLC sion is illustrated in FIGURE 4.31 . The pair of DNA a A conversion NOT FOR SALE OR DISTRIBUTION NOT FOR SALEResult: OR DISTRIBUTION duplexes across the top represents the DNA molecules of two alleles in a cell undergoing meiosis. One duplex contains a G2C base pair highlighted TCAAAT TCAAAT a a A G C T G G CAGT T TA CAGT T TA in color; this corresponds to the A allele. The other duplex contains an A2T base pair at the same © Jones & Bartlett Learning, LLC © Jones & Bartlett Learning, LL position, which corresponds to the a allele. In the FOR SALE OR DISTRIBUT NOT FOR SALE OR DISTRIBUTION Result: A aNOT conversion process of recombination, the participating DNA duplexes can exchange pairing partners. The result is shown in the second row. The exchange of pairTCGAAT TCAAAT a A A G A G G G CAGCT TA CAGT T TA ing partners creates a heteroduplex region in which bases & that are not identical in the LLC paren©any Jones Bartlett Learning, © Jones & Bartlett Learning, LLC tal duplexes become mismatched. In this example, NOT FOR SALE OR DISTRIBUTION NOTResult: FOR No SALE OR DISTRIBUTION conversion one heteroduplex contains a G2T base pair and the other an A2C base pair. At this point, the mismatch repair system comes into play and corrects TCGAAT A TCAAAT a T C G C T G CAGCT TA CAGT T TA the mismatches. Each mismatch can be repaired in Bartlett either of two ways, so LLC there are four possible © Jones & Learning, © Jones & Bartlett Learning, LLC ways in which the mismatches Result: No conversion NOT FOR SALE OR DISTRIBUTIONcan be repaired. NOT FOR SALE OR DISTRIBUTION One type of repair results in gene conversion of FIGURE 4.31 Mismatch repair resulting in gene conversion. Only a a to A, another results in gene conversion of A to small part of the heteroduplex region is shown. a, and the remaining two restore the sequences of the original duplexes and so do not result in gene conversion. © Jones & Bartlett Learning, LLC © Jones & Bartlett Learning, LL The crossovers needed for the FOR chiasmata to form NOT SALE OR DISTRIBUT NOT FOR SALE OR DISTRIBUTION are initiated by programmed double-stranded breaks 4.6 Recombination is initiated by a in DNA (FIGURE 4.32, part A). In a double-stranded double-stranded break in DNA. break, the size of the gap is usuallly increased by nuclease digestion of the broken ends, with greater In prophase I of meiosis, chiasmata are the physi© Jones & Bartlett Learning, LLC DNA & Bartlett Learning, LLC degradation©ofJones the 59 ends leaving overhanging 39 ends cal manifestations of crossing-over between as shown inNOT the illustration. These are repaired molecules. ­ These SALE structures between pairs of NOT FOR ORbridge DISTRIBUTION FOR SALE ORgaps DISTRIBUTION using the unbroken homologous DNA molecule as a sister chromatids in a bivalent and are important in the template, but in meiosis the repair process can result proper alignment of the bivalent at the metaphase plate in crossovers that yield chiasmata between nonsister in preparation for anaphase I. Bivalents that lack chiaschromatids. These crossovers are also the physical basis mata to help hold them together are prone to undergo © Jones & Bartlett Learning, LLC © Jones Learning, of what&isBartlett observed genetically as LLC recombination. nondisjunction.

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A

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B

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a

NOT Double-strand break (DSB)

Noncrossover pathway

Crossover pathway

Strand invasion © Jones & Bartlett Learning, LLC forms D loop. NOT FOR SALE OR DISTRIBUTION

© Jones & Bartlett Learning, LL NOT FOR SALE OR DISTRIBUT

Strand invasion forms D loop.

A

B

A

B

a

b

a

b

© Jones & Bartlett Learning, LLC Elongating strand ejected NOT FOR SALE OR DISTRIBUTION

© Jones & Bartlett Learning, LLC OR DISTRIBUTION

A

B

D loop NOT expands and SALE FOR serves as template for the remaining free 3 end.

a

b

A

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B

© Jones & Bartlett Learning, LLC a b NOT FOR SALE OR DISTRIBUTION

Ejected end pairs with correct partner.

Synthesis across gaps completes repair

Dark blue ©A Jones & Bartlett Learning, LLC A B and red NOT FOR SALE OR DISTRIBUTION strands a

b

B

a

break and rejoin.

b

blue & Bartlett Learning, LL ©Light Jones and pink NOT strandsFOR SALE OR DISTRIBUT break and rejoin. Holliday

New synthesis completesLLC © Jones & Bartlett Learning, repair of DSB; products NOT FOR SALE OR DISTRIBUTION are nonrecombinant for markers A and B. A

junctions Strand breakage and & Bartlett © Jones Learning, LLC rejoining separates duplexes; NOT FOR SALE OR DISTRIBUTION products are recombinant for markers A and B.

