1. 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

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

3. Meiosis without recombination Meiosis with recombination
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
ProphaseI
MetaphaseI
TelophaseI
MetaphaseII
Gametes
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?

4. Meiosis without recombination
A b
a B
A b
a B
ProphaseI
A b
a B
MetaphaseI
TelophaseI
A b
a B
If the frequency of meiosis without crossing-over is 60%, we have these gametes:
30% (½ del 60%) Ab
30% (½ del 60%) aB
MetaphaseII
A b
a B
Gametes
A b
A b
a B
a B

5. Meiosis with recombination
A b
a B
A b
a B
ProphaseI
A b
a B
A b
a B
MetaphaseI
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
TelophaseI
A b
A B
a B
a b
MetaphaseII
A b
A B
a b
a B
Gametes

6. Determine the gametes and the frequencies:
Meiosis without crossing over
Meiosis with crossing over
Total Ab
30%
10%
40% aB ab AB
80% Parental Gametes
20% Recombinant Gametes
If A and B were independent genes, which kind and with which frequencies are the gametes expected? A B a b
25% AB : 25% ab : 25% Ab : 25% aB

7. (position of the genes on the chromosome)
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 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.
L’associazione tra geni ci permette di costruire delle mappe genetiche ovvero di determinare la posizione dei geni sul cromosoma.
Unità di mappa= distanza
The map unit is the distance between two genes and correspond to 1 recombinant product on 100 meiosis products.
(recombination frequency 1%)

8. Zb zb p1 P1 x zb zb p1 p1 X Zb p1 zb P1
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
Zb zb p1 P1 x zb zb p1 p1 Zb p1 zb X P1
The cross is between a double heterozygous with a homozygous recessive (TEST-CROSS)

9. X Which are the parental classes? Which are the recombinant classes?
The heterozygous parent produces these gametes
Parental gametes Zb p1 e zb P1
Recombinant gametes Zb P e zb p1 Zb p1 zb X P1
95% X 5%
distance of 5 mu then frequency of recombinant gametes is 5%
Each recombinant gamete is expected with a frequency of 2,5%
The parental gametes are 95:2= 47,5% each
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)

10. X Determine the phenotypic classes of the progeny: Zb p1 zb P1
The homozygous recessive produces only gametes : zb p1
The final phenotypes are:
Frequency
Phenotype
47,5%
Zb zb p1 p1
Normal Leaf, pale pericarp
zb zb P1 p1
Striated Leaf, dark pericarp
2,5%
Zb zb P1p1
Normal Leaf, dark pericarp
zb zb- p1 p1
Striated Leaf, pale pericarp

11. P F1 Draw the map b+ vg+ b vg X 18
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.
Draw the map
b vg+ 18
b vg X P
b vg+
b vg 18 F1
F1 will be double heterozygous with genes in cis and dominant phenotype.

12. Gametes of the heterozygous parent are: Parental b+ vg+ e b vg
If an individual of F1 is crossed with an recessive homozygous, which phenotypical classes are expected in the progeny and with which frequencies?
Gametes of the heterozygous parent are:
Parental b+ vg+ e b vg
Recombinant b+ vg e b vg+ b+
vg+ b vg X
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% each
The 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
b -vg-
Black body, vestigial wings 9%
b+ -vg-
Brown body, vestigial wings
b- vg+-
Black body, normal wings

13. The second parent is double homozygous recessive
Phenotypes
TOTAL
AB X ab AB Ab aB ab 79
Progeny phenotypes
The second parent is double homozygous recessive
The first parent is heterozygous for both genes
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.

