That allows you to install the lattice of the Pennet. Some general methodological recommendations for solving problems of genetics

Objective: Development of skills to use the Pennet's grille, to determine the type of weight and genotypes of offspring.

Equipment: cards with tasks for students, collections of tasks in genetics for schoolchildren.

Progress

Exercise on the use of Pennet's lattice to designate the type of weight and genotypes.

Pennet's grille looks like a two-dimensional table, where a gamet of one parent is recorded in the upper part, and in the left side - vertically, the Gamets of the second parent. And in the cells of the table at the intersection of rows and columns, the genotypes of offspring are recorded in the form of combinations of these weights. Thus, it becomes very easy to determine the probabilities for each genotype in a certain crossing.

Solving problems on mono-librid crossing.

To solve this type of tasks, use the following solutions algorithm:

1. Carefully read the task condition, think about each word.

2. Make a brief task recording. Do not forget to put the appropriate characters.

3. Record the crossing scheme. Remember that the problem of the problem depends on the correct entry.

4. According to the crossing scheme, you can immediately answer the question of the problem, knowing the first and second laws of Mendel. Do it. If this is not possible - spend crossing.

5. If you came to the conclusion based on Mendel laws, you need to check your approval. Spend crossing.

Task 1. Tomato smooth skin of fruits dominates the pubescent. The homozygous form with smooth fruits is crossed with a plant having pubescent fruits. In F1 received 54 plants, in F2 - 736. Some questions: how many types of Games can form a plant with sutured fruits? How many plants F1 can be homozygous? How many F2 plants can have smooth fruits? How many plants F2 can have fetuses? How many different genotypes can be formed in F2?

A - smooth skin of fruits, and - pubescent fruit skin

Solution: 1. Record crossing scheme. The task says that they cross the homozygous plant with smooth seeds, which means its genotype AA, a pubescent plant - AA.

3. We carry out an analysis of crossing. F2 has a splitting: according to genotype - 1 (AA): 2 (AA): 1 (AA); Phenotype 3 (yellow-grained plants): 1 (green-rested plants).

4. We answer the questions of the task.

1) Plants with refrigerated fruits gives one type of weights, since its genotype homozygot for a recessive sign.

2) All plants F1 is heterozygous. Therefore, the number of homozygous plants with powdered fruits in F1 - 0.

3) in F2 - 736 plants. Plants with smooth fruits have genotype AA and AA. They make up 3/4 of the total number of plants - 736: 4 * 3 \u003d 552.

4) Plants with powered fruits make up? from the total number in F2, i.e. 736: 4 \u003d 184.

5) In F2, a splitting of genotype occurred in a ratio of 1: 2: 1, i.e. In F2 3 different genotypes.

Answers: 1) 1; twenty; 3) 552; 4) 184; 5) 3.

Task 2. Black bristles color in pigs dominates red. What offspring should be expected from crossing a black pig with ff genotype and black boar with FF genotype?

F - black bristles, F - red bristles color.

Answer: All offspring has black bristles.

Task 3. Normal rumor in humans is due to the dominant genome S, and the hereditary deafness is determined by the recessive genome S. From the marriage of a deaf woman with a normal man was born a deaf child. Determine the genotypes of parents.

S is a normal rumor, S is deafness hereditary.

P deaf x norm

The child manifested itself a recessive sign, which means its genotype SS. In the genotype of the child, one allele came from the parent organism, and the second is from the father. The mother under the condition manifested itself a recessive sign. Therefore, its SS genotype. The father has a normal hearing, which means one allele is dominant, and the other recessive, which he handed over to the child. If his father was homosigue for this sign, the child was born with normal hearing, but heterozygous (carrier of the deafness gene).

Answer: SS and SS parents genotypes.

Task 4. From the crossing of a comolet bull of the Ishirsk breed with horned cows in F1 received 18 calves (all comoles), in F2 - all 95 calves. What is the number of comolete calves in F2?

Sign: Having horns. D - comolet, D - horned.

95 3/4 \u003d 71,5 \u003d 72 comulom calves

Answer: 72 comulom calves in F2.

3. Solving problems of dihybrid crossing.

To successfully solve problems on the dihybrid crossing, you must be able to:

a) record genotypes of parental individuals. Remember, they are written in four letters: two pairs of alleles (two pairs of signs).

b) Record gametes that forms the parental. Remember, Gamets are recorded in two letters: one allele (gene) from each pair of signs.

For example, AAVA; AAVA; AAVA; AAVA

Gamets AV AV AV AB AB AV

1. Carefully read the task condition (maybe even several times), find out all moments causing any doubts.

2. Mark gene pairs with defined letters (you can take any letter of the Latin alphabet).

3. Briefly record the condition of the task.

4. Record the crossing scheme.

5. If both crossed individuals form more than one type of hemes, make the grid of the ponnet or decide the task to the algebraic method.

6. Record the results of crossing.

Give an answer to the question in the task. If necessary, make an analysis of the results obtained.

