Linkage Estimates from F2 Data Help
In organisms where the male is XY or XO, the male receives only the Y chromosome from the paternal parent (or no chromosome, homologous with the X in the case of XO sex determination). The Y contains, on its differential segment, no alleles homologous to those on the X chromosome received from the maternal parent. Thus, for completely sex-linked traits the parental and recombinant gametes formed by the female can be observed directly in the F2 males, regardless of the genotype of the F1 males.
If the original parental cross involves scute, vermilion females and wild-type males then both male and female progeny of the F2 can be used to estimate the percentage of crossing over. The student should verify that this expectation is valid.
An alternative to the testcross method for determining linkage and estimating distances is by allowing dihybrid F1 progeny to produce an F2 either by random mating among the F1 or, in the case of plants, by selfing the F1. However, this is a poor substitution. Such an F2 that obviously does not conform to the 9 : 3 : 3 : 1 ratio expected for genes assorting independently may be considered evidence for linkage. Two methods for estimating the degree of linkage from F2 data are presented below.
- Square-Root Method. The frequency of double-recessive phenotypes in the F2 may be used as an estimator of the frequency of noncrossover gametes when the F1 is in coupling phase, and as an estimator of the frequency of crossover gametes when the F1 is in repulsion phase.
- Product-Ratio Method. An estimate of the frequency of recombination from double-heterozygous (dihybrid) F1 parents can be ascertained from F2 phenotypes R-S, R-ss, rrS-, and rrss appearing in the frequencies a, b, c and d, respectively. The ratio of crossover to parental types, called the product ratio, is a function of recombination.
- For coupling data: x = bc/ad
- For repulsion data: x = ad/bc
F1 in coupling phase. AB/ab
F2: The frequency of ab gametes = of the frequency of all noncrossover gametes. If the crossover percentage is 20%, we would expect 80% noncrossover gametes (40% AB and 40% ab). The probability of two ab gametes uniting to form the double-recessive ab/ab = (0.4)2 = 0.16 or 16%. Now, if we do not know the crossover percentage, but the F2 data tell us that 16% are double recessive, then the percentage of noncrossover gametes = 2√freq: of double recessives = 2 √0:16 = 2(0.4) = 0.8 or 80%. If 80% are noncrossovers, the other 20% must be crossover types. Therefore the map distance between A and B is estimated at 20 units.
F1 in repulsion phase. Ab/aB
F2: The reasoning is similar to that in Example 6.20. With 20% crossing over we expect 10% of the gametes to be ab. The probability of 2 of these gametes uniting to form the double recessive ab/ab = (0.1)2 = 0.01 or 1%. Now, if we do not know the crossover percentage, but the F2 data tell us that 1% are double recessives, then the percentage of crossover gametes = 2√req: of double recessives = 2√0:01 = 2(0.1) = 0.2 or 20%.
The recombination fraction represented by the value of x may be read directly from a product-ratio table (Table 6-1). The product-ratio method utilizes all of the F2 data available and not just the double-recessive class as in the square-root method. The product-ratio method should therefore yield more accurate estimates of recombination than the square-root method.
Locating the value of x in the body of the coupling column (Table 6.1), we find that 0.1816 lies between the values 0.1777 and 0.1948, which corresponds to recombination fractions of 0.28 and 0.29, respectively. Therefore, without interpolation, recombination is approximately 28%.
Locating the value of x in the body of the repulsion column, we find that 0.3750 lies between the values 0.3643 and 0.3927, which corresponds to recombination fractions of 0.36 and 0.37, respectively. Therefore recombination is approximately 36%.
Practice problems for these concepts can be found at:
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