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Variations from Mendel’s Original Crosses •Multiple alleles •Polygenic inheritance •Linked dihybrid crosses Variations from Mendel’s work •Mendel’s original pea experiments studied the segregation of 7 different characteristics. •He looked at each of the 7 characteristics separately – he could do this because they were all on different chromosomes. In other words they were not linked together on the same chromosome. If they were on the same chromosome they could not segregate independently. •Single genes that are linked to the X chromosome - such as haemophilia and colour blindness - showed a variation to the usual Mendelian ratios because not all individuals have two X chromosomes. (see Beyond Mendel ppt in Blueprint of Life) •Another variation we have already studied is co-dominance – such as roan coat colour in cattle. (see Beyond Mendel ppt in Blueprint of Life) •Let us consider a few more variations from the standard… PART 1 Multiple Alleles Multiple alleles - ABO •With Mendel’s original crosses there were only Parents: AB x BO ever two alleles of the one gene to choose from; short or tall, angular or smooth, etc. Gametes: lA lB & lB •Sometimes there are more than two alleles to inherit even though we can still only inherit a total of 2 alleles (but 1 if the allele is linked to the X or Y chromosome). •More than two choices of alleles (and there can sometimes be hundreds – we won’t be doing any of those!!!) are known as multiple alleles. •Examples in humans are the ABO blood groups (3 alleles) and hair or eye colour. •Let’s have a look at the ABO alleles where allele A and B are co-dominant when inherited together while both are dominant if inherited with O. lWB lA lB lAlB lBlB l0 l0 lAl0 lBl0 W Phenotypes: 25%AB 50%B Note: 25% A l0l0 = O Multiple alleles - mice •Non-human examples occur in genes for coat patterns in many animals such as in mice. •In mice the following genotypes match the following phenotypes: Parents genotype: CyCa x CbCa Parents phenotype: Yellow x Black Gametes: Cy Ca & Cb Ca •CyCa = Yellow coat Cy •CbCa = Black coat •CaCa = Agouti (mixed) coat •CbCy – Black coat •Therefore black is the most dominant allele, the yellow allele is dominant over agouti and the agouti allele is recessive to both black and yellow for the coat colour gene. Ca b CyC CaCb C W b Ca Phenotypes: 50% Black 25% Yellow 25% Agouti y a a a CC CC PART 2 Polygenic Inheritance Polygenic Inheritance •Polygenic inheritance occurs when there is more than one gene involved in a particular phenotypic trait. •Each loci involved can also have multiple alleles. •Examples in humans include height, skin pigmentation, weight, cleft palate, neural tube defects, intelligence, the Rhesus factor and, most behavioural characteristics. •As there are several genes involved with polygenic inheritance it means there are several genes influencing the phenotype, and because of this the number of possible phenotypes tends to be large. The phenotypic characteristics blend from one to another and can appear continuous. •Polygenic inheritance usually shows what is called continuous variation, that is a trait that covers one end of a scale to another and can be shown with a bell shaped bar graph. •When only one gene influences a phenotypic trait, such as, yellow or green seed, or A/B blood group, the phenotype variation generally tends to be discontinuous. That is, either one character or another & nothing in between. Polygenic Inheritance •This graph shows continuous variation of the phenotype skin pigmentation. •Each of the 3 genes influencing skin pigmentation has two alleles, 1 for light skin and 1 for dark skin. The two allele show incomplete dominance patterns and show a blended effect. •There are 64 different combinations of the 2 alleles of the 3 genes. •An extremely pale skinned person receives 3 pale skin alleles from mum and 3 pale skin alleles from dad. •A dark skinned person receives a total of 4 or 5 dark skin alleles from http://www.emc.maricopa.edu/faculty/farabee/BIOBK/BioBookgeninteract.html mum and dad. Polygenic Inheritance http://www.emc.maricopa.edu/faculty/farabee/BIOBK/BioBookgeninteract.html Polygenic Inheritance A typical pedigrees showing traits with polygenic inheritance can appear like this one with distant relatives affected and no obvious pattern of inheritance. http://med.usd.edu/som/genetics/curriculum/2CINHER3.htm Polygenic Inheritance •A non-human example of polygenic inheritance is that of Chicken combs. •Two loci are considered http://www.ndsu.nodak.edu/instruct/mcclean/plsc431/mendel/mendel6.htm •R = rose while r = single •P = pea while p = single Rose Phenotype Pea Genotype Single rrpp Pea rrPP, rrPp Rose RRpp, Rrpp Walnut RRPP, RRPp, RrPP, RrPp Single Walnut PART 3 Linked dihybrid crosses Recall Mendel’s monohybrid ratio Parents: RR (round) x Gametes: R R & rr (angular) r r Parents: Rr (round) x Gametes: R r & R Rr (round) r Rr RR F1 R R r Rr Rr rr F2 R r R RR Rr r Rr rr Rr r Rr Rr Mendel’s monohybrid ratio states that in the F2 generation pure breeding parents will show a phenotypic ratio of 3:1 for the