TASKS AND METHODS OF GENETICS.
Genetics - a fairly young science.aboutits founder is the Austrian naturalist Gregor Mendel (1822–1884). In 1865, at a meeting of the society of science lovers in the city of Brno (Czech Republic), G. Mendel considered the mechanism of preservation of adaptive traits of the species among generations. In 1866, he published his work “Experiments on Plant Hybrids,” but this publication did not attract the attention of contemporaries. In the spring of 1900, three botanists - G.de Frieze in Holland, K. Chermak in Austria and K. Corrence in Germany - independently from each other, at completely different sites, discovered an important pattern of inheritance of characters in the offspring of hybrids. But it turned out that they simply “rediscovered” the laws of inheritance, considered by G. Mendel in 1865. Nevertheless, the official birth date of genetics is still considered 1900.
Genetics - The science of heredity and variability of living organisms.Heredity - This is the body’s ability to transmit its characteristics, the characteristics of development to future generations. The elementary unit of heredity are genes located on the chromosomes. The transmission of characters by inheritance is carried out in the process of reproduction. During sexual reproduction, the inheritance of signs and developmental features is carried out through germ cells. With asexual reproduction, inheritance is carried out through vegetative cells and spores, in which the material foundations of heredity are enclosed. The characteristic features of a species, breed, variety are preserved from generation to generation by plants, animals, microorganisms due to heredity. But with sexual reproduction, the similarities between parents and the new generation are less, as there is variability.
Variability - this is the property of the body to acquire new signs in the process of individual development. Variability provides material for selection activities and the evolutionary process. Due to the variability, individuals of one species differ from each other. The appearance of new traits in individuals of one species depends on a change in the material basis of the heredity of the organism and on the external conditions affecting the organism.
The totality of all the hereditary traits of the body (genes) is calledgenotype .
The totality of the signs of the organism (external and internal) manifested in the process of life was calledphenotype . Consequently, the phenotype is determined by the genotype, but the external conditions for the existence of organisms in which the genotype is realized can largely determine the manifestation of certain signs. Individuals of the same species having the same genotype may differ from each other depending on the conditions of existence and development. It can be concluded that the phenotype develops when the genotype and environmental conditions interact.
The main task of genetics is the study of such important problems as storage, transmission, sale and variability of hereditary information. The following methods are used to solve these problems.
The most widely used in genetics is the hybridological method of studying heredity.
The main features of the hybridological method:
1) G. Mendel did not take into account the entire diverse complex of characteristics of parents and their descendants, but singled out and analyzed inheritance by individual characteristics;
2) an accurate quantitative account was taken of the inheritance of each trait in a series of successive generations;
3) G. Mendel traced the nature of the offspring of each hybrid separately.
For his research, G. Mendel chose peas, since this plant has many well-distinguishing characteristics (form of seeds, color of seeds and flowers); pea is characterized by self-pollination, which allowed Mendel to analyze the offspring of each individual separately. For crossbreeding, G. Mendel selected plants that have pairs of alternative (mutually exclusive) characters.
Let us consider in detail monohybrid crosses. A classic example of monohybrid crossbreeding is the crossbreeding of pea plants with yellow and green seeds.
When crossing a variety of peas with yellow seeds and a plant with green seeds, all the offspring of the first generation turned out with yellow seeds. When crossing pea plants with wrinkled and smooth seeds, all offspring turned out to be with smooth seeds. The regularity discovered by G. Mendel was called the rule of uniformity of hybrids of the first generation, orthe law of dominance (I Mendel law).
The sign that appears in the first generation is calleddominant (yellow color of seeds, smooth surface of seeds), and a sign that has not developed (suppressed sign) -recessive (green color, wrinkled surface of the seeds). Mendel introduced genetic symbolism to record the results of the cross; P - parents; - female; - male; x is the sign of the cross; G - gametes; F - offspring; hybrids of the first, second and subsequent generations are indicated by the letter F with the number below - F 1 , F 2 , F 3 ...; Letters of the Latin alphabet A, a, B, B, C, C, D, d ... indicate individual hereditary characters, with dominant characters indicated by the capital letters A, B, C, D, ..., and recessive characters respectively a, b , s, d ...
When compiling a crossover scheme, it is necessary to remember that each somatic cell has a diploid set of chromosomes. All chromosomes are paired. Genes that determine the alternative development of the same trait and are located in identical regions of homologous chromosomes are calledallelic genes or alleles. There are always two allelic genes in the zygote, and the genotypic formula for any sign must be written in two letters. If any pair of alleles is represented by two dominant (AA) or two recessive (aa) genes, such an organism is calledhomozygous. If in one and the same allele one gene is dominant and the other recessive, then such an organism is calledheterozygous (Aa).
Genetic recording is as follows:
From the above cross-breeding scheme, we see that the hybrids of the first generation are uniform in their dominant character. This pattern, as already mentioned, is known asMendel’s Law: when two homozygous organisms are crossed that differ from each other by an alternative variant of the same trait, all hybrids of the first generation will be uniform both in phenotype and in genotype, and will carry traits of both parents in the genotype.
The experiments carried out by Mendel showed that the dominant gene appears in both the homozygous and heterozygous state, and the recessive gene only appears in the homozygous state.
Then G. Mendel crossed the hybrids of the first generation among themselves and obtained the following results: out of 8023 pea seeds received in the second generation, there were 6022 yellow and 2001 green. The same ratio was obtained with other variants of crossing between hybrids of the first generation. Based on this, Mendel came to the conclusion that in the second generation there is a splitting of characters in a ratio of 3: 1, that is, 75% of individuals carry dominant characters, and 25% are recessive.
Let's make a genetic record of this crossing:
We see that according to the phenotype, 3: 1 cleavage occurred, and according to the genotype 1AA: 2Aa: 1aa.
This pattern is calledsecond generation hydride cleavage rules, orII mendel’s law which is formulated as follows:when two heterozygous individuals (Aa hybrids), which have a pair of alternative variants of one trait, are crossed, the offspring splits on this trait in a ratio of 3: 1 by phenotype and 1: 2: 1 by genotype.
Recording crosses can be done in another way, using the so-called Pennet lattice, which was proposed by the English geneticist Pennet. The principle of constructing the lattice is simple: the gametes of the female are recorded along the horizontal line at the top, and the gametes of the male are written vertically to the left, and the genotype and phenotype of the descendants are determined at the intersection of the vertical and horizontal lines.
Let us make a genetic record of the considered examples using the Pennet lattice.
As a rule, genetic recording using the Pennet lattice is used in the analysis of more complex crosses. But it allows us to easily understand why the offspring of the first generation are uniform, and in the second generation there was a split.
Genes are known to be located on chromosomes. In the considered example, a pea plant with yellow seeds in a pair of homologous chromosomes carries a pair of yellow alleles. As a result of meiosis, homologous chromosomes diverge into various gametes, and with them allelic genes (yellow color of seeds). Since homozygotes (AA) have the same allelic genes, all gametes carry this gene. Similarly, with plants that have a green seed color (aa): both allelic genes are the same, therefore gametes carry the same gene. We conclude: a homozygous individual always gives one type of gamete.