Task 1. "In the family of carbohydrate right-handed parents were born by diverse twins, one of which is the carbonous left-handers, and the other blue-eyed right-hander. What is the probability of the birth of the next child, similar to his parents?"

The birth of the carrage-eyed parents of the blue-eyed child testifies to the recessivity of the blue painting of the eyes, respectively, the birth of the right-handed parents of the left-handed child indicates the recessiveness of the best ownership of the left hand compared to the right. We introduce the markings of alleles: a - brown eyes, and - blue eyes, in - right-handed, in - Lefty. We define genotypes of parents and children:

R Aavv x Aavv

F1 A_VV, AAAV_

A_VB - a phenotypic radical, which shows that this child with Left-handed with brown eyes. The genotype of this child can be AAVA, AAVR. If you look for an answer to the task, then we need a phenotypic radical - a_v_ - carbonous right-handed child.

Stressed 9 options for the descendants that we are interested.

Answer. Total possible options 16, so the likelihood of the birth of a child, similar to its parents is 56.25%.

Task 2. A woman with brown eyes and red hair married a man with not red-haired hair and blue eyes. It is known that the father of the woman's eyes were brown, and the mother had blue, both had red hair. The father of the man had no red hair and blue eyes, her mother - brown eyes and red hair. What are the genotypes of all these people? What could be the eyes and hair in children of these spouses?

Alllel gene, responsible for the manifestation of the karen eye of the eye, and (this is well known that the brown eye color dominates over blue), and the allel blue eye gene, respectively, will be a. Be sure to the same letter of the alphabet, as this is one sign - the color of the eyes. The allel gene is not red hair (the color of the hair is the second studied sign) denoted by, as it dominates the allele that is responsible for the manifestation of red hair color - b. The genotype of a woman with brown eyes and red hair we can first point out the incompleteness, and so A-BB. But since it is said that her father was carbonous with red hair, that is, with the genotype A-BB, and her mother was blue-eyed and also with red hair (AABB), then the second allele of a woman with and could be only a, that is, her The genotype will be AABB. The genotype of a blue-eyed man with non-red hair can be recorded first: aab-. But since his mother had a red hair, that is, BB, then the second allel gene with a man could only be b. Thus, the genotype of a man will be recorded by AABB. Gotypes of his parents: Father - Aab-; Mothers - A-BB.

P ♀Aabb × ♂aabb

Children from marriage analyzed spouses will be with equilibly genotypes or phenotype: carbonous not redhead, carbonous redheads, blue-eyed are not redhead, blue-eyed redheads in a ratio of 1: 1: 1: 1.

The task of 3.U Parents with a free urine urine and a triangular pamper on the chin was born a child with a surrounding urine of the ear and a smooth chin. Determine the genotypes of parents, the first child, phenotypes and genotypes of other possible descendants. Make a problem solving scheme. Signs are inherited independently.

A - a free ear of the ear, and - a fragile ear of the ear;

B is a triangular jam on the cheerfulness, in the smooth chin.

It turns out 16 possible genotypes with whom children would be born. The genotype of the child born is the one that is specified in the problem will be 1/16 or 6.25%, its genotype (AVAV), and the remaining 9/16 or 56.25% - a free duft and a triangular fossa, 3/16 or 18.75 % - a free domestic and smooth chin, 3/16 or 18.75% - a controversy died and triangular chin.

4. Solving tasks for analyzing crossing.

To successfully solve this type of task, it is necessary to remember: the analyzing crossing is the crossing of an individual of an unknown genotype with a person, homozygous by recessive alleles (features).

When solving tasks, use the following solutions algorithm:

Carefully read the task and mark the appropriate allele letters (signs, genes, genotypes).

2. Briefly write down the condition of the task.

3. Make an assumption about the genotype of an unknown individual using the laws of Mendel.

4. Confirm your assumption by crossing scheme.

5. Take the output based on the end result of the crossing.

Task 1. "The rooster with a pink ridge is crossed with two hebles, also having a pink comb. The first gave 14 chickens, all with a pink-shaped ridge, and the second - 9 chickens, of which 7 are pink and 2 with a sheet-shaped ridge. Sign. What are the genotypes of all three parents? "

Decision. Before determining the genotypes of parents, it is necessary to find out the nature of the inheritance of the form of the ridge of the ridge. When crossing the rooster with the second chicken, 2 chicken with a leaf ridge appeared. This is possible with the heterozygousness of the parents, therefore, it can be assumed that the rose-shaped crest in the chickens dominates the sheets. Thus, the genotypes of the rooster and the second chicken - AA.

When crossing the same rooster with the first chicken cleavage was not observed, therefore, the first chicken was homozygous - AA.

Task 2. Black fur painting in mink dominates blue. How to prove the chicbnation of two black mink acquired by the animal farm?

A - black coloring, and - blue painting

GENOTIP R-?

Solution: Black coloring can be both heterozygous, and homozygous individuals. But the purebred individuals are homozygous and do not give splitting in the offspring. In order to determine the genotype of individuals with dominant signs, the analyzing crossing is carried out - cross with individuals homozygous for recessive features.