B

a LLC © Jones & Bartlett Learning, NOT FOR SALE OR DISTRIBUTION

b

A

B

b © Jonesa & Bartlett Learning, LLC NOT FOR SALE OR DISTRIBUTION Crossover

FIGURE 4.32 (A) Double-strand break in a duplex DNA molecule with overhanging 39 ends facing the gap. Repair of the break makes use of the nonbroken homologous duplex. (B) Repair pathway that does not result in crossing-over, although the heteroduplex regions can undergo gene conversion. (C) Repair pathway that does result in crossing-over, also with possible gene conversion in heteroduplex regions. [Adapted from© D. Jones K. Bishop and D. Zickler, Cell 117 (2004): 9–15.] & Bartlett Learning, LLC © Jones & Bartlett Learning,

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LL NOT FOR SALE OR DISTRIBUT

plex region is the region where the light blue strand is A double-stranded break does not necessarily result paired with the red strand. Because it is a heteroduplex, in a crossover, however. Repair of the double-stranded any base-pair mismatches in this region could be corbreak by the noncrossover pathway is illustrated in Fig©4.32, Jones Learning, Jones & Bartlett Learning, rected by©mismatch repair in such a way as toLLC result ure part & B. Bartlett The first step in repair isLLC that a broken 39NOT end invades homologous ­unbroken DNA duplex, in gene NOT conversion. Such heteroduplex regions are FOR the SALE OR DISTRIBUTION FOR SALE OR DISTRIBUTION forming a short heteroduplex region with one strand typically only a few hundred base pairs in length. They and a looped-out region of the other strand called are much shorter than a gene and vastly shorter than a a D loop. (Specific proteins are required to mediate chromosome, and so gene conversions are rare events strand invasion; in E. coli the strand-invasion protein except for short regions very near the site of a doubleBartlett Learning, © Jones & Bartlett is known as RecA.) In LLC the illustration, the heterodustranded break. Learning, LLC

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the ­double-stranded break, but note that the resulting At one end of the heteroduplex, the free 39 end of structure includes two places where the strands have the broken DNA strand is extended (brown), but after exchanged pairing Learning, partners. Each of the structures a time it is ejected from the template, and the strands of © Jones & Bartlett LLC © Jones & Bartlett Learning, LLC where pairing partners are switched is called a Holthe unbroken duplex are able to come together again. NOT FOR SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION liday ­junction, named after Robin Holliday, who first At this point, the extension of the 39 end is long enough predicted that such structures would be involved in that pairing can take place with the complementary recombination. strand in the broken duplex. At the same time, this The problem with Holliday junctions is that they pairing provides a template for the 39 end of the other are from different duplexLearning, LL broken strand. Extension of the 39 ends across the © Jones & Bartlett Learning, LLCplaces where DNA strands © Jones & Bartlett m ­ olecules are interconnected. How the strands are DISTRIBUT remaining gaps completes the repair of the doubleNOT FOR SALE OR NOT FOR SALE OR DISTRIBUTION ­interconnected is shown for the DNA double helices stranded break. Note that although gene conversion in FIGURE 4.33, part A. Resolution of the ­ Holliday can occur in the noncrossover pathway, the resulting junctions is necessary for the DNA molecules to duplex DNA molecules are nonrecombinant. become free of one another. This requires breakage and The crossover pathway for repairing a doublerejoining of © one pair of & DNA strandsLearning, at each Holliday stranded break is illustrated in Figure 4.32, part C. © Jones & Bartlett Learning, LLC Jones Bartlett LLC junction. The breakage and rejoining is an enzymatic AgainNOT invasion of the unbroken duplex forms a D FOR SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION function carried out by an enzyme called the Holliday loop and a short heteroduplex region in which gene junction-resolving enzyme. conversion can occur. As in the noncrossover pathParts B and C in Figure 4.33 show two ways in way, the free 39 end of the broken DNA strand is which the Holliday structures can be resolved. Breakextended (brown), but in this case it continues until age and the strands LLC indicated by the red it displaces partner strand © Jones & Bartlettthe Learning, LLC(pink) of the template © Jones & rejoining BartlettofLearning, arrows results in a crossover at the site of the left-hand strand (red). The displaced strand can then serve as a NOT FOR SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION Holliday junction, whereas breakage and rejoining of template for the elongation of the 39 end of the other the strands indicated by the blue arrows results in a broken strand. Eventually, the extensions of the broken crossover at the site of the right-hand Holliday juncstrands become long enough that they can be attached tion. In both cases, the resulting DNA molecules have to the broken 59 ends. This completes the repair of

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Holliday

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© Jones & Bartlett Learning, LLC B NOT FOR SALE OR DISTRIBUTION

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Resolution of the Holliday junctions by breakage and reunion at the sites of the © Jones red arrows. A

a

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& Bartlett Learning, LLC NOT FOR SALE ORb DISTRIBUTION A

Site of crossover

B

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a

Recombinant products A b and a B

Resolution of the Holliday junctions by breakage and reunion at the sites of the blue arrows. © Jones &

Bartlett Learning, LL NOT FOR SALE OR DISTRIBUT b

Site of crossover

B

© Jones & Bartlett Learning, LLC NOT FOR SALE OR DISTRIBUTION Recombinant products A b and a B

FIGURE 4.33 (A) Two Holliday junctions in a pair of DNA molecules undergoing recombination (the parental chromosomes are AB and ab); (B and C) two modes of resolution depending on which strands are broken and rejoined. Part D is an electron micrograph showing a single Holliday junction between a pair of DNA molecules. [Illustration modified from B. Alberts. Essential Cell Biology. Garland Science, & Bartlett LLC Jones & Bartlett Learning, LLC Institute.] 1997. IllustrationLearning, reproduced with permission of Huntington Potter,©Johnnie B. Byrd, Sr., Alzheimer’s Center & Research

