In the diploid ancestor of our modern bread wheat ( Triticum aestivum ), resista
ID: 58925 • Letter: I
Question
In the diploid ancestor of our modern bread wheat (Triticum aestivum), resistance to a particular fungal pathogen is controlled by a dominant allele at a single locus. Assume that a population of this diploid wheat is in Hardy-Weinberg equilibrium and that the frequency of the dominant (resistance) allele is 0.51. In each of the following parts, we suggest carrying your answer to three decimal places.
a) What proportion (decimal value) of the population is phenotypically resistant to the pathogen?
b) Assume that all susceptible individuals in this population are eventually killed by fungal infections before they reproduce. What is the frequency of the recessive (susceptibility) allele among the survivors? (HINT: count the alleles)
c) Assume that the surviving plants cross-pollinate randomly and that each plant produces the same number of seeds. What proportion of the seeds produced will eventually become plants that are susceptible to the fungus?
Explanation / Answer
a) The organism is diploid (2N) and the resistance is DOMINANT. That means that either a homozygous dominant or heterozygous phenotypes will give the resistant phenotype.
The frequency of the dominant resistance allele is 0.51 = p.
Therefore the frequrncy of the recessive allele = 1 - 0.51 = 0.49 = q
( p + q = 1, according to Hardy-Weinberg's equilibrium)
Also, according to Hardy-Weinberg's Equilibrium, p2+2pq+q2 = 1
(where 2pq represents the heterozygous individuals)
So, 2pq = 1 - (0.51*0.51 + 0.49*0.49)
2pq = 0.4998
Thus, the proportion of individuals resistant are the dominant homozygous (p2) + the heterozygous (2pq) = 0.2601 + 0.4998 = 0.7599
b) All the homozygous recessive individuals are killed before they reproduce = q2 = 0.2401
Thus only the hetreozygous are left behing which still carry a recessive allele ( one dominant + one recissive allele in a heterozygote)
Therefore, the frequency of the recessive (susceptibility) allele among the survivors = half of the heterozygous frequency = half of 0.4998 = 0.2499
c) Now q = 0.2499 and p = 0.7501
The frequency of susceptible plants is q2 = 0.24992 = 0.625