If the studied individual is homozygous (pureborn) (AA), then the offspring from such a crossing will have black color and genotype AA:

If the studied part of the heterozygous (AA), then it forms two types of games and 50% of the offspring will have a black color and aa genotype, and 50%-tall color and genotype AA.

Task 3. Parents have I and II blood group. What blood groups should be expected from offspring? Write crossing scheme.

The blood group is controlled by an autosomal genome. The locus of this gene is denoted by the letter I, and its alleles are indicated as a, b and 0. Moreover, the alleles a and b are dominant, and the allele 0 is recessive. Of the three alleles in the genotype can only be two.

1 Var. P♀ i (a) i (a) x ♂ i (0) i (0)

2 Var. P♀ i (a) i (0) x i (0) i (0)

G i (a), i (0) and i (0)

F1 I (a) i (0) and I (0) i (0)

When crossing, if one parent is genotype I (a) I (a), all children will only have a second group of blood. If one parent I (a) i (0) has two options - I (a) i (a) - the second and (0) i (0) - the first blood group.

Task 4. In dogs, the black wool is dominated over brown, and short wool - over long. The hunter bought a black dog with a short wool and wants to make sure that he does not carry alleles of brown and long wool. What female need to pick up for crossing to check the genotype of purchased PSA?

Q - Black wool color, q - brown wool.

W - short wool, w - long wool.

P ♀ __ __ x ♂ q_ w_

P ♀ -? F1 -?

Decision. To determine the genotype of the PSA, it is necessary to cross it with a female, homozygous according to recessive alleles, i.e. Conduct an analyzing crossing. If puppies with brown and long wool appear in the offspring, the genotype of the crossed PSA will be heterozygous on both pairs of signs, or one of the pairs of signs.

P ♀ qq ww x ♂ qqww

G QW QW QW QW QW

Answer. Fenotype splitting in F1:

P black short-haired \u003d 0.25; 25%;

P black long-haired \u003d 0.25; 25%;

P brown short-haired \u003d 0.25; 25%;

P brown long-haired \u003d 0.25; 25%.

Another embodiment.

P ♀ qq ww x ♂ qqww

P brown short-haired \u003d 0.5; fifty%.

Such splitting of signs of puppies confirms that the cross-protected dog was heterozygous in the color of the wool.

The third embodiment.

P ♀ qq ww x ♂ qqww

Answer: Fenotype splitting in F1:

P black short-haired \u003d 0.5; fifty%;

P black long-haired \u003d 0.5; fifty%.

This result of crossing confirms that the cross-mounted dog was heterozygous on the basis of the length of the wool.

It is well known that the preparation of Pennet's lattices is widely used to solve genetic tasks in Mendelian genetics. The ability to properly draw up the lattice of Pennet is useful to schoolchildren and students in the lessons of biology. But professional genetics use these skills in their work. What is the grate of the Pennet?

The grate of the Pennet is the graphical method proposed by the British genetic Reginald Pennet in 1906, which in visual form demonstrates all possible combinations of various types of weights in specific crossings or in experiments on breeding (each gamet is a combination of one maternal and one father's alleles for each, studied in crossing, gene).

Drawing up a lattice of the Pennet in mono-librid crossing

In mono-librid crossing, the inheritance of one gene is investigated. In the classic mono-librid crossing, each gene has two alleles. For example, we will take the maternal and father's organisms with the same genotype - "GG". In genetics, capital letters are used to designate the dominant allele, and for recessive - lowercase. This genotype can only give two types of weights that contain either Allel "G" or Allel "G".

Our fenhead lattice will look like this:

	G.	g.
G.	GG.	GG.
g.	GG.	gG.

By summing the same genotypes in the Pennet's lattice for our offspring, we obtain the following genotype ratio: 1 (25%) GG: 2 (50%) GG: 1 (25%) GG is a typical ratio of genotypes (1:02:01) for monohybrid Crossing. The dominant allele will mask the recessive allele, which means that the organisms with the GG and GG genotypes have the same phenotype.

For example, if the allele "G" gives yellow and allele "G" gives green, then the GG genotype will have a green phenotype, and the GG and GG genotypes are yellow phenotype. By summing the values \u200b\u200bin the lattice, we will have a 3G- (yellow phenotype) and 1GG (green phenotype) - this is a typical ratio of phenotypes (3: 1) for mono-librid crossing. And the corresponding probabilities for the offspring will be 75% G-: 25% GG.

Pennet's grille and Mendelian inheritance

In the first these results were obtained in the experiments of Gregor Mendel with a plant - a pea garden (Pisum Sativum). Interpretizing the results of Mendel received the following conclusions:

Each sign of this organism is controlled by a pair of alleles.
If the body contains two different alleles for this feature, then one of them (dominant) can manifest itself, fully suppressing the manifestation of another (recessive).
In case of meiosis, each pair of alleles is divided (split) and each gamet receives one of each pair of alleles (splitting principle).

