Chapter 6 -- Linkage and Mapping in Eukaryotes

groucho 518 rough 471 groucho, rough 6 wildtype 5 ------- 1,000a) What is the linkage arrangement of these loci?

11 map distance = ------ x 100 = 1.1 m.u. or 1.1 cM 1000

This is a little complicated because we need to set-up a Punnett square but the four gametic genotypes will not have the same probability of occurrence of 0.25 each, as would expect if we had independent assortment. The first thing then is to estimate the probability of occurrence for each gametic genotype. To do this we must remember that a map distance of 1.1 m.u. means that a recombinant gamete will occur with a probability of 0.011. Thus the two recombinant gametes (

By summing the frequency of occurrence for each cell in the Punnett Square, we can determine the expected frequency for each phenotype in the F2.

grouchoBased upon the expected frequency of occurrence of the flies and assuming 1000 offspring, we would expect to see in the offspring of a dihybrid crossgro gro, + +0.24755gro gro, ro +0.00275 ------------------ total 0.2500 rough+ +, ro ro0.24755+ gro, ro ro0.00275 -------------------- total 0.2500 wildtypegro +, ro +0.24725gro +, ro +0.24725gro +, + +0.00275+ +, ro +0.00275 -------------------- total 0.5000

sons daughters abnormal, brown 219 197 abnormal 43 45 brown 37 35 wildtype 201 223What is the linkage arrangement of these loci?

abnormal brown 416 abnormal 88 brown 72 wildtype 424That two classes in highest frequency are the non-recombinants and the two classes in lowest frequency are the recombinants. To calculate the map distance between the two loci, we use the formula above such that

160 map distance = ------ x 100 = 16 cM 1000

To construct the data set, we need to setup the crosses and then calculate the expected occurrence of each phenotype in the F2 generation. We will cross an inflated, warty female with a wildtype male. This will produce in the F1 a heterozygous wildtype female and a hemizygous inflated, warty male. These two individual will then be mated to produce the F2 generation. As the male is hemizygous for recessive traits, this is essentially a testcross as the gamete from the F1 female will determine the phenotype of the F2 offspring. Also, there will be no difference in the frequency of expression of the traits between males and females in the F2 generation. For a map distance of 10, in the F2 generation 10% of the offspring must be recombinants. Thus from 1000 offspring, we expect 100 to be either inflated or warty as these are the recombinant conditions given that the P generation was either inflated and warty or wildtype. Of the 100 recombinant offspring, we will evenly divide them among inflated (50) and warty (50). The nonrecombinant offspring will be divided evenly among warty and inflated (450) and wildtype (450). As these traits are X-linked, we could also evenly divide each of the phenotypic classes evenly between males and females. This would give us a data set that is shown below.

sons daughters total inflated, warty 225 225 450 inflated 25 25 50 warty 25 25 50 wildtype 225 225 450If these traits were on an autosome, the difference would be that to get the F2 generation, the female would have to be backcrossed to a male that was homozygous inflated and warty and could not be crossed to her brother in the F1 generation. In the F2 generation, the data set would look like that shown under the total column above.

class 1) got-s got-s,amy-s amy-s,sdh-s sdh-s 441 class 2) got-f got-s,amy-f amy-s,shd-f sdh-s 421 class 3) got-f got-s,amy-s amy-s,sdh-s sdh-s 11 class 4) got-s got-s,amy-f amy-s,sdh-f sdh-s 14 class 5) got-f got-s,amy-f amy-s,sdh-s sdh-s 58 class 6) got-s got-s,amy-s amy-s,sdh-f sdh-s 53 class 7) got-f got-s,amy-s amy-s,sdh-f sdh-s 1 class 8) got-s got-s,amy-f amy-s,sdh-s sdh-s 1What are the linkage arrangements of these loci, including map units? If the three loci are linked, what is the coefficient of coincidence?

