Population Genetics is the study of allele and genotypic frequency within a population. It is concerned with how the frequency of alleles in a gene pool change over time.
In addition to understanding Mendelian genetics, the knowledge of Population Genetics helps breeders to understand changes in gene frequencies in populations as selection alters the gene frequency of breeding population.
Population Genetics model are useful both for statistical inference from DNA data and for proof or disproof of concept and lot more.
HARDY-WEINBERG EQUILIBRIUM
Hardy from England and Weinberg from Germany discovered that equilibrium between genes and genotypes is achieved in a large population. They showed that the frequency of genotypes in a population depends on the frequency in the preceding generation, not on the frequency of the genotypes.
This principle provides the solution to how variation is maintained in a population with Mendelian inheritance.
Basic Formulas
P = frequency of the dominant allele in the population
q = frequency of the recessive allele in the population
P2 = frequency of individual that is homozygous dominant in the population
2Pq = frequency of individual that is heterozygous in the population
q2 = frequency of individual that is homozygous recessive in the population
Q1: In a given population, 84% of the population can roll their tongue; determine how many people will be homozygous for tongue rolling if the population is at equilibrium. (Note: Tongue rolling is controlled by a dominant gene).
Solution1
Let represent gene for tongue rolling = R
Gene for non-tongue rolling = r
Therefore, tongue rollers in the population will be
(i) homozygous rollers = RR
(ii) heterozygous rollers= Rr
Because the ability to roll tongue is controlled by dominant gene, individual that are unable to roll their tongue = ‘rr’
From the question, 84% of the people are tongue rollers
Therefore, % of non-rollers 100-84 = 64% (non-rollers ‘rr’).
Remember rr = q2 (according to Hardy- Weinberg principle).
q2 = 16%
= 0.16
q = 0.4
If P + q = 1
P = 1-q
P = 1-0.4
P=0.6
Therefore, P = 0.6 and q = 0.4 (frequency of dominant and recessive allele in the population)
(i) number of people that is heterozygous roller = 2Pq
2Pq = 2 × 0.6 × 0.4
= 2 × 0.24
= 0.48
= 48% (0.48 × 100)
(ii) Number of people that is homozygous P2
P2 = 0.6 ×0.6
= 0.36
P2 = 36% (0.36 × 100)
On the other hand, we can decide to subtract the number of individual that is heterozygous tongue rollers (48%) from the total number of individual that is tongue roller (84%).
P2 = 84% - 48%
= 36%.
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