LINKAGE AND GENETIC MAPPING IN EUKARYOTES

genetic linkage map Based on recombination data What we will study now

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

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 5-4

What is crossing over? When non-sister chromatids of homologous chromosomes exchange DNA segments

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

(a) Without crossing over, linked alleles segregate together. Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. B B b b B b a a A a A A 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

Evidence for Linkage 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 in? You guessed it 5-11 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

His experiment involved three traits: Body color, Eye color, Wing length 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

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 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

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

Creighton and McClintock Experiment They demonstrated physical evidence of cross-overs. 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

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 5-42 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

A simplified genetic linkage map of Drosophila melanogaster 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

Genetic linkage map Based on recombination data to determine the relative position of genes on the chromosome

Physical Maps Use nucleotide sequences to map genes Will talk about this later

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

The first genetic map was constructed in 1911 by Alfred Sturtevant He was an undergraduate who worked in Morgan’s lab. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display 5-49

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.”

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

As the percentage of recombinant offspring approaches a value of 50 % 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 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.

If you need help, watch the Boseman video below. https://www.youtube.com/watch?v=TU44tR0hJ8A

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

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

Trihybrid Crosses 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. Order of genes not important here. 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 Step 3: Collect data for the F2 generation Phenotype Number of Observed Offspring Gray body, red eyes, normal wings + + + 411 parental Gray body, red eyes, vestigial wings + + vg 61 pr/vg Gray body, purple eyes, normal wings + pr + 2 b/pr and pr/vg Gray body, purple eyes, vestigial wings + pr vg 30 b/pr Black body, red eyes, normal wings b + + 28 b/pr Black body, red eyes, vestigial wings B + vg 1 b/pr and pr/vg Black body, purple eyes, normal wings B pr + 60 pr/vg Black body, purple eyes, vestigial wings B pr vg 412 parental 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 Notice pr is no longer with the b and vg And pr+ is not with the b+ and vg+. 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

Which are the double cross-overs? The ones with the least amount.

Step 4: Calculate the map distance between pairs of genes Number of recombs between pr and vg: 61 + 60+ 2 + 1 = 124 Number of recombs between b and pr: 30 + 28 + 2 + 1 = 61 Number of recombs between b and vg, all but double cross-overs: 61 + 60 + 30 + 28 = 178 5-65 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

Map Distance pr/vg = 124/1005 x 100 = 12.3 b/pr = 61/1005 x 100 = 6 b and vg = 179/1005 x 100 = 17.8 _____6____________12.3____________ b pr vg The distance between b and vg was found to be 17.8. The actual distance is 18.3 mu.

Interference The slightly smaller lower value was a small underestimate because we did not consider the double crossovers in the calculation between b and vg. The 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.

Need help with trihybrid linkage? You do not need to compute interference. https://www.ndsu.edu/pubweb/~mcclean/plsc431/linkage/linkage3.htm

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 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 In crosses of tan and black Neurospora cultures, the spores appear tan or black in a certain order. All black spores or all tan spores indicate no hybridization. 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 No hybridization 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.

Non-crossovers

Cross-overs

Non cross-overs Non cross-overs Cross-overs

(1/2) (Number of cross-over asci) Map distance = X 100 To calculate this distance, the experimenter must count the number of cross-over asci, as well as the total number of asci In cross-over asci, only half of the spores are actually the product of a crossover Therefore (1/2) (Number of cross-over asci) Map distance = X 100 Total number of asci 5-89 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

For the last questions on the worksheet about the fungi. 11. The distance between the gene and centromere is ½(number of recombs)/total asci counted. ½ (22+21+21+23)/993 = 4.4 mu 12. (a) Six types: pro-1 - I just listed as “1” 1111++++ ++++1111 ++11++11 11++11++ ++1111++ 11++++11 (b) Since you have 9.8 map units, remember that number was derived by dividing the percent recombs by ½, so there are actually 19.6% recombs or a total of 196 asci that are recombs. Divide that by four types and you get an expected number of 49 of each type of recomb and 402 of each parent types. 13. Ascomycetes produce spores in a sac in an ordered tetrad so you can see the products of meiosis and subsequently mitosis of those meiotic spores.

Linkage on human chromosomes Hampered by the inability to perform desired crosses and small number of progeny in most human families Geneticists mostly use pedigrees which are often incomplete.

Nail-Patella Syndrome One of the first documented demonstrations of linkage in humans Linkage between this syndrome and ABO blood types

Pedigree of Nail-Patella Syndrome

More on physical maps later……