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Chromosomes: The Physical Basis of Inheritance

Chromosomes: The Physical Basis of Inheritance. 1866 Mendel published his work 1875 Mitosis was first described 1890s Meiosis was described 1900 Mendel's work was rediscovered 1902 Walter Sutton, Theodore Boveri and others noted parallels between behavior of chromosomes and alleles.

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Chromosomes: The Physical Basis of Inheritance

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  1. Chromosomes:The Physical Basis of Inheritance • 1866 Mendel published his work • 1875 Mitosis was first described • 1890s Meiosis was described • 1900 Mendel's work was rediscovered • 1902 Walter Sutton, Theodore Boveri and others noted parallels between behavior of chromosomes and alleles.

  2. Physical basis for Mendel’s laws: Behavior of chromosomes in meiosis Physical basis for Mendel’s laws: Behavior of chromosomes in meiosis equal segregation of alleles into different haploid gametes random assortment of genes on different chromosomes into gametes

  3. Chromosomal Theory of Inheritance • Genes have specific loci on chromosomes. • Chromosomes undergo segregation (meiosis) and independent assortment • Thus alleles of genes are independently assorted.

  4. Eggs EN En eN en Sperm EN EENN EENn EeNN EeNn E N En EENn EEnn EeNn Eenn E n eN EeNN EeNn eeNN eeNn e e e E E N N N en EeNn Eenn n eeNn eenn n e n Independent Assortment As long as genes are on different chromosomes, they will assort independently

  5. Chromosomal Basis of Sex • X-Y system: females are homogametic (XX) and males are heterogametic (XY) with males and females have the same number of chromosomes. • Examples include humans and all mammals, Drosophila.

  6. X-O system: females are homogametic (XX) and males are heterogametic (XO) with males having one less chromosome than females. • Examples include grasshoppers (cavallette), crickets (grilli), and cockroaches (scarafaggi).

  7. Z-W system: males are homogametic (ZZ) and females are heterogametic (ZW) with males and females have the same number of chromosomes. • Examples include all birds, some fishes, butterflies (farfalle), moths (tarme), and wild strawberries (fragole).

  8. Haplo-diploidy. There are NO sex chromosomes. Fertilized eggs become diploid females. Unfertilized eggs become haploid males. Males are fatherless. • Examples are the social insects: bees (api), ants (formiche), termites (termiti).

  9. No sex determination system. Most plants and some animals are NOT dioecious organisms with separate sexes. • Most plants & some animals are monoecious, where the same individual produces both eggs and sperm. • Examples include earthworms (lombrichi), garden snails (lumache), pea plants (piselli) and corn (granturco).

  10. Thomas Hunt Morgan • First to associate a trait (gene) with a chromosome. • Worked with fruit flies (Drosophila melanogaster) • Why fruit flies? • Short generation time (9 days) • Survives and breeds well in the lab • Very large chromosomes in some cells • Many aspects of phenotype are genetically controlled.

  11. Drosophila Mutations

  12. More Drosophila Mutations Wild Type ++ ebony body ee white eyes ww

  13. Normal eye color in Drosophila is red. • Morgan’s wife discovered a male Drosophila with white eyes on the window in their lab. • In the parental generation, Morgan crossed this male with white eyes to several females with red eyes.

  14. X P F1 • The offspring of that cross (the F1) all had red eyes (both males and females). • Morgan concluded that white eyes was recessive to red eyes. • Morgan than cross the F1 males and females among themselves to produce the F2. A white-eyed male was discovered X

  15. 1/4 1/4 1/2 F2 • The phenotypic ratio in the F2 generation was ~ 75% red and 25% white, which confirmed that red was completely dominant to white. • However, when Morgan looked more carefully at the sexes of the flies, he found that all the females had red eyes while ½ the males had red and ½ had white eyes.

  16. P: red-eyed female X white-eyed male. • F1: all red-eyed females & males. • F2: 50% red females: 25% red males: 25% white males. • Conclusion eye color is controlled by gene on the X chromosome. • Males have one X chromosome, while females have two X chromosomes.

  17. X+ X+ Xw XwX+ XwX+ X P Y X+Y X+Y X X+ Xw F1 X+ X+ X+ XwX+ 1/4 1/4 1/2 Y X+Y XwY F2 Morgan’s Discovery Of An X-Linked Drosophila Gene A white-eyed male was discovered

  18. Notation for Alleles • White eyes is a recessive mutation. • When a mutation is recessive, lower case letters are used to denote alleles: • Let w = white eyes; w+ = red eyes. • The “+” symbol always denotes the normal (wild type) phenotype. • Therefore, female genotypes are: ww = white eyes; w +w + or w +w = red eyes.

  19. Notation for Alleles • Bar eyes is a dominant mutation in Drosophila. • When a mutation is dominant, upper case letters are used to denote alleles: • Let B = bar eyes; B+ = normal eyes. • Therefore, female genotypes are: BB or BB + = bar eyes; B+B + = normal eyes.

  20. Morgan’s Crosses - 1 • White eyes is recessive and on the X chromosome only. • P: w +w + (female) x wY (male). • F1: w +w (female) x w +Y (male). • F2: ¼ w +w +: ¼ w +w: ¼ w +Y: ¼ wY. • Females are either homozygous or heterozygous while males are hemizygous.

  21. Morgan’s Crosses - 2 • Morgan still wanted to obtain white-eyed females. To do this, he crossed the F1 females to P males: • w +w x wY, which produces in the offspring: • ¼ w +w : ¼ ww: ¼ w +Y: ¼ wY • 50% red : 50% white

  22. Note the difference between reciprocal crosses

  23. The Key To Morgan’s Discovery • The key to Morgan’s discovery was the observation that all the white eyed individuals in the F2 generation were males • Without this vital data on the association of white eyes with being male, the gene for white eyes could have been seen as a simple recessive trait on an autosome • This illustrates the importance of recording all the data possible and being alert to the possibility of interesting things being present in the data

  24. 0.05% Bridge’s experiment

  25. Unlinked versus Linked Genes • Unlinked genes are genes located on different chromosomes – assort independently of each other. • Linked genes are genes located on the same chromosome – always assort together unless crossing over occurs.

