1. Gene linkage seminar No 405 Heredity

2. Key words: complete and incomplete gene linkage, linkage group, Morgan´s laws, crossing-over, recombination, cis and trans linkage phases, strenght of linkage, Morgan´s number - p, centimorgan, gene map unit, Bateson´s number - c, gene mapping.

3. 3. Mendel´s principle of combination (independent assortment): Members of two and more allelic pairs segregate independently - there are as many types of gametes as possible combinations among maternal and paternal chromosomes (alleles). It is applied only in the case of different localization of genes on chromosomes (pairs of homologous chromosomes).

4. Dihybrid cross – genes on different chromosomes 2 allelic pairs – carried by different pairs of homologous chromosomes – segregate independently two types of parental crosses P: AABB x aabb AAbb x aaBB F1: AaBb double heterozygote genotypes of gametes AB Ab aB ab 1 : 1 : 1 : 1

5. ABAbaBab AB AABBAABbAaBBAaBb Ab AABbAAbbAaBbAabb aB AaBBAaBbaaBBaaBb ab AaBbAabbaaBbaabb Genotypes of gametes Punnett (combination) square for F2 generation genotype ratio 1 : 2 : 1 : 2 : 4 : 2 : 1 : 2 : 1 phenotype ratio 9 : 3 : 3: 1

6. Backcross - testcross B1 AaBb x aabb heterozygote x double recessive homozygote ABAbaBab AaBbAabbaaBbaabb Ratio of phenotypes in B1 1 : 1 : 1 : 1

7. Gene linkage 3. Mendel‘s law is not valid, if the crossed genes (A,B) are on the same chromosome (a pair of homologous chromosome) = They are in gene linkage Consequence: the alleles with non-recombinant (parental) combinations are more frequent, because they are passed together more frequently into gametes. The alleles with new recombinant combinations are less frequent – incomplete linkage

8. Incomplete linkage – dihybrid cross: Parental generation – genes on the same chromosome (a pair) not like in a situation of independent assortment F1 generation – AaBb = = incomplete linkage (cis) independent assortment In the case of incomplete linkage, alleles non-recombinant (AB, ab) are more frequently than the alleles new – recombinant (Ab, aB). A B a b A B a b AB ab

9. In the F2 generation are the same types of genotypes, but genotype and phenotype ratio is different – Non-recombinant (parental) alleles are formed more frequent than new recombinant alleles. In B1 generation are the same types of genotypes, phenotypes, but not in ratio 1:1:1:1 like in the case of independent assortment. Non-recombinant (parental) alleles are formed more frequent than new recombinant alleles e.g. 2:2:1:1

10. Individual AaBb in F1 forms for F2 and for B1 Alleles in cis phase in trans phase Non-recombinant AB ab Ab aB Recombinant Ab aB AB ab Incomplete linkage: AB ab AbaBAbaB

11. Proof of recombination: always in B1 testcross cis trans B1 AB>Ab aB ab AB and ab Ab and aB non-recombinant non-recombinant (parental) are more more frequent (parental) are more frequent Mechanism of recombination – crossing over Probability of c.o. depends on the distance between alleles

12. cis phase F1 trans phase F1 A B a b A b a B AB ab Ab aB Complete linkage - dihybrid cross Genes A an B are next to each other, there in no place for crossing over = recombination. Dihybrid forms only two types (non-recombinant/parental) of alleles with ratio 1:1 alleles AB ab Ab aB = AaBb =

13. P AB ab Ab aB x AB ab Ab aB F1 AB Ab ab aB x The whole dihybrid cross with complete gene linkage cis phase trans phase ABab ABABABabAB abABabababAbaBAbAbAbaBAb aBAbaBaBaB F2 cis trans

14. ABab ABABABabAB abABabababAbaBAbAbAbaBAb aBAbaBaBaB GP 1 : 2 : 1 1 : 2 : 1 FP 3 : 1 1 : 2 : 1 B1ABababABababab AbaBabAbabaBab 1 : 1 1 : 1

15. Morgan´s laws 1. 1.The genes on chromosomes are in linear order 2. 2.Number of linkage groups is equal to the number of pairs of homologous chromosomes

