LECTURE 6: LINKAGE

Linked genes, recombination, and chromosomal mapping Mendel's Law of Independent Assortment is a consequence of the fact that chromosomes segregate independently in meiosis Take two individuals AABB x aabb Take the heterozygous and cross with homozygous recessive AaBb x aabb ab AB AaBb 25% GrandParent phenotype aB aaBb 25% Aabb Ab 25% ab aabb 25% Grandparent phenotype

Chromosome alignment in MeiosisI These results are readily explained by the two alternative ways the chromosomes can line up on the metaphase plate during meiosis I: A a A a OR B b b B AB ab Ab aB Because the A and B genes assort independently, AaBb dihybrids constructed from different parental genotypes will behave the same. P AABB x aabb F1 AaBb P AAbb x aaBB F1 AaBb

AABB x aabb AAbb x aaBB AaBb A a B b OR Test Cross with aabb AaBb x aabb AaBb x aabb a b a b A B AaBb Parent 25% A B AaBb 25% a b aabb Parent a b aabb A b Aabb A b Aabb Parent a B aaBb Parent a B aaBb

Why 50:50 Why not 25:25:25:25 A hypothetical dihybrid cross involving the genes A and B produced the following results: · A= Tall a= short · B= Blue b = white Cross I: Cross II: Tall, Blue x short, white Tall, white x short, Blue AABB aabb AAbb aaBB Tall, Blue AaBb Tall, Blue AaBb X X short, white aabb short, white aabb 50% Tall, Blue 50% Tall, white 50% short, white 50% short, Blue Parent only Parent only Why 50:50? Why not 25:25:25:25?

In these crosses, independent assortment is not occurring. For example in the first cross, the alleles Tall and Blue behave as if they are linked to one another. They look like the Parental class only Similarly in the second cross the alleles Tall and white appear as if they are linked to one another. They look like the parental class only These results are readily explained if the genes A andBC lie next to one another on the same chromosome:

If the two genes are on the same chromosome then Cross: A= Tall a= short B= Blue b = white If the two genes are on the same chromosome then A-B Tall Blue a-b Short white X P A-b Tall White a-B Short Blue X P a-b X A-B a-b F1 a-b X A-b a-B F1 a-b a-b A-B a-b Tall Blue A-B A-B a-b a-b Short white a-b A-b a-b Tall white A-b A-b a-B a-b Short Blue a-B a-B

AABB x aabb AAbb x aaBB AaBb A a B b OR Test Cross with aabb AaBb x aabb AaBb x aabb a b a b A B AaBb Parent 25% A B AaBb 25% a b aabb Parent a b aabb A b Aabb A b Aabb Parent a B aaBb aaBb Parent a B

Separate versus Same chromosome AABB x aabb AABB x aabb AaBb AaBb AaBb x (aabb) Test Cross Test Cross AaBb x (aabb) A a B b a A b B A a b B OR a b a b A B AB ab AB ab 25% 50% A B a b ab ab a b A b Ab ab aB ab Ab a B A b 0% a B aB ab

Purple vestigial Morgan performed the following experiments in Drosophila to investigate the independent assortment of the genes pr and vg PR+ = normal red eyes pr = purple eyes VG+ = normal wings vg = vestigial wings P PR+VG+/PR+VG+ x prvg/prvg F1 PR+VG+/pr vg x prvg/prvg If they are on different chromosomes they should assort independently If they are next to one another on the same chromosome they should not assort independently pr vg pr vg PR+ VG+ pr vg Parent PR+ VG+ pr vg Parent PR+ VG+ 25% PR+ VG+ ~50% ~0% pr vg Parent pr vg Parent pr vg pr vg pr VG+ pr vg pr VG+ pr vg pr VG+ pr VG+ PR+ vg pr vg PR+ vg pr vg PR+ vg PR+ vg

When Morgan performed this cross, he obtained the following result: P PR+VG+/PR+VG+ x prvg/prvg F1 PR+VG+/pr vg x prvg/prvg pr vg PR+ VG+ pr vg Parent PR+ VG+ PR+ vg Pr VG+ pr vg 1005 ~44% ~6% pr vg Parent 968 Pr VG+ pr vg 143 PR+ vg pr vg 153 Although the non- parental classes are present, their frequencies are dramatically reduced from that expected from independent assortment. The two loci are linked !!! How do we explain the presence of non-parental classes?