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a crossover that yields reciprocal recombinant A b and attaches methyl (2CH­3) groups to the amino acid a B products. (In principle, resolution could also take lysine of an abundant protein called histone H3 that, at theLearning, red arrows inLLC one Holliday junction and the together with other histoneLLC proteins, binds with © Jones & Bartlett Learning, © Jones & place Bartlett blue arrows in the other, but these resolutions result in DNA to form a chain of small DNA-protein beads NOT FOR SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION noncrossover products. It is unclear how often these known as nucleosomes. DNA is usually packaged in noncrossover types of resolution take place.) the form of nucleosomes containing histone H3, and hence wherever the target 13-mer of PRDM9 occurs, nucleosomes containing histone H3 are also presRecombination tends to take place at ent. Evidently, methylation histone by PRDM9 © Jones Bartlett Learning, LLC © of Jones &H3 Bartlett Learning, LL preferred positions in&the genome. predispposes the DNA to undergo a double-stranded NOT FOR SALE OR DISTRIBUT NOT FOR SALE OR DISTRIBUTION In some organisms, including humans and other mambreak, but the molecular details of break formation mals, programmed double-stranded DNA breaks are still unknown. More than 30 alleles of the gene are much more likely to occur at certain ­ positions encoding PRDM9 have been identified, which differ in the genome than others. Crossovers resulting in in their propensity to recognize the 13-mer target ­recombination are much more likely to occur at these and to methylate histone H3. Variation in the LLC alleles © Jones & Bartlett Learning, LLC © Jones & Bartlett Learning, positions, which are referred to as hotspots of recomencoding PRDM9 accounts for the differing efficiency NOT FOR SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION bination. The human genome contains about 30 thouof PRDM9-associated hotspots. sand hotspots of recombination, spaced an average of Although recombination tends to be initiated by 100 kb apart. These hotspots are often located in the means of double-stranded breaks at hotspots of recomspaces between genes, and they differ greatly from one bination, it should be emphasized that there are so to the next Learning, in the likelihood of a double-stranded break. many to number © Jones & Bartlett LLC © Jones & hotspots Bartlettrelative Learning, LLCof crossover events One particular protein has been implicated in about that the sites of recombination in any particular meiotic NOT FOR SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION 40 percent of hotspots across the human genome. cell show a great deal of randomness. In the human The protein is known as PRDM9, and it is known to genome, for example, there are about 30 thousand bind with double-stranded DNA at sites that match or hotspots of recombination and about 60 crossover nearly match the 13-mer sequence events per meiosis. Roughly speaking, each crossover © Jones & Bartlett Learning, LLC Jones Bartlett LL could take place at any of©about 500&hotspots in Learning, the 5-CCGCCGTMWCCWC-3 vicinity. The relatively small number of crossovers per NOT FOR SALE OR DISTRIBUT NOT FOR SALE OR DISTRIBUTION where M means A or C, and W means T or A. The meiosis means that crossovers occur essentially at ranprotein does not cause the double-stranded break dom sites chosen from among the very large number of directly. PRDM9 is actually a methyl transferase that hotspots that occur across the genome.

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© Jones & Bartlett Learning, LLC CHAPTER SUMMARY NOT FOR SALE OR DISTRIBUTION

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n

Genes that are located in the same chromosome and that do n Variations in DNA sequence among individuals (polymornot show independent assortment are said to be linked. phisms) serve as genetic markers along the genome that are used for genetic mapping, tracing the genetic ancestry of n The alleles of linked genes present together in the same individuals, and many other purposes. ­chromosome tend to be inherited as a group. © Jones & Bartlett Learning, LLC © Jones & Bartlett Learning, LL n Among the most useful types of DNA polymorphisms are n Crossing-over between homologous chromosomes results NOT FORrestriction SALEfragment OR DISTRIBUT NOT FOR SALE OR DISTRIBUTION single-nucleotide polymorphisms (SNPs), in recombination, which breaks up combinations of linked length polymorphisms (RFLPs), simple-sequence repeats alleles. (SSRs), and copy-number polymorphisms (CNPs). n The frequency of recombination serves as a measure of disn Tetrads are sensitive indicators of linkage because they include tance between linked genes along a chromosome, providing all the products of meiosis. a genetic map of the relative positions of the genes. level, recombination is initiated by a doublen © The map distance between genes in a geneticLLC map is related to n At the DNA Jones & Bartlett Learning, © Jones & Bartlett Learning, LLC stranded break in a DNA molecule. Use of the homolothe frequency of crossing-over between the genes in meiosis. gous DNA molecule as aSALE templateOR for repair can result in a NOT FOR SALE OR DISTRIBUTION NOT FOR DISTRIBUTION n Physical distance along a chromosome is often, but not crossover, in which both strands of the participating DNA always, correlated with map distance. molecules are broken and rejoined.

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© Jones LEARNING & Bartlett Learning, OUTCOMESLLC NOT FOR SALE OR DISTRIBUTION n

Given a genetic map with two genes, predict the kinds and relative frequencies of gametes that would be produced by an individual of a specified genotype.

n

Explain how linkage is detected between two genes in an organism with unordered tetrads and estimate the distance in the genetic map between the genes.

n

Analyze the results of a genetic cross with three linked genes n Explain how linkage between a gene and its centromere is © ofJones & Bartlett LLC © Jones & Bartlett to deduce the genotypes the parents, the order ofLearning, the genes detected in an organism with ordered tetrads and estimate theLearning, LL FOR SALE OR DISTRIBUT NOT FOR SALE ORtheDISTRIBUTION along the chromosome, the map distances between genes, map distance between the gene NOT and its centromere. and the degree of interference between crossovers.

n

Distinguish between a single-nucleotide polymorphism (SNP), a restriction fragment length polymorphism (RFLP), and ­copy-number variation (CNV). Describe one experimental © Jones Bartlett Learning,canLLC method by which&each type of polymorphism be detected.

