Population Genetics: How to Solve HardyWeinberg Problems
A population is a group of organisms of the same species usually found in a clearly defined geographical area.
Population genetics: It is the study of alleles of genes in populations, and the forces which maintain or change the frequencies of particular alleles and genotype in populations.
HardyWeinberg equilibrium
The mathematical relation ship between the frequencies of genotypes and alleles in a populations was developed by G.H.Hardy and W. Weinberg in 1908. The relation ship , known as HardyWeinberg equilibrium, is based upon a principle which states that
"the frequency of dominant and recessive alleles in a population will remain constant from generation to generation provided certain conditions exist".
Conditions are:
 The population is large
 Mating is random
 No mutation occur
 No selection occurs
 Generations do not overlap
 There is no emigration or immigration from or into the population, that is, there is no gene flow between populations.
The HardyWeinberg Equation
This equation provides a simple mathematical model of genetic equilibrium can be maintained in a gene pool, its major application in population genetics is in calculating allele and genotype frequencies.
Starting with two homozygous organisms, one dominant allele 'A' and one recessive allele 'a', then all the offspring will be heterozyous (Aa).
When two individuals heterozyous (Aa) for the same gene are crossed (Aa x Aa), they segregate in a ratio 1AA: 2Aa:1aa. This ratio can be obtained by simple binomial expansion (A+a)^{2}= 1AA+2Aa+1aa or A^{2}+2Aa+a^{2}
Suppose p represents the frequency of allele 'A' and q represents the frequency of allele 'a',
then (p+q)^{2} =1pp+2pq+1qq or p^{2}+2pq+q^{2}
Sperms/Eggs
 A (p)
 a (q)


A (p)
 AA (p x p)
 Aa (p x q)

a (q)
 Aa (p x q)
 aa (q x q)

