Download presentation

Presentation is loading. Please wait.

Published bySimon Yam Modified over 2 years ago

1
Gene linkage seminar No 405 Heredity

3
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.

4
3. Mendel´s principle of combination (independent assortment) members of two and more allelic pairs segregate independently - there are as many types of gamets as possible combinations among maternal and paternal chromosomes (alleles) Genes on the same chromosome show linkage Genes on the same chromosome show linkage - they have tendency to be transmitted together through meiosis - they have tendency to be transmitted together through meiosis Genes on 1 chromosome = linkage group Genes on 1 chromosome = linkage group

5
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 Only the recombination of parental alleles results into the formation of gametes with the configuration of linked alleles different from the configuration of parent alleles

6
cis phase trans phase P AB ab Ab aB x AB ab Ab aB F1 AB Ab ab aB Complete gene linkage – dihybrid form only two types of gametes with ratio 1:1 A B a b A b a B x AB ab Ab aB

7
F2 cis transF2 cis trans ABab ABABABabAB abABabababAbaBAbAbAbaBAb aBAbaBaBaB GP 1 : 2 : 1 1 : 2 : 1 FP 3 : 1 1 : 2 : 1 B1ABababABababab AbaBabAbabaBab 1 : 1 1 : 1

8
Recombination – incomplete linkage A aA Bb B a b A b A a B b a B A B A aa b Bb A b A a BbBb a B cis trans Non-recombinant AB ab Ab aB Recombinant Ab aB AB ab

9
Proof of recombination: from B1 (testcross) cis trans cis trans B1 AB>Ab aB ab Mechanism of recombination – crossing over C.O. = reciprocal exchange of segments between nonsister chromatids of homologous chromosomes Probability of c.o. – dependent on the alleles distance 0 < P < 1 0 < P < 1 P = 0 complete linkage, P = 1 free (independent) combination

10
Crossing over - linked genes A/a, B/b

11
Strength of linkage p = fr + fnr fr from B 1 (testcross) fr – frequency of recombinant gametes, fnr = frequency of nonrecombinant gametes (phenotypes in B1) 0 < p < 0,5 p = Morgan´s number Unit: cM (centimorgan) = % of recombinants = expression of relative distance of alleles = map unit 1cM = rekombinant frequency 1% (0,01) p = 0 complete linkage p = 0,5 free (independent) combination

12
Bateson´s number c: expresses the ratio between recombined and non-recombined alleles (fromB1) Phase cis = Phase trans: c = Phase trans: c = a 1 + a 4 a 2 + a 3 a 1 + a 4 a 1 - frequency of gametes AB (zygotes AaBb from B1) a 2 - frequency of gametes Ab (zygot es Aabb from B1 a 3 - frequency of gametes aB (zygot es aaBb from B1) a 4 - frequency of gametes ab (zygotes aabb from B1) 1 < c < ∞ for independent combination c = 1

13
Phase cis: p = a 2 + a 3 a 1 + a 2 + a 3 + a 4 trans: p = a 1 + a 4 a 1 + a 2 + a 3 + a 4 Morgan´s number – from B1

14
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 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.

15
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

16
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. 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?

17
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?

18
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.

19
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?

20
Thompson and Thompson: Clinical genetics, chapter 10 : Human gene mapping, pg. 207-222, 7th edition

Similar presentations

OK

Heredity Unit 1 Test Review. 1. Another name for a sex cell. GAMETE.

Heredity Unit 1 Test Review. 1. Another name for a sex cell. GAMETE.

© 2018 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on dda line drawing algorithms Ppt on cartesian product example Download ppt on civil disobedience movement in the civil rights Ppt on study designs in epidemiology Ppt on trial and error dvd Ppt on classical economics vs keynesian Ppt on indian premier league Ppt on democratic and non democratic countries File type ppt on cybercrime articles Ppt on spiritual leadership in the home