PowerPoint Presentation Materials to accompany Genetics: Analysis and Principles Robert J. Brooker CHAPTER 5 LINKAGE AND GENETIC MAPPING IN EUKARYOTES Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

5.1 LINKAGE AND CROSSING OVER In eukaryotic species, each linear chromosome contains a long piece of DNA A typical chromosome contains many hundred or even a few thousand different genes The term linkage has two related meanings 1. Two or more genes can be located on the same chromosome 2. Genes that are close together tend to be transmitted as a unit 5-3 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

Chromosomes are called linkage groups They contain a group of genes that are linked together The number of linkage groups is the number of types of chromosomes of the species For example, in humans 22 autosomal linkage groups An X chromosome linkage group A Y chromosome linkage group Genes that are far apart on the same chromosome may independently assort from each other This is due to crossing-over 5-4 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

Crossing Over May Produce Recombinant Phenotypes In diploid eukaryotic species, linkage can be altered during meiosis as a result of crossing over Crossing over Occurs during prophase I of meiosis at the bivalent stage Non-sister chromatids of homologous chromosomes exchange DNA segments 5-5 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

Copyright © The McGraw-Hill Companies, Inc Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. b b B b B B a a A a A A Fig. 5.1(TE Art) Diploid cell after chromosome replication Diploid cell after chromosome replication Meiosis Meiosis B B B B A A A a b b b b a a a A Possible haploid cells Possible haploid cells (a) Without crossing over, linked alleles segregate together. (b) Crossing over can reassort linked alleles.

These haploid cells contain a combination of alleles NOT found in the original chromosomes This new combination of alleles is a result of genetic recombination These are termed parental or non-recombinant cells These are termed nonparental or recombinant cells Figure 5.1 5-7

Bateson and Punnett Discovered Two Traits That Did Not Assort Independently In 1905, William Bateson and Reginald Punnett conducted a cross in sweet pea involving two different traits Flower color and pollen shape This is a dihybrid cross that is expected to yield a 9:3:3:1 phenotypic ratio in the F2 generation However, Bateson and Punnett obtained surprising results 5-8 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

Figure 5.2 A much greater proportion of the two types found in the parental generation 5-9

Morgan Provided Evidence for the Linkage of Several X-linked Genes The first direct evidence of linkage came from studies of Thomas Hunt Morgan Morgan investigated several traits that followed an X-linked pattern of inheritance Figure 5.3 illustrates an experiment involving three traits Body color Eye color Wing length 5-11 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

yy ww mm y+y w+w m+m F1 generation x y w m Y y+ w+ m+ Y F1 generation contains wild-type females and yellow-bodied, white-eyed, miniature-winged males.

Morgan’s explanation: All three genes are located on the X chromosome P Males P Females Morgan observed a much higher proportion of the combinations of traits found in the parental generation Morgan’s explanation: All three genes are located on the X chromosome Therefore, they tend to be transmitted together as a unit 5-13 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

Morgan Provided Evidence for the Linkage of Several X-linked Genes 1. Why did the F2 generation have a significant number of nonparental combinations? 2. Why was there a quantitative difference between the various nonparental combinations? 5-14 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

White eyes, miniature wings 716 Red eyes, miniature wings 401 Let’s reorganize Morgan’s data by considering the pairs of genes separately Gray body, red eyes 1,159 Yellow body, white eyes 1,017 Gray body, white eyes 17 Yellow body, red eyes 12 Total 2,205 But this nonparental combination was rare Red eyes, normal wings 770 White eyes, miniature wings 716 Red eyes, miniature wings 401 White eyes, normal wings 318 Total 2,205 It was fairly common to get this nonparental combination 5-15 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

Morgan made three important hypotheses to explain his results 1. The genes for body color, eye color and wing length are all located on the X-chromosome They tend to be inherited together 2. Due to crossing over, the homologous X chromosomes (in the female) can exchange pieces of chromosomes This created new combination of alleles 3. The likelihood of crossing over depends on the distance between the two genes Crossing over is more likely to occur between two genes that are far apart from each other 5-17 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

Figure 5-5 Copyright © 2006 Pearson Prentice Hall, Inc. Figure 5-5 Two examples of a single crossover between two nonsister chromatids and the gametes subsequently produced. In (a) the exchange does not alter the linkage arrangement between the alleles of the two genes, only parental gametes are formed, and the exchange goes undetected. In (b) the exchange separates the alleles, resulting in recombinant gametes, which are detectable. Figure 5-5 Copyright © 2006 Pearson Prentice Hall, Inc.

