Week 9 Online Learning - Lecture notes EdX notes Week 9 PDF

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BIOL1020 Week 9 - EdX Notes 1. INTRODUCTION - RECOMBINATION AND LINKAGE MAPPING LEARNING OUTCOMES The outcomes of this week's online learning include being able to: 

Relate the observation of recombinant genotypes to phenomena arising from meiosis

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Gain familiarity with calculating recombination rates from an array of genetic markers

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Use recombination rates to infer genetic distances among loci

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Create simple linkage maps based on recombination distances among a few loci INTRODUCTION Linkage mapping is a cornerstone of modern genetics. Genetic maps are created by observing recombination frequencies among loci: loci that are close to each other on the same chromosome are said to be linked and only occasionally are their alleles separated by crossing-over events. Even before geneticists could isolate specific DNA loci, they were able to construct genetic maps based on elegant breeding experiments. Today, genetic maps are still very much in use, for instance, allowing phenotypic traits to be connected to contributing DNA loci and, in combination with physical maps based on basepair distances, to identify genomic regions of high and low recombination. Amazingly the basic methodology for using recombination frequencies to infer gene order was deduced by an undergraduate! Alfred Sturtevant loved puzzles and was working in the lab of the famous Drosophila geneticist Thomas Hunt Morgan. Morgan had proposed that genes are arrayed on chromosomes and that occasionally cross-over events would break up associations between alleles of linked genes. Sturtevant thought that perhaps observing recombination frequencies among multiple genes could be used to make a genetic map. He took home some data one night and returned to the lab the next day with a solution (which was later published: Sturtevant 1913). Not bad for an undergraduate project!

The online activities that follow will review why linkage occurs and how we can use the testcross breeding design to infer gene orders on chromosomes using the methodology developed by Sturtevant. This basic approach forms the basis for all modern genetic maps.

2. CHROMOSOMES AND LINKAGE In eukaryotes, DNA is organised into linear chromosomes. With each meiosis, crossing over occurs between non-sister chromatids. Figure 13.12 from your textbook shows an example of recombination arising from crossing over between one set of homologous chromosomes during a single meiosis On average, there are 2-3 crossing over events per pair of chromosomes. Crossing over is the physical process that leads to recombination, or the mixing of parental chromosomes. The resulting two chromosomes that contain stretches of DNA from both parents are recombinant chromosomes and the two chromosomes that are identical to the parents are

called parental type chromosomes.

Genomes  vast in length. For example: The human genome -> approximately 3.3 billion base pairs long. Drosophila fruit flies -> More compact genome is about 180 million base pairs long. Genetic Mapping:  How we, given the vast size of typical eukaryotic genomes, find the locations for genes that control specific phenotypic traits… Background: Genomes –> organised into linear chromosomes {important} During meiosis: -> chromosomes assort independently. ~ Loci on the same chromosome - may stay linked if they are close together on chromosome. However, if loci are far apart: -> likely that crossing over will create recombinant gametes. (seen in this diagram) The chromosomal theory of inheritance: -> developed primarily from work on Drosophila. Thomas Morgan: -> found chromosomes could explain patterns of inheritance and how phenotypic traits could be mapped to chromosomal locations. -> established that chromosomes contained the information underlying heredity.

Worked with colonies of Drosophila that they exposed to all sorts of horrible mutagens (including chemicals and radiation) Found some flies with odd characteristics: e.g: white eyes. -> called “mutants”

-> in most cases the mutant phenotype is recessive + a “knock out” of a fully functional gene. Experiment: - focused on two types of flies: 1. the typical “wild type” 2. double mutant with black vestigial wings. Aim: discover whether or not the loci encoding these two traits are linked. Procedure: 1. Two homozygous flies were crossed, one wild type, and one double mutant. -> created a doubly heterozygous fly. 2. F1 testcross design was used. Remember: o testcross indicates that an individual with an unknown genotype is being crossed against a homozygous recessive individual. o Dihybrid => fly is heterozygous at two loci. Unknown about F1 fly: whether or not the black and vestigial traits are linked. Strength of the testcross design: -> homozygous parent can only make one type of gamete – black and vestigial. -> dihybrid F1 parent on the other hand, can create 4 different types of gametes. Firstly: If the black and vestigial loci were on different chromosomes: -> expect even proportions of gamete types. -> + .: even proportions of offspring. Reality -> The loci for black and vestigial = on the same chromosome. (In these diagrams, lines represent chromosomes) Original parents = homozygous. ~ Their F1 offspring = heterozygous -> are mated against the tester… No recombination in the F1: -> Only two types of offspring are possible… => offspring match the original P generation phenotypes. Recombination in the F1: -> Two additional genotypes

=> offspring show a mix of traits from the two original parents (grey body+ vestigial wings / black body + normal wings. Results: Numbers of different testcross offspring phenotypes obtained in Morgan’s real experiment –  Not an even proportion of phenotypes. -

most common types resemble the original wild type and double mutant parents. -> call these the “parental types”.

 Other two categories show a mix of parental types… -> these are the “recombinant types”. Recombinant offspring: -> reflect crossing over that happened when the F1 was making gametes. Recombination frequency => take the number of recombinants and divide them by all the offspring. e.g: 17% of the offspring were recombinant .: recombination frequency of 17%. -> equivalent to a genetic map distance of 17 cM (unit of measure is named in Morgan’s honour) Important: keep in mind that where crossing over occurs on a chromosome is random. .: when we look at two loci (black body and vestigial wings in this case), different recombinant individuals will have different break points in their chromosomes even though they all have the same phenotype. These are some examples of what the recombinant genotypes could look like. Experiment: -> demonstrated that some loci are linked and .: these loci must be close to each other on the same chromosome.

