Mapping with Three point testcrosses PDF

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Mapping with Three point testcrosses...

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Mapping with Three point testcrosses we used a series of two-point test-crosses to map several genes, but a two-point test cross is not the most effective way to build genetic maps. By performing three-point test-crosses using three genes, geneticists can efficiently map three linked genes simultaneously. A cross between a trihybrid heterozygote and one homozygous recessive for all three genes can be used to determine the order of the three genes on the chromosome. Recombination of linked genes produces eight different gamete genotypes with unequal frequencies. The parental types will be observed most frequently, the recombinant types less frequently. You can see the difference between a single crossover between a and b, and a single crossover between b and c. They produce different gamete types. And both these single crossovers are different from a double crossover, one that occurs in both intervals, which you see at the bottom.

In this example, the trihybrid heterozygote does not have all three wildtype alleles on the same chromosome. Wildtype a and c are together, but wildtype b is on the homologue. Follow the figure through to see the parental games, the single crossovers, and the double crossovers.

Emerson mapped three genes in maize: one for seedling color (green is dominant to yellow), one for leaf texture (rough leaves are dominant to glossy leaves), and one for fertility (normal fertility is dominant to variable fertility).

He made trihybrid heterozygotes and test crossed these, then analyzed the progeny to map the genes. I want to point out the elegance of using the test cross to map genes in eukaryotes. So, if you look at the map genotypes on the right-hand side. you can see that every single phenotype has one chromosome that was received from the test cross parent, and whatever that was on that chromosome was not expressed. So, anything that is expressed comes from the trihybrid heterozygote. We can follow all of the crossover events in the progeny. You know you can count on the parents being the most common class of progeny. There are some other things you can rely on. When three genes are linked, each of the six recombinant gamete classes are produced at frequencies significantly lower than that predicted by chance. Within each crossover class (e.g., single crossover between genes a and b) both gamete types that result are equally frequent. Double crossover classes are the least frequent because both crossover events must occur to produce these.

1. Are the data consistent with the proposal of genetic linkage? Under independent assortment, eight genetically distinct gametes would be produced with equal frequency: 1/8 3 726 = 90.75 for each class. What we see here is a departure from this that distribution and this is consistent with genetic linkage; parental gametes are more frequent and recombinant gametes less frequent than predicted.

2. What are the alleles on the parental chromosomes? The genotypes of original parents are known in this case, so the alleles on the parental chromosomes are V Gl Va and v gl va. The frequencies of the F2 progeny also show this, as these are the most abundant F2 phenotypes.

3. What is the gene order on the chromosome? The double recombinants, or double crossover progeny, can be used to determine the gene order. To determine the order, genes can be listed in each of three possible orders and you can check the resulting double crossover progeny with your actual data. Alternatively, when you compare the parental alleles and those of the double crossover genotypes, the position of one allele will differ; this is the gene in the middle of the other two.

4. What are the recombination frequencies of gene pairs? Count the number of crossovers that occurred between the two members of each gene pair, including the double crossover classes. Given what we know about the parental genotypes and the gene order, we can predict the gametes that would result from a single crossover in each interval- between V and Gl , and between Gl and Va.

The classes for a single crossover between V and Gl are indicated with red arrows; those for between Gl and Va are indicated with purple arrows. Notice that the number of progeny within

each of these classes is similar. Go ahead and calculated the RF for each of these classes. Remember to include the double crossovers in each of your calculations because they count for a crossover in each of the intervals.

For V-Gl the frequency, r, is 60 + 62 + 4 + 7/726 = 0.183 or 18.3 cM For Gl-Va the frequency, r, is 48 + 40 + 4 + 7/726 = 0.136 or 13.6 cM To calculate the distance between V and Va, you can count the number of crossovers that occurred between the two most distant genes, including the double crossover classes. In fact, you have to cross the double crossover progeny twice, because each represents two crossvers between V and Va. This calculation should give you V-Va = 60 + 62 + 48 + m 40 + 4 + 7 + 4 + 7/726 = 0.320 or 32.0 cM BUT you can also just sum the two distances- this is actually more accurate.

5. Is the frequency of the double crossovers consistent with independence of the single crossovers? In most experiments, the number of observed double crossovers is actually less than what you expect. So here for Emerson’s data we would expect 9.1 double crossovers in each one of the progeny classes, but we only get 4 in one and 7 in the other. This is caused by an effect called interference (I): The degree to which a CO event in one interval interferes with a second CO; Interference identifies the double crossovers expected but not produced.

In Emerson’s data, we calculated the expected number of double crossovers to be (0.183)(0.136) 3 726 = 18.2, and we can see from the data table that the observed number of double crossovers is 11. The coefficient of coincidence, c, is the Observed double crossovers/Expected double crossovers (11/18.2 = 0.6). The interference, is then 1-c. Interference identifies the double crossovers expected but not produced In cases where I < 0, negative interference has produced more double crossovers than predicted.

We’ve answered our 5 questions, but let’s review what it means for us to do a 3 point test cross.  Determine nonrecombinant (or parental) and DCO offspring classes. This helps you figure out the gene order.  Pick out the SCO classes and calculate RF values for them and start your map. Put distances on your map- all three of them!!  Calculate expected number of double crossovers (DCO). Calculate C.O.C. and interference. Now try your hand at this process from the beginning. Here is a problem focusing on three mutations in Maltese Bippies. Extra nose, yellow eyes, and zebra stripes. Go ahead and create the genetic map for these three genes. The answer is on the next slide....