Technology of storage and primary processing of agricultural products

Polyduragay mixing

The purpose of the work. It is the study of the inheritance of traits in mixed offspring, taking into account two or more traits.
The content of the work. Mixing two organisms that differ from each other by two or more traits is called polyduragay mixing. For example, when an organism that is sharply different from each other is confused with three signs:
AAVVDD x aabbdd
gametes ABD abd
F ₁ AaBbDd
This AaVbDd genotype forms 8 different ABD, ABd, AbD, Abd, aBD, abD, aBd, abd gametes. When the F ₁ hybrid (AaBbDd) is self-crossed, 3 traits are observed to be passed on to offspring independently of each other in the F ₂ generation because these genes are located on separate homologous chromosomes. The parent forms each form 8 different gametes and are added in 64 combinations in the Pennet cell.
Monoduragay hybrid phenotype separation of class 2 - ¹ = 2, diduragayda 2 ² = 4, triduragayda 2 ³ = 8 and so on issue. In this way it is possible to calculate the types of gametes formed in a hybrid and the combination of gametes in F ₂ : in a monolayer two different gametes 2 ¹ = A / a; 4 different gametes from diduragay or 2 ² = AV / Av / aV / av; 8 different gametes or 2 ³ = AVS / AVs / AvS / Avs / aVS / aVs / avS / avs are formed from triduragay .
For example, two different male and female gametes are formed from the first joint monoduragay, and from their combination 4 combinations: 1 A / A; 2 A / a; 1 a / a i.e. 4 ¹ = 4 is formed. In Diduragay 4 ² = 16, in Triduragay 4 ³ = 64 gamete combinations are formed.
Thus, the number of gametes formed from F ₁ , the number of phenotype and genotype classes can be calculated while the number of genes in polydura hybridization is precise . The following table shows the combinations and phenotypic and genotypic separations that occur in different hybrids.
Table 5

Allel number of s	Number of gamete varieties	The number of gamete combinations	Number of phenotypic classes	Number of genotypic classes	The phenotypic formula of divorce
1	2 ^1= 2	4 ^1= 4	2 ^1= 2	3 ^1= 3	3: 1
2	2 ^2= 4	4 ^2= 16	2 ^2= 4	3 ^2= 9	9: 3: 3: 1
3	2 ^3= 8	4 ^3= 64	2 ^3= 8	2 ^3= 27	27: 9: 9: 9: 9: 3: 3: 3: 1
4	2 ^4= 16	4 ^4= 256	2 ^4= 16	3 ^4= 81	81:27:27: 27: 27: 9: 9: 9: 9: 9: 9: 9: 3: 3: 3: 3: 1
N	2 ^n	4 ^n	2 ^n	3 ^n	(3: 1) ^n

Issues to be solved independently
1. What gametes emerge from plants with the following genotypes ? a) AaVVSs; b) aaVbSs; v) AaVbSs; g) AaVbSsDD.
2. What phenotypic peas occur in the following hybrids ?
a) AaBbCC x aaBBCc b) AaBbCC x aabbcc
A - yellow, V - smooth, S - tall,
a - green, b - uncurled, s - short. 
3.Fill in the table below.
Table 6

Allel number of s	Number of gamete varieties	The number of gamete combinations	Number of phenotypic classes	Number of genotypic classes
5	2 ^5=	4 ^5=	2 ^5=	3 ^5=
6	2 ^6=	4 ^6=	2 ^6=	3 ^6=
7	2 ^7=	4 ^7=	2 ^7=	2 =
8	2 ^8=	4 ^8=	2 ^8=	3 ^8=