Figure 5.4 These parental phenotypes are the most common offspring These recombinant offspring are not uncommon 5-18 because the genes are far apart

5-19 Figure 5.4 These recombinant offspring are fairly uncommon because the genes are very close together These recombinant offspring are very unlikely 1 out of 2,205 5-19

Chi Square Analysis This method is frequently used to determine if the outcome of a dihybrid cross is consistent with linkage or independent assortment 5-20 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

Creighton and McClintock Experiment wx Normal chromosome 9 Parental chromosomes c Wx Abnormal chromosome 9 Crossing over c wx Knob Interchanged piece from chromosome 8 Nonparental chromosomes (a) Normal and abnormal chromosome 9 C Wx C = Colored c = colorless Wx = Starchy endosperm wx = waxy endosperm (b) Crossing over between normal and abnormal chromosome 9

Figure 5.6 5-30

Interpreting the Data Parent A Parent B C wx (nonrecombinant) C Wx (recombinant) c wx (recombinant) c Wx c wx By combining these gametes into a Punnett square, the following types of offspring can be produced 5-32 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

So let’s start by considering the unambiguous phenotypes Ambiguous phenotypes that could be produced whether or not recombination occurred in parent A So let’s start by considering the unambiguous phenotypes 5-33 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

The colored, waxy phenotype (Cc wxwx) can occur only if Recombination did not occur in parent A AND Parent A passed the knobbed, translocated chromosome to its offspring This was the case, as shown in the data table below 5-34 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

The colorless, waxy phenotype (cc wxwx) can occur only if Recombination did occur in parent A AND Parent A passed a chromosome 9 that had a translocation but was knobless This was the case, as shown in the data table below 5-35 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

The Data 5-31 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

As stated by Creighton and McClintock: These observations were consistent with the idea that a cross over occurred between the C and wx genes As stated by Creighton and McClintock: “Pairing chromosomes, heteromorphic in two regions, have been shown to exchange parts at the same time they exchange genes assigned to these regions.” 5-36

5.2 GENETIC MAPPING IN PLANTS AND ANIMALS Genetic mapping is also known as gene mapping or chromosome mapping Its purpose is to determine the linear order of linked genes along the same chromosome Figure 5.8 illustrates a simplified genetic linkage map of Drosophila melanogaster 5-42 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

Each gene has its own unique locus at a particular site within a chromosome Figure 5.8 5-43 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

Experimentally, the percentage of recombinant offspring is correlated with the distance between the two genes If the genes are far apart  many recombinant offspring If the genes are close  very few recombinant offspring Map distance = Number of recombinant offspring Total number of offspring X 100 The units of distance are called map units (mu) They are also referred to as centiMorgans (cM) One map unit is equivalent to 1% recombination frequency 5-45 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

Chromosomes are the product of a crossover during meiosis in the heterozygous parent Recombinant offspring are fewer in number than nonrecombinant offspring Figure 5.9 5-47

Number of recombinant offspring The data at the bottom of Figure 5.9 can be used to estimate the distance between the two genes Map distance = Number of recombinant offspring Total number of offspring X 100 76 + 75 = X 100 542 + 537 + 76 + 75 = 12.3 map units 5-48 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

Alfred Sturtevant’s Experiment The first genetic map was constructed in 1911 by Alfred Sturtevant He was an undergraduate who spent time in the laboratory of Thomas Hunt Morgan Sturtevant wrote: “In conversation with Morgan … I suddenly realized that the variations in the length of linkage, already attributed by Morgan to differences in the spatial orientation of the genes, offered the possibility of determining sequences [of different genes] in the linear dimension of the chromosome. I went home and spent most of the night (to the neglect of my undergraduate homework) in producing the first chromosome map, which included the sex-linked genes, y, w, v, m, and r, in the order and approximately the relative spacing that they still appear on the standard maps.” 5-49 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

Figure 5.10 5-52

The Data 5-53 Alleles Concerned Number Recombinant/ Total Number Percent Recombinant Offspring y and w/w-e 214/21,736 1.0 y and v 1,464/4,551 32.2 y and r 115/324 35.5 y and m 260/693 37.5 w/w-e and v 471/1,584 29.7 w/w-e and r 2,062/6,116 33.7 w/w-e and m 406/898 45.2 v and r 17/573 3.0 v and m 109/405 26.9 5-53 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

