Two varieties—X (susceptible) and Y (resistant)—of eggplant were used to charact
ID: 179057 • Letter: T
Question
Two varieties—X (susceptible) and Y (resistant)—of eggplant were used to characterize the genetics of resistance to a necrotizing plant pathogen. Eight thousand plants from each variety were inoculated, and the average number of necrotic lesions produced per leaf was recorded. Variety X averaged 52 lesions per leaf, whereas variety Y averaged six lesions per leaf. The F1 progeny from a cross between variety X and variety Y plants averaged 21 lesions per leaf, with a variance of 13.69. Of the F2 progeny, 8192 were again tested by inoculation, and a total of four plants were produced that were phenotypically similar to the original parental plants—two plants for each extreme (i.e., parental) phenotype. The F2 plants also averaged 21 lesions per leaf, with a variance of 56.25.
a) How many genes control resistance to this pathogen?
b) Determine the broad-sense heritability of resistance to this particular pathogen among the F2 plants.
Explanation / Answer
a) The number of genes can be found if the proportion of F2 individuals expressing the two parental phenotypes can be determined according to the following formula:
(1/4)^n=proportion of offspring either X or Y where n is the number of genes controlling a trait.
The numberof F2 progeny plants expressing either parental phenotype = 2/8000 = 1/4000. Using the formula (1/4)^n, we get (1/4)^n= 1/4000.
Since 4^6=4096, which is approximately equal to 4000, we can say that n=6, indicating that six gene pairs control resistance to this pathogen.
b) Heritability is an estimate of how much variability in a population is due to genetic factors, separate from environmental factors.
Heritability index (H2) is an analysis of variance among individuals of a known genetic relationship. H2 measures the degree to which phenotypic variance (VP) is due to genetic factors.
Mathematically, phenotypic variance (VP) is the sum of environmental variance (VE) genetic variance (VG) and the interaction of genetics and environment (VGE). The last is usually negligible, so is left out of the calculation.
VP = VE + VG
VG is the unknown variable we have to find. VG=VP-VE
Since the F1 plants have identical genotypes, we can use the variance in the F1 as an estimate of environmental variance. Thus, VE= 13.69
It is mentioned that the phenotypic variance in F2 generation, VP= 56.25
Substituting these values in the above equation and solving for VG,
VG= 56.25-13.69= 42.56
The heritability index (H2) is calculated by dividing the genetic variance by the phenotypic variance: H2=VG/VP.
Thus, H2= 42.56/56.25= 0.76
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