How to solve linkage map problems

How to Solve Linkage Map Problems

Determining Linkage/Independence

• Suppose you want to map the distance between genes "N" and "P." Each has two alleles (N and n, P and p). Both pairs of alleles have a complete dominance relationship; we will identify the phenotypes simply by the letters of the alleles.

• Your mating will be: NnPp x nnpp• If your genes are independent, then your first parent's four possible gametes (NP, Np,

nP, and np) will be produced in equal numbers, and thus you will get four offspring classes, also with equal numbers, like so:

NP offspring: 100Np offspring: 100nP offspring: 100np offspring: 100

• These equal numbers are the evidence that these two genes are independent. However, if you get numbers like:

NP offspring: 150Np offspring: 50nP offspring: 50np offspring: 150

Determining Linkage/Independence

• Then you have established that the genes are not independent. Moreover, you know that in your original heterozygous parent, one chromosome had N and P on it and the other had n and pon it (as opposed to N and p on one and n and P on the other). In other words, your parental linkage is N linked with P and n linked with p.

• In this example, the parental offspring are the NP and np offspring; the recombinanant offspring are the Np and nP offspring.

Three-point Cross I

• This time we are looking at the A, B, and C genes (alleles A and a, B and b, C and c). Again, complete dominance for all. Here's our mating:

• AaBbCc x aabbcc

• If the genes are independent, we should get eight classes of offspring, all pretty much equal in number. However, here's what we get:

ABC: 95 aBC: 50ABc: 5 aBc: 700AbC: 700 abC: 5Abc: 50 abc: 95

• Obviously, not all phenotypic classes are equal, so we definitely have linkage.

Determining Gene Order• To determine gene order, you must first identify the parental classes and the double crossover classes. Don't

fall into the trap of assuming that the parentals must always be ABC and abc. Parental classes are always the largest ones, since the connection between the genes will tend to preserve parental linkages above the expected level of independence. In our example, the AbC and aBc classes are the parental classes. Notice that these are reciprocals of each other--they have "opposite" phenotypes. This should make sense.

• The double crossover classes will always be the smallest classes, since they require a crossover between A and B and a crossover between B and C. So our double crossover classes are ABc and abC. Again, note that they are reciprocals.

• Now we compare the parentals with the double crossovers. The only gene which has recombined in a double crossover will be the middle one, because the chromatids will have crossed over between the first two genes, then crossed back between the second and third genes. Compare each double crossover with the parental which is most like it. For double crossover ABc, the most similar parental is aBc. (It's the same in two of the genes; the other parental is the same in only one gene.) Comparing these, we see that the only gene that is different is the A/a gene. If we compare the other double crossover to the other parental, we get the same result--only the A/a gene is different. So A must be the gene in the middle. Our actual gene order is B-A-C.

• To keep from getting confused, it's an excellent idea to rewrite the offspring chart above with the genes in the correct order, like this:

BAC: 95 BaC: 50BAc: 5 Bac: 700bAC: 700 baC: 5bAc: 50 bac: 95

Map Distance• To figure out the distances, we must identify our single crossover classes. We've got four of

them: BAC, bAc, BaC and bac. Note that this is two pairs of reciprocals (BAC and bac are reciprocals and bAc and BaC are also reciprocals). Again, we want to compare these to the parentals which are most like them to determine just where the crossover occurred to produce them. Comparing single crossover BAC to parental bAC, we see that it has recombined between the B and the A loci. Same for its reciprocal. Comparing the BaC single crossover to theBac parental, we see that it has recombined between the A and the C genes, as has its reciprocal. Note also that the numbers of offspring give you hints as to which crossover classes go together.

• Now we are ready to calculate distances. We have four offspring classes which have recombined between the B and the A genes. These are the single crossovers BAC (95) and bac (95) as well as the double crossovers Bac (5) and bAC (5). This gives us a total of 200 offspring who have crossed over between these two genes. There are a total of 1700 offspring, so we'd calculate this distance as (200/1700) x 100, which equals 11.8. So we've calculated a distance of 11.8 LMU between genes B and A.

• There are also four offspring classes which have recombined between the A and C genes. bAc (50), BaC (50) and again, our double crossover classes BAc (5) and baC (5), for a total of 110. Calculating: (110/1700) x 100 = 6.5 LMU between genes A and C.

• To calculate the distance between B and C, just add 11.8 and 6.5, to get 18.3 LMU.

Additive?

• Linkage map units don't actually correspond to a specific length of chromosome. The exact distance calculated between two genes depends upon how much information you have about what's going on in the region between the genes.

• For example, consider the problem above. If we'd been following only the two end genes (B and C), and disregarding the A gene, of the offspring in our example, only the single crossovers would have counted as recombinants. The only reason we know that the double crossovers recombined (twice) between B and C is because we see that the A gene is switched; the B and C genes are still in the parental formation. If we are ignoring A, they become parentals. Thus, if we calculate the B-C distance without regard to the information about the A gene, we get:

• Recombinants: BC {B(A)C [95] + B(a)C [50]} and bc {b(a)c [95] + b(A)c [50]} for a total of 290.

• All the rest would score as parentals. We we'd calculate the distance between B and C as: (290/1700) x 100 - 17.1 LMU. Compare this to the 18.3 LMU we calculated between the same genes when we were paying attention to the A locus.

• Keep in mind that when you are calculating linkage map distance, there will virtually always be many genes between each pair that you are considering--genes that you are ignoring in your calculations. So linkage map units are often not strictly additive, and you will often get different precise distances, depending upon the specifics of the calculation method.

