Linkage – Part II PDF

LINKAGE – Part II CHAPTER 6 Inheritance of Recombinant Offspring 2 how were they able to connect the feact...

LINKAGE – Part II CHAPTER 6 Inheritance of Recombinant Offspring 2 how were they able to connect the feact Barbra w transposable elements Harriet Creighton and Barbara McClintock, as well as Curt Stern performed crosses involving two linked genes to obtain direct evidence that genetic recombination is due to crossing over They wanted to see if there was a correlation between the occurrence of recombinant offspring and microscopically observable exchanges in segments of homologous chromosomes We will consider Curt Stern’s experiments next Stern’s Experiment Showing Correlation between Crossing Over and Recombination 1 3 shorter bc it has deletion this has a ____ shape color x chromosomes u (a) X chromosomes found in Stern’s flies and the result of them crossing over looking at color and shape of the eyes u Access the text alternative for slide images.  Stern’s Experiment Showing Correlation between Crossing Over and Recombination 2 4 y chromosome w no contribution parental like punnet square but more like a 2x4 parental prescence of the recombinent showed where crossing over occurs recombinant recombinant u (b) The results of Stern’s crosses u Access the text alternative for slide images. 6.3 Genetic Mapping in Plants and Animals 5 Genetic mapping (also known as gene mapping or chromosome mapping) is performed to determine the linear order and distance of separation of linked genes along the same chromosome Each gene has its own unique locus, the site where the gene is found within a particular chromosome if they know gene cauusing disease will give u a number and the minutes talking abt the distance Simplified Genetic Linkage Map of Drosophila melanogaster 6 WT u Access the text alternative for slide images. Uses of Genetic Maps 1 7 organization genes on chromosomes and if map then can 1. They allow us to understand the overall complexity and genetic organization of a particular species see if on same or diff and the bc can pin point then can go ahead and use primers, PCR, peice of distance 2. They can help molecular geneticists to clone genes genome isolate seq and make new primers and chromosome ____ (blocking)? gene can take distribution and construct 3. They improve our understanding of the evolutionary relationships among different species phylogenetic tree and show evolution of prot gene for disease and causing therapeutics for replaced mutated for WT chrispercasp syst etc 4. They can be used to diagnose, and perhaps, someday to treat inherited human diseases direction to 1 step closer to finding solution some disease we always study and only get to 1/2 way pt. and list of not studied v long and we still don't know musch; majrity studied r the ones affecting large group of ppl 5. They can help in predicting the likelihood that a couple will produce children with certain inherited diseases genetic counselors ^^ predict likelyhood and nav the results if have children that will have genetic diseases 6. They provide helpful information for improving agriculturally important strains through selective breeding programs tomatos diff hybrids not all tasty but some rly large some small seeds some longer shelf life etc hybrid varieties and come from which gene to change; non GMO and organic together hard to get; usually privedged to have these options and w advances the prices r varied; if anything is going under the ground pref to get organic or close to >> strawberries usually grown w pesticides and small Genetic Maps Allow Us to Estimate the Relative Distances Between Linked Genes 1 8 classical genetics; btwn genes and likely hood of cross over ___ u The relative distance between linked genes is based on the likelihood that a crossover will occur between them u Experimentally, the percentage of recombinant offspring is correlated with the distance between the two genes amt recombinant is prop to distance Genes that are far apart result in many recombinant offspring Close genes result in very few recombinant offspring Genetic Maps Allow Us to Estimate the Relative Distances Between Linked Genes 2 9 Number of recombinant offspring Map distance = ´ 100 Total number of offspring u The units of distance are called map units (mu); they are also referred to as centiMorgans (cM) u One mu is equivalent to 1% recombination frequency (RF) if distance is 10% then know its 10 mu units Testcross and Genetic Linkage Map 10 u Genetic mapping experiments are typically accomplished by a conducting a testcross, where an individual that is heterozygous for two or more genes is crossed to one that is homozygous recessive for the same genes u Example: testcross with two linked genes (s and e) affecting bristle length and body color in fruit flies s = short bristles and s+ = normal bristles e = ebony body color and e+ = gray body color Testcross to Distinguish 11 Recombinant and Nonrecombinant Offspring 1 p types higher in number u Access the text alternative for slide images. Testcross to Distinguish Recombinant and Nonrecombinant Offspring 2 12 u Testcross data can be used to estimate the distance between the two genes total # offspri recombinants u Map distance = ((75 + 76) ∕ (542 + 537 + 76 + 75 )) × 100 = 12.3 map units u The s and e genes are 12.3 map units apart from each other along the same chromosome on same chromos  Recombination and Map Distance 13 Multiple crossovers set a quantitative limit on measurable recombination frequencies as the physical distance increases A testcross is expected to yield a maximum of only 50% recombinant offspring w multiple cross the error rises bc the phys distance is large is best to consider when looking at test cross and if we get more than 50% we look for another gene u Access the text alternative for slide images. Three-Factor Crosses 14 example u Data from three-factor crosses can also yield additional information about map distance and gene order u The following experiment outlines a common strategy for using three-factor crosses to map genes u In this example, fruit flies that differ in body color, eye color and wing shape are looking at these 3 genes considered b = black body color b+ = gray body color pr = purple eye color pr+ = red eye color vg = vestigial wings vg+= long wings Step 1: Cross two true-breeding strains that differ with regard to three alleles 15 ebony bod purple eyes vestigial wings grey bod F red eyes long wings M u Access the text alternative for slide images. Step 2: Perform testcross by mating female F1 heterozygotes to male flies that are homozygous recessive for all three alleles (bb prpr vgvg). heterozyg w dom traits grey>ebony red>purpl long>vestigal ebony purple vestigal u Access the text alternative for slide images. Step 3: Collect Data for the F2 Generation 17 u Data from a Three-Factor Cross (see step 2) Phenotype Number of Observed Chromosome Inherited from Offspring (Males and F1 Female Females) Gray body, red eyes, long wings 411 An illustration shows 8 chromosomes inherited from an F1 female. parental both w grey red so add and get 472 Gray body, red eyes, vestigial wings 61 An illustration shows 8 chromosomes inherited from an F1 female. SCO Gray body, purple eyes, long wings DCO 2 An illustration shows 8 chromosomes inherited from an F1 female. eye color gene swapped w purple eye and must be double cross Gray body, purple eyes, vestigial wings 30 An illustration shows 8 chromosomes inherited from an F1 female. SCO Black body, red eyes, long wings SCO 28 An illustration shows 8 chromosomes inherited from an F1 female. Black body, red eyes, vestigial wings DCO 1 An illustration shows 8 chromosomes inherited from an F1 female. Black body, purple eyes, long wings 60 An illustration shows 8 chromosomes inherited from an F1 female. SCO black and purple add and get 472 Black body, purple eyes, vestigial wings 412 An illustration shows 8 chromosomes inherited from an F1 female. parental Total 1005 u Access the text alternative for slide images. Map the Genes 1 18 u Analysis of the u F2 generation flies allows you to map the three genes u The three genes exist as two alleles each Therefore, there are u 23 = 8 possible combinations of offspring u If the genes assorted independently, all eight combinations would occur in equal proportions not what u have It is obvious from the data that they are far from equal Map the Genes 2 19 u In the offspring of crosses involving linked genes Nonrecombinant phenotypes occur most frequently that's why we went thro parentals 1st Double crossover phenotypes occur least frequently freq 2 & 1 dco Single crossover phenotypes occur with “intermediate” frequency around 30 freq sco Double Crossover 1 20 u 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 u 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 bc there is based on the number of offsprings and can figure from data which is the gene in the middle Double Crossover 2 21 replacement of eye color ^^ each x is a cross over double crossover ebony purple vistigial ebony red vistigial u Access the text alternative for slide images. Step 4: Calculate the Map Distance Between Pairs of Genes 22 u One strategy is to regroup the data according to pairs of genes From the P generation, we know that the dominant alleles are linked, as are the recessive alleles u Nonrecombinant F2 offspring have a pair of dominant or a pair of recessive alleles u Recombinant F2 offspring have one dominant and one recessive allele The regrouped data will allow us to calculate the map distance between the two genes regroup data so consider them as pairs do body eye and body wings etc  Step 4: Calculating the Map Distance Between Pairs of Genes 1 23 Nonrecombinant