dominant characteristic. (See Mendel ppt in Blueprint of Life for more detail) Mendel was also curious to find out the ratios of offspring if two characteristics were considered rather than just one (such as seed shape shown above) Dihybrid crosses – F1 Let’s consider a cross between a pure breeding plant with round and yellow seed with one that is pure breeding with an angular/wrinkled and green seed F1 Parents genotype: RRYY x rryy F1 Parents phenotype: Round & Yellow F1 Gametes: RY RY & RRYY x Angular & Green rryy ry ry F1 RY RY ry RrYy RrYy F1 phenotype: 100% Yellow and Round RRYY F1 genotype: 100% RrYy ry RrYy RrYy Dihybrid crosses - F2 F2 Parents genotype: RrYy x RrYy F2 Parents genotype: Round and Yellow F2 Gametes: RY Ry rY ry & RY RrYy Ry rY x Round and Yellow RrYy ry F2 phenotype: F2 RY Ry rY RY Ry ry RRYY RRYy RrYY RrYy RRYy RRyy RrYy Rryy 9 round and yellow 3 round and green 3 angular and yellow rY ry RrYY RrYy RrYy Rryy rrYY rrYy rrYy 1 angular and green rryy Dihybrid crosses - F2 F2 RY RY RRYY Ry RRYy rY RrYY Ry rY RRYy RrYY RRyy RrYy ry RrYy RrYy Rryy rrYY rrYy F2 genotype: these need to be calculated separately for each phenotype, for example; The 9 round and yellow seeds can be either •RRYY •RrYy •RRYy ry RrYy Rryy rrYy rryy What are the possible genotypes for the yellow and angular seed? •RrYY Whereas the 1 angular and green seed can only be •rryy Non-linked genes segregating in a dihybrid cross without crossing over F1 Parents pure breeding: RRYY x rryy F2 Parents genotype: F2 Gametes: RY Ry Beginning of Meiosis RrYy rY ry Rr Yy Possible gametes after meiosis RY ry Ry rY Non-linked genes segregating in a dihybrid cross with crossing over Beginning of Meiosis after crossing over F1 Parents pure breeding: RRYY x rryy F2 Parents genotype: F2 Gametes: RY Ry RrYy rY ry R Possible gametes after meiosis are still the same 4 combinations RY rY Ry rR r Yy ry Ry ry RY rY Put either of these in a Punnett square and observe the normal 9:3:3:1 F2 ratio F2 RY Ry rY ry RY RRYY RRYy RrYY RrYy Ry RRYy RRyy RrYy Rryy rY RrYY RrYy rrYY rrYy ry RrYy Rryy rrYy rryy •Here we see one dominant gene and one recessive gene from different traits mixed together in the same individual organism, as well as two dominants from the two genes and two recessive genes in the same individual. Non-linked Vs. Linked •Located on different chromosomes. •Located on the same chromosome. •Sort independently at meiosis. • Do not allow typical Mendelian ratio of 9:3:3:1 in the F2 generation of a dihybrid cross. •Crossing over doesn’t alter the way the gene is segregated into gametes. •If two genes are linked they are close together on a particular chromosome. •Allow typical Mendelian ratio of 9:3:3:1 in the F2 generation of a dihybrid cross. •Do not sort independently at meiosis. •An X linked gene is one that is on the X chromosome. •If genes are closely linked, there is less chance of crossing over occurring between them at meiosis and altering the way they segregate at meiosis into gametes. •If genes are not closely linked then there is a high chance they may end upon different chromosomes after crossing over and therefore segregate differently. Linked genes segregating in a dihybrid cross without crossing over Beginning of Meiosis. Crossing over has occurred but the genes were are interested in have not recombined. F1 Parents pure breeding: AABB x aabb F2 Parents genotype: AaBb F2 Gametes: AB ab A a B b Possible gametes after meiosis AB ab AB ab Put these in a Punnett square and observe the 3:1 F2 ratio F2 AB AB ab Both dominant traits AABB AaBb Both recessive traits ab AaBb aabb •In this case, if genes are linked closely together on a chromosome and do not get the chance to cross over at meiosis, then we should never see both a dominant trait and recessive trait together in the same organism. •We will only see both dominant or both recessive traits in a 3:1 ratio and not the expected 9:3:3:1. They are therefore not sorting independently. •If the 2 genes are very closely linked, the ratio is the same as the monohybrid cross ratio in the F2 generation. Linked genes undergoing crossing over at meiosis Linked genes segregating in a dihybrid cross with crossing over Beginning of Meiosis after crossing over. The genes have ‘recombined’. F1 Parents pure breeding: AABB x aabb F2 Parents genotype: AaBb F2 Gametes: AB Ab aB ab Aa Possible gametes after meiosis – there are now 4 combinations rather than two. AB Ab bB aB ab Put these in a Punnett square and observe the 9:3:3:1 F2 ratio F2 AB Ab aB ab 9 Both A and B dominant 3 A dominant and B recessive aB AaBB AaBb aaBB aaBb 3 A recessive and B dominant ab AaBb Aabb 1 Both A and B recessive AB AABB AABb AaBB AaBb Ab AABb AAbb AaBb aaBb Aabb aabb If the genes are not closely linked then there is a high chance they will undergo crossing over. If this occurs then they segregate as if they were independent and show the typical 9:3:3:1 ratio in the F2 generation. Linked or non-linked dihybrid crosses? •Say we conducted some crosses with sweet pea plants. We crossed a pure breeding plant with long pollen grains and purple flowers (both dominant) with