For the parental classes these are: got-s,amy-s,sdh-s and got-f,amy-f,sdh-f For the double crossover classes these are: got-f,amy-s,sdh-f and got-s,amy-f,sdh-sA comparison of the gametes from the parental classes with the double crossover classes shows that the locus amy is in the middle.

number of recombinants between got-amy amy-sdh got-sdh --------------------------------------- 11 58 11 14 53 14 1 1 58 1 1 53 ---------------------------------------- totals 27 113 136To calculate the distance between each locus, we use the same formula as before: the number of recombinants between the loci divided by the total number of flies and then times 100.

27 distance (got-amy) = ------ x 100 = 2.7 m.u. 1000 113 distance (amy-sdh) = ------ x 100 = 11.3 m.u. 1000 distance (got-sdh) = 2.7 + 11.3 = 14.0

hotfoot, obese, waved 357 hotfoot, obese 74 waved 66 obese 79 wildtype 343 hotfoot, waved 61 obese, waved 11 hotfoot 9a) If the genes are linked, determine the relative order and the map distance between them.

b) What was the cis-trans allele arrangement in the trihybrid parent?

c) Is there any crossover interference? If yes, how much?

For the parental classes these are: h o wa and h+ o+ wa+ For the double crossover classes these are: h+ o wa and h o+ wa+A comparison of the gametes from the parental classes with the double crossover classes shows that the locus hotfoot (h) is in the middle. In addition, an examination of the nonrecombinant gametes shows that all widltype alleles are on the same chromosome in the trihybrid and all the recessive alleles are on the other chromosome. This is a cis arrangement of the alleles in the trihybrid.

number of recombinants between obese-hotfoot hotfoot-waved obese-waved ----------------------------------------------- 79 74 79 61 66 61 11 11 74 9 9 66 ----------------------------------------------- totals 160 160 160To calculate the distance between each locus, we use the same formula as before: the number of recombinants between the loci divided by the total number of flies and then times 100.

160 distance (obese-hotfoot) = ------ x 100 = 16.0 m.u. 1000 160 distance (hotfoot-waved) = ------ x 100 = 16.0 m.u. 1000 distance (obese-waved) = 16.0 + 16.0 = 32.0

The first step is to create the pedigree for this cross and label the chromosomes with as much inofrmation as is possible. The man with type AB blood had a father who was homozygous for fructose intolerance, thus the

Of the ten children 5 inherit the chromsome with the

# recombinants map distance = ---------------- x 100 total # 2 map distance = ---- x 100 = 20 cM 10To calculate the

prob(given 20 cM distance) = 0.40^5 * 0.40^3 * 0.10^2 = 0.0000065536

To get the probability of this family given independant assortment, we assign the probabiliy of any given phenotype the value of 0.25. Given a dihybrid testcross, each phenotype is equally likely. To get the probability of this family we then multiply the probabilities on each birth together.

prob(given independant assortment) = 0.25^10 = 0.0000009536743164062

To calculate the lod score we then take the logarithm of the ratio of these two probabilities.

Anylod= log ((prob given 20 cM distance)/prob given independant assortment)) 0.00000655360lod= log ------------- = log 6.872 = 0.83 0.00000095368

Trembling, Rex 42 Trembling, long-haired 105 normal, Rex 109 normal, long-haired 44a) Are the two genes linked? How do you know?

The cross is a dihybrid female testcrossed to a homozygous recessive male. If the two loci were not linked, then they would assort independently and you would see the four phenotypes in the offspring in a ratio of 1:1:1:1. In this case, two classes are in much higher frequency than the other two classes. This could be tested with a chi-square, which would reject the null hypothesis of independent assortment. Thus, we can see that the two loci do not assort independently and we must conclude that they are linked.

The two classes in highest frequency represent the parental or nonrecombinant classes. This means that the parents of the dihybrid female were Trembling, long-haired and normal, Rex. The chromosomes of these parents would be homozygous for these traits.

Trembling, long-haired normal, Rex Tr + + R ------------ x ---------- ------------ ---------- Tr + + RThe F1 dihybrid female (Trembling, Rex) would have received one of each chromosome from her parents, one chromosome from Trembling, long-haired and 1 from normal, Rex. Thus her chromosomes look like this

Tr + --------------- --------------- + RIn this case, Trembling is on one chromosome and Rex is on the homologue. As they are on opposite chromosomes, they are in the trans position.