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ISSUES AND IDEAS n Bartlett Distinguish between geneticLLC recombination and genetic © Jones & Learning, complementation. Is it possible for two mutant genes to NOT FOR SALE OR DISTRIBUTION show complementation but not recombination? Is it pos-

n What&is Bartlett meant by theLearning, term chromosome interference? © Jones LLC n In human genetics, why are molecular variations in DNA NOT FOR SALE OR DISTRIBUTION sequence, rather than phenotypes such as eye color or

sible for two mutant genes to show recombination but not blood-group differences, used for genetic analysis? complementation? n In genetic analysis, what is so special about the ability to n In genetic analysis, why is it important to know the position examine tetrads in certain fungi? of a gene along a chromosome? n Explain how tetratype tetrads demonstrate that recombinan What is the maximum frequency of recombination between tion takes place at the four-strand stage of meiosis and is Learning, LL © Jones & Bartlett Learning, LLC © Jones & Bartlett two genes? Is there a maximum map distance between reciprocal. two genes? NOT FOR SALE OR DISTRIBUT NOT FOR SALE OR DISTRIBUTION n Explain why the observation PD .. NPD with respect to n Why is the frequency of recombination over a long interval tetrads is a sensitive indicator of linkage. of a chromosome always smaller than the map distance over the same interval?

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Problem 1

Solution.  To determine which, if any, genes are linked, consider the mutants in pairs, and sum the data to find the In Drosophila, the recessive mutant allele spineless (ss) results in total number of parental types and recombinant types for each thin bristles, cinnabar (cn) results in bright red eyes, and ebony pair. For& ss Bartlett and cn, the parental types are 241 1 212 1 25 1 © Jones & Bartlett Learning, LLC © Jones Learning, LLC (e) results in a black body color. A cross is carried out between 35 5 513 and the recombinant types are 223 1 202 1 31 1 NOT FORfemales SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION of genotype ss cn e / 1 1 1 and males of genotype 31 5 487. These numbers are close enough to 500 : 500 that one ss cn e / ss cn e. In this type of symbolism, each 1 denotes the may infer that ss and cn are unlinked. Similarly, for cn and e, the nonmutant allele of the gene written at the corresponding posiparental types sum to 515 and the recombinant types to 485, tion. From this cross, the following 1000 progeny were obtained: hence cn and e are unlinked. For ss and e, however, the parental types sum to 878 and the recombinant types to 122, and this ss cn e / ss cn e 241 & LLCimplies that ss and e are linked. © Jones & Bartlett ss 1 e / ss cn©eJones 223Bartlett Learning, result The estimated frequency Learning, LL 1 cn 1 / ss cnNOT e FOR202 NOT FOR SALE OR DISTRIBUT SALE OR DISTRIBUTION of recombination is 122/1000 5 0.122 or 12.2 percent. 1 1 1 / ss cn e 212 ss cn 1 / ss cn e 25 ss 1 1 / ss cn e 31 1 cn e / ss cn e 31 1 1 e / ss cn e 35

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Determine which, if any, of the genes are linked, and for NOT FOR SALE OR DISTRIBUTION those that are linked, estimate the frequency of recombination between the genes.

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Solutions: Step by Step

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­recombinants with those of the nonrecombinants. If the gene © Jones & Bartlett Learning, LLC © Jones & Problem Bartlett2Learning, LLC order is correct, then it will require two recombination events In Drosophila, the recessive mutant allele cinnabar (cn) resultsNOT FOR OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION (one inSALE each interval) to derive the double-recombinant in bright red eyes, curved (c) results in curved wings, and plexus (px) results in extra wing veins. All three genes are linked. In a cross between cn c px / 1 1 1 females and cn c px / cn c px males, the following progeny were counted:

chromosomes from the nonrecombinants. If this is not the case, rearrange the order of the genes. (The “odd man out” in comparing the double recombinants with the nonrecombinants is always the gene in the middle.) In this particular © Jones Bartlett Learning, LL example, the gene order is correct as given.& Finally, with this cn c px / cn©c Jones px & Bartlett 296 Learning, LLC cn c 1 / cnNOT c px FOR preliminary bookkeeping done, we can proceed to tackle theDISTRIBUT NOT FOR SALE OR SALE63OR DISTRIBUTION cn 1 1 / cn c px 119 questions. (a) The frequency of recombination between cn cn 1 px / cn c px 10 and c is given by the totals of all classes of progeny show1 c px / cn c px 86 ing recombination in the cn 2 c interval, in this case (205 1 1 c 1 / cn c px 15 25)/1000 5 0.23. (b) The frequency of recombination between 1 1 1 / cn c px 329 c and px equals (145 1 25)/1000 5 0.17.Learning, (c) The frequency of ©1 Jones & cn Bartlett LLC © Jones & Bartlett LLC 1 px / c px Learning,82 recombination between cn and px equals (145 1 205)/1000 5   Total 1000 NOT FOR SALE OR DISTRIBUTION FOR OR DISTRIBUTION 0.35. (NoteNOT that the doubleSALE recombinants are not included in this total, because the double recombinants are not recom (a) What is the frequency of recombination between cn and c? bined for cn and px; their allele combinations for cn and px are (b) What is the frequency of recombination between c and px? the same as in the parents.) (d) The frequency of recombina (c) What is the frequency of recombination between cn and px? tion between cn and px (0.35) is smaller than the sum of that (d) Why is the frequency of recombination between cn and px © Jones & Bartlett Learning, LLC © Jones & Bartlett Learning, between cn and c and that between LLC c and px (0.23 1 0.17 5 smaller than the sum of that between cn and c and that between 0.40) because ofOR double crossovers. (e) The coefficient of coinNOT FOR SALE OR DISTRIBUTION NOT FOR SALE DISTRIBUTION c and px? cidence equals the observed number of double ­recombinants (e) What is the coefficient of coincidence across this region? divided by the expected number. The observed number is 25 What is the value of the interference? and the expected number is 0.23 3 0.17 3 1000 5 39.1; the (f) Draw a genetic map of the region, showing the locations of coefficient of coincidence therefore equals 25/39.1 5 0.64. cn, c, and px and the map distances between the genes. The interference equals 1 2 coefficient of coincidence, and so © Jones & Bartlett Learning, LLC © Jones & Bartlett Learning, LL Solution.  Do not try to hurry through linkage problems! You will be the interference equals 1 2 0.64 5 0.36. (f) The genetic map FOR SALE are ORin DISTRIBUT FOR SALE OR DISTRIBUTION rewarded by taking timeNOT to organize the information in the optimal is shown in the accompanyingNOT diagram. The distances manner. First, group the progeny types into reciprocal pairs—cn c px map units (centimorgans). However, the map distances of 23 with 1 1 1, cn c 1 with 1 1 px, and so forth—and make a new list and 17 map units are based on the 23 percent and 17 percent organized as shown here. (Ignore the cn c px chromosome from the recombination observed between cn and c and between c and father because it contributes no information about recombination.) px, respectively; the actual distances in map units are probably © Jones & Bartlett Learning, LLC © Jones &estimates Bartlett Learning, LLC a little greater than these because of a small amount cn c px 296 of double recombination within each of the intervals. 625 NOT SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION 1 329 1  1FOR



cn c 1 1  1 px

63 82

cn 1 1

119

1  c  1

15



 1  c px © Jones & Bartlett Learning,86 LLC NOT FOR SALE DISTRIBUTION cn OR 1 px 10

cn

145 205



25

  Total

1000

c

23

px

17

© Jones & Bartlett Learning, LLC Problem 3 NOT FOR SALE OR DISTRIBUTION In Neurospora, the gene arg12 encodes the enzyme needed

to convert ornithine to citrulline in the pathway of arginine biosynthesis. The gene was discovered in the experiments of Beadle and Tatum discussed in Chapter 1. In a cross between arg12 and ARG12 strains, where arg12 denotes the mutant In this tabulation, a space has been inserted between the pairs © Jones & Bartlett Learning, LLC © allele, Jones & Bartlett allele and ARG12 the nonmutant two-thirds of the Learning, LL of reciprocal products in order to keep the groups separate. NOT FOR SALE ORthis DISTRIBUT NOT FOR SALE OR DISTRIBUTION resulting asci show second-division segregation. What does The number next to each brace is the total number of chromoobservation imply about the map distance between arg12 and somes in the group. The most numerous group of reciprocal the centromere? chromosomes (in this case, cn c px and 1 1 1) consists of the nonrecombinants, and the least numerous group of reciprocal chromosomes (in this case, cn 1 px and 1 c 1) consists of Jones & Bartlett Learning, the©double recombinants. Rearrange the order LLC of the groups, if necessary, so that the nonrecombinants are at the top NOT FOR SALE OR DISTRIBUTION of the list and the double recombinants are at the bottom. (In the present example, rearrangement is not necessary.) At this point, also make sure that the order of the genes is correct as given, by comparing the genotypes of the double

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Solution.  Two-thirds is the maximum proportion of seconddivision segregation that can occur. This value is observed for © Jones & Bartlett Learning, LLC any gene that is so far from the centromere that one or more FOR SALE crossoversNOT are almost certain to takeOR placeDISTRIBUTION between the gene and the centromere. Hence, we can deduce that arg12 is at least 50 map units from the centromere. This map distance is a minimum, and the true map distance could be greater.

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© Jones & Bartlett Learning, LLC © JonesCONCEPTS & Bartlett Learning, IN ACTION:LLC PROBLEMS FOR SOLUTION NOT FOR SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION 4.1 A double heterozygote has the repulsion configuration

4.9 Two genes in chromosome 7 of corn are identified by the

recessive alleles gl (glossy), determining glossy leaves, and ra (ramosa), determining branching of ears. When a plant heterozygous for each of these alleles was crossed with a homozygous © Jones & Bartlett Learning, LLC © Jones & Bartlett recessive plant, the progeny consisted of the following geno- Learning, LL 4.2 What gametes, and in what frequencies, are produced types with the numbers of each indicated: NOT FOR SALE OR DISTRIBUT NOT FOR SALE OR DISTRIBUTION by a female Drosophila of genotype A B / a b when the genes Gl ra/gl ra 98 gl Ra/gl ra 91 are present in the same chromosome and the frequency of Gl Ra/gl ra 7 gl ra/gl ra 4 recombination between the genes is 8 percent? What gametes, A b / a B of two linked genes that have a frequency of recombination of 0.20. If a randomly chosen gamete carries A, what is the probability that it also carries B?

and in what frequencies, are produced by a male of the same genotype?