Interpreting the Data In some dihybrid crosses, the percentage of nonparental (recombinant) offspring was rather low For example, there’s only 1% recombinant offspring in the crosses involving the y and w or w-e alleles This suggests that these two genes are very close together Other dihybrid crosses showed a higher percentage of nonparental offspring For example, crosses between the v and m alleles produced 26.9% recombinant offspring This suggests that these two genes are farther apart 5-54 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

Sturtevant assumed that the map distances would be more accurate among genes that are closely linked Therefore, his map is based on the following distances y – w (1.0), w – v (29.7), v – r (3.0) and v – m (26.9) Sturtevant also considered map distances amongst gene pairs to deduce the order of genes Percentage of crossovers between w and r was 33.7 Percentage of crossovers between w and v was 29.7 Percentage of crossovers between v and r was 3.0 Therefore, the gene order is w – v – r Where v is closer to r than it is to w 5-55 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

Sturtevant began at the y gene and mapped the genes from left to right Sturtevant collectively considered all these data and proposed the following genetic map Sturtevant began at the y gene and mapped the genes from left to right 5-56 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

A close look at Sturtevant’s data reveals two points that do not agree very well with his genetic map The y and m dihybrid cross yielded 37.5% recombinants But the map distance is 57.6 The w and m dihybrid cross yielded 45.2% recombinants But the map distance is 56.6 So what’s up? As the percentage of recombinant offspring approaches a value of 50 % This value becomes a progressively more inaccurate measure of map distance Refer to Figure 5.11 5-57 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

When the distance between two genes is large Figure 5.11 When the distance between two genes is large The likelihood of multiple crossovers increases This causes the observed number of recombinant offspring to underestimate the distance between the two genes 5-58 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

Figure 5-12a Copyright © 2006 Pearson Prentice Hall, Inc. Figure 5-12a Three types of double exchanges that may occur between two genes. Two of them, (b) and (c), involve more than two chromatids. In each case, the detectable recombinant chromatids are bracketed. Figure 5-12a Copyright © 2006 Pearson Prentice Hall, Inc.

Trihybrid Crosses 5-59 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

Trihybrid Crosses Data from trihybrid crosses can also yield information about map distance and gene order The following experiment outlines a common strategy for using trihybrid crosses to map genes In this example, we will consider fruit flies that differ in body color, eye color and wing shape b = black body color b+ = gray body color pr = purple eye color pr+ = red eye color vg = vestigial wings vg+ = normal wings 5-59 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

Step 1: Cross two true-breeding strains that differ with regard to three alleles. Male is homozygous wildtype for all three traits Female is mutant for all three traits The goal in this step is to obtain aF1 individuals that are heterozygous for all three genes 5-60 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

Step 2: Perform a testcross by mating F1 female heterozygotes to male flies that are homozygous recessive for all three alleles During gametogenesis in the heterozygous female F1 flies, crossovers may produce new combinations of the 3 alleles 5-61 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

Number of Observed Offspring (males and females) Step 3: Collect data for the F2 generation Phenotype Number of Observed Offspring (males and females) Gray body, red eyes, normal wings 411 Gray body, red eyes, vestigial wings 61 Gray body, purple eyes, normal wings 2 Gray body, purple eyes, vestigial wings 30 Black body, red eyes, normal wings 28 Black body, red eyes, vestigial wings 1 Black body, purple eyes, normal wings 60 Black body, purple eyes, vestigial wings 412 5-62 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

The three genes exist as two alleles each Therefore, there are 23 = 8 possible combinations of F2 offspring If the genes assorted independently, all eight combinations would occur in equal proportions It is obvious that they are far from equal In the offspring of crosses involving linked genes, Parental phenotypes occur most frequently Double crossover phenotypes occur least frequently Single crossover phenotypes occur with “intermediate” frequency 5-63 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

The combination of traits in the double crossover tells us which gene is in the middle A double crossover separates the gene in the middle from the other two genes at either end In the double crossover categories, the recessive purple eye color is separated from the other two recessive alleles Thus, the gene for eye color lies between the genes for body color and wing shape 5-64 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