Morgan suggested that when homologous chromosomes pair during meiosis I, the chromosomes occasionally exchange parts PR+ VG+ pr vg P pr vg pr vg F1 PR+ VG+ PR+ VG+ pr VG+ pr vg Crossover chromosome PR+ vg PR+ VG+ The chromosomes that have gone through this crossover are known as crossover products or recombinants. The original chromosomes and those that have not undergone a crossover are known as parental. Evidence for the model that chromosomes physically exchange during meiosis is found in meiotic structures known as chiasmata. During meiosisI when homologs pair, non-sister chromatids appear to cross with each other. The resulting cross-shaped structure is known as a chiasmata.

Crossing-over through the microscope Duplicated homologous chromosomes Synapsis in meiosis I Crossing over between Non-sister chromatids Anaphase in meiosisI Segregation of homologous chromosomes Haploid gametes in meiosis II

Answer: How does one determine whether two genes reside on different chromosomes or reside on the same chromosome as linked genes? To explain this we need to define the terms parental and recombinant: Parents: AB/AB x ab/ab Gametes: AB ab F1: AB/ab Meiosis produces the following gametes: AB Parent ab Parent Ab Non Parent aB Non Parent Recombinant gametes are those with different allelic combinations than those gametes of the previous generation.

If genes A and B are on different chromosomes: P A B a b Gamete A a B b a b F1 diploid (tester) A B b 25% Parental a a b 25% Parental Test cross progeny A b a 25% Recombinant a B b 25% Recombinant

Genes A and B are linked on the same chromosome A-B a-b P A-B a-b Gamete A----B a----b a----b F1 diploid x (tester) A----B a----b > 25% Parental a----b > 25% Parental Test cross progeny A----b a----b < 25% Recombinant a----B a----b < 25% Recombinant

Total Recombination frequency is always less than 50% parental A B A b recombinant a b a B recombinant a b a b parental IF a crossover occurred between linked genes each time homologs paired, the recombinant frequency would be 50% This is because crossing-over involves only two of the four chromatids on the metaphase pair (each of the paired homologs consists of two sister chromatids). For example, the frequency of recombinant gametes between linked genes A and B is 50% if crossing-over occurred each time the homologs paired.

100 75 50 % Recombination 25 25 50 75 Distance (MU)

2 out of 16 gametes are recombinant for A and B

However there are many instances in which the homologs pair and crossing over does not occur between genes A and B. It occurs somewhere else (Recombination ALWAYS occurs between paired homologs but it might not occur between the two genes YOU are studying) Consequently the overall frequency of recombinants is significantly reduced from 50% A B a b C c A B a b parental Alternatively crossing over occurs twice between the gene pairs A B a b parental A B a b

2 out of 12 gametes are recombinant for A and B

Distance The larger the distance between two genes residing on the same chromosome, the higher the probability there is that a crossover event will occur between them. That is for any chromosome, there is a fixed probability per given distance on the chromosome that a crossover event will event. Sturtevant realized that this property could be used to map genes with respect to one another. For each pair of genes on a chromosome a recombination frequency can be determined. By determining the recombination frequency between many pairs of genes on a chromosome, the relative distance between genes and the order of the genes on the chromosome can be determined.