Without these basic laws, we will not be able to solve any genetic task. Having established the opportunity to predict the results on one pair of alternative features, the Mendel moved to the study of the inheritance of two pairs of such signs.

Drawing up a lattice of the Pennet in a dihybrid crossing

With dygibrid crossings, the inheritance of two genes is investigated. For dihybrid crossings, we can make a lattice of the Pennet only if the genes are inherited independently of each other - this means that in the formation of maternal and fatherly heams in each of them, any allele from one pair can get along with any other of another pair. This principle of independent distribution was opened by Mendel in experiments on dihybrid and polygibrid crossings.

We have two genes - forms and colors. For form: "R" is a dominant allele that determines the smooth shape and "w" is a recessive allele that gives a wrinkled form of peas. For color: "Y" is a dominant allele that determines the yellow color and "G" is a recessive allele giving green coloring peas. Male and female plants have the same genotype - "rwyg" (smooth, yellow).

At first it is necessary to determine all possible combinations of Games, for this you can also use the Pennet's grille:

	R.	w.
R.	RR	RW
w.	RW	wW.

Thus, heterozygous plants can give four types of hametts with all possible combinations: RY, RG, WY, WG. Now make a lattice of the Pennet for genotypes:

	RY	Rg.	wY.	wG.
RY	Rryy	RRYG.	Rwyy.	Rwyg.
Rg.	RRYG.	RRGG.	Rwyg.	RWGG.
wY.	Rwyy.	Rwyg.	wwyy.	wwyg.
wG.	Rwyg.	RWGG.	wwyg.	wWGG.

By summing the same genotypes in the Pennet's lattice for our offspring, we obtain the following ratio and probabilities by genotypes: 1 (6.25%) RRYY: 2 (12.5%) RWYY: 1 (6.25%) WWYY: 2 (12.5 %) RRYG: 4 (25%) RWYG: 2 (12.5%) WWYG: 1 (6.25%) RRGG: 2 (12.5%) RWGG: 1 (6.25%) WWGG. And since the dominant signs mask recessive, the ratio and probability of phenotypes we will get such: 9 (56.25%) Ry- (smooth, yellow): 3 (18.75%) R-GG (smooth, green): 3 (18.75%) WWY- (wrinkled, yellow): 1 (6.25%) WWGG (wrinkled, green). Such a ratio of phenotypes - 9: 3: 3: 1 is typical for digibrid crossing.

Drawing up a foam lattice in triumphant crossing.

Make a lattice of Pennet for crossing between two plants heterozygous in three genes will be more difficult. To solve this problem, we can take advantage of our knowledge of mathematics. To determine all possible combinations of Games for triumphant crossing, we must recall the solution of polynomials.

Make a polynomial for this crossing: (a + a) x (b + b) x (C + C).
I will multiply the expression in the first bracket on the expression in the second - we get: (AB + AB + AB + AB) X (C + C).
Now you will multiply this expression on the expression in the third bracket - we get: ABC + ABC + ABC + ABC + ABC + ABC + ABC + ABC.

Accordingly, they can give eight types of weights with all possible combinations. This solution can be illustrated using the Pennet's grille:

	A.	a.
B.	AB	aB
b.	AB	aB

	C.	c.
AB	ABC	ABC
AB	ABC	ABC
aB	aBC	aBC
aB	aBC	aBC

Now we will make a lattice of the Pennet for genotypes (the table will have 64 cells):

	ABC	aBC	ABC	aBC	ABC	aBC	ABC	aBC
ABC	Aabbcc.	Aabbcc.	Aabbcc.	Aabbcc.	Aabbcc.	Aabbcc.	Aabbcc.	Aabbcc.
aBC	Aabbcc.	aabbcc.	Aabbcc.	aabbcc.	Aabbcc.	aabbcc.	Aabbcc.	aabbcc.
ABC	Aabbcc.	Aabbcc.	Aabbcc.	Aabbcc.	Aabbcc.	Aabbcc.	Aabbcc.	Aabbcc.
aBC	Aabbcc.	aabbcc.	Aabbcc.	aabbcc.	Aabbcc.	aabbcc.	Aabbcc.	aabbcc.
ABC	Aabbcc.	Aabbcc.	Aabbcc.	Aabbcc.	Aabbcc.	Aabbcc.	Aabbcc.	Aabbcc.
aBC	Aabbcc.	aabbcc.	Aabbcc.	aabbcc.	Aabbcc.	aabbcc.	Aabbcc.	aabbcc.
ABC	Aabbcc.	Aabbcc.	Aabbcc.	Aabbcc.	Aabbcc.	Aabbcc.	Aabbcc.	Aabbcc.
aBC	Aabbcc.	aabbcc.	Aabbcc.	aabbcc.	Aabbcc.	aabbcc.	Aabbcc.	aabbcc.

Reginald Panel (1875-1967) as a tool representing a graphical entry to determine the combination of alleles from parental genotypes. Along one side of the square there are female games, along the other - men's. This makes it easier and visually present the genotypes obtained by crossing the parent weights.

