Hour 6

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Hour 6

  1. 1. LECTURE 6 <ul><li>2.3 Genetic Mapping </li></ul><ul><li>2.4 Pedigree Analysis </li></ul>
  2. 2. OBJECTIVES <ul><li>At the end of the lesson, students should </li></ul><ul><li>be able to : </li></ul><ul><li>Analyze the example on pedigree given. </li></ul><ul><li>Calculate and map a chromosome’s genetic loci using recombination data. </li></ul><ul><li>Determine the position of genes/loci along a chromosome based on recombinant frequency data. </li></ul><ul><li>Explain Pedigree analysis. </li></ul>
  3. 3. Genetic Mapping The proportion of recombinants resulting from a dihybrid test cross is used to calculate the cross over value (COV), a measure of linkage, and if linkage occurs, the distance between genes . COV is also known as recombination frequency.
  4. 4. Genetic Mapping Formula <ul><li>COV = total number of recombinant / </li></ul><ul><li>total number of offspring x 100 </li></ul>
  5. 5. <ul><li>If the genes are not linked, </li></ul><ul><li>the expected phenotype ratio of such a cross is 1:1:1:1 </li></ul><ul><li>and there is a 50 per cent chance that alleles on separate chromosomes will be inherited together, </li></ul><ul><li>the expected COV is 50 per cent; </li></ul><ul><li>the lower the value, the closer the genes. </li></ul>
  6. 6. <ul><li>Thus the COV can be used to locate the relative positions of genes on chromosomes, a process called chromosome mapping or genetic mapping. </li></ul><ul><li>By convention, one per cent COV is equivalent to one map unit . </li></ul><ul><li>Today the word centimorgan is often used. </li></ul>
  7. 7. Cross over (meiosis) Homologous chromosome in synapsis – 4 chromatid (tetrad) P Q p q kiasma r R
  8. 8. Genetic mapping of Drosophila melanogaster genes
  9. 9. Example P : CCSS x ccss P : CS/CS x cs/cs G : CS cs F1 : CS/cs x cs/cs G : CS cs Cs cS cs x co co CS/cs cs/cs cS/cs Cs/cs 480 480 20 20 Parent combination Recombinant
  10. 10. Genotype Individuals CS/cs 480 cs/cs 480 Cs/cs 20 cS/cs 20 Parent combination Recombinant (cross over) Eg : Test cross F1 : CS/cs x cs/cs Parent combination = (480 + 480)/1000 = 96% Recombinant = (20 + 20)/1000 = 4% (cov) If C/c dan S/s genes is far apart, cov > 4% Jumlah 1000 C S 4 map unit
  11. 11. EXERCISE 1 <ul><li>P, Q, R dan S genes are linked. </li></ul><ul><li>COV for pairs of genes is as follows: </li></ul><ul><li>P and Q = 35% </li></ul><ul><li>P and R = 5% </li></ul><ul><li>R and Q= 40% </li></ul><ul><li>Q and S = 10% </li></ul><ul><li>R and S = 30% </li></ul><ul><li>Map the chromosome. </li></ul><ul><li>Tips: Start with the furthest. (highest cov % ) </li></ul>
  12. 12. Solution : R and Q= 40% P and Q = 35% R and S = 30% Q and S = 10% P and R = 5% R Q 40 map unit 5 25 10 S P
  13. 13. Cross over value given for P and Q = 4% P and R = 1% State the cross over value possible for Q and R. P Q 4 1 3 Cross over value for Q and R = 3% P Q 4 1 Cross over value for Q and R = 5% EXERCISE 2 R R
  14. 14. EXERCISE 3 <ul><li>In tomatoes, </li></ul><ul><li>S = smooth s = wrinkled </li></ul><ul><li>M = red flower m= white flower </li></ul><ul><li>Test cross between smooth, red flower plant with wrinkled, white flower plant resulted in: </li></ul><ul><li>Smooth fruit ,red flower = 300 </li></ul><ul><li>Wrinkled fruit , white flower = 300 </li></ul><ul><li>Smooth fruit, white flower = 100 </li></ul><ul><li>Wrinkled fruit, red flower = 100 </li></ul><ul><li>Count the map unit for the distance between both S and M genes on its chromosome. </li></ul>
  15. 15. Total progeny = 800 Recombinant = 100 + 100 800 = 25% Map unit for both S and M genes = 25 m.u S M 25 <ul><li>SM/sm x sm/sm </li></ul><ul><li>Smooth fruit ,red flower = 300 (SM/sm) </li></ul><ul><li>Wrinkled fruit , white flower = 300 (sm/sm) </li></ul><ul><li>Smooth fruit, white flower = 100 (Sm/sm) </li></ul><ul><li>Wrinkled fruit, red flower = 100 (sM/sm) </li></ul>
  16. 16. Conclusion <ul><li>Through genetic mapping, we can determine : </li></ul><ul><ul><li>Relative distance between two genes. </li></ul></ul><ul><ul><li>Linear sequence of the linked genes on a chromosome. </li></ul></ul>
  17. 17. Pedigree Analysis <ul><li>A pedigree is a family tree that shows genetic interrelationship among parents and offspring across one or more generations. </li></ul><ul><li>Standardized symbols, methods and definitions are used to construct it. </li></ul><ul><li>Eg: Inherited disease in the British Royal family. </li></ul>
  18. 18. I II III Circle = female Square = male Roman = generation Numbers = individuals in generation Colored = different phenotype 1 2 1 1 2 4 3
  19. 19. Eg: “widow’s peak” trait (dominant allele) W Allele = widow’s peak w allele – no widow’s peak
  20. 20. I II III Widow’s peak pedigree Color = widow’s peak Ww ww ww WW @ Ww ww Ww Ww

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