Mapping using recombination frequency For example Sturtevant identified three recessive mutations that reside on the X chromosome of Drosophila W+ red eyes w- white eyes CV+ normal wings cv- crossveinless SN+ normal bristle sn- singed bristle Cen Tel w cv sn X chromosome By calculating recombination frequencies between each pair of genes we can begin to establish where these three genes reside on the X chromosome with respect to one another

To determine the distance between the w gene and the sn gene P w sn/w sn x W+ SN+/Y F1 w sn/W+ SN+ x w sn/Y F2 w sn y Red eye Normal bristle 102 Red eye Normal bristle 96 W+ SN+ Parental White eye Singed bristle 88 White eye Singed bristle 92 w sn White eye Normal bristle 20 White eye Normal bristle 27 w SN+ Recomb Red eye Singed bristle 24 Red eye Singed bristle 23 W+ sn Fill out the phenotypes- recombinants can be determined by phenotype analysis = 20+27+23+23 =94/472

Recombination frequency equals the number of recombinants over total number of progeny white eye red eye # recombinant progeny = normal bristle + singed bristle # total progeny # total progeny = 20+27+23+23 = 94/472 Therefore 19.92% or ~20cM or ~20 m.u. separate the W+ and SN+ genes. 1 map unit (m.u.) = 1% recombination frequency This is a relative distance- depends upon recombination between two genes. Not an absolute distance like bp In the above cross, we could have determined recombination frequency by counting only males (or only females)

W+ red eyes w- white eyes CV+ normal wings cv- crossveinless SN+ normal bristle sn- singed bristle W+---------SN+ = 20MU w----------sn = 20MU W+---------sn = 20MU w----------SN+ = 20MU What is the distance between SN+ and CV+?

To determine the distance between the cv gene and the sn gene P cv sn/cv sn x CV+ SN+/Y F1 cv sn/CV+ SN+ x cv sn/Y F2 cv sn y Normal vein Normal bristle 88 Normal vein Normal bristle 92 CV+ SN+ Parental crossvein Singed bristle 94 crossvein Singed bristle 86 cv sn crossvein Normal bristle 6 crossvein Normal bristle 8 cv SN+ Recomb Normal vein Singed bristle 7 Normal vein Singed bristle 7 CV+ sn Fill out the phenotypes- recombinants can be determined by phenotype analysis Distance between CV and SN is 6+8+7+7/388 = 7MU

The next issue is where does cv map with respect to w and sn: We find that there are 7 m.u. between cv and sn This means cv can map to either one of two positions:- left of Sn and right of Sn w______________20________________ sn_________7____ cv OR w _________13__________cv_______7___ sn These models can be distinguished by determining the map distance between w and cv. Is the distance 27 or 13?

To determine the distance between the cv gene and the w gene P cv w/cv w x CV+ W+/Y F1 cv w/CV+ W+ x cv w/Y F2 cv w y Normal vein Red eye 111 Normal vein Red eye 101 CV+ W+ Parental crossvein white eye 100 crossvein white eye 115 cv w crossvein Red eye 15 crossvein Red eye 19 cv W+ Recomb Normal vein white eye 17 Normal vein whiet eye 20 CV+ w Fill out the phenotypes- recombinants can be determined by phenotype analysis Distance between CV and W is 15+19+17+20/498 = 14MU

W+ red eyes w- white eyes CV+ normal wings cv- crossveinless SN+ normal bristle sn- singed bristle W+---------SN+ = 20MU CV+----SN+ = 7MU = SN+----CV+ What is the distance between W+ and CV+? Recombination analysis indicates 14MU between W+ and CV+. By determining the map distance between w and cv and the map distance between cv and sn, and the map distance between w and sn we can determine the distances and order of all three genes. w _________14__________cv_______7___ sn Small Internal Inconsistency- 14+7=21 not 20 Map gives you order of genes but not PRECISE distance

Genetic Map cv Crossveinless wing g Garnet eyes f forked bristles sc Scute Bristle ec Echinus eye ct Cut wing v vermilion eye 9.1 10.5 9.2 15.9 11.2 10.9 sc--- f = 9.1+10.5+9.2+15.9+11.2+10.9= 66.8 sc f ~50 sc--- g = 9.1+10.5+9.2+15.9+11.2= 55.9 sc g ~50