Allele and Genotype Frequency
In mathematical terms p+q=1 is the mathematical equation of probabality and p^{2}+2pq+ q^{2}=1 is the binomial expansion of that equation (p+q)^{2}.
‘p’ and ‘q’ represents Allele frequency
p= dominant allele frequency
q=recessive allele frequency
p^{2}, 2pq, and q^{2}are Genotype frequency
p^{2 }=homozyous dominant genotype
2pq =heterozyous genotype
q^{2 }= homozyous recessive genotype
The frequencies of alleles 'A' and 'a' in a population at HardyWeinberg equilibrium are 0.7 and 0.3, respectively. In random sample of 250 individuals taken from the population, how many are expected to be heterozygous?
Answer:
Frequency of allele A (p) = 0.7 or 70%
Frequency of allele a (q) = 0.3 or 30%
Total number of individuals(p^{2 }+ 2pq + q^{2}) =250
First we have to found out, 2pq (heterozygous genotype) = 2 x 0.7 x 0.3 = 0.42 or 42%
in 250 populations, 42% individuals are heterozygous i e,
250 x (42/100) or 250 x 0.42= 105
Calculating Gene Frequency
If in a given sample of N individuals of which D is dominant allele (A^{1}A^{2}), H is heterozygous (A^{1}A^{2}) and R is recessive allele (A^{2} A^{2}).
Then N=D+H+R
Each A^{1}A^{1} genotype has two A alleles. Heterozygotes (A^{1}A^{2}) have only one A^{1} allele.
Let‘p’ represents the frequency of the A allele and ‘q’ represents the frequency of the a allele
Then p= [2D+H]/2N = [D+1/2H]/N and q= [2R+H]/2N = [R+1/2H]/N
Frequency of gene = Frequency of homozygote for that gene + 1/2 frequency of heterozygotes
A population has the following genotypic distribution: 10BB 15Bb 25bb 1) Based on your calculated allele frequencies, what proportion of individuals would you expect to be homozygous dominant, heterozygous, and homozygous recessive if the population is in HW equilibrium? 2) Is this population in HW equilibrium?
(Can you show me how to do this hardyweinberg equation? asked by Lwelch on HubPages)
Answer to the question
Same as the above MN blood group problem, Here the genotype is 10BB 15Bb 25bb
1) Based on your calculated allele frequencies, what proportion of individuals would you expect to be homozygous dominant, heterozygous, and homozygous recessive if the population is in HW equilibrium?
Here
'BB' represents the homozygous dominant =10 individuals
'Bb' represents the heterozygous =15 individuals
'Bb' represents the homozygous recessive =25 individuals
Total number of individuals=10+15+25=50 individuals
10/50 individuals belongs to homozygous dominant (BB) i.e., 0.2 or 20%
15/50 individuals belongs to heterozygous (Bb) i.e., 0.3 or 30%
25/50 individuals belongs to homozygous recessive (bb) i.e., 0.5 or 50%
The proportion of individuals would you expect to be homozygous dominant, heterozygous, and homozygous recessive: 20%BB : 30%Bb : 50%bb
Frequency of B allele = (10 X 2) + 15
Frequency of B allele =20 +15
Frequency of B allele = 35
Frequency of b allele = (25 X 2) + 15
Frequency of b allele= 50+15
Frequency of b allele= 65
Then the frequency of gene B = B / (B+b)
B = 35 / (35 +65)
B = 35 / 100
Frequency of gene B=0.35
The frequency of gene b =b / (B+b)
b= 65 / (35 + 65)
b= 65 / 100
Frequency of gene b = 0.65
Frequency of gene B + Frequency of gene b = 0.35+ 0.65 =1
Or otherwise
Frequency of gene = Frequency of homozygote for that gene + 1/2 frequency of heterozygotes
Frequency of B gene = Frequency of homozygote for that gene (BB) + 1/2 frequency of heterozygotes (Bb)
Frequency of B = 10/50 BB + 1/2 (15/50) Bb
Frequency of B = 0.2 BB + 1/2 (0.3) Bb
Frequency of B = 0.2 +0.15
Frequency of B gene= 0.35
Frequency of b gene = Frequency of homozygote for that gene ( bb) + 1/2 frequency of heterozygotes (Bb)
Frequency of b = 25/50 bb + 1/2 (15/50)Bb
Frequency of b = 0. 5 bb + 1/2 (0.3) Bb
Frequency of b = 0.5 + 0.15
Frequency of b gene = 0. 65
2) Is this population in HW equilibrium?
yes
here
Frequency of B gene + frequency of b gene =1
0.35 +0.65 =1
Glossary
Allele: Any of the alternative forms of a given gene.
Allele frequency: The relative proportion of all alleles of a gene that are of a designated type.
Dominant allele: An allele associated with a phenotype that is expressed when the allele is either heterozygous or homozygous.
Gene: A segment of DNA, which specifies a single polypeptide or an RNA molecule.
Gene pool: The sum total of all the alleles present in the breeding or reproductive members of a random mating population.
Genotype: The genetic constitution of an organism.
Heterozygous: An individual having two different alleles for one or more genes
Homozygous: An individual having two copies of the same allele for one or more genes under consideration.
Random mating: System of mating in which mating pairs are formed independently of genotype and phenotype.
Recessive: Refers to an allele, or the corresponding phenotypic traits, expressed only in homozygotes.
For Sample Problems
 Problem: 1
Let's return to our population of pigs. Remember that the allele for black coat is recessive. We can use the HardyWeinberg equation to determine the percent of the pig population that is heterozygous for white coat.  Problem : 2
In a certain population of 1000 fruit flies, 640 have red eyes while the remainder have sepia eyes. The sepia eye trait is recessive to red eyes. How many individuals would you expect to be homozygous for red eye color?  Problem: 3
In the United States, one out of approximately 10,000 babies is born with the disorder. Approximately what percent of the population are heterozygous carriers of the recessive PKU allele?  Problem: 4
If you observe a population and find that 16% show the recessive trait, you know the frequency of the aa genotype. This means you know q2. What is q for this population?
Comments
Thank you so much for helping with this! I got the homework done, got the points, and got an A in my biology class.