Genetic linkage is the tendency of alleles that are located close together on a chromosome to be inherited together during the meiosis phase of sexual reproduction. Genes whose loci are nearer to each other are less likely to be separated onto different chromatids during chromosomal crossover, and are therefore said to be genetically linked. Linked Genes Violate the Law of Independent Assortment

RECOMBINATION OR CROSSING-OVER: The exchange of a segment of DNA between two homologous chromosomes during meiosis leading to a novel combination of genetic material in the offspring.Crossing over is the swapping of genetic material that occurs in the germ line

Crossing over results in a shuffling of genetic material and is an important cause of the genetic variation seen among offspring.

1 - Schematizza il processo di meiosi di un organismo diploide con n = 1, che porta due geni A e B sullo stesso cromosoma. L'individuo è eterozigote in trans per i geni A e B (i due alleli dominanti di A e B NON stanno sullo stesso cromosoma). meiosi in cui non meiosi in cui avviene ricombinazione avviene ricombinazione profase I metafase I telofase I metafase II gameti Supponendo che la frequenza di meiosi senza crossing-over sia del 60% e quella con crossing-over sia del 40%, indica i gameti prodotti e le rispettive frequenze. Quanti tipi di gameti e con quale frequenza sarebbero attesi se la cellula fosse doppia eterozigote per due geni indipendenti? Qual è la differenza tra le due situazioni?

Draw a scheme of the meiosis process of a diploid organism with n = 1, that has two genes, A and B, on the same chromosome. The individual is heterozygous in trans for the A and B genes ( the two dominant alleles A and B are NOT on the same chromosome).

Meiosis without recombination

Meiosis with recombination

Supposing that meiosis frequency without crossing-over is 60% and with crossing- over is 40%, show the resulting gametes and respective frequencies.How many kinds of gametes (and with which frequency) would be expected if the cell were double heterozygous for two independent genes?What’s the difference between these two situations?

ProphaseI

MetaphaseI

TelophaseI

MetaphaseII

Gametes

Meiosis without recombination

A b

a B

A b

a B

A b

a B

A b

A b a B

a B

A b

A b a B

a B

A b a BA b a B

If the frequency of meiosis without crossing-over is 60%, we have these gametes:30% (½ del 60%) Ab30% (½ del 60%) aB

ProphaseI

MetaphaseI

TelophaseI

MetaphaseII

Gametes

Meiosis with recombination

A b

a B

A b

a B

A b

a B

A b

A B a B

a b

A b

A B a B

a b

A b a bA B a B

If the frequency of meiosis with crossing-over is 40%, we have these gametes:10% Ab (1/4 del 40%)10% AB (1/4 del 40%)10% ab (1/4 del 40%)10% aB (1/4 del 40%)

A b

a B

A b

a B

ProphaseI

MetaphaseI

TelophaseI

MetaphaseII

Gametes

Determine the gametes and the frequencies:

Gametes Meiosis without crossing over

Meiosis with crossing over

Total

Ab 30% 10% 40%

aB 30% 10% 40%

ab 10% 10%

AB 10% 10%

80% Parental Gametes

20% Recombinant Gametes

If A and B were independent genes, which kind and with which frequencies are the gametes expected?

25% AB : 25% ab : 25% Ab : 25% aB

A

B

a

b

A

b

a

B

The recombination frequency allow us to figure out the distance between two genes and to build the genetic map.(position of the genes on the chromosome)

The map unit is the distance between two genes and correspond to 1 recombinant product on 100 meiosis products.(recombination frequency 1%)

The greater the distance between linked genes, the greater the chance that non-sister chromatids would cross over in the region between the genes.

Basing on the distance we can predict the frequency of recombination.

Zb

zb

P1

p1

Zb p1 zb p1

X zb P1 zb p1

The cross is between a double heterozygous with a homozygous recessive (TEST-CROSS)

Mais has Zb genes (Zebra crossband). The mutants has phenotype with striated leaf (Zb normal; zb striated). The gene P1 (pericarp color), determine the colour of pericarp dark red (P1 dark; p1 pale), These genes are on the same chromosome and they are 5 map units distant. Which phenotypical classes are expected in the progeny and with which frequencies?

Zb zb p1 P1 x zb zb p1 p1

Which are the parental classes? Which are the recombinant classes?