Mono-librid crossing

In this example, both organism have a BB genotype. They can produce gamets containing allel B or b (the first means dominance, the second is a recessiveness). The probability of a descendant with Genotype of BB is 25%, BB - 50%, BB - 25%.

		Maternal
		B.	b.
Food	B.	BB.	BB.
Food	b.	BB.	bB.

Phenotypes are obtained in combination 3: 1. Classic example - rat wool color: for example, B is black wool, B is white. In this case, 75% of the offspring will have black wool (BB or BB), then as only 25% will have a white (BB).

Digibrid crossing

The following example illustrates a dihybrid crossing between pea heterozygous plants. A represents the dominant allele on the basis of the form (round peas), A is a recessive allele (wrinkled peas). B represents a dominant allele on the basis of color (yellow peas), B is a recessive allele (green). If each plant has an AABB genotype, then, since alleles on the basis of form and color are independent, there may be four types of gaming with all possible combinations: AB, AB, AB and AB.

	AB	AB	aB	aB
AB	Aabb.	Aabb.	Aabb.	Aabb.
AB	Aabb.	Aabb.	Aabb.	Aabb.
aB	Aabb.	Aabb.	aabb.	aabb.
aB	Aabb.	Aabb.	aabb.	aabb.

It turns out 9 round yellow peas, 3 round green, 3 wrinkled yellow, 1 wrinkled green peas. Phenotypes in dihybrid crossing are combined in a ratio of 9: 3: 3: 1.

Czech researcher Gregor Mendel (1822-1884) is considered the founder of geneticsSince he is the first, even before this science was formed, formulated the basic laws of inheritance. Many scientists before Mendel, including an outstanding German hybridizer XVIII century. I. Kelreteer, noted that when crossing plants belonging to various varieties, a large variability was observed in hybrid offspring. However, to explain the complex splitting and, moreover, nobody managed to reduce it to the exact formulas due to the lack of a scientific method of hybridological analysis.

It was thanks to the development of a hybridological method, Mendel managed to avoid difficulties that entrusted earlier researchers. The results of their work of the city of Mendel reported in 1865 at a meeting of the Society of Naturalists in the city of Brynna. Work itself, called "experiments on plant hybrids" was later printed in the "writings" of this society, but did not receive a proper assessment of contemporaries and remained fornger for 35 years.

Being a monk, his classic experiments on the crossing of various varieties of the pea Mendel spent in the Monastery Garden in the city of Brynna. He selected 22 pea varieties that had clear alternative differences in seven signs: seeds are yellow and green, smooth and wrinkled, flowers are red and white, plants are high and low, etc. An important condition for the hybridological method was mandatory use as pure parents, i.e. Not splitting on the signs studied forms.

A great role in the success of Mendel studies has played a good choice of the object. Peas sowing - self-population. To obtain the hybrids of the first generation, Mendel castrated the flowers of the mother plant (removed the anthers) and made artificial pollution of the pestles of the pollen of a male parent. Upon receipt of the second generation hybrids, this procedure was no longer needed: he simply left the hybrids F 1 self-sexual, which made an experiment less laborious. The pea plants multiplied exclusively by sexually, so that no deviations could distort the results of experience. And finally, the pea Mentel found enough for analysis the number of vivid contrasting (alternative) and easily distinguishable pairs of features.

Mendel began analysis from the simplest type of crossing - monohybrid, in which parental individuals have differences in one pair of signs. The first regularity of the inheritance discovered by Mendel was that all the first generation hybrids had the same phenotype and inherited a sign of one of the parents. This sign of Mendel called dominant. Alternative to him a sign of another parent, not manifested with hybrids, was named recessive. Open regularity received names I law Mendel, or the law of uniformity of the hybrids of the i-th generation. In the course of the analysis of the second generation, the second pattern was established: the splitting of hybrids into two phenotypic class (with a dominant sign and with a recessive sign) in certain numerical relations. By counting the number of individuals in each phenotypic class, Mendel found that the splitting in the mono-librid crossing corresponds to the formula 3: 1 (for three plants with a dominant sign, one - with recessive). This pattern was named II of the law of Mendel, or the law of splitting. Open patterns were manifested in the analysis of all seven pairs of signs, on the basis of which the author came to the conclusion about their versatility. In case of self-pollination of hybrids F 2, Mendel received the following results. Plants with white flowers gave offspring only with white flowers. Plants with red flowers behaved differently. Only the third part of them gave a uniform offspring with red flowers. The offspring of the rest was split towards red and white color in the 3: 1 ratio.

Below is a diagram of the inheritance of pea flowers, illustrating I and II laws of Mendel.

When you try to explain the cytological foundations of open laws, Mendel formulated the idea of \u200b\u200bdiscrete hereditary deposits contained in gates and determining the development of paired alternative features. Each gamet carries one hereditary deposit, i.e. is "clean." After fertilization of the zygote, two hereditary deposits (one - from the mother, the other - from the father), which are not mixed and later in the formation of a hybrid gamets also fall into different games. This hypothesis of Mendel received the name of the rules of "Chimets". From a combination of hereditary deposits in the zygote, it depends on what a feature will have a hybrid. The deposit determining the development of the dominant sign, Mendel marked the title letter ( BUT), and recessive - capital ( but). Combination AA and AA The zygote determines the development of a hybrid of the dominant sign. Recessive sign manifest only during combination aA.