Step 4: Calculate the map distance between pairs of genes To do this, one strategy is to regroup the data according to pairs of genes From the parental generation, we know that the dominant alleles are linked, as are the recessive alleles This allows us to group pairs of genes into parental and nonparental combinations Parentals have a pair of dominant or a pair of recessive alleles Nonparentals have one dominant and one recessive allele The regrouped data will allow us to calculate the map distance between the two genes 5-65 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

The map distance between body color and eye color is Map distance = 61 Parental offspring Total Nonparental Offspring Gray body, red eyes (411 + 61) 472 Gray body, purple eyes (30 + 2) 32 Black body, purple eyes (412 + 60) Black body, red eyes (28 + 1) 29 944 61 The map distance between body color and eye color is Map distance = 61 X 100 = 6.1 map units 944 + 61 5-66

The map distance between body color and wing shape is Map distance = Parental offspring Total Nonparental Offspring Gray body, normal wings (411 + 2) 413 Gray body, vestigial wings (30 + 61) 91 Black body, vestigial wings (412 + 1) Black body, normal wings (28 + 60) 88 826 179 The map distance between body color and wing shape is Map distance = 179 826 + 179 X 100 = 17.8 map units 5-67

The map distance between eye color and wing shape is Map distance = Parental offspring Total Nonparental Offspring Red eyes, normal wings (411 + 28) 439 Red eyes, vestigial wings (61 + 1) 62 Purple eyes, vestigial wings (412 + 30) 442 Purple eyes, normal wings (60 + 2) 881 124 The map distance between eye color and wing shape is Map distance = 124 881 + 124 X 100 = 12.3 map units 5-68

Step 5: Construct the map Based on the map unit calculation the body color and wing shape genes are farthest apart The eye color gene is in the middle The data is also consistent with the map being drawn as vg – pr – b (from left to right) In detailed genetic maps, the locations of genes are mapped relative to the centromere 5-69 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

An alternative method does not require this manipulation To calculate map distance, we have gone through a method that involved the separation of data into pairs of genes (see step 4) An alternative method does not require this manipulation Rather, the trihybrid data is used directly This method is described next 5-70 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

Number of Observed Offspring Phenotype Number of Observed Offspring Gray body, purple eyes, vestigial wings 30 Black body, red eyes, normal wings 28 Gray body, red eyes, vestigial wings 61 Black body, purple eyes, normal wings 60 Gray body, purple eyes, normal wings 2 Black body, red eyes, vestigial wings 1 30 + 28 1,005 Single crossover between b and pr = 0.058 61 + 60 1,005 Single crossover between pr and vg = 0.120 1 + 2 1,005 Double crossover, between b and pr, and between pr and vg = 0.003 5-71 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

To calculate the distance between b and pr To determine the map distance between the genes, we need to consider both single and double crossovers To calculate the distance between b and pr Map distance = (0.058 + 0.003) X 100 = 6.1 mu To calculate the distance between pr and vg Map distance = (0.120 + 0.003) X 100 = 12.3 mu To calculate the distance between b and vg The double crossover frequency needs to be multiplied by two Because both crossovers are occurring between b and vg Map distance = (0.058 + 0.120 + 2[0.003]) X 100 = 18.4 mu 5-72 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

Alternatively, the distance between b and vg can be obtained by simply adding the map distances between b and pr, and between pr and vg Map distance = 6.1 + 12.3 = 18.4 mu Note that in the first method (grouping in pairs), the distance between b and vg was found to be 17.8 mu. This slightly lower value was a small underestimate because the first method does not consider the double crossovers in the calculation 5-73 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

Figure 5-8 Copyright © 2006 Pearson Prentice Hall, Inc. Figure 5-8 A three-point mapping cross involving the yellow (y or ), white (w or ), and echinus (ec or ) genes in Drosophila melanogaster. NCO, SCO, and DCO refer to noncrossover, single-crossover, and double-crossover groups, respectively. Because of the complexity of this and several of the ensuing figures, centromeres have not been included on the chromosomes, and only two nonsister chromatids are initially shown in the left-hand column. Figure 5-8 Copyright © 2006 Pearson Prentice Hall, Inc.

Figure 5-10 Copyright © 2006 Pearson Prentice Hall, Inc. Figure 5-10 (a) Some possible allele arrangements and gene sequences in a heterozygous female. The data from a three-point mapping cross, depicted in (b), where the female is testcrossed, provide the basis for determining which combination of arrangement and sequence is correct. [See Figure 5–11(d).] Figure 5-10 Copyright © 2006 Pearson Prentice Hall, Inc.

