Part B. Hardy-Weinberg Equilibrium Hardy-Weinberg Equilibrium, also referred to
ID: 149136 • Letter: P
Question
Part B. Hardy-Weinberg Equilibrium Hardy-Weinberg Equilibrium, also referred to as the Hardy-Weinberg principle, is used to must meet five rules in order to be considered "in equilibrium es in a given population over a period of time. A population of alleles I) No gene mutations may occur and therefore allele changes do not occur 2) There must be no migration of individuals either into or out of the population. 3) Random 4) No genetic drift, a chance change in allele frequency, may occur. 5) No natural selection, a change in allele frequency due to environment, may occur mating must occur, meaning individuals mate by chance. a change in allele frequency due to environment, Hardy-Weinberg Equilibrium never occurs in nature because there is always at least one rule being violated. Hardy-Weinberg Equilibrium is an ideal state that provides a baseline against which scientists measure gene evolution in a given population. The Hardy-Weinberg equations can be used for any population; the population does not need to be in equilibrium. There are two equations necessary to solve a Hardy-Weinberg Equilibrium question: p- the frequency of the dominant allele q the frequency of the recessive allele. p- the frequency of individuals with the homozygous dominant genotype. 2pq the frequency of individuals with the heterozygous genotype. q -the frequency of individuals with the homozygous recessive genotype. Steps: 1. Find q Individuals with recessive genotype/total population Square root of q p+q=1 2. Find q 3. Find p you know q, plug in and solve for p 4. Now that you have p and g (allele frequencies) you can solve for genotype frequencies by using p+ 2pg +q-1 5. Square p to find p 6. Multiply 2xpxq to get 2pq 7. You should now have all your values(p, q, p. 2pq, q2) 8. Multiply frequency of genotypes by total population to get number of individuals with that genotype Ex: p' x total population number of individuals with homozygous dominant 9. Repeat steps for next generation to see if populations fall within Hardy-Weinberg Equilibrium.Explanation / Answer
(1A) Add all the individuals of population = 1024 + 512 + 64
= 1600
Find q2 = Individuals with recessive genotype / Total Number of Individuals
= 64 / 1600
= 0.04
Take square root of q2 = 0.04
q = 0.2
Use Hardy Weinberg Equation
p + q =1
p + 0.2 = 1
p = 0.8
Frequency of dominant allele (p) = 0.8
Frequency of recessive allele (q) = 0.2
Frequency of individuals with dominant genotype (p2) GG = 0.64
Frequency of individuals with recessive genotype (q2) gg = 0.04
Frequency of individuals with Heterozygous genotype (2pq) Gg = 0.32
(1B) Solve for q2 = Individuals with recessive genotype / Total population
= 108 / 1092 + 108
= 0.09
Take square root of q2 = 0.09
q = 0.3
Because recessive allele frequency was changed. Therefore, population is NOT in a state of Hardy Weinberg equilibrium.
(2A) Add all the individuals of population = 104 + 549
= 653
Find q2 = Individuals with recessive genotype / Total Number of Individuals
= 104 / 653
= 0.16
Take square root of q2 = 0.16
q = 0.4
Use Hardy Weinberg Equation
p + q =1
p + 0.4 = 1
p = 0.6
Frequency of dominant allele (p) = 0.6
Frequency of recessive allele (q) = 0.4
Frequency of individuals with dominant genotype (p2) = 0.36
Frequency of individuals with recessive genotype (q2) = 0.16
Frequency of individuals with Heterozygous genotype (2pq) = 0.48
(2B) Solve for q2 = Individuals with recessive genotype / Total population
= 560 / 840 + 560
= 0.4
Take square root of q2 = 0.4
q = 0.63
Because recessive allele frequency was changed. Therefore, population is NOT in a state of Hardy Weinberg equilibrium.