Calculate the frequency of recombination between these genes.

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© Jonesmosquito, & Bartlett Learning, LLC 4.10 In the yellow-fever Aedes aegypti, a dominant gene DDT for NOT DDT resistance (DDT is dichlorodiphenyltrichloFOR SALE OR DISTRIBUTION roethane, a long-lasting insecticide) and a dominant gene Dl for tetrads undergoes a double crossover between two markers. Dieldrin resistance (Dieldrin is another long-lasting insecticide) If the ratio of 2-strand : 3-strand : 4-strand doubles is 1 : 2 : 1, are known to be in the same chromosome. A cross was carried what is the ratio of PD : TT : NPD tetrads? (PD stands for out between a DDT-resistant strain and a Dieldrin-resistant parental ditype tetrad, TT for tetratype, and NPD for nonparenstrain, and female progeny resistant to both insecticides were tal ditype.) © Jones & Bartlett Learning, LLC © Jones & Bartlett Learning, LLC testcrossed with wildtype males. Among the progeny, 99 were NOT FOR4.4 SALE OR DISTRIBUTION FORtoSALE OR DISTRIBUTION A coefficient of coincidence of 0.36 implies which one or NOTresistant both insecticides, 88 were resistant to DDT only, 89 more of the following statements are true: were resistant to Dieldrin only, and 106 were sensitive to both insecticides. (a) The frequency of double crossovers was 36 percent. (b) The frequency of double crossovers was 36 percent of the (a) Are DDT and Dl alleles of the same gene? How can you tell? number that would be expected if there were no interference. (b) Are DDT and Dl linked? & crossovers BartlettasLearning, LLC © Jones & Bartlett Learning, LL (c) There were 0.36 times© as Jones many single double (c) What can you deduce about the genetic positions of DDT NOT FOR SALE OR DISTRIBUTION crossovers. and Dl along the chromosome? NOT FOR SALE OR DISTRIBUT (d) There were 0.36 times as many single crossovers in one 4.11 Two pure-breeding strains of mice are crossed to produce region as there were in an adjacent region. F1 mice that are heterozygous for three linked genes with alleles (e) There were 0.36 times as many parental as recombinant Aa, Bb, and Dd. Numerous triply heterozygous F1 mice are progeny. testcrossed, and the genotypes and numbers of the resulting © Jones & Bartlett Learning, LLC © Jones & Bartlett Learning, LLC progeny are as follows: 4.5 A gene in Neurospora, a fungus with ordered tetrads, FOR SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION showsNOT 10% second-division segregation. What is the map dis A2 B2 D2 10 tance between the gene and the centromere? A2 B2 dd 350 A2 bb D2 100 4.6 In Drosophila pseudoobscura, the eye-color mutation A2 bb dd 40 purple (pr) and the wing mutation crossveinless (cv) are located aa B2 D2 60 in chromosome 3 at a distance of 18 map units. What phe© Jones & Bartlett Learning, LLC © Jones &aa B2 dd Bartlett Learning, 120LLC notypes, and in what proportions, would you expect in the aa bb D2 320 NOT FORprogeny SALEfrom OR FOR SALE OR DISTRIBUTION theDISTRIBUTION mating of pr+ cv+ / pr cv females with pr cv / NOT Total 1000 pr cv males? 4.3 ANOT cell undergoing meiosisOR in an DISTRIBUTION organism with unordered FOR SALE

(a) Which gene is in the middle? (b) Specify the genotype of the F1 triple heterozygote as ­following recombination frequencies between individual pairs of completely as possible, with the genes in the correct order and genes: r2c, 10; c2p, 12; p2r, s2c, 16; s2r, 8. You will discover © 3;Jones & Bartlett Learning, the LLC © Jones & Bartlett Learning, LL correct alleles on each chromosome. that the distances are not strictly additive. Why aren’t they? (c) Which two genes are closest together? NOT FOR SALE OR DISTRIBUT NOT FOR SALE OR DISTRIBUTION ( d) What is the map distance between the two closest genes? 4.8 A Drosophila cross is carried out with a female that is Assume that interference is complete between these two genes. heterozygous for both the y (yellow body) and bb (bobbed (e) If interference is not complete, how will the true map bristles) mutations. Both genes are located in the X chromodistance differ from the value you calculated in part (d)? Briefly some. Among 200 male progeny, there were 49 ­wildtype for explain why this is so. both traits, 51 with & yellow body, 41Learning, with bobbed bristles, © Jones Bartlett LLCand © Jones & Bartlett Learning, LLC

4.7 Construct a genetic map of a chromosome from the

59 mutant both genes. Do these show ­evidence for NOTforFOR SALE OR genes DISTRIBUTION linkage? [Note: The appropriate chi-square test is a test for a 1 : 1 ratio of parental : recombinant gametes.]

