Question

Plasmopara halstedii is a eukaryotic plant pathogen that infects sunflower, causing sunflower downy mildew. Sunflower has evolved genes that confer resistance to downy mildew, called Pl genes, and breeding to spread the disease-resistant phenotype is a major method for controlling the disease in sunflower crops. Breeding programs in France in the 1980s were initially effective at controlling downy mildew. However, by the 1990s and 2000s, populations of P. halstedii had overcome several of the resistance genes. Additionally, while only one pathotype of P. halstedii existed in France until 1988, twenty years later 14 pathotypes could be found throughout France.

Researchers collected DNA from 24 samples of P. halstedii isolated from infected sunflowers for population genetic analysis. They analyzed data from 12 genetic markers, including one called Pha43, which had two alleles, called R and r. The genotype data from Pha43are shown in the table below.

Genotype	Count
RR	13
rr	9
Rr	2
Total	24

Data: Delmotte et al. (2008). Infection, Genetics and Evolution.

Using the observed genotype data, answer the following questions.

Based on the frequency of the R and the r alleles, what is the expected frequency of the Rr genotype in Hardy-Weinberg equilibrium?

What is the observed frequency of the RR genotype in the French P. halstedii population?

What is the observed frequency of the rr genotype in the French P. halstedii population?

What is the observed frequency of the Rr genotype in the French P. halstedii population?

Do these data indicate that the population is in Hardy-Weinberg equilibrium for the Pha43 marker?

Answer 1

there are two equations to solve a question if it is in hardy weinberg equillibrium

p+q=1

p2+2pq+q2=1

p is the frequency of the doinant allele

q is the frequency of the recessive allele

p2 is the frequency of individuals with homozygous dominant genotype

2pq is the frequency of individuals with heterozygous genotype

q2 is the frequency of individuals with homozygous recessive genotype

in the given question

RR= 13

rr=9

Rr = 2

Total 24

now the frequency of rr can be calculated directy as the genotype corresponds to only one phenotype

frequency of individual =( individuals divided by total population)

=9 divided by 24

=0.375

frequency of homozygous recessive individual = 0.375 = q2 (expected)

q = square root of 0.375

= 0.61 (expected)

now we know p+q=1

so p= 1-q

= 1-0.61

=0.39 (expected)

p2= (0.39)2

= 0.15 (expected)

now we know p2+2pq+q2=1

2pq= 1-(p2+q2)

= 1-(0.15+ 0.375)

=1-0.525

= 0.475 (expected)

so if the alleles are in hardy weingberg equilibrium

frequency of the doinant allele (p) = 0.39

frequency of the recessive allele (q) = 0.61

frequency of individuals with homozygous dominant genotype (RR)( p2) =0.15

frequency of individuals with heterozygous genotype(Rr) (2pq) = 0.475

frequency of individuals with homozygous recessive genotype(rr) (q2) = 0.375

now coming to the observed frequencies of genotypes

observed rr genotype = 9 divided by 24 = 0.375

observed RR genotype = 13 divided by 24 = 0.54

observed Rr genotype = 2 divided 24 = 0.08

since the observed and expected genotype frequencies are different the data for the population for Pha43 marker is not in hardy weinberg equilibrium