Zb p1 zb p1

X zb P1 zb p1

distance of 5 mu then frequency of recombinant gametes is 5%

The heterozygous parent produces these gametes

Parental gametes Zb p1 e zb P1

Recombinant gametes Zb P1 e zb p1

Each recombinant gamete is expected with a frequency of 2,5%The parental gametes are 95:2= 47,5% each

X

Parental class= during the meiosis, recombination process did not happen then the gametes derived from the original structure.

Recombinant class= during the meiosis recombination process happens (crossing-over)

95%

5%

The homozygous recessive produces only gametes : zb p1

The final phenotypes are:

Determine the phenotypic classes of the progeny:

Frequency Phenotype

47,5% Zb zb p1 p1 Normal Leaf, pale pericarp

47,5% zb zb P1 p1 Striated Leaf, dark pericarp

2,5% Zb zb P1p1 Normal Leaf, dark pericarp

2,5% zb zb- p1 p1 Striated Leaf, pale pericarp

Zb p1 zb p1

X zb P1 zb p1

X

In drosophila gene b (black body) and gene vg (vestigial wings) are 18 mu distant. The dominant alleles are b+ (brown color) and vg+ (normal wings), the recessive alleles b (black color) and vg (vestigial wings). A individual b+b+ vg+vg+ is crossed with an individual bb vgvg.

b+ vg+

b+ vg+

18

b vg

b vg18X

Draw the map

F1 will be double heterozygous with genes in cis and dominant phenotype.

b+ vg+

b vg

18

P

F1

Gametes of the heterozygous parent are:Parental b+ vg+ e b vgRecombinant b+ vg e b vg+

The frequences dipend on the distance between 2 genes: we know that the distance is 18 mu. The recombination frequency is 18%The recombinant gametes are 18 : 2 = 9% eachThe parental gametes are 100 – 18 = 82 : 2 = 41% each

The recessive parent produces only b vg gametes Then the expected phenotype are:

Frequency Phenotype class

41% b+- vg+- Brown body, normal wings

41% b -vg- Black body, vestigial wings

9% b+ -vg- Brown body, vestigial wings

9% b- vg+- Black body, normal wings

b+ vg+

b vg

X

If an individual of F1 is crossed with an recessive homozygous, which phenotypical classes are expected in the progeny and with which frequencies?

ParentPhenotypes

TOTAL

AB X ab AB 23 Ab 21 aB 17 ab 18 79

1) The second parent is double homozygous recessive 2) The first parent is heterozygous for both genes 3) Genotypes Aa Bb x aa bb

2) Expected phenotype classes (if the genes are independent) ¼ AB : ¼ Ab : ¼ aB : ¼ ab Total of 79 individuals: we expected 19,75 individuals for each class.

Progeny phenotypes

Phenotypes Xo H Xa (Xo – Xa)2 : Xa c2

AB 23 ¼ 19,75 10,56 19,75 0,53

Ab 21 ¼ 19,75 1,56 19,75 0,08

aB 17 ¼ 19,75 7,56 19,75 0,38

ab 18 ¼ 19,75 3,06 19,75 0,15

Totale 79 4/4 79 1,14

GL = 4-1 = 3c2 = 1,14, PROBABILITY = 70% / 80%Conclusion: we accept the hypothesis of INDEPENDENT GENES

A

a

B

b

4) Scheme:

We can compare the expected progeny with the observed progeny and figure out the c2

Parent phenotypes

TOTAL

GH X gh GH 71 Gh 26 gH 32 gh 68 197

1) Second parent must be homozygous recessive for both characters.2) First parent must be double heterozygous (in progeny we have the

recessive phenotype)3) Genotypes Gg Hh x gg hh

4) Expected phenotipic classes ¼ GH : ¼ Gh : ¼ gH : ¼ gh total of 197, we expect 49,25 individual for each class

Phenotypes of progeny

Fenotipi Xo H Xa (Xo – Xa)2 : Xa c2

GH 71 ¼ 49,5 (21,5)2

Gh 26 ¼ 49,5 (–23,5)2

gH 32 ¼ 49,5 (–17,5)2

gh 68 ¼ 49,5 (18,5)2

Totale 197 4/4 197

We don’t performe the test because the values are very different. The genes are linked

3) We can establish how the genes is linked: the more frequent classes are GH e gh, then the genes are in cis. The distance is the frequency of the recombinants: (Gh, gH): 26 + 32 = 0,29 = 29%=29mu 197

We can compare the expected progeny with the observed progeny and figure out the c2

G H

g h

g h

g h

29 29X4)

Parentsphenotype

TOTAL

VZ X vz VZ 41 Vz 64 vZ 55 vz 38 198

Genotypes Vv Zz x vv zz

2) Expected phenotypes if the genes are independent ¼ VZ : ¼ Vz : ¼ vZ : ¼ vz Total of 198 individuals, we expected 49,5 individuals for each class.

