I\' struggling to figure out how to solve these Hardy Weinberg problems, my teac
ID: 72092 • Letter: I
Question
I' struggling to figure out how to solve these Hardy Weinberg problems, my teacher mentioned something about using punnet squares but I can't remember exactly. I'd appreciate any explanation you have!
A) Given the following genotype frequencies, explain and demonstrate whether the population is in Hardy-Weinberg equilibrium. fr(AA) = 0.49 fr(AB)=0.15 fr(BB)= 0.36
B) Dr. Microcephalus is studying a population of yellow-crowned whack-a-moles. At a key biallelic locus, these are the numbers of individuals for ach of the genotypes. Is this population in Hardy-Weinberg equilibrium?
A1A1=54 A2A2=32 A2A2=14
Explanation / Answer
Hardy and Weinberg mathematically proved that in a population, all dominant and recessive alleles comprise all alleles for that gene.
This was mathematically represented as p+ q = 1.0
Where,
p = frequency of dominant alleles
q = frequency of recessive alleles.
Hardy and Weinberg also described all the possible genotypes for a gene with two alleles. The binomial expansion representing this is, p2+ 2pq + q2= 1.0
Where,
p2 = proportion of homozygous dominant individuals
q2 = proportion of homozygous recessive individuals
2pq = proportion of heterozygotes.
A). The proportion of BB individuals in a population is, p2 = 0.36; p = 0.6
The proportion of AA individuals in the population is, q2 = 0.49; = 0.7
The population is not in HW equilibrium because p+q = 0.6+0.7 = 1.3 (This value must be equal to 1 to obey HWE)
B). The proportion of A1A1 individuals in the total population is, p2 = 54/ 100 = 0.54
The proportion of A1A2 individuals in the total population is, 2pq = 32/ 100 = 0.32
The proportion of A2A2 individuals in the total population is, q2 = 14/ 100 = 0.14
The frequency of A1 allele = p2 + 1/2 (2pq) = 0.54+0.16 = 0.7
The frequency of A2 allele = q2 + 1/2 (2pq) = 0.14+0.16 = 0.3
The population is in HW equilibrium because p+q = 0.7+0.3 = 1.
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