offspring Total Recombinant Offspring Total Gray body, red eyes Gray body, purple eyes body color and eye color (411 + 61) 472 (30 + 2) 32 Black body, purple eyes Black body, red eyes 72 (412 + 60) (28 + 1) 29 Total 944 Total 61 u The map distance between body color and eye color is u Map distance = ( 61 ∕ (944 + 61) ) × 100 u = 6.1 map units have to take 2nd set and 3rd  Step 4: Calculating the Map Distance Between Pairs of Genes 2 24 body color and wing shape Nonrecombinant offspring Total Recombinant Offspring Total Gray body, normal wings Gray body, vestigial wings (411 + 2) 413 (30 + 61) 91 Black body, vestigial wings 413 Black body, normal wings (412 + 1) (28 + 60) 88 Total 826 Total 179 wing shape 2(2+1)= 6 and add to 179 u The map distance between body color and eye color is 6.1 12.3 u Map distance = ( 179 ∕ (826 + 179) ) × 100 (185/(826+179)= 18.4 giving the correct amt given when body eyecolor wing adding the distances between body and eye color and eye u = 17.8 map units color and wing shape 17.8 u Note: This value is a little low because it doesn’t take double crossovers into consideration. that = 6.1+12.3 = 18.4 this is bc rhe 17.8 doesnt account for the dco Step 4: Calculating the Map Distance Between Pairs of Genes 3 25 eye color and wing shape Nonrecombinant offspring Total Recombinant Offspring Total Red eyes, normal wings Red eyes, vestigial wings (411 + 28) 439 (61 + 1) 62 Purple eyes, vestigial wings Purple eyes, normal wings 442 (412 + 30) (60 + 2) 62 Total 881 Total 124 wing type u The map distance between body color and eye color is u Map distance = (124 ∕ (881 + 124) ) × 100 u = 12.3 map units Step 5: Construct the Map u The body color and wing shape genes are farthest apart based on the map unit calculation The eye color gene is in the middle u The data are consistent with the map being drawn as u b − pr − vg (from left to right) 2 6  27 Interference 1 how often prob of dco u The product rule allows us to predict the likelihood of a double crossover from the individual probabilities of each single crossover by taking the prob of each individual sco P =P ´P ( double crossover ) ( single crossover between b and pr ) (between pr and vg ) single crossover.061.123 u = 0.061 × 0.123 = 0.0075 p(dco) multipl by total offspring and get exp value u Based on a total of 1,005 offspring The expected number of double crossover offspring is = 1,005 × 0.0075 = 7.5 but we saw 3; 2 + 1 and lower then expect Interference 2 28 u We would expect seven or eight offspring to be produced as a result of a double crossover u However, the observed number was only three Two with gray bodies, purple eyes, and normal wings One with black body, red eyes, and vestigial wings 1st cross over decre the possibility that a second one will occers and then it will give us a higher number u 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 Interference 3 29 u Interference (I) is expressed as I = 1 − C, where C is the coefficient of coincidence u I = 1 − C = 1 − 0.4 = 0.6 or 60% 60% of expected cross overs didn't occur u This means that 60% of the expected number of crossovers did not occur Observed number of double crossovers C= Expected number of double crossovers = 3 7.5 = 0.40 only 40% occured Interference 4 30 something called interence u Since I is a positive value, this interference is positive interference u The molecular mechanisms that cause interference are not completely understood when 1 cross over occurs decr a nearby cross over from occuring However, most organisms regulate the number of crossovers so that very few occur per chromosome # cross overs on chromosome r v small 6.5 Mitotic Recombination crossing over in mitosis way less common then in meiosis u Mitosis does not involve the homologous pairing of chromosomes to form bivalents Crossing over in mitosis is therefore expected to be much less likely than during meiosis u Crossing over during mitosis does occur on rare occasions In these cases, it may produce a pair of recombinant chromosomes that have a new combination of alleles when does there is exhcange of segments and new combo alleles This is known as mitotic recombination Crossing Over Occasionally occurs during Mitosis u If mitotic recombination occurs during an early stage of embryonic development The daughter cells containing the recombinant chromosomes continue to divide This may ultimately result in a patch of tissue with characteristics that are different from those of the rest of the organism embryo after fert and 1 cell nd mitotic recomb takes places cells that're derived where mitotic recombo took place were diff from the rest >>> and stern found in body of fruit fly u Curt Stern proposed that unusual patches on the bodies of certain Drosophila strains were due to mitotic recombination Mitotic Crossing Over in Drosophila 1 u Stern was working with strains carrying X-linked