a pure breeding plant with round pollen grains and red flowers. •The F1 generation would give 100% long pollen & purple plants whether or not the two genes were linked. •If we counted 1,600 plants in the F2 generation we could expect the following results… Phenotype Expected if not linked Long, Purple 900 Round, Purple 300 Long, Red Round, Red Expected if linked Expected if linked & x/over & not x/over 1,200 Actual results 900 1,105 0 300 78 300 0 300 78 100 400 100 339 Linked or non-linked dihybrid crosses? •The actual results do not match with any of the expected results. What can we make of this? •We can tell that the two genes of this dihybrid cross are linked because the actual numbers counted do not match the 9:3:3:1 ratio for unlinked genes. •The actual results counted for each trait are in between those for ‘linked and not crossed over’ and those for ‘linked and crossed over’. What does this mean? •It means that some of the plants have undergone recombination (crossing over) of the long/round pollen and red/purple genes and some haven’t. •If the actual results are closer to the ‘linked and not crossed over’ results, then we can say that they are so close there has been little chance of crossing over and therefore the genes must be closely linked (close together). •If the actual results are closer to the ‘linked and crossed over’ results then we can predict that the genes are quite a way apart on the chromosome and have a high chance of crossing over. Calculating linkage •Linkage can be calculated by working out the amount of plants that have undergone recombination (crossing over) compared to the total number of plants. This can be worked out as a percentage. •If there is a large percentage of crossing over then the genes are not closely linked, and if there is a small percentage of recombination then the genes are closely linked. •Let’s have a look at our example. 156 (78 + 78) plants showed recombination from a total of 1,600. Therefore, 156/1,600 = 0.0975 x 100 = 9.75% recombination. •A cross that is carried out in order to calculate recombination is a test cross. •Remember that recombination is random so the recombination % is not going to be exact every time. •By working out the recombination % of a few linked genes, geneticists can identify their relative position, or loci, on a chromosome. This is known as a chromosome map. Chromosome maps of linked genes •Say we have 3 genes (ABC) linked on a particular chromosome and we want to know which ones are more closely linked. •After conducting 3 sets of test crosses we observe the following results: Cross AB Results AB 20 BC 26 AC 17 Ab 2 Bc 1 Ac 1 aB 2 bC 1 aC 1 ab 6 bc 5 ac 7 Tot. 30 Cross BC Results Cross AC Tot. 33 •Recombination between AB is 4/30 x 100 = 13.3% •Recombination between BC is 2/33 x 100 = 6.1% •Recombination between AC is 2/26 x 100 = 7.7% Results Tot. 26 Which genes are the furthest apart? Which genes are closest on the chromosome? Chromosome maps •Recombination between AB is 13.3% •Recombination between BC is 6.1% •Recombination between AC is 7.7% A Segment of chromosome 13 7 Example of a map of chromosome 9 in a plant C 6 B http://users.rcn.com/jkimball.ma.ultranet/BiologyPages/L/Linkage.html Are the genes Se/se & Md/md linked? Uses of linkage •Linkage is used to identify the order of genes along a chromosome. •It is often difficult to locate actual positions due to the randomness of recombination, which can change (albeit slightly) all the time. •Scientists have tried using recombination maps to compare species. The more closely related the species, the more similar their chromosome maps should be. •This has created problems for taxonomist who may find that the new genetic maps do not concur with information already available. How do they make a decision whether or not to change the current classification of certain organisms? If they do decide to make the change, how do they notify the scientific community? How much would it cost to make the changes? •Scientists first working on the Human Genome project (to map the position of genes on the whole genome) used linkage maps, but these maps could not identify the exact position of genes they soon became redundant. References •Aubusson, P. and Kennedy, E. (2000) Biology in Context. The Spectrum of Life Oxford University Press, Melbourne, Australia. •Board of Studies (2002) STAGE 6 SYLLABUS Biology Board of Studies, NSW, Australia. •Farabee, M. J. (2001) Gene Interactions. Retrieved from the site http://www.emc.maricopa.edu/faculty/farabee/BIOBK/BioBookgeninteract.html July 2004. •Huskey, R.J. (2003) BIOL 121 Human Biology General Information. Retrieved from site http://www.people.virginia.edu/~rjh9u/geninfo.html July 2004. •Kimball, J.W. (2004) Kimball’s Biology Pages:Chromosome maps. Retrieved from site http://users.rcn.com/jkimball.ma.ultranet/BiologyPages/L/Linkage.html July 2004. •Krupp, D. (1999) Beyond Mendel. Retrieved from the site http://imiloa.wcc.hawaii.edu/krupp/BIOL101/present/byndmend/sld006.htm July 2004. •McLaughlin, L. & Hitchings, S. (2003) Biology Options. Genetics: the code broken? McGrawHill Australia Pty Ltd.