In 1902, V. Betson proposed to designate the phenomenon of the partie of the signs by the term "allelaorphism", and the signs themselves, respectively, "allelavorphic". According to its own proposal, organisms containing the same hereditary deposits began to be called homozygous, and containing different depositors - heterozygous. Later, the term "alleryorphism" was replaced by a more concise term "allelism" (Johansen, 1926), and hereditary deposits (genes) responsible for the development of alternative features were called "allele".

Hybridological analysis provides reciprocal crossing of parental forms, i.e. The use of the same individual is first as a parent parent (direct crossing), and then as a paternal (reverse crossing). If both crossings obtain the same results corresponding to the laws of Mendel, this suggests that the analyzed feature is determined by an autosomal genome. Otherwise, a sign of a sign with a floor due to the localization of the gene in the sex chromosome is occurring.

Letter notation: P - parental individual, F - hybrid individual, ♀ and ♂ - female or men's individual (or gameta),
capital letter (a) - dominant hereditary deposit (gene), lowercase letter (a) - recessive gene.

Among the hybrids of the second generation with yellow coloring seeds there are both dominant homozygotes and heterozygics. To determine the specific genotype of the hybrid, the Mendel proposed to conduct a hybrid crossing with a homozygous recessive form. It received the name of the analyzing. When crossing heterozygotes ( AA) With the line with the analyzer (AA), the splitting and genotype is observed, and according to the phenotype in the 1: 1 ratio.

If a homozygous recessive form is one of the parents, then the analyzing crossing simultaneously becomes a beccossom - a return crossing of a hybrid with the parent form. The offspring from such crossing is denoted F B..

The patterns found by Mendelem in the analysis of mono-librid crossing were also manifested in the dihybrid crossing, in which parents differed over two pairs of alternative features (for example, yellow and green seed painting, smooth and wrinkled form). However, the number of phenotypic classes in F 2 was twice as well, and the splitting formula on the phenotype was 9: 3: 3: 1 (on 9 individuals with two dominant signs, three individuals with one dominant and one recessive sign and one individual with two recessive signs ).

To facilitate the analysis of splitting in F 2, the English geneticist R. Pennet proposed its graphic image in the form of a grid, which began to be called by his name ( grid Pennet). On the left vertical in it there are female gamets of the hybrid F 1, on the right - men's. In the internal squares of the lattice fit combinations of genes, occurring during their merger, and corresponding to each genotype of the phenotype. If the gamets are located in the lattice in the sequence, which is presented in the diagram, then in the lattice, it is possible to notice order in the location of the genotypes: all homozygots are located on one diagonal, on the other - heterozygotes in two genes (digeterosigons). All other cells are occupied by monogenetherosigots (heterozygotes of one gene).

The splitting in F 2 can be represented using phenotypic radicals, i.e. Indicating not the entire genotype, but only genes that determine the phenotype. This entry looks like this:

Dates in radicals mean that the second allel genes can be both dominant and recessive, the phenotype will be the same.

Digibrid crossing scheme
(Pennet's grille)

	AB	AB	aB	aB
AB	Aabb. yellow. GL	Aabb. yellow. GL	Aabb. yellow. GL	Aabb. Yellow. GL
AB	Aabb. Yellow. GL	Aabb. Yellow. Forsk.	Aabb. Yellow. GL	Aabb. Yellow. Forsk.
aB	Aabb. Yellow. GL	Aabb. Yellow. GL	aabb. Green. GL	aabb. green. GL
aB	Aabb. yellow. GL	Aabb. Yellow. Forsk.	aabb. Green. GL	aabb. Green. Forsk.

The total number of genotypes F 2 in the lattice of the Pennet - 16, but different - 9, as some genotypes are repeated. The frequency of different genotypes is described by the Rule:

In F 2 of the dihybrid crossing, all homozygotes occur once, mono-produgotes - twice and digeterosigons - four times. In the lattice of the Pennet, 4 homozygots, 8 monogethterosigot and 4 digeterosigons are presented.

The splitting of genotype corresponds to the following formula:

1Avv: 2Avb: 1ABB: 2Avv: 4Avb: 2ABB: 1Avv: 2Avb: 1ABB.

Abbreviated - 1: 2: 1: 2: 4: 2: 1: 2: 1.

Among the hybrids F 2 only two genotypes repeat the genotypes of parental forms: AAVA and aABB.; In the rest there was a recombination of parental genes. It led to the emergence of two new phenotypic classes: yellow wrinkled seeds and green smooth.

After analyzing the results of dihybrid crossing for each pair of features separately, the mendel has established a third pattern: an independent nature of inheritance of different pairs of features ( III Law Mendel). Independence is expressed in the fact that the splitting for each pair of features corresponds to the formula of mono-librid crossing 3: 1. Thus, the dihybrid crossing can be represented as two simultaneously running monohybrid.