-> demonstrated proportions of recombinant offspring can be used to determine the recombination frequency, or map distance, between two loci (black body and vestigial wings).

Constructing full maps of chromosomes -> similar to that of Drosophila chromosome 2 Focus - loci black body, vestigial wings, and cinnabar eyes. Previously: -> map distance between black body and vestigial wings = 17 cM. -> map distance between black body and cinnabar eyes = 9 cM. => these loci are linked… but if you think about it carefully for a moment, there are two possible gene orders: Experiment: -> Dihybrid F1 testcross design for vestigial + cinnabar and, with this additional information, determine which order is correct (shown on the right). Having determined the order of these three focal loci -> use them to start building up a genetic map. Notice that the distances between adjacent loci: -> 9 and 9.5 sum to 18.5 and not 17 (the recombination rate between black and vestigial). => This phenomenon is due to double crossovers - occurs the farther two loci are from each other. ie: the recombination rate between black body and vestigial is more than 17%. For now, we will only estimate distance for two loci at a time. (noting that our estimates at larger distances are somewhat smaller than the real amount). In practice -> build up larger linkage groups by adding up shorter distances to build up a longer map.

For this example of Drosophila chromosome II : -> total length is 104.5 cM – that is, 104.5% recombination => clearly a rate that we cannot directly measure!

Experiment 2: -> Cross between maroon eyes at map position 16.5 + vestigial wings at map position 67.

Procedure: Standard dihybrid testcross experiment -> measuring the recombination rate between maroon eyes and vestigial wings. Firstly: -> calculated a map distance of 50 cM (pretty close match to the actual map distance of 50.5 cM) -> 50% progeny were recombinants + 50% were parental types => say that these two loci are unlinked even though they are on the same chromosome. Whenever we examine 2 loci, the maximum recombination distance we can get is 50 cM. Looking at two loci on the same chromosome: (even if all meioses involve crossing over) -> maximum of 50% of the gametes will be recombinant types. Looking at two loci on different chromosomes: (remember that under independent assortment we expect to get a maximum of 50% new combinations) -> Either way, the number of recombinants can only be about half of the total number of gametes

.: the maximum measurable map distance is 50 cM.

How modern eukaryote genetic maps are created using DNA markers -> based on entirely the same logic developed by Muller and Sturtevant a hundred years ago. In making a genetic map we have two basic goals: 1. Determine whether the two loci are linked 2. If they are linked, determine the recombination distance between them. You are already used to visualising loci as locations on linear chromosomes. If we imagine zooming in on those loci to the level of DNA then we could see the actual nucleotide sequence of each allele. And if our individual was heterozygous, we would see that at least at some basepair positions there were different nucleotides. Polymorphic sites -> called SNPs / single nucleotide polymorphisms. If we only concentrate on these two SNPs, we can think about the possible genotypes. First SNP -> have three possibilities representing our two possible homozygous genotypes + a heterozygous genotype. Second SNP

-> has three possible genotypes. SNPs are considered codominant loci: ie: can recognize both alleles in a heterozygote. Why?... Because we can resolve heterozygous genotypes from homozygous genotypes using modern SNP genotyping techniques. -> With SNPs, can still use a testcross design involving a heterozygous and a homozygous parent. => if we cross these individuals, can expect to recover parental type progeny that have genotypes identical to their parents. => if there’s crossing over in some meiosis, get recombinant progeny that have a mix of elements from the parental genotypes. Remember: if the two loci are quite distant on the same chromosome / actually on different chromosomes, then we will get approximately the same number of recombinant and parental progeny(offspring). Genotypic notation -> Use to refer to these progeny offspring where the underscore indicates linkage on the same chromosome copy. Common Question: what about recombination in the homozygous parent?” -> of course, recombination happens in the homozygous parent as well. But, cannot tell when it happens because the gametes are identical…(at least for loci we can examine). Working Backwards: Problem 1 Notice is that there is a common element shared among all these genotypes, the A_C element that is shown in blue. -> corresponds to the loci contributed by the homozygous tester parent – now we can immediately infer that parent’s genotype. Before we can directly address our two goals… -> have to figure out the genotype of the heterozygous parent. 1. Option 1: heterozygous parent had one chromosome with the same phase as the tester. 2. Option 2: heterozygous parent could have a different arrangement of linked alleles -> where the A and T alleles from SNPs 1 and 2 are in phase and the G and C alleles from SNPs 1 and 2 are in phase.

Different possible phases = called cis and trans -> relative to the phase of the tester parent. Because tester had A and C in phase on the same chromosome -> this represents the cis / matching configuration. To figure out if our heterozygous parent was cis or trans = look at the number of progeny: Remember: there will always be at least as many parental types as recombinants => can easily identify the parental types. Because A + C are in phase and G + T are in phase -> can recognise our heterozygous parent as being in the cis configuration. Now we can also recognise the recombinant genotypes… Because our parentals types are more frequent than the recombinants -> know that our two loci are linked => can answer the first question of our problem. Genetic distance -> take the number of recombinants and divide by the total number of progeny = 23 cM....