Chromosomes: The Physical Basis of Inheritance

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

26. 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.

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

28. 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.

29. assortment = 50% recombination Independent assortment = 50% recombination

30. X crossingover

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

32. Linked (non-independent) genes

33. Most offspring like parents mom dad

34. 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.

35. 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.

36. 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.

37. 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

38. 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).

39. 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

40. 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).

41. Recombination crossing-over

42. 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

43. 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

44. 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

47. 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 +/+ +/+

48. + 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

49. 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.

50. + 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