4.12 In corn,NOT the genes v (virescent pr (red aleuFOR SALEseedlings), OR DISTRIBUTION rone), and bm (brown midrib) are all on chromosome 5, but not necessarily in the order given. The cross v1 pr bm/v pr1 bm 1 3 v pr bm/v pr bm

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© Jones & Bartlett Learning, LLC NOT FOR SALE OR DISTRIBUTION Concepts in Action: Problems for Solution

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1000 progeny with the following phenotypes: © Jones & produces Bartlett Learning, LLC 1 v pr bm 226 NOT FOR SALE OR DISTRIBUTION 1 1

(c) Suggest a geneticLearning, hypothesis that can explain the data in © Jones & Bartlett LLC panel C. NOT FOR DISTRIBUTION (d) AreSALE the dataOR consistent with your hypothesis?

v pr bm 229 v1 pr bm1 153 v pr1 bm 185 1 v pr1 bm 59 1 v pr © bmJones 71 & Bartlett Learning, LLC v1 pr1 bm1 36 A NOT FOR SALE OR DISTRIBUTION v pr bm 41

(a) Determine the gene order, the recombination frequencies between adjacent genes, the coefficient of coincidence, and the interference. (b) Explain why, in this example, the recombination frequencies © Jones & Bartlett Learning, LLC are not good estimates of map distance.

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Number of each B

Phenotypes observed in population

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Phenotypes Phenotypes observed of parents in progeny

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4.15 The NOT following classesSALE and frequencies of ordered tetrads FOR OR DISTRIBUTION

4.13 The male I-2 in the accompanying pedigree is affected were obtained from the cross a1 b1 3 a b in Neurospora. (Only with Huntington disease, a type of neuromuscular degeneration one member of each pair of spores is shown.) What is the order caused by a rare autosomal dominant mutation HD with comof the genes in relation to the centromere? plete penetrance. The wildtype, nonmutant allele is denoted hd. Spore pair The woman Learning, II-1 is also affected. The RFLP alleles A and a © Jones & Bartlett LLC © Jones & Bartlett Learning, LLC Number of 1–2 3–4 5–6 7–8 asci yielding the bands in the gel are linked to the Huntington locus NOT FOR SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION with a recombination frequency of 10 percent. a1 b1 a1 b1 a b a b 1766 (a) Is the genotype of II-1 HD A/hd a or is it HD a/hd A? a1 b1 a b a1 b1 a b 220 (b) Given the pattern of bands in the gel, what is the a1 b1 a b1 a1 b a b 14 probability that III-1 will be affected? (c) Given the pattern of©bands in the gel, what is the probability Jones & Bartlett Learning, LLC © Jones & Bartlett 4.16 A portion of the linkage map of chromosome 2 in theLearning, LL that III-2 will be affected? tomato is illustrated here. The oblate phenotype is a flattened NOT FOR SALE OR DISTRIBUT NOT FOR SALE OR DISTRIBUTION fruit, the peach phenotype is hairy fruit (like a peach), and comI 2 1 pound influorescence means clustered flowers. II

1

o (oblate)

2

© Jones & Bartlett Learning, LLC III 1 2 NOT FOR SALE OR DISTRIBUTION

ci (compound influorescence)

p (peach)

©15Jones & Bartlett Learning, LLC cM 20 cM NOT FOR SALE OR DISTRIBUTION

Among 1000 gametes produced by a plant of genotype o ci 1/1 1 p, what types of gametes would be expected, and what number would be expected of each? Assume that the chromosome interference across this region is 80 percent but that interference within each region is complete.

A a

© Jones & Bartlett Learning, LLC the molecular variation in© Jones & Bartlett Learning, LLC 4.14 A human geneticist discovers 4.17 The yeast Saccharomyces cerevisiae has unordered tetrads. DNA sequence illustrated in the accompanying diagrams of NOT FOR SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION In a cross carried out to study the linkage relationships among

electrophoresis gels. In the human population as a whole, she three genes, the tetrads in the accompanying table were finds any of four phenotypes, shown in panel A. She believes obtained. The cross was between a strain of genotype 1 b c and that this may be a simple genetic polymorphism with three one of genotype a 1 1. alleles, like the ABO blood groups. There are two alleles that yield DNA fragments of sizes, (F) or slow (S) (a) From these data determine© which, if any,&ofBartlett the genes are ©different Jones & fast Bartlett Learning, LLC Jones Learning, LL migration, and a “null” allele (O) in which the DNA fragment is linked. NOT FOR SALE OR DISTRIBUT NOT FOR SALE OR DISTRIBUTION deleted. The genotypes in panel A would therefore be, from left (b) For any linked genes, determine the map distances. to right, FF or FO, SS or SO, FS, and OO. In the population as a Tetrad Genotypes of Number whole, the putative OO genotype is extremely common, and the type spores in tetrads of tetrads FS genotype is quite rare. To investigate this hypothesis further, 1 a 1 1 a 1 1 1 b c 1 b c 132 the geneticist studies offspring of matings between parents who 2 ©aJones b 1 a & b 1 1 1c Learning, 1 1c 124 © the Jones &FSBartlett Learning, LLC Bartlett LLC have putative genotype (panel B). The types of progeny, 3 a 1 1 a 1 c 1 b 1 1 b c 64 and their numbers, are shown in panel C. NOT FOR SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION 4 a b 1

( a) What result would be expected from the three-allele hypothesis? (b) Are the observed data consistent with this result? Why or why not?

