The document discusses genetic mapping and pedigree analysis. It defines genetic mapping as using recombination frequency data to determine the relative positions of genes on chromosomes. It provides an example of calculating recombination frequency and mapping genes in Drosophila. Pedigree analysis is defined as using standardized symbols in family trees to show genetic relationships and inheritance patterns across generations, like for inherited diseases.

1. LECTURE 6 2.3 Genetic Mapping 2.4 Pedigree Analysis

2. OBJECTIVES At the end of the lesson, students should be able to : Analyze the example on pedigree given. Calculate and map a chromosome’s genetic loci using recombination data. Determine the position of genes/loci along a chromosome based on recombinant frequency data. Explain Pedigree analysis.

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. Genetic Mapping Formula COV = total number of recombinant / total number of offspring x 100

5. If the genes are not linked, the expected phenotype ratio of such a cross is 1:1:1:1 and there is a 50 per cent chance that alleles on separate chromosomes will be inherited together, the expected COV is 50 per cent; the lower the value, the closer the genes.

6. Thus the COV can be used to locate the relative positions of genes on chromosomes, a process called chromosome mapping or genetic mapping. By convention, one per cent COV is equivalent to one map unit . Today the word centimorgan is often used.

7. Cross over (meiosis) Homologous chromosome in synapsis – 4 chromatid (tetrad) P Q p q kiasma r R

8. Genetic mapping of Drosophila melanogaster genes

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. 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. EXERCISE 1 P, Q, R dan S genes are linked. COV for pairs of genes is as follows: P and Q = 35% P and R = 5% R and Q= 40% Q and S = 10% R and S = 30% Map the chromosome. Tips: Start with the furthest. (highest cov % )

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. 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. EXERCISE 3 In tomatoes, S = smooth s = wrinkled M = red flower m= white flower Test cross between smooth, red flower plant with wrinkled, white flower plant resulted in: Smooth fruit ,red flower = 300 Wrinkled fruit , white flower = 300 Smooth fruit, white flower = 100 Wrinkled fruit, red flower = 100 Count the map unit for the distance between both S and M genes on its chromosome.

15. Total progeny = 800 Recombinant = 100 + 100 800 = 25% Map unit for both S and M genes = 25 m.u S M 25 SM/sm x sm/sm Smooth fruit ,red flower = 300 (SM/sm) Wrinkled fruit , white flower = 300 (sm/sm) Smooth fruit, white flower = 100 (Sm/sm) Wrinkled fruit, red flower = 100 (sM/sm)

16. Conclusion Through genetic mapping, we can determine : Relative distance between two genes. Linear sequence of the linked genes on a chromosome.

17. Pedigree Analysis A pedigree is a family tree that shows genetic interrelationship among parents and offspring across one or more generations. Standardized symbols, methods and definitions are used to construct it. Eg: Inherited disease in the British Royal family.

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. Eg: “widow’s peak” trait (dominant allele) W Allele = widow’s peak w allele – no widow’s peak

20. I II III Widow’s peak pedigree Color = widow’s peak Ww ww ww WW @ Ww ww Ww Ww