genes affecting body color and bristle morphology y = yellow body color y+ = gray body color sn = short body bristles (singed) sn+ = normal body bristles u Females that are y+y sn+sn are expected to have gray body and normal bristles expected ^^^ Mitotic Crossing Over in Drosophila 2 u However, when he microscopically observed these flies, he noticed places in which two adjacent regions were different from the rest of the body and from each other This is called a twin spot Stern proposed that twin spots are due to a single mitotic recombination within one cell during embryonic development Mitotic Recombination in Drosophila embryo larger v similar to fert egg 1 of purples xchange segments w blues rem: rare event homologous pairing only during meiosis twin spot and mitotic recomb in single somatic and happens after fertilization; meiotic recomb for meiosis in germ line cells u Access the text alternative for slide images. 6.4 Genetic Mapping in Haploid Eukaryotes 36 u Much of our earliest understanding of genetic recombination came from the genetic analyses of fungi u Fungi may be unicellular or multicellular organisms u They are typically haploid (1n) Haploid cells reproduce asexually Two haploid cells can fuse to form a diploid zygote (2n) which goes through meiosis to produce haploid spores The sac fungi (ascomycetes) have been particularly useful to geneticists because of their unique style of sexual reproduction Sexual Reproduction in Ascomycetes 1 37 Meiosis produces four haploid cells, termed spores A group of 4 spores is a tetrad The spores are enclosed in a sac termed an ascus (plural: asci) Some species form an octad of spores following a mitotic division Sexual Reproduction in Ascomycetes 2 38 u The cells of a tetrad or octad are contained within a sac (that is, the products of a single meiotic division are contained within one sac) u This is a key feature that dramatically differs from sexual reproduction in animals and plants u For example, in animals, oogenesis only produces a single functional egg and spermatogenesis produces sperm that are mixed with millions of other sperm u Using a microscope, researchers can dissect asci and study the traits of each haploid spore The analysis of these asci can be used to map genes Sexual Reproduction in Ascomycetes 3 39 u Access the text alternative for slide images. Types of Tetrads 40 u Upon completion of meiosis, three distinct types of tetrads are possible: 1. The tetrad will contain four haploid cells (spores) with nonrecombinant arrangements of alleles. This ascus is said to have the parental ditype (PD). 2. An ascus may have two nonrecombinant cells and two recombinant cells, which is called a tetratype (T). 3. An ascus with a nonparental ditype (NPD) contains four cells with recombinant genotypes. Possible Assortment of Two Genes in a Tetrad 41 u Access the text alternative for slide images. Types of Asci 42 u 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 u If the two genes are linked The type of crossover between them determines what type of ascus is produced u No crossovers yield the parental ditype u Single crossovers produce the tetratype u Double crossovers can yield any of the three types u The actual type produced depends on the combination of chromatids that are involved Relationship between Crossing Over and PD, T and NPD for Two Linked Genes 1 43 u (a) No crossing over u (b) A single cross over u Access the text alternative for slide images. Relationship between Crossing Over and PD, T and NPD for Two Linked Genes 2 44 u (c) Double crossovers u Access the text alternative for slide images. Data from a tetrad analysis can be used to calculate the map distance 45 between two linked genes 1 u As in conventional mapping, the map distance is calculated as the % of offspring that carry recombinant chromosomes NPD + (1/2) (T) Map distance = ´100 Total number of asci u 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 Data from a tetrad analysis can be used to calculate the map distance 46 between two linked genes 2 u A more precise way to calculate map distance is: Single crossover tetrads + (2) (Double cross over tetrads) Map distance = ´ 0.5 ´100 Total number of asci Mapping from Tetrad Analysis 1 47 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 Total number of DCO = 4NPD Mapping from Tetrad Analysis 2 48 u Considering single crossovers: Notice that T asci can result from SCO or DCO Since there are two kinds of T that are due to DCO u The actual number of T arising from DCO is 2NPD u So, T = SCO + 2NPD and number of SCO = T − 2NPD Mapping from Tetrad Analysis 3 49 u SCO and DCO can by accurately measured by: SCO = T − 2NPD DCO = 4NPD u Substitution of these values into our previous equation gives: (T - 2NPD) + (2) (4NPD) Map distance = ´ 0.5 ´ 100 Total number of asci T + 6NPD Map distance = ´ 0.5 ´ 100 Total number of asci