Technology of storage and primary processing of agricultural products

 

18- LABORATORY WORK:
Topic: Population genetics
Objectives: - To explain the laws of meeting genes and genotypes and phenotypes in populations by working on various issues and to analyze the F ₁ and F ₂ generations using the Hardy-Weinberg formula .
Methodical instructions : A population is a group of animals and plants that belong to the same species, spread in a certain area and reproduce in isolation from other populations. It is natural that every plant, animal population should be phenotypically polymorphic, genotypically heterozygous. Population genetics, which is a separate branch of genetics, studies the laws of meeting genes and genotypes as well as phenotypes in populations.
In general, self-interbreeding organisms are less heterozygous, more homozygous, and externally interbreeding organisms are the opposite. Accordingly, in self-interbreeding and externally interbreeding populations, traits are passed down from generation to generation to varying degrees. S.Wright, S.S.Chetverikov in the development of population genetics. The service of NP Romashov and others was great. Advances in population genetics help to understand the laws of evolution and at the same time play a major role in the study of animal and plant genetics in agriculture. The genetic structure of populations can be determined using the Hardy-Weinberg formula.
r ² AA + 2pqAa + q ² aa = 1
where r ^{2 is the} amount of homozygous dominant AA genes in the population,
q ² is the amount of homozygous recessive aa genes in the population ,
The amount of heterozygous Aa genes in the 2rq-population. In this case p + q = 1
For example, if 16 out of 100 flowers in a population are recessive white and the rest are dominant red, what is the structure of the population:

1. 
   Determine the ratio of phenotypes: white flowers (aa) = 16 * 100 = 16%
   
   

100 
red flowers (AA and Aa) = 100-16 = 84%

1. 
   We determine the ratio of alleles: we assume that the number of flowers in a population
   is equal to 1.
   

aa = q ² = 0.16; a = q = 0.16 = 0.4 = 40%
A = 1-a = 1-0.4 = 0.6 = 60%
3.Calculate the ratio of AA and Aa genotypes.
AA = r ² = 0.6 ² = 0.36 = 36% Aa = 2rq = 2 (0.4 * 0.6) = 0.48 = 48%; aa = 16% 

total AA 36% + Aa 48% + aa 16% = 100%

4. We check the amount of alleles.
a = aa + 1Aa = 16 + 48 = 16 + 24 = 40% 
2 2
 A = AA + 1 Aa = 36 + 48 = 36 + 24 = 60% 2 
2
4.  We examine the ratio of genotypes using the Pennet grid.
Table 12

♀ ♂	0.6A	0.4a
0.6A	0.36AA	0.24Aa
0.4a	0.24Aa	0.16 aa