16. Strength of gene linkage, p = Morgan‘s number fr + fnr fr from B 1 (testcross) fr – frequency of recombinant gametes, fnr = frequency of non- recombinant gametes (phenotypes in B1) Unit: cM (centimorgan) = % of recombinants = expression of relative distance of alleles = map unit 1cM = recombinant frequency 1% (0,01) 0 < p < 0,5 p = 0 complete linkage p = 0,5 free (independent) assortment p =p =

17. Phase cis: p = fAb + faB fAB + fAb + faB + fab Phase trans: p = fAB + fab fAB + fAb + faB + fab Morgan´s number – from B1

18. Bateson´s number c: expresses: how more frequent are inherited non-recombined (parental) sets of alleles than recombined sets of alleles (from B1) Phase cis : c = Phase trans: c = fAB + fab fAb + faB fAB + fab 1 < c < ∞ for independent assortment c = 1

19. 1. In B1 (testcross) generation of parental cross AB/AB x ab/ab 902 individuals of phenotype AB, 898 of ab, 98 of Ab and 102 of aB phenotypes was detected. Calculate the strength of linkage between genes A and B (number p). 2. How would you prove that two allelic pairs are linked or not? 3. Hybrid of parental cross Ab/Ab x aB/aB is crossed with individual ab/ab in testcross. What is the ratio of genotypes in progeny comprising 1850 individuals? The strength of the gene linkage between A and B genes is 40% of recombinations.

20. 4. Rabbits bearing the allele En are spotted, whereas en codes the absence of colour spots. Allele l codes for the long (Angora) hair, and allele L codes for the normal short rabbit hair. Both allele pairs are in linkage. a)You should cross homozygous spotted rabbits with the normal length of the hair with Angora non-spotted rabbits. Describe genotypes of the P and F1 generations. Determine the gene linkage phase b)During the testcross (F1 x recessive homozygote) the following offsprings were obtained: 83 spotted rabbits with normal lenght of hair 12 spotted rabbits with long hair 14 rabbits without spots and with normal length of hair 89 rabbits without spots and with long hair Write the course of the backcross. c) Determine the strength of the gene linkage between both genes, and find out what is the ratio between the amount of parental and recombined gametes produced by the F1 dihybrid. d)Calculate the phenotype and genotype ratio in the F2 generation

21. 5. Determine the order of genes on chromosome if you know that p=5% for genes A and B, p=3% for genes B and C and p=2% for genes A and C.

22. 6. Dominant allele D is coding for Rh+ factor, recessive genotype dd is coding for Rh- phenotype (absence of Rh factor on the surface of erythrocytes). Elliptic -oval shape of erythrocytes (eliptocytosis) is coded by the dominant allele E, whereas the homozygously recessive genotype ee codes for normal round shape of erythrocytes. Both genes are in genetic distance of 20 cM. There is a man with the eliptocytosis, whose mother had a normal shape of erythrocytes and was homozygous for Rh+ and whose father was Rh- and heterozygous in allelic pair for eliptocytosis. This man got married to a healthy woman bearing Rh- factor. a) What is the probability that the first child will be Rh- with eliptocytosis?

23. b) What is the probability that the first child will be eliptocytic and Rh+? c)If the first child is Rh+, what is the probability that it will suffer from the eliptocytosis too? 7. Determine the cross ratios in B1 of the hybrid AB/ab, if c=2 and the number of offsprings in the B1 generation is 1585. 8. What is the cross ratio of the hybrid Ab/aB in the F2 generation, if c=5?

24. 9.The woman (X DH /X dh ) is a carrier (heterozygote) of recessive alleles for a daltonism (colour-blindness) and a hemophilia. Relative distance of genes is 10 cM. a)In which percentual ratio the recombined gametes in the woman arise? Write their genotypes. b)What is the probability that woman will have a hemophilic and daltonic son? What is the probability that she will have a hemophilic but not daltonic son or a daltonic but not hemophilic son or a healthy son? You should assume that the father is X DH /Y. c) The same you should calculate for the case that the mother has got all the allelic pairs in the trans phase.

25. 10.The colour-blind man had married a phenotypically normal woman. Their first son is healthy, two sons suffer from the hemophilia one daughter is colour-blind, and the second daughter is healthy. a) Draw a pedigree and determine the genotypes of each family member. b) What is the probability that sons of both daughters will suffer from hemophilia if they will marry healthy partners?

26. Thompson and Thompson: Clinical genetics, chapter 10 : Human gene mapping, pg. 207-222, 7th edition