Thus, if the maternal individual (AA) gives one type of gametes (A) and the paternal individual (aa) gives one type of gametes (a), then only one combination of gametes - Aa is possible, that is, all hybrids of the first generation are uniform and are heterozygous for this symptom (seed color). Phenotypically, all plants will be with yellow seeds. When two heterozygotes (Aa) are crossed, gametes with the dominant gene (A) and recessive gene (a) are formed in equal numbers in each individual, from which four possible combinations of zygotes can be expected. An egg with a gene (A) can be fertilized by a sperm with both a dominant gene (A) and a recessive gene (a); in the same way, an egg with gene (a) can be equally likely to be fertilized by a sperm with gene (A) and gene (a). As a result, four zygotes are formed: AA: Aa: aA: aa. We see that according to the phenotype, 3 individuals with dominant characters and one individual with a recessive one, that is, a ratio of 3: 1, were obtained. According to the genotype, the ratio was 1AA: 2Aa: 1aa.aboutthis implies that if in the future, from each group of individuals of the second generation, offspring are obtained by self-pollination, then homozygous individuals AA and aa will only produce uniform offspring, without splitting, and the offspring of individuals with the genotype Aa (heterozygous) will continue to split further.
G. Mendel explained this by the fact that gametes are genetically pure, that is, they carry only one gene from an allelic pair. Based on this conclusion, G. Mendel formulatedgamete purity law: the pairs of alternative characters found in each organism do not mix and, when gametes are formed, pass one at a time in their pure form.
(Definition from the book: T. L. Bogdanova. Biology. Tasks and exercises. M., High School, 1991)
TASKS AND METHODS OF GENETICS.
FIRST AND SECOND LAW OF G. MENDEL
Task number 1.
How many types of gametes form individuals with the genotype BB; with the genotype BB; with the genotype of cc?
D a n about:Genotypes:
1) BB
2) BB
3) cc
Decision:
The number of expected gamete types is determined by the formula:x = 2 n , where n is the number of pairs of alternative signs of the organism being studied, andx - the number of gamete types.
To find:
the number of gamete types -?
1) BB - the genotype of the individual.
One pair of alternative symptoms.
Determine the number of combinations of gametes:x = 2 1 from herex \u003d 2 (B, c).
2) BB - the genotype of the individual; no alternative symptoms.
x = 2 0 from herex \u003d 1 (B).
3) cc - the genotype of the individual; no alternative symptoms.
Determine the number of combinations of gametes:x = 2 0 from herex \u003d 1 (c).
Answer: 2 types of gametes; 1 type of gametes; 1 type of gametes.
Task number 2.
How many types of gametes does an individual form: a) homozygous for the recessive gene? b) homozygous for the dominant gene? c) heterozygous?
D a n about:Genotypes:
1) aa
2) AA
3) aa
Decision:
a) aa - individual genotype
x = 2 0 \u003d 1 (a)
b) AA - the genotype of the individual
x = 2 0 \u003d 1 (A)
c) Aa - individual genotype
x = 2 1 \u003d 2 (A, a)
To find:
x -?
Answer: a) 1 type of gametes; b) 1 type of gametes; c) 2 types of gametes.
Task number 3.
Smooth coloring of watermelons is inherited as a recessive trait. What offspring will result from crossing two heterozygous plants with striped fruits?
Task number 4.
Find possible gamete variants for organisms with the following genotypes: AA, BB, Cc, DD.
D a n about:Genotypes:
AA, BB, SS, DD
Decision:
1) AA - a homozygous organism, forms one type of gametes: A.
2) BB - a heterozygous organism, forms two types of gametes: B and c.
3) CC - a heterozygous organism, forms two types of gametes: C and c.
To find:
possible gamete options -?
4) DD - a homozygous organism, forms one type of gametes: D.
Answer: 1) A; 2) B, c; 3) C, s; 4) D.
Task number 3.
Identify the genotypes and phenotypes of offspring from the marriage of brown-eyed heterozygous parents.
Note: if the task is about people, then the following designations of parents are introduced: ○ - women; □ - men.
Answer: 1AA: 2Aa: 1aa; 3 children with brown and one with blue eyes.
Task number 4.
A person’s ability to hold a predominantly right hand dominates the ability to hold a predominantly left hand. A right-handed man, whose mother was left-handed, married a right-handed woman who had three brothers and sisters, two of whom were left-handed.aboutlimit the possible genotypes of women and the likelihood that children born from this marriage will be left-handed.
Answer: if a woman is homozygous, then the probability of giving birth to left-handed people will be 0, if heterozygous, then left-handed people will be born 25%.
Task number 5.
When heterozygous red-fruited tomatoes were crossed with yellow-fruited, 352 plants with red fruits were obtained.aboutsteel plants had yellow fruits. Determine how many plants were yellow?
Answer: 352 plants.
Task number 6.
Myoplegia (periodic paralysis) is inherited as a dominant trait. Determine the likelihood of having children with abnormalities in a family where the father is heterozygous and the mother does not suffer from myoplegia.
Answer: the probability of having children with anomalies is 50%.
Task number 7.
In tomatoes, the gene that determines the red color of the fruit is dominant in relation to the yellow gene. 3021 tomato bush obtained from hybrid seeds was yellow in color, and 9114 was red.
Question: a) how many heterozygous plants are among hybrids? b) does this attribute (coloration of the fruit) relate to mendelating?
2) Calculate the number of heterozygous plants that make up 2/3 of the number of all red-fruited:
(9114: 3) · 2 \u003d 6742 plants.
3) The sign "color of the fruit" refers to mendeliruyuschih, as the ratio of bushes with yellow and red fruits is 1: 3, that is, obeys the second law of Mendel.
Answer: a) 6742 plants; b) applies.
Task number 8.
The black body gene of cattle dominates the red gene. What offspring can be expected from crossing: a) two heterozygous individuals? b) a red bull and a hybrid cow?
Answer: a) 75% of black calves, 25% of red calves;
b) 50% of black calves, 50% of red calves.
Hybrid Hybridization.
THIRD LAW OF THE MENDEL
Hybrid crossbreeding is called, in which individuals differing in two pairs of alleles participate. G. Mendel crossed two varieties of peas - with smooth yellow seeds and green wrinkled. The hybrids of the first generation all had smooth yellow seeds, since in the first generation only dominant characters always appear. When crossing the hybrids of the first generation among themselves, a splitting was revealed: 315 yellow smooth seeds, 101 - yellow wrinkled, 108 - green smooth, 32 - green wrinkled.
Let us analyze the results of crossing: as we see, when crossing homozygous forms, the hybrids of the first generation are uniform, as is the case with monohybrid crosses; in the second generation, the characters are split and four groups of individuals different in phenotype are formed (yellow smooth, yellow wrinkled, green smooth, green wrinkled), and the ratio of their phenotypes is 9: 3: 3: 1. We see that when the hybrid crosses, there is an increase the number of phenotypes doubled compared to monohybrid crosses. If we consider the ratio of each trait separately, we will see that it is 3: 1, as in the case of monohybrid crosses, that is, splitting for each trait occurs independently. Based on this, G. Mendel formulatedIII law - the law of independent inheritance of traits: splitting for each pair of signs is independent of other pairs of signs.
Consider the cytological basis of Mendel’s III law. Two homozygous pea plants with genotypesAABB (yellow smooth seeds) andaavv (green wrinkled seeds) form one type of gamete -AB andav . As a result of their crossing, the offspring will be uniform -AaBb (yellow smooth seeds). When crossing first-generation hybrids with each other, each plant forms four types of gametes, with only one of each pair of allelic genes entering the gamete. As a result of random chromosome discrepancies in the first division of meiosis, the geneA can fall into the same gamete with the geneIN forming a type of gameteAB , or with a genein (forming the second type of gametes -Av ), and the genea can integrate with the geneIN (forming the third type of gametes -aB ) or with the genein (forming the fourth type of gametes -av ):
During fertilization, each of the four types of gametes(AB, AB, AB, AB) one organism can meet any of the gametes(AB, AB, AB, AB) another organism. All possible combinations of male and female gametes are easy to establish using the Pennet grille.
