Firdaus et al (2012) described the screening of 35 different accessions for whitefly resistance. These accessions belonged to 12 different species. After free-choice testing the most resistant accession was taken. The plant Firdaus selected was used as father in a cross with Solanum lycopersicum. F1 plants were grown and selfed to obtain a substantial number of F2 seeds. Firdaus used this F2 population as segregating population for his linkage analyses. Consequently, he tested a population of 328 F2 plants individually in a no-choice whitefly resistance test. Simultaneously he collected leaf samples from all individual plants to isolate DNA from them and determined the segregations pattern of a large number of molecular markers (SNPs) in this population.
Questions segregating populations
4.1 Why did Firdaus use an F2 tomato population and not an F1 population? (Think about the breeding strategy en genomic composition of a tomato plant)
4.2 What segregation pattern do you expect in the F2 population if P1 is AA and P2 is aa?
4.3 What segregation pattern do you expect in a BC1 population that has been backcrossed to P2, if P1 is AA and P2 is aa?
As you can see from the answers in 3.2- 3.3, the segregation patterns are quite different for population type F2 and BC1. This influences the linkage analyses greatly.
This also means that you can determine what kind of mapping population is used, by determining the segregation patterns of the marker alleles in the offspring.
4.4. Can you determine what kind of mapping populations have been analyzed based on the segregation ratios?
On the next page the SNP data of 5 segregating populations have been given. Each segregating population is comprised of 24 plants and has five SNP markers determined
Examples of mapping populations are F1, F2, BC1, introgression lines, RILs.
(What are introgression lines and recombinant inbred lines (RILs ?)
Be aware that in molecular marker analysis the notation of the alleles is different from Mendel’s, since we do not know whether an allele is dominant or recessive with molecular markers
One homozygous genotype = a, other homozygous genotype = b, heterozygous genotype = h
(tip: for practice to use the molecular notation, determine the segregation pattern in question 3.2 and 3.3)
Table: Molecular marker segregation patterns in different segregating populations
Genetic linkage map
The SNP data in the segregating populations (also called mapping populations) can be used to make a genetic linkage map. A linkage map shows the position of known genes and/or genetic markers relative to each other on the different chromosomes. The distance between the genes/markers is given in terms of recombination frequencies, also denoted as centimorgans. These recombination frequencies are a result from crossovers between genetically linked genes/markers on homologous chromosomes during meiosis.
Genetic linkage is the phenomenon that two genes/markers on a chromosome stay linked during meiosis if no crossovers occur. However, in the case of a crossover between the original gene/marker alleles is lost (see figure below). This cross-over process is rare and the occurrence of recombinant offspring plant with this new combination of gene/marker alleles is very low. The frequency of these recombinant plants in the segregating/mapping population can be calculated and this gives a measure of the genetic distance between two genes/markers. Very closely linked genes/markers will have a low recombination frequency and genes/markers that are more widely apart will have a higher recombination frequency.
In summary, a genetic linkage map represents the distance between a lot of genes/markers along the chromosomes of an organism.
A segregating F2 population was made and the segregation pattern of in total 589 SNP markers were determined. Markers that were difficult to score of had identical segregation patterns were removed. From this dataset a genetic linkage map with molecular markers was calculated using JoinMap 4.1. Markers that were difficult to score were removed from the dataset and markers that showed an identical segregating pattern were considered as derived from the same place (locus) on the chromosome and only one was taken. Finally, 589 markers could be used to construct a genetic or molecular linkage map. An example of a linkage map of chromosome 6 of potato is given below.
linkage map of potato chromosome 6 is according to van Os et al. (2006). Click to enlarge
Questions linkage mapping
4.5 What segregation pattern do you expect (according to the rules of Mendel) in a BC1 population that has been backcrossed to P2, if P1 is AABB and P2 is aabb?
This rule applies to genes that follow independent assortment. That means that the genes are unlinked and located on different chromosomes.
4.6 What segregation pattern do you expect when the genes A and B are linked and located on the same chromosome?
4.7 Compare the different answers of question 3.5 and 3.6.
What do you notice? What genotypes are absent in 3.6? Sometimes these genotypes do appear yet in very low numbers, what are these genotypes called in this case?
Based on the number of recombinants in a population the recombination frequency between the two markers/genes can be calculated. This frequency tells you how far apart the two markers/genes are located on the chromosome. This distance is given in cM (centimorgan).
4.8 Can you determine the order of the following genes, L, R, S, based on their recombination frequencies?
- R-S: 29,4 %
- R-L: 35 %
- S-L: 7,8 %
4.9 What is the order and distance of the genes A, B, C, D and E based on the recombination frequencies?
- A,B 50 %
- A,C 15 %
- A,D 38 %
- A,E 8 %
- B,C 50 %
- B,D 13 %
- B,E 50 %
- C,D 50 %
- C,E 7 %
- D,E 45 %
Note: If the recombination frequency is 50% it means that the markers/genes are unlinked. This means they are located on different chromosomes or very far apart from each other on the same chromosome.
4.10 A genetic map consists usually of several chromosomes. How many chromosomes can you find in this dataset? What is the order and distance of the makers on these chromosomes.
- A,B 50 %
- A,C 17 %
- A,D 50 %
- A,E 50 %
- A,F 12 %
- A,G 3 %
- B,C 50 %
- B,D 2 %
- B,E 5 %
- B,F 50 %
- B,G 50 %
- C,D 50 %
- C,E 50 %
- C,F 7 %
- C,G 19 %
- D,E 7 %
- D,F 50 %
- D,G 50 %
- E,F 50 %
- E,G 50 %
- F,G 15 %
4.11 Now calculate the recombination frequency yourselves and construct a linkage map based on the data from the molecular dataset on the next page.
First count the recombinants between two markers and calculate the recombination frequency per marker combination. Do this for each marker combination possible. Finally draw a molecular linkage map based on the recombination frequencies.
From recombination frequency to cM (centimorgans)
In publications the distance between the genes/markers is give in in centimorgans (cM). This unit of genetic linkage is named after Thomas Hunt Morgan. A distance of 1cM between two markers means that the recombination frequence was 1%.
4.12 What population size do you need to discover a minimum distance between two genes/markers of 1cM? and 0,1 cM?
4.13 How many basepairs would that be on average? (for info see next page)
Comparison between genetic and physical maps
The chance for a crossover and thereby related estimate of the recombination frequency is not equal over the whole chromosome. In the lower figure on page 16 you see next to the genetic map (given in centimorgan) a physical map of the same chromosome given in basepairs), In the middle part of a chromosome (the centromere) there is hardly any recombination. Also, the frequency is much higher in the distal regions. This difference in recombination frequency is the reason that a genetic linkage map (based on recombination frequency) is different from a physical map (based on the number of bases between two recombination events).
Question comparison genetic and physical map
4.14 In the picture below the difference is shown between the genetic map (x-axis in cM) and the physical map (y-axis in Mb – Megabases). Can you calculate (estimate) the number of bases per cM for the first 50 cM, in the centromeric region and in the last 80 cM. How large is the region in cM and Mb where there is hardly any recombination? Many markers have been determined in this region but in the picture we see only one. It appears that there is only one marker because all identical segregating markers minus one have been removed from the data set.