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Lecture 4 Linkage and Recombination CAMPBELL BIOLOGY Chapter 9 Notes at: tcd.ie/Biology_Teaching_Centre/local/ junior-freshman/ ≠≠ by1101local This is an Irish family with an autosomal dominant disease mutation which causes blindness (the disease is called RP). A large family like this is powerful (statistically) & can be used to find out which human chromosome the disease gene is on. This lecture is about the basis of genetic linkage mapping which allows us to localise disease genes. Central to this is the concept of crossing/ over recombination during Prophase I of meiosis which we discussed in the last lecture (lecture 3). Let’s go thru’ some background first...... 1 Genes are located on Chromosomes • During the 19th century, cytologists identified chromosomes and described their behaviour during mitosis and meiosis • They did not realise that chromosomes were involved in the mechanism of inheritance • In the early 20th century, geneticists realised that chromosomes behaved just like the hypothetical unit factors proposed by Mendel to explain inheritance - they occur in pairs in diploid cells - their number is halved during gamete formation - fertilisation restores the diploid number • They eventually proved that genetic information (genes) is located on the chromosomes Mendel s studies • Mendel investigated the inheritance of 7 traits • Each trait showed independent assortment i.e. the inheritance of one trait did not influence the inheritance of the second trait • Because of this, he observed the following ratios in the offspring 1. Monohybrid cross (inheritance of a single trait) 3 : 1 (dominant : recessive) 2. Dihybrid cross (inheritance of two traits) 9:3:3:1 2 Is independent assortment always the case? No it depends on whether the genes are linked or not (a) Genes located on different chromosomes are not linked This allows independent assortment – in a di-hybrid cross the traits show the classic 9:3:3:1 inheritance pattern (b) Genes that are located very close together on the same chromosome may show complete linkage They may be so close to each other that they cannot be separated by recombination during meiosis (c) Genes located far apart on the same chromosome typically show incomplete (partial) linkage because they are easily separated by recombination Linkage Humans contain approximately 25,000 genes (haploid number) But there are only 23 chromosomes (diploid number 2n = 46) Clearly, there are hundreds / thousands of genes on a given chromosome All of the genes on a given chromosome are said to be linked or to belong to the same linkage group IMPLICATIONS OF LINKAGE Since chromosomes are the units that are distributed into gametes during meiosis, if 2 genes are located on the same chromosome they will tend to be inherited together whereas genes located on different chromosomes will show independent inheritance. 3 A di-hybrid cross between 2 fruit flies (Drosophila) We re using the fruit fly, Drosophila, as an example – the same rules apply to inheritance of two human traits. grey (wild-type) b+ DOMINANT black (mutant) b recessive large (wild-type) vg+ Trait 1 = body colour DOMINANT Trait 2 = wing size vestigial (mutant) vg recessive Genetic Map of Chromosome 2 of Drosophila b+ vg+ b vg 4 b+ vg+ b vg Sometimes a crossing over event during meiosis will occur giving new non-parental genotypes: b+ vg and b vg+ 5 Why so many parental types in the offspring? The 2 genes are not showing independent assortment because they are linked i.e. are on the same chromosome Mother Father Why are there any non-parental types (i.e. recombinants)? Answer: In this case because of recombination/crossing over in the female fly Cross overs between non-sister chromatids Meiosis II Types of gametes produced 6 http://www.uic.edu/classes/bms/bms655/lesson12.html Haemophilia A (Factor VIII) – bleeding disorder G6PD deficiency – hemolytic anemia Recap from Lecture 3 Meiosis Prophase I (Crossing-over / recombination) • Homologous chromosomes (each consisting of two sister chromatids) come together as pairs • The structure formed is called a tetrad • Chromosome segments are swapped between non-sister chromatids at cross-over points called chiasmata (= crossing-over) 7 8 Genetic maps or linkage maps Sturtevant proposed that recombination frequencies reflect the distances between genes on a chromosome It is assumed that the chances of crossing over is equal at all points on a chromosome If so, then the farther apart two genes are: (a) the higher the probability that a cross-over will occur between them and therefore (b) the higher the recombination frequency Why? Because the greater the distance between two genes the more points there are between them where crossing over can occur 9 Recombination frequencies can be used to construct a genetic map • Recombination frequencies are approximately additive • Recombination frequency between the body colour gene black (b) and the eye colour gene cinnabar (cn) = 9% • Recombination frequency between cinnabar and vestigial wing (vg) = 9.5% • Recombination frequency between b and vg = 17% 1% recombination equates to approx. 1 million base pairs of DNA (also termed 1 centiMorgan (cM) after the scientist Morgan) 10 Co-inheritance of a marker & disease gene helps localise the disease gene www.ncbi.nlm.nih.gov/bookshelf/br.fcgi?book=mboc4&part=A1618 11 www.csc.fi/english/csc/publications/cscnews/2009/2/twins_genetics_research ‘A genetic map of the genes affecting adult height. Genetic linkage analysis was used for locating genes affecting stature. This method utilizes genetic markers known to show variation between individuals. The markers are evenly distributed across the entire genome and they are determined from DNA samples. By using these markers it is possible to determine, by means of the linkage analysis, the chromosomal regions that are similar in family members showing a strong correlation in height. As a result, scientists can estimate the statistical probability of the region containing genes associated with adult stature. Statistical significance for the detected linkages must then be determined through computer simulation.’ 12