By genotype: 1AABB: 2AABB: 2AaBB: 4AaBb: 1Aacc: 2Aavv: 1aaVV: 2 aaVv: 1aavv.
As already mentioned, the obtained hybrids in the second generation have the following phenotype ratio: 9 parts - yellow smooth, 3 - green smooth, 3 - yellow wrinkled and 1 part green wrinkled. Moreover, the characters are inherited independently of each other and for each of them the usual splitting is observed - 3: 1.
Thus, in a hybrid hybrid crossing, each pair of characters, when split in the offspring, behaves in the same way as in a single hybrid hybridization, that is, independently of another pair of characters. Such splitting is obtained with complete dominance for each pair of characters. In case of polyhybrid crossbreeding (parental forms differ in several or many features), splitting for each character is the same.
TASKS FOR DIGBRID
AND ANALYZING CROSSING
Task number 1.
Write down the possible types of gametes produced by organisms with the following genotypes: a) AABB, b) SdDD, c) EeFf;
d) dhh (genes are inherited independently).
Answer: a) AB; b) SD, SD; c) EfEf, eF, ef; d) dh.
Task number 2.
Normal growth in oats dominates gigantism, and early ripeness dominates late ripeness. The genes of both traits are in different pairs of chromosomes. What signs will the hybrids obtained from crossing heterozygous parents by both characteristics have? What is the phenotype of parental individuals?
Otvet: P - normal, early ripe oats plants;
F 1 : 9 parts of normal growth, early ripening; 3 - normal growth, late ripening; 3 - giant early ripe; 1 part of the giant late-ripening.
Task number 3.
In Drosophila, the gray color of the body and the presence of bristles are dominant characters that are inherited independently. What offspring should be expected from a cross between a yellow female without bristles and a male heterozygous for both traits?
Otvet: 25% - gray, without bristles; 25% - gray, with bristles; 25% - yellow, with bristles; 25% are yellow, without bristles.
Task number 4.
When a black rooster without crest was crossed with a brown crested chicken, all offspring turned out to be black and crested. Identify the genotypes of parents and offspring. What signs are dominant? What percentage of brown non-crested chickens will result from the cross-breeding of first-generation hybrids?
3) Define the percentage of brown chickens without crest:
Answer: P 1 : aacc, aacenturies; F 1 : AaBv;
R 2 : AaBv, AaBv;
F 2 : 1AABB: 2AABB: 2AaBB: 4AaBb: 1Aacc: 2Aavv: 1aaVV: 2aaVv: 1aavv; dominant features - black plumage and the presence of crest; brown chickens without crest will turn out 6%.
Task number 5.
A pumpkin having yellow disc-shaped fruits was crossed with a pumpkin that had white spherical fruits. All hybrids from this cross had a white color and a disc-shaped fruit shape. What signs dominate? What are the genotypes of parents and offspring?
Answer: P: aacc, aacenturies; F 1 : AaBv; dominant features - white color and disc-shaped fruit.
Task number 6.
Polydactyly (multi-fingering) and the absence of small molars are transmitted as dominant features. The genes of these characters are in different pairs of chromosomes. What is the probability of having children without abnormalities in a family where both parents suffer from both diseases and are heterozygous for these gene pairs?
Answer: the probability of having children without anomalies is 1/16.
Task number 7.
In humans, some forms of myopia dominate over normal vision, and brown eyes are over blue. What offspring can be expected from a marriage of a short-sighted brown-eyed man with a blue-eyed woman with normal vision? Identify all possible genotypes of parents and offspring.
25% - normal vision brown-eyed
25% - normal vision blue-eyed
Answer: 1) R: aavv,aavv; F 1 : AaBv;
2) P: aavv,aacc; F 1 : AaBb, aainin;
3) P: aavv,aavin; F 1 : AaVv,aavv;
4) P: aavv,aainin; F 1 : AaVv,aavv, aaVv, aavv.
Task number 8.
Some forms of cataracts and deafness in humans are transmitted as recessive unconnected symptoms.
Questions about:
1. What is the probability of having children with two anomalies in a family where both parents are heterozygous for two pairs of genes?
2. What is the probability of having children with two anomalies in a family where one of the parents suffers from cataracts and deaf-mute, and the second spouse is heterozygous for these signs?
Answer: in the first case, the probability of having children with two anomalies is 1/16, or 6%, in the second¼, or 25%.
Task number 9.
Glaucoma (eye disease) has two forms: one form is determined by the dominant gene, and the other is recessive. Genes are located on different chromosomes. What is the probability of having a sick child in the family:
a) where both spouses suffer from different forms of glaucoma and are homozygous for both pairs of genes;
b) where are both spouses heterozygous for both pairs of genes?
Answer: a) 100% of sick children;
b) 13/16 sick children, or 81%.
Task number 10.
In snapdragons, the red color of the flower does not fully dominate the white. The hybrid plant is pink in color. The normal form of the flower completely dominates the pyloric. What offspring will result from crossing two diheterozygous plants?
Otvet: 3/16 - red flowers of normal shape;
6/16 - pink flowers of normal shape;
1/16 - red flowers of a pyloric form;
2/16 - pink flowers of a pyloric form;
3/16 - white flowers of normal shape;
1/16 - white flowers of a pyloric form.
Allelic genes. So, we found that heterozygous individuals have two genes in each cell - A and aresponsible for the development of the same trait. Genes that determine the alternative development of the same trait and are located in identical regions of homologous chromosomes are called allelic genes or alleles. Any diploid organism, whether it be a plant, animal or human, contains in each cell two alleles of any gene. The exception is sex cells - gametes. As a result of meiosis, the number of chromosomes in them decreases by 2 times, so each gamete has only one allelic gene. Alleles of one gene are located in one place of homologous chromosomes.
Schematically, a heterozygous individual is designated as follows:
Homozygous individuals with this designation look like this:
or, but they can be written and how AA and aa.
Phenotype and genotype. Considering the results of self-pollination of F 2 hybrids, we found that plants grown from yellow seeds, being externally similar, or, as they say in such cases, having the same phenotype, have a different combination of genes, which is usually called a genotype. Thus, the dominance phenomenon leads to the fact that, with the same phenotype, individuals can have different genotypes. The concepts of “genotype” and “phenotype” are very important in genetics. The totality of all the genes of the body is its genotype. The totality of all the signs of the body, starting with external and ending with the features of the structure and functioning of cells and organs, makes up the phenotype. The phenotype is formed under the influence of the genotype and environmental conditions.
Analyzing Crossing. By the phenotype of an individual, it is far from always possible to determine its genotype. In self-pollinating plants, the genotype can be determined in the next generation. For cross-breeding species use the so-called analyzing crosses. When analyzing crossing, an individual whose genotype should be determined is crossed with individuals homozygous for the recessive gene, i.e., having the genotype aa. Consider analyzing crossing by example. Let individuals with genotypes AA and Aa have the same phenotype. Then, when crossing with an individual recessive by a definable trait and having a genotype aaThe following results are obtained:
From these examples it is seen that individuals homozygous for the dominant gene do not produce cleavage in F 1, and heterozygous individuals when crossed with a homozygous individual give cleavage already in F 1.
Incomplete dominance. Not always heterozygous organisms according to the phenotype exactly correspond to the parent homozygous for the dominant gene. Often heterozygous offspring have an intermediate phenotype, in such cases they speak of incomplete dominance (Fig. 36). For example, when a plant is crossed, a nocturnal beauty with white flowers (aa) and a plant with red flowers (AA), all F1 hybrids have pink flowers (Aa). When crossing hybrids with pink color of flowers among themselves in F 2 splitting occurs in the ratio 1 (red): 2 (pink): 1 (white).