Three Point Cross II

• Mating: DdFfGg x ddffgg

Where:

D = Calm personalityd = Dithery personalityF = Five toedf = Four toedG = Smooth fur, andg = Grizzled fur.

• Offspring:

Calm, Five, Smooth - 620 Dithery, Five, Smooth - 5Calm, Five, Grizzled - 100 Dithery, Five, Grizzled - 75Calm, Four, Smooth - 75 Dithery, Four, Smooth - 100Calm, Four, Grizzled - 5 Dithery, Four, Grizzled - 620

• Once again, assumption of independence would predict all phenotypic classes be about the same size. The only parent influencing the diversity of the offspring is the heterozygous one. She can produce eight kinds of gametes (all eight possible combinations of her alleles), and all should be produced in equal numbers if the genes are independent. Since we obviously do not have equal numbers in our classes, we do not have independence. Our genes are linked.

Three Point Cross II• How do you tell what the gene order is?

This task requires that you identify the parental classes and the double crossover classes among the offspring. The parental classes (there should be two, and they should be reciprocals) will be the largest classes. This is because the linkage makes the parental connections tend to remain intact more often than they would if assortment were independent. The double crossover classes (again, two and reciprocal) should be the smallest classes. This is because a double crossover requires two crossover events between the end genes, which is statistically significantly less likely than a single crossover.

Double crossover offspring result from the occurrence of two crossover events, one between the first two genes in the sequence, and another between the second and third genes in the sequence. In effect, this means that only the middle gene has been recombined, and the two end genes remain in the parental configuration. (The first crossover swaps the chromatids, then the second swaps them back.)

To determine which gene is in the middle, you compare each double crossover class to the parental which is most like it. Whichever gene is different is the middle gene. The two double crossover classes should give the same result.

Parental Classes: Calm, Five, Smooth (620) and Dithery, Four Grizzled (620)

Double Crossover Classes: Dithery, Five, Smooth (5) and Calm, Four, Grizzled (5).

Comparing, it is evident that the middle gene is the personality gene (Calm/Dithery).

Rewriting the genes in the correct order:

Five, Calm, Smooth - 620 Five, Dithery, Smooth - 5Five, Calm, Grizzled - 100 Five, Dithery, Grizzled - 75Four, Calm, Smooth - 75 Four, Dithery, Smooth - 100Four, Calm, Grizzled - 5 Four, Dithery, Grizzled - 620

Three Point Cross II• How do your calculate the map distances?Before you can calculate map distance, you have to figure out which single crossover classes represent which region of crossing over. This is done by identifying the reciprocal pairs of single crossovers, then once again comparing each to the parental class most like it. Noting which of the three genes has been switched will tell you where the crossover occurred.

The single crossovers between the toes and personality genes are Four, Calm, Smooth (75) and Five, Dithery, Grizzled (75). Note that both of these are different from the parentals only in the first trait.

The single crossovers between the personality and fur genes are Five, Calm, Grizzled (100) and Four, Dithery, Smooth (100). these differ from the parentals by only the final trait.

Linkage map distance is calculated just as for the two point cross, by calculating the percentage of the offspring showing recombination between the two genes in question.

Between the toes and personality genes, our two single crossover classes total 150 offspring (75 + 75). But recall that the double crossovers also recombined in this region, and thus need to be added to the single crossovers. There are 10 of them (5 + 5). So the total number of offspring which have recombined in this region is 160.

So the percent recombination is (160/1600)x100, or 10%; the linkage map distance between these two genes is 10 LMU.

Repeat the process with the second and third genes, again remembering that the double crossovers also recombined in this region, so the total number of recombinants is 210 (100 + 100 + 5 + 5).

Percent recombination is (210/1600)x100 = 13.125, or 13.125 LMU.

Finally, the distance between the end genes is 10 + 13.125 or 23.125 LMU.

Three Point Cross III

• For example, we have a mating: scsc ecec vgvg x sc+sc+ ec+ec+ vg+vg+

Three Point Cross III• If these genes were on separate chromosomes, they should be assorting

independently and all the classes should be equally frequent.• To map them, we simply examine the pair-wise combinations and

identify the parental and recombinant classes:• For example to determine the distance between sc and vg:

sc to vg # recombinant/Total progeny = 506/ 1008= 50%.

Therefore sc abd vg are not linked

Three Point Cross III

• What about sc and ec? • Identify the parentals and the recombinants. What is the map distance?

• From these observations what is the expected map distance between ec and vg?

Three Point Cross IV

Another example:


1. Identify the two parental combinations of alleles2. The two rarest classes usually represent the product of double crossover With this knowledge, you can establish a gene order in which a double cross producing the allelic combination observed in the rarest class

The parental chromosomes in fourth example are:v cv+ ct+ and v+ cv ct.

The products of the double recombinants were: v cv+ ct and v+ cv ct+


There are three possible gene order for the parental combination (basically we want to know which of the three is in the middle)


3. Once the parental chromosomes are identified and the order is established, the non-recombinants, single recombinants and double recombinants can be identified.



Now the non-recombinants, single recombinants, and double recombinants are readily identified Parental input:

• With this information one can easily determine the map distance between any of the three genes:

Where SCO is single crossover and DCO is double crossover

Big thanks to

http://www.cod.edu/people/faculty/fancher/genetics/LinkageMappingExamples.htm

and

http://bio.classes.ucsc.edu/bio105/winter%2008/Bio105_W08/Lectures/Lecture6/Lecture6.html






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How to solve linkage map problems