Progeny phenotypes

Phenotypes Xo H Xa (Xo – Xa)2 : Xa c2

VZ 41 ¼ 49,5 72,5 49,5 1,45

Vz 64 ¼ 49,5 210,25 49,5 4,2

vZ 55 ¼ 49,5 30,25 49,5 0,61

vz 38 ¼ 49,5 132,25 49,5 2,67

Totale 198 4/4 198 8,93

GL = 4-1 = 3c2 = 8,93, P= 1% and 5%Conclusion: The GENEs are not independent, they ARE LINKED

3) The most frequent classes are Vz e vZ, then the genes are in trans The distance is figured out from the frequency of recombination (VZ, vz): 41 + 38 = 0,39 = 39% 198

V z

v Z

v z

v z

39 39X4)

Parent phenotypes TOTAL

DE X de DE 0 De 78 dE 81 de 0 159

Genotypes Dd Ee x dd ee2) Expected genotype classes: ¼ DE : ¼ De : ¼ dE : ¼ de Totale of 159 individuals: expected 39,75 individuals in each class.

Progeny phenotypes

Expected class and observed class are very differentTHE GENES ARE LINKED

3)The most frequent classes are De e dE, then the genes are in transThe distance is : 0 = 0,00 = 0% 159The distance is NULL

D e

d E

d e

d e

0 0X4)

Parents phenotypes TOTAL

SF X Sf SF 92 Sf 87 sF 30 sf 31 240

Genotypes Ss Ff x Ss ff2) Expected Phenotypical classes: 3/8 SF : 3/8 Sf : 1/8 sF : 1/8 sf The Total is 240 individuals, we expect 90 : 90 : 30 : 30 individulas for each class

Progeny phenotypes

Fenotipi Xo H Xa (Xo – Xa)2 : Xa c2

SF 92 3/8 90 4 90 0,04

Sf 87 3/8 90 9 90 0,1

sF 30 1/8 30 0 30 0

sf 31 1/8 30 1 30 0,03

Totale 240 8/8 240 0,17

GL = 4-1 = 3c2 = 0,17 P= 90% -100%INDIPENDENT GENES

F

f

S

s4)

Parent phenotypes TOTAL

QW X QW QW 178 Qw 58 qW 61 qw 17 314

Genotypes Qq Ww x Qq Ww2) Expected Phenotypes classes: 9/16 QW : 3/16 Qw : 3/16 cW : 1/16 qw Total 314, expected 176,6 : 58,9 : 58,9 : 19,6

Progeny phenotypes

Phenotypes Xo H Xa (Xo – Xa)2 : Xa c2

QW 178 9/16 176,6 1,96 176,6 0,01

Qw 58 3/16 58,9 0,81 58,9 0,01

qW 61 3/16 58,9 4,41 58,9 0,07

qw 17 1/16 19,6 6,76 19,6 0,34

Totale 314 16/16 314 0,43

GL = 4-1 = 3c2 = 0,43 P = 90% -100%INDIPENDENT GENES

W

w

Q

q4)

Drow the mapGenes in cis M N

m n

10

E

e

For the genes M and N we expected 4 different gametes

MN 45%mn 45%Mn 5%mN 5%

For the gene E we have : ½ E, ½ e.

Final results with the gene E: 8 different gametesMNE 22,5% MNe 22,5% mnE 22,5% mne 22,5%MnE 2,5% Mne 2,5%mNE 2,5% mNe 2,5%

Genes M and N are concatenated at 10 mu. The E gene is located on another chromosome. Which gametes are produced by an individual: MmNnEe if genes MN are in cis?

For the genes M and N we expected 4 different gametes:

Mn 45%mN 45%MN 5%mn 5%

Final results with the gene E: 8 different gametes

MnE 22,5% Mne 22,5%mNE 22,5% mNe 22,5%MNE 2,5% MNe 2,5%mnE 2,5% mne 2,5%

M n

m N

10

E

e

Which gametes are produced if the genes M and N are in trans?

Drow the mapGenes in cis