Fig. 36. Intermediate inheritance in a night beauty
The principle of gamete purity. Hybrids, as we know, combine different alleles introduced into the zygote by parental gametes. It is important to note that different alleles that find themselves in the same zygote and, therefore, in the body that developed from it, do not affect each other. Therefore, the properties of the alleles remain constant regardless of what zygote they have been to before. Each gamete always contains only one allele of any gene.
The cytological basis of the principle of gamete purity and the law of cleavage is that homologous chromosomes and the allelic genes located in them are distributed in different gametes in meiosis, and then, when fertilized, reunite in a zygote. In the processes of discrepancy in gametes and association into zygothallele genes, they behave as independent, integral units.
- Will the definition be correct: is a phenotype a set of external signs of an organism?
- What is the purpose of analyzing crosses?
- What, in your opinion, is the knowledge of the genotype and phenotype practical?
- Compare the types of inheritance of genetic traits in crosses with the behavior of chromosomes during meiosis and fertilization.
- When crossing gray and black mice, 30 descendants were obtained, of which 14 were black. It is known that gray color dominates over black. What is the genotype of mice of the parental generation? See the solution at the end of the tutorial.
- The blue-eyed man, whose two parents had brown eyes, married a brown-eyed woman whose father had brown eyes and his mother blue. From this marriage a blue-eyed son was born. Identify the genotypes of all the individuals mentioned.
Within the population gene pool, the proportion of genotypes containing different alleles of the same gene; subject to certain conditions, it does not change from generation to generation. These conditions are described by the basic law of population genetics formulated in 1908 by the English mathematician J. Hardy and the German geneticist G. Weinberg. "In a population of an infinitely large number of freely interbreeding individuals in the absence of mutations, selective migration of organisms with different genotypes and the pressure of natural selection, the original allele frequencies are preserved from generation to generation."
Hardy-Weinberg equation in solving genetic problems
It is well known that this law is applicable only to ideal populations: a sufficiently high number of individuals in a population; the population should be panmix when there is no restriction on the free choice of a sexual partner; virtually no mutation of the studied trait; there is no influx and outflow of genes and no natural selection.
Hardy-Weinberg's law is formulated as follows:
in an ideal population, the ratio of the frequencies of alleles of genes and genotypes from generation to generation is constant and corresponds to the equation:
p 2 + 2pq + q 2 \u003d 1
Where p 2 - the proportion of homozygotes for one of the alleles; p is the frequency of this allele; q 2 is the proportion of homozygotes for the alternative allele; q is the frequency of the corresponding allele; 2pq is the proportion of heterozygotes.
What does the “ratio of the frequencies of gene alleles” and “the ratio of genotypes” mean - are they constant? What are these values \u200b\u200bequal to?
Let the frequency of a gene in the dominant state (A) be equal to p, and the recessive allele (a) of the same gene is q (it is possible and vice versa, but it is possible in general with one letter, expressing one notation from another) and understanding that the sum of the frequencies of the dominant and recessive alleles of one gene in a population is 1, we obtain the first equation:
1) p + q \u003d 1
Where does the Hardy-Weinberg equation come from? You remember that when monohybrid crosses of heterozygous organisms with genes Aa x Aa according to the second Mendel law in the offspring, we observe the appearance of different genotypes in the ratio 1AA: 2 Aa: 1aa.
Since the frequency of occurrence of the dominant allelic gene A in our case is denoted by the letter p, and the recessive allele by the letter q, the sum of the frequencies of occurrence of the very genotypes of organisms (AA, 2Aa and aa) that have the same allelic genes A and a will also be 1, then :
2) p 2 AA + 2pqAa + q 2 aa \u003d1
In tasks in population genetics, as a rule, it is required:
a) find the frequency of occurrence of each of the allelic genes according to the known ratio of frequencies of individual genotypes;
B) or vice versa, to find the frequency of occurrence of any of the individual genotypes by the known frequency of occurrence of the dominant or recessive allele of the studied trait.
So, substituting the known frequency of occurrence of one of the gene alleles in the first formula and finding the frequency of occurrence of the second allele, we can always find the frequency of occurrence of the various offspring genes themselves using the Hardy-Weinberg equation.
Usually, some actions (due to their obviousness) are decided in the mind. But in order to make it clear what is already obvious, one must well understand what the letter designations in the Hardy-Weinberg formula are.
The provisions of the Hardy-Weinberg law apply to multiple alleles. So, if an autosomal gene is represented by three alleles (A, a1 and a2), then the formulas of the law take the following form:
RA + qa1 + ra2 \u003d 1;
P 2 AA + q 2 a1a1 + r 2 a2a2 + 2pqAa1 + 2rrAa2 + 2qra1a2 \u003d 1.
"In a population of an infinitely large number of freely mating individuals in lack of mutations, selective migration organisms with different genotypes and natural selection pressures the original allele frequencies are preserved from generation to generation. "
Suppose that in a gene pool of a population that satisfies the described conditions, a certain gene is represented by the alleles A 1 and A 2 detected with a frequency of p and q. Since there are no other alleles in this gene pool, then p + q \u003d 1. Moreover, q \u003d 1 – p.
Accordingly, individuals of this population form p gametes with A 1 allele and q gametes with A 2 allele. If the crosses occur randomly, then the proportion of germ cells connecting to gametes A 1 is p, and the proportion of germ cells connecting to gametes A 2 is q. Generation F 1 resulting from the described breeding cycle is formed by the genotypes A l A 1, A 1 A 2, A 2 A 2, the number of which is correlated as (p + q) (p + q) \u003d p 2 + 2pq + q 2 (Fig. . 10.2). Upon reaching puberty, individuals AlAi and AgA2 form one type of gamete — A 1 or A 2 — with a frequency proportional to the number of organisms of the indicated genotypes (p and q). Individuals A 1 A 2 form both types of gametes with an equal frequency of 2pq / 2.
Fig. The logical distribution of genotypes in a series of generations depending on the frequency of formation of gametes of different types (Hardy-Weinberg law)
Thus, the proportion of gametes A 1 in the generation F 1 will be p 2 + 2pq / 2 \u003d p 2 + p (1 – p) \u003d p, and the proportion of gametes A 2 will be q 2 + 2pq / 2 \u003d q 2 + q (l -q) \u003d q.
Since the frequencies of gametes with different alleles in the fi generation are not changed in comparison with the parent generation, the F 2 generation will be represented by organisms with the genotypes A l A 1, A 1 A 2 and A 2 A 2 in the same ratio p 2 + 2pq + q 2 . Due to this, the next cycle of reproduction will occur in the presence of p gametes A 1 and q gametes A 2. Similar calculations can be performed for loci with any number of alleles. The conservation of allele frequencies is based on the statistical laws of random events in large samples.
The Hardy – Weinberg equation in the form in which it is considered above is valid for autosomal genes. For genes linked to sex, the equilibrium frequencies of the genotypes A l A 1, A 1 A 2 and A 2 A 2 coincide with those for autosomal genes: p 2 + 2pq + q 2. For males (in the case of a heterogametic sex), due to their hemizygosity, only two genotypes A 1 - or A 2 - are possible, which reproduce at a frequency equal to the frequency of the corresponding alleles in females in the previous generation: p and q. From this it follows that phenotypes determined by recessive alleles of X-linked genes are more common in males than in females.
So, with a hemophilia allele frequency of 0.0001, this disease is observed 10,000 times more often in men of this population than in women (1 in 10 thousand in the first and 1 in 100 million in the second).