As was found later, the independent type of inheritance is due to the localization of genes in different pairs of homologous chromosomes. The cytological basis of Mendelian splitting is the behavior of chromosomes in the process of cellular division and the subsequent merge of the weights during fertilization. In Profase I of the reducing division of MEIOS, homologous chromosomes are conjugated, and then in Anaphze I diverge to different poles, so that allele genes cannot get into one game. Negomological chromosomes during discrepancy are freely combined with each other and move to the poles in different combinations. This causes the genetic heterogeneity of the genital cells, and after their confluence in the process of fertilization - the genetic heterogeneity of the zygota, and, as a result, the genotypic and phenotypic diversity of the offspring.

An independent inheritance of different pairs of features makes it easy to calculate the splitting formula in di- and polygibrid crossings, as they are based on simple formulas for monohybrid crossing. When calculating, the probability law is used (the probability of occurrence of two and more phenomena is simultaneously equal to the product of their probabilities). Dogibrid crossing can be decomposed on two, triumphant - on three independent mono-librid crossings, in each of which the probability of the manifestation of two different features in F 2 is 3: 1. Therefore, the splitting formula on the phenotype in F 2 of the dihybrid crossing will be:

(3: 1) 2 = 9: 3: 3: 1,

triumphant (3: 1) 3 \u003d 27: 9: 9: 9: 3: 3: 3: 1, etc.

The number of phenotypes in F 2 polygibrid crossing is 2 N, where N is the number of pairs of signs that parental individuals differ.

Formulas for calculating other hybrids characteristics are presented in Table 1.

Table 1. Quantitative patterns of splitting by hybrid offspring
With different types of crossings

Quantitative characteristic	Type of crossing
Quantitative characteristic	mono-librid	digibrid	polygibrid
The number of types of weights formed by hybrida F 1	2	2 2	2 N.
The number of Games combinations when forming F 2	4	4 2	4 N.
Number of phenotypes F 2	2	2 2	2 N.
Number of genotypes F 2	3	3 2	3
Fenotype splitting in F 2	3: 1	(3: 1) 2	(3: 1) n
Gogotype splitting in F 2	1: 2: 1	(1: 2: 1) 2	(1: 2: 1) n

The manifestation of the patterns of inheritance, open by Mendel, is possible only under certain conditions (independent of the experimenter). They are:

Equalious formation of a hybrid of all varieties of Games.
All sorts of combination of weights in the process of fertilization.
The same vitality of all varieties of the zygota.

If these conditions are not implemented, the character of splitting in hybrid offspring changes.

The first condition may be violated due to the unvisability of a type of weight, possible due to various causes, for example, the negative effect of another gene manifested in the hametical level.

The second condition is broken in the case of selective fertilization, in which the preferred fusion of certain grades is observed. In this case, the goveta with the same genome can behave in the process of fertilization in different ways, depending on whether it is a female or male.

The third condition is usually disturbed if the dominant gene has a fatal effect in a homozygous state. In this case, in F 2 mono-librid crossing as a result of the death of dominant homozygotes AA Instead of splitting 3: 1, a splitting of 2 is observed: 1. An example of such genes is: Fur Platinum Painting gene in foxes, gray-colored wool colored sheep shearages. (Read more in the next lecture.)

The reason for the deviation from Mendelian splitting formulas may also become an incomplete manifestation of a trait. The degree of manifestation of the genes in the phenotype is indicated by the term expressiveness. In some genes, it is unstable and strongly depends on external conditions. An example is the recessive gene of the black color of the body in Drozophila (Ebony mutation), the expressness of which depends on the temperature, as a result of which the generozygous individuals can have a dark color.

The opening of the laws of inheritance for more than three decades was ahead of the development of genetics. The work published by the author "Experience with plant hybrids" was not understood and appreciated by contemporaries, including C. Darwin. The main reason for this is that by the time of the publication of the work of Mendel, chromosomes were not yet discovered and the cell division process was not yet described, which was mentioned above, the cytological basis of Mendelevian laws. In addition, Mendel himself doubted the versatility of the discovered patterns when, at the Council, K. Nemeli began to check the results obtained on another object - hawk. Not knowing that the hawk is multiplied by parthenogenetically and, therefore, it cannot be obtained by hybrids, the mendel was completely discouraged by the results of the experiments, in no way fit into the framework of his laws. Under the influence of failure, he abandoned his research.

Recognition came to Mendel at the very beginning of the twentieth century, when in 1900, three researchers - D. De Friz, K. Korrens and E. Chermak - independently published the results of their research, reproducing Mendel's experiments, and confirmed the correctness of its conclusions. . Since by this time, Mitosis was fully described, almost completely Meiosis (its full description was completed in 1905), as well as the process of fertilization, scientists were able to associate the behavior of Mendelian hereditary factors with chromosome behavior in the process of cellular division. The reserveing \u200b\u200bof the laws of Mendel and became the starting point for the development of genetics.