We know that Trembling, long-haired and normal, Rex are the parentals thus, Trembling, Rex and normal, long-haired must be recombinants. To calculate the map distance, we use the formula

number of recombinants map distance = --------------------------- x 100 total number individuals 86 map distance = ----- x 100 = 28.7 m.u. 300

Tunicate, liguleless, Glossy 58 Tunicate, liguleless, nonglossy 15 Tunicate, Liguled, Glossy 55 Tunicate, Liguled, nonglossy 13 nontunicate, Liguled, Glossy 16 nontunicate, Liguled, nonglossy 53 nontunicate, liguleless, Glossy 14 nontunicate, liguleless, nonglossy 59a. Determine which genes are linked.

Ignore the phenotype at the glossy locus. Tunicate, liguless 58 + 15 = 73 Tunicate, Liguled 55 + 13 = 68 nontunicate, liguless 14 + 59 = 73 nontunicate, Liguled 16 + 53 = 69These four phenotypes are in about equal frequency, which indicates independent assortment between the tunicate and liguled loci, and thus these two loci are not linked.

Tunicate, Glossy 58 + 55 = 113 Tunicate, nonglossy 15 + 13 = 28 nontunicate, Glossy 16 + 14 = 30 nontunicate, nonglossy 53 + 59 = 112These four phenotypes are not in equal frequency, in fact two classes are at much higher frequency (these are nonrecombinants) and two classes are at much lower frequency (these are recombinants). Thus, these data indicate that these two loci are linked. As we know that the liguled locus is not linked to the tunicate locus and that the tunicate locus is linked to the glossy locus, then we can assume that the liguled locus is not linked to the glossy locus. We could show this by the above method, if necessary.

T G L ---------------- ---------- ---------------- ---------- t g lThe

Tunicate, Glossy 58 + 55 = 113 Tunicate, nonglossy 15 + 13 = 28 nontunicate, Glossy 16 + 14 = 30 nontunicate, nonglossy 53 + 59 = 112From this table, the number of recombinants is 58 and the total numbers of individuals is 283.

58 map distance = ----- x 100 = 20.5 m.u. 283

880 kidney, cardinal 887 ebony 64 kidney, ebony 67 cardinal 49 kidney 46 ebony, cardinal 3 kidney, ebony, cardinal 4 wildtypea. Determine the chromosomal composition of the F1 females.

For the parental classes these are: kidney, cardinal and ebony For the double crossover classes these are: kidney, ebony, cardinal and wildtypeA comparison of the phenotypes from the parental classes(for example, "kidney, cardinal") with the double crossover classes (for example "kidney, ebony, cardinal") shows that the locus ebony is in the middle, because it is the one locus that does not match.

number of recombinants between kidney-ebony ebony-cardinal kidney-cardinal ----------------------------------------------------- 64 49 49 67 46 46 3 3 64 4 4 67 ------------------------------------------------------ totals 138 102 136To calculate the distance between each locus, we use the same formula as before: the number of recombinants between the loci divided by the total number of flies and then times 100.

138 distance (kidney-ebony) = ------ x 100 = 6.9 m.u. 2000 102 distance (ebony-cardinal) = ------ x 100 = 5.1 m.u. 2000 distance (kidney-cardinal) = 6.9 + 5.1 = 12.0The map is this.

k e cd ---+------------+------------+------- 6.9 m.u. 5.1 m.u.

Estimate the distance between the two genes.

see your book for the large table (Table 1) Table 2. Clone --------------------- human enzyme A B C D E ------------------------------------------------- gluthathione reductase + + - - - malate dehydrogenase - + - - - adenosine deaminase - + - + + galactokinase - + + - - hexosaminidase + - - + - -------------------------------------------------

Table 1. Human Chromosomes ------------------------------- Clone 3 7 9 11 15 18 20 --------------------------------------------- X - + - + + - + Y + + - + - + - Z - + + - - + + --------------------------------------------- Table 2. Enzyme -------------------- Clone A B C D E ---------------------------------- X + + - - + Y + - + + + Z - - + - + ----------------------------------

Last updated on 22 August 1996.