Another general consequence is that in the case of inequality in the frequency of the allele in males and females, the difference between the frequencies in the next generation is halved, and the sign of this difference changes. Usually, several generations are required in order for an equilibrium state of frequencies to arise for both sexes. The indicated condition for autosomal genes is achieved in one generation.
Hardy-Weinberg Law describes conditions genetic stability of the population. A population whose gene pool does not change in a series of generations is called mendelian. The genetic stability of Mendelian populations puts them outside the evolutionary process, since under such conditions the effect of natural selection is suspended. The isolation of Mendelian populations is of purely theoretical significance. In nature, these populations do not occur. The Hardy-Weinberg Act lists conditions that naturally change the gene pools of populations. The indicated result is led, for example, by the factors limiting free crossbreeding (panmixia), such as the finite number of organisms in a population, isolation barriers that prevent the random selection of mating pairs. Genetic inertness is also overcome by mutations, the influx into the population or outflow of individuals from it with certain genotypes, and selection.
Examples of solutions to some problems using the Hardy-Weinberg equation.
Task 1. In the human population, the number of individuals with brown eye color is 51%, and with blue - 49%. Determine the percentage of dominant homozygotes in a given population.
The difficulty of solving such tasks in their apparent simplicity. Since there is so little data, the solution should be as if very short. It turns out not very.
Under the condition of such tasks, we are usually given information about the total number of phenotypes of individuals in the population. Since phenotypes of individuals in a population with dominant traits can be represented by both AA homozygous individuals and heterozygous Aa, to determine the frequency of occurrence of specific genotypes of individuals in this population, it is necessary to first calculate the frequency of occurrence of alleles of gene A and a separately .
How should we reason when solving this problem?
Since it is known that brown eye color dominates over blue, we denote the allele responsible for the manifestation of the sign of brown eyes A, and the allelic gene responsible for the manifestation of blue eyes, respectively, a. Then, brown-eyed in the studied population will be people with both the AA genotype (dominant homozygotes, the proportion of which must be found by the condition of the problem) and the Aa heterozygotes), and blue-eyed - only aa (recessive homozygotes).
By the terms of the task, we know that the number of people with the AA and Aa genotypes is 51%, and the number of people with the AA type genotype is 49%. How, based on these statistics (a large sample should be representative), can we calculate the percentage of brown-eyed people with only the AA genotype?
To do this, we calculate the frequency of occurrence of each of the allelic genes A and A in a given population of people. Hardy-Weinberg's law, applied to large freely crossing populations, will allow us to do this.
Denoting the frequency of occurrence of allele A in a given population with the letter q, we have the frequency of occurrence of the allele gene a \u003d 1 - q. (One could designate the frequency of occurrence of the allelic gene by a separate letter, as in the text above - this is more convenient for someone). Then the Hardy-Weinberg formula itself for calculating the frequencies of genotypes in monohybrid crosses with the complete dominance of one allelic gene over another will look like this:
q 2 AA + 2q (1 - q) Aa + (1 - q) 2 aa \u003d 1.
Well, now everything is simple, you probably all guessed that we know in this equation, and what should be found?
(1 - q) 2 \u003d 0.49 is the frequency of occurrence of people with blue eyes.
We find the q value: 1 - q \u003d square root of 0.49 \u003d 0.7; q \u003d 1 - 0.7 \u003d 0.3, then q2 \u003d 0.09.
This means that the frequency of brown-eyed homozygous AA individuals in this population will be 0.09 or their proportion will be 9%.
Problem 2. In clover meadow, late ripeness dominates early maturity and is inherited monogenously. During testing, it was found that 4% of plants belong to the early ripening type of clover, which part of the late ripening plants are heterozygotes?
In this context, testing means evaluating the purity of a variety. And what, unless the variety is a clean line like Mendel’s pea variety, for example. Theoretically, “yes”, but in practice (the fields are large - these are not the experienced plots of the brilliant Mendel) in each production variety there may also be some “garbage” alleles of genes in each production variety.
In this case, with a late-ripening clover variety, if the variety were pure, only plants with the AA genotype would be present. But the variety was not very clean at the time of testing (approbation), since 4% of individuals were early ripe plants with the aa genotype. This means that the alleles of a were “dangling” into this variety.
So, since they are "hesitated", then individuals must be present in this variety, although the phenotype is late ripe, but heterozygous with the Aa genotype - do we need to determine their number?
By the condition of the problem, 4% of individuals with the aa genotype will be 0.04 part of the entire variety. In fact, this is q 2, which means that the frequency of occurrence of the recessive allele a is q \u003d 0.2. Then the frequency of occurrence of the dominant allele A is p \u003d 1 - 0.2 \u003d 0.8.
Hence the number of late-ripening homozygotes p2 \u003d 0.64 or 64%. Then the number of Aa heterozygotes will be 100% - 4% - 64% \u003d 32%. Since the total number of late-ripening plants is 96%, the proportion of heterozygotes among them will be: 32 x 100: 96 \u003d 33.3%.
Problem 3. Using the Hardy-Weinberg formula with incomplete dominance
When examining the population of Karakul sheep, 729 long-eared individuals (AA), 111 short-eared (Aa) and 4 an earless (aa) were identified. Calculate the observed phenotype frequencies, allele frequencies, and the expected genotype frequencies using the Hardy-Weinberg formula.
This is a task of incomplete dominance, therefore, the frequency distribution of genotypes and phenotypes are the same and they could be determined based on available data. To do this, you just need to find the sum of all individuals in the population (it is 844), first find the proportion of long-eared, short-eared and earless in percent (86.37, 13.15 and 0.47, respectively) and in fractions of frequencies (0.8637, 0.1315 and 0.00474).
But the task says to use the Hardy-Weinberg formula for calculating genotypes and phenotypes and, moreover, calculate the frequencies of alleles of genes A and a. So, to calculate the frequencies of the alleles of genes themselves, one cannot do without the Hardy-Weinberg formula.
Please note that in this problem, unlike the previous one, to designate the frequencies of allelic genes, we will use the notation technique not as in the first task, but as discussed above in the text. It is clear that the result of this will not change, but you will be entitled in the future to use any of these methods of notation, which seems more convenient for you to understand and conduct the calculations themselves.
We denote the frequency of occurrence of the allele A in all gametes of the sheep population by the letter p, and the frequency of occurrence of the allele a by the letter q. Remember that the sum of the frequencies of allelic genes is p + q \u003d 1.
Since, according to the Hardy-Weinberg formula p 2 AA + 2pqAa + q 2 aa \u003d 1, we have that the frequency of occurrence of dry q2 is 0.00474, then taking the square root from the number 0.00474 we find the frequency of occurrence of the recessive allele a. It is equal to 0.06884.
From here we can find the frequency of occurrence and the dominant allele A. It is 1 - 0.06884 \u003d 0.93116.
Now, according to the formula, we can again calculate the occurrence frequencies of long-eared (AA), earless (aa) and short-eared (Aa) individuals. Long-eared with genotype AA will be p 2 \u003d 0.931162 \u003d 0.86706, long-eared with genotype aa will be q 2 \u003d 0.00474 and short-eared with genotype Aa will be 2pq \u003d 0.12820. (The newly obtained numbers calculated by the formula almost coincide with those calculated initially, which indicates the validity of the Hardy-Weinberg law).
Task 4. Why is the proportion of albinos in populations so small
In a sample of 84,000 rye plants, 210 plants were albinos, as their recessive genes are in a homozygous state. Determine the frequencies of alleles A and a, as well as the frequency of heterozygous plants.