14. Conclusion: we accept the hypothesis of INDEPENDENT GENES
We can compare the expected progeny with the observed progeny and figure out the c2
Phenotypes Xo H Xa
(Xo – Xa)2
: Xa c2 AB 23
19,75
10,56
0,53 Ab 21
1,56
0,08 aB 17
7,56
0,38 ab 18
3,06
0,15
Totale 79
4/4
1,14
GL = 4-1 = 3
c2 = 1,14, PROBABILITY = 70% / 80%
Conclusion: we accept the hypothesis of INDEPENDENT GENES A a B b
4) Scheme:

15. Second parent must be homozygous recessive for both characters.
phenotypes
TOTAL
GH X gh GH Gh gH gh
197
Phenotypes of progeny
Second parent must be homozygous recessive for both characters.
First parent must be double heterozygous (in progeny we have the recessive phenotype)
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

16. We can compare the expected progeny with the observed progeny and figure out the c2
Fenotipi Xo H Xa
(Xo – Xa)2
: Xa c2 GH 71
49,5
(21,5)2 Gh 26
(–23,5)2 gH 32
(–17,5)2 gh 68
(18,5)2
Totale
197
4/4
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): = 0,29 = 29%=29mu
197
G H
g h 29 X 4)

17. Genotypes Vv Zz x vv zz Parents phenotype TOTAL VZ X vz VZ 41 Vz 64
198
Progeny phenotypes
Genotypes Vv Zz x vv zz
Phenotypes Xo H Xa
(Xo – Xa)2
: Xa c2 VZ 41
49,5
72,5
1,45 Vz 64
210,25
4,2 vZ 55
30,25
0,61 vz 38
132,25
2,67
Totale
198
4/4
8,93
GL = 4-1 = 3
c2 = 8,93, P= 1% and 5%
Conclusion: The GENEs are not independent, they ARE LINKED
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.
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): = 0,39 = 39%
198
V z
v Z
v z 39 X 4)

18. 2) Expected genotype classes: ¼ DE : ¼ De : ¼ dE : ¼ de
Parent phenotypes
TOTAL
DE X de
DE 0 De dE
de 0
159
Progeny phenotypes
Genotypes Dd Ee x dd ee
2) Expected genotype classes:
¼ DE : ¼ De : ¼ dE : ¼ de
Totale of 159 individuals: expected 39,75 individuals in each class.
Expected class and observed class are very different
THE GENES ARE LINKED
3)The most frequent classes are De e dE, then the genes are in trans
The distance is : = 0,00 = 0%
159
The distance is NULL
D e
d E
d e X 4)

19. 2) Expected Phenotypical classes: 3/8 SF : 3/8 Sf : 1/8 sF : 1/8 sf
Parents phenotypes
TOTAL
SF X Sf
SF 92 Sf sF sf
240
Progeny phenotypes
Genotypes Ss Ff x Ss ff
2) 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
Fenotipi Xo H Xa
(Xo – Xa)2
: Xa c2 SF 92
3/8 90 4
0,04 Sf 87 9
0,1 sF 30
1/8 sf 31 1
0,03
Totale
240
8/8
0,17
GL = 4-1 = 3
c2 = 0,17 P= 90% -100%
INDIPENDENT GENES F f S s 4)

20. 2) Expected Phenotypes classes: 9/16 QW : 3/16 Qw : 3/16 cW : 1/16 qw
Parent phenotypes
TOTAL
QW X QW QW
Qw 58 qW qw
314
Progeny phenotypes
Genotypes Qq Ww x Qq Ww
2) 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
Phenotypes Xo H Xa
(Xo – Xa)2
: Xa c2 QW
178
9/16
176,6
1,96
0,01 Qw 58
3/16
58,9
0,81 qW 61
4,41
0,07 qw 17
1/16
19,6
6,76
0,34
Totale
314
16/16
0,43
GL = 4-1 = 3
c2 = 0,43 P = 90% -100%
INDIPENDENT GENES W w Q q 4)

21. For the genes M and N we expected 4 different gametes
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?
Drow the map
Genes 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%
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%
For the gene E we have : ½ E, ½ e.

22. For the genes M and N we expected 4 different gametes:
Which gametes are produced if the genes M and N are in trans?
Drow the map
Genes 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%
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 ,5%
mnE ,5% mne ,5%