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Population Genetics: How to Solve Hardy-Weinberg Problems

Updated on December 17, 2012

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.

Hardy-Weinberg 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 Hardy-Weinberg 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:

  1. The population is large
  2. Mating is random
  3. No mutation occur
  4. No selection occurs
  5. Generations do not overlap
  6. There is no emigration or immigration from or into the population, that is, there is no gene flow between populations.

Hardy-Weinberg Equation
Hardy-Weinberg Equation

The Hardy-Weinberg 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 A2+2Aa+a2

Suppose p represents the frequency of allele 'A' and q represents the frequency of allele 'a',

then (p+q)2 =1pp+2pq+1qq or p2+2pq+q2

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 p2+2pq+ q2=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

p2, 2pq, and q2are Genotype frequency

p2 =homozyous dominant genotype

2pq =heterozyous genotype

q2 = homozyous recessive genotype

The frequencies of alleles 'A' and 'a' in a population at Hardy-Weinberg 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(p2 + 2pq + q2) =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 (A1A2), H is heterozygous (A1A2) and R is recessive allele (A2 A2).

Then N=D+H+R

Each A1A1 genotype has two A alleles. Heterozygotes (A1A2) have only one A1 allele.

Letp’ 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 H-W equilibrium? 2) Is this population in H-W equilibrium?

(Can you show me how to do this hardy-weinberg 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 H-W 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 H-W 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.

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    • Lovely 7 profile image
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      Lovely 7 4 years ago

      Lwelch,

      I am extremely happy to know that this hub worked for you.

      Thank you very much for your visit and comment.

    • Lwelch profile image

      Lena Welch 4 years ago from USA

      Thank you so much for helping with this! I got the homework done, got the points, and got an A in my biology class.