We denote the frequency of occurrence of the dominant allelic gene A by the letter p, and the recessive a by the letter q. Then, what can the Hardy-Weinberg formula p 2 AA + 2pqAa + q 2 aa \u003d 1 give us for applying it to this problem?
Since the total number of all individuals in this rye population is known to 84,000 plants, and in parts this is 1, the proportion of homozygous albino individuals with aa genotype equal to q2, of which 210 in total, will be q2 \u003d 210: 84000 \u003d 0.0025, then q \u003d 0.05; p \u003d 1 - q \u003d 0.95 and then 2pq \u003d 0.095.
Answer: the frequency of the allele a is 0.05; Allele A frequency: 0.95; the frequency of heterozygous plants with the Aa genotype will be 0.095.
Problem 5. Chinchillas were raised rabbits, and got a marriage in the form of albinos
In rabbits, the hair color of the “chinchilla” (Cch gene) dominates albinism (Ca gene). CchCa heterozygotes are light gray in color. Albino appeared on the rabbit farm among young chinchilla rabbits. Of the 5,400 rabbits, 17 were albinos. Using the Hardy-Weinberg formula, determine how many homozygous rabbits were obtained with chinchilla coloring.
And what do you think, the obtained sample in the population of rabbits in the amount of 5400 copies can allow us to use the Hardy-Weinberg formula? Yes, the sample is significant, the population is isolated (rabbit farm) and it is really possible to use the Hardy-Weinberg formula in the calculations. To use it correctly, you need to clearly understand what is given to us and what needs to be found.
For convenience of design only, we denote the genotype of chinchillas AA (we will need to determine the number of them), the genotype of albinos aa, then the genotype of heterozygous seroushes will be designated Aa.
If we “add” all the rabbits with different genotypes in the studied population: AA + Aa + aa, then this will be a total of 5400 individuals.
Moreover, we know that there were 17 rabbits with the aa genotype. How do we now, not knowing how many heterozygous gray rabbits with the Aa genotype were, determine how many chinchillas with the AA genotype are in this population?
As we can see, this task is almost a “copy” of the first, only there we were given the results of counting% of brown-eyed and blue-eyed individuals in the population of people, and here we actually know the number of albino rabbits 17 pieces and all homozygous chinchillas and heterozygous seryoshki in total : 5400-17 \u003d 5383
We take 5400 pieces of all rabbits for 100%, then 5383 rabbits (the sum of genotypes AA and Aa) will be 99.685% or in parts it will be 0.99685.
Q 2 + 2q (1 - q) \u003d 0.99685 is the frequency of occurrence of all chinchillas, both homozygous (AA) and heterozygous (Aa).
Then from the Hardy-Weinberg equation: q2 AA + 2q (1 - q) Aa + (1 - q) 2aa \u003d 1, we find
(1 - q) 2 \u003d 1 - 0.99685 \u003d 0.00315 is the frequency of occurrence of albino rabbits with the aa genotype. We find what the value 1 - q is equal to. This is the square root of 0.00315 \u003d 0.056. And q then equals 0.944.
Q 2 is 0.891, and this is the proportion of homozygous chinchillas with genotype AA. Since this value in% will be 89.1% of 5400 individuals, the number of homozygous chinchillas will be 4811 pieces.
Task 6. Determination of the frequency of occurrence of heterozygous individuals from the known frequency of occurrence of recessive homozygotes
One of the forms of glucosuria is inherited as an autosomal recessive trait and occurs with a frequency of 7: 1,000,000. Determine the frequency of occurrence of heterozygotes in the population.
We denote the allelic gene responsible for the manifestation of glucosuria a, since it is said that this disease is inherited as a recessive trait. Then the allelic dominant gene responsible for the absence of the disease is A.
Healthy individuals in the human population have genotypes AA and Aa; diseased individuals have only aa genotype.
We denote the frequency of occurrence of the recessive allele a by the letter q, and the dominant allele A by the letter p.
Since we know that the frequency of occurrence of sick people with the aa genotype (which means q 2) is 0.000007, then q \u003d 0.00264575
Since p + q \u003d 1, then p \u003d 1 - q \u003d 0.9973543, and p2 \u003d 0.9947155
Now substituting the values \u200b\u200bof p and q in the formula:
P2AA + 2pqAa + q2aa \u003d 1,
Let us find the frequency of occurrence of 2pq heterozygous individuals in a human population:
2pq \u003d 1 - p 2 - q 2 \u003d 1 - 0.9947155 - 0.000007 \u003d 0.0052775.
Problem 7. Like the previous task, but about albinism
General albinism (milky white skin, lack of melanin in the skin, hair follicles and retinal epithelium) is inherited as a recessive autosomal trait. The disease occurs with a frequency of 1: 20 000 (K. Stern, 1965). Determine the percentage of heterozygous gene carriers.
Since this trait is recessive, diseased organisms will have the aa genotype - their frequency is 1: 20,000 or 0.00005.
The frequency of the allele a will make the square root of this number, that is, 0.0071. The frequency of the allele A will be 1 - 0.0071 \u003d 0.9929, and the frequency of healthy AA homozygotes will be 0.9859.
The frequency of all heterozygotes is 2Aa \u003d 1 - (AA + aa) \u003d 0.014 or 1.4%.
Task 8. It seems how simple it is when you know how to solve
The population of Europeans according to the rhesus blood group system contains 85% Rhesus positive individuals. Determine the saturation of a population with a recessive allele.
We know that the allelic gene responsible for the manifestation of Rh positive blood is dominant R (denote the frequency of its occurrence by the letter p), and the Rh negative is recessive r (denote the frequency of occurrence by its letter q).
Since the problem says that p 2 RR + 2pqRr accounts for 85% of people, it means that rhesus-negative phenotypes q 2 rr will account for 15% or the frequency of their occurrence will be 0.15 of all people in the European population.
Then the frequency of allele occurrence r or “population saturation with the recessive allele” (indicated by the letter q) will be the square root of 0.15 \u003d 0.39 or 39%.
Task 9. The main thing to know what penetrance is.
Congenital hip dislocation is inherited dominantly. The average penetrance is 25%. The disease occurs with a frequency of 6: 10000. Determine the number of homozygous individuals in a population by recessive trait.
Penetrance is a quantitative indicator of the phenotypic variability of gene expression..
Penetrance is measured as a percentage of the number of individuals in which a given gene appeared in the phenotype to the total number of individuals in the genotype of which this gene is present in the state necessary for its manifestation (homozygous in the case of recessive genes or heterozygous in the case of dominant genes). The manifestation of the gene in 100% of individuals with the corresponding genotype is called complete penetrance, and in other cases, incomplete penetrance.
The dominant allele is responsible for the trait under study, we will designate it A. Therefore, organisms having this disease have genotypes AA and Aa.
It is known that phenotypic hip dislocation is detected in 6 organisms from the entire population (10,000 examined), but this is only one fourth of all people who have genotypes AA and Aa (since it is said that penetrance is 25%).
So, in fact, people with genotypes AA and Aa are 4 times more, that is, 24 out of 10,000 or 0.0024 parts. Then people with the aa genotype will be 1 - 0.0024 \u003d 0.9976 part or 9976 people out of 10,000.
Task 10. If only men are ill
Gout occurs in 2% of people and is caused by an autosomal dominant gene. In women, the gout gene does not appear; in men, its penetrance is 20% (V.P. Efroimson, 1968). Determine the genetic structure of the population by the analyzed characteristic, based on these data.
Since gout is detected in 2% of men, that is, in 2 people out of 100 with a penetrance of 20%, then in reality carriers of gout genes are 5 times more than men, that is, 10 people out of 100.