  26. Meiotic “crossing over” • During meiosis, sister chromosomes of homologous pairs “close pair” • Adjacent arms can transfer identical regions of their genes

  27. How to Determine Whether Genes are Linked or Not • Testcross the F1 individual. • If the parental types equal the recombinant types (1:1:1:1), genes are NOT linked. • If the parental types are significantly greater than recombinant types, genes ARE linked.

  28. assortment = 50% recombination Independent assortment = 50% recombination

  29. X crossingover

  30. P PP LL x pp ll purple flower, long pollen red flower, short pollen F PpLl 1 purple flower, long pollen

  31. Linked (non-independent) genes

  32. Most offspring like parents mom dad

  33. Example of Linked Genes - 3 • Recombinants are so much rarer than parental types when genes are linked because they are due to crossing over. • Crossing over occurs rarely between the same two genes, so the frequency of recombinants is less than the frequency of parental types.

  34. w A c x B d y 1 2 3 4 Linkage Groups - 1 • A linkage group is a chromosome. • Consider the 4 linkage groups below: Genes w, x, and y are linked. Genes c and d are linked. Genes A and B are not linked, nor is gene A or gene B linked to the genes in the other linkage groups (chromosomes). Genes c and d are not linked to genes w, x, and y.

  35. Linkage Groups - 2 • If genes are unlinked, they are said to be assorting independently. • If genes are linked they may be: • Partially linked if crossing over is possible, or • Completely linked if crossing over is not possible.

  36. Mendel studied loci on separate chromosomes • Gene loci on the same chromosome are “linked” • Their alleles tend to be inherited together • Crossing over causes linked alleles to be unlinked • Crossing over is very common somewhere among most eukaryotes’ chromosomes (low rate) Therefore there are many more possible outcomes

  37. Chromosome Maps - 1 • In 1917, Alfred Sturtevant, a student of Morgan, reasoned that the frequency of crossing over is directely related to the distance between the genes on the chromosome. • He used recombination frequencies to position genes in correct order on a chromosome (genetic map).

  38. Chromosome Maps - 2 • Sturtevant defined 1 map unit as equivalent to 1% recombination (centiMorgan): • Map Distance (MD) = 100 X [(# recombinants)/(total # of offspring)]. • Example: black body and vestigial wings • MD = [(206 + 185)/(965 + 944 + 206 + 185)]*100 • MD = 17.0 map units

  39. Chromosome Maps - 3 • Recombination frequencies provide information on the relative distance between genes along a chromosome. • From this information, it is possible to determine the sequence of genes along a chromosome (order and distance between each consecutive pair).

  40. Recombination crossing-over

  41. parental eye color : pr+ (red) and pr (purple) wing length: vg+ (normal) and vg (vestigial) P pr pr vg vg x pr+ pr+ vg+ vg+ F pr pr+ vg vg+ 1 pr pr+ vg vg+ x pr pr vg vg pr+ vg+ 1339 pr vg 1195 pr+ vg 151 pr vg+ 154

  42. P pr pr vg vg x pr+ pr+ vg+ vg+ pr pr+ vg vg+ F 1 pr pr+ vg vg+ x pr pr vg vg pr+ vg+ 1339 parental pr vg 1195 pr+ vg 151 pr vg+ 154 305 recombinants = 10.7% 2839 progeny

  43. parental P pr+ pr+ vg vg x pr pr vg+ vg+ pr pr+ vg vg+ F 1 pr pr+ vg vg+ x pr pr vg vg pr+ vg+ 130 pr vg 121 pr+ vg 990 pr vg+ 1094 251 recombinants = 10.7% 2839 progeny

  44. TEST A 3 PUNTI 3 mutanti recessivi di Drosophila vocchi vermilion cvsenza crossvein ct ali a lancia P +/+ cv/cv ct/ct X v/v +/+ +/+

  45. + cv ct + cv ct v + + v + + X P + cv ct v + + F 1 v cv ct v cv ct Cross to triply recessive tester v + + 580 Parentali + cv ct 592 v cv + 45 Ordine: v ct cv + + ct 40 v cv ct 89 + + + 94 v + ct 3 Doppi scambi + cv + 5 1448

  46. 1°- Individuare le classi più frequenti. Perchè? Sono i genotipi Parentali in cui non ci sono stati scambi. 2°- Individuare le classi meno frequenti. Perchè? Il verificarsi simultaneo di 2 crossing over, uno nel primo intervallo e uno nel secondo, essendo 2 eventi indipendenti, avrà una probabilità pari al prodotto delle singole probabilità e quindi avrà un valore inferiore ai singoli. Le classi meno frequenti rappresentano quindi i doppi crossing over ( un doppio crossing over ha come effetto di cambiare la posizione del solo marcatore centrale) Noto l’assetto parentale e quello dei doppi scambi, si capisce subito l’ordine dei geni, cioè quale sta nel mezzo.

  47. + ct cv + ct cv v + + v + + Riscriviamo X P + ct cv v + + F 1 89+94+45+40/1448 268/1448 = 18.5 v + + 580 Parentali v - cv + ct cv 592 89+94+3+5=191/1448=13.2 Scambio tra v e ct v ct cv 89 + + + 94 45+40+3+5=93/1448=6.4 v + cv 45 Scambio tra ct e cv + ct + 40 v ct + 3 Doppi scambi + + cv 5 1448

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