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.

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?

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

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

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

(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%)

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)

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 P1 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)

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

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.

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

The second parent is double homozygous recessive Phenotypes TOTAL AB X ab AB 23 Ab 21 aB 17 ab 18 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.

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 ¼ 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:

Second parent must be homozygous recessive for both characters. phenotypes TOTAL GH X gh GH 71 Gh 26 gH 32 gh 68 197 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

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 ¼ 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): 26 + 32 = 0,29 = 29%=29mu 197 G H g h 29 X 4)

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 ¼ 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): 41 + 38 = 0,39 = 39% 198 V z v Z v z 39 X 4)

2) Expected genotype classes: ¼ DE : ¼ De : ¼ dE : ¼ de Parent phenotypes TOTAL DE X de DE 0 De 78 dE 81 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 = 0,00 = 0% 159 The distance is NULL D e d E d e X 4)

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 87 sF 30 sf 31 240 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)

2) Expected Phenotypes classes: 9/16 QW : 3/16 Qw : 3/16 cW : 1/16 qw Parent phenotypes TOTAL QW X QW QW 178 Qw 58 qW 61 qw 17 314 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)

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.

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 2,5% mnE 2,5% mne 2,5%