Выбрать язык / Select language:
Ukranian
English
French
German
Japanese
Italian
Portuguese
Spanish
Danish
Chinese
Korean
Arabic
Czech
Estonian
Belarusian
Latvian
Greek
Finnish
Serbian
Bulgarian
Turkish
body { background-color: #E5E5E5; scrollbar-face-color: #DEE3E7; scrollbar-highlight-color: #FFFFFF; scrollbar-shadow-color: #DEE3E7; scrollbar-3dlight-color: #D1D7DC; scrollbar-arrow-color: #006699; scrollbar-track-color: #EFEFEF; scrollbar-darkshadow-color: #98AAB1;}
Суббота, 21.12.2024, 21:37Главная | Регистрация | Вход

Меню сайта

Форма входа

Приветствую Вас Гость!

Друзья сайта

Мини-чат

Наш опрос

Оцените мой сайт
Всего ответов: 356

Статистика

Examples of genetics topics. - Форум / ForumExamples of genetics topics. - Форум / Forum
[ Новые сообщения · Участники · Правила форума · Поиск · RSS ]
  • Страница 1 из 1
  • 1
Examples of genetics topics.
genetika-guppyДата: Четверг, 28.05.2009, 01:24 | Сообщение # 1
Admin
Группа: Администраторы
Сообщений: 2794
Награды: 1
Репутация: 5
Статус: Offline
For recording of genetics problems we have simple rules, but adherence to them will greatly facilitate the understanding of genetics of the studied traits.

TOPIC:

When breeding females guppi with a black tail and the males who had a red tail in all F1 fish were in the red-tailed black. In the F2 was splitting into 4 phenotypic classes, one (blue) was new.

Explain and define the splitting of parental genotypes.

ANALYSIS:

1. Parents are recorded as follows:

First written sign PP (parents), then the signs of a female, the "x" and symptoms male. In doing so, the words male and female fall, because it is clear that the female on the left and right - male.

For example,

PP black red tail x tail

So the female with a black tail crossed with red tail male.

If the known genes, we can write the following designation of the genes, for example:

PP black red tail x tail
aaVV and AAbb

In this example, the color of each tail is indicated by two genes - A and B, but different alleles - dominant (AA and BB - it is genotype) and recessive (aa and bb).

We assume that these are autosomal genes and will not be the difference in phenotype between males and females. Thus, in both sexes have the same phenotype in the F1 and the same splitting in the F2.

After recording signs parents can bring pictures of fish (if available).

2. Progeny of first generation is recorded as a F1.

In our case:

F1 red-black
Aa Bb

3. The second generation (hybrid F1xF1) is written as F2, etc. (F3, F4 ...).

Consequently, to draw the lattice Pinneta (see below), we obtain the following splitting of F2:

F2 AABB AABb AAbb AaBB AaBb Aabb aaBB aaBb aabb,

which is made 4 color-type tail

(AABB AABb AaBB AaBb) - red-black

(aaBB aaBb) - black tail

(AAbb Aabb) - red tail

aabb - blue (new phenotype)

genotypes in the ratio 4: 2: 2: 1 and phenotypes 9: 3: 3: 1.

Now, explain the option, when these genes are sex chromosomes - X and Y.

Record condition of the problem we need in another way:

PP black red tail x tail
XAXA Xa YB

Get

F1 XAXa (female) XAYB (males)
red-tail black black tail

In the F2 cleavage will be, respectively, as follows:

F2 (XAXA XAXa)
all females with a black tail

XAYB
males with red and black tail

[B] X [/ b] aYB
males with a red tail

When analyzing a crossover (crossover with the homozygous recessive form, which is the analyzer, since only one type of gamete with the recessive allele) - written Fa (analysis), back when (when crossing the progeny with one parent) - Fr (return).

 
pshaddockДата: Четверг, 28.05.2009, 04:51 | Сообщение # 2
Новичок
Группа: Пользователи
Сообщений: 6
Награды: 0
Репутация: 1
Статус: Offline
Gene Notation
Here is a useful shorthand for describing the genetic makeup of guppies.

  • Letters are used to designate the name of a mutation, like “b” for blond or Cp for Caudal Pigmentierte.
  • Letters are case sensitive to indicate a gene’s dominant (first letter is uppercase) or recessive (first letter is lowercase) relationship to its allele.

Sex Linked Genes
Sex linked genes are shown beside the chromosome they are found on.

XYSb

Autosomal Genes

Autosomal genes are shown in pairs.

gg

Crosses

The capitalized letter “X” is used to indicate that two guppies are “crossed”.

Example: Gg X gg

By convention the guppy on the left of the “X” is male and the guppy on the right is female.

Сообщение отредактировал pshaddock - Суббота, 30.05.2009, 04:10
 
genetika-guppyДата: Пятница, 29.05.2009, 00:46 | Сообщение # 3
Admin
Группа: Администраторы
Сообщений: 2794
Награды: 1
Репутация: 5
Статус: Offline
Thanks, Philip!

My rules more simple.

But I want to say that these rules are needed for GENETIC ANALYSIS understanding.

For example, you can to know all rules but not understanding rules of genetic analysis.

