Gene linkageGene linkageseminarseminar

No No 405405 Heredity Heredity

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 Mendel´s principle of combinationprinciple of combination

(independent assortment)(independent assortment)

members of two and more allelic pairs segregate members of two and more allelic pairs segregate

independently - there are as many types of gamets as independently - there are as many types of gamets as

possible combinations among maternal and paternal possible combinations among maternal and paternal

chromosomes (alleles)chromosomes (alleles)

Genes on the same chromosome show linkageGenes on the same chromosome show linkage

- they have tendency to be transmitted together through - they have tendency to be transmitted together through

meiosismeiosis

Genes on 1 chromosome = linkage groupGenes on 1 chromosome = linkage group

Morgan´s lawsMorgan´s laws1. The genes on chromosomes are in linear order

2. Number of linkage groups is equal to the

number of pairs of homologous chromosomes

Only the recombination of parental alleles results into the Only the recombination of parental alleles results into the

formation of gametes with the configuration of linked alleles formation of gametes with the configuration of linked alleles

different from the configuration of parent alleles different from the configuration of parent alleles

cis phase trans phasecis 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

• F2 cis trans F2 cis trans

ABAB abab

ABABABABABAB

ababABAB

ababABABabab

abababab

AbAb aBaB

AbAbAbAbAbAb

aBaBAbAb

aBaBAbAbaBaB

aBaBaBaB

GP 1 : 2 : 1 1 : 2 : 1

FP 3 : 1 1 : 2 : 1

B1 ABAB abab

ababABABabab

abababab

AbAb aBaB

ababAbAbabab

aBaBabab

1 : 1 1 : 1

RRecombination ecombination –– incomplete linkage incomplete linkage

A aA

B b B

a

b

A

b

A a

B b

a

B

A

B

A a a

b B b

A

b

A a

B b

a

B

cis trans

Non-recombinant AB ab Ab aB

Recombinant Ab aB AB ab

Proof of recombination: Proof of recombination: from B1 (testcross)from B1 (testcross)

cis cis trans trans

B1 ABB1 AB>>Ab aB Ab aB << ab AB ab AB << Ab aB Ab aB >> ab ab

Mechanism of recombination Mechanism of recombination –– crossing over crossing over

C.O. = reciprocal exchange of segments between C.O. = reciprocal exchange of segments between

nonsister chromatids of homologous chromosomesnonsister chromatids of homologous chromosomes

Probability of c.o. Probability of c.o. –– dependent on the alleles distance dependent on the alleles distance

0 0 << PP << 1 1P = 0 complete linkage, P = 1 free (independent)

combination

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

Strength of linkage pStrength 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 numberUnit: 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

• Bateson´s number c: expresses the ratio between recombined and non-recombined alleles (fromB1)

Phase cis =

Phase trans: c = Phase trans: c =

a1 + a4

a2 + a3

a2 + a3

a1 + a4

a1- frequency of gametes AB (zygotes AaBb from B1)

a2 - frequency of gametes Ab (zygot es Aabb from B1

a3 - frequency of gametes aB (zygot es aaBb from B1)

a4 - frequency of gametes ab (zygotes aabb from B1)

1 < c < ∞ for independent combination c = 1

Phase cis: p = a2 + a3

a1 + a2 + a3 + a4

trans: p = a1 + a4

a1 + a2 + a3 + a4

Morgan´s number – from B1

1. In B1 (testcross) generation of parental cross AB/AB x ab/ab902 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.

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 hairWrite 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

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?

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?

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.

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?

Thompson and Thompson: Clinical genetics, chapter 10 :

Human gene mapping , pg. 207-222, 7th edition