Figure 5-11 Copyright © 2006 Pearson Prentice Hall, Inc. Figure 5-11 Producing a map of the three genes in the cross in Figure 5–10, where neither the arrangement of alleles nor the sequence of genes in the heterozygous female parent is known. Figure 5-11 Copyright © 2006 Pearson Prentice Hall, Inc.

Interference P (double crossover) = The product rule allows us to predict the likelihood of a double crossover from the individual probabilities of each single crossover P (double crossover) = P (single crossover between b and pr) P (single crossover between pr and vg) X = 0.061 X 0.123 = 0.0075 Based on a total of 1,005 offspring The expected number of double crossover offspring is = 1,005 X 0.0075 = 7.5 5-74 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

Interference Therefore, we would expect seven or eight offspring to be produced as a result of a double crossover However, the observed number was only three! Two with gray bodies, purple eyes, and normal sings One with black body, red eyes, and vestigial wings This lower-than-expected value is due to a common genetic phenomenon, termed positive interference The first crossover decreases the probability that a second crossover will occur nearby 5-75 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

Interference (I) is expressed as I = 1 – C where C is the coefficient of coincidence Observed number of double crossovers Expected number of double crossovers C = 3 7.5 C = = 0.40 I = 1 – C = 1 – 0.4 = 0.6 or 60% This means that 60% of the expected number of crossovers did not occur 5-76 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

Since I is positive, this interference is positive interference Rarely, the outcome of a testcross yields a negative value for interference This suggests that a first crossover enhances the rate of a second crossover The molecular mechanisms that cause interference are not completely understood However, most organisms regulate the number of crossovers so that very few occur per chromosome 5-77 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

5.3 GENETIC MAPPING IN HAPLOID EUKARYOTES Much of our earliest understanding of genetic recombination came from the genetic analyses of fungi Fungi may be unicellular or multicellular organisms They are typically haploid (1n) They reproduce asexually and, in many cases, sexually The sac fungi (ascomycetes) have been particularly useful to geneticists because of their unique style of sexual reproduction 5-78 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

5-79 Figure 5.12 Meiosis produces four haploid cells, termed spores These are enclosed in a sac termed an ascus Figure 5.12 5-79

The cells of a tetrad or octad are contained within a sac In other words, the products of a single meiotic division are contained within one sac This is a key feature that dramatically differs from sexual reproduction in animals and plants In animals, for example Oogenesis only produces a single functional egg Spermatogenesis produces sperm that are mixed with millions of other sperm Using a microscope, researchers can dissect asci and study the traits of each haploid spore 5-80 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

Types of Tetrads or Octads The arrangement of spores within an ascus varies from species to species Unordered tetrads or octads Ascus provides enough space for the spores to randomly mix together Ordered tetrads or octads Ascus is very tight, thereby preventing spores from randomly moving around 5-81 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

5-82 Tight ascus prevents mixing of spores Ascus provides space for spores to randomly mix together Mold Yeast Unicellular alga Figure 5.13 5-82 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

Ordered Tetrad Analysis Ordered tetrads or octads have the following key feature The position and order of spores within the ascus is determined by the divisions of meiosis and mitosis This idea is schematically shown in Figure 5.13b The example depicts ordered octad formation in Neurospora crassa Spores that carry the A allele show orange pigmentation Spores that carry the a (albino) allele are white 5-83 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

Pairs of daughter cells are located next to each other All eight cells are arranged in a linear, ordered fashion Pairs of daughter cells are located next to each other Figure 5.13 5-84 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

020 020. A linear, cylindrical, eight-spored ascus of Sordaria fimicola. Note ascosporogenesis (sexual reproduction in the Ascomycotina) involves the formation of ascospores, typically eight, by "free cell formation" within an ascus.