I don't know internet links and books with good knoledge of real genetical analysis of guppy yet!!!

There are simple genetical sites only. We have partial information. I know it becouse I am biologist.

Because they do not exist and I hope this site will be one of first but not last.

I think that the best site about guppy genetics today is http://www.guppydesigner.com/.

But this site more club and database than genetic school and scientific portal.

I think this site will be first international professional genetics school with genetics analysis elements and

international Enternet-portal for connection russian guppybreeders with other world.

I hope Philip and other professional guppybreeders help me and Vladimir with it.

 
pshaddockДата: Суббота, 30.05.2009, 04:15 | Сообщение # 4
Новичок
Группа: Пользователи
Сообщений: 6
Награды: 0
Репутация: 1
Статус: Offline
Sergey,
I simplified my suggestion above.

I don't think you understand who the audience is for my site and what I am trying to do with it. But that is okay. I welcome a site that claims to be strictly scientific. I hope to learn from it. Since I am not a scientist, I am not sure how I can contribute to it.

Philip

 
genetika-guppyДата: Суббота, 30.05.2009, 22:29 | Сообщение # 5
Admin
Группа: Администраторы
Сообщений: 2794
Награды: 1
Репутация: 5
Статус: Offline
I am sorry Philip if it is mistake. But it was my opinion only.
Your site one of the best in guppy genetics.

We will have friedship. smile

 
pshaddockДата: Воскресенье, 31.05.2009, 02:12 | Сообщение # 6
Новичок
Группа: Пользователи
Сообщений: 6
Награды: 0
Репутация: 1
Статус: Offline
No problem. I am relieved of the responsibility to make my site the most scientific. happy

Philip

 
genetika-guppyДата: Воскресенье, 31.05.2009, 23:18 | Сообщение # 7
Admin
Группа: Администраторы
Сообщений: 2794
Награды: 1
Репутация: 5
Статус: Offline
smile smile smile
 
genetika-guppyДата: Четверг, 13.08.2009, 15:08 | Сообщение # 8
Admin
Группа: Администраторы
Сообщений: 2794
Награды: 1
Репутация: 5
Статус: Offline
So, I suggest you to look at the task of selecting of guppy.

The first task that we will be dedicated to explain the base body color of guppy.

Task:

"Blue guppy was cross with blond guppy, the F1 all the gray fish was reached.
In the F2 cleavage occurred: 55 gray, 17 blue, 19 blond and 6 whites.
When mating of F1 gray with white from F2 were 17 gray, 20 blue, 15 blond and 19 whites.

1. What is the genotype of parents and how these traits are inherited?

2. What happens if homozygous gray guppy will cross with homozygous white guppy?"

So to solve this problem use of genetic analysis:

1. Blue guppy was cross with blond guppy, in all F1 fish was reached gray
So the original parents were homozygous, as in F1, all fish received the same.
Nevertheless, all fish were gray - that is, do not resemble their parents and did not have the intermediate phenotype - it means we are dealing with the interaction of several genes, each of which is in the homozygous form.

2. "In the F2 cleavage occurred: 55 gray, 17 blue, 19 blond and 6 white."
The second generation, we see a splitting into 4 classes, usually inherent digennomu inheritance (when the alleles of two genes interact).

Verify this hypothesis.

For this demitted number of fish in all 4 classes:

55 + 17 + 19 + 6 = 97

Divide the resulting number by 16 (4 in the second degree = number of possible genotypes in the second generation of the cleavage of 2 genes):

97: 16 = 6.06 or 6 for level account (although a better share to a number).

Next, divide the number of fish in each class by the number 6:

55: 6 = 9,2
17: 6 = 2,8
19: 6 = 3,2
6: 6 = 1

We can see the splitting of close to 9:3:3:1 - confirmed the hypothesis (of course it is necessary to check its statistical methods, but more on that, we'll talk later - let us first with the genetic analysis investigate).

3. "In the mating of F1 of gray with white from F2 were 17 gray, 20 blue, 15 light and 19 whites."

Let us test our hypothesis:

PP Bb Rr (gray) X bb rr (white)

F1 BbRr (17) + Bbrr (20) + bb Rr (15) + bb rr (19)

Given the double homozygous one of the parents we have to get 1:1:1:1.

Checking:

Only F1 was obtained 71 fish - divide by the number of classes (4 classes):

71:4 = 17.75

Now divide the number of fish in each class at 17.75:

17: 17,75 = 0.95

20: 17,75 = 1.12

15: 17, 75 = 0.84

19: 17,75 = 1.07

As can be seen from the results we got close to the splitting of 1:1:1:1.

4. BB RR (gray) X bb rr (white)

F1 Bb Rr (gray)

All fishes will gray.

Thus:

1. Genotype of parents - BB rr (blue) and bb RR (blond)

F1: Bb Rr (gray)

F2: bb rr (white), BB rr and Bb rr (blue), bb RR and bb Rr (blond), Bb Rr or BB RR (gray)

2. All fishes will gray.

 
  • Страница 1 из 1
  • 1
Поиск:

Copyright MyCorp © 2024 | Используются технологии uCoz