But, since men make up only half of the population, there will be 5 out of 100 people with AA + 2Aa genotypes in the population, which means that 95 out of 100 will be with aa genotype.
If the frequency of occurrence of organisms with aa genotypes is 0.95, then the frequency of occurrence of the recessive allele a in this population is equal to the square root of 0.95 \u003d 0.975. Then the frequency of occurrence of the dominant allele “A” in this population is 1 - 0.975 \u003d 0.005.
Task 11. How few people are resistant to HIV infection
Resistance to HIV infection is associated with the presence of some recessive genes in the genotype, for example, CCR and SRF. The frequency of the recessive CCR-5 allele in the Russian population is 0.25%, and the SRF allele is 0.05%. In the Kazakh population, the frequency of these alleles is 0.12% and 0.1%, respectively. Calculate the frequencies of organisms with increased resistance to HIV infection in each population.
It is clear that only homozygous organisms with aa genotypes will have increased resistance to HIV infection. Organisms with the genotypes AA (homozygotes) or Aa (heterozygotes) are not resistant to HIV infection.
In the Russian population of resistant organisms, the CCR allelic gene will be O, 25% squared \u003d 0.0625%, and the SRF allelic gene 0.05% squared \u003d 0.0025%.
In the Kazakh population of resistant organisms, the CCR allelic gene will be O, 12% squared \u003d 0.0144%, and the SRF allelic gene 0.1% squared \u003d 0.01%.
Hardy – Weinberg Law
We will consider mendelian populations:
- individuals are diploid;
- reproduce sexually;
the population is infinitely large; and panmictic populationswhere random free crossing of individuals proceeds in the absence of selection.
Consider a single autosomal gene in the population, represented by two alleles A and a.
We introduce the following notation:
N is the total number of individuals in the population
D is the number of dominant homozygotes ( AA)
H is the number of heterozygotes ( Aa)
R is the number of recessive homozygotes ( aa)
Then: D + H + R \u003d N.
Since individuals are diploid, the number of all alleles for the gene in question will be 2N.
Total number of alleles A and a:
A \u003d 2D + H;
a \u003d H + 2R.
Denote the fraction (or frequency) of the allele A through p, and the allele a - through g, then:
Because the gene can be represented by alleles A or a and no others, then p + g \u003d 1.
The state of population equilibrium by a mathematical formula was described in 1908 independently by mathematician J. Hardy in England and doctor W. Weinberg in Germany (Hardy – Weinberg law).
If p is the gene frequency A, and g is the frequency of the gene a, using the Pennet lattice, we can present in a generalized form the nature of the distribution of alleles in a population:
The ratio of genotypes in the described population:
p 2 AA : 2pg Aa : g 2 aah.
Hardy – Weinberg law in its simplest form:
p 2 AA + 2pg Aa + g 2 aa = 1.
Problem number 36
The population contains 400 individuals, of which genotypes AA – 20, Aa - 120 and aa - 260 individuals. Determine gene frequencies A and a.
Given: |
Decision: |
|
| N \u003d 400 D \u003d 20 H \u003d 120 R \u003d 260 p -? g -? |
 |
Answer: gene frequency A - 0.2; gene a – 0,8.
Task number 37
In shorthorn cattle, the red color dominates over white. Hybrids from crossing red and white - roan suit. In an area that specializes in shorthorn breeding, 4,169 red animals, 3,780 roan and 756 white animals are registered. Determine the frequency of red and white cattle genes in this area.
Answer: frequency of the gene for red coloring - 0.7; white - 0, 3.
Task number 38
In a sample of 84,000 rye plants, 210 plants were albinos, as their recessive genes are in a homozygous state. Determine allele frequencies A and aas well as the frequency of heterozygous plants.
Answer: gene frequency A and a – 0,5.
Task number 40
Allele frequencies p \u003d 0.8 and g \u003d 0.2 are known in the population. Determine the frequency of genotypes.
Answer: allele frequency A - 0.45; allele a – 0,55.
Task number 42
In the herd of cattle, 49% of animals of red color (recessive) and 51% of black color (dominant). How many percent of homo- and heterozygous animals in this herd?
Answer: in a population of 81% of individuals with a genotype AA, 18% with genotp Aa and 1% with the genotype aa.
Interesting genetic tasks
Task number 44. "The Tale of Dragons"
The researcher had 4 dragons: fire-breathing and fire-fighting females, fire-breathing and fire-breathing males. In order to determine the fire-breathing ability of these dragons, various crosses were carried out.
1. Fire-breathing parents - all fire-fighting offspring.
2. Non-breathable parents - all offspring are non-breathable.
3. A fire-breathing male and a non-fire-breathing female - in the offspring are approximately equal to fire-breathing and fire-breathing dragons.
4. Non-breathable male and fire-breathing female - all offspring are non-breathable.
Considering that the trait is determined by the autosomal gene, establish the dominant allele and write down the genotypes of the parents.
Decision:
By crossing number 4 we determine: A - non-breathable, a - fire-breathing; fire-breathing - ♀ aa and ♂ aa; fireproof male - ♂ AA;
Crossbreeding No. 3: Fireproof Female - - Aa.
Task number 45. "Consultant of the company" Cocktail ".
Imagine that you are a consultant to a small Cocktail firm, which literally translates from English as “cock tail”. The company breeds exotic breeds of roosters for the sake of tail feathers, which are readily purchased by owners of hat stores around the world. The length of the feathers is determined by the gene. A (long) and a (short) color: IN - black b - red, width: FROM - wide from - narrow. Genes are not linked. There are many different roosters and chickens on the farm with all possible genotypes, the data of which are entered into the computer. Next year, an increased demand for hats with long black narrow feathers is expected. What crossings need to be carried out in order to obtain the maximum number of birds with fashionable feathers in the offspring? Cross pairs with exactly the same genotypes and phenotypes is not worth it.
Decision:
F 1: A*IN*cc
1. P: ♀ AAVVss × ♂ aabbss
2. P: ♀ AAVVss × ♂ AAbbss
3. P: ♀ AAbbss × ♂ aaВВсс etc.
Task number 46. "Smuggler".
In the small state of Lysland, foxes have been bred for several centuries. The fur is exported, and the money from its sale forms the basis of the country's economy. Silver foxes are especially appreciated. They are considered a national treasure, and it is strictly forbidden to transport them across the border. A cunning smuggler, a good student at school, wants to trick customs. He knows the basics of genetics and suggests that the silver color of foxes is determined by two recessive alleles of the hair coloring gene. Foxes with at least one dominant allele are red. What needs to be done to get silver foxes in the smuggler’s homeland without violating the laws of Lysland?
Decision:
Carry out an analysis crossing and find out which red foxes are heterozygous for the color alleles, transport them across the border;
in the smuggler’s homeland they cross each other, and 1/4 of the descendants will be silver in color.
Task number 47. “Will Prince Uno's wedding be upset?”
The only Crown Prince Uno is about to marry the beautiful Princess Beatrice. Uno's parents learned that there were cases of hemophilia in the Beatrice family. Beatrice has no siblings. Aunt Beatrice has two sons growing up - healthy strong men. Uncle Beatrice goes hunting all day and feels great. The second uncle died as a boy from blood loss, which was caused by a deep scratch. Uncles, aunt and mother Beatrice are children of the same parents. How likely is the disease to be transmitted through Beatrice to the royal family of her fiancé?
Decision:
By constructing the proposed genealogical tree, it can be proved that the hemophilia gene was in one of Beatrice's X chromosomes; Beatrice's mother could have received it with a probability of 0.5; Beatrice herself - with a probability of 0.25.
Task number 48. "Royal Dynasties."