The genetic content of spores in ordered tetrads can be determined This allows experimenters to map the distance between a single gene and the centromere The logic of this mapping technique is based on the following features of meiosis Centromeres of homologous chromosomes separate during meiosis I Centromeres of sister chromatids separate during meiosis II 5-85 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

5-86 Figure 5.14 (a) No crossing over Octad contains a linear arrangement of 4 haploid cells with the A allele which are adjacent to 4 with the a allele This 4:4 arrangement of spores within the ascus is termed a first-division segregation (FDS) or an M1 pattern Because the A and a alleles have segregated from each other after meiosis I Figure 5.14 (a) No crossing over 5-86 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

5-87 Figure 5.14 (b) Single crossing over These arrangement of spores are termed a second-division segregation (SDS) or M2 patterns The A and a alleles do not segregate until meiosis II Figure 5.14 (b) Single crossing over 5-87

The percentage of M2 asci can be used to calculate the map distance between the centromere and the gene of interest 5-88 Figure 5.15

(1/2) (Number of SDS asci) X 100 Map distance = Total number of asci Therefore the chances of getting a 2:2:2:2 or 2:4:2 pattern depend on the distance between the gene of interest and the centromere To calculate this distance, the experimenter must count the number of SDS asci, as well as the total number of asci In SDS asci, only half of the spores are actually the product of a crossover Therefore (1/2) (Number of SDS asci) X 100 Map distance = Total number of asci 5-89 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

Unordered Tetrad Analysis Unicellular algae Figure 5-18 The life cycle of Chlamydomonas. The diploid zygote (in the center) undergoes meiosis, producing [[p]][[p]] or [[p]][[p]] haploid cells that undergo mitosis, yielding vegetative colonies. Unfavorable conditions stimulate them to form isogametes, which fuse in fertilization, producing a zygote that repeats the cycle. Vegetative colonies are illustrated photographically. Figure 5-18 Copyright © 2006 Pearson Prentice Hall, Inc.

Unordered Tetrad Analysis Unordered tetrads contain randomly arranged groups of spores An experimenter can do a dihybrid cross and then determine the phenotypes of the spores Such an analysis can determine if two genes are linked or assort independently It can also be used to compute distance between two linked genes 5-90 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

Unordered Tetrad Analysis Consider a diploid yeast zygote with the genotype ura+ura-2 arg+arg-3 ura+ and arg+ = Normal alleles required for uracil and arginine biosynthesis, respectively ura-2 and arg-3 = Defective alleles Result in strains that require uracil and arginine in their growth medium Figure 5.16 illustrates the assortment of the two genes in the unordered tetrad 5-91 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

5-92 Figure 5.16 PD ascus: contains 100% parental cells T ascus: contains 50% parental cells and 50% recombinant cells NPD ascus: contains 100% recombinant cells 5-92

If the two genes assort independently The number of asci with a parental ditype is expected to equal the number with a nonparental ditype Thus, 50% recombinant spores are produced If the two genes are linked The type of crossover between them determines what type of ascus is produced No crossovers yield the parental ditype Single crossovers produce the tetratype Double crossovers can yield any of the three types The actual type produced depends on the combination of chromatids that are involved 5-93 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

Figure 5.17 5-94 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

Figure 5.17 5-95

A more precise way to calculate map distance As in conventional mapping, the map distance is calculated as the % of offspring that carry recombinant chromosomes NPD + (1/2) (T) Map distance = X 100 Total number of asci This calculation is fairly reliable over a short distance However, over long distances it is not Because it does not adequately account for double crossovers A more precise way to calculate map distance Single crossover tetrads + (2) (Double crossover tetrads) Map distance = X 0.5 X 100 Total number of asci Crossover tetrads also contain 50% nonrecombinant chromosomes 5-96 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

So let’s take another look at Figure 5.17 For the equation to be useful, it needs to be related to the number of various types obtained by experimentation So let’s take another look at Figure 5.17 The parental ditype (PD) and tetratype (T) are ambiguous They can each be derived in two different ways The nonparental ditype (NPD), however, is unambiguous It can only be produced from a double crossover (DCO) 1/4 of all the double crossovers are nonparental ditypes Therefore, DCO = 4 X NPD But what about single crossovers (SCO)? Notice that T asci can result from SCO or DCO Since there are two kinds of T that are due to DCO The actual number of T arising from DCO is 2NPD So, T = SCO + 2NPD Therefore, SCO = T – 2NPD 5-97 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

Now we have accurate measures of both SCO and DCO SCO = T – 2NPD and DCO = 4NPD So, let’s substitute these values into our previous equation Single crossover tetrads + (2) (Double crossover tetrads) Total number of asci X 0.5 X 100 Map distance = (T – 2NPD) + (2) (4NPD) Map distance = Total number of asci X 0.5 X 100 T + 6NPD Map distance = Total number of asci X 0.5 X 100 A more accurate measure of map distance because the equation considers both single- and double-crossovers 5-98 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display