Genotyping Example, Step 1 - Calculating What Is Expected
Following along with our tomato bacterial spot case, let’s next walk through the steps in calculating a chi-square analysis for a genotyping study. In the first step, we will focus on one DNA marker and hypothesize what banding patterns are expected and the number of individual plants for each pattern in the same F2 population we were working with for the phenotyping.
The gel photo below (Fig. 6) is a single DNA marker, CosOH57, being tested on our original 2 parents (lanes 2 and 3) and the resulting F1 individuals. The first lane is the ladder (giving a reference point for the size of each band). The next 2 lanes are the marker's banding patterns seen in each of the tomato parents. The remaining lanes are DNA samples from the F1 plants. Notice the F1s all show the same three band pattern, with a combination of one band like susceptible parent, OH88119 (216 bases long) and a banding pattern like the resistant 6.6068 parent (2 bands, one 145 base pairs long and one 71 base pairs long). Therefore, we can characterize each of the the F1 as heterozygous, because those plants have a banding pattern like each of the parents. In genotypic terms this would be like a RR x rr giving all F1 Rr genotypes.
Figure 6. Sample gel photo of CAPS marker CosOH57 in an F1 population. F1 Plants, OH88119 and 6.8068 are compared. OH88119 has a band at 216. 6.8068 has two bands, one at 145 and the other at 71. Modified for educational purposes. Photo credit: Matthew Robbins, The Ohio State University.
Based upon our banding pattern observations in the F1, we can hypothesize that CosOH57 is co-dominant, with alleles from each parent easily identifiable in the F1. We decide to test this hypothesis that molecular DNA marker, CosOH57, will segregate in a typical 1:2:1 pattern in the F2, as expected from Mendel’s law of segregation (with ratio of 1 homozygous for parent OH88119 allele : 2 heterozygous: 1 homozygous for parent 6.8068 allele). Therefore, for the 197 plants in our F2generation of selfing the F1s from the 6.8068 x OH88119 cross, we expected 49.25 tomato plants to show a banding pattern like OH88119, 98.50 plants to show a heterozygous pattern (like the F1s) and 49.25 plants to show a banding pattern like the parent, 6.8068.
Click on the video below to watch a tutorial further explaining how to calculate the expected frequencies and numbers of plants expected in each F2 genotype class.