The first decade of the twentieth century. It became a period of the triumphal march of Mendelamin. The laws opened by Mendel were confirmed when studying various signs of both plant and animal objects. There was an idea of \u200b\u200bthe versatility of Mendel's laws. At the same time, the facts that did not fit into the framework of these laws began to accumulate. But it was the hybridological method that allowed us to find out the nature of these deviations and confirm the correctness of the conclusions of Mendel.

All pairs of signs that were used by Mendel were inherited by the type of complete domination. In this case, the recessive gene in heterosigot does not act, and the phenotype of heterozygotes is determined by an exclusively dominant genome. However, a large number of signs in plants and animals are inherited by type of incomplete domination. In this case, the hybrid F 1 completely does not reproduce a sign of one or another parent. The expression of the attribute is intermediate, with a large or smaller dodging in one direction or the other.

An example of incomplete domination can be an intermediate pink color of flowers in the hybrids of the night beauty obtained when crossing plants with a dominant red and recessive white color (see scheme).

Scheme of incomplete domination while inheritance of flowers coloring near night beauties

As can be seen from the scheme, the law uniformity of the first generation hybrids is in crossroads. All hybrids have the same coloring - pink - as a result of incomplete domination of the gene BUT. In the second generation, different genotypes have the same frequency as in the experiment of Mendel, and only the splitting formula on the phenotype changes. It coincides with the glotype splitting formula - 1: 2: 1, since each genotype corresponds to its sign. This circumstance facilitates the analysis, as there is no need for the analyzing crossing.

There is another type of behavior of allele genes in heterozygot. It is called codomination and is described when studying the inheritance of blood groups in humans and a number of pets. In this case, a hybrid, in the genotype of which there are both allele genes, and both alternative features are exercising. Codomination is observed in the inheritance of blood groups of the system A, B, 0 in humans. People with a group AU (IV Group) There are two different antigen in the blood, the synthesis of which is controlled by two allelic genes.

Oracle in Generation F 1 is the result of crossing two parental organisms: male and female. Each of them can form a certain number of types of weights. Each goveta of one organism with the same probability can meet with any games of another organism in fertilization. Therefore, the total number of possible zygotes can be calculated by mailing all types of weights of both organisms.

Mono-librid crossing

Example 7.1. Record the genotype of the first generation of individuals when crossing two individuals: homozygous for the dominant gene and homozygous over a recessive gene.

We will write the letter designation of genotypes of parental couples and formed by them gears.

R AA X AA

Gameta A A.

In this case, each of the organisms form gamets of one type, therefore, with the merchant merges, individuals with the AA genotype will always be formed. The hybrid individuals developed from such weights will be uniformly not only in genotype, but also a phenotype: all individuals will carry a dominant sign (according to the first law of Mendel about the uniformity of the first generation).

To facilitate the recording of the genotypes of the offspring, the Games meeting is not allowed to designate an arrow or a straight line connecting the gametes of the male and female organism.

Example 7.2. Determine and write down the genotypes of the first generation of individuals when crossing two heterozygous individuals analyzed by one feature.

R AA X AA

Gameta A; A A; but

F 1 AA; AA AA; AA

Each parent individual is formed by the gamets of two types. The arrows showed that any of the two Games of the female individual can meet with any of the two Games male individuals. Therefore, four options are possible and the individuals are formed in the progement with the following genotypes: AA, AA, AA, AA.

Example 7.3. Hair can be light and dark color. Dominant dark color gene. The marriage joined a heterozygous woman and a homozygous man with dark hair color. What genotypes should be expected in children of i-th generation?

sign: Gen.

dark color: a

light color: a

R AA X AA

dark dark

Gameta A; A.

dark dark

Digibrid crossing

The number and types of zygotes during dihybrid crossing depends on how non-allele genes are located.

If the non-allest genes responsible for different signs are located in a single pair of homologous chromosomes, the number of types of heams in the digerozygous organism with the GMG genotype will be two: AB and AV. When crossing two such organisms, fertilization will lead to the formation of four zygotes. Recording the results of such a crossing will look like this:

P avav x avav

Gameta AV; AVA; AU

F 1 avav; Avav; Avav; avav

DiGaterozygous organisms containing non-allele genes in non-homologous chromosomes have a AAVA genotype and form four types of weights.

When crossing two such individuals, the combinations of them will give 4x4 \u003d 16 of the options for genotypes. The genotype of the formed individuals can be recorded consistently after each other, as we did during mono-librid crossing. However, such a line record will be too cumbersome and difficult for subsequent analysis. The English geneticist Pennet proposed to record the result of crossing in the form of a table, which is named by the name of the scientist - the lattice of the Pennet.

Initially, the genotypes of parental pairs are recorded as usual and their types of them, then the lattice is drawn, in which the number of vertical and horizontal columns corresponds to the number of types of gaming parents. Horizontally, the female individuals are discharged at the top, the vertical on the left is the men's individuals. In places of intersection of imaginary vertical and horizontal lines coming from Games parents, genotypes of descendants are recorded.