Suppose Emperor Alexander I had a rare mutation on the Y chromosome. Could this mutation be in:
a) Ivan the Terrible;
b) Peter I;
c) Catherine II;
d) Nicholas II?
Decision:
Immediately cross out Catherine II, in view of her belonging to the female sex.
Cross out Ivan the Terrible - he is a representative of the Rurikovich clan and did not belong to the Romanov dynasty.
Answer: Nicholas II could have a mutation.
Task number 49. "Leafing through the novel" War and Peace. "
Suppose that on the X chromosome, Prince Nikolai Andreyevich Bolkonsky had a rare mutation. Pierre Bezukhov had the same mutation. What is the probability of this mutation in:
a) Natasha Rostova;
b) the son of Natasha Rostova;
c) the son of Nikolai Rostov;
d) the author of War and Peace?
Answer:
Andrei Bolkonsky did not receive the X chromosome from his father. His wife was not a relative of either the Bolkonsky or the Bezukhovs. Therefore, the son of Prince Andrew does not have a mutation.
Natasha Rostova married Pierre Bezukhov. Pierre passed the X chromosome to his daughters, but not his sons. Therefore, the daughters of Natasha Rostova received a mutation, but the sons did not.
The son of Nikolai Rostov received his X-chromosome from his mother, the daughter of the old Prince Bolkonsky (of the 2 chromosomes of Princess Mary, there was only one mutation, therefore, she transferred the X-chromosome to her son with a 50% probability).
Lev Nikolaevich: the action of the novel ends a few years before the birth of Tolstoy, the author himself does not appear on the pages of the novel. But the writer’s father was a retired officer Count Nikolai Ilyich Tolstoy, and his mother was nee Volkonskaya, i.e. the writer's parents were the prototypes of Nikolai Rostov and his wife, nee Maria Bolkonskaya. Their future son Leo will get a mutation with a probability of 50%.
Task number 50. "The dispute between Bender and Panikovsky."
There was a dispute between Bender and Panikovsky: how is the color of the budgerigars inherited? Bender believes that the color of parrots is determined by one gene that has 3 alleles: With about - recessive in relation to the other two, S r and With w codominant, therefore, in parrots with a genotype C about C about - White color, C g C gand C r C o - blue C w C w and C w C about - yellow and C g C g - green color. But Panikovsky believes that the color is formed under the influence of two interacting genes A and IN. Therefore, parrots with a genotype A * B * - green A * bb - blue aaB * –– yellow aabb - white.
They made up 3 pedigrees:
1. P: Z × B |
F1: B, B F1: B F1: F, F, F, F, F, F, F, F, F |
Which pedigrees could have been compiled by Bender, which by Panikovsky?
Answer: genealogies 1 and 2 could be composed by Panikovsky, and genealogy 3 - by Bender.
LIST OF USED LITERATURE:
1. Bagotsky S.V.“Cool” problems in genetics // Biology for schoolchildren. 2005. No. 4.
2. Gulyaev G.V.Genetics problem book. - M .: Kolos, 1980.
3. Zhdanov N.V. Problem solving in the study of the topic: “Genetics of populations”. - Kirov: publishing house Ped. Institute, 1995.
4. Tasks on genetics for entering universities. - Volgograd: Teacher, 1995.
5. Kochergin B.N., Kochergina N.A.Tasks in molecular biology and genetics. - Minsk: Narodnaya asveta, 1982.
6. A brief collection of genetic problems. - Izhevsk, 1993.
7. Methodological development for students of the biological department of VZMSh at Moscow State University "Mendel Laws". - M., 1981.
8. Guidelines for self-preparation for practical exercises in general genetics. - Perm: publishing house Med. Institute, 1986.
9. Murtazin G.M.Tasks and exercises in general biology. - M .: Education, 1981.
10. Orlova N.N.Small workshop on general genetics (collection of tasks). - M .: publishing house of Moscow State University, 1985.
11. A collection of tasks in biology (teaching aid). - Kirov, 1998.
12. Sokolovskaya B.Kh.One hundred tasks in molecular biology and genetics. - Novosibirsk: Science, 1971.
13. Fridman M.V. Tasks on genetics at the School Olympiad of Moscow State University // Biology for schoolchildren. 2003. No. 2.
14. Shcheglov N.I.Collection of tasks and exercises in genetics. - M .: Ecoinvest, 1991.
The gene pool of a population can be described either by the frequencies of genes or by the frequencies of genotypes. Imagine that there are N diploid individuals in a population that differ in one pair of alleles (A and a); D - means the number of homozygotes for the dominant allele (AA); P is the number of homozygotes for the recessive allele (aa); H is the number of heterozygotes (Aa). Thus, in the population there will be three types of individuals with corresponding genotypes AA, Aa, aa. Since each individual with the AA genotype has two A alleles, and each Aa individual has one A allele, the total number of A alleles will be 2D + H. Then p is the frequency of occurrence of the dominant allele A is equal to:
The frequency of the recessive allele (a) is usually denoted by q. The sum of the frequencies of genes A and a is equal to unity, p + q \u003d 1, hence q \u003d 1-p. If a gene is represented by only two alleles (A and a) with a frequency of p and q, then what will be the frequencies of the three possible genotypes?
The Hardy-Weinberg law gives an answer to this question. At first glance, it may seem that individuals with a dominant phenotype will occur more often than with a recessive one. However, the 3: 1 ratio is observed only in the offspring of two individuals heterozygous for the same allele. Mendel’s laws say nothing about the frequencies of genotypes and phenotypes in populations. They are referred to in the named law. It was formulated independently by mathematician J. Hardy in England and doctor Wilhelm Weinberg in Germany. To understand the meaning of this law, suppose that males and females in a population cross randomly, or, which is the same thing, gametes of males and females will combine randomly to form zygotes. In the zygote, the maternal and paternal chromosomes are combined, each of the homologous chromosomes carries one allele from this pair. The formation of individuals with the AA genotype is due to the probability of receiving the A allele from the mother and the A allele from the father, i.e. pxr \u003d p2.
Similarly, the emergence of the aa genotype, the frequency of occurrence of which is q2. The Aa genotype can arise in two ways: the body receives the A allele from the mother, the a allele from the father or, conversely, the A allele from the father, and the a allele from the mother. The probability of both events is pq, and the total probability of the occurrence of the Aa genotype is 2pq. Thus, the frequency of the three possible genotypes can be expressed by the equation: (p + q) 2 \u003d p2 + 2pq + q2 \u003d 1
From the equation it follows that if the crossing is random, then the frequencies of the genotypes are associated with the frequencies of the alleles by simple relations according to the Newton binomial formula.
Let us examine an example when the frequencies of alleys of a given gene in a population are 0.1A; 0.3a (the geometric expression of the Hardy-Weinberg law for this case is shown in Fig. 21). In the progeny of 100 zygotes, there will be 49 homozygotes AA, 9 homozygotes aa and 42 heterozygotes Aa, i.e. this corresponds to the ratio of genotypes already known to us - p2 (AA): 2pq (Ad): q2 (aa).
Interestingly, in the next generation, gametes with the A allele will occur at a frequency of 0.7 (0.49 from AA homozygotes + 0.21 from Aa heterozygotes). This ratio will continue in the future. The frequencies of genes and, accordingly, genotypes remain unchanged from generation to generation - this is one of the main provisions of the Hardy-Weinberg law. However, the named law is probabilistic in nature and therefore is implemented in an infinitely large population. At the same time, gene frequencies remain unchanged if: there is unlimited panmixia; no natural selection; no new mutations of the same genes arise; there is no migration of individuals with other genotypes from neighboring populations.