1. Genetic recombination in Eukaryotes: crossing over, part 1 I.Genes found on the same chromosome = linked genes II.Linkage and crossing over III.Crossing over & chromosome mapping

2. I. Genes found on the same chromosome = linked genes Conflicting cytological evidence, only a few dozen chromosomes/individual – so must be several genes per chromosome Conflicting cytological evidence, only a few dozen chromosomes/individual – so must be several genes per chromosome cytological studies revealed only a few dozen chromosomes present, yet each species has thousands of genes cytological studies revealed only a few dozen chromosomes present, yet each species has thousands of genes highly likely that each chromosome would carry many hundreds/thousands of genes highly likely that each chromosome would carry many hundreds/thousands of genes therefore not all genes could assort independently therefore not all genes could assort independently Testcross experiments revealed gene linkage – observed deviations from the expected 1:1:1:1 ratio based upon independent assortment Testcross experiments revealed gene linkage – observed deviations from the expected 1:1:1:1 ratio based upon independent assortment

3. If a testcross is done and the genes are on separate chromosomes: Aa/Bb x aa/bb Aa/Bb aa/bb Aa/bb aa/Bb 1:1:1:1 observed 2 genes, located on different chromosomes, will segregate independently.

4. Chromosome is the unit of transmission, not the gene Linkage = two or more genes located on the same chromosome Linked genes are not free to undergo independent assortment Instead, the alleles at all loci of one chromosome, should in theory, be transmitted as a unit during gamete formation.

5. When two genes are compeletely linked, no crossing over occurs therefore, each gamete receives the alleles present on one chromatid or the other: AB or ab

6. II. Linkage and crossing over A.Crossing over – breakage and rejoining process between 2 NONSISTER chromatids Crossing over produces recombinants The % of recombinant gametes varies, dependent upon location of the loci. The closer the genes are, the less likely recombination will occur NR= parental R = recombinant

7. Breakage and rejoining process between two homologous non-sister chromatids -there can be one or more cross-overs -the cross over can occur anywhere

8. RECOMBINANT Crossover gamete Recombination Frequency = the # of recombinants/total progeny Parental gamete Parental gamete

9. 1). In the case of unlinked genes, independent assortment holds true… Testcross: Heterozygous x homozygous mutant AaBb x aabb Offspring: the # of recombinants = the # of parental types Recombination Frequency = (RF) = 1/2 or 50% B. Recombination Frequency, unlinked genes v. linked genes

10. 2). In the case of linked genes, no independent assortment Offspring: the # of recombinants < the # of parental types Recombination Frequency = the # of recombinants/total progeny  (RF) < 1/2 or 50%

11.  Crossing between adjacent non sister chromatids generates recombinants  The two chromatids not involved in the exchange result in non- parental gametes  The closer two loci are, the lower the RF (<1/2)

12.  We can compare the RF to what one would expect with independent assortment… RF Range – 0% to 50% RF significantly < 50% - Linkage RF = 50% - not linked

13. Recombination by Crossing Over – points to keep in mind: 1.CO’s can occur between any two nonsister chromatids 2.If there is NO crossing over, only parental types will be observed 3.If there IS crossing over, RF will increase up to 50% 4.when the loci of two linked genes are very far apart, the RF approaches 50%, 1:1:1:1 ratio observed, thus transmission of the linked genes is indistinguishable from that of two unlinked genes

14. Morgan noted the proportion of recombinant progeny varied depending on which linked genes were being examined… Testcross F1 results: RF = 11% RF = 1.4% As Morgan studied more linked genes, he saw that the proportion of recombinant progeny varied considerably. pr + vg + 1339 pr vg 1195 pr + vg 151 pr vg + 154 y w 43 y + w 2146 y w + 2302 y + w + 22 pr + pr vg + vg x pr pr vg vg y + y w + w x yy ww

15. III. Crossing over & chromosome mapping Morgan thought the variations in RF might indicate the actual distances separating genes on the chromosomes. Sturtevant (Morgan’s student) compiled data on recombination between genes in Drosophila test crosses  The closer the two linked genes, the lower the recombination frequency- thus RF may be correlated with the map distance between two loci on a chromosome

16. A. Linkage Maps Linkage of genes can be represented in the form of a genetic map, which shows the linear order of genes along a chromosome. Can also determine the distance between the genes. The % recombinant offspring is correlated w/the distance between the two genes Can also determine the distance between the genes. The % recombinant offspring is correlated w/the distance between the two genes Variations in recombination frequency indicate actual distances separating the genes on chromosomes Variations in recombination frequency indicate actual distances separating the genes on chromosomes

17. B. Map Units Map Unit (m.u.) = the distance between genes for which one product of meiosis out of 100 is recombinant [RF of 1% = 1 m.u. or 1 cM] e.g. if RF 12% between A & B, and 28% between B & C: A B C 12 mu 28 mu

19. F1: F2 RF – 1.3%, therefore y is 1.3mu from w males females

20. F1 F2 males females w 37.2 mu from m

21. A plant of genotype: AB ab Is test-crossed to a b If the two loci are 10 m.u. apart, what proportion of progeny will be A B / a b?

22. In the garden pea, orange pods (orp) are recessive to normal pods (Orp), and sensitivity to pea mosaic virus (mo) is recessive to resistance to the virus (Mo). A plant with orange pods and sensitivity is crossed to a true-breeding plant with normal pods and resistance. The F1 plants were then test-crossed to plants with orange pods and sensitivity. The following results were obtained: 160 orange pods/sensitive 165 normal pods/resistant 36 orange pods/resistant 39 normal pods/sensitive calculate the map distance between the two genes 36 + 39 =.1875 x 100 = 18.8 m.u. 400

23. C. Mapping multiple genes – Three- point mapping & Alfred’s research Hypothesis = when multiple genes are located on the same chromosome, the distance between the genes can be estimated from the proportion of recombinant offspring. Sturtevant’s First Genetic Map

24. A.Sturtevant’s First Genetic Map The linear order of these genes can be determined using testcross data Examined 5 different genes: y, w, v, m, r  All alleles were found to be recessive and X linked.  Crossed the double heterozygote female with hemizygous male recessive for the same alleles. Example:y+y w+w x yw y+w+ yw y+w yw+ RF = 214/21,736 = 0.0098 w+w r+r x wr w+r+ wr w+r wr+ RF = 2,062/6116 = 0.337 1 mu between y & w, 33.7 mu between w & r

25. genes are arranged on the chromosome in a linear order- which can be determined…

26. Alleles # R./total# RF y and w 214/21,7361% y and v 1,464/4,55132.2% y and r 115/32435.5% y and m 260/69337.5% w and v 471/1,58429.7% w and r 2,062/611633.7% w and m 406/89845.2% v and r 17/5733% w and m 109/40526.9% The Complete Data:

27. y-w = 1 m.u. v-r = 3 m.u. y-m = 37.5 m.u. w-r = 33.7 m.u. w-v = 29.7 m.u. Suggesting that v is between r & w, but closer to r y w v r m 57.6 1 29.7 3 23.9 Map distances more accurate between genes that are closer together, as the RF approaches 50%, the value becomes more inaccurate as a measure of map distance… More accurate