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a b c

1 1 1 1 1c 80

Total

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CHAPTER 4  Gene Linkage and Genetic Mapping

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A small portion of the genetic map of Neurospora crassa © Jones & Bartlett Learning, LLC © Jones 4.18 & Bartlett Learning, LLC chromosome VI is illustrated here. The cys-1 mutation blocks A A MOMENT TO THINK NOT FOR SALE OR DISTRIBUTION NOT FORcysteine SALE OR DISTRIBUTION synthesis, and the pan-2 mutation blocks pantothenic

Answer to Problem: The coincidence is equal to the observed number of double crossovers divided by the expected number, but the observed and expected fre­ quencies (proportions) can be used as well, because in the conversion to frequencies, both numerator and denominator (a) First-division segregation of cys-1 and first-division © Jones & Bartlett Learning, LLC © Jones & Bartlett Learning, LL are divided by the same number. Set r1 5 0.32 and r2 5 segregation of pan-2. NOT FORof SALE NOT FOR SALE OR DISTRIBUTION 0.25. We know that r includes the proportion gametes OR DISTRIBUT 1 (b) First-division segregation of cys-1 and second-division with single recombination in region 1 (call this s1) plus the segregation of pan-2. proportion of gametes with double recombination (call this d   ). (c) Second-division segregation of cys-1 and first-division Hence r1 5 s1 1 d. Similarly, r2 5 s2 1 d, where s2 is the segregation of pan-2. proportion of gametes with single recombination in region 2. (d) Second-division segregation of cys-1 and second-division In the two-point cross between w and r, the observed © Jones Jones &0.45, Bartlett segregation of pan-2.& Bartlett Learning, LLC frequency of © recombination, equals s1Learning, 1 s2 because LLC (e) Parental ditype tetrads. none of the double are detected. Now we can NOT ditype, FORtetratype, SALE and ORnonparental DISTRIBUTION NOT recombinants FOR SALE OR DISTRIBUTION solve for d because r1 1 r2 5 (s1 1 d   ) 1 (s2 1 d   ) 5 s1 1 s2 1 cys-1 pan-2 2d 5 0.57, whereas s1 1 s2 5 0.45. Subtracting, we obtain 2d 5 0.57 2 0.45 5 0.12, or d 5 0.06. This is the “observed” (inferred in this case) frequency of double recombinants. The 7 cM 3 cM expected frequency of double recombinants equals r1 3 r2 5 © Jones & Bartlett Learning, LLC shows the positions of © Jones Bartlett Learning, LLC 0.32 3& 0.25 5 0.08. Thus the coincidence across the region 4.19 The accompanying gel diagram is 0.06/0.08 5 0.75, and the interference equals 1 2 0.75 5 bandsOR associated with the A, a and B, b allele pairs for NOT FORDNA SALE DISTRIBUTION NOT FOR SALE OR DISTRIBUTION 0.25. In other words, there is about a 25% deficit in double two linked genes. On the left are the phenotypes of the parents, recombinants from the frequency that would be expected and on the right are the phenotypes of the progeny and the with independence.

acid synthesis. Assuming complete chromosome interference, determine the expected frequencies of the following types of asci in a cross of cys-1 pan-2 3 CYS-1 PAN-2.

number of each observed. Is the linkage phase of A and B in the doubly heterozygous parent coupling or repulsion? What is the frequency of recombination genes? ©between Jonesthese & Bartlett Learning, Parents

LLC NOT FOR SALE OR DISTRIBUTION Progeny 155

44

36

165

© Jones & Bartlett Learning, LL NOT FOR SALE OR DISTRIBUT GENETICS on the web GeNETics on the Web will introduce you to some of the most important sites for finding genetics information on the Internet. To explore these sites, visit the Jones & Bartlett © Jones & Bartlett Learning, LLC companion site to accompany Essential Genetics: A Genomic NOT OR DISTRIBUTION Perspective, Sixth FOR Edition,SALE at http://biology.jbpub.com/ Hartl/EssentialGenetics.

A a

© Jones & Bartlett Learning, LLC B NOTb FOR SALE OR DISTRIBUTION 4.20 Janet is performing a test cross to determine the linkage relationships between three Drosophila genes, dpy, unc and dor.

There you will find a chapter-by-chapter list of highlighted keywords. When you select one of the keywords, you will be linked to a Web site containing information related to that Jones & Bartlett Learning, LLC keyword.

HerBartlett entire gradeLearning, depends on getting © Jones & LLCthis right! She testcrosses © heterozygous for the recessive alleles: NOT FORfemales SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUTION dpy (dumpy body) / dpy1 (normal body) unc (uncoordinated) / unc1 (coordinated) dor (deep orange eye) / dor1 (red eye). The cross yields the following results:

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normal body, red eye, coordinated NOT FOR SALE normal body, red eye, uncoordinated normal body, deep orange eye, uncoordinated dumpy body, red eye, coordinated dumpy body, deep orange eye, coordinated dumpy body, deep orange eye, uncoordinated

348 96 110 306 65

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(a) What is the genotype of theOR F1 heterozygous female? NOT FOR SALE DISTRIBUTION (b) Construct a map of the region indicating the order of the genes and the distances (in map units) between them. (c) What is the interference in this region?

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© Jones & Bartlett Learning, LL NOT FOR SALE OR DISTRIBUT

© Jones & Bartlett Learning, LLC NOT FOR SALE OR DISTRIBUTION Concepts in Action: Problems for Solution

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